DIFFERENTIAL GEOMETRY WITH APPLICATIONS TO MECHANICS AND PHYSICS Yves Talpaert Ouagadougou University Ouagadougou, Burkina Faso
M A R C E L
Library of Congress Cataloging-in-Publication Data
Talpaert, Yves. Differential geometry : with applications to mechanics and physics / Yves Talpaert. p. cm. - (Monographs and textbooks in pure and applied mathematics ; 237) Includes bibliographical references and index. ISBN: 0-8247-0385-5 (alk. paper) 1. Geometry, Differential. I. Title II.Series
French edition published by CepaduCs Editions: Yves Talpaert, Leqons et Applications de Ghmitrie Diffirentiale et de Mkcanique Analytique, 1993. ISBN 2-85428-325-9.
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PREFACE
Differential geometry is a mathematical discipline which in a decisive manner contributes to modem developments of theoretical physics and mechanics; many books relating to these are either too abstract since aimed at mathematicians, too quickly applied to particular physics branches when aimed at physicists. Most of the text comes from Master's-level courses I taught at several African universities and aims to make differential geometry accessible to physics and engineering majors. The first seven lectures rather faithfully translate lessons of my French book "GComCtrie DiffCrentielle et M6canique Analytique," but contain additional examples. The last three lectures have been completely revised and several new subjects exceed the Master's degree. The text sets out, for an eclectic audience, a methodology paving the road to analytical mechanics, fluid-dynamics, special relativity, general relativity, thermodynamics, cosmology, electromagnetism, stellar dynamics, and quantum physics. The theory and the 133 solved exercises will be of interest to other disciplines and will also allow mathematicians to find many examples and concepts. The introduced notions should be known by students when beginning a Ph.D. in mathematics applied to theoretical physics and mechanics. The chapters illustrate the imaginative and unifying characters of differential geometry. A measured and logical progression towards (sometimes tricky) ideas, gives this book its originality. All the proofs and exercises are detailed. The important propositions and the formulae to be framed are shown by * and Two introduced methods (in fluid-mechanics and calculus of variations) deserve further study. There is no doubt that engineers could overcome difficulties by using differential geometry methods to meet technological challenges.
Acknowledgements. I am grateful to Professor Michel N. Boyom (Montpellier University) who allowed me to improve the French version and to Professor Emeritus Raymond Coutrez (Brussels University) who taught me advanced mathematical methods of mechanics and astronomy. Many thanks to my former African students who let me expound on the material that resulted in this book. AH my affection to Moira who drew the figures.
vi
Preface
I wish to express my gratitude to Marcel Dekker, Inc. for helpful remarks and suggestions. Yves Talpaert
CONTENTS .............................................................................................. v
Preface
.
Lecture 0
TOPOLOGY AND DIF'F'ERENTIAL CALCULUS REQUIREMENTS
.
1
........................................................................... 1 ................................ Topological space ...................... . .
1 Topology
1.1 1.2 1.3 1.4 1.5 1.6
Topological space basis .................................................... Haussdorff space ............................... . ........................... Homeomorphism .............................................................. Connected spaces .................................................... Compact spaces .................................. . .......................... 1.7 Partition of unity ............................................................. 2 Differential calculus in Banach spaces ............................... .......
.
2.1 2.2 2.3 2.4 2.5
Banach space .................................................................8 Differential calculus in Banach spaces ................................... 10 Differentiation of R " into Banach ........................................ 17 Differentiation of R into R .............................................19 Differentiation of R into R .......................................... 22
.
3 Exercises ..........................................................................30
.
Lecture 1
MANIFOLDS
37
Introduction ................................ . . ....................................... 37
.
1
1.1 1.2 1.3
Differentiable manifolds
............................ . .........................40
Chart and local coordinates ................................................ 40 Differentiablemanifold structure ......................................... 41 Differentiable manifolds ................................................... 43
.
2 Differentiable mappings ........................................................ 50 2.1 2.2 2.3
.
Generalities on differentiable mappings ................................. 50 Particular differentiable mappings ........................................ 55 Pull-back of function ........................................................57
3 Submanifolds
3.1 3.2
.
4
..................................................................59
Submanifolds of R ........................................................ 59 Submanifold of manifold ...................................................64 Exercises .......................................................................... 65
viii
Contents
Lecture 2. TANGENT VECTOR SPACE
.
71
...................................................................71 Tangent curves ................................, . .......................... 71 Tangent vector .............................................................. 74
1 Tangent vector 1.1 1.2
.
....................................................................80 Definition of a tangent space ............................................. 80 Basis of tangent space ................................ .................... 81 Change ofbasis ........................................................... 82
2 Tangent space
2.1 2.2 2.3
.
3 Differential at a point ........................................................ 83 3.1
3.2 3.3
.
4
.
Lecture 3
Definitions ................................................................. 84 The image in local coordinates .......................................... 85 Diferential of a function ................................................... 86
Exercises ..........................................................................
87
TANGENT BUNDLE.VECTOR FIELD.ONE-PARAMETER GROUP LIE ALGEBRA 91 Introduction ........................................................................... 91
.
1 Tangent bundle
................................................................. 93
1.1 1.2
Natural manifold TM ................................................... 93 94 Extension and commutative diagram .................................
.
.......................................................96 Definitions ................................................................... 96 Properties of vector fields ............................................... 96
2 Vector field on manifold 2.1 2.2
.
............................................................97 ........................................................................97
3 Lie algebra structure 3.1 3.2 3.3
.
4
4.1 4.2
.
Bracket Lie algebra .......................... . . ...................................100 Lie derivative ............................................................... 101
One-parameter group of diffeomorphisms
Differential equations in Banach ........................................102 One-parameter group of diffeomorphisms ............................ 104
5 Exercises
.
Lecture 4
................................ 102
.........................................................................111
COTANGENT BUNDLE -VECTOR BUNDLE OF TENSORS
.
125
....................................... 125 .......................................................................125
1 Cotangent bundle and covector field 1.1 1.2 1.3
1-form Cotangent bundle .......................................................... 129 Field of covectors ......................................................... 130
Contents
.
IX
.................................................................. 130 Tensor at a point and tensor algebra .................................... 130
2 Tensor algebra 2.1 2.2
Tensor fields and tensor algebra ......................................... 137
3. Exercises .......................................................................... 144 Lecture 5. EXTERIOR DIFFERENTIAL FORMS
.
153
........................................................153 Definition of a p-form .................................................... 153 Exterior product of 1-forms ..............................................155 Expression of a p-form ................................................... 156
1 Exterior form at a point
1.1 1.2 1.3 1.4 1.5
Exterior product of forms ................................................ 158 Exterior algebra ...................................... .....................159
2. Differential forms on a manifold ............................
. .............
162
2.1 2.2
Exterior algebra (Grassmann algebra) .................................. 162 Change of basis .......................................................... 165
.
.......................... . . . . ..........167
3 Pull-back of a differentia1 form
3.1 3.2
.
4
4.1 4.2
. 6.
Definition and representation ............................................. 167 Pull-back properties ....................................................... 168 Exterior differentiation
..................... ..............................170
Definition ................................................................... 170 Exterior differential and pull-back ....................................... 173
............................ . ............................174 ........................................................................... 178
5 Orientable manifolds Exercises
.
Lecture 6 LIE DERIVATIVE.LIE GROUP
185
1 Lie derivative ......................................................................186 1.1 1.2
First presentation of Lie derivative ...................................... 186 Alternative interpretation of Lie derivative .............................195
.
.............................................199 Definition and properties ..................... . . ...................... 199 Fundamental theorem ..................................................... 201
2 Inner product and Lie derivative 2.1 2.2
.
3
.
Frobenius theorem
.............................................................. 204
4 Exterior differential systems 4.1 4.2
.................................................
207
Generalities ............................................................... 207 Pfaff systems and Frobenius theorem ...................................208
Contents
1.5 1.6 1.7 1.8 1.9
XI
Killing vector field ..................................................... 274 Volume ...................................................................... 275 The Hodge operator and adjoint ......................................... 277 Special relativity and Maxwell equations ..............................280 Induced metric and isometry ............................................. 283
.
2 Affine connection ................................................................ 285 2.1 2.2 2.3 2.4 2.5 2.6
. ........... 285
Affine connection definition ................................
Christoffel symbols ....................................................... 286
Interpretation of the covariant derivative .............................. 288 Torsion ..................................................................... 291 Levi-Civita (or Riemannian) connection ...............................291 Gradient - Divergence - Laplace operators ........................... 293
.
3 Geodesic and Euler equation
..............................................
300
4. Curvatures.Ricci tensor.Bianchi identity.Einstein equations ... 302 4.1 4.2 4.3 4.4
.
5
Curvature tensor ...........................................................302 Ricci tensor ................................................................ 305 Bianchi identity ........................................................... 308 Einstein equations ......................................................... 309
Exercises
.........................................................................
Lecture 9 LAGRANGE AND HAMILTON MECHANICS
310 325
.
1 Classical mechanics spaces and metric ....................................325 1.1 Generalized coordinates and spaces .................................... 325 1.2 Kinetic energy and Riemannian manifold ............................ 327
.
................. 329 .................................................................329
2 Hamilton principle. Motion equations. Phase space 2.1 2.2 2.3 2.4 2.5
.
Lagrangian Principle of least action .................... . . . ..................... 329 Lagrange equations ....................................................... 331 Canonical equations of Hamilton ....................................... 332 Phase space .................................................................337
3 D'Alembert-Lagrange principle.Lagrange equations
3.1 3.2 3.3 3.4
.
................. 338
D' Alembert-Lagrange principle .......................................... 338 Lagrange equations ............................... ... ....................340
Euler-Noether theorem .................................................... 341 Motion equations on Riemannian manifolds ........................ 343
4 Canonical transformations and integral invariants ...................... 344 4.1 Diffeomorphisms on phase spacetime .................................. 344 4.2 Integral invariants ........................................................ 346 4.3 Integral invariants and canonical transformations ..................... 348
Contents
4.4
Liouville theorem
........................................................
.
352
.......... 352 ........................... 353
5 The N-body problem and a problem of statistical mechanics 5.1
5.2
N-body problem and fundamental equations A problem of statistical mechanics ...................................... 358
.
6 Isolating integrals
6. I 6.2 6.3 6.4 6.5
.............................................................. 369
Definition and examples ..................................................369 Jeans theorem .............................................................. 372 Stellar trajectories in the galaxy .........................................373 The third integral ....................................................... 375 Invariant curve and third integral existence ............................ 379
7. Exercises
.........................................................................
.
Lecture 10 SYMPLECTIC GEOMETRY-Hamilton-JacobiMechanics preliminaries
381 385
.......................................................................... 385
.
1 Symplectic geometry ............................................................ 388 1.1 1.2 1.3 1.4 1.5
Darboux theorem and symplectic matrix ...............................388 Canonical isomorphism .................................................. 391 Poisson bracket of one-forms ............................................393 Poisson bracket of functions ............................................. 396 Syrnplectic mapping and canonical transformation ................... 399
.
..................................404 Hamilton vector field .....................................................404 Canonical transformations- Lagrange brackets ...................... 408 Generating functions ......................................................412
2 Canonical transformations in mechanics 2.1 2.2 2.3
.
3 Hamilton-Jacobi equation
.....................................................
415
3.1 3.2
Hamilton-Jacobi equation and Jacobi theorem ........................415 Separability ................................................................. 419
.
.......................... 422 Variational principle (with one degree of freedom) ..................423 Variational principle (with n degrees of freedom) ................... 427
4 A variational principle of analytical mechanics 4.1 4.2
.
...........................................................................429 Bibliography ......................................................................................... 443 5 Exercises
Glossary ...................
....................................................................... 445 . .
. . . . . . . . . . . . . . . . .Index ......
Contents
xiii
449
LECTURE 0
TOPOLOGY AND DIFFERENTIAL CALCULUS REQUIREMENTS
1. TOPOLOGY This section presents the required basic notions of topology. 1.1 TOPOLOGICAL SPACE
D
A topological space S is a set with a topology. A toplogy on S is a collection 0 of subsets, called open sets such that:
- the union of any collection of open sets is open,
-
Let
',
any finite intersection of open sets is open, space S and empty space 0 are open sets. be an open set.
D
The complement of
with respect to
D
An open neighborhood of a point containing x.
is said to be a closed set, namely:
in a topological space S is an open set U
Afterwards, any open neighborhood will be simply called neighborhood. Let P be a subset of space S, D
D
The relative topology on P is defined by
A point
is a contact point of P if every neighborhood of x contains at least a
point of P. Afterwards, "open set" will often be simply called "open. "
Lecture 0
2
D
A point is an accumulation point or limit point of P if every neighborhood of x contains (at least) one point of different from x.
D
A point is an isolated point of P if x has a neighborhood which does not contain any point of P different from x.
In other words, an isolated point of P is a point of P, which is no accumulation point. PR1
Every accumulation point is a contact point, the opposite is evidently false.
PR2
Any contact point is either an isolated point or an accumulation point.
D
The closure of P,denoted
PR3
The closure of P is a closed set, it's the smallest closed set containing P
D
A subset P of S is (everywhere) dense in S if =
is the set of contact points of P.
S.
The definition means every point of S is a contact point of P
1.2 TOPOLOGICAL SPACE BASIS
Let S be a topological space. 1.2.1
Definition
D
A basis B for the topology on S is a collection of open sets such that every open set of S is a union of elements of B. In other words: is a basis of open sets of if every open set of S is I ). J
1.2.2
D
Example of the metric space
A metric space M is a set provided with a distance. A distance on M is a function
satisfying the following conditions:
Example. The standard distance on
is defined by
Topology and Differential Calculus Requirements wherex = (x', ...,
and y = (y' ,..., y n ) .
We introduce a topology on M related to the distance notion and called metric topology.
An open sphere about
D
in metric space M is
A subset P of a metric space M is an open set of M if either 0 or P: P (r .
D
-
PR4
Every open sphere of a metric space M is an open set of
Proof. The open sphere is an open set. If the case is not trivial, we prove the existence of an open sphere included in B(a,r), namely:
B(x,r-d(a,x)) c B(a,r). Indeed, every point y~B(x,r-d(a,x)) is so that: d(x,y)< r-d(a,x)
which implies: then any point y E B(x,r-d(a,x)) necessarily belongs to B(a,r). To conclude, there exists an open sphere B(x,r-d(a,x))rB(a,r)and, by definition, B(a,r)is an open set of M
The previous proposition implies every union of open spheres of a metric space M is open. Reciprocally, every open set of M is the union of open spheres. That follows from the open set definition. Then, we can express:
PR5 The open spheres of a metric space make up a basis for a topology called metric space topology. Let S and T be topological spaces.
D
The product topology on SxT is the collection of subsets that are unions of opens as U x V, such that U and V are opens respectively in S and T.
Thus open rectangles form a basis for the topology. 1.2.3
Separable space
D
A topological space S is said to be a space with countable baris if there is (at least)
one basis in S consisting of a countable number of elements, countable meaning finite or denumerable.
Lecture 0
4
D
A topological space S containing a (everywhere) dense countable set is called separable.
PR6
Every topological space S with countable basis is separable.
ProoJ Let B = (I4
D
={
xi E V,
I
I EN
I
i
d ) be a countable basis of
S. We provethe space S contains a set
) (everywhere) dense in S.
Every neighborhood V, of y E S includes an open set containing y. This open set is the union of a certain number of V,. Thus
UY, I
V,nD#0 because, in particular, this set contains x i € VimThis conclusion is true for every point YES, thus every pointy is a contact point of D and hence D is dense in S. Example. The space R" is separable with the topology defined by
B= (B(x,r) I x e Q n , ~ E f Q where Q is the set of rational numbers. 1.3
HAUSSDORFF SPACE
D
A topological space S is called a type TI space if for any two distinct points x and y of S exist a neighborhood U, of x with y not belonging to U, and a neighborhood U, with x not belonging to U,.
D
A topological space S is called a type Ta space or a Huussdoflspace if for any two distinct points x and y of S exist a neighborhood U, of x and a neighborhood U, of y such that:
u,nu,
=0
Example. The real straight line with two origins Haussdorff space.
01 62
I
01
and
02
is a type TI space but not a
Indeed, a topology is defined using: - the usual open intervals in R (for the semi-lines), - ~~~,)UI-E,O)U(O,E)) (for 011, (for 0 3 ) . - { { oz1U(-&',O) U(0,s')1
Figure 1.
In the last two cases, the intersection of two open sets (of 01 and 02)is necessarily not empty. Thus it's not a Haussdorff space example. Later, Haussdorff spaces (any two distinct points x and y of the space have disjoint neighborhoods) will play a fundamental part.
PR7
Every metric space is a HaussdorE space.
5
Topology and Differential Calculus Requirements
ProoJ Let x and y be two points of a metric space M and let r be the distance d(x,y).ClearIy, the open spheres B(x,r/z) and BCy,r/2) are disjoint.
PR8
Every subspace of a HaussdorfTspace is a Haussdorff space.
This proposition is immediate. 1.4
HOMEOMORPHISM
Let S and T be topological spaces.
D
A mapping
f:S+ T : x ~ y is said to be continuous atpoint XESif, for every neighborhood V, of f ( x ) , f is a neighborhood of x in S.
-'(vY)
D
A mapping f of S into T is continuous on S if it is continuous at each point of S. In an equivalent manner: A mapping f of S into T is corotimious on S if, for every open set W in T, f -'(W) is an open set in S.
PR9
A mapping f of S into T is continuous on S if, for every closed set A in T, f -'(A) is a closed set in S.
Proof: The explanation is immediate sincef -'(CA) =
c f'( A ).
Notation. The set of continuous mappings of S into T is denoted
CO(S;T). We know that two topoIogica1 spaces S and T are homeomorphic if there is a bijection f : S + T "exchanging" the open sets, i.e. to each open V in S corresponds an open f( V ) in T and to each open Win T corresponds an open f -'(W) in S.
D
A homeamo~hismf of S onto T is a bicontinuous bijection, namely a bijection are continuous. such that fand
f''
This definition is logical because iff and f -'are continuous, then the inverse image of every open set of T is open and the image of every open set of S is open.
Examples.
I. The open sphere { xaR"
I
n
C ( x 1 ) ' < r , r d + ) is homeomorphic to
r.
i=l
2. The space { x e F
I
n
a<
Z(x1)' < b ; a , b ~ R }+ is not homeomorphic t o r . i-l
Lecture 0
1.5
CONNECTED SPACES
D
A topological space S is connected if every partition of S into two open sets A and 3 implies A = 0 or B= 0. In other words:
i f J(A,B)EO~O: [ A + @ , B#ld; A ( 1 B = 0 and AUB=S
1.
This definition means 0 and S are the only subsets of S that are both open and closed.
D
A topological space S is focally connected ai point x ES if x has a basis of connected neighborhoods,
D
A topological space S is focally connected if it is locally connected at each point.
D
A topologicaI space S is arcwise connected if for every two points a and b in S there is a continuous mapping f of a closed interval [a,P]cR into S such that f(a)= a and f (P)= b .
That is: If every two points in S can be joined by an arc in S.
1.6
COMPACT SPACES Let 5' and T be topological spaces.
D
A topological space S is compart if for every covering of S by open sets (that is Uui= S ) there is a finite subcovering. I
The reader will easily prove the following proposition.
PRlO Any closed subset of a compact space is compact. PR 1 1 Any compact subset of a Haussdorff space S is closed in S.
Proof: Let A be a compact subset of 5'. Since S is a HaussdorfT space, there are disjoint neighborhoods of X E A and of ~ E C AIn. addition A is compact and thus there are disjoint neighborhoods of y and A. Therefore CQ is open. PR12 Any continuous mapping f of a compact space S into a Haussdorff space T implies the subset f(S) of T is compact. Proof: First observe that the subspace f (S) of T is Haussdorff because T is a HaussdoriT space. Secondly, let {U,),,I be an arbitrary open covering of f(S). Since f is continuous, ( f - ' ~ , ) i e l is an open covering of S. But the space S is compact, then from (f' ' l J ) i G I there is a finite subcovering Since f (f -'u,) c U,then (U,)i,J is also a covering of f (S). This implies f ( S ) is a compact space. 1
In this definition, the Haussdorff separation condition could be added in order to avoid the spaces with trivial topology {as}.
Topology and Differential Calculus Requirements
7
PR13 Every continuous bijection f of a compact space S onto a Haussdorff space T is a homeomorphism.
-'
Pro*$ It is sufficient to prove thatf is continuous on Tor in a similar manner (from PR9) that: [ V A closed in S 3 f (A) closed in T ] .
We have the following sequence: [ PRlO ] 3 [ A closed in S 3 A compact ] [ PR12 ] [ f(A) in Haussdorff T 3 f(A) compact ] [ P R l l ] 3 [ f ( A ) closedinT].
D
An open covering (U,IiEr of a topological space S is called a refinement of a covering {Cljd if for every Ur there is (at lest) an open ?$ such that Il,c 4.
D
An open covering (U,) of S is called locally finite if each point XES has a neighborhood Vxwhich intersects only a finite number of U,:
#{u, 1 ~ n y # 0 ) < ~ . D
A topologicaI space S is called paracompact if (i) S is Haussdorff, every open covering (CT,) of S has a locally finite refinement { V,}. (ii)
D
A topological space is called locally c o q a c t if each point has a neighborhood whose closure is compact.
1.7 PARTITION OF UNITY
The partition of unity notion is important in differential geometry because it allows to reduce the study of global problems to local problems as seen later in the integration context. Let S be a topological space.
D
The support of a real-valued function g:S+R is the closure of the set of points XES such as g(x)s 0.
We denote suppg
D
=
c ~ { x € S( g(x)zO).
Apartition of unity on S is a family {g,}of continuous functions (even of class CQ) g, : S + R, : X I + g , ( x ) such that: (i) {supp gi) is a locally finite covering of S , (ii) VXES: c g i ( x ) = l .
Lecture 0
8
Remark. Since {supp gi) is locally finite then every point X E S has a neighborhood intersecting only one finite number of supp g,. Therefore the previous sum is well defined because it can hold only a finite number of terms for each x, 1
D
A partition of unity { g i ) is subordinate to an open covering (U,) of S if for every g, there is at least an open U, such that supp g, c U/ . In other words: if {supp g j } is a refinement of { Cr,) .
OnIy the required topics of topology have been introduced. Every complementary notion and proof will be found in the numerous books related to topology, for example in "Topologie" (G. C hoquet), "General topology (J. Kelley). "
2. DJTFERENTIAL CALCULUS IN BANACH SPACES 2.1 BANACEI SPACE
2.1.1 Norm and normed vector space
Let E be a real vector space. D
A norm on E is a mapping
I I :EHR+:xH
1x1
such that V x c E : Rx(l=O a x = O Q X ~ ~ EIIx+yl E : I lxl+lyl Q X EE , V ~ R E : IkxI = lkl IxII
D
Two norms on a vector space E are equivalent if they induce the same topology on E.
D
A normed vector space is a vector space supplied with a norm.
PR14 A normed vector space is a metric (therefore topological) space
Proof: A norm on a normed vector space E automatically defines a distance. The mapping E x E + R + : (xy)~+Ix-yI is evidently a distance because: d(x,y)= Y x-y # = I(-I)&-x) i = I( y-x X = d@,x) d(x,y)= Ix-yY
=O
x-y=O
C- x = y
d(x,z)= I x-z tl = I (x-y) + (y-z)1 < d(x,y)+ d(y,z).
2.1.2 Banach space
Let (x,) be a sequence of points of a metric space E. Furthermore every supp g, will be assumed compact.
Topology and Differential Calculus Requirements The sequence ( x , ) is called a Cauchy sequence if V E > 0 , 3 v E N : [ Vp,q>v : I xp-xq I< E ] .
D
PR15 Every convergent sequence is a Cauchy sequence. Proof: If l represents the limit of the sequence ( x , ), we have:
The converse is not necessarily true! If it is true, we define: D
A metric space E is called complete if every Cauchy sequence converges.
D
A Banach space is a complete normed vector space (complete for the induced metric).
Let E and F be normed (real) vector spaces,
LfE; F) be the (normed vector) space of all continuous linear mappings from E into F. The following proposition can be proved:
PR16 If F is a Banach space, then L(E;F)is a Banach. 2.1.3
Isomorphism of normed vector spaces Let E and F be normed vector spaces.
D
A mapping f :E -+ F is an isomorphism if: (i)
f is a continuous linear mapping,
(ii) there is a continuous linear mapping g : F +E such that
gof=idE
and
fog=idF.
The requirements in the isomorphism definition imply f is a bijection of E onto F (g is the inverse). The bijection g is also linear. However, take care: a continuous linear bijection f does not imply that the inverse linear bijection g is continuous! This last remark leads us to introduce an equivalent definition of isomorphism between normed vector spaces (the following definition specifying the continuity of inverse mapping).
D
A mapping f of E onto F is an isomrphism if it is a linear homeomorphism (between topological spaces).
The reader will demonstrate the following Banach theorem: PR17 Every continuous linear bijection between Banach spaces f : E isomorphism.
+F
is an
Lecture 0
10
We specify this proposition means f
-' is continuous.
We remark that:
PR18 If E is a finite-dimensional normed vector space, then every linear mapping of the (Banach) space E into a norrned vector space F is continuous. Lastly, PR19 The set of isomorphisms between two Banach spaces E and F, denoted Isom(E;F), is an open subset of L(E;F).
2.2 DIFFERENTIAL CALCULUS IN BANACH SPACES Let E, F be Banach spaces, U be a non-empty open subset of E. 2.2.1 Tangent mapping
Let f and g be two continuous mappings of U into F.
D
The mappings f and g are tangent at no E U if lim
[If (XI- g w l l -0
o
-
111-sl
PR20 The notion of tangent mappings at a point defines an equivalence relation. Proof The two first properties of an equivalence are immediately verified. Prove that two mappings f, g tangent to a third h, at xo, are tangent at this point. Seeing that
lim llf
(XI - Mx)l[ =
Il+ -xoll
and
lim
llg(x) - 4 x ) ) l = o ,
o
lix - xo
1I
then the equality
If
(XI-
ntx)ll < -
Ik- xo l
Ilf (x) - W X ) ~+ C(X)- g(x]I llx - I Ilx - xoll XO
implies the third equivalence property. 2.2.2.
Differentiable mapping at a point
D
A mapping f : U ( c E) -+ F : x Hf (x) is differentiable at point xo of U if there is a continuous linear mapping I : u + F : x ~ +e ( ~ ) such that the mapping U ( C E ) + F : X H f(x,) + P ( x - x , ) is tangent to f a t xo.
Topology and Differential Calculus Requirements
Let us introduce a definition that we are going to show to be equivalent to the previous. Let x = x o + h . A mapping f : U ( c E) + F : x wJ(X)is differentiable at xo if
D
there is a continuous linear mapping called dflerential of f at xo
e : U(C E) -+F : h H~ ( h ) such that
B
J(xo+h)-j(xo)= C(h)+ h ]I 17 (h) lim q(h)= 0 h+O
(h = x-xo)
The differential o f f at xo is denoted:
dl,: U + F : h w df,(h)=P(h) with df, EL(E;F).
PR21 The two previous definitions are equivalent. Proof: If, hypothetically, the mapping
U -+ F : x Hf(XO) + [(x-XO) is tangent to f at xo,we necessarily have: lim Ilf~x)-ftxo)-c(x-xo)# -0 Xo llx - X O
1
=>
~ ~ >=-a(x-xo) f i )-t I h I ~ ( h ) lim q(h)= 0 . h-30
Reciprocally, if the mappings ! and q are such that: Ax~+h) -AXO) = [(h)+
lim ~ ( h=)o h 4
1 h 1 v (h)
then the mapping g : u -+F : x H g ( ~=8xo) ) + ~(x-x,)
is tangent to f at xo because lim
In conclusion, we can define: D
*
A mapping f : U ( c E) -,F : x t+ f (x) is differentiabIe at xo i f there is a continuous linear mapping called dflerentid o f f at xo:
df, : U ( cE) -,F : h H dfxo(h) i.e. such that :
Lecture 0
In an equivalent manner, we can express: D
A mapping is differenfiubfeat xo if there is a mapping dfxo called dwrerential of f at xo such that
where
Remark 1. It must be specified that the previous definition can be written: IIf(*o+h)-f(x0)-hft(xo)(l=o(lhl) where f '(xo)EL(E;F)is the den'vutiw o f f at xo, also denoted Dflx, ) . Remark 2. If E and Fare R, the definition of a function R rediscovered:
+ R differentiable at a point
is
f(x0 +A)-f(x0) = hCf'(xo)+rl(h)l = df, (4+ h rl(h). Remark 3. If E is R and F is a Banach space, then we find again the differentiable aspect of vector valued function at point G:
7
7(x0 + h) - j ( x o ) = h + &)I = d j ( r o) + h ;(h) where
is the derivative of
7: R -+ F
2.2.3 Differentiable mapping D
A mapping f : U(c E) + F : x I+ f ( x ) is dvjerentiabte on U if it is differentiable at each point of U.
D
A mapping f : U ( cE) +F : x H f ( x ) is digererttrhble on U if
there is a (linear) mapping called dzflerenlYtz1 of f in U df :U(c E) - + L ( E ; F ) : x H ~ ~ , .
Topology and Differential Calculus Requirements Differentiable composite mapping theorem
Let E, F, G be Banach spaces, U be an open of E, V be an open of F.
PR22 If f : U ( cE) +F is differentiable at xo€ U, g : V ( c F') + G is differentiable at f ( x ~ ) EV, then g of : f '(v) (c U) + G is differentiable at xo and
2.2.4
(?diffeomorphism
(q21)
Recall that the differential off at xo,namely dfxOE L(E;F), is related to a point.
Iff is differentiable at each X D
The den'varive of the mapping f : U -+F is the mapping
U(c E) +L(E;F): x
f' :
also denoted Df. D
U,we recall the following definition.
~ E
H
f' ( x )
'
A mapping f of U into F is continuously diflerentiable or of class C' if: it is differentiable on Uand if its derivative mapping f ' is continuous namely: f E cO(U;L(E, F)).
In an equivalent manner:
D
A mapping f of U into F is of class C' if the differential o f f is continuous on U.
Notation. The set of mappings U +F of class C' is denoted
c'(u;F). Let U be an open set of E, V be an open of F. D
A mapping f : U V is a C'diffeom~~hisrn if f is a bijection U -+ F of class C' (9 (ii) f : V +E is of class c'.
-'
We can remark that: FR23 A homeomorphism fof U onto V of class C' is not necessarily a C' diffeomorphism because f is not necessarily of class c'.
-'
Evidently, the images underf ' andf do not belong to the same space.
Lecture 0
ProoJ: Let the following mapping be ~ : R + R : X H X ~
which is a homeomorphism of class c'.However, the inverse mapping g:X H x ' ' ~ is not differentiable at the origin because we obtain the following absurd result from the composite mapping theorem; 1 = ( g f )'(O) = g'(f (0)) f'(0) = g'(0) f '(0) = 0 what implies the nonexistence of g'(0).
D
A mapping of class C' f : ycE)t, V ( c F ) is a local diffeonwrphism at xo if f ' ( x , )
E
lsom(E;F )
A mapping is a local difeomorphism on U if it is a local chffeomorphism at each point of C!
PR24 The homeomorphism of class C' f : U ( cE) + V ( cF )
is a C' diffeomorphsrn if f is a local diffeomorphism at each point of U. D
A cbfferentiable mapping, f: U(cE)+F is twice d~ferentiableatpoint xo of U if its derivative f ' is differentiable at point xo. The second derivative atpoint xo, denoted f "(x,) , is the derivative of f ' at xo.
Let us specify that
f Yx,
) E L(E;L(E;F ) )
D
A mapping of U into F is twice differentiable on U if it is twice differentiable at each point of U In an equivalent manner: if f' is differentiable on U.
D
The second derivative mapping of f is: f " : U(c E ) + L(E;L(E;F)) . A mapping f of U into F is of class @ if it is twice differentiable on U and if its second derivative f a is continuous on U, or if the derivative f ' is of class C'on U, or if the differential df is of class c'.
Topology and Differential Calculus Requirements
The previous definitions will be extended to higher orders, for instance:
D
A mapping f o f U into F is of class C? if it is q times differentiable on U and if its derivative of order q:
f 'q' : U + Lq(E;F ) = L(E;L(E;L(E; ...L(E;F )...))) is continuous on U.' In an equivalent manner: if the differential df is of class P-' . We can generalize PR22:
PR25 I f f : U +E and g : V + F are of class @, then g 0 f is of class @. Notation. The set of mappings of class CQ on U is denoted @(U;F) .
Let U be an open of E, V be an open of F.
D
A mapping f of U onto F is a t? difleomorphism if (i)
(ii) 2.2.5
f is a bijection U+ F of class C?, f -': V -+ E is of class P.
Inverse mapping and implicit function theorems
Inverse mapping theorem Let U be an open of E, V be an open of F. PR26 If a mapping f : U + V of class C' is a local diffeomorphisrn at point xo [ f J ( r o )E I S O ~ ( FE ); ] , then f is a C' diffeomorphism of some neighborhood of xo onto some neighborhood of JTxo). L2(E;F) [resp. L4(E;F>] denotes the space of continuous bilinear [resp. multilinear] mappings from ExE [resp. . . . x E (q copies)] into F: [resp. L,(EJ) = L(E,.. . JJ)]. &(E;F)= L(E,E;F)
Ex
We leave to the reader that there is a natural isomorphism: L(E;LqE;F)) = L,+l(E;F). The reader will easily prove that the following mappings (derivatives of order q+l):
g€L(E;Lq(E;F)) and
2 E Lq+,(E;F ) are isomorphic
defined by
,..., e , , = g(e,+,I (,,..,,e q )
Lecture 0
16
With the same notations, assuming f is not only of class C' but of class C4 (q>l), the previous theorem becomes: PR27 Iff is of class ?r onto v,,,, .
(q>l),then the restriction o f f to V, is a C? diffeomorphism of V,
Remark. Let us insist on the "local" character of the previous theorem. For example, in polar coordinates, the everywhere local isomorphism
f: R, x R -+ R ~ - ( o: )(r,B)I+ (rcos8,r sine) is not even injective (one-to-one). Implicit function theorem PR28 Let E,F,G be Banach spaces, VE,VF,VG be opens of respectively each space, f : VEXVF+G : (x$) H X X ~ be ) a mapping of class c', (xoj0)be a point of YExYF.
Iffixo,yo)=O and d,f(x,,,y,)~ Isom(F;G), then there is an open neighborhood of xo, namely, Vxoc E, an open neighborhood of yo, namely V, c V, and a mapping of class c1 g : V,(cE)+F such that [ X E < ~ , Y E V :~f ( x , y ) = O I [ x ~ : vy =~g ( x~ ) l .
-
In other words, this proposition means that, in a neighborhood of (xofi), the solutions of the equation f ( x j ) = 0 are given by y = g (x) where g is of class C' on V,.
Remark that f(xo~lo)=O and it is immediately proved that:
2.2.6
g(xo)=yo
Tangent mapping
Let U be an open subset of a vector space E.
D
The tangent of f : U ( c E) + F is the mapping Tf : U x E
+ F x F : ( x , e ) ~Tf(x,e)
such that Tf(x,e>= ( f ( x X f '(x1.e)
where the second element of the pair is f'(x) (linearly) appIied to e G E.
Topology and Differential Calculus Requirements We can view Tf (x,e) as the image of a vector with its origin. The differential composite mapping theorem is written: (g O f Y(x1.e = g'(f (x)).(f'(x).e) .
We say that T is a covariant functor. Remark. The reader will see that T(g 0 f )
= Tg
0
Tf.
2.2.7
immersioni and submersion
D
at x is a mapping f : U ( c E) + F of class C? such that f ' ( x ) is injective (one-to-one). A submersion at x is a mapping f : U(c E ) + F of class C? such that f l ( x ) is surjective.
2.3
DIFFERENTIATION OF
An i-mion
R" INTO BANACH
Let F be a Banach space, U be an open of R n , xo be a point of U, f be a hfferentiable mapping U(cR n ) +F : x I+ f (x) The space R", with its vector structure, is a Banach space with the norm
D
The differential o f f at x,
IS the linear mapping
dfxo : R n- , F : h e d J X D ( h ) such that
Make explicit this mapping and first remember that f(x) and df,(h) are vectors of Banach F. Let (e ... ,en) be the canonical basis of R".
,,
Considering the special vector o f R"; h' e,
also denoted Any linear mapping of a finitedimensional real normed vector space (Banach) into a normed vector space is continuous. The canonical basis vectors are such that Vj :e, = (0,..,1,..,0) where the#h entry is unity md the remaining are zeros.
Lecture 0
18
we have:
(h',... ,o), f(xi
2 + h l , x,,...,
x i ) - f(xk,x; ,.,.,x:)
I I/
= h1df, (el + lh' JIe1 tl(hLel).
But the following lim P7(h1el )= 0
~'HO
implies
Consequently, at x,, f possesses a partial derivative with respect to xl:
and generally
af
d S , ( e , ) = -(x,). dxJ
Let
The differential of J; at xo,is the (linear) mapping
We denote the image of h:
More especially, if we choose R instead of F, we can express: PR29 Every projection p J : R" + R : x H p ( x ) = x J is differentiable and f
Pro$
The obvious equality J
p ( x + h)=x' + h j
implies P'(X + h ) - p J ( x )= h f . There is a differential (linear mapping) of
because we do have
pj
at x:
Topology and Differential Calculus Requirements
From this proposition, eq. (0-2) is written:
Remark. It is obvious that p = dpi . J
We finally give the well-known expression of the differential of f a t xo. Instead of f , we choose the n successive "projectors":
pi : R n+R :
( X I , ..., X " ) H
j = ~..., , n.
xi
The formula (0-2) is successively written in these special cases: dx' (h) = h'
Hence, the differential o f f at XO is
In conclusion
Remark. In particular, we specify that df% (e,) =
"
af
j=1
-+x, ax
DIFFERENTIATION OF
R"
af
) d x (e, ) = -(xo) ax1 J
seeing that
2.4
INTO R
Let xobe a point of R".
2.4.1 Directional derivative Let u be a unit vector of R" f be a mapping of R" into R defined on a neighborhood of xo.
Let us consider the function
20
Lecture 0
D
The directional derivative of f at x, in the direction of u is the limit (if it exists):
D,f ( x , ) = lim f(x0 + m - f ( x 0 ) h
h+O
In particular, the partial derivative with respect to x i is the directional derivative in the direction of the vector el of the canomcal basis (e,,... ,en):
af
% f ( x , ) = a i f ( x o )= -(x,) ax'
2.4.2 Theorem of differemtiation
PR30 If a mapping f of R" into R is differentiable at xo, then f possesses at this point a partial derivative in every direction and
D"f (4,) = df,
(u)
.
The converse is false as it is proved by the function f :R
+R
defined by
and
PR31 A mapping of R" into R, having at a point derivatives in all directions, is not necessarily differentiable at this point(and not even continuous). A fortiori: PR32 A mapping of R" into R, having at a point partial derivatives is not necessarily continuous and differentiable at th~spoint.
On the other hand: PR33 A mapping of R" into R, defined on a neighborhood of a point of R" and having all continuous partial derivatives in this neighborhood, is differentiable at this point. It is called of class C' or continuously dlflerenriable.
2.4.3 Linear differential forms on R" First let us remember the linear form notion.
D
A linear form is a linear function of a vector space (here R") into R: f:R"-+R.
A function defined on an open U of R" is called linearform if it coincides with the restriction of a linear form (on R")to U, Let us establish its expression in a basis. Given the canonical basis (e,) of LT a vector of R" is denoted ' 1 The Einstein summation convention, i.e, summation is implied when an index is repeated on upper and lower levels, generaIly wifI be used.
Topology and Differential Calculus Requirements
The image of h under f is
and putting = JTei),
it is
f ( h )= S,h i . The dual basis (e" ) of the ordered basis (e,), such that eWi(e ) = 6'. J
J
allows to write e"(k) = hi
f ( h )= f , e*' (h). Consequently, PR34 The linear form f i n the basis (e") is written:
Remember that the space of linear forms on a vector space E is a vector space called dual space of E. It is denoted E' . Now we introduce the notion of diflerentid orte$orm (or linear dflerential form) on R" or on an open U of R". Let (ei)be the canonical basis of R", i h = h e, be a vector of R*, w, : R n + R bealinearformatxo~U.
We have previously mentioned the differential of the projection p' at x was such that: dpl 5 dx' . The function x I+ is identified to the image xi of point x.
>
Problem. Give the expression of the linear form w, in the dual basis (dx') of the basis (ei) which is such that: dxf(ej= ) 6;,. Answer. The image of h under w, is
Lecture 0 = o,(x,,)h
[ putting the real wi(x, ) = w, (el ) 1.
i
From
dx i ( h )= hi it follows:
o, (A) = m, (xo ) dxl(h). A difSentia1 vne-fonn on U is a mapping w which links each point X E U with a linear form w, defined on F,that is:
D
w : U ( c R n ) + (R")' : x HW ,
In other words and more explicitly: D
* A digerenth1one-form on U is a mapping w : U ( c R n ) 4L(Rn;R ) = (Rn)' : x I+ w,
such that
o,:Rn+R:hw~,(x)dx'(h). The real numbers w,( x ) are calIed components or coordinates of the differential one form o relative to the basis (dx').
Notation. A differential one-form on U is denoted w = W , dx' . PR35 The differential of a real function at any point x is an example of differential one-form. Proof. Let us consider the real-valued function
f:U(cRn)+R. The differential off at any point x df*:Rn+R has the following recalled expression:
d f , = di f ( x ) dx' where
dif (4= d f . ( e ,1. Remark. A differential one-form is not necessarily the differential of a differentiable function. But, we recall:
D
A differential one-form is met if it is the differential of a differentiable function.
2.5
DIFFERENTIATION OF
R"
INTO
fl
Let U be an open of R", xo be a point of U, h = (h',..., h n ) be a point of U, f : U(cR") + R m : x Hf (x) be a differentiable mapping at no.
Topology and Differential Calculus Requirements 2.5.1
Differential and Jacobian matrix
Let US consider the differential of f at xo d S , :U(cR n ) + R m : h t + d f J h )
such that
f (43 + h )- f ( x o ) = d / , (4+ llhli tt(h) limq(h) = 0 0
It is obvious that the previous equality between vectors of R'",which defines the differential dfxo,is equivalent to rn equalities ktween the components of these vectors relative to the canonical basis of R",namely:
The ith component of the differential df, is
The differential d/L of the ith component f i (- pi 0 f : R" + R) of the mapping f is: dLo = d(p'
O f
), .
We know (see section 2.4) this differential is
dfio: R" -+ R : h n dXo(h)= d, f '(xo)hJ
Let us use the obvious equality
(df,)'
= df; .
So, the differential d f , : R n + R m is defined by its rn components: (dLo)' : R" -+ R : h H (dLo)'(h) = dfj0 (h) = d, f'(x0)hJ
(this last notation indicates a summation on j). The differential d f , is represented by matrices, such that
D
The m by n matrix representing the differential df, is called Jacobian mat& of f at point xo.
It is indiscn'minateIy denoted:
Lecture 0
By introducing again the n linear mappings dp! denoted dxJ:Rn-+R:h~hJ, we can denote the differential o f f at xo by
(df,)' dfxo=
f
)j
.
.
dx'
[idLo [iIfrn.. i3,JjX[ixj =
.
In conclusion, the differential of f at xo is the linear mapping of lp" into Ily" defined by the following Jacobian matrix
Remark. It could be written Vi = l,..., rn : ( j = 1, ..., n).
2.5.2 Image (in
r)of a basis vector (of X") under dfxa
Let xo be a point of U(cR n ), (el,... ,en) be the canonical basis of R", (u,,... ,urn)be the canonical basis of R"s. PR36 The differential df, is such that V j = 1,. . . ,n :
ProoJ: We have:
where 1 is thejth element; the proposition is so proved. 2.5.3 Theorems Let x, be a point of R"
Topology and Differential Calculus Requirements
25
PR37 If a mapping of R" into
defined on a neighborhood of xo has continuous partial derivatives on this neighborhood, then it is differentiable at XO. (It is called of class c').
PR38 A mapping f : U(cR")+ R m is of class P at X,,E U fll the functions that definef have 9th-order partial derivatives (with respect to n real variables x') continuous on a neighborhood of x,. 2
2.5.4 Diffeomorphism and Jacobian Previous definitions can be used again. So, let U,V be opens of R"
D
A C'd~fleornorphismof U onto V is a bijection of U onto V such that f and f of class C'.
-' are
The reader will easily generalize to the C? diffeomorphism notion. D
A mapping f : U ( c R") + R" of class C' is a local tirsrfeonwrphism at point xoof U if dfxDis an isomorphism. It is a local difleomorphism on U if it is a local diffeomorphism at each point of U.
Remark. In the context of finite-dimensional spaces, the isomorphism notion implies the same dimension for corresponding spaces,
D
The rank of a differential at a point is the rank of the Jacobian matrix defining this mapping.
D
The rank of a differentiable mapping at a point is the rank of its differential at this point.
D
The Jacobian of a differentiable mapping of R" into R" at a point is the determinant of the Jacobian matrix at this point.
It is indiscriminately denoted
or simply J. PR39 In order that a mapping f of class C' between opens of R" be a C' diffeomorphism it is necessary the Jacobian of f be different of zero at every point.
This necessary condition is evidently not sufficient!
Indeed, by giving
' Iff means "if and only if"
Every partial derivative of lower or equal order than q is defined and continuous at xo.
Lecture 0
then the following mapping D,
-+D, : ( x , ~ ) H ( x= xZ - Y 2 , y
=2xy)
is such that
and J = 4(x2 -+ y 2 ) > 0.
The previous mapping is of class C' and of rank 2. It is nevertheless not a diffeomorphism because it is not an injection (not one-to-one) and thus not a bijection. On the other hand, we can say: PR40 If f is a bijection of class C' between opens of R"
(f-'
exists), then the condition of nonzero Jacobian is necessary and sufficient in order that f be a C' diffeomorphism between opens of R".
In a similar manner: PR41 A mapping f : U ( c R n )+ R n of class C' is a local diffeomorphism at xo of U ~ f l the Jacobian off at xo is different from zero.
2.5.5
Inverse mapping theorem
The reader will refer to the inverse mapping theorem in the case where E=R" and F=R" [f ' ( x o ) E Isom(E;F ) implies finite dimension of spaces having the same dimension]. The mapping
f : U(cR n ) + R" is defined by n functions of n real variables:
f '(XI,...,x n )
l l i s n
of class C' on open U.
The mapping f'(xo) E L(Rn;R n ) is so defined by the Jacobian matrix at point x,:
In the local inversion theorem, the assumption
f ' ( x o )E Isom(R";R n ) is equivalent to the following condition for the Jacobian:
Topology and Differential Calculus Requirements
PR42 If
f : U ( c R " ) -+ R n is a mapping of class C' defined by n functions f 'of real
variables x' and if the Iacobian
D(fi
D(xi )
is nonzero at xo of U,
then there is an open neighborhood Vxoof xo, and an open neighborhood Vf(,, of f (x,) such that f is a C1diffeomorphism of Vxoonto V',,,. The inverse mapping f -'is then defined by n functions (f + I ) ' of class C' on V,,,,
&"
and the differential of f-' is defined by the inverse of the Jacobian matrix representing df , i.e. (4.f -l),c,, = (dLo1-I . (0-8)
2-56 Implicit function theorem
The implicit function theorem wiJl be expressed under the assumpon of finitedimensional spaces: E = R ", F = G = R * (the two last spaces having the same dimension). Let VE be an open of R", VF be an open of Rm.
PR43 If the mapping f : V, x V, + R m is defined by rn functions f ' of class C' of n real variables, 1 if the point (x,, ..., x,";~;,...,y,") of V, x V, verifies the system of m following equations f !(XI ,..., x ~..., y"); = 0 ,~ ~ ~ if the lacobian
D(fl) is different fiom zero at point
(x;
,..., x,";y : ,..., yo"),
D(Y' )
then, in a neighborhood of ( x i ,..., x,") and in a neighborhood of system of equations f i ( ~ ,..., l x n ; $ ,..., y r n )= 0 is equivalent to the system ( 1 1i 4 m ) yf = g i ( x ' , . . . , ~ n ) where the various g' are of class C' on a neighborhood of ( x : ,. .., x,").
(yA ,. .., y,"), the
2.5.7 Differentiable composite mapping theorem
PR44 If f : U(cR") + Rm: x H f ( x ) is a differentiable mapping at point x,, if g : f ( U x c R")+ Rr : y H g(y) is a differentiable mapping at point f( x , ) , then the mapping g 0 f of R ninto Rr is differentiable at xo and O
f 1,
= dg,,,,
o
dfxo.
(0-9)
Lecture 0
2.5.8 Constant rank theorem
Let R be an open of R n .
PR45 If f : n(cR")+ R m is a mapping of class @ (q2l), if, for every xo of a,the rank o f f at point nois r Iinqn, m), then there are: an open neighborhood U, (cQ) of xo,
,
an open neighborhood Vf,( of f (x, ) E R m, an open neighborhood Uo of O ( c R"), an open neighborhood Vo of O ( c P), a C? di ffeomorphlsm 4 : Uxo+ U ,, a C? diffeomorphism y/ : V,,%, + Vo such that: y 0 fa#-' : uo+ v, : (x',..., X n ) H( X I ,.., X',O ,...,0). 2.5.9 Immersion - Submersion Let U be an open of R", q f : U(cR n ) + R m a mapping of class C
D
A mapping f of class C? is an immersion at a point x of U if its differential df, is injective.
This requires n 5 m. D
A mapping f of class C? is a submersion at a point x of U if its differential df, is
surjective. This requires n 2m.
D
Imntersion and submersion are defined if the two previous definitions are valid at every point of
PR46 A mapping f : U ( c R n ) -+ R m of class C? is an ittunemion at x rff the rank of its Jacobian matrix is n. It is a submersion at x ~ f fthe rank of its Jacobian matrix at x is m.
PR47 If f : U ( c R n ) + R m of class C? is an immersion at point x of U, then there are R" c U , an open Vx of R" x Rm-"containing x such that an open Vf(,, of R" containing f (x),
~,fl
a C? diffeomorphism g : V, -+ Vf,(
having the same restriction to V,
n
R" as$
Topology and Differential Calculus Requirements Example. n=l,m=2
Figure 2.
PR48 If f : U ( c R")+R m of class C? is a submersion at point x of U, then there are an open U, containing x, an open Uf,,, of R m x Rn-"containing f ( x ) and a @ diffeomorphism g : U, + U,,,,such that the restriction of f to U,is II o g where ll denotes the canonical projection of R m x Rn-"onto R". Example. n=2,m=1
Figure 3.
In conclusion, we have introduced the essential and necessary requirements of elementary differential calculus for the understanding of the following lessons. Proofs and other theorems of classical analysis will be found among an abundant literature. Standard possible books are "Foundations of modem analysis" (J. DieudonnQ),"Calcul diffkrentiel" (H. Cartan).
Lecture 0
3. EXERCISES Exercise 1. Find the differential of the linear mapping
f :Rn+Rm:xt+Ax. What is the differential in the case of f is the mapping i d : R n + R n : x ~ x?
Answer. The condition of "d&erenttability " at point xo of R" is A X - Ax, = d f x o ( x - x , ) + ~ ~ x - x , ( l q ( x - x o )
limq(x - x , ) = 0 x,
Putting h=x-xo
and, since A is linear, we obtain: Ah -
( h ) = llhll v(h)
and df, ( h )= Ah.
The mapping f being linear, then df is necessary f. '
In the particular case of identity, we have:
A - I and n = m , thus the differential is the identity.
Exercise 2. Find the differential of the mapping
at point xo. R
Answer The differential dfxoof any vector y =
y'ei
of Rr is the following real:
I-1
@ x o ( x ~ ' e i Cyidf,(ei) )= = I
In conclusion,
1
CY'~ax?,f( x . 1 1
Topology and Differential Calculus Requirements
~ : R : + R : ~ H E ( ~ ~ ) . llxll
Exercise 3.
Find the differential mapping of
f :R~ + R ~: ( r , B ) ~ ( x = r c o s 8 , y = r s i n d ) with ~ E R, 8: ~ [ 0 , 2 r [ . Answer. We have:
The differential
df : R~ + R~ : (dr,dt?)H (dx,d y )
is defined by
Exercise 4.
Find the differential mapping of
f :R2 + R 2 : ( x , y ) ~ ( =Xx + y , Y
=x-y).
Answer. If ul and uz designate the vectors of canonical basis, we have:
The differential
Lecture 0
d f : R'
+ R~: ( d x , d y ) I-+
(dX,dY)
is defined by
Exercise 5. Find, if there is one, the differential of
where GL(n,R) is the general linear group of all non-singular nxn matrices with real elements.
Answer. The definition off at A, is such that
I A'
- 4'- df4 ( A- A,
1 = o([\A- A,II) .
Putting A- A, =B, we have:
I/(Ao
+ B)' - 4'- df4
(B)I
=o(ll~ll)
By analogy with
1 (a,+ P)-' = --
ao+P
1
P
1 "
--C
a , ( l + -P ) = a0
0
(-I)"(-)" a 0
a 0
under conditions P/a, < 1 and azO , we deduce what follows: ( A , + B)-I
91
=
[C(-l)n(~-l~)"]&-' = [I-&-I-'B+(&-~B)~ -.....14-] 0
with
which describes a neighborhood of A,. The definition of differential entails
dfA,( B ) = 4-'BA,' because the reader will immediately prove that:
I[(A;'B)' A;'
- (A;' B)' 4' + ...
. I1 = o(ll~1l).
Topology and Differential Calculus Requirements
Moreover, we have:
Exercise 6. Find the differential of the mapping Answer. The mapping det is not linear and then the previous classical process won't get
anywhere. Let us use the formula (0-1)
that is here d dx,
dfi(e,) = -(det
d axg
A) = - ( x x i k
,
A,)
(minor). If we have that establishes the differential of mapping f = det. Exercise 7. Given
verify the differentiationformula for g o x Answer. Since t--f3(1,t~)A(t+t~,t-f~)
we have: g o f : ~ - +: ~t ~2 (
t + t ~ , t - t ~ ) .
On the one hand, e bein the vector of the canonical basis of R and ul, u2 being the vectors of the canonical basis of then the formula (0-7) leads to
2
Lecture 0
On the other hand, we have:
and
Thus we have:
Exercise 8. Given a mapping of class C' f : R~ 3 R : ( x ' , x 2 , x 3 ) H f ( x 1 , x 2 , x 3 ) and a function g : R + R : X H f(x,-x,x),
prove that the function g is of class c2and calculate g t ( x ) , g W ( x )from 8,f,d, f,8,f
Answer. Introduce the linear ( thus C" ) following mapping
The composite mapping g =fo h o f R into R is then of class c2(the class o f f ). From the derivative composite mapping theorem, the derivative of g at x is:
In the same way, we have:
gp(x)=a,,f(x,-x7x)-2a,,f(x,-x7x)+2a,,f(x,-x,x)
'
a22f
(xY-x,x)-2a2,f
Exercise 9.
Given the following mappings
(x,-xyx)+a,f
(xy-x,x).
Topology and Differential CalcuIus Requirements
f : R' + R~ : (x,y) H (excosy,exsiny,x2 + y 2 ) g : I('
-+ R'
: (x, y,r) H ( , / x f , x y z , x ) ,
check the composition theorem for Jacobian matrices:
( ~)(*,Y,. f
( ~ ( Ogf )X,Y, = ( D ~ ) I I , , , Answer. On the one hand, we immediately have:
g o f : (x, y) H (ex,+(x2+ y2)e2"sin2y, e x cosy)
which implies
i
e"
0
( D ( 0~f))(,,, = xe2"sin 2y + (x excosy
2
ye2"sin 2y + (x 2 + y2)eZx cos 2y
+ y2)e2"sin 2y
- e x sin y
On the other hand, we successively have:
e x cosy
.]
- e x sin y
(.).x,y)
9
cosy
sin y
0
0
+e2'sin2y 0
(~'+~')e~sin~ 1
Thus we finally have: (~g)fx,,,c~fXx,Y, ex (x
2
+ yZ)eZX sin 2y + xe2"sin 2y excosy
Exercise 10.
Let us consider the following system
0 (x
2
+ y2)e2"(cos2 - e x sin y
Lecture 0
36
From the implicit function theorem show that y and z can be made explicit in function of x in the neighborhood of point (1,1,1). Calculate the first and second derivatives of y and z at x = 1. Answer. Denote f ' ( ~ ; ~ , z ) = 3+xY~2i-2z2- 6 = O f2(~;y,z)=xZ+y2+z2-3=~. ThcJacobian
D(fl)
atpoint(1,1,1),where y l = y and g = z , i s
D(Y'
The implicit function theorem tau ht us there is an open neighborhood Vl of point 1 of R and an open neighborhood V(t,1,1) of R such that:
E
(
~
7
2) E ~
3
7I.1.:1f)( x , Y , ~=) 0 I
#
[ x E V , :Y=Y(x),z=z(x)~.
By deriving the system equations, we obtain: 6x + 2 y(x) yf(x)+ 4z(x) zl(x) = 0 2x + 2y(x) yl(x) + 2 z ( x ) z'(x) = 0. At point x =I, since y(1)
= z(1) =1,
we deduce:
y'(1) = 1
z'(1) = -2 .
Once again we derive VXEVl :
Knowing that y(1)
= z(1) = 1
, y'(1) = 1, z'(1) = -2,
LECTURE
1
INTRODUCTION This lecture presents the manifold concept that allows overcoming difficuIties notably encountered in undergraduate studies of the definition of surfaces. The 2-sphere S is a particularly illustrative example.
1. Coordinates on S*
Introducing latitude A. and longitude 4 coordinates we can write:
If we consider
then the mapping : D + R~ : (a,@) I+ (x(&@),~ ( 4 4 1z(A,#)) ,
is not injective because
7r
VqJ E [0,2n[ : A(--,@) = (0,0,1) 2
(north pole) .
So, the latitude A and the longitude # are not "good global coordinates on S 2. But global coordinates do not exist on s2!
PR1 The differential of the differentiable mapping h is not of (maximum) rank 2 everywhere.
ProoJ: The following Jacobian matrix
Lecture 1
1
- cos A sin # cos R cos # 0 ?r
is of rank 2 on D except at points (- -,() and (1, #) where it is of rank 1. 2 2 At every other point there is a reversible 2x2 matrix and the determinants are respectively: (zero if R = 0) X 3~ (zero if # = - -) 2' 2 (zero if 4 = 0).
cos2 A sin # 2. Stereographic projection
Let us to overcome the pole problem due to the global parametrization of To cover J let us introduce two coordinate systems. Let us choose: (i,= 1(2
{ p E s 2 : . 7 ( ~ ) > - 1 / 2}
= { pE
s2: Z@) < 112
)
R2= ( q R ~3 : Z ( ~ ) = O } . The stereographic projection from the south pole onto the meridian plane is:
(since
Figure 4.
s'.
Manifolds The stereographic projection from the north pole onto the meridian plane is
5 2 772 (since - -= 1.
x
1-2
y
Each of the so defined parametrization systems is valid for more than half the sphere where Ul is without south pole and U2without north pole. The parametrization systems cover the sphere: UI Uu, =s2. PR2
The composite mapping 42
04;' : R 2 --)R2 :(61,11>~(6z~772)
is a differentiable bijection and its differential is of maximum rank at every point.
Proof Let the sequence be
+
x = (1 z)&
Y = (1+z)771
l-~~=x~+~~=(l+z)~(&;+~;)
a
l-z=(l+z)({;++;)
The coordinate change ( & , ? ? l ) ~ ( & , l l ) on u 1 n ~ 2 = ~ p ~ ~ ~ : - t < ~ is( p ) < ~ 1 defined by
This mapping #2 o 4;' is a differentiable bijection and its differential is of maximum rank at every point because:
Plan We are going next to provide a topological space M with a structure of manifold from the concept that (local) coordinates are attached to every element of M and that if a point of M belongs to two different coordinate systems then the coordinate change must be "agreeable." Let us thus introduce rigorously the manifold notion which generalizes the one of surface.
Lecture 1
40
1. DIFFERENTIABLE MANIFOLDS Let M be a set of elements called points, F be a finite-dimensional real normed space. 1.1
CHART AND LOCAL COORDINATES
Provide M with a topology by considering that every point of M belongs at least to an open U, of M (covering!). 1.1.1
Chart
D
A Iocal chart on M is the pair (&p) consisting of:
- anopenUofM, - a homeomorphism p of G onto an open subset p(U,) of F. The open U is called domain of the chart I .
An arbitrary point of M can belong to two distinct opens, for instance Cr, and Uk..The corresponding distinct charts are (U,,qj)and (Un,~lk).
The homeomorphisms gr, and a being different, we link the opens qz(l/,) and yln(Uk) of F by introducing the following definition. Let us denote pJ;u, fluk the restriction of pi1to the open pj(U,
Figure 5 .
nU,)
of F.
&F
Afterwards, the space F will be only 12". So to each point M is associated a chart (U,p) such that p(U) is an open of R". The term "local" will generally be omitted.
Manifolds D
41
Two charts (U,,g3) and (Uk, ~ ] k )on M, such that U j nU, # 0, are called C-compatible ( q 2 l ) if the overIap mapping pbj = pk "
GJnu,
is a d diffeomorphism between the opens pj (U,nU,) and p, (U, nU, ) of R".
'
1.1.2 Local coordinates
D
The local coordinates xi of a point p belonging to the domain U o f a chart ( Q p ) of M are the coordinates of point p(p)of R:
We denote by
... ,Xn) the ordered n-tuple of real numbers linked to pointp. (XI,
Figure 6. The bijection g, assigns to any point p of U c M the system (x', ... pn). Reciprocally p" assigns to every ordered n-tuple of real numbers a point of C.! So, to coordinate lines of coordinate system.
1.2
R" are associated coordinate lines on M and a chart defines a
DIFFERENTIABLE MANIFOLD STRUCTURE
1.2.1 Atlas
D * An atlas of class i? on M is a family A of charts (U,,g)such that: ( i ) the domains I/, of charts make up a covering of M
uut (ii) any two charts (U,,p,), (q,q) of A, with U if) U, # #, are Cg-compatible. 7
t ;1
Remark. An atlas of class CQ generates an atlas of class (?such I
Remember a diffwmorphisrn of class C4 is called @ diffeomorphism.
that p
I q.
Imal
Lecture 1 Example. Atlas of sphere.
Let the unit 2-sphere be:
Consider the mapping @, stereographic projection from the north pole n onto the plane { g E R) : x 3 ( ~=) It is a bijection between s'-(n) and this plane. Similarly, the stereo aphic projection g, from the south pole s onto the previous plane is a !- { s) and the plane. bijection between S ? Because of poles the sphere cannot be covered by only one chart: no single homeomorphism y,can be used between s2and the plane. On s2,with the topology induced by the one of R~and refering to the introduction, we know l ) compatible. The 2-sphere atlas is composed of that two arbitrary charts (U,,@)and ( U 2 , ~ )are (at least) two charts.
01.
The reader will immediately generalize to the n-sphere:
He will consider two homeomorphisms: the stereographic projections from respectively the north pole and the south pole, i.e. two mappings of S n - { n ) (resp. Sn-fs)) onto the hypersurface of equation xn+'= 0.With the usual topology on S n as a subset of R"", then an atlas with two charts will be defined.
D
A chart (U,v)is compatible with the atlas
~u, D
((u, ,p,),,, 1 or is uMssible
if the union ~ ((u, ) ) ,uq,), ) is again an atlas; in other words if it is a chart of the atlas.
Two atlases of class C? are equivalent or compatible if their union is still an atlas.
1.2.2 Differentiable manifold structure
D
The maximal atlas 2 , associated with an atlas A is the atlas being composed of all (equivalent) charts compatible with A.
We say that: D
A maximal atlas on M provides M with a di/ferentiable manifold structure
In practice, a differentiable manifold structure is defined from an atlas representative of its equivalence class (all the equivalent atlases defining the same differentiable manifold structure). The definition of a differentiable manifold structure requires that: (I;) The opens of local charts cover M. (i!, Two any charts (U,,p,), (U,,e) such that U, f Ul j z 0 are C4-compatible. Make more explicit the second requirement with the aid of a change of local coordinates.
Manifolds 1.2.3
43
Change of charts
Let p be a point belonging to the intersection U, fluj of domains of distinct charts (U,,@) and (Cl,,e). I li
Figure 7.
Let us consider two local coordinate systems. The definition of an atlas of class C? means the coordinates x", ...,xfn of p with respect to a local coordinate system are functions of class C4 of coordinates x',. . of p with respect to the other system of local coordinates.
.x
In this case, we define:
D
The change of charts ( q ,and ~ )(U,,@)or local coordinate transformation of pointp is admissible if there is a C? diffeomorphism between opens of R": o v);' : R n -# R n : (xi ,..., x n ) H ,..., x'") , that is if the functions f defining the coordinate transformation (XI'
x" =f '(xl,. . .q"). . . . . x'"
=fil,. .. $1
have continuous 9th-order partial derivatives with respect to variables xk
1.3
DIFFERENTIABLE MANIFOLDS
1.3.1
Definitions
D
A diflerentiable man~ofd of class CQ is a pair consisting of a topological space and a maximal atlas:
( M ,2). We will henceforward assume the basis for the topology defined by chart domains is countable; so we will assume M is separable.
Lecture I
44
If the topological space M is assuredly of type T I ,on the other hand it is not certain it is a Haussdorff space as the following example proves (see Gkomktrie dafkrentielle, Berger and Gostiaux, ed. P.U.F). Example. Let the subset of R~ be:
E =kx,0)1 x < o ) u ~ x , o )I ~ O } U ~ X J ) /
X>O)
Let us provide E with a manifold structure from the charts (U,,p,)and (Uz,@)defined by:
Evidently we have: U, UU,
=
E.
The mappings p, and fi are homeomorph~smssuch as the following sets =
~?(U~)=R n (, u l n u 2 ) = n c u l n u 2 ) = { x ~ x < o l
are open. The mapping
P, 09;~ : v 2 ( u ,n u , ) + ~ , c u ,nu2> is the identity X H (x,O)
t-9 X .
This mapping ]- co,O [ +] - w,O [ is of class C" In conclusion, the space E is provided with a manifold structure (of class C"). However, E is not a Haussdorff space because the points (0,O)and (0,l) have no disjoint U, l , then the neighborhoods. Indeed, let U be an open of E containing (0,O). Since (0,O) E U f open q,(U flU ,) of R contains 0 and then includes I-&,.$. Therefore we necessarily have:
In the same manner, given an open U'of E containing (0,I) it is proved that
{ (x,o) 1
< 0) C u' So the opens Wand Cr are not disjoint. -'El<
J
Spaces such as E will be removed because in the future every manifold will be assumed of Haussdorff type. In practice, we define a differentiable manifold from an atlas on M We say:
'
D
w A difJ;entiable manifold is a pair consisting of a Haussdorff space with countable
basis and an atlas. A differentiable manifold is an n-manifufd if for every point of space there exists an admissible local chart (U,q) such that (U,p)c R".
' Unless 1hear to the contrary, the differentiable manifolds will be of class C:
Manifolds Notation. Its dimension being n, the manifold (M, ,A) will be denoted M, or simply M Remark. A manifold is such as the topological space is locally compact and locally connected (see solved problem). 1.3.2
Product manifold
Let M, (or simply M) be a manifold of class C4 defined by an atlas
Let N, (or simply N ) be a manifold of class C? defined by an atlas ~ = ( ( v j , w , ) ~l
D
E J } .
A"
The product atlas A x is ( ( ~ i x ~ j , ~ l x x( p~ j , )~l ) E I x J ) where
ui xvj = ( ( ~ / , y ~~~ ul ~€vj} , y ~ 9, x W j : U,x v, + R" x R" : ( x , y ) I-+ (P,(~),Y,(Y)) Remark. The structure of manifold MxN foIlows from the structures of manifolds M and h! Proof: If we denote
then the following set
w,= u,x vj
w;= u ; x v ; ,
w,n q = (uin u : ) x ( v j nv;)
is an open of MxN. If we define R, such as P&(x>Y) =(P~(X),Y~(Y)) and analogically for pi, then the (Tdiffeomorphism between opens of R"'" is:
v;
O
pi' = (p,!0 p,:') x (y;0 y ; l ) .
Indeed (P; 04$)(x2y) = Q:J(~ ; l ( x ) , y ~ ' ( y )
v;
= ( P,'(P;'(~))~ (V;'{Y)))
= ( (P;O P ; ' ) ( ~ ) ~ ( VO ;v;')(Y) , 1
Consequently, we can naturally say:
D
The produd manifold M,xN, product atlas of M, and N,.
of two manifolds is the manifold defined from the
Its dimension is the sum of dimensions of each manifold.
Lecture 1
46
1.3.3 Examples of manifolds I. Space Rn is a manifold such that the atlas {@",id)) is made up of only one chart (R",id).
'
'.
2. Circle S Looking at S as a subset of R ~then ,
S'
=
{ ( x 1 , x 2 )Jt2 ~ : ( x ' ) ~+ ( x 2 1 2 = 1)
is provided with the induced topology.
The circle is evidently not homeomorphic to R. Therefore, S' cannot be covered by only one chart. On the other hand, let us check an atlas of S' defined by two charts (U1,ylland ) (U2,qa). Let these following charts be:
*J- (
u, = { ( X 1 , X 2 )E S' : X'
< 1)
1
pl : U,(CS ) +]0,2x[:(xl = cos0,x2 = sine) I+ B
u, = ( ( X 1 , X 2 )E S'
: X I >-I}
p2: U ~ ( C S ' ) - + I - x , a [ : ( x l = c o s 8 , x 2 =sin#)t+O
These charts evidently cover S Figure 8
Considering
uinti,= si- ((l,o),(-1,o)) v;' is a diffeomorphlsm between the opens
let us prove that the mapping pl and q,(U, n u 2 ) of R.
Figure 9.
0
n
qz, (U, U , )
Manifolds If then we have:
0 E 1 0 , ~[
~f we have:
e €1-Z, o [ q,(q;18)= pl (cos B, sin 8) = e + 2~ .
The last equality is obvious. The reader will put it in concrete form taking for instance
In conclusion, p, 0 pi1 is really a diffeorno~hismbetween the above mentioned opens of R. Remark the only one-dimensional connected manifolds are R and S '. 3. Sphere S ". In R ""I
let us consider the n-sphere
To provide S n with a differentiable manifold structure we define an atlas consisting of 2n+2 charts ( I li l n+l ): U: = ( x c s n x i >o}
I
U ; = { X E S ~ ' I ~ I < O } .
n
The sphere S is really covered with such charts. Now, we must construct transformations between charts (changes of charts) which are ( C m ) diffeomorphisms.
Let us consider QI: : U:
+ R n: x = (xl ,..., xntl)I-+(XI,..., i ' ,...,xn+')
Lecture 1
48
where the symbol A means the ith coordinate is removed. It is in a way the orthogonal projection of the "positive hemisphere" onto the corresponding equatorial "plane." That is really a bicontinuous bijection. Analogically we define p,: :
ur -+ Rn : x H (x' ,..., i t,..., xn+').
For instance, let us consider any point x of U: positive.
nU,'
such that the rth and jth coordinates are
The following mapping between opens of R":
P; o(P;)-~ :p;(u; nu;)+p;(u,?
nu;):
is actually a diffeomorphism. A difficulty could have occurred because of the square root but the expression under the radical sign is always positive. 4. Torus. The 2-torus T' = S ' X S is the product of two manifolds S ' .
In the same manner, the n-dimensional torus is the product of n circles: T" = { ( e * , ..., e * ) ) Cylinder. The cylinder S' x R , provided with the product manifold structure is a two dimensional manifold.
5.
The product manifold S n x R is called the "n+ 1 dimensional cylinder. " 1.3.4
Orientable manifolds
Let ( x ' ) and (y') be two coordinate systems of an open U of M. D
A differentiable manifold is orientable if there is one atlas ( ( u , , ~ ~such ) ) , that ~ , in the common domain of any two charts the orientations are the same; in other words
Let us note the orientations associated with each coordinate system (in the common domain) are opposite if, at every point of the domain:
D
A differentiable manifold is orientable if there is one atlas
Jacobian of every coordinate transformation pi o
(U,,V,)~,, such as the
is positive at every point.
Manifolds We immediately deduce from the definition: PR3
The product manifold of orientable manifolds is orientable.
PR4
Any open of an orientable manifold is an orientable manifold.
Example 1. The manifold S" ( n 2 1 ) is orientable. The atlas, which has permitted defining the differentiable manifold structure of S n, does not allow using the definition of orientation; therefore we are going to choose another atlas. Consider the opens
B and the poles N = (0,..., 0,l) and S = (0 ,... ,0,-1). Let g, -be the stereographic projection from the north pole N onto the plane of equation
g+'-o. Let pzbe the stereographic projection from the south pole S onto the previous plane Let y be the symmetry with respect to the plane of equation x' = 0. The atlas consisting of charts (U1,*) and (U2,y0 e)is in accordance with the definition. Indeed the coordinate transform ( Y ~ v ~ ) o=PY~ ' 0 ~ ; ' )
is the composition of an inversion (mapping presenting an always negative Jacobian) with a symmetry with respect to a plane. Thus it is a diffeomorphisrn with positive Jacobian. Therefore the sphere S" is orientable. Example 2. Any torus or cylinder is orientable. This is an obvious consequence of PR3 and of the fact that S n and R" are orientable. To prove a manifold is not orientable it is easy to consider the next proposition. PR5
In order that a differentiable manifold M be orientable it is necessary, for any pair of connected charts (Up) and (V, y/), the Jacobian of y l o p-' to have a constant sign on P(U v).
n
Example. The Mobius strip is not an orientable manifold. Indeed in lZ2let us consider the strip defined for example by {(x1,x2)ERZ: 1x11< 4) . An equivalence relation can be defined by
( x " , x ' ~ ) ~ ( x ~ , xe ~ )[xfl =xl and xfZ= x z ]
A Mobius strip is diagrammatically represented by a once twisted strip, of length 8 (for instance) and whose extreme parts are superposed and stuck for a length 1 (for instance).
Lecture 1
50
To define a differentiable manifold, we choose, for example, charts whose domains are U,={(x1,x2) : - 4 < x 1 < l , x 2 E R}
u, = {(x1,x2) :
- I < x 1 < 4 , x2 E R ] .
We choose the canonical projection as a homeomorphism, more precisely its restriction pl to the set of x1 E ] - 4,1[ and its restriction f i to the set of x i E 1 1,4 [. Every coordinate transformation (chart change) 0 p, on the next intersection of domains
-
U , n W 2= ( ( x 1 , r 2 ) : - 1 < x 1 < I ,
X I E R ]
is such that x'l
=
Xf2
= X2
.
3
but, after a complete trip round the strip, it is
We immediately see the Jacobian sign is not constant and thus the Mobius band is not orientable.
2. DIFFERENTIABLE MAPPINGS Let us introduce the notion of mapping of class (P between differentiable manifolds from the chart notion.
GENERALITIES ON DIFFERENTIABLE MAPPINGS
2.1
Let M, N, be manifolds of class 0, f be a continuous mapping of M, into N,, x be a point of M,.
2.1.1
Differentiable mapping between manifolds
D
A mapping f of M, into N, is of class C ( q l p ) at point x of M H if, for each chart (U,q)such as XE Uand each chart ( V, y/) such as y =f ( X ) EV, the mapping called "local representative" of f fPY z v o f o p - ' : p ( ~ n f - l ( V ) > c R n + R m is of class C?. A mapping of class C9 between manifolds is also called C-mrphism.
Manifolds
5I
The previous definition is translated in the language of coordinate systems in the following manner. Let x',. . ., x" be the local coordinates of x in (U, p), y', ...,y " be the local coordinates of y =f (x) in (V, y). D
A mapping f of Mninto Nm is of class C? at point x of Mnif the m local coordinates y' of point y =f ( x ) are, in the neighborhood of x, the m functions of class C4 j = 1, ..., n y i = f'(xJ) of n coordinates x' of x.
I . . C
fw
I ,
f'
Figure 12.
Remark 1. The continuity (class C?) between topological spaces is presupposed; so in fl(V) is open. particular the set U f
-'
Remark 2. Let us specify the local representative f,, is only defined on a part of q(U) because: (x', ..., x n ) E p(U) f(p-'(XI,..., x n ) ) E V
s
rp-'(xl
,..., x") E f -'(v)
and, rp : U -,p(U) being a bijection, we have:
D
A mapping f of M,, into Nm is a mapping of class CQ of Mn into N, if, for every x in M,, to any (admissible) chart (V, I@)on N, is associated a chart (U,p)on M, such as XE U, AX)EAU)C V and also f,, y o f oq-' : p ( U ) c R N+ R m
is of class P. Notation. C?(M,;N,) denotes the set of mappings of class C? of M, into N,; C(Mn;Nm)denotes the set of differentiable mappings of M, into N, (understood of class C").
Lecture 1
52
Remark that it can be proved:
PR6
A mapping f of M, into Nm is a mapping of class @ r f l for each x of M, there exists one chart (U,p)with XXEU and one chart (V, yi) with J l x ) V~ such that f ( U )c V and
f,,
E Cq(~(U);Rm).
Now, we consider the following proposition. PR7
The canonical projections are differentiable mappings of the product differentiable manifold M, x N, into the respective manifolds M, and N,.
ProoJ: Let us consider the canonical projection
p : M , x N , +M,. It is sufficient to prove that there are a chart (UxV,pyi) on M,xNm "at" each ( x ,y) E M, x N,,, and a chart (U', p') on M, "at" x = p(xy) such as p(U x V) c U' and tp' 0 p o ( q x v/)-' is a mapping of class C" of ( p x y ) ( U x V ) into R".
L
R" Figure 13
Let (U x V , p x ly) be a chart at (xa) on the product manifold and (U,p) be the corresponding chart on M,at x. We have: p(Ux V )= U . The following mapping
is of class e".
Manifolds
53
2.1.2
Properties of differentiable manifolds
PR8
The composition of mappings of class C ] between manifolds is a mapping of class C?.
Proof: Let M, M', M" be differentiable manifolds,
f g
q
EC EC
q
( M ;M'),
(M';M").
Prove that
q
g 0 f E C ( M ; M ") .
Consider XEM
y = f ( x ) E M'
= g ( f ( X I ) E M"
On the one hand, we know by hypothesis there is a chart (Wf,p')on M'at y and a chart (WRYp")on M" at z such that g(U1)c U" and pa g tpl-' E C4(p'(U');p"(U")). 0
0
Figure 14. On the other hand, for an arbitrary chart ( U, p) on M at x we have:
But the composition of mappings of class P: v"o(gO f)op-' =(Go"ogop'-l)o(p'o f
op-l)
is of class C?. Therefore, for each x of M, there is a chart (W flf - ' ( U 1 ) , p )on Mat x and a chart (Uw,q")on M uat z such that ( g O f )(U) c M" and pn f c g ( ~ (nuf-i(uf));pw(u~)). In conclusion, g 0 f is a mapping of class C? of M into M a .
Lecture 1
54
PR9
Let M,, N,, Pr be differentiable manifolds, pl be the canonical projection of M,, x N, onto M,, pz be the canonical projection of M, x N,,, onto N,. The mapping f :P,+M,,xN,,, is of class C? 18 the coordinate functions P
are of class
I
e.
O
:P,+ M ,
~
~
2
0
:P, f +Nm
Proof The necessary condition is immediately verified. Indeed, sincepl andpz are of class C? and f too by assumption, then PR8 implies p, 0 f and p, 0 f are of class C?. Let us prove the sufficient condition. Let us suppose p, o f and p, 0 f are mappings of class Cl of Pr respectively into M, and into N,.
Figure 15.
Consider ZEP,,f ( z )M ~ , x N , , ( p I o . f ) ( ~ ) =( F ~ ,Z ~ ~ ) ( ~ ) = Y -
The mapping p, 0 f of class CQ implies there is a chart ( W I , ~ ,at) z and a chart (U,p) at x such that (p,of)(W,)cU and po(pI 0 f ) o e ; ' E C4(4(&);Rn). The mapping p, 0 f of class C? implies there is a chart (W2,02) at z and a chart (V, y) at y such that ( P , f )(W2 1 c v and wo(p2 0 f ) o @ i 1 E C4(@2(W2);Rrn).
Manifolds
Let us restrict to
w=w,nw,
8= Q l / w
To prove that f is a mapping of class C4 it is sufficient to note there is a chart ( W,', aat z and a chart (U x V, q x ly) on M nx N , at (x, y ) = ( ( p , f )(z),(p2 f )(z)) such that:
f (W)c (PI
O
S ) ( W )x (
f )(W) r U x V
~ a2
and ( U z ~ v ) o f o 8 -~1c ~ ( e ( w ) ; R ~ + ~ ) . This last mapping ( t p x y ) f~ 06-' = ( p o p l o f o 8 - ' ) ~ ( ~ 0of~of?-') , is actually of class P. Indeed, 8, 0 8-'is of class C? (because change of chart). Furthermore, we know that P O ( P , 0f)0@,-~E C ~ ( Q , ( C ~ : ) ; R " )
thus p o p lo f 08-I = p o ( p ,a f)oe;' o(8106-')
is of class C9. It is the same for
y/
0
p, o f o 8-' and thus the proposition is proved.
The reader will demonstrate the next proposition.
PRlO A mapping f : M,
+N ,
is of class (? 1 f l there is an (open) covering (U,),,r of Mn
such that f / ,is of class C? for every ie I.
2.2
PARTICULAR DIFFERENTIABLE MAPPINGS
2.2.1
Diffeornorphism and local diffeomorphism
Let Mn and N, be differentiable manifolds of same dimension.
D A mapping f of Mn onto Nn is a e difleomorphism of M,, onto N, iff is a bijection of @(Mn;Nn)and f -' E Cq(N,;M,). Notation. Let DlffQ(Mn;Nn) denote the set of CQ diffeomorphisms of Mn onto N,, D~f(Mnfl~)denote the set of (e") diffeomorphisms of M, onto Nn. PRl 1 If M, is a differentiable manifold then Dlff(M,;N,) composition of mappings. (see PR8). D
is a group with respect to the
A differentiable mapping between manifolds (of same dimension) f : M, + N, is a local dz~eoomorphismat a point x of Mnif the rank of fat x is n. It is a local difleomorphism on M, if it is a local diffeomorphism at every point of M,.
Lecture 1
56
The reader will refer to PR39, PR40 and PR41 of lecture 0. PR12 A mapping of class C? of Mn onto Nn is a CQ diffeomorphism zff it is bijective and is a local diffeomorphism on M,. Let us emphasize the importance of the bijective assumption. PR13 A bijectionf of Mn onto N, is a diffeomorphism of M, onto Nn 1 f i in local coordinates xi, the n differentiable functions 1 = I, ...,El, f'(xi) that definefl show a non-zero Jacobian.
2.2.2 Immersion - Submersion - Embedding Let M,, N, be differentiable manifolds. D
A differentiable mapping f : M,
+ N,,, is an imm~rslonat point x of M,
if the rank
off is equal to the dimension of M,. It is called immersion of M,, into Nm if it is an immersion at every point of M,.
It is necessary that n Irn
D
A differentiable mapping f : M n+ N , is a submersion at point x of M,, if the rank off is equal to the dimension of N,. It is called submenion of M, into N, if it is a submersion at every point of Mn.
It is necessary that n 2 m .
D
-+
N,,, is an embedding of Mn into Nm iff is an injective immersion and a homeomorphism of Nm onto JTM.) (for the induced topology). A differentiable mapping f : M,
The reader will prove the following propositions by having in mind the constant rank theorem.
PR14 If f : M, + Nmis an immersion (resp. submersion) at point x of M, , a chart (U,p) on M, containing x and a chart (V, y/) on N, exist such that
f(U>c v and
falv = Y f 0
0
Q-'
: p(U) + Rm : (x',...,x n ) H ( x l ,..., xn,O,...,O)
[ resp. (x',. ..,xn) H (xl,. ..,x m ) , n L m].
PR15 If f : Mn + N,,, is a mapping of class CQ of constant rank r on Mn then, for every x of M,, a chart (U,y>)on M, containing x and a chart (V, y) on N, exist such that:
f (u)c v and
Manifolds
PR16 Iff is an embedding of Mn into N, then the set AM,) is provided with a differentiable manifold structure (induced by the embedding).
1
~ r o o jIf {(u,, vt),€, is an atlas of M, let us prove that
{(/(u, X P ~o
.f-l
),€I
)
is an atlas of
fW n ) .
Figure 16. Every pointfix) of f i u ) corresponding to x E U , has a neighborhood which is homeomorphic to an open of R"; and the opens f(U,),,, cover f(Mn); moreover the image of any
n
x E U , U , is a point
f (x) E f (U, ) nf (U, ).
The mapping (P, o f -l)"(P,
o f
is a diffeomorphism between the opens (of R") q3(
4)and
@(Ut)because:
and that (I%*)and (U,,9)are charts of atlas on M,.
2.3
PULLBACK OF FUNCTION Let M,and N, be differentiable manifolds.
2.3.1 Real-valued function on manifold Let g :M ,
D
-+ R : x t,g(x)
be a function on M,.
A function g on M, is of class C' at point x of M, if there exists a chart (U,tp) containing x such that : R n+ R is a function of class C? on the open q(U) of R". This function gap-I is called 'yunction g expressed with respect to local coordinates " or function g "read on the chart ".
Lecture 1
58
The real (g o p-')(xl,..., x n ) will be denoted g(xl,...,x n ) or simply g(xi) with i = I ,.. .,n.
Figure 17.
Now let us prove that for a hnction the notion of class P is independent of the chart choice. In other words: PR17 A function g of class P in a local coordinate system is automatically of class CQ in any other admissible coordinate system.
PrmJ Remember that a change of local coordinates is admissible if there exists a @ diffeomorphism, for instance p o ly-I, between opens of R". Let g be a function of class "read" on (U,p). Since = (gop-l)o(qoly-')
and because the functions p o ry-' and g o p-' are (at least) of class P, then (PR5) the @ composite mapping theorem implies the function g 0 yl-l is of class C!
Figure 18 D
A function g is of class C on M,,if it is of class (?at every point of M,.
Manifolds 2.3.2 Pull-back of function under differentiable mapping
Let f be a differentiable mapping of M, into A,' h be a differentiable real-valued function on N,.
D * The pull-back of the function h by f is f 9 h = h of . The mapping f is so defined:
Cm(Nm;R) + C m ( M , ; R )h: Hf *h Therefore, from a mapping f : M,, + N,,, we have constructed an induced mapping
$* : Cm(Nm; R)+ Cm(M,;R).
3, SUBMANWOLDS
A subset V of R" is a subman&iold of R",of dimension rn (5 n ) and of class P if, for every XE V, there exists an open U, of Rn containing x and a P diffeomorphism g of U, onto the open g(U,) of R" such that
Figure 20.
Lecture 1
60
Let us prove propositions that allow to find submanifolds of R". Let U be an open of R".
PR18 Let f : U ( c R n ) -+ R m be a mapping of class C, y be a point of R", V =f-'(y). If f is a submersion at every point of V, then V is an (n-m>dimensional submanifold of R". ProoJ Let x be a point of Y. The PR48 of lecture 0 is applicable and implies that there is an open U,(of U) containing x, an open U,of R* x Rn-"containing y and a C4diffeomorphism g : U , +Uy : X ' H g(xr)
such that for each x' of U, :
where
l7 is the canonical projection of R" x Rn-"onto R".
We have the following sequence for any x'
x'~U,flV
#
[ x ' E U , and x ' E V ] [ X'
[XIE
#
and f(xl) = y ] ux and n ( g ( ~ I )=)~1
E Ux
[ xr E Ux and g(xl)E ( yf x Rn-"]
and thus
The submanifold definition means definitely that V is an (n-m)-dimensional submanifold of
R".
Figure 21 From the previous proposition we can deduct another one very usefkl in practice:
Manifolds PR19 Let f' : U ( c R") + R be m functions of class C4, v = { x = ( x l ,...,x n ) e R":~'(x' ,..., x n ) = 0 , y i (I~,..., m)}. If for every x of V, the rank of the Jacobian matrix
is m (In), then V is an (n-m)dimensional submanifold of R", Proof We recall $is a submersion at x r@ its Jacobian matrix at x is of rank m. The proposition immediately follows from PR18 where f is defined by the following element of R": f (xl ,..., xn) = (fl ( x l ,... , x R ,..., ) f yxl,. ..,x"))
and where the p i n t y is (0,. .. ,O)E R m . It should be noted the submanifold is so defined as the intersection of hypersurfaces.
Example. An ellipse is a one-dimensional subset of R ~ . Indeed, consider the ellipse
The following Jacobian matrix
has rank 2 at each point of V since, for example, the determinant
is different from zero at each point of V.
A special case of the previous proposition is the following
PR20 Let f : U ( c R n ) + R be a function of class @, V = ( X E Rf (~x )~= ~ J . If, for every x of V, one of partial derivatives of f is nonzero (nonzero gradient of f ), then V is an (n-1)-dimensional submanifold of R". Examples.
1. The hyperboloid(one sheet) of equation
is a two-dimensional submanifold of R ~ .
Lecture 1
62
Indeed,f is of class e" on R) and
2y -22 qf =, ( 2x , is different from zero at every point 2 ,T ) a
b
of the hyperboloid.
2. It is immediate that a sphere of equation x 2 + y2 + z2 - r 2 = 0, an ellipsoid of equation 2 x2 y2 z 2 y2 z 2 +-2 + --i-- 1 = 0, a hyperboloid (two sheets) of equation -x7- - --1 = 0, an elliptic c a b Z c2 a b x2 y2 x2 y2 paraboloid of equation z = 7+ 7 and a hyperbolic paraboloid of equation z = - a b a b2 are two-dimensional submanifolds of R).
,
3. The cone(with vertex o and axis oz) of equation z Z= r2 + yZ is not a submanifold of R ~ . It is necessary to remove the origin (singularity) to obtain a manifold. In this last case G ' = (2x,2y,-22) is nonzero on the open R~- (0) and thus the previous proposition is applicable, Another interesting proposition allows to conclude to the existence of submanifolds of R":
PR21 If f : U ( c R m ) + Rn is an injective immersion (m 5 n), if f -' : V = f (U) + U is a continuous mapping, then V is an m-dimensional submanifold of R". ProoJ Let us schematize the situation in the case m = 1, n = 2.
t x J u
Figure 22. Let y be a point of V such that y = f ( x ) , x being a point of the open U
The PR47 of lecture 0 means there exists: - an open V, of R" x R"-" containing x such that V, nR m c U , - an open Vf(,,of R" containing y, - a d diffeomorphism g : V, + Vf,,, having same restriction as f to V,
-'
The continuity of f at y implies there exists an open V;,,, of included in Vf(,, such that
~ ~ ~ ~ v ; ( , ) n vf - l ( y f ) E ~ x .
nRm
R" containing
y and
Manifolds
The restriction of
g-l to
Vj,,, is a C? diffeomorphism of V;,,, onto an open V: c V'
The restriction of g to V,' is a C? diffeomorphism denoted g,v; : v: + v;(x)*
.
(iJ On the one hand, because g and f have same restriction to V: , we have:
vt
v: n R" : g,,: (t) = f ( t )
E
v;,,, n v .
(id On the other hand, if we have: Y' E v;(~) v [af - ' ( Y ' ) E V, I m then there exists t G Vx',n R such that f (t) = y' .
n
But
Y' E V.(x) thus t~v:nR"'.
From (i) and (id, we deduce that V;(,,
This definitely means that v: = g;(v;r,).
V
nV is the image of
is an
Vi fl R m under giv:, that is
m-dimensional submanifold of
R"
since
Example
then f (U) is a two-dimensional submanifold of R" Indeed, the three requirements of the last proposition are fulfilled, namely:
- In the introduction, we have showed the Jacobia. matrix of f - sin1 cos# - cos A sin #
[
-sinRsin4 COSA
cosRcos# 0
is
1
and is of (maximum) rank 2. So,f is an immersion.
- The mapping f is injective because 'v'(~,#)y(A'y+') u : f (A,#) = f (A1,#') = (A,#) = Vf,4'). Indeed, from sin A = sin A' we deduce A = A', from the equalities cos A cos 4 = cos A cos 4' and c o d sin# = cos A sin 4' we deduce cos # = cos #' and sin # = sin 4' (because cos A # 0) and thus # = #' .
-
Y. The mapping f -' : f (U) + U is continuous since R = arc sin z and # = arc tg X
Consequently, f (U) is a submanifold of R3.
Lecture 1
64
3.2
SUBMANIFOLD OF MANIFOLD
D
A subset W of a manifold Mnis an m-dimensional subman~ildof M, (m I n) if for each x E W there is a chart (U,p)in M, containing x such that: ~ ( w n w=) ~ ( u ) n R ~ .
n
A chart (U,p) such that p(U W) is the set of points (x',. .. f ) of p(U) fulfilling i+' = ...=x" = 0 is said to be "adapted" to K
Figure 23.
~ ( nuw )= P ~ Wn ) R~ =((I1,X2,X3)Ep0(LI)
I
.X3
=O}.
The reader will refer to propositions expressed in $2 in order to prove the existence of submanifolds. So, for example, PR15 leads to: PR22 Given two differentiable manifolds M, and N,, if f : M, -+ N , is of class ?i and of
(constant) rank r on M,, then, for each y E f ( M , ) , f -'(y) is an (n-r)-dimensional submanifold of M,. PR23 If f : M ,
+N ,
is a mapping of class CQ between differentiable manifolds, if y is a
point of f(M.) and if f is a submersion at each point of f '(y), then f'('y) is an (n-m)-dimensional submanifold of Mn. In another manner: Let f i : M, mapping
+R : x t+ f : M,
f '(x) be m differentiable functions on M, defining a differentiable
+ R m : x H (f1(x),...,f n ( x ) ) .
Manifolds
We can say:
PR24 A subset W of M,, defined by rn equationsf'(x) = 0 and f having rank rn at each point of W, is an (n-m)-dimensional differentiable submanifold of M,,. Prooj: Let (U,p) be a chart of Mncontaining a point x, E W . Putting (fo #-')I (XI,..., xn)= ?(xi) i = 1, ...,n , the assumption of rank m amounts to saying that the matrix
(i, j = I, ...,m) different from zero.
has a Jacobian D(x'
Then there exists an open U:(c U) containing x, such that xll = fI(x') , ...., dm= f "(x') , X I ~ + I = xm+l,...., X l n -xn is an admissible coordinate change, that is
-
The chart (U', p') defined in this manner is such that
Such charts exist at every point of Wand, by changing the numbering of coordinates x", we find again ,'(W, U')= (xfrn+' ,...,x r n ,..,, , 0) meaning that W is an (n-m)-dimensional submanifold.
4. EXERCISES Exercise 1.
Prove that a manifold M is a locally compact topological space. Answer. Let us show every point x of M has a compact neighborhood. Let (U,p) be a chart containing x where p is a homeomorphism U + p(U) such that q(U) is a neighborhood of &x) in R" locally compact. Therefore there is a compact K of R" such that p(x) E K c p(U) . But, p-I being continuous and M being Haussdorff, we can conclude p-'(K) is a compact neighborhood containing x in accordance with PR12 of lecture 0.
Exercise 2.
Prove that a manifold M is a locally connected topological space.
Lecture 1
66
Answer. We proceed as in exercise 1. So q(U) is a neighborhood of d x ) in R" containing a connected neighborhood C of q(x). Then p-' ( C ) is a connected neighborhood in M containing x. Exercise 3.
Prove that every open Uon an n-manifold M is a same dimensional submanifold. Answer. First we use the following theorem, the proof of which is left to the reader: If ,p?) ( i E I ) is an atlas on Mythen a subset W of M is an open f l
k~,
vi I ,
pi(~nu,>
is an open in R. Next,let
A = ~ u , , ~ ) ~ ~ E beanatlasonM. I )
It is sufTicient to check
k~
U,,pi,
)
/
j
i E I is an atlas on W.
Let us remark the opens U nU , cover U because Mis covered by the U, ( i E I). The image of W nU ,under the homeomorphism qrlanu, , denoted P',, , is the open
v,, ( u n u , ] = p , ( u n u , ) of R" (see theorem). Since U nU, ( j E I ) is an open in M, the recalled theorem implies p, (U nU, 0 U , ) is an open of R. Therefore, the mapping
a,,
0
pi1, restriction of p, 0 p;' to 9,(U
nUi n(l,1. is
a C' diffeomorphism between the opens piI((LI (IU, nU,) and p,, (U fl U, fl u,). Exercise 4.
Given a function f E C w (R n ; R) and knowing that f is a submersion at each point of M = f (0) , make explicit the differentiable structure of the manifold M.
-'
Answer. The PR18 indicates that M is an (n-1)-dimensional submanifold of R". We specify by assumption that the rank of df, : Rn -+R is 1 at each point x~ of M. One of the partial
derivatives, for instance
is nonzero. Let R, be an open on F-', Q2 be an open on R. If the point (xi ,..., i,k,...,x:; X: ) E R, x R2fulfils the following equation f(xi ,..., i k x n ; x k ) = O
with
af -(xo)#O axk
then the implicit function theorem means that: Given a neighborhood U,(ca,)on R"-'and a neighborhood V,(c R,) on R there exists a function g : U, + V, of class C" such that the equation x k = g(x 1 ,..., x- k ,..., x " )
Manifolds is equivalent to f ( X I ,...,g(xl ,..,, jk,..., x n ),..., x n ) = 0
whatever (xi,..., i k,..., xn)E Ux. Let us introduce a chart (U,,pk ) defined by
U , = (U, x V x ) f l M and (Dk : (UX xVJnM
+ ux: (xl,...)xk7...,x") I+ (x', ,,.,P,..., x")
such that
q7i1(xI ,.**, ik,...,x n ) = (x1 ,...,g(x I ,..., x- k ,..., x*),..., x"). We can likewise proceed at every point x so that we cover Mwith opens (U, x V x )f M l We must prove every change of chart is admissible (e"):
It is definitely a diffeomorphism (of class e")between opens of of class e",
R"-'because g is a function
Exercise 5. Show that SL(n;R) has a natural structure of differentiable manifold where SL(n,R) designates the special group of all inversible n x n matrices of determinant +1. Answer.
Let
(I;
be the set of matrices
U, be the set of matrices
( aV)
having a positive minor,
( a p ) having a negative minor
So we have domains of charts covering SL(n,R). There exists a function f :SL(n;R)+R:A ~ d e t A - 1 which is zero in the neighborhood of a point A, E SL(n;R). The image of every matnx A E SL(n;R), i.e. f (A) = det A - 1, is definitely zero because det A = I in the neighborhood of Ao. The differential is a linear form R" + R of which the components d j , ( e , ) = are (see exercise 6 of lecture 0) the partial derivatives of f , namely:
(a),,
a (det A ) = A, axti
where the various xu pte the variable coefficients of matrices (u,). In order to be allowed to apply the implicit function theorem and to make any variable explicit in function of others, one of the partial derivatives of f must be nonzero. It is
Lecture 1
68
because every matrix of determinant 1 has at least a nonzero minor, which allows making the corresponding variable explicit. Let us use U i , domain where A, > 0 . Let R1 be an open on R"'-' , R2 be an open on R. 4E R, x R2 be the point verifying f (A,)
= det A,
- 1= 0.
The implicit function theorem means the folIowing: In a neighborhood U(cR,) and a neighborhood V ( c Q, ) there is a function of class e"
such that
I... XI!...
det
.... g ,...
] l o
We choose as a chart domain the open
u,;' = (ux v)nSL(n; R ) and the homeomorphism
To remove xii amounts to a "projection." Let us "pull back up" by making a variable explicit (implicit function theorem!), namely:
*
We consider another chart (U2,pL) such that Ui.+f Ulr 0. The change of chart
P; O(P;)-I is defined by
:P;(u;
~ u ; ) + P ; ( nu;) u;
Manifolds
The change of chart is of class differentiable.
C" since
g i is so; the manifold SL(n;R) is actually
Exercise 6. If a function h on N, and a mapping f of M, into N, are of class C4 prove that the pullback of h by f is of class C? (q 2 1).
Answer. The hnction (h o f )
0 QI-'
of R" into R can be written:
(by associative law).
Thus it is the composition of two mappings
f oq-':An+ R m with
~ow-':R"+R. The first mapping is of class CQ since f is so, the second is also of class C? since h is so.
The differentiable composite mapping theorem implies that the function f'hoq-'
: R n- + R
Lecture 1
70
is of class CQ. Therefore, the function f ' h on M, is of class C4 in accordance with the definition of C? differentiable function. Exercise 7.
Let z = ln define a surface in R' (i) What revolving curve generates this surface? (id Is this surface a submanifold of R~? The answer of this last question will be first given by using the submanifold definition and secondly from gradient notion. Answer. (I;) The surface is generated by revolving the curve of equation z = lnlxl [in the
plane ( X J ) ] about the axis oz. (ii) A surface w c R3 is such that for every point p E W there is a chart (U,p)on R3 such that
The surface of equation r = l n d m is defined on R'
- (0,0).
The manifold R3 being differentiable, then each point belongs to at least a chart such that the homeomorphism
~ : u n +dunw): w (x,Y,z)H(x,Y,o) maps every point ( x y , z ) into the open R2-(o,o). We are going to use the PR20. Let us consider f(x,y,z)= z - l n J W
.
The gradient
being nonzero at every point of the surface, then this last is actually a submanifold of R ~ .
LECTURE
2
TANGENT VECTOR SPACE
To each point of a differentiable manifold M we are going to associate an n-dimensional vector space: the tangent space to Mat this point. A decisive progress in differential geometry occurred when tangent space was defined as a manifold without reference to R". Different techniques can be used, for example the algebraic approach using the notion of ideal, but we have chosen the method which is the most used by engineers and physical scientists.
1. TANGENT VECTOR 1.1
TANGENT CURVES
Let M be a differentiable manifold, po be a point of M, I be an open interval in R containing 0 1.1.1 Curve
D
'
A (diflerentntlable)curwe , passing through po, in M is a differentiable mapping
c:I-+M:tHc(t)
such that 4 0 ) = Po
So, the previous mapping makes a correspondence from each real t to a point of M (image of t). One says it is an arc of parameter t. We will assume the mapping c is at least of class c'. Remark. Refering to the mechanics, then the real parameter t can be called "moment, " but a priori t is any parameter.
1.1.2 "Reading" of a curve
Let c be a curve of M, (U,p) be an admissible chart. I
Strictly speaking it is a matter of an arc
Lecture 2
72
D
The reading of the curve c in the chart (U,p)is the mapping p o c : I ( c R ) + R n : t ~ ~ ( c ( ~ ) ) = ( ..., x 'x(nt( )f ),) .
Figure 25.
Notation. The parametric equations of the curve c are excessively denoted xi = x i ( t )
i = 1, ..., n.
Example. On the sphere 9,the equations of any curve are x i = O(t) x2 = # ( f ) t€l where x' and 2 are respectively the colatitude and the longitude. So, a parallel is defined by ( x' being constant ) x2 = #(t) and a meridian is defined by x1 = 8(t> 1.1.3
( xZ being constant ).
Tangent curves
Let (U,p)be an admissible chart of M,
Po
u,
ci
cl, cz be two curves of Mpassing throughpo.
D
The curves cl and c2 are tangent atpo with respect to g, if the mappings (I + R n) 9 0 c, and p 0 c, are tangent at point 0 E I (or at moment 0):
Let us show the choice of p doesn't matter, which will allow introducing the notion of tangent curves.
Tangent Vector Space
Let (Ui,rpi),z1,2 be two admissible charts containing
PR1 Two curves cl and c2are tangent at p~ with respect to p
fl they are tangent at po with
respect to @,
Proof. Let us demonstrate the necessary condition that is
Taking restrictions (if necessary) we let U, = U ,. In this case, we have: ~2 O C I
i = 1,2.
= ( P ~ o v ~ ' ) o (ocr) v~
By applying the differentiable composite mapping theorem, we see the following
The converse is analogically proved. Therefore, two curves cl and c2are tangent at p, with respect to any local chart.
EM
(at moment 0 ) if they are tangent, at po,
This proposition means that the tangency of curves at a point is independent of the used chart and this notion is well-defined. So we say:
D
Two curves cl and c2are tangent at point p, if c, (0) = c2 ( 0 )
In local coordinates we denote: (q O cl )(t)= ( x l ( t ) , . . . , x " ( t ) )
(p0c2)(t= ) tyl(r) ,..., y f l ( t ) ) . We can "read the curves by using: poc, : R
pot,:
+ R n : t t + ( x l ( t ) ,..., x n ( t ) )
R + R" : t k + ( y L ( t...,yn(t)). ),
More accurately, the chart change is admissible
Lecture 2
74
The pointpo being common to two curves it is evident that (x1(0),.. . Sc"(0)) = Cy1(0),. .. lJI(0))
which is denoted
(4)=(Y:)
i = 1, ...,n.
The definition (2- I ) is expressed in the local coordinate context as follows.
The previous curves cl and
c2 tangent at point
p, in Mare such that V i = I, ...,n
Example. In M, the curves defined by the three following types of parametric equations ( i = l , ...,n): VC(;,ER , k ( i ) ~ R : XI
= C(,)f
i
3
u = k(,,t
+ c(,,t
yt = c(,,( t 2+ t )
go through the same point at t = 0. Moreover they have the same tangent vector because
TANGENT VECTOR
1.2
Let po be a point of differentiable manifold M Notation. C(pO) denotes the collection of functions of class e" on every open neighborhood of PO. Let c be a curve in M, g, l.2~CM(p0). It is said that g and h have same germ at po if there is an open neighborhood U of po such that gl, = h,, .
D
We so obtain an equivalence relation in e"(po). We say:
D
A differentiable germ of function g, at po, is the equivalence class of differentiable h c t i o n s coinciding in an open neighborhood U ofp,.
Notation The set of differentiable function germs in an open neighborhood U of po is denoted by O(U). 1.2.1
Fint definition of tangent vector
Let po belong to domain U of a chart ( U , p ).
Tangent Vector Space
75
By referring to the tangent mapping notion defined in lecture 0, the reader will easily check that the tangent curve notion (at a point) introduces an equivalence relation among curves (same tangent at point!). An equivalence class of tangent curves at point p,
E
A4 is denoted
[cJ, where c is a representative of the cIass.
A tangent vector to M, at po, is an equivalence class of tangent curves at po.
D
One says that "the curves of the equivalence class have same tangent vector at p," This vector is also denoted i(0)
by referring to the mechanics and more particularly to the velocity vector of a curve on the surface.
1.2.2
Function along a curve and tangency
Consider p, D
E
U ( cM ) .
Thefunction g along the curve c is the function goc:R - + R : t ~ ( g o c ) ( t ) .
PR2
Two curves cl and cz have the same tangent vector at point p,
EV
8
Vg E O(U);
in other words: iff the derivative of g along cl ai p, is equal to the derivative ofg along c2 at po.
Lecture 2 Proof: Let the function along c be
Its derivative at po is
Consequently, the following equivalence
proves the theorem. In condusion, we can express:
PR3
All the curves c, tangent at po of M, 1.e. having same tangent vector at po , are characterized by a same value:
All these curves will excessively be designated by c. The function g read on a chart is the following function expressed in local coordinates by g o p-I: R n
-+
R : (XI,..., x " ) H ( g 0 q - L ) ( ~,...,' x n ) =g(xi) i = 1, ...,n.
We immediately express: PR4
The h c t i o n g along c g o ~ = ( g o ~ - l ) o ( ~ o c )
represents the "reading" of the curve c in the chart followed by the "reading" of g on the chart.
PR5 A tangent vector to M, at po, i.e. an equivalence class of curves c, presents the same : value Vg E O(U)
'
I
The notation xo represents the coordinate system (xo,..., x: ) of po
Tangent Vector Space Prooff From
and by using the rule of composite function derivation, we deduce
Remark. Make clear that curves c are tangent at a point if, in a chart, they lead to the same vdue
in other words, if they present the same successive values
for each of curves. 1.2.3
Derivation in the Leibniz sense
Let us interpret a tangent vector at a point po as a real-valued mapping of derivation defined on the set O(U) of differentiable function germs in an open neighborhood U of po.
D CZ- A (linear) mapping
X, : O(U)-,R : g H X,(g, is said to be a &fiation mapping
D
' if
Va,b E R,Vg, h E O(U):
The derivative of g, at p,, in one tangency direction associated with XPo is the real
for an equivalence class of curves [c].
In local coordinates, the derivative of g in one tangency direction is next expressed
Remark I. Considering the differentiable function atpo X':M+R:XHX'
we have
' In the Leibnix sense
Lecture 2
also denoted Xi Thus, the derivative of g, at po, in one tangency direction is
Remark 2. For any constant function I, we have:
X, ( I ) = 0. Indeed, introducing the identity function i, we have: xpo(1) = 1xpo(i) =1xpo(i. i) = 21 that is necessarily zero.
xpo(i)
Remark 3. From above, we deduce for every constant h c t i o n I that
1.2.4
Second definition of a tangent vector
PR6
A tangent vector is a derivation (in the Leibniz sense).
Prooof: Let a tangent vector be an equivalence class [c], such that the value of the derivative of g along c, at po, is
The mapping
is
-
linear : Vk E R, Vgl, g2 E O(U):
(see definition with the aid of a chart), - of derivation: Vg,, g , E O(U):
Reciprocally, we can prove the following theorem.
Tangent Vector Space PR7
Every red-valued (linear) mapping of derivation in O{U) is a tangent vector atpo.
Answer. Let
X, denote this mapping.
Let (x: ,. . .,x,") be a local coordinate system of a point p, E U , (xl,. .., xn) be a local coordinate system of a neighbouring point p E U , Remember that the real i = 1, ..., n ax') designates the image of p under g 0 q-' . Let us prove the following lemma: Given g E O(U), there exists n functions g,
E O(U)
such that:
]=I
Indeed, in R" let us consider the following diagram.
Figure 27
As a general rule, we have: g"(y' ,..., y n)- E(0,... ,O) = [Z(uyl,.-.,uyn1;1:
Putting
1 itYag
gj(yl,.., y n ) = +uyl
,..., uy") du,
0
we have
So, the change of variable yJ t,xJ defined by (y-' = 0 for x = xi) Y~ = X ~ leads to J
Lecture 2
Now let us prove the theorem. The functions being represented by the images of points on which they "operate" and by remembering the properties of linearity and derivation of the operator X , : O(U)+ R :g
Xpo(g),
we write the following in accordance with the usage that identifies the image of any point with the function: n
=
xpo[p(xA,...,x,")+ x, [(i - 4 ).p,(x',.. .,xR)] j=l
Otherwise
leads to
Therefore, (1) is written
Using again the habitual writing of functions, we obtain
We have definitely found again the expression (2-4) of tangent vector atp,. Consequently we can express a second definition equivalent to the first: D
A tangent vector to M, at PO,is a (linear) mapping of derivation:
2. TANGENT SPACE Let po be a point of a differentiable manifold M. 2.1
DEFINITION OF A TANGENT SPACE
D * The tangent (vector) space of M, at po, is the set of equivalence classes of tangent curves at po. It is the set of tangent vectors at p,
E
M
The reader will immediately check that linearity and derivation properties of any tangent vector show the (R) vector structure of the tangent space.
Tangent Vector Space Notation. The tangent space at po E M is denoted TpoM or simply Tpo.
In short: T,M = {[cl,j=
1x,J
where X , is a representative of tangent vectors at po (with components
2.2
dx' xi = -(O) dt
).
BASIS OF TANGENT SPACE
Writing convention. We denote X , g the derivative of (germ) g in X tangency direction,
atpo. Let us recall it is the real
ag = a(g where henceforth the upper line in ax1
v-' ) is
axi
Tangent vector expression. A tangent vector, a t p , in the direction corresponding to X' has the following expression
'
This immediately follows from (2-7) where Xpo"operates" on every germ g.
PR8 The tangent vectors at a point of a manifold M form an n-dimensional vector space Prooj: The expression (2-8) shows every tangent vector Xpo is a linear combination of at the most n tangent vectors at po of M which are defined by
These vectors form a system of tangent space generators. Therefore, the dimension of the tangent space, at po, is at the most n. Furthermore, the vectors
are linearly independent. Indeed, if they were dependent
then there would be reals a,,not all zero, such that
But such a,do not exist because for any fimction x i : x M x J, we have:
' Simplified writing in accordance with the Einstein summation convention
Lecture 2
To conclude dim TpoM = n. D
The n operators
make up a basis of TpoM associated to a local coordinate system. It is said they form a nahrral busis of TpoM with respect to the coordinate system
(xi1. The n componeprls of vector XpOwith respect to the basis
(gl0
are the reds X i .
We denote
Example. Let us illustrate with 9 that every tangent vector X , E T,M is a linear combination of tangent vectors e,. Indeed, the coordinate lines, i.e. the meridians and the parallels are respectively: x' = B(t) x 2 = #(t) = # (constant) 6 = 0, (constant). By omitting the reference index of point po, every tangent vector to a curve of SZ,at this point, is written:
that is explicitly:
dt? a d# d -+-- . dt a0
dtd#
So, the sum of a tangent vector to a parallel (0constant): d# a -dt d# with a tangent vector to a meridian ( constant): dB a -dt ae
+
is a tangent vector to SZ. 2.3
CHANGE OF BASIS
Every vector of TpoM can be expressed in another basis than the one associated to local coordinate system ( x i ) .
Tangent Vector Space
Let us establish the formulas giving the transformation law for components. M , let n basis vectors ei be linear combinations of In TpO
The matrices a and
(s),.
P being inverse, we respectively have:
but
but
thus
thus
More simply, let a change of natural basis be
(s), -(&lo.
(6)(&), " [s)
The classic rule of partial derivatives calculus implies:
=
thus
thus
but
but
0
therefore, we obtain the following formulas:
To sum up, every vector X, E TpoMis expressed in the form of linear combination of n tangent vectors to coordinate lines sometimes denoted by (a,), .
We say that Xo is tangent atpoto the curve t H x i ( r )
(1
= 1,. .., n).
3. DIFFERENTIAL AT A POINT The notion of tangent (vector) space at a point of a manifold allows defining the differential regardless of local coordinates.
Lecture 2
84
Let M and N be differentiable manifolds, f be a differentiable mapping of M into N, Xobe a tangent vector to Mat point x, E M, zo be the image of xo under fin N, W be an open of N containing 2 0 .
DEFINITIONS
3.1
In section 2 of lecture 1, we defined the mapping f*:Cm(N;R)+Cm(M;R):h~ f'h that associates to any function h on N the pull-back of h byf D
*
The diflerenflaal ((mapping) o f f at point xo E M is the linear mapping df, : T , M + T , N : X o t+df,X,
such that, V h E O(U):
w The vector d
dL,Xo(h) = Xo(f '4 -
(2-10)
m tangent to N at .a, is called the image of tangent vector Xounder f:
If the image of X,under f is .&,, then (2- 10) is written:
Zo(h) = X*(f*h)
Remark 1. The linearity of differential, namely Va,b E R,VXk,X," E T X o M : dLo(aXL + bXz) = a dfxoX;
+ b dLoX: ,
Tangent Vector Space follows immediately from the linearity of tangent vector Xo. Indeed, Vh E O(W): df, (a& + bX,")(h)= (ax;+ bX,")(f *h) = uX;( f *h)+ bX,"(f 'h)
= adL0X;(h)+bdLoX,"(h) = ( adLD Xi + b df,XXh).
Remark 2. We leave the proof of the following propositions to the reader.
(9
A differentiable mapping f : M -,N is an immersion (resp. a submersion) at xoof M ~ f l df, is injective (resp. sujective).
(io A differentiable mapping f : M -+ N is a local diffeomorphism at x,
EM
~ f ldfxois an
isomorphism.
Remark 3. The differential notion is well defined because the image independent of choice of curves c defining Xo. So, the differential o f f at xobeing
dLo : qoM -,C O N IcIso H[f cl O
,,(
2, = d&X0 is
) >
let us prove that if two curves cl and cz are tangent at xo, then the curves f o c, and f o c, are tangent at Axo). Indeed, let f U,p) and (U', q') be charts respectively of Mand N, We have p r o f oc, = ( v l o f o ~ - ' ) o ( P o E ~ ) p r o f oc* = (p'o f op-l)o(po~Z).
But the curves cl and ~2 being tangent, it follows Hence the differentiable composite mapping theorem (c'at least) implies
This expresses the tangency of curves represented by f 0 c, and f 0 c, .
3.2
THE IMAGE IN LOCAL COORDINATES
Let xi be the n coordinates of xoin a chart of A4 The differentiable mapping f makes to any chart of M correspond a chart of N in accordance with z' = f'(xi) where
zJare m
local coordinates of the point .zo=f(xo).
Let us make explicit in local coordinates the definition of the differential df, . The equality zo(h) = Xo(hof ) ,
Lecture 2
86
having permitted to introduce the image & of tangent vector, is written by using (2-7):
. ah af =xr-I=o-\ j
82'
ax'
".
So, the components of tangent vector & (image of Xo under dS,) are
Let us mention the matrix
(1
representing d/, is the lacobian matrix defining the
Xo
differential of the corresponding mapping R n
+Rm .
In conclusion, the image of Xo is expressed in the natural basis of TzoN in the following form:
3.3
DIFFERENTIAL OF A FUNCTION Let U be an open of M containing xo.
Let us apply the preceding notion to the case of a h c t i o n g E O(U).
Figure 29 The differential of g at x,
EM
is the (linear) mapping
dg, : T x o M +,,T(,
such that, Vh E Cm(g(xo)):
R : Xo t-,dg,X,
Tangent Vector Space
&,XO ( h ) = Xo( g ' h )= X,(h o g )
In conclusion, we have obtained the following result: (2-13)
Since dg, Xo is a vector of R (and has thus only one component) it is identified with its component that is the real Xog. Therefore, we can say the following. PR9
The derivative of the fbnction g in the direction of vector Xois the image of Xo under the differential of g at xo.
4. EXERCISES Exercise 1. 7
En R ~express , the tangent vector at point (xoyo)("instant ' x = xocost - y, sint
f = 0)to the curve of equations
y = x , s i n t + yocost.
Answer. In this example, where x' = x and x2 = y, we have:
and hence
Exercise 2, Given the following differentiable mapping
f : R -,R' : t I-+ (r,t 2 ), d calculate the image of tangent vector - to R, at point t, under the differentiable mapping j dt The expression of this tangent vector will be given in the natural basis of T , ( , ) R ~ .
Answer. The differentia1 of f at point i is the linear mapping df, that associates to tangent d . . vector - its image underf at point f (t) = (t,t 2 ). dl
Lecture 2
88
Let us name
d d and - the natural basis vectors of tangent plane to R~ at At). & ay
We immediately have: d d t a dt2 a df -=--+--=-+2t-
a
a
QY
ax
ay
dt
dt ax
dt
Exercise 3. Given two differentiable mappings
f,:M+Mr
f,:M'+MR
and
between differentiable manifolds; prove that ( f , o s , > *=f;*of2'.
6%'
Answer. kfh E Cm(M") : (f2
o f i ) * h = ho(f2 o A ) = ( h o f 2 ) o f ; =fi*(hofi)=fi*(f;h) = (A* 0f;)h.
Exercise 4. Consider the group of isornetries of Euclidean plane R2 with a natural structure of manifold (diffeomorphic to R 2x s').Let (x, y , 8 ) be the composite of a rotation about the origin through an angle 8 with a translation of vector xe, + ye, where (e,,e,) is the standard basis of R ~ . (i) Make explicit the composite mapping of two isometries. (iij Calculate the differential of the following mapping
f : (u,v,#)
(X,Y,~)O(U,V>~)
at point (u, v, 4). Answer.
(d Let us compose two isometries, that is (a, b, 8 ) (a', b',O1). 0
The first isometry is such that to any
(3 G
R~ corresponds
and the second is such that
Therefore we have: coso sin8
- sin8 cos8' cos8
X
sine'
7;)
- sin cos8
+
8 - sin 01;:) sin0 cost3
(COS
(;;- IT')(;)
+
(6)
cos(8+ 8') - sin(& 8'))r) + sin cost9 b' sin($ + 8') cos(8+ 8') y The last two terms represent a translation and the first is a roMon of angle (8 + 8'). +
Tangent Vector Space (ii) The mapping f is explicitly written from the previous result as follows:
The differential is calculated at a point from the formula (2-12), that is
where x' = u, x2 = v and x3 = 4. We have thus:
d&,,,,)
e, =cosOu, + s i n B u , + Ou,
df, ,,,, ,e2 = - sin 0 u, + cos 0 u2 +0 u, 4U.v.l)e3
= OU,+ 0 % + 113
[I] I':" I:[ r:'J I:[ I:[
or in an equivalent form:
&,,,()
0 = sin 0
df,,,,, 1 =
cosQ
qu,v,o 0 = 0.
Exercise 5. In the basis
($)
of Jt3 express the vectors of the natural basis of *-torus T~ c
these vectors orthogonal ? Answer. Let ox I x 2 x 3 be a fixed system of reference of R ~ , c(o, R,) be a horizontal circle of center o and radius R,, c(0,R, ) be a circle of center 0 E c(o,R,) situated in a vertical plane through a.
Figure 30
Are
Lecture 2
90
Let 0 ~ be the ~ system ~ of ~ reference 7 ensuing ~ from a rotation of angle 0 about axis ox3 1 where 8 is the angle between the axes ox and ox1,the axis 0X3 being parallel to ox3. are The coordinates of any point of vertical circle c(0,R,) in
where 4 designates the angle locating the point on the circle 40,R,). The coordinates of this point with respect to the fixed system of reference are the following:
R, sin # where 8 E [ 0 , 2 ~and ) I$ E [0,2r)are the locd coordinates on T *.
a and a with respect to the At point (x 1,x 2 ,x 3 ) d 3 , the components of tangent vectors a0 84 fixed basis are respectively: (-(R, + R, cos 4) sin 0, (R, + R, cos#) cos 8 ,0) and (-R, sin# cost9, - R, sin4sin0, R, cosb). A tangent vector to the manifold T at (x1,x 2 J3) is expressed with respect to the basis
where
x1= (R, + R, cos #)(- sin 8)8+ (-R2 sin #)d cos 8 X'
= (R]
x 3= R
+ 4 cos /) cos 0 8+ (-R2 sin q5)d sin 8
~ C O S ~ ~ .
So, a tangent vector to the coordinate fine 8= t
(4 constant) namely
(-(R, + R, cos 4) sin r , (R, + R, cos #)cost ,0) d is really parallel to - .
ae
In the same manner, a tangent vector to the coordinate line
4 = t (0 constant) namely
(-hsint cos8 ,-R,sintsin6 ,&cost) d is really parallel to - . 34
a as
The tangent vectors -and
d
-are orthogonal vectors because
a# - (4 + 4 cosq5)sin 8 (-4sin#cos8) + (R, + 4 cos+)cosB (-4sinq5sinO) = 0
LECTURE 3
TANGENT BUNDLE VECTOR FIELD -
ONE-PARAMETER GROUP LIE ALGEBRA
Later it will be necessary to consider ("simultaneously") the set of a11 tangent vectors at all the manifold points. So, simply speaking, we are going to introduce a manifold with a vector space attached to each point p,q,... and that will be called: "tangent bundle. " Such a situation can easily be schematized in the case of a one-dimensional manifold MI. In Fig. 32, it must be understood that the lines T,M, T p . .. are boundless.
Figure 31 The intersections of tangent spaces having no meaning, it will be advisable to represent these spaces by parallel lines (vertical for instance). Make clear the suggested vector bundle notion.
INTRODUCTION First, we give some general definitions about vector bundles that will be quickly particularized.
Let E and F be (finite-dimensional) vector spaces, U be an open of E.
Lecture 3
92
D
The Cartesian product U x F is called a local vector b u d e and U is the base space.
D
The projection of U x F is the mapping
n:u~F+u:(x,y)~n(~,y)=~
by specifying that the open U x F of E x F is a local manifold. For each x tz U , the fiber over x is (x)x F , it is n - ' ( x ).
D
Let U x F, U' x F' be local vector bundles, $h,:U+Lr' #2 : U + L(F;F').
A mapping and
D
# : U x F -+ U'x F' is a local vector bundle mapping
if
4 is of class C"
C(X,Y> = ( ~ W > C ~ ( X ) . Y ) . diffeornorphism #, such that #2(x) is a (linear) isomorphism for each
Moreover, a x E U , is called a local vector bundle isomorphism.
# is a bijection of U(cS )
D
A local bundle chart of a set S is a pair (U,& such that onto a local bundle U' x F' .
D
A local bundle atlas on S is a family B of local bundle charts (Ui,#i)such that: the domains (I,make up a covering ( ) of S,
-
-
i d
'v'(U,,h),(U, ,#, ) E B , Wi flU, +0 , the overlap mappings local vector bundle isomorphisms.
4,
= #j
0
hic
A vector bundle structure on S is an equivalence class of vector bundle atlases.
D D
Uui
*
A vector bundle is a pair consisting of S and a vector bundle structure on S.
are
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra The vector bundle structure induces a differentiable structure on S. Moreover, we assume that a vector bundle is a HausdorfTspace with countable topology. Every vector bundle will be denoted by its only underlying set. Let S and S' be vector bundles.
A mapping f : S + S' is a vector bundle mapping if for each x E S and each (admissible) local bundle chart (V,#) of Sf,f (x) E V, there is an admissible local is bundle chart (U,#), f (U) c V, such that the local representative fM = # I o f o a local vector bundle mapping.
D
+-'
1. TANGENT BUNDLE Let M be an n-dimensional differentiable mapping.
NATURAL MANIFOLD TM
1.
We are going to view the union of tangent spaces to M "attached" to each point of M. More precisely, let us consider the set
I
TM=((~,x,) XEM,X,ET,M}. D
*
The tungeni bundle of M is the manifold
We prove in detail (see exercise 1) the following:
PR1
The tangent bundle TM is a natural differentiable manifold of dimension 2n.
In short, we can consider that: Given a manifold M with an atlas A of admissible charts (I/,, pi ), a vector bundle atlas of TM is the natural atlas TA = {(TU,T P ) (Ia PI A where Tp is the tangent of q, notion having been defined in lecture 0.
1
1
Indeed, firstly since the union of chart domains of A is M then TM is covered by the domains of TU. Secondly, let us consider any overlap mapping pi o ~ ~ ., The ~ , corresponding ~ ,
~91;'
T(pi o P;') = Tpi o is a local bundle isomorphism. This is obvious because if f : U ( c E ) + V ( c F)is a diffeomorphism, then Tf : U x E -+ V x F is a local vector bundle isomorphism. D
The space M is called the base space of the tangent bundle.
94
Lecture 3
D
The (canonical)projection of tangent bundle TM is the (C) surjective mapping
n,
:TM-+M:(x,x,)HX,
that is:
V ( X , X , ) E T M: n M ( x , X X ) = x . Remark that the tangent bundle projection Il, is of rank n.
D
The fiber over (x) is
D
A C9 section of tangent bundle TM is a mapping s of class @ of M into TM such that the composition of s with the tangent bundle projection is the identity on M:
nM o s = i d l M .
Notation. We denote s E
Pa), where rq&) designates the set of C
q
sections of TM.
Remark Each point of some fiber specifies one particular tangent vector. Example What is the tangent bundle of the configuration space (i.e. the space of positions) of a material point in a Keplerian force field ? The particles move in the position space which is the manifold R~- (0) = R: . If 7 and v' respectively designate the position vector and the velocity vector of a point p, then the tangent bundle of configuration space R: is T R =((F,J) ~
1.2
1
~ E R : G , ER')=R:XR'.
EXTENSION AND COMMUTATIVE DIAGRAM Let M and N be differentiable manifolds of respective dimensions n and m y f : x nz = f(x) be a differentiable mapping of M into N, X, be any vector of T I M , Zf,,, be the image of X, under f , that is K X , .
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
We can "extend" the differentiable mapping f and define:
D
A linear mapping of TM into TN is called the tangent of f ( or digereenrial of f) denoted Tf or df if to any element ( x , X x ) of TM corresponds the element (f (x),Zf(,))of TN and such that:
6) the following diagram is commutative, that is f
0
JIM= Il, 0 df
(ii) the restriction of df to each tangent space, at a point, is the differential at this
point, that is: d4T&, = df,.
Remark. Let us introduce the local coordinate systems (x i ) and Cfi) defined by charts of which the respective domains contain x and z. We can say that the mapping df associates to each pair (xJ,) the pair (z,Z,) such that the tangent vector image is
PR2
If f ,: M -,N and f, : N -+P are C' mappings between differentiable manifolds, then f, 0 f; : M + P is of class C' and
ProoJ: This C' composite mapping theorem will be later proved (see exercise 3).
PR3
id,, : M -+ M is the identity mapping on the differentiable manifold M, then d idlM: TM +TM is the identity mapping on the tangent bundle TM.
If
ProoJ This proposition is an obvious consequence of the definition of the differential
PR4 If f : M -+ N is a diffeomorphism between differentiable manifolds then !$L
: TM
+TN
is a bijection and
Proof: See exercise 4. The previous properties of Tf : TM
-+
TN allow the statement T is a covariantfunctor.
Lecture 3
96
2. VECTOR FIELD ON MANIFOLD Let M be an n-dimensional manifold, TM be the tangent bundle of M. 2.1
DEFINITIONS
D * A (tangent) vectorfield on M is a mapping X : M + T M : x k + X(x)=(x,X,) which assigns to each point x E M a pair composed of a point and a tangent vector at point x.
Terminology, We will use the expression "vectorfield" instead of "tangent vector field." Remark A vector field X on M is a section of TM. In other words, it is a mapping
X:M+TM such that noX:M+M:x~n(X(x))=x
(idl,)
where ll is the projection of TM.
D
A vector field is diflerentiable if the mapping that defines it is of class C" .
Notation. The collection of differentiable vector fields on M is denoted Therefore, we denote that X is a differentiable vector fieId on M by
X E X(M) or
x E rm(nM) by referring to the notion of a section. 2.2
PROPERTIES OF VECTOR FIELDS
'
A differentiable vector field X on M assigns to each point x
E
M a vector of T,M
ki where the coefficients X'=- are the (local) components of vector field. df Because the vector field is differentiable these components are differentiable functions of n coordinates x i (of point x).
In the following, the term "differentiable" will be generally omitted.
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
Notation. The ring Cm(M; R) is simply denoted by C" ( M ). PR5
A vector field on M defines a derivation on C m ( M )that is
X ; C m ( M )+ C m ( M :) g
I-+
X(g)
such that
Proof: The linearity property:
and the derivation rule:
are fulfilled at every point of M
We Ieave this to the reader who can use the local coordinates. PR6
The set X ( M ) of vector fields on M is a module on (?(Ad) defined by the following operations: - the addition: X ( M ) x X ( M ) + X ( M ): ( X , Y )H X + Y such that VX,Y E X ( M ) , V ~ E C * ( M: () X + Y ) g = Xg+Yg ,
-
the multiplication by a differentiable function: C m ( M ) x % ( M )+ X ( M ) : (h, X ) M hX such that VX E X(M),Vg, h E C m ( M ) : ( k Y ) g = h Xg Let us notice that the dot between function and vector is not denoted.
Remark. We consider the restriction %(U)of vector field to a domain of a local coordinate system (xi) of points x E U c M . d made up of n vector fields 7 . A basis of module is for instance the natural basis ax
3. LIE ALGEBRA STRUCTURE 3.1
BRACKET Consider X, Y E X(M).
Aflerwards, we denote Xg instead of X ( g ) .
Lecture 3
98
3.1.1
Vector field product
D
Theproduct of vectorfiefdsX and Y is an operator defined by
V g E C Q ( M ): ( n ) g = X(Yg). The product of vector fields is not a (differentiable) vector field.
PR7
Prooj The derivation property is not satisfied because V g , h E C" ( M ):
( r n ) ( g h )= X(Y(gh)) = X(gYh+ hYg) = g X(Yh) + Xg Yh + h X(Yg) + Xh Yg
is not
g ( X Y ) h+ h ( X Y ) g = g X ( n ) + h X(Yg).
PR8 The operator XY is a differential operator of second order. d
ProoJ By introducing local coordinates and putting di = -, we have:
ax'
3.1.2
Operation "bracketn
We can nevertheless construct an operator of derivation from product of vector fields.
D * The bracket is the mapping [ ] : Z ( M ) x X ( M ) + X ( M ) : ( X ,Y ) H [ X , Y ]= XY
- YX
The bracket of vector fields Xand Y "operates" on C m ( M ) , that is
[ X , Y ]: C m ( M ) + C m ( M ): g
PR9
H [ X , Y ] g=
X(Yg)- Y ( X g )
The bracket of differentiable vector fields is a differentiable vector field too.
Pro05 The linearity property m
Va,b E R,V g , h E C ( M ) :
[X,Y](ag+ bh) = a [ X , Y ] g+ b [ X ,Y ]h is immediately proved. Check the derivation rule:
Vg,h E C m M ) : [X,Y](gh)= g [ X , Y ] h+ h[X,Y]g. We successively have:
(3-2)
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
Bracket expression in local coordinates. From the expressions of (XY)g and ( Y a g in local coordinates, we have Vg E C m ( M ) :
This also allows checking that the bracket of vector fields is an operator of derivation on C m( M ) . In local coordinates, the expression of bracket [X,Y] is
In particular: [d,,djl = 0 . In such a case, one says ''coordinate (induced basis. " 3.1.3 Important theorem
Let M and N be same dimension manifolds, f be a (C") diffeomorphism between M and N, X and Y be vector fields on M.
Figure 34 Before expressing the theorem, let us point out X ( f e h )E C W ( M )
and define the following: D
The image of vectorJ5eld X under f is the vector field Z on N defined by Z : C m ( N ) + C " ( N ) : h ~ Z ( h ) = ~ ( f *f-'. h)o
We denote Z(h) = (dfX)(h) -
Lecture 3
100
PRlO Given differentiable vector fields X and Yon M, we have:
ProoJ By definition, we have for the vector field [X,Y], V h E C " ( N ): (df [ X ,YlXh)= [ X ,Y I ( f ' h ) o f - I = X ( Y ( f * h ) of - ' - Y ( X ( f ' h ) ) o f - ' = X(Y(h0 f ) o f - ' 0 f ) o f - I - Y ( X ( h 0 f ) o f-lo f ) o f-' = X((Y(h0 f ) o
f - ' ) o f ) o f-'-~ ( ( ~ (f )hoof - ' ) o f ) o f-'
but Y(h0 f ) o f-'= (df Y)(h)
X(h O f )
f -' = ( d f X ) ( h ) 7
thus, omitting parentheses, we continue:
3.2
LIE ALGEBRA
PRll The module # ( M ) provided with the (inner) composition law, namely the bracket law, is an R-algebra.
'
Proof: The module X ( M ) is evidently a real vector space.
In addition, the bracket is a bilinear law:
+x~,x,I =[x,,x3I+[x~,x~I 'Jx,,X2,x3 E x ( M ) : [x~,x2 +x3l= ~ X I , X ~ ] + [ X I ~ X ~ ~
vx,,x,,x3E
: [XI
Vk,,k, E R, V X , , X , E % ( M I :[klX1,k,X,l= klk2[Xt,X,l The previous properties are proved by way of exercises. PR12 The R-algebra X ( M ) is an algebra such that the bracket law is anticommutative and
the Jacobi identity is verified.
'
Given a commutative field K , we recall that an algebra on K or K-algebra is a vector space E on K provided with a bilinear mapping +: Ex E + E. In other words, it is a vector space K,E,+ provided with an (inner) law such that: 'dx, y, z E E : (x+y)+z = x+z + y+z Vx, y, z E E : x*(y+z) = x * y + X + E V k l ,R2 E K , Vx, y E E : &,x) + ( k 2 y )= klR2 ( x * ~ ).
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra Proof: First, the law bracket is anticommutative:
V X ,Y
E
X ( M ) : [ X ,Y ] = -[Y, XI
since V g E CP( M ) : [ X ,Y J g= X(Yg)- Y ( X g )= -[Y, X ] g . Let us mention that [ X ,X] = 0 . Finally, the Jacobi identity is V X ,Y,Z E X(M):
(see exercise 8). Let us introduce a fundamental definition.
D * A Lie algebra is an algebra for which the inner law is anticommutative and satisfies the Jacobi identity. From PR12, we can say: PR13 The R-algebra X ( M ) is a Lie algebra.
PR14 The bracket law is not associative.
3.3
LIE DERIVATIVE
Bracket allows the introduction of the Lie derivative notion that will be discussed in general in lecture 6.
D 6;r
The Lie den'vatratrve of vector field X with respect to vector field Y is the vector field defined by
LYX = [Y,xj.
(3-6)
In local coordinates, let us consider the two vector fields
The Lie derivative of X with respect to Y is the following vector field:
L,X = [ Y , X ]= (Y'B,x'
-
x-'~,Y') a, .
(3 -7)
PR15 If f : M -+ N is a diffeornorphism, then 'dX E S(A4): L, : % ( M )+ X(M) is natural with respect to following diagram is commutative: (i) the mapping
that is the
Lecture 3
(iQ L, is natural with respect to restrictions, that is for any open following diagram is commutative:
Proof The first assertion is obvious since VY
E
U c M the
X(M):
L 4 x d f Y = [ d f X , d f Y ] = d f [ X 7 Y ] = dLxY f The second assertion, namely VY E Z ( M ):
L,," YIU = (L*Y)Iu
is also clear from the equality d ( f 1 ~=)d f l following ~ from the definition of d. The Lie derivative notion plays a fundamental role in mechanics; we will go back to it.
4. ONE-PARAMETER GROUP OF DIFFEOMORPHLSMS What follows is very important (notably in physics) because, for instance, the various states of a material system can be described by a moving point in a position and momentum space. A problem of major interest is to study a "flow" obtained from the solution of differential equations.
4.1
DIFFERENTIAL EQUATIONS IN BANACH
Let E be a Banach space, fl be an open of R x E, J be an interval of R, f : R + E : (t,x) Hf ( t , x ) be a mapping that will be specified later.
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra 4.1.1
Integral curve
D
A curve z : J + E : t ~ i ( t ) is an integral curve of the differential equation
V t E J : ( t , X(t)) E R
and
dz
( t )= f ( t ,Z ( f ) )
The existence of an integral curve means that a tangent vector f ( t , x ) is assigned at each point x of the curve. Remark. If the vector space E is R", then the vector differential equation is equivalent to a system of n differential equations:
4.1.2
Existence and uniqueness of solution
D
A mapping
f : n + E : ( t , x ) f~( t , x ) is locally L&schitz in x if, for every neighborhood V of R, we have: W j x , ) , ( f , x 2 E) V ,3 k E R+ :
I/f ( t , ~ ,-) f ( t , x 2 ) /5/ kllx, - x, 11.
PR16 Given a continuous and locally Lipschitz in x mapping f : n + E : ( t , x ) f~( t j x ) then for every point ( t o , x o )of there is a neighborhood of xo in E and an interval [to- a , t o +a]of R such that the equation
has only one differentiable solution F : t I+ T(t) defined on [to - a ,to + a ]and such that x ( t o )= xo. 4.1.3
Differential equation and vector field Recall that to each "instant" t corresponds a point of a manifold M J
+M :t H x ( t )
and that a vector field on M is a mapping from M to TM which assigns to each point x of M one vector in T,M .
Lecture 3
104
To every vector field on Mis associated a first-order differential equation
= x(x(t)>
with x(0) = x,
and conversely. 4.2
ONE-PARAMETER GROUP OF DJPFEOMORPEISMS
Let M be a differentiable manifold, J be an intervai of R, X be a vector field on M. Question. Given a vector field, does there exist a curve that each tangent vector is a vector of this field ?
4.2.1 Local transformation of M
D
An integral curve of field X on M is a curve
c:J+M:twc(t) such that dc dt
-( t ) =
X(c(t)).
Re-examine the theorem of local existence and uniqueness of differential equation solutions.' PR17 Considering X E X ( M ) and a point of M, there exists: - a neighborhood U c M of this point,
- aninterval I , = ( ~ E R : I ~ ~ < E , . F E R + ) ,
-
a differentiable mapping #:I,xU+M:(t,x)~fit,x)=#,(x) such that: I, +M : t H 4t(x) is an integral curve of X with 4,,(x) = x . We can introduce the following definition.
D
A local transformation of Mgenerated by vector field Xis a difieomorphism between neighborhoods of M
~,:U(CM)+M:XH~~~(X) verifying the following differential equation
and
Manifolds, fields and mappings are assumed of class C".
Tangent Bundle, Vector Field, Oneparameter Group, Lie Algebra
105
For each t E I,, the diffeomorphism 4, maps U to some other open #,(U) following the integral curves (respectively for each point). The following drawing suggests the idea of flow at given time t.
From the previous context, we define:
D
A flow box of X at x E M is a triple (U,E,#).
We specify that for each point x E U the mapping 4 leads to a locally unique integral curve. More precisely:
Uniqueness theorem of flow boxes :
If (U,E , #) and (U', E',#') are flow boxes at x E M , then # = 4' on ( I , flI,,) x (U fl U') . ProoJ Putting I = I, f l l ,we have vx
UnU'
# / I x ( , )=
4b,(,)
because for each x E U there is an integral curve of X (at x), namely: c, : I
+ M : t H C, ( t )= 4(t,x)
and it can also be proved if c, and c: are two integral curves of X , then c, = c: on the intersection of their domains (see e.g. Founciations of Mechanics, R. Abraham and J. Marsden, 1978).
In this reference book, the following theorem is easily proved: Existence theorem of flow boxes:
If X is a vector field on a manifold M, then there is a flow box of X at each x E M. 4.2.2 One parameter (local) group of diffeomorphisms
PR18 Every vector field on A4 generates a one-parameter (local) group of diffeomorphisms on M. Proof: Let U be an open of M, r={t:ltl<E,E>o),
i2 be a neighborhood of (0)x M .
Lecture 3
Consider a local transformation
U ( c M )+ M : x H#,, ( x ) with t , s, t + s E l This transformation satisfies, like # [ ( x ) ,the differential equation associated with field X Let us use the uniqueness theorem. The following integral curves associated to X
having for t = 0 the same value
#s
(x) [because
40(4s(x)) = #s (x)] , are such Vt,s,t + s E I :
Therefore, the local transformation of Mare such that
Vt,s,t
+sE I :
In particular, we deduce
4-3 = 4;' because (4-* O
)(XI= #-s+s (4= 4" (XI.
In conclusion, the local transformations of M have a group structure with composition law, 4" being the identity element; we say:
D
The pair
defines a one-parameter (local) group of &!eonwrphisms on M
Remark. We must emphasize the previous diffeomorphisms are local and not global on M. The local transformations 4, are defined in any neighborhood of each point xo in M for
"instants" of the interval defined by
Iri < ~ ( x , ) .
4.2.3 One-parameter (global) group of diffeomorphisms We are going to consider globally the flow of vector field, extended as far as possible in the parameter t.
Let us consider
-
CI=RxM
and @:RxM+M:(t,x)~$,(x)
where the diffeomorphisms of M ( V t E R ) :
#f : M -+ M : X I + # ~ ( X ) are such that: (i/
4,, is identity on M,
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
Under those conditions, we define: D
(a,@)
We say or more simply diffeomorphisrnson M
#, is a
one-parameter (global) grorrp of
Example. For M and every t E M , the mapping @ such that @(t7X ) = e'x
defines a one-parameter (global) group. It is the one-parameter group of homotheties. In the case of global diffeomorphisms on M, let us notice that PR18 is not necessary verified. so, PR19 Every vector field on Mdoes not necessarily generate a one-parameter (global) group of diffeomorphisms. Proof: Let us give the following counterexample.
vector field on R Consider the (e")
The associated differential equation is -'h = x 2, dl
Since 1 d (- -) = dt X
then, for the initial condition x ( 0 ) = x, (# 0 ) , the integral curve equation is xo . x(t) = -
1 - tx,
1
This soIution is not defined for r = - . xo
So, the mapping
#',
Let Ex be the set of (t, x ) E R x M such that there is an integral curve c : I with t E I.
D
1
such that the point -corresponds to x,, is not defined for t = -. 1- tx,, Xo
-+
M of X at x,
The vector field X is complete if Ex(cR x M) is R x M; in other words: if the vector field generates a one-parameter (global) group of diffeomorphisms on M.
Lecture 3
108
So, a vector field X being complete gff each integral curve can be extended so that its domain becomes (-a,*), the reader will conclude that the vector field of previous counterexample is not complete. Every integral curve is not defined at every "instant." Remark. In mechanics, most of (Harniltonian) vector fields are not complete. Problems with singularities in elasticity, in general relativity, and so on, are incompleteness examples unlike endlessly persistent dynamics problems.
From the uniqueness theorem of integral curves, we express the following definition.
D
We call integral of X the unique mapping 4, : Ex +M such that, for all x E M , the mapping t H @x (t, x) is an integral curve at x called maximal integral curve.
Linked this way, we say:
D
If X is complete,
#, is called the flow of X and
( M , W , #)~a flow box.
Applications. Every vector field on a compact manifold is complete. It necessarily generates a group of global diffeomorphisms.
More generally, the reader will prove what follows: If X is a Ck vector field ( k 2 1 ), if c : I +M : t H c(t) is a maximal integral curve of X and if c((a,b)) lies in a compact subset of M for every open finite interval (a,b)in the domain of c, then c is defined for all t 6 R. It is sufficient to prove that b E I (and analogically a E I). The reader will consider from t , (E (a,b))+ b a sequence c(t,) which converges to a point x E M (because compactness) and then a neighborhood of (0,x) E R x M . He will prove that c is extended to a time greater than b. Consider as it were the converse of PRI9. PR20 To each one-parameter group of diffeornorphisms on M corresponds a vector field X called a generatingfield of group. ProoJ: Let 4 be a one-parameter (global) group of diffeomorphisms on M, c be a curve of M through xo:
Xo be the tangent vector to the curve at xo, that is
Let us prove at any point x E c there is one tangent vector X(x) also denoted X,. . By hypothesis, the diffeomorphisms being elements of a one-parameter group, we have: 4,+s(xo)= 4,(#3(x0)) and thus
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
If t = 0,this result becomes:
It is perfectly well that the tangent vector a x ) to curve c at point x = #s (x,).
Figure 36
D
An orbit of one-parameter group is an integral curve of vector field, that is a curve tangent to vectors of field X
To sum up: sa (Tangent) vector field
a One-parameter (global) group
3 one-parameter (local) group 3 one-parameter (global) group 3 generating field of group.
4.2.4 Second order taogent bundle This concerns the tangent bundle of a tangent bundle. Let us see how we shall associate with a vector field on TM a second-order differential equation. Let TTM be the tangent bundle of a 2n-dimensional tangent bundle TM. Any element of TM is a pair (x, X,) such that in a local coordinate system, the tangent vector at point (x') E M is denoted y r = -dx' . ax dr Following a normal usage, we denote any element of TM by its 2n coordinates (xi,y'). In the same manner, let
X , = r y 8y
with
Lecture 3
110
be a tangent vector at any point ( x ' , ~ ' )of TM; it is the second element of a pair of TTM. Following the normal usage, we denote an element of TTM by the following 4n coordinates (x',y';u',vi). The 2n first-order differential equations
are equivalent to n second-order differential equations, namely: d 2x i ah' vi(xl,-), dr df which are obtained provided that we put yi = ~ ' ( X J , ~ ] ) . Introduce the following commutative diagram
Consider the vector field
z : TM -,T T M : ( x ' , ~ ' H ) (x',Y';u',v') and the mapping ll,: TTM
4 TM : (x',yi;u',v') I+
(xf,u').
We have
(n**z)(xl,y') = ~ I * ( X ~ , ~ ' ; =U(xt,ui). ~,V'~ In other words, putting ui = y i as in the case of second-order equation existence, we have:
So, we can express:
PR21 A second-order differential equation on M is associated to a vector field Z on TM n,oZ=id\, .
iff
(3-9)
Remark 1. At point (x i ), the tangent vector of a pair of TM is expressed as
because
(24' =
y').
Tangent Bundle, Vector Field, Ooeparameter Group, Lie Algebra
I11
Remark 2. A solution of a second-order differential equation on Mis a differentiable curve
c:J-,M:t~c(?)
such that L.: J + T M :t wL.(t)
is an integral curve of X .
5. EXERCISES Exercise 1.
Show that the tangent bundle TMof an n-dimensional manifold has a natural structure of a 2n-dimensional differentiable manifold. Answer. 1O. Let us define a chart on TM.
Let
&u,,pa),e, 1 be a set of charts making up an atlas of M, where tpa is a homeomorphism
pa : Ua4 R ' : x H (XI,...,x n ) . So the opens in M are homeomorphic to opens in R". In the same way, for each x E M , we can define a second diffeomorphism: : T f l + Rn ; X, H(x', ...,X") where the X' are the components of X, relative to the basis of T,M associated to ( x i ) . To domains U , covering M correspond the following domains covering TM : T% =((x,x,) : X E U , , X , E T , M ) . These are the IT-'(u,) which actually cover TM because the opens U , cover A4 and ll: TM +M is surjective. Finally, we naturally define the homeomorphism ya of TU, into R& from the product of homeomorphisms qa and pi, namely:
So, the local coordinates of a point (x, X,) x', ... $$,. ..
E TM,
in a chart (TU,,(v, ) are the 2n reals
,xn.
24 Now let us examine the differential manifold structure of TM
Figure 37
Lecture 3
112
M being a differentiable manifold, remember that tp,aq;'
: R n+ R n
is differentiable. Next, the formulas
imply that
pi o ,pi-' : R" + R"
is differentiable. Finally, let Uaand
(la be
two opens in M(chart domains) with nonempty intersection.
Let V, = K 1 ( U a ) and Vg = II-'( U D )be the corresponding domains of charts in TM
Consider the corresponding homeomorphisms Y , : Va + R ~ :"(x, X,)H (xi,X J )
va: Vs + R'" : (x, X,)
H
(F~,X')
of TM into R ~ . The mapping Yp " Ya- l :
v a ( ~ - l ( ~~a U
~ ) ) + V ~ ( H -nufl)): ~(U,
is a ((P") diffeomorphism. 39 The tangent bundle TM is a Hausdodf space. Consider two tangent vectors, one at x the other at y. There are two cases are to view: - First, the two points are distinct ( x # y) . The opens U x and U ,, regarded as chart domains in Mare such that U ,
nU,,= 0 .
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra These two neighborhoods go back into TM under the empty set:
(0 mapping
IT1and we have the
rr-'(~,)fln-~(u,) =0 . - The two points are not distinct (x = y). Let X, and Y, be two distinct tangent vectors of T,M. Let ( U , q ) be a chart such that x E U . The 2n-tuples of R ~ * ( x l ,..., x n , x1,..., X n ) for ( x , X , ) 1 (X ,..., xn,yl ,..., Y n ) for tx,Y,) n are different since ( X I ,. .., X" ) # ( Y ' ,..., Y ). Let q ( U ) be an open of R" containing (xl,. .. $), #?, be an open of R" containing (A?, ... X), #?,, be an open of R" containing (Y',... ,Y), V / : T U + R * ' : ( X , X , ) ~ ) (,..., X x~" , x l,...,X n ) . In TU we have the empty set:
a.
~ - l ( ~ ( ~ ) x ~ ~ ) n ~ - l ( ~= cu)xp,)
4". The tangent bundle is with countable basis. It's easy to prove by considering y-'.
Exercise 2. Make explicit the differentiable manifold structure of the tangent bundle of S'
'
Answer. Let (U,,p, ) and (U,,p, ) be charts of an atlas on S such that:
U,= ((x',x2) E S' : XI < I} pl : Ul +]0,2n[ : (cos e,sin e)H 8
The following mapping between opens in R:
is a diffeomorphism such that
'
Any point of S can be defined by e'B"'.The tangent vector, at
r = 0, is defined by
)' (0)= i 8 ' ( O ) ei*'").
(ele
I
The tangent bundle of circle is the set kx, x,) x E s',X,
'
E
113
T,s').
The change of charts p, 0 pi1on 5' gives rise to a change of charts on the bundle.
Lecture 3
114
So, we have seen that p, op;' (everywhere) the identity.
is a diffeomorphism between opens in R which is not
For tangent vectors, the following change of local coordinates between opens in R pi 0fpi-l : X' I+ Y '
is such that
and
The tangent vector components are the same and ?he differentiability is assured. The differentiable manifold structure of TS' is so made explicit, this tangent bundle being evidently of dimension 2.
Exercise 3.
If f, : M + M' and f2 : M' 4 M u are mappings of class C' between differentiable manifolds, show that f, 0 f ; is a mapping of class C' and d(f2o.#i)=df20ddfi .
Answer. Let (U,p), (U', p')and (Uw,p") be respective charts of M, M' and M" such that
f,(W c u' The local representative of f, o A is
and
f2
(U') c U".
The C' composite mapping theorem implies that the previous expression is of class c'. By definition we have
d(f, J;Xx,tcI,)
=
((A
f ; ) ( ~ ) y E f i
)
oJ; a ~ l c ~ , o ~ x x ,
and (df, a ddJ;)(x,[cf,)= #2tf,f;(x) =
3
[A
cl,,,,)
((52 O J;)(x), [ f 2 A O
CI(,~.~XX)
thus d(f2 0 f i ) = d f 2 0 4.
Exercise 4. If f : M + N is a diffeomorphism between differentiable manifolds, prove that df : TM 4 TN is a bijection between the corresponding tangent bundles and (cay)-' = df - I .
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra Answer. Since f is a diffeomorphlsm, the compositions
f-'0 f = idlM
f o
f-'=idlN
lead (referring PR2) to d(fl)odf = d(fU'o f ) = d i d ( , . The PR3 implies that d(f-')odf =idl,. In the same way, we have:
qod(f-')=idl,
.
As a general rule, remember that #:A+B is a bijection z f l there is
ry:B+A
such that /oy=idl,
and
yo/=id],
Then, we can assert df is a bijection of TM onto TN .
The mapping df is evidently the inverse of d( f -') .
Exercise 5. Prove that
VX,Y , Z
E
X ( M ) : [ X + Y ,Z] = [ X ,Z] + [Y,Z] .
Answer. Vg E C" ( M ): [ X + y, Zlg = (X+ Y)CZg) - Z( ( X + YIg = X ( Z g ) + Y(Zg)- Z(X-.) - Z(Yg) = [X, Zlg + [ Y ,Zlg.
Exercise 6. Prove that V X ,Y E X(M), V a E R : [ax,Y] = a [ X ,Y ] .
Answer. Qg E C w ( M ): [ a X , Y ] g = a X(Yg) - Y ( aXg) = a X(Yg)- a Y ( X g )= a[X,Y]g .
Exercise 7. Prove that VX,Y
E
X(M), Vg E CQ(M): [gX, Y] = g[X, Y] - Yg X.
Answer. Vh E Cm(M) :
[gX, Y]h = gX (Yh) - Y ( g X h ) = gX(Yh)- gY ( X h )- Yg.Xh = g[X, Y jh - Yg
m.
115
Lecture 3 Exercise 8.
Establish the Jacobi identity VX,Y,Z E X ( M ) : [X,[Y,Z]]+[Y,[z,x]j+[z,[x,yl] =0
and deduce: L[X,,]Z= CL, LY 12. 3
Answer. Vg E C" (M) :
Let us add the second members: X(Y(Zg) - w . g ) ) + Y(-wKg)- X ( Z g ) ) + WV'g) -CY,ZI(Xg) - [~,XI(Yg) - CX,YI(Zg) = 0.
-Y(Xg))
So the sum of first members shows the Jacobi identity. This identity [X,[Y,z11+[Y,[Z, XI1 = [[X, Y1,ZI is written ~ , [ y ~ ~ l + ~ , [L~,,,y,, xzl = that is L I x , , ] Z =L x L y Z - L y L x Z = [ L , , L , ] Z Exercise 9. In R 2 let
(-$)
Calculate the bracket
= (dx,a,,)
be the natural basis (with respect to Cartesian coordinates).
[i,,iej of unit vectors in the polar coordinate system.
Answer. The vector fields tangent to respective coordinate lines are -1, =cos@-+sined a (0constant) ax 3Y
-I,
The components of field Y' =-sine, Y' = cos8. The bracket of fields
a
= -sin 8 -
dx
ir are
ir and i8 is:
a
+ cos 9 @
( r constant).
X 1 = cos8, X' = sin8
and the ones of
&
are
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
117
Exercise 10.
Show that the vector field on lZ2
gives rise to a one-parameter (global) group of diffeomorphisms that will be specified. Answer. The vector field may be interpreted as the right-hand side of the following system:
Given the initial conditions x(O)=x,
and
y(O)=y,,
the differential system admits an integral curve c:R
+ R2: t H( x ( t ) , y ( t ) )
defined by x(t) = x, e'
~ ( t=)Y O e'
and through ( x o j 0 )at initial "instant." The field is complete because each integral curve is defined to each "instant." The integral curves are semi-straight line without origin. The oneparameter group is made up of the following diffeomorphisms: : R'
-+ R'
: ( x ~ , Y ~ )4, ( x o , ~ , =) ( x ( t ) , ~ ( t = ) ) (xoe',~oe').
Since 4 ' ( ~ 0 , ~=0e)1 ( ~ o , Y 0 ) , we conclude that
4' is a homothety (center o, ratio e').
Exercise 11.
Does the mapping
a:R
X R -+ ~ R~ : ( t , ( ~ , ~ ) ) ~ ( ~ + t , ~ - 3 t )
define a one-parameter group? Answer. Putting the following element of R
Lecture 3
118
Q(t, u) = 41(4
9
we have:
(iii) On the one hand, we have
#_,(x,y)= (X-S,Y +3s), on the other hand )(x,Y) = #;I(x + S.Y - 3s) = (x,y) (4;' thus #il(x, y) = (x - S,y + 3s). Consequently 4-, (x,Y)= #;'(x,Y)*
9
A one-parameter group is well defined.
Exercise 12. Does the mapping @ : Rx
R Z+ R 2: (t,(x,y)) H(tx,y-x)
define a one-parameter group? Answer. Because #O(X?Y)=(O,y-x)+(x,v), 0 doesn't define a one-parameter group.
Exercise 13.
Does the mapping @: R X R 2
+ R2: (t>(x,y))H ( ~ ~ 0 ~ 2 t - y s i n 2 t , x s i n 2 t + y c o+s t2)t
define a one-parameter group? Answer. We successive~yhave: (i)
9
(I
40(x,y) = (x>Y)> #$+,(x,y ) = (XCOS~(S + t) - ysin 2(s + t),xsin2(s + t ) -tycos2(s + t) + s + t) 04~(x,y) = ((xcos2t -ysin2t)cos2s-(xsin2t + ycos2t +t)sin2s, (xcos2t - ysin2t)sinZ.s +(xsin2t + y c o ~ 2 t + r ) c ~ s 2 s )+ r = (x(cos 2s cos 2f - sin 2s sin 2t) - y(cos 2ssin 2t + sin 2scos 2t) - t sin 2s, x(sin 2s cos 2t + cos 2s sin 2t) + y(cos 2s cos 2t - sin 2s sin 2t) + t cos 2s + s ).
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra We clearly have @s+r
f
4
41
and it is not a one-parameter group. Exercise 14. Prove the mapping
a:R
X R+ ~ R2 :(t,(x,y))w (XCOSI-ysint,xsint+ycost)
defines a one-parameter group. What is this group? Answer. We successively have:
(0
~O(X,Y) = (x,Y)
=((xcost -ysint)coss-(xsint +ycost)sins, (xcost- ysinr)sins+(xsint + ycost)coss)
+
= (xcos(s + t) - y sin(s + t), xsin(s + t) ycos(s + t) )
= #$+, (4Y). (iig
4-,
(x, y) = ( X COS(-S)- y sin(-s) ,x sin(-s) + y cos(-s) ) = (xcoss+ysins,-xsins+ ycoss),
but #s(x,y) = (cOss sins hence $4i1(., ). =
- sin $) coss
('""
sins) -sins coss
();
(;]
= (xcoss+ysins,-xsins
+ ycoss)
= #-,(X,Y).
So, the one-parameter group is the plane ratation group. Exercise 15. Show that the vector field on R
a
a
is complete. SpecifL the one-parameter group
Lecture 3
120
Answer. This field is associated to the differential system:
dr
@=x. dt
-= - y
dt
Given the initial conditions x(0) = x, ~ ( 0=) Yo the differential system admits an integral curve:
9
c : R -+R*: t H ( ~ ( t )y(t)) ,
such that x(t) = xocost - yo sin t
y(t) = x, sint + yocost
and passing through (xoj0)at the "initial instant." The field is complete. The integral curves are concentric circles. The one-parameter group is made up diffeornorphisms
4' : R2 -+ R2 (xO>YO I-) 4 t ( ~ o r ~=o )( ~ ( t ) , ~ ( t. ) ) Since #,(x0,y,)=(x, cost-yo sint,xo sint + yo cost) we can conclude #, is an angle t rotation. The rotations form a group. In this problem, the use of complexes is more profitable:
x(t) + j ( t ) = (x,
+ iyo)ei'
4,(~09Yo)= ~~'(xOIYO). This result means rotations:
#,
is an angle t rotation and the set of diffeomorphisrns forms the group of
{#,
1
~ E R } =
SO(2;R).
Exercise 16. The configuration space Mof a particlep moving in a Keplerian force field, such that
10)
Show that Newton's equations of motion confirm PR20;that is more precisely, a is R 3 second-order differential equation is associated to a vector field Z on TM. Show that the vector field Z is not complete.
Answer. We have TM = ( l Z 3 - @ ) ) x l t 3 = ((7,~))
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra
where 7 and u' are respectively the position vector and the (tangent) velocity vector of p. Consider Z : TM
and
-+
TTM : (?,ii) H (r',u',v',G)
n, :TTM + TM : (7, G,?, w)H (F, c).
The differential system associated to Z is du' d? - v-rn-=w. dt
dt
Remember that the condition
leads to a second-order differential equation:
and
The vector field Z is not complete because there is (at least) an incomplete integral curve, namely the energy:
~=irnu~+v where
k
~ = j f . r ~ = - r
(with an additive constant).
We have lmu2 2
--k -- c r
(constant).
The velocity of p is not limited if r -a oo. The velocity becoming infinite after a finite lapse of time implies Z is not complete (singularity!). Exercise 17. (Important!) With the help of diffeornorphisms give an interpretation of the Lie derivative of a vector field X with respect to vector field Yon a manifold M. Given one-parameter groups of diffeomorphisms #, and v,, of which X and Y are the respective generating fields, show that the curve
is differentiable at t = 0 and admits [X, Y] = L, Y as a corresponding tangent vector. Answer. (i) Let xobe a point of M (at t = 0 ), #, be a one-parameter group of diffeomorphisms, X being the generating field, Y, be a one-parameter group of diffeomorphsms, Y being the generating field.
Lecture 3
122
Every point x E M in the neighborhood of xo admits an expansion about t = 0 :
So, the point x,, image of xo under 4l is: X, = [ I + txfai+ +t2xia,(xja,)+ o(t3)]xlo
Y
Figure 39 Also, for the point x, = y / , ( # l x o ) , we have:
In the same way, for the point
x, = l y , x o , we have:
and, for x4 = 4, (cy,xo), we have:
Let us compare xz with x4 and remember x2 is obtained "following ' the successive orbits of X and Y, and x4 is obtained "following" the successive orbits of Y and X: 7
X,
- X,
+ - ( I + txia,+ +r2xiai(xja,) + tvrar + r2x1a,(vrar) + f t 2 ~ r a , ( ~ s+ao,()t 3 ) ) ~ l o = t 2 ( Y r a r x t a-ix i a i y r a , )XI, + o ( t 3 ) = (1 + t~ '8, + f t z ~ r a r ( ~ +txial " a , ) +t2yra,(x*a,)jt2x'a,(xJaj) + o(r3))x),
=t2(yiaixr x i a i y r)ar =
X/
, + o(t3)
t 2 [ y , x l x 0+ ~ ( t ~ ) .
Taking the limit as t
0 and the previous expression being legitimate for every xo
+ D
have:
Iim ( v I
I+Of
o
4, )X t- ( A
YI
)X
= Iy, XI, = (L, X ) , .
E
M, we
123
Tangent Bundle, Vector Field, One-parameter Group, Lie Algebra The reader will immediately interpret this result, by "following" the orbits in the right order (that is imposed by the previous formula). (it;)
The following limit expressing the tangent vector (at f = 0 ) at point x
lim ( 4 - J 7 0 ~ - f i 0 @ f i ~ ~ J i . ) x - ~ exists r f l
3 lim (4-1 O ly-, O 412 O
wax - x
t
1+0+
l f 3 lim
O V : ) X - ( Y 0, 4 t ) x )
4.t O V - , ( ( # :
1+0+
tZ
8 3 lim (4, ,-to+
v, )x - (Y, o t
M:
( because #o = yo= i d )
t
r+o+
E
#I
)X
2
It follows from (i) this limit, at point x, is:
LECTURE
4
COTANGENT BUNDLE VECTOR BUNDLE OF TENSORS
1. COTANGENT BUNDLE AND COVECTOR FIELD
In lecture 0, the definition of a differential one-form on R n was recalled. It is immediately transposable to tangent vector spaces. Let M be an n-dimensional manifold.
1.1.1 Definition
D
'
A l-form or covector at point x E M is a linear form on T,M.
Such elements of
L,(Tfl; R) are denoted wx,Px,ax...
So, the definition of a 1-form or covector w, is expressed as follows: w,:TXM+R:X~wx(X) with
V a , b R,VX,Y ~ ET,M : w , ( a X + b Y ) = a w , ( X ) + b w , ( Y ) . PRl
The duality between 1-forms and vectors is expressed by
W A X >= X(w,) . Denoting by ( , ) this natural "pairing" or contraction between vectors and 1-forrns, the previous equality is written: 6-w
x)= w, (X)= X(w, ).
(ox,
Also called dierentid one-fonn.
(4- 1 )
Lecture 4
126
Example 1. The row vectors are 1-forms. With the multiplication of matrices, a row vector (linearly) associates a real to each column vector. For instance
Example 2. In quantum mechanics, the I-forms called bras complexes
(6,y l )
to vectors called kets
(1
(linearly) associate
Iv ).
D * The cotangent vector space, at x, of M is the dual of tangent space TxM . We denote
T:M this space of covectors (or 1-forms) on T x M . The vector structure of T;M is immediately verified since:
Dual terminology. The vectors of TxM are sometimes called contravariant vectors and the ones of T,'M are called covariant vectors. This old terminology refers to components of these vectors. Let us introduce bases. 1.1.2 Expression of a 1-form Let us remember the basis of T,Massociated to coordinate (xi) is made up of n tangent vectors d ei =ax' '
the index x being omitted. What is the corresponding basis in the cotangent space T,'M? D
The cobasis of T i M or dual basis of basis (ei) is composed of n basis vectors denoted h i each , linear form dr' being the differential of the projection x t-,x' and such that
where 6;= 1 if i = j and
= 0 otherwise.
Cotangent Bundle, Vector Bundle of Tensors We denote
PR2 The expression of a 1-form oxrelative to the dual basis (du')is 0, = Oi dxi
where the reals 0, = o , ( e , )= ( o x , e , )
are called components of o relative to (dx*). Proof On the one hand, we have V X E T,M :
m i ( x =) ( d r l , x )= xi since
h i ( X )= &'(xje,) = xj4
2 :
X' .
On the other hand, we have:
(a,,x)= q ( X )= w , x t since
w , ( X ) = w X ( x i e ,=) w , ( e , ) x i . From (4-5)and ( 4 4 , we deduce the real
cu, ( X ) = widxi( X ) that implies W x = w,drl.
Remark I. The formula (4-5) following from the duality expressed by (4-2) shows that dr' associates to X the ith component X', Remark 2. According to convention, the components of vectors show an upper index and the components of I-forms a Iower index. Recall that the Einstein summation convention about indexes is systematically used and that any basis vector ei is distinguished by a lower index while any one-form dri of cobasis is characterized by an upper index. Remark 3. The differential of a function g at point x E M, namely
d . , = d,g(x) dx' is an example of 1-form. 1.1.3
Change of cobasis
As a general rule and with recognized notations, we know that
Lecture 4
e;. = a,i e,
s
ef*j
= p; e*i i
e*' = aje
where a and
(1)
I*/
(21
fl are inverse matrices. a
Denote by (0') the dual basis of (e;) = (a;) . ax'
The relation (1) is written
The relation (2) is written
6" = p/dx'.
dr' =a: 8'.
We have = m;. 8' =
0: fl/ dr '
but Ox= U' dri
then, the formulae expressing the component changes are
w
0;.=
w, = /?: a;.
a; a,
More particularly, let a cobasis change be ( k t++ ) (dr").
It's obvious that:
Relative to the "new" cobasis (du"), the 1-form i s written:
but W,
= mi dxi
then
Remark Let us remember the following formulas i
e: = aJ . eI
= afmi
x'j= f l ,x'
We note that the law of component change for a 1-form is the one of change of basis vectors. It's not the case for a vector: the matrix is inverse. That's the reason why, initially, every vector of tangent vector space was called contravariant vector and every 1-form was named covariant vector. This terminology is logically given up because vectors and forms exist as their own entities regardless of any basis change.
Cotangent Bundle, Vector Bundle of Tensors 1.2
COTANGENT BUNDLE
The cotangent bundle plays an important role in various applications. In mechanics, we shall see the cotangent bundle is the phase space of conservative systems. Let us consider the union
U T ~ofMall cotangent vector spaces of M at every point of M. x€hf
In other words, let us view the set denoted
T'M. More precisely,
D * The cotangent bundle to M is the manifold
Indeed, we prove (see exercise I ) the following PR3
The cotangent bundle TOM has a natural structure of 2n-dimensional differentiable manifold.
Example. The cotangent bundle to the configuration space M = R~ - (0)
of a particle
moving in a Keplerian force field is T' M = ( R-~{o)) x(R~)*, D
The basis of the cotangent bundle is M
D
The (canonica1)projectionof cotangent bundle T'M is the (C") surjective mapping
n$ : T*M + M :(x,w,) H X. In other words Q(x,o,) E T*M : II;(x,w,)
D
The cotangent fiber over {x>is (H;)-'(x)
D
= x.
= ( X ) X T:M.
A C? section of cotangent bundle T*M is a mapping s of class CQ of basis M into T * M such that the composition of s with the cotangent bundle projection is the identity on M
Lecture 4
1.3
FIELD OF COVECTORS
D
A field of covectors (or of I-forms) on M is a mapping o : M + T ' M : x w o ( x ) = (x,oI) which assigns to each point x E M a pair composed of a point and a covector oxat x.
Remark A covector field of 1-forms on M is a section of T'M In other words, it is a mapping o:M+TSM such that n~ow:~-,M:x~~~(m(x))=x (that is id/, ). D
A covector field is diflerentiable if the mapping which defines it is of class e".
So, given a (C")differentiable vector field X, if o (X) is a (C") differentiable function then w is differentiable.
Notation. A covector field or field of I-forms on M is an element of %*(M)= R' (M). We denote w E Q'(M) or w E rm(n;)).
The properties of covector fields are similar to the ones of vector fieIds. A differentiable covector field o on M assigns to each point x E M a l-form 0, = w, '35' defined with respect to cobasis (&) by components w, which are differentiable functions of n coordinates of x.
We leave the following proposition to the reader. PR4
The set Q ' ( M ) of differentiable covector fields on M is a module on the ringCm(M). It is the module of l-forms on M
2. TENSOR ALGEBRG 2.1
TENSOR AT A POINT AND TENSOR ALGEBRA We ask the reader for a very elementary knowledge of tensor calculus.
Let M be an n-dimensional manifold.
Cotangent Bundle, Vector Bundle of Tensors
131.
It's pointless insisting on the considerable importance that the tensors have gained through developments of exact and applied sciences in the 2 0 ~century, more especially in Riemannian geometry, quantum mechanics, analytical mechanics, fluid dynamics, cosmology, electricity, special and general relativities, differential geometry, meteorology, electromagnetism and so on. We are going to make up a vector space consisting of tensors on a vector space. In the present context, this will be done from a tangent space at any point x. Finally, we will consider vector bundles of a manifold. 2.1.1
Definition and examples
D W= A tensor of type
(i) at point
x E M is a (p-t-q)-linear function defined on the
Cartesian product of p spaces TxM and q spaces TxaMwhich is denoted:
( T ' ) Px (T**W9, the order of product factors being to be specified. We denote (omitting index x): t E L,+,((TxM)Px (TX.M)';R) or simply f E T,: where T,: designates the set of tensors of type contravariant oforder q and covuriant of order p.
(i); these
tensors are also named
Let us remember the n 1-fonns B J , making up the dual basis of a basis (ei), make corresponding to every vector X = X' ei E TIM the following reals
Example 1. A vector X of T,M is a tensor of type (:). It is a linear form on Tx*M.
The space T,M of vectors at x G M is also denoted T,: and we write
X E T,:. Example 2. A covector or 1-form u, at x E M is a tensor of type ( y ).
The space TXwMof covectors at x E M is also denoted T=: and we write:
o = u,, # j
E T,,
0
.
The duality imposes: (e,~ = r)n , ~ ~ ( # ' , e ,=) a r , ~ ' G : = o , x i .
Example 3. A tensor of type on TxM x TIM.
( 9 ) (or covariant of order 2) at x E M is a bilinear form defined
The inertia tensor of the rigid body is an interesting example of a tensor of type (: ).
Lecture 4
132
Let us consider the following n2 tensor products Qi@8': T X M x ~ , M + R : ( X , Y ) ~@6"(X,Y) B' = e i ( x ) B J ( Y=) x i y J The set Tx*M63 TxoMof tensors of type (i)is also denoted T,: . The reader will easily verify that (8' C?J 8') is a basis of T,: . The expression of a tensor of type (:) relative to the previous basis is t = t , e i @oJ where the reals t , = t(ei,e, ) are the components of t. Indeed, the tensor t associates to vectors X,Y of T'M the reaIs: t ( X , Y )= t(X i e,,Y e j ) = X i Y'r(e i ,e j ) J
= t,XiYJ
[by putting t(ei,ej)= t,]
=
rg ei(x)8'(y) = tg 8' €3B J ( x , Y ) .
Example 4. A tensor of type
(1) (or covariant of order p) at
x E M is a p-linear form
defined on (T,M) P. We denote the vector space of these vectors by BPTx*MITx;.
The expression of a tensor of type (:) at x E M is t = t . . 8" @... 8 6 ' ' tl...lp
with respect to the basis (Oh €3.. . €3 6 ')~of T.:. Indeed, VX,, ,...,X,,, E TxM :
,
r(X,,,,..., X,,,) = t ( X 4 e ,,..., X ' P ~ =, ~ )~ ~ . . . ~,..., ' ~e itP( )e , , = r i l , , , i p X i l .= , .r~&, i ~O"(X,,) ...eip(X,,,)
,,,.
,,,
= t ipOilC 3 . . . 0 e i p ( x ..., X,,,).
Example 5. A tensor of type
(i) (or contravariant of order 2) at
xE
M is a bilinear form
defined on TX8Mx T'M. Let us consider n linear forms ei making up a basis of T,M : ei : T ~ M + R : w ~ e , ( w ) = w ~ 2 and also the n tensor products o e, @ e j :T x MxT x *M + R : ( o , p ) ~ e@ej(w,,u) , such that e, @ e,(o7 P ) = e i ( @ e , ( ~ = ) WiP,.
If (ei),(e;.)are bases of
T'M
and (Bk),(t9'") the respective dual bases, let us prove the
ei B e; compose a basis of T,: = T,M G3 TxM, vector space of dimension n2. Indeed, on the one hand, the ei 63 e; are linearly independent because
Cotangent Bundle, Vector Bundle of Tensors
On the other hand, every tensor t of type (i) can be written as a linear combination of the different elements e, 63e; . Indeed, let us find its components tu. Since t is a bilinear form, we have r(Ok,8'") = r u e , @e;(8",9'") = r h . Thus, any tensor r of type (i) is written t = t(8',8")e1 me; where its components are t(Bi,9").
Example 6. A tensor of type (,4) (or contravariant of order q) at x E M is a q-linear form defined on (Tx*M)q. We denote the vector space of these tensors by T,;. It's the space L, (Tx'M;R) dual of aqT iM. @' T,M
The reader will easily prove that every tensor of type (,4) is expressed by
r =tl"'."e,, O... Qe, q
where the n products e,, @... C3 a, defined on (T:M)q form a basis of T,: and the reais t"...t= t(Q" ,..., oq ' )
are the components of t relative to this basis. Example 7. A tensor of type (f) at x is a bilinear form defined on T,'M x T,M or on T,M x Tx*M. Let us consider the cobasis I-forms 1 9 ' : T f l + R :X H ( B ' , x ) =X' and the basis vectors ~,:T=*M+ ~ : s ~ ( e , , w ) = m , . 2 We define the n tensor products B16e, : T x M x T ~ M + R : ( X , w ) ~ O i @ e 1 ( X , w ) by
8' @e,(X,m) = x l m j . It is easy to prove the 8' @Je, compose a basis of the vector space T:M @ TxM.
In particular, the Kronecker delta is the tensor 6 E T,: such that V X E TIM, V w E TM : : S(m,X) = 6(w,Qi,x J e j ) = w,Xj6(O1,e,) = S J w , X 1
x).
= w,x1 = (0,
Lecture 4
134
Example 8. According to convention, a tensor of type basis!).
t)is a scalar (independent of the
Change of basis
2.1.2
Let us consider T,M and a change of basis defined by i
e;
Let
= a,ei.
(dr') be the dual basis of
(ei),
(8') be the dual basis of (e;.). We recall that the change of cobasis is defined by
eJ= p , j d r i . The formulas of change of vector and covector components are well-known. Remember that we have X" = p;xi
w; = O 1j O i
and that, in the case of a change of particular bases associated to coordinates x i and x" , we have
These formulas can be generalized to tensors of t y p (:) . Remember that, in a language full of imagery, the tensors are "intrinsic mathematical beings," that is independent of choice of coordinate systems. In other words, each real defined by a multilinear form t (tensor) is not "altered" by any change of basis. For example, a vector X or tensor of type
(k)
is such that 'dm= a,@' = w>B''
x ( w , ~=) x(o;B1')
that is
atxi= w ; x t l .
More generally, given some elements of T,'M : il
oil
=O;~~Y .....J ,I W, , eiq = ~ ; . ~ e ~ ~ q ,
a tensor of type (,4) is such that t(m;lol~l ,..., ~ ; ~ e l = *q t(wiI@ ) ,..., wi e i q ) . .
41 thus
- ail.
f~J~.-.Jq
jq
f
i
1
= p; ..pc '3;
1 ...lo . t I " I
J9
j -- f f ...piqq t i .i
ttj~...jq
and also 4
',
I '
rtJ'1.4
= Jg
In a change of coordinate system ( x i ) t,( x ' j ), we immediately have:
E
Tx*M:
Cotangent Bundle, Vector Bundle of Tensors
1
J
-
axfjl
axiq
axil
t il
In the same manner, we obtain and
The reader will write the formula of component change in the case of a tensor of type
(4,).
Example. Consider a tensor of type (i) t = t l j2I3 4 j5
0'1C 3 e, @ e, @ Qi4 @ ei5
A change of basis (and of cobasis) implies: and
2.1.3 Tensor algebra
We recall that an inner law, namely addition, can be defined on the set of same type tensors. We say: D
The sum of two tensors of which the nptQcomponents are respectively 4 ... ' Jp '11
and
is the tensor of type (:) of which the components are i1...tq fl..iq f j l...jp+' jl...j,.
The addition of two tensors of type T,;xT,; +T,;
( z ) at x is
:(t,u)~-,t+u
where t + u is the tensor sum. Define now the multiplication by a scalar. D
The product of a tensor at x, with components which the components are
k t"",ig
, by a scalar k is the tensor of
.
JI ,..Ip
The multiplication of a tensor of type (: ) by a scalar k is RxT,: +T,:
:(k,t)~kt
Lecture 4 where k t is the product of t by k.
D
The tensor product space of vector spaces T,: and T,: is the vector space of tensors of type ("p':).) denoted:
T,; car,: = T,; Let us remember a tensor of type
.
(4,) is a (p+q)-linear form such that:
being understood that the order of spaces can be modified to obtain another tensor of type
(3. D
The tensor mulcipIIcation of any tensor t of type (;) and any tensor u of type (:) is the mapping
the tensorproduct t C3 u being such that
@~(~~~),.~.,~(q),~~q+~~*~.~s~~q+~~>X(~>,~.~ = t ( ~ ~ l ) ~ . . . , ~ (, + ~ .). , X X(~p~1)()@) ( q + 1 ) ~ . . , ~ ( q + s ) ~ X ~ p + \ ) * " . ~ X ( p + r ) ) , f
This law verifies the following properties:
PI. The tensor multiplication is bilinear:
These properties are immediately checked from the definition of tensor product.
P2. The tensor multiplication is associative: Y~,U,SET,;: ( ~ @ u ) @=sI @ ( u @ s ) = ~ @ u @ s .
P3. The tensor multiplication is not commutative. One of exercises shows a counter-example. D
W-
The tensor algebra, at x, is the infinitedimensional vector space:
r, = R ~ T , M @ T ; M ~ B T ~ ~ @ T @~ -:. .@@ T ,,' ,~@ . - -, direct sum of vector spaces of which the dimensions are higher and higher, and where R represents the tensors of type (:) (also called scalars).
Cotangent Bundle, Vector Bundle of Tensors This space is provided with a bilinear inner law: the tensor multiplication. Therefore, we can express: PR5
The tensor algebra I, is associative, non-commutative and of infinite dimension.
2.1.4
Contraction Let t be a tensor of type
D
(4,) such that p,q 2 1 .
The contraction of a tensor is the operation which consists in choosing a contravariance index and a covariance index, in equalling these and in summing with respect to the repeated index.
For example, let us consider the tensor tq1
emB e , 8akk E T,:
Contracting i n p and k, we obtain a tensor of T,: whose components are
It's a tensor having lost a contravariance and a covariance because the component change is as follows:
,-= s;~'.,
axrmaxrpaxr = 8; a ~ 'a~' ax"'
r
axrrnaxrk axr a ~ 'ax' ayk
=-
This is in accordance with the rule of change of contravariant vector components.
We can express: PR6
Every contraction of a tensor removes one contravariance and one covariance.
PR7
ARer q contractions a tensor of type (:) is reduced to a tensor of type (:) principle q! in number).
(in
Note that the tensor multiplication with contraction can be called the "contracted multiplication. "
2.2
TENSOR FIELDS AND TENSOR ALGEBRA
2.2.1
Vector bundle of tensors
To each point x E M are associated various types of vector spaces that are tensor products of several spaces T& and T,M.
Lecture 4
138
By analogy with the tangent and cotangent bundles of M, we can define on M the vector bundle of tensors of type ) :
(z
D
The vector bundle of tensors of type (;) is
PR8 The vector bundle T,PM is a natural differentiable manifold. Pmof Let ( ( ~ ~ , q be ~ )a )family of charts composing an atlas on M where pp is a homeomorphism
q, : Ug -+ R" : x H (xl ,..., x ) , n
x i being the local coordinates of x.
We can define a vector bundle atlas. How? To domains Up covering M correspond the domains ~ ; u , = { ( x , t , ) : ~ E u , ,~ , E T , ; J which cover TlM. At each point x E M there is a homeomorphism
Therefore, we define in a natural manner the following homeomorphism:
So, the local coordinates of a point i
x , t;
,:;
(x,tx) of
T,'M in a local chart (T,PUp,y,) are the reds
.
In particular, the tangent bundle is
T,'M=TM and the cotangent bundle is T,'M = T'M 2.2.2
Pull-back of a tensor of type (:) Let f be a differentiable mapping of M into N, y E N be the image f ( x ) of x E M.
139
Cotangent Bundle, Vector Bundle of Tensors
Let us give an important definition that we will later reintroduce, more particularly, in the study of exterior differential forms.
he pull-back
D
of ternor t ,
E
T,: by f is a tensor of T.; denoted (f ' t ) ,, such
that V X,,,,..., X,,, E T , M :
( S e t ) x( X ( I ) , . . . , X ( p=)t)y (df,X(,p.-. , d f , X f p J.
w
(4- 14)
Remember that dfxXI,, E T, N is the image of tangent vector XI,, under f and so on. The previous definition means theplinear form f 't assigns to (X(,,,..., XI,,) the same value as the plinear form t,, assigns to images of successive vectors X Recall that each component of a tensor u, of type (:) defined by ' 1 ,...t, = 'x(eii ,...,eip )
relative to a basis (ei) of TxM is
then, the pull-back ( f ' t ) , has components such that
Later we will essentially view the pull-back of completely antisymmetric tensors of type nevertheless we consider briefly the case of any tensor.
(1);
2.2.3 Covariant functor Tz
With (4-14),we have introduced, in particular, a mapping f'
E
L(T;N;T:M), namely
f * : T ; N . + T ~ * M : w Hf * ~ such that VX E T p : ( f * w ) , X = wY(dfX ) .
We can generalize to higher orders and say: D
If df
E
L(T=:;T~;)is an isomorphism, we define T,'f = f; e L(Tx:;Ty;>
such that Vt E Tx;,Va)lk,E T ~ :Vqt, ,
E
I
Tyo :
(f;t)(o( ,,,...,o(,,,Y(,),..., Y,,,)= t ( f *q ,. ., , u
Inaead of vectors).
4' a.could have simply denote f' by viewing f
-'
1
f*w,,,.(df )Y(,, ,...,(df -')Y(,,).
E L(T,;;T,:)
(the W o n being tmgrnt
Lecture 4
In particular,
fd = df and
Lo =(df -' )' This last maps forward as df (unlike f *) and is called push-fomrd mapping, also denoted
Question. Would the expression PR9
f,O have a sense?
If df : T,M -,T,N and dg : T,N
+ T,P are isomorphisms, then
(11
(gO f 14, = g;
{ii)
if i is the identity an T,: , then i: : Tx: + Tx: is id 5: ,
Pi,)
the mapping f; : T,:
f;, + T,:
is an isomorphism such that (f;)-'=(f -I):.
Proox {i)
(id
,...,@(,
ET:,
VZ(1),... Y z ( p ) E c , vfETx;
The second assertion immediately follows from the definition of
because i* = idl? and di-'= idl, . Therefore, :i = idlSz . (iii)
The third assertion follows from (i) and (ii) . We have: f; o (f');
and
= (fo Ti);=
= id/
TY;
Cotangent Bundle, Vector Bundle of Tensors
The reader will be persuaded that T,4 is a covariant functor from the following:
PRlO If Tf : TM + TN and Tg : 77V + TP are vector bundle mappings that are isomorphisms on each fiber, we have:
ti)
(g"f)Q,=g;of;,
(ill
if i is the identity on TM, then :i : TJM -t TJM is idl,:, ,
(iig
the mapping f: : T,9M + TJN is an isomorphism such that (f;)" = (f
-I):.
(see PR9).
Exercise. The reader can prove the natural differentiable manifold structure of T;M. Information. Considering a tangent bundle projection Il, : TM +M, he will prove that the set of all natural charts on T ' M (with ll;M : T'M -+ M ) is a vector bundle atlas. The axiom of covering is obvious. Moreover, any two overlapping naturaI charts will be (T:U, 9;) and ( T jV, y;). A local vector bundle isomorphism o p-' implies (W 0 p-I); =
y/i
0
(9;)-' is a local vector bundle isomorphism. (The reader will prove that if
g, is a local vector bundle mapping then so is pi). The previous natural charts compose a natural atlas generating a vector bundle structure. 2.2.4 Tensor field and algebra
Let T,QMbe a vector bundle of tensors. If we remember that a section of a bundle assigns to each base point x an element in the fiber over x which is a tensor in this case, we can express:
D
A tensorfield of type (: ) on M is a (e") section of TJM. More explicitly, it's a mapping t : M -+ TJM : x H r ( x ) = (x,t,)
which assigns to each point x E M a pair composed of the point and a tensor at this point, A tensor field is d~flerentiableif the mapping t is of class e".
Notation. The collection of differentiable tensor fields of type
T".
(B,)
on M is denoted
Lecture 4
142
Therefore, we express t is a differentiable tensor field of type (;) on M by tE
7p4M
or, if we think of section, we denote f E
where (II,);
rrn((nhf);
is the projection mapping T,4M +M : (x,t,)
I+
x , that is such that:
(IT,);o t : M -+ M : x H (nM); (t(x)) = x .
( idl~f1.
We recall that P ( M ) is the ring of C real-valued functions from M into R, with
The following definitions are natural: {
Vt E 7;M, we define f t
E
7;M by
f t : M - + T , 4 M : x ~ f(x)t(x). (ii)
V X E X ( M ) , V d E X8(M), we define the following element of C"(M):
t(a(,,,...7a~q)3X~1)7...,X~p,) : M + R : x I-+ ~x(a~,l(x),...,X~p,(x)) (iii)
Vt E 7iM, Vu E CsM, we define the following element of 7z:M
A tensor multiplication ( t , u ) r~@ u is so defined.
The rnuitiplication by a scalarfunction : V X E X(M), V g E Cm(M): C* (M) x X(M) + X ( M ) : ( g ,X)t+
is well-defined from above and X(M) is a e"(M)-module. We could also consider the e"(M) linear mappings: L ( X ( M ) ; C m(M)) = f '(M)
and the T(M) multilinear mappings:
L,+,(X(M),...,f*( M ) ; C W ( M ) ) .
gX
:
Cotangent Bundle, Vector Bundle of Tensors
D * The
tensor algebra of M is the (real) infinite-dimensional vector space
I M = & ~ M $ I , ' M $ ~ O M $ ~ ~ M ,@ . . .
direct sum where
&OM
= C"(M) and with tensor product 63 such as in particular
f a t = ft.
We say also "bigraded Cm(A+algebra. " Through the following proposition, let us make explicit the components of the pull-back of a tensor field of type (0,). PRl1 The pull-back of a differentiable tensor field of type (: ) by rr differentiable mapping f (M
-+
N : x I+ z = f ( x ) ) is a differentiable tensor field of type
(1).
This proposition is proved with the help of local coordinates. Recall that the components of the image Z, of a vector X, E T,M under f are [ see (2-12) 3:
ProoJ
So, in particular, if X, is the kth basis vector e, , then the only nonzero component is the kth and its value is 1. Therefore, thejth component of the image of e, is (df,e,)/
dfj
s
=dx11x6',
ay-j
and the vector d f , e , with respect to the natural basis (4)of T, N is dfxegk
i3fJ
Ej .
ax"
In conclusion, from (4- 14), we have: V t z E T=:,V(
f * t l xE T ~ :
d f 'I af" ((f*t).)i,..,tp= t z ( -dx1] lx 4),-.>-Ix axlp af~~ - - ...-
Consider a differentiable field
tE
Q"~P
ax" I.
axlp
Ix
f ' t on M are differentiable since f is
ConsequentIy the tensor field f ' t is differentiable. Vf't E
7 : ~:
tjl ..jp .
N.
The components (f of tensor field differentiable at every point of M.
Moreover,
Elp)
Lecture 4
Simply, iff is defined in local coordinates by z' = zi(x'), we have:
This expression shows the character (:) of the pull-back of t
3. EXERCISES Exercise 1
Show that the cotangent bundle T'M of an n-dimensional manifold has a natural structure of 2n-dimensional differentiable manifold. Answer. We proceed as in the tangent bundle case, but now, we introduce the projection
na:T*M-tM.
Let U, and Up be chart domains in M. These two neighborhoods are going back into T'M under . Let us consider the corresponding domains of charts in T'M, namely
W,= n'-'(Ua) c T'M
Figure 40
and
Wg= ~ * - I ( U ~ ) c T*M .
Cotangent Bundle, Vector Bundle of Tensors
Let the homeomorphisms of T'M into R
'"be:
qa : W, -+ R'" : ( . T , o (M ~ )()x ' , ~ , ) pp :
W' +
: ( x , q , , ) H ( x v J , 4 ).
In the chart (Wa,pa),a basis of T i M is ( h i )and in the chart (WP,qp) it is (dr'j)). Let us consider the following mapping, Vx E U, flU p:
The expressions of every 1-form o(,,of T,'M in one and other charts are:
but
thus
diffeomorphism In conclusion, we have specified the following (C) 4p'
O
P o-I
:va(n*-l(ua n r / , ) ) + ~ ~ c n * -nu,)): ~(u, (X
I
axf , m i )H (xvi,w;,= -ai). ax' j
An atlas of cotangent bundle and the chart changes are so defined.
Exercise 2. Show that the Kronecker delta defined by
is a tensor of type (f). Answer. The proof is immediate because
This tensor is very particular since its components are unaltered under every change of (natural) basis. Indeed, the equality term on the left is 6:. The reader will easily verify that the Kronecker symbols 6"and JVare not tensors.
Lecture 4
Exercise 3.
In a change of (natural) basis, prove that r'P
,-[, ax'
j
ax'
t i= - t P P .
ax'
Answer. We have:
Exercise 4.
If ,udesignates a I-form on M, prove that, given a change of basis e:
= a:e,,we have
V X ' ~E, T,M, 'd,u,driE T,'M : pi = a:p, and pi X 1 = &X". Answer. We have: b;.= p(eJ) = a;p(e,) = a:pi
and
X'p, = y ( X t e i )= p(Xf*e;)= X1'&.
Exercise 5. By using the contracted tensor product, prove that a force is a covector. We recall that contravariance and covariance are indistinguishable in the special case of the elementary mechanics.
Answer. Since the work dz of a force f is an (intrinsic) scalar and any infinitesimal displacement dx is a vector, then the contracted product dr = f,dxi implies that S, are the components of a tensor of type ( y ). Indeed, the contraction of tensor product off with vector dx removes one wntravariance and one covariance to lead to a scalar d r .
Exercise 6. Make explicit, at least one way among twelve, two successive contractions of a tensor of components t i j k m .
:
Answer. From this tensor of type ( ) we obtain a (i) tensor by contracting in the first and last indexes:
Cotangent Bundle, Vector Bundle of Tensors
This new tensor has components denoted u j k , such that:
A new contraction amounts to putting the first and second indexes and summing on them:
Finally, we obtain: axn
k Uk n
=
axn
t
axrs
,
k k ni,
that are the components of a tensor of type ( p). Exercise 7.
Solve the equation
4x'X'xk- &x'xkxh= 0. Answer. Explicitly, we have the equation:
with the solutions:
Exercise 8.
Given a change of coordinates making Cartesian coordinates (x, y, z ) into spherical coordinates (r,0,4), calculate the "new" component A{ of ( ) tensor of components A, = 2.y
Answer. Let by:
US
A2 = 2x+(y)'
designate the "new" coordinates ( r d a l distance, colatitude and longitude) Xrl
=r
y2= 0
The Cartesian coordinates are such that sin x'' cosxr3 x Z = r sin 8 sin 4 = x" sin X" sin xr3 XI
= r sin0cos# = x"
x3 = r cos8 = x" COSX'~.
Since in a general manner we have
then
A3 = xz .
x r 3= 4 .
Lecture 4
that is 2r 2 sin3 @cos24 sin#+r(sin 2 ~ s i n ~ ) ( 2 c o s ~ + r s i n@~)s+irn2sin8cos2 ~ @cos#.
Exercise 9.
Show that the tensor product is generally not commutative. Answer. Let us give the following obvious counter-example of tensor product of vectors. We have:
X@Y=Y@X c=,
XiY'ei@ej=~-'X'ej@e,
that is I@ the two vectors are parallel. Exercise 10.
Prove that a tensor r of type (: ), such that V X E T,M : tx(X, X)= 0 , is antisymmetric. Answer, We have 'dY E TxM:
3 3
t, ( X , X ) = 0 O=t,(X+Y,X+Y)=tx(X,X)+t,(X,Y)+t,(Y,X)+tx(Y,Y) fX(X,Y) = -tx(Y,X).
Exercise 11.
In R2a basis (E,,E,) is rotated relative to the basis (e, = dl,e, = a,) through an angle a(t). Let a vector be X = X I E ,+ X 2 E 2 and a linear form be f (X) = X' cosna + x 2sinna (n E N). Express the basis (6' ,02) dual of (E,, E , ) in according to basis (&',dx2 ) dual of ( ( e , ,e2 ). (iiJ Make $ explicit in the basis (dr',dr2)and calculate df . (iii) Calculate df (X).
Answer. (i) From cosa
( ( - s i n we deduce
sina el
caso)(J
Cotangent Bundle, Vector Bundle of Tensors
) (
coso = i na
sin
asa
(ii) The linear formf is written:
f = cosna 0' isinna 6' = cos(n + 1)adr' + sin(n + 1)a dx2 and its differential is df = ( n + I)(- sin(n+ 1)cr dx = (n
' + cos(n + I)a d x 2) da
n a 6'+ cosna 8 ' ) d a .
+ l)(-sin
(iiiJ From above, we deduce
+x
df (X)=(n + I)(-X' sin na
cosna) d a .
Exercise 12.
a tangent to parallel relative to the basis (1 On S' check that the expression of vector 84
(a,, d, ,d, ) of R~ is
a a a -= -y - + x- , this last vector being also denoted HZ. a4 ax a~
(ii) Having analogically written !fX and f l y ,calculate the brackets [hx,F.ly ],
[ M y ,r-1, 1 and
[Hz,r-1, ] according to previous vectors.
pi4
-4
Calculate Hx(r),Wy(r)and N,(r) where r =
We define the 1-form gradient of r by:
also denoted
Give the expression of the 1-form dr and deduce, for instance, the real (dr,M,). Was the result predictable? (iv) Defining the operator
N2= LH,LH,+ LW LH + LW,LH* Y
Y
and using the property [L, ,L, ] = q,,,l (see exercise 8 of lecture 3), calculate
[fi2,LM* I and comment upon the result.
' The gradient is a remarkable example of I-form. This notion will be reintroduced in Riemannian geometry
Lecture 4
150
a a
(v) Express kx,W, and h, with respect to the basis (-- -).
aeya+
Given a function g, show that
N 2g = - - 1
1 aZg a (sin@-)ag +-
sine d o
ae
sin28ad2
where Band # are spherical coordinates. Answers. (i) From x = sinBcos~
y = sin8sin#
z = cosB
we deduce the vector
also denoted
kf, .
(ii) Analogically, we have:
H, = -2-
and
a +za . H, =- xdz
ax
We immediately obtain:
and analogically
tkf,,h,I= (iii) From the expressions of
-kfx
[Hz,hrl= - b y .
MI,N, and Fr, we immediately deduce:
H, (r) = hy(r) = N, ( r ) = 0. The I-form d r = ( x 2 f Y 2 +2z) -% ( x & + y d y + z & )
implies that { d r , ~ , ) (=I' + y 2 + z ' ) - ~(x(-y)+yx+O)=O.
This result was predictable since the 1-form dr represents, in a way, a set of surfaces (equations r = constants) and h, is tangent to the sphere of equation r = 1. The intersection number of vector h, with inner surfaces, represented by ( d r , ~ , )is, zero. (iv) We successively have:
Cotangent Bundle, Vector Bundle of Tensors
and also
In conclusion, H 2 commutes with Lhx,LMYand LMX . (vj From
a
ae a a# a ayae+z$
-= --
an,
we deduce:
(obvious)
and
a a e a +-a4a a~ a~ ae az agl
-=--
LECTURE
5
EXTERIOR DIFFERENTIAL FORMS
The contribution of differential forms to recent developments in mathematics and physics is significant as the next lectures will partly prove. The introduction of these forrns lets us unify, generalize and view better previous notions seen in other disciplines such as elementary geometry, calculus, thermodynamics, fluiddynamics, electromagnetism, analytical dynamics,... The works of Elie Cartan relating to exterior differential forms were at the root of a theory whose contribution to modern physics was decisive. First, we are going to consider a vector subspace of the tensor algebra at a point. In the second section, we will define the vector bundle of p-forms and the algebra of exterior diflerential forms. Next, the pull-back of a p-form, exterior differentiation and orientation will be introduced.
I. EXTERIOR FORM AT A POINT Let M be a differentiable manifold, t , be a p-linear form at x E M , (8')be a basis of Tx*M, dual of the basis (e,) of T p . 1.1
DEFINITION OF A p-FORM
D
A p-linear form is called skew-symmetric or completely antisymmetric if, for any permutation a making (I,.. ., p ) + (a(l),. . .,a ( p ) ), we have:
where D
E,
(or signa) is +1 or -1 according to a being an even or an odd permutation.
The alternation mapping or antisymmetrization is a (linear) mapping A, : T,
0
0
-+ Txp: tx H
such that V X,,..., X, E T ' :
Aptx
Lecture 5
where u is the permutation (1,..., p) + (a(l),..., a ( p ) ) , is the sum over all the permutations of the sequence (I,...,p) P
This sum is also denoted
where Sp is the symmetric p u p (orderp!). ass,
Note the presence of conventional factor
1
- and the sum
over all p! elements of S , .
P! Remark 1. The linearity of the antisymmetrization is obvious. Remark 2. The antisymmetrization A, is the identity on the vector space of skew-symmetric p-linear mappings:
(since S , has order p!).
= fx(Xlr...,Xp)
D * A p-form (ar exterior fonn of degree p) at
x E Mis the image of a p-linear form by
antisymmetrrzation. In other words, a p-form at x is a skew-symmetric tensor of type (: ). Notation. Thep-form Apt, being well-defined, we denote: W, = A p t x .
PRl
The set ofpforins at
R,P(M) Example 1.
'dl,
E
xEM
is a vectot subspace of T,: denoted:
or simply Q,P.
T,:, VX,Y
E
T,M :
~,(X~Y)=A,~,(X,Y)=~(~,(X,Y)-~,(Y,X)). Example 2.
V t , E T,: , Ve,,e,
E
T,M :
w , ( e , , e , ) = ~ , t , ( e , , e ,= ) $(tg -ti,).
We evidently have the skew-symmetry property, namely: W,(ei, e, ) = -w(e,, ei )
also denoted
Exterior Differential Forms
1.2
155
EXTERIOR PRODUCT OF 1-FORMS Given the 1-forms 0' making up a basis of T,:, we have: A,(B~I@ - - @ e i p ) (,,..., x x,)
~ ~ (@0 B ~' ) ( xY, ) =
f(ei @ e J ( x , y ) - e 1@B-'(Y,x))
=
L x ~ v@...@eip @ 6 (xo(,) ,..., xn(,)) P!
u
G e j -0' B e i ) ( x , y )
=+(el
=
fs;ol
. .
= 4s+f. P . I,...I , ell ~ . . . @ e " ( x ..., ,,
@eJ(x,y)
x,)
@'f I.. are:
where the symbols 6; are:
where the symbols
0 if (IJ)is not a permutation of ( i j )
0 if ( I ,... I,) is not a permutation of ( i , ...i,) ,
1 if (IJ) is an even permutation of ( i j ) -1 i f ( I 4 is an odd permutation of (ij).
1 if ( I ,...I,) is an even permutation of (i, ...i,), -I if (I,...Ip)isan odd permutation of (1, ...i, ).
p
In conclusion, we have: A , ( ~ IB . . . B ~ ~ P+si1-'~ )= '@...@~IP p,
Remark.
I,..,Ip
s;sl @ e J ( x , y ) = ( e me i -eJ @ e i ) ( x , y )= s i ( x p j ( ~ ) - e J ( x ) e i ( ~ ) J
D * The erteriorproductofp linear forms is thep-form '
eilA . . . A O * P =$:::;el1@ - - @ s l p . Remark. The factor
is not present in this definition. So we have conventionally chosen:
Several other conventions exist, but as the reader will see in the next, we have adopted the one that eliminates the most constants. Examples.
e3
=e3me1- 0 ' @ 8 =-el ~
8' ~ 8 ~8~ '
' The symbol
A
=s;;el @eJ@ e K
can be called (< wedge D or cr hat )).
156
Lecture 5
= e i @ e J mek+ e i @ek @ e i +ek638' @ e J -81 B e i mek-a i @ e k @ e j -ek @ e i @ e i . These examples show the importance of the order of exterior product terms. Later we will treat that more generally. , = P, 0 ' is Exercise I proves the exterior product of two I -forms a = a, 8'and B
1.3
EXPRESSION OF A pFORM
1.3.1 Expression of a 2-form
Consider o E Q : ( M ) , x beIonging to an n-dimensional manifold M. PR2
The C: products 8'
A
8' ( i < j ) form a basis of the vector space R : ( M ) .
ProoJ: On the one hand, every 2-form is a linear combination of C: products of 1-forms
@'A@(i< j). Let us express indeed the value of w on vectors X and Y of T,M , more precisely the image of ( X y ) bym: m ( X , Y ) = m ( ~ ' e , , ~ '=eX~i )Y j w ( e , , e j ) = wgXrYJ
[putting m, = m(e,,e,) = A,t,(ei,ej) = t!, = - t p ] = w,,x'y2+o,,xZyl + W , , X ' Y ~+0,,x3~' +...
So, the products 6' A 8' ( i < j ) generate every 2-form.
On the other hand, these products are linearly independent, namely:
xe,@~9j=O t<j
because ' d r , s {~I , ..., n):
V i , j ~ { ..., l ,n): my = O
Exterior Differential Forms Therefore, the expression of a 2-forin relative to the basis (8' A 8' ),,,is
This sum contains C i =
n! terns. 2!(n - 2)!
The components of a 2-form w relative to the basis (8' A 8j),j are called strict components and are denoted . So the expression of a 2-form relative to (6' A =~
is
( ~A e, j e ~
by knowing that the sum is onIy over i < j.
Obviously, we have o=&w,8'
A@'
Example. Let M be a 3-dimensional manifold. The exterior product of two 1-forms a,P E RL(M) is a 2-form w with C$strict components, namely: ~ ( 1 2 =) a 1 8 2 - a 2 P 1 ~ { 1 3= ) -a3A @(23) = a 2 P 3 - f f 3 p 2 3
these being the components relative to the basis (8' A BJ),, of i2:(M). 1.3.2
Expression of a p-form
D
A p-form is decomposable if there exists p I -forms a, such that @ = a ,A . . . A ~ , .
PR3
The C,P decomposable forms Oil A ... A Brp (i, < ... < i,)
make up a basis of the vector
space 85(M) . This proposition can be proved as before or in a rather different manner (see exercise 4). We say that: il...tP 81, (8' h . . . h ~ ~ ' ) (,,..., e , e,) = 6,1:':(0'~8 . . . 8 8 1 p ) ( e,..., j , e,) = 6,1,,,p , ... As previously, we can obtain the expression of ap-form, that is
where
I
,,..., c {I,...,n 1. lp
We have also:
w
,,, 8" A , . AB" .
o = ;q1
,,.
=&yP A
.
Lecture 5
158
(1)
It is essential to distinguish the n P components of a tensor of trpe from the C,P components of the p-form obtained from antisymmetrization. Consequently, we define:
D
relative to a basis (Oh called st& components and denoted The components of a p-form
n,
A ... A
eip
of R: (M) are
),!,
(il < . . * < I , ) .
~ ( i , . . . ~ ~ ~
So the expression of ap-form is W
CO
0" A ...A 8'
=
by knowing that the sum is only over il < ... < i, .
Remark I. Expression of ap-form relative to the basis (8" B...C3 0'' ) of T,:. Let *I
@ . . . @ o i p
= til...tp
be a tensor of T,:. We have: ~ e i l @-..@I t 8'' ~
=~ , t = , ~
,
~
~
=[, , aP(ei1 @...@e")=ti1,.,;s;:,::.,e" I' p
@ . . - @ e l p
= IP! t ti...lp
.
el1A . . . A o ~ ~
The next to last equality allows to write:
In particular, we find again
Remark 2. According to convention we call 0-form any scalar; that is an element of (M) (of dimension C: = 1 ).
PR4 If the degree of a p-form is higher than the dimension n of the manifold, then the form is necessarily zero; that is V p > n : R,P( M )= (0). Proof: If p > n then two indices are necessarily equal and thep-form is thus zero.
1.4
EXTERIOR PRODUCT OF FORMS
D
The extenbrproduct of a p-form wand a q-form p is a ( p + q) - form defined by
&"
, -
A
(~+47)!
p!q!
p+q
(w €3PI
3
(5-7)
Exterior Differential Forms
that is an element of vector space
(M) having C r qdimensions.
Problem. What is the image of p + q vectors X , ,..., X,,,
by the exterior product w A p ?
From the definition of antisymmetrization, we have V X ,,. .., X,,,
E T,M
:
Remark. In particular, we find again the definition of exterior product of two 1-forms:
8' A@' = & ~ , ( 8 ' 6 8 ' ) = $ ( 8 ' @ 8 ~ - 8 ' @ 8 ' ) .
1.5
EXTERIOR ALGEBRA
Let us show that at (some)x E M are associated vector spaces whose direct sum is an algebra of finite dimension. Let us recall that the following (internal) law can be defined in the set of exterior forms of same type: D
The addition of twopforms at x E M is the law where o + p is the p-form sum, the sum of twop-forms with respective (strict) components
and p(,,,,,ip, being
thep-form with the C,P following components: @(il
... ip)+ / 1 ( i
)
.
We can also define a second law: D
The multiplication of apform, at x E M, by a scalar k is an (external) law: RxRC +R,P : ( k , w )k~~ where kw is the product of w by k, the product of the pforin with (strict) components
, p ,
by the scalar k being the
p-form with the C ,P components
k#(,,.J p ) . D
The aten'or product space of vector spaces ST: (M) and njE(M) is the vector space of @q)-forms, that is R,P x l2: = R r 4 ,
Now, we can define a third law that is bilinear:
Lecture 5
160
D
The exterior multiplicatr'on of a p-form wand a q-form A is the mapping
the areriorproduct w A p being such that VX,,...,X,+,E T,M :
The following properties are verified: (i)
The exterior multiplication is bilinear:
ProoJ: The first equality (the others being demonstrated in the same way) is immediate:
w ~ ( p + v ) = 2 ! A , ( w @ ( p + v ) ) =2!(w@p+w@v) =2!A2(w@p)+2!A,(w@v)=w~p+w~v.
(id
The exterior multiplication is associative:
but
thus
At this level we recall the notion of signature of a permutation. We can view the permutation ( i,.....i,) -+( j ,.....j,) as the composite of two other permutations:
(
. ..i,] = (k,. * * J1...jm
.?) [ il-.--im ]
J~,..bm
kl*.-km '
but the signature of the composite of two permutations being the product of their signatures, then we have: So the permutations imply
Exterior Differential Forms
imply jl - j p + gip+q+l-.-ip+q+r - S ! . . . ~ p + q~ ~ +..;p+q+r ~ + l fil...p+q p + q + ~ , , . p + q + r - 11 .,.ip+g;pq+l...ip+q+r
-- i , , . JI, i pJp+g +q I
i l - . j ~iPtg+l...ip+q+r +~
4....................... p +4+r
.ip+qip+q+~.-ip+qcr .................... p + , .
11
Therefore, the sum relative to the indexes i ,...i,+,, in [(wA p) A v ] ( X ,,..., X,+,+,) is such that:
Finally, [(w A P ) A vI(X1 ,...rXp+q+t
In the same way, we can show this expression is
that proves the associative property. (iir;)
The exterior multiplication is not comrnutafiveand V w E R,P,V p E 0; :
ProoJ: On the one hand, we have: (m A ,uXXl
,...,x p>xp+17...>xp+q)
and on the other hand:
(~~~)(X,r...r~p,Xp+l,...rXp+p) =
. . . . lp+,...Ip+qI ] ...1 , ...,..,,,,...p+q i~...i~+~
C4
P ( X , ~, +. . . ~, x ; p + q ) 4 X i , ,..., x i p 1.
The permutation (i,+, ... iP+, i, ...i,) + (i,...I ,i,, .. .i,+,) corresponds to pq transpositions; thus we have: o ~ p = ( - l ) ~ p ~ o . Special cases: If w ~ f l , Pandifpisodd,then w ~ w = O .
If~,~~flf,,tw h en np = - p ~ a , and w ~ w = O
D
The exterior algebra at x is the vector space
direct sum of vector spaces which is necessarily finite. This last precision foIlows fiom the fact there is no exterior form of degreep higher than n.
Lecture 5
162
The previous vector space is provided with a bilinear law: the exterior multiplication. We can express:
PR5
The exterior algebra Q x ( M )is associative, noncommutative and of finite dimension.
The exterior algebra is a vector subspace of the tensor algebra 7, whose the dimension is d i m R ; ( ~ ) + . . . + d i r n R : ( ~ ) +Cf, = C :+...+Cz = 2 " .
Specify that the product of an element of degree p with an element of degree q is of degree p + 4 (graduation!).
2. DIFFERENTIAL FORMS ON A MANIFOLD If necessary, the following notions could be restricted to an open U of M 2.1
EXTERIOR ALGEBRA (GRASSMANN ALGEBRA)
2.1.1 Differential form
D
The vector bundle of p-fomts (or differential forms of degree p) on M is
If we remember that a section of a bundle assigns to each base point x an element in the fiber over x , we can express: D
A pgorrn on M is a C msection of w P(M). Explicitly, it is a mapping
that assigns to each point x E M a pair composed of the point and a p-form at this point.
Notation. The collection of differentiable p-forms on M is denoted
R P( M ) . Therefore, we express that w is a differential p-form on M by w E QP(M). If we think of a section, we denote:
w
E
rrn(w;)
where rn; is the projection mapping w P (M) + M : (x,cv,)
I+ x
that is such that
Exterior Differential Forms
So, a differential form of degree p on M is a differentiable mapping that associates to each point x E M a p-form at this point:
where :M
mi...,p
-,R :x
mi
,..., (x)
Thus a differential form of degree p on M associates to every point x E M a p-form whose value on p vectors X, ..... X, is
Exercise. The reader will show the natural differentiable manifold structure of co ' ( M ).
InSormation. Considering a tangent bundle projection IT, : TM set of all natural charts on w P(M)is a vector bundle atlas.
+ M , he will prove that the
Remark 1. We denote the collection of 1 -forms on M by Q'(M)=<'M = X * ( M ) .
Remark 2. According to convention, a real-valued function on M is a 0-form on M We denote: no( M )= cC" ( M ). Example 1. Let xi be the local coordinates of any point x of a 3-dimensiona1 manifold M and ( d r ' ) be the dual basis of (ai). A 1-form on M, namely a E R'(M), is a mapping a:M-+wl(~):x~a(x)
defined by 3
a ( x ) = x a i( x )dxi( x ) i=l
where a,= a ( 8 , ) . The differential form of degree 1 on M is denoted
Example 2. Let x' be the local coordinates of any point x of a 3-dimensional manifold M and (dri)be the dual basis of (d,).
Lecture 5
164
A 2-form on M, namely w E Q*( M ) is a mapping
w : A4 + u 2 ( ~x)H : ~(x)
defined by W(X)= w12(x)dX1AG!x~(x)+ mI3(x)dr'~ d x ~ ( x ) + w , , ( x ) dr\dx3(x) x~ where w, = a?(Bi,d,).
The differential form of degree 2 on M is denoted
Example 3. A 3-form on a 3-dimensional manifold M is a mapping w : M -,u~(M):xH W ( X ) defined by W(X)= W(X) A dr2 A dr3(x) and the value of which is known for three vectors
(XI,
x',x3),(Y',Y 2 , ~ 3 ) and (2',z2,Z3)
namely:
x'x 2x3 G(x) Y1 y 2 Y 3 . Z' zZz3
We denote the previous differential form of degree 3 on M by @=I%&'
A&*
A
~
X
~
.
Example 4. If g is a real-valued function and w a differential form of degree p on M, then gw is a differential form of degree p such that:
g w : M + u P ( M ) : x - gw(x) where gw(x) = g ( x ) m( , ,p,(x) drij A ... A d;'(x). 2.1.2
Algebra of exterior differential forms
The addition of tensor fields and the multiplication of a tensor field by a real-valued function being previously defined, the addition of differential forms (of same degree) on M and the multiplication of differential Forms on it4 by a real-valued function are thus operations well-defined. We specify that skew-symmetry is evidently safeguarded with regard to the two previous operations.
Viewing ?:A4 as a module on Cm(M), we immediately see that SZP(M)is a submodule.
Exterior Differential Forms
We have: V ~ , ~ E R ~ ( M ) , V ~ E C f~ @ ( M~ )+: , €u R P ( M )
because
f 6 w + ,u E
is such that f €3w ( x ) -t- p ( x ) is an element of the vector subspace of skew-symmetric p-linear mappings on M. This submodule is provided with an algebra structure by the introduction of a third law, namely the exterior multiplication: A : R " x R +~ R P ' q : ( w , y ) ~ ~ ~ p
defined by
M
+ w ~ ' ~ ( :Mx )H (W A p)(x)= w ( x )A p(x).
It is bilinear; that is, with evident notations: ~ A ( P +, + 2 ) = w v , + @ A P 2 (W, +6.l,)Ap=wl A / l + w 2 A p g(o~,u)=gw~,u+w~g/r.
It is associative and anticommutative.
D
The set of differential forms of any degree on M provided with the three previous operations is the following direct sum called algebra of exterior dvferential f o r m of M or Grassmann algebra of M :
Remark. We have included ! ~ O ( M = ) C m ( M )in this sum, the real-valued functions being such that f A o = f @ w = fw , and we often consider this algebra as a vector space rather than a module on C" (M).
2.2
CHANGE OF BASIS
The cobasis (&I) and another ( 8 ' ) ( such that 0' = P:dni or dKi = a j Q J ) that were previously introduced at a point, are assumed defined at every point of a chart domain of M The cobases are thus bases for differential forms on a chart domain of M. Let us consider p forms of a differentiable field on such a domain. 2.2.1
Differentiable form of degree 2
We have: w = qg>dxiA dr' = o(u)aiBP~ a i 8 ~ =a+,(a;O1 + a i B 2 + . . . ) ~ ( a { e+aid2 l +...) =
qu,(a6a: - a:a:)QP A Bq
Since the expression of a 2-form relative to (B P
A Bq) is
(P < 4 )
Lecture 5
the change of strict components is such that
In particular, let us consider a change of cobasis defined by
A 2-form is such that
Since the expression of a %formrelative to (dr'P A dxfq) is
then the change of strict components is
So, given a change of cobasis, a differential form of degree 2 is written:
2.2.2
Differential form of degree p
In a similar manner, the following formulas are established: a;,.......ai! JP
.jp)
- W(*, , , , t p ) -
............... a; ......
alp JP
Exterior Differential Forms
167
Those fomulas must be compared with the ones (4-13) for basis changes in the case of tensors of type (i) .
We evidently have:
3. PULL-BACK OF A DIFFERENTIAL FORM Let M, N be differentiable manifolds, f : M -+ N : x H y = f(x) be a differentiable mapping such that y' = yj(xk), o be a differential form of degree p on N. 3.1
DEFINITION AND REPRESENTATION
(i)
The definition (4-14) of the pull-back of a tensor of type by f is valuable, in particular for differential forms of degree p, the skew-symmetry being safeguarded. Thus we express:
D M
The pull-back of w E RP(N)by f is the differential form f 'w such that, Vx E M, V X,,..., X p E T ~ M :
(f'fi),(X~~...,X~) = (4X~,.d''Xp).
(5-1 1)
Note that f' is a linear mapping R P (N) + SZB(M).
Remark. Before continuing, the reader will be aware of the following:
D
The push-forward of p E R P (M) by f is the differential form f,p such that V ~ E N , V ,..., K Yp E T ~ N :
From the comparison of (4-13) with (5-9), the recalled formula (4-15):
implies
Lecture 5
168
Important rule PR6
The representation of the pull-back f 'o in local coordinates on M is obtained from the locaI representation of o expressing the dy as differentials of functions yJ (xi). J
ProoJ W e have:
--
(Y' (xi1)
fi(jl...jp)
Y hilA ...A & I p D(xq,..., x" ) 7
.
~
'
~
( i , < ... < i p ),
the expressions of components of f 'o being recognized. This rule is used in the exercises. To simplify, let us use J,, the Jacobian determinant appearing in (5-12). PR7
Only one constant Jfexists such that the particular pull-back J* : Q:( M )+ R: (M)
defined by (5-1 1) satisfies f *o= J, o for every w. Moreover, for any other pull-back g' : C2:(M) -+ Rt(M), we have:
f (19 (iid
J,,*=J,Jg3 J , =1 , Jf., = ( J ~ ) - if' J t 0 (that is i f f is an isomorphism).
ProoJ 0) Given the standard basis wo = a!xl such that w = kw, where k is a constant.
A ... A
dxn of Cl:(M) then every n-form w is
Hence f8m0= Jfwo implies f'w = fgko0= kJf oo= J, w . Next, the proof of (9 is the following (see section 3.2): J , . , o = ( f o g ) * w = ( g * of ' ) w = g * J f o = J f J g w . (ir;) is obvious.
(iii) is also obvious because
I = J,, = Jf -, = J, J,., .
3.2
PULLBACK PROPERTIES
PI
Let f : M + N : x I+ y = f ( x ) and g : N mappings between manifolds. We have: ( g o f ) * =f 8 0 g * .
Proof. V V E C ~ ~ ( P ) , V X E M :
+P : y H z = g(y)
be differentiable (5-13)
Exterior Differential Forms
P2
Let f : M -, N : x Hy = f (x) be a differentiable mapping. q Vw E nP(N)V , p E R ( N ): f * ( w ~ p ) f= ' w ~f * p .
(5-1 4 )
Proof: This formula can be proved by introducing the skew-symmetry operator A, (simple
exercise). Let us give the proof by using the previous important rule. Consider w = w(,,,.,p,dyJ1A ... A dy jp
On the one hand, we have: @ I,'
1
f ' ( ~ A P= )a)(jl-.jpI&jp+I,,~+q~(ax'." '
@,J'" +A'.)
A...A(-
ax"
hit +.,.1
and on the other hand, we have:
ay jP f * w = q , l , , , j p ,;tyil ( - d r h +-..),+,...A(--dr"...)
axti
axil
ayj#+l
@,JP+~
f ' r = P ( , ,., p+o,(--hil +-. )A...A(ax ax" i1
hil+...)
Therefore
f'w
A
f *r= f ' ( 0
A ,U) .
In particular, if we consider the real-valued function g E ZZ'(N),we have: f * ( g A o ) = f ' g A f * w = f'g f * U . The above is also
We can immediately verify that Vg E C w(N), b'w
E
R P( N ):
f ' ( g 4 = f'g J'w . Indeed, Qx E M, VXl,..., X, E T&f :
(f' ( g o ) ) ,( X , , . . . , X , )= ( g 4 , (df,X17...>dfxXp) = g(.Y) @,(dLX1Y...,dLXP)= g ( f ( x ) ) ( f'w), ix,, . . . , X p ) = f * g ( x )( f 'o), ( X ,,--.,X,), The following property is evident.
P3
If f :M -+ N is the identity, then f
P4
We have: ($*)-I
= ( f - I ) * = fC.
: ! J P ( N )-+ Q P ( M ) is the identity.
Lecture 5
170
Indeed, the first equality is obvious because f 0 f -' = id a (f f -I)' = id =(f -I)* f * . The second follows from:
( ( f ' ) ' o ) ,,..., (~ Y p )= (o),(df;l~, ,...,df;'Y,) = (LW),(Y,,...,Y,).
4. EXTERIOR DIFFERENTIATION 4.1
DEFINITION The differential dg of a 0-form g : M + R at x E M has been defined in lecture 2.
Having so defined the differential d g : TM
+TR = R : ( x , X , ) t ; r dg,X = Xxg,
where dgxX is the derivative of g in the direction X at point x, we introduce the following mapping: D
The exterior d~flerentiationis a mapping
~ : ~ P ( M ) + R ~ + ' w( M ) : satisfying the following properties:
PI. If g is a real-valued function, then dg is the differential of g that is a differential form of degree 1 (see lecture 4). P2. Linearity: Vm,,o, E R P ( M ) : d(w, + m2) = dml f dm2. P3, Zero square: d o d = 0. p Vw E R ( M ) :d(dw)= 0 .
P4. Antiderivation: Qw E Q P ( M ) , V pE n q ( M ) : d(w~,u)=dwnp+(-l)~~~dp. P4 '. In particular, for every real-valued function g E Q'(M) we have d ( g P ) = d g n P + gd ~ .
D
The differential form do is called erterior di'erential or exterior derivadive of w.
Remark 1. The operator d is natural with respect to restrictions; that is, if U c V are opens of M, then for each w E nP(V) we have d(miu)= (dm)," or the following diagram is commutative: .raP(V) ' i ra'(U) d
.l
QP+'
.l d
( v )--p OP+' (U)
Exterior Differential Forms Remark 2. Considering local coordinates, the previous PI means that
because
Uniqueness. Let us prove the uniqueness of d : OP( M )
np*'(M) . For that, let us establish the lemma:
d(dxiiA... A d x ' ~= ) 0.
(Inductive reasoning). From PI and P2 we deduce that The lemma being proved for p = 1, assume that it is true forp-l and show it is true forp. Indeed, from d(xil( A t 2A . .. A dxtp)) =&'I A(&'' ~ . . . ~ d x ' ~ ) + x ' ~ d ~( d. .x, '~~ d x ' p ) (since P4 ') =Ai1A...Adxi~, we deduce from P3: d(dxi' A... ndxip)= d(d(xi1dxi2A , . . r\dxip))= 0. Using this lemma, P2 and P4 ', we immediately conclude that to each differential form of degree p
corresponds one differential form of degree p+ 1, namely
So, the four properties defining d give a well-defined differential form of degree p+ l from a differential form of degree p. This form is denoted:
d o = do,
,,,,p,
A
drl A . ..A kip
To prove the existence of such a mapping d, let us show that, in a chart, the mapping leading to (5-15) satisfies the four properties. Indeed, the two first properties are immediately fulfilled. Relating to P3, the following expression Var E LIP( M ):
Lecture 5
172
is zero because permuting summation indexes k and r, and by symmetry of mixed partial derivatives, we have:
Finally, let us show that P4 follows fiom the anticommutative property of exterior law Considering 0= , , A ...A &kip j.l
A.
= p( jq,dX" A . . . A dXJ4,
we have
3
d(@A P ) = (dm(i,,.,iP,j.l(jI.. jq, + ~ i l . . . c , d f i .j. lJ q ) )
=dm(,
,,
A G ! X ~ A . . . A d r i p A p(,,,&jl
+(-l)'&,, =dunp
,, hi'
A...hd*lp
A
A
kip A
A
... A kJq
A... A&*
~ d j l ( ~Adx'l ~ , , A, .~. . hki ~
+ (-T)Pu~d,u.
To establish the existence of d, we must also prove that the expression (5-15) of dm does not depend on the chart. Let y j = y ( x i ) be a change of local coordinates (at least of class C ). The exterior differential of A ... A hi'(y') . w = qiI-. (x'(Y')) dxil (y') J
p )
is d~ = d o , , ,
A
dr" ( y ' ) A ... A kip (y')
The formula (5-15) is recognized; that proves the definition of dm is independent of local coordinate choice.
Example. Given a 3-dimensional manifold, calculate the exterior derivative of W = W(p,dXi A&' = fwydx' A & / .
Answer. We have:
Exterior Differential Forms or also = $akm, hkA h1 A kLri
A form w on a manifold M is closed if its exterior derivative is zero:
D
w E LIP(M) closed if dw = 0.
EXTERIOR DIFFERENTIAL AND PIJLCBACK
4.2
Let M and N be differentiable manifolds, f be a eO differentiable mapping of M into N defined by y' w be a differential form of degree p on N.
= y'(xi),
PR8 The exterior differential of the pull-back of a pform is the pull-back of the exterior differential of the differential form: V w E S ~ ~ ( Nd () f:* w )= J'dw. Proof: Consider 0 = 'u(u,l,p)dy'l A ... A
4'' .
On the one hand, we have
and on the other hand
So, d is said to be "natural with respect to mappings" or the following diagram is commutative; R P( N )
d
OP(M)
.l-
3. d
~'+I(N)
,.
>ClP+'(M)
Let f : M + N be a diffeomorphism. We leave the following commutative diagram to the reader: QP(M)
d
J-
QP+I
since
ft =(f - I ) * .
QP(N)
.I- d ( M )y 2 ! p+' (N)
Lecture 5
174
5.
ORIENTABLE MANIFOLDS
Before dealing with the manifold orientation, let us recall the practical calculation of a pull-back of a differential form. If the differentiable mapping f : M + N is defined by y1= f l(xI ,..., xn),. . ... , ym = f y x l,..., xn)
then
Let M be an n-dimensional manifold. D
A volume on M is a nonzero differential form of degree n at every point of
M
*
In other words, a volume R E R n (M) is such that Vx E M : n(x) 0 . We recall that the definition of an orientable manifold has been introduced in lecture 1. PR9
A manifold M is orientable
#I
there is a volume on M
Proof: Note that a volume !2 attributes an orientation to each fiber of TM, because there are
nonzero elements of any one-dimensional space Q Z M ) . First, prove that if there is an atlas ((~,,p,)),,such that every coordinate change shows a (strictly) positive Jacobian of the overlap mappings at every point of M, then there is a volume on M. Consider the change of chart p,
op,-'
: Rn
-+R n : (xl ,...,x n ) H ( y ' ,..., y n )
and let
n, = p,*(drl A ... A d r n )E n n ( u , ) be a volume (element) on I/,
Figure 41
Exterior Differential Forms
Let 51,
E
Rn(M) be the volume element such that vx E U , : Rj(x) = n j ( x ) VXE u, : n j ( x ) = 0 .
Let us consider the differential form of degree n:
where {g,) is a partition of unity on M.
From local representatives and since the previous sum is finite in any neighborhood of each point, we see that i2 E On( M ) . Since = pj*(dyl A ... ~ d y "=) q$(p[')g(dy1 A . .A dyn)
a,
then we have
Because the various g,(x) and Jacobian are (strictly) positive by assumption, then the n-form nis nonzero at every point x E M , that is a volume. Conversely, if there is a volume on M, let us prove there is an atlas every coordinate change shows a (strictly) positive Jacobian.
{(u,,~~)),, such that
Let (U, ,pi ) (U,, pj) be charts of the atlas, R be a volume element on M, h, : ~I,(U,)-+R:(X' ,..., x " ) n h , ( x l,..., x n ) ),
Considering the chart (U, ,pl ),very generally speaking we have
where h, ( x k ) is (strictly) positive on p, (U,) If h,(xh) was (strictly) negative we would change the chart by permuting x,-, and xn In the same manner, considering the chart (U,,qj), we have:
where hj lyk) is (strictly) positive on pj(U, ) . We successively have:
Lecture 5
ht(xk)dxlA... n u!xn= (pi1)*t2= ( p , 0 p['Y(p~1)*i2 = h,(p, o p,-Yxk 1)
,
D ( Y ' ~ . . . , Y ",...) ~ ' &", ~ ( x ..., ' ,x")
In concIusion, the Jacobian
is (strictly) positive. PR 10 An n-dimensional manifold M is orientable $?'I dimensional (one generator!).
SZn(M),module on Cm(M) is one-
ProoJ: Firstly, let i2 be a volume on M. In local representation, we have:
VxeM:
R(X)=~,,~~&'~A...A~!~~.
Consider any other R' E R n ( M ). Each fiber of Rn(M) is one-dimensional and we define a real-valued function g on M such that From
R1(x)= m;, ," (x) drJ1n... A dr*' ( x ) ,,
and since vx E M :
(4 0
~il.,.t~ f
we see that
is clearly an element of C m ( M ) The converse is obvious since each fiber is one-dimensional and the (generator) n-form i-2 E LId(A4) is nonzero for every x E M . Since there is a unique g E C m ( M )such that D
a' = g R we can express:
Volumes R, and i2, on an orientable manifold M are equivalent if there is a function g E C: ( M ) such that R, = go,.
An orientafion o f M is an equivalence class of volumes on M, denoted [a]. An odented manifold is an orientable manifold with an orientation [n]on M. The reverse orlentation of an orientation [a]is the orientation [-a].
Remark. The so-defined equivalence relation is natural with respect to mappings and diffeomorphisms.
Exterior Differential Forms Indeed, given the C ' mapping f : M + N, if O Nand then f *a',is equivalent to the volume f *aN on M. Indeed, we have [see (5- 14')] :
Moreover, f being a diffeomorphism, if R, andR; f,W, is equivalent to the volume j;Q, on N.
$2;
177
are equivalent volumes on N,
are equivalent volumes on M, then
Indeed, we have: f*igS2)= i g o
f
-'If*Q
PR11 An orientable manifold M is connected I@ M has exactly two orientations. Proof: Firstly, let R and Q' be volumes such that L?' = gR, g being a nonzero function on M If M is connected then either g(x) is strictly positive for every x E M or strictly negative for every x E M. Thus R' E [R] or R' E [-R] .
Conversely, by assumption M has exactly two orientations. If M was not connected, then there should exist a subset (different from 0 or M) which would be both open and closed. Hence, from a volume fl on M, the reader could construct a volume on M that wouldn't belong either to [R] or [XI]. D
A chart ( U , p ) on an orientable manifold M is positively oriented
equivalent to tp*(drl A ... A dr" ) where dx' volume.
A.
if RI, is
..A dxnE SLn ( q ( U ) ) is the standard
This notion is well-defined from the previous remark. Remark. If M is orientable, there is an atlas with all positively oriented charts; if a chart had a negative orientation we should change its orientation by an index pennutation.
Given volumes a, and R, of respective orientable manifolds M and N, we say: D
A C mapping f : M + N is called orientationpreserving if f $' 2, orientation reversing if f '0, E [-QM I.
E
[R, ] and
PR12 Given a C mapping f : M + N then f *S2, is a volume iff for every x E M there is a neighborhood U such that U +f (U) is a local diffeomorphism. ProoJ: First, if f '0,is a volume then the determinant of the derivative of the Iocal
representative is not zero and thus the derivative is an isomorphism. The inverse function theorem allows the conclusion. Conversely, if f is a local diffeomorphism then f *RN( x ) # 0 (because PR8). We can aIso introduce the determinant notion.
Lecture 5
178
D
The
detednunt
of a
e" mapping f : M + N is
the (unique) function
det(QM.n,,f E cm(MI such that:
f '0,= (det(nM,n,,f
(5-17)
.
In the case of a mapping f : M -,M we denote det(,M,nM, f by det,
PR13 A rmapping f : M + iV is a local diffeomorphism det,QM,n,,f
fi
f.
for all x of M:
* 0.
PR14 A eD mapping f : M + N is orientation preserving (resp. orientation reversing) r f l for allx of M: (resp. det,QM,*,,f (x) < 0 ). det,,,nN, f (4> 0 To end, we say: D
A
C mapping
So, it is fr
f : M -+ N is called volumepreserving if f 'a, = R, .
det(QM,nNl $=I.
6. EXERCISES Exercise 1. Given two 1-forms a = a,6' and
fl = flj 0' , show various expressions of
a np
Answer. The tensor product a @ f l = a , & e i 80'
is a (:)tensor such that the n 2 components relative to (0' 8 6') of T,; are a,& The exterior product a A P is an antisymmetric tensor of T,: : ar\fl=a,~8'~ = a8, P ~ ,6i@0J-a;flj6J86' = ( d i p j- ajP,)Bi 8 Qj
< where ej = ai aj
8 , P,l
To sum up: a A P= qflj19' ~ 6= xj ( a t f l j -ajfli)O1AB' icj
)'
Exterior Differential Forms
Exercise 2. Given m I-forms 8 ' at point x E M which are linear combinations of m forms h i , calculate 0' A ... A em. Answer. Considering 6' = fl: dx'
p: E R , i = 1,...,m ,
with
we have:
8'A . . . ~ 8 "= f l i d x i h A...AP%~* ' , = fl:
...c dr"
A... A him
= p; ...pz'5:;;2drl,.. A h r n = det(p,!)
dxl
A ... A dxm.
Exercise 3. Find the value of the m-form dxil A... A dxim applied to m vectors X,= x?aIl, .. .,
x,,, = xis,, . Answer. We have; dx" A ... A him (xl,..., X m )= 6j"') I". dr'l @ ...@ &Im(xl ,..., X,) ,I
= s;::;
x;...x:
x;... X f - . . . . xi... xi Exercise 4. Prove that the C : products dril A ... A dr
ip
( i , <... < i, ) form a basis of S2,P( M ) .
Answer. We have previously seen to eachp-form was associated ap-form "generated" by C,P
elements 8' A ... A Oip (or dxh A ... A dxip) such that i, < ... < ip .
Thus, it is sufficient to show these C,P elements are linearly independent, that is: (jl < ... < i,) . dxh ~ . . . ~ d x="O f3il,,,ip= O
xrul,,
i,<..
Let (j, < ... < j, ) be a sequence which is completed in order to (j, ,..., j , , j,,, ,..., j , ) be (1,. .., n). The order of terms will be changed if necessary. The assumption implies x t W i ., dx" A . . . A d X i p hdXh+' A,.,AdxJ' = O . it<.
I'
P
In this sum, only hJ1 r\...r\hJp A h J " ' A... Adxh
Lecture 5
180
is nonzero, the other terms having necessarily two equal factors. The previous sum is thus reduced to only one term, namely: &'I A , , ,A & j p A&jp+' A , . .A h J n OJI . ;tp
(JI
<Jp)
which necessarily is zero.
So, we conclude:
Exercise 5. Show that the oriented area S of a parallelogram constructed from vectors X and Y is an example of 2-form. Answer. The area of the parallelogram defined by vectors X ( X ' ,x 2 ) and Y(Y Y ') is actually the following image of (X, Y) under S:
Exercise 6. Make explicit the exterior product of a Zform w with a 1-form ,u defined on R ~ . Answer. This 3-form is such that b'(x, y , z ) E R3: (W A P)(x, Y , Z )= [P(x,Y , z ) @ + Q ( x , y , z ) d z A& + R(x,y7z ) h A dyJ A[A(x,y,z)dx+ B ( x , y , z ) d y + C ( x , y , z ) ~ I
that is = [ P ( x ,Y , z )
Y , z ) + Q(x, Y , 2 ) B(x7 Y , 2 ) + R(x5 Y , z ) C ( x , y , z ) l h A 4 Adz.
Exercise 7. Make explicit the exterior product of three 1-forms defined on R3 Answer. Consider the three 1-forms mi = A,& + 3,dy+ C,dz ( i = 172,3) . The exterior product of three 1-forms defined on It3 is the following 3-form on R3 : A,(x,y,z) B,(x,y,z) C,(x,y,z) Az(x,y*z) B , ( x , y , z ) C , ( x , y , z ) & A b y ~ d . . A3 ( x , Y , Z ) B3(x, Y ,2) C,(x, Y7 2 )
Exercise 8. Calculate the pull-back of the 3-form dx A dy A & by the mapping f : [o,m) x [0,2~)x ( - a , m ) -+ R3 : ( r , O , z ) H(x,Y,z) where the cylindrical coordinates r,8,z are such that
Exterior Differential Forms
Answer. The pull-back of the 3-form by f is
f (dx A dy A dz) = d(r cos 8 ) A d(r sin 8) A d i =(cos8dr-rsin~9d0)~(sin~dr+rcosBd~)~dz = r c o s 2 Qd r ~ d e ~ d z - r s i n 2d8 0 ~ d r ~ d z =rdr~dB~dz, this result being expected since the Jacobian is r Exercise 9.
Calculate the pull-back of the 3-form dr A d y A dz by the mapping
f : [&a)x [O,Z]X [0,2ir)+ Jt3: (
~ ~ 8 ~ (x,Y,z) 4)
where the sphericd coordinates r,8,4 are such that x=rsin8cos#
y=rsinBsin#
z=rcosQ.
Answer. It is easy to prove that
f * ( & ~ d ~ ~ d z ) = r ~dsr innd8B ~ d 4 .
Exercise 10. Let U be an open of a two-dimensional manifold M and f : U + R 3 be a mapping defined by i = 1,2 and j = 1,2,3. zJ = zJ(x')
Express the pull-back of a 2-form on R~by f: Answer. Given the 2-form
the pull-back of w by f is:
This result could be directly obtained.
Lecture 5
182
Exercise 11. Curl Let w = o j ( x i )d X j be a differential form on R" such that the functions w j are of class c'.Give the expression of the exterior differential do. Answer. The following equalities
mean that the coefficients of dw are the (strict) components of curl of covector field w. We also have dm = +(a,@, -a,@,)dri A akj .
Remark that the curl is an element of exterior algebra. But it can be viewed as an element of tensor algebra as follows: i
dm= x ( a , ~ -, ajq)(mi@ d r J-mj @ m ) f<j
= ( a , @ ,- a 2 ~ , ) d x@hZ 1 +(a,@, - a 1 ~ , ) a k @dxl 2 = ( a i w j-a,wi)dui B ~ J .
+,..
In conclusion, do = (aimj - a j w i ) d x i @hJ' = a,oj dx'
A ~ X '
Exercise 12. Dlveqefice
Let
m, dr' A akJ be a 2-form on
a,=
R" such that the functions rug are of class c'.
i <j
Calculate the exterior differential d o . Answer. We have:
The coefficient of dm is called divergence of field of antisymmetric tensors of type
We can also write
The preceding exposition can be generalized to degree p. The exterior differential of
(i).
Exterior Differential Forms
,,,,, dr" A ... A drip = f
w=
drh A ... A dr
ip
i,<..
is then
the minus signs in the last expression folIowing from the permutation of the indexes of dx and w, permutation implying factors (-1) (-1) = -1.
Exercise 13, In R', let X be a vector field of components X , , X,, X,and n, = X,dx + X,dy + X,& be the associated 1-form. Calculate d w n w and find an important class of vectors (in mechanics) making zero this exterior product.
Answer. From
dw=d!~dx+aY,,~dy+d!~a'z = ( a x x ,- a , x , ) d x ~ ~ ~ + ( a , ~ , - a , ~ , ~ d ~ ~ d ~ + ( ~ , ~ , - a , ~ , ) d ~ ~ d x we deduce d ~ ~ w = [ x , ( a ,-a,x,)+x,(a,x,-a,x,~+x,(a,x, x, -a,x,)]drAdyAdz. In vector notation, the previous expression is zero Therefore, in particular
d w ~ w = O if
tff 2.(9A 2)= 0.
curlX=O
that is if there exists a function V (calledporential ) such that X = -grad V .
Exercise 14. Curl andgradient Given a 1-form w = w j ci!xJ and a function g , establish the formula
curl(gw) = g curlo -tgrad g
A w.
Amwer. We have
curl(gw)= [ai(gw,) - 8 , (gw,)I dri8 drj = ai(gw,)dutA drl = ga,w,
A
drj +a,gdrlA q m J
=gdw+dgt,w. Exercise 15.
Consider
$2 = dr'
A , ., A
dn" and a vector field X
(
Introduce the divergence notion from $ 2 ( X ) .
(id
Given w = h(x1,.,.,x")i2,prove that div,X = - a,(hX1).
(iiQ
Show the divergence expression in spherical coordinates r, 8,4
I
h
Answer. (i) From Q we immediately deduce:
Lecture 5
184
and thus ~ ( R ( x ) )= ( a , x l + a2x2
+..--t- a , x n ) d r l A dn2 A ... A dxn = ( a , x i ) o
= (div, X)R . (iz)
From wherehisafimctionof(xl ,..., x n ) ,
w = h d r l r \ ... A&"
it follows
&(X)=h ( ~dr2 ' / \ . . . A & " --..) and thus
1
= a,(hxi)i;m
that is
(iid In spherical coordinates (r, 8, z) , we know that
and the divergence of the particular vector field
X = X r d , +xed,+ X4da,
(comprehensible writing)
is
1 div,X = ,r d r ( r 2 ~ ' ) sin + 28 a , ( s i n d ~ @ ) + a , ~ ' .
Exercise 16. Given an orientable manifold M with volume R prove that: (i) ( I
(iii)
Iffandgaremappings M+M,tthendet,(fog)=((det,f)og)(det,g). If id denotes the identity on M, then det, id = 1, If f is a diffeomorphism M +M , then det, f = (det, f
-'
Answers. (i) det,( f 0 g)R = (f
g ) ' n = g9(f*Q) = g8((det, f1 0 ) = (det, f o g ) g * R = (det, f 0 gxdet, g)R .
0
f -' ) - I .
0
( since (5-17) ) ( since (5-14') )
LECTURE
6
LIE DERIVATIVE LIE GROUP
The reader will refer to the notion of a one-parameter group of diffeomorphisms developed in lecture 3 (paragraph 4).
W e recall that every (differentiable) vector field X on a manifold M generates a oneparameter (local) group of diffeomorphisms #, between neighborhoods of M : Every local transformation #, verifies the differential equation
with &o(x)= x Recall also that the orbit of a one-parameter group through a point x, (= +oxo) is the integral curve: R
-+ M : r I+
~(t)=#~x,
that is tangent to vectors X(##x0)of vector field at every point #,x, Thus we have a collection of curves associated to field X filling M (or some part of M ) and such that through each point passes only one curve without intersection.
Lecture 6
186
We point out that any diffeomorphism notion leads to a derivative definition.
4, defines a mapping between opens of M .
This
I. LIE DERIVATIVE Unlike Euclidean spaces, the manifold notion doesn't let us simply introduce the derivative notion. Indeed, how shall we compare, for example, vectors at various points and how shall we define the derivative of a vector field at a point? A first answer is supplied with the notion of orbits of a one-parameter group. 1.1
FIRST PRESENTATION OF LIE DERIVATIVE
1.1.1 Lie derivative of a function Let g be a differentiable function on M . The tangent vector, at point
xo, to the orbit of diffeomorphisms
#, is
We recall that the derivative of (germ) g in X tangency direction, at x,, is the real d xog = -(g04,)(~0)/,=0 dt In a chart, if the n x i ( t ) designate local coordinates of
4,x0 = x and
xi the ones of xo, then
we know that:
D
The Lie derivative of a function g with respect to X, at point xo, is the derivative of g in the direction X :
~~~g= Xog= lim g(hx0 1- g(x0 ) r-o t More precisely, we compare, at x, , the value g8(x0)= g(#lxo)of g obtained at point q5,(x0) with the value g(x,). Next we divide by the variation of parameter t and take the limit t + 0 . We go back to xo along the orbit! D * The Lie derivative of a function g with respect to X is the function L,g: M
+R : X I +
such that L,g(x) = X , g ( x ) .
(see PR9 in lecture 2).
L,g(x)
Lie Derivative, Lie Group
187
In short, omitting the bracket:
In local coordinates, the Lie derivative of g with respect to X is expressed by ~ , g= X I aig = aig &(x*a,) = dg(x)
denoted in short form:
L,g = d g X
&;r
Remark. In exercise I , we introduce the gradient of g denoted dg and such that: ( d g , ~=} Lxg.
&a
Let f : M -,N be a diffeomorphism, X be a vector field on M, Lx be a differentiable Lie operator on C m(M). PRI
The Lie operator L, is (9 natural with respect to pull-back by
cm(~)
C ~ M )
I
LQI m
C (N)
that is the following diagram is commutative
Lx
7 Cm(M)
(id natural with respect to restrictions; that is the folIowing diagram is commutative (for any open U of M )
Proof: 0) The image of X under f (see lecture 3.83) is the vector field df X on N such that V h E C m(N) : @'"X(h)= X ( f * h ) 0 f-' that implies L,(f*h)= X ( f * h ) = d f X ( h )o f = f*L,,h. (id The second assertion
L,," (hlU) = (Lxh)IU is obvious because
d(hlu)= (dh)lu.
Lecture 6 1.1.2 Lie derivative of vector field Let
4, : M +M
be diffeomorphisms.
Let X be the (generating) field of tangent vectors to the orbit of a group of diffeornorphisms 4, passing through xo.
Let Y be a vector field associated to a diffeomorphism point x, = 4txo.
v, and Y,,
be the tangent vector at
Figure 43 We recall that the image of this last vector under the diffeomorphism 4,-' = 4-, is dK1Y#,%
where d4,-) : T#,xo M
+ Tk M
"Going backwards " to no along an orbit and comparing the previous image with vector Y, , we define: D
* The Lie derivative of
vectorfield Y with respect to X, at
x, , is
1 L, Y = lim - (db;'~, - Y, ) I=O t
Important remark. Zero Lie derivative
From the previous definition, if d4t-1~+,, = Y, then the Lie derivative is zero.
Lie Derivative, Lie Group This particular case is illustrated as follows:
Figure 44
In a general way, there's no reason for such an equality apart from when the image of orbit (with tangent vector X ) passing through xo under ry, is the orbit corresponding to X passing through v,xo. It's the case if (see exercise 17 of lecture 3) the points x, = ry,(#tx,) and x, = #,(y,x0) are the same. Let us now introduce the important proposition that follows.
PR2 * The Lie derivative of vector field Y with respect to X is the Lie bracket of X and Y . Prooj First let us point out the following remark.
Let g : I x U + R be a function defined on I x U c R x M . There is a function h : I x U + R of class C1such that: and The function h such that 1
h(t,x) = laIg(tu, x) du 0
fits the requirements.
Indeed, from the change of variable v = tu, we deduce:
and also
Now let us prove the proposition.
Lecture 6
190
The following comparison between vectors d#,-' Y - Y
leads to 1
1 t
-1
L,Y(g) = lirn - ( d + ; ' ~- Y X g )= lim - d4, (Y - d+,Y)(g) t=o
t
t=o
1
= lim - (Y - d#tY)(g) t=o 6
[this will be used later]
this last equality following from the definition of the image of vector field Y under lecture 3.43), namely: (d41Y)f g ) = Y(4t.g) 4;'
4, (see
The previous remark bas shown there was a b c t i o n h(r,x) such that (g O 4, )(x) = g(x) + t h(r,x)
where h(0,x) = a ( g 0 4 t ) ( ~ , x=) at
xg,
the last equality following from the definition of the directional derivation of g along X, (point x being not mentioned). Then, by using the expression of L,Y(g) , we have:
because 1im4~-'= id. 150
From the directional derivative of a function Y ( g ) along X :
and since h(0,x) = X g
we deduce:
L,Y(R) = X(Y(g)) - Y ( X ( g ) )= [X,YIg.
Exercise. Prove that [L, ,L, ] is a derivation on the algebra Cm(M) and LIx,yl= [L, ,L, 1. (solved 9). Properties Remember that the vector space X ( M ) provided with the bracket of vector fields is a Lie algebra. The law [ ] implies the following properties (see lecture 3):
Lie Derivative, Lie Group PI.
R-bilineariiy. V X ,Y , Z
P2.
Anticornmutative property. V X ,Y E X ( M ):
E
X ( M ) , Va,b E R
So the properties P2 and P3 show the algebra is a Lie algebra. The last equality proves L, is a Lie bracket derivation. Remark. In lecture 3, exercise 7 has shown that Vg E C m ( M )
We will note the presence of the last term. From this exercise, we deduce L x ( g Y ) = [ X , g Y ] = - [ g Y , X= ] -gL,X+XgY
that is the well-known derivation rule:
equality showing that L , is a derivation on (Cm( M ) , X ( M ) ) 1.1.3 Lie derivative of tensor fields and forms Similarly to the definition of the Lie derivative of a vector field, we say:
D
The Lie deriv&ianveof
( z ) tensorjkfd T with respect to X, at point xx,, is
1 LXu T = lim ( ~ !=O
t
E ' T -~T,, ),
Before defining the Lie derivative of a differential form we give the definition of the Lie derivative of (:)tensor field. "Going backwards" to x, along the orbit associated to X and denoting 4,*T#,, the pull-back of tensor T#,, at x, , we define:
Lecture 6
192
D
The Lie derivdive of (: ) tensorjield T with respect to X, at point xo, is
We are going to use again this definition within the framework of completely antisymmetric tensors of type
(1).
Let w be a differential: form of degree p on M, b:w,,, be the pull-back of p-form w+,, at point xo. D
The Lie derivative of differeential form w with respect to X, at xo,is
w
Remark 1.
Remark 2. The definition (6-5) leads again to the formula (6-1) in the context of real-valued functions:
PR3
The operator d is natural with respect to L, ; that is the following diagram is commutative:
aP(~)-bn*(~) d
-1 P+I
M
(
,
-1
d
,flP+l(M)
In other words, V w E Q P ( M ): dL,w = L,dw. Proof:
d d d d ~ =~d -4f*wI,70 w = - d41*mII=0= - 4t*d~II=0 = &dm, dr dt dt
(see PR8 of lecture 5 on the commutative property of d and
Lie Derivative, Lie Group
193
Now, we introduce the notion of a differential operator on the tensor algebra and then Lie derivative with respect to X .
D
A differential operator on the tensor algebra 7M is a set of mappings: D(U) : 7,4U -+:7 (U)
(for every open U c M and p, q 2 0)
such that: (i) D is a tensor derivation, that is linear and VT E TpqM,VS E 7,SM :
D(T@S)=DT@S+ T@DS, (ii) D is natural with respect to restrictions, that is for opens U c V c M and VT E 7'V :
(DT)(U=
D(TIU)
E
cqu.
In other words, the following diagram is commutative:
7;vA7pJ
JD
DL
rp'v77+7;u (iii) If S E I;'u
designates the Kronecker delta, then D6 = 0.
It must be mentioned the various indexes p,q... have been removed from D to simplify the writing.
Let f : M -+ N be a diffeomorphism, X be a vector field on M. PR4
The Lie derivative is natural with respect to pull-back by f ; that is the following diagram is commutative:
that is f*oLqx
= L,
0
f*.
Proof: First, the following Wi?IImorelemma can be proved (using local coordinates):
L
If for every open U c M , following operators E, : Cm(U)+ Cm(U) and F, : X(U) + X(U) are tensor derivations and natural with respect to restrictions; that is Vg, h E Cm(U),VX E X(U), VY G X ( M ):
Lecture 6
E,(g@h)= E,g@h+g@E,h E, ( g l u )= (EM g)(u F u ( g @ X ) = E u g @ X+ g @ F u X
F,(YIV= (F, Y) I U
7
then there is a unique differential operator D on 7M that coincides with E, on Cm(W)and with F, on %(W). Considering in particular the Lie derivative L,,, in place of E, and F,, we immediately know the Willmore's lemma is applicable. Then we say: D
The Lie derivative with respect to X is the unique differential operator on 7 A4 denoted Lx, such that Lx coincides with Lx previously defined on C W ( M )and
%(MI. Now, let us go back to the proposition. We must prove: V T E ~ : N : L , ( f e T ) = f*L,,T.
Given any open W c N, we define the operator
where the restriction
IW
are not written.
We know, from PRI of this lesson and from PR15 in lecture 3 (written in "pull-back language), that the operator D coincides with the Lie derivative on C m ( W and ) X(W). So, if we prove that D is a differential operator on 7 N , then the Willmore's lemma will let us conclude there is a unique operator L,, on T N with respect to pull-back by f . Indeed, first V T ,S E T,4N :
Second:
o~lw = ( ( f * ) - l ~ x)Iw f * = ~ (f)-I
(L, ~ * T ) ( =w(/*)-I
= D(TJw). Finally:
Ds=(f*)-'Lxf*6=(f*)-'Lxd=O. In conclusion, DT is LgxT and the proposition is proved.
So, we have the following equality: f 'I.RT=
f'T)
L,,~~*TIW
Lie Derivative, Lie Group 1.2
ALTERNATIVE INTERPRETATION OF LIE DERIVATIVE
1.2.1 An alternative definition
Let T be a tensor field on M, To be the tensor of field at point xo Let us define the ''transpursed fensot" under a local g.ansformation coordinates. In a local chart, the coordinates of the point x = +,x,-, are x' = +:x,.
#, by using local
The diffeomorphism defining the mapping such that x, I+ x can be considered as a change of point coordinates. So a "new" coordinate system is defined such that the coordinate lines ate the images of coordinate lines of the initial system under 4, .
D
The transported tensor of a tensor To under 4, is the tensor defined at point x such that its components are the ones of T, expressed in the "new" coordinate system.
It is denoted
7; To Example. Let us illustrate this notion by a tensor at x,
expressed in the "old" basis. The transported tensor is
a
(I, To)', -@&" dxr
a ax:,ax axr
BdX" = ~ ' , k --@-drs r
axk
ax;
ax
s
axu
since the respective lectures 2 and 4 teach us that
a -hJ a -ax;
ax; axi
and
&A
hA=-drj, ax"
In the "new" basis, the components of the transported tensor are:
ax' a;ax,' ,,*.
(( &)rm = 7s u ax, ax ax
Next, at point x = +lxo, we compare the tensor Tp,+ of field T with the transported tensor 7; To at point #p,. Then, we say: D
The Lie derivative of a tensorfield T with respect to X, at point x, is a tensor of the same type as T such that:
Lecture 6
1.2.2 Lie derivative of vector field Let Y be a vector field,
a
Yo = Y,' - be the vector of field at xo. ax;
To define the Lie derivative of Y with respect X, we introduce what foIIows. Any point x in the neighborhood of xo admits an expansion about x, :
On the one hand, the transported vector of Yo under 4 , at x , is
ax*
axi
( J ~ ~ ) ' = ~ q = q ( S ; . + ( - - -+--)) ~. t ax,/ dxl On the other hand, the components of vector field Y, at x, are:
We make the subtraction
and, having divided by r , we take the limit as t
-+ 0 .
Then, the Lie derivative of Y with respect to X, at point x , is the vector such that
We find again (see lecture 3) the following expression: 6b"
L,Y
= [ X , Y I=
( x / a 1 y i- y J a j x i ) a ,.
1.2.3 Lie derivative of covector field
Let w be a covector field w, = w/Odr,' be the covector of field at xo. On the one band, the transported covector of o,under
at x, is
Lie Derivative, Lie Group
On the other hand, the components of covector, at x, are:
We make the subtraction
ax-' + ...
do.
w , ( x ) - ( < a , ) , = (---XJ + W j ; - ) O t axJ ax
and, having divided by t, we take the following limit:
lim @'(x)- ( < @ O ) i t=o
=
+ &]
axJ
t
ax'
.,
ax
In conclusion, the Lie derivative of w with respect to X is the covector field such that ( L , ~ ) ,= X / ~ , O + I ,OI, a,x .
(6- lo)
J
1.2.4
Lie derivative of tensor field of types (:)
Let T be a tensor field of type (i), To = T: dx: 8 dx,' be the tensor of field at xo. On the one hand, the transported tensor of To is axk a x r (
xi
) - (ax r ax
- x,'t
...)(Sj'--tax; ax
-
- ...
dxi = T ' ( 6 , - - tax: axt
=T;
-
- ...)
.-.)
ax; -6;-tax; + ...) (6:q-g-t a ~ j axf
=q-q,O-ax; t - Tb0 axi
On the other hand, the components of
st +
...
ax'
(i)tensor, at x, are
aT, $(x)=T,O +(-) axk
x,Lt + ... O
We make the subtraction
aT, T , ( X ) - ( C T , ) , =((,),x: ax
,, ax;
ax," + ...
+T,-+T;-)~ axJ ax1
Lecture 6
198
and take the following limit
ax
r
lim
axk
t
t=o
dxj
In conclusion, the Lie derivative of tensor field T of type field of type (i) such that: (L,T), = x ka,q, + T~ aixk+
1.2.5
axk
;TTh-
ax'
10
(i) with respect X is the tensor
qrajxr.
Lie derivative of tensors of type (:) and (:)
Given a tensor field of type
(i) ;
and a vector field X, we immediately have:
w
(L,T),I .,I,-. = xkakq1 + T, ?..., ailxL+ . . + ql,,,ka,p xk. p
,p
Given a tensor field of type
(6-i1)
(i) :
and a vector field X,we immediately have: ( , r X ~ ) b -i qX k a k ~ i ~ -- -~ i ~q *4 1 atxh - ... - ~ f l . . ka,xc 62'
1.2.6 Lie derivative of a tensor field of type
Given a tensor field T of type
(6- 12)
(4,)
(4, ), we have:
Remark. The Lie derivative is said of "zero degree" that is L,T
is a tensor of the same
type as T and
VT,,T2E T , : M : L x ( T , 0 T 2 ) = L x T , @ T 2 + T , @ L x T 2 .
Lie Derivative, Lie Group 1.2.7 Lie derivative of a pform
Consider a p-form
o=
o, t , , , dr'
h
... A drip
t,<.,.
Its Lie derivative is also completely antisymmetric and the expression of the Lie derivative of a (:) tensor implies:
To end this section, we notice the antisymmetrization of Lx(q @T,) implies V o E R P ( M ) , V y E Rq(M):
=
L,T, €3T, + ir; 8 L,T,
2. INNER PRODUCT AND LIE DERIVATIVE Let X be a vector field on M, w = ~ o l l dx' , , i Ap..hG be a differential form of degree p on M .
DEFINITION AND PROPERTIES
2.1 D
*
The hner muItipIicatr*on of w by X is the linear mapping i, :
R P ( M ) + RP-'(M) :w
t-+ iXw
defined by the innerproduct i, w of w and X such that VX,,.. ., Xp-, E X ( M ): ix+e(X,,..., X p - , )= m ( X , X,,... , X p - , ) .
The reader will check that the expression w
ixw = ( P - I ~ w! k i 2 . , i p xkdrbA . . . A ~ x ~ P
is actually the definition of o z ,
in local coordinates.
The expression (6-15) of the inner product is equivalent to
Lecture 6
the inner product of a function (O-form) and vector field being zero. Solved exercise. Check the equivalence of expressions (6-15) and (6-16) if, for example, w is a differential form of degree 3 on a 4-dimensional manifold.
Remark. In the case of a l-form w, we have ~,W=I~(W,&*)=~ = (~, ,x X ) .~ Moreover, it is a r ( x ) = w , d x i ( x ) = o i ~ ' 6 ;=&,Xi=(w,x). In this particular case, we find again (4-6). PR5
The inner product is an antiderivation on the exterior algebra, that is:
(0
ix
1s
R-linear, q
(iq Vw E R P (M), Vp E R (M) :
i , ( w ~ p ) = i , w ~ p+ ( - l ) P ~ ~ i x p . Proof: The linearity is obvious. An easy proof of (ii) is obtained by using the formula (6-16). So the reader will write i, (w A p ) . He will next consider ' X ~ A P
x i 1 'fy = O(I,. tp )&iP+] .,imql
- w(tI...lP P(ip+, ...ip+q 1
x i 2
,...
&4
Ah
i p
hip+]
, .. A ,
, &4 ,... , alyiP+4
h i p v
+
...
must appear in and in the same manner w A i,p. He will immediately conclude that the equation (6-16) because the signs of the concerned terms are governed by the degree of w.
In another manner, we see that V X , E X ( M ) :
PR6
The inner product is nilpotent : Vw E R P (M) : i : ~= ix(ix&)= 0,
Lie Derivative, Lie Group
that is an operator of zero square. Pro05 It's obvious from the definition in the context of local coordinates.
wrjl llx'
Exercise. Check this last property if w =
A dr' A
drk
i<j
Answer. We have: i
ixm = w(w,X drJ A drk -
qW, X J hiA&'
+ qt,x k dri A&'
There are many other properties and the following are evident from the definition, V w E nP( M ) ,Vk E R, 'ifg E C 0 ( M ):
+
i x + r= ~ ixw i,w ,
i,w= k i x w and iX iY o = - i Y j x u .
igxw=gi,o ,
PR7 The inner product is natural with respect to diffeomorphisms. In other words:
If f : M + N is a diffeomorphism between manifolds, then the following diagram is commutative: Q P ( N )A LIP(M)
ix
.l
0'-I (N)?
$ if.x W ( M )
Proof. We must establish that V w E STP(N):
if., f e w = f *i,ro. V X l,..., Xp-l E TM : lf.X
2.2
f* o ( X ~ , . . . , X = ~ -f~*)m ( f*Xy XI ,.'.'Xp-l) = o ( X , d f Xl ,..., dSXP-,) = iXw(dfX I ,..., df X P - , ) = f'ixw(X1,. ..,X,-,) .
FUNDAMENTAL THEOREM
The following theorem shows a very helpful expression of the Lie derivative.
since (5-1 1)
Lecture 6
202
PR8 The Lie derivative of a differential form with respect to a vector field is a mapping such that:
w
V X E % ( M ) : L, = d o i x
+
i,od.
(6-17)
ProoJ:
- First, we introduce three lemmas (with simplified notation).
Consider 1, = d i x + t x d .
Lemma 1. The operators d and I, are commutative. Indeed, we have:
Ixd = di,d
+ i,dd
+ di,) = dl,.
= di,d = d(ixd
Lemma 2. The operators Lx and I, coincide on C m ( M ) .
Indeed, in local coordinates, we have Vg E C m ( M:) ag Lxg = X' -ax'
Lemma 3. Vw E n P ( M ) ,V p E flq(M): The equality being true for L, , let us prove it for 1, ; that is:
The left-hand member is written:
The second and seventh terms are destroyed and also the third and sixth terms. Thus we have: dix (m A p ) + ixd(m A p )
= d i X w A p+ i , d w ~ p+ w n d i x p + u ~ i , d p = (di, + i,d)u A p + w A (di, + i x d ) p .
- Now we prove the theorem, that is: VW E Q P ( M ): (LX- l X ) w= 0 . By considering W =
lemma 3 implies:
q,, -,p,dr'J A...Adxip,
Lie Derivative, Lie Group
Lemma 2 implies: (Lx - 1, )w(l,.lp 1 = 0 and lemmas 1 and 2 imply (L, -Ix)&' =d(Lx - 1 , ) ~ ' = O . Thus we have: V W E RP(M) : (L,
- l,)w = 0
and so L, = I x .
Remark 1. We immediately find again the definition of the Lie derivative of a hnction g : Lxg = ixdg = ( d g , x )= x t d i g .
Remark 2. The previous theorem lets us simplify calculus. So,
Remark 3. Also we can easily prove that the following diagrams are commutative:
Indeed, L,
o
i, = ix
0
Lx because
Lxix = (ixd+ di,)i, = ixdix = ix(dix+ ixd)= ixLx
and
Lx 0 d = d 0 Lx
because
Lxd = (ixd+ di,)d = dixd = d(i,d
+ dix)= dL, .
Remark 4. The fundamental theorem allows to prove readily that Lx+,w= L,w
+ L,w
because LX+,,m= d i , + , ~ ~ + i , + = , dd(i,w+i,w)+i,dw+i,do= ~ L,w+L,w. Before starting on the very important Frobenius' theorem, we end this section by defining the notions of derivation and antiderivation.
Lecture 6
We say, V w E n P ( M ) p,
D
q
E
R ( M ):
An algebra endomorphism is a derivation D if D is linear and if D(wnp)=Dwnp+ w ~ D p .
On the other hand, it is an antiderivation A if A ( w ~ p )A =wnp
+(-l)P~~Ap.
Examples. The exterior differentiation is an antiderivation of degree +1 since dw E R ~ " ( M ) + The antiderivation ix is of degree -1 since zxoE W ' ( M ) . The Lie derivation is of degree 0 because L,w E n P ( M ) .
3. FROBENIUS TmOREM Initially we establish a proposition of which the converse is the Frobenius' theorem (relative to vector fields). Let M be an n-dimensional manifoid, S be an mdimensional submanifold of M, with the induced topology.
PR9
Given any two vector fields on S, such that
Vx E S : X ( x ) , Y ( x ) E T,S , then
[ X ( x ) ,Y(x)l E T.S , Prooj First we show the following lemma :
L
The bracket of any two vector fields which are linear combinations of m commutative vector fields is a iinear combination of these m vector fields.
Let
be two Iinear combinations of m vector fields, J; and gi being real-valued functions. The properties of Lie brackets imply:
Since the vector fields are commutative, then the brackets [ X , , X , ] are zero; that proves the lemma. Now we prove the proposition.
Lie Derivative, Lie Group
a
We first observe the basis vectors -- E T,S commute: axJ
a
a
axJ
axk
[-,-I
= 0 . We recall that such a
basis (8,) is called "coordinate basis. " Since every vector field on S is a linear combination of vector fields (3,) which are commutative, then the lemma implies any two vector fields tangent to S have a Lie bracket which is tangent to S. Now set out the following important Frobenius' theorem. PRlO Consider m ( l n) vector fields X, on M such that, at each point x E M , the vectors X i ( x ) are linearly independent in TIM and generate a vector subspace P(x).Then the condition Vx E M : [ X , ,X , ] ( x ) E P ( x ) is necessary and sufficient in order that, through every point x, one submanifold S of M exists such that Vy E S : T,S = P(y).
In fact, there is a family of such submanifolds of dimensions which are at most rn. This family of submanifolds fills an open domain U of M and builds up a foliarion such that each submanifold is a leaf of the foliation. No leaf intersects with any other one. Thus, in an open domain of M we consider m linearly independent1 vector fields X , such that: where c: are differentiable functions; then the integral curves of the fields give rise to a foliation of M . ProoJ We are going to establish the theorem by induction.
Leave the obvious case ( m = 1 ) with only one vector field X where each integral curve is a one-dimensional submanifold of the manifold M . Especially, if all the m vector fields commute (zero Lie brackets!) then they evidently define a coordinate system and the requisite family of submanifolds follows. Generally speaking, we are going to prove the family of manifolds exists if the Lie brackets of rn linearly independent vector fields X j are linearly dependent on the fields. d dz
We choose one vector field in {X,), for instance X, = - . Let Vi be m - 1 vector fields which are linear combinations of X j , j E (1,. . . ,m),and defined by ( d r , ~ =, )L K z= 0 ( Vi E { l,..., m-1)). We have:
The case of dependent vector fields must not be considered since we would remove enough fields.
Lecture 6 m-1
~ x ~ , v , l = C c , v+, d , X m j-1
where a,, , b,, c,, d , are functions on domain U. However, bo = d, = 0 , indeed we have:
m-l
Since all the previous Lvkare zero, we have
dz Since the first members of previous equations are zero and L,= z = - = 1 , we deduce: dz
Therefore, we have: m-l
m-1
k=l
j-1
[ ~ , y ] = x aand~ ~ [~X , ~, , . Y , I = ~ C , V , The induction supposes that any set of m- 1 vector fields on which all Lie brackets are linearly
dependent generates an (n-1)-dimensional submanifold. The fields 6 are suitable for that. Thus, we have a family of (m-1)-dimensional submanifolds filling U . On U, we define vector fields &, I E { I,. .. ,m-1 ), that in particular form a coordinate basis (zero brackets!) on one submanifold S' and such that, in U, by means of the "Lie transport law": LxmY, = [X,, Y,] = 0, we transport the vector fields Y, along every orbit relative to X . passing through S '. Let us recall that we must prove the existence of a set {X,,K) on U, where all the fields commute and form a coordinate basis of an n-dimensional submanifold. = 0 is a (unique) linear We are going to verify that each field Y, satisfying [Xm,Y,] combination of only Vj, j E { 1,. . . ,m-1 }. Let us show that 1=1
determines a unique solution of [X,, Y, ] = 0. Indeed, we have: nt-1
[X.,Y,I=
LXJC
j-l
m-l
m-1
A, v,)=C ( L X m f & )+C ~ , fjLXJ, j-1 j=l
Permuting the indices j and k in the second sum and since the fields Vk,k E (1,. .. ,m-l), are linearly independent, we obtain the following system of differentiable equations:
Lie Derivative, Lie Group
207
The initial conditions corresponding to the fields Y,on S' lead to a unique solution and thus the fields Yi are linear combinations of fields yi . Therefore, we can conclude because we have constructed rn fields X, and Y, forming a coordinate basis for an m-dimensional submanifold. Note we must emphasize the "Lie transport law" safeguards the commutative property since: [Xm,[Y,,Yj31+[Yi~[Yj>XmlI + [Yj,[Xm,Ytll= 0 where the last two terms are zero, and thus [Y,,Y,I= 0. The proof is so established.
4. EXTERIOR DIFFERENTIAL SYSTEMS 4.1
GENERALITIES
D
An exterior diflerentialsystem of rank r on M is a system composed of r exterior forms equaling zero, namely Vx E M, V X E TN : f E {I, ...,r ] . wl(x,x) = 0
Let f : W +M be a differentiable mapping between manifolds.
D
A differentiable manifold W is called the integral manifold of an exterior differential system ( w' = 0 ) if the pull-back of each w' E QP(M) by f is zero: f *@I = 0 1 E { I , ...,m).
Cbssical example. By refering to exercise 10 of lecture 5, given an open U of the differentiable manifold R~ and a mapping f : U -+R~ defined by z = z (x i) with i = 1,2 and j = 1,2,3, let us consider the exterior differential equation on R~: J
J
There is an integral surface, defined by z ~ = z ~ ( x ~ , x ~z )~ = z ~ ( x ~ , x z~ ~) = 3z (1x ,2 x ) if b'(x1,x2)E U :
D
The ideal of an exterior algebraQP(M) generated by {a'}, denoted I, is the set of sums Cw' (€1) I
defined for any p'
E RP(M).
Lecture 6
208
Note that every solution of an exterior differential system is a zero element of I and conversely.
D
Exterior differential systems are equivalent if they generate the same ideal (i.e. have the same solutions).
D
The closure of an exterior differential system { w' = 0 ) is the exterior differential system {wl=O, dmI = 0 ) ( / = I ,..., r).
PRl 1 An exterior differential system and its cIosure have the same integral manifold.
ProoJ: It's obvious since d(f ' w ' ) = f *dwl = 0. D
An exterior differential system is closed if it is equivalent to its closure, ( d l c l ).
4.2
PFAFF SYSTEMS A N D FROBENIUS THEOREM
More particularly, consider an exterior differential system composed of r independent differential forms of degree 1 (sometimes called w a f l system)namely:
al,a:&i=o
( i = 1, ..., r ) .
(6- 18)
We have, Vx E M, V X E T , M , the following system of linear equations: a l ( x ) = 0 , also written a:Xt = O . Consider the r-dimensional subspace generated by the equations w' = 0 at any point of the n-dimensional manifold M. At any such point, called a "generic point," the equations (6-18) define an (n-r)-dimensional vector subspace P, of TM .
Classic example. Given an open U of the differentiable manifold R~, we consider the oneform equation ( i = 1,2,3 ) . a=a,dxi=O If there is an integral manifold then it is two-dimensional because one restriction exists on the vectors, namely a,X i = O . There is an integral manifold defined by xi = xf(u,v) with (U,V)EUCR~ if
Lie Derivative, Lie Group
209
Generally, there is no solution x' = xl(u,v) . By using next integrability conditions, we will show that solutions exist in this example. D
An exterior differential system { a'
= a: h' = 0 ) of rank
r is called completely
integrable on an open U if there are r independent differentiable functions y' defined on U such that {a' = 0 j is equivalent to the exact differential system { dyl= 0 ). The functions y1 are called first integrals of the exterior differential
system. We remark that a' = 5; dy' where the various Z; are differentiable functions on U
If the exterior differential systems { af dx' = 0 } and { dy' = 0 } are equivalent, then the r 1-forms a' and r 1-forms dy' define at each point x the same r-dimensional cotangent subspace Pi (M) of T=*(M). Then the mapping M + R r : x t+ (y',...,y r ) is of rank r. PR24 of lecture 1 proves there is one (n-r)-dimensiona1 differentiable submanifold through one generic point x, which is defined by equations vl(x>= y'(x0).
Meaning of the Frobenius theorem The Frobenius theorem signifies that, given a vector subspace P(x), the condition
VX(x), Y (x) E P(x): [X,Y)(x) E P(x) is necessary and sufficient in order that one integral manifold through each generic point of
M exists. Let X and Y be vector fields of a submanifold P and a E R being generated by {a',I = 1,..., r }. We have:
be a differential form of degree 1,
(a,X )= a(X)= a(Y)= 0 and V[X,Y]E P , V a ~ ,nwehave;
a ( [ X , Y ] )= 0. But generally speaking (see exercise 3), given some differential form P of degree 1 and any two vector fields Xand Y,we have:
dP(X,Y)= LxP(Y)- LYP(X)+ P([X, YI). Thus we have
VX,YE P,Va E R : da(X,Y)= 0 . So, d a E I where I is the ideal generated by R.
(6-19)
Lecture 6
210
Therefore, the exterior system {a' = 0, I = 1, ...,r ) is closed and the Frobenius theorem means in the differential system language that there is a family of integral submanifolds (foliation), namely the Pfaff differential system is completely integrable #I" the system is closed.
Integrability criterion PR12 An exterior system of r differential forms a' of degree 1 is completely integrable j f l
v l € { l ,...,r ] : d a ' ~ a A'
. . . A ~0 ~
.
(6-20)
Proof: First we are going to prove that an exterior differential form 8 belongs to the ideal I generated by r independent 1-forms a' in the neighborhood of a generic point
The necessary condition is obvious. Indeed, if 8 belongs to the ideal I generated by ( a') then the previous identity is verified. Conversely, let us prove that 8 A a' A ... a' = 0 is sufficient so that 8 E I . In a basis(a',...,a r,.. .) consider the I-form 8= a,, .,,, a" A .,. A a"
C
4 <..
Thus the condition {I,. ..,r) = 0 ; that is 0 E I .
implies that the functions ah..,are zero if { i , , . . . ,i,)
The integrability criterion follows from (6-21) where 6 is any da' (belonging to I ). PR13 Pfaff systems of ranks r or r-1 are necessarily integrable in the neighborhood of a generic point of an rdimensional space.
Proof: First we have:
a'
E
a\&' = 0
i,l
E {I,-.,,r}
dri=o.
Secondly, r-1 forms a' being independent, we obtain a basis by adding a rth 1-form a'. Since each da' is a hnction of any form aj ( j= 1,..., r - I), the condition
da' ~ aA' . . . r\arL' = 0 is automatically fulfilled.
Classic example. By refering to exercise 13 of lecture 5, we consider the following Pfaff equation in R ~ : a=u,drl=O
i
=
1,2,3.
This exterior differential equation is completely integrable on a neighborhood of x,
E
R~:
Lie Derivative, Lie Group
In this case, there is only one integral manifold passing through each generic point x, defined (in a neighborhood of x, ) by y ( x ' , x 2 , x 3 )= Y ( x ; , ~ ; , ~ ; ) .
Remark. Generally speaking, first integrals do not exist globally on a manifold. We say:
D
A Pfaff system is completely integtable on a naanifold M if there is a covering of M by opens where the system is completely integrable.
5. INVARIANCE OF
TENSOR FIELDS
The Lie derivative expresses without difficulty the invariance of a tensor field under any transformation. Let
4, be a diffeomorphism on an n-dimensional manifold M DEFINITIONS
5.1
D
*
A contravariant tensor field T on M is invariant under 4, if V x E M :
d#tq = T(@tx).
A field of differential forms w on M is invariant under #tgm(#tx> =
D
4 x 1-
(6-22)
#, if
Vx E M :
(6-23)
A tensor field on a manifold is invariant under a om-parameter (local) group of diffeomorphisms if it is invariant under every diffeomorphism of the group.
Let us give an immediate necessary and sufficient condition to have an invariant field. It is often taken as a definition. PR14 A tensor field is invariant under a one-parameter (local) group of diffeomorphisms #I its Lie derivative with respect to the generating field of the group is zero.
Briefly, V T E f;M, VX E X(M) :
w
T invariant ~ f l L,T = 0.
(6-24)
We also say that T is invariant under a vector field X . Proof:
The necessary condition is obvious from definitions of vector field invariance and Lie derivative.
Lecture 6
212
To prove the sufficient condition [L,T following lemma :
L
=o 3
T invariant], the reader will verify the
In a neighborhood U of a point x, E M , we can choose local coorrllnates x i such that a vector field X, nonzero at x, , has components
xk= q k , the index numeration being eventually modified. Such local coordinates are called "adapted. This lemma means that the curves such that: x' variable 1 x = c(,) E R ( for i = 2, ..., n ), are the orbits tangent to vector field X in U. "
Now, the sufficient condition can be proved because, for example, we have for some tensor field of type (: ):
,,.
0=
=
+ T,i*.,ipai,xk + +~ ; ~ . . . k a , ~ ~
= a,T,,,.
=
TI.
.Ip
.
(~2,*p.>~n)
Since the di ffeomorphism is such that & ( x ) = xi + t ( because X' = 1 ) +,!(x) = xi E R ( for i = 2, ...,n ) , 2 then the components T,I '..,p ( x ,..., x" ) are constants in U. Since #( 0 4$= q+t+, , the invariance of T is thus proved for every diffmorphism. In particular, we can say: PR15 Every vector field Y is invariant under a (local) one-parameter group of diffeomorphisms 4t (of generating field X ) fl LxY = 0.
Exercise. Prove the previous proposition from the Lie derivative definition (see exercise 17).
5.2
INVARIANCE OF DIFFERENTIAL FORMS
Let w be a differential form of degree p on M. We say:
D w A differentia1form w E R P(M) is invariant of vectorfield X or invariant under X i f
L,o=O.
PR16 The exterior product of two invariant forms is an invariant form.
(6-24')
Lie Derivative, Lie Group Proof: If L,w = 0 and if L X p= 0 then
Lx(w~p)=Lxw~p+~~L,p=O.
PR17 The exterior differential of an invariant form of vector field X is an invariant form.
Proof. LXw=O
Remark. L,w is closed
iff
s
Lxd~=dL,o=O.
d o is an invariant form.
Now, let us introduce a definition of invariance of forms with regard to differential systems.
We know that to vector field X on M correspond curves
:
R + M : ~ H ~ ~ X
such that
In local coordinates ( x i ) , the curves are solutions of equations
Instead of considering these equations, it is sometimes better to view a differential system composed of n-1 differentia1 equations, namely Vg E C m ( M :)
and we say:
D
A differential form w is invariant under the vector field X associated to a differential system if it is invariant under every one-parameter (local) group of generator gX where g is a nonzero function on M, that is
L@w=O.
(6-25)
PR18 A differential form o is invariant under the vector field X associated to a differential system
zfl
the respective inner products i,w and i,dw are zero.
Proof: We have
L,o = diSm
+ ifldw = d(gi,m) + g i,dm
= dg~i~w+gdi,w+gi,dw
thus L@o=O
Remark
a
[i,o=O and
Lflw=dg~ixw + gLxw.
ixdw=O].
Lecture 6
214
PR19 A closed differential form w is invariant under the vector field X associated to a differential system Iff the inner product of w and X is zero. Proof: The necessary condition is obvious because PR18 and dm = 0 imply i,m = 0 Conversely, if i,u = 0 and do = 0 = i,dw, then PR18 implies that w is invariant.
5.3
LIE ALGEBRA
The study of tensor field invariance is significant in mechanics (tensors being vectors as angular momentum, functions as energy, metric tensor and so on); we will meet with the last case in Riemannian geometry by introducing the Killing vector. The knowledge of the generating field under which a tensor field is invariant is important in physics. So, for instance, considering an axially symmetric system, i.e. invariant under rotations in some plane, then the associated vector field is invariant under the group of rotations. The angular momentum relative to the fixed axis (or projection of angular momentum) is conserved. It's a particular illustration of the following general proposition: "The invariance under a group of transformations leads to a conservative law." This notion will be later dealt. PR20 The set of vector fields under which the tensor fields of a set IT,) are invariant shows a Lie algebra structure. Proof: If a field of tensors 7'' is invariant under any two vector fields X and Y,then T, is invariant under every linear combination uX + bY (a,b E R) because we have:
So, the set of all such vector fields has a vector space structure. In addition, we have: LxTk=LyTk=O
=\
4x,r1T,=[Lx,Ly]Tk=0
(see exercise 8 relative to forms). Therefore, every field [ X , Y ]making up a Lie algebra answers the question.
6. LIE GROUP AND ALGEBRA Only one restrictive presentation of Lie groups is given since, besides this geometrical surrounding, the Lie groups are developed in special courses.
The Lie groups and Lie algebra play an important role in physics as illustrated with the examples of the General Linear group GL(n;R),the Special Orthogonal group SO(n), the Special Unitary group SU(n) and so on.
Let us introduce the Lie group notion which associates the notions of manifold and group. This theory founded by Sophus Lie was later developed by Elie Cartan; nowadays it is an important area of research.
Lie Derivative, Lie Group 6.1
LIE GROUP DEFINITION
D
A Lie group G is a finite-dimensional manifold of class C" provided with a group structure for which the group operations
.:GxG+G:(g,h)~g.h -': G + G : g I 4 g-' are of class C" . The dot wit1 be omitted. D
A lefi translation by g
EG
is a diffeomorphisrn defined by
L, : G - + G : ~ HL , h = g h . A right translation by g E G is a diffeomorphism defined by
$:G+G:hr-,R,h=hg. Remark 1. Generally, the group G is not commutative ( g h Remark 2. From L, 0 L,. = L., , we deduce
= Lg.,
+ h g ).
and likewise R;' = Rg-,.
Remark 3. Both L, and Rg are diffeomorphisms such that Lg o R,, = R,, o L, .
6.2
LIE ALGEBRA OF LIE GROUP
Let e be the identity element of G. Any neighborhood of e is mapped by a left translation Lg onto a neighborhood of ge = g .
We can introduce the differential mapping
dL, :T,G+TgG and so define invariant vector fields on G.
Lecture 6
216
D
A vector field X is leftt-invatiant if Vg E G :
It is equivalent to express V h E G : dL,X(h)= X ( g h ) .
Indeed, the implication (6-26') a (6-26) is obvious and the converse follows from the equalities: X(gh) = d & ( e ) = d(Lg o Lh)X(e)= dL,dLhX(e) = dL,X(h).
In the same manner, we define a right-invariant vector field by
dR,X(h) = X(R,h) .
PR21 The left (resp. right)-invariant vector fields on an n-dimensional Lie group G form an n-dimensional vector space.
Pro@ This proposition is obvious because (6-26) implies that the sum of any two invariant vector fields and the product of some invariant vector field by a real are also invariant vector fields. In addition, the set gL(G)of left-invariant vector fields on G and T,G are isomorphic vector spaces: the mappings J; : %,(G) + T,G: X I+ X ( e )
fi : KG + %(GI : 4 H ( g I+
XJg) = dL,5(e) 1
satisfy
A of,
= ~ d l . , ~ and
h0f;=idlzL,,,.
Only one left-invariant vector field corresponds to any vector of T,G and gL(G) has n dimensions. PR22 The bracket of any two left (resp. right)-invariant vector fields is a left (resp. right)-
invariant vector field. Proof: We use (3-4):
dS[X,YI = [dfX,dfYI where f is a diffeomorphism, that is here: dL,[X,Y]
= [dL,X,dL,Y].
From dLgX ( e ) = X ( g)
and
dL,Y (e) = Y(g),
we deduce:
dL,[X,Yl(e) = [dL,X,dl;,YI(e) = CX,YI(g), which means the bracket of any two fields X and Y is left-invariant (resp. for right-invariant).
Lie Derivative, Lie Group
217
In conclusion, the set X, (G) of left-invariant vector fields is provided with a structure of real vector space and with a composition law having the bilinear and anticommutative properties of vector field brackets on G and verifying the Jacobi identity. This set has a Lie algebra structure and we say:
D
The Lie algebra of Liegroup G is the Lie algebra of left-invariant vector fields.
It is denoted L(G)
.
The Lie algebra structure of XL(G) can be transferred to T,G (isomorphism!); so the vector space T,G is the Lie algebra of G by defining a Lie bracket as foliows:
that is also
(e)
Constants of structure
Let ( X i) ( i = 1, ..., n) be a basis of the Lie algebra L(G) . Since any bracket [X,, X, is an element of vector space L(G), we have:
where the reals c;l are called constants ofstmcfure of G which completely characterize the Lie algebra L(G). Property I. Under a change of basis, the reds c i are transformed as components of a tensor
of type (i ) owing to the bracket biiinearity and the antisymmetric property: c i . = -ci ~k
C'
Property 2. From the Jacobi identity, we immediately remark that c;,c;
+c;c;
+c;c,; = o .
Remark. Each Lie algebra of a Lie group has a unique tensor of structure. If two Lie algebra are isomorphic (namely there is a vector space isomorphism in accordance with the bracket), then their constants of structure are the same by choosing good bases. Conversely, is a Lie group G determined from its Lie algebra? The following can be proved: Given constants verifling the two previous equalities, there is only one simply-connected Lie group having these constants of structure. We must emphasize the local character of structure constants which generally do not globally determine G.
6.3
INVARIANT DIFFERENTIAL FORMS ON G
D
A differential form w on a Lie group G is lefl-invariant if Yg E G :
Lecture 6
The lea-invariance of o is simply denoted 'dg E G :
o=Lio.
Simple example. Let dt be a left-invariant form where t is a parameter such that t = 0 is the identity e on G. The product p(s, t) = s.t of points of respective coordinates s and t [ with p(s,O) = s and p(0,t) = t ] defines the following transfornation of G : L, : r Hp(s, t ) . The left invariance condition of the one-form dt , namely L:(dt) = dt , is
that implies p(s,t) = t + f( 8 ) =s+t since s = p(s,O) = f (s) . So, the group operation is the addition and G is commutative. Remark. If w is a lefi-invariant differential form on G, then it's the same for d o because we have Vg E G : .Lido = d q o = d o . The preceding can be said for tight-invariant forms. Structure equations of Maurer-Cartan
Let (Xi) ( z = I, ...,n) be a basis of the Lie algebra L(G). Let us consider the (dual) basis (a') of left-invariant differential forms of degree 1 such that al(X,) = 6;. From (6-1 9), we have in this particular case V j ,k E { 1,. .. ,n): d a i ( x , , x k )= ~'([x,,x,]) = cJkai(x,)= cJk and thus we have obtained the equations of Maurer-Cartan:
which let us know the structure of the Lie algebra of G from differential forms of degree 1.
6.4
ONE-PARAMETER SUBGROUP OF A LEE GROW Let G be a Lie group.
D A one-pamme&r subgroup of G is a ( C" ) differentiable homomorphism of the (additive) group R into G
Lie Derivative, Lie Group Thus, we have
g(s + t) = g(s)g(t) g(0) = e . This subgroup is evidently abelian.
PR23 The one-parameter subgroups of G are the integral curves (or orbits) of left-invariant vector fields on G passing through e ( at t = 0 ). Proof: First, we are going to prove that a one-parameter subgroup G defines an orbit of a left- invariant vector field passing through e, X(e) being the tangent vector at point e. d
Remember the equations -g(t) = X(g(t)) and (6-26'). dt The equality (6-30) is written: + t ) = Lg(,)g(t)= L#(,)~(S)
If s=O,nameIy g(O)=e and then
1-
where X (e) = dg(s) ds
dL, = i d ,
,,,is the tangent vector, at e, to the curve defined by g.
The equality (6-31) means the curve t I+ g(t) is an orbit of a left-invariant vector field with tangent vector X(e) at point e. Conversely, we must prove that a (unique) one-parameter subgroup of G corresponds to any tangent vector X(e) of G. By denoting
and if
[s)
L, : ( h i )H ( ~ , h '=) ( p'(g',hr) )
denotes the matrix defining dLg (see also exercise 181
then the equation (6-31) is written in local coordinates:
where the various ht = 0 and X r are respectiveIy the coordinates of e and X(e) . There is a unique solution g(t) (It( < E ) such that g(0) = e and that is a one-parameter subgroup of G. In other words, the curves t k-3 €s)(!
g(t)
and
t
H g(s +t)
verify the same differential equation and the initial condition g(s)g(O) = g(s) = g(s + 0).
Lecture 6
220
d Indeed, the tangent vector -(g(s)g(t)) to the curve t I+ g(s)g(t) is the image of the dt tangent vector to the curve t H g(t) under the diffeomorphism Lg(,, , that is
that is also
There is a one-to-one correspondence between the one-parameter subgroups of G and the elements of the Lie algebra of G : to each vector of T,G is associated a unique one-parameter subgroup (curve through e ). Exponential mapping We define a diffeomorphism of G onto itself as follows. The points on an integral curve of a left-invariant vector field X (trough el at t = 0 ) are defined by g, (t) : x H exp(tX)x .
D
The mapping g, (t) = exp(rX) is called exponential mappinggenerated by the vector fierdx:
This mapping has the property of a one-parameter subgroup of G :
Therefore, for every X
E
KG,the integral curve of X passing through e at t = 0 is
This mapping g, exists for any real f (the flow is complete). This smooth (i.e. C" ) homomorphism is a one-parameter subgroup of G .
D
The exponentz'ai mapping ofthe Lie algebra of G into G is exp, : T,G + G : X
H g,(l)
=expX.
Example 1. The vector space R" which is a manifold and a group under vector addition is the most obvious example of Lie group. The one-parameter subgroups are the straight lines through the origin. The left-invariant vector fields are parallel to these lines and these commutative fields have zero brackets. In conclusion, the Lie algebra of Lie group R" is the vector space ToR" = R" provided with zero bracket. The exponential mapping exp : Rn+ Rn is the identity.
22 1
Lie Derivative, Lie Group
Example 2. The group of ( n x n ) real nonsingular matrices is called general linear group and denoted GL(n, R) .
If the space of ( n x n) real matrices is identified with R"' , then the general linear group is defined by a nonzero determinant. identified with the open submanifold of Atn2 Let A = ( 0 : ) be some element of GL(n,R) with det(a/) + 0 . There is an open subset of R*', GL(n,R) is provided with a differentiable manifold structure. A neighborhood of A is composed of matrices 3 = ( b / ) such that b,! - 0; < 6
1
1
where the real E is chosen mall enough so that det(b,!)t 0 . In this neighborhood, the coordinates are defined by the n 2 reals xi' = b: - a;. GL(n, R) is a group with the multiplication law of class C m: GL(n, R) x GL(n, R) + GL(n, R) : (A, B) I+ AB where AB = ( a / bi ). Besides, GL(n,R) + GL(n,R) : A H A-' is of class C" , so G is a Lie group. Since R"' is identical to the tangent space at any of its points, the tangent space at identity point e of GL(n,R) is naturally identified with Rn2: any tangent vector is a (n x n)reaI matrix. Consider a one-parameter subgroup generated by any matrix A E L(R n ,R n ) that is an integral curve of left-invariant vector field passing through e (at t = 0) and represented by the matrix ( t )=(
)
with
A=
[y) 0
Since g~ (f + At) = gA (f ) g~ (At)
7
we easily obtain:
g, (t) = exp(tA),
so,
is a one-parameter subgroup with gA(0)= I . The exponential mapping is A"
exp: L ( R n , ~ "+) G L ( ~ , R ) :A H ~ , ( I ) =~ n=O n! Finally, we find the Lie algebra of GL(n,R) as follows. For every A E L(Rn,R n ) , the left-invariant vector fields on GL(n, R) are defined by X, : GL(n,R) + L(Rn, Rn): Y H YA . Here, the Lie bracket is defined by IA,BI = [XA, X,l(I) .
Lecture 6
222
If we consider the one-parameter subgroups or integral curves of left-invariant vector fields g, and g,, and if we refer to the exercise 17 of lecture 3 giving an interpretation of Lie bracket with the help of diffeomorphisms, we immediately have:
1
= lim - ( ( l + t A + - ~ - - ) ( 1 + ! B + . ~ ~ ) - ( l + r B + - ~ . ) ( l + f A + ~ ~ ~ ) ) t=0
t2
So, at point e we obtain: [A, B] = AB - BA and the Lie bracket of any two left-invariant vector fields at e is the usual commutator of the two matrices "generating" the fields. The left-invariant vector field generated by this commutator belongs to the Lie algebra of GL(n, R). L(R n ,R n ) is the Lie algebra of GL(n, R), the Lie bracket being the matrix commutator.
We remark that, given C E GL(n, R) , the mapping C g, C-' : R + GL(n,R): t I-+ C g, ( t )C-'
is an integral curve of vector field Xu,-,passing through I Indeed, we have: cg,(O)C-' = I and
In addition, we see that gm-I = C g ,
c-~
or ~x~(cAc-'= ) C exp A C-' .
PR24 If h : G + H
is a C" homomorphism of Lie group G into Lie group H , then dh, : L(G) + L ( H ) is a Lie algebra homomorphism.
Proof We have 'd 4
, E~T,G : dhe[5,V1= dh, ( [ X g' X,l)(e) = [dh,Xc,dh&.X,I(e, = Ex4i,x~,,l(e~)
( homomorphism!)
= Idhe5,dh,rll
PR25 If h : G + H is a C" homomorphism of Lie groups, then Mexp, ProoJ: The mapping
5)= exp, (dh,C).
V{
E
L(G)
Lie Derivative, Lie Group
is a one-parameter subgroup of H . Thus, we have d dh.C = ;i;g(t)l ,=o = v and g(t)= exp, trl.
That implies: h(exp,
5) = g(l) = exP, V = e x P (dhe5). ~
The adjoint transformation For example, the matrix group GI,(n,R) has been considered through a faithful representation as a matrix transformation of n-dimensional vector space (a representation is termed faithful if it is one-to-one). Besides such a type of representation there is the adjoint representation that we are going to introduce. First let us consider the inner automorphism of G associated with g E G , that is
This mapping of G into itself is Cmand is a homomorphism because Vh,1 E G :
In particular, the identity e is mapped by any I, into e. So, each I , induces a mapping of TeG into itself and we say: D
The adjoint tranrformation associated with g defined by Ad, = (dl,), = d(Rg-,L,)(e).
Remark 1. PR25 lets write tlg E G, V {
E
G is the mapping TeG -+ TeG
E T,G :
exP(Ad,5) = exp((d1, ),
5
= 1, (exp5 = g exp5 g-'
Remark 2. Considering a one-parameter subgroup of G defined by t t-, exp(tX), let h and 1 be any two points of this integral curve of X passing through e at t = 0 . The respective images under I, of the previous points, that are h'= g h g - l , I'= gig-' and
e = geg-l, define another curve passing through e such that the tangent vector field is ~d 8 x (this follows from I,(hl) = I,h ],I). Thus we denote I,(exp(tX)) = exp(t Ad, X).
This section relating to Lie groups and Lie algebras has left the door half-opened to a fundamental domain of theoretical physics.
Lecture 6
7. EXERCISES Exercise 1. Verify that V X 'X E+(M),Vg E C m ( M :) (dg, X ) = L,g Answer. We have:
( d g , x )= d g ( x t a , )= a j g d * j ( x t a i )= a j g x t d : = a , g x i - - dg = Xg. df The 1-form dg is called the gradient of g. Exercise 2. From the Lie derivative definition, find again the expression (6-14') of L,w in the particular case of a one-form o,dr' :
Answer. Let us consider the diffeomorphisrn
,..., xn)l+ (4; ,..., 4;).
:(XI
The important rule of pull-back (see lecture 5. $3) implies:
and the Lie derivative definition is written:
(since
4; = xi )
d4; ~,=,&'=-&J. ax 11-0 &I dt dxj l Let us note this result is obvious because from the definition of the Lie derivative of a function we immediately have:
-dad --d ax.i
d*i-
t
a
= d x J a j x r= d ~ J 6 ;=.dxl= aJ.xi&'.
L,&'
Therefore, the announced formula immediately follows:
L,
(0,A')= L,D,
dx'+ U , L,&'
= x l a j w tA'
+ mia,xi drl
= ( x k d , w ,+ w , a , ~ ~ ) d x ' .
Exercise 3. Given some differential form j of degree 1 and any two vector fields X and Y , prove the following equality:
d P ( X ,Y )= L, Answer. We have
P V )- Lu A X ) + P ( [ X ,Y1).
Lie Derivative, Lie Group
and thus
Exercise 4.
Check the fundamental theorem in the case of a 1-form. Answer. Consider w=w,&. On the one hand, we have: i,w=w,X1 = ( w , x ) di,w = (d j w,)X i dxJ + w , ( d j ~ ' ) d x J . On the other, we have: dm = a,w, dxJ A dxi i,do = ( d j w l ) X dxi -(i3,w,)Xi d r j . J
Thus, we have: (di, + i,d@ = (ajw,)xJdx' + w , ( d J ~ ' ) d x J =(xk akml-t- w,
a,x")dxj
=Lxw. Exercise 5.
Check the fundamental theorem in the case of a 2-form. Answer. Consider a=@,,,&' A & ' = + ~ , & ' A d r '
On the one hand, we have: !,a, =
m M x hdkj = n, &J
On the other hand, we have: d o = f (aha,+ a l @ , + a j w h , ) h hA&'
where
RJ = W,,.xh
A&
ixdo=$(dhw,+a,rup+djo,,)Xh&t~drJ. Thus we have: (i,d
+ O @ , X +~ q,djxh)dxi A
-t- dix )ol= & ( x ~ L $ , ~ ,
&j
= &m.
Lecture 6
Exercise 6. Check the fundamental theorem in the case of ap-form. Answer. Consider 1
t3 = --itDil ,,%
P
A ... A d
i
p
.
On the one hand, following from exercise 12 of lecture 5, we have: d o = --L(akrul.,ip @+I)!
-a,,@,
i,dw = $(akw, , , -a,mk
,.
,,- ..-- a, q,.)drk - - - -aj u
, , ,
A hh A... A h
ip
)xkdthA . . . A & ' P .
On the other hand, we have: i,w=-
a, X k d x t .2. . A~& ~ P (p-I)! Y .JP
Therefore, we have the following expressions of di,m :
=S(xka,,ob*.,,p+Oh a,xk+ xka,*a?,,,,+ ~ 2.,,
+...+xkaipm,,,ip-l + uk,,,a,+,xk)
k3.., l i,ai*xk
A . . .A..,dXi~
Because the last p terms of iXdoare opposed to first p terms of di,w , we have:
Exercise 7. Given two vector fields X, Y and a function g, check the following equalities:
Answer. (i) We have: [Lx,rY]g= L X i Y g - i Y L X g= 0
and i[x.ulg= 0 .
(ir;) We have:
Lie Derivative, Lie Group
but
thus [L, ,iy ]dg = Lx (Yg) - ir ( d ix
+ ixd)dg = X(Yg)- i, d(i, dg)
= X(Yg>- iyd(Xg)= X(Yg) - Y ( X g )= LIx,Jlg
= i[*,Y
-
Exercise 8. If w E IZP(M),prove that
Answer. We will proceed by induction in order to demonstrate that
First, this equality is checked on some 1 -form g dh . We have: il,,Ylgdh = g i[,,,]dh = gL[X,Y,h= g [ X , Ylh. From Lx iy( g dh) = Lx( g Yh) = ( L , g)Yh + gLx (Yh) = Xg Yh + g X(Yh) and iyLx ( g d h ) = i, ( (Lxg ) dh + g Lxdh) = iy( X g dh + g L, dh) = XgYh+gi,di,dh = XgYh + g i r d ( % ) = x g Yh + g Y ( X h ) , we deduce: L,i,(gdh)- iyLx(gdh)= g X ( Y h ) - g Y ( X h ) = g [ X , Y ] h= i,,,,,gdh.
Now, if we assume the formula is verified for my exterior form of some degree less than or equal to p-1, then we are going to prove it is true for exterior forms of degree p. are less than or e q d to So, given a p-form o = a A p such that the degrees of a and p- 1, we have: $r.ula = ELx ,ir la l [ ~ B=CLx,iylB. .r~ Thus, we have:
Lecture 6
228
Exercise 9.
If X and Y are vector fields on M, prove that: (
[L,, L,] is a derivation on the algebra C" (M) and
(19
V w E h P ( M ) : L , X , r l=~[ L X ,L,]w.
4,,,, = [L, ,Lr
,
Answer.
(9 If we remember that Lx : Cm( M ) + C m ( M ) is a derivation on C" ( M ), we write Wf,g E Cm(M): rLx SLY I(fg)= Lx L y ( f g ) - Ly Lx (fg)
=Lx(fLYg+gLYf)-LY(fL,g+gL,f) = fLxLyg+L,gL,f +gLxLyf +Lyf Lxg - f L , L x g - L x g L r f - g L , L x f - L x f Lyg =f[~x.LYlg+gC~x,Lrlf. The second assertion is immediate because V f
ECm(M)
(ii) We successively have V o E RP( M ) :
Lix,rl@ = di[x,q@+ ikx,rlda = dCL, , i , ] w + [ L , ,iyldm = dI,,irw-di,Lxo+ L x i y d w - i y L x d w = L,di,w+L,i,dw-di,L,w -i,dLxw = L x L , o - L,Lxw = [ L x , L y ] w . Exercise 10. Verify the equivalence of expressions (6- 15) and (6- 16) in the case of a differential form
of degree 3 in Md. Answer. We have:
Lie Derivative, Lie Group
Exercise 11.
If o = drl A ...drn designates the (canonical) volume n-form on R" and if X is a differentiable vector field on R" : fiq (zii)
Calculate d(i, o). Calculate div, X if g = g a!!' A ... A dxn. Find again the expression of the divergence of X in spherical coordinates.
Answer. (
We immediately have: ixw-
X ' ~ X ~ A , . . A C L-Xx"2 d x 1A
=(
Ei3,xj)o
~ AX . . . A~& "
+...
= (div,X)m.
J
('I&!
Wehave:
1
div,X = -dj(gXj)
( summation on j).
g
(iii)
Since we have in local coordinates x' = r, x2 = 8, x3 = 4 : d u ~ d y ~ & = r ~ s di nr 8~ d B ~ d 4
then the divergence of a vector field X = ~ ' d+, x2a6 + x3a,is written
Exercise 12.
d If w designates the volume 3-form of R band if X = X i i is a differentiable vector
ax
field on R 3, give the expression id,@
by using the linear mapping
Lecture 6
230
Answer. In R 3 , as we will later see, the types of variance are indistinguishable, and we have on the one hand: curIX = ( d 2 x 3- d 3 x 2 ) d ,-I-(a,xl- alx3)a, + (a1xZ -a , ~ ' ) d ,
and thus z,rlx
= ( a , x 3 - d,x2)dr2A dr3 - ( a 3 x 1- a I x 3 ) d r 1A k3+ (d,x2- 8,~')dw' A d x 2 .
On the other hand:
and thus
In conclusion, we have established that
Exercise 13.
Given an invariable mass rn, calculate the Lie derivative of density p defined by dm= p d x l ~ . . . ~ d = x "p'dx" A...A&'".
Answer. We have:
~ d x A...A&" ' = pdet
)-
A .
A
din=
Anrn .
So, under a change of coordinates, we have:
Let us calculate the Lie derivative
On the one hand, the density at point x is:
On the other hand, the transported tensor of po is the density p, expressed in the "new" coordinate system:
but
Lie Derivative, Lie Group thus
and
ax' 0
dxl t
210
....
ax2
1-?lot ax
The only interesting terms are the ones of power I of t :
Thus we successively have:
, t
( summation on k )
Lx P = (diplo X i + p0(dkxk10
Exercise 14. Choosing a local coordinate system, consider the Lie derivative of a tensor field T with d respect to vector field --r. Find ( L ~ ~ T ) " jq" .Give ' ~ ~ , .the result in the particular case of a ax
vector field Y. Answer. In the formula (6-13), if we replace
X*with 8: for any k, then we have:
( L , , T ) ~ " =a,~'"" ~ ' ~ ~ ~ . ~ .~ More particularly, if d[ takes the place of X' in LxY = (x-'djyi - ~ j d , ~ ' ) d , then we obtain ( L a Y ) ' =a,yi. 7
ax'
Exercise 15. In R ~given , a curve defined by equations p = p(t),q = q(t) and a vector field
calculate L, (dp A dq) with the help of the fundamental theorem and by taking into account of canonical equations, namely:
Lecture 6
Answer. We have:
L, (dp A dq) = di, (dp A dq) = di
, , (dp .
p-+q-
aP
*
A dq)
Exercise 16. Consider a differentiable mapping of A4 into N and Prove the following equalities V X E X(M) : (i) f*(i@,w)=i,f"w (ii)
fa
: R P ( N ) +Q P ( M ): w H f *w.
f*L8,w=Lxf*w.
Answer. We successively have VY, ,..., Yp-, E X ( M ) :
{ii)
f*L4,w=f*dimw+f*i,,a'o=df*i~,w+~*iQ~d~ = di, f 'w
+ i,f*dw
( because fi) )
= L xf*w. Exercise 17.
From the Lie derivative definition, prove that any vector field Y is invariant under the (local) one-parameter group of diffeomorphisms 4, (of generator X ) t f l L,Y = 0. Answer. The necessary condition being obvious, let us prove the sufficient condition. We have tft E R:
The first equality follows from the lecture 3. $4 where we have established that
that is
The third equality follows from the writing of the Lie derivative definition. In conclusion, if L, Y = 0, the image d+-'~#,+ under the diffeomorphism is constant for every r. The vector field is invariant.
Lie Derivative, Lie Group
Exercise 18.
Prove that a Lie group G is locally isomorphic to R" r f l its structure constants are zero. Answer. First, if G is locally isomorphic to R", then there is a diffeomorphisrn between neighborhood of e E G and neighborhood of the origin of Rn, L, ; h HL,h, such that the
local coordinates of L,h = g.h (E G) are g' + h i . Therefore, the differential of L, is defined by the matrix
and thus, locally, a basis of the left-invariant vector field space is (8,) and any bracket of such fields is zero. Conversely, let us consider some left-invariant differential form a' of degree 1. The constants of structure being zero, from d a i = 0 we deduce there is locally a differentiable function g' such that a' = dg'. If such n independent functions g' are chosen as local coordinates, then the diffeomorphism L, is expressed by L, : (h') H (L,h1)= (p'(gJ(g),h') where the various p' are the coorhnates of the product of elements of which the coordinates are g-' and h r , with pi(g-',O) = gi and ~7'(O,h')= hi.
api r = dh'. The invariance of l-forms under the left-translation L, is expressed by -dh dh' The differential dLgis defined by its m d x
($1
that is (6;) since, by hypothesis, the
forms a' are invariant. Therefore, we have: p'(gJ, h') = h'
+ f '(g').
We choose the functions g J such that, for the identity element e of G, we have gi(e) = 0 . Then pi(g',0) = f Y g j ) ( coordinates of the product g.e ). = gi
Finally, we have pi = h'
+ gi and
G is locally isomorphic to Rn.
LECTURE
7
INTEGRATION OF FORMS STOKES' THEOREM, COHOMOLOGY AND INTEGRAL INVARIANTS
In this lecture, we are going to define the integral of an n-form on an n-manifold. Next, the important Stokesytheorem will be expounded and finally the cohomology theory will be set out briefly as well as the integral invariants.
1. n-FORM INTEGRATION ON n-MANIFOLD 1.1
INTEGRATION DEFINITION
1.1.1 n-form being zero outside a compact Let M be an oriented n-manifold, U be a chart domain of M, w be a differential form of degree n on M, zero outside a compact included in U. We choose a Iocal coordinate system which is compatible with the orientation of M; we have:
D
The integral of w on M is the Riemann multiple integral of the function w, ,. on R":
jMru,,n(~) di' A . . . A & "
=
o=
C-..Ew
( x ) &'...dxfl,
the function a,,,,, being zero outside the compact.
We denote:
1w
= rm...J-:a1,,"
&I...&.
(7-1)
This integral exists if the function w,,,,, is locally integrable, in particular if it is continuous.
Lecture 7
236
Remark 1. We say w has compact support if suppo is compact in M, where suppcl, is CZ(X E M I W(X)+ 0). Remark 2. The definition (7-1) is obviously well-founded because it is independent of the chart choice. l and we Indeed, given two any charts (Uf,g ~), and (U, ,p, ), we consider supp w c U, fU, view the chart change p, 0 q,:' that is the change of local coordinates x' t,2'. The formula of change of variables in the case of Riemann multiple integrals is:
The equality
lets us conclude the definition is not dependent of local coordinates provided that the chart change safeguards the orientation (that is positive Jacobian). 1.1.2
Differential form of degree n on M
Let M be an oriented n-manifold, w be a differentiable form of degree n on M, (g,) be a partition of unity.
We have Vx E M :
g, (x)ar(x) = e ( x ) . 1
D
The integral of w on M is
for any partition of unity.
Remark 1. The sum has only a finite number of nonzero terms. For any x E M , there is a neighborhood U such that a finite number of g, are nonzero on U.By compactness of supp w , there is a finite number of such neighborhoods which cover suppu. Thus there is only a finite number of nonzero gion the union of these covering opens U. Remark 2. The above definition is independent of the partition of unity (with positively oriented charts). Indeed, if ( g ) 3 is another partition of unity subordinate to another covering { U; } of M, then the functions g,g; are such that Vx E M : g,g;.(x) = 0 (except for a finite number of (i, j))
and
g,g; (x) = 1 . Thus we have: (since
xgl = I ) J
Integration of Forms
Now, we are going to define the integral of a differential form of degree n on a subset of M.
D
The integral of a form o of degree n on a subset S of M is (if existence) the integral of 40, where 4 is the characteristic function of S that is VIES: 9 5 ( ~ ) = 1 VX E s : #(x) = 0.
From the definition, we immediately have: {g the linearity, that is:
@+P=L&+~~P
\ I O , ~ E ~ ~ jM (M): Vk,ksR:
I kru= k] M
M
0,
the additivity with respect to integration domains, that is: if S, and S, are two complementary subsets of S with respective characteristic fhnctions 4, and 4), then we have: (ii)
1.2
PULL-BACK OF A FORM AND INTEGRAL EVALUATION Let M and N be oriented n-manifolds.
PRI
I f f is an orientation preserving diffeomorphism of M onto N and if w E Rn(N)has compact support, then:
Proof: First, we specify that supp f'w = fl(suppw ) is compact. Second, we recall that the image under f of a chart (Ui,pi) of an atlas of M is a chart (f(U,),#,0 f-')of an atlas of N.
Remember (see lecture 5.53) that if a = o n ( y j ) d y l~ . . . ~ d y ~ is an n-form on N, then its pull-back (in M) is
,..,
Next, in a chart domain, the definition of the integral of a form, namely
and the introduction of a partition of unity lead to the proof We summarize this proposition by the following diagram:
Lecture 7
A subset S of M is of m e w r e zero if S is the countable union of sets Si such that
D
each Si is included in a local chart ( U , , q i )and every q , ( S , ) is of measure zero in
R".'
PR2
A differential form o of degree n, having compact support, is integrable ~ f lthe set of discontinuity points of o is of measure zero.
This is justified from the definition of the integral and from Riemann integral properties
Proof
In practice, we can: subdivide a manifold into a sum of sets such that each of them is included in a chart domain (except for sets of measure zero), introduce a diffeomorphism between M and a set of Rn (except for a set of measure
-
zero).
Standard example. Integrate a differential form of degree 2 on the sphere
Answer.
s2.
s2is defined by x2 = sin 0 sin #
x' = sin@cos#
x 3 = coso
with 0 < 8 < 1 r , 0 < # < 2 z .
So, the pull-back of a form on
s2:
~=P&~A~X~+Q&~A&'+R&'A&~
is (in R') :
f *w = P(cos6 sin# dB + sin0 cos# d#) A (-sin @)dB
+ Q(-sin8)dB A (cos8 cos# dB - sine sin# d#) + R(cos B cos# do - sin 8 sin# d#) A (cos 8 sin # dB + sin 8 cos# d#)
Therefore, S' being provided with induced orientation (B,$) of R ~we, have:
A subset E of R nis of measure zero if there is a mumable sequence of closed intervals P( such that:
Integration of Forms
2. INTEGRAL OVER A CHAIN
Let o be a differential form of degreep on M, P = ( (xi) E R P I a' < x1 S b' ;a ' , bi E R ) be a closed pdimensional interval. 2.1
INTEGRAL OVER A CEAIN ELEMENT
D
A pdimensional chain element on M or element oJ a p-chain is a pair, denoted c, = (P,0 ), composed of - a closed interval P which is oriented from numeration of coordinates, - a differentiable mapping @ of an open U c P into M .
Figure
D
The support of the chain element c, is the subset @(U)of M
D
The integral of a form o of degree p over an element of a pchain ci is
D
Chain elements c, and cj are equal (resp. opposite) if, for every form o,we have:
Remark. To opposite orientations of P correspond opposite chain elements.
2.2
INTEGRAL OVER A CHAIN
D
A chain on M is a (formal) linear combination of chain elements, denoted
D
A chain is jinite if it is composed of a finite number of elements.
D
The integral of a form u, of degree p over a chain c is
Lecture 7
240
Remark. We will suppose the previous sum is finite. It is the case if for example a, has
compact support and if the chain support is locally fmite.
D
Chains c and c' are equal if for every form a, of degree p we have:
Now, let us prove an important proposition. Let P be a closed interval of R P , f be a differentiabje mapping of M into N , 0 be a differentiable mapping of P into M, w be a differentiable form of degreep on N, p being the dimension of P. PR3
If f : M + N is a differentiable mapping between compacts mahng locally finite chains c and f (c) correspond, then for every form w (having compact support) we have:
ProoJ: The image of cl = (P,@) under f is the following chain element on N:
f ( c , > = ( P , fo
m .
The definition of the integral of w over the chain element f ( c i ) is:
From (f .@)*&I = (@* 0 f*)w =@*(f80)
we deduce
j,(f
o @ ) ~ = S p c D * ( f * w ) ==I Jf*w. k , q an M and every compact of N being the
By considering the locally finite chain c = i
image of a compact of M under /: then the chain f (c) =
k,f (c,) is locally finite on N. 1
From the previous result about an element of chain and from the integral linearity, we deduce the important formula (7-6).
3. STOKES' THF,OREM 3.1
STOKES' FORMULA FOR A CLOSED p-INTERVAL q
Assume the form w of class C (q 2 1)
Integration of Forms
24 I
p
D The boundary of a closed p-interval P of R is the set of 2p closed (p-1)-intervals of RP-'that are the faces x i = a' and x i = b' of P.
It is denoted dP. Some point of a face is defined by p-1 coordinates ( x l ,..., i',.. ., x P ) such that a' 5 x 5 b where j = 1,...,i",.. .p. I J
An orientation is associated to each face, namely:
D
An orlentadon of a face is the one of coordinates determined by the order i i (xl,...,i', , . , , x P ) for the face x = b' if i is odd and for the face x = a' if i is even. Any odd permutation of coordinates gives the opposite orientation.
So, in the example of a cube, the orientation of the face x 3 = b 3 is the one of coordinates determined by the order ( x ' , x 2 ) and the orientation of the face x 3 = a 3 is determined by (x2,x1).
Consider a (p-1)-form on P : 13=
q,,;.,p h' A . . . A d ; '
A . . . ndrP
and thus
The definition of the integral of this p-form on P, oriented by the natural numeration is:
(with summation over the repeated index i )
It's a sum of integraIs of o over the faces x i = bi and x i = a' of P, on condition that the previous faces be oriented by the numeration 1,..., i*,...p of Atp-' for the face xi = b i if i is odd and for the face x i = a' if i is even; in the opposite sense in the other cases. There are the chosen conventions to define the boundary 8D of P. The reader will view the cube as an example. Finally, we have the Stokes'formula:
' A "hat" denotes a missing term.
Lecture 7
242
3.2
STOKES' FORMULA FOR A CHIAN Let 0 be a differentiable mapping of P into M
D
The boundary of an element of a p-chain c, = (P,@) in M is the element of the (p - 1) -chain (dP,Qi) where aP is defined by a set of 2p oriented closed (p - 1) intervals of R ~ -, '
It is denoted dc,.
D
The boundary of a chain c =
k,c, is i
Let us prove the Stokes' formula for a chain. Let o be a (p - 1) -form on M The integral of dm over a chain element
ci
is (definition of the integral of dm)
k d o = jP @ d o =
d (a*@)
=I =Lo
@*u
Finally, for a chain c =
( since @'dm = d@*w)
(in accordance with Stokes' formula for P) ( from previous definition of integral ).
k,c,, in accordance with the definitions of chain boundary and i
integral, we have:
Therefore, we have the Stokes'formula for a chain:
To end, let us remark:
PR4
The boundary of a chain boundary is zero, that is VC: a 2 c = o .
ProoJ: For every form w, we have:
Integration of Forms
4. AN INTRODUCTION TO COHOMOLOGY THEORY 4.1
CLOSED AND EXACT FORMS
D * A form
w on M is said
- COHOMOLOGY
to be closed if its exterior differential is zero
o E R ~ - ' ( M ) closed if d o = 0 . D * A form w is =act (or is a coboundary ) if there is a form v of which the exterior differential is o: o E R P ( M ) exact
If 3 v E R ~ - ' ( M: )d v = w .
We immediately say:
PR5
Every exact form is closed.
Proof: It's clear because d 0 d = 0 . The converse is not true. But, we can prove every closed form w is locally exact, that is at each point x, E M there exists an open neighborhood U (of x,) such that the restriction a,,is exact. More precisely we will demonstrate if U is an open sphere of R" and w is a closed differential form of degree p (21) defined on U,then there is a ( p - 1) -form v such that w = d v . We must specify the (p- 1) -form v is not unique because any (p - 1) -form of type v + d p is perfectly suitable. Every exact form being closed, the vector space BP(M) of exact pfoms on M is a subspace of real vector space F P (M) of closed p-forms on M. Therefore, we are going to introduce the quotient space of FP by BP , denoted F pI B P, from an equivalence relation. Let us show equivalence classes splitting up F p ( M .)
D
Any two closed forms are cohomlogic if their difference is exact:
VW,O'EF P ( M ) , ~
V
0E' - ' ( M ) : o
- O'
&-a'= dv.
In particular, D
A cohomologic to zero form is nothing else but an exact form: V ~ E W ( M ) w: - 0
D
a w=dv.
The cohomology class of the closed form w is the set of closed forms which are cohomologic to o.
PR5 The cohomology classes of closed forms of degreep on M make up a vector space.
Lecture 7
The set of these classes is the quotient space denoted H P ( M ;R) = F P ( M ) I B P ( M ) or simply
H P ( M ). For every p-form and real k, we have: 0-wf,p-p'
3
w+p-wl+p'
D
The space H P is called the pth de Rham colromology space of M
D
The dimension of H P is called the pth Betti number of M, denoted b P .
It is proved that b P is finite in the compact manifold case. 4.2
POINCARE)
PR7
Every closed p-form on an open homeomorphic to P is exact.
LEMMA
Proof: We know there is a homeomorphism of a neighborhood U (of some x E M) onto an open ball of R" about 0. 1O. The proof is immediate in the case of monomial forms. In an open of R", let ~=~dr'A...Adw~ be a closed form where g is a differentiable function in R" that we must specifl. We have:
that implies
Therefore g is only a function of x' ,..., x P . We recall it is sufficient to prove the existence of a form v such that o = dv . We choose v = h d w 2 r\...r\dxP ah where h is a function of xl, ...,xP and such that - = g . ax1 Thus we have:
ah
d v = d h ~ d x ~ n . . . =~- ddrrl ~ ~ d r ~ . . . =@. ~ d r ~ ax1
2'. If the closed fonn w is not a monomial, to prove the Poincark lemma we are going to construct an R-linear mapping: I : Q P ( U )+ np-'(u)
Integration o f Forms
such that 1 0 d+ d o l = I
[ identity on RP(U) 1.
As soon as this will be established, the Poincare lemma will be proved because w=(Iod
+ doI)w=(doI)m
which will show the existence of a form v = Iw such that w = d v Let @(x)= iflf,, t p &'IA .,, A hip
be a p-form of R P ( U ) . We know (see exercise 12 of lecture 5) that dm(x) = &(akmil..ip
- ailmki
2 .i p
-...
-
a, .co,,,)&*
ALWI
A...AU~P.
On the one hand, we introduce the following contraction i,diu(x) = ~ x ~ ~ , w i , -, ,ai1oksi, ,p -
and at point tx:
di mr,r)dxilA ... A dr
ip
,,
i,dm(tx) = j ~ ~ t ( a , ~ ,(tx) , , ,- a , , ~ , , (tx) -.-. - a , ~ , ,(tx))dxig ^ . . . k i p
On the other hand, we have: 'xu(') =
@.Ci2..i,
( x ) x dXi2 ~ A... /\dXip
but the formula (5-19) implies that this ( p- 1) -form *Qi2.,,#
ht2 h...Ah"
has the following exterior differential:
qarni,,, PI + a,ni,,, + .+ a, n,, , )mil A ... A mi. = ~ ( a , ~ n +ai2n , ,3,id, +...+a, n , , )mi?A ... A&'' P
PI
l..p
).-
P
P - ~
IIZ - P I
A & I
.
By taking this result into account, di,w(tx) is written: ~ P! [ ~ ~ t ( a , ~ , , ~ , , , ~ ( t ~ (fx)+--+aipw,.., ) + a , ~ , . , ~ ~ ~( I ~ ) ) + ~ ~ k i , , i ~ ( ~ A~ . ). .~Ai ~, ' ~~ .~ l ~ "
We have thus: i,dw(tx)
+ di,o(tx)
a~, ~ , (a) ~ c pmki2..ip (tx) a,,xk)mil = qP!t x k a k ~ i l ,( ,ti ~ p+)p ~ , , (tx))dxh ,~ A..h = qP!
t
k
Choosing the linear mapping I such that: Iw(x) =
we have:
A ... A hip
,,,p
6 tp-I i p ( t x ) d t , 1
i
p
.
Lecture 7
Remark 1. Let us insist that a closed form is locally exact. It is not always true globally! The Poincare lemma is no more true if the form w is not differentiable at some points of R"; this lemma doesn't apply to a manifold defined from R" without those points. Consider the classic example of the closed differential form
defined on U = R~- (0).Is it exact? In other words, is there a function g such that w = dg ? The answer is in the negative, because if there was such a differentiable function on U we would have: X Y 2 g(x, y) = arc tan - + h(x) X
But, arc tan y / x is not a function on U since there is an increase in 2n after one turn around the origin. We could have been misled by the use of polar coordinates which leads to w = dB, but w is not the differential of a differentiable function on R ' - (0) and thus o is not exact!
Remark 2. Since every closed p-form on an open homeomorphic to R" is cohomologic to zero, we write: H ' ( U )= 0 ~ 2 1 , for every open U homeomorphic to R" . Let us point out that, for p = 0, F O ( M ) denotes the set of closed forms of degree zero and thus the set of functions g : M + R such that dg = 0. Such functions are thus (locally) constant on each connected component of M. Therefore, we write FO(M) = R~ where k is the number of connected components of M. Let us specify B'(M) is reduced to the only zero-function. The equivalence relation is the classic equality between constant functions. Therefore, we have: HO(M) = F O t (0) = FO(M) = R ~ .
Integration o f Forms
Exercise. Prove a manifold M is simply connected
iff
H' ( M ) = 0 .
Answer. The reader will use the following property. For every closed curve in M, if V w E R'(M): fa,= 0 then there is g E Q0(A4) such that w = dg .
-
BOUNDARY
- HOMOLOGY
4.3
CYCLE
D
A p-cycle or closed p-chain on M is a finite p-chain whose boundary vanishes:
p-cycle c
D
u
dc = 0 .
Finite chains are homologic if their difference is the boundary of another chain: C-C'
G
C-cf=dy.
A chain homologic to zero is a boundary: c-0
D
c3
c=dy.
The set of finite p-chains hornolog~cto a p-cycle is called homology class of order p.
PR8 The homology classes of orderp of M form a vector space. This space is denoted H,(M) . It's the set of classes of cycles different from one other by a boundary. In short, H , ( M )= (p-cycles)l{boundaries~.
The dimension of the quotient space H , (M) is also the pth Beni number, PR9
The integral of closed form over a boundary vanishes.
Proof We have: Ba)=ldw=0
PRIO The integral of a closed form has the same value over homologic chains, that is: [e-cfanddw=O] Proof: The hypotheses
c - c' = ay
jw=d,w.
and dw = 0 let us write:
PRI 1 The integral of a coboundary over a cycle vanishes. Proof: Given a coboundary w = dv and a cycle c that is dc = 0, we immediately have:
Lecture 7
We can immediately express: PR12 A cycle is homologic to zero r f f the integral of every closed form over c vanishes.
PR13 A closedp-form is exact
#I
its integral over everyp-cycle vanishes.
Let us sum up a few analogies: Closed form or cocycle o dw = 0
Closed chain or cycle c ac = 0
Exact form or coboundary dv
Boundary 8~
Every coboundary is closed d(dv) = 0
Every boundary is a closed chain a(ay)= o
Cohomologic forms o - w ' c=> w - w ' = d v
Homologic chains
Coboundary = cohomology to zero w- 0 o=dv
Boundary = homology to zero
Cohomology class
Homology class
Quotient space H P ( M ) , b,
Quotient space H,(M), b,
c-C'
C=O
a e
c-ct=ay c=ay
5. INTEGRAL INVARIANTS The inte g a l invariants play an important role in mechanics. 5.1
ABSOLUTE INTEGRAL INVARIANT
Let ci = (P,@) be an element of ap-chain on M, where P t R P is a closed interval and @is a differentiable mapping of an open U c P into M k,c,be a corresponding chain.
Let c = 1
We know to a vector field X correspond curves (called orbits): that are solutions of the differential system:
Integration o f Forms
One says these orbits form a "tube.
"
More precisely, given an open @(U)of M, the mapping
I ( c R) x @(U)+ M : ( t , x ) I+ y,x assigns to any point x of the support c a point through yl, (x) = x .
D
yl,x
of the curve (orbit)
f H y,x
passing
Every element c, = (P,@)of a chain is "deformed" along a tube of orbits associated to a vector field X.So, to each t E I there is a deformed element of chain denoted c; = (P,y10 a).
So, cj is defined by the same closed interval P as the one defining c, and by the composition of the mappings 0 and y, , this last mapping being defined on @(U). Obviously, the previous notion becomes widespread to a deformed chain c' =
kicj. f
D * A differential form w of degree p is an absolute integral invariant concerning the differential system (7-11) if its integral over a pchain c is invariable over every deformed pchain c' , that is
Example. If the components of a force field are f,, then the work of this covector field along a curve C (support of a chain) is defined by
where the tilde indicates that the one-form is related to the curve C.
Figure 47 The one-parameter (local) group of diffeomorphisms yl, assigns to each point x E C a point y,x. Given t, the "deformed curve" of C is C' = y/,C .
Lecture 7
250
So, the work is an absolute integral invariant ~ f lthe integral defining the work over a chain of support C is invariable, that is:
Jxi = jC, A ;xi
(for every C' ).
It is proved (see exercise 3) this condttion is equivalent to ( L x f ) ,=o,
the orbit of one-parameter group being the curve tangent to vectors of field X. PR14 A differential form o of degree p is an absolute integral invariant concerning the system (7-11) 5tjr w is invariant under every one-parameter (local) group of diffeomorphisms y, associated to system (7-11). Proof: The necessary condition amounts to prove that if for any chain element c and any deformed chain element c' we have: I m = then y : =~w for every one-
lo,
parameter (local) group of diffeomorphisms y , Indeed, we immediately have:
By hypothesis, these integrals being equal for any c and c' (thus for any P), we have:
This equation being verified for any 0, we have Vyl, : y:w = w . Therefore, w is invariant under y, . Conversely, it is sufficient to prove that if for every one-parameter (local) group of diffeomorphisms v, associated to system (7-11) we have 60= w , then for any chain element c and deformed chain element c' . Indeed, we have: jw = = k * y ; w = jp(wr 0 @)*e = C
Io*~ P
lfl)
In another manner, we can say:
PR15 A differential form w of degree p is an absolute integral invariant concerning the differential system (7-11) or with respect to the generating field X of group fz o fulfils the following conditions [i,w=O
and i,dw=O]
or t f l w is an invariant of field X
( L , o = 0 ).
Integration of Forms
25 I
Prooj: The first necessary and sufficient condition immediately follows from the PRIS of lecture 6. The second condition is also obvious because L,o = i , d o + d i , o . We deduce from this proposition the following corollary.
PR16 A closed differential form w is an absolute integral invariant with respect to X
rff
i,w = 0.
PR17 A differential form o is an absolute integral invariant with respect to X respect to gX [Vg E C m(M) 1.
fi
it is with
Proof. We have: i,w=gi,w=O
and
i,,dw=gi,dw=O.
PR18 If o is an absolute integral invariant with respect to X, then it is for i, o Prooj: We have:
Let us give a Jacobi theorem.
PR19 A volume element r,~= A4(x)dw1A...AG?x" is an absolute integral invariant with respect to X if the function M(x) is the "last multiplier"; that is:
ProoJ: The expression (6-14) of the Lie derivative of a form implies L=yq= ( x k a k M+ M d l x l +...+~ d , ~ " ) d / \u. . .'A & "
which is zero fr
d, (A&')
=0
We deduce from this "last multiplier theorem" the following corollary: PR20 The quotient of the last two multipliers is a first integral
Pro@ We have:
Lecture 7
252
5.2
RELATIVE INTEGRAL INVAIUANT
We consider ( p +1) -chains with boundaries (of dirnensionp).
D w A differential form of degree p is a relative Integral invariant if, for any boundary dc and boundary dc' of a deformed chain of c, we have:
From this definition, the reader will immediately deduce that every absolute integral invariant is relative. PR21 A differential form of degreep is a relative integral invariant l f l its exterior differential is an absolute integral invariant (L,dw = 0). Proof: We have:
Let X be a generating field of group. We can say:
PR22 A differential form o is a relative integral invariant
fi
i,dw = O . Proof: We have:
[ o relative integral invariant ]
lff
[ du absolute integral invariant ] [ i,do=O
and i,ddw=O]
(since PR21) (since PR15)
lff
i,dw = 0 .
This proposition implies again an absolute integral invariant is necessarily a relative integral invariant.
Remark. A relative integral invariant w is such that i,dw is a closed form. It's obvious because
L,dw=O=di,dw.
Integration of Forms 6. EXERCISES
Exercise I. Find again the Green-Riemann, Ostrogradski and Stokes formulae in classic calculus. Answer. (I) In R~ consider a domain D and its boundary
JD
A differential form of degree 1 being denoted
w = Pdx+Qdy, we have
dw = ( a x e - d , P ) d x ~ d y . The boundary dD being positively oriented (domain always on the left), the Stokes' formula is written: Pdr+Qdy (a.Q-a,P)d.~ dy = I,(a,Q -8,P)drdy.
,.J
=ID
It's the classic Green-Riemannformula. (ii) In R ~consider , a domain V and its boundary 8V
A differential form of degree 2 being denoted w=Pdy~dz+Qdz~d,+R&~dy,
we have
dm=(d,P+i?,Q+i?,R)dx~dy~dz. The Stokes' formula is written
The boundary dV being positively oriented (normal vector to V pointing outward), we find again the usual Ostrogradski formula:
meaning that the (outward) flux of vector field X(P,Q, R) through dV is the triple integral of the divergence of X over V. (iig In R 3 , consider a domain
~ = { ( ~ , Y , Z ) :=X R = X~( ~ > V ) ~=Y ~ ( 5 ~ f 7 ) 3 ~ = ~ ( 5 ~ 7 7 ) ~ ( 4 3 7 7 ) ~ A ) delimited by a boundary aS, A r R 2. A differential form of degree 1 being denoted w =Pdx+Qdy+R&,
we have:
dm=(a,R-d,Q)dy~dz+(d,P-d,R)dz~&+(a,Q-d,P)dxndy. The Stokes fonnula is written:
Lecture 7
254
The boundary dS being positively oriented ( from the positive orientation of the boundary of A ), we find again the classic Stokesformula. The circulation of vector field (P,Q,R) around 8s is the flux of curl of X through S. Exercise 2.
On a 3-manifold M, we consider the chain S defined by the mapping: U = { ( ~ J , Q ) EORi p~
such that x=asinBwsg,
y=bsinesinq,
z=ccos@
(a,b,c R ) ,
the orientation being given by ((a, 8).
Make explicit the tangent vectors to the ellipsoid S with respect to the Cartesian basis ) of R ~ . (ii) Calculate
(iid Check the Stokes formula:
where the parametric equations of
3s (0= x/2 ) are
Answer. fi) The tangent vectors to the ellipsoid, namely:
are linearly independent provided that sin t9 be nonzero: 8 e { 0, R ). (ii) From
dr = a(cosBcosq, do-sin8sinp d p ) , we deduce: ydr = absinBsinp(cosBcosy,dB - sinesinp dp)
(iii)
The parametric equations of the given ellipse imply: yak = -absin 2 p d p .
The boundary orientation corresponding to choice (p,B) being indirect (or left-handed), the integral along aS is
Integration of Forms
Exercise 3. Prove the work of a force field of components f, is an absolute integral invariant over a chain of support c r f l ( L x f ), = 0 &' where X is the generator field associated to differential system =X ' dt Answer. By definition, the work off along c, that is
I I; k',is an integral invariant 8
where the tilde indicates the one-form is related to the curve and ly,c represents some deformed curve of c along a "tube" of orbits associated with X. This condition is equivalent to what follows:
(Lxf),
=o
(for every i).
LECTURE
8
RIEMANNIAN GEOMETRY
The works of Riemann about the non-Euclidean spaces were at the origin of the manifold differential geometry towards the mid-19" century. By the intrinsic consideration of manifolds withn this new context, the Riemannian geometry contributes a great deal to modern developments of physics and the introduction of a metric is essential.
1. RIEMANNIAN MANIFOLDS Let M be an n-dimensional C P differentiable manifold, T,M be the tangent vector space at x E M . 1.1
METRIC TENSOR AND MANIFOLDS
1.1.1 Pseudo-Riemannian and Riemannian manifolds D
A manifold M is provided with a pseudo-Riemannian structure by giving a tensor field of type ( ) on M (class c P-' ): g : ~ + ~ ; ~ : ~ ~ g x such that: 6) For every x E M the tensor g, is symmetric. (ii) For every x E M the bilinear form g , : T,M x TxM -+ R : ( X , Y ) n g , ( X , Y ) is nondegenerate, that is V X e T X M : g,(X,Y)=O a Y=O In addition, we will generally assume the structure is Riemannian, that is: (iig The bilinear form which is symmetric with respect to all pairs of X E T,M, called mdric form, is positive-definite: vx+o: g,(X,X)>O.
Lecture 8
258
The tensor field g is called metric and g , is the metric tensor defined on T,M x T,M
D
.
D * A Riemannian man&ofd is a pair (M,g) where g fits the requirements 01, (ii) and (iil;), g being the
Riemannian metric.
Let us introduce local coordinates. Considering the vectors of basis
(5)
also denoted (ei) and putting g, (e,,e, ) = g, , the
condition (I;) means that g .Y . = g J'.. .
#
(8- 1)
The condition (iQ means that the matrix (g!,) is nonsingular. Indeed, we have: VX = xie,: g y ~ ' ~= 0 '
e
g,yj = o j : yi = 0
provided that det(g,)
+0.
The condition (iiij means that
vx # 0 :
g , ~ i >~0 J
Let ( h i be ) the cobasis, dual of (el) .
D
*
The metric element of M is the (intrinsic) scalar ds2 = g, d r i d x ' .
It's also called line element. Introducing another cobasis ( d l ) , we will denote the corresponding metric element by ds2 = g ; e l e / .
1.1.2
(8-2')
Metric signature
Given a basis change e; = ajei (and 8' = f i d r ' ) , we know the formula of component change of a (i) tensor is g:, = a : a i g g . If (g) designates the matrix (g,), we denote these equalities in the matrix context as follows:
( g ' )= 'a ( g )a where 'a = (a:) is the transpose of a = (a:). In order to present ( g' ) as a diagonal matrix, we view a as a = OD where 0 is some
Riemannian Geometry
orthogonal matrix and D some diagonal matrix:
We can transform the matrix (g) into diagonaI form by a good choice of orthogonal matrix 0, that is:
Thus we have: (8')=
lgnl-'
If we choose d , = for every i, then ( g ' ) is a diagonal matrix with elements + 1 or -1, the nonsingularity preventing some zero. Note that the choice of d,, cannot after the signs of diagonaI elements. For instance, the matrix 0 can be selected such that the +1 are in front. The metric tensor can be written: g = Q ' @ 0 ' +. . . + Q P @ Q P - Q p + I @ Q P t l -...- QP+Q @ O P t ' l
with p+q = n . Several definitions of the signature of g have been given. Let us make the following choice.
D
The signahrre of g is the pair ( p , q ) of natural numbers such that p + q = n .
This is independent of the choice of basis.
If q = 0 then the metric tensor is positive-definite, p = 0 then the metric tensor is negative-definite. In the other cases, the metric tensor is called "indefinite." An important example, with p=l, is given by the spacetime of special relativity that we will later consider. 1.1.3 Scalar product
At each point x E M the metric tensor defines a scalar product.
D * The scalar product of two vectors X and Y of T,M is the real defined by the nondegenerate symmetric bilinear mapping: g, : T f l x T f l + R :( X , Y ) H g,(X,Y).
Lecture 8
It's denoted by
(x,y) = g,(X,Y). D
One says that the metric tensor g, provides the vector space T,M with a Euclidean vector space structure.
In local coordinates, the scalar product of two basis vectors is written:
w
g,(e,,e,) = (e,,e,} = gg.
These are the components of the
(i ) tensor g, .
The scalar product of two vectors of T,M is written:
Note that it is a scalar ensuing from the double contraction of a tensor g, by two vectors X and Y .
Any two vectors X and Yare orthogonal if their scalar product is zero:
D
X 1Y if 1.1.4
(x,Y) = 0 .
Norm and angle
Given a vector X of T f l , let us consider the real
(x, x)= g,, X'X' . It is positive in the case of a Riemannian manifold.
D
The norm of X is the real
Remark. If the bilinear form (or metric) is positive-definite, that is the manifold is Riemannian, then the previous definition corresponds to the usual definition of the Euclidean norm. is not a norm in the On the contrary, if the manifold is pseudo-Riemannian, then
,/m
usual sense because it can be a positive real, an imaginary number or zero and called pseudo-raorm
D
The angle a between two vectors X and Y of T f l is defined by
J(x,X)
is
Riemannian Geometry
26 1
Example 1. At each point x of Rn we define the scalar product of tangent vectors (origin x ) by A canonical metric is so defined at every point x E R n from gx. Thus, knowing the following metric tensor field g : ~+ nT : R " : X H ~ ~
then (8" , g ) is a determined Riemannian manifold. Example 2. In R"" let Sn be the sphere
The canonical or standard metric on Sn is induced by the one of Rn" . Indeed, Iet us express such that: the metric tensor in a chart (UA1,pn+,)
U k I= { x E S n: xntL(x)>O) pntl : Un++l+ R" : x = (xl,. .., xn+l) H Every point x E U,',, is
Let
:(
l
,. ..,xn).
(XI
n
be the natural basis of S at x.
a
In R"", the basis vector - has the following components: ax'
and in the chart, we have:
The so defined pair (S",g) is a Riemannian manifold. Example 3. Given two Riemannian manifolds ( M I ,g , ) and ( M , , g , ) , let us provide the product manifold M , xMZ = W , , x 2 ) :X I E M , , x ~ EM^ 1 with a Riemannian structure. How?
h the Euclidean vector space at (x,, x 2 ) tangent to the product manifold denoted by
Lecture 8
262
From the preceding, the reader will be convinced that the torus T Z= S' x S' is provided with a Riemannian manifold structure if he is aware of the Riemann structure of S' . The generalization to T n = (x)" S' is immediate.
CANONICAL ISOMORPHISM AND CONJUGATE TENSOR
1.2
The metric tensor characterizes the duality between T& and TXaM 1.2.1
Canonical isomorphism existence
Let us establish the (canonical) isomorphism existing between PR1
T'M and T'*M. How?
The bilinear form g being nondegenerate, there is a (canonical) isomorphism between TxMand T:M defined by the flat mapping :
'
such that to every vector X is associated a 1-form defined by g(X, ) : T J 4
-,R :Y
H
g ( X , Y ) = (x,Y).
ProoJ The flat mapping is linear since:
It's injective (one-to-one) since:
Xb= Y, 3
( X - Y ) , = g ( X - Y , )=0
X=Y
since the bilinear form is nondegenerate.
It's surjective (onto) because we are going to prove that to every 1-form associated one vector X such that w = X,.
w E T:M
is
Indeed, if (dxJ) designates the cobasis, dual of the basis (e, ) of TIM, we have: where w = w(e ) . In addition, given the bilinear form components g,, the vector answering the question has its components X' such that w=wjdxJ
gvX1= W,
(X,= w ).
Therefore, there is a (unique) solution since the matrix ( g v )is nonsingular. The proposition is so proved.
Also called lowering mapping ("bemol" in French)
Riemannian Geometry
263
D
The radical of TIM is the kernel of the flat mapping, that is the set of vectors of T f l such that each image of which (under ) is zero.
PR2
The bilinear form g is nondegenerate if the radical of T,M only contains the zero vector.
Proof: In the case of a nondegenerate bilinear form, the conchtion structure definition means the 1-form
g ( X , ) : T$4
+ R : Y I+
(ig
of a Riemannian
g(X,Y)
is zero for the only vector Y = 0 . 1.2.2 D
*
Conjugate tensor The inverse mapping of a flat mapping is called a sharp or raising mapping :
'
# : T,'M-+T,M:~Hco#.
Symbolically, we denote: 6-2"
(#)-I
=
b.
Make explicit the components of any vector w' dual of X, by introducing the conjugate tensor of metric tensor g . Under the previously defined isomorphism, to the components xiof X correspond the following components of covector X , :
Denote by (g4') the inverse matrix of (g,), it exists because det(ge) # 0.It is such that:
From equalities g",k
we deduce:
=4
gq, = g g g i k ~ k = '$xk= X J
,
The duality between vector and covector is expressed by the following formulae about components: w
XI = g , x J
The components X,and
Called "d12se" in French.
xj
= gvx,.
xi of X are respectively called "covariant" and
03-71
"contravariant."
Lecture 8
264
D
The gU are the n2 components of a tensor of type tensor g. It is denoted g-I .
(i ) called
conjugate tensor of
The metric and conjugate tensors are respectively: g : T N x T , A 4 + R:(X,Y)w g(X,Y)
and g-' : T,'Mx T:M -+ R : ( g ( X , ),g(Y, 1)
g-'(g(X, ),g(Y, 1)
with g - l ( g ( ~),, g ( ~),) = g - 1 ( ~ i 8 1 , ~ = j egJ-)l ( e l , e J ) ~ , ~ j = gl' X,Yj.
The bilinearity of g-' is certain because this last is a function of 1-forms g(X, ) and g(Y, ) such that g-'(g(x, ),g(Y, )) = g(xYy) that is explicitly: ~ " X ' Y ,= g , X'YS = gmgirXigJsyj = glrg,gJiXt~,
with g , gjS= 8;.
Remark. The tensor character of g-' is also revealed from a criterion of tensor calculus. Indeed, the second formda (8-7) expresses a contraction between the g" and the components Xi of a tensor of type (:). Since this contraction leads to a (L) tensor of components X', a criterion relating to tensors immediately shows the g"' are components of a (i) tensor.
Generalization. The previous developments relating to vectors and covectors can be extended to tensors of any type. So we write: gSP ASq, = A P g r
gqsAPT = APT . Specifjl also that repeated contractions with metric tensor g let canonically a (:) tensor correspond with a (: ) tensor, so we have: ...jq til.Aq - gil .*.giQ jQ f *'1-,f., - g l l ~ , ...giqfq 51...)., .
(8-8)
(8-9)
1.23 Calculation of metric and conjugate tensors The components of metric and conjugate tensors can be calculated either directly from , from the metric element ds2 = g, dr*dr'. the basis vectors and go = ( e , , e j )or
Riemannian Geometry
265
Example 1. Express the metric tensor and its conjugate in relation to the cylindrical a d d coordinate basis (-,-,-) of R 3 . ar as a~ Answer. Every point x of R~ is ( X I = rcos,9,x2 = rsin$,x3 = Z ) . With respect to the basis d a d d (-,-,-) the components of basis vector e, = - are (cos9,sin 9,0), the ones of ax @ az ar
a
e - - are (- r sin $7cos 9,O) and the ones of -
a9
the other g,,(x)
(1 #
a
az are
-
(0,0,1). We immediately deduce:
j) being zero.
Remark that the basis (e,) is not normed. The metric tensor and its conjugate are: 1
0
0
0
0
0
0
1
0
1
A second method consists in writing the metric element h2= ggdr'drl = (dr1)2+ (dr2)2 + (dr3)*
by using the cylindrical coordinates, namely:
ds2 = dr2 + rzd92+ dzZ. This metric element immediately implies:
- - - ) that is
Observe that with respect to the orthonormal basis ( 11,1,,
the metric tensor has its diagonal elements equal to 1:
the other elements being zero.
Example 2. Express the metric tensor and its conjugate in relation to the spherical coordinate system of R ~ .
Lecture 8
266
Answer. Every point x of R~ is
(XI
= r sin 9 cos4 ,x2 = r sin 9sin 4 ,x 3 = r cos 9)where r, 9
and 4 are respectively the radlal distance, the colatitude and the longitude. By proceeding as in the previous example, we immediately find: 0
1.3
0
0
0
ORTHONORMAL BASES
1.3.1 Orthonorma1 bases
D
A basis of a Riemannian manifold is orthonormal if
s, = 6, where 6,is the Kronecker symbol.
In the Euclidean case: (go) = I .
The following proposition is obvious. PR3
The components of a vector X with respect to an orthonormal basis are such that
xt= X
i
, the variance being without significance. The metric tensor not appearing any more, the Einstein summation convention is not applicable any longer and it is necessary to introduce the summation sign. So, for instance, we write:
Remark. If the bases are not orthonormal, then the components of a covector are not necessarily the ones of the corresponding vector! So in the spacetime of special relativity we have:
x,= go, xf l = x0 x3= -x3. XI = g l B ~ f l = -xl, x, = -xZ, The gradient is another evidence of this (see exercise 9). 1.3.2
Orthogonal group
PR4
The changes of orthonormal bases at a point of a Riemannian manifold form a subgroup of linear group GL(n;R) called "Orthogonal group " and denoted O(n;R).
Riemannian Geometry ProoJ The matrices of change of orthonormal basis are orthogonal, that is: V A G GL(n;R): A 'A = I .
Let (e,), (e;) be orthonormal bases. For any basis change denoted by e: = a; e, or follows the expression
So we have
el = a; e,
n(n + 1) -following independent equalities defining a 2
submanifoId of Gi,(n;R) :
If we introduce the transposed matrix 'A = '(a:) = (a: ) , then the previous independent equalities are written: A 'A=(u,!) ' ( ~ : ) = ( 6= ~I ,) this shows 'A is the inverse matrix of A. The special orthogonal group or direct rotation group is the subgroup of orthogonal matrices of determinant +1.
D
It is denoted SO(n.R)= { A E O(n;R )
(
det A = 1).
1.4
HYPERBOLIC MANIFOLD AND SPECIAL RELATIVITY
1.4.1
Minkowski spacetime
At the beginning of the lecture, we introduced the signature @,q) of the metric tensor field
g = 8' @ 8' + ... + 8 P 8 @ P - e p + l 8 o p t ' -... - ,gn @ 0''. The relativistic (or Lorentz) manifold is a fundamental example of a hyperbolic manifold. More precisely, the manifold R:, also called Minkowski's spacetime is a fourdimensional manifold with the metric element ds2 = gas a!xad.x'
= (dr0)2-
- (dxZ)'- (h3)'
(8- 10)
where x0 = ct (c being the light velocity) and where Greek indices are usually introduced.
In the case of a pseudo-Riemannian structure, for example a Minkowski spacetime, the square norm of a vector X
/ I x=J(x, ~ X)= g, can be positive, negative or zero.
X~X'
Lecture 8
268
An isotropic vector is a vector of zero norm. At every point x E define a cone of equation g,# xB= 0,that is:
e4the isotropic vectors
xa
This cone is called "isotropic cone " or "light cone relative to an event.
"
~ > 0) The vectors inside the cone (such that g , xaxB are said to be of time type, the vectors outside the cone (such that gapxaxp< 0 ) are said to be of spatial type, the isotropic vectors (on the cone) are called light vectors.
Figure 48. Example of spacetime
(such that x3 is constant).
The light cone relative to an event (cT,T',F~,F~) is composed of two parts: a past light cone inside of which are the events fiom which light can reach ( c f , i ' , ~ ~ , and x ~ )a future light cone inside of which are the events that can be reached by light emitted fiom ( c i , ~ i2, ' , i3).
In the Minkowski's spacetime, a universe trajectory of a particle is defined by x0 = ct
x' = xl(t)
x2 = x 2 ( f )
x3 = x 3 ( t ) .
Tangent vectors to universe trajectory are defined by
x = (c,xL,x2,x3), In special relativity, the vector X is only of time type or isotropic because the light velocity is maximum, that is c2- (i1)2 - (i2)2 - (x3)' 1 0. The only universe trajectories of zero mass particles show isotropic tangent vectors. In particular, it's the case of light (tangent to the light cone). The nonzero mass pamcles show universe trajectories inside the light cone.
1.4.2 Special relativity and special Lorentz transforms In classical mechanics, there is no unique (privileged) coordinate system (or fiame of reference) in which the classical mechanics postulates are valid. All the inertial coordinate systems, also called Galilean (in uniformly rectilinear translation), are equivalent. The classical mechanics laws have the same expression in every inertial coordinate system. In other words, these laws are invariant under the Galilean transformations denoted by
Riemannian Geometry
- Vt
t' = t where v is the "velocity of a coordinate system" (x') with respect to system absolute time (independent of observers). X'l
=
X'3
= x2
= x3
(x)
and r is the
In special relativity, Einstein's postulate asserts that all physical laws must be such that if they hold in any coordinate system then they hold in any other coordinate system moving at a constant velocity with respect to the previous. There is a set of inertial coordinate systems (moving uniformly with respect to one another). This postulate is a generalization of the Galilean relativity principle since it takes invariance of electromagnetism laws into account besides the ones of mechanics.
In addition, Einstein has postulated that the speed of light c is a fixed universal constant relative to every coordinate system: c is an invariant under every change of Galilean coordinate system. In this context, we emphasize space and time are not only linked but space and time are also connected to the coordinate system to which they refer. Now, we are going to show the Lorenfz transformafions under which the physical equations are invariant in relation to the Galilean coordinate system. Let (G') be a Galilean coordinate system moving with uniform velocity relative to another Galilean coordinate system (G), Consider two infinitely neighboring events. The first event has coordinates (ct,x',x2x3) with respect to (G) and coordinates (ct ',x" ,x", xf3) with respect to (G') . The second event has coordinates (c(t + dt),xl + dX1,x2+ h2 ,x3+ dr3) with respect to (G) and coordinates ( ~ ( t+' dt'), x'l + dx",xf2+ h I 2 ,xr3+ dd3)with respect to (G') . In (G), the metric element is expressed by ds2 = g,,dr'drv = (drO)' -
3
(dri)'
(xO= c t )
i=1
and in (GI) by
In special relativity, the required condition
is equivalent to ds2 ht2
The Lorentz transformations leave invariant the metric element. To find the relations between the coordinates x a and x ' ~we are going to simplify the problem without restricting the general nature of special relativity developments. This follows from space isotropy concerning the axis choice, and from space homogeneity and time uniformity concerning the choice of origins of Galilean coordinate systems. Because of the nonrestricting simplification, the Lorentz transformation is called "special. " So, the frame of reference (G') moves parallel to x1 with velocity ii with respect to the frame of reference (G) . We suppose the origins of coordinate systems coincide at t' = t = 0 .
Lecture 8
270
Therefore, we have the following coordinate transformation which must be linear (since every uniformly rectilinear motion in (G) must still be in (G') ): xtO= AX' + BX' = cXR + DXL xf2= XZ = x3 where the coefficients are hnctions of the velocity of (GI) with respect to (G) . The previous equation system is denoted by where the matrix L = (La,) goes from "unprirned" to primed. We inversely denote x' = M Y B
with Lap M p r = 6;. Using basis vectors, we immediately have:
Introducing the metric elements
and making equal these metric elements, we obtain: gr v = g> ~
~
,
~
f
l
~
that is, by using the matrix notation:
1; I=;: :: :]IP-31: :: :] A C O
0
A B O O
0
1
0
0
0
This implies:
- C 2 = 1 (1)
AB-CD=O
(2)
0
0
1
B~ - D 2 = - 1
(3).
By putting
then the equations (1) and (3) are written:
~ ~ ( 1 -=p1 ~ ) Since A and D are reals, then
~ ' ( 1 - p 2=)1
p 5 1 and we have:
Note that if the frames of reference (G') and (G) are neighboring (that is u +0), then xtO + x0 and x" -+ x' , that is A +1, B + 0,C +0, D +1.
Riemannian Geometry
Therefore, we necessarily have: A=D=-
I
C-P'
We can easily find
C-B=-
P . Indeed, let (ct,x',O,O) be the origin of (G') with respect to (G)
x 0 . This event is (ctr,O,O,O)with respect to (G'),that is: where x 1 = ut = u C
The special Lorentz transformation is thus:
that is
and
We introduce the hyperbolic rotation angle y such that
-
p thfy and thus
Lecture 8
272
Therefore, the special Lorentz matrix is defined by
Remark. If u is small in comparison with c then we find again the Galilean transformations
of classical mechanics. 1.4.3
Lorentz group
In a more general manner, we have: -1 =
det(g*) = det 'L det(g) det I, = - ( d e t ~ ) ~
which implies detL=fl. The corresponding Lorentz group is formed of linear isometries of the Minkowski spacetime manifold with the invariant metric element (dxO)'
-
3
(hi)'. It is a subgroup of GL(4,R) i=1
defined by
where
It is immediate that: The proper Lorentz transformations, that is such that detL = +1, form a subgroup of the Lorentz group. The Lorentz group has four connected components:
Riemannian Geometry chy
shy
[;v-iy
; 0
0
The first component which corresponds to the special relativity does not reverse the time and is "proper " ( det = +1 ). To conclude, we say the physical laws are invariant under the transformations of Poincard's group which are denoted xla
= LapxP + Aa
(where the four A" are the coordinates of the origin o in the primed coordinate systems); that is the laws have the same expression in all the inertial systems. 1.4.4 Time dilation and length contraction (r) Consider a particle moving in a Galilean frame of reference (G) with coordinates
(ct,x,y,z) with respect to (G) . Let (GI) be a Galilean frame of reference such that the particle is motionless at time t . It is sometimes called '>roperframe of reference. " A local clock in (GI) indicates t' at this time and is called the "roper rime. " We choose the axes and origins in accordance with special Lorentz transformation obtaining. Consider two neighboring events defined by (ct,x,O,O) and (c(t + dt),x + dr,O,O) in (G) and the same events respectively defined in (G') by (ctr,x',O,O) and (c(tl+ dtl),x',O,O) (since motionless). We have: ds2 = dx2 - c 2dt 2 = (v2 - c 2 )dt 2 in (G) : d.~"= -c2 dlr2. in (G'): The equality ds2 = dsr2 immediately implies: dt = y dt' Since dt > dt' ( y > l), we conclude the (proper) moving clock runs more slowly than the other. This result is also obtained by writing
ct = y (ct' + @') and thus
c(t + dt) = y(c(t' + dr')+ P(x' + dx'))
cdt = y(cdtr+ pdwf).
Since the two events stay in the same position in (G') (that is dr' = 0 ) at different instants (that is dt' $ 0 ), then the already shown relation of time interval dilation is found again.
Lecture 8
274
(ii) In the same manner, we consider two neighboring events in (G) and (G') . By subtracting X' from x' + dx' , we obtain:
If the two events are the measures of two limits x and x + dx of a space length in (G) at the same time t , then we put dt = 0 which implies: dr ' &=---.
Y The (parallel to x) length element being fixed in (G') means that &'is the ')roper length element. " The previous formula shows dx < dx' that is the contraction in the length of a moving rigid body. (iii) A point moving along the axis ox, we consider that two events are the positions of this point at instants t and t + dt (clock in (G)). From dx = y(dx' + PCdt') cdt = y(Qdx' + cdt'), we obtain the important relation between velocities: dx' dx'
-+PC dt'
-P&+l c dt'
where
GY
-+II
dt'
u dx' -c2 dtr
+1
dx and - are the respective velocities of the point with respect to (G') and (G)
dt dt In particular, if u is small in comparison with c we find again the classic formula of velocity addition: v = v' + u .
1.5
KILLING VECTOR FIELD
There is now the question of the invariance of a metric tensor field g D
A vector field X such that L,g = 0
is called a Killing vectorfield. In particular, we can choose local coordinates x i such that
xk= 4k which implies:
If this expression is zero then the metric tensor is independent of the coordinate x1 Conversely, if there is a chart in which the metric tensor is independent of any coordinate, then a Killing vector field automatically exists. Example 1. In 3-dimensional Euclidean space, the metric tensor field is such that the g, (here a!,) are independent of x, y and z. Thus d,,d,,d, are Killing vector fields.
Riemannian Geometry
275
Example 2. The expression of metric tensor field components in relation to cylindrical coordinates shows that 8, and 8, are Killing vectors.
1.6
VOLUME
Unless otherwise indicated the Riemannian manifolds are assumed to be of finite dimension.
1.6.1 Completely antisymmetric tensors Consider the example of a completely skew symmetric tensor of type (!). In the case of a 3-dimensional manifold, we recall that such a tensor has one strict component. Denote this by D . The components of the (i)tensor relative to basis (dx' @J dx2 C3 h 3are )
D. & = E uk D where sVk is the Levi-Civita symbol equal to 1 if (iJ;b) is an even pennutation of (123), equal to -1 if (iJk) is an odd permutation of (123) and equal to zero if (at least) two indices are equal. Every component change is clearly denoted
The strict component change is thus
The previous formula can be generalized to completely antisymmetric tensors of type ( ) on an n-dimensional manifold. The physicists call D "tensor density. " The components of a completely antisymmetric tensor of type (:) relative to basis
(dxil @ . . . @ d r t mare )
Exercise. The reader will immediately verify that a completely antisymmetric tensor C of type (,") on an n-dimensional manifold shows the following change of strict component
C' = (det - ) C . )r:(
-I
He will prove the components of such a tensor relative to basis (e,, @ ... @J e, ) are
Lecture 8
where
E'~...'~is the Levi-Civita symbol.
1.6.2
Volume form on Riemannian manifold
Under any basis change defined by
'
ax'
e.=J
&I,
ei
we have:
We denote detg = det(gy)
The rule of determinants products is applied: detg' = (deta)2 detg
On the oriented manifold, the detg and detg' being supposed positive (direct basis change), we have:
ddetg'= deta fi
&
So, is a tensor density and is the strict component of a completely antisymmetric tensor of type (0, ). Hence, we say:
D 6 v
The volume form on A4 (associated to the Riemannian metric g ) is the following completely antisymmetric n-form:
q=
&dr'n...
(8- 1 5 )
A&".
The components of the tensor pg relative to basis (dr" O... €3him ) are
a
a .
Notice that the volume spanned by vectors -,..., - 1s ax1
ax*
Remark. The following change of strict component
lets us introduce a completely antisymmetric tensor of type (:), dual of volume form and having the following components relative to basis (eil @... C3 e, ) :
Riemannian Geometry
Note that (P~)"...~* ,.,,,* = n!
since the previous sum contains n! equal terms such as
1.7
THE HODGE OPERATOR AND ADJOINT
From the p-form notion the duality allows introducing the q-vector definition and the notion of adjoint.
D
A q-vector is a completely antisymmetric tensor of type (: ).
The reader will transpose the developments about pforrns. For example, he will show that the q-vectors form a Cz -dimensional vector space. Let t be a completely antisymmetric tensor of type (:). The corresponding q-vector of components t ' ~ "is' ~such that t'l...i4= g i l A .,.g'qJ' t
,
J I .-Il
D
'
relative to volume form pg is the (n - q) -form
The od/ont of q-vector denoted * r such that
The operator * defines an (n-9)-form from a q-vector and conversely.
We also denote (*t)lp+,.. in -
1 - -(&
P!
. .
Given a p-form w and w"
D
"'"
ti1..+
= gi'" .. .gipJp a r,.,j p , we say:
The (n-p)-form denoted *oand called adjoinedform (or dualform) of p-form w is defined by 1
i * 4 , p +.,"l = - ( ~ g ) i ~ . . . l ~ail-.ip P!
Lecture 8
So, given an oriented orthonormal basis (el,...,en) of
D
T'M , we say:
The (previous) star operator such that (*@I(ep+,,.*.,eV) = is called Hudge star operator.
a
1 The adjoint of a completely antisymmetric tensor of type (:) or of type (,") is a
scalar.
L,
Remark 2. The adjoint of a completely antisymmetric tensor of type ( "i' ) ( resp. ( ) ) is a covector (resp. vector). This will let us find again mathematical notions of elementary vector geometry on R~ such as vector product, curl and divergence (see exercises 4,5 and 6). Remark 3. For n-Riemannian manifolds, we have V w E R,P(M):
* that is We also have Vf
= (-1)P("-~1
*-I = (-l)p(n-P) * E
T,: :
* *f = ( - ~ ) q ( ~ - q )f
.
Proof: Let us first verify the preceding in the case of functions (or O-forms).
By definition, the adjoint of a function f is an n-form of the following components
(*f)i
]..,in
=(
~ g l i .,i n
f
*
The adjoint of adjoint off is a O-form such that
We denote: **f = f .
Now, we prove the formula (8-22a). The adjoint of ap-form w is an (n-p)-vector of components
and thus
Riemannian Geometry
-
1
( P ~ p!(n- p)!
...i,, jl...jp
(Pg
)il
@i l . . . i p
P(~-P)
p!(n- p)!
(Pg)Ip+,
...i"jI ...jp(&)lp+l i"' -ip W i t . ip
since An-p) transpositions are necessary to change from the sequence ( i ,...ipi,+,...i,) to the sequence (i,,, .. , i , i , .. . i p ). So, in particular, we have:
But from (8-19) it is easy to verify what follows:
Thus, we have:
but (pg l p + l . . . n t l . . ~ p
m,,.Ap = P!(
pcl ~
g
..nl . . . p
" 1 ...p
and therefore (** 4
1 . p
-(-l)~(n-~) -
.
0 1,..p
Since this result holds for every new index designation then the formula (8-22a) is proved.
Remark. If the metric is not positivedefinite, the reader will be careful to sign (see later, in special relativity). Example 1. In the Minkowski spacetime, compute + o if w = a h 0 , a E R . Answer. The component of the 3-form * w is
that is =
gipm p = EOlz3goow, = a .
Therefore, we have: *w = a d r ~ d y ~ d z .
Example 2. Given a one-form w = dr' + dr2 + dr3, find the expression of dimensional Euclidean space. Answer. The components of the 2-form * w are:
* o in the 3-
Lecture 8
Explicitly, we have: II
g CL), = 0,= 1 (*lo),, = -0, = -1
(*w),,
=El,,
( * w )=~W~3= 1. So, we have: *CL)=dX2AdX3--dr' A d r 3 + d x 1 ~ d X 2
Notice that in Euclidean space, we have g, = 6, and thus the index position does not matter. 1.8
SPECIAL RELATJYmY AND MAXWELL EQUATIONS
1.8.1
Faraday t f o r m and its adjoint
Let us recall that in relativistic dynamics the elementary notion of 4-vector is introduced. In particular, the 4-velocity of a particle has components Ua
dxa =-
dt' where t' is the proper time relative to the moving particle (own clock!).
The components of the (usual) velocity relative to an inertial (or Galilean) 3-dimensional frame being denoted
and
we know the 4-velocity is explicitly related to the corresponding 3-velocity as follows:
or
u1=
vi
J'r'
c I- ,
In special relativity, the change of momentum relative to the proper time t' is considered. So, given an electric field E , a magnetic field E , a particle momentum jj = mC and a particle charge e, we know the classical Lorentz force law is
Riemannian Geometry and in this context we consider the change of momentum relative to the proper time:
and the change of mechanical energy
These equations also written: dp O dt'
-=
--
eE.U
are denoted in tensor notation as follows:
dp" = e F p B -
( p , P = 0,1,2,3 1.
dt'
We make explicit these equations:
-dpO - e(FOoU 0+ F O I U' + F02 U 2+ F
O ~(i3)
dt'
dp1= ~ ( F 'Uo0+ F'I W' + F'ZU 2 + F ~ ~ u ~ ) dt'
= e(E, U 0+ B3 u2- B,
u3)
and so on. After identification we deduce that the electromagneticfield is represented by
Since
FmP= gap F " a , the following matrix
- E l 0 - B 3 B2 - E 3 - B 2 B'
0
Lecture 8
282
lets us introduce the following Faraday 2gorrn F :
where the Ei and B, are respectively the three components of electric and magnetic fields. Now we are going to find the expression of
* F.
Since 1
11
=?&ha gCglsFw = ~%IZ3gmg"FO1 +igIO23g g
(*F)23
00 6 0
= -Fa1 = El
(*F)13= FO2= -E2 (*F)12= -h3= E3
(*F)~l = F23
= B~
and so on,
we obtain:
Remark. The reader will verify given a p-form m
* *W = (-l)p("-~~s~n(~) 0 and in particular
**F=-F. 1.8.2
Maxwell equations
We are going to consider the exterior derivative of the Faraday 2-form and Maxwell equations.
PR5
The Faraday 2-form is closed.
Proof: We immediately have: d~ = (alBl+ a 2 ~+2a 3 ~ 3 ) A ~ AXd y 3~ + (d,B, +d,E, -d,~~)du* A dX2/\dr3
+ (aOB2f d3E1- dlE3)dYOA dx3 A dY1 + (d,B, + dlE2- d2E1)dxOA dyl A du2. The well-known Maxwell equations divg =
lead to the result dF = 0
c.B= 0
Riemannian Geometry
283
Now we are going to consider the exterior derivative of the adjoint * F . Given the 4-current vector J such that J O = cp where p is the electric charge density and J ' , J ~J 3, are the components of the electric current density (in the 3-dimensional space), we can say:
PR6
The exterior derivative of the adjoint of Faraday 2-form F is related to the charge current 3-form * J by
Proof We immediately obtain: d * F = (a,E, + d2E2+ a,E3)dxl A dx 2 A dx 3 + (doE3+ a2B1- d,B,) a!xoA a!x' A dr2 + (d,E, + d,B, - a,&) dXOA dr2 A dx3 + (ao& 4- d,B, - a,B,) dxOA dX3A dr' Two remaining Maxwell equations being the (scalar) electrostatic
G . R= 4zP and the (vector) electrodynamic equation
we have:
The adjoint of the 4-current vector J is the charge-current 3-form (*J),, = J I such that:
*J
with components
The proposition is so proved. In conclusion, the Maxwell equations are summarized in the context of forms by dF = 0
w
4a
d*F=-
*J.
C
1.9
INDUCED METRIC AND ISOMETRY
Let M,, N, be differentiable manifolds, f : M, + N,,, : x H f (x) be a differentiable mapping.
(8-23)
Lecture 8
284
The reader will refer to formula (4-14) that defines the pull-back of
PR7
(0,) tensor by f
If (Nm,g)is a Riemannian manifold and if f is an immersion, then the tensor field f * g provides M,,with a Riemannian structure called induced from the structure of Nm*
ProoJ: The assumption of immersion and thus of constant rank implies the image of vector X E T&fn 1s dS,X=O X=O [see (2-11 )].
The bilinear form f 'g such that for every nonzero vector X:
is positive-definite (since g is positive-definite by hypothesis). The Riemannian structure condition for manifold M , is thus hlfilled. This structure is induced byf from the one of Nm. Example. Consider a differentiable mapping
f :R - + R~ : X H f(x)=(fl(x), f2(x))
If the manifold R' is provided with the metric defined by vz = ( z i , z Z )E Tf(,) R~: g,,(, (z,z) = (2')' + (z2)', is then the manifold R provided with a Riemannian structure? Answer, The image of every vector X E TxR is defined by
=r
8f' z'=--;x'=~"(x)x ax
and
af2
z ~ I axi -x~=~'~(~)x.
So the form
(f*g),(X, X)= ((f" ( x ) ) ~+ (f' 2 ( ~ ) ))2( x I Z is positive-definite (and defines a metric on R ) if the derivatives f f ' and f 2 are not simultaneously zero. f
To end we introduce a bijective mapping between M, and Nm preserving the metric, namely:
'dx,yE T N , , : ( X , Y ) =~ ( @ X , ~ J Y ) ~ ~ . D
Riemannian manifolds ( M , , g M ) and (Nm,g,) are isometric if there is a diffeomorphism f between the manifolds, called isomtry, such that:
f*
g =~g ~ .
(8-24)
Riemannian Geometry Let En be a Euclidean space with the metric element dsZ = f ; ( dx i )z . i=l
PR8 The group of isometries of an n-dimensional Euclidean space is isomorphic to afine group R n x O(n;R). Information. The space En is defined by one chart with the previous metric element. An isometry f defined by n differentiable functions Yi= f l ( x l , ..., x") is such that
The space being Euclidean, we have:
I::[
which means that - is an orthogonal matrix. Next by writing the diffeomorphisms of En the reader will be able to conclude.
2. AFFINE CONNECTION Let us introduce a notion "enriching" the manifold structure and which is very useful in mechanics and physics. In Euclidean space, comparison and operations between vectors are immediate. On the other hand, coordinate systems of tangent spaces at different points of some manifold cannot be a priori associated. But we are going to provide the manifold with an additional notion namely: the afflne connection (also called linear connection). This branch of modern geometry uses the notions of covariant derivative, parallel transport, geodesic, curvature, torsion, etc. In addition, a metric is added in Riemannian geometry.
Let M be an n-dimensional manifold of class C" . AFFINE CONNECTION DEFINITION
2.1
Let us show a mapping which to every pair of Cmvectorfields on M assigns another C" vector field. D
*
An amne connection on M is a mapping
V : %(M)x % ( M )-+ X ( M ) : ( X , Y ) H V,yY such that
Vf,h E Cm(M), V X , Y , Z E Z ( M ):
Lecture 8
286
w
PI. P2. P3.
Vfl+,,Z = f VxZ + h V,Z
Vx(Y+Z)=V,Y + v x z V x ( j Y ) = f V,Y + (Xf)Y
The vector field V,Y
where Xf=Lxf .
is called covan'ant derivative of Y along X.
We will later present a geometric interpretation of the covariant derivative of field Y along a
curve (tangent to vector field X). 2.2
CHRISTOFFEL SYMBOLS
Let us show a covariant derivative is completely determined from n3 functions on any local chart domain U denoted:
Let us express the components of vector fieId V,Y with respect to a local coordinate system ( x i ). Given vectors
x = X ' a , = x'e,
Y =y'dj=YJe,,
we have: V,Y = vX1q(Y'e,) =xiV,(~'ej)
( because PI )
this last equality following from P2, P3 and from L , Y J = 6 : a,YJ
D
= diy'
.
The Christoffel symbols G~of the affine connection (or connection coeffcients) are the components of the following covariant derivative with respect to the natural basis: Ve,ej =
43,"
k
ek.
Thus we have:
SO, we can say: PR9
The components of vector field VxY are in a local coordinate system (x') :
(8-26)
Riemannian Geometry
If X is in particular e j , then we denote the covariant derivative of Y along ej by V,Y = V,Y . We obviously have:
D
The covariant diflerential of Y , at point x, denoted VY , is a tensor of type (:) such that the inner product of VY and vector X is the covariant derivative of Y along X
In a local chart, this definition means that
also denoted
Remark. The connection is called Euclidean if there are coordinates such all the various
ri are zero or
aYi
= - . In this case, where M = R n, we have: (v,Y)'
ax
J
dY' = -X '.
axr
Another terminology can be introduced. Given a parameter t, we say: D
The absolute derbarlve of vector Y, at t, is the covariant derivative vector field defined by
In particular, gwen a moving point p E M such that its coordinates x i are functions of time parameter t, we express: The velociv vector of p has components
and the acceleration vector has the following components
Lecture 8
,
Dvi dt
a =--- -
dv' &' +I-;vdt dt
PRlO The Christoffel symbols am not the components of a tensor.
Proof: We successively have:
that is
The second term of the right-hand member shows the Christoffel symbols are not the components of a (: ) tensor. Notice these symbols are tensors if the affine (or linear) changes of coordinates xi = x
2.3
i
dZx'
(Yi,..., y n ) are such that -are zero for every i,p,q. aupdyq INTERPRETATION OF THE COVARIANT DERJVATTVE
Let c : [a, b] + M : f I-+c ( t ) be a differentiable curve of the Riemannian manifold such that c(0)= x, . Let
be the vector field tangent to the curve c. Let Y be a field of vectors Y,,,, tangent to M at every point c(t) of the curve. In particular, the vector Y,(,, E TxoM is denoted Yo.
Riemannian Geometry
289
To give a geometric interpretation of the covariant derivative, we are going to use a method closed to the one that has been introduced to define the Lie derivative; but we emphasize the vector fields are here defined along a curve. We denote by (T-lyc(t))o
the "backwards transported parallel vector" of field Y, along the curve c, at x, . The previous expression means the vector is transported backwards in accordance with the requirement
D
A differentiable vector field Y is parallel along c if 'dt E [a,b]: (V,Y),(,, = 0
(8-33)
where X is the vector field tangent to the curve c. Defining the covarianl derivative of Y along c by V,,,,Y which is the field of well defined vectors (V,Y)(x) at every x E c , we say in an equivalent manner: A differentiable vector field Y is parallel along c if
V,,,,Y = 0
DY ( also denoted -= 0 )
dt
In a local chart, the previous equation is written:
PRl 1 The covariant derivative of field Y along the curve tangent to field X is
Pi-& The vector Y,,,, is backwards transported parallel along c to xo. Denote by corresponding vector field which must verify (8-34).
For instance, for the ith component, we have thus:
the
Lecture 8
Let (YC{,,) or have:
(c,,)
be the components of vector field Y at point c(t). We immediately
E,,,
We deduce T i(x,) from ( 1 ) = (2). Since is backwards transported parallel along c to x, , we must take (8-36) into account to write the components of the vector at x,, : dY' = Y; +tx,'-(x,)+tr;(o)x;~; ax3
+~(t*)
Therefore, we have: ayl
~-'C,,,(X,)= [Yd + t ( X J - g ) * + ~ ~ ; ( O ) X O / Y ~
a
+ o ( t ' ax) ~ ~ I , (8-37)
and the formula (8-27) leads to: I
(V,Y), = lim - [(T"Y~,,,),- Yo]. r-+o t Finally, we introduce the geodesic notion that will be developed later.
D
Vector field X is autoparallel along c if
D
A curve c is called geodesic if c is autoparallel alongc.
In other words, from (8-36),we say c is a geodesic if and only if, in any coordinate system, the following well-known geodesic equations exist
Given c(0) and C(O), there is a unique geodesic c defined on some interval [a,b].In other words, there is a unique autoparallel field X along c such that X ( 0 ) = X , and X ( t ) E T,,,,M. D
Given two points
and cz of the curve c, there is a linear isomorphism, called the
parallel translation dong c: <El
: Tc*M+ Tc2M
such that tangent vector field is autoparallel along c.
Riemannian Geometry
From uniqueness of solutions of differential equations, we immediately have: fCIc3
and
c,, is the identity.
2.4
TORSION
O CIC2
=
CC3
Let M be a pseudo-Riemannian manifold. A vector field of type (1) is associated in a natural manner to the connection V D
The torsion is the tensor field of type ( ) such that VX, Y E X(M) :
w
(8-38)
T( ,X,Y) = V,Y - V , X - [ X , Y ] where the blank space is reserved for some differential form of degree 1.
The torsion is also denoted by Tor(X, Y) . We immediately show that T is a tensor. So, the multilinearity is obvious, for example Vf E CW(M): T( , f l , Y ) = VY , - V,(fl) - [jX,YI = fV,Y- fV,X-XL,f - f[X,Y]+XL,f = ST( ,X,Y). The expression (8-38) clearly proves the skew-symmetric character of the tensor of torsion.
Now, make explicit the torsion components in a local chart (x k).
a
Given basis vectors e, = - and ax
a
ej = ----, since
ax'
T ( ,ei,ej)=V,e, -V,,e,
[e,, e j ] = 0 we have:
=(qf-r;)ek
=cek
where we have put 6w
k
(T( ,ei,ej))k = q,
=ri-q:.
Let us specify the T,k and q: are not tensor components, but their differences components. The subtraction defining the torsion vanishes the annoying terms. 2.5
(8-39)
qf are tensor ,
LEVI-CIVITA (OR RIEMANNIAN) CONNECTION
Here we consider the case of a zero torsion for which we have the following proposition. PR12 The Christoffel symbols are symmetric in relation to their lower indices:
q kj = 0
0
c;=q;.
Lecture 8
292
By adding a metric we now "enter" Riemannian geometry; but we recall parallel transport, covariant derivative, connection coefficients, curvature, torsion, etc. exist without a metric! PR13 On a (pseudo) Riemannian manifold ( M , g ) there is a unique connection V associated to g such that the torsion is zero and W , Y , Z E % ( M ):
Prooj First we specify that X g ( Y , 2)= L,g(Y, Z ) = V,g(Y ,Z). Let us prove that given the assumptions there is a unique vector field V,Y defined by
Indeed we have: ~ ( V X Yz), = x g ( y ,z)- g(y, VXZ) = x g ( y , z ) - g ( y , v ~ x+ t x , z ] ) = X g ( Y , 2)- g(Y,[X, 21)- g(Y, V , X )
(because (8-41)) (because T = 0 )
but
this last equality following from (8-41). In addition, we have g(V,X,Z)
=
g(V,Y
- [X,Y],Z) =
g(V,Y, Z) - g ( [ X ,Y1,Z)
and thus g(V,Y,Z) = X g ( Y , Z ) - g(Y,CX,ZI) - Zg(Y,X) + g(X,[Z,YI) + Yg(X,Z) - g(V,Y,Z) + g([x,yl,z)*
This last expression leads to (8-42) which shows VxY is uniquely determined by the metric. Conversely, if the vector field V,Y is defined by the formula (8-42) then the so defined scalar product verifies the three properties of an afine connection, the zero torsion property, that is V,Y - V,X [X,Y] , and the equality (8-41); it's sufficient to reverse the steps.
-
From the previous proposition (or "Fundamental theorem of Riemannian geometry") we say:
D
'
A Lwi-Civita connection is a zero torsion connection such that V X ,Y,Z
This last condition is explicitly written in a local chart ( x k ) as follows:
' Also called Riemannian connection.
E
Z ( M ):
Riemannian Geometry
Remark. The condition (8-41) of the Levi-Civita connection, also written
means parallel transport along the c w e of tangent vector field X is an isometry (that is preserves the scalar product). So, for vector fields Y and Z defined along this curve, we have:
Christoffel formulas
Since g(V,,e,,ek) = g(qrer,ek)= qF g(er,ek)=
crg,
=
rM
then the equality (8-42) is, in this particular case, the (first) Christoflel formula:
the (second) Christofel formula being:
Ricci identity From the condition (8-41) of the Levi-Civita connection, we deduce: Le,g(e,,e, = g(V,e,,e, 1+ g(e,,V,e, which is nothing else the Ricci identity:
ag
jk
=
rhj + rjik
or = g,
2.6
GRADIENT
2.6.1
Gradient
-
I;; +gjr G,'.
DmRGENCE
-
LAPLACE OPERATORS
Introduce again the differential of a function f E C m ( M,) a concept already defined at the end of lecture 2. Here M is a Riemannian manifold and we have:
Lecture 8
df, being an element of T:M
The PR9 of lecture 2 means the image of Xx under dS, is the derivative Xxf of the function f in the direction of vector X x.
In the context of Riemannian geometry, we define the following fudamental differential form of degree 1: D
The gradient of a functionf, at x E M , is the 1-form gradxf : T N
+ 1P : X, H grad, f (X,)
such that
(grad,f
,x,)= df,(X,).
The gradient off on M, denoted gradf , is the differential form of degree 1 defined by
w
V X E X ( M ) : (gradf,X ) = df ( X ) .
(8-45)
We also denote the gradient off by df and we simply write:
w
(df
,x)= Xf
The reader will immediately observe that it's a question of inner product: w
i,df =(df,X ) .
(8-45")
In particular, we find again the expression of gradient of r seen in exercise 12 of lecture 4, namely:
In a local chart, we have:
i
= (grad f),X .
The components of gradf are thus (gradf 1, = 3if
and df = d , f & ' .
Riemannian Geometry
295
Remark 1. We will mention the difference between gradf that is a 1-form and the associated vector under the (canonical) isomorphism: # : TX9M
-+ T I M :(8,f)w (grad f)'
= (g",
f)
i,j =
1,.,.,n.
Remark 2. In the particular case of an open of R" we find again the usual gradient definition. 2.6.2
Divergence
D
The divergence of a vector field X is the function following from the contraction of (: ) tensor field VX (covariant differential).
In coordinates, we denote:
w
divX = V i x i
Make explicit this notion if a metric is introduced. The formulas (8-29) and (8-43) imply:
v j x i= ajx8+ r ; x k= a,xi + g i h r , x k
A contraction leads to the following
(i)tensor field
By changing the names of summation indices, we obtain:
gha,gMx k= ghr dhglkxk= gAahgitx k which implies:
Denoting the cofactor of g , by G" and the determinant of ( g , ) by detg , we have:
From the determinant theory, we know that
8,detg = G" d,g,
But we have
3
d, detg -- gihdk&h
3
V , X ' = d , x i+ - X '
detg
1
2
dkdetg detg
Lecture 8
3,J ldetgl
1 2 detg a'detg =
-/,
and thus
Jm dm) 1
divX = --- a,( X i
In the particular case of an open of Rn we find again the usual definition of the divergence: divX = a i X i .
Denoting by
no= dyl A . . . A & " the standard volume on R", we have:
LxR, = L X h 1A&'
+ + dul A ... A L,drn A , . . A ~ P +... + dul A . . . A ~ ~ X ~ ~ X '
A ... A & "
= a,xtdrj = (div X ) 0, .
We can generalize and define: Given a vector field X on an orientable manifold with volume a,the divergence of X is the (unique) function denoted div,X such that
D
L, n= (div,X)R . A vector field is called incompressible if its divergence is zero.
Remark. A vector field X on M with volume volume preserving.
is incompressible
~ fevery l flow box of X is
Indeed, if X is incompressible then L,C! = 0; that is R is invariant under the vector field X. In other words, given a diffeomorphism 4,, we have:
We say
6, is vo1ume preserving.
Conversely, if we have V X E M : #,*Q(x) 2.6.3
=
R(x) , then X is incompressible.
Laplace operators
Let f be a differentiable function on a Riemannian manifold.
Riemannian Geometry
D
The Laplaciizn off is the divergence (function) of vector field gradient. In short: Lap = div.grad .
We specify every vector of this vector field gradient is the image of a 1-form gradient under the sharp mapping. The Laplacian off is expressed in local coordinates as follows:
This formula lets us find again the classicd expressions of Laplace operators in curvilinear coordinates (polar, cylindrical, spherical, elliptical...). Example. Compute the Laplacian of a function f in cylindrical coordinates r,B,z
Answer. Introduce the local coordinates U
1
3
u2=e
=r
u
=z
The components of gradf are
From the following conjugate tensor fl
0
01
we deduce the corresponding contravariant components gu d jf :
So, since
\/ ldetg = r , we obtain: 1 a af a raf a Lapf =-(-(r-)+-(--)+-(r-)) r dr 8r a8 r 2 a @ az 2 --+--+-- a2f i af 1 a f +-a2f ar2 r d r r 2 a 0 2 dz2
af dz
Let us now define new operators. D
The operator
6 = (-l)n~+n+'* d * ,
associating a (p-1)-form to any p-form, is called the codin;ereential operator. Note Vf
E
C m ( M:) 6f = 0 , since the differential of an n-form on an n-manifold is zero.
Lecture 8
298
D
The operator RP(M) + Q P ( M ): A=dS+Sd
is called Laplace-De Rham operator.
D
A differential form w is harmonic if Am = 0 .
To end this section, we consider any two p-forms a and B at x E M . The n-form a A *p is necessarily proportional to volume form pE(at x) which we express as
It's also
p A *a as seen later on.
Given differential forms of degree p on an orientable manifold M and assuming compact supports, we define the following inner product:
If necessary, the compactness notion will be used in the following applications. Application 1. Give the expression of factor (a, B) in Iocal coordinates and deduce:
Answer. Consider two any p-forms
C a,,,
a=
dril A . . . A&''
W i , dril A . . . A h C .
fl= il <. .
We immediately have:
and
a A */?=
,
So, from (a,p) = lP!a , , deduce:
hail
.,,,
Jm
pilip
d*' A ... A d*'
and by introducing g and go sufficient in number, we
a~*p=fl~*a and thus {a,B) = ( P P ) . Application 2. Show the operators d and 6 are adjoint, that is V a E Rp(M),BE Q ~ * ' ( M: ) ( d a ,/?) = (a,@.
Riemannian Geometry
Answer. We have:
I, Since fl
E
d a ~ * f l + ( - l ) ~a L~ d * p = dL( a ~ * f l ) =wI a ~ * f l = O
fiP+'(M), we have:
Application 3. Prove that: 0) V ~ E ~ ~ ~ ( M ) : W A * W ~ O ~ ~ ~ ( ~ , ~ ) = O ~ J T ~ = O
fig
A differential form w is harmonic
d o = 0 and Sw = 0 .
#I
Answer. (i) We have: I
I
W A * WP.= ~11 --Ww. ' " P p 2g 0 p
and thus
(w,w)=o
1fS
w=o.
(ii) If dm = Sw = 0 , then w is harmonic.
Conversely, if by hypothesis Aw = 0 then the inner product (AD,w ) = (
d b ,w )+ ( a m ,w ) = (220, SW) + (dw,dm)
is zero. The terms of this sum being nonnegative, we deduce that
and hence
dw=SW=O Application 4. Show the operator A of Laplace-De Rham is self-adjoint, that is:
V a , p ~ n ~ ( M( )~: a , f l ) = ( a , ~ ~ ) . Answer. Each member of the previous equality is
a)
( d a ydP)+ (sa, .
Note that the operator A is positive definite; that is ( A a , a ) is positive and is zero if Aa = 0 . To end this section we mention that the Laplace-De Rham operator plays an important role in the Hodge-De Rham theory. So, Hodge has proved that any pform a, has a unique decomposition: w=da+Sp+y with Ay = 0.
Lecture 8
300
3. GEODESIC AND EULER EQUATION Consider the following curve c on a Riemannian manifold ( M , g ):
[a,b] + M : t H ~ ( t ) such that, given the subdivision a = to < t, < ... < t , = b of [ a , b ] , the continuous h c t i o n t I+ c(t) is differentiable on each [ti,t,+,]for every i E [0, p - 11. First, we recall the following definition.
D
Thelengthofarccisthereal
(b > a ) . The integral exists since the integrand is defined and continuous for every t ( # t i ) belonging to [a,b ] . In local coordinates ( x i ) , the length of an arc c is
and is also denoted
D
A geodesic c is a curve such that its tangent vector field X is autoparallel along c:
V,X =0.
(8-5 1)
In local coordinates, this geodesic equation is written under the form of the n following differential equations:
Let us find again the geodesic equations from calculus of variations which consists in obtaining extremal curves (basic calculus of variations!). In this context, the geodesic distance between two points of an open set U is the lower bound (if it exists) of curve lengths joining the points and lying in U. The geodesic arc is the arc corresponding to this lower bound. Let us recall a fundamental result of the variational calculus. Consider (class c') curves joining two points xo and x,. To each curve is associated a previously defined length, namely;
Riemannian Geometry
where
A (necessary) condition of extremum of I, is obtained if the system of n Euler equations:
dr '
(where the various f' are -) is verified. dt Remark. A geodesic doesn't necessarily lead to a minimum of length. Furthermore, the 2sphere example proves there are several geodesics joining two diametrically opposite points (an infinity).
Now we establish the geodesic equations from the calculus of variations. On a manifold, a geodesic joining two points xu and x, is a curve such that
the calculus of variations means it is a curve "satisfying" the Euler equations (8-53):
Substituting the curvilinear parameters for t, the geodesic equations are
By multiplying by g' , we obtain the geodesic equations (8-52): d2xr -+r ds2
dxi hk --=o
' d s ds
where s is proportional to t. In lecture 9 we will prove that the geodesic equations express that the acceleration vector of a point moving dong geodesic is zero. So, a geodesic is a curve followed by a point of zero acceleration.
Lecture 8
302
4. CURVATURES
-
RICCI TENSOR ElNSTEIN EQUATIONS
4.1
- BIANCHI IDENTITY -
CURVATURE TENSOR
4.1.1 Operator R(X,Y)
PR14 The Cmmapping %(M)x X(M) x X(M) + F(M) :
(X,Y,Z) t+ (V,Vy
- VYVX - V[X,y])Z = R(X,Y)Z
is linear in each X, Y, Z separately and defines a tensor R ( X , Y ) of type ( ). ProoJ: Having
R(X,Y)Z = -R(Y,X)Z ,
it is sufficient to prove that V h E C m ( M ) :
We have:
and also
So, R(X,Y) is a tensor and is of type (i)because the "contracted product of R ( X , Y ) and a tensor of type (i)is a tensor of type (b). By introducing the commutator
[V,
,vyI = v,vy - v y v x ,
we denote: 62'
R(X,Y) = CV,,VYl-V,x,Y,.
Riemannian Geometry 4.1.2
Curvature tensor or Riemann-Christoffel tensor
Given any vector fields X,Y,Z E X ( M ) we are going to define a new tensor called curvature tensor which operating on X,Y, Z leads to (b ) tensor R(X,Y)Z . This curvature
:
tensor is of type ( ). Make explicit the components of this tensor in a local chart. From
we deduce the following element of X ( M ) :
a
a
a
ax1
ad
axk
R ( X , Y ) Z = x ' y ' Z k R(--,-)-. In particular, make explicit
W e successively have:
D
The curvature tensor or Riemann-Christoffel tensor is the tensor of type (i ) such that its components are
w The mapping
R : , ~= a i q i - 8 , ~ ;+qic:-4;q;.
X ( M ) x X ( M ) x X ( M ) + X(M) : ( X ,Y , Z )
is so well defined by
and, in particular, by the following expressions where VI,l,,,ld, = 0:
The following proposition is obvious:
(8-55)
R ( X ,Y ) Z
Lecture 8
304
PR15 The curvature tensor field of components R:;, is skew-symmetric in i and j . A geometrical interpretation of curvature tensor Let X be a vector field tangent to a curve V be a tangent vector field on M
c,
of M,
Let ( 7 - l ~ be ~ )the ~ vector of the field V which is backwards transported parallel along c, from q top; in general, it is different from V ( p ) . From (8-37) we immediately deduce (index 0 corresponding top):
I c1
Figure 50
Let us consider a set of orbits c associated to vector field X as it was introduced at the time of the Lie derivative study (one-parameter group!) and also another set of orbits r associated to a vector field Y. We assume they are such that
[ X , Y ]= 0 . About that the important remark of lecture 6 (in 8 I. 1.2) will be seen again. Let r be the intersection point of two curves c,and TIof the respective sets of curves. The vector V(r)can be backwards transported parallel along rl to point q. Next, the so transported vector is backwards transported parallel along c, top. The vector V ( r ) can also be backwards transported parallel t o p successively along cl and TO.
In the first case (r -,q + p), since the "backwards transported parallel vector" at q is then the "transported is, at p:
Riemannian Geometry
In the second case ( r + s + p ), the "transported"is, at p:
The difference between the two results, namely:
amounts to
+ o(t". This last expression makes appear R(X,Y)V since [ X , Y ]= 0. t 2[V,,V,IV(P)
In conclusion, the previous difference "measures" the changes following fiorn successive parallel transports along the closed curve pqrsp and it is expressed by the vector
Flat space D
A flat space is a manifold with zero curvature tensor.
In the case of a flat space, the parallelism is a global notion. The previous geometrical interpretation lets us assert that a vector is parallel (or not), at every point, to a vector at point p. The parallel transport is independent of the path choice.
The metric of flat spaces is such that
The geodesics are straight lines. The Minkowski spacetime of special relativity is an example of flat space. 4.2
RICCI TENSOR
4.2.1
(: ) curvature tensor
D * The curvature tensor of type (: ) is defined by R ( X , Y , Z , T ) = g ( R ( X , Y ) Z ,T ) . In local coordinates, given X = X' a, , Y = Y 8, , Z = Z' 3, ,T = T ma, : J
~(R(x,Y)z,T) = X ' Y ~ Z * TR ~; , g(a,,a,) =x
' Y ~ z ~ TK&" ~ ,
Lecture 8
The components of this tensor of type ( ) are
~ ( a ~ , a ~ , a , , a=, )g(~(d,,a,)a,,a,) = Rho.
(8-58)
The curvature tensor of type (: ) is skew-symmetric in its last two indices:
PI.
R(X,Y,Z,T)=-R(Y,X,Z,T). I This property is obvious because R;,!,= - Rk, ,. We have proved: RhiJ = -Rknr,,, .
The curvature tensor of type ( 0,) is skew-symmetric in its first two indices:
P2.
Rh,lj. = -4th .Ii .
'
Indeed in exercise 16 we proved:
R(X,Y,Z,T) =-R(X,Y,T,Z). The curvature tensor of type (: ) is syrnmetnc under a pair exchange of indices:
P3.
(8-6 1)
Rstsr/ = 4,,h. Indeed in exercise 18 we proved:
R(X,Y,Z,T) = R(Z,T,X,Y).
Indeed in exercise 17 we proved:
R(X,Y,Z,T)-t- R(Y,Z,X,T)+ R(Z,X,Y,T) = 0 which is written:
Rh, + R,,,,, + R,,
a
4.2.2
=0
Rh,$+'~k,lrn+%~,lk=~ Rh,o+Rpi,Jrn+'~,mt=O.
Ricci tensor
Can new tensors be deduced by contractions of the Riemann-Christoffel curvature tensor? The components R,,, being antisymmetric in i and j and also in k and m, every contraction between two indices of a same "set" leads to a zero tensor. So, for instance, by contracting
G., = g'"~,,, in k and I, we obtain:
Riemannian Geometry h
R:., = g R,,,
= gd
44,
= -g
ht
R,, = 0.
On the other hand, if there is contraction in any index of the first "set" and any index of the second "set," then we obtain a new tensor of type (i) or its opposite. Indeed consider the following successive contractions in indices k and k: (2 and 3):
&=&!,kj -- g
(1 and 4):
g"
(land 3):
g kl Rhj,&= g k l R i h , j k = -g
(2 and 4)
gM
D
Rhi,jh
=g
&h,@
=g
Rih,&
=
hk
M
h
Rjk.hi
= Rj,hi = Rji
4 1k
Rib,, = - g hk R,,&= -R,
4h.U
= -Ri,
.
The Rkci (curvature) tensor is a symmetric tensor of type (; ) with components
obtained by the contraction of Riemann-Chistoffel tensor in the second and third inhces. In other words, considering R(X,Y) where XY are fixed, then the trace of the linear mapping
R ( X , Y ) : T,M
+ T,M
: Z J+ R(X,Y)Z
is the Ricci tensor.
The component notation of Ricci tensor is
R, =I& =a&; -a,r;+r;r;-r;r; 4.2.3
Scalar curvature
From the Ricci tensor
D
T'M x T,M + R , we say:
The scalar curvature on a manifold is the trace of the Ricci tensor, namely the sodenoted scalar R=&'=~~R, (8-65) obtained from the Ricci tensor contraction.
Example. A cone is an example of flat space. Show that the scalar curvature is zero. Answer. Parametric equations of a cone are
x=utanOcos~ where u is the altitude,
y=utanBsin6
6 the longitude and 0 the half angle at vertex o.
We successively have: ds2 = (1 + tan2 B)du2 + u2 tan2 B d#2
z=u
Lecture 8
and from (8-64): &=O
4.3
+
=
R*=O
R=O.
BIANCHI IDENTITY
4.3.1 Operator (VxR)(Y,2 )
We define the operator ( V xR) (Y,2 ) by
(VxR)(Y,Z)T=Vx(R(Y,Z)T)-R(VxY,Z)T-R(Y,VxZ)T-~(Y~Z)VxT~ (8-66) 4.3.2
Bianchi identity
PR16 Given vector fields X,Y,Z there is the Bianchi identi@:
w
( V x R ) ( Y y Z+) ( V y R ) ( Z , X )+ ( V z R ) ( X , Y )= 0 .
(8-67) , we must prove
ProoJ If we denote the sum on the cyclic permutations of (X,Y,Z) by X.Y.Z
the following identity:
From (8-66),we have:
and thus we have:
But cyclic permutations do not alter X,Y,Z
But we have:
VZ (Vvxy - VVUX) = V z V ( v x ~ - v ,=x )V Z ~ I X . ~ ] 3
Riemannian Geometry
and thus
Finally, the sum on the cyclic permutations
is zero since the previous summation of the first two terms is zero and it is so for the two following. For the last term the sum is also zero because the Jacobi identity means that
In local coordinates and by introducing the Riemann-Christoffel tensor, the Bianchi identities are written:
~ ~ + v,.%,, g + v,, < ,~ ~= 0 . ~
w
4.4
EINSTEIN EQUATIONS In the Bianchi identities, a contraction in I and k entails
v,R:, + v,~:, + v k ~ : , u= 0 0
-ViR,
+VjR,
+v,R~,, =O.
A new contraction in m and j implies:
-vi gnJiR,,#+ v jg"gR, G
- v , R + v ~ R / +v,R;
tS
2Vk girR;
e
V, ( R * - i g b R ) = 0.
- V,
+ v , gw~:,, = 0
=O
g "= ~ 0
So the symmetric tensor G with components G*=R*-' *R 2g verifies the contracted Bianchi identities:
w
V,G* = o
.
The tensor G is called the Einstein (curvature) tensor.
(8-68)
Lecture 8
310
By searching a spacetime metric such that every particle follows a geodesic, Einstein had to consider a nonzero curvature Riemannian space. The previous equations play a fundamental role in the relativistic theory of Gravitation. Indeed, let us recall the Einstein equations in this theory are based on the fact that the gravitational potentials generalize the notions of Laplace and Poisson equations and establish relations between tensors in a Riemannian spacmme Md . In the relativistic theory of Gravitation, the curvature of this manifold M4gives an account of gravitational phenomena. The coefficients gy of the metric (called gravitational potentials) are supposed to verify the Einstein equations describing the state of energetic distribution: G* =) R* - f g d ~ = x ~ *
(8-70)
where X is a constant linked to the gravitational constant and T is the stress-energy tensor. The stress-energy tensor describes the state of energy distribution at each point and "generalizes" the right-hand member of the Poisson equation. More precisely, the components a of this tensor T are such that TOO represents the density of mass-energy, ~ " momentum density, T O j an energy flux and T~ a stress. In 1917, Einstein modified his field equations such that G* + A ~ = * XT*
where A was called the cosrnofogicafconstant (making the universe open or closed). Later he left this addition. The constant A has a hdamental influence on universe evolution models. Without energy the Einstein equations are
G* E) R* - + g * ~= 0 The G* verify the four equations (8-69). Make clear that the G~ being symmetric, there are $n(n + 1) = 10 in number. The four equations (8-69) reduce the Einstein equations to six independent equations.
Notice that searches are also turned through universe models showing torsion. The reader will refer to the general relativity literature.
5. EXERCISES Exercise 1.
Show the expression of the 1-form g ( X , ) in a local chart and find again the expression of g ( X , Y ) for every Y E TIM.
a ax
Answer. From X = Xi - we deduce 1
Riemannian Geometry X , = g ( X , )=x,dr'=ggx'dri
and X J Y ) = g ( X , Y ) = g,Xidri(Y) a , a = g,X'Yi ax
Exercise 2. On the hypersurface defined by
construct the Riemannian metric induced by the following bilinear form (on R"" ): 132
Is the corresponding bilinear form positive definite? Answer.
The hypersurface M, = { x : ( X I ) '
-
n+l
( x i ) ' = 1 ) is i=2
generalizing the hyperboloid (two sheets). From
we deduce two domains of charts of Mn namely:
and
A curve on the manifold is of type:
A tangent vector at point c(0)is
with
E
= 1 or - 1 .
a submanifold of R"+'
Lecture 8
By putting i i ( 0 )= X i , we have:
Any other vector of T,M, is written:
The pseudo-metric induced by the scalar product on R"" is defined by
Remark that the field g : Mn -,T . M :~x H g, is of class Cm. The bilinear form is not positive definite. Indeed 1
n+l
st1
ntl
/-2
1=2
x*xi)(C x'x')- (xi)'
Therefore, the value of the bilinear form g , ( X , X) is less than or equal to
and thus it is an always strictly negative real because x1 cannot vanish and at least one of components of tangent vector X is nonzero. Exercise 3. Prove that on any paracompact manifold Mn here exists a Riemannian metric. Answer. Let { (U,,pi) ) be a covering of the paracompact manifold, ( h, ) be a partition of unity subordinate to this covering.
Let g('' be the (i) tensor on U, with components g:' = S,, in a local coordinate system. The (i) tensor
h, g"' on M , provides M,, with a Riemannian structure, the corresponding i
bilinear form being positive definite. For every i the pull-back p; induces (from Rn) a Riemannian structure on Ui.
Riemannian Geometry
3 13
Exercise 4.
Find the adjoint of exterior product of two vectors on a 3-dimensional Riemannian manifold and give its meaning in an orthonormal basis of 3dimensional Euclidean space,
Answer. The exterior product of two vectors X and Yon a manifold M3 is written:
The adjoint of this (i) tensor is the covector of components
*
(*T), = $ ( y ) r ( ~ ' ~ -' X'Y i ) =
+
and explicitly:
&*
(XiY' - X'Y')
(*T), = Jdetg ( x 2 y 3- X ~ 2,Y (*T)2= Jdetg ( x 3 y 1- x'Y') (*T), =
(x'Y'- X'Y' ) .
By considering an orthonormal basis of Euclidean space, we know that detg = 1 and variance is indifferent. We find again the components of vector product of vectors Xand Y: x2y3
- x3y2
x3q
- x,&
x,Y2 - X J .
Exercise 5.
Find the adjoint of curl on Mj and find again the classical expression of curl on R ~ . Given f E C m ( M 3 prove ) the curl of grad f is zero. Answer. Exercise 1 1 of lecture 5 has shown:
The adjoint of curl, that is * d m , is the vector of components
curlk# or explicitly:
1 ijk = Z ( P g ) ( aI. mJ. - a [email protected] =
Eijk
*(ai?
- aJu 1. )
Lecture 8 1 curl 3 w = -
In an orthonormal basis of 3-Euclidean space, detg = 1 and variance is indistinguishable. We find again the curl notion of elementary geometry. The curl of grad f is zero because
* d(df) = *(ddf) = 0 Exercise 6.
Find again the usual definition of divergence on R~from the notion of adjoint. Answer. Let
x=x1a,+x2a2+x3a3
be a vector field. Its adjoint is the differential form of degree 2:
and thus
The divergence of field X is the coefficient of the exterior derivative of the adjoint ofX. Exercise 7. Calculate the Christoffei symbols in spherical coordinates. Answer. Given the spherical coordinates u1 = r
u2 = B
u3 =#,
we know the only nonzero gU are g,, = 1
g22 = r 2
g33= r 2sin 2 8,
the metric element being ds2 = d r 2 +r2d8' + r 2 s i n 2 8d+b2.
Riemannian Geometry Therefore
Exercise 8. Find the first Christoffel formula from the Ricci identity.
Answer. The following Ricci identities, successively obtained from cyclic permutations,
r, + r,, d,gH = Cjk + r@, 4gjk =
-
= - q k i - Jkj
are added and from the symmetry of connection coefficients (in last two indices), we have: d,g,k +ajgk,-',gu
=rhj +&
=2rhj,
This proves the first Christoffel formula.
Exercise 9. Given a Riemannian metric g, calculate the components of the gradient (1-forin) of a fbnction f and the ones of the associated vector (under g # ) (i) relative to basis (a,,a,,d,) in a cylindrical coordinates system, (izj relative to basis (1,,yo ,a,) of R~ such that
Answer.
fi) The components (grad f), are: and the components (gradf )' are
because gl' = g33= 1 and g22= -1
r2
(ii) Since
'
the components (gradf ), relative to basis 1
(arf7-8,f r
(5,t,a,> are the ones of Xf :
J,f ).
They are the components (gradf )' because ( i , & , d , ) is an orthonormal basis (gy = 6').
Covariance and contravariance are indistinguishable unlike the case of nonorthogonal basis! Exercise 10.
Calculate the Laplacian of a function f in spherical coordinates r76,4 where 0 is the colatitude. Answer. The components of covector grad f are
From
we deduce the corresponding contravariant components goajf :
Since
&= r
2
sin t3 we have thus
Exercise 11. Given a subset S c M we suppose a p-volume exists on S defined by
where (6')are local coordinates on S and g, is the metric tensor on S.
Riemannian Geometry (i) Find again the formula of area of a surface defined by = Z(X,Y)
t f ( x , y ) ~U c
R'
(id Find again the area of a spherical surface (radius R).
Answer. (i) By putting
4 = ayz , the metric element ds2= dX2 + dy2 + dz2 immediately leads to metric tensor p = a,z
and detg, = 1 + p 2 + q 2 .
We find again the well-known expression of area:
(ii) From spherical coordinates B and # , such that in R~: x' = RsinB cos4
x 2 = Rsin 8 sin4
x3 = RCOSQ,
we deduce the following vectors tangent to the sphere:
The metric tensor g, is
and detg, = R~sin2 8 . The area of the sphere is
Exercise 12.
From expressions of tangent vectors to torus T' c IZ3 (see exercise 5 of lecture 2) obtain the area of T . Answer, We know the tangent vectors to T~ are
Lecture 8
The metric tensor g,, is such that:
and (~+$COS#)~
0
detg,, =
=
0
(&A&+ 4 c o w 2-
(4)2
The area of T' is
Exercise 13. Given a vector field X on an orientable manifold Mwith volume a , prove that 1 V ~ ( # O ) E C " ( M )d:i v m X = d i v , X + - L x g . g
Answer. Since the following equality is obvious L A @ ) = gLxR+ QLxg
and since g R is a volume and divergence definition:
L, ( g Q )= (div,X) @, we deduce that
and the exercise is proved since Lx CJ = (div, X)a.
Exercise 14. Given a vector field X on an orientabIe manifold M with volume R , prove that:
VgcCWCI(M):d i v n ( g X ) = g d i v n X + L X g .
Riemannian Geometry Answer. The PR18 of lecture 6 has established the following equality
In the following expression
we know that dg AS^ = 0 since dg A R E a n t l ( M ) . Therefore, we have: L,f2=i,a'g~Q+gL,CJ
that is L,R=L,~ASZ+~L,R
and thus
R div,(gX) = RLxg + gRdiv,X
The proof is so produced. Exercise 15.
Given a surface S c R n , prove that any geodesic in S is such that, at each point of it, the osculating plane is orthogonal to the surface. Answer. Let us recall that every osculating plane equation is defined by the first and second
derivative vectors; geodesic equations are:
,, d2
gv
akk dx' 1 dr' d X k + dkgr-- -ajg,k -- = 0
~ '
ds ds
2
ds ds
But
a2
a
(because d,e, = -- -= die, ), axk8xf axfaxk
Since the summation indices play the same role, the geodesic equations are then simplified: d2x' ( e l , e j ) -ds2 +(ei,a,ej)--
dx' dXk ds a's = 0
Lecture 8
320
Now, we consider the following vector:
and thus 2
d --
--d2xJ
ds2
dXJ
dyk
dsze'+dsdr
aae,
Therefore the geodesic equations (1) mean the second derivative vector and tangent vector are perpendicular at each point of the geodesic. The osculating plane is thus orthogonal to the surface. Exercise 16. Prove the (0, ) curvature tensor is antisymmetric in its first two indices. Answer. Since
R(X,Y,Z,T) = g((VxVy - VyV,
- V[x,yJ)Z, T)
and (8-41): we have:
Therefore we have proved:
lib.,C= -Rnk,y. Exercise 17. Prove the identity R(X,Y,Z,T)+R(Y,Z,X,T)+R(Z,X,Y,T) = 0 Answer. This follows from the next zero sum:
(VxV, --VYVX-VIX,Y1)Z + (VYVZ- VZVY - V[Y.Z])X+ (VZVX - V x V z - V[Z,X,)Y
Riemannian Geometry
321
Exercise 18. Prove the equality R(X,Y,Z,T)=R(Z,T,X,Y). Answer. By adding the following identities:
R(X,Y,Z,T)+R(Y,Z,X,T)i-R(Z,X,Y,T) = 0 R(Y,Z,T,X)+R(Z,T,Y,X)+R(T,Y,Z,X)=O R(Z,T,X,Y)+R(T,X,Z,Y)+R(X,Z,T,Y)= 0 R ( T , X , Y , Z ) + R ( X , Y , T , Z ) + R ( Y , T , X , Z )= 0 , we obtain (from P2 of curvature tensor):
R ( Z , X , Y , T ) + R ( T , Y , Z , X ) + R ( X , Z , T , Y ) +R ( Y , T , X , Z ) = 0 and (from PI and P2 o f curvature tensor) this identity becomes 2 R ( Z 7 X , Y , T+ ) 2R(T,Y,Z,X ) = 0 that is 2R(X,Y,Z,T) = -2R(T,Z,X,Y) = 2R(Z,T,X,Y). We have thus proved the symmetry under exchange of the first and last pairs of inches Rh,ij = Rij.km.
Exercise 19. Find the independent components of the (: ) Riemannian vector on a manifold M, . Make explicit these components if the 3-dimensional manifold is a Lobatchevsky space. Answer. From properties of the ( y ) curvature tensor, but (8-62) being useless, we immediately see that only six components are independent, namely: R12,12
3
'13,137
R23,23
5
R12,13
9
'12,23
9
'13,23
.
Let x1 = x, x2 = y, x3 = z be local coordinates in Lobatchevsky space (metric element: 1 ds2 = I ( d r 2 + dy2+ h2). We have: z
Lecture 8
and the ChristoffeI symbols are
q ; = & l , = ~ ;2 = I & = I& ; = - z
From (8-55) and since R,,,/ = g,,RL,,. , we obtain:
Exercise 20.
Find the Ricci tensor, the scalar curvature and geodesic equations for 2-torus T 2 c Answer. In exercise 5 of lecture 2, we have introduced parametric equations of T~:
with latitude # and longitude B , R being the radius of circle that follows the center of the vertical circle (with radius r). Since ds2 = ( 2+ r c o s & ~ do2 ~ + r2 d4' formulas (8-43) imply:
e r sin 4 rw=-R +rcos+
From (8-64) we obtain:
"
r cos 4
R -
R + rcos4
R,
( R+ t cos+)cosh
=
r
R=
Geodesic equations are
r 2 d 2+ ( R + r c o ~ # ) ~= 8I .~ These three equations are not independent. The second equation leads to a first integral (~+rcos#)~0=k.
2 cos 4
-
r(R + r cos 4)
Riemannian Geometry
323
From the third we deduce:
( ( R+ r cos 4)' - k2)'" = r(k+ r c o s ~
The initial condition
is obtained from the third equation where
d2 . d2 is replaced by #fz
The integral is elliptic. Remark. Since g, = (W+ r c o s 0 2 and gll = r (all the other g, being zero), it's easy to 2
verify that the various G,
=
Ri;. - ig,R
are zero.
Exercise 21.
We have considered several surfaces in R~ and obtained the following results. 1 ". Cylinder. Equations: x = a c o s u
ypasinu
z-v
(~ER,).
The geodesics ( ii = i; = 0 ) are helixes because
and thus
z=bu,
the pitch of the helix being h = 2 x b . 29 Helicoid.
Equations: x = rcosu ds 2 = dr2 + (r 2 + h 2 ) du2.
y = r sinu
z = hu
( h : nonzero real)
Lecture 8
324
Geodesic equations are 2r with i 2+ ( r Z + hZ)ti2= 1 ii + -----(2) 2 2 i u= 0 r +h 2 2 Since (2) implies (r + h )u = k and deducing i from (3), we obtain:
i;-ru2 = O
=
(1)
,/(c2
kr
65 = u - u, 2 + hZ)(g2+ h - k 2 )
(k)
The integral is elliptic. 3 9 Hyperboloid (one sheeo. y = acha sin fl
Equations: x = acha cos p
z = b sha
with a, b E R , latitude a and longitude p . ds2 = (a 2 sh 2 a + b 2 ch 2 a)da 2 +a2ch2ad p 2 . 2 2 ra = (a + 2b )sha2 cha 2
ra -Bs -
a2sh a + h ch a
pa
a2 sha cha a2sh2a+ b2 chZa
sha cha
=-
'
Geodesic equations: d +
( a 2 + b 2 )sha cha a 2 sh 2 a + b2ch2a
-
a 2 sha cha a 2 sh 2 a + b2chZa
B2 G O
sha cha
b+2---hb=0
(a2sh2tx + b2ch2a)b2+ a2chZa
with
= 1.
Since (2) implies chZa = A (constant) and deducing d from (3), we obtain: d:
da
cha
(3).
LECTURE
9
LAGRANGIAN AND HAMILTONIAN MECHANICS
In this lecture we recall several notions of Lagrangian and Hamiltonian mechanics in order to prepare the reader for the next lecture. So we give the classical principles of Lagrange and Hamilton in analytical mechanics. The equations of motion are deduced. The canonical transformations and integral invariants are also introduced. Moreover, a fluid-dynamical method that I have perfected is shown. To conclude comments on isolating integrals and ergodicity are made.
1. CLASSICAL MECHANICS SPACES AND METRIC 1.1
GENERALIZED COORDINATES AND SPACES
Consider a system of particles with a finite number n of degrees of freedom. This system is called rheonomic if the position vectors of particles depend explicitly on time; it is called scleronomic in the opposite case. Classical mechanics teaches us that the motion of a system of particles is described at any time by n variables called generalized coordinates denoted by q'.
We must specify that there are other systems of generalized coordinates such that
with
These conditions should be always present in mind. Recall also the evolution of a (material) system with n degrees of freedom can be described in spaces of n coordinates and we say:
D * The confi$pration space of a (material) system with n degrees of freedom is the differentiable n-manifold of points (q',..., q n ) .
Lecture 9 It is denoted by Q.
Example. The configuration space of a system composed of N points of R3 is the manifold R ~every ~ point , having 3 degrees of freedom. Generalized trajectory. A point of the configuration space determines the "position" of the material system at a given instant; in other words this (descriptive) point gives the configuration of the system. The evolution of the material system in the course of time makes the point (ql,...,qn)describe a curve in the configuration space called the generalized trajectory of the system. The notion of configuration space of a system subject to constraints is developed in elementary mechanics. So, for example, a set of N particles in R 3 with Cartesian coordinates x, y, t is subject to constraints defined by m equations f,(x, y,z) = 0. This (material) system shows n = 3N - m degrees of freedom and the configuration space is a (3N - m) -manifold. Example 1. What is the configuration space of a set of two particles following a curve defined by { (x, y, z) E R3: f(x, y, Z ) = 0 ,g(x, y, Z ) = 0 }?
Answer. A priori two particles of R3 have 3N = 6 degrees of freedom. But the constraint equations are m = 4 in number because the coordinates of particles (xl,y,, z, ) and (x, ,y,, 2,) must verify the following equations: f ( x l > ~ , > z=, 0)
f ( ~ 2 , ~ 2 ,=~ 02 )
g(x,,y,,z,) = 0
g ( ~ Z r ~ 2 7 ~= 2 )0.
The configuration space is thus a 2-dimensional manifold (since 3N - 4 = 2 degrees of fieedom). More quickly, the curvilinear coordinates of every particle let us see that the configuration ' such that q' and q2 are the respective curvilinear coordinates s, and s,. space is R
Example 2. What is the configuration space of a system of two particles linked together by a rod of negligible mass and moving in the space? Answer. A priori the two particles have 3N = 6 degrees of freedom. But an equation of constraint expressing that the distance of two particles (x,, y,, z,) and (x,, y,, 2,) is constant, that is:
reduces to 5 the number of degrees of freedom. The dimension of the configuration space is 5. Generalized coordinates are the three coordinates of the mass center G and the two angles giving the direction of the rod. The configuration space is R3 x
s2where q1= x,,
q 2 ' Y G , ~3
-- Z
4-0,q5=4.
G , ~-
Lagrangian and Hamiltonian Mechanics
327
Example 3. A rigid body with a fixed point in space has 3 degrees of freedom. Generalized coordinates are the three Euler angles. The configuration space has the manifold structure of SO(3).
Example 4. The configuration space of a double pendulum in a plane is a 2-torus, cartesian product of two circles S' . D
*
D
* The coordinates (or components) of a tangent vector to a trajectory of equation
The conJguration spacetime is the Cartesian product Q x R of the configuration space Q and the time axis R .
q = q(t) [ where q(t) = (q1(t),...,qn(t))] in Q are the reals q'(t) and are called generalized velocihres 9'. We denote: (q' ,..., 4") E TgQ.
D
The velocityphase space is the tangent bundle of the configuration space Q, that is
It is the set of 2n-tuples (qi,q') defining a coordinate system on the tangent bundle.
1.2
KINETIC ENERGY AND FUEMANNIAN MANIFOLD
1.2.1
Kinetic energy
Consider a system of N particles of R ~ . The Cartesian coordinates of the N particles are respectively denoted: ( x , , y , , z , )= (x',x2.x3)
( x ~yN,zN) , =
x3N)
The kinetic energy of the system is written:
where the common coeficient m,= m, = m, represents the mass of the first particle and so on. Since the x i are functions of q and t, we have: J
Lecture 9
The kinetic energy is composed of three terms:
which is independent of generalized velocities,
which is linear in the generalized velocities,
which is quadratic in the generalized velocities. In the case of scleronomic systems (constraints independent of time!), we have:
and the kinetic energy is the following quadratic form T = -2Ia .Jk (q ' ) q ' q k . 1.2.2
Riemannian manifold
PR1 The configuration space of a scleronomic system is a Riemannian manifold. Proof: The kinetic energy being quadratic in the generalized velocities, T = t a r @ ' ) 4'gk,
and introducing a Riemannian metric
( ) on Q, we denote W = ( g l ,..., 9")s T,Q :
T=~(x,x). Thus the metric element is evidently ds2 = ajk(ql)dq'dqk= 2T dt2.
Example. The configuration space of a system of two mass points linked together by a rod of negligible mass and moving in a plane is a manifold R' x S1. The metric is determined by the kinetic energy: T = im,( ( 4 ) '
+(LIP)
+f s ( ( i 2 ) 2 + (
1
~ 2 7
Lagrangian and Hamiltonian Mechanics
given two pints ( x i ,y, ) and (x,,y2) of respective masses m, and m,. The reader will easily express the kinetic energy from the generalized velocities xG,yGand 6 (usual notation).
2. HAMILTON PRINCIPLE, MOTION EQUATIONS AND PHASE SPACE 2.1
LAGRANGIAN
By considering the space TQ x R , we say:
D
The Lagrangim is the differentiable function
L : TQ x R + R : (qi,q',t)H ~ ( q ' , ~ ' , t ) of the n generalized coordinates q i , the n generalized velocities qi and time t such that 6 v
L(q',qi,t)= T(q',q i ,t)- V(q i ,t).
Explicitly:
~ ( ~ ' , g '=, iol(q',t)cjicj' t) + ~ 7 , ( ~ ' , t )+ 4 'o ( s i , t ) - v ( q ' , t )
(9-2)
We recall the variable t doesn't always explicitly appear in the kinetic energy T and potential energy V.
Remark. A rheonomic system can present a Lagrangian which is not explicitly dependent on time. Indeed, consider a pearl following a circle c(o;R) . If this last rotates about a vertical axis with a constant angular velocity o,then this example ilIustrates t h s remark. The generalized coordinate is the angle 0 (in the plane of the circle) locating the pearl from the horizontal plane of equation x3 = 0 . The position of the circle is located by the angle o t in the horizontal plane and so, is explicitly dependent on time. The components of the position vector of the pearl in R~ are The Lagrangian L=m ( ~ '+ 8 Rim2 ~ cos28 )- nag R sin B 2
is not explicitly dependent on t 2.2
PRINCIPLE OF LEAST ACTION
According to Maupertuis, a succinct and rough expression of principle of least action is that nature chooses the simplest way.
Lecture 9
330
We recail that the variational principles have been helpful in the mechanics developments. Two fundamental ways are possible in classical mechanics:
(4 The motion equations of Newton, Lagrange or Hamilton are deduced from variational principles as theorems. (id From the postulated motion equations we derive variational principles as theorems. Firstly, we introduce a variational principle for which we show that the extremals' verify the motion equations of Newton.
We recall the d'Alembert-lagrange principle and the (infinitesimal and instantaneous) virtual displacements show a differential character and let us obtain the Lagrange equations. These last equations as well as the canonical equations of Hamilton can be deduced from an integral principle. In order to present this last variational principle, which is valid for any (material) system, we introduce an essential definition. Given two fixed points q, and q, on the manifold Q , we consider the following set of curves c of class C' : C(q,,q,,C~l~t,l)= (c:[t,,t,l c= J2 -, 4 1 ) = qPc(t2) = 9,).
e;
Hence we say:
D
The function S : C(qllq2,Etl,t21) + R
is defined by the action integral
The action integral is also denoted by
s=
L(q',dl,t)dt . 1 '
Remark. Consider a curve c : r I+ c(t) of C(q,,q,,[tl ,t2]) and any curve C(Y,,q2,PI,t2]) about c, (= c) and such that c, (t,) = gl, c,(t,) = g, .
For each t , any tangent vector
d dll
c;(O) = (-c,)(~)
c, : t H c,(t)
is historically denoted by &(t)
E
of
T,(,,Q
such that &(tl) = &(t,) = 0
Variational principle of Hamilton
PR2
Among all the possible trajectories of a (material) system from an instant t, to an instant t,. the natural trajectory (extremal!) is such that the (first) variation of the action integral vanishes:
In the calculus of variations, every curve solution of the Euler equation is called an exrremal.
Lagrangian and Hamiltonian Mechanics
33 1
The general nature of this variational principle (which is a postulate) widens the field of theory compared Newton laws of motion.
PR3
The extremals relating to variational principle of Hamilton verify the Newton equations.
Prooj. Consider a system of N points of masses m, of position vectors rj and without constraints. The extremals make the following integral extremum:
written with the notations of
5 1.2
and m, = m, = m, = M I ,. . . , m,,-, = in,,-,
= % , = M,
.
The extremals are the solutions of 3N Euler equations:
these last, grouped together 3 by 3, are really the Newton equations for each of N points: M , ~ ; ' + ~ , v = o..., , M , Y ~ + ~ , V = O .
2.3
LAGRANGE EQUATIONS
Firstly, in the context of the manifold Q, we say:
D
A motion in the configuration space Q is a mapping
x : R + Q : t H ~ ( t=) (q l(t), ...,q n (t)) making extremum the action integral (L being the Lagrangian):
S=
I" L(x(t),i(t),t) dt. 11
Historically, the Lagrange equations were established before the Hamilton principle, but these equations are also deduced from this principle. In a general way, we are going to prove what follows. PR4
The motion of a (descriptive) point takes place in the configuration space Iff the generalized coordinates and generalized velocities of this point satisfy the Lagrange equations.
Proof. We first recall the Hamilton principle:
simply denoted:
Lecture 9
where the variations are synchronous (that is St = 0). In addition, we recall that the variations are zero at limits, that is by denoting (cL(0))' = 69' : 6gi(t,)= 6q1(t2)= 0 . We notice also that
d Sq' = -6q
i
dr
Now, the proof is easy:
=jtz (---a= '1
aq'
aL)sqidt. dt aqi
From a lemma of calculus of variations1 and since the Sqi are arbitrarily independent, then the previous integral is zero #I Vz E {I, ..., n):
They are the n Lagrange equations in the case of a (material) system of n degrees of freedom, that are n differential equations of second order also called Euler equations in variational calculus.
2.4
CANONICAL EQUATIONS OF HAMILTON
2.4.1
Legendre transformation
The Legendre transformation lets us establish the canonical equations of Hamilton from Lagrange equations (first met in the thermodynamics context). the
Figure 51
' Proof for instance in our book Gkornetrie dzflkrentielle eiMkcanique analpique.
Lagrangian and Hamiltonian Mechanics Let f : R
-+ R : x H f ( x ) be a convex function
(f"(x)> 0).
Given a real p, we denote by x(p) the common abscissa of the most distant points of the curve y = f ( x ) and the straight line y = px .
So, the function xr-, p x - f ( x )
shows a maximum for x(p) D
The Legendre transformation off (with respect to x) is the function g : R - + R : P H ~ ( P ) p=x ( p ) - f ( x ( p ) ) .
This definition becomes immediately widespread to convex functions of a vector variable x = (xl ,..., 3 ) .So for p = ( p ,,..., p,) the Legendre transformation is obtained by
Example. What is the Legendre transformation of the function f ( x ) =
2
x2 ?
m P px - - x2 shows a maximum at x ( p ) = - . The Legendre 2 m p2 transformation is the function defined by g ( p )= 2m
Answer. The function x
I-+
2.4.2 Canonical equations of Hamilton
We are going to show that the Legendre transformation associates to n second order differential equations of Lagrange a system of 2n first order differential equations.
We recall that
L : R" x R" x R + R : (qt,q',t) H L(qf,q',t)
is the Lagrangian function T - V (at least of class C' ). D
The generalized momentum or canonically conjugate momntum (to q' ) is
) D * The Hamiltonian is the Legendre transformation of Lagrangian ~ ( q ' , q ' , twith respect to q = ( q l ...,qn) , , that is
w
H(q*,pi,t)= p,9' - L(q',(i'(qi,~ i , t ) , t )
where the q' are expressed from the p,
(9-7)
Lecture 9
334
This transformation converts the formalism and equations of Lagrange to formalism and equations of Hamilton. PR5
A system of 2n first order differential equations called system ofcanonical equations of Hamilton and ensuing from the system of n Lagrange equations is written:
Proof: Since
dL (since p, = -) aq'
and
we deduce:
q. * =-aH api
dL = --aH 89' dq'
dH at
dL at
-= --
The system of n Lagrange equations:
leads to the system of 2n equations of Hamilton:
4'
dH = --
apt
p.
=--
dH aq'
describing the evolution of the material system
Example. The canonical equations of a point of mass m moving under a central force are
since
Remark. The equations
Lagrangian and Hamiltonian Mechanics
introduce the complementary equation:
which takes the following form in Lagrangian formalism:
where L* = piq'
-L
is called adjoint Lagrangian.
Evidently we have: H(qi,pi,r) = ~*(q',g'(p,),t).
First integrals and cyclic coordinates
2.4.3
In the Lagrangian context we obtain interesting conclusions about the existence of first integrals (see also the calculus of variations). We recall the "first integral" terminology designates, in mechanics, a function of positions, velocities and time which is constant during the motion. Here a first integral is a function of q',p,and t which remains constant along any "phase orbit." We adopt this usual terminology but we must emphasize that H. Poincare used the more suitable term "invariant" to designate this notion. So, in the case of conservative systems (with potential V ) and scleronomic (d,L = 0), the complementary equation
means that H (or
C)is constant during the motion and in the Lagrangian context we say:
PR6 Every scleronomic and conservative system has the Harniltonian (or adjoint Lagrangian) as a first integral also called the Jacobi integral or Painlev&integral. This first integral is the mechanical energy.
We prove this last assertion. Let
'
be a constant.
The Euler theorem about the homogeneous functions applied to the homogeneous quadratic function Tof the generalized velocities q' (scleronomic case) is written:
Lecture 9
and thus
By transposing this result into Harniltonian formalism, we can say:
PR7
Every scleronomic and conservative system has the Hamiltonian (or mechanical energy) as a first integral:
H=E. D
A generalized coordinate q' is cyclic if it doesn't occur explicitly in the Lagrangian or in the Hamiltonian.
For every cyclic coordinate qi we clearly have since H = p, q' - L .
PR8
The generalized momentum that is conjugate to a cyclic coordinate is a first integral of the equations of motion.
Prmj In the Lagrangian formalism, we have:
3
pi = 0
and in the Hamiltonian formalism we write:
Relevance of cyclic coordinates
If a coordinate (e.g. q") is cyclic, then the conjugate momentum is a first integral ( p , = c ) . So the Hamiltonian is a function of other q i , of other pi,of an arbitrary constant c (determined by the initial conditions) and o f t , namely: So the problem comprises n - 1 generalized coordinates. Its solution will let us determine the
cyclic coordinate by integrating q"
aH dc
= -.
Lagraogian and Hamiltonian Mechanics
337
Remark 1. To integrate the canonical equations of Hamilton is to search the independent first integrals. If all these were known the solution of the problem would be reduced to a system of 2n finite equations. Remark 2. The first integrals are deduced from the canonical equations. So in the example of a central force field, we have:
which implies that
is a constant during the motion. We find again a well-known result since the coordinate 6' is cyclic. Remark 3. It turns out to be necessary to choose a coordinate system that gves the maximum of cyclic coordinates, this choice making the most of symmetries of the problem.
In the previous example, no cartesian coordinate would be cyclic.
2.5 PHASE SPACE
D
Q=
A cotangent vector (or covector) to the configuration space Q at point q = ( q i , . . . , q n ) is a 1-form of T,'Q such that its coordinates (or components) are the generalized
momenta p,. By remembering that the cotangent space at point q is the space of cotangent vectors to Q at point q, we say:
D
*
The momturnphase space is the cotangent bundle of the configuration space Q:
Consider n local coordinates qi on Q and n components pi of a 1-form in a natural basis
(4'). The 2n coordinates q' and p, being independent and for own pedagogical reasons, we denote the momentum phase space :
w
T*Q={(p,,qi): (4' ,..., q n ) ~ Q , ( p,..., , p.>€Tq*Q1.
Remark. For each value of the mechanical energy IY, the equation H = E defines a hypersurface in the momentum phase space. In this case, the representative point of the
Lecture 9
338
D
The momentum phase spacetlme or sfate space is the cartesian product
T * Q x R = { ( p , q , t ) :~ E Q , P E T , ' Q , ~1E R
w
where t belongs to the time axis.
3. D'ALEMBERT - LAGRANGE PRINCIPLE LAGRANGE EQUATIONS 3.1
-
D'ALEMBERT-LAGRANGE PRINCIPLE
The study of constraints is introduced in the courses of analytical mechanics that present the d'Alembert-Lagrange method from the notion of virtual displacements consistent with the constraints. We recaIl this method obtaining the Lagrange equations and we begin with two examples.
3.1.1 Particle constrained to move on a surface Consider a point of mass m moving on a surface Q, and let
be a motion between two configurations
x ( t , ) = x,
and
x(t,) = x, .
The idea of d'Alembert has been to make Dynamics similar to Statics which is presently
written: f-mx=O
where f is the resultant of applied forces, namely f = F + L is the vector sum of known forces F and forces L of constraints. Later the concept of virtual displacement was used by Lagrange. This displacement is a priori an arbitrary vector denoted S x . It is called virtual displacement because the time is fixed (st = 0).
So, we write:
(f- m i ) . S x = O
In addition, the constraints are supposed perfect; that is the virtual displacements, tangent to Q,,do not contribute to the virtual work, namely the virtual work of forces of constraints vanishes: L . 6 x = 0.
So, the virtual displacements are judiciously chosen and we can say:
Lagrangian and Harniltonian Mechanics PR9
339
The d'Alembert-lagrange principle consists in choosing virtual displacements Sx E Tq([)Qz for which the (unknown) constraint forces do not virtually work.
From this important principle we deduce that (F-m?).Sx=O.
The unknown forces of constraint do not appear any more! We are going to prove the d'Alembert-Lagrange and Hamilton principles are equivalent. We recall that a configuration space trajectory between two instants t, and t, is an extremal concerning the following action integral
(we consider the problems with potentials). PRlO A trajectory x : R -,Q, : t I+ x(t) satisfies the Hamilton principle iff it verifies the d' Alembert-Lagrange principle.
We consider the curves x and x -t- b x . For any &(ti)= &(t,) = 0 we have:
ProoJ:
Sx(t) E T,(,)Q2
and by integrating by parts, we have:
The principle of Hamilton:
is equivalent to the d'Alembert-Lagrange principle:
(F-mx).bx=O
av where F = -ax
represents the resultant of the known applied forces.
Indeed, from the above-mentioned lemma of calculus of variations, we have:
3.1.2 Systems of particles with constraints Consider a system of N particles ph with constraints.
such that
Lecture 9
340
If the number of constraints is k, then the material system shows n = 3N - k degrees of freedom. Given any virtual displacement of point ph, denoted by
and fulfilling the d'Alembert-Lagrange principle, we have:
where and F,"' respectively represent the resultants of external and internal forces applied to ph. The equivalence of the Hamilton principle and the d'Alembert-Lagrange principle will be proved as before.
3.2
LAGRANGE EQUATIONS
The Lagrange equations are deduced from the principle of d' Alembert-Lagrange. Indeed, the equations of this principle are equivalent to the system of equations
because the Sq' are arbitrarily independent variations. This system of equations is also written:
where each
Qi
is the genemlizedforce defined by
Finally, by considering the kinetic energy
obvious developments of elementary analytical mechanics lead to the well-known Lagrange equations:
By introducing the Lagrangian, the Lagrange equations are written:
Lagrangian and Hamiltonha Mechanics
PRI 1 The Lagrange equations are invariable under changes of generalized coordinates. Proof Consider a change of generalized coordinates denoted by
If we denote the variational derivative of L with respect to q i:
and if we put
then it is easily proved ':
Remark also that the variational derivative is a tensor of type ( ). The proposition follows from the preceding, namely:
6L -0 6qi 3.3
- SL' =O. Sq"
EULER-NOETHER THEOREM Courses in mechanics for senior undergraduate students show the foIlowing principle: Law of
Invariance under
s any group of transformations
conservation.
For instance: Invariance under a group of
Conservation of angular
rotations leaving an axis fixed
momentum about the axis.
We are going to introduce a general theorem which plays a fundamental role in the Lagrangian theory and that Euler used evidently under another form. This so-called Euler-
For instance in our book: "Mecanique generale et analytique."
Lecture 9
342
Noether theorem lets us associate a first integral of Lagrange equations to any one-parameter
group of diffeomorphisms (on the configuration space) which conserves the Lagrangian. Let Q be a configuration space, L : TQ + R : ( q i , q i ) t,~ ( ~ ' , c j ' )be a (differentiable) Lagrangian.
D
A diffeomorphism 4 : Q -,Q is admissible for the space Q provided with L if
V(x, X,)
E
TQ : L(d4 X,.) = L(X,).
Example. Given
Q2={(x,y): ~ E R , Y E R )
then a translation ofy , namely Vs E R :
A ; Q2
+Q2
:(X,Y>H(~,Y+~)
is admissible. PR12 If a one-parameter group of diffeomorphisms is admissible for a configuration manifold Q, then there is, at least, a first integral of the Lagrangian equations.
ProoJ: Let
be a one-parameter group of diffeomorphisms, c : R + Q : t H c(6) be an integral curve of Lagrange equations.
#s
Since 4s is admissible for Q (with L), then every (transformed) curve is also solution of the Lagrange equations.
Consider
a> : R x R +Q :(s, t) H @(s,t) = #s(c(t)).
The mapping 4soc:R+Q:f
verifies the Lagrange equations. By putting 9 = @(5,f)
we have:
H$~(c(~))
Lagrangian and Hamiltonian Mechanics The ({ admissibility )) of diffeomorphisms 4simplies:
and thus
So, at point c ( t ) , the first integral is
The Euter-Noether theorem is so proved.
3.4
MOTION EQUATIONS ON RIEMANNIAN MANIFOLDS
In the Riemannian context, we are going to establish the motion equations of mechanical systems and find again the geodesic equations.
Let Q be the configuration space of a mechanical probIem. In a general coordinate system (q') a metric ( , ) is defined by V v = (9',...,qn)E TqQ:
We denote 4 = (4' ,-, 4 " )
q = ( 4I
,...,an).
A Lagrangian
L :TQ -+ R : ( 9 4 ) H L(q,q) is defined by
L=T-V where V:Q+R:qwV(q) is the potential. PR13 An integral curve c is satisfying Lagrange equations
iff
V t i (t) = -grad V (c(t)).
Proof: Let c : 1 + Q : t H ~ ( t=)(q'(t),...,qn(t))
be an integral curve of the Lagrange equations:
Lecture 9
0
guqJ+("-'-
aqk
ag,k)'j,qkE
dV
dq'
a4'
By multiplying by gri, we obtain the n general equations of motion: w
qr +I-;
gjqk
= -gHdlv.
(9-16)
The right-hand member represents the rth component a' of the gravitational acceleration. In particular, if the gravitational acceleration vanishes, then these equations are the wellknown equations of geodesics. This confirms the result of lecture 8; namely: the geodesics are the curves followed by points of zero acceleration.
4. CANONICAL TRANSFORMATIONS AND INTEGRAL INVARIANTS 4.1
DIFFEOMORPFITSMS ON PEASE SPACETIME
For readers who are not familiar with developments in mechanics, this section is devoted to notions which will later be introduced within the modern context of symplectic geometry. The Hamiltonian mechanics dates from 191h century. The canonical equations of Hamilton play a fundamental role particularly in celestial mechanics and in quantum theory. In classical mechanics, the Hamiltonian theory doesn't greatly simpliQ the differential equations of a problem in comparison with the Lagrangian version. Nevertheless, the Hamiltonian formalism introduces the following advantage. The Lagrange equations can be simplified by transformations concerned by the only variables q i ; on the contrary the Hamilton equations can be simplified by transformations concerned by the variables pi, q'. We also specifL that we'iI have to search for transformations preserving the canonical form of Hamilton equations which is not the case with the Lagrange equations (see PRl1). This and other comments will be developed in the ulterior context of symplectic manifolds. In this section, integral invariants will be considered with respect to the state space T'Q x R , but also with respect to the phase space T'Q (at a given instant). PR14 Any diffeomorphism on the phase spacetime doesn't preserve the form of Hamilton equations.
ProoJ: Consider the canonical system of Hamilton equations
Lagrangian and Hamiltonian Mechanics
and the following diffeornorphism
e1= Q ' ( P ~ , ~ ~ J )
4 = 4(p,,qi,t)
In the "new" coordinates the Hamiltonian H is written as foilows: K(P,,Q1,t>= ~(p~(~,,Q',t),q'(~,,~',t>,t). Firstly, we introduce the definition of the Poisson bracket.
D
*
The Poisson bracket of functions f (pi,q', f ) and g(pi,qi,t) is the function defined
by
Under the diffeornorphism the Hamilton equations take the following form:
In the same manner we obtain:
However the canonical form of Hamilton equations is preserved after transformation rf there is a function p(4 ,Q1,t) such that
and we notice it is not always the case, which proves the proposition.
Now, the question is to know under what conditions the canonical form is preserved. PRI 5 A diffeomorphism defined by P, = P, (P,, q i , f ) and Q' = canonical form of Hamilton equations ifl
{ Q ' , Q ~>={M 3
=
{Q1,4 >=a: and if there is a function ~ ( 4 Q' ,,f )
,qi,t) preserves the
0
(9 - 18) ( 0 (P,,QJ>=-6;
such that
Lecture 9
Proof In this case, we actually have:
with the following transformed Harniltonian
We will go back to this question in the lecture dealing with symplectic diffeomorphisms. The integral invariants were introduced in lesson 7 and we are going to show their sigmficance in mechanics by refering to H. Poincark.
4.2
INTEGRAL INVARIANTS We first recall the notion of universe velociry.
Given a system of n local coordinates ( x ' ) on a manifold, we know that to a vector field X are associated integral curves of the following system:
In the nonstationary (also called rheonomic) case, time t explicitly appears and putting the real x0 = t , we have a similar system, namely:
So, next to the spatial velocity vector of components
'
there is the universe velocity of components :
' Let us recall that in special relativity, we have considered the metric element dsZ and and so
that is
dra -=2 ds
rush that gd
0'0 ' = 1 (dependence o f universe velocities).
Lagrangian and Hamiltonian Mechanics
and we denote
u = (U") = (l,X i j . Example. The Lie derivative of a function g with respect to U is written
Remark, A system of local coordinates universe velocity field has components:
(2")
can always be introduced that is such that the
which implies the existence of n first integrals zi. Now, we recall that a differential form w is an absolute integral invariant concerning the differential system
D
An integral invariant is called complete if the term dt appears; without dt it is called truncated.
PR16 If the form
ru=qbiP,dr4 /\...A&'
is a truncated absolute integral invariant, then the associated form
0 = m(il.,.ipl (k''- Xildt)A ... A (hip - xiudt) is a complete absolute integral invariant.
ProoJ: Firstly, we establish that R=w-dt~i,w.
Indeed, =
ip
(dx' - X i' dt) A (dr" - X l2 dt) A ... A (dr - x '.dl)
Lecture 9 Now we calculate &R . Since = &w - L,(dt = -dt
and since
A ixm) =
-L,(dt
A hixu
A i,w)
( because &dt = d &t = dl = 0 )
&(ixw) = (dlx+ ixd)i,o = i,di,o = -i,i,dw
( because (i,d
+ di, )w = 0 )
=o, we can conclude that
4.3
INTEGRAL INVARIANTS AND CANONICAL TRANSFORMATIONS
PR17 The Hamilton equations show a truncated absolute integral invariant, namely:
w
w =dpi~dqi
(9-20)
which is invariant with regard to the only canonical system of Hamilton equations.
ProoJ: Firstly we must prove that the Lie derivative of w with respect to vector field X (tangent to integral curves of Hamilton equations) is zero.
It is sufficient to prove that since dm = 0 and L,w = ixdw+ d i x w . Consider the canonical equations of Hamilton:
with the Harniltonian function H : T'Q
+R : (pi,qiH ) H(p,,q ). i
The vector field X is
wherep and q represent the respective systems of local coordinates ( p i ) and ( q i ) of points of phase space. So, the components of X constitute the 2n-tuple:
also denoted (X,,X'>
Lagrangian and Hamiltonian Mechanics So, we obtain: t,m = I, (dp, A ndq') = -
8H
aq
dH dq' --dp, a
~
i
This implies d1,o = 0.
Secondly, given w = dp, A dqi , is there a system of differential equations of type:
for which o is an absolute integral invariant? Yes, we are going to establish that the truncated absolute integral invariant is the one for the only canonical system of Hamilton. We must thus prove if a field X is such that Lxw = 0,then there is a function H(p,,ql) verifying the canonical equations of Hamilton (along the integral curves). We have: L,W = dixm = 0 and, the conditions of Poincare lemma being hlfilled in the phase space, there is a function H such that i X 0 = -dH . (9-2 1) From equalities ix w = 4 dq - Q dpi i
i
we deduce:
where His the Hamiltonian. The proposition is so proved. D
*
A vector field X verifying i,w
=
-dH is called Hamiltonian veciorfiefd.
PR18 The differential form i2=dpi~ d ~ ' - d H ~ d t is the complete absolute integral invariant for the canonical system of Hamilton. Proof: This immediately follows from PR16 which asserts that to the truncated absolute
integral invariant is associated the following complete absolute integral invariant: 0 = (dp,- Xi&) A (dq i - X'dt)
Lecture 9
350
Exercise. The reader will easily prove that
is a complete absolute integral invariant for canonical equations of Hamilton, the vector field aH aH U being (X,,xi,l), that is (--,I). d4' ' 'pi
He will look at the following: &,R = d i , 8 = d(0)= 0 .
PR 19 The differential form
A = pi dq'
6~
(9-23)
called Liouville form, is a truncated relative integral invariant for the canonical system of Hamilton. Proof: The differential dA = w is an absolute integral invariant and PR21 of lecture 7 leads
to that conclusion. The following proposition is easily proved. PR20 The differential form
A = pidql - ~ d t is a complete relative integral invariant of canonical system of Hamilton. Remark. Let us situate in the Lagrangian context the integral invariant A = p, dqi - Hdt
We have
Along a trajectory c of equations qi = q'(t) such that dq' = ¶'dt (that is solution of Lagrange equations), we find again the action integral:
D
*
A canonical transformation on the phase space T'Q is a differentiable mapping
T'Q
+ T'Q,
denoted by
4 = 4(p,,gi) which preserves the 2-form w = dp, A dq' .
Q' = Q ' ( P , ~ ~ ' )
Lagrangian and Hamiltonian Mechanics
So, given a differentiable mapping
we have the truncated absolute integral invariant concerning the canonical system of Hamilton:
We similarly consider the following differential transformation T'Q x R + T'Q x R defined by xrA= xfA(xB,t) t' = t such that the complete absolute integral invariant is
More explicitly, the previous transformation (sometimes called contact transformation)
denoted by
Q' = Q'(p,,ql,t)
6 = P,(pi,qiYt)
I = 1, ...,TI
is such that 0 = dpi ~ d q * - ~d ~d= & d ~A ,~ Q '-dKr\dt
where
H
is the transformed Hamiltonian.
Important remark. equations.
Every canonical transformation preserves the shape of Hamilton
Indeed, let us consider a transformation
4 = P,(pt,qfJ)
Q'
=
QI(P~,~~,~)
and the corresponding complete absolute integral invariant
a = d &~ d ~ ' - d p ~ d t . y
The associated vector field being X = (x,, Y',I) in the "new ' coordinates, we have: 0 = iXfl = i(Xr,Y I , , ) (dP, A d ~ -' dR A dt) dH dH - I(x, ,yr , l ) (dp, A dQ1- -dP, A dt - -dQ1 A dt)
34
So the shape of Hamilton equations is preserved.
aQ'
1= 1, ...,TI
Lecture 9
352
On the other hand, every diffeomorphism preserving the canonical shape of Hamilton equations is not necessarily a canonical transformation.
So, the diffeomorphism defined by P=p
Q=29
preserves the canonical shape of Hamilton equations (with transformation.
H = 2H ), but is not a canonical
The important canonical transformation concept will be largely developed in the next lesson.
4.4
LIOUVILLE THEOREM The absolute integraI invariant
for the canonical system of Hamilton means the integral of w over an eIementary domain of the phase space is constant when this domain loses its shape along the phase orbits induced by the differential system of Hamilton.
The condition L,w
=o
is the expression of the Liouville theorem which is written in undergraduate courses as follows:
This expresses the invariance of the volume of the phase space domain. The phase space is sometimes called incompressible.
5. THE N-BODY PROBLEM
AND
A PROBLEM OF STATISTICAL MECHANICS Let us use the stellar dynamics context to show the N -body problem and to gve an original method allowing the evolution study of densely populated systems (N > 106). Stellar dynamics is the branch of mathematical astronomy which attempts to establish the laws of structure, motion and evolution of stellar systems, such as galaxies and clusters.
Star motions are essentially governed by the gravitational forces; others such as radiation pressure, electromagnetic fields, etc, are generally neglected in stellar dynamics. The gravitational forces are essentially caused by the masses of the system (internal forces). In addition, the star dimensions being small by comparison with interstellar distances, we can compare the stars to particles.
Lagrangian and Hamiltonian Mechanics 5.1
N-BODY PROBLEM AND FUNDAMENTAL EQUATIONS
5.1.1
N-body problem
The N-body problem consists in determining the trajectories of N particles interacting in accordance with the gravitational law of Newton. Let us show the equations. In an inertial frame of reference oxyz, we consider N points P, of masses m,, and coordinates (X~SY~>Z~). Each point 4 is attracted by the N - 1 other points P, with a force
where
We evidently have:
The projections onto the coordinate axes lead to 3N differential equations, namely: dt
b;;
((xi -
Gmimj(x, - x,) + ( y , - yi)2+ (z, -
and others in y and z. The force function is written N
u =G2- C {>ii
mi mj
((x,
- Xi12 + ( y , - y,)l
+(z, - z , ) ~ ) I / ~
(9-2 5 )
where the sum is concerned with N ( N - 1) arrangements without repetition taken 2 by 2 (e.g. if N = 4 , then 12, 13, 14 ; 21,23,24; 31,32,34; 41,42,43). Since for each value of k we have:
then the equations of motion are
PR22 If the forces are conservative (that is derivable from a force function U ) then the differential equation of the dynamics can be put in a canonical form.
Lecture 9
3 54
Proof. Consider a system of N points subject to the gravitational law in It3 and let the respective Cartesian coordinates denote
The kinetic energy of the system is
where the common coefficient m, = m, = m, represents the mass of the first point and so on.
By putting y
,k
k
(no summation of course!)
=mkX
we obtain k-l
mk
So, the system of motion equations (k
= 1, ...,3 N ):
is equivalent to the following system
y R = - a(T - U ) ax'
'
If we introduce the hnction
F=T-U
then the system of motion equations takes the canonical form:
by recalling that the position of indices has no significance, it is only a notation to designate 3N coordinates. Sometimes F is a called characseristicfunction and
are the conjugate variables.
We have obtained a system of 6N differential equations of first order and there are 6N unknowns: the positions and velocities of points. A priori 6N integrations are thus necessary, but some first integrals can be known.
Lagrangian and Hamiltonian Mechanics 5.1.2
Poisson equation
The previous developments show the beginning of celestial mechanics theory. In stellar dynamics, the potential function V is used instead of U : v=-U. Let ?(t) and C(t) be respectively the position vector and the velocity vector of a star in the gravitational field of N stars located by position vectors c(t) . The motion equations of this star are T=C
where
-
? = -V Vefl(F, T)
i = 1, ..., N
denotes the usual gradient and Vg(r',f) is the "efTective" potential per unit mass:
Since the theory of Dirac distributions shows the following Laplacian
then the "effective" potential satisfies the Poisson equation:
The stellar systems are discontinuous. The density p (mass per unit volume) and the potential are not continuous functions.
Figure 52
Lecture 9
356
The figure shows that the (real) density is nonzero at each stellar point and elsewhere vanishes. A denseIy populated stellar system can be viewed as a "cloud" of points for which can be introduced a smooth density and a mean potential (e.g. N = lo9 for globular clusters and N = lo1' for galaxies). So the smooth density is illustrated by a continuous and decreasing curve. This process is justified if N is large because it can be proved the (real) potential is a "random" fimction such that the first order moment is the mean potential and the higher order moments can be neglected for large N. Therefore, we can use the notions of the statiktical mechanics. So, we consider a distribution function f (F,G,t) in the phase space. The number of particles included in the infinitesimal cell of the phase space, such that T E (F,F + d7) and i j E (3,? + &) is evidently (at instant t ) : with the total number of particles: IJfdifi=~. The (mass) density is
j
p(r;r)= rn f ( ~ , ~ , t ) &
where m is the (average) individual mass. The total mass of points is M = / ~ & .
The gravitationa1 potential
must verify the Poisson equation: &"
A V ( 7 , t ) = 4nGp(F,t)
with the boundary condition:
lim I/(?, t ) = 0
1q + m
( p(r',t)= 0 to infinity).
We mention that realistic models of material systems must take into account the various masses and thus the introduction of a suitable mass function is an additional difficulty; fortunately, consideration of a few mass groups is sufficient.
Lagraagian and Hamiltonian Mechanics 5.1.3
Liouville-Boltzmann (or continuity) equation
Consider the representative point of a star in the phase space. In this 6-dimensional phase space we view a "cloud" of points in the neighborhood of the previous representative point. In the phase space, the points of the cloud represent particles (stars of the stellar system) of which respective positions and respective velocities are close together.
Figure 53
Each representative point of the cloud moves along a phase trajectory according to the canonical equations of Hamilton. Let f ( p ,, q i ,I ) be the frequency hnction of representative points of the cloud and dp, A dq' be the volume of the cloud.
In different situations we can consider that the number of representative points of a cell moving in the phase space is invariable. It is the case if there is no capture and no evasion in a moving phase cell or if the captures compensate the escapes; then, the system is said "in dynamic equilibrium. "
The number of points of the moving domain
f (~i>q'>t)dpi ~'9' being invariant during the evolution of the domain, we have:
where X is the well-known vector field tangent to integral curves of the Hamilton canonical system. The Liouville theorem Lx(dpi A dq')= 0
implies
Lecture 9
It is the Liouville-Boltzmann equation (by refering to the conservation of phase volume). This equation will be studied in the next section.
5.2
A PROBLEM OF STATISTICAL MECHANICS
5.2.1 Fundamental equations
One of the most fundamental problems of physics developed in statistical mechanics is to resolve the Boltzmann equation coupled with the Poisson equation. We recall it. The phase density or (velocity and position) distribution function f in the phase space is classically connected to the number of particles by: = f (q,v,t)
[ with q = + dq ;
+
We say that f (q,
and v these particles being included in an infinitesimal cell of the phase space at the moment t.
is the number of particles per unit phase volume.
If the potential of a system of particles is smooth then the distribution function f is continually varying and the infinitesimal cell is continuously deforming along the phase space trajectories. Given a metric element ds2 =
where
=
dqi
these trajectories are defined by [see
- are the components of the gravitational acceleration.
In a phase cell the number of points varies under the action of "regular forces" in the gravitational field, but also under the action of "irregular forces" due to encounters between particles. Encounter means that close particles interact through their gravitational attraction without contact, this last phenomenon being rare since star dimensions are small compared to their mutual distances. The literature unfortunately uses "encounter"" indeed "collision, instead of "interaction. " We denote by (c - e) or by (-)
the number of points which are captured or have escaped
with regard to a moving cell during time
and due to encounters.
Lagrangian and Hamiltonian Mechanics
359
Taking the two previous phenomena into account, one must set the total rate of cell density change is equal to the density change due to encounters. The Boltzmann equation is then obtained:
It shows the variation of phase points, that is explicitly: -
+
f
+a
f (a i -
In particular, if the encounter effect can be neglected and by considering the Hamiltonian flow then the distribution function f
verifies the following equation:
that is the Liouville-Boltzmannequation:
which is also called collisionless Boltzmannequation [ see (9-34) ]. It is a fundamental equation of stellar dynamics and it can be helpful in other domains like biology. In 1915, Jeans already showed that the effect of binary encounters in our Galaxy could be neglected; only the forces exerted by the main body of our Galaxy were important. He concluded it was "Boltzmann's well-known equation with the collisions left out" that could be employed. Jeans indicated the potential V was connected to the density = f by the Poisson equation: The last three equations form the well-known equations of the Jeans approach. Rigorous!y, the precedent fundamental problem must be treated with different masses; the total distribution function is
where each distribution function must verify an integrodifferential system:
Lecture 9
60
The Liouville-Boltrmann equation use is correct in the dynamic study of our Galaxy, on the contrary it isn't in the (globular) star clusters where the encounters modify the content of the moving cell. Nevertheless, in the cluster core, where the most of matter is, one can consider that the cumulative effect of encounters into any infinitesimal cell is zero. The evasions and captures are statistically compensated. A quasi-equilibrium state is attained which very slowly changes. Then the Liouville-Boltzmann equation can be also employed. In 1938, in plasma physics, Vlasov obtained equations which are structurally equivalent to the three equations of the Jeans approach, the gravitational field being replaced by the etectromagnetic field. The physics of a system of particles interacting according to Coulomb's law is not similar to the theory of gases. In a gas, short-range (intermolecular) forces are found, while in plasma physics (like in galactic dynamics) the forces decrease slowly with the distance (long-range forces -1/r2 ); isolated binary encounters do not describe the system evolution. Each particle interacts with the whole particle set. Vlasov has also shown the term
should be removed. In plasma physics, the collisionless Boltzmann equation associated with Maxwell equations constitute together the Vlasov approach. The two preceding approaches are formally similar!
5.2.2
An original fluid-dynamical approach
When I was studying populated spherical star systems, I introduced the Fourier transform of the Boltzmann equation to obtain a general hydrodynamical system. -
A general system of moment equations
Let =
f
be the Fourier transform of the distribution function f on the velocity space and where represents the "kinemaricfrequency. "
We are going to calculate the Fourier transform of (9-35), namely:
We successively have:
Lagrangian and Hamiltonian Mechanics
From these results and by putting: and
,
-
,
then the Fourier transform of the Boltzmann equation is:
By introducing the successive local moments: f [ in particular:
= f (q,t )
we can replace
by a series:
with
In the same manner, by considering the Fourier transform =
and by defining:
we replace
by
Here, = respective masses.
if we consider that every encounter between particles doesn't change the
Lecture 9
362
In addition, iy' "elastic": +
=
=
+
=
if we consider that every encounter between particles is (see Madelung, 1957).
By successive derivatives with respect to of terms of (9-37) and putting all the obtain a sequence of local moment equations. The first equation is (for
=
we
=0 ):
that is
The second
is for
=
The third
+
+
+
at
is for
=0: at
The fourth
+
+
a'
+
is for
+ at
ak
+a r
+a m
+
and so on.
As a general rule the previous equations are written:
where (p) represents the set of circular permutations. In order to make the most of these equations we are going to introduce two important definitions.
Lagrangian and Hamiltonian Mechanics
D
The velocity conditional distribution is defined (
t)
) by:
where the marginal distribution
represents the number of particles per unit spatial volume. Note that the normalization condition is fulfilled concerning
:
=
By denoting the "metric volume" :
D
=
, we say:
The number density of particles is the number of particles per unit "metric volume," that is:
Because the (velocity and position) distribution function f is
and by introducing the moments of the velocity conditional distribution also named moments around the origin: =
...vm hq(v, t ) d(v)
= (VJ vk...vm),
then we immediately see that: r~k.,.m =
So, knowing that V,
where
fi
u ~ k
a f i = 0 and &= 0 , the system of equations (9-39) becomes: at
Lecture 9
364
In these equations it is more judicious to directly express the moments ukl-" fiom successive moments around the mean which are:
also denoted: = ((vk- uk)(vl- u')...(v m - urn))
We introduce the residual velocities: Wk
= vk
-U
k
and since (d)= 0 and vi = wi + u' , we immediately obtain:
and as a general rule (since nk= 0 ):
So the hierarchy of moment equations (9-43) is in a general writing: d,p + V j ( pu J )= 0
(9-45a)
and so on. The first equation is the continuity equation, the second expresses the fluid's movement submitted to internal tensions pa&,the other equations relate other higher moments.
Lagrangian and Hamiltonian Mechanics
365
So the hydrodynamical approach lets us know the evolution of the most important physical parameters of sets of particles such as number density, mean radial velocity, etc., which give all the information about the evolution.
-
Closure of the hydrodynamical system
The fluid-dynamical approach always presents more unknowns (moments) than equations. Therefore, it is necessary to introduce supplementary equations to close the hydrodynamical system. A hydrodynamical approach was developed and exploited by R.B. Larson in the case of spherical systems. The Larson method imposes the velocity deviates by only a small amount from a Maxwellian distribution. In this case of a nearly Maxwellian velocity distribution developed with a Legendre polynomials series it is possible to close the hydrodynamical system.
Unlike the Larson method, our proposed method is general and takes the two third-order moments (not only the radial one) and all the fourth-order moments into account. It is necessary to limit the hydrodynamical system to a finite number of equations without neglecting a fundamental physical effect. Because the evolution of a system of particles is largely due to the effect of an energy flux (heat flow in a gas), then the radial third-order moment representing the radial energy flow plays a very important role in the evolution. Thus one cannot cut the system with the equations governing the evolution of the third-order moments. It is necessary to limit the system of moment equations at a higher order.
We propose the following closure. Let z be a volume limited by a surface S and driven in the general motion of (star) system with mean velocity. The trajectories of individual particles, entering z, produce a variation of the internal tensions by the flow of the fifth-order mechanical tensions through the surface S:
By remembering the continuity equation expressing the integral invariance of the barycentric movement, we suggest to write an analogous relation for higher orders such that
that is, the variation rate of fourth-order tensions creates a fifth-order tension flow. The derivative
d
- is taken following the barycentric movement.
dt
The coefficient K relates the fourth and fifth-order tensions. A priori, this coefficient of exchange of tensions depends on the k, 1, m, p, positions and time values. The previous relation is written:
Lecture 9
and by considering
v,(pn-)
= lim 7-0
-I s pnkhq dS, 1
then the closure equation is
If K could be known, this relation would close the hydrodynamical system. Unfortunately K cannot be determined. It can be postdated that K is a constant parameter during the evolution of a chosen model; that is a way we can get out of it. It's the only restriction of the theory. This remark suggests to cut the hierarchy of moment equations to higher orders in order to take the maximum number of physical parameters into account.
- Moment equations in spherical coordinates Using spherical coordinates r, 8, 4 ( resp. radial distance, latitude and longitude), the mean velocity components are u' = ( u ) ( r )
u 2 = (v)= 0
u3 =(w) = 0
u=i
v=r@
w = r#sin8 .
where For spherical symmetry systems the components (v) and (w) of the mean velocity in transverse direction are zero. This is also true for the moments of odd order in v and w. Because the two transverse coordinates play equivalent roles, the value of moments is unaltered by permuting v and w . The mean square random velocities in radial and transverse directions are
and the tensor of dispersions has the diagonal form:
To construct the moment equations, the well-known expressions of coordinates are used as well as
vkpi= piik= pi,
+r;pJ.
VkPU = p @ ; k = pu,k + r A pmJ + r; pin
ri
in spherical
Lagrangian and Hamiltonian Mechanics Putting the only nonzero moments of higher orders
((
n t l l = & =U -
(413)
and
and
I have finally obtained the general following hydrodynamical system:
Lecture 9
Indeed, the equations (9-45a) and (9-45b) immediately lead to equations (9-48a) and (9-48b). For k = I = 1 and k = I = 2 the equation (9-45c) leads to respective equations (9-48c) and (9-4 8d).
-
From (9-45d) and after several calculations, we obtain the equations (9-48e) for k = I = rn = 1 and (9-48f) for k = 1 = 2 and rn 1 . The last equation (9-45e) is exploited with the following respective values k=l=m=p=l;
k=l=l,m=p=2; k=l=m=p=2, So, rather hard calculations lead to the last three equations (9-48). The second members of the moment equations describing the encounter effects (relaxation terms) are usually calculated from the Fokker-Planck equation. They were calculated by the Larson method for which the choice of a nearly Maxwellian velocity distribution is correct because the encounters are essentially present in the core (central part). So, from the Larson results and our calculations, the encounter terms are:
where T is the classical relaxation time
Lagrangian and Hamiltonian Mechanics
369
Dm,,denoting the dimensions of the regon where the relaxation effects are important and m the average mass. Boundary and initial conditions must be again chosen. If one of the most delicate points of the fluid-dynamical method is to cut the equations system, another is to replace partial derivatives by finite differences. The reader interested in numerical treatments of fluid-dynamical methods will refer for instance to papers of R.B. Larson (and others) in Monthly Notices Roy. Astron. Soc., Astronomy and Astrophysics, Astrophys. Space Sci., . ..
To conclude, I obtained fragmentary results with this method. I hope that different values of K do not lead to very important changes in the behavior of any chosen physical system. The problem is open and I hope the most important physical parameters such as density, mean radial velocity, square of the radial and transverse velocity dispersions, outward energy flux, etc, are not too sensitive to the choice of K. In that case, it should be necessary to carry on considering higher order moments. The method is all the more interesting since it would be usable in other scientific fields.
6. ISOLATING INTEGRALS To end this lecture the notion of the isolating integral is introduced. It is important to study star trajectories in our Galaxy. Ergodicity appears. 6.1
DEFINJTION AND EXAMPLES
Thefirst integral of a differential system of order 2n is a well-known notion. We recall that H. PoincarC used the most suitable name "invariant" since this suggests a function constant along each integral curve. , equations being of second order, there is a first So, in mechanics, in R ~differential integral I zfl
(for all solutions of motion equations). The Liouville-Boltzmannequation
means that the distribution function f is a first integral of motion equation (in phase space), that is the equation
f (p,q,t) = c
(cER)
defines, in the phase space, a hypersurface (fixed or in motion) to which the descriptive point (p,q) belongs at every time t ,
Lecture 9
370
Since 2n variables p,q define a descriptive point at every r, there are 2n hypersurfaces intersecting at a point, one of them being necessarily not fixed There are at the most 2n independent first integrals of motion equations and thus only 2n-1 explicitly independent of t : Make clear that this time-independent nature implies the corresponding hypersurfaces are fixed in the phase space. If m first integrals of this last type are known then the trajectory of the representative point in the phase space belongs to the intersection of m fixed surfaces in this 2n-dimensional space. The hypersurfaces which intersect at several points are not considered.
By considering a subset R of the phase space, we say:
D
A first integral I(p,q) is an i&olaa'ngintegrul in R if there are at Ieast two sets S and S' such that V XE S, ' d ~E' S' : J ( X )f
](X').
An isolating integral defines a hypersurface I = c ( c E R ) which splits the phase space into each other's inaccessible domains. If 2n isolating integrals are known: f , ( ~ 2 4 , t= ) c,
with
then the previous 2n equations can be solved in order to obtain the descriptive point orbit in the phase space by considering the following equations
where the various cj are the "constants of motion." So there is only one orbit of the descriptive point through a point x,(p,,qo) which is defined by
Conversely, given p, and conditions that are
at time to then the constants of motion are determined by 2n
c, = & ( P O , ~ O , ~ O ) .
Therefore, the knowledge of initial conditions is equivalent to the one of constants of motion. From one of the 2n- 1 first integrals I , ( p , q ) = c, (obtained by elimination of time) a variable can be made explicit, for instance: 4' 'f (P~,...,P~,~~,...,Q~,C~) .
Lagrangian and Hamiltonian Mechanics
371
By definition, if such a solution shows a limit point (accumulation at finite distance) then the first integral is not isolating. In particular, from the Bolzano-Weierstrass theorem, this solution q1will have at least one accumulation point at a finite distance if there is an infinite number of solutions in a finite domain. So a nonisolating integral is "a first integral which, if solved for one variable as a function of the others, gives an infinite number of values for this variable in a finite interval, for a range of the other variables." The hypersurface representing a nonisolating integral is composed of an infinity of sheets densely filling the phase space. Such a condition I = c doesn't give any physical condition and thus is not useful. On the other hand in the case of an isolating integral, there is only a finite number of sheets of the integral surface in any finite region; this type of integral is the only one taken into account. Example. In order to illustrate the nonisolating character of a first integral, let us consider a two-dimensional harmonic oscillator for which the equations of motion around the point of equilibrium (x = 0,z = 0) are
The general solution is
where a, 6, p,,p2 are independent integration constants. We evidently have:
B= a4Fc0s(fit+~,)
i = b&cos(,@t
+p2).
(2)
The phase space is four-dimensional and so there are at most three time-independent first integrals. Two are immediately:
which define 3-dimensional hypersurfaces in the phase space. From (3) we find with ( 1 ) and (2) the dependence of E, and E, on the integration constants:
The remaining time-independent first integral is obtained by elimination o f t in the general solution: 1
x
I3 = -arcsin--a fi
1
JZ; -
=
4'1
P2
arcsin- =- -b f i &
The first integral shows the variation of z in function of x .
7
372
Lecture 9
By putting
from the previous first integral we deduce:
is irrational, to any value of 5 The presence of orcsin is interesting. So if corresponds an infinite number of values 6 dense in [-1,+1]. The first integral I, is nonisolating. It doesn't impose supplementary constraint on the (spatial) trajectory of the oscillator which can fill the space delimited by the only first integrals E, and E, , namely:
On the other hand, if ,/?@ is rational, to any value of 5. corresponds a finite number of solutions in a finite domain. The first integral I, is isolating. It imposes a supplementary constraint. The spatial trajectory doesn't fill, in the (x,z)-space, the whole of the rectangular region specified by values of E, and E,; on the contrary it is a closed Lissajous figure corresponding to a periodic oscillation. Since, in the phase space, a nonisolating integral represents a hypersurface with an infinite number of sheets, it is not useful for a physicist (no information or condition!). The first integrals E, and E, are obviously isolating. We define:
D
A first integral 'j(ps9) = c,
is isolating in a subspace R, of the phase space if a finite number of values can be deduced (in every finitedimension domain of Q,) for each of variables pi,qi in function of cj E [c: ,c;] c R and other variables. The term isolating follows from this: an isolating integral can be recognized by its property of isoIating points on the solution curve from neighboring points in the phase space.
6.2
JEANS THEOREM
The distribution function f ( p , q , r ) is a first integral of canonical equations since the Liouville-Boltmann equation
Lagrangian and Hamiltonian Mechanics
3 73
is verified along the phase space orbits. In other words, the function f is an invariant function along these orbits. Consequently, the Jeans theorem is PR23 In the 6-dimensional phase space, the frequency distribution is a function of six first integrals of equations of motion:
f D
('19'..9'6)
'
A dynamical system is scleronomic or in a steady (or stacioreury) state if
a,f = o . In this case, f is a distribution function of only first integrals which do not depend explicitly on time (sometimes called scleronornicfirst integrals). The distribution function having a physical significance D. Lynden-Bell has proved that the only isolating integrals are to be taken into account.
6.3
STELLAR TRAJECTORIES JN THE GALAXY
Besides their own interest, the stellar trajectories allow asserting the eventual existence of isolating integrals. Let us consider stellar systems with axial symmetry like galaxies. It can be proved that such = 0 ). systems have necessarily, at equilibrium, a plane of symmetry (considering limp m Let us choose a cylindrical coordinate system (r,B,z) where r is the distance from the axis, 19 is the angle around this axis and z the distance from the plane. The equations of motion in this system are readily obtained from the following Lagrangian (per unit mass):
they are the Lagrange equations:
or the HamiIton canonical equations
In this scleronomic problem (stationary potential), the Lagrangian is explicitly independent of t and so the equations of motion always show a scleronomic first integral, namely the energy (Harniltonian) per unit mass:
Lecture 9
3 74
The Euler-Noether theorem gives an immediate proof of this, the time translations, #s(qi,t), being admissible. Another first integral follows immediately from equations of Lagrange or Hamilton (6 is a cyclic coordinate):
It is twice the area "swept out" by the radius vector (r,B,o) in the plane of symmetry, that is the z-component of the angular momentum or "constant of areas" also denoted by A (see exercise 3).
In 1916, Jeans supposed that the distribution function would be a function of only two (analytical and isolating) integrals I, and I,, improperly denoted by However all the observations show the dispersions of motions in r- and zdirections of the stars at any point (in a spatial element) are not equal. So the discrepancy between Jeans theorem and observations implies a giving up of Jeans function. The idea of the existence of a third isolating integral was put forward. The 1960s introduced a "third" integral, "third" because it has been derived as an integral in addition to the energy and angular momentum (see G. Contopoulos et al.). Before starting on this notion, let us show that the axial symmetry of our Galaxy permits to replace the study of 3-dmensional stellar orbits by 2-dimensional orbits in a moving plane which contains the star and the z-axis of rotation at each instant. The Lagrange equation
shows that the meridian plane which accompanies the star is not uniformly revolving. So, in the 3-dimensional configuration space the particle describes a 3-dimensional trajectory, but, in the meridian plane in rotation, this particle describes a meridian trajectory.
If we introduce the reducedpotential:
then the remaining equations of Lagrange
are written:
-
i: = -arv
The Hamiltonian (per unit mass)is
-
i' = -a,v.
Lagrangian and Hamiltonian Mechanics = * ( p ; + p;)
+ V(r,z).
In the meridian plane, the force components are
7is a supplementary (centripetal) force resulting from the use of a noninertial where r system and where the terms - a,V and - d,V are defined for instance from a simple function approximating the galactic potential near the Sun. The consideration of meridian orbits simplifies the problem. Then by knowing the initial position of a star, we determine its orbit in the 3-dimensional configuration space as follows: - its meridian trajectory is calculated from equations (9-50) by having chosen values of isolating integrals E and A ; - the angular position of the star is obtained from the following integration:
Numerical studies show there are several types of orbits. These orbits stay inside a torus which exceptionally may open and let stars escape.
Figure 54
/
So, for instance, some orbits fill a "torus box" and the corresponding meridian orbits densely fill a "curvilinear parallelogram" and are topologically equivalent to Lissajous figures.
6.4
THE THIRD INTEGRAL
To construct the "third" integral which designates the third isolating integral, G. Contopoulos has used the foliowing nonexistence theorem of H.Poincark:
Lecture 9
376
If H = H , +&Hi + c2H~ -I--.. is the Hamiltonian of a canonical system such that: Ho depends only on variables q i ,its Hessian with respect to q' is not zero and H , , H 2 ... , are ( 2 ~-periodic ) functions of pi, then it is impossible that the series ~=#,,+E#~+E~#~+... be integral of canonical equations, the 4 j ( p i , q i )being (2~)-periodicfunctions of p, and 4 being analytic uniform with respect to reals p,. So, under certain conditions it can be asserted that there is no third integral. However, such conditions are not always fulfilled! For instance, the Poincark theorem is not applicable to the following example.
Contopoulos model In the case of a galaxy with axis of symmetry, the following potential is chosen
where E is an arbitrary constant and x = r - r,, r, being the initial radial distance. This potential is obtained by truncating a series expansion of type:
The Hamiltonian per unit mass is written:
It is the energy integral or Hamiltonian of the canonical system:
z. = - dfl
pz =--
dz
~ P Z
also written:
N
=
dH
+ pf + px2+ QZ') + E(-xz2) = H,,+ E H, .
Under the following transformation p, = @
sin p,
p, =
sin p, ,
Lagrangian and Hamiltonian Mechanics
the Hamiltonian becomes
+aq'
with respect to q' and q2 being always zero, the The Hessian of H,= fiql Poincard theorem is not usable and thus nothing stands in the way of the formal construction of a third integral. We search for an integral # of the canonical system (9-51) of type where the various 4j are polynomials (in x, z and corresponding momenta) which are determined by the condition
{4.H
I = d,4
apxH+a,$ apzH-ap,4 a,H-aPz4 a,H = o
meaning that 4 must be a scleronomic first integral. Since H
=
Ho + E HI the condition becomes
This equation must be satisfied for a range of values of E (sufficiently small); thus each of the terms must vanish, that is:
(W,HoI=O
..... C4",Hll+{4n+,,Ho)=0
The first equation of the hierarchy means that unperturbed system ( E = 0)
4,,
.....
is a first integral of the following
These are the equations of motion of a two-dimensional harmonic oscillator around the equilibrium position (0,O).
The corresponding characteristic system
admits two immediate isolating integrals which are Ho and
40= $(p: + px2)
[ o r Ho -#o = +(p:+ Q z 2 ) ] . We notice that if the auxiliary variable t is considered then the general solution is given by
Lecture 9
and px =
cos \iji(t
- to )
Previously, we have specified that if elimination of time:
$@
1
p, =
JmJZit cos
(9-52)
.
is irrational then the first integral obtained by
26 x T,= -orc sin ,/P
1
fi
arcsin i2WK-4,)
is not isolating. From the second equation of the hierarchy:
we deduce:
dr - -=-dr dpx ----dfi - dB, -Px P, -Px -Qz -p,z
2
- dl
(with the three known first integrals H,, #o and To). We have also:
Il = J - P , z 2
dt
that is [from (9-52) 1:
--
1
4Q- P
((P - 2e)xzZ- 2xp3 + 2pXpzz).
So Contopoulos obtained the third integral
The adelphic integral has been obtained (before the " third integral) by E.T. Whittaker when searching a first integral of the canonical system with the following Hamiltonian in the case of two degrees of freedom:
where s,,s, are constants and the various Hi are such that: m
n
for i = rn + n : H,,, = (ql)? (q2)2cos(ap, + b p z ) where a,b are integers,rn - la1 and n - (bl are zero or even.
Lagrangian and Hamiltonian Mechanics
From the condition { @ , H ) increasing powers of
where each
@i
=
3 79
0 , Whittaker has obtained a first integral developable in
@ and @ , and periodic with respect to p, and p,
such that:
has the same properties as H,.
Whittaker has proved that the adelphic integral (so called because it is quite similar to the energy) is associated to the (infinitesimal) adelphic transformation which changes any trajectory into a close other such that every periodic solution is transformed into a close periodic solution of the same period (and same energy). It is straightforwardly proved that the Contopoulos integral is an adelphic integral.
6.5
INVARIANT CURVE AND THIRlD XNTEGRAL EXISTENCE
In the study of 3-dimensional orbits of stars in a galaxy with an axisymrnetric mass distribution, we know that (per unit mass) the energy E and the z-component A of the angular momentum vector are isolating. The question of the existence of a third isolating integral is open. By using the isolating integral A , the 3-dimensional problem is reduced to one describing the motion of stars in a (meridian) plane. In the corresponding Cdimensional phase space if the time does not appear explicitly, then there are at most three independent scleronomic integrals of the motion (different from A ). One of them is the well-known isolating integral:
and the problem is to know if there exists an additional isolating integral. Let us introduce a method that can determine this. The existence of the isolating integral E implies it is sufficient to know three coordinates of the representative point in the phase space to obtain (from E ) the fourth 2 for example. The trajectory of this point can be followed in a 3-dimensional subspace of coordinates ( x , x , z ). Since i2 2 0 then the obvious condition
defines a bounded domain R If there is no isolating integral different from E, we say: D
If the trajectory fills the previous bounded domain 0 , then the trajectory is called ergodic .
In the phase space of points ( x , z , f , i . ) , the 3-dimensional hypersurfaces of the following equations: H(x,z,x,z) = E @(x,z,x, i) = C respectively "contain" all the trajectories of same energy E and all the ones having the same isolating "third" integral C .
380
Lecture 9
In the phase space, the intersection of the hypersurfaces H = E and # = C is a surface on which lie the trajectories characterized by E and C and whose respective initial points are on it. Its equation is obtained by solving the equation H = E for one of the variables, for example: and by inserting this z in the equation # = C . Then the surface equation is
This really defines a surface in the space of coordinates ( x , z , i ) . Different studies characterizing such a surface lead to a better knowledge of 3-dimensional orbits and ergodicity. The melhod of the surJace of section doesn't consist in viewing the trajectory in the phase space but precisely the successive intersections of the trajectory with certain surfaces. This simplification safeguards the essential properties of the phase space trajectory following from the study of the sequence of points so obtained. So, in particular, by intersecting the surface H fl4 by the "surface of section" which is for example the plane z = 0,we obtain a curve of equations (choosing y > 0 ):
This curve which is the locus of successive points pi (intersections of the "phase trajectory" by the "plane of section" ) is called the invariant curve. It is so called because it remains invariant under the transformation M that, in the plane of section, brings any previous point p,-, to a next point p, (consequent of p,-, ):
Lagrangian and Hamiltonian Mechanics
Generally speaking, for an infinite time there will be an infinite sequence of points p, Therefore there is a simple criterion to prove the possible existence of a third integral 4 , namely: If the successive points of intersection piof the phase trajectory with the plane of section ( z = 0) lie on an invariant curve, then there exists a third isolating integral. In the opposite case, by following the phase trajectory with a known energy E during a longer and longer time these points p, will fill an area determined by the condition:
which represents the intersection of domain R by the plane of section. The reader interested by this subject will be aware that the main authors of analytical works in this domain are PoincarC, Birkhoff, Siege1 Moser, Arnold, etc.
The third integral studied by Contopoulos and his collaborators (and others) provides a typical example of a numerical study of the above-mentioned method. The reader will refer to astronomical reviews (among others) such as Astronomical Journal, Astrophysical Journal, Astronomy and Asfropkysics. It is clearly established that a "dissolution" of the invariant curves appears as the energy increases. This phenomenon of transition from "isolating case" to "ergodic case" can be justified by the fact that the third integral is a series which would stop converging for no Ionger small perturbations. Then the third integral would be without usefulness. For a preliminary study at this level, the reader can refer to the advanced course "Dynamical structure and evolution of stellar systems " where Contopoulos shows the bases of topological methods and where the third integral breakdown and the dissolution of invariant curves are justified from the phenomenon of resonance interaction.
7. EXERCISES Exercise 1.
Deduce the canonical equations of Hamilton from the principle of least action. Answer. The principle of Hamilton is successively written:
0 = 6 r (p,q' - H)dt I
=
% (&,tj' + p,6rji - aq.H 6pi - i3,H b j ~ ~ ) d t
Lecture 9
Since this integral vanishes and the various variations canonical equations of Hamilton are obtained.
@j
and Sijl are independent, the 2n
Exercise 2.
Prove that a function f of time and spatial coordinates x' is a first integral of the system
dua -= ua( x a ) dt
[
(U") being the universe velocity ]
t f l its differential is a linear combination of hi- xidf .
Answer. Since
then $ is a first integral 18
Exercise 3.
Prove that if rotations about an axis oz are admissible (invariance under a group of rotations leaving the axis points fixed), then the angular momentum about the axis, denoted Hzis constant.
I:!
Answer. Let rl, = y be the position vector of a point ph For any rotation
+o
of angle 0 about oz, we obtain:
The Euler-Noether theorem leads to the following first integral:
Exercise 4.
Find the expression of the Liouville-Boltzmann equation in cylindrical coordinates by using the Harniltonian formalism.
Lagrangian and Hamiltonian Mechanics Answer. By using the cylindrical coordinates of some particle:
q1 = r
q2
=6'
93
=Z,
then the corresponding Lagrangian (per unit mass) is
L = 3 ( i 2- 1 - r ~ +8 j~2 ) - V(r,O,z,t). Since the generalized momentum components are p, = i.
' p s = r 2 13
p, = i
the Hamiltonian (per unit mass) is written:
H=p,qi-L=f(p~+p~/r2+p~)+~(r,B,z,t)
and the canonical equations of motion are
The Liouville-Boltzmann equation is thus
Exercise 5 Consider a potential
where A is the (constant) 2-component of the angular momentum vector, x = r - r, [ r and r, being the respective distances of the star and the sun from the gaIactic center (r = 0,z = 0) 1, P and Q are well-determined constants and are panmeters small enough for the two last terms be considered as perturbation terms. Construct the third integral. Answer. The method showed in this lecture (see Contopoulos model) leads to the following series
Lecture 9
384
That's the question of knowing whether the third integral is convergent. It's a difficult problem because of the presence of divisors that may become arbitrarily small. Several numerical results are a point in favor of convergence. Exercise 6.
Establish the equations of invariants curves in the case of a potential field of the form
previously considered for irrational values of Answer. The unperturbed case will be studied first. We consider the unperturbed Hamiltonian that is the first integral of energy 2 ,y - 12(x . 2 ++x2 +i + e z 2 ) .
The first integrals are q50 = +(x"
Ho -q& = f ( i 2 + Q Z ' ) ,
+ x 2)
the "energies in x and z " are constant. Notice that in the (x,i)-plane of equation r = 0 of the (x,x,z) -space there is an invariant curve for each value of = f ( i 2+ P x 2 ) = C .
+,,
Therefore, from the zero order approximation ( E = 0) of the formal series 4 = #o+ E 64 + ... , we deduce that the invariant curves do = C are concentric ellipses which are circles in the ( f i x , x) -plane of section.
Secondly, by eliminating i in the perturbed Hamiltonian expression, we obtain:
that defines a domain such that in the plane of section of equation z = 0 all the invariant curves are necessarily inside the ellipse (or circle) of equation
The theory of Moser and Arnold proves the analytical existence of (closed) invariant curves in perturbed problems close to an unperturbed problem, thus there are invariant curves in the neighborhood of invariant circles of the unperturbed problem. To find the equations of perturbed invariant curves, it is sufficient to eliminate i between the is irrational the following Hand # expressions and to set z = 0.So, in the case where invariant curves are obtained:
2d0 = x 2 + + x 2 = x i + + x i
+ ~ E ~ ~ ( x - x ~ ) -. . . -
Whereas for zero order approximation the invariant curves are ellipses defined by px2+k2=p~;+i;
,
on the other hand a higher order approximation Ieads to a deformation of ellipses.
LECTURE
10
SYMPLECTIC GEOMETRY HAMILTON-JACOB1 MECHANICS
This lecture shows the most powerful process to integrate differential equations of dynamics. The classical theory of Jacobi and Hamilton is based on the canonical transformation notion, but here a modern presentation is given.
PRELIMINARIES Let us recall elementary notions in a finitedimensional real vector space E. D
A bilinear form o : E x E + R is nondegenerate if
[ V Y E E : w(X,Y)=O]
X=O
Remark that if (e,) is an (ordered) basis of E, then the matrix (wv) of o is nonsingular. If we denote (8') the dual basis, we recall that o=wv8'@8J where we know that 0, = w(e,,ej).
Here also we can introduce a (canonical) isomorphism between E and E* . We say:
D
'
The lowering mapping associated to w is the isomorphism ~:E+E*:XHX,=O(X,) such that, VY E E ; X , ( Y ) = @(X,Y ) also denoted X,.Y.
Remark that the matrix of o is such that
' Called Mmol in French.
Lecture 10
PR1
If w is antisymmetric ( ' w = - w ) , then its rank is even, for instance 2n, and there is a dual basis (8.') of basis (e,) of E such that
ProoJ: For o # 0 there are vectors e, and en,, of E such that o ( e l ,en,,) # 0 , for example o ( e ,, en+,)= 1 by "dividing" el by a good constant. Given the skew symmetry of w that implies &(el,el) = "(em,,en+,) = 0 then, with respect to the plane PI generated by (e,,en, ), the corresponding matrix is (-OI
Q
Let us denote by P,, the w-orthogonal complement of PI, that is
I
P,,=(XEE VYE~:O(X,Y)=O). We immediately know that P, nP,,= (0) and also E = P, + P,, because we obviously verify that V u E E : U - 4% en,, )el + w(u,e, )en+,E 4 1. We can conclude that E=P,@P,,.
We begin again but, this time, on P , , by choosing two vectors ez and en+, such that w(e,,en+,) = 1 , and so on. Finally, we view the matrix w relative to the basis (e,) is
where I is the (n x n ) identity matrix. I#
To end, we show the form p =
ZB' ok+"is w . A
k=l
Indeed, from p ( e i ) ,that is (e,), with respect to p , we see:
This means that V i , j I n : p(ei.ej) = 0, p(ej,em+,) = 1, ~ ( e,en+j) j = 0, ~ ( e n + i , e= i )-1. and therefore the matrix of p is the one of o . n
~(e,+,,en+j) = O
Remark. Given o = CB'n 8"" , a reasoning by induction proves the exterior product of n i=l
factors o is
Symplectic Geometry, Hamilton-Jacobi Mechanics
on
=
Coil,,@I+. ,.-. n
#,h
,,@I+.
i,, ..ik=l
n(n-I)
= n!(-l)Z@'
A . . . A ~ * ~ ,
the exponent of (-1) can be replaced by the largest integer in n/2. Clearly, o n is a volume and an orientation is so defined. Let us end the preliminaries by the sympIectic notion that will be afterwards essential. D
A symplectic vector space is a pair denoted (E, w ) such that w is a nondegenerate 2-form on E.
D
Given symplectic vector spaces (E,w) and (F,,u), we say a linear mapping f : E + F is symplectic if f * , u = w .
D
The symplecticgroup denoted Sp(E,o) is the set of symplectic mappings f : E forming a group under composition.
+E
It's really a group, since f belonging to the group GL(E,E) , it is sufficient to see that ( f 0 g ) * o= g*(f 'o)= g * o = a and also ( f - ' ) * o = ( f n ) -f'*o=w .
D
Given f E Sp(E,w), the matrix of the nondegenerate 2-form u E R'(E) is called symplectic.
This (2n x 2n) matrix is denoted by
Remark 1. We evidently notice that J-' =
'J =-
J
and
J 2 =-I. Remark 2. If f E L(E,E) with corresponding matrix @ , then the symplectic condition f 'w = w is written: *@J@=J. R
Indeed, from e; = f (e,) = j=1
that is, in matrix notation:
f;i e, , we immediately deduce that
Lecture 10 J f = '0 J a. Thus, the symplectic condition f 'w(e,,e,) = o(e:,e;.) = o(e,,e,) leads to the result
'@J@=J.
1. SYMPLECTIC GEOMETRY Let M be a manifold of even dimension 2n.
D
A syrnplectic structure on A4 consists in giving a nondegenerate closed 2-form at each point of M called symplectic form w . A symplectic manifold ( M , w ) is a manifold M provided with a symplectic form w on M.
1.1
DARBOUX THEOREM AND SYMPLECTIC MATRIX
1.1.1
Moser lemma
L
Given symplectic forms w, (supposed differentiable for every t E [O,l]), for every x E M there are a neighborhood U of x and a family of local transformations p, : U + U such that pi = id and p,*w, = w,.
Proof: The problem is to know if there are vector fields X I on U such that
with u),*o, = o,, .
By refering to (6-6) with here w,(t, x), that is by introducing the notion of the Lie derivative L, linked to the flow, we have:
a,), = 9;( a , ~+, but the form d,o, is closed since dd,o, = d,dw, = 0; therefore, from the Poincark lemma, there is a neighborhood U of x on M such that the form d,w, is exact and we can write:
d,w1 = 4 4
and thus
Sy mplectic Geometry, Hamilton-Jacobi Mechanics
To look for p,' such that
amounts to search for X, such that
~,*d(~cr + lx,w,) = 0 or such that i X , q = -p,
+
In a local chart, we write: a+ = & ~ ~ ( t , x ) d X ~ ~ d X ~( A,B = 1, ...,2 n ) C'l
3
= PA (t,x)
i,w, = m m x A d y B.
So, the problem amounts to solve ( for x A ( t , x ) ) the following system m , ( t , x ) x A ( f , x ) + ~ s ( t , x ) = 0,
that has a unique solution because the forms o,are nondegenerate. The field X, is thus determined and consequently also the forms o,such that p j w , = w,, the lemma is so proved.
Let us note the nonlinear problem of the determination of 9, has been reduced to a linear problem consisting of finding X I . 1.1.2
Darboux theorem
Given a nondegenerate 2-form on Mand a chart (U,tp) at each x E M such that
&x)=O
and
VUEU: 1
n
p(u) = (x ,.. ., x , x
n+l
2n
,..., x ), then we obviously know the following 2-form on U 0
= &l
, &"+I +
.,, + &", &2"
is closed.
Conversely, let us establish the following important Darboun theorem.
TH
Q=
Let ( M , m ) be a symplectic manifold. For each x E M there is a chart ( U , q ) such that q ( x ) = 0 and there are 2n differentiable functions x' ,.. ., x2" (local coordinates) defined on U such that:
Such charts are called symplectic and the functions x' are called canonical coordinates Proof We h o w there is, at x , a basis of T,M such that
Lecture 10
By remembering PR1, we consider a chart on which the exterior form
is the following constant %-form Consider the family of exterior forms = t m + ( f -[)a,
t E [OYIl which are closed and nondegenerate on a neighborhood U of x . Indeed, at x, the 2-form W,
0, (x) = wo( x ) = O ( X )
is nondegenerate and by continuity there is a neighborhood U of x on which w, is nondegenerate. Therefore, the exterior forms of the family { o r )being symplectic, the Moser lemma implies there is a diffeomorphism 9,such that:
pi= id
and
p;m, = o,.
In addition, we specie that for t =1 : w,= w and p;,'w = w, which means it is possible to produce a coordinate change such that:
1.1.3
Volume n
The exterior form w c R 2(M) is nondegenerate (the rank of o = xdr' A &"+'is in) 1=1
~
A4 is an even-dimensional manifold (dim M = 2n). From wn we define the standard volume ( where [n/2]is the largest integer in n/2 ):
which defines an orientation on M. 1.1.4
Symplectic matrix
Let ( x A) = ( x l ,...,x n , xn+',...,x 2 " ) be an ordered set of 2n reals. For instance, they are the 2n coordinates on a phase space in mechanics : i = 1, ...,n. ( x A )= (pi,ql) Make explicit the matrix (0,) from w =+'3,dXA A h B (10-2) that in mechanics is dpt A dq' .
We have VA,3 E (1,..., 2 4 , Vi, j E (I ,..., n ): w=+w,dxA
*ahB
= f (fly&'A hi+ u,~+,&' A dxn*'
+ on+' jdTn+i
dr'
+
&n+lnt,
hnci A hn+J)
Symplectie Geometry, Hamilton-Jacobi Mechanics The symplectic matrix
(0, )
is
that is, in the previous sum, the terms w, are zero (in mechanics there is no term of type dpi A dp ); the terms wn+i,,+' are also zero (in mechanics there is no term of type dq' A dq' ). The only nonzero terms are
V ~ E { I...,, n): mi,,+, =-w,,,, = I . We recall also that; '48
(@,)-I
= ( o )='(aAB)=-(om)
and w,oBC =SAC.
1.2
CANONICAL ISOMORPHISM
The canonical isomorphism notion has been previously introduced within the context of Riemannian structure defined by a bilinear form g. A symplectic structure defined by a nondegenerate dosed 2-form on a manifold establishes also a vector bundle isomorphism between TM and T'M .
M
First, it can be proved what follows: PR2
Given a differential form of degree 2 on M ( in particular o E R'(M)), we can say: (I;) The mapping TM +T'M : ( x , X ) w ( x ,X , ) = w,(X, ) such that Vx E M , VX,Y E TIM : m,(X,Y) = w ( X , Y ) ( x ) is a vector bundle mapping. (ig The mapping -x(M)+ x*(M) : x H X , = w ( x , m is C (M)-linear. If (I, is nondegenerate, then (iil) TM + T'M :( x , X ) H w,(X, ) is a vector bundle isomorphism.
ProoJ In brief: (i) The local representative of the mapping TM + T'M with respect to a chart ( U , q ) of M, with U' = p(U) c E , being defined by U'x E + U' x E* : ( y ,Y ) t+ ( y ,o,,(Y, )) , the reader will prove it is of class Cm. (id We have immediately V X ,Y E X ( M ): w ( X + Y , )=cL)(X, )+w(Y, ) on each fiber. (iiij If w is nondegenerate, then the mapping TM +T'M is a bijection on each fiber.
Lecture 10
392
Indeed, it is "one-to-one": w ( X , )=w(Y, ) w ( X - Y , )=O what implies w is nondegenerate. It is "onto" because to every a G T,'M is associated one X E T,M such that a = w ( X , ). (Take again the proof in lecture 8 5 1.2). So, we have an isomorphism on each fiber. Let M be a 2n-manifold, for example in mechanics the phase space of a configuration space Q,that is the cotangent bundle M = T'Q . We introduce a syrnplectic form on M, namely: n
w = c d r i ~dr"+l
(or
+W,&~
r\dxB)
(or
xA-dA
).
I=1
and also a tangent vector field
x = xJa,+ xn+jan+,
ax
We say: Dw
The symplectic form w E R ~ ( M )being nondegenerate, there is an isomorphism called flat mapping, denoted by b :~E(M)+x*(M):xHx~ such that to every vector X is associated a 1-form defined by X6 = i X w
6iP
=4x,1 = U b( X ) .
Make explicit (10-6): VY E X ( M ) : w ( X , Y ) = i;,w(Y) by introducing local coordinates.
On the one hand, we have: w=+wABdXA*akB ixu= + ( W , . , ~ X -~ Q ~ XI ~~ X ~ ~uX A B~ x)A = drB s
on the other hand: o ( X , Y ) = 3w,drA
A drB( X , Y )
Symplectic Geometry, Hamilton-Jacobi Mechanics
D * The sharp mapping denoted by
# is the inverse of the flat mapping b :
So, we have V X E X ( M ) ,V a E x * ( M ):
(x,)" x (a#), = a . Make this notion explicit. Consider the flat mapping such that X H X , = o ( X , ) and where X,= i,w = w , x ~ ~ x ~ . To every 1-form
a = rn,xAdrB = C(-xn+'&' +xi&"+')
(lo-%) ( see (10-7a) )
1
(by putting a, = m , X A )
= aBdxB
is associated the following vector (under the sharp mapping):
and so we denote
a# = X , . Indeed, we find again a = (a')a = iatu= an+,a!xn+' - (-ai)&'
Prove the following formulas: V a , f l ~ i 2 ' ( ~ )i,.a,=J :
P(a#> = -a(PU1 Indeed, we have: i
B
= w ( J n , ) = (P*)*= /3
and
P(a9 = (Byb(aii) = ~ ( / 3 ' , a= ' )- w ( a " P B " )= - a ( p P ) . 1.3
POISSON BRACKET OF ONE-FORMS Let ( M , w ) be a symplectic manifold.
In mechanics, the symplectic structure of the phase space authorizes a very important operation: the Poisson bracket. To any one-form a E % ' ( M ) corresponds the vector field a'
E
X ( M ) such that:
Lecture 10
394
To any one-form a E X * (M) corresponds the vector field anE X ( M ) such that: i,,o = a .
D
The Poisson bracket of one-forms a and by: {a,~ =@ ) ([~",P'I,1
'
P on M is the one-form of M defined
-
- 'La',pqo
that is
( a , ~= )
P'I~
where [a',P'] is the (Lie) bracket of vector fields a$and
p'
It follows: t a , ~ )=#
[aN7P#1
So, we immediately construct the commutative diagram: %xi%
& x'xx*
[ I ,
3- 17
bxb
PR3
X
{ ] +
z*
The space of one-forms on M provided with the Poisson bracket is a Lie algebra.
Proof: The following properties are easily checked.
~ a , p , Ey C l l ( M ) : PI. Linearity: Vk,,k, E R : (a,k,,B+k,y)= k , { a , P ) + k , ~ a 9 y ) . P2. Skew symmetry: ( a , p ) =- ( ~ , a l .
The previous properties obviously follow from the Poisson bracket definition. P3. Jacobi identi&: ~ . { P , Y > > {+ ~ , ( r , a j{)r+, b . ~ ) 3 0. = Indeed, we have: ( a 7 ( P 3 ~(Po=r ~ [ ~] =~ ~~a '~ ~~[ Pl ' r, ~ # l l r . By obtaining analogous results for the two last terms of the Jacobi sum, then the Jacobi identity relative to the Lie bracket proves the Jacobi identity relative to the Poisson bracket.
Other Poisson bracket notations exist.
Symplectic Geometry, Hamilton-Jacobi Mechanics = g { a , P ) + P X , ( g )*
Indeed, we have: b , g p ) = [a",gS"], = (La.(gB")),
but L , I ( S B ~ )
= g ~ , n B #+ P U ~ = n g = g~a",P+
+/3#a*(g)
thus (a,gpJ=&[a"pPYlb+ P a x ( g ) .
PR4 The closed differential forms of degree 1 form a Lie subalgebra of the Lie algebra of differential forms of degree 1. Proof We are going to prove the Poisson bracket of any two closed one-forms is closed. For that, let us prove the Poisson bracket of any two closed one-forms a, /? E a'( M ) is exact:
(.,PI
=
J,.,$.,@
( see exercise 8 of lecture 6 )
= [La,.is" Iw = Lanip" - i
Pn
La,w
= La,B-ip.diflu
( because (10-9)and dm = 0 )
= di,p- i,,da
( because dp = 0 )
=d$*P = dl ,l ,w a P = -dw(an,
p') .
This last equality is immediately verified because if we denote:
then we have: n
The proposition is established since we have proved:
~ ap E, R'(M): (a, B ) = -dm(.', fl"
>.
Remark 1. The equality (1) is also written:
(a,p)=L a . p - Lp.ia.w+diP.ianu (a,~)=~~,~-~~,a-di,.i~.w So, if n and pare closed we find again that (a,p ) is exact (equal to - dip.ia. a, ).
(1)
Lecture 10
396
Remark 2. The sets of closed one-forms and exact one-forms on M,respectively denoted by SZ; ( M )and Qj( M ), are real vector subspaces of R' (M).
Since every exact form is closed, the algebra @ ( M ) of exact one-forms is a subalgebra of Q: (MI.
To summarize: V ~ , P E Q ; ( M ) :{ ~ , ~ ) E Q ' , ( M ) C R ; ( M ) and evidently ' ~ ~ , P E Q ; ( M()~: , P ) E Q : ( M ) .
Let us observe that we find again the notion of quotient space of the algebra of closed oneforms by the algebra of exact one-forms.
Any two one-forms a and fl on a symplectic manifold (M, w ) are in involution or
D
involutive if
@(a" p" = 0 . PR5
If a and pare closed and involutive one-forms on a symplectic manifold, then their Poisson bracket is zero.
The proof is obvious from the equality (10-1 1).
1.4
POISSON BRACKET OF FUNCTIONS
Let (M,w ) be a symplectic manifold, f be a function of Cm(M). Let us refer to (10-8). In particular, we consider the following vector field
(df)# = x, also denoted =
Xf.
So, we have:
and in particular:
Given any two functions f ,g E C m( M ), we say: D
*
The Poisson bracket off and g is the function
(f,gf = -~(X,,X*)
Symplectic Geometry, Hamilton-Jacobi Mechanics that is, from (10-6):
(f, g j = - i x / o ( X g ) . PR6
Given any two functions f and g we have
Proof The following equalities are obvious: x f ( g ) = ~ x / g = i x , d g = ~ x / ( x=i,,ixpw g), = -w(X, , X g ) .
We can also verifjl(10-13b)by using local coordinates:
{f
=- ~ x , d1 x~ = -i XI
Notation. From
(f g 1= L,/ 9
we nicely denote
(pid r n " ) ( x g) A
(g)= d g ( X , ) = - d f ( X , )
{f,g 1= -df . X g = dg.X, .
In local coordinates:
and we find again the expression of the Poisson bracket in accordance with the one met in the ninth lecture.
Properties. Given any two functions f , g E C m( M ) , we have: PI. Indeed,
d ( f , g ) = (df,dg) d { f ,g )= - d d X f X g = -d@((df 9
= {df , dg
P2. Indeed,
1
(dg)#) ( since (10-1 I) ).
X{f.g,= [ X f J g ] X { , , , = ( d ( f , g ) Y =(df,dg)" =[(df)*,(dg)*l = EX,,X,I.
Lecture 10
398
PR7 The space of functions C m ( M )together with the Poisson bracket forms a Lie algebra. ProoJ: The following properties are verified Vf, g, h E Cm(M) :
PI. Bilinearity : vk,,k,ER:
(f,k,g+k,h)=k,{f,g)+k,{f,h)
since
{f,k,g + k2hJ=X#,g + k2h) = k,Xf(g) + k,X/(h) P2. Anticornmutative law: (f, g ) = -{g,f since {f,g)= -o(X,,X,) It is clear that {flf 0 -
=w(Xg,Xf)= -{g,f I.
3=
P3. Jacobi identity : {f,{g,hf)+{g,(h,f l)+th,tf,gj)= 0. Indeed,
(f4g,hlj+ tg,fh,f 11= X,(X,(h)) + X,(X,(f 1) = [X,,X,l(h) and (h,{f,g))= -Iff,gl,h3= -X{,,,](h)
=
-[X,,Xgl(h).
In addition, we can establish: { f 7 g h j =g{f,h)+h{f,g).
P4.
Indeed,
{f gh 3
D
= X,(gh) = d(ghXXt ) = g dh(X,) + h &(X( )
Any two functions f , g E C" (M) are in involution or involutive differentials are so, that is if
if their
In an equivalent manner if their Poisson bracket is zero: {f,gj=-w(~,,~,)=~. PR8
Given a function 1,the mapping CYM) + CYM): g H (f, g } is a derivation.
Proof: The mapping defined by PR9
u, }
g that is by L , g is obviously a derivation.
A function f E Cm(M)is constant along the orbits of X, ~ f f i and g are involutive.
Symplectic Geometry, Hamilton-Jacobi Mechanics We note that f is called a first integral of field X,. Proox Let p, be the flow of X , .
By remembering the formula (6-6) which shows the Lie derivative of differential form w with respect to X, we know that in the case of some 0-form f :
which vanishes
rff (f, g )= 0 .
We can notice that the same conclusion exists ~fg is constant along the orbits of Xf because if W, indicates the flow of X, we have: d -(v/fg)=O # (f$)=o. dt
Poisson theorem If a Cmfunction f is in involution with Cmfunctions g and h , then f is in involution with the Poisson bracket h 1.
TH
lg,
Proof. By hypothesis, we have ( f , g ) = ( f , k) = 0 and the Jacobi identity allows us to
1.5
SYMPLECTIC MAPPING AND CANONICAL TRANSFORMATION Let (M,w ) and (N, p ) be symplectic 2n-manifolds.
D
G-
A C" mapping f : M
+N
f * p =W
is called a symplectic or canonical t r a ~ o r m a t i o nif
.
By referring to the last part of lecture 5, the reader will be convinced by the following: PRlO If f : M + N is a symplectic mapping, then: (i) f is volume preserving, (4 det,,=. n p f = 1 , (iig f is a local diffeomorphism.
,
PRll Every symplectic diffeomorphism orientation.
Proof By recalling the volume form is
f
E
Dif(M2,,M2,)
preserves volume and
Lecture 10 n(n-1) a n= ( - 1 ) n ! d X I A... A & " , it is immediate the volume is preserved under f since
and a symplectic orientable manifold preserves the orientation. This proposition is a modern presentation of the famous Liouville theorem that will be again introduced in the next Hamiltonian mechanics section.
PR12 The cotangent bundle of a manifold is a sympIectic manifold (and is thus orientable). ProoJ: We are going to consider the cotangent bundle T'M of an n-manifold M. Let IJ*: T'M + M be the projection of a cotangent bundle, &:I TT'M * +TM be the tangent mapping of II* .
To make the proof easier, we draw the following diagram:
R
TT'M
d"'
"
'a,
(xux1= ax(xx)
9TM
nL
L.
17
Clearly, we have the projection ~*:T*M-+M:(x,~)Hx. We denote (x,a) by ax.
In the same manner, we simply denote dl*: TT'M +TM : X,,+i
X,.
So, from a chart ( U , q ) on M, by introducing a basis (e,) of R" and the dual basis T'R n, we have successively:
n
p* : T'M
+ R2": a, H p * ( a , )= ~
( x ' +xnffei) E ~
r=l
dp' : TT'M
+ TR"
5
R2": X,,
I+
dp*(Xax) .
Let us specify the vector Xu* tangent to T'M is expressed in (U,p) by n
dp*( X , ) =
C(A,E' + Biei) i=l
where A, and Biare the local coordinates of Xu .
(6')
of
Symplectic Geometry, Hamiltonian-Jacobi Mechanics
To the tangent vector X, E T,,T'M is associated the one-form Am, : TaxM+ R defined by
iz,=(X,,=
Q,
( X m )) x
= a, ( X , ).
So, a differential form A of degree 1 on T'M (that to each a, makes AaKto correspond) is defined by: A : T'M + T'T'M : a , H Lax such that ( ~ ( a , )X,, ) = (a,.rn*(X, )). In a local chart, the previous equality leads to:
We immediately see that
Indeed, by viewing
we find again the expression of Aax(XaF) :
So, from the form
we obtain the symplectic form
Consequently, there is a volume element and hence the cotangent bundle of any manifold is orientable. D * The forms R. and dR. are called the canonicalforms on M
PR13 The canonical 1-form d on T*M is the (unique) 1-form such that for any 1-form a on M the pull-back of A by a is a'A = a . Proof: We consider
A : T'M + T*T*M a:M+T*M a*: T*TBM+ T'M d a : TM + T T * M .
Lecture 10
402
The definition of the pull-back ( see (5-11) ) means that VX E TxM:
(a:A),(x) = a(a.)(da,.X)= (aax .da.~) = (a, , & * d n , ~ )= (a,,d(II*a,)x) = (ax,x)
since lIva(x)= x .
Because the result is a,.X the proposition is proved. The following theorem shows how symplectic diffeomorphisms on T*M are generated from diffeornorphisms on M. Theorem of canonical extensions TH
If f : M + N is a diffeomorphism then diffeomorphism.
f * : T 4 M 4 T'M is a symplectic
Proof: First, remember the pull-back mapping f*: T * M -,T'M : a, Hf ' a , is such that V X E T,M :
( f * a ) , ( X= > af(,,(df
.
We are going to prove that (f*)*w=w.
For that, it is sufficient to establish such an equality for the canonical one-form: f-n=n. Indeed, we have VXar E Tax(T'M):
(f**l)(ax )(xax )=( ( f
)*AG )xu, = nf.,a,(df '.xa,
= f*(aX).(dn8df*(Xa, )) = a,(df m ' d f *(Xas 1) = ax( d ( f o n* f >Ixa, ) but
fan* O S *
=
n*
since
(f n* f * ) ( a , )= f (II'(f*a,>)= f ( f " ' ( x ) )= x = n*(a,). 0
Therefore, we have
(f'*n)(a,>(x,, = a,(rn*)(X,,) = a a , ) ( X a ,) that is
f-a = n
and thus ,f**w = 0 .
Remark. The following diagram is commutative
Symplectic Geometry, Hamiltonian-Jacobi Mechanics +T*M
T'M
n*4
3. n*
M
y
M
since
The diffeomorphism f * js sometimes called lif"t off
Jacobi theorem TH Given symplectic manifolds (M,w )and (N, p ), a diffeomorphism f : M + N symplectic lfl V h E Cm( N ): df -'x, = x,,
is
We also denote: f 'X,= X,.,
.
Proof: First, we remark that f is symplectic r f l V a E R 1 ( M ): (f 'a)' = df - ' ( a ' ) , expression also simply denoted by f *a'. Indeed, by putting X,= a we must prove that (f8x,)"df-'x. This is true since VY E X ( M ): f * X , ( Y ) = X,(dfY) = w ( X , d f Y )= f 'w(df -'x,Y) = w(df -'x,Y) [ rff f is symplectic j = (df-'x), ( Y ) . Now we can prove the Jacobi theorem. I f f is symplectic then we have: df -'x,= @-'(dh)# =(f 'dh)' = (df'h)' =
x,.,.
Conversely, the hypothesis being f ' X , = X,., , we have, on the one hand: d( f ' h )= f '(dh);= f '(X,), = f *ixhp =i
( see PR7 of lecture 6 ),
f8p rx*
on the other hand: d ( f ' h ) = ( d (f *h)):= (X,.,), = ix w = 1, .,y,
0,
l'h
So, the hypothesis of the reciprocal and the comparison between the two previous results (for any h and every X h f ) imply f 'p = w, that is the diffeomorphism f is symplectic.
Lecture 10
404
In particular, this theorem is applicable to the case of a diffeomorphism f : M
+ M.
To end this section, let us show an important link between symplectic diffeomorphisms and Poisson brackets. PR14 A diffeomorphism f : M + N is symplectic #I f preserves the Poisson brackets of functions, that is: v g , h E c ^ ( M ) : { f 8 g , f * h ) =f'(g,h}. ProoJ: We have:
f8ig,h)= f'X,h= f'LXgh (see PR4 of lecture 6)
= L,.& f'h
=L
f'h
xf-,
(this last equality following from the Jacobi theorem = X,.g f' h
f
= (/*g, *A
#I
f is a symplectic diffeornorphism)
I.
The reader will immediately verify this proposition hoIds in the case of one-forms. Remark that f is a Lie algebra isomorphism on C m(M) and 0' (M)
2. CANONICAL TRANSFORMATIONS TN MECHANICS In mechanics, we recall there are two fundamental spaces: the configuration space Q and the phase space which is the cotangent bundle M = T'Q . This momentum phase space is thus provided with a symplectic form. By recalling the 2n coordinates of T'Q are (pi,q') then, from previous developments, the canonical forms are w n = p, dqp (10-16') 6&" w =dpi ~ d q ~ . ( 10-17') The form 2.1
A is sometimes called Liouville fonn .
HAMILTONIAN VECTOR FLELD
Let ( M , w ) be a symplectic 2n-manifold (cotangent bundle), H : M + R be a function of class C ", cailed Homiltonian. D Q=
A Hamiltonian vectorfield is a vector field, denoted X, or simply X , such that
Symplectic Geometry, Fitrmiltonian-Jacobi Mechanics It is also denoted by
(x),= -dH
g/
So, for every vector fiefd Y , we have: w(X,Y) = -dH(Y) also denoted
w D
w(X,Y) = -dH.Y
A Hamiltonian system is a triple ( M ,w,X ) .
We immediately notice that X = ( X ) : = -(dH)# . Remark 1. Several authors choose the canonical 2-form to be w = dqi
A
dp,
and introduce the Hamiltonian vector field X such that:
w(X,Y)= dH.Y . Remark 2. Any two Hamiltonians H and H ' for the same Hamiltonian vector field (X= X') differ by a constant since O=jxm-ix.m=d(H1-H). Remark 3. The existence of Hamiltonian vector field is guaranteed by the nondegeneracy of m.
PR15 If (M,w,X ) is a Hamiltonian system with H E C W ( M )and flow @,, then d dt
Vg E Cm(M):- ( g o + * ) =
{ ~ O + ~ , H ) .
ProoJ We have:
PR16 Given a Hamiltonian vector field X then t I+ lff
that is gff the 2n Hamilton equations hold.
(p(r),q(t))is an integral curve of X
Lecture 10
406
Proof: By introducing the 2n canonical coordinates obtain: ( X ) , = i,w = - dH
such that w = dp, A dq', we
Then, the expression of the Hamiltonian vector field is
simply denoted by
x = (-7aH ,-)
dH
a9
PI
This expression of Hamiltonian vector field X is confirmed since ixw = i,(dp,
i
A d q ) = i,dpi A
dqi - dp, A i,dql
is truly - dH .
Theorem of energy conservation
TH
The one-parameter group generated by X preserves the HamiItonian function. In other words: Given an integral curve c(t) for X , then H(c(t)) is constant in t : bt*H = H .
ProoJ We have:
but
[ moreover the previous expression is L, H = dl?.(-dH)' = w( ( d ~ ) (' d , ~ ) '1)
thus #,*H = #;El = H .
We evidently find again a welI-known property, namely:
The Hamiltonian H is a first integral (energy) of motion equations if t is not explicitly in H:
since this is dH(c(t)).Y t )= dH(c(t)). X(c(t))= o(X(c(f)), X(c(t))= 0
Symplectic Geometry, Hamiltonian-Jacobi Mechanics Liouville theorem
Let us show the modem expression of the Liouville theorem showing the flows of X are canonical transformations.
TH
Given a Harniltonian system (M,w, X ) , if a flow flow preserves the phase volume a,.
4,
of X is syrnplectic then this
Proof. We must prove there is the following constant in t : +/*,lo= w . Indeed, d
-+,*w = #/*LXw= #/*(iXd+ dl,)@ = -#,*ddH = 0, dt and thus the constant 4,*0 is o (= # i o ) . Since L ~ = WW~A L ~+WL X w ~ ~ = O , . . . and so on, we conclude that the phase space volume is preserved: #,*an = ( 4 , ' ~ )=~wn. PR17 The Harniltonian vector fields form a Lie subalgebra of the Lie algebra of vector fields.
ProoJ We must prove that if X, and X , are Hamiltonian vector fields, then [X,,X,] is a Harniltonian vector field. It's true because
This Lie bracket is a Hamiltonian vector field, the one associated to Hamiltonian { H
,GI.
Now, let us introduce the notion of a locally Hamiltonian vector field. D
A vector field X on a symplectic manifold ( M , w ) is locally Hamiltonian if for every x E M , there is a neighborhood U of x on which X is Hamiltonian.
PR18 A vector field X is locally Harniltonian r f l ixu is closed , rfS L,w=O, fr its flow is composed of symplectic mappings. Proof: We have: [ X locally Hamiltonian, that is on U : X = - ( d ~ )]#
fl !#-
[ ixw locally exact, that is on U : ixw = -dH dixw = 0
Moreover the expression
1
( by the Poincare lemma).
Lecture 10
d dt
-(bt*u = 4; (L,w) = (b; (d1,w)
vanishes ( @,*u = w ) Iff X is locally Hamiltonian and so the flow of locally Hamiltonian vectors is "symplectic." Remark. If a Hamiltonian vector field is necessariIy locally Hamiltonian, the converse is not necessarily true because closed one-forms are not necessarily exact [ see e.g. R. Abraham and J. Marsden, p. 189, 1978 1.
2.2
CANONICAL TRANSFORMATIONS
-
LAGRANGE BRACKETS
In the previous section, we have showed the symplectic structure of the phase space and also the Lie algebra structure of Harniltonian vector fields from Poisson bracket operation. We know that every Hamiltonian vector field defines a "Harniltonian flow," that is a oneparameter group of symplectic diffeomorphisms which preserves the phase space structure. We forge ahead with the Lagrange bracket and we end with the first integral notion.
In local coordinates xA = (p,,qi) we first notice that the 2n Hamilton equations are denoted
with A,B E (1,..., 2 n ) . Lndeed, they are:
dH
dH dxn+' 84' dH dH
dpt = dx* = a@-= dt dt axB dqi =-= dnCi W dt dt
=
t3xB
with i = 1, ..., n. Let us write again the Poisson bracket and Hamilton equations.
Let f be a function f (1, P,,q i ) . The expression of the Poisson bracket of functions f and H is
that is
In matrix notation this is written:
Symplectic Geometry, Eamiltonian-Jacobi Mechanics
The Hamj1ton equations have the following elegant expression:
kA = ( x A , H } or
w
x A= ( X ^ , H ) ,
Indeed, since
we can conclude from (10-20). Before introducing the notion of Lagrange bracket, we are going to use the matrix notation to prove the important following proposition.
PR19 A chart (U, q) is symplectic @, by denoting
~ ( x=) (p,(x),
'( x ) ) , we have on
U:
ProoJ: Firstly, given a symplectic chart characterized by w = dp, A 4', then the Poisson brackets of canonical coordinates are truly the previous, for instance:
This is also obtained from (10-13c):
(see also exercise 2). Conversely, let (U. p) be a chan with ( p i ,p,)= 0,( q f ,q ~ }= 0 and We immediately see that the inverse matrix
is J-' . Indeed,
thus S = I , and so on.
(qi,p,
)= 6:
Lecture 10
410
Consequently, the matrix (u,) of w is J. Now, let us view the canonical transformations from the notion of the Lagrange bracket. Let ( M ,u) be a symplectic manifold, (U, p) be a symplectic chart on My f : M + M : x H y be a diffeomorphism denoted by yA= yA(xB)where the different values ofA and B belong to f l,..., 2n).
D
*
The Lagrange bracket of any two functions yC and
yD,
denoted (y C,yD ), is
The Lagrange bracket expression Gwen the charts (U,p) and (V,y/) such that p(x) = ( p ,,..., q n ) , y(y) = (4,..., Q n ) and given the diffeomorphism f such that (4,..,Q") = f ( p ,. . . , q n ) , then for any P and any Q of (8, ...,pnJ, we have:
Indeed, the previous definition leads to
and the expression of (P,Q) ensues from the following equalities
Note that the Lagrange bracket (P,Q) in the "new" canonical coordinates is expressed from the " old canonical coordinates which are expressed as functions of the "new" ones. So, the Lagrange bracket is the function expressed by
For example, on the phase space, we have:
Relation between Poisson and Lagrange brackets There is an immediate relation between Poisson and Lagrange brackets:
Symplectic Geometry, Eamiltonian-Jacobi Mechanics
Indeed, we have: 2n
D=I
axA arE
I*, g p
ayD ayB -, ;, a
PR20 A diffeomorphism on M is a canonical transformation fz the matrix associated to the Lagrange bracket is the symplectic matrix J, Iff the matrix associated to the Poisson bracket is J-' . Proof: yA= yA( x B) defines a canonical transformation
sff
{yC,yD}=eCD
since (10-26).
So, on the phase space, we know that the existence criterion of a canonical transformation is
Exercise. With the help of coordinates, prove that a transformation is canonical preserves the Poisson brackets. (See exercise 8).
fi
it
Remark, Once more we recall that a canonical transformation 4, is such that
that is ip,*w = 4;w =
This invariance was clearly shown in lecture 9. We emphasize that
w=+w,'hA is a (truncated) absolute integral invariant of canonical equations
and conversely if we search a differential system such that w is a (truncated) absolute integral invariant then we obtain the canonical system of Hamilton.
Lecture 10
412
Exercise. By using (10-28) show again the symplectic form w is an absolute integral invariant concerning the canonical system of Hamilton. (See exercise 3).
First integral existence The existence condition of a first integral of canonical equations of Hamilton is
Poisson theorem TH If f and g are first integrals of Hamilton equations, then their Poisson bracket is a first integral. The proof is provided with exercise 6. This theorem introduces a process to obtain first integrals from two which are known; however, these so obtained first integrals are not necessarily new.
23
GENERATING FUNCTIONS Given a canonical transformation on T'Q x R defined by P~= P ~ ( ~ , , q ' , r )
Q'
=Q1<~i,qi>t) we recall that the complete relative integral invariant is (9-24):
tl=t,
A = p,dq i - Hdt = P , ~ Q' Hdt+d~ where dS (such that obviously ddS = 0) is a (classically exact) differential of a function S of 4n+ 1 variables pi,q' ,4,Q' ,t . D
The previous function S ( p j ,q',P, ,Q' ,t) such that
pidqi- P , ~ Q '= (H - g ) d t +dS is called generatingfunction of the canonical transformation. The generating hnction S is a priori dependent on 4n+ 1 variables. However 2n+ 1 variables are independent in view of the following 2n relations: =e(~i,q~>')
Q'
=~ ' ( ~ j ~ q l 9 ~ ) .
Consequently, any generating function is one of the following types: S,(qiyQ',t) , S2(4,qiyt) , S3(~iyQ'yt), S4(p,,4,t).
Symplectic Geometry, Hamiltonian-Jacobi Mechanics (i) Generating function S,(q',Qi , I ) .
We suppose det
$ 0 in such a way that q' and Q' are independent coordinates.
a(pi,qr)
We have:
as,
as,
as,
+-del aQi
pldqi - ~ d =t e d ~ -Hdt+-dt+--;dqi ' at aq
as,
-
=--
as
H = ~ + H . at
aQi
These relations define the canonical transformation (nonsingular if det[&)
r 8).
Example. The generating function
generates the following canonical transformation PI = Q' P, = -qi
(H=H).
This transformation is assuredly canonical because: dpl A dqi = de
A dQ
i
.
Exercise. From (10-31) prove again the invariance of Hamilton equations under canonical transformations. Answer. By considering the canonical equations p. =-- aH
' aqt and the canonical transformation P, = P,(p,,qi,t) we have:
. i aH q =y 89
Qi = Q ' ( ~ , . q ' , t ) ,
[ since (10-31) ] .
But, the Poisson brackets are invariant under the canonical transformations; therefore we have:
and thus
[ since (10-31) 1.
Lecture 10
414
The invariance of the other Hamilton equations is proved in the same way.
(ii) Generating function S, (4,ql ,r) . Define S, from S,. From
pidqi
- ~ d =t p i d g i
-
Kit + d s , ( g l , ~ ' , t )
and since
we choose
s,(q1,Q',t> =S,(P,,qi,t)-P,Q1
and hence d s , = pidqJ- ~ d+ Q t ' ~ P+, Kdt . The identification leads to
with the condition
(i$'k.)
*
det - O .
(iii) Generating function S,(pl ,Q1,t) Define S, from S, Since
we choose
Sl(qi,Q',t)= S , ( P ~ ~ Q ' f, i~g)i +. Differentiation leads to
with det
-
*O.
(iv) Generating function S,(p,,C,t).
Define S4 from S, :
s , ( p i , 4 , t )= s , ( ~ ' , Q ~ , ~ ) + P -PA' ,Q' Differentiation leads
as4 4' =-apt
as,
Q1=-
aP,
- = as H 4 + H . at
Symplectic Geometry, Hamilton-Jacobi Mechanics
3. HAMILTON-JACOB1 EQUATION The analytical integration method of Jacobi is famous for solving insoluble problems by Lagrangian or Hamiltonian formalisms. This method based on Hamilton works is very effective in celestial mechanics, in perturbation problems and other areas. 3.1
HAMILTON-JACOB1 EQUATION AND JACOB1 THEOREM
The canonical transformations are particularly useful insofar as the transformed Hamiltonian K is simpler than the original Hamiltonian H.A clever man as Jacobi thought to choose a zero transformed Harniltonian in order that the solutions piand Q i of "new" canonical equations
be constants. Therefore there are 2n (independent) first integrals of motion of the representative point in the phase space. The Hamilton-Jacobi equation is obtained from such a judicious canonical transformation. If the transformed Hamiltonian H is zero then the existence criterion of a canonical transformation for a generating function of type S ( q i , Q i , t ) ( i = 1, ...,n ) is
pidqi-
and the famous nonlinear Hamilton-Jacobi equation
In the Jacobi theory, the 2n equations of Hamilton being
i: = o
Q~=o,
then the 2n (independent) first integrals of motion of the representative point in the phase space are
where ai and bi are 2n arbitrary constants. Given an orbit of a descriptive point in the phase space, the values of parameters a, and bi are the motion constants. In mechanics, only the search of a complete integral of the Hamilton-Jacobi partial differential equation is interesting (and not the general solution).
Lecture 10
416
D
A complete irttegral of the Hamilton-Jacobi equation is a solution which depends on as many arbitrary independent integration constants as coordinates q' and t, on q i and r.
However, the Hamilton-Jacobi equation shows S through partial derivatives. Therefore, one of the arbitrary constants is necessarily additive (since it doesn't alter the partial derivatives). This additive constant plays no role in the canonical transformation (only the partial derivatives of S are present) and we say precisely:
D
A complete integral of the Hamilton-Jacobi equation is a solution of the equation depending on as many nonaddtive independent constants as coordinates q', on q' and t.
It is denoted by
S(qi,bi,r) by taking the "new" coordinates Qi as arbitrary constants. Now, the n relations
at the initial instant are n equations linking the n constants bt and initial values ofp, and q i together. They so allow to calculate the integration constants from particular initial conditions. Next, the n following equations
provide the various initial values of q' .
ai
from the initial conditions since the right-hand member is known from
Therefore, the problem is solved because the last equations let us obtain the coordinates q' from initial conditions and time:
qi
=
qt(a,,b',t) .
It's the powerful integration method of Jacobi. Remark 1. Don't forget the existence condition of the canonical transformation:
a2s
*
det (-)
#
0
Remark 2. The notation of a generating function by S(qi,Qi,t)is fully justified because we have (Q' being constants):
Symplectic Geometry, Hamilton-Jacobi Mechanics
417
Thus S (corresponding to a complete integral of the Hamilton-Jacobi equation) is nothing else than the action integral S = ~ d. To t each complete integral is associated s class of orbits of a representative point in the phase space fitting the least action principle.
1
Jacobi theorem
PR21 If S(q1,b',t) is a complete integral of the Hamilton-Jacobi equation, then the functions p,(a,,bi, t ) and ql(ai,bi,r)obtained by solving the 2n equations
constitute the general integral of Hamilton canonical equations. d
ProoJ On the one hand, by considering pf = ---yS(qf, b , t) , the derivative of f
a!?
with respect to parameters b' leads to
a2s +--=o. af? a2s
dtdb'
dp, dqJ db'
On the other hand, the total derivative of ai with respect to time, namely a, = 0 , is written:
(since b' = 0). Subtracting (1) and (2), we obtain:
and the canonical transformation condition det(-)
a2s agJ abr
#
0 leads to n Hamilton canonical
equations:
Now, on the one hand, the derivative of the Hamilton-Jacobi equation with respect to coordinates q' leads to
Lecture 10
418
as
On the other hand, the total derivative of pi = - with respect to time is aqt
Subtracting (4) and (3), we obtain:
which implies the n other canonical equations:
Remark. We have shown a canonical transformation (a,,b i) t,(p l ,q r) and the general integral of the Hamilton canonical equations: q1= ql(ai,b',t).
P, = pi(a,, b i , 0 Any other general integral
doesn't define a canonical transformation and doesn't alIow obtaining a complete integral of the Hamilton-Jacobi equation. However, there is an exception, namely: If the constants c,and d'are respectively the various initial values pp and qi, then the previous general integral shows a canonical transformation (c, = p;,dr = q:) o (p,,q i ) :
Indeed, the general integral (5) with D(pi'qi) D(c,,di)
+ 0 leads to the system
of 2n first integrals of Hamilton canonical equations and thus the following first integrals: { ci,c, ), { ci,d 1 and { d t,d ). J
J
We immediately have:
{ ~ , , d ' } ,=-6: ~
{ c , , ~ , ) , ~= O
{d i ,d')
,o
=0
and since the Poisson brackets are first integrals we deduce (for every time): {c,,d' ) = -6;
{ci,cj)= 0
(dt,d'} = 0.
Therefore, the general integral ( 5 ) , where c, and d' are respectively the initial values of various p, and q' , is a canonical transformation.
Symplectic Geometry, Hamilton-Jacobi Mechanics
3.2
SEPARABILITY
Obtaining a complete integral of the Hamilton-Jacobi equation is often a fastidious even unsolved problem; nevertheless, searching this integral is sometimes made easy. We restrain our study to the (frequent) case of conservative material systems with scleronomic Hamiltonian. In the case of a scleronomic Harniltonian
the Hamilton-Jacobi equation
has a complete integral of type S
where
= -E
t + S(ql,..., q",b',..., bn-',E )
a,,? = 0 .
Therefore, the Hamilton-Jacobi equation becomes:
The Jacobi theorem immediately leads to the general solution of Hamilton canonical equations:
-
(by putting
dS
to = - ).
dE
Example. In the case of the simple harmonic oscillator, the Hamilton-Jacobi equation is
as
obtained by replacingp by - in the Hamiltonian, that is: 89
where q is the displacement and k the stiffness of the spring. This Hamilton-Jacobi equation is easily solved by considering S = S(q,b)- bt
Lecture 10
420
where b is the energy of the simple harmonic oscillator. The Hamilton-Jacobi equation
is here
The complete integral is thus
We have:
The constants a and 6 are linked to initial conditions. At initial instant ( r = 0) the oscillating point of mass m is relaxed at equilibrium position go with p, = 0 . Given the initial conditions, the Hamilton-Jacobi equation
implies
-
-
and we find again an obviousness: the parameter b is the (initial) mechanical energy - m b = -k q 2, --w 2 2
2
q,2
=
E.
Therefore, we have: q = q, cos w(t - a)
and the initial condition q(0) = q, implies the constant a must vanish. In conclusion, S is the generating function of a canonical transformation and is expressed by
s = m u I J=dq (given the initial conditions).
-dqmzq;t
Symplectic Geometry, Hamilton-Jacobi Mechanics
42 1
PR22 A scleronomic mechanical system with a cyclic generalized coordinate, e.g. q' , shows a complete integral of the Hamilton-Jacobi equation of type S = - Et +alq' +S(q2,..., q n , a , , ~ , ,..., b Zhn-')
(10-35)
where a, is an arbitrary constant. Proof: The scleronomic character explains the first term. In addition, if q' is a cyclic coordinate we have:
as
The generalized momentum p, = - conjugate to q' is an arbitrary constant a,. 84'
The complete integral is thus of type (10-35). Since it must verify the Hamilton-Jacobi equation, then the function must satisfjr:
In the complete integral (10-35) the variables t and q' are separated from the other variables.
The reader will generalize if t , q l , ..,qk . are cyclic coordinates, with H ( ~ ~,..., + qn, ' ,,.. ., p,) . In this case: f-l
D
A scleronomic conservative mechanical system is called separable if the corresponding Hamilton-Jacobi equation shows a complete integral S of type:
This simplified writing shows only the separable variables.
Make clear that the separability existence is dependent on the choice of generalized coordinates. Remark. Solving the Hamilton-Jacobi partial differential equation is not a priori more simple than solving Hamilton canonical equations, but separability leads to quadraturcs! Example. In the central force problem, find a complete integral of the Hamilton-Jacobi equation and the general solution of motion equations. The Hamiltonian is well-known:
H = & ( P : + P i / r 2 ) + Fk . The HamiIton-Jacobi equation is
Lecture 10
We search a complete integral of type S =-E6+S,(r)+S2(Q)
which implies
that is
Since the two members are hnctions of respective 8 and r, we obtain:
This expected result, S2= AB, implies:
The general solution is such that
For a trajectory through (r,,B,,) at instant t,, there are thus the well-known equations:
8-8, =
+ A "S r 4 2 m -~~drr n k /-r ~ ~ / r ' 2
giving the positions (by B ( r ) ) and position instants.
4. A VARIATIONAL PRINCIPLE OF ANALYTICAL MECHANICS We suggest the following presentation of a variational principle inspired by the quantum mechanics and the last chapter of "ClmsicalMechanics " witten by H. Goldstein. It is also developed in the case of n degrees of freedom.
Symplectic Geometry, Hamilton-Jacabi Mechanics INTRODUCTION
In classical notations, with Sf = 0 and instants (and phase positions) denoted by 1 and 2, the principle of least action is
o
6 j'I (pdq - ~ ( p , q , t ) d t =) 0
where p = ( p ,.. , ., p,) and q = ( q1 ,...,q n1. But
and, since the square bracket vanishes, the principle of least action is briefly written: @dq-@dp- 8,H@dt- 8,Hsqdt = O . We find again the Hamilton canonical equations: dp = -aqH dt
dq = d,H dt
In addition, the (complete) relative integral invariant P dq - ~ ( ~ , q ,dft = ) P d~ - H ( q , ~ , p , t ) d+t ds(q,e,t)
(1 0-36)
implies P = -8,s
p = 3,s
8,s + ~ ( d , s , g , t =) H(g,e,p,t).
The principle of least action, expressed by G I ~ ( P -~ Q~+ ds) a = f ( h ~ d ~ - i Q da~, ~- & d t- a Q ~ @ d-r a , H ~ d t ) + [ 6 ~ ]+[P&?]: : =O
where the last two terms are zero, implies: dQ = d,Hl dt
d~ = -dQHdt
~,H=o.
2
The principle 61 L dr = 0 (implying that H is not dependent on q ) is going to be given up for the following principle. The consideration of individual trajectories will be abandoned for a "field mechanics." 4.1
VARIATIONAL PRINCIPLE (with one degree of freedom) Let i
w = exp(%S)
Lecture 10
424
be a complex function where A is the Planck constant h divided by 2n and S is a real-valued fwnction. From this fiction y(q,Q,t) of dimension [ L - ' ' ~ ]we , define a new (real) Lagrangian of dimension [ M T - ~: ]
where
y/*
is the conjugate of y
Remark 1. The Lagrangian L is a priori a function of
U ,I,
y * ,d4y,d@*, dQv, dQ+v*, d,ry and
a,w* Remark 2. For L to be real, we have made
K
symmetrical with respect to complexes.
Remark 3. From d v = ijjdry* dS = -ifi -
v
V/
we deduce 8,s y
= -it)
a4v/
this last equation being also written:
Piy = ih d , ~ . Remark 4. The "new" Hamiltonian
2 is evidently dependent on
y, y', d,y,...
Principle Recalling that there is no variation in q and 1 (4= & = 01, we say:
PR23 For variations Sy and Sy' which are zero at of the configuration variable q, we admit that:
t,
and t, and also at limits of a domain
The dimensions of this are [ A 4 L 2 ~ -]' Notation. To simplify, we denote the partial derivatives 3, L , daqVL ,.. . respectively by L,,Lyq ,...
Consequently, the variational principle (10-39) leads to the Euler equation:
Symplectic Geometry, Hamiltonian-Jacobi Mechanics
and also to its conjugate.
Hamiltonian presentation If we introduce the generalized momenta
n = L~~
n*=~. YQ
b
respectively conjugated to y(q,t) and y (q,t),then the Hamiltonian is written:
By viewing again S, + H = H and by noticing that the contribution of S, = -iA-
v, = ititvf Y
W
to L = y* H y is immediately written: 1
ifi (vv,'- ~ * r y , )7
we see that:
H,, = h y
H
. = - 1 - Vh2.
Wt
By specifying that the following integrals are zero since the variations are zero at the limits for the respective variables q and t :
we are going to prove the following:
PR24 The principle (10-39) leads to the following Harniltonian equations: ~q
yi = H,.
=Hn
n, = -H, + d
, ~ ,
(10- 41)
n:, = -H,. + a , ~ ~ ;
Proof: We have: (2)
(2)
0 = 8 ~ I ~ d ~ d t = 6 1 ~ ( n ~ , + n * y ~ - ~ ) d ~ d t (1)
(1)
Lecture 10
426
But
n w , + n * s =~ a,(nsy+n*~y*)-n,w-n:,fiv* ;
and also (since H,
= ll - L ,
= 0 = H .): (P;p
- 6H = -H, 6l3- H,. 6II' - H, 6 ~- H,. SY* - H,, 6 ~- Hv; , 6 ~ :
where the last two terms are such that
H , 6y, - H v':. Gy* = d,(H, 6 y + Hvt. 6 y 0 -) a,H,, Gyr - a,H,,. 6 ~ ' . Therefore
the third and last terms vanishing, the equations (10-41) are thus shown.
Example of the harmonic oscillator.
Since
and
then with usual notations relative to the harmonic oscil1ator (remembering the corresponding Hamiltonian expression), we have:
The corresponding Eulerian equation, namely
is the Schrodinger equation: m 2 2 -tr2 2 m Vqq+ j o 9 ~ = i ' W t
9
the conjugate equation having the same significance. We can find again this equation as follows:
H = l l y q + I I * y ~ - L = L , ~ wq q. + y ;L- L
Symplectic Geometry, Hamiltonian-Jacobi Mechanics
and from (1 0-41) we obtain Ir/ 4
, ah n *
that is the Schrodinger equation.
If we introduce ~ / ( q , t=) f (q, t)e's(q3')'" in the equation (10-42), we obtain two fundamental equations:
-
the Hamilton-Jacobi equation complemented with a second member:
-
the equation
~ ; + t / , s , + & =f Os ~ ~ which is the continuity equation, that is
a,f2+&a,(f2a,s)=o .
(10-44)
So f behaves Iike a conservative density in the configuration space (Ehrenfest theorem).
Remark. From a transformation such that Ldq=ndy+II*dy*-Hdq =ndly dw' -r?dq +&(y, y*,y,v*,q)
+n*
a "new" Jacobi theorem, for
= 0, could be deduced extending the analytical mechanics, this last folIowing from the heuristic theory of quantum mechanics.
VARlATIONAL PRINCIPLE (with n degrees of freedom)
4.2
By refering to section 1 of lecture 9, we consider the following Riemannian metric
and the Lagrangian L = -2Ia g q l q J+ a i q l + a , - V
with L., = a,.qJ+a, = p, . 4
Lecture 10
428
Let us consider again the case P = 0 = y/,. From the expression of the (complete) relative integral invariant: Ldt =
f ( p , +p;)dqJ- H ( q r , p . , p ~ , t ) d ~
we first deduce
where yl, is denoted W, Second, the comparison of terms in t leads to
Principle PR25 For variations Sy/ and Sy/* which are zero at t, and t, and also at limits of a domain of the configuration space, we admit that:
where
& d ( q ) is the volume form
and
~ = ~ a ~ ~ ( i h ~ ; - n , ~ ' ) y( )- +i l( V ~ -- a ,~ ) ~ y2JZi *A + i - - ( ~ ~ ~ - y , ~ * ) .
PR26 The variational principle leads to the following Euler equation
---I
'
fi 2
a,(&aJk
= ih(d, - a
akY)+ ((Y- a,) + +a, d - i L ( a , & - %(a' 247
J
(1 0-46)
3,)~.
Proof: The Euler equation being
(L&)*. - d , ( ~ & ) ~ ;- B,(L&)*; and since
=0
Symplectic Geometry, Hamittoniau-Jacobi Mechanics
then the proposition is proved.
Example. In classical notations, given two points of masses m, and m, , and putting:
then, in a central field, the Lagrangian determining the metric is written:
L
MA^ + + m i 2 - ~ ( I I i l l ).
The Euler equation is written:
from which two Schrodinger equations are obtained, the one for the mass center: it2 --(C a:, ) y , ( S , t ) - j h a , ~ , ( R ,=t )k , c R o 2m
and the second for the couple of reduced mass:
So, if k is nonzero, the barycentric movement influences the couple state.
EXERCISES Exercise 1. If a closed one-form a is in involution with closed one-forms /?and involution with the Poisson bracket { j9,y ). Answer. We must prove:
fl(a"{~,r I#) = o . We have (by playing):
4 a u {P,r , I#)= w ( ~ # , [ P n , r U l )
prove a is in
Lecture 10 = i,lw([P",r")
= a([/l#,r'l)
= i , ,a
since a ( X ) = (a,X ) = i,a
= Lpliy,a+iy,Lp,a
since [L, ,i,] = il,,,,
[B
.Y
1
+
= Lprir*a i ,di ,a! , r B
but
i7.a = a ( f )= ~ ( a ' , ~ ' ) lp"a= m(a#,p#) and thus the involution assumptions: w ( a t Y r # =) w(a#,PU) =0
lead to the proof
Exercise 2.
If (x*) are 2n local coordinates on a symplectic manifold and w = zdr' A drm', prove i=I
that: ( x l , x J )= O
,
{ y + t , X l ~ + 1j = 0
,
(x',xn+' } = -S/ .
Answer. For the coordinate functions, we have:
X,,= 3
&+I
=
dl
{ x ~ , 1x=~- - W ( X ~ , = , X- ~~( -)a ~ + , , - a , + , = ) o
,xx.+, ) = -w(a, ,a,) = o
{ x ~ + ~ ,) = x -m(xxn+, ~ + ~
{X~,X= ~ -+~ ~( x x , , x x . + ,=) -w(-a,,,ai)
= -4.
These expressions are also obtained from (9-17) with local coordinates ( x A ) = (x',xn+');for example:
Exercise 3. Prove that the symplectic form w is an absolute integral invariant of Hamilton canonical equations XA
=
aH ax"
Answer. The well-known expression i,w = w,
x Aa!xB is written:
Symplectic Geometry, Hamiltonian-Jacobi Mechanics
Exercise 4.
With the aid of d Bprove , the Jacobi identity b'f , g , h E Cm(M2,) : {V;g>*h)+{{g,h),fI+{{hJI,g)=0 . Answer. We have:
a
w" -(W"
axA
af ag )- ah + 0." - -a- ; i ( ~ c D -&c
axD ax*
ax
ag
ah )af+.,.
-- B axc axD ax
The second and third terns cancel and so on. Indeed, by changing the names of summation indexes A -+ C, C -+ A and next C + L), D + B, B + C , then the third term
becomes :
(the second term opposite).
Exercise 5. If (M,w) and (N, p) are symplectic manifolds and if 4 : M + M' and y : N + N' are diffeomorphisms, then the Cmmapping f : M + N is symplectic fr the mapping ly 0 f 0 4-' of (Mf,4,0) into (N', y,p) is symplectic. Answer. First, we recall that if (M,w) is a symplectic manifold and # : M + M f is a diffeomorphism, then we know that (Mf,$*w) is a symplectic manifold and #is symplectic. Now, if f : M + N is syrnplectic then we have:
Conversely, if
y/ o f o
4-' is symplectic then we successively have:
Lecture 10
432
In addition, we have so proved that f is symplectic symplectic.
#I
every local representative off is
Exercise 6 (Poisson theorem). I f f and g are first integrals of Hamilton canonicai equations, then their Poisson bracket is a first integral.
Answer. From
8,f + { f , H
>=o
d,g + { g , H 1= O
and { H , { f , g J ) + { f , I g , H ) l + { g , { H , fl ) = O ,
we have:
that is
Exercise 7. In the case of a conservative system (H = E), prove that the existence of a first integral
f (p,, q l ,t) implies the successive dnf are first integrals. at" Answer. The Poisson theorem lets us assert { f , H } is a first integral. But f is a first integral
!IT 3,f + ( f , H
1= 0
and thus d,f is a first integral. Then, the Poisson theorem means { a,f , H ) is a first integral. But 8,f is a first integral r f l
a2.f is a first integral, and so on. thus we conclude df Exercise 8. Prove that a transformation is canonical #I it preserves the Poisson bracket. Answer. Given a transformation y A= y A (x B ) , we have:
Symplectic Geometry, Hamilton-Jacobi Mechanics
433
but
and thus If,g1y=(frgIr
Q
uAB = ( Y ~ , Y ~ ~
that is gff the transformation is canonical.
Exercise 9. By considering a generating function S,(ql,Q') , prove that a canonical transformation on the phase space f: ( p i q') , + (4,Q1) verifies f'w = w . Answer. With the generating function S,, the transformation
implies
and
The first term of the right-hand member of
f'w = dp, ~ d q='-dq' aqi %I
A dq'
+* dQj A dq' a p aqi
is zero because
For the same reason, we immediately have:
and thus f ' o
= w.
Exercise 10. Establish f in order that the transformation
4 = f (qi)cosp, be canonical.
Q' = f (q' 1sin pi
Lecture 10
Answer. Since
f ~ i n p , d p ~ )asA ( ~ ~ i I I p ,fdcosp, y ' + dpi) aq
d < ~ d Q =(7cospidq'' ?If aq
the transformation is canonical ~ f l Vi E ( I,. ..,n ) :
that is (f(q1))2 = -2qr + -2
( c i ~ R ) .
Exercise 11. By considering the phase space, is the diffeomorphism
Q = ~ n ( ~sin - 'p)
P = qcotp
a canonical transformation? Answer. Yes, because ~ P A = ~ ( -Q- d p42 + c o t p d q ) ~ ( - d p -cos q - 'pd q ) sin p sin p
Exercise 12.
Specify the type of constant matrix on the phase space
( a v ) making
canonical the following diffeomo~phism
Q' = ql.
4 = pi +a,qJ
In this case find the generating function of type S, (4,qf). Answer. From
d c A dQ1 = (dp, + aa,dqJ)A dqi = dp, A dqi we deduce i
(av - a,,) dq'
avdqJ A dq = 0 = jqi
thus the matrix in question is symmetric.
The canonical transformation condition :
A
dq'
Symplectic Geometry, Hamilton-Jacobi Mechanics
leads to
and thus S, = -$a,qiqJ
Exercise 13.
Find the motion equation of a simple harmonic oscillator by introducing the generating function
where q is the displacement and m2 = k / m Answer. We immediately have:
as, -- mwq cotQ p=dq
This last relation leads to
and thus p=j
G z
cosg
The Hamiltonian is invariable under the transformation because the time doesn't explicitly appear in the generating function. Since
and
the Harniltonian is written:
Since the coordinate Q is cyclic, we conclude that conjugate momentum P is constant, it is
Elm.
Lecture 10
The motion equation is
Q=wt+c where c is an integration constant (fixed by the initial conditions). We find the well-known displacement
Exercise 14. A particle moves in a vertical plane under the action of its weight (mg) without friction. This plane is constrained to rotate about a vertical axis with constant angular velocity w. z() Calculate the Hamiltonian. Find the Hamilton-Jacobi equation and a complete integral. (id Determine the general solution of Hamilton canonical equations. (110
Answer. (r) By introducing the cylindrical coordinates r7B7zand knowing that is written:
d = w , the Lagrangian
There are two degrees of freedom, the generalized coordinates are r and z. Since the generalized momenta are
the Hamiltonian is immediately:
(id The Hamilton-Jacobi equation is
By separating the variables t ,r , z : S = -El
+ S,( r )+ S2(z) ,
the Hamilton-Jacobi equation becomes:
Syrnplectic Geometry, Hamilton-Jacobi Mechanics
So, the first term only depends on variable r and the second term only on variable z; they are thus constants. By putting
with E l + E 2 = E , we immediately have a complete integral:
{zii) The general solution of motion equations follows from
that is m dr
m dz
Exercise 15.
Find a complete integral of the Hamilton-Jacobi equation of a spherical pendulum of length R. Establish the general solution of Hamilton canonical equations. Answer. In this problem of two degrees of freedom the Lagrangian is written
where the two generalized coordinates are the colatitude 0 and the longitude Hamiltonian is
I$,
and the
The Hamiltonian-Jacobi equation is written:
In this scleronomic problem where # is a cyclic coordinate, we search a complete integral of type:
S = -Et+c@+0(0). The Hamilton-Jacobi equation that is
Lecture 10
lets us obtain 8. A complete integral is thus
The general solution of motion equations follows from
Remark 1. The first part of the general solution shows the "horary law," the second leads to #(@,E,c,a,). Remark 2. We can immediately verify that: 2
det(-)
d S = da,,dqJ
-
0
d dO -dB
1
ado -dc dB
z0.
Exercise 16.
In spherical coordinates r , 8,4, let
be the Hamiltonian of a particle of charge e subject to a central electric field [potential V ( r )] and to a constant magnetic field = b fi, (along pole axis), c being the velocity of light, (Q Deduce there is no separable solution of the Hamilton-Jacobi equation. (ii) If the term ( e b / ~ )is~neglected, show the Hamilton-Jacobi is separable. Find a complete integral and the general solution of Hamilton canonical equations. Answer. (i) The coordinate 4 being cyclic, let us try a separable solution of type S = -E t + H(r) + 0(6)+ A#
where the constant A is the generalized momentum associated with 4. The Hamilton-Jacobi equation is written:
Symplectic Geometry, Hamilton-Jacobi Mechanics
Two successive partial derivatives lead to the following absurd result:
There is thus no separable solution. eb 2 (id If (-) = 0,then the Hamilton-Jacobi equation is written: C
where the first term depends only on r and the second term only on 9; there are two (constant) opposite terms:
Therefore, a complete integral is
The general solution of motion equations is
m dr
Exercise 17,
Given the unperturbed Hamiltonian
+
H,(P,~) = p2 + c q
- dt?
+
4.
Lecture 10
and the perturbed Hamiltonian H ( P * ~=) H , ( P , ~ )- el2
EER*.
(r;l Find a complete integral of the Hamifton-Jacobi equation and show the generd solution of motion equations of the unperturbed problem. Deduce a canonical transformation cancelling H , (with "new" variables E and a). (ii) With these ones, determine the Hamilton canonical equations of perturbed motion. (iir;) Determine the perturbed motion equation. Show the general solution and the particular solution with the following initial conditions q(0) = 0 and q(0) = q, .
Answer. (9 Let us search a complete integral of type S = -Er + S,(q) The Hamilton-Jacobi equation
admits a complete integral
*I,/-
S(q,E,r) = - ~ t
dq.
The general solution of motion equations is: - - a =as --=-[&
aE
that is
with
The system of equations (1) and (2) is equivalent to
Therefore, the "new" canonical variables are two integration constants: Q=E
P=a
such that
In conclusion, a canonical transformation cancelling H , is
(id By refering to the canonical transformation cancelling H , , the perturbed Hamiltonian H is written:
Symplectic Geometry, Hamilton-Jacobi Mechanics
The Hamilton canonical equations are
Given E and well determined values Eo and a,, we have:
E = -2&(t- ao)Eo+ &c2(t- a0)3 and thus cL
E = - ~ ( t - a , ) ' ~ ~ f4- ( t - a ~ ) ~ + ~ From ( t = 0 ) c2 4
E, = -EO; E~+ &-a:
+K
we deduce: cZ 4
E = -&(t -a,)' E, + & ( t- a,)'
f
In the same manner, we obtain:
(iii) The Hamilton canonical equations
lead to the following equation q-2&q=-c
of which the general solution is = Ae&'
+
Be-&' +C 2&
From initial conditions ( t = 0 ):
we deduce the particular solution
By developing around
E
=0
and putting T = a
c2
Eo + E U ~-E&-ao ~
t , we have:
4
4
Lecture 10
From series expansions
we obtain (after obvious simpIifications):
BIBLIOGRAPHY
R.ABRAHAM and J. MARSDEN, 1978, Foundations of mechanics, Benjamin. V. ARNOLD, 1976, Les methodes mathhatiques de la mecanique ciassique, Mir. L. AUSLANDER and RE. MACmNZIE, 1963, Introduction to differentiable manifolds, Mc Graw-Hill. M. AUDZN, 1991, The topology of toms actions on symplectic manifolds, Birkhaiiser. M. BERGER and B. GOSTIAUX, 1992, Giornchie differentielle, P.U.F..
G. BREDON, 1994, Topology and geometry, Springer.
E. CARTAN, 1946, Leqons sur la gbrndtrie des espaces de Riemann, Gauthier-Villars. E. CARTAN, 1958, Lepns sur les invariants integraux, Hermann. H. CARTAN, 1967, Calcul diffkentiel, Hermann.
C. CHEVALLEY, 1946, Theory of Lie groups, Princeton University Press. G. CHOQUET, 1984, Topologie, Masson. Y. CHOQUET-BRUHAT, 1968, Geornktrie diffihentielleet systemes extkrieurs, Dunod. Y. CHOQUET-BRUHAT, C. DEWITT-MORETTE and DILLARD-BLEICK, 1977, Analysis, mmifolds and physics, North-Holland, Amsterdam.
G. CONTOPOULOS, M. HENON and D. LYNDEN-BELL, 1973, Dynamical structure and evolution of stellar systems, Geneva observatory, Switzerland.
Th. DE DONDER, 1%7, Theone des invariants integraux, Gauthier-Villars. J. DIEUDONNE, 1960, Foundations of modem analysis, Academic Press, New York.
H. FLANDERS, 1963, Differential forms with applications to the physical sciences, Academic Press. S. GALLOT, D. HULIN and J. LAFONTAINE, 1990, Riemannian Geometry, Springer. C. GODBILLON, 1969, G h e t r i e differentielle et mecanique analytique, Hermann. H. GOLDSTEIN, 1959, Classical mechanics, Addison-Wesley, Reading, Mass.
S. HEGALSON, 1978, Differential geometry, Lie groups and symmetric spaces, Academic Press. J. KELLEY, 1975, General topology, Van Nostrand, Princeton, N.J..
L. LANDAU and E LIFSHITZ, 1960, Mechanics, Addison-Wesley, Reading, Mass.
Bibliography
S. LANG, 1962, Introduction to differentiable manifolds, WiIey, New York. A. LICHNEROWICZ, 1964, E l h e n t s de calcul tensoriel, A. Colin. E. MADELUNG, 1957, Die mathematische Hilfsmittel des physikers, 6" auflage, Springa. J. MILNOR, 1974, Morse theory, Princeton university press.
CLW.MISNER, K S . THORNE and J.A. WHEELER, 1973, Gravitation, Freeman and CO, San Francisco. A. ONISHCHIK and E. VINBERG, 1993, Lie groups and algebraic groups, Springer
S. KOBAYASHI and K. NOMIZU, 1963 and 1969, Foundations of diffaential geometry, 2 vol., Interscience, N.Y. PHAM MAU QUAN, 1969, Introduction a la gborndtrie des varietks diffbentiables, Dunod. H. POINCARE, 1957, Methodes nouvelles de la mkanique dleste, 3 vol., Dover Publications. J. ROELS, 1985, La gBom6trie des systemes dynarniques harniltoniens, CICAO.
Y. TALPAERT, 1991, MBcanique gknerale et analytique, Cdpadues, Toulouse. Y. TALPAERT, 1993, Geomdtrie diffkentielle et mecanique analytrque, Cepaduks, Toulouse.
M. SPIVAK, 1979, Differential geometry, Publish or Perish, Berkeley. A. WINTNER, 1947, The analytical foundations of celestial mechanics, Princeton University Press.
GLOSSARY List of a few successively encountered symbols S, T...
topological spaces
E, F...
finite-dimensional real vector spaces
U, V.. .
open sets (in E, ... or on a manifold)
f l g . . . :U(C E ) + F
C " (or smooth) mapping
Isum(E ,F)
set of isomorphisms between Banach spaces E and F
L(E $3
continuous linear mappings of E to F
df, E L ( E ; F )
differential off at x
C4(U;F )
set of mappings of class C on
L, ( E ;F )
space of q-linear mappings from ( x ) E ~ into F
Tf : U x E + F x F
tangent off
(U,PI
local chart
M , N...
C "manifolds
w,...
submanifold of manifold
f:M+N
mapping of manifolds
m"YM,, N,)
set of Cq-diffeomorphisms of M, onto N,
g , h... E C m ( M ;R ) = C m ( M )
smooth (or C" ) real-valued functions
f b h = h of
pull-back of function h byf
f' : C a ( N ; R )+ C m ( M ; R )k: H f ' h
pull-back of fwlctions
EE q
UcE
derivation or tangent vector at p,
xPo
tangent space at po E M basis of TpoM tangent bundle of M projection of tangent bundle 7M ( x ) : TM
--+ M
fiber over {x)
~¶m)
set of Cqsections of TM
X,Y ... E X(M) = r m r n 4 )
vector fields
E
M
Glossary
446
[
1 : X ( M ) x X ( M ) -+ X ( M )
I.,Y
(Lie) bracket Lie derivative of Y with respect to X
= [X,Y]
#, : W ( c M ) + M
local transformation
ax,mx,Px,...
one-forms or covectors at x
(w,,x)= u x ( X )= X ( 0 , )
contraction between vectors and 1-forms
T:M
cotangent space at x E M
(dr')[or (@)I
cobasis of TiM or dual basis of
E
M
(a,) [ or of (ei) ]
cotangent bundle projection of cotangent bundle T*M cotangent fiber over {x) set of Cqsections of T'M one-forms or covector fields on M tensors of type ( ) at x E M tensor multiplication tensor algebra at x E M bundle of tensors of ty-pe tensor fields tensor algebra p-forms at x E M exterior multiplication
exterior algebra at x bundle of p-forms p-form on A4
Q ( M )=
60'( M )
Grassmann algebra
r=O
pull-back of p-forms
d : OP( M ) -+
aP+' (M)
exterior differentiation
( 5)
exterior differential (or derivative) of p-form o push-forward volume
div, X
divergence of a vector field Lie derivative (operator)
i, : C2
( M )+$2P-' ( M )
inner multiplication
G
Lie Group
L(G)
Lie algebra of Lie group G
exponential mapping of L(G) boundary of a chain c
metric tensor scalar product of two basis vectors
metric element of M
g, (X, y) =
(x, Y)
scalar product of X and Y
TM +T:M:XHX,=~(X, ) I
flat (or lowering) mapping
:T,*M+T,M:~HCV#
sharp (or raising) mapping
b' #
p, =
ddetg drl
( * t ) i q + !..,in =
A,.. A&"
$ (pgj i l...in ti1-i9
( * @ ) l p + , . . in = $!(/Jg
)t, ,,in
loi1..iF
VxY
volume form adjoint of q-vector adjoined form (or dual form) covariant derivative of Y along X
V : X ( M ) x X(M) + X ( M ) : (X, Y) H VXY affme connection T( , X , Y ) = V,Y - V , X - [ X , Y ]
torsion
grad, f : TIM + R
gradient off
Lap = div.grad
Laplacian
6 = (-l)w+n+' * d *
codifferentid operator
4,~
(: ) Riemannian-Christoffel tensor components
R ~ , y
(I1 ) curvaturetensor components Ricci curvature tensor components
scalar curvature
Glossary
448
configuration space generalized coordinates q' generalized velocities
9'
velocity phase space Lagrangian
generalized momenta pi momentum phase space Hamiltonian Poisson bracket of functions Liouville form or canonical 1-form canonical 2-form symplectic form flat mapping sharp mapping Poisson bracket of 1-forms Hamiltonian vector field Hamilton canonical equations
Lagrange bracket of functions
action integral
8s dS hy7,q1,t) at aq
--I-
=0
Hamilton-Jacobi equation
INDEX
absolute derivative, 287 absolute integral invariant, 249 acceleration vector, 287 action integral, 330,350,417 accumulation point, 2 adapted chart, 64 adapted coordinates, 212 addition ofp-forms, 159 addition of tensors, 135 adelphic integral, 378 adelphic transformation, 379 adjoined form, 277 adjoint Lagrangian, 335 adjoint operators, 298 adjoint of q-vector, 277 adjoint transformation, 223 admissible change of chart, 43 admssible chart, 42 admissible diffeomorphism, 342 affme connection, 285 a f i e group, 285 alternation mapping, 153 R-algebra, 100 algebra of exterior differential forms, 165 angle, 260 antiderivation, 200,204 antisymmetrization, 153 arcwise connected, 6 atlas, 4 1 autoparallel vector field, 290,300
backwards transported parallel vector, 289 B m h space, 9 Banach theorem, 9 base space, 92 basis for topology, 2 pth Betti number, 244 Bianchi identity, 308 bigraded algebra, 143 N-body problem, 353 Boltzmann equation, 359 boundary, 241,242,247 bracket, 98
canonical coordinates, 389 canonical extension, 402 canonical forms, 40 1 canonical isomorphism, 262,391 canonical projection, 29, 52,60 canonical transformation, 350,399 (canonically) conjugate momentum, 333 Cauchy sequence, 9 central force problem, 42 1 chain, 239 change of cobasis, 128,134 characteristic function, 354 chart, 40 Christoffel formulas, 293 Christoffel symbols, 286-288 circle, 46 class Cq 50-51, 58 closed exterior differential system, 208 closed form, 173 closure, 2 closure of an exterior dfferential system, 208 cobasis, I26 coboundary, 243 cocycle, 248 codifferential operator, 297 cohomologic to zero, 243 cohomology class, 243 collisionless Boltzmann equation, 359 commutative diagram, 102, 110, 170, 173, 187, 192, 193,201,203,402 compact space, 6 compact support, 236 compatible atlases, 42 compatible charts, 41 complementary equation, 335 complete absolute integral invariant, 349 complete integral, 416 complete integral invariant, 347 complete relative integral invariant, 350 complete space, 9 complete vector field, 107 completely antisymmetric form,153 completely integrable fext. diff. syst.), 209-211 components of a one-form, 127 components of a tensor, 132-135
Index components of a vector, 82 components of a vector field, 96 configuration space, 325 configuration spacetime, 327 conjugate momentum, 333 conjugate tensor, 264 connected, 6 connection coefficients, 286 conservative force, 353 conservation law, 341 conservation of energy, 406 conservation of angular momentum, 34 1 constant of structure, 217 constant rank theorem, 28 constraint equations, 326 contact point, 1 contact transformation, 351 continuity equation, 357, 364 continuous mapping, 5 continuously differentiable mapping, 13,20 Contopoulos model, 376 Contracted multiplication, 137 contraction, 125, 137 contravariant, 126, 128 contravariant component, 263 coordinate (induced) basis, 99, 205 cosrnologicd constant, 310 cotangent bundle, 129, 144 cotangent vector, 337 cotangent vector space, 126 countable basis, 3 covariant, 126, 128, covariant component, 263 covariant derivative, 286,289 covariant differential, 287 covariant functor, 17,95, 141 covector, 125, 337 covector field, 130 covering, 40-41 curl, 181, 183,313 curvature tensor, 303 (: ) curvature tensor, 305 curve, 7 1 p-cycle, 247 cyclic coordinate, 336 cylinder, 48, 323 cylindrical coordinate basis, 265
d' Alembert-Lagrange principle, 339 Darbow theorem, 389 decomposable form, 157 deformed element of chain, 249
dense, 2 pth De Rham cohomology space, 244 derivation, 77,78,97,204 derivative, 12, 13 derivative in one tangency direction, 77 determinant, 178 diffeomorphism, 13,25,41, 55-56 differentiablecomposite map theorem, 13,27 differentiable covector field, 130 differentiable manifold, 43,44 differentiable manifold structure, 42 differentiable mapping, 10, 12 differentiable tensor field, 141 differentiable vector field, % differential, 11, 12, 17-19,23, 84,95 differential one-form, 22 differential operator, I93 direct rotation group, 267 directional derivative, 20, 77,87 distance, 2 divergence, 182,295,296,314 domain of chart, 40 dud basis, 126 dual form of fom,277 dual space, 2 1 dynamic equilibrium, 357
ergodic trajectory, 379 Ehrenfest theorem, 427 Einstein curvature tensor, 309 Einstein equation, 310 Einstein summation convention, 20 electromagnetic field, 28 1 element of a p-chain, 239 embedding, 56 encounter, 358 energy conservation theorem, 406 equal chain elements, 239 equivalence class of tangent curves, 75 equivalent atlases, 42 equivalent exterior differential systems, 208 equivalent norms, 8 equivalent volumes, 176 ergodic case, 381 Euclidean connection, 287 Euclidean vector space structure, 260 Euler equations, 332,425,428 Euler-Noether theorem, 342 exact form, 243 exact one-form, 22 exponential mapping, 220 exterior algebra, 16 I
Index exterior derivative, 170 exterior differential, 170, 173 exterior differential system, 207 exterior differentiation, 170 exterior multiplication, 160 exterior product, 155, 158 exterior product space, 159 extremal, 300,330
Faraday 2-form, 282 fiber, 92,94, 129 field of covectors, 130 finite chain, 239 fust integral, 209, 369,399,412 flat mapping, 262,392 flat space, 305 flow, 108 flow box, 105,108 0-form, 158, 163 p-form, 154, 162 foliation, 205 Fourier transform of Boltzmann eq.,361 Frobenius theorem, 205,209 function along a curve, 75 fbture light cone, 268
general linear group, 221
g e n d i z e d coordinate, 32 5 generalized force, 340 generalized momentum, 333 generalized trajectory, 326 generalized velocity, 327 generating field of group, 108 generating hction, 412-4 I4 geodesic, 290,300,344 geodesic equations, 301 germ, 74 gradient, 149, 183, 187,294,295 Grassmann algebra, 165 gravitational acceleration, 344 gravitational potentials, 312 Green-Riemann formula, 253
Hamilton canonical equations, 334,409 Hamilton-Jacobi equation, 415,419 Hamilton principle, 330 Hamiltonian, 333
Hamiltonian flow, 408 Hamiltonian system, 405 Hamiltonian vector field, 349, 404 harmonic form, 298 harmonic oscillator, 4 19,426 Haussdorff space, 4 helicoid, 323 Hodge decomposition, 299 Hodge star operator, 278 homeomorphism, 5 homologic chains, 247 homology, 247 homology class, 247 hydrodynamical system, 367 hyperboloid (one sheet), 324
ideal, 207 image of tangent vector, 84, 86 image of vector field, 99 immersion, 17,28, 56 implicit fimction theorem, 16,27 incompressible vector field, 296 induced Riemann structure, 284 inner automorphism, 223 inner product, 199 inner multiplication, 199 integral, 108 integral curve, 103, 104 integral manifold, 207 integral of a form, 235-237 invariant curve, 380 invariant field of forms, 21 I invariant tensor field, 2 11 invariant under X, 212 inverse mapping theorem, 15,26 involutive fimctions, 398 involutive one-forms, 396 isolated point, 2 isolating integral, 370, 3 72 isometric Riernann manifolds, 284 isometry, 284 isomorphic vector spaces, 216 isomorphism, 9 isotropic cone, 268 isotropic vector, 268
Jacobi identity, 101, 191,394, 398 Jacobi integral, 335 Jacobi (last multiplier) theorem, 251 Jacobi theorem (H-J equation), 417
Index Jacobi theorem (symlplectic diffeo.), 403 Jacobian, 25 Jacobi matrix, 23 Jeans approach, 359 J m s theorem, 373
Killing vector field, 274 kinematics frequency, 360 kinetic energy, 327-328
Lagrange bracket, 410 Lagrange equations, 332 Lagrangian, 329 Laplace-de Rharn operator, 298 Laplacian, 297 leaf of foliation, 205 left-invariant, 2 16-217 left translation, 215 Legendre transfonnation, 333 Leibniz derivation rule, 77, 97 length of arc, 300 Levi-Civita connection, 292 Levi-Civita symbol, 275 Lie algebra, 101,214,398 Lie algebra of Lie group, 217 Lie bracket, 98 Lie derivative, 101, 121, 194 of covector field, I96 of diffaential form, 192 of function, 186 ofp-form, 199 oftensor field, 191,195,198 of vector field, 188,196 Lie group, 215 Lie subalgebra, 395 Lift off; 403 light cone, 268 light vector, 268 limit point, 2 line element, 258 linear form, 20 Liouville-Boltanam equation, 358-359 Liouville form, 350,404 Liouville theorem, 352, 357,407 Lobatchevsky space, 321 local bundle atlas, 92 local bundle chart, 92 local coordinate system, 4 1 local coordinates, 40-4 1 local diffeomorphism, 14,25, 55-56
local moment, 361 local representative, 50 local transformation, 104 local vector bundle, 92 local vector bundle isomorphism,, 92 local vector bundle mapping 92 locally compact, 7 locally connected, 6 locally finite covering, 7 locally Hamiltonian, 407 locally Lipschitz, 103 Lorentz group, 272 Lorentz transfonnation, 269 lowering mapping 262,385,392
M manifold, 43 n-manifold, 44 Maupertui s principle, 329 Maurer-Cartan equations, 218 maximal atlas, 42 maximal integral curve, 108 Maxwell equations, 282 meridian trajectoq, 374 metric, 258 metric element, 258 metric form, 257 metric space, 2 metric space topology, 3 metric tensor, 258,264-266 Minkowski spacetime, 267 Mobius strip, 49 module, 97 moment around the mean, 364 moment around the origin, 363 moment equation system, 364 momentum phase space, 337 momentum phase spacetime, 33 8 C '-morphism, 50 Moser lemma, 388 motion in configuration space, 33 1 multiplication of a fonn by a scalar, 159 muItiplication of a tensor by a scalar, 135
natual basis, 82 natural with respect to diffeomorphisms, 20 1 natural with respect to Lx , 192 natural with respect to mappings, 173 natural with respect to restrictions, 170, 193 Newton equations, 33 1 nilpotent, 200
Index nondegenerate bilinear form, 385 norm, 8,260 normed space, 8 number density, 363
one-form one-parameter (global) group, 107, 109 one-parameter (local) group, 105-106, 109 one-parameter subgroup, 218 open (set), 3 open sphere, 3 operator (V, R)(Y,2 ),308 operator R(X,Y), 302 opposite chain elements, 239 opposite orientations, 48 orbit of one-parameter group, 109 orientable manifold, 48, 174 orientation, 176,241 orientation preserving, 177 orientation reversing, 177 oriented manifold, 176 orthogonal group, 266 orthogonal vectors, 260 orthonormal basis, 266 Ostrogradski formula, 253 overlap mapping, 4 1
Painleve integral, 335 paracompact, 7 parallel translation, 290 parallel vector field, 289 partition of unity,7 past light cone, 268 perfect constraint, 338 Pfaff system, 208 phase density, 358 Poincard group, 273 Poincare lemma, 244-246 Poincare non-existence theorem, 375 Poisson bracket, 345,394,396 Poisson equation, 355,356 Poisson theorem, 399,4 12 positively oriented chart, 177 potential, 183 principle of least action, 329 product atlas, 45 product manifold, 45,52 product of vector fields, 98 product topology, 3 projection, 92, 94, 129
proper length element, 274 proper Lorentz transformation, 272 proper time, 273 pseudo-norm, 260 pseudo-Riemannian structure, 257 pull-back, 59 pull-back of differential form, 167,237 pull-back of tensor, 139, 143 push-forward, 140, 169 push-forward of differential form, 167
Q quotient space, 246-248
radical, 263 raising mapping, 263 rank, 25 "read on a chart" function, 57 "reading" of a curve, 72, 76 reduced potential, 374 refinement, 7 relative integral invariant, 252 relative topology, 1 reverse orientation, 176 rhwnomic system, 325,346 Ricci (curvature) tensor, 307 Ricci identity, 293 Riernann-Christoffel tensor, 303 Riemann connection, 292 Riemannian geometry, 25 7 Riemannian manifold, 258 Riemannian metric, 258 Riemannian structure, 257 right-invariant, 216 right translation, 215
same orientation, 48 scalar curvature, 307 scalar product, 259,260 Schrijdinger equation, 427 scleronornic system, 325, 373 second derivative, 14 second order tangent bundle, 109 section, 94,129, 169 self-adjoint operator, 299 separable system, 421 separable topological space, 4 sharp mapping, 263,393 signature of g, 259
Index skew-symmetric form, 153 spatial type vector, 268 special Lorentz transformation, 269-271 special orthogonal group, 267 special relativity, 280 sphere Sn,47, 49 spherical coordinate system, 265 standard volume, 390 state space, 338 statistical mechanics, 356 steady state, 373 stellar dynamics, 352 stellar orbits, 373-375 stereographic projection, 38,49 Stokes formula, 241-242,254 strict component, 158,275 submanifold of manifold, 64 submanifold of R ",59 submersion, 17, 28, 56 subordinate, 8 support, 7 support of a chain, 219 surface of section, 380 symplectic chart, 389 symplectic form, 388 symplectic group, 387 symplectic (linear) mapping, 387,399 symplectic manifold, 388 symplectic matrix, 387 symplectic structure, 388 symplectic transformation, 399 symplectic vector space, 387
tangent bundle, 93, 111 tangent curves, 73 tangent of a mapping, 16,95 tangent mappings, 10 tangent space, 80 tangent vector, 75 tensor, 13 1 tensor algebra, 136, 142 tensor density, 275 tensor derivation, 193 tensor field, 141 tensor multiplication, 136 tensor product, 136 tensor product space, 136
third integral, 374-38 1 time type vector, 268 topological space, 1 topology, 1 torsion, 291,3 10 torus, 48 transported tensor, 195 truncated absolute integral invariant, 348 truncated integral invariant, 347 truncated relative integral invariant, 350 tube, 249 twice differentiable, 14 two-dimensional harmonic oscillator, 371
uniqueness of flow boxes, 105 universe trajectory, 268 universe velocity, 346
variational derivative, 341 variational principle in field mechanics, 424, 428
variational principle of Hamilton, 330 q-vector, 277 vector bundle, 92 vector bundle isomorphism, 391 vector bundle mapping, 93 vector bundle ofp-form, 162 vector bundle of tensors, 138 vector fidd, 96 velocity conditional distribution, 363 velocity phase space, 327 velocity vector, 287 virtual displacement, 338 Vlasov approach, 360 volume, 174 volume form, 276 volume preserving, 1 78
Willmore lemma, 193 work, 249
