The Geometry of Submanifolds

+ Ljl sin' ,p = LAk,

L,1 = (nir;,,,-,) = (LA, - L,1) cosp sin tip+ Lk, cos2p = 0.

L = (nir",,;,) = LA,t sin' cp + LL sin 2cp + 4 cos'- p = L. If L2 = L.2 holds, then we consider the second fundamental form with respect to n3,

and so on. If Lkk above

L, then we use the fact that TT2Iki /.R1(Lkk

0. Similarly to that stated

- 4) = 0.

This implies LRt = 0 for k 34 j and for any r. Thus, the submanifold with at normal connection admits n principal directions. Conversely, suppose that F" admits n principal directions at each point. Choose the coordinate system so that at the fixed point Po one has gik = L' = 0 for i 0 k. Then (I) implies T,,,,Ifk = 0. Therefore, the suhmanifold admitting n principal directions at each point has a flat normal connection. 4.8 On Non-Immersibility of n-Dimensional Compact Manifolds of Non-Positive Curvature It is well known that every compact surface in E3 has a point with positive

Gaussian curvature. That is why there exists no regular isometric immersion of a

SUBMANIFOLDS IN EUCLIDEAN SPACE

81

two-dimensional closed Riemannian manifold of non-positive curvature into E3. In this section we state theorems which are much more general than the one above. Theorem 1 (Chern, Kuiper) Let M" be an n-dimensional compact submanifold in Euclidean space E"+P. There exists at least one point Po E M" such that there are no real asymptotic directions on M" at Po.

Recall that the direction dr on the submanifold is asymptotic if 11°(dr) = 0 for each second fundamental form II'. Let r be a position-vector of M" and p = r2. Then dp = (r, dr), d2p = (dr)2 + (rd2r). For M" compact p reaches its maximum at some point Po. Introduce the coordinates u1..... u" about this point. Then d2r =

&2r

Ol Oui

du' du' = (1 r. r,,, + L ',,n,) du' du1.

At point Po we have dp = 0 and d 2p < 0. It follows then that (r, r"k) = 0. Therefore dZp = (dr)2 + (rn,,)IIh'.

If the direction dr is asymptotic at Po, then llP = 0 and hence d2p = (dr)2 > 0, which contradicts d2p < 0. The theorem is proved.

Theorem 2 (Otsuki) to X':

The following system of quadratic equations with respect

A`;X'X' =0.

a= l.....N.

(1)

i, j = 1, .... n satisfying conditions N < n and

AhAJk -A`-A)h)Xi}"XkY* <0

(2)

a=1

for any X' and Y', has a non-trivial real solution.

Consider X1,. .. , X" as the Cartesian coordinates of a point X in n-dimensional Euclidean space. Denote

a= 1.....N. Define the function

H(X) _ E+G. Let Xo = {Xo} be a point in the unit sphere S"-1: &.,X'Xi = I such that H(X)Is._, reaches its minimal value A2 at Xo. We are going to show that A = 0. Suppose that A A 0. Consider qj,,,(X) as the components of some vector in N-dimensional space.

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82

Then H(X) is the square of its length. If u are the elements of an orthogonal matrix N

of order N and

<%(X),

(X) _ 3=1

then

N

H(X) _

;(X))2.

n=1

Taking appropriate a we can satisfy the conditions iP,*,(X) = A and 0 (Xo) = 0, a = 2, ... , N. We may assume A > 0. Otherwise we just replace ij'1 with -01. Further on, the symbol * is omitted.

At Xo, the conditions dH = 0, d2H > 0 must hold. Set dXj = P. The equation dH = 0 obtains the form A1iXXY'=0. (3) Since X is the unit vector, the equality b;,XXY1 = 0 holds. If it is a constant then

(AjXX-µb;,Xx)Yr=0 for any Y orthogonal to X. In the case µ = A this equality is true for all Y. Therefore the next relation holds:

A'jXo=Ab,X. Considcr d2H:

(4)

N

d2H = 2 [d.1 )2 +

+

=2

+

=2 Ad2?V1+t(dp,,)2 >0.

.

X (5)

n=2

At Xo we have

d2t1 =

dO,, = 2A`;,

yk

(6)

Take N vectors A' in n-space of the form

A° = {A-Xo},

cr= l,._N.

By hypothesis N > n. So there exists at least one vector Yo = { Ya} orthogonal to all A", i.e. the vector satisfying

A`;.XXYj=O, a= I,...,N.

(7)

Because (4) holds and A 96 0 we have b;, Xo Y = 0. So, we can substitute the com-

ponents of Yo into (6). Then do,, = 0 for all a = I , ... , N. By virtue of (5) d2,i1 = 2A YYYY > 0.

SUBMANIFOLDS IN EUCLIDEAN SPACE

83

On the other hand, by virtue of (2) and (7)

Therefore A Y Y o = 0. Let H(Yo) = /l2. By virtue of (7) for a > 2 we get t / (Xo cos 9 + Y o sin g) = A (Xh cos 9 + Yo' sin 0)(Xa cos 9 + Ya sin 0)

= sine

Yo)

Further tpi (Xo cos 9 + Yo sin 0) = cos, 9x.

Hence H(Xo cos 9 + Yo sin 9) = cos4 9a2 + sin4 9µ2.

Denote the right-hand side above by f (9). The derivatives off (9) at 0 = 0 satisfy the conditions f'(0) = 0, f'(9) > 0. The direct calculation gives f"(0) = 4 cos 20(-A2 cos' 0 + tie sin20) + 4 sin 29(x2 + µ2).

It follows then that f"(0) = -4X2 < 0 which contradicts the assumption of Xo being the minimum point for H(X). So, .\ = 0. The theorem is proved. As a corollary of the theorem just proved we obtain the non-immersibility theorem (Chern, Kuiper for q = 2.3. Otsuki [2] for the general case). Theorem 3 Suppose at each point of the compact Rientannian manifold M" there exists q-dimensional tangent subspace such that the sectional curvature of M"

for planes contained in this subspace is non-positive. Then M" can not be isometrically immersed into

E"+q-1

Suppose M" is isometrically immersed into E""'- 1. Apply the previous theorem to

the situation when X and Y belong to the subspace Tq of the hypothesis and N = q - I < q. Consider Ay as the coefficients of the second fundamental form. By virtue of Gauss' equation they have to satisfy (2). In accordance with theorem 2 there exists an asymptotic vector (it belongs to Tq). But this fact contradicts theorem 1. In a particular case, if M" is a compact manifold of non-positive curvature, then there exists no isometrical immersion of M" into E2n-'. Observe that 2n - I is the lowest dimension because there exist immersions of :dimensional compact flat manifolds into E2' (Clifford torus, for instance). 4.9 Extrinsic and Intrinsic Nullity Indices Let L' be the coefficients of the second fundamental form of M" in E"+P with respect

to the normal vector n at a point P. Denote by v(P) the dimension of the largest subspace T' in the tangent space T" satisfying

L' X'=0 for every X = {X'} E T", r = 1,... , p.

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84

The number v(P) is called the relative index of nullity or the extrinsic nullity. The subspace T" is the subspace of relative nullity (or extrinsic nullity). Denote by A(X, Y) the second fundamental form operator. It associates to each pair of tangent vectors X, Y E T" the normal one with components L'y . X' P. The definition above of v(P) can be reformulated as follows: v(P) is the dimension of the largest subspace T" C T" such that if X E T then A(X, Y) = 0 for every Y E T". If we take X = Y, then A(X, X) = 0, i.e. each X E T" is an asymptotic one.

Consider the Riemannian curvature operator Rxy, which takes Z E T" to R(X, Y)Z. Denote by p(P) the dimension of the largest subspace TO' in the tangent space T" satisfying Rxy Z = 0 for each X E TO' and all Y, Z E T". The number p(P) is called the nullity index or intrinsic nullity index. We shall write simply v and p assuming that the point is the same. The subspace T" is called the space of intrinsic nullity. The definition implies that the sectional curvature in TN is equal to zero. Theorem The extrinsic and intrinsic nullity indices (v and p respectively) of the submanifold M" C E"+P satisfy inequalities

v
(R(X, Y)Z, W) = (A(X,Z)A(Y, W)) - (A(X, W)A(Y,Z)). , ) means the scalar product in tangent or normal spaces, X, Y, Z. W E T". Suppose X E T". Then A(X, Z) = A(X, W) = 0 for all Z, W. By virtue of Gauss' equation (R(X, Y)Z. W) = 0. i.e. R(X, Y)Z = 0 for all Y. Z. Thus X belongs to the intrinsic nullity subspace TO'. Then T" c TO' and v < p.

where (

Prove now that u:5 p + v. Suppose X E TO'. Then (R(X, Y)Z, W) = 0. The Gauss equations imply

(R(X,Z)A(Y, W)) - (A(X. W)A(Y,Z)) = 0. Take the vector Z satisfying A(X, Z) = 0. Then (A(X, W)A(Y, Z )) = 0. If p = v the inequality holds. So. assume v < p. Take the vector X E TO' such that X T". Then there exists Wo E T" satisfying A(X, Wo) 34 0. Observe that A(X, WO) is orthogonal to all A(Y, Z ). Let TN-" be the orthonormal complement to T" in TO' and its basis. Take any vector e E T" and consider the system of vectors A(e, ei ), ... , A(e, ec,-")

Suppose the system above is linearly independent. Then the number of vectors in the

system is not greater than p. Hence p - v < p and the theorem is proved. Suppose now that only part of the system above is linearly independent. Taking if necessary the linear combination of e; we can achieve the situation when A(e, ei ), ... , A(e, eq) are linearly independent and A(e, eq+,) = ... = A(e, e,,_") = 0. Set Z = e, X = eq+, .

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85

As A(eq+i. e) = 0 and eq+i E Tr', there exists Wo such that A(eq+l, WO) t 0 and in addition A(eq+l, Wo) is orthogonal to A(e, el ), ... , A(e, eq). Consider the system of vectors A(Wo,ej),.... A(Wo,eq). The system A (e, ei ), .... A (e, eq). A(Wo, eq+i) is linearly independent. For sufficiently

small a the system

A(e,ei)+FA(Wo,el),

A(e,e2)+eA(Wo,e2),...,eA(Wo,eq+1)

is linearly independent as well. Set e = e + eWO. Then the latter system is equivalent to

A(e,ej),...,A(e,eq+i) Thus the number of linearly independent vectors in normal space has been increased

by I. Continue the process by induction. Eventually we come to the linear independent system

which is the maximal one. Thus p - v < p. The theorem is proved. 4.10 Saddle Submanifolds

The surface in three-dimensional Euclidean space with Gaussian curvature K < 0 is

called saddle. You are never able to cut "a cap" of the surface and its second fundamental form can always be reduced to the form a2dr2 - b2dv'-.

It would be interesting to generalize the notion of saddle surface for multidimensional spaces and some generalizations of this kind have been proposed by Sheffel [3-4). We call an image of a circle under a continuous map a loop or a contour. We begin with the following definition: We say that the suhmanifold contains no two-dimensional caps if each of'its points has a neighborhood U such t/rat.for any flat section E' (r > 2) any contour in U n E' is contractahle in this intersection to the point.

Completeness of the submanifold is not necessary here. The complete n-dimensional (n > 2) .suhmanifold in E' is called a saddle if one can span the two-dimensional simply connected surface F2 C F" n E' for any closed rectifiable contour L in the section F" n E' (2 < r < m) which is contractable in F" to the point. The contractability of contour L in F" is essential. The hyperboloid of one sheet x, + x" -.%23 = I is saddle surface F2 C E3. The last condition of the definition above is not fulfilled for the section x3 = 0. The section x3 = 0 gives the contour L which is not contractable in F2 n E2 to the point. For the case of F'- C E3 the definition above means that if L is a flat contour in F2 n E2, contractable in F2 to the point, then the entire flat domain in E2 bounded by L is in F2 (see Figure 10).

86

THE GEOMETRY OF SUBMANIFOLDS

FIGURE 9

FIGURE 10

If the section of saddle submanifold F" with some E3 is a surface F2, then the latter cannot be locally convex. Suppose we are able to cut "a cap" with some plane E2. Let L c F2 n E2 be the closed curve. Then L is contractable in F2 (hence, in F" )

to the point. On the other hand, there exists no two-dimensional surface in F2 n E2 = F" n E2 with L as its boundary. This fact contradicts the definition of saddle submanifold.

The notion of saddle submanifold can be expressed in terms of second fundamental forms. The following theorem holds: Theorem 1 The class of C2 saddle subntanifolds coincides with the class of submanifolds satisfying the following property: for each normal vector v at x E F" the corresponding second fundamental form 11(v) at the same point has, after diagonalization, the form 11(v) = Ap dan + Ay &c2,

with ApAq < 0.

SUBMANIFOLDS IN EUCLIDEAN SPACE

87

See the proof in [3]. Theorem 1 implies that the metric of a saddle submanifold has

a non-positive sectional curvature. Indeed, consider the expression L°L; - (Ljj)2, where Ljj are the coefficients of the second fundamental form with respect to n,, in the coordinate system u11... , u". Suppose xI , ... , x" is the coordinate system in which the form above has diagonal expression A. d + Ay dx2-. As Lii are tensor components. " = A° ar" Lij

aa

5W AU-1 +

Ay

a.Xq axg

ani -5;j-

Set ax°/su' = -y°;. We get

L L r - (L ) z= [A,,(7p1)2 + Ay(7yi)2]l r[A°('Y°i)z + Ae(7v)z - [App°r7p1 + Ag7y,'Yyt] z

A°Aq('YpPYv - 7p1Yyi)2 < 0.

Thus L°Li - (Lijp)2< 0. It follows from Gauss' equations that the sectional curvature is non-positive. It is easy now to give an example of a saddle submanifold F3 C E5. Let XI, x2, x3, yl,y2 be Cartesian coordinates in E5. Define the submanifold by equations J'I = x - a;,

yz = (xI + xz - x3)2 - (xI + x2 + x3)2.

-oo<xi
i=1.2,3.

However, the hypersurface z = x1 + .r2 - x3 is not a saddle in E 4 by the definition above, because its second fundamental form 11= 2(dx? + d ri - dx?) contains three squares of differential instead of not more than two according to Theorem 1. It shows that the given above definition of saddle submanifold narrows the class of submanifolds too much. Sheffel's second generalization of the saddle surface notion uses the notion of relative homology group. In this way we get a more wide class of submanifolds. A complete n-dimensional submanifold F" C Em is called k-saddle if for any rplane Et (k < r) the (k + I)-group of relative homology with integer coefficients vanishes: HA.,I (F", F" fl E'') = 0. The saddle submanifolds defined by the first definition are 1-saddle submanifolds

according to the second definition. Glazyrin [6,7] stated the theorem, which generalizes Theorem 1. Theorem 2

Suppose Fis an n-dimensional C22-submanifold in El". The submanifold F

is oj'class k-saddle if and only if for each point x E F and each unit normal v at x the corresponding second fundamental form can be reduced by coordinate change to the form

Q;dxi +

+ail dxl

-bjdxj,

where i, j < k.

The hypersurface in E"+', defined by

Z=x1 +...+.1"J - _,C+I

C+2

j

THE GEOMETRY OF SUBMANIFOLDS

88

is k-saddle submanifold with k = maxi, j ). We already proved that the sectional curvature of a I-saddle submanifold is nonpositive.

For the case of k-saddle submanifolds (k > 1) the curvatures for some 2-planes can be positive. But necessarily there exists 2-planes with non-positive sectional curvature. The following theorem was proved in [5]. Theorem 3

Suppose F is an n-dimensional k-saddle C3 submanifold in VIP. Then

for each x E F" and every (k + I)-tangent subspace Tk+I and every e E Tk+I there exists r E Tk+l such that the sectional curvature K(e. T) < 0. Definition Let A be a symmetric matrix having n+ positive and n_ negative eigenvalues. The type of matrix A is the number tip A = max (n+, n-). When A is the matrix of the second fundamental form of the submanifold with respect to v, then tippv is the A type.

Lemma Let F be an n-dimensional submanifold in E"+p. If there exists el E T" such that for any e E T" K(e, el) > 0, holds then one can point out the normal vector v with tipF v = n.

Let et,. .. , e" be the basis of T". Choose the orthonormal system of normals vi,. .. , vp in such a way that v, is directed along the vector of normal curvature corresponding to e, . Then for coefficients of the second fundamental form we obtain: Li I = = L = 0, L11 360. We may assume Li I >0. If tipF v < n, then there exists the vector e producing a non-positive value of second fundamental form. Set e2 = e. Then L22 < 0 and p

KF(el,e2) = L11IL22 - (L12)2_ E(Li2)2<

0

:=2

which contradicts the hypothesis. Now prove Theorem 3. Suppose that there exist Ek+I C T" and e E Ek+I such that KF(e,f) > 0 for every f E Ek+l. Endow E"+p with Cartesian coordinates with the origin at x and axes x l , ... , xn, y,.,. .. , yp parallel to e, , ... , e" E T" and v, , ... , vp E NP respectively. Moreover, we choose the first e1..... ek+I in Consider a (k + 1)-dimensional submanifold S in E"+p, defined by Ek+l.

k+I

r 1.%= I

Xk+r = 0,...,X. = O.

The F-normals v,, ... , vp are the S-normals as well and L', i, j < k + I are the coefficients of S second fundamental forms. Observe that ek+2, -..,e. are normal to S in E"+p as well, but the corresponding second fundamental forms vanish.

Therefore the inequality Ks(e,f) > 0 is satisfied for some e E Ek+I and any f E Ek+I. By the lemma, there exists the vector v in the normal space to S with

SUBMANIFOLDS IN EUCLIDEAN SPACE

89

tips v = k + 1. It is clear that v = E; a,v;. Let and Ly(v) be the coefficients of the second fundamental forms of F and S respectively. Then

>a,L',j, L;,1(v) _ >a,L` , i,j < k+ 1. Therefore L41(v) = L,1(v) for i, j < k + 1. The matrix II the matrix IIL;,(v)II is of higher order n. Hence

II is of order k + 1 while

tiPFV = tip II L;J(v)II > tip IILy(v)II = tipsv = k + I.

The last equation contradicts the hypothesis of F being k-saddle. Theorem 3 is proved.

Note that the Mathematical Encyclopedia V.4 [13] contains a slightly different definition of k-saddle submanifold, given by Toponogov and Sheffel: The submanifold F"' in E" is said to be k-saddle if at any point for any normal f the number of eigenvalues of the same sign of the second fundamental form with respect to does not exceed k - 1, where 2 < k < m. It was proved that if F" is complete and saddle in the meaning above, then its homologies satisfy N;(Fm) = 0, i > k.

5 Submanifolds in Riemannian Space 5.1 Submanifolds in Riemannian Space and Their Second Fundamental Forms

Let V' be an m-dimensional Riemannian manifold endowed with curvilinear coordinates y' , ... , y"' and the first fundamental form ds2 = a"edy° dy3.

The system of functions

y° =f°(x...... x"), a = 1,...,m defines a regular n-dimensional submanifold F" C VI if f" are regular of class Ck and the rank lla ,I is equal to n. If the length of any curve in F" is evaluated with the metric of V", then we say that the metric on F" is induced by the metric of ambient space. Let us be given an n-dimensional Riemannian manifold R" endowed with the metric d12 = gj dx' dxj. We say that the Riemannian manifold is isometrically immersed into V"' as a submanifold F" if the length of any curve in F" being evaluated with the metric d12 is equal to the length of the same curve being evaluated with the metric ds2 of ambient space. This means that for any infinitesimal motion, the differential dx' and corresponding differentials dy" are related by a"sdy" dyI

= gy dx'dx1.

Since y" are the functions of x', then dy" = a dx'. Comparing the coefficients of two forms we obtain 8y" 8y' a°o

8x' 8.rj 91

= gij.

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92

As y« are the functions on P. their usual derivatives are covariant one. Hence the latter equation can be rewritten as a«,3

Yn.i

d Yj =

gij.

with i fixed form the components of vectors on

As dy« = ez, dx', the values

lar;J VI" tangent to coordinate curves x' in F". Their linear combinations form the tangent space Ts. Let p be the components of a vector normal to F", i.e. anriY.r

- 0. ;r =

Choose the system of (m - n) unit orthonormal vectors ", o = I,. .. , m - n, which are orthogonal to F". Then evidently a08

Sri =

a«3Y.; v_ o - 0.

Differentiate the equality

a«,Byy =g,; covariantly with respect to uk in the metric gj. We get 8a«p y,° y j Y 'k + a0 y a 11 k Y,; ay7

+ a«#Y. Y = 01

(1)

Change the places of subscripts i, k and then j, k. We obtain two equations:

ay Y,ky y',+a«fly 1Y

+a00JAYk =0,

8r Y,°YkY,j+a«BY.jjyk+aOfl)'a.iYkj=0. d 8

Adding the last two equations and subtracting (1), we obtain (after suitable change of summation indices)

(8y + byp - 8y )Y i YY k + Denote by the form

k

= 0.

(2)

the Christoffel symbols of the metric of ambient space. Then (2) has

a«RY, ji y

k

+ r«J0,' Y' y jj Y, k = 0.

We can write the second term as l'$ Y,", Y,;Y k = a^/] i ,,Y' Y,JY k

SUBMANIFOLDS IN RIEMANNIAN SPACE

93

Therefore. equation (2) finally takes the form

ryy)

o

a,,3 Y

=

Thus the expression in parentheses gives the components of a vector which is normal to F". On the other hand, this vector has to be linear combination of normals , with some coefficients L,. Thus we obtain the Gauss decomposition

ya+ra

(3) v

The values LOij' form the components of a symmetric tensor. The form L! du' dul is called the second f undamental form of the subnranifold with respect to normal vector G.

Exercises

(1) Let Vbe endowed with the metric ds2 I H?(4')2. Prove that the second fundamental form of hypersurface y"' = const has the following coefficients L

H; 8H; L; = 0 for i l n=-l1M49y,",-34

(2) Let the metric of V3 be ds2 =gii (d1'1)2+2g12dyIdy2+g22(dy2)2 +dr2. where g;; depends on y I , Y 2. t. Prove that the second fundamental form of the surface

t = const is 1 dg; clu`du'.

2 Ot 5.2 Gauss-Weingarten Decompositions for the Submanifolds in Riemannian Space

Let X be the vector field on V'. If cp is the scalar function on V', then the derivative of cp in the direction of X is by definition

vxp = v"wX", where means the covariant derivative with respect to _v" and X' means the v-th component of X. For any vector fields Y. X on the Riemannian manifold V'", define the covariant derivative of Y with respect to X by

VxY= (0"Y)X", where rt

vI,

= ay""

Y'

is the covariant derivative in V'" of Y with respect to y

THE GEOMETRY OF SUBMANIFOLDS

94

In local coordinates, aa

Y"

VxY" = Y" X° + r" Y'`X". If Z is once more a vector field, then we can find the derivative of the scalar product (YZ) by the following rule:

'x(YZ) = (OxY,Z) + (Y,VxZ). Denote by rj =

V } the vector fields tangent to coordinate curves in the sub-

5717

manifold F". Find the expression for the covariant derivative of r,; with respect to rj

in the metric of V'. We have

r Or r"1 = Caf.v` + I' "P+' ar °

r

02)'"

!j' +

rv = J

OU'auJ

+

ay'' ay" =

its, au' auJ

Vrr" J

(I )

We see that'Vrri = V,,rj. Now find the expressions for the second covariant derivatives of v" in F": 'J

(2)

'J all A

au' auj

where I are the Christoffel symbols of the metric on P. Combining (1) and (2) we get y',; + I',ja

au A

av" ay " + I,p" atr' auj .

Using the Gauss decomposition (3) from Section 1, we obtain Vr,rj = ICI, r`a + LA"'jj.

(3)

We can regard this equation as another form of the Gauss decomposition. Therefore, we can define the coefficients of the second fundamental form as L" = (Vr,ri

where the parentheses mean the scalar product on V'. Using (1) we have Ly = L;,. Take now the normal vector field " as Y and rj as X. Then

V, n =

n rn 1

Lp n

.j

Define the torsion coefficient by

A'./j = vp r,S") b,," we have ju.1j = -µ,,/j. Decompose the covariant derivative into tangent and normal to F" components. Let V" AiA r. A. + Bp"li As

SUBMANIFOLDS IN RIEMANNIAN SPACE

95

Just like in Sections 2 and 3 of Chapter 4 we find

A.k = -Ljtgtk,

Bn./j =

µv"/j'

Thus we come to the following Weingarten decomposition

Vr,&" = -Ljto

k

r, k + ltnrr/! p.

the last decomposition can be rewritten

Taking into account the expression for as

auf -

(4)

Ljtgk);k -+Uar/jG

(5)

5.3 Gauss-Codazzi-Ricci Equations for the Submanifolds in Riemannian Space

To obtain the Gauss-Codazzi-Ricci equations one can use the Gauss-Weingarten decompositions. The method is the same for Euclidean space. We produce here only the final results. The Gauss equations: Rijk, =

(L,A Ljt

- L I Lk) +

Y.; Y Y'k )'i,

(I )

where Rfh is the Riemannian tensor of V"'. The Coda:_i equations:

Lij.k - R. =lam/k Lo' - lim/i L A + R,..nh '° k .

(2)

The Ricci equations:

I1m/j.k - lkm/k.j +

gu, (LT

(llrrli PrM/k - Ilpr/k Raa,,,.

.

1:R

T = 0.

Lnk - Lik

l (3)

If the ambient space is of constant curvature, the terms in (2) and (3) containing Rr,,Hh vanish.

For the case of general Riemannian manifold the analogue of the Bonnet theorem does not exist. Indeed, the formulation of a similar theorem is impossible because the

equations (1)-(3) enclose members r.°, which depend on a position-vector of submanifolds. The Bonnet theorem can be formulated only for spaces of constant curvature. The proof is similar to that for Euclidean space, see Section 5 Chapter 4. Consider the Gauss equation for a two-dimensional surface F2 in Riemannian space V'". Set i = k = 1, j = I = 2. Then divide the both sides by 911922 - g12. On the left-hand side we obtain the intrinsic curvature of F2. Denote it by K;: Ki =

R1212

911922- g12

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96

Introduce also the extrinsic curvature K,, of F2 in V'", setting

FPa= L° IIL°22 - (L- )2) 12

1

Ke =

911922- g12

Write the expression for the sectional curvature of VI along the plane, tangent to F22:

Kr - =

(a,-, a,* -

)), v; v y 2

Consider the denominator of the expression above. As the metric on F2 is induced from V", we have

(a, a,M - a.6 a.h

.1

2 .1

'

= 911922 - 9_12'

Thus, the Gauss equation for F2 C V"' can be written as K; = K,. + KY,-,

(4)

i.e. the intrinsic curvature of F2 is equal to the sum of the extrinsic curvature of F2 and the sectional curvature of V"'. Exercise

The asymptotic direction r on F" C E"11' is such that all the sectional fundamental forms vanish on r. Prove that for any 2-plane containing r the sectional curvature of F" is non-positive.

5.4 Relations of Covariant Derivatives in Ambient Space and in Submanifolds

Denote by V and V the covariant derivatives in Vand F" respectively. Let X, Y E T,(F") be vector fields tangential to F". We are going to prove that

VXY= (VXy)T

(1)

where ( ) T means the projection into the tangent space of F". Recall the definition of covariant derivative in V"'. In a space T,(V"') there is the basis e,, of vectors tangent

to coordinate curves y" of V"'. Any vector in T,(Vcan be expressed as a linear combination of these vectors. In the special case, Vx Y = ('JX Y" )e"

The components are 'x Y°

= (D,Y°)X°

_ ay "

X'', av' +r- Yl'

where YN and X° are components of X and Y with respect to the basis e,,. In the same way we define the covariant derivation V in F".

SUBMANIFOLDS IN RIEMANNIAN SPACE

97

Let r; = {L } (i = 1, ... , n) be the basis in the tangent space of F". Decompose X ou,

and Y via this basis:

Y=a'r,,

X=blr;.

Then the covariant derivative in F" of Y is of the form

vY

(2)

Y"=dr"+

V= Mr", 1

where r" are the components of rl in ambient space. Therefore

VxY" =

a r" Ibkr k

-'r' bkrA+a(aVV+I',,,ri')rkbk = oyp u r°bk+hkclrVrir,. Recall Gauss decomposition (3) of Section 2: ' Dpi "r ; =a I',k

r +" L.

Then

Ox Y° =

jd;

r,

bk+bkdI'

Change superscripts in the second term as A

Using (2) we can write

ra+a'bkLu t.o. i

j. This gives

//

f7xY=VxY+dbiLf

.

The second term on the right-hand side belongs to the normal space while VxY is in T(F"). Therefore (1xY)T = Ox Y. Q.E.D.

Observe that the projection of 'xY into the normal space of F" has the form (VxY)1 = L°ya'bJ G.

The operator h(X, Y) = (OxY)£ is called the operator of the second fundamental forms. It takes two vectors X, Y which are tangential to F" into the vector which is normal to F".

9K

THE GEOMETRY OF SUBMANIFOLDS

5.5 Totally Geodesic Submanifolds

The submanifold F" in Riemannian space VI is called totally geodesic if each geodesic y in the submanifold F" is geodesic in the Riemannian space V"'. The simplest example gives the plane E2 c E3. Geodesics in E2 are the straight lines only. At the same time they are geodesics in ambient space E. So, the plane is a totally geodesic submanifold in E3. Is it true that planes are only totally geodesic submanifolds in E39 Further on we shall see that the answer is positive. For the sake of comparison let us consider the cylindrical surface. It is well known that geodesics in it are the straight line elements or circles orthogonal to elements or helices. It is easy to see that some geodesics in a cylinder are geodesics in ambient space (straight line elements) while others are not (circles and helices). Hence the cylinder is not a totally geodesic submanifold in E3.

FIGURE I I

FIGURE 12

SUBMANIFOLDS IN RIEMANNIAN SPACE

99

FIGURE 13

We are going to state the criterion for the surface to be totally geodesic. Let T be a unit tangent vector to the curve ry in F". Decompose its curvature vector OTT in V' into tangent to and normal to F" components: V,T = (QrT)T + (VrT)N

From what was proved above. (Vrr)T = V,-r. Hence Vrr = VTT + (QTT)N.

If -y is geodesic in F", then VTT = 0. Therefore 'y is geodesic in V' if and only if (QTT)N = 0. We have

QrT = (Or,a`ri)"a' = a`a1(Orrj)N = a'a'L°

The normals G are linearly independent. Hence afaJL4 = 0 for all o = 1, ... , p. If this supposed for any r, then LO, = 0. So we conclude the following: Theorem A submanifold F" is totally geodesic if and only if all its second fundamental forms are identically zero. The totally geodesic submanifolds are multidimensional analogs of geodesic lines. It is known that a geodesic line is uniquely defined by the point and the direction at this point. The generalization of this statement in the two-dimensional case would be

as follows: given the point and the 2-plane at that point, does there exist a twodimensional totally geodesic surface tangent to the given plane and passing through the given point? Is it unique? Suppose the required surface exists. Denote it by F2. We are going to show that we can obtain the surface by construction. Choose the direction r in a given plane T. Draw in F2 the geodesic in the r-direction. Let it be 7. By assumption F2 is totally geodesic. Hence ry is the geodesic line in V'". As only the unique geodesic passes through the given point in the given direction, we can construct F2 in the following manner: in any direction T E T, draw the geodesic of VI starting at x. The union of all geodesics forms the surface F2 passing through x and tangent to T, (see Figure 14). Thus, if the required totally geodesic surface exists it necessarily coincides with

THE GEOMETRY OF SUBMANIFOLDS

100

FIGURE 14

F2, i.e it is unique. But the surface F2 is not totally geodesic in general. Indeed, if we take two points A and B in it and join them by a geodesic line in F2, then this latter geodesic is not, in general, the geodesic in V"'. However, at the point x for any direction r the condition a'a'L,1 = 0 is satisfied. In other words, F2 is totally geodesic at x.

Apply the formula K; = Ke + Kv,. to the case of F2 at x. Because of K, = 0 we have Ki = Kv-. Therefore the sectional curvature V' on the 2 -plane T is equal at x to the intrinsic curvature of the totally geodesic surface F2. In just this way Riemann introduced the notion of curvature for multidimensional space in his famous lecture. A Riemannian manifold containing totally geodesic submanifolds cannot be ar-

bitrary. Ricci [1] found the system of differential equations that the Riemannian manifold has to satisfy to admit the totally geodesic submanifolds. Note that for the case of a two-dimensional totally geodesic surface in V3, at each point the following condition has been satisfied: R y i y 6 = 0, where y; and O are the components of tangent and normal vectors respectively. As an example, let us consider the totally geodesic submanifolds in the unit sphere

S"'. Geodesic lines in it are great circles. Therefore every totally geodesic submanifold in S' can be obtained in the following manner. Take the tangent subspace T" at x E S' and then draw all geodesics from x tangent to T". The union of the points of all those geodesics forms the totally geodesic submanifold - the great sphere of dimension n. Evidently the inverse is also true, i.e. every great sphere S" is totally geodesic in S", because every geodesic in S" is the great circle in S" and hence is geodesic in S"' Exercises

(1) Let V3 be the hypersurface in E3 defined by equation

xi+x2+x3-x4=1. Prove that the two-dimensional sections of V3 by hypersurfaces xi = 0 are totally geodesic surfaces in V3.

SUBMANIFOLDS IN It1EMANNIAN SPACE

101

(2) Let the metric of V'" be of the form ds2

m

n

= >aijdyidyj+ E anydyndy"y, n3=n+I

i.j=1

,y"'

where aij are independent of y"+1, Prove that submanifolds y" = const a = n + I , ... , m are totally geodesic in V'. (3)

Consider the complex space C"T1 with points (tvl, ... , N'n+1) where wi is a with iv, complex number. Take two collections (w1, ... , wn+1) and (wi, ... , and w; not all equal to zero. We say that two collections represent the same point of the complex projective space CP", if there exists a complex number A 0 such

that ii = Aivi. The real dimension of CP" is equal to 2n. Complex numbers ivj (i = 1, ... , n + 1) are called homogeneous coordinates in CP". It is possible to endow CP" with the Riemannian metric (the so-called Fubini-Study metric) which has the following form with respect to homogeneous coordinates dw, ds2

=

n+1

n[[+I

n+1

-4

lilt'i =I

/=1

tj dwj

4'=11+1

In+I

r-t

2

i

The curvature of this metric on 2-planes (the sectional curvature) takes its values in the segment [,1-t, I]. The metric of CP2 in non-homogeneous coordinates, i.e. when iv; = I. IvI = uI + iv2, W2 = u2 + iv2, can be represented in the form ds2

l

= W t (du2 + dl )Ai + 2B(dui due + dvl di'2) + 2C(dul dv2 + due dvl) }, ci_1

JJ

where 2

W=1+E(u;+1A,=W-Ir,' -vZ, i=

B = -(it, it, + vI v2),

C = -(III v2 - VI u2).

Prove that the second fundamental forms of surface u2 = 1+2 = 0 are identically zero, i.e. this surface is totally geodesic. Hire Take it,, vI as coordinates in this surface. Then its position vector is v = (ul , s,,, 0, 0). Direct the normals 1:1, E2 along the tangents to u2 and v2 curves respectively. Then prove that L,, = r , L = T j.

5.6 On the Intersection of Two Totally Geodesic Submanifolds

In this section we are going to prove the theorem "in the large" on totally geodesic submanifolds [2].

102

THE GEOMETRY OF SUBMANIFOLDS

Theorem (Frankel) Let M" be a complete connected Riemannian manifold of positive sectional curvature. Let V' and W' be compact totally geodesic submanifolds in M". If r + s > n then V' and W' have a non-empty intersection.

Proof Suppose first that V' and W' are arbitrary compact regular submanifolds. Suppose they have no intersections. Then there exists the shortest geodesic a(t) of

length t > 0. joining V' to Ws. Set a(0) = P E V', a(l) = Q E W'. As a(t) is the shortest, it is perpendicular to V' and W' at P and Q respectively. We will prove that there exists the variation X of a(t) such that b2LX < 0, i.e. a(t) is not the shortest. This contradiction implies the statement of the theorem. Let Vo be the tangent space to V' at P. By parallel transport along a(t) we obtain

the subspace V, of the space Mg which is tangent to M at Q E W. Since Vo is perpendicular to a(t) at P, VI is perpendicular to a(t) at Q. Let WQ be the tangent space of WS at Q. Then V, and WQ are two linear subspaces of MQ. Moreover, both are perpendicular to a(t). Therefore the dimension of their intersection satisfies

dim(V,fWQ)> r+.s-(n- 1) > 1. Thus V, and WQ have at least a one-dimensional common subspace. This means that there exists the unit vector X0 tangent to V at P such that after parallel transport it becomes tangent to W` at Q. Let X = X(t) be the corresponding parallel vector field

along a(t) with X(0) = Xo. The second variation of arc-length, represented by the integral r

- J K(T,X)dt. 0

where T is a tangent vector to a(t) and K(T.X) the curvature of M" on the plane (T, X), is strictly negative due to the hypothesis. Up to this point V' and W' were arbitrary. Suppose now that V' and W` are totally geodesic. This makes it possible for the boundary terms to vanish in the second variation. Take X(t) as a variation field. To obtain the "bend" we take the geodesic in M" going in the direction of X(t) for each t. As V' is totally geodesic and X0 is tangent to V0 at P the geodesic going in the X0-direction is contained in V'. For the same reason the geodesic going in the

X,-direction will be contained in W`. Take on each geodesic going in the X(l) direction the arc-segment of length m Then each curve ct = const will have its endpoints in V' and W' in accordance with the variation requirements. But since X0 and X are tangent vectors to geodesics the equation VxX = 0 holds, at P and Q. Hence the boundary terms in the second arc-length variation formula are equal to zero. Therefore

62L.y(0) _ - J K(T, X) dt < 0. 0

Q.E.D.

SUBMANIFOLDS IN RIEMANNIAN SPACE

103

Observe that the intersection of two totally geodesic submanifolds is totally geodesic, too. 5.7 Totally Geodesic Surfaces in the Hypersurface of Revolution in E4

Define the hypersurface of revolution F3 in E4 by the equation

y'-+~'+w2=R2(x),

(1)

where R(x) is some positive function of x. The x-axis is the axis of symmetry for the surface above. Let F2 be the totally geodesic surface in F3 with curvilinear coordinates u1, u2.

Then x, ... , r are some functions of 111,112- Substitute these functions into (1). Differentiating the equation with respect to ui we find i = 1.2.

-RR'x,,, + y,,, + zzu, + wwu, = 0, Therefore one of the normals to F22 is b1 =

{-RR',y,z,w}.

Note that S, is normal to F3, as well. We find the second normal to F2 by taking the vector product of three vectors in E4. We obtain e1

e2

e3

xu,

.ru,

zu,

xu;

y,,2

=u;

-RR'

>

C'4

1Vu,

1r

By definition the coefficients of the second fundamental form of F2 in F3 are scalar

products of VrJ,,, and the normal of F2 in F3, that is 2: L,j = (Or,,ru,2).

As the metric on F3 is an induced one from the ambient space E4, the covariant derivative ru, in F3 differs from the covariant derivative in E3 only on the component normal to F3. Since i 2 is tangent to F3,

Lij =

If F2 is totally geodesic, then Lij - 0. Thus we get the system of three differential equations which define the totally geodesic surface F2 in F-1: xu,

yu,

Zu,

K'u,

X2

)'u:

zu:

Ivin

-RR'

y

z

w

= 0,

i, j = 1, 2.

(2)

THE GEOMETRY OF SUBMANIFOLDS

104

In addition, we have to take into account equation (I). Parametrize F3 as y = R(x) cos W cos 0. z = R(x) sin V cos 0, w = R(x) sin 0.

First, consider the surface x = const. Take cp = u1, 0 = 142 as its parameters. Then

r,, _

I

-

0

cos cp

R cosh,

0

0

cos 9

0

-cosy -cos y,sin0 re = R r., = R cos O -sin -sin k sing '

0

The equation (r,,,r ,ro£I) = 0 implies R' = 0. Therefore the surface x = const can only be totally geodesic when R' = 0. In expressions for r;, and re the first components are zero. Hence the first components of r,,,,,, ro and r, are zero, too. If R' = 0. then the first components of 1;I is zero, as well. In this case equation (2) is satisfied. Thus, the surface x = const is totally geodesic when R' = 0. We call surfaces of this kind a critical sphere.

Now consider the surface p = const. This kind of surface is analogous to meridians of surfaces of revolution in E3. That is why we call them meridian-like surfaces. Take x = 111, 0 = U2 as the parameters in this surface. Introduce four mutually orthogonal unit vectors by e_

('). 0 0 0

a-_

0

cos p cos 0 sin p cos 0

h-

sin0

0

- cos'p sin 0 - sin sin 0 cos0

c=

cosp 0

Then r,,, = e + R'a,

r,,, = Rh.

where (') means derivative with respect to x. Further, we have

= R"a,

= R'b,

r",", = -Ra.

Therefore

(R"a,e+ R'a,Rb,-R'Re+ Ra) _ -RR'R"(a; e + R'a, Rb, e) = 0, (R'h,e+R'a,Rb.1;I) = 0. r 2 {I) _ (- Ra, e + R'a. Rh, -R' Re + Ra) = 0. Equations (2) are completely satisfied. Hence the meridian-like surfaces are totally geodesic in F3. They have the same type of topology as cylinders. Each surface is

contained in E3, passing through the axis of rotation and the plane V = const. Moreover, it is the surface of revolution in that E3.

SUBMANIFOLDS IN RIEMANNIAN SPACE

105

Turn now to the general case. Then we can take x = u1, cp = u2 as the coordinates in F2, while 0 is the function of (x, cp), that is 9 = 9(x, (p). Then r", = e + R'a + R9.1b,

Re cos 0 + R0 ,b,

£1 = -RR'e + Ra.

Taking into account that c., = -a cos 9 + b sin 9, r",", _ (R" - R0Y)a + (2R'B_,. + R9.1Y)b,

r,,,,,_ _ - R9_Y9,,a + (R'9 + R9.1y,)b + (R' cos 0 - R91 sin 9)c,

r,,,,,, _ - R(cos20+ 02. )a + R(sin 0 cos 0 + 9,,,,)b - 2R sin 0 0,,c. Find the mixed product J = (Aa + Bb + Cc, r,,, , r",, CI), where A, B, C are some coefficients which will be substituted later using the expression for r,,,,y. For now we have

J= (Aa+Bb+Cc,e+R'a+R91b,Re cos0+R0,,b,-RR'e+Ra) = R2(abce){B cos0(1 + (R')2) - C9,,(I + (R')2) - R'ROT cos9A}. For the vector r,,,,,, the coefficient C = 0. So we obtain 2R'

-

RJR" g1 +

R'R

93

= 0.

(3)

Introduce the functions of x, setting

R'R" R'R a= 2R' R (l+(R')2)(1+(R')2) Then (3) takes the form

911+ct9,.+I30 =0.

(4)

Using the expression for coefficients A, B and C from decompositions of r,,,", and we obtain two more equations:

01,;+010,; tan0+0029,,=0,

(5)

0.,,,+sin0cos0+(2 tan0+001)02 +0 cost 001=0.

(6)

Rewrite equations (4) and (5) in the form

aIn9, OX

+a+091

0,

aIn0"

+tan9B,+0919,, =0.

Using the equality of mixed derivatives of In 0 and applying (4), (5) we come to the equation 019,,(0' - 0a + 0 = 0. (7)

106

THE GEOMETRY OF SUBMANIFOLDS

If 0' - /3a + I = 0, by integration we obtain R 2 = -x 2 + CI X + c2, where c; = const. Therefore F3 is the three-dimensional sphere. If 0,, = 0, equations (4) and (6) again give /3' - ,Oa + I = 0. Let 0, = 0. Then (6) implies 0,P,o + sin 0 cos 0 + 2 tan 0022 = 0. This is the equation of a great circle in a sphere in spherical coordinates because this is nothing but the equation of a geodesic line of the form 0 = 0(ep). By a change of

coordinates we can write the circle in the form p = const. Thus we come to the conclusion: the totally geodesic surfaces in the hypersurface of revolution in E4 which differ from a 3-sphere are only critical 2-spheres and meridian-like surfaces.

5.8 The Relation of Curvatures of Surfaces in Lobachevski and Euclidean Spaces The Lobachevski space L3 can be represented as a ball D of unit radius in Euclidean space E3 with the metric ds2L

_(rdr)2+(I -r2)dr-, (I -r2)2

(I)

where r is a position vector from the ball center of the point in E3. The surface F2 of

this ball can be considered both as a surface in Lobachevski space L3 and as a surface in V. It happens that there exists a simple relation between the surface geometrical characteristics with respect to both considerations. This relation allows the transfer of the intuitive geometrical perception and moreover the theorems and other assertions from one space to another. The first theorem, stating such a relation, was obtained by Sidorov [6]. Theorem I If K: is the external curvature of the surface F22 C L3 and KF is the curvature of the same surface F2 C E3 and n is the unit normal to F2 C E3 then

i -r2

K;! = KE

, (2)

- (rn)-

In the paper by II'ina and Kogan [7] the relations of other geometrical quantities were obtained. We state them in theorems 2 and 3. Assume that the surface in both L3 and E3 is represented with respect to the same coordinate system a 1, u 22. Let ds 2 = g;; du' du' be the metric of F2 c E3 and dl2 = a;, du` du' the metric of F2 C L3. The following theorem holds. Theorem 2

The ratio of metric tensor determinants has the form al IQ22 -

a1,,

gI1g22 - gf2

=

1 - (rn)2 (1

-

(rn)2

)

3

(3)

Denote by a;, and a;. the coefficients of the second fundamental forms of the surface F2 in E3 and F2 in L3 respectively. There exists a very simple relation between them.

SUBMANIFOLDS IN RIEMANNIAN SPACE

107

Theorem 3 The coefficients of the second fundamental forms of F2 C E3 and F2 C L3 are proportional: j = µ,\ ,1,

(4)

(rn)2)-1/1.

inhere p = (I - r2)(1 To prove the Sidorov formula and Theorems 2 and 3 we apply the method from [7]. First, consider the metric coefficients of dl2 as induced by ds2

(I

a;, =

(I -r2)

r2)(r r,,,) 2

Therefore, the determinant of the metric tensor a;, is equal to alla22 - a12 =

(A + (I - r2)2(r2,r2, 1. U

r2)-3,

(5)

where A = ru,(rr,,:) It is possible to transform A as

A=(rru:)(r,ru:(r,r'r,, Apply the well-known formula from analytic geometry, [[ab]c] = (ac)b - a(bc). Set g = jgyj Then A -g(rn[nr])

= g(r2 - (rn)). Noting that r2,r2, -

r,,:)2 = g, after substituting A into (5) we get (3). Denote by < > the scalar product in the Lobachevski space metric while () as above denotes the scalar product in the Euclidean metric. Every vector a of contravariant components (a') at the point of Lobachevski space corresponds to some vector of the same contravariant components at the corresponding point (which coincide with each other in the model) of Euclidean space. We denote both of them by

the same letter a and will find the relation between the scalar products < > and ( ). Introduce in E3 the Cartesian coordinates x1, i = 1.2.3 with the origin at the center of the ball D. We will also use the same coordinates in P. Let c,, be the metric coefficients of dsi with respect to this coordinate system. We have

(I -r-)b;,+x'x' (1 -r2) where b;, are Kronecker symbols.

Therefore

=

(ar)(br) + (I - r2)(ab) (I - r2)2

- (ab) - ([ar][br]) (1 - r2)2

(6)

THE GEOMETRY OF SUBMANIFOLDS

108

where [ ] means the vector product in E3. In the sequel we shall use the following:

Volkov formula The following relation of scalar products holds in L3 and in E3: (ab) < ab' > _ -r2,

I

where b' = b - r(br).

Indeed, using (6) we obtain directly

_ _ - <ar> (br)

- (ab) - ([ar] [br]) _ (br) (ar) - ([ar] [rr] ) (I -r2) _

(1 -r2)2

(ab) -r2.

I

The normals to F2 in E3 and to F2 in L3 we denote by n and nL respectively.

Lemma The relation holds between the normals: 1

-r

2

2

1 - (nr)'

(n - r(nr)).

Indeed, let r,,. be the tangent vectors to both of the surfaces. Verify that n - r(nr) is orthogonal to F22 c L3. By the Volkov formula we have

_ (_r =0.

1-r2

The norm of a = n - r(rn) in the Lobachevski metric is

_ (aa)(I - r2) + (ar)` (I -r2)2 Further, a2 = I - 2(rn)2 + r2(nr)2, (ar) _ (nr)(l - r2). Noting the expressions just obtained, we have

_

1 - (rn)2

I - r=

So, the lemma is proved. The proof of Theorem 3 is now quick.

Taking into account the Volkov formula, we have the following expression for the second fundamental form coefficients of the surface:

CI-r

A _ < r,,,,,,nL > _ <

r

1

1 - (nr)2

_

1 - (nr)2

- µA,J.

1/2

SUBMANIFOLDS IN RIEMANNIAN SPACE

109

The following step immediately leads to the Sidorov formula. Indeed, using the proportion of second fundamental forms, for extrinsic curvature of F2 C V we have KL K L _ ,

A1 1A22 - (A 112)

= 2Ai1As - (A, 12 )2 (I - r2)3 14

a11a22 - ail

_

KE

911922 - g12

1 - (rn)2

1-r2

Thus, Theorem I is proved. By means of (3) we conclude the simple corollary. We shall try to describe the form which has the Lobachevski plane L2 of curvature -1 isometrically imbedded into P. Evidently, the extrinsic curvature of the surface which realizes this imbedding must be equal to zero: K, = 0. Hence, by formula (3), KF = 0: that is. in a Euclidean space this surface is developable. It is well known that such a surface is either a cylinder or cone or the ruled surface formed by tangent lines to an arbitrary spatial curve. If the singular points of such a surface are located beyond the ball D, then the surface in L3 which realizes the isometrical imbedding will be regular and complete. 5.9 IsometrIcal Immersions of the Euclidean Plane into Lobachevski Space

Volkov and Vladimirova stated in [8] that the Euclidean plane can be immersed into a three-dimensional Lobachevski space in two ways: (1) in the form of a horosphere; (2) in the form of the surface of revolution of some equidistant line around its basic geodesic line. The second surface was called the equidistant cylinder by the authors. The proof of this assertion was conducted by applying Jorgens' theorem [10]. Indeed let d3.2 = dug + du, be the metric of a plane immersed into V. 11=all dug + 2A12 du1 due + A22 dug the second fundamental form of immersion. The Gauss-Codazzi equations are very simple for this case: OA12

OA11

t9UI

812

= 0,,

dA12

all

0U2

0'

AI IA22 - A1, = 1.

From the two first equations it follows that there exist functions All = (Pu,.

But it must be the case that

A12 = 'A,,,

A12

1\12 = Viu, -

Therefore there exists the function 0 such that 0u, = 4P,

Hence All

A22 = +Gu_ ,

and 0 such that

Bu2 =

A22 = 0U.,. From the Gauss equation we find

0..,,, = I. By Jorgens' theorem [9], any solution of this equation over all the plane (u1, u2) is a polynomial of the second order. Hence, du;u, =const. By rotation of coordinate axes we can achieve 0. Then AI I = I /A,2 = a = const. In

THE GEOMETRY OF SUBMANIFOLDS

110

Lobachevski space of curvature K = - l the principal curvatures of the horosphere are k1 = k2 = I. Introducing the coordinates u1, u2 into the horosphere, we get I = 11= du? + dug; that is, the horosphere is isometric to the Euclidean plane. Consider the positionvector of the surface F2 obtained by the rotation of an equidistant line around its basic curve. Find the extrinsic curvature K, of this equidistant surface as a product of the principal curvatures. To do this, intersect the surface with the Lobachevski plane L2 which is perpendicular to the basic line. We obtain the circle of radius u in the section. Its normal curvature k1 on F2 coincides with geodesic curvature on Li. To evaluate it, introduce in L2 the polar coordinate system with the origin at the point on the basic line. The metric of Li is then ds2 = du2 + sinh2 The geodesic curvature of the circle of radius u, if the chosen normal is external, is k1

_

1

G.

Pg,

2G

_

2 sinh u cosh u = 2 sinh2 u

- cosh u.

The second principal curvature k2 coincides with the geodesic curvature F2 section with Lobachevski plane L22 through the basic line. Introduce in L22 the semi-geodesic system of coordinates (u, v) such that the family of u-curves would consist of geodesics of this plane perpendicular to the basic curve and the v-family would consist of

the basic curve and its equidistants. Then the metric of L2 takes the form ds2 = du 2 + cosh2 udv2. The geodesic curvature of coordinate curve u = const is evaluated by

_ k2

I

pg.,

_

-G 2G

2 cosh u sinh u 2 cosh2 u

= - tanh u.

Therefore the extrinsic curvature Ke of the equidistant surface of revolution is equal to

Ke = k1k2 = tank u cosh u = 1.

By the intrinsic curvature formula for the surface in Riemannian space we have K1 = K,. + KR. Hence K1 = 0. The equidistant surface fundamental forms are I = dui + du,, !I = a du? + ! duZ.

a

Since these forms coincide with the fundamental forms of the required surface, then,

provided that lal 0 1, the equidistant surface gives the solution of the problem. If jal = 1, then the horosphere gives the solution too. Here we use the Bonnet theorem for a space with constant curvature.

6 Two-Dimensional Surfaces in E4 6.1 Normal Curvature Ellipse In this section we will study in a more detailed way the properties of two-dimensional surfaces in four-dimensional Euclidean space.

Choose an orthonormal pair ni,n2 of normals on F2 and consider the corresponding second fundamental forms 11°

= L11 (dug )2 + 2Li, du' du2 + L,°2 (du2)2,

v = 1, 2.

We also have the unique torsion form Y1211dd _ (n1n2,)dil

To each point p E F2 and each direction r E T. there corresponds the vector kN(-r) of normal curvature in the normal space N,,. Put the initial point of kN(r) to p. Then its end-point is the point M of the normal curvature indicatrix. If r rotates in the tangent plane T. then M traces some closed curve. For the arbitrary direction r of infinitesimal components (du',du2) we have 111

112

kN(r) =2ni + 2n2

(1)

Introduce in to the normal plane N., Cartesian coordinates x1, x2 with the origin at p and basis vectors n 1, n2. Then the coordinates of M are

III

THE GEOMETRY OF SUBMANIFOLDS

112

Choose the coordinate system u'. u2 in F2 in such a way that g,, = by at p; then at p we have ds2 = (du')- + (du2)2. Set du'

coscp=

Sinp=

(dul )2 + (

due (th,l )2 + (dU2)2

2)2

Then V is the angle between r and the ul coordinate curve. For the coordinates of indicatrix points we obtain the expression cos'- V + 2Li2 cos p sin p + L,, sin2 ;p.

Transform the expression above to the form

xl =L' 2+ j'2i +L1

L'22

2

L', , sin 2;p.

cos

Set

L11

2

L22

=n

Lit

L222

2

Since gy = b;J at p, in our system a and 0 are the components of mean curvature vector H, i.e.

H=ant+13n2. Let q be a point in N. of coordinates (c.$). Translate the origin to q. In the new coordinate system k, the indicatrix equations have the form

i' = L11

L22

2

cos 2W + Li, sin

i = 1, 2.

This is the equation of an ellipse with the center at q. Rotate coordinate axes on F2 in such a way that the end-point of the normal curvature vector corresponding to the u'

direction would take a position at one of the normal curvature ellipse vertices. Moreover, choose n1,n2 to be parallel to the axes of the normal curvature ellipse.

FIGURE 15

TWO-DIMENSIONAL SURFACES IN E4

113

If = 0 then .Y' = 0. Hence L = L22. Moreover, as p = 0 is an extremal point of dz1 /dip = 0 at V = 0. Hence L12 = 0. Therefore the function .YI = L11 2 L2', cos

k22 = L2,, sin

Denote by a and b the semiaxes of the normal curvature ellipse. Then L 112- L22 -= a,

Iz =

So, we can find the expressions for the coefficients of the second fundamental forms: L11 = « + u,

Lz11 = 0,

Ll, = 0,

L2

1

12 = b,

(2)

/ =11. L22=a-a, Lf1 Now express the Gaussian curvature K of the surface in terms of a. b. a,13. We have at p:

K=

81212 911922

= Li 1

, - (Liz)2 + Li L?, - (Li2)2 1

_ (a+a)(a-a)+fi2-b2=a'-+32-a'-b'. Thus, the Cartan formula for Gaussian curvature is obtained:

K = a 2 +, 32-a2-b22.

(3)

By definition, the Gaussian torsion rct = 2ab. It is proportional to the area of normal curvature ellipse. For the case of F2 in some three-dimensional hyperplane E3 the normal curvature ellipse degenerates into the intercept of a straight line passing through p. Indeed, in this case for any tangent direction at p the normal curvature vector kN is parallel to the normal n of F2. Therefore, the kN end-point is in the straight line passing through p in the n-direction. Prove the inverse statement [1]. Theorem Suppose that at each point p the normal curvature indicatrix or F2 C E4 is degenerated into the intercept of a straight line passing through p. If the Gaussian curvature satisfies K 0 then F2 is in some hyperplane E3.

Observe that examples show the necessity of the K # 0 condition. Proof Choose the normal fields n1, n2 in the following manner. Taken, along the normal curvature intercept, while n2 is in the perpendicular direction. Let L°y du' du1 be the second fundamental forms. Then L2 0. The Codazzi equations List - L ki =1-112/A L1 - µ12// L k

114

THE GEOMETRY OF SUBMANIFOLDS

imply the system with respect to 1112/1: 1412/1

L;, -1t12/241 = 0,

/L12/IL12

- 1 I2/2LI, I = 0.

As its determinant satisfies 2

(L1,) -L1IL _ -K(g11g22 -gj2) 0 0, 1

1

1tI2/i = 0, that is (n2,ni,F) = 0, i = 1,2. Take the curve y: u' = u'(t) in F2 and consider the hyperplane E3(t) spanned on r,, n1. Let y(t) be any smooth vector field along 7 such that y(1) E E3(t). The conditions µ12/i = 0, L'f = 0 are equivalent to 0.

(n,,r,n1) = 0,

Consider the vector field 112 = P12(111, u2). Its derivative is orthogonal to n2. Therefore,

it decomposes into a linear combination of ra, . rn.. n1: n2,,; = Al ru, + A2r,,, + A3n1.

By virtue of the conditions mentioned above, A, = 0. Hence, n2,,, = 0 and n2 is a constant vector field on F2. Moreover, as n2 is a normal, then (n2r,,;) = 0,

i = 1.2.

As n2 is a constant vector field, integrating, we get (n2r) = c.

where r = const. Thus, the components of the F2 position vector satisfy the equation of a hyperplane E3, so F2 is in some E3. Consider now the special class of surfaces F2 c E3 with K =- 0.

(1) Let -y be a regular curve in E4 which is not contained in any E3. Let F2 be the surface formed by all tangent lines to y. Let us find the normal curvature indicatrix and Gaussian curvature of that surface. Its position vector is r(u, v) = p(u) + Pus.

where p(u) is a position vector of y and u is the arc-length parameter. We have r = Pu + PuuV,

re=Pu,

Pan + Puauti.

rnr=Pull

r., = 0. By virtue of the Frenet formulas Par, = k12,

TWO-DIMENSIONAL SURFACES IN E°

115

where 2 is the principal normal to y, k, is the first curvature of -y. Therefore the tangent space to F2 is a span off 1 and {2 and the normals are f;, {4. Find the second fundamental forms with respect to S3, 4. We have

+kk)

=klu6

P,nn, = (k1C,)u =

Hence

(r1,t4)=0,

(r,,,,t3)=k1k2v.

(r,. 3) =

0,

0.

(rre I;a) = 0.

Thus, K - 0 and the normal curvature indicatrix is the intercept parallel to Because -y is not in any E3, the surface F2 is not in any E3, either. (2) Consider the Clifford torus. Its position vector and derivatives are

r=

a1 cos u1

-a, sin 111

a, sin it,

a, cos 111

a2 COS u2

rui

0

-a, cos u, -a, sin 111

0 0

0 ru=

0

a2 sin u2

ruiui =

0

-a2 sin u2 a2 Cos u2

0

r u,en

0

-a2 c os u2 -a2 sin u2

FIGURE 16

'

)

ruiu:=0.

116

THE GEOMETRY OF SUBMANIFOLDS

As the unit normals we take cos u,

nt _

0 0

_ n2 _

sin u, 0 0

COS u2

sin u2

The coefficients of the second fundamental forms are

Lit =-at. L1,=0. L;,=0, Lit=0. L12=0. L;,_-a2. Therefore the normal curvature vector is of the form

kN(r) =

- (at l dui, nt + a2 dtr, n, J ds -

ds -

where ds'- = ai duf + az dtDenote by yt and y2 the coordinate system in the d normal plane. The coordinates of the kN()z end-point at x are y;_-a;dO Hence they satisfy the line equation al)?, + a2y2 = -1.

Thus, the normal curvature indicatrix of a Clifford torus is the intercept [Figure 16]. For definiteness we take a; > 0. 6.2 Knotted Spheres in E4 We present two ways to construct the knotted spheres in Artin [2].

E4. These were proposed by

The firs: way Let xt. x2, x3, x4 be Cartesian coordinates in E4. Define the hyperplane by x4 = 0. Take in E3 the closed knotted curve C without self-intersections. In the x4-axis take two points a(0, 0.0, I) and b(0, 0.0. -1). Join all points of C to a and b with intercepts. Denote by S the set of all these intercepts [Figure 17). This set is called a suspend. Let us show that the suspend has no self-intersections. Indeed, let P1, P2 E C be distinct points. Suppose the intercepts aP, and aP2 have it common point Q. As only one straight line passes through a and Q, the lines containing aP1 and aP2 coincide. These lines intersect E; at Pt and P2. Hence P, and P2 coincide, which contradicts the assumption. The set S can be endowed with the natural system of neighborhoods each of which is homeomorphic to a disk (excluding, perhaps, the neighborhoods of p and q). Let us show that the set of intercepts joining the points of C to a is homeomorphic to a disk. To do this, suppose that 1 is the length of C. Take in E2 a disk D with the center at 0 and boundary St of length 1. Define the one-to-one correspondence between points of this St and C by equality of arc-lengths. Suppose P E C corresponds to Q E S1. Let p(P) be the length of intercept aP. Define the one-to-one correspondence

TWO-DIMENSIONAL SURFACES IN E4

117

CCE6

FIGURE 17

between points of the radius OQ of the disk D and points of aP as follows: map the point of D in the circle with the center at 0 radius r and the point in aP at a distance of p(P)r27r/1 from a. When r -+ 0, the points of S tend to a. The center 0 corresponds to a. Thus, S is homeomorphic to the pair of disks glued by their common boundary C; that is, S is homeomorphic to S2. Let us show that the surface F3 is knotted. We use the fundamental group of the complementary domain: 7r1 (E4\F-) to demonstrate this. Consider, first, the knots in

E3. If y is the standard circle in E3 then ir,(E3\y) = Z, where Z is the group of integers. If y is a knotted curve then iri(E3\y) # Z. If S2 is the standard sphere S2 C E3 then 7r, (E 4\S2) = Z. Let r be a closed curve in E4\F2 with marked point

A E E3. We may consider r as a loop, also. Without loss of generality, we can suppose that t lies in the region - I + e < x4 < l - e, where e > 0 is a small number. Let P be a point of r which lies in a half-space x4 > 0. Draw the ray aP up to the intersection with a hyperplane E3 at some point P' E E3. This gives us the projection

of the upper part of r into E3. By projection from the point b we construct the mapping of the lower part t onto some curve in E3. As a result we obtain some closed curve I" C E3\-y. Moreover, F' is homotopic to r in the domain E4\F2. If I' is contractible in E3\y to some point A then r is contractible in E4\F2 to the same point A and vice versa. This correspondence between loops r and I" generates the

homomorphism of 7ri (E4\F2) into ir1(E3\y). Moreover, the unit element of the second group is the image of the unit element only of the first one. Therefore, 7ri (E4 \ F2) is isomorphic to its image. But ir, (E3\y) is a subgroup of irl (E4\F2).

THE GEOMETRY OF SUBMANIFOLDS

118

Hence, a, (E4 \ F2) = 1r1 (E3 \'y) # Z. As a result we conclude that the surface F2 is knotted. Note that F2 is an irregular surface, because y is an edge, a and b are conic points.

The surface singularity at -y is smoothable but the singularities at a and b seem essential. The second way In a half-space E+(0) defined by x4 = 0, x3 > 0, take the arc y with the endpoints in the plane x4 = 0, x3 = 0. Denote that plane by 11 (see Figure 18). Rotate the half-space E+(0) around H. The latter transformation means that the point with coordinates xl,x2,x3,x4 passes into the point with coordinates 9j by xI = X1,

x2 = x2,

X3 = x3 COS cp

x4 sin (p.

X4 = x3 sin Sp + x4 COS yp.

The rotation of E+(0) by an angle v is denoted by E{ (cp). In rotation by 360° the points of -y, being in E3 (gyp), form the set S homeomorphic to S2. Let y be smooth with tangent vectors at P, and P2 orthogonal to II. Then S is a smooth submanifold. The latter is evident at all points except P1 and P2. Let us show that the tangent planes exist at P, and P2. Let r be a tangent vector to -y at P1. It is directed along the x3-axis. If EE(0) rotates around II then P, stays fixed, but r rotates in the plane which is parallel to e3 and e4. After a complete rotation we get the tangent plane to S generated by T. Join P, and P2 with an intercept. Denote by j the closed curve which is the union of y and the intercept above. We are going to show that rr, (E4\S) is isomorphic to a, (E3\1). The condition 7r, (E4\S) 0 0 means that S is a knotted sphere in E4. Each element of ir,(E4\S) can be represented by a piecewise linear path s in E4 \ (S U ry). Set s = sl s2 ... s,,, where si is the straight path from pa_ I to p;. pn = po.

Pot

FIGURE 18

TWO-DIMENSIONAL SURFACES IN E4

119

Suppose P is in n but not in y. Draw the three-dimensional half-space E3 (cp) through the point pi and H. Then join Pi to P with the path v, in this half-space such that v; does not intersect S U 5. The path s' = s, v1 vi 1s2v2v21 . snvnvn 1 is homotopie to s. Let us consider a closed curve vi_11siv;. It has no intersections with S U ry. By rotation around II, transfer it into some arc in E (0). More precisely, rotate arcs v,_1, and v1 as rigid bodies. Map a point from E+ (0) to Q E Si in the following way: draw

the circle through Q in the plane parallel to e3, e4 with the center in H (it is clear that the circle has no intersections with S, otherwise the intercept si would intersect S

and the entire circle would be in S); let this circle intersect E+(0) at Q'; Q' corresponding to Q. Thus, the arc v,!1s;v; is homotopic in E3\(SU 11) to some closed curve in E+\ Therefore, to a closed curve s in E4\S there corresponds some closed path s" in

E3\1. Two homotopic paths s and t correspond to paths s" and t" which are homotopic to each other in E3-,\!. Thus we have h:

ir,(E4\S)

a1(E'3\1')

On the other hand, every closed path in E3 which does not intersect 5 will not intersect S; that is, it will represent some element of the homotopy group ir, (E4\S ). To summarise we have the inclusion mapping i.: 7r,(E3\7) -+ R,(E4\S). Besides,

by construction of h the homeomorphisms i.h and hi. are identical mappings. It follows that a, (E4\S) is isomorphic to 7r, (E4\y ). Let us find the second fundamental forms of S. Project y into H and let t be the arc-length parameter of the image. The position vector of S can be represented as: x1(1)

\

X2(1)

r =

J(t) cos V f (t) sin p The curve r(t, 0) coincides with -y. Hence f (t) represents the x3-coordinate of points in -y. Further, we have xt

0

'

r, =

x2

cos V

f' sin cp

r=

0

f sin p /'cos

where (t) means differentiating in t. It is easy to find one of the normals:

nI =

ds2 = (I +f,2) d,2 +f2 dV2.

THE GEOMETRY OF SUBMANIFOLDS

120

The other normal n2 we can find by means of the vector product of r,, r,,,, n1: e,

e2

e3

e4

xi

xi

f' cos.p

f' sin p

0

0

fsinyp

n2

f 1

The second derivatives of the position vector are x,'

r=

0

0

/I

r'`

f" COS,p

f" sinV

-f' sin y7

,"

-f COSH

['['cos,

f sin ;o

Therefore, the second fundamental forms are

II' = kdt2,

f

112 = 1

die +

i f'2

dtip2.

1 + f'-

where k is the curvature of projection of -y in H. As Li, 0. the directions of the coordinate curves are the principal ones. Hence, the normal connection is flat and Gaussian torsion Kr = 0. The Gaussian curvature is

K=

.f"

f(l +.f,2)'

The latter expression of the Gaussian curvature is the same as for the surface of revolution in E3 around the axis with parameter t and meridian equation f =f(t). 6.3 Codazzi-Rieci Equations for a Two-Dimensional Surface with Respect to a Special System of Coordinates

Codazzi equations for a two-dimensional surface F2 in E4 with respect to a general coordinate system have the following form r n o L+ik - LA. i = 11,1k 1 rit - µr7/i L;k.

With respect to the usual derivatives we can write 8414. chug

8Ui

+ I'j La

- I'ik L _ {Mv/k Lr

µ",/i L

TWO-DIMENSIONAL SURFACES IN E''

121

Set 1112/r = v;. The Codazzi equations are formed with four equations. Two of them contain derivatives of 11(n1): dLl l 8L1'' + (r'1 r212)L12 LI - 1712L, + I'2 LI 1II 22 = v LZ_ ,, - v 2L211, 07212 - u1 1

1

011,

'' + ML11 + (r22 - I'12)Lj,

1

I'j,

= v,Lj, - vl L;,.

The other two contain derivatives of 11(n,). Further on, we set u1 = u, u'- = v. For now, choose coordinates u, v in the surface and coordinates in the normal plane in a special way. First, we get the family of u-curves in such a way that at each point x the normal curvature vector for the u-direction with its initial point at x would have its end-point at one of the vertices of the normal curvature ellipse. Direct the normal n1 parallel to the ellipse axis corresponding to the vertex above. The v-family of coordinate curves we get as an orthogonal family to the u-curves. Choose the normal n, orthogonal to n1. Then the metric has the form ds2 = E((1u)2 + G (dv)2.

Put the origin of the coordinate system in Nl put at x. Then the second fundamental forms have the following coefficients:

Lj'1 = (a+a)E.

Li1 = BE,

Li, = 0.

Li, = h EG.

L;, = (n - a)G,

L22 = f9G.

It is easy to find the Christoffel symbols:

rII

- EE,

r2 I - - G,

r12

2G'

r2-2 _ -2E'

EE, r22 =

2G'

If we replace the Christoffel symbols and the coefficients of the second fundamental forms in the Codazzi equation with the expressions obtained above then we get

Gb EG-E.b

EG-

(1)

=0,

(2)

f3Ev2 - bv' vI = 0,

(3)

-(aG) + a. G + i3Gv1 - b EGv2 = 0.

(4)

(aE), +

THE GEOMETRY OF SUBMANIFOLDS

122

For the case of F2 C E4 there is only one Ricci equation. With respect to a general system of coordinates we write it as 1212/1.2 - µ12/2.1

+9m(Li1Lhh2

- L12Lhl) = 0.

With respect to a special system of coordinates: (L1 L2 n

h2

22L) = 2abvE . 11 - S - L L2) - L2,(SHHL1 12

1-

h1

2z

Thus, the Ricci equation with respect to the special coordinates above gives the form

v2.1 - v1.2 = 2ab EG. It follows that the fundamental system of immersion for the case of F2 C E4 consists of six equations. Now we turn to the special cases of (I H4). Let (u1, u2) be coordinates in F2 such

that the second fundamental forms are diagonal, that is L12 = Li, = 0. Those coordinates can always be endowed with a hyperbolic Grassmann image in the surface.

Note that the classification of surfaces in terms of their Grassmann image will be presented in Chapter 8, Section 6. Set

L11 = Acos01,

L2, = Bcos02,

Lit =A sin01,

L222 = Bsin02.

(5)

Write the next pair of Codazzi equations:

a2 + r i1 L22 - r: , Li = 1

µ21/2 L211+

2

W2 + I'll L22 - r12Li1 = µl2/2LI I Here µ21/;

( 6)

(2)

In the latter equations, replace L11, L22, L11, and L22 with their

expressions from (5). Then multiply the first by sin 01, the second by - cos B1 and add. We get

- a Z + 1'11 B sin (01 - B2) = µ21/2. 2

(8)

Multiply the first equation by 2A cos 01 and the second by 2A sin 01 and add. Then we get 042 8u2

+ 2r 1 AB cos (01 - 02) - 2r12A2 = 0.

In the same way, using another pair of Codazzi equations j9Lr au12+r22Li1

-Iz1Li2 =A.T1ILi2,

r= 1,2

(9)

TWO-DIMENSIONAL SURFACES IN V

123

we get

Bii+I'22gsin(02-01)=µ21/I, OBI

(10)

(Il)

+2I';2ABcos(0I -02)-2r2'B2=0.

Note that (9) and (11) do not contain torsion coefficients. The Gauss curvature and torsion have the forms K = AB

cos (01 - 02) 2

911922-912

sinw

rr

gl 1922 - g12

We can write the Ricci ^equation as

l

/

Kr v Lalall2

\1

- a42 1 F22 B sinw I - al (r'il A sinw IJ,

where w = 01 - 02. This expression for wr is similar to the Liouville one for K.

6.4

Condition for a Surface to be in a Three-Dimensional Sphere

The Clifford torus is in the sphere S3. At each torus point x the normal curvature indicatrix degenerates into the intercept at a fixed distance from x. The latter condition is common for every surface in S3. Theorem If F2 is in S3 of radius R. then at each point x the normal curvature ellipse is the intercept at the distance O = 1 /R from x.

Conversely, if at each point x the normal curvature ellipse degenerates into the intercept at the fired distance /3 from x and the Gaussian curvature K -A,32, then F2 is in a sphere S 3 of radius 1/0.

+x4 = R2, or r2 = R2 in Proof Let F2 be in a sphere S3 of the equation x,+ vector notation. Differentiating the latter equation in ui, we get (rr,,,) = 0,

(r,,,r ,) + (rra.,) = 0.

From the first expression it follows that r is one of the normals to F2. From the second expression we obtain the coefficients of the second fundamental form with respect to that normal. Set n2 = r/R. Let nl be the normal orthogonal to n2. Then

gi z---R L" Therefore, the normal curvature of F2 with respect to n2 is

L du' dui g, du' dui

1

R'

124

THE GEOMETRY OF SUBMANIFOLDS

i.e. it is constant. Hence the normal curvature indicatrix degenerates into the intercept at the fixed distance I/R from the corresponding point in the surface. Conversely, at each point x E F2 let the normal curvature indicatrix degenerate into the intercept of the straight line at the fixed distance from x. Direct n1 and n2 parallel and orthogonal to that intercept respectively. Using the same system of coordinates as at the beginning of Section 3, we have

LII=(a+a)E,

L11=I3E,

L12 = 0,

L12 = 0,

L2'2 = (a - a) G.

L', = 3G.

Use the Codazzi equations (l)-(2) Section 3. As 3 = const and b = 0,

(a - a)v1 = 0.

(a + a)v2 = 0,

By hypothesis K # 32. Hence a2 54 a-' and vi = 0. In this case the Weingarten decomposition for derivatives of 112 is of the form n2,

Lei

r;

Therefore

112=c-3r, where c is a constant vector. As n2 is the unit vector, the position vector of the surface satisfies the equation of a sphere

of radius 1/3. If ,i = 0 then the sphere degenerates into Euclidean 3-space.

6.5 Whitney Invariant and Gaussian Torsion In this section we consider a two-dimensional closed surface F2 C E4. Locally there always exists a regular normal vector field on F2. Indeed, r,,, (i = I, 2) are tangent

vectors to P. The vector n = {Z;,} is normal to F'- if the components satisfy the following system: XIu,SI +XZU,i;2 +X3u,6 +X4,,,{4 = 0, X1u9I + X2en6 + X3a,e3 + X4u, 4 = 0.

As r,,, and r,,, are linearly independent, one of the second-order minors of the matrix IIx,, is not equal to zero. Suppose, for instance, that in some neighborhood of P the

TWO-DIMENSIONAL SURFACES IN E4

12$

minor x,,,, x,,,; - x,x2,,, # 0. Taking arbitrary s and 44 we can find E, and 2. The space of solutions has dimension two. So we get two unit linearly independent normal vector fields in some neighborhood of P. But it may occur that there exists no global continuous normal vector field. Remember that the surface, being a two-dimensional manifold, is called orientable if there exists its partition into curvilinear triangles oriented in such a way that every two with a common side induce opposite orientations in this side. The orientability notion above is an intrinsic one. On the other hand, in three-dimensional Euclidean space there exists the extrinsic notion of orientability: the surface F22 is orientable if

there exists a continuous normal vector field on it. In E; both definitions are equivalent. For the surface F22 in E4 a new phenomenon arises. There exist closed orientable (in the intrinsic sense) surfaces F22 with no regular normal vector fields. The Whitney

invariant gives the answer to the question of whether or not there exists a global regular unit normal vector field on F2 in E4. Recall that orientation in E" means the choice of a basis positively oriented by assumption. Any other basis related to the chosen one with some non-degenerate matrix - the transformation matrix - will be positively (negatively) oriented if the determinant of the transformation matrix is positive (negative). Let E' be an oriented space and F2 an oriented surface in

E4. Then the orientation

in the tangent plane Tp defines the orientation in the normal plane Np in the following way: let e l, e2 be a positively oriented basis in T,. Then n,, n2 define a positive orientation in Np if (el. e2, n, , n2) is a positively oriented basis in E4.

Let n(P) be a normal vector field on F2 with singularities at a finite number of points P,..... Pk. Let a;(P) be a regular normal field defined in some neighborhood of Pi. Surround P; with the closed curve I'i in Let Y; be a positive angle between n and a;(P) with respect to orientation in N,. Denote by Aye; the increment of the angle above in an F2 positive bypass along Ti. Introduce the index of point P; with respect to the normal vector field n as F2.

Ind,, (P,)

Ate`

(1)

Further on, it will be stated that the index does not depend on the choice of a.(P) and

is the characteristic of a singular point. The Whitney invariant v for the closed surface F2 is the sum of indices of all singular points the of normal vector field n(P). The following theorem holds. Theorem torsion

The Whitney invariant v.lor closed surface F2 C E4 depends on Gaussian

v

2n J

ter dS.

(2)

The latter formula implies that the Whitney invariant does not depend on the choice of n(P).

THE GEOMETRY OF SUBMANIFOLDS

126

Let n1 and n2 be unit mutually orthogonal normal vector fields on F2, regular except for a finite number of points. Assume, in addition, that n1, n2 define positive

orientation in N. Then ti1 _ (nlui n,, ) - (n11 n2u )

(3)

V91

To prove the latter formula we use the Weingarten equations for F2 C E4: nrui = -LLA K I r,r + (n;,., n,, )n

We obtain (nlut n,,,:) - (n2ui nl,.2) = (Llr L22A

- L 2i

LIR)g'

(4)

The expression on the right-hand side of (3) is invariant with respect to the rotation of n1. n, in NP; that is, if cp is some function on F2 and

a1 = cospnl +sin;Pn2, (5)

a2 = -sin'Pnl +cospn2, then

(nlu, n2.r2) - (n2ui III,,!) = (alni a,,,) - (alu= a,ui ).

Moreover, it is easy to verify that the expression on the right-hand side of (3) is invariant with respect to coordinate change. Therefore to evaluate (3) at the fixed point P choose coordinates ul, 1/2 and normal fields a1,a2 in such a way that at P coefficients L with respect to chosen normals would have the form of (2) Section 1. Then we get 2L

2

-LIiL2A )gA =2ah=I:l,

which completes the proof of (3). We can suppose that n1 and n2 have singularities at the same points because one of them and the orientation in Np uniquely define another. Let U;. be e-neighborhoods of each point P; and r'; the boundary curves of these neighborhoods. Set D = F2\ E, U1 . Using (3) we can write

J D

xr dS =

'f D

5 a .i [(nl n2i,) - (n2nI, )]

0

((nl n,ui) - (n2nlu' )]

l

dnl di2 1 (6)

(n1dn,)-(n2dn1). -r,

The orientation of -t; is opposite to the orientation of t;.

TWO-DIMENSIONAL SURFACES IN E4

127

Let a,, a2 be regular mutually orthogonal unit vector fields defined in the neighborhood of a singular point positively oriented in normal planes. They are related with n, and n2 by (5). By direct calculation we get (n, dn2) - (n2 dn,) = (a, dal) - (a2 dal) - 2 dep.

(7)

Contract r i to a singular point. Taking into account the regularity of ak we get lim

J

(ak dal) = 0.

r,

From the definition of index of a singular point, for n, we have

Indn, (Pi) = I I dep. r;

Therefore, passing to the limit while e

J

0 in (6) we obtain

tcr dS = 2ir

Ind., (P;),

which completes the proof of the theorem. Note that from (7) follows the independence of the singularity point index on the choice of regular field a(P). Indeed, if n, (P) is another regular vector field, then in bypassing I'; the increment of angle p between n, (P) and a(P) is equal to zero. Let us consider the surface F2, which is homeomorphic to the sphere S2, with zero Whitney invariant: v = 0. We will show that on F2 there exists a continuous normal vector field. Decompose F2 into two disks D, and D2. Let O; be points in D;. We can suppose that D2 is a small geodesic disk with its center at 02. In each disk we define the regular normal field a; (for instance, by means of parallel transport in a normal bundle of arbitrarily taken normal vectors at O; along the radial curves which start from these points and have no mutual intersections). On the boundary curve of both disks, which we will denote by y, we have two continuous vector fields a, I., and a21,. As the Whitney invariant v = 0, these fields are homotopic to each other. Indeed, if we project each vector of ail, into normal space No, then we obtain two mappings f, andf2 of y into the unit circle SI in the No, plane with the center at 02. Both of these mappings are homotopic to the constant mapping and hence. homotopic to each other. Therefore, on y there exists the homotopy a24.,, 0 < r < I of normal vector fields class, such that amp, = a2 I,, a2, I, =a, L. Now we shall use the properties of fiber bundles. We apply Section 3 of Theorem 3.1 from [15, p. 91[ on the fulfilling of the axiom of the covering homotopy expansion for the pair: n-dsimensional disk D" and its boundary sphere S"-I. There exists the homotopy a2, of a2 in the class of normal vector fields onto D2. which coincides with a211, on y. Thus we get the continuous normal field on the whole surface F2. which coincides with a,, on D, and with a, on D1.

THE GEOMETRY OF SUBMANIFOLDS

128

FIGURE 19

FIGURE 20

Now we are going to construct the closed regular surface F22 in E4 with non-zero Whitney invariant.

At first, we construct the non-regular surface c2 which is homeomorphic to a sphere and has the self-intersection point. Take two unit disks KI and K2 in the planes of (x1, x2) and (x:. x4) respectively with common center 0 (see Figure 19). Let -y; be their boundary circles, A I a point in -yI of coordinates (cos ep, sin ,p, 0, 0). A2 a point in '72 of coordinates (0.0, cos cp, sin P). Join A 1 to A2 with the intercept AIA2. While ;p varies, the intercept AIA2 forms some

surface. Let us show that this surface is regular (see Figure 20). The vector AI A2 has the following components:

= (-cos,p,-sin;p,cosp,sin p). We write the position vector of the surface as the vector function of two parameters AIA2

cp and t:

pr(cp,t) 0<;,<27r, 0
cos cp(1 - 1)

= 0AI +toIA2 -

(sin(1 - t) tcos(p

tsinp

TWO-DIMENSIONAL SURFACES IN E1

129

We have

-

- sin cp(1 - t) r-0

_

coscp(l - t) -tsincp

-since 's 'p '

tcoscp

cos cp sin cp

The metric coefficients are

E=r,= I -2t+2t2> I

,

F=(r,rr)=0, G=r, =2.

Therefore, the surface generated by Al A2 is regular. It has two boundary contours ry1 and -12. Glue the disks K1 and K2 to them. We obtain the closed surface V with one

self-intersection point 0 (see Figure 20). This point we regard as two coincident points 01 and 02 such that 0; belongs to Kj and its neighborhood is the intersection of the neighborhood of 0 with K,. The surface 'F2 is non-regular at the points of yt. However, by means of smoothing we can get the regular surface. We do the smoothing in the following way. We draw the plane E2(cp) through the intercepts 0AI and OA2. This plane contains the intercept A1A2. Consider the right-angled triangle 0AIA2 with the right angle at 0. The angles at AI and A2 are smooth with sufficiently small arcs A, A" and AZ A2 (see Figure 21). OA;, A" AZ are intercepts. We

obtain the closed curve which we denote by F. When an angle p varies, the curve F makes the closed regular surface F22 with 0 as the point of selfintersection. Evidently, F22 is homeomorphic to 4D2. Now we will show that 'F2 is homeomorphic

to a sphere. In E3 we take a circular cylinder of unit radius and with the height of f. Let D1 and D2 be the bases of the cylinder, QI and Q2 be their centers. P1 and P2 be the points in the boundary circles such that the intercepts QIP1 and Q2P2 are parallel. Gluing these bases to the lateral surface area, we get the surface which is homeomorphic to a sphere (see Figure 22). Now set the homeomorphic (moreover,

FIGURE 21

130

THE GEOMETRY OF SUBMANIFOLDS

FIGURE 22

the isometrical) correspondence between K, and D, which takes O; into O;, the point in 'yl with coordinates (1, 0, 0, 0) into P1, the point in y2 with coordinates (0, 0. 1.0) into P2. It is required as well that the intercepts Q. (A1) i = 1.2 be parallel to each other. For the point B in AI A2 take a corresponding point C in f (A1).f(A2) in such a way that the length off (A1)C is equal to the length of AI B. Recall that the length of AI A2 is equal to f. The mapping just obtained maps 4)2 onto a cylindrical surface with glued bases homeomorphically.

Now construct on F2 the normal vector field n with two singular points 0I and 02. The vector r, of the "lateral surface area" of F2 is orthogonal to E2(,). Indeed, it is easy to see that it is orthogonal to intercepts OA;. Hence, if in E 2(:P) we construct the normal vector field n on I' then at each point of .4,A2 the vector n will be orthogonal to r,,, i.e. n will be the normal vector field on F2. The normal field n on the intercept 01 Ai is a constant one and it is directed along 01 A1. The normal field n

on the intercept 02A' is constant too and is directed along 01A1. Thus, the vector field n is defined at every point of F2 except 01 and 02. At these points n is not defined because the directions of O;A1 vary while the angle cp varies.

Let us find the indices of singular points of is. Let e1, .... e4 be the unit basis vectors in E4. In the neighborhood of 01, by the rule stated above, define the regular

normal vector fields a and b in such a way that the basis a, b defines a positive orientation in the normal plane. Let the basis el. e2 define the positive orientation in the disk K1. The basis el, e2, e3, e4 defines the positive orientation in V. Therefore in the neighborhood of 01 we set any a = e3, b = e4 and we shall produce the bypass around 01 along the small circle in the direction from el to e2, that is in the positive varying direction of the angle cp. With this bypass, the increment of the angle between n = OA and e3 is equal to 21r. Therefore Indo, n = 1.

Now we find the index of 02. The direction of the bypass around 02 must be compatible with the direction of the bypass around 01 in such a way that both of these directions would define the same orientation in F2. As F2 is homeomorphic to the cylinder with glued bases, it follows that in the neighborhood of 02 the bypass must be produced from e4 to e3, that is in the negative direction of the varying angle

TWO-DIMENSIONAL SURFACES IN Ej

131

ep. The basis e4, e3, e2, e, defines the same orientation in E4 as e,, e2, e3, e4. Therefore, in the neighborhood of 02 we can set a = e2 , b = e1. The vector field n = 01 A, rotates from e2 to e1 in the bypass around 02 in the direction from e4 to e3.

The angle between n and e, gives the increment 2ir. Hence Indo, n = I. Thus, the Whitney invariant of F2 is equal to 2. 6.6 Normal Torsion of a Two-Dimensional Surface in E4

The normal torsion of a two-dimensional surface has been defined in [3]. In [4], [6], [7] the properties of surfaces with zero normal torsion were considered.

Let r be some tangent direction on F2 C E4 at the point x. Draw the threedimensional space E3(r) through r, x and the normal plane N. It intersects our surface by the curve i(r). The torsion of ry at x we shall call the normal torsion of F2 for the 7--direction. Let us find the expression for the normal torsion of the surface

y= Z = :(x1,x2),

at the point 0 of coordinates x, = x2 = 0. y = z = 0 in terms of y, z Taylor expansion coefficients up to the third order inclusively under the assumption that the tangent plane at 0 is parallel to the plane of x, . x2. We have 1

y = 2(a0xi + 2a,x,x2 + u2x2 ,) + 31(c0x + 3clx x2 + 3c2x,); + c3x3) + -,

= Let cp

2(boxj + 2b,x1x2 + b2x2 ,) + T! (403, + 2d,x?x2 + 3d2x,x; + d3xlj) +

-

be the angle which r makes with the x1-axis. Set x, = t cos gyp,

x2 = t sin 5p.

In consequence we set cos cp = p, sin p = q. Then the representation of the curve 1, which is the section of F2 with E3(r), has the form x1 = t cos cp,

x2 = t sin cp,

y = y(t cos p, t sin cp),

z = z(t cos p, i sin gyp).

Recall the formulas from Section 1 Chapter 1 for curvatures k and ti of a given curve,

k = I[ririill

=

Ilrrrrrrrrrll

irI(rir:J12

where brackets [ ] denote the ordinary multivector. At 0 we have r, _ (coscp,sincp,0,0),

rtr

(0, 0, R, S),

(1)

THE GEOMETRY OF SUBMANIFOLDS

132

where P = aop2 + 2a1pq + a2g2,

Q = bop2 + 2bipq + b2g2, R = cop3 + 3cip2q + 3c2pg2 + C3Y, S = dop' + 3dlp2q + 3d2pq2 + d3g3.

Using formulae (1) we find

k=/2+Q2.

K=±PS-RQ

P2+Q 2 2

Write the expression PS - RQ explicitly: PS - RQ = (a2d3 - h2c3)gs + (2ai(4+ 3a2d2 - 2b1c3 - 3b,c,)g4p + (aod3 + 6aid2 + 3a2d1 - b0c3 - 6b1c2 - 3b2c1)g3p2

+ (3aod2 + 6a1d1 + a2do - 3b0c2 - 6b1c1 - b2co)g2p3 + (3aodi + 2a1d0 - 3b0c1 - 2b1c0)gp4 + (aodo - boco)pi.

Thus at the fixed point of the surface either there exist no more than five directions with zero normal torsion or for each direction the normal torsion is zero. We have examples of surfaces with the number of zero normal torsion directions equal to any

of 1.2,3,4,5. If for every tangent direction the normal torsion is zero, then the expansion coefficients of y, z have to satisfy six conditions: ao

CO

C3

a2 b2

0,

b0

do

3

ao

C2

d3

d2 + 6

a1

c'3

d3

c3

+ 3 a2

C2

61

d3

b2

d2

c1

b1

d1 +

bo

ao b0

at

2

0,

1612

CO

b2

do

+6 a1 C2 + 3 a2 d2

b1

b2

= 0,

0, I

1"

d1I

0.

In [6] V. Fomenko obtained the invariant expression for normal torsion. Consider the second fundamental forms with respect to normals it, and n2: 11 t = hlj du' dug,

112

= cj dui duj.

and the torsion form

w=µidu, where pi = (nin2d). Let r = (tt , t2) be a unit tangent vector. Then the normal torsion for the direction T is equal to

K(T)=

-

pit'+(bijVkClm"cgVkbb,,)titjtktltm

(bllt itj)2

+ (Ci,ljtj),

2

TWO-DIMENSIONAL SURFACES IN E'

133

where Vk means covariant derivative. To prove this, consider du'

du'

r5 = r,, T. ,

rw,,

d2u'

UJ W + r,,

By means of the Gauss decomposition, for ru we have rdzu'

ruI, = ds +

;

dua ds

22

du' \ ds

re + b,

du' du' du' du' nI + c; W ds ds ds n2

Let n be the unit tangent to F2 vector field which is orthogonal to r. Let kg be geodesic curvature of -y and k, , k2 be normal curvatures of F2 in the r-directions with respect to ni, n2. Then ru = kg>) + king + k2n2.

Since -y is in E3(r) and ,j is orthogonal to E3(r) at 0, then kg(O) = 0. Hence at this point ki +k2. For the normals n,,n2 we have dn,

=A,r+BIn2+C10,

ds

dn2

ds = A2r + B2n2 + C2;7.

Find the third derivatives of the 7 position vector:

r"' = [+icuci +k2C2+kg + I dsI

+ k2 B21 it, + [

+[k1A1 +k2A2[r

ds + k, B1, n2.

Since r' is in E3(r) and kg(O) = 0, the first two terms in the expression for r' are zero. To evaluate the normal torsion we use the formula ti(T)

--

(r 'r"r"') k2

So, we find

(r) K(r)

ki k_, - k2ki5 + k1 B1 - k2B2 (3)

k1 + k2

Note that B, = -B2 = -w(r) = -µ;t'. Thus, we obtain the formula for c(T) which has an even more geometrical form than (2):

K(r) = -w(r) + as arctan k .

134

THE GEOMETRY OF SUBMANIFOLDS

Consider now the derivatives of curvatures k, and k2:

k, = bjt'ti,

k15

= b,Jkt`tjtk + 2bjt'ItAtJ.

The vector is the geodesic curvature vector of ry. Therefore at 0 we have k1 = b/jktftjtk. In an analogous manner we obtain the expression for k2.,. We have klk2s - k2" Is = (b/j Clmk - Crj bimk )t

ltjt ltmtk

Substituting the latter into (3) we get (2). 6.7 Pozniak Theorem on Isometric Immersion of a Two-Dimensional Metric into E4

It is well known that an arbitrary two-dimensional regular metric can not be isometrically immersed into V. For instance, the flat metric defined on a torus can not be isometrically immersed into E3 as a regular surface. The Lobachevski plane, as Hilbert proved, cannot be isometrically immersed into E3 either, however, every geodesic disk can be realized in that space. Give the simple example of a metric in a given domain with a boundary which can

not be isometrically immersed into E3. Let us consider the torus with flat metric without disk of small radius r. If r is sufficiently small then this domain can not be isometrically immersed into E3. Indeed, suppose the torus has been obtained by identifying the points of opposite sides of a square. Assume that the length of sides is a. Cut out from this square the small disk of radius r with its center at the center of square. Let p be a point on the boundary circle. As the flat metric in E3 always has realization as a developable surface, the straight line element passes through each point of the surface. Let -y be the straight line element passing through p. Since 7 is a geodesic of Euclidean space, it is the geodesic for the flat metric. Let q be another point of intersection (which always exists) of that geodesic with the boundary circle.

The distance in space between p and q is greater than or equal to (a - 2r), while the arc distance between p and q is less then irr. If rrr < (a - 2r) then the intrinsic distance from p to q is less than space one, which is impossible. Naturally the question arises of the realization of a two-dimensional metric in E4. It is natural to suppose that any regular two-dimensional metric has a regular immersion into E4. For this formulation the question is still open. A sufficiently general result has been obtained by Pozniak. Theorem (Pozniak) Let as be given the metric of class in the plane. Then an1, compact part of the given metric can be immersed into V. The surface which realizes this metric is of class

We can suppose that the metric is defined in some domain S2 of the .r, isotermic form: ds2 = A2(d.Y2 + d)'2),

where A E

Introduce in E4 the bipolar coordinate system p, a. u. v:

in

TWO-DIMENSIONAL SURFACES IN E4 1

x = EP cos

u-,

2

x = Ep sinu-,

cosv-,

x3 = Eo

£

E

135

x = Eo

E

sin-v, £

where c is a positive parameter. The main purpose of the introduction of the small parameter e is that we can represent the required surface in the form of tightly torsioned surface in a thin tube. The metric of E4 with respect to these coordinates has the form = a >(dx')2 = p2du2 + c dv2 + e2(dp2 + dv2).

ds2

=1

Suppose the immersion is defined as

p=Q=

u = u(x,Y),

eat-VA,

v = v(x,Y)

Then the following equality has to hold: e2", (du- + dv'-) + 2e2 e22", (dw)2 = A2 (dx2 + d)2).

We rewrite this equation in such a manner that on the right-hand side is the metric of the plane: A2 e-2"- (dx 2

+ dY-) - 2E2

due + dv2.

(1)

So we get the differential equation with respect to one unknown function it,. It turns out that this equation is of second order with respect to derivatives of w. Set b = 2e2, A = \2 a-2w

W= E

E=A-bn',',

.

The Gaussian curvature can be evaluated by the general formula

K=

-I4 F

E E,

E,.

F, G GY

F.

2W{I

G,

-F,) + "

W

G, W \

'1

(2) Y

The determinant on the right-hand side of (2) is the polynomial with respect to 6, A, Y, ... , 1',.,.. Denote it by D. It has b as a multiplier. We have (E' WF,,)

= 2(1 -

bA)-1 /2 [-11',.

1l

+ (In A),, + 2

11'Y11'YI )A 'J

where A = (11.2 + x-,)/A. We obtain a similar expression for (GY - F,.)/W. After multiplying by 2W,'the equation K = 0 has the form 2

I - bA )3/2 +

.,,

2(In

t'Y.Y - N'Y11'xy.)A,A-2 - b(1'Yiv

+bf-2w,.+2(ln\)1. {

2b t+'Y,YK,.,. -

.-

1',.11'Y).)A,YA-21(1

J1 b(11'1.u'.rx-NpKxl)A-IIA,.

+b[-2w,Y+2(lnA).Y+b(w n',.).-w,.iv .)A-' A,Y=0.

- *A)2

w2 .r ,. )A-'

THE GEOMETRY OF SUBMANIFOLDS

136

Note that the third derivatives of w vanish, which is characteristic of the Darboux method. In addition, (in A).r., + (In A), = -2KA2,

where K is the Gaussian curvature of a given metric. Multiplying both sides by (I - 6A)312, we get

(2.w+ 2A2K)(I - 6A) = 60(6. A is the Laplace operator. From this we get Ow + A2K = 6[A(Ow + A2K) + 0] = 6'I'(6, A, A.,,...

.,...,

where' depends on 6, a", ss....... w,.,. polynomially. Let wo be the solution of An- + A2K = 0,

1+'I,,tl = 0.

Since K has an expression in terms of the first and second derivatives of A and A is of regularity class there exists the constant M such that the norm IIws'oll2." <- M. Setting w = U + wo we get the following equation with respect to U:

AU = 6f(67 A_., U'...' U'.."), where the right-hand side contains the small parameter 6. Set the boundary condition Ulan = 0. Now we use the Shauder estimate II UIl2.,, < 6NIIf II,,, where the constant N depends only on the size of domain fl. Since the function .f is polynomial-like, there exists a continuous non-negative function F such that III II < F(6, IIAII1a, I1UII2.,,)

If o < 6 < 60, llAII2." < A0, 11 UII2." < vo then F is bounded from above F < M. Let us such that the be given the sequence {

DU = 6f(6,A,...,A,.,.,....

Unlnst = 0.

Let II2," < vo. Choose 61 in such a way that 61 < 60 and 6, NM < vo. Using the Shauder estimate and the estimates above, we get II U II2." < vo. The sequence J U" } is

formed by functions which, together with their derivatives up to the second order inclusively, are uniformly bounded and equicontinuous. By the Arcel theorem, some subsequence { U,; } of a given sequence converges together with derivatives up to the second order to some function U E which is the solution of the problem. Since the constructed solutions U are uniformly bounded in the whole domain 0 and for

all 6 < 61, there exists 62 < 61 such that for all 6 < 62 the discriminant of the quadratic form (I) is positive. After that, making use of the Darboux method, we can find functions u and v. The theorem is proved.

TWO-DIMENSIONAL SURFACES IN E4

137

Exercises

(1) Prove that the indicatrix of normal curvature of a surface F2 C E" is an ellipse located in some two-dimensional plane. (2) Let a surface F2 C E4 have the explicit representation u = u(x1, x2 ), v = v(x1,x2). Prove that the orthonormal basis n1 = {{;}, n2 = {q,} of the normal space can be given in the form 6 = -1116 - V16,

771 = -111113 - 171714,

2 = -112e3 -

712 = -112713 - V2714,

3 = -B C4 =

cos cp

V=

+E

sin cp

7A

713 = -B

,

VIA- cos V

'4 =

09-

sin cp

-e

cos ip ,

v/A- sin V 109-

where A = I + (grad u 12, B = (grad u grad v), C =1 + ]grad v12, g = I + ]grad u12+ (3)

12,

E = f 1, u; = er Prove that the Gauss curvature K and Gauss torsion kr for the surface from (grad v12 + I [grad It grad v]

exercise (2) have the following expressions:

K={(uI1u22-ir2)(1+Vj+1 )-(u11v22-2u12v12+u22v11)(u,v,+u2v2) +(VI I v22 - 1'

1+u

2

Kr={(ulIv,2-1112VII)(1+!2+v2)-(u,1v22-11111'11)(111112+V1Y2) +(u12v22 - 1'121122)(1 +uj + 11)}

',.

Prove that the surface It = x2 + y-, v = x2 - y2 has K = tcr = 0. (4) Prove that the Gauss curvature K coincides with the Gauss torsion Kr if u v = fi.r2' where 4 is some function of x1, _X2. (5) Let the metric of surface F2 C E4 have the form ds2 = du2 + B2(u) dv2. The surface is called helicoidal if the coefficients of the second fundamental forms

and the torsion form do not depend on v. Prove that the first integral is T2 + B,1, = const, where T is some function of coefficients of dr2, If' and the torsion form. Prove that the Lobachevski plane can not be isometrically immersed into E4 as a helicoidal surface. (Rosendorn [141). (6) Prove that the surface F2 C E4:

xI =u22-v2,

x2=UV,

x3=1W, x4= VH',

where u2 + v22 + 1r2 = 1 is regular at each point and is homeomorphic to the

projective plane. (7) Let the surface F2 C E4 be given in the form: 4)(x1.... , x4) = const,

fi(xl .......

const.

THE GEOMETRY OF SUBMANIFOLDS

138

Let o, =

grad W - %PiJ grad 4', where subscripts denote the second derivatives

of 4' and %Y. Let q'' = eA'04T, - 4'I'Pk) and Tj,Jk = (ayaki) - (,7&t,). Prove that the Gauss curvature has the following expression:

E E T;,jkq"9rt

Aminov Yu.

(8) Let the surface F2 C E4 be represented in the form 4

1

E a;.xr = d1. `i=1

4

> h;x; = d?. i-1

where ai, bi, di are some constants. Let for i 0 j, ci; = a1!J - a1h; and c,; = dlhi - dzai. Prove that the Gauss curvature of F2 can be calculated by the following formula: 4

K=

E ;.I

+<]

[X1 1 c )` ,

7 Minimal Submanifolds 7.1 Minimal Submanifolds in Em

Let F" be n-dimensional submanifold in Euclidean space Em. We say F" is a minimal

submanifold if for any sufficiently small domain U in F" the variation with fixed boundary of its n-dimensional volume is equal to zero. We are going to find the necessary condition for the manifold to be minimal. As shown before (see Section 2 Chapter 2), the volume of F" can be represented as the integral

V = J f du' ... diu". where g is the determinant of the F" metric tensor. Consider the variation of F" which is non-zero only in some neighborhood of the fixed point PO E U. In a sufficiently small domain U we can always construct the fields of unit normals c,,.. . ,

which are mutually orthogonal and depend smoothly on coordinates in F". We consider the special variations of the form

F=r+E where e is constant, w(P) is some regular function of P E F" being zero on the U boundary. In these variations the shift of the point of a submanif'old happens in the direction of some of the normal vector fields &,. We have

r,,, =ry,+Ew Now we use the Weingarten decomposition for L.,. Then

r,,, + ctv(-L grk r,,k - pp/j(n) + e w,,, 139

THE GEOMETRY OF SUBMANIFOLDS

140

So, we can write the expression for the metric tensor g; of the surface F" represented by the position vector F: Sii = Fw Fu1 =Sri - 2weL; + ... .

where we restrict ourselves to terms of first order with respect to 6. To evaluate the determinant g we decompose g into the sum of determinants and restrict ourselves to the terms of first order with respect to e:

g=g-2wegLyg`f+.... Therefore

f = f(l -ewL' g)+ The first variation of volume is dV

rr

JwL,;g"dV.

If it is zero, then

L, g' = 0.

(1)

The direction of variation can be chosen arbitrarily. Therefore the latter equalities

hold for every a = 1.... ,p. We can represent these equations in the following geometrical way. We introduce the vector H in normal space as

H=>L;jgi1(2) r

1

n-1

It is called the vector of mean curvature. Note that in Riemannian space the mean curvature vector has been introduced with the same formula. We ought to be sure that the right-hand side of (2) is independent of the choice of normal vectors &,,, i.e. if we choose another basis of normals then we obtain the same vector.

Let f v = I..... ni - k be another basis in normal space and

f. = a,,, , is the orthogonal matrix. Let Lt be the coefficients of second fundamental forms with respect to f,,. We have where ja,.r, I

L = (J,r" ) = a,, L;i It is easy to see. now, the invariant meaning of H:

H=kgOf,=au,Lyg'jav.iSli=hi:)LgH. Let us return to the equalities (1). The following theorem holds.

MINIMAL SUBMANIFOLDS

Theorem

141

The mean curvature vector of a minimal submanifold is zero: H = 0.

Now we would like to state some properties of the position-vector of a minimal submanifold in Euclidean space. If p is the scalar function on the Riemannian manifold, then the Laplace-Beltrami operator is defined by: V2cp = p,yg'ij. In P we can introduce the coordinate system in such a way that at a fixed point PO holds g'f = bi, and r . = 0. Then the covariant derivatives at Po have the form of ordinary ones. The Laplace-Beltrami operator atP0 is V- )W = 2 I

+ ... +

If we denote by 8/8s; the differentiating operators with respect to the arc-length parameter on the geodesic, which is tangent to the coordinate curve ul at PO, then at PO

v2p= " a2V

Eat =1

In accordance with the Gauss decomposition, we have

Contracting with g'J , we get

V2r=Lg'JnH.

(3)

Hence for the minimal submanifold

V2r=0. that is, each component of the position-vector satisfies the Laplace equation. If we take the function p = i r2, then P.r = (rr,,),

p,;i = (r;rj) + (rr.j) = g;J + (rH )n. Therefore

V2p=n(l+(rH)).

(4)

Thus, if F" is a minimal submanifold then

V2P=n.

(5)

It follows that a minimal submanifold in Euclidean space E'" can not be closed. Indeed, if the submanifold F" is closed then there exists a point at which p takes its maximum. At this point V2p < 0, which contradicts (5). The theory of minimal surfaces in Euclidean space Em is closely related to the theory of analytic functions. To show this, introduce in a minimal surface F2 the isometric coordinate system u1, u2 (it is possible in a simply connected domain), with respect to which dr2 = A2((du1)2 + (du2)2). Introduce the complex variable

THE GEOMETRY OF SUBMANIFOLDS

142

Z = ul + iu2. Let xk be the components of a minimal surface position-vector. From what was proved above, we know that xk are harmonic functions of uI, u2. Introduce functions 'pk of the complex variable z as axk

axk

(6)

Vk = 8u1 - 15U-2

We state that pk are analytic functions of z. The Cauchy-Riemann conditions for the real and imaginary parts of Spk are 0 2xk

49 2xk

0 2xk

0 2xk

au12 + au22 -

°'

au1 au2 = au2aul

.

They are satisfied due to the harmonicity of xk. As u1, u2 are isothermal coordinates, then a,

_ aYk

k=I au

I

2

_ =A2. k

2

= A-I u2 M 09y

axk axk

0. k=I au1 au'- =

It follows that the analytic functions Wk on a minimal surface F2 satisfy the equation (7) k=1

Conversely, if Vk(z) are given analytic functions, which satisfy (7), in some domain G

of complex variable z then by use of these functions one can construct the minimal surface in E"'. Indeed, as Wk(z) are analytic, then there is the definite integral P

Pk (=) dz PO

along any curve 7 in G joining Po to P1. If two curves joining P0 to P1 are homotopic, then this integral has the same value. By definition (6), Vk = axk/az. Therefore, for the required minimal surface we have P

X

k

= Re

fpk(z) dz + Ck,

Ck = const.

PO

Thus, every minimal surface F2 in E"' can be described in terms of analytic functions Vk, satisfying (7). If m = 3, equation (7) can be solved by means of two analytic functions j and g. Rewrite it in the form (VI - i'p2)('PI + icp2) = -'p3

(8)

Set

'PI - i 'P2

'p3

Cpl - i42

= g.

(9)

MINIMAL SUBMANIFOLDS

143

By means of (8) we find P1 + iW2 = -f92-

Therefore, equations (9) and (10) imply the expressions for Wk:

V1=10-92),

(l +g2),

W3 =fg

2

Conversely, if Wk are given by the formulae above then they satisfy (7). Take the

function g as an argument. Set F(g) = f dz/dg. Then we obtain the classical Weierstrass-Enneper representation ofrra minimal surface in E3:

x1 = Re IJF(l -g2)dg+c1, r'- = Re

2JiF(l +g2)dg+c2,

x3 = Re

JFgd + C3.

(10)

The function F(g) is called a Weierstrass function. Thus, the minimal surface in E3 can be defined by means of one analytic function.

An analogous representation for a minimal surface has been obtained by Eisenhart. We write equation (7) for m = 4 as W1 + AP2

W3 - iW4

'P3 + 4P4

VI - IV2

= u,

(I 1)

where the common value of the ratio is denoted by analytic functions u. Put API - i'P2 = llg,

W I + iW2 = U f

where f and g are some analytic functions. Then from (11) we get

'P3 + ip4 =f

'P3 - AN = -u2g

Therefore, for the functions cok we obtain the representation W1 = 2 u(f+

g),

'P3 = (f - u2g),

'P2 = - 2 (f - g), ' P 4 = 2t (f+ u2g)

The analogous representation for m > 2 has been obtained by Beckenbach [2]. We will distinguish the cases of even or odd values of m. Let in = 2k. Introduce the analytic functions F1,. .. , F2k-1, setting F2.- I =

'2t- I + iW2s

'23-1 - if

F2k-I = WP2k-I + i1P2k

'2k-1 - i'P2k F2s = V7s-1 - iV2r.

1 < s:5 k - I,

(12)

THE GEOMETRY OF SUBMANIFOLDS

144

2,_ I FS = cpjs_ I +,p22. Put It is evident that F,,-,FS' k-1

TF2,_1F;, .f=1

Equation (7) implies T

`P2* -I - 402k = -

4P2k -1 + Y2k

Using the definition of F2k_1, we find 472A-1 - icc2k =

(13)

FU I

V

Relations (12) and (13) permit us to express cp; in terms of functions F1 , ... , F2 - _ I. Therefore, for the minimal surface F2 C E2" there is the representation 1

x 2.v - I = Re 2J (1 +Fz<_1)Fz,dz+c2T_,.

I <s
x2i=Rei !(1-F2,-1)Fsdr+c2,,.

xl "1 = Re 2 J (1 +F;k_1) T dz+ru xv = Re2J (1 - F,k

1)

-F.T dz+c2, 1

If in is odd, i.e. to = 2k - 1. then take the components xi of the position vector for 1 < i < 2k - 2 as above, while for x2k'11 we get from (7) the representation

x`F-1 =ReJv- dz+c2k

1

Let us consider the example of a minimal surface in E4. Let f = f (z) be an analytic function of the complex variable z = x' + ix2. Let F2 be a two-dimensional surface in E4 defined in explicit form as

x; = Ref(z).

x4

= Im,f(z)

Set Ref (z) = u, Im f (z) = v. We are going to prove that F2 is a minimal surface. To do this, we find the mean curvature vector. The tangent vectors to F2, which are tangent to coordinate curves, are 0

1

I=

0

,

kv I i

ra : =

I

ux! VY:

MINIMAL SUBMANIFOLDS

145

Since f(.) is an analytic function, u,, = v,2, uxx = -V u. Therefore

g1, =r2, =u;1+V21+I =u2,+VY2 :+I =$22, V_ V

11

g12 = (rYlrx:) = uXIuX: + VYI VY: = 0.

That is, the coordinate system x1, x2 in the surface is orthogonal and isothermal Further r.Y, Y, =

0 0 ,

u.Y.xj

i,j=l,2.

VYIx,

Recall the definition of the mean curvature vector I ij H=2Lyg nn,

where n,, are unit normals. We have

Lii

t..nn )=gii(r Y c +r Y=x'+nn )

r,

Since u and v are harmonic functions, uYIY, + uY:_Y: = 0, vxIY1 + VY:Y: = 0 and, as a consequence, r, I, I + r.Y: Y: = 0. Thus, the mean curvature vector H = 0. We now consider many-dimensional minimal submanifolds. The following class of

minimal submanifolds is generated by analytic functions of several complex variables. Let us consider the Euclidean space E2" of dimension 2n with Cartesian coordinates x I, ... , x'-". Introduce n complex variables z" = x" + it"" For convenience, set .r"+" = r Let us be given k analytic functions ft........ ."), j = I , ... , k of n complex variables z"...' z" in some domain in E'". By definition of an analytic function, 8f;/dzi = 0. Let uj and v be real and imaginary parts of the analytic function fi respectively, that is f = uj + ivj. Define a 2n-dimensional submanifold F2" in E("+A'), having a position vector and its derivatives of the form

10

xi

0 I

r(xI...... t,,l) =

3

ui

ri =

0

0

vk., .V

V'A'

\ VA'X')

Since f is analytic with respect to every .", then uj and v1 satisfy the CauchyRiemann conditions ujx" = JI.',

Up' = -Vjx'.

THE GEOMETRY OF SUBMANIFOLDS

I46

Let us consider a four-dimensional submanifold of E°, defined as

xs = Ref(-1,z2), xb = Im f(z1,z2),

where f (z 1, z2) is the analytic function of two complex variables z, = x, + ix2, Z2 = x3 + ix4. Set It = Ref, v = Im f. These functions satisfy Cauchy-Riemann conditions with respect to each pair of variables x1, x2 and x3, x4: !!A, = "AJ,

U31 = VX2

-V"'

U"

U"

-V".

In addition, the pluriharmonic conditions (see, for instance, [3]) are satisfied, as a consequences of Cauchy-Riemann conditions 0. r, r, + P T2,, - U` r,, = 0.

where instead of (p we can take any of u, v. The position vector and its derivatives has the form X1

0 0 0

.r7

r=

X3

X4

rz, =

it

0

U#

!!

V

v1i

V!

Here the subscripts of u and v denote the partial derivatives. The metric coefficients are

g11=I+!!iz +V2!,

g12=!lIU2+VI1'2

Using the Cauchy-Riemann conditions, we get 91 1 = 922,

912 = 0

In an analogous manner, we get 933 = gqq, g34 = 0. Further, we have gi3 = UI U3

+

v1 v; = V2V4 + vt v3.

g14 =UIU4+VIV4=-1'2V3+VIVq. g23=U2U3+V V3=-v1 V4+V2V3. g24 = U2U4 + V2V4 = VI V3 + V2V4.

Comparing these expressions, we get g13 = g24,

g14 = -923-

MINIMAL SUBMANIFOLDS

147

Thus. the matrix of metric tensor has the form

Ilg,111=

911

0 911

g13 g23

-g23

0 913

g23

g33

0

-g23

913

0

g33

913

a 0 c

c -d

0

ad db

-d c

0

c 0 b

Now we verify that for the components of the inverse metric tensor, the relations analogous to the relations for gy hold. We have

g11-1

adc d b 0I =' (ab2 -bc2 -d2b),

glc 0

g

b

a

c

-d

c

b

0

-d 0

b

_ I (ab2 - d-b - c2b).

Therefore, g11 = gu. Analogously, we set g33 = g44. Further,

g12 =

g13 =

0

dc

c

h

0

-d

0

h

-I

g

1

g

0

a

c

d 0 _ 1 (c3 + d2c - abc),

c

-d

c

b

a

0

c

g- =g C d b -d c 0 ,4

1

= - - (dbc - dbc) = 0,

=

(c;+d2(- -abc).

Therefore. g 122 = 0. g 13 = g24. In an analogous manner, we get g14 = -g223. gas = 0.

Now we find the vector Only the two last components are non-zero. Taking into account the conditions on g'', we have Uil glf - (IIII /+ u22)gl1 + (u33 + U44)g33 + 2(1113 + u24)g13 + 2(1114 - 1/23)g14.

Since the function 11 satisfies the conditions of harmonicity and pluriharmonicity, 11#9`j = 0. In an analogous manner, we get ryg') = 0. Thus, ryg'' = 0. Therefore, the mean curvature vector H E.(rug'l,n,,)n,, = 0 and our submanifold F4 is minimal. In an analogous way we can prove that the submanifold. which is constructed by means of k analytic functions z,) i = I, ... , k, is a minimal submanifold in Euclidean space of dimension 2(1 + k).

148

THE GEOMETRY OF SUBMANIFOLDS

7.2 Minimal Submanifolds in Riemannian Space

For a given submanifold in Riemannian space we define the mean vector H in the same way as in Euclidean space:

H=

n

g'' n,

The submanifold is called minimal, if its mean curvature vector vanishes: H = 0. To obtain interesting examples we consider the following quite general situation. Denote by V, V and V some manifolds. Suppose that V C V C V. Denote by T the tangent space to V. Let H be the mean curvature vector of V C V and H the mean curvature vector of V C V. Then

H = (H)T, where ( )T means the projection into T. To prove this, we state the analogous relation on normal curvature vectors of V C V, with respect to the same direction r in tangent space to V. Let k,v(r) be the normal curvature vector of V C V. By definition

kN(r) = (Vrr)', where N is the normal space to V C V. Let N be the normal space to V C V. Then the normal space to V C V is the direct sum N + IV. Let kN(r) be the vector of normal curvature of V C V. We have

kN(r) =

(Vrr)."" N,

(I )

where p is the covariant differentiation in V. We can represent the right-hand side of (I) as a sum of two projections, i.e.

kN(r) = (vrr)N + (D:r)N.

(2)

Note that N c T. Therefore, the first term on the right-hand side of (2) is in T and (Vrr)N = (7rr)N = k,v(r), while the second term is orthogonal to T. We rewrite equation (2) as kN(T) = kN(r) + (7tr)'N Therefore

kN(r) = (kN(r))T

(3)

Let L be the coefficients of the second fundamental forms of the submanifold V C V and &, the unit normal basis. Then, from the definition, we have H = L° g'i1;o, where n is the dimension of V. Choose the coordinate system in V in such a way that at some fixed point P will be g,, = bj,. Then H = > , L,,° t;,,. Denote

MINIMAL SUBMANIFOLDS

149

by k.,vi the vector of normal curvature of V C V with respect to the u' coordinate curve direction. By definition of normal curvature vector we have Ila k,v

Hence, in the u' coordinate curve direction we have kN, = Li'i &,. Therefore

H=-ll l_ k,,;. I

In the same way, we get

where kNI is the normal curvature vector of V c V with respect to the direction of coordinate curve u'. Applying (3) we obtain k,v; = (k,v;)t for all i. Hence H = (H)t`. From this it follows, for instance, that if F" C VP" c VA and F" is a minimal submanifold in Vk then F" is minimal in V'. Indeed, H = 0 and hence H = (H)* = 0. Let us consider the Clifford torus T22 as a surface in a three-dimensional sphere S3

of radius f and on the other hand as a surface in V. So we have the situation considered above: T2 C S3 c E4. The position vector of T2 and its derivatives have the following expressions:

- sinx,

cosx1

_ r

sin x,

cos x2

,

rt,

_

0

_

cos x1

rr.

0 0

sin x2

0

- sin x2 COS x2

- cos x,

_ rY,

-sin x,

rr,x, =

rY, .,= 0.

0 0

We can find the second fundamental forms of T2 in E4 by formulas from Section 14:

L° - 1 02r Ox'8xi " Normal vectors t" are already found in Section 1 Chapter 6. We have

L:,=-I.

L112=0,

Lu = 0,

Lit = 0,

Lie = 0,

Liz = -1.

150

THE GEOMETRY OF SUBMANIFOLDS

Therefore, the mean curvature vector H of T2 in E° is

To find the mean curvature vector H of T2 in S3 we need to find the projection of H

into the tangent space to S3. But r is orthogonal to S3. Therefore, H - 0. So. the Clifford torus T 2 is a minimal surface in V. To construct further interesting examples of minimal surfaces in spheres we are going to state the lemma on the Laplace-Beltrami operator representation. Let H be the mean curvature vector of some submanifold F" in Riemannian space V"'. From a point x E F" draw m geodesics 1; of ambient space VI in such a way that l; are mutually orthogonal at x and 1; are tangent to F" at x for i = I,... , n, while 1;

are perpendicular to F" for i = n + 1, ... , m. Denote by 0/01, the differentiating operation along /1 with respect to arc-length parameter. Denote by grad f the gradient off in V"'.

Lemma Let V2 and V2 be Laplace-Beltrami operators in VI and F" respectively. Let f be some regular function of class C2. Then

V2f= V2 f-n(Hgradf)+ E at?.

(4)

i=n+ I

Take n geodesics in F" from x which are tangent to l1, i = I , ... , n. Denote by a2/as? the second derivatives with respect to the /1 arc-length parameter. It is easy to see that

02f 0112

a2fasil

(k, gradf )

i=

where ki is the curvature vector of the geodesic in F" as a curve in V"'. Summing these expressions from i = I up to i = n and observing that E ki = nH, we get (4). Consider from this point of view the sphere S2 of radius p in Euclidean space V. Let x 1, x2, x3 be Cartesian coordinates in E3 with the origin at the center of S2. If v is the outward normal vector to S2 then H = -P v. In this case, (4) has the form

02f=02 f+2af+a2f, P&

(5)

where a/ap is the derivative with respect to the radius of the sphere. x1,x2.x3 satisLet f be a homogeneous polynomial of degree k with respect to '2 f = 0. Suppose this polynomial has the form fying the Laplace equation i,+1,+1,=k

where ai,;,;, are constants. It is easy to see that homogeneity off implies

ofkf ap

02f k(k- 1) p,

ape =

p2

f

MINIMAL SUBMANIFOLDS

151

Therefore, from (17) it follows that f Is, is the eigenfunction of the Laplace-Beltrami operator on S2: k(

V2.f=

PZ

1)f

(6)

with the eigenvalue Ak = k f. So, f is a spherical harmonic. P Define the surface in E5 by a position vector of the form _

r-{

f' f' f'

CIx2

X2X3

X1X3

(XI)2

-

(x1)2 + (X2)2 - 2(x3)2

(x2)2

2f

'

6

provided that x + x2 + x3 = 3. This surface is called a Veronese surface. Denote by i = 1, ... , 5, the components of r. Each f is a polynomial of second degree with respect to x1, x2, x3 satisfying the Laplace equation in E3: 02 f = 0. Therefore, on the sphere xi + x2 + x3 = 3 these polynomials satisfy the equation

v2f=-2f, i=1,...,5.

(7)

It is easy to show that F22 is in S4. Indeed, take the sum of component squares

-x2)2/12+(1 -x3)2/4

( I2 X$2

=X3(x +x2)/3+(x1 +x2)2/12+(1-.x3)2/4 = x3(3 - x3)/3 + (3 - x3)2/l2 + (l - 2.r3 + ,r3)/4 = 1.

We can find the first fundamental form of F2 as (dr)2 =3{&ixi x2)2 + (dx2x3)2 + (dxi X3)22+4(d(xi -x2))2} +4(d(1

- x3))2

= 3 j (x2 dxi + XI dC2)2 + (x3 dx2 + X2 &3 )2 + (X3 dXl + Xl dx3)2

+(xi dxI -X2dx2)2+3x3dr31 =3

{(x1 + x2 + x3)(dr1 + dx2 + dx3) + 2x3 dr3(Xi dxl + x2dr2 + x3dx3)}

=dxi +dY2+d.3. Thus, F2 is isometric to S2 of radius

We can rewrite system (7) in vector notations

as V2r = -r. As V2r = 2H, where H is the mean curvature vector of F2 C E5, then

2H = -r. Let us consider F2 as a submanifold in S4 C E5. Let H be the mean curvature vector of F2 C S4 and T be the tangent space to S4. Then H = (H)t. But r is orthogonal to T. Hence H = 0. Therefore, the Veronese surface is minimal in S4 . The Veronese surface was one of the first non-trivial examples of minimal surfaces in a

sphere. Other examples have been constructed by Calabi, do Carmo and others. Their investigations helped other mathematicians to solve the Bernstein problem on the existence of a complete minimal hypersurface defined over the whole n-dimensional space.

THE GEOMETRY OF SUBMANIFOLDS

152

By means of spherical harmonics of order k on S22, we can obtain minimal isometric immersions of S2 into S24. It is known that on the sphere S2 of curvature p'the eigenspace of V2 with respect to Ak is of dimension (2k + I). It turns out that we can choose the orthogonal basis .ft, ... ,f2k+1 of this space in such a way that 2k+1 f2

2k+I

=I

r=I

(df)2 = (dx1)2 + (dx2)2 + (dr;)'.

-1

Therefore. the surface of position vector [A .. . . ,f2k+l } is in Stk. It is isometric to S2 and is minimal in Stk.

7.3 Stable and Unstable Minimal Surfaces A minimal submanifold is defined by the condition that the first volume variation is

zero: 6 V = 0. For the case of a two-dimensional surface in a three-dimensional Euclidean space we say area variation. This condition leads to the equation with respect to the mean curvature vector: H = 0. Minimal surfaces in three-dimensional Euclidean space can be realized as soap films. If you immerse the closed wire contour

into dissolved soap then, after you take it out, you obtain the film based on this contour with zero mean curvature H = 0. The la place formula is known: if some surface is a common boundary of two media with pressures P and p respectively then

the remainder of P and p is proportional to the mean curvature

P - p=Ha, where a is a numerical multiplier. It is constant on the surface and is called the surface tension. In the case of a soap film, based on a wire contour, the pressure is the same from both sides. Hence H = 0. If the film, maybe together with another rigid surface, forms the closed surface in and out of which the gas medium has constant pressures P and p respectively, then the remainder of the pressures are constant and hence H = const. However, it happens that Nature realizes only stable minimal surfaces. We now consider this notion. Some domain G in a minimal surface is called stable if its area is the smallest one among close surfaces with the sane boundary. We can find large unstable domains in some minimal surfaces. In a physical experiment these domains are either not realizable or soon collapse. To state, whether there is a stable or unstable given domain in a minimal surface, it is natural to consider the second area variation of a minimal surface. We consider the minimal surface F2 in E3 represented by the position vector r and the variation F = r + ewn,

where n is the unit normal to the surface, e is the small parameter, x' is the piecewise smooth function which is zero over the boundary. The varied area is a function of e,

MINIMAL SUBMANIFOLDS

153

i.e. S = S(e). This function takes its minimal value on F22 if dS/de = 0 and

d2S/de2>0fore=0. Therefore, we can state the criterion of stability of a domain in the surface: the domain is stable if its second area variation 62S is positive for all non-trivial variations reducing to zero over the boundary. The domain is unstable if there exist variations such that b2S. Investigations of minimal surface stability in terms of area second variations were carried out by Schwarz in the nineteenth century. He stated the area second variation

formula. To derive that formula, we suppose that a given domain is endowed with lines of a curvature coordinate system (u, v). Then Rodrigues' formula holds: ny

_

- TI r,,

I Trr,

nr

where R, are the principal radii of curvature. We consider variations in the normal direction because it can be shown that more general variations with fixed boundary lead to the same result (see [4)).

On the surface. corresponding to the e-parameter, the vectors tangent to coordinate curves are of the form R

rr = l 1 - R cw) rv. + ew, n. 2

The metric coefficients are

P= / (I

2

-

R1

ew E+e21vu,

G= I I-RZewIG+e2w2. Omitting the infinitesimals of order greater than e'-, we get

E6 -F2 = I_E_G

+Ke2w2+`2Ew2+Gss,2 I

2EG

)

Denote by V1v the Beltrami first differential parameter on the minimal surface F2, i.e. the square of this function gradient:

EG-F'Then we obtain the expression for the second-order area variation of G c F2:

625= J {2Kw2+Vw}dS.

(1)

THE GEOMETRY OF SUBMANIFOLDS

154

Denote by w the area element of spherical image. Then dw/dS = J A1. Let dn2 = e du'- + 2 f du dv + g dv 2 be the spherical image metric with respect to (u, v) coordinates. It is related to the metric ds2 of F'- by dn2 = (q dS2. Let V'w be the first differential parameter of it, in the spherical image metric. We can express it in terms of the first differential parameter of a minimal surface as gw

+

Vw

H', eg _f 22

IK12EG

IK1

Therefore, the expression of second area variation can be rewritten solely in terms of spherical image: b2S = 2

f

w2} dw,

(2)

where is the spherical mapping of a minimal surface. Set G' = P(G). The formula just obtained was found by Schwarz. From this formula it follows that the stability property of a domain in the minimal surface is characterized by its spherical image, i.e. by the domain G'. In some cases the domain G' in a unit sphere can be found easily. If it is sufficiently simple. it is possible to determine the sign of 62S. It turns out that the question of stability can be reduced to the question of the first eigenvalue of the Laplace-Beltrami operator V of the unit sphere metric in G'. Indeed, since ' is zero over the boundary of G', the/p following integral relation holds:

-G'

f

w Vz t dw.

G,

Therefore, we can rewrite the second area variation in the form

b2S=- f (V

(3)

G.

Let us consider the eigenvalue problem in G':

V w4-taw=0 provided that I,x;. = 0. From the general theory of elliptic differential equations it follows that there exists the orthonormal system of eigenfunctions WL corresponding to the eigenvalues 0 < Al < A2 .... Let (p1 be the first eigenfunction in G' corresponding to eigenvalue A,. Suppose al < I. Represent the surface variation in the n. Then form F = r +

62S= -f (-2A, w + 21v)iv dw < 0. U,

Thus, in this case we can point out the variation decreasing the area. It is the domain instability case in a minimal surface.

I»

MINIMAL SUBMANIFOLDS

Suppose \1 > 1. Then \, > I for all i. Any surface variation directed along the normal and vanishing over the boundary we can represent in the form of a linear combination with constant coefficients of eigenfunctions w = E', akcpk. Then V;w = -2 k 1 ak,\kcpk. We have x

r ac

b2S =

- G.J

a,

E{-2akAkcpk +2ak+pk} I: ai'pidw = 2>ak(Ak - 1) > 0. ka1

i=1

k=1

Here we used the orthogonal property of tp1,... , Wok.... . Thus, the following theorem holds.

Theorem (Schwarz) The domain in a minimal surface is stable if the first eigenvalue of the Laplace-Beltrami operator of the domain spherical image Al > I and unstable if Al < 1.

Here Ai are the eigenvalues of the spherical image Laplace-Beltrami operator, i.e. the number for which the equation V;1v = -2A;1v has a non-trivial solution vanishing over the boundary. 7.4 Pogorelov Theorem on the Instability of Large Domains in a Minimal Surface in E3

In the previous section we considered the description of stability or instability properties of a domain in a minimal surface in terms of spherical image. But it is possible to obtain the criterion of instability starting solely from intrinsic properties of the minimal surface metric. In [5] the following theorem has been stated. Theorem (Pogorelov) A simply connected nrinimal.cu{jace is unstable if it contains the geodesic disk of radius p satisfying even one of the following conditions:

(1) Ict(g)I >

P

'tp1 In ip ) for

some < p;

(2) S(p) > 47rp2

Here S(p) is the disk area, l(p) is the length of the circle of radius p, w({) is the integrated curvature over the disk of radius C with the same center. Define the polar coordinate system (u, v) in the considered geodesic disk. It is possible because K < 0. Let ds 2 = dug +g2 dv2 be the metric of the surface with respect to that coordinate system. We consider deformations r = r + ewn such that the function w depends only on the distance from the disk center, i.e. on u. Then the second area variation can be expressed as 2w

62S=

if 00

+1 v )gdudv.

THE GEOMETRY OF SUBMANIFOLDS

156

Suppose that the first condition of the theorem is satisfied. Set ' = I for it < . In the segment { < it < p introduce the new variable a defined by means of equation 2-

du =

1(tt)

where 1(u) =

,

J

gdv.

0

i.e. 1(u) is the length of the geodesic circle of radius u. Let v be a linear function with respect to a over the remaining part of the disk and also equal to I for it = and

to zero for u = u(p). Such a function has the form (u) -

u - u(P)

it(.) - ft(P)

For the function w we have

,+tt(P)

u(t,)

2::

b2S< Jf 2 Kgdudv+ 0 0

J

- tt(0

u(F)

/f tp

=2

+

I

du

1(u).

The function g(u. v) is convex with respect to it because of ,,,/, = -K > 0. Hence 1(u) is also convex downwards. Note that 1(0) = 0. Since the graph of 1(it) lies under

the straight line through the origin and the point (p, 1(p)), then td(p)/p> 1(u) (see Figure 23). Therefore p

1( ) >

In 1(p)

For the second area variation we obtain an estimate: 1(P)

6 2S <- 2w(E) +

p Ine

FIGURE 23

MINIMAL SUBMANIFOLDS

157

By virtue of condition (1) of the theorem, we get b2S < 0, that is the deformation is an area-decreasing one and the surface is unstable. Suppose that the second condition of the theorem is satisfied, i.e. 2ir - - s < 0. The expression for the second area variation can be rewritten in terms of the function g: 2a P

2T

62s= -Jf 2g"w2 dudv + 00

if

wrugdudv.

00

Integrating by parts twice and taking into account the conditions g(0, 0) = 0, g"(0,0) = 1, r(p) = 0, 1, we get 2-

27rrP r!P

62S = 4a - 4

f

3f

JJ 00

w11'2gdudv.

00

Set v = I for u < E (e is small). Let r be linear for e < u < p. Smooth this function in the C2 class about the point u = E. For such a function x and sufficiently small E the first integral in the expression above is however small because g(0) = 0. The second integral in the same expression differs however small from S. Since

i >r

2a-3S <0, b2S<0. Thectheorem is proved.

A complete simply connected minimal surface is called stable if any bounded domain in it is stable. In E3 this property possesses the plane. Do Carmo and Peng [6] stated the following result. Theorem 2

A plane is a unique complete stable minimal surface in E3.

The simple proof of this theorem has been obtained by Pogorelov with the help of Theorem I. Suppose that a surface is stable and its curvature is not zero identically. Take the geodesic disks with centers at some point 0. If the disk radius is sufficiently large then 0 0. Consider the ratio l(p)/p which tends to a finite or infinite limit while p - oe. If 1(p)/p < const < oc while p - oo then for fixed and p oc the ratio l(p)/2p in (,e) 0. Therefore, when p is sufficiently large the first condition of the theorem is satisfied. The surface is unstable. If 1(p)/p - cc while p oc then P

S

P,

=

I

P,

f

J 1(Pi) dP. -- oc. 0

It follows that for sufficiently large p the second condition of the theorem is satisfied. The surface is unstable.

Thus, the Gaussian curvature K - 0. Since H - 0, the surface is a plane. Let us note that Theorem 2 is the generalization of the Bernstein theorem because a minimal surface z = --(x, j,), by virtue of the Schwarz theorem (see [4]), is stable (it can be included in the field of extremals).

158

THE GEOMETRY OF SUBMANIFOLDS

7.5 One Problem on Minimal Surfaces with Free Boundary

In Minimal surfaces (Yu. Aminov 1978) - as part of the lecture course I gave to fourth-year students in the mechanics and mathematics department in Kharkov State University - I posed the following problem which, as far as I know, has never been posed before. Take some non-closed wire contour r. Join its endpoints A and B with flexible but inextensible thread y. If the length of the thread is not large then in experiments we obtain a soap film with a closed boundary formed by the contour F and the thread y. The latter takes a quite definite position in space under the effect of surface tension forces. In the film surface the thread takes the position of a curve with constant geodesic curvature. The problem is to prone the existence of, and to find, this minimal surface (see Figure 24). We can formulate the solvability criterion of that problem: let L be the length of

wire contour and / the length of thread. The problem is solvable for the class of regular surfaces if / < L. Many authors have considered the problem where a part of a minimal surface boundary moves on a prescribed surface. However, this is fundamentally another problem.

A modification of the problem stated above is that where the boundary obtains maximally possible freedom. Let us be given n points A1.... , A,, joined sequentially

with threads in such a way that we obtain a closed contour. The lengths of the threads are given and equal to /;. The problem is to find the surface of minimal area with a boundary formed by inextensible threads y; and points Ai. It is important, as Pogorelov pointed out, that threads 7i have to coincide with asymptotic lines of the required surface.

Here we have a unique, perhaps, and wonderful example, when Nature alone creates, and demonstrates to us, asymptotic lines.

To prove this. we consider the conditional extremum problem - to find the surface of minimal area S provided that one part of the boundary is fixed while

FIGURE 24

MINIMAL SUBMANIFOLDS

159

As FIGURE 25

another is formed by an inextensible thread of fixed length. Let us consider the functional

fJ E

duds+aJ ds, .

F=

where ) is a Lagrange multiplier. In this case we have to consider the general variations. i.e. not only the normally directed and non-vanishing variation on y;. If we take, first, variations vanishing on the boundary and normally directed then we come to the condition H = 0. The first surface area variation amounts to the contour integral (see [4], p. 12)

f(6x/) ds, where fix is any admissible displacement in the boundary, v is tangent to the F2 unit normal of y;. Therefore, the following condition has to hold:

(6xv)ds+AbJds=0. h,

ry,

For the fixed endpoint variations of the curve length the following formula holds (see [7]):

6 /ds=-J(kbx)(Ls, N

'Y8

where k is the curvature vector of y;. Hence on y; the following relation has to be satisfied:

f(oxz,-)k)ds=0.

THE GEOMETRY OF SUBMANIFOLDS

160

Since the thread is free at all its points except the end-points, then bx can be chosen arbitrarily everywhere except at the end-points where bx = 0. Therefore

v = \k. From this it follows that the curvature vector of -yi is in the tangent plane of the surface, i.e. -yi is the asymptotic line in a minimal surface. Since v is unity and A is constant, the geodesic curvature of the boundary is constant. As the normal curvature of -y; is equal to zero, -yi is a curve of constant curvature. 7.6 Second Area Variation of a Submanifold in Euclidean Space

Let us consider the variation of a minimal submanifold F" in Euclidean space E"+r directed along one of the normals np of the normal basis nl, ... , np: F = r + ES,

where r; _ rn,,.

To simplify notations we set r; = r,,,, i = w,,,, n,,,1 = n,,,,.. Then F; = r, + eivin,, +

Consider the metric coefficients of a variated surface F up to terms of order E2: r;) + E-{w,w, + w2(npl;n,,ij)}.

gij = (F,Fj) = gij + ew(nl; rj) +

Introduce the notations rij = (npl, rj) + (nalj ri), aij = w;wi + w2((np,, n,,lj). Then

gij = gij + EiVrij + E2aij.

Further on, it will be convenient to use the following columns: 91j

rl1 Tj =

gj =

(g.j) Now we find the determinant of the metric tensor gij:

g = detgij = [g1 +

e2al, $2 + 61172 + E'-a2, ...]

g+E".E[g1...r;...g"]+EZ i

i<j

Note that due to F" being minimal the coefficient of c is zero.

i=1

MINIMAL SUBMANIFOLDS

161

Expanding every determinant [g, ... of ... gn] with respect to the i-th column, we get n i%

i=1

Therefore gn]T. Il

(1)

i<j

Applying the Weingarten decomposition formulas, we find a,j $'/ = wjwjgll + w2(n.li nlj)$ij

= (grad

WI2 +11.2 {L,

Lipgkrghi + 1lan/i11v-/jgl1J

Now we find the second term in braces (1). Let P be any point of a minimal submanifold. It is always possible to choose the coordinate system about P in such a way that gij = bij at P and coordinate directions at P would be principal with respect to nP. In this case, at P we have L = 0 for i 54j. Then

[$I ...Tj...Tj...$n] =TiiTp - (Tij)z We have r1, = 2(n,,lj rj) = -2L1jj = -2k", where kf is the normal curvature of a minimal submanifold with respect to n. in the direction of the coordinate curve iii. Besides, rij = 2(nljrj) = 0 for i # j. Moreover, at P we have n

Lik Lnpg'tpgj% = > r=1

n

(L,)2= E(kv)z i=1

Thus, we can write

rn l >(k°)2+4>kjlkj +>(14,ii)2

g+e2{ IS

i<j

e-1

1.

0.1

k° = 0. Hence, Ej<j(k°)2+2 Fi kf kj1 - 0. Due to F" being minimal, we have With the help of the latter relation, transform the expression in braces. Thus, the second volume variation of a minimal submanifold in Euclidean space has the form 62S =

J

{rai W12 + w2

i

(A li)2+2W2>2k°kj dS, n,j

i<j

(2)

where dS is the n-dimensional volume (area) element of a submanifold. The field of variation S we took in the form wn.. The field ( defines each sum in (2) uniquely because they are independent of the orthogonal to C basis rotations. We can write the

sum of the first two terms in invariant form as (V(V ), where V is a covariant

162

THE GEOMETRY OF SUBMANIFOLDS

derivative in a normal bundle. Besides, note that Ei<j k f kJ = KK(n.) is the second symmetric function of principal curvatures of F" with respect to np. 7.7 Duschek Formula for Second Volume Variation of Minimal Submanifolds in Riemannian Space

We consider now a more general situation. Let FA be a minimal submanifold in the Riemannian space M"+P. Let V be a normal field of variation on Fk. Through each point x E Fk with V 0 0, draw the geodesics of M"+P in the V-direction. The set of all points of these geodesics form some (k + 1)-dimensional manifold M containing Fk as a hypersurface. Consider the variation of Fk in M. Introduce the coordinate system in M. Take as t the arc-length of a geodesic starting at the point of Fk. Then the metric of M has the form k

ds'`

_

g,

du'du'+dt2.

i.l=l

where (u' ..... u k ) are coordinates in F" and gl depend on face, obtained from Fk by variation, can be represented in ft as

uA, t). The sur-

t = e11'(u'.....Ilk),

where w is some smooth function of class Cl. Varying a we obtain a one-parameter family of surfaces. The surface corresponding toe has metric coefficients of the form 9i1 = g,1 + e21v", w,,, .

In the sequel we set w; We are going to find g = det Ig;;I and its derivatives of first and second order with respect to a for s = 0. Introduce the following columns: till

91i

gi=

a=

Uj

I

Then we can write 2 g = [g1 +6 wla,...gk + e2H'ka) A

_ [gl...g&]+e2 E[gl...a...gk]w;. i=1

Expanding all the determinants [gl ... a ... gk] with respect to the i-th column and summing them over i, we obtain the coefficient of e2: iv; w1g''g = g V, w. Here we use

the standard notation V1w for the square of the gradient off. As gi, depends on t, then g = [g, ...gkj depends on e. We have

dw... 191

atgk]+2egVln'+

MINIMAL SUBMANIFOLDS

163

where dots mean the terms of order higher than e. Note that for e = 0 1 8g;

-

2 8t where L, are the coefficients of second fundamental form Fk C M. If we introduce the column

Ll; L; _

,

Lki

= L;. Differentiating (I) we get the expression of the second then we can write derivative for E = 0 as d 2g __ 2

dE2

1 2 ... gk1 v` + 2g01 w. gi ... agr ... agi ... gk w +kL I ... a'gr at 2 at 8t 19 I

(2)

Introduce in Fk the coordinate system in such a way that at a chosen point .yo would be g; = S;, and coordinate curves would be tangent to the principal directions of the L, du' dui form. If n is the unit vector directed along V then at a chosen point L1; is equal to the normal curvature k; with respect to n. By the very definition, this value is the same either whether we consider Fk in ft or in Mk+P (note that n is tangent to Al). Then

{gi.........]4Lgi...Li...Li...1=4(Liilii_L).z4Jciki. dt

at

(3)

The second term in (2) at xo has the form 492

A

ar-

IV-

Consider the curvature tensor of M at the points of Fk. Denote by Rtgp its components. For convenience, set k + l = v. The metric of M is rewritten in the form

ds2=E j'Igy th' di, where we set x = t, x' = u1 for I

k. The general formula for Riemannian tensor components has the form a2gjj _ a2gk; _ 82g,, ax'8x/ + 8xkav dx'8xp axkaxl

I (82gkp RhJ°

+g1,n(r,,R rhp, - rip, r,JJ), where I'd,,, are Christoffel symbols of the M metric. From this formula it follows that 82gu + g11

2

[(rv;,1)2-rW.1 r;; r,

.

THE GEOMETRY OF SUBMANIFOLDS

164

The Christoffel symbols at x0 are rvi / = I 09-.1 2

ui

0Sul

OS;,

+&J

2 i31

_

- Lii,

r." = 0.

Thus, at x0 10 2 gii + Lz

Rn,;,,

(4)

8 Taking into account (3) and (4), we can rewrite (2) as

Id=g_ ,

_

= 1V

4

2

1: k l + E k? - k ;,,;,, i o

i
+ v 1 w.

(5)

i

Now express the components of the Riemannian tensor of M in terms of the curvature of M"+P and the normal basis torsion coefficients of Fk. Let a 1, ... , ek be unit vectors tangent to coordinate curves in Fk. Denote by k(e,, n) and K(e,, n) the sectional curvatures of ft and M"P respectively for the 2plane spanned on e; and n. In addition, let K,(ei, n) be the extrinsic curvature of M. Then at x0 we have the relation (see Section 5 Chapter 5): k(ei, n) = Ke(e,. n) + K(ei, n).

At xo we have P-1

Riviv = K(e1,n),

E(L,L°

Ke(ei,n) _

- (Laj) ).

e=1

where L' are coefficients of the second fundamental form of M with respect to normals no, o = I,... ,p - 1. Note that these normals are also normals to Fk. Now denote by tip the unit vector parallel to V earlier denoted by n, i.e. express the variation field in the form V = iiw,. We have LW = toV"Pnp

,

where V is the covariant derivative in M"P. As tip is tangent to a geodesic, then 0. Hence L,, = 0. Further, L0

where u,,Ii is the normal basis torsion coefficient of the submanifold Fk C Mk+P Thus P-1

1: /l;P/i + K(ei, n). n=1

Now consider the covariant derivative of V = wnp in the Fk normal bundle: De; V = W, tip + W1Lap/i no.

MINIMAL SUBMANIFOLDS

165

The square of this derivative is k

p-I

(VV VV) = v1w+w2y:EpQpl,. 1=I o=I

If we substitute the expression of R,,,,,, into (5) then w2 E pc';,pi together with V, w form in (5) the expression for (V VV V). Besides, as Fk is minimal, we have

4 E kik; + > k2 = 2 >2 k,kj. i<j

i

i<j

Substitute the relations just obtained into (5). To find d2 fg/de2 it is sufficient to find

d2g/de' because dg/de = 0 for e = 0 if Fk is minimal. Thus, we have found the expression for d22 f /de2 at the point xO under the special choice of coordinate system. With respect to an arbitrary coordinate system this expression gets the multiplier f. Integrating over the whole P. we obtain the expression for the second volume variation of minimal submanifold in Riemannian space.

fits=

F,Li<1 J

k

{(vv VV)+It,

}dS. i-I

JJJ

By another method this formula has been stated by Duschek [8). For minimal submanifold the inequality Ei
To prove this, take as a submanifold variation the family of submanifolds parallel

to F'I obtained from F" by the shift of each point in the it direction by the same distance, i.e. let V = n. As Vn = 0, the integrand in the second volume variation expression satisfies 2 >i<, k,ki - K(T, n) < 0, which implies the instability of Fk. Here T is the tangent space of P. For the case of a hypersurface, the second condition is satisfied automatically. K(T, n) is the ordinary Ricci curvature with respect to it. Therefore, every closed orientable minimal hypersurface in a Riemannian space of positive Ricci curvature is unstable.

THE GEOMETRY OF SUBMANIFOLDS

166

7.8 The Sum of Two Second Volume Variations of Two-Dimensional Surfaces In this section, we will study the stability of a domain D in the minimal surface F2 as well as the stability of a closed minimal surface F2. We consider two variations VI = wn1 and V2 = wn2, where iv is some C2 regular function on F2, nI and n2 are unit mutually orthogonal normal vector fields on F2.

Take the sum 62S(VI)+b2S(V2) and reduce it to the form convenient for our purpose. Consider, first, 62S(VI ). We suppose that on F2 there is introduced the

isothermal coordinate system with respect to which the metric has the form ds- = A(du2j + du;). We shall discuss in another section the possibility of introducing that system of coordinates. Expand (V VI V VI ), separating torsion coefficients µI2/1 and 1412/2: n-2

(V VI V VI) = A

114' + Hu, + H'2 [(µ1z/1)2 + (1412/2)2] + H'2

[(ill//t)2 + (iJij12)2} j=3

=A

(VI-, + 1yµ12/2)2 + (wu, - 1412/1)2 + (IV21412/1)nz - (w21412/2)u.

n-2 + w2({112/2.1 - 1412/1.2) + w2

[(111j/I )2 + (141//2)2]

(I )

j=3

Here 611j/k are torsion coefficients with respect to the basis of normals Replace 1412/2.1 - 1412/1.2 in (1) with their expression from (3) of Section 3 Chapter 5: lml/2.1 - µW11,2 = Krnl + O'rnr + Tml

where the right-hand-side values have expressions in terms of the second fundamental form coefficients Lk# of F2, torsion coefficients and Riemannian tensor components

given by Kml = (L11 L12 - L1 2L11 + LZI L;, - L°;Li1)/A,

n-2/ O'nrl = E(61mj/I µl1/2 - 61mj/21 /j/I ),

j-1 Tmr = R.jx,,,,n44 ni y'i J''2 /A.

n

These values depend on the plane through nand nl and on the orientation in it. Rewrite (I) as (V VI V VI) =

A I (H"' + y141z/2)2 + (lc , -will./1)2 n-2

+ w2(h,12 + T12) + H'2 E(611v) 6121/2 - 144/2 ,021/1) p-3 n-2

+ IV2 J=3

c1(plj/t

)2 + (µ1i/2)2]

MINIMAL SUBMANIFOLDS

167

In an analogous way transform (V V2 V V2). We obtain an expression different from

the one above only in the last term which has the form n-2

[(pall )2 + (14/2)2] . r-3

Now we can find the sum >,`_I 6-S( Vi). Note that n-2

t{(µ1j/1)2 + (µIj/2)2 + 2(µ'1,i/1 /421/2 - AIj/2 i22J/I) + (p2j/I )2 + ({22j/2)21 j=3

n-2

= F, [(µij/I +lhj/2)2 +(PIj/2

-142//{)2

j=3

Therefore, we can write 2

E(V V1 V Vi) = i=1

IV /112/2)2 + 2(w, - wjz12/1)

A

n-2 + 11'2

j-3

[(141j/I + 1L2j/2)2 + (µ1j/2 -

+ 2w2INp2 + T121 +

A

[(1V2µ12/I)w

2

Iii/1)2

- (II°2{212/2)..]

(2)

The integral of the last term over any domain D C F2 can be turned into the integral over the boundary of this domain:

J[(V2VV1)-(ViVV2)1.

(3)

t1D

The latter integral is written in invariant form. It is equal to zero if variation fields VI

and V2 turn into zeros on the boundary. If F2 is closed then integration of the divergent terms on the right-hand side of (2) produces zero. Applying the second volume variation formula for the surface in Riemannian space, we get

22

i-I

62S(Vi) = f

A

[20".. + t'µ12/2)2 + 2(w,, -

2] VVIA 12/1

F2

w2 n__

+

A

(µl1/1 + µ2j/2)2 + (1411/2 - µ2j/1)2 + IV2 2n.17 + 2T12 J=3

-2

I Kr(n,)I i

1

-

2

K(ei, nj)J } dS. i,j=1

JJ

(4)

168

THE GEOMETRY OF SUBMANIFOLDS

This formula will be used in the sequel. Here el and e2 are mutually orthogonal unit

vectors in the tangent space of F2, Ke(n;) is the second symmetric function of principal curvatures of F2 with respect to the normal ni. 7.9 Theorems on Instability of Minimal Surfaces in Riemannian Space of Positive Curvature

By means of the theory of geodesic lines, which attracted the attention of geometers and was developed by them in many directions, in the papers of Rauch [43], Berger (21], Toponogov [22], Klingenberg [23] and others, deep results on the topology of Riemannian manifolds have been obtained, in particular the theorem on the sphere

has been proved. However, for some cases the methods of this theory are insufficient. The investigation of the homotopy groups of manifolds have been associated with

the theory of the stability of minimal submanifolds. The hypothesis on this subject was stated in 1975 [26]. In this paper and the paper published the following year [27] i proved the theorems on the instability of closed minimal surface in Riemannian space of positive curvature. Using the existence theorems, Yau and Schoen in 1979 [9], and Micallef and Moore in 1984 and subsequent years [101, [ 1 I ], [ 12], [13] applied

this approach to investigations of the topology of manifolds. In the work [13] the authors introduced the notion of curvature for totally isotropic two-planes and proved the following theorem. Theorem (Micallef, Moore) Let M be a compact simply conntected n-dimensional Riemannian manifold which has positive curvature on totally isotropic two-planes, where n > 4. Then M is homeomorphic to a sphere.

Also, note that, applying the minimal surfaces theory, Burago and Toponogov in 1973 [51] obtained an interesting result in the theory of geodesics. They stated that in a three-dimensional closed simply connected Riemannian manifold of positive Ricci curvature the length I of any closed geodesic is bounded from below. More precisely, I > 12 a 6/Ri, where Ri is the minimum of Ricci curvature. That paper is related to Toponogov's favorite (by his own admission) problem: prove that if for arty 2-plane cr the sectional curvature K, of a simply connected closed Riemannian manifold satisfies the inequalities 0 < b < K < I then there exists the positive-valued function f (b) such that the length I of any closed geodesic satL /les the inequality l > A b) . In this section and the following section we prove the following result: Theorem (Aminov) Let F2 be a minimal surface homeomorphic to a sphere in a complete simply connected Riemannian manifold R" of curvature from the interval (4 , 1]. Then F2 is an unstable minimal surface.

Note that in [14] Lawson and Simons considered minimal flows in Riemannian space and stated the hypothesis on their instability in a complete simply connected Riemannian space R" of curvature from the interval (4-1. I].

MINIMAL SUBMANIFOLDS

169

One may consider the surface F2 in Riemannian space R" with the boundary defined in the neighborhood of F2. In this case it is necessary to pose the condition of

F2 normal bundle triviality, i.e. of the existence on the surface of the continuous basis normal fields n1,...,n"_2. The following holds: Theorem 2 (Aminov) Let F2 be a minimal surface, homeomorphic to a sphere, in the Riemannian space M" of curvature from the interval (4 1, 1]. Suppose the normal bundle of F2 is trivial. Then F2 is an unstable minimal surface.

For the surface in four-dimensional Riemannian space there hold more strong theorems. But we shall state and prove them in the next section. First we prove Theorem 2. Let M" be a C4 Riemannian space and components of

the position vector of F2 be of class as functions of parameters. As F2 is homeomorphic to S2, it is possible to introduce isothermal coordinates u1, u2 singular at one point (see [15], [18]). The metric of F2 with respect to that coordinate system has the form ds2 = A(dui + dug), where A 0 as 1 if we set z = ul + iu2. Select the variation fields V, and V2 in such a way that they satisfy the following systems simultaneously: (a) '+'u, + i'µ12/2 = 0,

x'", - i'µ12/1 = 0,

(b) Ali/1 + µ2112 = 0, µU/2 - 112j/I = 0.

2.

(1)

Observe that E 3 L({L,j/, + /z2j/2)2 + 011#2 - 112j/1)2] depends on only the plane distribution spanned on V, and V2. From the triviality of the normal bundle on F2 it follows that there exists the field of frames C1, ... -G-2 of normal space N.Y such that Ci are continuous mutually orthogonal unit vector fields on F2. Moreover, if F22 is of the Cm regularity class then Ci can be chosen from the C'"-I class. We can use the Pontriagin method to prove this (see [16] Section 7). In this paper he considered framed manifolds in Euclidean space. By the Nash theorem [17] the Riemannian space M" together with F2 can be regularly and isometrically immersed into some Euclidean space. The method from [16] of constructing the regular framing F2 homotopic to the given continuous framing is applicable to our case. Thus, we can suppose that if F2 is of the C2." regularity class then i E Cl-". Write the decomposition

k = 1..... n - 2.

V k = link =

As I V, = I V21 = iv and V, 1 V2, then K'2=Eair=>a2-,

>a,Ta2T=0

(2)

It is easy to see that the torsion coefficients pjk/i can be expressed in terms of torsion coefficients pjk/i of the 1;i basis as 1t'2 pjk/i = air

8akr

8a- + afr aA,, µm/i,

(3)

THE GEOMETRY OF SUBMANIFOLDS

170

Using (3), the equations of system (1)(b) can be written as

a air

aair aui

raa, r 1\ 0142

+

8a2,

+ an(aI, jro/1 + ago

8142

- 8827 1 8141 /1

0,

(4)

+ atr(a1, lam/2 -agol1) = 0.

Since w2 = Eai,, the equation w,,, + n7112/2 = 0 after multiplying by w gives the form

alr

0a1r 814 1

+aIr

aa2r 814 2

+alra2v{k,o/2 =0,

i.e., taking into account that air aio µro/1 = 0, we conclude that the latter equation can be obtained from (4) if we set j = 1. In an analogous manner we state that (4) produces the equations of system (1)(a) if we set j = 1, 2. Thus, all equations of (1) can be written in the form of (4) if we set j = 1, ... , n - 2. Let ak1 be components of the matrix inverse to IIa;iII Multiplying (4) by ak/ and contracting over j, we get as

as

8141

+ 8142 + a, ILkm/I + a2m Pkn,/2 = 0,

aaik aa2k 5142 - 514i + as,,, Jkm/2 - a2m µkm/l = 0,

(5)

where k = 1,2,...,n- 2. Let U be

a complex vector with components Uk = a1k - ia2k and B the matrix with components Bmk = µmk/1 + iµmk/2. Then we can write (5) as 2

aUk

of

(6)

= Bk,,,

The solution of (6) is the generalized analytical vector function (the definition in tend to zero as [18]). Since (i are regular of class CI-0, the coefficients B4,,, E 1 /IzI2, when z --# oc. Indeed, the torsion coefficients µkm/i have the form {kA-m/; = (nm, nk.i) + IOV+ O1 Yi k Sm,

where [av, I 1 are Christoffel symbols of the M" metric. We can write '

t"

where alas; means the derivative with respect to the arc-length parameter. By virtue of the regularity of k on F2 we have that Ia"I is bounded on F2. Besides, it is evident

that ICI < const. Since v(A- -. 0 as I/IzI2 when z -, oo then µk/; -. 0 as From the conditions I V1 I = I V2I and V, 1 V2 it follows that U2 _ U; = 0. 1/IzI2

Prove that (6) has the solution U defined over the whole z-plane including the infinite

point oo and such that U2 = 0.

MINIMAL SUBMANIFOLDS

171

Since BTU = -B1 for any two solutions X and Y of (6) the functions X2, Y2 and (XY) are holomorphic functions of z. Indeed, for instance, we have 8U2

a_U

I

Un,(Bkm + Bmk) = O.

aZ = 2 Uk In the case of bounded X and Y the holomorphic functions X2, Y2 and (XY) are 2 U aZ

constant complex numbers. If there exist two linearly independent solutions X and Y defined over the whole plane then, taking a linear combination U = al X + a2 Y such

that the equation a X2 + 2a1a2(XY) + a2 Y2 is satisfied, we obtain a non-trivial solution over the whole plane and such that U2 = 0. Show that the above mentioned two solutions X and Y exist. If C is a constant vector and U is a solution of (6) then W = U - C satisfies the inhomogeneous equation 2

=BW+BC.

-

Let B° be the space of continuous vector functions with the usual metric, defined over the whole z-plane and tending to zero when z -' oo. Introduce in that space the linear operator S as

SV=

1

a

f/ BV(t) dE, z - t

E

where E means the complex plane. Since Bj 0 as 1/I:I2 when z oo then, in the same way as in the Pogorelov monograph, ([19], p. 428) for a one-dimensional system, we prove: (1) if V C B° then SV C B°; 2) S is totally continuous. Suppose (6) has the non-trivial solution U tending to zero while z - oc. Since U2 is a bounded holomorphic function, then U2 = 0 and, as a consequence, we suppose below that condition (3) is satisfied and the homogeneous equation has only trivial solutions. Then for an arbitrary C the integral equation W + SW + SC = 0 has the solution W such that W 0 while z - oc. Selecting two linearly independent vectors C, and C2 we get two linearly independent solutions X = W, + C1, Y = W2 + C2 such that X C1, Y -. C2 while z oc. Therefore, there exists the solution U of (6) such that U2 = 0. This fact implies the solvability of (1) in the regularity class. Select the variation fields V, and V2 in such a way that (1) would be satisfied. Below we use the Berger inequalities on curvature tensors (we shall prove them in Section 12): If for any 2-plane o, in the tangent space to M" the sectional curvature K satisfies 6 < K, < I then the components of a Riemannian tensor with respect to the orthogonal coordinate system satisfy

(I - b),

I R,Jk,I < z

j4k

IR+fk1I 5

(1 - b).

i i,j,k0l.

(7)

From this it follows that I T121 < 3 (1 - b). It is easy to show that for a minimal surface F2 Ik12I 5 IKK(nl)I + IKe(n2)I

(8)

THE GEOMETRY OF SUBMANIFOLDS

172

Taking into account (7) and (8), we have

Eb'-S(Vi)=Jsv2 [2K12 -2EIK,.(ni)I+2T,2i-1

K(ei,ni)JdS i.i

i

F_

(

< rtv2L3(I F'

9)

-h)-46JdS<0.

L

because by the theorem hypothesis b > 1/4. Therefore, F2 is an unstable minimal surface. Theorem 2 is proved. To prove Theorem I we consider the case of n = 4 separately.

7.10 Minimal Surfaces in Four-Dimensional Riemannian Space

The case of n = 4 is rather distinctive. The instability theorem holds under weaker conditions. It is not necessary to require the completeness of M4 and normal bundle triviality of F2. Theorem (Aminov) Let F2 he a minimal surface homeomorphic to a sphere in an oriented Riemannian space M 4 oJ' curvature froth the interval (,1-t . 11. Then F2 is tatstable.

If there exists the regular unit normals vector field on F2, then the statement on instability is true under the assumption that the curvature of M4 belongs to the closed interval (a . 11. At the same time, in complex projective space P2C with the Fubini-Studi metric of curvature from the interval (,1-t. I) there exists the totally geodesic surface which produces the area minimum among close surfaces. But this surface does not admit the regular field of unit normals. In section 5 of Chapter 6 we considered the Whitney invariant of a surface in fourdimensional Euclidean space. It was defined as a sum of indices of normal vector field singular points. We proved that in the case of a closed oriented surface the Whitney invariant can be evaluated by means of integrated Gauss torsion Kr:

V =dS. JKI F=

In the case of the surface F2 in four-dimensional Riemannian Space M4 a more general formula, stated by Chern, holds: I = 2 V

(t:12 + Tie) dS. F:

This formula is a particular case of the Whitney invariant formula, stated for a closed submanifold FA in Riemannian space M21 (37].

As above, we suppose that F2 is homeomorphic to the sphere. Construct the normal vector field on F2 with only one singularity of index v at = = oc. To do this,

MINIMAL SUBMANIFOLDS

173

take in some neighborhood of z = oc the regular mutually orthogonal normal vector fields SI and S2 which generate a positive orientation in the normal space NT. Set z = IzI e'a. Now define fields i with singularity at z = oc as SI = 1 cos vcp - f2 sin i'

42 = I sin vcp + 6 cos vtp.

,

Since the torsion coefficients it12jj of the (FI.e2) basis are related to the torsion coefficients µ,2/i of the

basis as PI2/i = µ12/i + v aul ,

then {i the singularity index at z = oo is equal to v. Because of it being equal to the Whitney invariant of F2, the fields li have the regular prolongations on the complement to the chosen neighborhood of z = oc. By definition, we have

o = arctan

12

III

,

_ dul

-

U2 1z =

1`

a 0

111

121

Since i are regular about z = oo , then µ12/i -. 0 as 1/1212 while z -+ 00. Therefore, µ,2/i - 0 as 1 /1212. This speed of decrease is not sufficient, in general, for the regular solvability of (6) Section 9, but in the case of v > 0, due to the special choice of {i, we prove that the solution exists. Moreover, IUI 0 while z - 00. Therefore, at z = oc

the variation fields vanish and, as a consequence, the variations will be defined everywhere on F2. Next we shall consider the case of v < 0. The system (6) from Section 9 for F2 C M4 has the form

2a =BI2U2

2aa2=B21U1,

where B12 = µI211 + iµ1212. Besides, it must satisfy U2 + UZ = 0. Set U2 = 1UI. Then

we obtain only one equation

2aa'

=iB12U1.

(1)

In some neighborhood of z = oc we can write iB12 = -A12/2 + iiII2/I = -P12/2 + 1µ12/2 -

V

-.

Suppose that v > 0. Set W = U, (I + Iz1°). Then near z = oc we have

8W out aZ=aZ

(1 + ICI

+v2UI

(Z)e_I

ZY

v

= - 2lµ12/2 + 2µI2/2 - 22(1 +

Izlv), W.

Therefore, W satisfies the equation with a regular coefficient everywhere in the complex plane, where this coefficient tends to zero not slower than 1/1=12 while z -' oc. Then there exists a regular solution W which is bounded in infinity together with the derivative. Then the required solution satisfies

THE GEOMETRY OF SUBMANIFOLDS

174

W

U, =

while z -oo.

If the Whitney invariant v < 0 then we consider the system

2 aU = -BU.

(2)

The point is that the sum of second variations can be transformed to l2S(Vi) =

J1

IVA1212)2 +

" (wu, + IVAA1211)2]/A

F2

+w2[-211.12 -2T12+...]}dS. The latter expression differs from that obtained in Section 8 in the signs before p12/j, ,,I2 and T12. Then the coefficient of U1 on the right of (1) changes to

Ic]2/2 - i,1211 +-;. In this case, set W = UI(1 + Iz]-°). Applying the procedure, analogous to the case of v > 0, we find the solution W. Finally, observe that the contour integral (3) of Section 8, which can be regarded as taken over the circle IzI = const, tends to zero while z -y oc. Theorem 3 is proved. For the surface F2 with zero Whitney invariant we prove a stronger assertion.

Theorem 4 If F2 is homeomorphic to a sphere minimal surface with zero Whitney invariant then it is unstable provided that the curvature of M4 takes its values i n [4 ,

I].

The following holds. Lemma If F'- is homeomorphic to a sphere minimal surface in a Riemannian space M4 of positive curvature and the local imbedding invariant n, n'Z e' e; is sign-preserving on F2 then F2 is an unstable surface. T1 2 =

Here ei and ee are components of mutually orthogonal unit vectors e1 and e2 tangent to F2, ni and n; are components of mutually orthogonal unit normals on F22, R,,m are component of an M4 Riemannian tensor. Since v = 0, there exists on F2 the regular normal basis field (n1,n2). There exist the solutions of (I) and (2). From the two pairs of variations defined by these solutions, select the pair for which the T12 would be included into F_, 62S(VV) with a

negative sign. For that pair E.b2S(Vi) < 0 and, as a consequence, the surface is unstable. The Lemma is proved. Turn now to the proof of Theorem 4. Consider one of the pairs of variations, say that which is defined by system (1). Since the curvature of M4 is from the interval [4, 1], then Eib2S(Vj) < 0 where the equality to zero occurs only in the case when T12 at each point takes its maximum value, i.e. T12 = i (1 - 4) = z Therefore T12 preserves the sign. By the previous lemma, F2 is unstable.

MINIMAL SUBMANIFOLDS

175

By analogous methods, one can prove the instability of a totally geodesic surface homeomorphic to a sphere with zero Whitney invariant in an oriented symmetric space M4 of positive curvature [25]. In this case, from the total geodesity of F2 and symmetry of M4 it follows that T12 = const. Moreover, the condition v = 0 implies T12=0. 7.11 On the Instability of Minimal Surfaces in a Complete Riemannian Space

Here we continue the proof of Theorem l from Section 9 based on the result of the two previous sections. Let M" be a complete simply connected orientable space of curvature from the interval (1, 1]. Under this condition the following theorem holds: Theorem If M" is a complete simply connected orientable Riemannian space of' curvature satisfying inequalities I < K. < 1 then M" is homeomorphic to the sphere S".

The proof of this theorem was begun in the paper by Rauch [43] and was then continued in papers by Berger [21], Toponogov [22], and Klingenberg [23]. It is known - due to Milnor's wonderful result - that on an n-dimensional manifold, which is homeomorphic to a sphere S" for definite values of n, there can be defined differential structures different from the differential structure of the usual sphere. Those spheres are called exotic. There were attempts to prove that in the theorem above M" is diffeomorphic to the usual sphere S". Provided that 0.87 < K < I, Sugimoto and Shiogama [24] proved that M" is diffeomorphic to the usual sphere S", but if a < Ko < 1 it was proved that M" is diffeomorphic to either the usual or the exotic sphere. The exotic sphere E" is the surface obtained from two balls D" and D2 by gluing by means of boundary diffeomorphisms [38]. If one cut out from E" the ball D", however small, then the complement would be diffeomorphic to the Euclidean ball Dl-). As F2 is compact, there exists the point in M" and its sufficiently small neighborhood not intersecting F2. Cut out Di from that neighborhood. Let D. be the complement of D1. In papers [39] and [40] it was proved that the normal bundle of immersed spheref(S2) C E" for n > 5 is trivial. The regular homotopy between f (S2) and the standardly immersed sphere S2 C E3 exists. We can suppose that it is realized in some sufficiently large ball homeomorphic to D". Therefore, if n > 5 then the normal bundle F2 C M" is trivial. In this case we can apply Theorem 2 from Section 9. For n = 4 the instability is stated

by Theorem 3. Thus for any n the surface F2 is unstable. Theorem 1 is proved. In the case of locally conformally flat space, the condition on curvature of the space can be weakened to the condition of curvature positivity. Theorem 5 Let F2 be a minimal homeomorphic to sphere surface in an orientable locally conformally flat space Mn of positive curvature. Suppose the normal bundle F2 is trivial. Then F2 is an unstable minimal surface. If n = 3 then the surface F3 is the minimal hypersurface. In this case the instability

occurs only provided that the curvature of ambient space is positive. Therefore, consider the cases of n > 4.

THE GEOMETRY OF SUBMANIFOLDS

176

Remember the expression for the conformal curvature tensor: Chijk = Rh;jk + n

2 (bhjR;k - bhk R;j + gik Rhj - g,jRhk )

R

+ (n - 1)(n - 2) (bhkgij - 6hj9ik ). As e1,e2, n1 and 112 are mutually orthogonal, by the expression above we find

Therefore, for the case of a surface in locally conformally flat space we have T12 = 0. Then in the integrand in (9) from Section 9 we can replace the number (I - b) with 3 0. Therefore, in this case F_, 6 2S(V,) < 0 which means instability.

7.12 Berger Estimates of Tensor Curvature Components

In previous sections we used Berger inequalities on tensor curvature components. They are useful in many other situations if the curvature K, of the investigated Riemannian manifold satisfies definite conditions over all 2-planes a in the tangent space. Riemannian space is called b-pinched if there exists the constant b such that over all 2-planes the curvature Ko satisfies the inequalities b < Ko < 1.

(1)

For the case of 6-pinched Riemannian space Berger stated the inequalities on Riemannian curvature tensor components R;jA, with respect to some orthonormal basis. They are I Ryk,I

2 (1 - b)

for

i 96 j

IRijkil53(l-b) for (i,j)

k.

(k,!),

(2)

i<j,

k
(3)

Denote by K(X, Y) the curvature K? of the 2-plane a generated by the vectors X, Y. If Xi and Y' denote the components of these vectors then by definition

Y) = K(X,

R;jk,XiYiXkY!

(XX)(YY) - (XY)2'

where parentheses in the denominator mean the scalar product. Denote by e1, ..., e the orthonormal basis in the tangent space. We also use the notation K;j = K(e;,ej) _

R.

Inequalities (1) imply b < K(e;, aej + bek) < 1,

where a and b are arbitrary real numbers not equal to zero simultaneously. From this it follows that

MINIMAL SUBMANIFOLDS

177

a2(K4 - b) - 2abRij;k + b2(Kik - b) > 0,

a2(l - Kij) + 2abRjik + b2(1 - K&) > 0. The discriminant of these two quadratic polynomials with respect to a and b must be non-positive. Hence (Rijk)2 < (Kij - b) (Kik

- 6),

(Rijik)2 < (I - K;1)(I - Kik) As the geometric mean is less than the arithmetic mean, from the above we conclude

RjuI < (K;j+K;k - 2b). Real :5

(2 - Ky - Kk).

Adding the corresponding sides of inequalities just obtained, we get the first inequality under the proof: JR;jkil < 1(I - b), where i # j 36 k. Obtaining the second estimate (3) is more complicated. Consider the inequalities 6 < K(aei + bek, eel + dej)

for any set of basis vectors ei, ej, ek, ei and any real numbers a, b. c, d. Transfer b to the right-hand side and set

F(a,b,c,d,i,k.!,j) = K(ae1+bek,cei+dej) - b.

Let aei + bek, cei + dej be unit and orthogonal to each other. Then a22 + b2 = I , 2 + d2 = 1. We can consider the function Fas a polynomial of degree 4 with respect to a, b, c, d while with respect to each of a, b, c, d it is a polynomial of degree 2. Consider the polynomial C

G(a,b,c,d,i,k,!,j) = 2 [F(a,b,c,d,i,k,1,j) + F(a,-b,c,-d,i,k,1,j)]. It is easy to find that K(ae, + bek, cei + dej) = K,1a2c2 + Kk;b2c2 + K;ja2d22 + Kkjb2d2

- 2Riikiabc2 - 2Rijkjabd2 - 2R;j0a2dc - 2Rkikjb2cd - 2(Riikj + Rijki)abcd.

If we change the signs of b and d, then we obtain an analogous expression which contains the terms in the second line of the above with the sign "+". Hence the function G(a,...) contains terms of the form a2c22, a2d2, b'-c2, b2d2 and abed. Now introduce a polynomial

H(a,b,c,d,i,k,1,j) = G(a,b,c,d,i,k,!,j) +G(-a,b,c.d,i,1,j,k). Observe that the second term differs from the first not only because a is replaced with -a, but also in last the three arguments. That is, the replacement: k --F I -+ j , k has been made. Thus, for the function H we get the expression

THE GEOMETRY OF SUI3MANIFOLDS

178

H=(Kit +K,1)a2C2+(KkJ+K,)b2C2+(K,1+K;a)a2d2+(Kkj+K,k)b2d2 - 2(R,,kj + Rfjk, - R1ik - R;ky)abcd - 26.

We can simplify the coefficients of abcd using the Bianchi identity: R;jk1 + R&1j + R;1jk = 0. We obtain that this coefficient is equal to -GR;jk1. Introduce the notations

A=K;j+Ki1-26, B=K;j+K;k-2b, C=Ku+K1j-26, D=Kkj+Kk,-2b, E = 3R;jk1.

The function H has the form H = Aa2c2 + Ba2d2 + Cb2c2 + Db2d2 - 2Eabcd = (Ac2 + Bd2)a2 - 2Eabcd + (Cc2 + Dd2)b2. Since F > 0, H > 0. If we consider Has a quadratic expression with respect to a and b

then we conclude that the discriminant of this expression must be non-positive. Hence

ACc4 + (AD+BC- E2)c2d2 +BDd4 > 0. From this we get

(AD+BC-E2)22-4ABCD<0. We can rewrite the latter inequality as

E4-2E2(AD+BC)+(AD-BC)2 co. Therefore E2 is less then the largest root of the left-hand-side expression, that is

E22
6[A+D+B+ C].

(4)

By an analogous method we get IR;jk/I 5

3

[(AD)I/2 +(B'C')1j2] <6(A'+D'+B'+C'],

where

A'=2-K;j-K;1, B'=2-K;j-Ku, C'=2-K1k-K1j, D'=2-&j-Kk,. We can rewrite the inequalities (4) and (6) in the form

(5)

MINIMAL SURMANIFOLDS 1

179

12K,,+2Kkt+Kit+K;k +Kjt+Kjk -8b],

IRiiktl <1

IRik,I1[8-2Kii-2Kkt-Kit-K,k-Kp-Kjk]. Adding the corresponding sides of the inequalities above, we obtain IRijktl S

(1 - b),

for (i, f) # (k. 1).

Thus, the Berger inequalities are3 stated. 7.13 Minimal Surfaces and Scalar Curvature of Ambient Space

Schoen and Yau [19] applied the notion of minimal submanifold stability to the investigation of topology of three-dimensional Riemannian manifolds. They observed that if a closed minimal surface in three-dimensional Riemannian space N3 has positive genus then its instability follows from a less heavy condition on N3 than

in the Simons theorem proved in Section 7. It is sufficient to require the scalar curvature to be positive. Namely the theorem holds: Theorem 1 (Schoen, Yau) A closed minimal surface F2 of positive genus p in a Riemannian space N3 of positive scalar curvature is unstable.

In the considered case the second variation formula has the form 62S = J {VIII, + w22(2K,. - K(ej,n) - K(e2,n))} dS, Fz

where it is the normal on F2. Set ir =const. The scalar curvature R of N3 is equal to R = 2{K(e1,n) + K(e2,n) + K(e1,e2)}. Then

2Ke - K(el i n) - K(e2. n) = Ke + K., + K(e1, e2) -

R 2

In the expression on the right-hand side, only K(ej, e2) has an indefinite sign. But summing it with Ke - the extrinsic curvature of F2 - we get the intrinsic curvature K,, i.e. Ke + K(ej, e2) = Ki. Since by hypothesis the surface is of positive genus,

JKIdS=21r(I -p)<_O. F2

Therefore, the second area variation is non-positive:

62S=J;v2(Ke+K,-2 f dS<0, F=

/

where equality to zero occurs only for v = 0. Thus, the surface F2 is unstable.

180

THE GEOMETRY OF SUBMANIFOLDS

Note that in [14] there was obtained the expression for the second derivative of area of the surface from the family of parallel surfaces in three-dimensional Riemannian space: 2

dtST J

(ic + K, -2dS,

where t is the arc-length parameter along the normals to the surface of the family. The application of Theorem I to the investigation of the topology of Riemannian spaces is based on the theorem on the existence of closed minimal surfaces of prescribed genus p > 1. Here is a reminder of some preliminary definitions. Let F be a two-dimensional closed manifold and y1, ... , y, closed curves in F with a common point xo. Assume

that they are representatives of different classes from the fundamental group 7r1(F, xo) and generate this group. Let f be a continuous mapping of F into a closed Riemannian manifold N. Every

continuous mapping of one manifold into another induces the mapping of the fundamental group of F into the fundamental group of N. Denote this induced mapping by.f.:

f.: 7r1(F,xo) -* 71(N,f(xo)).

We say that two continuous mappings W and fare conjugate with respect to the fundamental group if there exists the path T in N joining f(xo) to cp(xo) such that every loop r 1
the mapping f of class L. This class is convenient to the proof of existence of a mapping which gives the minimum of an energy functional. Rather complicated theorems, such as the Nash theorem on isometric immersion of a Riemannian manifold into a Euclidean space, and the Mumford theorem on the compactness of some set in the space of modules and others, have been used. That is, the chain of arguments takes a dangerous size. Theorem 2 (Schoen, Yau)

Suppose that N is a compact Riemannian manifold, F N is the injective snapping on fundamental groups. There exists a conformal structure on F and a branching minimal

is a compact surface of genus p > I and W: F

immersion f: F- N with action on 7r1 which is conjugate to that of V and which minimizes the area over all such maps.

In the case of p = I a stronger theorem is proved: any Abelian subgroup of rank 2 in 7r1(N) can be realized as the image of ir1(T2) under minimal immersion of a twodimensional torus T2. If N is three-dimensional then from the results of Ossermann, Gulliver, Hildebrandt, Kaul and Vidman follows the regularity of the area minimizing surface.

MINIMAL SVBMAN4FOLQS

1Kl

Theorem 3 (Schoen. Yatt) Let N he a compact Riemanniat manifold. Suppose that rri(N) contains the finitely generated non-ire/lea! Ahelian subgroup. Then there ext3ts the branched minimal inntrer ion f: T' - N such that the induced mapping oil 7ri has rank 2 for the image in rr! (N ). Moreover, / minimi_e' area over all Harps g of toms T2 into N irlitch are conjugate to /: From Theorems 1-3 follows the theorem on the non-existence of the metric of positive scalar curvature on the manifold of prescribed topological type.

Theorem 4 (Schoen. Yau1 Let N3 be a compact orientahle Riemannian space. Suppose that one of the following conditions is satisfied:

(1) rri (N3) contains a finitely generated non-cirlic Abelicnt subgroup: (2) r1(N3) contains the subgroup Lsotnorpliic to the fimdimrental group of genus

p > I surface. Then N3 does not admit a metric of'positiie scalar curvature.

Moreover, applying the results of Kazdan and Warner, it was stated that the metric of non-negative curvature on these manifolds is necessarily flat. 7.14 Two-Dimensional Minimal Surfaces in Four-Dimensional Riemannian Spaces of Constant Curvature

For a surface in Riemannian space the normal curvature indicatrix is defined in the same way as for the surface in Euclidean space (see Section I Chapter G). For a twodimensional space it is an ellipse in normal space. If F2 is a minimal surface then at each point x E F2 the center of that ellipse coincides with Y. It is known that the metric of a minimal surface in E3 is not arbitrary. There exists the necessary and sufficient condition on the metric to be the metric of a minimal surface. That condition has been found by Ricci: in the first place, the Gaussian

curvature K:5 0. in the second place. the Gaussian curvature of the metric

doe ='I

the is zero. We can represent the Ricci condition in the following form

0_ In $k=K,

(1)

where V2 is the Laplace--Beltrami operator in the metric ds'. To show that (1) follows from the Ricci conditions, represent the metric ds2 with respect to isothermal coordinates: dc2 = A(du12 + dug`) and find the Gaussian curvature of

da- = %/-K*? + du;). We have

1i4._-2v

I

1

=

t?=

8= W_

In

1

182

THE GEOMETRY OF SUBMANIFOLDS

Since by the Ricci condition K = 0, we obtain equation (1). In this section we state the necessary and sufficient, for some cases, conditions on the minimal surface metric. Theorem I If K is the Gaussian curvature of the metric on the minimal surface F2 in Riemannian space R4 of constant curvature K0, then the following inequality is satisfied:

V2 In '/XD - K - K > -ZKo.

(2)

2

In the particular case, when F2 is in Euclidean space E°, this condition expresses the fact that the Gaussian curvature Ko of the metric do-2=V-__K_ ds2

is non-positive: Ko < 0. If Ko = 0 then ds2 is immersible into E° as the metric of minimal surface with the circle as the normal curvature indicatrix. The equality in (2) holds for minimal surfaces with the normal curvature indicatrix being the circle.

From inequality (2) follows the assumption: if F2 is homeomorphic to a sphere minimal surface in the space of constant curvature Ko = I and the curvature of its metric satisfies K < 1, then its area S _> 121t. An analogous condition has been stated by Calabi [27] for a homeomorphic to a sphere minimal surface in an n-dimensional sphere of unit radius.

Now pass to the proof of the theorem. Denote by a and b the half-axes of the ellipse - the normal curvature indicatrix. Let e;1 and {2 be unit normal vectors directed along the principal axes of the ellipse. Consider, first, the domain where a 54 b. The second fundamental forms define the mapping from the set of directions

in tangent space Tx into the normal curvature ellipse; moreover, the orthogonal directions in T, pass into diametrically opposite points of the ellipse. At each point x E F2 select two mutually orthogonal directions Tj and T2 which pass into points in one of the principal axes of the ellipse, say in the axis of the {I direction. Lines, which are tangent to Ti and r2, form the orthogonal net on F2. Take it as a coordinate net. Suppose the metric with respect to these coordinates has the form ds2 = Edu2 + Gdv2.

Then the second fundamental forms with respect to normals 1 and i;2 have the simple expression

Ill = aEdu2

- aGdv2,

II2 = 2bvdudv,

because the ellipse center coordinates are a = Q = 0. Set v1 = (S1S2u), V2 = (.1t2,) We can represent the Codazzi equations in the form of two systems: (aE), = v1bVrE_G,

(aG)u = -v26 EG,

(3)

MINIMAL SUBMANIFOLDS

vl EG

G

183

- EGb = aEv2, (4)

E,b V

- EGb , = -aG v1.

E

In addition, the Gauss equation

K-Ko=-a22-b2

(5)

i/2, - v,, - tab EG = 0

(6)

and the Voss-Ricci equation

hold.

Multiply the first equation of (4) by b/E and the second equation of (3) by a/ EG. Subtracting, we have

J(bhu_aau)+(b2_a)_7=0.

(7)

Without loss of generality we can suppose that b > a. From (7) we find

_ _ (In b2 -

G.

EG

(8)

EG

Further, multiplying the first equation of (3) by a/ EG and the second equation of (4) by bIG and subtracting, we obtain E,

(In

EG

b'- - a2),E VEG

(9)

From (8) and (9) it follows that the chosen coordinate net is isothermal. We set

E=G=

1 62-a2.

By means of (8) and (9) we find the Gaussian curvature

K= V2ln 'b'--a2

(10)

Applying (6), we can find the second equation on a and b. To do this, by means of (3)

and (4) find v, and v2. Set a = r cos 0, b = r sin 0. From (3) we have

v, - (aE), = cot0 b EG

E (in EG

r) - 0,E +cot9E, EG

EG

Replacing E, by (9), we find

v, _ - I 1

2

in

tang- 1} E tan 0+ 1)

EG

(11)

THE GEOMETRY OF SUBMANIFOLDS

184

In an analogous way, by means of the second equation of (3) and (8), we find

_1 1/2

2

In

tang-1 tan 0 + I

G (12)

u

EG

Then (6), (11) and (12) imply

2V2 In

a--+b= -tab.

(13)

It is easy to find 1

2 V2 In

a+b

a-b

_

2jgrad 012 sin 29 V20 cost 29 cos 29 +

Using the relation just obtained, we can consider equations (10) and (13) as a system: V20

+ 21grad 012 sin 20

cos 0 +

cost 20

V2 In r - tan 20 V29 -

21grad 012

- _2ab, = 2K.

cos'- 20

We exclude V20 from the first equation and substitute it into the second. We get V2 In r + 2ab sin 20 = 2K + 21grad 012.

Note that r2 = a2 + b2 = Ko - K. Also

2absin20
V2 In

Ko - K + Ko - 3K > 21grad 012 > 0,

which completes the proof of Theorem I for the points of a 36 b. Suppose now that in some domain D C F2 the normal curvature ellipse is a circle, i.e. a == b. Define some orthogonal net on F2. From (3) we get

vi = [E,. + E(In a)J/ EG,

(14)

v2 =

Substituting these expressions into the Voss-Ricci equation (6), we get

K-2V2In a=a2. Since 2a2 = Ko - K, from the latter equality we find the necessary condition on dc2 to be immersible into F3 as a minimal surface with a circle as a normal curvature indicatrix. Namely V2 In

' Ko-K+ 20-2K=0.

(15)

MINIMAL SUBMANIFOLDS

1115

This condition is sufficient for the immersibility of cis'- into R4. Indeed, choose in an abstract two-dimensional manifold. endowed with the metric ds2, any orthogonal coordinate system and define two quadratic forms:

Ill = a(E(lu= - Gds-2).

112 = 2ic'cluch'.

where 2,t2 = Ku - K. Define I'I and 1l_ by(14). Then the Gauss-Codazzi equations are satisfied automatically, while the Voss-Ricci equation reduces to (15) which is satisfied due to conditions on the metric ds--. Thus, 111, Jl', ill and rm satisfy the fundamental equations of immersion and the surface F2 is defined uniquely up to rigid motion. In the particular case when the sphere S2 is endowed with the metric of curvature

K = 1/3, then for K) = I equation (15) is satisfied. The corresponding minimal immersion of S2 into S4 is known as a i iroaese surface. We alread) considered it in Section 2, Almgren [29] proved that a homeomorphic to the sphere S2 minimal surface in

spherical space S3 is the only great sphere. The condition on the surface to be homeomorphic to S2 is essential because Lawson has constructed the immersions of closed surfaces of various topological types into Ss as a minimal surface. Now we state one generalization of the Alnigren theorem

Theorem 2 A nrinhnul hoaweamorphic to a sphere sur/arc' F2 in spherical ,pace S4 either has it ,run-trivial I intner inrcrriant r' # 0 or is the great two-dinie,isio al sphere in S4.

Both cases are realizable because a Veronese surface has a non-trivial Whitney invariant.

Proof' Since F2 is homeomorphic to a sphere we can introduce on F2 the iso-

thermal coordinates ul. u: with respect to which the metric has the form ds= = A(diw + duz). Suppose that the Whitney invariant v = 0. Then there exist on F2 regular unit normal vector fields nl and a_. Since F2 is minimal, the second fundamental forms with respect to these fields are of the form 111 = A(ailui + 2,3dut (hr_ - aclu4 ). 112

= A(rih +2hthi1 d-

ed;).

The functions a, 13, r and b are geometrical values related to the normal curvature ellipse parameters. Note that aril + cn2 is the normal curvature vector of it, curve and dnt + bn2 is the normal curvature vector of the curve making the angle of r/4 with the it, -curve. Therefore a, ,3,e and b are regular bounded functions on F2. The

extrinsic curvature of F2 in Sa is K. = -a' - .12 - h2 - c-=. If we introduce the Cartesian coordinates x. y in normal space then we can represent the normal curvature ellipse point coordinates x, J1 as

x= itcosp+.3siny:. Y = e cosy' + h sin y:.

THE GEOMETRY OF SUBMANIFOLDS

186

If we set

w1=(a-b-i(Q+c))A, w2=(a+b-i(f3-c))A, then we can express the Codazzi equation for F2 as 8w1

1

8z =2'v1(u2 -iv,), a2=-1W2(v2-iv1),

where v, are the torsion coefficients of the n1,n2-basis, i.e. v; = (n1n2i,,). Since nj are regular, the functions vi are bounded and regular. Moreover, vi = (n1,nzs) 'fig tends to zero as 1/Iz12 while z oo. Therefore wk can be found in the form (see [181, p. 159):

vt = 4'1 (z) e(),

W2 = 4'2(z) a-.A:),

where 4'k are analytic functions and W(z)

I j v2(() - iv' (9 2a E

C-z

dE(.

Here E is the whole complex plane. oc, wk As A - 0 while z 0. By the theorem from [18] (p. 156, Theorem 3.11) we have wk = 0. Thus

a=/3=b=c=0, i.e. F2 is a totally geodesic surface. The equation v1 2 - v2,,, holds. By virtue of this equation and the fact that F2 is simply connected, we can construct two vector fields parallel in the normal bundle (see Section 6 Chapter 4). Since F2 is a totally geodesic surface, these vector fields are constant on F2. Thus F2 is a great sphere.

7.15 Surfaces of Constant Mean Curvature in E3

In this section we consider the surface in three-dimensional Euclidean space of constant non-zero mean curvature. The simplest surface of.that kind is a sphere. There exist surfaces of revolution which are not spheres but they are non-closed. Libman proved that the sphere is the only closed convex surface of constant mean

curvature. A more general theorem has been proved by Hopf [30]. It avoids the convexity condition.

Theorem (Hopf) If F2 is a regular class C3 closed homeomorphic to a sphere surface in E3 of constant mean curvature then F2 is a sphere.

MINIMAL SUBMANIFOLDS

187

Introduce about some point of the surface the isothermal coordinate system u, v with respect to which the metric has the form dc2 = E(du2 + dv2). Then the mean curvature H and the Gaussian curvature K can be expressed as

2EH=L+N, E2K = LN

- M2,

(1) (2)

where L, M and N are the coefficients of the second fundamental form. With respect to isothermal coordinates the lines of curvature satisfy the equation

M(du2 - dv2) - (L - N) dude = 0.

(3)

Consider the Codazzi equations:

2E(L, - Mu) = EV(L + N),

(4)

2E(M, - Nu) _ -E,,(L + N). Differentiating (1) in u, we get

2E,,H+2EH,,=L,,+N,,. Using the second equation of (4), find

L. - Nu + 2M, = 2EH,,.

(5)

Differentiating (1) in v and using the first equation of (4), we get the following equation:

(L - N), - 2M,, = -2EH,..

(6)

Introduce the complex variable w = u + iv and the complex function 0 _ (L - N) -iM of this variable. Then

I4I2 = 4(L - N)2 + M2 = 4(L2 +2LN+N2) - (LN- M2) 2

=E2(H`2-K)=

(k1-k2)2,

where k, are the principal curvatures of the surface. Therefore, 214'1 = E Ikl - k21. It

follows that if at some point 4' = 0 then this point is an umbilic one: k1 = k2. Consider the real and the imaginary parts of 4'dw2. We have 4' dw2 = { 1 (L - N) - iM } (du + i dv)2

= [(L_N)(du2_dv2)+2Maudv] - i[M(du2 - dv2) + (L - N) dudv]. Comparing with (3), we find that the condition that the imaginary part of 0 dw2 is zero coincides with the curvature lines equation. Thus, the equation of lines of curvature can be written as Im (4' dw2) = 0.

THE GEOMETRY OF SUBMANIFOLDS

188

On the other hand, if the imaginary part of a complex number is zero then its argument is divisible by 7r. Note that the argument of the product of two complex numbers is equal to the sum of the multiplicand arguments: arg (4' dw 2) = arg 4) + 2argdw.

Thus if dw corresponds to the principal direction then

arg4'+2argdw=ka,

(7)

where k is an integer. Rewrite (5) and (6), which were obtained from the Codazzi equations, in terms of a complex function (D. We have

4,; _ 4D+i

r

[2I

11

=

[(2EH-2M,.)-iM,, +I E(H -iH,) = EH,,..

L

Thus the complex equation holds: (8)

0. From (8) it follows that If the surface has a constant mean curvature H then 4' is an analytic function of w. If 4 # 0 then its zeros are isolated. If 4' is identically zero then all points of the surface are umbilic, i.e. the surface is either a sphere or a plane. By the hypothesis of the Theorem, the second possibility is excluded. Now suppose that the surface has only isolated umbilic points. Let 0 be one of these points. Consider the vector field of principal directions on the surface. The umbilic points are singular for this field. Let ry be a closed positively oriented curve in the surface surrounding 0. If u, v are local coordinates about 0 then 7 is the image of some curve y in the u, v plane. Also, we can consider directions r, and r2 on this plane corresponding to principal directions on the surface. Let y be parametrized by t E [0, 1]. Assume that ri = ri(t). Let us analyze the behavior of r, while y moves in the positive direction. There are three possible cases: ( I ) TI(I) = 'n (0),

(2) r, (l) _ -r, (0), (3) r1(l) = ±r2(0). Assume that r, is the principal direction field such that the corresponding principal curvature k, satisfies k, > k2. Note that the field r, never "mixes" with 72, i.e. it never occurs that after moving along y the field r, continuously varies into the field r2. Indeed, any two points P, and P2 on the surface can be joined to each other with the curve not passing through the umbilic points. Along such a curve one always has k1 > k2. Therefore the direction 7-1 is defined (up to the opposite direction). If after moving along y the field r, passed into ± then at some point of this curve there

MINIMAL SUBMANIFOLDS

189

FIGURE 26

FIGURE 27

would be a point such that k, = k2 (Figure 26), which contradicts the choice of y. Thus (3) does not hold. Figures 26 and 27 show examples of curvature line nets with singularities at the origin. The increment after the motion along y of the angle which makes n and the positive direction of the u-coordinate curve is divisible by 7r. Denote

by A the increments of the complex number argument after passing around the singular point. We say that 0 is an isolated umbilic point of index j if i(argdw) =rr. From (7) it follows that A(arg 4) + 20(argdw) = 0.

loft

TI[E 6I0\1I iR) 01 Si'B\IANII-01 DS

I IGL KF 21i

Hence the singular point index I function 4, icro at 0 Then

Let it the be intill iplicity of analytic

2rn 2T

Thus. the index of an} umhihc point in the surface of constant mean curvature is negatite. B% the Potneare theorem. the sum of indices of all actor field singular pointy on a closed surface does not depend on it specially chosen field but is defined h,, the genus of the surli ce The reader can find the proof of this statement in [31] (pp. 5(1 64). if g is the ,genus of the surl:ice and !h are the indices of'singular points

then rt Ix = 4t I

1 f3} the hypothesis. F2 is homeomorphic to a sphere, that is g = W. Hence T ."A - 4 0. But from hat ne have just proved it follows that ra ix : U. This contradiction completes the proof In series of papers [32]. published in 1956 1958 Ale.andro% proved, based on the minimum principal for elliptic equations. that a closed h%persurface of constant mean cursature and of am topological structure but without self-uuersections in itdimensional Euclidean space is a sphere. For a long time the problem. posed b% Hopf. was open: does there exist the closed surlitce of it positive genus in Et (with self -intersections neecssurih) of constant mean cur%ature? The solution of the Hopf problem has been obtained by Wente [34). He constructed a closed surface in Ei of constant mean curvature Other examples are mentioned in Walter's paper [35] (see Figure 28 from that paper). In [33] the theorem was proved stating that there exist infinitely many unmersions of the form of genertlired surfaces of resolution al' the type ViA I - VIA t into E=t which are difieomorphie to S''t i and have prescribed mean curvature. Let us gne in general the method of construction of-a three-dimensional h%persurfuce in El %%ith

MINIMAL SUBMANIFOLDS

191

H = Ho = const. We take the hypersurface from the example of Section 13 Chapter 3, where we indicated the principal curvatures of this hypersurface. So the condition H = Ho gives us the ordinary differential equation

xyit -yx"=3Ho+-y x r'

v'

If we take as the parameter t the length of the arc of the curve {x(t), y(t)} and if a is

the angle between the tangent to this curve and the x-axis, the equation can be transformed to the following form:

da=3Ho+cosa-sin-. ds

v

(9)

x

where

x= J cosads+xo, y=J sinads+yo There exists the following theorem: Theorem For an arbitrary given positive value Ho and a positive integer n, there exists an analytic solution curve of (9) which starts at the x-axis and ends at the y-axis and whose total change of direction is

Da=2n,r+2. The solution curve which intersects the x-axis and y-axis with angle equal to i gives the regular and closed hypersurface with H = cont.

8 Grassmann Image of a Submanifold 8.1 Spherical Image Analogs

In studying two-dimensional surfaces in three-dimensional Euclidean space, both local and global aspects, it is very useful to invoke the concepts of spherical image and spherical mapping. By Gauss's theorem the limit of the ratio of the spherical

image area An, of the domain G in F2 to the area AS of this domain while G contracts in F2 to x is equal to the modulus of the Gaussian curvature at x: lim OS = IK(x)I. An important role in the theory of minimal surfaces is played by the conformality of a spherical image. The first fundamental form of the spherical image dn2 is proportional to the first fundamental form ds2 of the surface: dn2 = IKI ds2. There are

many important papers where the authors either use various properties of the spherical image or formulate theorems in terms of the spherical image. For instance, the metric of unit sphere dn2 has been used by Efimov as an ancillary metric on the surface of negative curvature.

We would like to have a similar sufficiently simple geometrical analog of a spherical mapping for a submanifold. Let F" be a regular submanifold in

E"+*. Of course, it is possible to take some unit

normal vector n(x) at each point x in F" and make it begin at a fixed point 0 in E"+k. While x varies in F", the endpoint of n(x) describes some submanifold in the S"+k-I. A construction like this holds for hypersurfaces, where, after fixing sphere the orientation in F", it is possible to select n(x) uniquely. However, if the codimension of a submanifold k > 2 then the choice of n(x) is ambiguous. We would like to have a geometrical object with points which would be 193

194

THE GEOMETRY OF SUBMANIFOLDS

H. GRASSMAN

related invariantly to the points of a given submanifold just like in the case of a spherical image.

To construct it we use not the unit vector field n(x) but the normal spaces N, themselves. The dimension of NT is k. Since N, depends on x, then it is possible to say that N, is a space function on P. although this notion is not commonly used.

Draw k-dimensional space N through the fixed point 0 in E"+k such that N is parallel to N,. N belongs to the set of all k-dimensional subspaces which pass through the fixed point 0 in Euclidean space E"+k. The set above of k-dimensional planes is called a Grassmann manifold. In Klein's book [2] on the history of mathematics in the nineteenth century there are many pages devoted to Grassmann. He did not attend lectures in mathematics but, through his original and distinctive education, he made an essential contribution to the development of mathematical science. His fundamental geometrical work, "The study of extension as a new branch of mathematics", was published twice in his lifetime - in 1844 and in 1862. He was a school teacher in a provincial town, but his ideas featured in University courses and, as Klein wrote, he "exerted an original and strange influence still felt today."

In a preface to the second edition of "The study of extension ..." Grassmann names the mathematicians who influenced him: MBbius with his barycentric calculus

and Belavitis, who in 1835 gave the geometrical algebra of intercepts in complete

GRASSMANN IMAGE OF A SUBMANIFOLD

195

generality. Also, Grassmann was stimulated by one of Gauss's remarks in the preface to a collection of his works (1831). Only the geometrical interpretation of complex numbers leads to the geometrical calculus of intercepts in a plane. Grassmann introduced the concept of the geometrical product of intercepts, invented complex numbers of higher order which have multiplication which may be not commutative and proposed no fewer than 16 types of complex multiplication, one of which - the combinatorial product - he used in the theory of linear extension to the notion of determinant. He also introduced new geometrical objects - multivectors. After the publication of his book in 1845, a paper by Sen-Vennan on the geometrical multiplication of intercepts was published. This paper repeated Grassmann's ideas and he sent two copies of his book to Cauchy with the request that one copy be sent to Sen-Vennan. Surprisingly, O. Cauchy published a paper in Comptes Rendus (1853) with methods based on symbolic quantities. One of these methods was the same, in Grassmann's opinion, as his method published in the book he had sent to Cauchy. The situation was delicate, but although Grassmann did not want to accuse the famous scientist of plagiarism, he sent his claim of priority to the Paris Academy. The commission did not come to any conclusion, but Cauchy published nothing more on this subject. It seems that Grassmann never considered the manifold named after him "in the large" but he studied the objects which are the points of this manifold. It is hard to say now who first gave the name of Grassmann to that manifold. Severi cited the papers of Grassmann in 1909 whereas Shubert did not mention them in 1886. So, the k-dimensional plane in Euclidean space E"+t which passes through the fixed point in En+k is called the point in a Grassmann manifold. The word "plane" has been replaced by another word - "point". One of the mathematical objects the plane - obtained another name - the point. And nothing more. But after this there arose some associations, some questions. For instance, we can also ask: what

"points" are close to each other? How can we measure the distance between points? If we define a neighborhood of these points in such a way that they would satisfy corresponding axioms then we obtain a manifold. At first it seems too abstract and

unusual. Then a question arises: how we can imagine it or model it (in modern language)? Its topological structure is particularly confused and complicated. It is difficult to ascertain what is more abstract - the notion of number, straight line and other objects of so-called elementary geometry or such a modern mathematical notion as the Grassmann manifold. However, the objects of elementary geometry are habitual and have been examined for thousands of years. It seems that the notion of number is a priori habitual to the intellect and nothing to speak about.

Therefore it is necessary to make new mathematical concepts habitual and use them if they are fruitful. From this arises the question: where do we meet more abstraction - in the notion of number and the method of number representation or in the generation of modern mathematical constructions? Endowed with the system of neighborhoods, the set of all k-dimensional planes in the (n + k)-dimensional Euclidean space E"+k that pass through the fixed point 0

1%

THE 41:OM1ETRS OF Si. KMAN8TOLDS

becomes a topological space. Moreover, this topological space forms a manifold. i.e.

for each point there exists a neighborhood which is homeomorphic to the ball in some Euclidean space E' of dimension 1. for which I can be expressed via ri and k. The standard notation for this manifold is GA . &. Denote a point in it by P. We shall study the Grassmann manifold in more detail in the next section. Now turn back to the correspondence x A' as at the beginning of this section. Now we can consider it as a correspondence x -- P. Denote it by 1P. Since the image of 'If is located in a Grassmann manifold. 'I' is naturally called a Grassmwnn nwppnrg (in analogy to spherical) and the image'P(F") is called a Grassinuwm image. Other names such as generalized spherical image or tangential image are used. It seems that this mapping was introduced in the nineteenth century. Such a mapping was widely used in topology, for instance in Pontryagin's papers on characteristic classes [3]. However, from the geometrical point of view the properties of the Grassmann image received scant attention and remain poorly studied still.

S.2 Grassmann Image of a Two-Dimensional Surfaces F2 in E4

Leaving aside for a while the general case, consider the Grassmann image of a two-dimensional surface in E4. Let x' be Cartesian coordinates in El. The twodimensional plane N can be represented by a pair

i1) of mutually orthogonal unit

vectors in this plane. Consider the ordinary bivector p = [ i] with Plucker coordinates p

it/

tit ,

where a and q' are components of 1; and q. There are six Plucker coordinates. P1 '-. p23, p22a, p34. It is easy to check that p'' stay unchanged under a rotation of p13, p'''. the basis k, , in N. Values p'J are related to each other by two relations. First, taking

into account the bivector scalar product formula, we have P"

= Q41111410

1r,

(1)

Also, p12p34

+P I i pa_ + p14p21= o.

Consider P":

We can write the left-hand side of (2) as £ I (q p34 + rt3 p i_ + 1;4p= 3) -Tai (42 p 34 + ' pr' + 4 p23) .

(2)

GRASSMANN IMAGE OF A SUBMANIFOLD

197

The coefficient of 1 can be represented as

Ir

q4

73

2 3 f4 = 0.

it

q3

'T

Analogously, the coefficient of q' is zero, too. Equality (2) can be interpreted geometrically. Consider a bivector q defined by the plane complementary to p. If T, v are mutually orthogonal and orthogonal to N unit vectors such that , i?, T, v form a positively oriented basis in E4 then q = [Tv]. The scalar product of p and q is zero. Indeed, (pq) = ([frl][TV]) = (CT)(r!v)

-

0.

On the other hand, the components q'' of the bivector q can be represented in terms of p'". Namely q'2 = P34,

q13 = _p24

q14 = p23,

q24 = _p1a,

q:3 = pI4,

q34

= Ptz.

(3)

To prove (3), form the matrix of , rl, T, v components: r>z

r

r ill

T1

T2

T3

T4

vl

vz

v3

v4

A=

As A is an orthogonal matrix, the inverse matrix coincides with its transpose: A' = A'. Use the formula for minors of an inverse matrix (see [4], p. 31). Let

/

BIk

1 be a minor of a matrix B consisting of elements of the i-th and j-th rows, k-

th and 1-filth columns. Suppose that il, i2, i3, i4 form a permutation of 1,2,3,4 such that i, < fz, i3 < i4. The same conditions are imposed on k,, k2, k3, k4. Taking into account that the determinant of A is 1, we can write A

i,

iz

k,

k,

_ (-1} i, «i,tk, i k; A

k3 i3

k4 i4

Set, for instance, i, = 1, iz = 2, k, = 1, kz = 2. Then T3

F4

v3

y4

i.e. p12 = q34. Set i, = 1, iz = 3, k, = 1, kz = 2. Then i3 = 2, i4 = 4, k3 = 3, k4 = 4. So we have n3 C3

I-

(_I}4f3I Vr21

74

0 I.

THI (JEOMI I R) 01 St SA14NII'OLDS

198

ti(il RE 2')

Hence. pl' = -e. In an analogous way we get other q". We can now represent the condition (pq) = 0 as

(pNl= E

pugs=)(p12p34+pl pit+p'4p2z)=0.

fr

which coincides with (2).

Consider the six-dimensional space E. Then the endpoint of the vector p with components (p12.....p'4) attached to the origin determines the point in El. Various points in E4 obtained in that way and the coordinates of which satisfy (1) and (2) are located in some algebraic four-dimensional submanifold in E. That submanifold is the Grassmann manifold 624 immersed into E'. In other words. this is its model. That submanifold we also denote by G24Let N be a plane parallel to the normal plane N, of F2. If it,, it, form the basis of

normals then the corresponding bivector p is of the form p = [n1ai2). Since u, = F11 (ill u2) are vector functions of it', u2, p = p(rri. it'). Therefore. the image of a Grassmann mapping * is, in general, some two-dimensional surface r2. That surface r2 is located in the submanifold G_,i (see Figure 29). As an example. consider the Grusrnuriui image of the Clifford trues xi = Cos ri.

X2 = sin o,

N' = cos d.

x4 = sin d.

Unit normals of that surface are

it, = (cosri.sin't.0,0), it, = (0,0,cos,d.sin:3). Hence, the components of the bivector p are of the form p12=0.

p 13 = cosre cos,d, pit = cosh rind,

p21 = sin rt cos d

p=; = sin ri sin .3.

p14 = 0.

GRASSMANN IMAGE OF A SUBMANIFOLD

199

As p12 = p34 = 0, in this case r2 C It is easy to state that r2 is a Clifford torus as well. Study some geometric properties of a Grassmann image of a general surface. Since G2,4 is represented as a submanifold in E6, the ambient space induced Riemannian metric on that submanifold is E4.

do,- = E(dp'1)2.

kj

As t- is a subset in G2,4, the metric arises on t'- as well. Denote it by dp2. State the relation that arises under Grassmann mapping between the metric dp2 and the external geometry of F2. We have dp = [dn1 n2] + [n1 dn2].

This formula has a symbolic sense. It arises from the PlOcker coordinates expression and the way of differentiating determinants. Apply the Weingarten decomposition to differentials of normal vectors dnr, = r'_ L" gll rl +

no du',

where L. are coefficients of the `second fundamental forms of F2, gil are components of inverse metric tensor of P. Using the decomposition rule of a bivector into a sum of bivectors, we get

-L1g''[rln2]du'.

[dnln2] = Analogously we proceed with [n1 dn2j. Thus

dp = - (-L gi'[rl n2] + L2 gil [nl r,J) du'.

To Find the square of dp use the formula for the scalar product of bivectors:

([ab][cdj) = (ac)(bd) - (ad)(be). Note that by virtue of the orthogonality of it, and n2 the relation ([rln2J[nlrk]) = 0 holds. Now write

(Ljg1l[rln2]du,)' = L' Lrlq guIgq",du'duP([rln2][rmn2J)

= L1 L' g" gtm'(rlrn,)du'duP = In the same way 2

(Lfgu'[nIrl]du') = L LMg'ldu'duP.

Thus, we can represent the metric of the Grassmann image r2 as dtt' du'. dp2 = (L;,t Ljl + L2ik L2)gkl j/

THE (d OMLI ItY or S11INIANIFOLUS

2(H)

So. the abstract thing - the Grassmann image I'2 - gives quite real features. On 172 we are able to crawl, to measure distances. angles and so on. Also. we obtain some information on the external geometry of F2. We related the internal geometry of a

Grassmann image to the geometry of F2. We can express the components of clp2 = G ahi char. where G'a - _

(LE

LIra

A

) gAa 2 + L,A L'11 ,

with respect to a special system of coordinates in terms of parameters of the normal curvature indicatrix of F2. Choose uI. ar in F2 and normals in a special way which was described in Section I of Chapter 6. Then at some fixed point r E F2.

Ljj=n+a.

L{2=0.

L;, =n -ca.

L';,=d

L2jI =rii.

Substitution into G,, produces [(LiA)2+(L24)2] _ (c:+(a)2+112+h'.

GPI = A_I

Go =

-I

(L'14 LA +LTAL_A) = 2;3h.

A), =tat-u)-+r12+h-.

A-I

Write the metric of r2 at W (1 ):

dp' = [(u + ca)' + 32 + h' (du' )2+4dhdaa' clan- + (ct

- a)2 +J2 + h2 (daa-')2.

This form of metric has been stated in [7]. Consider the volume element of the Grassmann image dSt =

G12clra' cltr'.

Using the Cartan formula from Section I of Chapter 6, we have ' +' j2 + ca' + h- + ?nra)(ct

GI

((1' + +i=

-2

i-

+ cat + h2 - _ciu) - 432/'2

- ca- - h' )' +4((k21,2 + ri2ra2 )

ti'- +4(n2h' + area' ).

where K is the Gaussian curvature of F2. With respect to the chosen system of coordinates, the volume element of F2 at r,I is dS = dul daa2. So, we have the following generalization of Gauss's theorem: the ratio of the Grassmann image volume element to the volume element of the original is of the form

At dS

K2 +4(n2h' +;i:1(t

()4

GRASSMANN IMAGE OF A SUBMAN[FOLD

201

While this equality has been stated with respect to a special system of coordinates. due to the invariant meaning of both sides of (4). it stays valid with respect to an arbitrary system of coordinates.

From (4) it follows that " > SK I. If F2 is minimal then n = 3 = 0 and dp2 = (K I dS2. Therefore, in this case the Grassmann mapping T is conformal. 8.3 Curvature Tensor of Grassmann Manifold G2 4

Denote by p12. p1z. p1;. p''. p-;, p3 the Cartesian coordinates in E". Standard immersion of G24 into E' is defined with two equations:

F2

(1)

1,

F1

=P12P - -P13P-

+P 14r = 0.

(2)

Gradients of F1 and F2 determine two normals to the submanifold: I

grad Fl = {p12, p1' p14, p' P. p34} = p.

2 grad F, = {p4.

p_4,

P2:4 p14, -PI;, P' } = q.

As (pq) = 0, these normals are mutually orthogonal and unit because P` = q-' = 1. Equation (1) means that 62 4 is located in a five-dimensional sphere Ss: also q is the normal to G34 C S` Let f 1, t2, r , y4 be some curvilinear coordinates about a point in G2 .. A position

vector of the considered submanifold can be represented as a vector function p = p(p1.... , v ). Let drr= = a ! 40 dr i be a metric of G_ 4 induced from ambient ) space Eb with respect to these coordinates. Then a,, . We would like to -0. find the Riemannian tensor components R #,0. To do this. use the Gauss equations

which relate the Riemannian tensor to the second fundamental forms. Let S2,", i dig" dp;, o = 1.2 be the second fundamental forms of G.24 c E° with respect to

normals p and q. Setting a = I corresponds to p. o = 2 corresponds to q. Then sill,

a=6

kV I

P}

0P n { = f7f)r

11

q)

.

Since p and q are normals, these coefficients can be represented as

VO)

SZ,';i=-k q{J.

THE GEOMETRY OF SUBMANIFOLDS

202

In terms of differentials dp and dq, the second fundamental forms of G2,4 c E6 can be represented as

III = -(dp)2,

112

= -(dpdq)

(3)

Up to a sign the form !!I coincides with the submanifold metric. Consider the Gauss equation of immersion of G2,4 into E6:

L(na, a=1

?A - W., TO

a,,, a,m - a,,s ah +

(Lp Lq ) (Lp Lq

ar Or

( ap _q ( Op Oq ).

OYI a?

Or P) Way" Or

Let X = {X" } and Y = { Y" } be mutually orthogonal unit vectors in the tangent space of G2.4. Denote by a a 2-plane spanned on X and Y. Then the curvature of G2.4

with respect to this plane is x(a) = R"chn X° Y.X, Ye

= I + (VxpVxq)(VypDyq) - (OrpVyq)(Vypvxq), where Vx means the covariant derivative in E6 with respect to X. In the last term of the above we have (Vxp Vyq) = (Vyp Vxq). Since G2.4 is in S5 and p is normal to

this sphere, Vxp = -X, Vyp = - Y. Therefore, the curvature of G2,4 can be represented as

K(a) = I + (XVxq)(YVyq) - (XD),q)(YDxq)

(5)

Vector Vxp is unity, i.e. (Vxp)2 = 1. Taking into account that Vxq is formed with the same components as Vxp but taken in a different order, we have (Vxq)2 = 1. In the same way, (Vyq)2 = 1. Vector q is normal to G2.4. Hence (Vxpq) = 0. From this we find (pVxq) = 0. Also note that q is unity. Hence Vxq has decomposition via vectors of tangent space of G2.4. Let X, Y, Z and W be mutually orthogonal unit vectors from that tangent space. Consider decompositions

Vxq =a, X+a2Y+ a3Z+a4W, Vyq = b1X+b2Y+b3Z+b4W.

As Vxq and Vyq are unity, E; a = rib? = 1. As (XVyq) = (YVxq), a2 = b1. So K(a) = 1 + a1 b2 - a;.

Find k(a) variation boundaries. Consider in the plane a pair of vectors with components (a,,a2), (b1,b2). Their lengths are not greater than I. Their vector product modulus Iaib2 - ail is also not greater than 1. Therefore, 0 5 K(a) < 2. Show that

GRASSMANN IMAGE OF A SUBMANIFOLD

203

boundary values are attainable. Take the pair of three-dimensional vectors is = {p'} and v = {v'} of components =P12-p34,

µl =P12+p34, µ2 = p13 +p42,

VI

µ3 = p14 +p23,

V3 = p14 - p23.

V2 = p13 - p4-,

By virtue of (1) and (2), each of them is unity, i.e p22 = 1 and v2 = 1. We can rewrite the forms dp2 and (dpdq) in terms of these vectors: 2dp2 = dµ2 + dv2,

2(dpdq) = dµ2 - dv2.

The normal curvature k of G2,4 with respect to q is

k-_

(dpdq)

- dv2 - dµ2

(dp)2

dv2 + dµ'

From this it follows that the normal curvatures of hypersurface G2.4 C S5 vary in [-1,11 with attainable boundaries. Ask changes its sign, the hypersurface is a saddle. With respect to tangent direction dv = 0 we have k = - I ; with respect to tangent direction dp = 0 we have k = 1. The components of p E G2,4 can be represented via µ' and V. Namely p = (pl' p13 p14 p23 p24 p34)

=2(µl+v1,µ2+v-, µ' t +v -1, 'U3 -v3,

I) -µ +V-1 µ - v. 2

1

Thus, G2.4 is isometric to the direct product of two two-dimensional spheres Si X S

of the same radius I/f. Keeping in mind this correspondence, we shall represent the points of G2.4 as µ x v. Let F, = S1 x vo be a submanifold of p-endpoints while µ

moves along the sphere µ2 = 1 and v is fixed v = vo. Let F2 = µo x SZ be a submanifold corresponding to v-variation while µ is fixed: p = M. Denote by T, and T2 the tangent planes to the respective submanifolds. At every point p = PO x there exist both of these planes. Each vector of the T; plane is principal; also, the normal curvature with respect to directions from T, is - I while from T2 it is 1. Let X and Y be mutually orthogonal in T1. Then Vxq = X,

V}'q = Y.

From (5) we find that k(a) = 2. Note that T, and T2 are perpendicular to each other. Indeed, if X E T, and Y E T2 then X = (dµ 1, dµ2, dµ3, dµ3, -4112,4111),

Y = (dv,, dv2, dv3, -dv3, dv2, -dv, ).

Their scalar product (XY) = 0. Moreover

Vxq=X, Vy=-Y. Therefore, from (5) we see that for the X, Y-plane the curvature of the Grassmann manifold is k(a) = 0.

N4

THh UEOMC1 RY Of S1'HM \Ni1OLD.S

8.4 Curvature of G2 4 with Respect to the Tangent Planes of the Grassmann Image of a Surface

Let F2 be a regular surface in E4. r2 its Grassmann image and p a point in r2 In the

tangent space of G2; consider the plane a tangent to I'= at p. State the relation between the curvature K(cr) and the geometry of F'. Represent I'= in parametric form:

r =}'''(u.r), r+= 1.2.3.4. where it. r were taken out from F2 by Grassmann mapping. Then 11 are tangent to 1`2 but not orthogonal to each other. in general. Use the formula for the G24 curvature tensor in the form equivalent to (4) of the previous section:

+

Jp

W)

CeJt r?,

02P

p

)-(

allrJt,W

atd,,. carrW

"p q

l1r 1c)r'

The curvature A(rr) can be found by

I+ (p

nu W){p ,t

q}-(p

oil

W)`

(1)

Pj,P - l pupr)2.

Consider (dp q) on 1' Find the tangent vectors of t=. They are derivatives of the vector-function p = p(u. r): pff = -G R gA1 [r, F_] -

gc1 [f1

where u1 = it. tr' = 1.

To simplify notations, set L;1 = L' g"1. Bisectors [rr fir] are located in the tangent space of G2 To prove this. consider their scalar products with p = (41 and q = [ri r2] f f (here g is the determinant of the F2 metric tensor). Then

(p[r,£f(} _ ( in,)( ,) -

4.

(r1r,)(r2 ,) - (rlf r)(r_r??) _ 0.

Using Gauss and Weingarten decompositions with respect to F2. find the second derivatives of p: 01P 011,01i

1t

__

-t

1

-Ir11r',-.L. L2 1iti,r

+(_ 1)L'

!

011,

-Lrgr4 +L;U11!1 rI].

,I-1

[,r

f1

f

[E1r

GRASSMANN IMAGE OF A SUBMANIFOLD

205

On the right-hand side of the equality above, the first two terms are linear combinations of [{i r.1. So they are tangent to G2,4. The third is directed along p. The fourth is directed along q because of [r1r2] = vgq. Hence (d 2P q) _ -

ij

f L("11 L2121 dui duj

= 2 {(Li2Li1

vg

- L12Li,)(dul )2 + (Lil

t - Li1L 2) du' due

+ (142"21 - L2'1L42)(du2)2}.

With respect to the special system of coordinates in F2 mentioned above (so that (2) from Section 1 Chapter 6 holds), we obtain

(d2Pq) = -2{-b(a + a)(du' )2 - 2af du1 du2 + b(a - a)(du2)2}.

(2)

Observe that the determinant of the F2 metric tensor matrix (as stated in Section 2) is

P.Pv

-

K2 +4(a2b2 +/32a2)

(3)

du dv + (p,,,q) dv2 with Comparing the expression of (d'-p q) = (p,,,,q)du2 + (2), we find the scalar products (p,,,,,,q) with respect to the chosen system of coordinates:

2af,

(p,, q) = 2b(a + a),

-2b(a - a).

Therefore 4(a2b2 - (b2a2 +a2132))

(4)

Substituting (3) and (4) into (1), we get k(or). The result just obtained is presented as a theorem. Theorem The curvature K(c) of the Grassmann manifold G2,4 with respect to the plane a tangent to the Grassmann image r2 of a surface F2 C E4 has the form:

K(°)

_

K2 + 4a'-b2 K22 + 4(a2#2 + b2a2)

This theorem was proved in [7).

Find the characteristic equation on the principal normal curvature of 1'2 with respect to the normal q. Remember that for this case the principal curvatures are where dp varies in the plane tangent to extremal values of ratio Let p correspond to x. The metric of f'2 with respect to the chosen coordinates has the following form at p: r2.

dp2 = [(a + a)2 + Q2 + b2](dul )2 + 4Qbdul du2 + [(a - a)2 +Q2 + b2](du2)2.

THE GEOMFTRI (N SIIRNIANII.01 DS

_'tx+

Then, the characteristic equation can be written as

[(ri' +d=-a=-h2)'+4(rt`h-+a'rj')] +4). h[n2 +12 -cc2 -h=] +4(h=u'- - h=ri- - a=,3=) = 0. Using the Carton formula on the Gaussian curvature K of T' and the formula on K just obtained we have

!i=+4(nW +,12a2) Then, the characteristic equation can be written as A=+2A

_ct h l;

K2+4(n h=+d (1'-)

+A-I =0.

Let A, be the principal normal curvatures of 1''- with respect to y. Denote by R, and A the symmetric functions of A,. Then

_ Ai + A, 2

-2ahK

R, = A1A: = K - 1.

R' +4(ci-h' + 132cr-) '

(5)

If the principal normal curvatures A, are of the same sign. i.e. the form (d=pq) = -(dpdq) is of fixed sign. then k, = k - I > 0 If A, are of different signs. i.e. the form (dpq) is of alternating signs. then k < 1. By means of (5) we find

h

E= +4a:h: -2obK

Therefore 2ah K

f /it ' - 4N,.

2A. Thus, the Grassmann image r2 of F2 C E4 determines the ratio of Gauss torsion to Gauss curvature; 8.5 An Estimate for the Area of Closed Surfaces in E4 In this section we find an upper bound for the area of it closed surface in terms of the surface enclosing ball radius and the area of the Grassmann image. The following holds. Theorem Suppose that a c loseel on nrahle sw/ac , F2 C E; of class Ca u *h Gaussian curvature K of constant signs. Eider characteristic x and area S is con-

tubred in a ball of radius R. Let the Grasstrrann image area be equal to SZ. Then

S
.

(I)

GRASSMANN IMAGE OF A Sl1BMANIFOLD

2117

To prove the theorem. state the equation on p = r2/2 analogous to the Darboux equation for the case of a surface in P. The equation we need to obtain is (2)

V2-,p +t r

where g = det I1g 11. a22 is the Monge-Ampere operator, V2 is the Laplace-Beltrami operator. We have P+<, = (r'r'ra)

1-

P,,,,,, _ (r,,,"r) +g = 114p", + L,(rnt) +g,.

to be the second covariant derivatives of p. Then

Set p,,r

p,, = g,r + L,(rn,).

From this we rind

detjp,-g,,I =(pu p,

(P1ug22-2pIZg12+P22$11)+g

_ (L 1 L,, - 42

Taking into account the definitions of Vz, and 022 = (P I1 P 22 - Pi&)/gi

V:!:

V2P = P rig",

we obtain (2). Now estimate its right-hand side from the above. Select the normals it, and n2 in such a way that n, is orthogonal to r at each point of the surface. Then on the righthand side of (2) we obtain

[Li, L;2 - (Li2)-J/g. Set

(Li IL;, - (L-112))/g. Note that K = K(nl) + K(n2). Then K(n1) =

2+K(n1)-K(n2)2

Find the upper bound for IK(n1) - K(n2)1. Introduce the system of coordinates in F2 in such a way that at a fixed point P the components of the metric tensor would have the following values: g, I = g22 = 1, $12 = 0. Consider the following quantities;

A = K(nl) - K(n2). B = L11 LL, -2Li,Li1 +L;,L11.

If we rotate nl and n2 in the normal plane by angle w then the corresponding expressions for A and B with respect to rotated normals are

A=cos2ipA+sin2V B,

B=- sin2+A+cos2lpB.

THE GEOh111"TRY Or SL'BMANIrOLDS

20$

P + B- is a local invariant of imbedding F2 into E'i. Evidently. Hence. JK(rrl) - K(n2 )I < 742 -+B2. Consider A2 + B2 in more detail. Select normals 11, and introduce coordinates in the surface in a special way as in Section 1 Chapter 6. Then the coefficients of the second fundamental forms have a specific form at the fixed point. So

A-+ 8" _ ((LiiL;, - (LI,)2 - L11L22 +

(LiiLea -'-L' L,, +L;,L10-

[W

[n- +,12 - rr' - h-J- + 4(n* +,t-rr-) K2 + 4(n`h` + 32a2).

In Section 2 the expression for the ratio of the Grassmann image F'-' area element SS2 to the area element of F2 was found: (A2

rt

=

/-K= V

+ 4{r12h2 + 32a ).

Thus. dig/dS = 42 + B-. Now apply formula (7) Section 13 Chapter 3 to a closed surface F2:

f

C'22 p dS

=1

J

KI grad p12 1/.5.

F=

J

Observe that Igrad pI < 2p < R2. The integral of Cep over a closed surface is zero. We have 1

I KIgradp12dS+S= 1 r'

From this we find .S <

r` Moll)

!

2

I K(rui

r,

2

r

rtS - J

rK

I'+

K(n11 - K(n2)

Klgradp1= 2

r A grad p1' ,.,

(rill)2sS

IL.

r!S +

42 (rill) JS

c!S

QR2

2

If K < 0 then the first term on the right-hand side is negative. In this case

S< flR2+R- r ASS = If the surface is homeomorphic to a torus and K = 0 then S S (1R2/2. If the surface is homeomorphic to a sphere and K > 0 then the second term on the right-hand side is negative. Thus the area estimate is of the form S:5 (11+3zr)R2. Show that in the

GRASSMANN IMAGE OF A SUBMANIFOLU

2t$

case of the Clifford torus the inequality S:5 S2R2/2 turns into equality. Consider a position vector r(u, r) and basis tangent vectors of a Clifford torus:

fcosu r(u, v)

.

sin u

cosr

- sin if

r

sine

cosu 0

0

r

0 ,

sin 1,

cosr

0

Then ds= = du' + dr=. Then torus area S = 4rr'. It is easy to check that cosu

_

sin if

0 0

"-

0 0 cos r

sinr

are normals to this surface.

Therefore, the components p" of the position vector of the torus Grassmann image I'- are as follows

P12-0. p1' = cosu cosr, p'4 = cosu sin r. p23 = sin a cosr. p=4 = sin a sin r. p-4=0. The surface 1.2 is located in four-dimensional space and. moreover, in a threedimensional unit sphere. By direct evaluation we find the metric of r don = du= + dv-.

Therefore the area of the Grassmann image 0 = 47r. Since r2 = 2. the Clifford torus is contained in a sphere of radius vC1. So. for this case S = 32R2/2. 9.6 Three Types of Grassmann Image of Two-Dimensional Surfaces in E4

Let F" be a regular submanifold in (n + nn)-dimensional Euclidean space E"+",. We

shall suppose that the Grassmann image I'" of F" is a regular submanifold of dimension n in G,1,,+,,, and that the Jacobian of a Grassmann mapping 'P is non-zero. Consider the following problem:,from a given regular sulnnanituld f'" C G,,, +,,, find a szthnranifold F" in Euclidean space E+ 'm such that I" fts Grassmann image. In this and subsequent sections we consider the solution of this problem for if) = it = 2. We

state that the following theorem holds. Theorem Each component .v, of a position reclor of the required surface satisfies one differential equation of second order with the coefficients and, car a consequence, the type determined by the Grassmann image. If the curvature of G24 on the plane tangent to r2 satisfies k > I then the equation

be of elliptic type; if k < I the equation is of hyperbolic type; if K = 1 then the equation either degenerates or is of parabolic type.

THE GEOMETRY OF SUBMANIFOLDS

210

In accordance with the values of k, separate the set of surfaces F2 in E4 into three types: the surfaces with elliptic Grassmann image: -K > 1; with parabolic Grassmann image: K = I ; with hyperbolic Grassmann image: K < 1. The problem is reformulated as follows: let us be given a parametric representa-

tion of r2 in the Grassmann manifold as p = p(u1,u2). In each two-dimensional plane - the point of r2 - select two mutually orthogonal four-dimensional unit vectors l;l and 2. Since a point of 1,2 depends on u1 and u2, {, also depends on u1 and u2. Thus, we can suppose that there are defined two vector fields ,(u1i 112). Denote by

{;' a component of {, with respect to an orthonormal basis of E4. Denote by r = r(u1, u2) a position vector of the required surface F2. A component x,(u1, 142), i = 1, ... , 4 of this position vector must satisfy the system 4

=0, a= 1,2;

1: k=1

i= 1, 2,

(1)

which expresses the condition on the tangent vectors of the required surface to be orthogonal to the vector fields ,. Set A'# = p'f/p34. Consider some neighborhood of po E 1,2. Without loss of generality we may assume that p34 = 1 at po and p34 34 0 in the neighborhood. Rewrite (1), separating xcc3 and X4. Set a = 1, i = 1, 2. Then from (1) we have Xlu, SI + X2.,, dI + X3u,

CI3

4

+ X4u1 CI = 0,

X1u.Q2+X2uiC2+X3u,C2+2 =0. Multiply the first equation by C21, the second by C4 and then subtract the second from

the first. We get 24 X1ui P + Xzui P + .X3u,34 p =- 0. 14

From this exclude X3u, .In an analogous way we can find expressions for other derivatives of x3 and .X4. Thus we can write 2

X3u,

2

XiuoA'4,

x4u,

_-

Xiu,A3i

(2)

The equality conditions for mixed derivatives of x3 and x4 yield two equations which relate the first derivatives of x1 and x2: 2

2

1(x,u,A'_ -Xiu2A'u) = 0,

Denote by c and d the columns I \\\

24 X32

-x,".Auj) = 0.

(3)

I and ( X31) respectively. Three cases are

possible: (1) the equations in (3) are li/nearly dependent; (2) both minors [cu cu2 ], [du du2] are zero but the equations are linearly independent; (3) the general case one of the minors [c,,,, cu2], [d,,,, is non-zero. Here brackets mean a determinant of the matrix with derivatives of c and d as columns.

GRASSMANN IMAGE OF A SUBMANIFOLD

211

Consider case (3). Suppose, for instance, that [du du:[ 54 0. Then (3) can be solved with respect to derivatives of x1: x1u, = Ax2u, - Bx2,,2,

(4)

xIu2 = -Cx2u, + Dx2u,,

where coefficients A,. .. , D are determined by the given surface I'2. If we set t = [d,,,du:] then

A=-[du,cu,]lt, B=-[duicui]/t,

D=[du:cuJlt.

C=[d,,cu2]1t,

Thus, we reduce the problem to finding a solution of (4). Now state the dependence of the type of (4) on the structure of F2. To do this, clarify the geometrical sense of t

and of

-Bdui + (D - A) dul due + Cdu2. Let T3 be three-dimensional surface in E6 represented by equations p12 = p23 = p24 = 0. Let r = (p14,p13,p34) be a projection of the position vector p of r2 into T3 and v = (q14, q13, q34) a projection of G2.4 normal q into T3. Since the components of q can be expressed in terms of p'), we can write: v = (p23, -p24,p12). The equality

(pq) = 0 stated above obtains the form (rv) = 0. Denote by f2 a projection of I'2 into T3. Let r(u1iu2) be a position vector of f2. We have ),14

t = [du,du.J =

u

A31 u,

A31

14

I

(p34)2

14 34

),14

u,

(P 34) 4

u2

Pu2

P31

P31

Pla

P34

14

Pill

(P,), P"

p14

p31

p34

Pu,14

P31 u,

Pu134

14

31

34

I (p34)3

Pu2

Pu2

14 34

I Pu, p34_ P Pu,

1

34 31P34-P31Pu,

Pu,

P

Pua

P",

I

14 34 3

Pu,1P34-P31P34 p.14

2

14p34

Pu, P- P Pu1

(P )3IPu"

p14 +'

I

(rru,rut) (P34)3

(5) ,

Pu2

i.e. up to a numerical multiplier t coincides with a support function of fl. Further 14 rr,

[du'ell,

14I= u,

A31

A32

it,

u,

14 34 724 34 24 34 P - P Pu, J"' P -P Pu,

14 34 1

(-34)4 - 1-34 _ p31 pu pu2 p34 - -32pu4 '4

p14

P.1a

31

32

Pu,

P1a

P24

p Pu, p3I

p32

pea

p34 34

3a

31

Pu,

P

32

3a

u,

U/

p24 P32

}

(6)

THE GCOb1F1 R? 01 St IRMANH -OLDS

212

Using (pq) = 0, write 24

14 II

(7)

h21 =pl2p3;

p3i1

Write the first determinant in braces in expanded form. then substract and add P ' 1' Remember that r = (p14. p11 p.34). 1, = (p'; -p'; pt=). So, we get ptr,.

p1,,, 4

ptl24

I3

P,1,4

12

31

u I'

-1

1.t

1

p,f - p p' pu ptt. + Fit. pt,.

P,1

(8)

pu;

Transform the second term in braces as follows 14

, u P"

ur

p'

I

14 14

i.y

14

%4

la

p p' - pe. pn - pt,, P;" + pu p,t

p- _

v4Ma

11

+ p p- R, p,t, ' litt

11 +P'" P". A411)

p,,,

Write the third determinant in braces in expanded form. Now write (6) taking into account (7) (9):

rv

'z

(p14))+(

+

r; s

(prsp1'

p;`'

PK;

pr,;

p13

p''1

I p,;

(p_ zJ)

pt"l

I

p1e4

ii

1 t,

- p12 14 +n''p'i _pl4pl2 +p31pr'.I

The expression in parentheses is zero, because of (pq) = 0. Consider it vector product in T3:

;I.- p

Pt,

pt,,

P0.

34 P e,

[r,,, r,,,]

=

The third component of

Al -1 ,,

14

tt.

14

p;to.

34

pt

1z

14,,

P,,,

34

is of the form 14

'1 p,t

p IIM I4

P

u

+ 24 p

14

Pit.

11

pit

pit,

rat

11

ptt

1

Using the latter relation, we can write (r,,: u

)r1.i

+ [i'fr,,,rrr,])

,

(pu) where the brackets on the right mean the vector product in T3 and superscript 3 means its third component. It is possible to puss from the expression for (r,,, r,,,) in

GRASSMANN IMAGE OF A SUBMANIFOLD

213

terms of projection of r2 to the analogous expression in terms of -2. Namely, with the help of the coordinate representation, it is not difficult to find that

(Pt,,qu) +(pa,qu) =2{(ru,vu) +(r,,,v.,)}.

(11)

Set p = l /(rr,,,T,,). Using (5), (10) and (11) we have

B=

P34

P( Put q..),

C = - P P(Puqu,),

D - A = - 2 P34P{ (P,,, qu:) + (Pu2 q.,)).

Thus the following equality of quadratic forms holds:

-Bdu, + (D - A) du, due + Cdu2 = - 2 p3'4p(dpdq).

So, the type of (4) is determined by the form (dpdq). This form is of fixed sign if k > I on the plane tangent to t'-. In this case (4) is of elliptic type. If K < I then (4) is of hyperbolic type. From system (4) we can pass to a single equation with respect to x2. Denote x2 by W. Then, writing the equality conditions for mixed derivatives of x1, we obtain

Cpu,,,,+(A-D),Pu,u,-B*u,u,+Wu.(A,+C,,,)-4u,(B,,+D..)=0.

(12)

Solving this equation, we find x2 first. Then from (4) find x, and from (2) find x3 and X4.

For the sake of a perfect picture, observe that instead of (2) there can be found the systems on each of functions x3 and x4 representing their relations with x2 just as in the case of a system on x1. We can express these three systems in a vector form.

Consider a three-dimensional subspace E3 in E4 passing through the origin and perpendicular to e2. Denote by i = (-x3i -x4, x,) a position vector in this subspace. We shall assume that T3 and E3 coincide in such a way that p14 corresponds to -x3, p13 corresponds to -x4, p34 corresponds to x1. Then a position vector of F2 satisfies the following vector system: xu. = Wu.P{T(Tu.Vu:) + [v(Tr.Tu:U} - T-,Pr(Tr.Vu.)r (13)

xu, =

- ,,.P{T(Tu.Vu.) +

[V[Tu,ru.I]}.

The coefficients of (13) are regular functions if the projection of 1,2 into T3 is a regular surface 1 2 and a support function of I 2 with respect to the center of G2.4 is non-zero.

The conditions of equality of mixed derivatives of each component of i produce three scalar equations on x2. But it happens that all of them are proportional to the already derived equation (12) on x2. Thus, (12) gives the compatability condition to (13).

We shall not produce a long analytic discussion here. We refer the reader to [8] because we already reached to the corresponding conclusion in another way.

THE (R-ON] P I R1 01'S1 RM1t.Nii OLDS

214

8.7 Analysis of Specific Cases

In this section we consider specific cases (1) and (2). First we find an expression for h,.. As the normal curvature k of G24 C S' is determined by the ratio of (il2pq) to

(dp)2, we can write the external curvature k, of a hypersurface G_4 C V with respect to the plane tangent to r2 as (Pr,, qr1,) (P,,. q,,,)a- [(i'

(Pr1. yrr, )]=

Pl,,PHi - (P.,Ptr0)2

At PI0 we have p;; = 1. p,,4 = P,',' = 0. Therefore

(Pu.gr,.) =?(pail';' 3 4"

(p,,,q.. ) + (.p,,.q,,,) = 2(Pn;P

;

+P.IPi +P ;Pi:i +pn24p, ).

Thus, the expression for k, at p can be transformed to It

1='

It

, Pal

PU,

P Put

14 1' 11_

P1r1.4

l

'I (/(P,r,Pir: -(P

+2 P,13

Pil

It

Ii

l,,3

P,r:

P,r,

(1)

J

Consider system (3) from the previous section for cases (I) and (2). Suppose that at po we have (e [drr,du+] = 0. Then p's P; = MR. where A and p are some numbers, i = 1.2. Thus , (P-A),

-

In case (1) the equations of (3), Section 6 are linearly dependent and hence A = p and

vice-versa. Then E, = 0 and A = 1. Surfaces like this are excluded from the considerations of Section 8, where we state the existence theorems. and [c,, c,,] arc equal to zero, but the equations In case (2) both the minors of (3) Section 6 are linearly independent. So. we can suppose. for instance, that 1] # 0.

For some numbers 0 and , it must be that d,1, = ed,,, ,

pI i = 0 one has the result:

Since r2 is regular, at p1, with pU = 1. X13

rank Therefore, d

VI,: = '} c

14

,_tI

111

X13

X14

1, a14

1

A2

N

A_a

2

,r

1. System (3) Section 6 can be written in the form

GRASSMANN IMAGE OF A SUBMANIFOLD

215

Since (d,,, c,,, ( j4 0, the following system must be satisfied: xlu, = Ox1u_,

x211, = 1'x2u,.

Define on ul a coordinate curve with two functions x1(uI,O) and x2(uI,O) of nonzero derivatives. From the latter system find x, and x2. Then find x3 and x4. Since the rank of matrix C, u,

xIu;

x2ui

x1

is not equal to zero, the surface F2 is regular in a sufficiently small neighborhood ofpo.

8.8 Existence Theorems for Two-Dimensional Surfaces in E4 with Prescribed Grassmann Image

In this section we use the system obtained above to prove two theorems - one local and one global. Theorem 1 (Aminov) Let 1,2 be a two-dimensional regular surface in G2,4 of class C4. Suppose that the curvature k of the Grassmann manifold G2,4 with respect to the plane tangent to I'2 at po satisfies k 96 1. Then there exists some neighborhood of po in I'22 which is the Grassmann image of the C2 regular surface p2 C E4.

To formulate the second theorem we need some preliminary definitions. Denote by S2(e) a submanifold in G2,4 consisting of two-dimensional planes in E4 passing through the origin 0 perpendicular to e E E4. This submanifold is isometric to the standard two-dimensional sphere in three-dimensional subspace which is perpendicular to e. In the standard imbedding of G2.4 into E6 some three-dimensional subspace passes through S2(e). Denote it by T3(e). This space passes through the

origin in V. The following theorem holds: Theorem 2 (Aminov) Let D be a simply connected domain in a regular surface r'2 in G2,4 with univalent and regular projection into some subspace T3(e), with a support function of projection which never turns into zero. Suppose that at all points of D the curvature K > 1. Let P1. P2, P3 be fixed points in the boundary of D.

Let E2 be a two-dimensional plane in E4 passing through e and 0 a simply connected domain in it with fixed points QI, Q2, Q; in a boundary of A. Then there exists a C''-" regular surface F2 in E4 with A as its projection into E2 and D as its Grassmann image; moreover, these points of F2 which are projected into Qi have P, as their Grassmann image. To prove Theorem 1 we need the following Lemma Let T2 be a plane tangent to G2,4 at Po such that k(T2) 76 1. Then there exists a three-dimensional subspace T3(e) passing through Po such that the projection of T2 into T3 is a 2-plane.

THE GEOMETRY OF SUBMANIFOLDS

216

Through any point P E G2,4 there goes a three-dimensional subspace T3(e), with a manifold of all such subspaces being one-dimensional. Indeed, let E2 C E4 be an inverse image of P and e be some vector perpendicular to E2. Then both S2(e) and T3(e) pass through P. The manifold of all T3(e) passing through P is in one-to-one correspondence with the manifold of unit vectors e orthogonal to E2 and, as a consequence, is one-dimensional. Let T4 be a tangent space of G2,4 at Po. Find those 2-planes T2 C T4 which have no regular projections into any T3(e) passing through P0, that is, those planes T2 in which for any subspace T3(e) there exists the vector which is orthogonal to T3(e). Consider in E4 the basis {e1}, where e3, e4 E E2, and

the basis (e,) which is obtained from lei) by rotation of the angle cp in the plane perpendicular to E2. Take as e the vector e, = cos cp el + sin cp e2. Then an arbitrary 2-plane which is perpendicular to e and passing through the origin can be defined by a pair of mutually orthogonal vectors 4

4

Y=>1'e'.

X=Ea'e,,

i=2

i=2

If we make a substitution e2 = - sin cp e, + cos cp e2 then we can see that Plucker coordinates of this plane with respect to the basis {e,} are of the form p13

p12 = 0,

= x(3(321 sin cp,

p23 =f3(3a-J cos w,

P24 =

)314

a

P14 = a[412) sin W, 21

cos c',

Pi4 = x[3/341

(I )

Suppose that the vector product of X and Yin the subspace spanned on e2, e.1, e4 has the following components with respect to this basis: [XY] = {sin 9,cos9cos-y,cos9sin-y}.

On the other hand, we can find the same vector product in terms of a' and 0': [XYJ = {aI314I, a[4a2J, x[2-/33[ I.

Comparing, we see al3(341 = sin O,

a(402I = cos 9 cos ry,

x[2/33[ = cos 9 sin ^y.

Then the Plucker coordinates (1) of the X, Y-plane have the form p12 = 0,

pl3 = -cos0 sin-y since,

p23 = cos 9 sin y cos cp,

p14 = cos0 cosy since,

p24 = - cos 9 cosy cos cp,

p34 = sin 9.

The plane E2 corresponds to 9 = 7r/2. Find tangent vectors of the submanifold S2(e) at Po, i.e. for 9 = it/2. The introduced parametrization (9, -y) of S2(e) is singular at 9 = it/2, but in this case the set of all 8p/89 determines, while -y changes, a twodimensional plane - the tangent plane of S2(e) at Po. Take as a basis of this plane the following pair: 1, = (0, sin cp, 0, - Cos cp, 0, 0),

.

j2 = (0, 0, - sin gyp, 0, Cos cp, 0).

GRASSMANN IMAGE OF A SUBMANIFOLD

217

The plane off, andf2 depends on V. We can now formulate our question as follows: what kinds of 2-planes T2 C T4 at Po can not be projected onto any fl, f2-planes? Let us assume that there exists such a plane and k, l form the basis in it. As T4 is tangent to G2.4, p13, p14, p'3, p24 form a system of coordinates in T4. Let k and l have the following components in T4: k = (k,,k2,k3,k4),

1=(11,/2J1,14)-

For any V there exist numbers CI (cp) and C2(cp) not equal to zero simultaneously such that a linear combination C, k + C21 is orthogonal to both f, and f2. Therefore, the following equations hold: C, (k, sin 1P - k3 cos,p) + C2(11 sin lp -13 cos cp) = 0,

C,(-k2 sinlp+k4 coscp)+C2(-!2 Sin ce+l4 cos) ,p=0. For any cp the determinant of the system above with respect to C, and C2 is zero. As ,p is arbitrary, we obtain three equations: k, k2

1'

I = 0,

12

I

k3 k4

4,1 14

kI

= 0,

-!3 I

k2

I

14

1,

+ I -k3 k4

1 = 0.

121

(2)

From the first two equations of (2) it follows that there exist two numbers A and µ such that k1=A11,

k2=A12,

k3=µl3, k4=µ14-

(3)

The numbers A and µ are not equal to each other because k and I are linearly independent. The third yields 1114-1213=0.

(4)

Now find the curvature K of G2.4 with respect to the k, /-plane. To do this, find the extrinsic curvature & of hypersurface G2.4 C S` by (1) Section 7 with replacements 24 -

''3

14

P, = k2,

pN, = k3,

...

...

D24

k4,

= !d.

Then the numerator of K, formula gives the form k

-) k4

11

14

12

Ik3 13I

+2I

k3

13

k1

11I1-k4

k2

12

2I kk2

1

-14

+

1

12 11

3

II kk4

13

-14

(5)

Make use of the relations (3) on the components of k and F. By virtue of them, the last term in (5) vanishes. So (5) has the form -(A - p)21; l4 - (A -µ)2l;1; + 2(A - µ)211 12 13 14 = -(A - µ)2(l,14 -1213)'`.

By virtue of (4) we get zero. Thus KP = 0. Therefore, as k,. = K - 1, the curvature of a Grassmann manifold k = 1. The lemma is proved.

THE GEOMETRY OF SUBMANIFOLDS

218

From the lemma we conclude that only the two-dimensional surface r2 passing through Po with a tangent plane of property k I is regular projectable in some neighborhood of P0 into some T3(e) in the form of a regular surface r2. The projection of the tangent plane of r2 is the tangent plane of f2. But the tangent plane of F2 at P0 coincides with some jl,f2-plane in 7°. Hence, a support function of f2 at Po with respect to the origin in T3(e) is equal to 1. So, it differs from zero in some neighborhood of P0. Choose Cartesian coordinates in E° in such a way that e2 = e. Then T3(e) coincides with the subspace introduced in Section 6: T3 = T3(e2). From what has been stated above, it follows that in some sufficiently small neighborhood of Po, system (12) Section 6 has regular coefficients. Let gi be a metric tensor of the required surface F2. Find g = det IIg1II Let F2 be a

projection of F2 into the subspace E3 perpendicular to the basis vector e2, i its position vector, gi its metric tensor. Then, evidently, g+i = 9# + X2u, x2u,,

g = g +gl l Xiu, - 2812 X2u, X2a2 +g22 2u, = I[Xui

i..'112

+ (Xu, X212 - Xu, X2u,)

.

By means of (4) Section 6 and condition (r, v) = 0 we get [Xu, Xu] = -vpl,

x1, X2u2 - xu, X21, = -TPI,

where we set

I = X22141(7842

V12) - X2ui X2u, ((T1, V12) + (rue Vu, )) + X2u2 (Tu, Lu, )

Taking into account that Ir12 + IVI2 = 1, we find g = 12p2.

(6)

Suppose that all over r2 the curvature K < 1. Use the differential equation on x2. It is of hyperbolic type for this case. The characteristics r;(ul, u2) = c and n (ul, u2) = c for this equation are characteristics for the (d2q) form. Pass the TI-coordinates to t. Then (rrvf) = (r,,vl) = 0. Give values for x2 on the characteristics in such a way that at the initial point XX2q 54 0, which guarantees 154 0 at this point. In some neighborhood of the initial point there exists a solution of the differential equation under consideration (see [31], p. 458). With the help of (2) Section 6 we find other components of the F2 position vector, each of which is found up to an additive constant. As 196 0 in some neighborhood of the initial point, by virtue of (6) the surface is regular in that neighborhood.

In the case of k > I a stronger assertion holds - Theorem 2. It has been formulated already at the beginning of this section.

Now we apply the theory of quasi-conformal mappings. Lavrent'ev has introduced the concept of a quasi-conformal mapping associated with a given system of partial differential equations. A homeomorphism of domain D onto the domain A as u = u(X,Y),

v = v(X,Y)

GRASSMANN IMAGE OF A SUBMANIFOLD

219

is called the quasi-conformal mapping associated vith an elliptic system

8u !2 1

ax , 8y = 0,

i = 1, 2,

if u(x,y) and v(x,y) satisfy that system. The geometrical notion of a strongly elliptic system has been introduced equivalent to the usual notion of ellipticity for the case of linear system. In [9) Lavrent'ev proved a general theorem on the existence and uniqueness of a quasi-conformal mapping associated with a strong elliptic system such that it maps given domains ,6 and 0 bounded with piecewise smooth curves one onto another and a given triple of points z1 i z2, z3 in the boundary of D onto given triple of points w1, W2, w3 in the boundary of A. Since k > 0, our system is of elliptic type. We apply the Lavrent'ev theorem to the

simply connected domain of parameters b corresponding to D and the simply connected domain i in the E2-plane passing through e, and e2. Using (3) Section 6 we find, first, x, and x2; then x3 and x4 with definite components up to a shift along the plane perpendicular to E2. To solve system (3) Section 6 it was possible to apply the Vekua theory, as well. Consider the question of the regularity of F2. From (6) and k > I it follows that F2 is regular if x2u. + x22,,, 54 0. In [10] Polozij stated the following: Theorem In order that functions u(x,y) and v(x,y), which are continuously differentiable solutions of au,,, + bu, - v,. = 0,

d satisfying the Holder condition at zo = xo + iyo, map a one-sheeted neighborhood of zo in the plane z = x + iy onto a one-sheeted neighborhood of vo = f (zo) in the plane w = u + iv, it is necessary and sufficient that the Jacobian uCv,. - v,.u,,. 54 0 at zo.

Zeros of the Jacobian urv,. - v_ru, coincide with zeros of the function u_t + iu,., i.e. with zeros of x2u, + ix2u, in our case. We can also apply, the Levi theorem from [11].

Since a quasi-conformal mapping from the Lavrent'ev theorem is a homeomorphism, by the Polozij theorem x2i, + ix2u, # 0. Therefore, the constructed surface F2 is regular. On 1'22 it is sufficient to require the regularity of C2 Then F2 will be regular of class C2"", a < 1. Theorem 2 is proved. Note that for the hyperbolic case some theorems on the reconstruction of a surface by its Grassmann image have been proved by Kizbickenov [12). Another approach to the problems above has been proposed by Weiner [ 13], [14].

8.9 On Local Projections of Two-Dimensional Surfaces in E4

The local behavior of F2 C E4 essentially depends on the type of Grassmann image.

Consider the projection of F2 c E4 into three-dimensional space E3(r) passing

1H1: C,FA)MLIR\ OF SUBMANII OLDS

2211

through the point x, the normal plane N, and the tangent vector rat x. Let el and e2 be an orthonormal basis in the tangent plane such that e, is a tangent to the u'-curve. Coordinates are consistent with the normal curvature ellipse. Let c'u e4 be normals to

F' parallel to the principal axes. Set r = cosdcl +sinOe2; x' are Cartesian coordinates with respect to the basis e1,--..c4 with the origin Y. Then up to Intinitesimals of 3-d order with respect to 1' (i = 1, 2). a position vector of F2 projection into E'(r) has the form f = r(cos Ox' - sin Ox) -F ea [(ci + (1)(11)2 + (ei -11)(.x)2 + ea

1d(.x' )2 + 2bv' x 2 +3(x2)2 ]

Denote the projection above by P. We suppose, also, that n,13. it. h are not equal to zero simultaneously. Depending on r, the point x in F2 may be either regular or singular. We can find the direction r with respect to which F-(-r) has no singularity at x from the following equation

cth+(1.3sin 20-nhcos20=0.

(1)

Let 0, he the roots. If all coefficients of the equation above are zero then F2(r) is a plane for all T. 2 Further on. we exclude that possibility. Equation (1) has no roots if (u32) + (ah) - (ah) -< 0. i.e. if the Grassmann image is elliptic. In addition. for any r the surface F2 is singular at t (see Figure 30). The structure of F2 (r) is similar to a two-sheeted Riemannian surface. The surface is tangent to z at x. The surface section with a plane through r and it e N, produces a pair of parabolas tangent to r and with branches in distinct half-planes. In some of the sections one of the parabolas degenerates into a half-line which is a line of self-intersections. If (ah) 2 - (ad)2 - (nh) 2 < 0, i.e. if K < I and 0 d, then

FICiLIRE 30

GRASSMANN IMAGE OF A SUBMANIFOLD

221

P2(r) is also singular at v. But in contrast to the previous case. P2(r) is included in some two-sheeted angle (see Figure 31). The section of P with the plane through r

and it E N, is either one point x or the pair of parabolas tangent to r and with branches in the same half-plane. In a particular case one of the parabolas degenerates into the half-line. If 0 = Q, then the projection is a regular parabolic cylinder. In the parabolic case, i.e. when K = 1. generally there exists only one subspace

Ea(rl) such that P has no singularity at x. If r 0 rI then P has the structure as in Figure 32. The section of P with the plane through r and n E N, consists of the parabola and the half-line. For some sections this parabola degenerates into the halfline which is the line of surface self-intersection.

FIGURE 31

FIGURE 32

222

THE GEOMETRY OF SUBMANIFOLDS

8.10 Reconstruction of Two-Dimensional Surfaces in n-Dimensional Euclidean Space by Grassmann Image

Consider the two-dimensional surface F2 in n-dimensional Euclidean space E" for n > 5. Its Grassmann image 1'2 is located in a Grassmann manifold Gn_2," denoted further on by G. As dim G > n for n > 4, the arbitrariness in determination of the two-dimensional surface in G is greater than in E". Hence, an arbitrary surface in G

can not be a Grassmann image of the surface in E". It is possible to state the necessary and sufficient condition for r2 to be the Grassmann image (see [17]). In this section we state the theorem on uniqueness in the general n-dimensional case and then consider in more detail the case of n = 5, where we state the necessary and sufficient condition mentioned above.

Let Ei-4 be an (n - 4)-dimensional Euclidean space passing through 0. Set E= En-4. Denote by G2,4(E) a manifold of all (n - 4)-dimensional subspaces E`2 in E" passing through E. Then G2,4(E) is a Grassmann manifold G2,4 imbedded in G as a submanifold. We say that I'2 is tangent to G2.4(E) if the tangent plane of r2 at x E r2 is in tangent space of G2,4(E) at x. We shall call 1'2 the surface of a general

kind if both F2 is not tangent to G2.4(E) and any tangent to r2 vector is not simultaneously tangent to the pair of G2,4(E) at any point x E 1'2. The following theorem of uniqueness holds:

Theorem (Aminov) A two-dimensional surface in n-dimensional Euclidean space E". n > 5. is determined by its Grassmann image of a general kind uniquely up to homothety and parallel transport.

The proof and also the necessary and sufficient condition mentioned above we present in the case of n = 5. Use a representation of points in the Grassmann image in the form of matrices. Select in E5 an orthonormal basis et,... , e5. Let Eo be a three-dimensional subspace passing through e3, e4, es. Take the normals {,,, v = 3, 4, 5 of F2 in such a way that each & would be of the form 2

= e + E a ,e;,

v=3,4,5.

(1)

The matrix Z = Ila,II uniquely determines E3 - the normal space of F2 - if it is one-to-one projectable onto Ea and vice versa. If UI, u2 are curvilinear coordinates in a regular surface I'22 c G then the components of Z are regular functions of u1, u2. So, Z = Z(ui, U2). Denote by a' the columns of Z. The matrix Z consists of two columns and two rows. By [ ] we denote a determinant. Introduce the quantities f' _ [d",au au,],

i jAk;

i, k = 1, 2.

Set A = f11 f 2 - f2 fi-. From the lemma (see below) we conclude that 0 a general kind.

0 for r,2 of

GRASSMANN IMAGE OF A SUBMANIFOLD

223

Theorem 2 (Aminov) In order that C3-regular surface r2 C G3,5 of general kind is a Grassmann image of C3-regular surface F2 C E5 it is necessary and sufficient that the following equation is satisfied.

a

au,

I

f;u: -f'2.,

A_ a

flu, -f22ui

f2

1

IPq -A.,

f;

art2 ° f2, -f2u, f,2

-

0

(2)

Let r2 be a Grassmann image of some surface F2 of the position vector r = {x1}. Let E3 be a normal space of F2. Suppose that &, = {t j,} are the normals and hence 5

k=1,2.

Exj,,,V,,=0, v=3,4,5;

(3)

j=I

By virtue of decompositions (1), equation (3) can be represented as 2

xy,,, = -

x;,,, a,,.

(4)

;= I

From the equality conditions on the mixed derivatives of x we get three equations on x,, x2: 2

E(x;,,, a',,,,. - x;,,_ a,,,,.) = 0,

v=3,4,5.

(5)

;= I

There exists a function A(u,, u2) such that

x;,,, = of ,

where f =

k, i = 1, 2

(6)

i 0 k, k = 1,2. From the equality conditions on the mixed

derivatives of x, and x2, we obtain a system of two linear equations with respect to the function 'I' = In A: 'P u,f; -'I',,, Al' =fiu: -fz,,, I (7)

wu, f - pu, f12

=fi2-f;,,.

The determinant of the system above is A. By the lemma below, A 4 0. So, we have

4"'. _ ,

1

0

IPU2

-fi,, A

f fl',,, -fiu,

o j2 f,us -fZ, I

(8) .

f,u_ -fiu, fl'- I' The equality conditions on mixed products of produces equation (2). In a sequence of systems (4), (6), (8) each subsequent system is a compatibility condition for the previous one, while (2) is a compatibility condition for the latter system. If (2) is satisfied then 'I' is determined up to an additive constant. Thus, the function A and, as a consequence, the right-hand sides of (6) will be determined up to the constant multiplier. The variation of this constant multiplier means the transformation of F2 into a homothetic one. If the multiplier is fixed then from (6), (4) the function x1 is determined uniquely up to additive constants. The choice of these constants means

224

THE GEOMETRY OF SUBMANIFOLDS

the choice of parallel shift of F2. The surface F2 of position vector r = {x1} is regular because its metric tensor determinant

g = .402 (1 + (a' )2 + (a2)2 + I[aIa2)12)

(9)

is not equal to zero.

Lemma Suppose that r2 is not tangent to G2,4(E). Then 0 = 0 if and only if there exists tangent to r2 a vector such that it is simultaneously tangent to the pair of G2.4(E) For the case of n = 5 the space E is of dimension one. Denote by e the unit vector in E. Suppose that some v = cos a Z,,, + sin a Z,, which is tangent to r2, is tangent to the pair of G2,4(e). Represent the equation on G2,4(e) in terms of the matrix Z. Let e be of the form S

5

j=1

1-3

e = Eb 1e; = > c'

,.,

where {,,, v = 3,4, 5 determine a three-dimensional space E3 which contains e. Using

(1) we get b° = c", b' = E5_3a1c°, v = 3,4,5; i = 1,2. Hence, each of the submanifolds G2.4(e) is determined by the equation µZ = n, where µ = (b3, b4, b5), 17 =(b',b2) are constant vectors. Therefore, if tv,i are curvilinear coordinates in G2,4(e) then the equations UZH., = 0, $ = I,... , 4 are satisfied. Thus, for the vector

W there exists the vector p such that p W = 0. This means that cos a (pa,,. ) + sin a (pa,,) = 0, where () means the scalar product in three-dimensional space. Non-zero /s is orthogonal to the pair of vectors: Ti = cos a ail + sin a au:, i = 1, 2. Since W is tangent to the pair of G2,4(e) then we have a pair of vectors of u type orthogonal to the pair (r', r2). Hence r1 and r2 are parallel to each other. So, r' is in the plane of vectors a;,,, au,, k = i. Hence, the mixed products are

[r'a, a;:]=0, k,i=1,2;k

i.

(10)

From this it follows that rows of determinant

6=

an,

[a

a2. 2

, , an.

2]

'"a" l a"

,

i

au, au, au,

r2

I,]l

a"'a' a'

are proportional, that is 0 = 0. Conversely, suppose that 0 = 0. If for some fixed i the vectors a., and a,,_ are parallel to each other then there exists a number a such that r' = 0. Take as µ any vector in the plane perpendicular to rk, k 0 i. Then the vector W tangent to I'2 satisfies an equation pW = 0, i.e. it is tangent to the pair of G2.4(e) simultaneously. For this case, the lemma is proved. So, suppose that each pair of a',, a, consists of two non-parallel vectors. Then they determine a plane. If four of the vectors a;,, i, k = 1, 2 are in the same plane then there exists the vector p orthogonal to that plane such that (pa,,t) = 0, that is p 0, k = 1, 2. This means that the tangent plane of r2 is in the tangent space of G2.4(e) which contradicts the lemma hypothesis.

GRASSMANN IMAGE OF A SUBMANIFOLD

225

Thus, we suppose that planes of pairs a,,. a,, and a are distinct. From L = 0 it follows that there exists a number a such that the mixed products (10) are zeroes. As the planes of pairs a,F, are distinct then rl and r2 are collinear. There exists a twodimensional plane of vectors p perpendicular to r', i.e. pW = 0. Therefore, W is tangent to the pair of G2,4(e) simultaneously. The lemma is proved. We note that Gorkavy [34]-[36] has proved very interesting theorems on the reconstruction of multidimensional submanifolds by their Grassmann image. 8.11 On Hyperplanar Sections of Two-Dimensional Surfaces In E4 With Elliptic Grassmann Image

Borisenko remarked that a two-dimensional surface F2 C E4 has an elliptic Grassmann image if and only if at each point P E F2 the normal curvature ellipse is not degenerated and P is located in an ellipse bounded domain. This means that for any normal n there exists a system of coordinates x, y such that the second fundamental form with respect to that normal has the form a2 dx 22 - b2 dy2. where a and b are not

equal to zero. As the normal curvature ellipse of the surface with an elliptic Grassmann image is not degenerated at each point, any domain on the surface can not be located in a hyperplane V. Moreover, the following is true. Theorem Let F22 be a C2-regular surface in E4 tt'ith an elliptic Grassmann image. Then the boundary r of any connected compact domain D C F2 can not be located in E3.

Suppose that r is in some hyperplane E3. Introduce Cartesian coordinates X1, x2, x3, x4 in E4 in such a way that the xi, x,, x3 axes are located in E3. Suppose that x4 of P E F2 is some function u(P). Then u = 0 on F. Let us assume that u 0 0. As D is a compact domain and u is continuous, then there exists a point Po in D of maximal or minimal u-value. Introduce about Po in F2 a system of coordinates x,y. Then at P0 the following conditions must be satisfied: ux = 0, u,. = 0, u,,u,.,. > 0. The vector v = (0, 0, 1, 0) is the normal one to F2 at P0 and the second fundamental form with respect to v is 11(v) = u,, dx2 + 2uy dxdy +

dye.

Since F2 has an elliptic Grassmann image, then by a coordinate change 11(v) can be reduced to the form u_Tr dx2 + u,.y dy2, where u,r_ruy1. < 0. This contradicts the extremal property of Po: u,,u,., > 0. Hence, r can not be located in E3. From this theorem it follows that any domain in F2 of a boundary formed by an intersection of F2 and a hyperplane is necessarily non-compact. 8.12 Representation of Points of a General Grassaraan Manifold in Terms of Plticker Coordinates and Matrices

In the previous section the points in the Grassmann manifold G2,4 were represented by Pliicker coordinates. Evidently, the Pucker coordinates can be introduced in any

_.h

THE (DEOME rttl Of SUBUMAMFOLD5

Grassmann manifold GA,, IA. Also. we shall consider the representation of points in terms of matrix elements.

Let NA be the k-dimensional subspace in E"" passing through 0. We shall consider NA taking into account the orientation. Choose in NA an orthonormal frame C),..-CA. With respect to the ambient space basis a .....cl,,,A. each of , obtains some components 4`,. The Plucker coordinates p'f ' of Ni" are the set of values f,

=

?A

p11

Si

..

S)

..

.

f,

. !q

In the other words, the Plucker coordinates of NA are the Plucker coordinates of the

ordinary multivcctor p = .......4J. Order them lexicographically and select the components with distinct strings of i)....,iA. Consider the Euclidean space E" of dimension v = C; (the binomial coefficient). Put the initial point of the vector with components (p) A.. ) at the origin of E''. Then each point p E GA ,, , A corresponds to the unique point in E''. In this standard way the Grassmann manifold GA ,,,A can be immersed into E''. Plucker coordinates of an ordinary hivector are subject to some relations. First 61)=k2l .. The equality above can be represented as (p'

a)=

= I.

=1.

(1)

4

11

Hence, a Grassmann manifold GA,, A is immersed into a unit sphere S" ). Also, we have the set of relations

p" ",p,f 14I=0.

(2)

where [ ] means the cyclic permutation of superscripts, The latter equality can be represented as A

E(-1)'p', f,

I hpM, ,,

1

),

1

1, =0.

(3)

1=1)

where we set ix =J(). The term with a = 0 is a product p" 'kph '/. Denote by .41' a cofactor to e,,4 from the last column of the determinant

SAl

...

SAC

Then for each Plucker coordinate we have Pee

It 11 = At'),

GRASSMANN IMAGE OF A SVBMANIFOLI

229

So. the left-hand side of (3) obtains the form tr++4

r,

.. en

'

cr+11+

...

u

... .. ... ... A ...

4h

L

t4

4

E1

X41+

...

S4111

Each determinant in the latter expression is zero because it contains two identical rows. Thus, (3) is proved. In terms of the Phkker coordinates no GA +A it is possible to introduce the metric clrr=

induced hr the immersion into E":

d4r' = > (dp'+ a )14

Now we describe another way to introduce coordinates in a Grassmann manifold. They will be independent of each other. i.e. the proper coordinates. In this way we shall find a dimension of the Grassmann manifold. Let N, be some k-dimensional fixed subspace in E"lA passing through 0. Define the neighborhood of N,, as a set of all k-dimensional subspaces NA passing through 0 which contain no vectors except zero orthogonal to N,. Let NA be an arbitrary k-dimensional subspace from the neighborhood above. Show that for any e E N, there exists only one f E NA such that a projection off into N coincides with e. Indeed, project all of the vectors from NA into N,4,. If t and 11

have the same image then 4 - q is orthogonal to N, because of the linearity of projection. But NA is from the neighborhood. hence E = n1.

If the projection of N, does not cover N,, then, by virtue of the dimension equality, there exist two vectors in NA with the same projection, which is impossible. Let e,, .... en, k be the orthonormal frame in E"+A, where eI, ... , eA form the basis in N;. In NA take a frame fi, ....ft (non-orthonormal, in general) such that e, is the

projection off into N,4,. Consider the decomposition off with respect to basis of E"+A: +++A

f,=e,+>2a,e,, i=1,...,k. r-A+i

Thus, to each k-dimensional plane NA from the chosen neighborhood of N, there corresponds the matrix Z = IIaaII of k rows and n columns. Conversely, to each

matrix Ila,II there corresponds some frame f,,...,ft and, as a consequence. a kdimensional oriented subspace Nk spanned on that frame, i.e. the point in GA n+A. The correspondence above between points of a Grassmann manifold and matrices defines a homeomorphic mapping of the NQ neighborhood onto a neighborhood of origin in kit-dimensional space EAn, where coordinates are generated by elements of matrices 11u,11. In contrast to PlUcker coordinates, a, are independent. Thus, the dimension of a Grassmann manifold GA n+A is equal to irk.

THE t;LOMFTRt I)} S1 K MANIM)I CIS

22$

8.13 Metric of a Grassmann Image of Arbitrary Submanifold Let F" be a regular submanifold in E" A parametrized by u' , .... ii". Find the metric of the Grassmann image r' of the submanifold, assuming that l"" is parametrized by a' , ... , u" by a Grassmann mapping. Apply the same method as in Section 2.

Consider a differential of an ordinary multivector. Using the Weingarten de-

l

composition. we have dp=(I

A

...i] _

u...

M-

[=;I

ctic'

I ... r,

j

t ]cliff'.

where r, = Or/au'. The square of dp produces the metric of the Grassmann image(dp), -

--r'

r, ...E4]

L;Lm,g"g"

14I

To find the scalar product of ordinary multivectors in the formula above we use (5) Section 1 Chapter 1. where this product is expressed in terms of scalar products of vector components of a multivector, The matrix of these scalar products for rt # ri consists of units in the main diagonal. excluding two places with zeros. Besides, the (ri, '3) position take (r,r,) All the other elements are zero. So, 0

0

1

.

0

0

d 0 0 0

s

...

1

0

...

0

0

0

-

0

...

0

1r,r,)

...

0

...

0

0

...

0

0

.

...

0

0

I

If a 0;1 then the fi-th row and rl-th column consists of zeros. In this case the determinant equals zero. If ct = c3 then the matrices have diagonal form. where (r,r, )

takes the (n, a) position. Thus,

(fr' ... P,.

=(r,r,)=g,%

Tlwre/ore, the metric of the Grassmanni image t" of a suhinani/old has the form R

(clp}= -

LU L`,s g" rlur dit'. t,_1

GRASSMANN IMAGE OF A SUBMANIFOLD

229

Denote by G1; the components of the metric tensor of a Grassmann image. Then k

G11=

Fe, L g.t

The expression above for (dp)2 has been found in [18], [32], [33].

8.14 Angles Between Planes

Let us be given a pair of k-dimensional planes T and E passing through the origin 0. Find the angles that they make with each other. Let u be a non-zero vector in T and v be its orthogonal projection into E. The stationary values of angle 9 between u and v while u rotates in T we shall call the angles between T and E. The direction u1 which corresponds to the stationary value of 9 we shall call the angle direction of T with respect to E. We want to show that if v1 is a projection of ui into E and vi 0 0 then vi is the angle direction of E with respect to T. Let ft,. .. , fk be some basis in T while el, ... , ek is some basis in E. Represent an arbitrary unit vector u E T as k

where E cost a1 = 1.

u = E cos a1 f ,

i=I

i=I

Represent the projection v of u into E as

v = E(ue;)e;. j=I

If 9 is the angle between u and v then L

I.

Cos20 = E(ue;)2 =

(cos ai (f e;)J 2.

(1)

!=1

1=I

We can consider the expression on the right as a quadratic form with respect to cos ai. The matrix of this form consists of k

au = E(.fe1)(fiei)

(2)

1=1

Since the matrix is symmetric, there exist k stationary values of cos2 u and k principal directions ul, ... , uk, which we called the angle directions of T with respect to E. Let

us set f = ui. Then the quadratic form matrix (2) is diagonal, that is k

Df e,)(f e,) = 0, /=I

for

i `i.

(3)

THE GEOMETRY OF SUBMANIFOLDS

230

Let vi be the projection of u; into E. Then k

vi = E(Iel)ei i=I

For i 11-j the vectors vi and vJ are mutually orthogonal. Indeed, by virtue of (3), we have

(vivi) = E(.fei)(fiei) = 0. t-1

Direct et along vi. Then 0 for i 1. Find the angle directions of E with respect to T. Let e be the unit vector in E k

e = E cos Qi ei. i=

Denote by V an angle which e makes with T. Then k

rr

k

k

cos f3 (ef,)=

cos'-

i=1 Li_I

I

cos'- (3J (ei.f')'.

i--I

The expression on the right is a quadratic form with respect to cos(i; moreover, its matrix already has diagonal form. Hence, cos 2 W attains stationary values for e = ei. So, vi are the angle directions of E with respect to T. The plane which passes through ui and vi we shall call the angle plane. It is well defined if vi # 0 and ui 0 vi. Denote by Bi an angle between ui and vi. We shall call it the angle between T and E. So, we can formulate the Wong theorem from [23]. Two k-dimensional planes T and E in E" I k make k angles among which no more than r = min(n, k) can be non-zero.

8.15 Geodesic Unes In a Grassmann Manifold

Let T and E be k-dimensional planes in El" k passing through the origin. Consider them as points p and q in a Grassmann manifold. Find a geodesic passing through these points. Let p and q be represented in terms of ordinary multivectors: p = Jul ... uk],

q = [vI ... vk],

where u, are the angle directions of T with respect to E, vi are the projections of it, into E. Let 0, .... OA be the angles between T and E. Consider a curve -y of parameter t E [0, 1) in a Grassmann manifold p(t) = [al (t) ... ak(1)

where each of ai(t) is a vector function of unit length. Moreover, if ui and vi are distinct then ai(t) is located in the ui, vi-plane, otherwise ai(t) = ui. Let us require that

GRASSMANN IMAGE OF A SUBMANIFOLD

231

a,(0) = it, a,(l) = vv, and ('h) = c-, i.e. the speed of rotation of any a,(t) would be constant. It is evident that either c, = 8, or c, = 9, + 2,rk where k, is integer. Prove that p(t) is a position vector of a geodesic in a Grossmann manifold. We have dt = dp

al ... A ,_I

da, aA . J

dt

(1)

Taking into account the formula for the scalar product of multivectors, we get [171

...

[a[

dt7... aA]

...

+ ... aA

= 0,

for

i

0j.

because the components of the first multivector are orthogonal to a,. If i = j then the corresponding scalar product is {%}-= O. Thus

(5,)2 0

dt

Since I`C'I is constant, the parameter i differs from the arc-length parameter of I by a

+92t. constant multiplier. Namely, if a is the arc length of ry then e = 82 + To show that ry is a geodesic on a Grassmann manifold we verify that the second derivative of p(t) is normal. Differentiating both sides of (1). we get ci-p -it-2

f

l

d -a,

E Lat... 76-2

a*J +

La1

dct,

du,

[it -

Tit

... aLl

(3)

As each of the a,(t) endpoints passes a circle, moreover with constant speed, then d2a,/dt'- = -8, a,. Hence, the first sum on the right-hand side of (3) is -(0? + + 8k')p. Since a Grossmann manifold is located in the unit sphere of center in the origin, then p is one of its normals. Show that each of the muftivectors ..ad from the second sum is orthogonal to the tangent plane of a Grossmann manifold. Every tangent vector of a Grossmann manifold at p(t) can be represented as [0,

b

by=Eful ...ba,,...ctAJ, n=1

where b means differentiation in some direction. For i

(fa1...ba,,...aA] [a

n\, i 0 j we have

da,. tea+..aA])

=0.

because a, from the first multivector is orthogonal to each of the vectors from the second. Note that (a, w) = 0 because a,(t) is in the plane orthogonal to a, for all t. If a = i then a, from the first is orthogonal to each of the vectors from the second.

THE GEOMETRY OF SUBMANIFOLDS

232

So, the second sum on the right-hand side of (3) is perpendicular to the tangent space of a Grassmann manifold. It follows, then, that y is geodesic. Using (2), we can find the distance between planes T and E as an integral of dQ over the shortest geodesic y: I

0

Thus, the square of distance between two planes is equal to the sum of the squares of angles between them.

8.16 Geodesics in a Grassmann Manifold C2,4

Let p be a position vector of submanifold GIA C E6. Normals of this submanifold are mutually orthogonal vectors p and q, where q is a bivector complementary top. If y is a geodesic in G2.4 of a natural parameter s then

p=ap+bq, where a and b are some unknown coefficients. Both sides of the equation above multiply by p. As p and q are mutually orthogonal, we have (d2p

p

','2)

_ d

dp dp ds(pds)-(ds) =-I =a. ''

So,

dip = -p+bq.

(1)

From this it follows that the curvature of -y in E6 is equal to 073Y and b is the normal curvature of G2.4 C SS in the -y-direction. Expand (1) into components and use the expression of components of q in terms of the components of p: dzdpr2l2

u 2p 13

= _p12 + bp-A = d

= _P 13 + b pat =

ds2

d 2ga2

_ _q3a + bp3s

=

-q42 + b pat

ds2

So, the following holds:

d2q=-q+bp. ds2

(2)

GRASSMANN IMAGE OF A SUBMANIFOLD

233

Multiply both sides of (1) by q, taking into account that (q I) = 0:

(q

ds

(q

ds)

- \ds dc)

- (ds ds)

(3)

.

Multiply both sides of (1) by dq/ds, (2) by dp/ds and then add. We obtain d

(ds ds2) + (ds

d sq

)

ds

( d s 1) = 0

because p and q are normals to G2.4. Hence, b = const. So, we come to the conclusion: a normal curvature of G2,4 with respect to q in a geodesic direction stays constant. Find a position vector of a geodesic in explicit form. Set b = k. Then we have

d,pds2 _q

d

-p+kq,

= -q + kp.

Adding and subtracting these equations, we obtain

d2(p q) _ -(I -k)(p+q), d2(p q) dy2

-(1 +k)(p-q)

Since k is constant, the solutions of the latter equations are

p+q=Acos 1-ks+B sin 1-k s,

(4)

p - q=Ccos l +ks+D sin l +ks,

where A, ... , D are constant vectors in six-dimensional space such that A I B, C -L D, JAI _ ... = IDI = vf2-. Also, as all components of q can be expressed in terms of components of p, the components of A, ... , D are not arbitrary. Namely, the three

last components of A and B replicate the first three. For instance, the first and last components of p + q are p12 +p34. Therefore, AI = A6, BI = B6 and so on. The vectors p ± q are of constant length vll While s varies, both of their endpoints trace a circle. Adding, we find p V-1

ks.

Since p2 = q2, then setting s = 0 in above, we get A 1 C. In which case is the geodesic closed? Suppose that there exists a number I such that

1 -kI= 2irm,

1 +kI= 2irn,

where m and n are integers. Then

k= n2-m2

n222+m22*

THE GEOMETRY OF SUBMANIFOLDS

234

From this we find

m2

1-k

i.e. I - k/ I + k is a square of a rational number. Note that IkI < 1 always. The value of k depends on the choice of starting direction of a geodesic. If k is not of the form above then the geodesic is not closed. Project the geodesic -y into spheres Si and S, whose direct product generates the Grassmann manifold G2.4. Denote by y; the results of projections. Consider the three-dimensional position vector p of sphere S? (see Section 3): p = (P12

+P34,P13 +p42,p14 +P13).

Its components coincide with the first three components of p + q. Denote by j and h the three-dimensional vectors consisting of the first three components of A and B respectively. Then, using the representation (4) of p + q, we can find p at points of yl as

µ(s)=AcosvTs+Bsin VI -ks. In an analogous way we find a position vector v(s) of y2

v(.r)=Ccos 1+ks+Dsin l+ks. As A and h are unit mutually orthogonal, the endpoint of p(s) traces a great circle in S2. The same conclusion is valid with respect to y2. So, the projections of geodesic in G2.4 into SZ are the great circles. The points in these circles are related to each other with a parameter s. Let t; be the arc-length parameter of y;. Then we have

dµ=(-A sin 1-ks+Bcos 1-ks) 71-1

I

-kds. dtl

I --k s. Analogously, t2 = 1 -+k s. Hence, the arc-lengths of y; are related to each other linearly: tl = V -R 12 . So, we can give another way to describe a geodesic in G2.4. Take great circles y, in S;2 and set point-to-point correspondence between them such that the ratio of lengths of corresponding arcs is constant. The pairs of corresponding points of y; represent a point in G2,4. Therefore, 11 =

8.17 On the Canonical Form of a Matrix Let Z be a matrix of k columns and n rows. Consider the transformation Z generated by an orthogonal matrix S of order k and an orthogonal matrix Q of order n of the following kind:

2 = QZS.

GRASSMANN IMAGE OF A SUBMANIFOLD

235

Show that by a specific choice of S and Q the matrix Z can be reduced to the simplest form (we shall call it the canonical form) such that 2, = 0 for i 9Q. Let Eo and Eo be mutually perpendicular subspaces in E"+k passing through the fixed point O. Let a, , ... , an and a,,+ I , ... , an+k be orthonormal bases in Ep and Eo

respectively. Then it is possible to map the n-dimensional subspace E" in E"+k to each matrix Z and vice versa (see Section 12). Let e,, . . . , e,, be the angular directions of Eo with respect to E". Let f,, ... f" be the vectors in E" such that a projection of j into Eon coincides with e;. From what was proved above, e; form orthogonal basis in Eo andf form the orthogonal basis in E".

Let f,, ... , fr, r < n be a maximal subset of {f,} with the property of not being located in Eon. Through each pair (e;, ,) i < r draw a two-dimensional plane E,. From what was proved, these planes are mutually orthogonal. In each plane E,2 i = 1, . . . , r orthogonal to ei. We obtain the system of r orselect the vectors thonormal vectors en+i, . , e,+, E E01. If r < k then enlarge this system up to a basis

en+1, ,en+k in E. Write a decompositions of fit. ..,fn as ei+

z,en+I,

e2+

-

-+

Z2en+2,

+

...

er+

Zr

r en+r,

f,, = en.

Matrix Z with respect to this choice of bases has the canonical form

Z=

0

10o

f2

.........

0

...

...

zr

0

...

0

0

0 0

0

We can take the matrix of basis change from a,, ... , a" to et,. .. , en as Q, and the matrix of basis change from a,,1,... , an+k to en+,, ... , en+k as S. 8.18 The Equations of Isometric Immersion of a Grassmann Manifold and Second Fundamental Forms

Our next objective is to find the curvature tensor of a Grassmann manifold. We present two ways to obtain it. The first (which is rather long) exploits equations already found of isometric immersion of a Grassmann manifold into Euclidean space. Therefore, the expression is complete and conceptually simple. In addition, we obtain some lemmas of independent interest. For instance, we state a minimal system

of equations for the standard immersion of a Grassmann manifold into Euclidean

THE (OMETR1 UI' Sti3Ai.tNlt(TLDS

23h

space. The second method which is short and elegant, we shall expound in Section 22. It belongs to Leichtweiss and is based on Cartan's method of external forms. As we already know, the Plucker coordinates p'" in GA,, A satisfy equations 11

pl,

E (p"

;l: pi, III = p.

P

) 2= 1.

(1)

11

But not all of them are independent of another. We are going to find a minimal subsystem of the above equations which is sufficient to represent the immersion. The manifold G4,, , A is imbedded into the Euclidean space E t of dimension A'= C,4, A. Since the dimension of GA ,,,A is equal to kn, to represent the immersion it

is necessary and sufficient to make a system of

lit =Ciit -kit

(2)

independent equations. Consider a point Pi, of Pliicker coordinate p" A = 1. while other coordinates are zero. Represent the immersion with the following system of equations The first of them is (3)

(ii + 1... . n + k). Consider the

Introduce two sets of integers: 11 = (1..... it), equation of the form , Pn

P,,

I i A =0,

(4)

where it..... i A E li . Next, consider the equations f o r superscripts it , .... iA _ I E and it E12:

pnh

ip"

Si

11

(5)

Repeating this process. we obtain n

i..,,

pn

1,

p

I,,

., p .

,r

I

,,,.r:p"hi

Si

A = (),

A ={).

(6)

In the latter equation ii .... , IA 2 E !, and only it, is E 11. The total number Al of equations in (3)-(6) is

A1=&+kC'A I+...+(A'CA +...+CA

(7)

where CI are binomial coefficients. The following formula is known (see [22]. problem 59,b p. 28): A

A

C i,

+

A

CACrI

If ,- {,

(8)

GRASSMANN IMAGE OF A SUBM ANIFOLD

237

where C,',' = C, = 1 is assumed. The right-hand side expression in (7) differs from the kn. Hence, the total number of right-hand side expression in (8) only by equations (3) -(6) is

Af = C44, - kit.

which coincides with (2). Thos, an immersion of GG,,,.A can he represented with

(3)-(6). Each equation from (4)-(6) is determined by a generalized index Denote by F11, the quadratic expressions on the righthand sides of (4)-(6). Then (4)-(6) can be represented as PD =

F,1, = 0. (9)

F,,,,=0. Denote by t,,, the unit normals of a submanifold. To simplify the notations we omit c} in subsequent expressions. The unit normals can be found in terms of gradients as ,, = grad F,,/grad F,,I. Prove that at P these vectors are mutually orthogonal. To do this we represent grad F,, as the vector in E I with coordinates 1/1 ' . Introduce generalized indices

u=(il...f

ii...!A 1 it+13)

t=(i&-, n+2...it+k), I _ (n+ l ...it+k). Let e,,,, be an odd or even number, depending on the c and h sets. Then F,, can be represented as

E(-1)'"'p'v=Pop- +(-1) rIfp"+... Represent the vectors p and grad F,, as

p= {...P"...e...p°... grad F,, _ {... p".. - (-1):- pr... At Pu all of the PlOcker coordinates except p° are zero. Hence, at Pi,

gradF,, _ {...p' ...} _ {...1...}, where the position of 1 corresponds to the position of p. At P11 then IgradF,,I = 1.

For different F the I takes different positions. Therefore, at P the gradients are mutually orthogonal and orthogonal to p. Now find the coefficients of the second fundamental forms with respect to C,, at P0. We have

f 0-.p L'r = 18u,8u, "1

- r7u;

8111)

-8

8u,

grad F,,I .

THE GEOMETRY OF SUBMANIFOLDS

238

The structure of the components of p and grad F, imply

L Lemma

(ap app+(_ly au; aui

ap°apb+apbaps+ap,a.-

au; au;

au1 au;

au; au1)

(10)

If the Plucker superscripts it ... ik contain two or more elements of 11

then at Pa ap;,...;A

au;

=0.

Let us represent p as a multivector [a I.... ak], where a; are En" vector fields which depend on the coordinates u1i ... , ukn in a Grassmann manifold. Let e1, ... , en, en+1,... , e"+k be the orthonormal basis in E"+k We suppose that if the values of parameters uI,... , ukn correspond to P0 then a; = en,;, i = I,. .. , k are the vector fields which correspond to u1i... , ukn. Then for p = [a1,. .. , ak] we have the Plucker coordinate p1 = 1. Further, 5wi

=L[ R_1

.k

The Plucker coordinate °1y of -,° is equal to the sum of the Plucker coordinates of

t,

ak] . It is sufficient to prove that the corresponding Plucker multivectors [a1 coordinates are zero for them. We shall denote corresponding Plucker coordinate by superscripts in brackets. Let be Cartesian components of aR. They are functions

of uI,... ,ukn. We have S;1l

...

elk

as[al...1...ak]41*=L ...

Sk

_ k = 0. Therefore in (11) there are two columns of all zeros except 0(n /au; in the first and aQ; /au; in the second. These columns are proportional. So, the corresponding Plucker coordinate of [al oul° ak] is zero. The lemma is proved. Suppose that i 1 i i2 E 11. Then at Po are C',' = fk =

Consider all equations of (9) for a < 2. Their left-hand sides consist of products of p°pb. One of the generalized indices a, 6 contains two or more elements of 11. Hence,

the second fundamental forms with respect to p are zero by virtue of the lemma. It remains to consider equations of the form pi,...i*_2iili=pn+i...n+kl = 0.

(12)

Superscripts j1 ... A-2 belong to the set of n + 1, ... , n + k. Suppose that among ./l, ... ,jk_2 there are no n + a, n +)3. By permutations of superscripts, represent the

GRASSMANN IMAGE OF A SUBMANIFOLD

239

Plucker coordinate p J' Jk-2'1'2 in such a form that i, and i2 take the a and 0 positions R U

We can represent equation (12) as

respectively:

pn+I...i,...ti2In+,9+I...n+kIpn+1...n+kt = 0,

(13)

where the commutation is taken over i2i n + I,. .. , n +k while superscripts in I I stay immovable. Introduce the generalized superscripts, each of which consists of the k usual superscripts: (I

l+

p= (n+ I ...i, ...i2...n+k), ry = (n+ ...n+a...n+Q...n+k), a, = (n+ ... it ...n+a...n+k), p, _ (n+ 1 ...i2...n+,O...n+k), a2 = (n+ l ...i, ...n+,0...n+k), p2 = (n+ 1 ...n+or ...i2...n+k). In addition, introduce two permutations of (k + 1) elements of the form

... n+a ... n+a n+ 1 n+2 ... i2 ... n+ I n+2 ... n+a ... A2 _ 1n+,0 n+ I n+2 ... n+a ... i2

A

i

=

n+ i n+2

12

... n+k n+/3 ... n+kJ n+/3

,

n+/3 ... n+k

... n+k)-

i2

Let e(A,) be the Kronecker symbol of A,. In terms of generalized superscripts, equation (13) can be represented in a brief form: pPp7 + (-1)`l\')p°'pP' + (_1)t

2)p°2pP2

= 0.

In a, interchange the places of it and n + a, i.e. put i, in the f-th position and n + a in the a-th position. Set a3 = (n + 1... n + a ... i, ... n + k). Then 2K1 Out' .Note au, that the parity of A, and A2 is the same. Indeed, A, is obtained from A2 by two interchanges: first, interchange the places of n + /3 and n + a; then interchange the places of n +,0 and i2. Thus, e(A,) = e(A2). So, PPP" + (-1)E(A') (_p°, pv- + p°2 pot) = 0.

We find the coefficients of the second fundamental form with respect to {P by (10):

Lj=(-1)e(a')[-Cam'a

au, 8uj

'+8p°,-app,

atlj aui

)+( &i l7uj +aanj

r

)J

The vectors Op/&i and Op/&j tangent to the Grassmann manifold at Po are represented in terms of matrices X = Y = IIY°11 of k columns and n rows. The element of X in the a-th column and i,-th row is n+k

aui

THE GEOMETRY OF SIiBMANIFOLDS

240

We have analogous expressions for elements of Y. We can represent coefficients L° as

p= 3,. y;2 + Y"; ll

,9

xa

;i -

<*

'

r;

where p = (n + 1 ... ii ... i2 ... n + k).

8.19 Leichtweiss-Wong Formula for the Curvature of a Grassmann Manifold

Leichtweiss and then Wong found a convenient formula for the sectional curvature of a Grassmann manifold GA,,, {.k. Let X, Y be tangent vectors represented in terms of Y) as matrices of n rows and k columns. Introduce the matrices A =

A1=XY"-YX', A2=X'Y-Y'X, where * means transposition. Then the curvature k(X, Y) of a Grassmann manifold can be found by the Leichtweiss-Wong formula: IC(X, Y) =

2Tr[A1A, +A2A2]

4Tr(XX')Tr(YY') - [Tr(XY' + YX')]2

(1)

Set for definiteness that k < n. To prove (1) we use the fact that one of the matrices, say X, can be transformed into a standard form: 1 xl

10

0 x

0 0

X= 0

...

0

xk

0

...

...

0

Denote by a,, , bp, the elements of A1, A2 respectively. It is easy to find that apq = xn y p - x q y ,

a

=-xgyr,

for p, q < k,

for pk,

a y = xnyy,

for p > k,q < k,

apq = 0,

for p, q > k.

The elements of the k-order matrix A2 are ft>(xfy9 - x7yf) = x,yo - xqy'°. bR4 =1=1

GRASSMANN IMAGE OF A SUBMANIFOLD

241

The numerator in (1) can be represented as

Tr[AIA +A2A;] =

(XPi'vP -xq:P)v` A (2)

R

+

(x r - xa rP)-.

(xglr )`' + 1_4_u.d. P=n

This formula has been cited in [24) without proof. But it was remarked that it was derived with the help of a specific matrix form of a system of equations of geodesics. We shall obtain this formula in another way using the Gauss equation of immersion into a Euclidean space E -v. The curvature tensor of GA +6 can be expressed in terms of coefficients of the second fundamental forms of the immersion. The second fundamental form with respect to normal p coincides with the first fundamental form. Let Lf; be the coefficients of the second fundamental form with respect to 4,,. Let X and I be the tangent vectors to coordinate curves it, and it, respectively. In the previous section we saw that "

,

of,,

Take into account that matrix X has the standard form. Then

it-` n v

rv

for p.q
L;' = O.

LJ = xpiq.

q > k. q > k.

for p <_ k.

Let A,,,, be the components of a Riemann tensor of GA .+k. Then

Rvv=Pa,pii -(pu'ur)2+Ls-(Lj) I,

pupa,+4'VP PV

P-4 q'.A

Substituting the expression for p;,,..., (pupa). we obtain a

a p

,=1 y=1

-2

vq}gxPl p `-1, q. A

y=1

1,

4 1 q- A

THE GEOMETRY OF SUBMANIFOLDS

242

-4

E

XPXg

Pgygyp9 P

- F(xPpyqg)2 - E(YP"Pegq)2 Pq

-2

(x,)2(yq)2

E xPvgyPxq p.q

pk

[(Xn yq - xgyq )2 + (xppyp - Xq yq)21 +

(Xpy9)2. i#q. p
p, q
The latter expression coincides with the right-hand side of (2). Therefore IZi1ij = Tr(A1A j +A2A:).

Dividing both sides by p pug - (per,

we obtain the formula (1).

8.20 Boundaries for Variation of Curvature of Grassmann Manifold In this section we find point-wise boundaries of curvature of a Grassmann manifold when it is not a sphere.

The curvature of a Grassmann manifold Gnr+k for n, k 0 1

Theorem (Wong) varies in [0, 2).

From (2) of the previous section it follows that k > 0. Moreover, there exists a 2-plane with K = 0. Indeed, consider a 2-plane of vectors X

0 - 0

1

0

0

0

0

0

0

0

1

0

0

0

0

0

0

0

...

Y

Then A,=A2=0andK=O. Show that the upper boundary is 2. Consider, first, G2,4. In this case the tangent vectors are represented by 2-d order matrices X, Y. One of them, say X, can be taken in a diagonal form. So,

X=

1A,

Ol

0

- y v11. _

Yi

Then

A, _

0 _a12

a12

0

),

0

A2 = 1-b12

h12

0

where a12 = alY2 - A ', b12 = alyi - A2W. Denote by A the numerator in the formula for K. Then

A=2Tr(AIA,+A2A;)=a212 +b12 _ (AIY2 - A2Y1)2 + (Aiy1 - )2y2)2.

GRASSMANN IMAGE OF A SUBMANIFOLD

243

Consider B = Tr (XX') Tr (YY*) - (Tr(XY'))2. We have XX`

(Y

-(0

fix)2=

0),

2

(AI 0

A2

1(y11)2

y2/ =

(A

Y,

0

A,

(i,2)2

/ y; Y

YH

(V21 )2 +

0

(A,

XY`

vi

all

0

*

+(Y2)2

A*i

y2)=1

2

A2Y2

Y1

Therefore, (Y22)2]

B = (A2 + A) [(Y1)2 + (Y2)2 + (),1)2 +

)2

- (A,y

So, we have k = A/B. Consider the difference:

2B-A= 2(A +a2)[(Y:)2+(Y2)2+(Yi)2+(Y2)2] - (A1Yi

+,\.)-Y 2)2

z

- (Aiy2

- ayi

2)2

- (auyi -

A2Li,1)x + Jai(Y1)x + 2ai(y; )2 + a2(Yi )'- + A2(y,)2 + 2A (Yi)2 + 2A1A2yzyi + 2A A2y2Y2' - 4A,)xyiy, 2

(1)

= 2(a,y2 - A2Y,)- + (Alyx + A2y1)2 + (,\,y1 + A2Y2)2 > 0

Thus, we see that k < 2. It is easy to rind a 2-plane with K = 2. Set, for instance,

/0

X= (

1 f,

\\is

Y=

(1

0l).

Therefore.

Then the right-hand side in (1) zero. K = 2 with respect to this 2-plane. Turn now to the general case. Represent X in a standard form. Then

X =

Al

0 . 0

0

...

0

A2 ...

Ak

0

0 Y1' ...

0 ...

0

...

0

Yk

...

Y1

Y;ii ...

Y1

...

...

... ... Yk 4 , 1 ...

..

.

The matrix A, = XY' - YX' is of order k and has the following form:

r 0

A2yk - AkY2 I.

... y"

THE GEOMETRY OF SUBMANIFOLDS

244

Therefore, Z TrAIA, = F, I<;<j
A2 =

AIYI

AIYI - A2Y2

0

A2YI - AIyi

-

AkYk

A2Yz -

+1

...

AIYi

A2Yz+1

...

A, Y,

AIyi

AkYk

0

-A1Y

If p, q > k then Cp, = 0. So, k

n

TrA2A; = E (A ;Y; - AIy )2 + 2

A; (Y; )2. ;=1 j=k+I

1
Denote by A, as above, the numerator in the Leichtweiss-Wong formula. Then k

A=

{(x14-A;Y;)'-+(A;Y -A;Yj)2}+ I<;<j
n

A?(y;)'-. ;=1 j=kit

Note that in all terms above, the subscript i coincides with either p or q. It is easy to find the denominator: k

n

B=(Ai+...+A2k) E E(y,!)'

-(AIYi+...+AkYJ )2.

i=1 j=1

Consider 2B - A. Split the sum Ej(y;i)2 in the B expression into two parts of j < k and j > k + 1. The second part annihilates the third term in the A expression. Expand the parenthesis in the A expression and then annihilate ('o' )2, (AjyJ)2 (A;y;')', (Ajy1')2 with the corresponding terms in the 2B expression. After that the expressions of the form A?Q)- and A (y; )'- remain in 2B - A which we arrange into groups with 2A;A. y/ from A. In the B expression expand the second term - (z; A;y )2. Then A?(y)2 is annihilate of together with the corresponding terms in the first summand while the products 2A;Ajy yJ arrange themselves into groups with (A;yJ)'-, i 96 j from the first summand. Thus we get n

k

2B-A=

if

k

it

Ai(Yi)2+2 j=1 1=1 j=Ail

t=1 j=k+1

+2

{A (Yj )2

A(Yi)2

2AiAjY;)'j + Aj (Y)p }


2 + 2A;Ajyjy; + A?(Y`)-}

+ I
+

{Ai(Yi)2 + 2AiAjyjy; + Ai (Y;)2} +2

i(y!)2.

1.1,1=I i.i.j#

GRASSMANN IMAGE OF A SUBMANIFOLD

245

The expressions in braces are perfect squares. Hence we can write k

2B-A=

k

re

n

A2(,i)2+2E E Ai(Yi)2 i.l=11=k+I

i=1 j=k f 1

E

k

+ 2 1: A (y;)Z + 2

(),1j';

I
i. j.1=1

+ 1:

- Ajy')2

{(A;Y+A,Y)2+(Aiyi+Aj4)2}>0.

I
Therefore, k < 2. The upper bound 2 is involved. If we take X and Y in the form X

1

0

0

I

0

0.0

0

0

... ... ...

Y-

0

0

-1

-1

0

0

0 0

0

... ... ...

0

0

then the following system will be satisfied

AiY/ =0, j>k+l, 1 <1
Ai. l - Aj),, = 0,

1,j, i < k,

i,j, i #,

Ai)' + AjY,! = 0,

A, y// + A,) j = 0,

which guarantees k = 2. The theorem just proved was formulated in Wong's paper [24].

8.21 Curvature of a Grassmann Manifold with Respect to 2-Planes Tangent to a Grassmann Image of a Submanifold. Classification of Submanifolds

Consider the n-dimensional submanifold V" in Euclidean space El". Let r = r(u1i... , u") be its position vector. Let be the orthonormal basis of normals on V" and ri 1 < i < n, the tangent basis. We can assume that at the fixed point xo all of the vectors above are mutually orthogonal. The unit basis vectors in E"+"' can be chosen in such a way that ri = giiei, i < n, = e"+, at :co. The multivector p of normals is represented in the form

p=16...G'J The correspondence x - p defines the Grassmann mapping. The image is some set of points, in general a submanifold, in G.,",.,", which we denote by r. By Grassmann mapping, the coordinates in V" parametrize 1'" as well, i.e. the multivector p is a function of these coordinates. The tangent basis of r" form m

49 n=1

4

m

I...a , =-ELkigij r=1

THE GEOMETRY OF SUBMANIFOLDS

246

Here we used the Weingarten decomposition for Op/auk. To use the LeichtweissWong formula we need to represent this basis in matrix form. As we stated, the derivatives of components of multivector p are zeroes except those for which only one of the superscripts I i , . . . , im belongs to the 11, n) set, while the others belong to the [ n + 1, n + m) set. We have apii..;,,

auk

m

-

-

p,

Cf4 g iiS1

...Ti

p=1

Suppose that among n + 1, ... , n + m the number n + p is missing. Set l is one of p I, ... , n. Introduce a generalized superscript j(1, p) = (n + 1, ... ,1, ... , n + m). We have t[ni l

p I(l.p) = b

... ... ... ... ... Tl ...

...

..

TI J

tn+ml m

= rJ n+I ,

I

Fn+l m

.

.

.

.

.

.

.

.

.

+m

... T/n+m

to+m m

Therefore

api(l,P)

p

Lkl

allk

I

gll

Represent the tangent basis ap/auk of rn as matrices Xk of m columns and n rows. In the p-th column and 1-th row position in the matrix Xk is the element api(4)) auk

_

LPI

Lkl SIl

Hence L" I

K,,

Find the formula for curvature of G0,,,,+m, with respect to any 2-plane of vectors X, Y

tangent to the Grassmann image. Set X = a'X;, Y = b'X,, q'' = a'bi - aib', A,,;i = Xi). Denote the elements of A1ij and A2;; by a7' and bid,, respectively. We have

Al (X,Y)=XY'-YX'=(a'bi-b'ai)X;Y1 = 2A,(X1,Xi)q''. It is easy to find the elements of using the expression of X; in terms of the second fundamental forms of a submanifold:

a1 _

n

1=1

LIP Litl

1,

gll

A4

L1.Lli

b;Jlrs = P=1

grrgxs

GRASSMANN IMAGE OF A SUBMANIFOLD

247

Consider the numerator of the k formula: I

2

n=1

(a,'

2

(bijlra

p. y=1

r. s=1

Substitute at'y and bijlrc with their expressions. Using the identity Tr(X;Xj') = Tr(XjX, ), find the denominator:

> (L LL q")2__

4Tr(XX')Tr(YY') - [Tr(XY' + YX')]2 = 2

grrgss .

r. s= I p, q=1

Therefore, the curvature k of a Grassmann manifold Gm.n+m with respect to a 2 -plane of vectors X, Y tangent to the Grassmann image of a submanifold V" is of the form

r

y E; K(X' '/-

P

=1

2

m

n

(F1=1 L LjjE) +Er.s=1 cn

9

P

P

P

(Ep=1

ij2

j2

1

1i

1)

I

EP-' 4[.r.=1 `p4=1(Lirl'jsq) 9.9-

This formula has been obtained in [9] of Chapter 10. A simple formula holds in the case of a Grassmann image of a submanifold V" with n principal directions. In this case LPjj = 0 for i 36 l and hence the first term in the

numerator of (1) is zero. The remaining expression can be simplified. So, we get ,,rs 2

(Ep n

En

m

=1

2

Ep.y=1(Ln L ) (qrs)

<< 1.

2

(2)

R BRn

From the latter inequality it follows: Theorem Let V" C E"+" be regular submanifolds with n principal directions and a regular Grassmann image r". Then the curvature k of Gn.n+m with respect to 2 -plane tangent to rn takes values in 10, 1].

We already considered the classification of surfaces in E° in terms of Grassmann image. We can use the same approach to classify many-dimensional submanifolds. In [3] we proposed the following classification of types of Grassmann image: if for any 2-plane tangent to t" the curvature k of a Grassmann manifold is (1) K < 1; (2) K = 1; (3) K > 1 then we say that a Grassmann image type is (1) hyperbolic; (2) parabolic; (3) elliptic. Moreover, if there exist 2-planes of k > I and K = I or K < 1 and K = 1 then we say that the Grassmann image type is mixed. The latter case can be separated into three subcases:

(a) hyper-parabolic, if k < 1 for any 2-plane; (b) elliptic-parabolic, if K > 1 for any 2-plane; (c) strictly mixed type, if there exists a 2-plane of k < 1 and a 2-plane of K > 1.

THr C,FUhiFTRi Ul SUONIAN11-OLIAS

248

In papers of Borisenko [28] and Borisenko, Nikolaevsky [271. [29] another more detailed classification was proposed. Borisenko proposed the following hypothesis. Let Ft c Er' i" with I '> 3. p ? I he a regular of class C' suhmanifiild with a non-degenerate Grascnuunt image. Then at any point of a Grassmamr image there exists it tangent 2-plane of curvature 9:5 1: if at some point k ? I for all tangent 2-planes then for an), 2-plane K = I and p = I at that point. Nikolaevsky [26] and Savel'ev [37] proved this hypothesis for all hp except some finite (not very large) set of i.,p values.

Muto [18] considered the case when, for all 2-planes tangent to a Grassmann image, the curvature k takes its minimal value k = 0. It is possible if and only if Ft is a flat submanifold with flat normal connection in Et' P, and in addition p > I. The case when K =- 2 was considered by Borisenko and Nikolaevsky [2S]. It was

proved that the surface is necessarily two-dimensional (i.e 1= 2) and minimal. Moreover, the normal curvature ellipse at any point is a circle with the center in the surface. In the, particular case of p = 2. the surface is a complex curve in Ea. Sheffel' [38] considered the classification of points on submanifolds with the help of the stability property with respect to some group of space transformations. He called this group geometric if it includes the similarity group as a subgroup, Only groups of affine and conformal transformations are geometrical groups preserving the local geometrical properties of submanifolds [39]. The dimension of space of the second quadratic forms of F" c E"' at the point P E F" one calls the point-wise codimension of the submanifold at this point. Borisenko [28] found all cases of pairs (dimension, point codimension) with a finite number of affine equivalent classes. Theorem Let F" C E' be it regular it-dimensional submanifold in in-diinenrional Eucidean space. Then it is possible to allocate the points of the submentlfold among the ,finite number of equivalent classes only in the cases:

(I) for taro-clbnensional submanifolds iritit arbitrary point-it'ise cochmens:on: (2) for htpersurfaces F" C E"' and for the case rritlr the point-wise codinw;is un

p (3) .for submanifolds F" C E"' with the minimal point-wise codimension p = 0 and with the maximal pouu-wise codimension p = ""'2 1 1

(4) for three-diniewsional submanifolds F3 C E" with the paint-wise codimension

p-2 or 4. Thus, in the general case there is the infinite number of affine equivalent classes.

8.22 The Leichtweiss Way to Find the Curvature Tensor of a Grassmann Manifold

In this section we represent a simpler way to find the curvature tensor of the Grassmann manifold G,, "Irk proposed by Leichtweiss. It involves the external forms method of Cartan and structural equations of Riemannian space.

GRASSMANN IMAGE Of A SUBMANIFOLI?

249

Let {e,} be the orthonormal basis in the n-dimensional plane E" C E"'4. Let be the orthonormal basis in the k-dimensional plane Ex C E"" which is perpendicular to Leichtweiss obtained the expression for the metric of a Grassmann manifold in the following form: E".

"

d

1=1 ,.=1

As independent coordinates in a Grassmann manifold we take the elements of matrix Z (see Section 12). Denote by it,,, the elements of Z. The pair (ki) is considered below as a single generalized subscript - the symbol which distinguishes the coordinates. So. the metric (1) is represented in the form 11

do,'

_E

4

g11n11)it(ua-)do,,, du,,.

1/-In 1-1

where the metric coefficients are Slrnltltl

c9tt,

i

e-

att,

We can write the differentials of the basis vector fields e, and e as

dr,=Fdi,, "

I.

ae,

-1 i=1

IOU,

ae,.

de

r-t -1Ou,t Introduce three kinds of linear forms with a single generalized subscript in parentheses as W 1".1 = (en de1)

FF r-I

aeau, _(e,.

t=1

\ du, s. J

;

wu,1 = (e,de1)

wp. i) =

Then the metric of a Grassmann manifold can be represented as "

c

do- _

(wtlnt )-.

,=I n=1

which As Chern proved, there exists the set of two-indexed Pfaff forms satisfy WI,t( ,, f) = -wtI such that the external differential of w,,, I is of the form n

dWtn11 =

k

EW11n1(,4) Awt(2)

/_) .i=1

THE GEOMETRY OF SUBMANIFOLDS

250

where n means the external product of forms. On the other hand, for the external differential of new forms w(in)(Jii) we have k

n

l

k

[om

dw(in)(J;i) +

L W(in)(h) A w(h)(Jd) = 2 1=1 7=1

Q(in)(J:i)U7)(hs) (w(h) A W(h6)),

(3)

h,l=I -),6=I

where the four-tensor Q(in)(Jri)(17)(he) is the Riemannian curvature tensor if at a fixed

point of the manifold we take w(in) = dui,,. The equality (3) allows us to obtain the formulas for the curvature tensor. The external differential of w(in) is the form of the second order and can be transformed to

-/

k

n

(i9e n8ei(in)

8tq, C7tlj0

J,1=1 ii,ry=1

du, n dujd

= (den A dei).

(4)

Represent differentials dei and den as a linear combination of eJ and e,i: n

k

dei = E w(.J)eJ + L w(is)ed,

(5)

d=I

J=1 n

k

den = - F,w(J,)e) + W(nd)ed

(6)

:i=I

J=1

Substituting (5) and (6) into (4), we get k

n

dw(fn) = EW(M) Aw(,J) + LW(n;i) Aw(ii) :i=1

J=1 n

k

_ E E(W(iu)bn,i + biJw(na)) AW(J;i). J=1 d=I

Comparing the right-hand sides of the above and (2) we find the expression of W(in)(J;J) in terms of a single generalized subscript: (W(il)bo,i + bijW(n,i))

W(in)(Ja)

Set A to be the left-hand side of (3). Then n

A = d&(i(j)(J/i) +

k

W(in1(11) A 1=1

=I k

_ (de1 A deJ)bnd + (den A deq)biJ + E 1=1 7=I

+ biiw(n7)) A (w(,J)b7ri + biJw( L) ).

GRASSMANN IMAGE OF A SUBMANIFOLD

251

Replace dei, ... , de;, with their expressions from (5), (6). We obtain

+

A

`/=1

+ 1=1

n

+

w(i,)e,, >W(jr)e, +

w(j,)e,

I

k

n

k

=1

r=1

=1

bij

k

/ E(W(il)

AW(lj)bilb7;3

1=1 ,=1

+W(ie) AW(,:3)bn,bej +W(n7) Aw(,3)b/ebej

(r,

k

n

W(n)

=1

AW(n))& +bij

,=1 n

w(rn) Aw(l,i) l=1

k

_

AW(hb).

(bilbjkb,bbn;3 + 1,h=1 i 6=1

Thus, at a fixed point the Riemannian tensor components have the form R(i(k)(j 3)(h)(hb) = (bilbjh - bihbjl)bnub,b + 6jblh(bn,6 4 - b06b;37)

Find the sectional curvature of a Grassmann manifold for the 2-plane generated by a = 11X1011 and b = IIYi1I Note that at our point w(i0) = duin and as a consequence 8(in)(j;i) = bijb0;,. Transform the numerator and denominator of the sectional curvature formula separately. Split the numerator into two summands. The first is k

n

E(bi/bjh - bihbjl)bn/36,bXi,,Yj.-iXhYhb =

- XjnYin) E ((xinYin n=I

,

1,1=1

and the second k

L.

bijblh(bn,b;ib - 6,, &tv)XicYj;JXl,Yhb =

(n 1

2

DXinyi;3 - Xi/3Yin)) i=1

It is easy to find the expression for the denominator as n

(8(in)(h)$(j, 00.1) - $(in)(hb)g(h)(j;J)

k

(Xinyji - XjiiYin)

inY%liXhYhb =

i.j=10.8=I

Thus, the curvature of a Grassmann manifold with respect to a 2 -plane of vectors a and b has the form z

k(a, b)

_

v"j=1

(E0=1(XiOYjn

+En.13=1 ( 1(Xinyi/i - Xi3Yk+))2 - XjnYin)) .!c rij=1 Ln,/3=I (XiaYj;3 - Xj1 yin)2

THE GEOMETRY OF SUBMANIFOLDS

252

It also is easy to find that the Ricci tensor of the Grassmann manifold is proportional to the metric tensor: R(,3)(r,) = mg(;;;)(h,), where m = 2 - n - k. So, the Grassmann manifold is the Einstein one. We remark here that the case n = 2, k = 2 is exceptional. On the manifold G2,4 there exists the whole family of invariant metrics d[r2

=

'',,CC

\[(171dd1)2 + (,)I A2)2 + (Y2d 1)2 + (1)2&2)2J

+2)c4(mdd1)(ihdf2) - t]1dy2)lfidS1)1,

where A, it be real constants and A > 1pl.

So. I state that there Ls no another solid, besides five mentioned solids, mhich is contained between equilateral and equiangular polygons equal to each other. Euclid, "Origins

9 Regular Polyhedra in E4 and EN There are rive regular polyhedra in E3. In Euclid's "Origins" the 13th book is devoted to the theory of regular polyhedra. which are often called Platonic solids because in a famous dialog "Thimey" the four of them were described: tetrahedron, icosahedron, octahedron and cube. But there is the background to suppose that the

dodecahedron was unknown to Plato at that time. Morduchaj-Boltowski wrote about this in his comments to Euclid's 13th book. It seems that the dodecahedron was discovered considerably later than the other four regular solids. The science historians (see, for instance, Zmud' "Pythagoras and his school") relate the discovery of the dodecahedron by Hyppas, the Pythagorean, and by the octahedron and the icosahedron by Thiatet. It is possible to construct the regular polyhedra in E4. A polyhedron in E4 is called regular if all its faces of dimension 1,2 and 3 are congruent to each other and the neighborhoods of each vertex are congruent to each other. Their total number is six. Some of them have very complicated structures. These polyhedra are called:

simplex, cube, 16-hedron, 24-hedron, 120-hedron, 600-hedron.

All of them were discovered around 1850 [41 by the Swiss mathematician, physicist, astronomer, and professor of Bern University L. Schlafli (1814-1895). For a long time his paper on regular polyhedra was forgotten and the discovery was attributed to another scientist. Schlafli's mathematical gift was apparent early in his school years. Although he studied in the theological department of Bern University since 1829, he aspired to

study science. From 1837, he taught mathematics and physics in Thun. For 10 years he was a successful teacher, but all his free time was devoted to the fundamental study of high-level mathematics. He also studied botany and was particularly interested in the positions of the leaves on a stem. He tried to relate 253

254

THE GEOMETRY OF SUBMANIFOLDS

LUDWIG SCHLAFLI 1814-1895

this to mathematical law of the Fibonacci series. At this time he met J. Steiner who

worked in Berlin and Steiner recommended the young Schlafli to the German scientists Jacobi and Dirichlet as a translator for their trip to Italy during the winter of 1843-1844. Schlafli was 29 years old and knew the Italian language, and also had brilliant mathematical skills and abilities. He benefited greately from these

journeys with the famous mathematicians. Dirichlet educated him in number theory and Schlafli's later paper on quadratic forms originated from these discussions. Also, he translated two of Steiner's papers and two of Jacobi's papers into Italian and during these journeys Schlafli received many stimulating ideas for his subsequent papers. In 1848 he moved to Bern to work in the department of mathematics in Bern University as an associate professor. He wanted to devote himself to mathematics

but this opportunity proved very expensive because of famine that year. In his memoirs he wrote "...limited by salary of only 400 francs I had to live in relative poverty. However, I went through with this experience with joy. because I had all-helping science". His supernumerary professor assignment in 1853 was not

REGULAR POLYHEDRONS IN E4 AND EN

255

helpful and between 1854 and 1860 he had to earn a salary by performing calculations for the Swiss national insurance office. At the same time (1850) Schldfli discovered the regular polyhedra in E4 and

E". Also, he successfully developed some problems of the Italic geometrical school. He stated the conditions on Riemannian manifold to be of constant curvature: the geodesics in it with respect to an appropriate system of coordinates must be of straight line form. The assertion that an analytic Riemannian space

of dimension n allows the local analytic immersion into Euclidean space of dimension "ice is known as Schlaf is theorem. However, the strict proof of this theorem (rather hypothesis) was given later by Janet and Burstain. SchlAtli also obtained the formula for the volume of a simplex in Lobachevski space.

From 1856 onwards Schlaf i corresponded with the English mathematician A. Cayley. That correspondence lasted, with some breaks, up to 1871 and played an important role for the theory of surfaces of third order, as well as for other branches of mathematics. Schlalli's 70th birthday jubilee was celebrated at Bern University and colleagues from Zurich wrote to him: "This century distinguished due to life of the incomparable Euler, and no other Swiss master of mathematics has such a broad knowledge as you. Now you are known to contemporary mathematicians of all cantons." In 1870 Schldfli received the Steiner award for his geometrical achievements, the French institution presented him with Cauchy's papers, the Norwegians presented him with Abel's papers and the Berlin Academy presented him with Dirichlet's papers. For detailed information on his life see biography by J. J. Burchardt, Ludwig Schldf li, 1814-1895.

9.1 The Four-Dimensional Simplex and Cube

Consider, first, the two simplest polyhedra in E4: the simplex and the cube. Take in some three-dimensional hyperplane E3 a tetrahedron with edge of length a. Through its center draw the straight line I perpendicular to E3. If M is a point

in I then it is equidistant from all vertices of T3. Since the distance from 0 to vertices of T3 is less then a, there exists the point Mo in / such that if M tends in I to infinity then the distance from Mo to the vertices will be equal to a. The set of all points of intercepts joining Mo to every point of T3 forms the four-dimensional simplex.

Consider the four-dimensional cube. Define it as a set of points in E4 whose coordinates satisfy the inequalities: 0 < xi < 1, i = 1, ... , 4. The set of points for which one of the coordinates xi satisfies either r, = 0 or x, = I is called the face. The faces of xi = 0 and .r, = I are called opposite. Denote them by A, and B; respectively. The faces A, and B, have no intersections because they are in parallel hyperplanes. There are four A,-faces and four B,-faces because i takes only four values: 1,2,3,4. Therefore the four-dimensional cube is the 8-hedron. The cube vertex coordinates are formed with 0 and 1. Enumerate all of the vertices. First, consider the vertex with

256

THE GEOMETRY OF SUBMANIFOLDS

all coordinates zero: (0, 0, 0, 0); then all coordinates with one being 1, and so on. We obtain the set of vertices as (0000)

(1000)

(1100)

(1110)

(0100)

(1010)

(1101)

(0010)

(1001)

(1011)

(0001)

(0110)

(0111)

(1111)

(0101) (0011)

Thus, the cube has 16 vertices. In spite of the fact that this is a figure in the manydimensional space, the four-dimensional cube can be represented in a plane as follows. Take four straight lines from one point 01. These straight fines we use in the capacity of coordinate axes in E. Separate off equal intercepts in them. The cube edges are parallel to e1 i e2, e3. So, we obtain eight vertices. Then construct A3, A1, A2

and mark their vertices. Thus we get 15 vertices. The 16th vertex 016 can not be obtained by the procedure above because A, faces do not contain all of the cube vertices, namely (1, 1, 1, 1) does not belong to any of A,. The vertex 016 belongs to one of the B,-faces. To find it, construct a cube in the face x4 = 1. To do this, consider the straight lines passing through 09 in the e4-axis which are parallel to el, e2, e3. All of the B4-vertices are already marked except 016. Drawing the straight lines parallel to corresponding edges, we find 016. The cube A4 of vertices 01, 02, 03. 04, OS, Os, 06 07 is opposite to the cube B4 of vertices O9, Olo, 012, O11, 013, 015, 016. 014 Consider the intersection of faces A, and A. It is defined by the system of equa-

tions xi = 0, x, = 0 and inequalities 0 < xk < 1, 0 < x, < 1, where k and I are subscripts complementary to i and j. Therefore, the intersection of faces Ai and Al is the square. In an analogous way we find that the intersection of Ai and B, faces for i 54j is also a square. Thus, any two faces, except opposite ones, intersect each other by the square (see Figure 33). The intersection of three faces A, n A, n Ak or A; n B; n Bk or Bi n Bk n B,, where all subscripts Q k k are distinct, produces an intercept - the cube edge. How many

edges has the cube? How many vertices is each vertex of four-dimensional cube joined to? From OI there go out four edges. The same number of edges go out from any other vertex. The total number of vertices is 16. Therefore, the total number of edges is 16.4/2 = 32. The total number of two-dimensional faces is equal to the product of the total number of three-dimensional cubes and half the number of faces in each three-dimensional cube, that is 8.6/2 = 24. The boundary of a four-dimensional cube can be represented in another way - by its surface development into three-dimensional space. Consider, first, the development of the surface of the usual cube onto a plane. If we cut it appropriately and develop it onto a plane then we get a cross-like domain. Using the method of edge

gluing, we are able to make the cube surface by an inverse gluing process (see Figure 34).

REGULAR POLYHEDRONS IN E4 AND E" Of$

FIGURE 33

FIGURE 34

044

257

255

THE GFOAIF rRl OF SUHMANIF01 I>S

FI((tRE 31

We shall proceed in an analogous manner to get the three-dimensional surface development of a four-dimensional cube surface. Construct the cube A., in threedimensional space of basis e, . e2. ea. Glue to it three more cubes A,, i = 1.3.3. The AI-cube intersects A. by a two-dimensional face x, = 0. Therefore, the face P1RP:P4 is their common one. Glue A, to A.4 along this face. In an analogous way glue A, to the P, PKPP,-face and A, to the P, P5PSP.,-face. It is possible to pass from the A4-face to the opposite B3-face along any other face,

say B1. In the face chain A4B,B4 each two units intersect each other by a twodimensional face. The B,-face intersects A4 by the square PKPSP,,P7. Glue the cube B, to it. In an analogous way we glue B2 and B; cubes. Note that A, and B, are glued to A4 by the opposite two-dimensional faces. It remains to glue B4 and to show how

to identify free two-dimensional faces. B, and B4 have the common face x, = I. .r; = I which is opposite to the face of gluing B, to 44 because these two-dimensional faces are in parallel planes. Therefore, glue B., to B, by the P,P,aP1, P,2-face. From the development just obtained it is possible to get the four-dimensional cube surface by an ideal identifying process (see figure 35). Now we show how to identify the faces of the cubes A, and B,. If some edge is

common to the faces of two cubes then these faces ought to be identified. For

REGULAR POLYHEDRONS IN F.' AND E'

359

instance. POP-, is the common edge of faces P6PPjPIIPill and P,,P5PI3P1I. Thus, we identify them. Moreover, we see that the edge PIIIPI I is identified with PI.IPI t. Hence, the faces P14PI.P11PI6 and PIQPIIP17PIK ought to be identified, too. In the figure, correspondence between identified faces is marked with arrows.

Thus, all free cube faces are identified to each other. As a result, we obtain the closed three-dimensional manifold - the surface of a four-dimensional cube. From a topological point of view this manifold does not differ from the surface of a fourdimensional ball, i.e. it is homeomorphic to the three-dimensional sphere S3. 9.2 The Analog of an Octahedron: 16-Hedron

Now we proceed with consideration of the next regular polyhedron which is the analog of an octahedron. It is called the 16-hedron. To construct it, take in the space of x1, x+. x3 an octahedron with vertices Q1, Q2, .... Q,,. Suppose the octahedron edges are of length a. Through the octahedron symmetry center 0 draw the straight line parallel to the Y4 axis. Take two points Q7 and QII in this line which are symmetric with respect to the hyperplane x.I = 0. All the distances from Q7 to the octahedron vertices Q; are equal to each other. Since the distance from the octahedron symmetry center 0 to its vertices is a/vl < a, then there exists a point Q, such that Q7Q, is equal to a. The polyhedron just obtained with vertices QI .... , QK is regular. Its three-dimensional face is formed by the set of intercepts joining Q7 or QK to the two-dimensional face of an octahedron. For instance, one such face is QIQ2Q3Q7 - the tetrahedron. The total number of faces through Qi is equal to the number of octahedron faces. i.e. 8. Since every threedimensional face passes through either Q, or QK, the total number of three-dimen-

sional faces is equal to 8 .2 = 16. The number of edges which go out from each vertex of the 16-hedron is equal to the number of octahedron vertices, i.e. to 6. As each edge joins two vertices, the total number of edges is 6.8/2 = 24. Now find the number of two-dimensional faces. We can separate all such faces into three disjoint classes: those passing through Q7, those passing through Qs, those passing through neither Qr nor Q. The number of faces passing through Q7 is equal

to the number of octahedron edges, i.e. to 12. The two-dimensional face passing through neither Q7 nor QK is a two-dimensional octahedron face. There are eight of them. Thus, the total number of two-dimensional faces of a l6-hedron is equal to

12+ 12+8 = 32. For convex polyhedrons in E4 the following analog of the Euler formula is valid. Set PII is the tofu! number of rertices, P, Is the total number of edges, P3 and P3 are the total numbers qt tt o- and three-dimensional faces respectirelr. Then these numbers are related as

Po-PI +P3-P3=0 which is the particular case of the Poincar6 formula. Check it for a 16-hedron:

8-24+32-16=0.

260

THE GEOMETRY OF SUBMANIFOLDS

In an analogous way it is possible to construct the regular polyhedra in n-dimensional Euclidean space E": the simplex, the cube and the 2"-hedron. It happens that for n > 5 there are no other regular polyhedra. Consequently we shall need the concept of director polyhedron. Let A be the vertex of the regular polyhedron M 4. Consider the ends of all edges from A. They are

equidistant from A, symmetric with respect to an axis of symmetry through A and situated in a common three-dimensional plane. The intersection of this plane with M4 is also the regular polyhedron in E3. It is called the director polyhedron. We shall denote it by N3.

9.3 24-Hedron

Take as a director polyhedron the cube N3 in some three-dimensional space E3. Suppose that this E3 is defined by x4 = 0. Let QI,... , QR the cube vertices and a the edge length. Draw the straight line parallel to x4 through the center 0 of the cube N3. Take on this line a point Q9 in such a way that the distance of Q9 to the cube vertices would be equal to a. It is possible to do so because the half-length of the diagonal satisfies avr/2 < a. Draw the three-dimensional space through the twodimensional cube face, say QIQ2Q3Q4, and the point Q9. Construct in this space the octahedron with vertices QI, Q2, Q3, Q4 and the point Q9. Denote by Q1o the vertex which is opposite to Q9. The analogous octahedra we shall construct with respect to other cube faces. Thus, we get six octahedra with the common vertex Q9 and upper faces glued to each other. Let Q10, Q1 1, Q12, Q13, Q14, Q15 be the octahedron vertices

opposite to Q9. All of them are equidistant from Q9 and located in a hyperplane E? parallel to E3. The lower octahedron faces are not glued to each other. Reflect the figure just obtained symmetrically with respect to E, . Let Q16 be the point symmetrical to Q9. Denote by Q17, ... , Q74 the points which are symmetrical to the vertices of the cube N3. Thus, we obtain 24 points which are the vertices of the required polyhedron. In figure 36 we see the schematic picture of this intermediate construction step: we depicted E2 instead of E3 and the square instead of N3. We constructed 12 boundary octahedra. Some of the faces of these octahedra are left

free. Paste the holes with the auxiliary octahedrons. In figure 37 you can see the spatial picture of free octahedron faces. The pyramids QIQ2Q6Q5QI1 and Q6Q2Q3Q7QI5 form the lower parts of an octahedron. They have the common face Q6Q2. The faces QIIQIBQI7 and Q18Q17Q15 belong to unglued octahedron parts which arise after symmetry. Observe that Q1sQ17 is the edge of the cube which was obtained from N3 by symmetry. It is parallel to Q2Q6. loin Q6 to QIS and Q2 to Q17 with edges. The polyhedron QIIQ6Q2QI7QI8QI5 has the edge net of an octahedron. To prove that this is indeed an octahedron, we need to state that the distance between the straight lines Q2Q6 and Q17Q18 is equal to a. The latter is equivalent to the statement that the distance between Q6 and Q1s is equal to a. These points are symmetrical with respect to the plane El. Therefore the distance between them is equal to the distance between E3 and the plane symmetrical to it,

REGULAR POLYHEDRONS IN E4 AND

261

FIGURE 36

4

FIGURE 37

THE GEOMETRY OF SUBMANII.OLDS

262

Q9

FIGURE 38

that is to double the distance from QI I to E3. The distance from Q, I to E3 is equal to the distance from Q9 to E3 (see figure 38). Let h be this latter distance. Then

h=

a---4a---=2a

.

Thus, the distance between Q6 and Q18 is equal to a. The quadrangle Q6QISQI7Q2 is the square and the figure Q6QISQI7Q2QISQII is the octahedron. Since the cube has

12 edges, to paste the holes we need 12 octahedra. Thus, the boundary of a constructed polyhedron consists of 24 octahedrons. The number of vertices is 24. The number of edges is 8.24/2 = 96.

Another way can be given to construct the 24-hedron by pointing out all its vertices and making the convex hull. All the vertices of the 24-hedron can be obtained from the points (±a ± a00) by various coordinate permutations. Separate them into two subsets (a) and (3):

(±a ±a (a) (±a (±a

0

±a

0). 0),

0

0

±a),

0

(0 (/3) (0

(0

±a ±a ±a 0

0),

±a), ±a ±a). 0

From this we see that all vertices are located in the hyperplanes x, = a, x, = -a. Also, the first and last hyperplanes contain six vertices each and they form the (a)set. The hyperplane xI = 0 contains 12 vertices and they form the (i3)-set. Six vertices in the xI = a hyperplane form the boundary octahedron. The (Q)-set makes the vertex set of an icosahedron which forms something like a belt of a 24-hedron. Both of the ways described above produce the same figure. Let 0 be the center of symmetry of a 24-hedron and A be the center of some boundary octahedron, say

Q9Q2Q6Q7Q3QIS. Denote it by T1. The point A is the center of the square

REGULAR POLYHEDRONS IN E' AND E'

263

Q2Q6Q7Q3. Introduce the system of coordinates in E4 in such a way that the x1 axis coincides with OA and the origin coincides with O. Let N be the hyperplane containing T1. The straight line OA is perpendicular to N. Direct the other coordinate axes x2i x3, x4 along the three diagonals of the boundary octahedron T1. Then its

vertices obtain coordinates of the form (a ± a 0 0), (a 0 f a 0), (a 0 0 ± a). Let K be the square in a hyperplane EE' (see figure 38) obtained by projection of the square QIQ4QgQ5 into E2. The boundary octahedron T2, which is formed with Q16 and K, is the image of T, under parallel transport along OA. It can be shown that all of 12 other vertices of the 24-hedron are located in a hyperplane which is parallel to N and passes through O. 9.4 The Symbol and Theorem of Schlifli

Assign to each regular polyhedron P in E" the ordered set of numbers { r 1 , . . ., r,,_ 1). It is called the Schldfli symbol. The numbers for this set are defined inductively. The first number r, is the number of edges of a two-dimensional face P

I) is the Schlaf i symbol for director polyhedron TP. Remember that TP is regular too. Consider the Schlafli symbols for regular polyhedra in E3. They consist of two and { {r2, r 2, . . . ,

numbers {r,,r2}. For the cube we have r, = 2 and the triangle as the director polyhedron (polygon). Hence, the Schlifli symbol of the cube is {4, 3}. This symbol for the tetrahedron is, evidently, {3, 3}. For the octahedron we have r, = 3 and the pentagon as the director polygon. So, the Schlafli symbol is {3, 5}. The dodecahedron is dual to the icosahedron. Its faces are pentagons and the director polygon is a triangle. Therefore, the Schlgfli symbol is {5, 3). The Schli fli symbol for the regular polyhedron in E4 consists of three numbers {ri, r2, r3 }. Evidently, this is 13, 3, 3, } for the tetrahedron, {4, 3, 3} for the cube,

{3,3,4} for the 16-hedron and {3,4,3) for the 24-hedron. For the other regular polyhedra we calculate these symbols later. Let a be the edge length of a regular polyhedron P and r be the radius of the sphere circumscribed around P. Introduce the number p(P) = 4r,. If r"_, } is the Schla"fli symbol for P then the following formula holds

AM = 1 -

2

p(TP) '

(1)

where TP is the director polyhedron for P. Let 0 and O' be the centers of spheres of radii r and r' circumscribed around P and TP respectively. Let A be the vertex of P with AB and AC as the going out edges. Their length is a. The intercept BC is the edge of the director polyhedron. Denote its length by a' (see Figure 39). The points B, A and C are consequent vertices of the two-dimensional face of P. This face has r, sides. Let 0" and r" be the center and radius of the circumscribed circle around this regular polygon (see Figure 40). The

264

THE GEOMETRY OF SUBMANIFOLDS

FIGURE 39

FIGURE 40

angle BO"A is equal to 21r/ri. Therefore, by the cosine theorem a2 = 2r"222ri2 cos. Hence

a=2r"sin-.

(2)

Express a' in terms of a. From the triangle BO"C we see

a'=2r"sin

-. r, 21r

(3)

Therefore, taking into account (2) and (3) we get

a'=2acos-.

(4)

r,

Draw a plane through the points A, B and O. Then draw in it the circle of radius r with the center at O. Let AD be the diameter. Let W be the value of the angle BDA.

RLGULAR POLYHEDRONS IN E' AND E"

265

A

a FIGURE 41

Then the central angle BOA is equal to ABD we find a = 2r sin p. Then

(see Figure 41). From the right triangle

ca -

P(P) =

4r' =

sm- V.

(5)

The point O' is in intercept AO; it is also the projection of B into this intercept. From the right triangle BO'O we find r' = r sin tsp. Hence

r'=acosv.

(6)

Taking into account (4) and (6) we get

a'- 4a2 cos=, p(TP) = 4r,_ = 4u= cos

cos=yo'

(7l

The latter relation together with (5) gives (1).

Since p(P) does not change under a transformation like that, it is possible to represent it as a function of the Schlgfli symbol. So. (I) can be written as

P(rl,...,r,,.l)=1 -

cos' _

P(r,,...,r. I),

This formula allows one to pass from the function of the (n - 1) variable to the function of (n - 2) variables. In addition, we see that p(P) decreases steadily while rl increases. As r> > 3 , then cos=; >t. Write the inequality L COs,-,

d.

From this it follows that p(a .... r 1) > 1/4. The latter inequality allows us to analyze the set of system

IHE(i1:OMEIR1 Of SLIM NIIOLL)S

2t(3

We begin with consideration of the values of 1(P) for regular polygons. The Schlafli symbol consists of one number ri. So,

1(ri) = 4r, = sin- r . In more detailed form: p(3)3.

p(4)p(5)5 Sam

for nyb p(n)=sin'1r

4 From this it follows that only the triangle, the square or the pentagon can be taken as the director polygon for the regular polyhedra in V. For the case of the tetrahedron, the Schlafli symbol is J3.31. Then

p(3.3) = I _ cos=- _ I _ 1/4 3/4 MA)

-2

The Schlafli symbol for the cube is {4.3}. Therefore 1(4.3)=I-cos-=1-1/2

3/4

1(3)

=1

3'

For the case of the octahedron the Schlafli symbol is {3.4} and

p(3.4) = I -

cos-- ii = P(4)

I

- 1 /4 = 1

.

1/2

For the cases of the icosahedron and dodecahedron the Schlittli symbols are 13. 51 and {5.3} respectively. Hence

J)(3.5) = I -

cos' 1(5) cos'

P(5.3) = 1 -

p(3)

5-f 10

3 - f

I

< 4'

6

From the latter inequaht) it follows that the dodecahedron can not be taken as a director polyhedron for the regular polyhedron in E4 Consider the Schlafli symbols of regular polyhedra in E`l. They can be of the forms: (a) 4r1.3,3}. (h) {r,,4.3}, (c) {r1.3,4}. (d) {ri,3.5}. In case (a), rI takes values of 3.4 or 5 only. Indeed, if rI ? 6 then

3/4 1( 3,3)

2/3

9 =1-<0. i

This contradicts the positivity of p(P). Thus, only {3.3.3}. 14.3.3.1.15 3. 3 } arc the admissible Schlafli symbols in case (a). So.

p(3.3,3) = 1- (3,

1(3.3)

=1_1/4 =5. 2/3

REGULAR POLI HFDRONS IN L4 AND E%

p(4,3,3)=I_

cos-4

1/21.

p(3,3

2/3

- 7-3f P(5,3,3)=1-P(3,3)-

4

cos'-s

16

267

1

<

Therefore, the polyhedron of the {5.3, 3} symbol can not be taken as a director for the regular polyhedron in ES. In cases (bHd). r, takes the value of 3 only. Indeed. P(4,4,3)=1-cos=;

p(4,3)

1/3

p(43,4) = I - 3 = 1 - 1/2 =0. p(4,3,5)=I-cos',=1-

5 - 73

p(3, 5)

<0.

Thus, the Schllifli symbols of the regular polyhedron in E4 can be only of the forms

{3.3.3}. {4,3,3}. (5.3, 3}, {3,4,3}, {3,3,4}. {3.3.5}. So, there are no more that six regular polyhedra in E;. This conclusion is known as the Schllitli theorem. We constructed four of these six polyhedra above. Now we construct the other two. Consider p(P) for the last three symbols:

p(3.4,3) = 1 -

CQS,

p(4,3)

p(3.3,4) = I -

cos p(3 4)

= 1 _ I/4

= I - 1/4 1/2

cos=;

p(3,3,5)=1-p(3

l

1/3'4 1

2'

1/4

=

1

5)

3 - NF5

w

8

I <4.

Therefore, only two polyhedra of the symbols {3.3, 3} and {3.3.4} can be taken as director for the regular polyhedra in E. Hence, the Schlafli symbols of the regular polyhedrons in E5 can be only of the forms: (a) {r, , 3.3.3). (b) {r,. 3.3, 4}. In case (a), r, takes values of 3 or 4. In case (b), r, = 3. Indeed,

p(5,3,3,3)= I-

cosp(3, 3, 3

1-

,

co s

5%(5-

< 0.

s

P(4,3,3,4)=I--(3,3,4)=1-V2_0. So, there are only three regular polyhedra in E5: the tetrahedron, the cube and the octahedron analog. The same assertion is true for all n ? 5. The Schlafli symbols of

regular polyhedra in E" for n > 5 are only J3.3--..3). {4.3.....3}, {3.3.....4).

THE GEOMETRY OF SUBMANIFOLDS

268

Consider the value of p(P) for the five-dimensional cube of symbol {4, 3, 3, 3). We have co")

1 /2 p(4,3,3,3) = I - p(3,3,3)

5/8

I

1

=

1

5 < 4

Therefore, the cube in E4 can not be taken as a director polyhedron. For n > 5, the only tetrahedron or the octahedron analog can be taken as a director polyhedron. 9.5 The Regular 600-Hedron It was a Diamond! As large as a plover's egg. The light that streamed from it was like the light of the harvest moon. When you looked down into the stone you looked into a

deep yellow that drew your eyes into it so that they saw nothing else. It seemed unfathomable as the heavens themselves. We set it in the sun, and then shut the light out of the room, and it shone awfully out of the depths of its own brightness, with a moony

gleam, in the dark. Wilkie Collins The Moonstone.

Consider a polyhedron of symbol {3, 3, 5}. Its director polyhedron is a {3, 5} polyhedron, i.e. the icosahedron. To represent this polyhedron we use the coordinate method of Coxter and Berger [2]. It is simple to construct this polyhedron using the coordinates of its vertices. Denote by r a positive number satisfying r2 = r+ 1, i.e. T = J 2}'. All vertices of the icosahedron and the dodecahedron can be represented by means of this number. Namely, twelve vertices of the icosahedron are of the form (see section 7):

(0,±T,±l),

(±1,0,±r),

(±r,±1,0).

(1)

They are obtained by even permutation of 0, ±r, ± 1. All twelve vertices of the dodecahedron have coordinates as

(f1,fl,tl), (0,fr-',fr), (fr,0,fr-'),

(fr

±'r' 0).

(2)

Various sections of the 600-hedron by hyperplane represent the polyhedra with the icosahedron and dodecahedron among them. The set of 600-hedron vertices consists of the points (±2,0,0,0),

(0,±2,0,0),

(0,0,±2,O),

(0,0,0,±2),

(t1, fl, fl, tl) and the points obtained by even permutations of coordinates

(± r,±1,±r'',O).

(3)

REGULAR POLYHEDRONS IN L AND E"

269

Set a = ± r, b = ±1, c = ±'r-1, d = 0. Then the latter set of points is of the form (d, b, a, c),

(b, d, c, a),

(a,c,d,b),

(b, a, d, c), (d, a,c,b),

(b,c,a,d),

(d,c,b,a),

(a, d, b, c),

(c, a, b, d),

(c, d, a, b),

(c, b, d, a).

(a, b, c, d),

(4)

The set of the polyhedron vertices consists of 24 vertices of (3) and 96 vertices of (4), i.e. 120 vertices.

The origin is the polyhedron symmetry center. The vertices are located in nine hyperplanes x, = const. Mark in the xi-axis the points that the hyperplanes containing these vertices pass through. Also, subscript the corresponding numbers of vertices in each hyperplane I

I

I

-2 1

12

20

12

I

I

I

0

T'

30

12

1

T

2

20

12

1

In the x, = 2 hyperplane there is only one vertex A (2, 0, 0, 0). Consider the vertices in the x, = T hyperplane. They belong to the set of vertices (4) with a in the first place:

(T,±I,±T-1,0), (T,O,±1,±T-1),

(T,±T ',0,±1).

(5)

In the x I = T hyperplane the variable coordinates are x2, x3, x4. With respect to these coordinates the vertices of (5) can be represented as

T-'(±T,±1,0), r '(O,±T,±1),

r-'(±1,0,±T).

(6)

Comparing with (1), we conclude that (6) represent the vertices of the icosahedron which is obtained from (1) by similitude of the r--' similarity ratio. Join the point A to every vertex of an icosahedron with the edges. Since the lengths of these edges are equal to the icosahedron edge length, all the two-dimensional faces are triangles and three-dimensional face is a tetrahedron. In the hyperplane x, = I there are 8 vertices of set (3):

(1,±1,±1,±1)

(7)

and 12 vertices of set (4) with b in the first place:

(1,±T,0,±T-'), (l,±T-',±T,O),

(1,0.±r',±T).

(8)

Comparing with (2) we conclude that 20 vertices (7)-(8) make the set of dodecahedron vertices. The point A takes a symmetrical position with respect to the icosahedron (5) and dodecahedron (7H8) vertices. The sets (5) and (7)-(8) are invariant under the action of group of the icosahedron. In the hyperplane x, = T-1 there are 12 vertices:

(r',±T,±1,0),

(T-',O,±T,±1),

(r',±1,0,±T),

rHF

2711

OP SLI,)NMANii'()LDS

which coincide with the vertices of the icosahedron (1). Finally, consider the hyperplane xi = 0. It contains 30 vertices: (0.±2,0.0).

(0.0.±2,0),

(0,0,0.±2).

(0.±r.±r '.±l). (0.±1,±r.±r''). (0.±r'',±I,±r).

(9)

The number of these vertices conicides with the number of icosahedron vertices. In the hyperplane v, = 0 consider the icosahedron obtained from (1) by the homothetic transformation of ratio 2r 1. The coordinates of its vertices are

(0,±2.±2r '). (±2r '.O,±2). (*2,±2r '.0)

(10)

Find the midpoints of the edges of this polyhedron. The midpoint of the edge of

vertices (2,2r ',0). (2.-2r 1,0) is (2,0.0). The midpoint of the edge of vertices (2.2r , 0), (0.2.2r ') is (1. r + 1, r 1) = (1. r, r 1). Comparing this with (9). we 1

%ee that these points belong to the set (9). In an analogous way we can state that the other points of (9) are the midpoints of the edges of an icosahedron (10). Therefore. this point set is invariant under the action of the group of the icosahedron Thus, the set of polyhedron vertices is invariant with respect to rotation around the x, -axis which takes the director icosahedron into itself. Therefore, the neighborhood vertices of polyhedra in each separately considered hyperplane .r, = const are congruent to each other.

Note that the polyhedron edges join the iertices located either in the same hyperplane or in neighboring hyperplanes. Mark two points: A(2.0.0.0) and B(I, -1. -1 -1). Draw the hyperplane through the midpoint and perpendicular to the intercept AB. Perform the symmetry transformation with respect to this plane. Let it be the unit normal of the hyperplane. The r = r - 2(rvt)n is a position vector of the symmetric point to the point of position vector r. In coordinate form, taking into account that it = ( . ; . 1 ), this transformation can be expressed as .v,=

(2x, - vi - x' - .ri - .v4) !2.

1 = I ..... 4

Under this transformation the set of vertices (3) passes into itself. For instance. (-2.0, 0.0) passes into (-1.1,1,1). The point symmetric to (u. h, d) has the following coordinates

(cr-h-c h- ct-c

-cc-h-rl

To show that these coordinates can be expressed in terms of it. h, r. d, it is sufficient to state the following equalities-

r+1+r

-r+l+r0 r-l+r.,, r+1

taking into account that I + r = r. Hence, the coordinates in (11) can be expressed via a, h, c, d. For instance. (r. 1, r -1, 0) passes into (0, -r 1. -1. -r). which has the form (d, h, u) and, therefore, belongs to (4).

REGULAR POLYHEDRONS IN E4 AND E"

271

The set of points (4) is invariant under symmetry transformation. Therefore the neighborhoods of A and B are congruent. Conducting the same considerations with respect to B, we conclude that the neighborhood of the vertex A is congruent to the neighborhood of the vertex in the hyperplane x, = r. From this it is easy to state that the neighborhoods of each vertex are congruent to each other. So, the polyhedron is regular.

Each vertex adjoins 20 faces - the tetrahedrons. Since the total number of polyhedron vertices is 120 and the number of tetrahedron vertices is 4, the total number of polyhedron faces is 120.20/4 = 600. 9.6 The Regular 120-Hedron Consider now the polyhedron of the {5, 3, 3} Schlaffli symbol. It is dual to the 600hedron considered above, that is, its vertices coincide with the centers of faces of the

600-hedron. Evidently, the polyhedron dual to the regular one is regular too. Therefore, the polyhedron under consideration has 600 vertices. We shall show that it has 120 faces. The symbol of the director polyhedron is (3, 3). So, it is a tetrahedron. The faces of the 120-hedron are dodecahedrons. Indeed, as {5, 3, 3) is the 120hedron symbol, the number of sides of a two-dimensional face is 5. Among the regular polyhedra in E3 only the dodecahedron has pentagons as faces. This fact can be also proven geometrically. Consider the centers of 600-hedron faces with a common vertex A (2, 0, 0, 0). Each of these faces corresponds to the 2dimensional icosahedron face in the x1 = r hyperplane. Find, for instance, the center of the 3-face of vertices {2,0,0,0}, 17-,,r, 1, 01, Jr, 0, r,1 }, Jr. 0, r, -1 }. We obtain { "r 11, a 2 - y} . In an analogous way we can find the centers of the other three faces. All of them lie in the same hyperplane x, = . Therefore they are vertices of the same face of the 120-hedron. If we draw rays from A (2, 0, 0, 0) through the centers of 3-faces then they will pass through the center of 2-faces of the director icosahedron, i.e. to each 2-face of an icosahedron there corresponds one vertex of a 120-hedron face. Therefore, the faces of a 120-hedron are dodecahedra. Denote by D, the dodecahedron constructed with respect to vertex A, see Figure 42. Two neighboring dodecahedra are glued to each other along the pentagon. The number of dodecahedron faces is 12 and the number of 600-hedron vertices in the hyperplane x, = r is 12. For each of these vertices there is a corresponding dodecahedron D2,..., D1.1. Each of them has a common edge with D1. So, we can imagine

that the dodecahedron DI has 12 dodecahedra glued around it. In the x, = I hyperplane there are 20 vertices of the 600-hedron. T h e r e are 20 corresponding dodecahedra D14, ... , D33 which are glued to the previous layer D2,..., D13. There are 12 vertices of the 600-hedron in the hyperplane x, = r -I. Thus we have 12 more dodecahedra. Finally, 30 dodecahedra correspond to the vertices in the hyperplane x, = 0. The dodecahedra gluing process continues for x1 < 0 also. Thus, the polyhedron consists of nine dodecahedron layers and their total number is I + 12 + 20 + 12 + 30 + 12 + 20 + 12 + I = 120.

272

THE GEOMETRY OF SUBMANIFOLDS

FIGURE 42

Coxeter represented the vertices of a 120-hedron in the following convenient form: these are all points obtained by permutation of coordinates of the points

(±2,±2,0,0),

(±T,±T,±T,±T 2),

(ff,fl,fl,fl), (±T2,±T-',±T-',±T-'), and also the points obtained by even permutations of coordinates of the points ((±T2,±7--2, ±1,O),

(±v7,±7-',±T,±O),

9.7 A Simple Way to Construct the Icosahedron

In this section we shall state a simple and descriptive way to construct the icosahedron. Inscribe into the circle of center 0 two regular pentagons 11, and 112 of vertices A,, . . . , A5 and B1, . . . , B5 respectively in such a way that each B; would be the midpoint of arc A;A141. Let a be the pentagon side length and r the circle radius.

We shall regard these pentagons as located in two coinciding planes. Draw the straight line I perpendicular to the pentagon planes through the circle center 0. Shift these planes along / in such a way that they stay perpendicular to /, the pentagons take symmetrical positions with respect to I and lA;Bjl = a. This is possible because in the starting position IA;Bjl < a and we infinitely increase the distance lA;BjI by the distance increase between planes. Join B ; to A; and A; }, with edges. Then construct two caps for pentagons n, and 112. Mark the point Q, in I such that l Q 1 A j = a. Such a point exists. Indeed, shift Q along I starting from the center of 11,. In the starting position I QA,I = r = 2 si n x/5 As sins > 1 then r < a. If Q tends to infinity along ! then IQAjI increases to infinity. So, there are two required points symmetrically located with respect to the n1-plane. This plane splits the space into two half-spaces. Select Q, in that half-space which contains no 112. Join Q, to Aj with edges. In an

REGULAR POLYHEDRONS IN E4 AND E?'

273

analogous way find Q2 and join it to B; vertices with edges, too. Thus, we obtain the required icosahedron. Denote by C; the vertices of the icosahedron. To get the simple coordinate form of icosahedron vertices we ought to introduce a convenient system of coordinates. Let 0 be the center of symmetry of the icosahedron. Consider the edge C, C2. Draw the

straight line OK perpendicular to C1 C2. Suppose that IOKI = r and the length of edges is equal to 2. Then IKC1 I = 1. Draw a plane through the points 0, C1, C2. This plane contains two more icosahedron vertices C12 and C11 which are symmetrical to

C1 and C2. If we join C, to C12 and C2 to C1 then we obtain the rectangle C1 C2C1I C12. To the edge C, C2 there are adjoined in a symmetrical way the icosahedron faces - the regular triangles C, C2C3 and C, C2C4. The straight line C3C4 is evidently perpendicular to the plane of the rectangle C1 C2C11 C12. Let us now define the system of coordinates. Put the origin into the center of symmetry O. Direct the axes x3 along the straight line OK, the axes x, parallel to C2C1 and the axes x2 along

C4C3. With respect to this system of coordinates we have C, (l, 0, r), C2(-1, 0,,r). Let C3 and C4 be represented as C3 (0, a, b), C4(0, -a, b). Let C10 and G) be the points 0-symmetric to C3 and C4 respectively. They are also the icosahedron vertices. Thus, we know eight of the icosahedron vertices. Let C. C6, C7, C8 be the other four vertices. Suppose that C5 and C6 are joined with an edge. As we saw above, if we draw the plane through 0 and the edge then it contains four icosahe-

dron vertices. These vertices are disposed symmetrically with respect to OK. Therefore, the planes C1 C2C11 CI2 and C5C6C7C8 are perpendicular to each other; also the axis x, is their intersection while the sides of the second rectangle are parallel to the x, and x2 axis. The distance from 0 to the edge C5C6 has to be equal to r and

IMC5I = 1. Hence, we have C5(r,1,0), C6(r, -1,0). The intercept C3Cq is the icosahedron edge. The points C3 and C9 are disposed symmetrically with respect to the plane C5C6C7C8. Therefore C3C9 is perpendicular to this plane. So, the distance from C3 to the plane x1, x2 is equal to 1. Note that the distances from C3 to C1 and C5 are equal to 1. Thus we have two conditions: 12

IC3C1

= I + a2 + (r -

1)2 = 4.

IC3C5I2 = 1+r2+(a- 1)2 =4.

From this we find a = r. Substituting a into the first equation, we obtain the equation with respect to r. The required number is the positive root of the equation r2 = r + 1. After that it is easy to find the coordinates of all icosahedron vertices. We already expressed them above, see (1) Section 5. We can take the centers of icosahedron faces as the vertices of a dodecahedron.

10 Isometric Immersions of Lobachevski Space into Euclidean Space This chapter introduces some background information and provides an individual perspective on work developing from the theory of isometric immersions of the Lobachevski plane into Euclidean space. Section 10.1 gives an account of the author's visit to Nickolina Gora with Professor Efimov and Professor Pozniak and also gives some interesting insights into the work of Professor Efimov and Professor Pozniak and colleagues at Moscow State University. It introduces the conference on Non-Euclidean Geometry held in Kazan in 1976, dedicated to the 150th anniversary of Lobachevski's discovery. Section 10.2 gives an account of the fire in Kazan in 1842. Lobachevski was a professor at the University of Kazan and this section describes his work, achievements and discoveries and the significance of his work in classical mathematics. 10.1 Meeting at Nickolina Gora

The Kharkov-Moscow train arrived in Moscow early in the morning while the meeting with my scientific advisor - Professor Nickolai Vlad mirovich Efimov was scheduled for 5 p.m. Many things depended on that meeting. I hoped to discuss the forthcoming plans and perspectives of my scientific work and I wanted to use my time in Moscow to the best effect. I went to Lenin's library to get copies of scientific

papers not available in Kharkov. At that period I was studying the geometry of vector fields and minimal surfaces and I was interested in the interpretation of mass as a curvature of space. The book by J. Willer and others, The theory of gravitation and gravity collapse, contained a curious formula: in Einstein space the density of mass-energy p is equal to the sum of the scalar curvature of a space-like hypersurface and its second symmetric function of principal curvatures, namely 4ap = R + Ke. 275

276

THE GEOMETRY OF SUBMANIFOLDS

This formula directly relates geometry to physics. In one of my papers I introduced the concept of the curvature source power of a vector field singular point and found

the interpretation of this concept in terms of general relativity as a mass source power. So, I was interested in the reference, where the uncommon hypothesis that the centers of galaxies can be the sources of mass was suggested. I collected a copy of this for further study. At last, the time for the meeting with Professor Efimov arrived and I took the elevator to the 14th floor of the Moscow State University building, to the department of mechanics and mathematical. N. V. Efimov was there and it was good to see him. He had been ill but had recovered and was, as always, friendly, joyful and benevolent. It was late June and the weather was hot and sunny. On the office wall there was a

portrait of Gauss and "the king of mathematicians" gazed at me with a stern, penetrating and slightly mocking look. Efimov was an unusual dean in the history of the mechanics and mathematics department. He was able to pay attention to students, to fascinate them with science and he was respected by professors and students. Consequently he took the position of dean for two terms, i.e. for eight years. He was always attentive to everyone he met regardless of their position. He asked me about my work in science and about the life-style in Kharkov and told me that he was waiting for Edward Genrichovich Pozniak. They planned to go to the cottage at Nickolina Gora to stay in the country and to get some fresh air and he invited me to

go with them. Pozniak arrived and approved the "cottage idea". This was very flattering and unexpected for me, although I knew that academician P. S. Alexandrov had often taken his students out of the city. Sometimes, Pavel Sergeevich came to the student's hostel with a portable record-player and listened to classical music records with them. The rector, Ivan Georgievich Petrovski, also visited students and discussed aspects and perspectives of University-life with them. It was good to be in the company of Efimov again, the famous geometer, Lenin and Lobachevski Prizes laureate. In 1983, the famous French geometer Marcel Berger dedicated the Russian translation of "Geometry" to Prof. Efimov. In a preface he wrote: "In May 1968 Efimov was the guest of Paris VII University and everyone who simply talked to him or discussed mathematical problems was profoundly impressed by him. I would like to express here my deep respect for this extraordinary man." Efimov was not a fisherman, a hunter or a collector. He liked, it seems, only to read books. One day, I went to him for a tutorial consultation and I noticed a book "Dear My Man" by Yu. German on his table; he loved books and once he told that he had reread Tolstoy's War and Peace. I remember also the seminar in geometry when he gave us his impression of Bulgakov's "Master and Margarita" just published in Moscow magazine. "It is something tremendous", he said, and recited the beginning of the first scene with Pontius Pilate. I remember with pleasure and some sadness that hour and a half journey by car to Nickolina Gora with Efimov, "the geometry clan's oldest father" (to use M. Berger's expression), driving with E. Pozniak beside him. The fresh soft air of the evening and the scent of the forests near Moscow generated optimism and the joy of living.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

277

The most recent of my results which I had reported earlier in Moscow were concerned with the instability property of closed minimal surface in a Riemannian space of positive curvature. I hoped that it would provide an approach to the Hopf problem

on the metric on S2 x V. Efimov and Pozniak recommended that I presented a report at the P. K. Rashevski seminar after summer, in the presence of A. Fomenko who studied minimal submanifolds at that time. Then the conversation turned to the importance of geometry. Sometimes geometry was viewed with scepticism although it was accepted that it was an interesting subject. Efimov said: "You know, we. the geometers, studied very deeply the equation w = sin w, which arises in the theory of isometric immersions of Lobachevski plane into Euclidean space. Recently. Sergej Petrovich Novikov told me that this equation occurs in many physical phenomenona. for instance in superconductivity, and it is developing the theory of soliton solutions for it. At the beginning of the century, in geometry the method allowing new solutions to be obtaining by the Backlund transformation was developed, and this method also has applications, So. geometrical approaches justify themselves, maybe not at once but after some time. You must believe in the your work."

At last, the car turned off highway and entered the yard of the dacha which Efimov and his family rented for the summer. A table was placed in the yard which was shaded by trees and we continued the conversation. Efimov told me that it had been proposed to stage a conference in geometry in 1976 in Kazan devoted to the

150th anniversary of Lobachevski s discovery. Efimov said: "Do you know the reason for the obstruction and delay of that discovery? The obstacle was not only the solution of a mathematical problem, but also in overcoming the inertia of secular traditions. It had to be understood that all 'the new' was true although 'the old' was also true. It had to be understood in what sense 'the old' and 'the new' can be true and also mutually exclusive. Before Lobachevski, the geometrical idea was tied to Euclidean space, and Lobachevski opened and released unbounded possibilities of

fruitful generalizations. It was he who got modern geometry off the ground" LQbachevski was regarded as a scientific hero by Efimov. The conversation made an impression on me and I planned to develop the theory of isometric immersion of Lobachevski space into Euclidean spaces.

I remember Efimov's suggestion: "Time is ... nobody knows. It would be very interesting and important to penetrate the secret of time. Only one saying can be expressed definitely - it is inexorable." In 1976, at the All-Union conference in non-Euclidean geometry in Kazan I presented a report: "Lobachevski spaces as a submanifolds in Euclidean spaces" That was my first visit to Kazan and together with other participants in the conference, I visited the monument to Lobachevski. 10.2 Fire in Kazan

In the late summer of 1942, Kazan was hit with a tremendous disaster - fire - and most of the city was destroyed. There was a strong wind. and the fire spread quickly from house to house and the connecting streets up to Kazanka river. The ringing of

278

THE GEOMETRY of SUBMAN1FOLDS

N I LOBACHE% SKI

fire-bells, explosions. people shouting and crying. the unbearable heat the noise of

storm ... People left their houses and tried to save their property. They were in despair. As night fell the dark blue sky was painted with reflections of the fire. The disaster

was exacerbated by the force of the storm and the lire spread to areas nine miles from the city. N. I. Lobache%ski was the rector of the Unnersity and with Musin-Pushkin. the supervisor. organized a plan to protect and sa'e the University from fire. The rector's steel) willpower was pitted against the temf)ing power of the fire. The main University buildings were sa%cd and the students and officials worked hard to protect the library. the anatomy theater, the main University and adjacent building.

The instruments of Astronomic Observatory was carried out into Unicersit} square. The student's library books and equipment from the mineralogy. botany. and zoology departments and other offices were also removed for safety. From the main University library only rarities were carried out while the other books remained. In spite of strong fire around the library building the books remained undamaged. The next morning the fire abated and the people of Kazan were shocked by the

devastation it had casued. About 1300 houses were destroyed together with the University Astronomy Observatory and University residences. including Lobachevski's flat. U. M. Simonov, famous citizen Kazan, was abroad when he learnt of the fire in Kazan. He was a brave and experienced man but tears filled his eyes when he read

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

279

about this disaster in the newspapers. In scientific circles Simonov was more famous than Lobachevski due to his participation, in the capacity of scientist-astronomer, in

the historic Russian round-the-world expedition to the mysterious South Sea in 1819-1821. That expedition of the "Vostok" and "Mirnyj" ships was guided by Bellinsgausen and Lazarev. During the voyage the Antarctic Continent was discovered, in 1820. One of the islands from the Fiji group in the Pacific Ocean was named after Simonov. After that expedition his reputation was enhanced. He corresponded with foreign scientists, regularly visited France, Britain, Austria, and Belgium and became a corresponding member of the Academy of Sciences. Lobachevski did not leave Kazan at that time. He reconstructed the University buildings, gave lectures in various courses, and achieved the discovery of a new geometrical "continent" which remained unrecognized at that time. The expanded account of that discovery, "On origins of geometry", was published in "Kazan University Notices", 1829. Later, he wrote five papers on that topic and

in 1837 his book was translated into French. In 1840, the book Geometric investigations in the theory of parallel lines was published in German. There was also translations into English. That activity played an important role in the international recognition of new ideas. Nevertheless, the respect accorded to Lobachevski was

for different reasons, for other scientific papers and for his term as University rector. His main scientific activity and achievements remained unrecognized by the Academy. However, his papers were carefully studied Gauss and in 1841 he wrote in one of his letters: "I begin to read in Russian with some fluency and benefit. Mr. Knorr send me a short paper in Russian by Lobachevski (Kazan) on parallel lines. Due to this paper and its German translation I am interested to read more papers from this ingenious mathematician." In November, 1842, Gauss proposed that Lobachevski should be awarded the Corresponding Member Diploma of the Konig Gottingen Society. The Diploma

stated that the Society had elected him as a friend and co-member in scientific association, and expressed a hope that he would work collaboratively and share discoveries, notices and thoughts with Society.

In 1843, Lobachevski responded to Gauss and thanked the Konig Gottingen Society for the honor and expressed the desire that all his scientific works be worthy

of the excellent Society. In addition, he wrote: "Please excuse the delay in responding. My health and my personal circumstances have suffered as a result of the fire in Kazan and it was has created many problems for the University." In 1846, Simonov replaced Lobachevski in position of rector, and the government awarded him St. Ann Order first. Lobachevski was rewarded with three Orders much later. But all of the recognition and misunderstandings that Lobachevski suffered in the last years of his life must not undermine the main thing --- the joy of creation that he felt. And time put everything into perspective. It should be remarked that the fate of the Hungarian mathematician Janos Bolyai,

who worked on the same ideas some years later, was much more unhappy. But currently the Hungarian Mathematical Society is named after him.

280

THE GEOMETRY OF SUI3MANIFOLDS

103 Eflmov's Theorem on Non-Immersibility of the Lobacbevski Half-Plane into E3

It is well known that a complete Lobachevski plane is not immersible into E3 as a regular surface. This was proved by Hilbert in 1901. Efimov generalized this result in two ways: (I) he proved that a half-plane (i.e. an infinite domain with a complete geodesic line as a side) is not immersible into E3; (2) he stated theorems on nonimmersibility of complete two-dimensional metrics with variable negative Gaussian curvature bounded away from zero: K < -a2 < 0, a = const. In this section we consider Efimov's theorem of the first kind. Theorem I For any a there exists no band in E3 of constant width a located along the complete asymptotic line in a C4-regular surface of constant negative curvature. Suppose that the band from the hypothesis exists. Let -y be a complete asymptotic

line in it. Let y, be a curve in this band of distance a from y. The distance from a point in y, to -y is the length of the shortest line joining the point in y1 to the point in y. Each asymptotic line in the surface belongs to one of two families. In addition, the curve of the first family is distinct from the curve of the second family by the sign of torsion on a surface of Gaussian curvature K; for the first family the torsion is

for the second family the torsion is - vr--K. Let -y be of the second family of asymptotic lines. Emit from 0 E -y the asymptotic of the first family and fix a point A in it such that the length of asymptotic fine arc OA would be smaller than e: OA < C.

Emit from A the asymptotic of the second family. Then it is located in the band. Indeed, suppose that it intersects yI at a point B (see Figure 43). Emit from B an asymptotic of the first family. Suppose that it intersects y at a point C. The arcs OA and OB are opposite sides of an asymptotic quadrangle. Since the net of asymptotic lines

in a surface of constant negative curvature is Chebyshev's one, then

OA = CB < E. However, the length of CB is greater than a because the distance from B to -y is equal to e. So, we come to a contradiction.

FIGURE 43

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

2KI

By means of asymptotic lines, introduce in the surface the coordinates (ir. r) with respect to which the metric has the form d k2 = dl r' + 2 cos w cIu dr + ch=. The Gauss-

Codazzi equations then reduce to a single equation:

' = sin.;:

(1)

The angle w between asymptotic lines must satisfy 0 < w+ < n by means of surface regularity. The first family of asymptotics is represented by u-curves, the second by

r-curves. From what was proved above we can suppose that o, <_ u < -x. < r < +oc. As sin' > 0, there exists a point at which either w > 0 or w,, < 0 It is sufficient to consider only one of these possibilities because (I) is invariant with respect to simultaneous reorientation of the coordinate axes. Moreover, we can assume that', > a > 0 in some segment n < it < .I. r = ro = const where a = const. Integrating (1) in r. we get L,, > a for a < ti ,,5 1. r > rp. Let ((i_, .i_) be a segment in (a. d). Let na <- it < ,32. Integrating in it from n up to it, we get to

w du > cr(ir - n) > a(02 - (1) = wu > 0.

w(u, r) - W(ct, r)

Therefore w(u, r) > wy, > 0.

where ' = coast,

(2)

Integrating in it from it up to .3 we get

w,thc>a(d-u)>_a(.1>0 u

where w, = cont. Therefore W(tr, r) < w(3, r) - wl < 7r - wI.

(3)

From (2) and (3) it follows that in a half-band a:! < it < i_, r > rp the inequality sin w > h > 0, where h = const, is satisfied. Integrating both sides of (1) in r. we get

r) -',(ir. rU) =

J

sin'dr > Mr - r0).

Therefore, for any ni there exists r such that w,(u, tip) > in, where a2:5 t t:5 :3±. In this case 0 < w < Tr fails if nt is sufficiently large. The theorem is proved. If we consider a metric ds2 = Edit' + 2Edu dr + G di'2 then the geodesic curvatures

c, of coordinate curves u and r are, respectively, 1

EG - F2

F

E

\T),-I -

EG-F'-Vfir+VGr}

282

THE (GEOMETRY OF SUBMAN[t-OLDS

Therefore, the geodesic curvature of asymptotic lines in a surface of curvature K _ -1 can be expressed in terms of derivatives of net angle j with respect to natural parameters. Namely, NJ = -Vj,,,

n: = W,,.

The integrated geodesic curvature of any asymptotic line are can be estimated from above by ir:

J

h',(iS,I

=

J4idszZ7r.

(4)

We shall use the remark above to prove the next Efimov theorem from [1].

Theorem 2 The Lohaehe rski half-plane does not admit isometric and regular miniersiun into V. Here the regularity assumed is of class Ca while in [21 this condition is lowered to C2. Some lemmas are stated. Lemma I

Let G,, he a simply connected curvilinear polygon of boundary I',, in a

surface of constant negative curvature K= -1. Let r,, consist of smooth arcs Al ...... ,,, where each of the a, is an arc of either an asymptotic or geodesic line. Then the area S(G,,) satisfies the following inequality: S(G,,) <- 2(n + 1) 7r.

(5)

Apply the Gauss Bonnet formula to G,,:

fKJS= -SW

ds f+27r+Eni

where - is the geodesic curvature of A,, n, is the angle between A, and .4+r Since the integral on the geodesic curvature of each asymptotic line is less than ;r and In,I < it

for each ri,. we get the upper bound (5).

Suppose that the Lobachevski half plan TI is immersed into El as at regular surface. Find in II a domain G,, of infinitely large area but with the number of sides n <4. Then we shall come to a contradiction with (5). Denote by (li) and (12) the first and the second families of asymptotic lines as above. Since each family is a family of integral curves of an autonomous system of differential equations without singular points, then every curve of the families is a topological straight line and thus it separates II into two domains. It is not excluded that the curve comes to the side of II with one or both its "infinities". Let .4 he a point inside 1I. Emit from A the asymptotic ray li of the first family, supposing that it

exists. Here the "ray" means a curve with one endpoint .1 and a monotone

sequence of points Bi..... BA.... divergent in li Let a be some smooth arc from .4 with no more intersections with li. Suppose that the union of li and a separates II into two domains. Denote one of them by NED. In this case %%c include the points of li

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

283

FIGURE 44

and a into E. So, E is a closed set in II. Let B be an arbitrary point. but not A. in the ray It and 12 be an asymptotic ray of the second family which passes into E. We consider 1, as maximally prolongated. Three cases are possible: 12 comes to the arc a. 1, comes to infinity. L, comes to the side of 11. In any case 1, separates E into two parts such that one of them contains A in the boundary. Denote this part by T(B), considering T(B) as a closed subset in II (see Figure 44). Lemma 2 If E is a complete metric space as a subset in II then

elim T(B) = E. Here the limit of T(B) while B - x means the closure of the set-theoretical sum of closed sets T(BI ), .... T(Bt),.... where BI , B2.... is a monotone divergent sequence of points in 1I .

Set T' = UT(Bj). Let T' be a closure of T'. Show that T' = E. Suppose that T' is only a part of E. Then there exists a point in E but not in P. From this it follows that there exists a point Q which is both the interior point to E and the cluster set point to T'. Denote by h a quadrangle of an asymptotic net which contains Q and is interior to E. Through the point Q draw an arc If of the first family asymptotic line such the li is located in h. Since Q is a cluster set of V. the there exists an integer N such that for any k > N the intersection of T(BL) and It is not empty. For k > N there exists an arc 12 from Bk intersecting 11 at some point B. Consider the asymptotic net quadrangle where k. ni > N (see Figure 45). Fix k while ni -+ oo. As the sequence BI .... , BL.... is divergent in lI and E is complete, the length of the arc BL Bm tends to infinity while ni -' x. However, the length of the arc BABA, is bounded. The arcs B&Bm and B'B,,, are the opposite sides of an asymptotic

quadrangle. By Chebyshev's net property, they must be equal So. we come to a contradiction. Lemma 2 is proved. Let r be a side geodesic line of the half-plane H. Draw in R a complete geodesic line 7 which has a common perpendicular with r. By the property of the Lobachevski plane geodesic lines, the smallest distance between the points of r and , is

THE GFOML1R) (W SUBMANIFOLDS

214

FIGURE. 45

the length of this common perpendicular. So. There exists -- > 0 such that the disk of

radius r and with its center at any point of I is in If. Denote by 1I, a half-plane with , as a side Take an arbitrary closed disk D which is strictly interior to 111. Through the each point of D draw the maximal in III asymptotic of the family (11). In this case all possibilities are exhausted by the following:

(1) the set of all drawn arcs is located in some compact part of the closed half-plane IIi U,)--

(2) among the drawn arcs there exists the arc outgoing to infinity in III U' in both directions; (3) among the drawn arcs there exists the arc outgoing both to infinity in III U -, in one of its directions and to some point A E I in another direction. If the first possibility occurs then there exists a domain G of G type. where n = 2. In this case D C G. The second possibility never occurs because Theorem l implies that 11, would contain an --band along a complete asymptotic line. So. our consideration reduces to the third case. Apply Lemma 2. taking as a one of two rays of the geodesic ; with A as a starting point Denote it by I(,. Moreover, select it in such a way that corresponding to this choice the domain E would include not less than half of the area of disk D. Let S'1 be

a union of the interior of E and all of the points of the ray 1,. Define a mapping ,1': E, - 11 as follows. If x is interior to E then by Lemma 2 it belongs to some set T(B), where B is some point in 11 be some segment [A,y] in the ray 1,. Set 1`(x) = v. If

x E lI then set f (x) = x. The mapping 1 is continuous on Si. Denote by DI the intersection D U E,. As E is closed. D, is a compact set. The mapping f is continuous on D, and the image 1(DI) is a compact set in the ray 11.

From the construction of E it follows that the image of the E interior is not a compact set in 1,. This implies that there exists a point B, E 1i such that D, C T(BI ).

ISOMETRIC IMMERSIONS OF LOBACREVSKI SPACE INTO EUCLIDEAN SPACE

285

FIGURE 46

If T(BI) is a compact set then it is a curvilinear polygon G of G. type, where n = 3,

and in addition D, C G. Suppose that T(B1) is not a compact set. Now consider T(BI) as E, B, as A, /2 from the second family of asymptotic lines as /I. Take an arbitrary point C in 12 and then draw the asymptotic !' of (11) family through C into T(B1). The arc !,, being maximally prolongated in T(BI), must come to rya. This follows from Theorem 1 because if we prolong /1' in the other direction, then across 12 it does not meet either /1 or /2 for the second time and, hence, it goes out to infinity.

Denote by T'(C) a closed part of T(BI) cut off by /' (see Figure 46). Applying Lemma 2 again, we conclude that there exists a point C E 12, such that D, C T'(C). The set T'(C) is a compact curvilinear polygon of type G,,, where n < 4. It contains either a disk D or the part of D, namely D1, of area not less than half the area of D. Thus we come to a contradiction. The theorem is proved. In connection with this theorem the natural problem arises of describing those domains in the Lobachevski plane which can be regularly and isometrically im-

mersed into E3. In Pozniak's paper [3], devoted to the 150th anniversary of Lobachevski geometry, isometrical immersions into E3 of infinite polygons in Lobachevski plane with finite or infinite number of sides but not containing a halfplane were constructed. Here the "infinite polygon" means the intersection of the finite or countable set of closed half-planes with sides - the straight lines in the Lobachevski plane - having no common points. It was proved that any infinite polygon with a finite number of sides can be isometrically immersed into E3. Also, two classes of infinite polygons with infinite numbers of sides were introduced, which can be immersed as well.

Note that the Yugoslavian Mathematician D. Blanusha embedded the Lobachevski plane into E6 while E. Rozendorn immersed it into E5 (see [33], [34]).

THE GEOMETRY OF SUBMANIFOLDS

286

10.4 Local Non-Immersibility of n-Dimensional Lobachevski Space L" into E2 -1

Cartan in papers of 1919-1920 [4] and independently Liber in 1938 [5] proved the following theorem. Theorem The Lobachevski space L" of dimension n can not be immersed even E21-2. locally into (2n - 2)-dimensional Euclidean space

(By the way, it ought be noted that I have never seen [4] and give the citation from [61).

The proof follows from the analysis of the Gauss equations. In the case of space of constant curvature the Riemannian tensor is of the form Rolfld = K(gM g,.* - gnb gih ),

where K means the curvature of space. For the case of L" the curvature K < 0. Consider the solvability problem of the Gauss system P

E(LnLI16 - L.'hLIj,) = K(g.-,g;ra

(I )

n=1

with respect to the p symmetrical tensor a = I , ... , p. The theorem is proved if we show as in [5] that this system has no solutions for

p
Let {e, }, ..., (e,,) be a system of n unit n-dimensional vectors which form an orthonormal basis in metric i.e. which satisfy

ei ej gnJ=b,j. With the help of the tensor g,,;3 we are able to drop a superscript, setting a

g1,. Pass from the solvability problem of (I) with respect to an arbitrary basis to the solvability problem of an analogous system with respect to the orthonormal basis. Multiply both sides of (1) by e°e] ek e1 and set Loge," eJ = b°,.

Then we get the following system:

E(b)bk,-bi,b7R)=K(brfai-,-6,,bki).

(2)

a

Lemma l

If the system (1) is solvable for K = Ko 36 O then it is solvable for any K.

Indeed, if L;, satisfy (I) for K = Ko then L°

K1 KK0

satisfies (2) for K = K1.

rt

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

287

Leninia 2 Let an arbitrary vector I in the in-dimensional plane E'". nz <- n be a principal direction of a symmetrical tensor T,,,,. Then T,,,, has tit eigenvalues equal to each other.

Let T be an operator of matrix T,,,,. Let ei and e2 be linearly independent vectors in Em of eigcnvalues al. A2 respectively. Then

Tel = Aiei,

Tee, = \,e2.

Let A12 be an eigenvalue for el +e2 E E. Then T(el + e2) = A12(ei + e,).

As the operator is linear. T(el +e2) = T(el) + T(e2) = Aiei + A2e2.

Thus, we have Aiei + A,e2 = Ai2ei + A12e2.

Since ei and e2 are linearly independent, we conclude that

AI =.\12

A2=ail.

Therefore. every vector in E"' corresponds to the same eigenvalue. Leniinu 3 Suppose that (1) is solvable for n = no and p = po. Suppose that E"'. in < n is a subspace such that every vector in it is principal for r of tensors L,'+.. Then the system is solvable for n = in and p = po - r.

Without loss of generality suppose that L; ...... L;, satisfies the hypothesis. For any 1= {1"} E E"' the resuslt holds:

L l' = A-/,,. where A' is the eigenvalue corresponding to L',,,. Choose in E" an orthonormal basis {e`,;,} and contract (1) with e°e; e, e16, where i. j, k.1= L.-in. We get

b°h, - blb =K(b,1bA,-b,,b14).

(3)

As h+,', = A'b,, for 1 < a < r. the equality (3) can be rewritten as ,

f (hr"NA)-h,',h)

= K-F (Aq)1(blbxl-6+16!1

.

(4)

y=1

+=++1

Thus, we obtain the system analogous to (2), where it = in, p = po - r. Since K - E{Ay)2 0 0, by lemma (2) the system (4) is solvable for any coefficient of (S 1,41- b,, 6A). The lemma is proved.

Denote by IIB"'II an orthogonal matrix of order p. Then we can pass from the system of p tensors L,,,, to the system of p tensors 4, by

11 =B'"L"

THL GrLOMETR' 01-

288

DS

If LI,,, satisfies (1) then L',, satisfies the same system. Indeed. the properties of an orthogonal matrix imply

jr L rr i= B'"L"1"Br"L,I"_ A"I L"pb Lp''' = L"lit-L"n f' to

Let us be given a system of p tiectors r1..... rp in p-dimensional space and we know that precisely in of them are linearly independent. Then there exists the orthogonal transformation with matrix B = IIB"rII such that among fr = Bmr' there are precisely in linearly independent while others VI are zero. Indeed. as precisely in of r1..... rp are linearly independent then there exist p - in linear relations on them: (.11r1j

} ctprr'=0:

Cr-mit.i+ . .+cp 1"pr1' =t).

where the rank of Ilr" fll is p - m. We can consider the rows of this matrix as pdimensional sectors c'. r = I.....p - in. As they are linearly independent then by means of linear operations we can turn them into the system of mutually orthogonal

unit vectors. Save the previous notation cr for them. Complete the system of c',

r = I..... p up to an orthogonal basis er. r = I..... p in p-dimensional space. Suppose that cr are components of je-J. Set B T-7 = e,m

0 for r = l..... p - in The vectors r = Br"r° for p - in + I < p are Then linearly independent because are subjected to only p -m relations. A tvto-valent tensor of components u r, which are the products of components of two unnalent tensors u and r t see shall call a dyad. Lemma 4 The metric tensor g,, f of n-dimensional Riemannian space can not be represented as a sum of k dyads if k < ,,.

Suppose that

Consider det

ill. The columns of this determinant are linear combinations of

u er- I..

I"=

.A

u,E

with r7 as the coefficients. Since the number of columns is less then it. then, expanding det II ill into the sum of determinants, we find that e%ery term of the expansion contains at least two identical columns. Therefore, the expansion of each determinant is zero and us a consequence det IIg,, ill = 0, which is impossible.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

Lemma 5

289

If K< 0 and p = n - 2 then the system (I) is unsolvable for all n > I.

Evidently, the lemma is true for n = 2. Suppose that La. forms a solution of (1) if n = no and p = po = no - 2. Take any n-dimensional non-zero vector 1 of components 1'1. Set v° = La 1". The vectors va..... vQ are not zero simultaneously because in the opposite case it follows from (1) that 1° ga - g°,1N = 0. The latter implies that 1° = 0 for all a, which is impossible.

Suppose that the system of vectors vrt..... vo contains precisely m which are linearly independent. As we just proved, m > 1. Since po < n - 2, then m < n - 2. By means of the appropriate orthogonal transformation B'", this system can be transformed into a system such that the first p - m vectors will be zero while the other m

are linearly independent. Save the previous notation for the vectors of the new system. So, we obtain the system of vectors with the following properties:

v° =LAP =0 for a= 1,...,po-m v° = L° I' are linearly independent for a > po - m. Consider the (no - m)-dimensional subspace E""-'" in E"° which is orthogonal to each of v°. Let be an arbitrary vector in E"0-". Since v°{° = 0, then contracting (1) with 111f0 we get

E PO

00

Q. t6 - g.6 1.6

(5)

°=Po-m+I

where ub = L' CR. If

96 0 then (5) implies that the metric tensor gb is re-

presented as a sum of m +I dyads. This contradicts Lemma 4 because m + I < no - I + I = no. Hence, we must set 10 CO = 0. Then (5) has the form PO

v°° ub° = K1° s r-Po-m+l

(6)

Let r be an arbitrary vector orthogonal to . Contracting both sides of (6) with T6, we get v° uh Tb = 0.

Since v° are linearly independent, u, Tb = 0, a = po - m + I,. .. ,po. Since T is arbitrary and orthogonal to , ub = This means that

Lxe = A°G,

a =Po - m+ 1,...,m

i.e. f is the eigenvector for each of L. Since 1; is an arbitrary vector from E"°-"', then by Lemma 3 the system (1) is solvable for n = no - m and

p=po-m= (no-m)-2. Setn, =no-m. Since I <m<no-2,2
Thus, the system (1) is solvable for n = n, < no - I and p = n, - 2. Repeating our arguments, after a finite number of steps we come to the conclusion that system (1) is

THE GEOMETRY 01- SUBMANMFOti)S

29(1

solvable for ii = 2. which is false. Lemma 5 and as a consequence the theorem are proved. 10.5 On the Existence of Principal Directions on a Domain of L Immersed into E2" t ous

a%te r tin

' fcui a ', et

roes

c

_1

It's jeunes. "

J. Dieudonne.

This section presents the theorem of E. Cartan, and provides an introduction to the life and achievements of this brilliant mathematician.

E. Cartan is often called the architect of modern mathematics, In the paper published in the Bulletin o/' the American Mathematical Soc'wir v. 58. 1952. S: S. Chern and C. Chevalley wrote: "Undoubtedly. one of the greatest mathematicians of this century. his career was characterized by a rare harmony of genius and modesty." His long life extended from the end of the nineteenth century up to the middle of the twentieth. It was a period of great turmoil and change, including the First and Second World Wars, and he experienced many difficulties. Dolomieu - a village in the south of France -- was Cartan family home and Elie Cartan born in 1859. His father was a blacksmith and his mother a fine needlewoman He always returned to Dolomieu for holidays from Paris. where he was a student of "Evoke iVorntale Superitire". and sometimes he would help his father in the smithy. Later on. the Carton family traditionally gathered in a large family house for vacations.

Cartan's study looked out towards the mountains and provided the right atmosphere conducive to scientific research. But he was also surrounded by the young members of his family and enjoyed their company. In that intellectual home with so much thirst for knowledge nobody could suppose that the War would bring so much suffering to this family.

Louis Cartan, his son, was a talented physicist. He worked in the group investigating the problem of precise measurement of nuclear mass. The War forced him to stop his scientific research and in 1942 he entered the Resistance Organization and became a member of its staff A careless action permitted the German and French police to discover the group and in 1942 Louis was imprisoned. At the beginning of 1943 he was moved to Germany and following his departure to Germany there was no information about him. All the efforts of the Red Cross and French and foreign scientists to find his location were unsuccessful. Louis Cartan was murdered in December 1943, but his family received this information only in 1945. Cartan's outstanding contribution to world science could not protect his son. Another son, Jack, was a very talented composer. and his daughter Helen was interested in mathematics, but both died young. His elder son, Henry. became a famous mathematician, and contributed to the development of a mathematical book series by Nicolas Bourbaki. Biographers said that Bourbaki as a young man, visited Kharkov and there produced papers which had been lost in war-time. Perhaps this occurred because of Cartan's influence.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

291

E. CARTAN

In 1930 E. Cartan visited Kharkov, the capital of Soviet Ukraine at that time, in spite of the uncertain situation. Together with other French mathematicians, J. Hadamard, A. Danjua and P. Monten, he left Paris to participate in the first AllUnion Congress of Soviet mathematicians. Perhaps Cartan, as a worker's son was interested in visiting a country where new principles, and new standards of life were

being established. Also, the visiting mathematicians were an impressive bunch. Cartan met many famous and emerging mathematicians, such as S. N. Bernstain, W. Blaschke, L. Lihtenstain, A. Kolmogorov, 0. Schmidt, D. Sintsov, V. Kagan, L. Pontryagin, P. Novikov and others. In the lecture "Projective geometry and Riemannian geometry", Cartan said: "As a matter of fact, the non-Euclidean Lobachevski geometry, which immortalized the name of the great Russian geometer, can be considered from the projective geometry viewpoint." In 1937, Cartan was awarded the N. Lobachevski prize for his series of papers on geometry. In 1931, he was elected to the French Academy of Science and later he became a foreign member of the US National Academy of Science and a foreign Fellow of the Royal Society. His contemporaries noted that Cartan had been an exceptional teacher.

Remember that the principal directions on a submanifold are those which are principal for all second fundamental forms simultaneously. Cartan stated the following result.

THE GEOMETRY OF SUBMANIFOLDS

292

Theorem If a domain in L" is immersed into E2i-1 then at each its points there exist n principal directions.

Before giving the proof in the general case, consider n = 3, when the theorem follows from elementary reasons. Introduce a coordinate system u1, u'-, u3 such that at some fixed point Po one has g,1= Sid. At Po the Gauss equations are

- (L112)2 + LiIL`zz - (Li,)2 = -1+

Li1Liz

LiILa3 - (Li32 +LiILa3 - (L13)2 = -1,

- (4)2 + L;,L23 - (Li3)2 = -1+

L 2Li3

LiILi3 - Li2Li3 + LiILi3 - Li2 L13 = 0,

Li2Li3 - Li2Li3 + Li2Li3 - LiILi3 = 0. L33L21 - Ls2L31 + L33L221

-

LJL31 =

0.

Moreover, the system of coordinates can be chosen in such a way that at P0 one of the second fundamental forms, say III. would be diagonal: Lv = 0 for i 0 j. Then the system above has the form I

z

I

z

L11L22+LIIL22-(L12)-=-1+ LiILa3+Li1L333-(L13)2=-1+ LbL33 + L;,Li3 (L23)2 = -1, ,2 L212 L213+ L222 L213 - - L2 12L2 23+ L2 33"21 =

-

,2 ,2 II

2

-

L2 1 L231 .

Set Did = LuLJ - (L,)2 for i # j. Suppose that all L j j = 0. Then LI I L22 = - I, L11La3 = -1, L212L33 = -1. Therefore, (L11L L33)2 = --I, which is absurd. Hence, at least one of A,j is non-zero. Set that A12 0 0. Multiply the both sides of the fourth equation of the system above by Lie and use the fifth equation. We get L12 1'-22 3L22

= L22LI2 2'- 13 = (L12)21-,3 2

Therefore (L11 L22 - (L12)2)L223 = 0. As 012 # 0, a3 = 0. The fourth equation of the Gauss system implies either L12 = 0 or L13 = 0. We may set that L12 = 0. The fifth

implies that either L13 = 0 or L22 = 0. If L13 = 0 then Lie = L13 = L;3 = 0 and, hence, the matrix of II2 is diagonal, too, which completes the proof. Therefore, suppose that L22 = 0. Then we have L11Li2=-1,

LLL33=-1, LiIL33+LiILi3-(Li3)2=-I.

From L11Li = -1, L;,L33 = -1

it follows that L11 = L. Then (L11)2+ LiILi3 - (L13)2 = -1. Let us take it that Po corresponds to u; = 0. Introduce a new coordinate system v1i v2, v3 related to the old one by u1 =COSWVI+SincPV3,

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

293

u3 = -sin,pv1 +cosWV3, u2 = V2,

where p is a constant selected in such a way that L-23 = 0 with respect to the new system of coordinates. We have 8W au;

2 8u1 8uj &I OV;

al'] OV3

Since L22=L12=L;3=0, thenL12=L23=0and L2 13 =

2

(LI

8ul + LZ OU3 OuI (L2 Out + L233 (9u3) C7u3 19 8v3 I3 a},3 X I + X.3 C7V3 8Vl

\

= 2 L1 I - LIB) sin 2,p + L13 cos 2V.

Hence, there always exists the angle ,p such that L13 = 0. Consider III. Since L', = LI33+ I

I - L33) sin gyp + L13 cos

L13 = I (L I

(n1r,,,,,,)

L12 =

8u2

0,

= 0.

I

Analogously we state that L23 = 0. Thus, by means of orthogonal rotations in tangent space, both matrices III and 1122 obtain a diagonal form. So. the theorem is proved for n = 3. Turn now to the n-dimensional case. In this case the system of Gauss equations has a solution L°. -- the coefficients of the second fundamental forms A', a = I , ... , n - 1. Since the curvature of Lobachevski space K < 0, we can introduce one more form An = vKds and rewrite the Gauss system in the form "

E (A°(X, Y)A°(Z, W) - A°(X, W )A°(Z, Y)) = 0,

(1)

n=1

where X, Y, Z, Ware vectors in an n-dimensional vector space V. These forms we call externally orthogonal.

Further on, we shall follow the proof given by Moore in an interesting and important paper [6]. Lemma Let A',..., A" be the set of n externally orthogonal symmetric bilinear forms in n-dimensional real vector space with the following property: if x E V satisfies A"(X, Y) = 0 for all Y E V and all A = 1, 2,....n then X = 0. Then there exists a real orthogonal matrix 11a,,11 and n linear functionals cp1, ... , V" such that

Aaaa1
119

THE GEOMETRY OF SUIfMANIFOLDS

Clarify the latter notations. If

_ E it, [hr' then

The matrix of this bilinear form is

[riu

uj

«3lI

[r The lemma means that the matrices of forms Aa are diagonal with respect to basis 'Pi

Relate to each matrix AA the linear operator Aa on 1'. Denote by VA V a set of bivectors [XY], where X. YE V. To each ordinary bivector [XY] we can map the bivector [A'XA"Y]. Therefore. the linear operator A" induces the linear mapping AA A A- of VA V into itself.

Show that Aa A A' = 0 if A' = ±9,k where y^'' is a linear form. Indeed. suppose that A" = fy: c- p', where p' = it, (la'. Then al

A"X = (aX )a.

where a =

lr Hence [A".%7.4"Y] = (aX)(aY)[aa] = 0 Conversely. let A" AAA=O, i.e. [A "XA" Y] = 0 for any X. Y. This means that AAX and .4 S Y are collinear. Therefore.

the image of operator AA is a one-dimensional space and A4 = ((iX)h. where a. h are vectors. The matrix of Aa is a1hi

a:hi

aih2

[r2h2

al h

As Aa is a symmetric operator, a,b, = a,h,. i.e. [abj = 0. So. a and h are collinear: h = pa. Therefore

A'X=p(aX)a If p < 0 If Is > 0 then introduce a linear form .p' _ 11i a, drr. Then .4,' = rA then introduce a linear form pA = V--it a, the. Then A = y7'. The hypothesis of the theorem can be reformulated as follows: let us be given the

linear operators A'.....A" on an n-dimensional vector space I' such that (AIXY) = (XA`)") We can write the condition of external orthogonality as

A'AA"=0. a-I

(2)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

295

Indeed, consider the value of AAA AA on bivectors [e,e;] generated by the basis vectors e;, ej. We have (A' AA')[e,ei] = [A'e;AAej].

The components of the latter bivector (we mark them with a subscript near the brackets) are [AAe; AAei]kl

_

(AAe;)k I

(AAe,),

(Aae!)k (AAei),

I

L'a

1'Jk

Lu

Lill

Therefore, the system of Gauss equations (1) can really be written as one bivector equation (2). Further, the Lemma hypothesis means that the intersection of kernels of bilinear forms A', ... , A" consists of zero vectors:

nKerAA=0. A

Now show that by means of an orthogonal matrix ([aA.11 we can pass from the system

of operators AA to the system of operators B" of the form

B" =>a. AA. p= 1,...,n, such that B" A Bµ = 0 for each p. Suppose that there exists X E V such that the system of vectors A' (X), ... , A"(X) is linearly independent. Prove it by contradiction. Let X be a fixed vector. Denote by U(X) a space generated by A'(X). A = l .... , n. Let p be a maximum of dimensions of U(X), when X varies in V. Suppose that p < n. If Xo is such a vector that dim U(Xo) = p then without loss of generality we can assume that the subsystem

A'(Xo),...,AP(Xo) is linearly independent while AP" (Xo) = 0, ..., A"(Xo) = 0. If Y is any other vector from V then (2) and the conditions on Xo imply P

EA^(Xo)AA"(Y)=0.

(3)

u=I

If Xo and Y are fixed then we can put the vectors A"(Xo) and A"( Y) we can put into correspondence with the linear forms. The result holds: Lemma of Cartan In order that given linearly independent forms f , ... , fP (the rank matrix of coefficients of these forms is p) satisfy the identity P

A=1

296

THE GEOMETRY OF SUBMANIFOLDS

it is necessary and sufficient that each form gyp' is a linear sum of the forms f k with a symmetric matrix of coefficients:

EC,

k

k

k

See the proof in [7] p. 20. So, we have P

A°(Y) = EcnA3(Xo).

(4)

0=1

Thus any vector of the form A° (Y ), a = I,... , p, Y E V, can be expressed in terms

of A1(Xo), 0 = 1, ... p. If we denote by W a subspace generated by A"(Y ), Q = 1, ... , p, Y E V then (4) shows that the dimension of W is p. As p < n, there exists Z E V which is orthogonal to W, i.e. A°(Y, Z) = 0,

a = 1, ... ,p.

But from the properties of A 1, ... , A" it follows that there exist A and N such that AA(N, Z) 0 0,

where, evidently, A > p + 1. If e > 0 is sufficiently small then the vectors A° (cos a Xo + sine N ), I < a < p, generate W. On the other hand Aa (cos a Xo+ sine N) is located beyond W. Indeed, ZA'(coseXo+sin eN) = sin eAA(Z,N) 34 0. Therefore, the subspace generated by

A"(cos e Xo + sine N), µ= 1,...,n is at least (p + 1)-dimensional. This contradicts the definition of p. So, p = n. Let v1,.. . , v,, be a basis in V such that the system

A'(vl),...,A"(v1) is linearly independent. In application to the equation A"(vl) A A"(v') = 0 1i=1

Cartan's lemma implies that there exists a symmetric matrix C(i) = flc(i ) 11 such that

AA(v,) _ Ec(i)NA"(v1),

I < A < n,

1A

where C(1) is a unit matrix. Use the equation A"(vi) A A\(vi) = 0. a=1

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

297

Substituting now the expressions above for AA(v,) and AA(vj), we have

>(c(i )N c(j )"

v
- c(j)VA c(i)N)A1'(v1) A A`(v1) = 0.

As the system 41 (vi ), .... A"(vi) is linearly independent, the coefficients of bivectors A!'(vj) A A"(vi) are zero, that is

Dc(i) c(j ) - c(j )

e(1)N)

= 0.

(5)

Note that C(i) are symmetric matrices. Equation (5) and the symmetricity property of C(i) mean that C(i) and C(j) commute. From the linear algebra it follows that there exists an orthogonal matrix A = 11aA J) such that each of the matrices AC(i)A',

I < i < n is diagonal. Here A' means transported A. Set 0

BA = >2 aaA".

(6)

a=I

Introduce a diagonal matrix

Consider BA(v1)

fri

0

AC(i)A'. Since A' = A'1, AC(i) = T(i)A.

T(i) 0

Tn

In more detail: al

AC(i) _

...

an

... ... ... ai ... a, )

...

Tn

cn

... ... ... ( ... c, cm,

0 ) (all ...

T 0

Cl...

a "I

a a,",

Therefore, aA, c'= TA a,A,. We have

BA(vi) = >2TAavA"(vl) = TABA(vl)

So, every BA(v,) is proportional to BA(vi). Therefore, the bivectors generated by BA(vj)i = 1,...,n are zero: BA(vr) A BA(vj) = 0.

This implies BA A BA = 0.

As was proved above, in this case there exists a one-dimensional form 0A such that e = ftpA (9 V%.

THE GEOMETRY OF SUBMANIFOLDS

298

From (6) we find As:

As = E d BI' _

fag; GPs 0 ips,

where Ild 11 is the matrix inverse to A.

Take ps at the fixed point of an immersed domain as the differentials of new coordinates, i.e. set dria = yoa.

Then each second fundamental form L du' duj is diagonal with respect to a new system of coordinates. Then the theorem is proved. Note that Moore [6] proved the existence of asymptotic coordinates on the immersed domain L" C E2"- 1. 10.6 Lemma on Principle Directions on the Submanifolds of Negative Curvature to be Holonomic

Lemma Let R' be a submanifold of negative curvature in Euclidean space EN. Suppose that at each point in Rthere exist m principal directions. Then they are holonomic.

Denote by T,, a tangent space at x E R"' and by np the normals to RI. Denote by a comma the covariant differentiation in E"'. By definition, the prinicipal direction X E T, is that which satisfies

np.,X'=AX+ayny

(1)

for any np. Denote by X I , .... Xthe field of nr unit principal directions. We say they are holononic if for any X1 there exists a family of hyper-surfaces orthogonal to X,. This hypersurface, then. can be taken as a coordinate surface. To prove the lemma it is necessary and sufficient to prove that for any pair Xk, X; of vectors orthogonal to Xj. the bracket [XkX1] is orthogonal to X. Consider, for instance, [XI, X2] and show that it can be expressed via X, and X2. We have np.r Xj' = Apt X, + > pvp/I nq.

(2)

Let K(XI. Xj) = K, be the curvature of RI with respect to the 2-plane generated by X;, X/. We have K(Xi. Xj) = F, Ap+Apj'

(3)

P

Multiply (2) by Ape and sum over p. Taking into account (3), we obtain Apt np.i X, = KI2 XI + Ape E l4ep/ I ny.

(4)

H

Differentiate equation (4) covariantly in the X2 direction. Thus, we have Ap211p.ij XIX2 + Ap211p.l XI,i XZ = K12 X1.i X4 + ....

(5)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

299

Here and subsequently we denote by dots the linear combinations of X1, X2 and ny. Since np,,, means second covariant derivatives of the vector field np in V. np,, _ ,tp',. So. we can write

n

Xi, _ (11,X2) I X1 -11,X2 r Xi _ (Ap,XX +

,yp.2nq) I

i - np, Y; r X1,

(6)

=Ap2X2,X1, -III, X±IX[+. Substituting (6) into (5). we obtain an equation which we shall use below. Interchanging the roles of XI and X. we get the second equation. Denote by [XI X2}' the j- th component of brackets and set %2 = E ,,(A,,)2. Then the system of equations just obtained can be represented as

K,2XI,X_-X-X',) =EAp2)tpr[XlX2J'+..-. p (7)

K12X2,Xi-X,X;"riEAr, lip, [Y2X,J'+. p

Using system (7). we find that

(Ki,--, - )[XIX2)=(K12-ii)EA, np,[XjX2J, r

+(KI:- )

)EAplnrr[XIX2}'+....

18)

I,

Since RIO' is an integrable submanifold. f X 1 X2J can be expressed in the form of a linear combination of tangent bases. Suppose that [X1 X2[ has a non-zero projection onto XA, k * 1.2. that is [X, X2] = PXA +...,

where p 96 0 and the dots mean the terms without X,1. As XA is a principal direction Ap+np,[X,X2}'=PExMxpr,Z.+...

=pKAXA+ ---

M

Project (8) onto XA. Taking into account (9) and dividing out p y6 0, we get (Ki2 -

i illi ) = K2k(Ki2 -'}i) + KIA(K12 - "r2).

If the curvature of R' With respect to any 2-plane is negative then the right-hand side of the latter equality is positive while the left-hand side is non-positive. Hence, p = 0 and the statement is proved.

THE GEOMETRY OF SUBMANIFOLDS

300

10.7 Metric Form with Respect to Curvature Coordinates From the lemma stated in the previous section it follows that in an immersed domain we can introduce an orthogonal system of coordinates uI, ... , u" such that the coordinate curves are tangent to the principal directions. It happens that after scaling the coordinate curves the metric form has a specific form. The following theorem, stated in [8], [9], holds:

Theorem (Aminov) Let a domain in L" be isometrically immersed into E2" Then in this domain we can introduce local coordinates uI..... u such that coordinate curves are tangent to principal directions and the metric obtains the form I

"

sin2 o; d:r.

d52 = r= I

where IT

(I)

sin2o-, = I. r=1

Observe that in E3 the orthogonal systems of coordinates with condition (1) were considered by Gishar and Darboux [10] and in E" by Bianchi [11]. Since at each point of an immersed domain there exist n principal directions, the normal connection is flat. In this case we can choose the normal basis of normal 0. Let vector fields parallel in the normal bundle. With respect to this basis Lp be the coefficients of the second fundamental form with respect to n,,. Since the coordinate curves ui are tangent to the principal directions, Lf = 0 for i 34j. Use, then, the equations of Codazzi for the immersion of L" into E21-I. We have

L°.A-L°ti=0.

(2)

Observe that Lea, = 0 for i 0 k because of the choice of a system of coordinates. Consider (2) in more detail, setting i = j A k: 8Lpa

_ I'LL°,+I';;Lkk =0.

(3)

where rfk are Christoffel symbols. If g;; are metric coefficients then I',k =

2gi; duk'

rk

-

2gkk (JUL

Multiply (3) by Lj° and sum over p. On the left-hand side we obtain

8 duk

E(' " P

)2-

1

agii (Le)2+gjiagii=0. ,i

gii &,,A-

P

(7!!k

(4)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

101

If we set

h

cot- a,=

() ] /g,,,

_

2

P

then (4) has the form O(ln g,, - In sing a,)

#i

-0

Therefore, by means of a suitable choice of parameter in the it, coordinate curve, we get g,, = sin` a,. Then

E(LF)2 = cos'- a, sin' a,. P

Now we prove (1). Consider the set of g, p = I..... it - 1. where i is fixed, as some vectors of length cos a, sin a, in normal space and project it onto coordinate axes in normal space. Set

p=I.....tt-I; t=1.....n.

(5)

It is evident that the angles y,,n must satisfy n-I

cos3 OW _

(6)

n=1

for any I. Since the curvature of L'" is -1. for any 196 j we have

L! = -g g,, = - stn" a, sin2 a,.

(7)

Using (5) and (7) we obtain n-i

Cosyor Costp,, _ -tang, tan a,.

(8)

P=I

The equalities (6) and (8) express the condition on the n-matrix A to be orthogonal sin al

sin an

COS Sot I COS al

...

cos tat,,-1 cos al

...

Cos.p

cos

cos an cos an

Namely, the product of two distinct columns is zero due to (8) and the square of each column is I due to (6). The orthogonality conditions for rows produce (1).

THE GEOMETRY OF SUBMANIFOLDS

302

10.8 The Fundamental System of Equations for the Immersion of Ln into E2n-' (System "LE")

In consequence we use the notation Hi = sin aj. As ds2 = E" , Hi du; is the metric of the Lobachevski space of curvature -l, the coefficients Hj must satisfy some special system of equations. Let Riikj be the components of the Riemannian tensor of the metric ds2 = E" , H2 du;. It is well known that if dsz = gjjdu; duj is the metric of a Riemannian space of constant curvature Ko then the Riemannian tensor satisfies the relations Rjjk1= Ko (gik gij - gil gik)

For our case we can write these conditions as

Rkyi=0,

i34j;

Ri;ii=Hj2H2,

k3&i.

(1)

In addition, we have (2)

Rewrite (I) in an expanded form while replacing (2) with a system of differential equations. To do this, we use the expressions for a Riemannian tensor with respect to orthogonal coordinates (see (37.4)-(37.6) from Eisenhart's "Riemannian geometry"). In terms of Darboux symbols /3; j = N- 2, i 0j, [3;; = 0, equations(I)can be written as aQ'j+80jj+EQkhOki

auj

auj

k=1

ajki=Q

ji

4 k#i i1 96

kj$j;,

auj

= H,H;, .

Differentiating (2) in uj, we get aHj auj

E

Q%%Hq,

( 3)

q=1.q#J

From the definition of Darboux symbols: aHj = $,; H,, auj

i0j

( 4)

Write the compatibility condition of the system (3)-(4). Differentiating (3) in uj and dividing out Hj # 0, we have a2Hj aujauj

E (10A QiH q

q=1.q#i

+Q. aHg 811i

)

n

(lQi,Qjq H4 + QhQw Hj) - --" Hi + Oilk#i E QtkHk q=1.q#j.i

auj

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

303

In the last term the subscript k can be equal to j. We can write 1l

01r>2/3dH4 = it

>2

/3,rgHH)

v=1.yr!

R#t

Hence r7-H,

= - 1:

d,gdwH, -

du,Oa,

Old''

H, +

(5)

Differentiating (4) in a, implies _q=Hl 011,01t,

= t3r3u if, +

6)

011,

Subtracting (5) from (6). we obtain V3

tad

r3Nd,,, =0.

t1t,,all, +

q-I 4+ !

Thus. we passed from the system (1). (2) to the system with respect to functions If, and 13,,:

(u)

13i,Ht,

r Out 011A

-

(h)

rr

,3xyHy.

q 96 k.

q=I w

{c)

X13 11

u,

= i3 ljA,

(d)

13

Ou14t

+ Ou, +

0.

(7)

,=1

8r3A, _ H,Hh, {e} P++>24s3,t auk 1

where Q k k are distinct from the range 1.... , n. We shall call this system the fundamental ,s wtent of isonnetriak immersion of L" into

E14-1, or in shortened from: the system "LE" (Lobachevsky, Euclid). The former integral of this system is the relation Er,=1 H, = const. Indeed, mul-

tiply (a) by H, and sum over all I. 10 k. Then multiply (b) by Hk and add the expressions obtained in that way. We get

28ak ( !=I

H')

=EJ3k,HkH,->}3tgHtHq =0. 1

q

Due to the choice of initial conditions we can take 1 as the constant in this former integral. Therefore, we can state that (7) is equivalent to the system (1). (2). On the other hand. the fundamental system of equations of isometric immersion of one manifold into another is formed with the equations of Gauss, Codazzi and Ricci. As at each point of the immersed domain there exist tt principal directions. the

304

THE GEOMETRY OF SUBMANIFOLDS

normal connection is flat. Therefore, we can choose the normal basis field parallel in

a normal bundle. In this case the torsion coefficients p,,, , = 0. In addition, with respect to the curvature coordinates the coefficients of the second fundamental forms satisfy L f = 0, i 0 j. Therefore, the Ricci equation is satisfied identically. The Gauss equations are of the form n-1

Rijki = 1(L°&Ljpl - L lLjk).

(8)

P=1

It is known that with respect to the orthogonal coordinate system the Riemannian tensor component is zero if all subscripts Q, k,1 are distinct. On the other hand, the right-hand side of (8) is equal to zero too, because L = 0 for i 36 j. So, if all Q, k, I are distinct then (8) is satisfied. Suppose, now, that two of them, say j and k, coincide while i, j and / are distinct. The right-hand side of (8) stays zero. The equation has the form Rkjii = 0, i.e. it coincides with the second equation of (1). If among Q, k, I there are two pairs of equal subscripts, say i = 1, j = k and i 96 j, then (8) has the form n-1

Rijij = - > Lii4 = sine ai sine aj. p=1

Here we applied (7) from Section 7. Thus, the Gauss equations have the form of system (1). Set'I,,p = cos aj cos Ipjp. Then the expressions from Section 7 for L , can The orthogonal matrix A from Section 7 has the form be written as L, = H1 H1

Hn

A= 4II

I

'In-1

t.-1

...

...

Hn

LI11

LInn

HI

H.

Ln-1

Lnnn-1

11

Hn

HI

We can write the Codazzi equations

6`k-rkL,+r:Lkk =0, i &k as

dip -

=f3ki4Pkp,

k#1.

(9)

Uk

The orthogonality of matrix A implies Ej 1 Hj4bjp = 0. Differentiating the latter relation in u, and using (a), (b) of (7) and (9), we get Qi1Hitlp+IQaHit2p+...+tip(- rl(figHgl

LGg

+HI Oil dip + H2a124bip +

+ Hi

ui

+...

//J1

`0 + -

= 0.

ISOMETRIC IMMERSIONS OF LORACHEYSKI SPACE INTO EUCLIDEAN SPACE

305

Therefore

=0. A

Since for regular immersion H, 94 0, the expression in parentheses above is zero. Thus, fi,,, satisfies the following system: ;1t, 'Pip.

k -A i,

(10)

c

all,

Denote the i-th column of A by a,: H,

oil

u, -

Then (a). (b) of (7) and (10), (11) can be written in the following form OR,

(12)

aq.

+

The compatibility conditions of this system are the equations (c). (d) of (7). Indeed, if k A j 10 k. Then ;3+lr i3!,

8ut8u! Hence.

M= I ,',,,

(tA + A, ,3

Now let i

a,r

02u`

!t

011,0uu

!

r

tt

/. We have

a, - i,!

`d aq +

lII, t!w

!

+

E 13ry aY -

'_ - ` Differentiating

i! J !r! + ;3r r

4

Y.Y#j

;3" uy + 13!r Y Yfr

r

a, + rlrr i3 u,.

E3'y;iry +

=$a, in it,, we get 6Pa, 013#

01110u,

Ou,

U,+l1'lrail u,

Therefore tL

du,J!!,

oil,

+

a, = 0. i,

a)

THE GEOMETRY OF SUBMANIFOLDS

306

by virtue of d) of system (7). Thus, the system (10)-(l 1) together with the Codazzi system has (7) as the compatibility conditions. The system of Gauss-Codazzi-Ricci has (7) as a compatibility condition. Therefore, we shall call system (7) the.fundamental system of isometric immersion of L" into E2"-1. In Section 20 we shall give the matrix form of this system. System (7) serves for the determination of functions H,. If they are found then from (10){11) we find $,p. Then the coefficients of second fundamental forms are defined as LPi = H1 ,p, L,j = 0 i 31j. The initial conditions on d jp can not be taken arbitrarily. Since the system (10)--(l l) contains all the partial derivatives of and moreover the equations are resolved with respect to them, we may take the initial conditions on tip at a single point Po. Show that if the matrix A at Po is orthogonal

then it is orthogonal in all the domain of solvability. Introduce the functions Tjk = (ajak) - 8,k, where (ajak) is a scalar product of aj and ak. If Q, k are distinct then BTjk 0-U,

= Qij(arak) + /3rk(aja,) = f3ijTik + $3,4. Tj,.

Set i = j 34 k. Then BT;k

ou'

- E Qkq (agak) + Aka? _ q

f3Iq Tqk + /3;k TI q#k

Finally, set all subscripts to coincide. Then BT;,

r

_ 8u , = -2 `'f3,q(aiaq) q

2 E l3igTiq q

Thus, we see that Tjk satisfies the homogeneous system of linear differential equations IM BTjk_ Bu,

Bjkr

Tu.,

i+j, k

where B; are the functions of f3,.. If the regular solution of (7) is known then Btj; are definite and regular of the corresponding class. From the theorem of uniqueness for the solution of the system of linear differential equations it follows that if Tjk = 0 at

the initial point then T,. =_ 0. Hence, if A is orthogonal at one point P0 then it is orthogonal in all the domain of solvability of (7), (10)-(l l). 10.9 Gauss and Weingarten Decomposition

A position vector of a submanifold, which corresponds to a given solution of the Gauss-Codazzi-Ricci system, can be found from the system of equations which consists of Gauss decompositions and Weingarten decompositions: k

r,,,,,, = r,j r"k + Ljj no, n,,,, _ -Lijng A r,,,, .

(1)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

307

Remember that torsion coefficients are zero. Consider this system in our case. We have I'u r,,, + L° n,,.

rwu, = I; r,,; +

(2)

kk#i

The Christoffel symbols are

rk=-Hk$ki

ri;=Hji

Let Ti _ as

be unit vectors tangent to coordinate curves. Then (2) can be represented R

8r;

a=-1:OkiTk+IN,n,.

(3)

k=1

Here the summation over a is assumed. For i $ j, we have rau

=rjjru,+rjr.,+E

PUk

k#i.j

But r = 0 for distinct Q, k. In addition

18H; H, Buj '

jTIOHj Hj 0u;

I',j

Therefore, /3;j Tj, i # j. The Weingarten decomposition has the simple form n,,,,, = --Dtt,T. Finally, the system of Gauss and Weingarten decompositions is of the form

Sr;_ 8u;

-

Oki Tk +,k, n,,, k

(4)

uj = A j Ti,

n,

Ti.

The coefficients (3k;, -ti, are known functions of u1.... , up. The compatibility condition for this system gives the system of Gauss-Codazzi-Ricci equations. If we find the solution ri, no of that system then the position vector of the submanifold can be found as r(ui , ... ,

r(Po) + > I

J T;HI du;.

Note that at the initial point Po all the vectors ri(Po), no(Po) are chosen in such a way that they are unit and mutually orthogonal.

THE GEOMETRY OF SUBMANIFOLDS

308

10.10 Local Analytic Immersions

Find the arbitrariness in determining the local analytic solution of (7) Section 8. This system belongs to the class of Bourlet-type systems (in Bianchi terminology, see [I I]):

aU, axk = T;k

aUi

UI,...,

Here the system of functions UI, ... , U, forms the required solution. The variable Xk is called main to U; if the system is resolved with respect to All the other variables are called parametric to U,. It is supposed that the right-hand sides contain, maybe, the derivatives aU;/ax, in a parametric variable x, that enters linearly if I > k and if I = k then j > i. We suppose that by virtue of the system the integrability condition r>:u, CPU' x = sk sk is valid with respect to main variables.

If the assumption that'P;k are analytic functions is made the existence theorem holds [2]: in some neighborhood of an initial point (x0,,..., xx) there exists a unique system of analytic functions Ui, ... , U, such that if main variables take the initial values then U, is an arbitrary given analytic function of parametric variables. Apply this theorem to our system. Verify that compatibility conditions are satisfied. System (7) Section 8 is resolved with respect to all derivatives of Hi. The equality of mixed derivatives holds because of (c) and (d). Order the functions /3,;

lexicographically and set /312 = U1, /313 = U2..... The following rule holds: if

(3i;=Uo,/3k,=U1then a
O$ik

O$ki

au,

auk

+ A,k ,

0% au,

0Qlk

auk

+ B,k,

where A& and B,k are functions of /3,; and H. According to our notations we have

aU,

au,

au,

auk

+ Aik,

0U,

OU,

auk + B;k.

au,

The right-hand sides are linear with respect to derivatives of the required functions; also k > i. Therefore the solvability condition of (7) Section 8 is satisfied. Moreover, each function U, satisfies the equation = p,;/3;k. Therefore, for the function U, = $,k the variables u3, u,, where j 54 i, k, are main variables while uk is the parametric one. Verify the compatibility condition a2 U,

02 U,

aukau; - au;au,

092 U,

aA,k

aukau; + au;

a au;

1iak

Taking the second derivative of U interchange the order of differentiation on the right-hand side, i.e. we do it as 0-2

a2 U,

/3k; . U' a a-k' a;, + a-'' aukau; = au;auk = auk WOW = auk auk

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

309

Taking into account that A;k = - Er 13a/3Pk + H;Hk, we reduce the compatibility condition to

k+a

+>[3Pj33Pk-HkHj

ij

(

)

l3PrQPi -

L

HrH

= 0,

which is satisfied by virtue of (e). In an analogous way we verify the equality of mixed derivatives 7wr and with respect to main variables. Thus, (7) Section 8 is totally integrable. Separate the initial conditions on (3, j into two groups: 01'2 (142),

013(u3),

...,

13

I203(u3),

...,

Ys(un),

n(40,

132'1 (U2),

...,

#. (un),

0 3 ' 2( U 3 ) ,---?.

N.2 (un),

031(u3).

I

An0-In(un),

where Ji°(uj) is the initial value of #j j on the coordinate curve uj (the other arguments are fixed). The number of initial functions is n(n - 1). Each initial function depends

on one argument and must be analytic. The number of functions on distinct coordinate curves is distinct. On the ul-curve there are no functions, on the u2-curve there are two functions, on the u3-curve there are four functions, and so on. 10.11 The Principal Vectors of Normal Curvature and the Indicatrix of Normal Curvature

By definition, the i-th principal vector of normal curvature is n-I LP

k, = E n nr P=1 grr

Since L, = cos ar sin ar cos IPrP, 8a = sin2 a:, n-1

k; =cota;ECosv;PnP. P=1

The length of k; is cot O. The scalar product of two vectors of normal curvature is n-1

(krkj) = cot at cot aj E cos W;, cos pjp = -1 P=1

due to the orthogonality of matrix A. Consider the indicatrix of normal curvature at a point x of a submanifold. By definition, this is the subset in normal space formed by the endpoints of all normal curvature vectors from x with respect to all tangent directions at x. Let y,. .. , yrt-1 be the coordinate system with respect to the basis

nl,... , nn_ I in N. For any unit tangent direction X= {`}, the components of the

310

THE GEOMETRY OF SUBMANIFOLDS

FIGURE 47

normal curvature vector with respect to this direction are of the form y,, = )2 Set ), In a space of parameters A1,... , A., the equation Ei Aig = I and the inequalities A, > 0 represent the simplex. The normal curvature indicatrix is the image of this simplex under the linear mapping y,, = >; Lsa,. It is easy to see that this mapping is not degenerate. Thus, the normal curvature indicatrix of L" C is the non-degenerate simplex of dimension (n - 1), see Figure 47. The asymptotic directions on the submanifold are mapped into x. The vectors from x to the simplex E2"-1

vertices are the vectors of principal normal curvatures k,. The product of two principal normal curvature vectors is (k,kj) = -1. The simplex face opposite to the i-th vertex consists of linear combinations of kj - k,, j 96 i,13& 1. Since (kj - k,, k,) = 0, k, is orthogonal to this face. Therefore, the point x is the common point of simplex altitudes. Consider the case of n = 3. There are only two unit normals n1i n2 and therefore instead of cos Vi1 and cos V,2 it is convenient to use cos cp, and sin cp,, setting gyp,, ='p,.

We represent the coefficients of the second fundamental forms as Lr, = cos a, sin a, cos gyp,,

L = cos a, sin a, sin gyp,.

The Gauss equations produce tan a, tan aj = - cos(cp, - pj), i 0 j. Under the assumption that det A = I (note that A is an orthogonal matrix), we rind sin (vi -yp2) _ sin (

sin a3

- coSa1 cosa2 ' sin a1

- W3) _ - cos a2 cos a3 '

sin (V3 - V0 _

(I )

sin a2

- COs 01 cos a3

Since for n = 3 the normal space is a plane. the normal curvature indicatrix is a triangle. Denote its vertices by A, B, C. The image of unit sphere S22 C Tr under the mapping generated by the normal curvature vectors covers this triangle eight times.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

311

The normal curvature vectors have the form k; = cot 0j (cos Vj n, + sin Vj n2). Therefore, the angle between kj and kj is equal to tpj - tpj. The vectors kj go from x the common point of heights of triangle ABC - to the vertices A, B, C. Find some elements of triangle ABC. Set I BCI = a. From triangle xBC we find a2

= kl +k2

- 21k111k21 cos(V2 - WI)

= cot2 a, + cot2 a2 + 2 cot al cot a2 tan a, tan 02

=cotta,+cot2a2+2=

2a+

sin

sin a2

,

2OS2 a32

=

1

sin a, sin a2

Therefore, a = cos a3/ sin a, sin a2. We find the analogous expressions for b = I ACI and c = I ABI: b

=

cos al

sin 02 sin a3

,

c=

COS 02 (2)

sin a, sin 03

Note that the length of each side is greater than f. Find the angles of triangle ABC. Let a, Q, ry be the values of angles about vertices A, B, C respectively. We have a2 = b2 + c2 - 2bc cos a. Therefore co__ =

b2 + c2 - a2 The

_

cos2 a, sin2 al + COS2 a2 sin2 a2 - COS2 a3 sin2 03

2 sin or, sin a2 cos 0, cos a2 sin2 a, sin2 a2 = tan al tan 02. sin or, sin 02 cos or, cos 02

Using the relation tan a, tan a2 = - cos(Vl - V2), we have a = x - (9 - VI). In a similar manner we state that Q = it - (tp3 - V2), -y = 7r - M - .p, ). Find the altitude hc from the vertex C: he = bI sin al = b1 sin(ir - ( p 2 - Bpi ))I = bI sin 1('p2 - API )I =

2

sin 202

Here we use (1)-(2). Thus, he = 2 /sin 202. The area of triangle ABC is I

SAeC = 2

hcc

_

COSa2

I

= sin 202 sin or, sin a3

I

3v

2 sin or, sin a2 sin a3 - 2

Now state a consequence of the Codazzi equations which we shall need in Section 16. Apply (10) Section 8. In our case they have the form 8 cos cpj cos aj

- cos 0k cos (Ok 8 Sin aj Sin ak

8Uk

8 sin cpj cos aj 8Uk

8Uk

cos ak sin cpk 8 Sin aj sin ak 8Uk

By means of these equations we find

a

kk=

Bnok sin(tpk-`,j) k.

THE GEOMETRY OF SUBMANIFOLDS

312

If we use a notation 3i j =; H; then j = cos ak sin(vk - pj)f3kj.

(3)

COS aj

auk

10.12 Asymptotic Lines and the Chebyshev Property

The asymptotic directions are related to the principal directions in a simple way. Let v be a tangent direction defined by

_ =c,, where a1. i=1,...,n. & Therefore, v = Efe; r,,;. Suppose that the direction v is asymptotic. For any normal du;

np we have do ds

_ 8n du;

8u ds -

-L11g"rN,

du; TS

= -top eir,,,IH,.

If r = r(s) is a parametric representation of the asymptotic line then for any p (np r,,,) = 0 must hold, or, equivalently, (np, r,(s)) = 0. Taking v as r we get by virtue of the orthogonality of matrix A

(vnp:) = - E Hi = 0. Hence, v is the asymptotic direction. The number of asymptotic directions is 2' Taking n vector fields, constructed by means of asymptotic directions, we can construct a system of coordinates with asymptotic lines as the coordinate curves. These coordinates are called asymptotic. Since the choice of n from 2"-1 directions is ambiguous, the choice of asymptotic system of coordinates is ambiguous. With respect to an asymptotic coordinate system the metric of a submanifold has a

Chebyshev form. From this it follows that the lengths of opposite edges of a curvilinear coordinate parallelepiped are the same (see Figure 48). Take, for instance, as coordinate asymptotic directions v, the following vectors:

Denote the asymptotic coordinates by a1. Their relations to the curvature coordinates are realized by means of a matrix of components ±1: ul U2

u"

-I 1

1

-1

... ...

1

...

1

1

al

I 11a2

-l

a"

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

G it

313

2144

FIGURE 48

Therefore, the metric obtains the form

ds'FH,'rltr; n H; dn +? p

Since

F N,' - H;' - H, dap dnq. ,topq

r-1

H,2 = 1, the metric can be represented as ds2

Fdor2,+2 p

(1 -2H -2H4)dapdaq. p
Since the coefficient of each dal is equal to 1, the length of each edge of the coordinate parallelepipeds is the same, i.e. the coordinate net is the Chebyshev one. In the case n = 3 we have

ds2 = Flop+2 cosw,da,dok, where Q, k are distinct and the angles 1w, I < sr.

From here we obtain two generalizations of inequality (4) Section 3. One generalization is linear. The integral geodesic curvature of asymptotic lines on a twodimensional surface constructed by two families of asymptotic lines is bounded: Jidoi < sr.

314

THE GEOMETRY OF SUBMANIFOLDS

The second generalization (two-dimensional) is the following. The integral inner curvature of this surface is bounded also: fKtdSI <27r. =III

For example. for the surface o3

coast this inequality follows from the equation

KJ __

012

1

W3

2 sin w3 801C%2' -

Indeed, the metric of this surface has the form: dh2f + 2 cosw3 dal dal + da. 10.13 Grassmann Image of Immersion of L" into E2"-'

Denote a Grassmann image by r". If a submanifold F" C E`" has flat normal connection and ul, ... , uk are the coordinates of curvature then the metric of t" is 2

r"

2 duk

p

,r"

(

r°-EE\LkA) k=I p=1

gkk

As Lkk = cos ok Sin Ok COS tpkp, gkk = sin2 ak, the metric of the Grassmann image r"'

of immersion of L" into E2"-1 has the form It

dew _ E cost a; du?. Since the metric of L" with respect to the curvature coordinates is ds2 = the following result is true.

HH duj ,

Theorem 1 The sum of the metrics of a Lobachevski space and its Grassmann image is the flat metric dc2+dsl2,, =due

From this it follows that parameters u; have a geometrical meaning. Theorem 2

For any k-dimensional submanifold Fk, k _> 2. in an immersed domain

of L" C E2i-1, it > 2, the k-dimensional volume of its Grassmann image is greater than the volume of the original in that submanifold. The volume of the image ought to be found taking into account the multiplicity of covering. Let a k-dimensional submanifold Fk in L" be represented by

ui = ui(xl,...,xk).

i= 1,....n.

The coefficients of the induced metric on Fk are gki

aui au, _ L sin2 ai dxk , axi

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

315

Set

gl,...lk

= 8xk

8xk

It is easy to find the determinant of the metric tensor gkj:

g=

sinz a;,

... sin2 a;k (gi,...lk)z.

11 <... <1k

Using the form of the metric of the Grassmann image of L", find the determinant g of the W (Fk) metric tensor:

,k= E cost 0;, ... cost a,

(ql,...1* )2.

h< ...
Note that if n > 2 then cos a; > sin aj, i 54 j. Indeed, (1)

cost a1 = I - sin 2 al =

sin- a, > sin2 ail

where (i) over the sum sign means that the subscript i is omitted. From the latter inequality we find that g > g if k < 2, n > 2. If n = 2 the inequality turns into equality. From Theorem 2 follows the proposition on non-immersibility of a complete Lobachevski space: there exists no regular isometric immersion of complete L" into E2"-1 such that the Grassmann image r" is located in a closed n-dimensional submanifold and the Grassmann mapping of L" onto Imo" is finite-sheeted.

The length of a curve taken arbitrarily in L" may be decreased under Grassmann mapping. The exclusion consists of asymptotic lines: the ratio of the length of the

asymptotic line arc Grassmann image to the length of this arc is

n --I. This

statement easily follows from the equality E, cost a; = n - 1. The surface created by two families of asymptotic lines we shall call asymptotic. The following more precise theorem holds: the ratio of ' image area of an asymptotic surface to the area of the original is greater than 2(n - 2). Suppose, for instance, that an asymptotic surface is tangent to a pair of asymptotic directions of components with respect to u;. Let all 02 be the parameters in this surface, namely, the arc lengths of asymptotic

lines. Then the Plucker coordinates of the tangent plane of the surface are

q12=q1j=0, -q"=q21=2,i,j=3,...,n.

Therefore, the area element of the surface is f doI dal = 2[(sin20,1 + sin 2 a2)(sin2 03 +... + sin 2 a")]

112

do1 dal.

THE GEOMETRY OF SUBMANIFOLDS

316

The area element of the Grassmann image of the surface is f dal dal = 2 [(COS2 a, + Cost 0'2)(COS2 a3 +... + Cost an)]

I/2

dal dal.

Consider the ratio

ff -

1(2 I

- sine or, - sine a2)(n - 3 + sine or, + sine a2)

1 /2

(1 - sine a, -sin 2 a2)(sin2 or, + sin2a2)

Note that

n-3+sin2a,+sin2a2>n-2

2-sin2a, -sinea2>2 1 - sin2 a, - sin2a2

sine a, + Sine a2

The latter inequalities imply f > 2(n - 2)g. The theorem is proved. Let X and Y be tangent vectors to T", k(X, Y) the curvature of a Grassmann manifold with respect to the plane of (X, Y). By means of formula (2) Section 21 Chapter 8 and taking into account the expressions of the second fundamental forms, we can write n

K(X,Y)=

E

sine ar sin2

E

COS2 Or COS2 as(grs)e

r ,n1

r,r-I

Since the submanifold is of flat normal connection, then by a Theorem from Section 21 Chapter 8 this curvature satisfies 0 < K(X, Y) < 1. In the case of immersion of L" into E24-1 the extremal values are not reachable: 0 < IC(X, Y) < I. From the expression above it follows that K(X, Y) is equal to the ratio of the bivector area element in L" to the area element of the corresponding bivector in r". Therefore, the area S of any two-dimensional submanifold F2 in L" is expressed by

S= *(F2)

where dw is the area element of %p(F2), X, Y span the tangent plane to W(F2).

10.14 The Analogy of a Pseudosphere

The isometric immersion of part of three-dimensional Lobachevski space L3 into five-dimensional Euclidean space E5 as a pseudosphere analog was given first in Brill's paper [13] in 1885. In the following year Schur expanded this result to ndimensional space [14]. Let a,, I < i < n - I be non-zero real numbers such that metric representation of the immersion is as follows: x2i_, = a1 ey- cos (y;/a;),

x21 = a; ey- sin (yi/a;),

aj = 1. The para-

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

r0

I=-J

1 - e2" du,

317

-00
Y.

where I < i < n - 1. Here xJ are Cartesian coordinates in EZ"-I, yA, are curvilinear

coordinates in the submanifold. Note that y < 0. Find the metric of this submanifold. We have 1

o

-ey'^ sin (yilai)

i
e)'A cos (y,/a,)

ryq=

ale" cos(y)/al) ale`'' sin(yl/al)

0

I - e2y

0

From the expressions it follows that y; form the orthogonal coordinate net in the submanifold and 8i; =r2 =e2y..

-I =eZr,1: a?+(1 -e2`'^) = 1.

8,",

1= I

Therefore, the metric of the submanifold has the form

dy

ds2 = e'-y'" (dyr +... +

.

This is the metric of the space of constant negative curvature - 1. The coordinate curves y,, are geodesics. Changing constants a,. but saving the condition E,a? = 1, we obtain a continuous bending of the submanifold. The metric reduces to the form 1=1 H2 du, with Er_I H? = I if we set dy; _

dui

d)' =

1,

I - e''Y1. du",

ey°

Find Hj as a function of us,.. . ,

r J

We have

dy,,

1-e2.

=c-Inle-y.+ e-2y'a-l

The constant c may be chosen arbitrarily. So, set c = 0. Then

e_

1

cosh h u '

1

sinh u

H;- cosh u h -1 ' H"cosh

I
THE GEOMETRY OF SUBMANIFOLDS

318

Thus, the metric of L° with respect to the curvature coordinates has the form 2

_du?

+(n - 1 ) sinh2u,, du,-,' cosh2

1)

In the case of n = 3 the unit normals can be taken of the form

nl =

a1 cos(yl/al)

-a2 Cos(y1/al)

a1 sin (y1 /a1)

-a2 sin (),I /aI )

a2 cos (y2/a2)

1 - e2j7

,

n2 =

a2 sin (y2/a2)

-ey'/

at cos (y2/a2)

a1 sin (y2/a2) 0

1 - e'2y,

Introduce the new parameters zl, z2, z3 by setting zl

al

z2

a2

z3 = arc cosh a

Then the fundamental forms of the submanifold will be represented as ds- _

ai dzi + a2 dz2 - sinh2 z3 dz'l I

cosh- Z3

II = 1

sinh z3

,

z (a, dzIz + a2 dz2 - dz3),

cosh` z3 aIa2 (dz1 II2 = cosh z3

- *2).

It is easy to verify that the torsion coefficients 0i2/1 = (nln2,,,) _- 0. Hence, the normal basis n1, n2 is parallel in the normal bundle. In the case of n = 3 consider a projection of the submanifold just constructed into the hyperplane of coordinates XI, X2, x3, .r4. The projection position vector has the form

u3al cos(y)/al) u3a1 sin(yl/al) u3a2 cos (y2/a2) u3a2 sin (y2/a2)

where U3 = ey'. This is a three-dimensional hypersurface F3 in E4 with two components of boundary. The first component corresponds to u3 = 0 and consists of one point - the origin; the second component corresponds to u3 = I and consists of the Clifford torus. The hypersurface is covered with a family of Clifford tori which corresponds to distinct values of Ira. The metric of this hypersurface is ds2 = u3 (dy? + dy.,) + du3.

The curvature of this metric with respect to the 2-plane tangent to the surface U3 = coast is equal to -1 /u3 . The other curvatures are zero. Therefore, this is the

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

319

surface of non-positive curvature. The hypersurface is homeomorphic to the domain obtained from the sphere S3 by cutting out one isolated point and the interior of the torus. The submanifold just constructed, i.e. the the immersion of a domain of L3 into E5, is defined by means of values of the fifth coordinate over the points of F3, where this fifth coordinate tends to infinity when the point tends to the isolated one.

10.15 The Immersion of L3 into E5 with a Single Family of Lines of Curvature which are Geodesic

Consider the immersion of V into E5 such that the fines of curvature of the uIfamily are geodesics in V. Find the arbitrariness in the determination of such an immersion and show that in this case the fundamental system of immersion reduces to the sine-Gordon equation. The theorem holds: Theorem Let some domain in V be isometrically immersed into E5 such that the lines of curvature of the ul family are geodesics in L3 and the metric of L3 has the

form ds e = sine a du, + cost o (sine y duZ + cost y du2). ).

(1)

Then a depends only on uI and is the inversion of the elliptic integral u1 =

f

da

costo--

,

a, 00,

(2)

---2cp

where co, cI are constants and -y is a sum of two functions rl and w of uI and u2, u3 respectively: y = T(uO) + w(u2i u3), where w satisfies the sine-Gordon equation: wwu, - w,,,,,, = -c sin 2w,

where c = V;OT+ C,

and q is expressed via a: 2ri = arccos

+ CI

c).

(TC-OCOTO-

Conversely, if a and y are given and satisfy (2)-(4) then there exists the isometric immersion of L3 into E5 such that the lines of curvature of the uI family are geodesics in L3 and the metric with respect to curvature coordinates has the form (1) with given a and y. The arbitrariness for the solution of the equation on w consists of two functions of one argument. Let us prove this theorem. As uI is geodesic, then aw'° = 0, i=2,3, i.e. a is a function only of uI. Consider the equations Ruph = 0 for 176 k. Use the expressions for Riemannian tensor components with respect to orthogonal coordinates (see [1] to

THE GEOMETRY OF SUBMANIFOLDS

320

Chapter 3). Then taking into account that H, = H, (u,) the equations R1223 = 0 and R1332 = 0 mean that

aH2 0 In H3

02H2

01410143 = 0143

au,

aH3 0 In H2 au,

a2H3 '

01410142 = 0842

Rewrite them in terms of Darboux symbols: 0032 = 031012,

aul

'9(323

0ul

= 021013

From this it follows that 032 = - /13 and 023 =

/H2 are functions of only u2 and 113. Substituting the expressions of H2 and H3, we find YU UI = 0,

"YU2UI = 0.

Therefore, y = y(ul) + ry(u2, u3 ), where y' and ry are some functions. Note that the equation R2113 = 0 is satisfied identically. Consider now the Gauss equations R1221 = sin2 a, sin2 a2,

R1331 = sin2 al sin2 03.

They have the form

a

1

13 cos a sin y

au,

sin or

0ul

0

aul

1

a cos a cos y

sin or

au,

= sin a cos a sin -y,

= sin a cos a cos y.

If we multiply the first by sin y, the second by cos -y and add, we obtain

du = -cot ayu, -sin a cos a.

(3)

1

If we multiply the first by cos y, the second by - sin y and add, we obtain 2 a all In (csina y"') = 0.

From the latter equation we find y,,, = co sin a/ cos2 a, where co is constant. Substitute this expression into (3) and multiply by 2a,,,. Integrating, we come to the equation of first order:

a,, =cos2 a-

co

cos2 a

-2c1.

From this follows (2). Note that a,,, $ 0 because in the opposite case (3) implies sin a cos a - 0, which is a contradiction of the regularity condition of immersion. Now analyze the equation R2332 = sin2 a2 sin2 a3. In expanded form that equation is

0 au2

aH3 H2 5u2 I

I a I aH2 OH2 OH3 _ +au3 H3 5u3) + Hi aul aul

H2H3-

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO

SPACE

321

The substitution of H, gives EP

8u3

o, + 8u:"

1

8 cos a sin y

sin- a

8111

0 cos a cos -8u1

3 COS a cos

sin ').

Since 7 = - (uI) +'j(iu2, a:). we can represent this equation as iuur, - 'lrnx = 7 sin 2't

I

+ COS`t a + COO ell7 t 1 + Cos 21 cot a"}ur au,.

Here we came to an absolutely wonderful fact -- while the coefficients of sin 21 and

cos try in the right-hand side are variable, the sum of their squares is constant. Replace c with the expression obtained above. Then we have 1-11

hurt - yu:a = 2 sin 2y [cots alai + CoS a + 2CI ] + cos 21 cot.ayrr, qu, = sin 21

[cot' a-)+ CI] +cos 2-) cot

tai au, .

(4)

Set

A = cot-(nU, +cI. B= Coto'),, aai. Write

A2 +B-= I_' °, +2c11 + L COS- a

°,

COS'

a

f cos-a-

\

, -?c11

COS- a

Thus, (4) has the form A sin 21 + B Cos 2-),

(5)

where A2 + B3 = t'3 is constant. Set B R cos9=-. sine=-.

Prove that due to the choice of 0 and the following equations on - and a: yar =

-

co sin a

,

cos- a ,

1

-a-

COS,

aut - U

c

a - 2t't . cos- tr

the equality (2y + 0),, = 0 holds. We have sine = co

cos- a -

co

cos- a

- 2c1

/2 cos a.

Differentiating in uI, we get co

co stns

1

co sina

cosSea = - cos a cos a sin a + Cas- o/ c- + C cos'- a a- 2,7u,

("+Ci)/C.

322

THE GEOMETRY OF SUBMANIFOLDS

Consider the expression cos 6 = ( + c1) /c from another point of view. Using the previous relation, we can write cos 89,,, = -2y,,, cos 0. Assuming that cos 0 $_ 0, we get

the required relation 0,,, + 2rya, = 0. The case 0 =_ 0 is excluded because it implies a = const which is impossible for regular immersion. Equation (5) has the form y`-n,,, - 'tN,e; = c sin(2ry + c2).

The latter reduces to the sine-Gordon equation if we set w = y + 2. Also, set rt = I - c2. Then y = Tt + w. The case co = c, = 0 is also possible. Then the equation for w has the form of a wave equation, y = const and a., = f cos a. The converse statement on the existence of immersion follows from the fact that the fundamental system of immersion has been reduced to the equations from the hypothesis of the theorem. 10.16 Local Immersions of L3 into E5 with Hyperplanar Grassmann Image

The Grassmann image r3 of the regular immersion of a domain in L3 is a regular submanifold in a Grassmann manifold G2,5. Map to each point in G2,5 of the Plucker coordinates pU the vector p in E10 of coordinates p'2...., p45. This correspondence defines the standard imbedding of G2.5 into E10. The Grassmann image is called hyperplanar if it is located in a hyperplane of E10, that is, in some subspace of E9 (see Figure 49). The P10cker coordinates of the point in r3 satisfy the linear equation

Ea,jpli+a=0,

(1)

i
where aid and a are constants. State the criterion on the immersion of V into E5 to have a hyperplanar Grassmann image.

Theorem I The Grassmann image r3 of immersed domain of L3 into E3 is hyperplanar if there exist the constant C1 not equal to zero simultaneously and the constant a such that CIQ,!

- CJ$lr = aHkEYk,

where eUk are the Kronecker symbols.

FIGURE 49

(2)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

323

FIGURE 50

Moreover, note that this immersion has the following remarkable property: the submanifold admits the motion by itself along the curves of constant curvature. Write (1) in terms of the scalar product in E3: (pa) + a = 0. where a is a constant 10-dimensional vector with components (a,,}. Differentiating this relation, we obtain that for any shift dp along r3 the scalar product (dp a) = 0. Hence, the constant vector a is a linear combination of normal basis vectors of v. We shall construct this basis later. Let ri, r2, rr be the unit vectors along the principal directions at a point x E V. F, be a unit vector in the normal plane N along the i-th principal vector of normal curvature while 17, is a unit vector in N, orthogonal to , (see Figure 50, where the triangle means the normal curvature indicatrix). If ni.n2 are the fields of unit normals parallel in a normal bundle and mutually orthogonal, then & and ri, can be represented as , = cos V, nl + sin,, n2,

17, _ - sin y:, n i + cos p, n2.

(3)

Denote by [ ] the bivector generated by the vectors in V. The lemma holds;

Lemma The system of vectors in 00 1= 1, 2, 3,

(r,-. ,n],

[r3, rill,

Ira, r$],

p

(4)

form an orthogonal basis of normals of the submanifold r3 c E10. The unit vectors [r n,], 1= 1, 2; 3 are tangent to P. Observe that the normals jr, ty,] to 173 are located in the tangent space to G2.5 while [r T.] are the normals to G2,S. First, it is easy to see that all of the vectors have unit length and are mutually orthogonal. The following formula holds (see Section 1 Chapter 1).

([abl[cd]) = (ac)(bd) - (ad)(bc).

(5)

By direct substitution we obtain the orthonormal property of basis (4). Find the vectors tangent to r3. We can represent the position vector of r3 as (nln21. Let L, be the coefficients of the second fundamental forms of the immersion of V with respect

THE GEOMETRY OF SUBMANIFOLDS

324

to nk. Using Weingarten decompositions for the derivatives of nk, we find the tangent

vectors to r3: r

Lu[nir,,,I)H;-

[nin2] _ -(Li[r,n2] +

In notations H, = sin ai, h; = cos a; the coefficients of the second fundamental forms can be represented as

L = Hihi cos Vi,

Li2i = H;h; sin Vi.

If we use the relation (3) then ap = -[ri ii]h;, ui

i = 1, 2, 3.

Using (5) it is easy to verify that the vectors (4) are orthogonal to p,,,, i.e. they are normals to the submanifold 173. Show, for instance, that p,,, is orthogonal to 1741. We have

([rilli[[rjV[) = (ri7j)(r1iEj) - (ritj)(q,rj) Since j is normal to L3, (r1 t j) = 0. If i = j then (fit e) = 0. If i 0 j then (T;rj) = 0.

Hence the expression on the right is zero. So, the lemma is proved. Now prove the theorem. Represent the constant vector a as a linear combination of basis vectors (4): a = E a, [r, 4,1 + E bk ei/k [rirjj - ap.

(6)

i<J

i

The coefficient of p is -a because (ap) _ -a by (1). The other coefficients are still indefinite. The condition on a to be of constant length produces the system of differential equations which we obtain by differentiating both sides of (6) in ui, u2, u3. To find this system, take some preliminary actions. Write the Gauss decomposition

for the second derivatives of position vector r of immersion of V into ES, which have been obtained in (4) Section 9, in the form 8r;

j

8u

= J3ij rj,

car

su,

= -I3ji rj - Oki rk + hi tti.

(7)

By means of Weingarten decompositions write the derivatives of normal vectors: 6(1

auk

3pi H;Hk = r7i auk + rk, h,

i

8u,

ft - h,r,.

= r)i M1

(8)

Introduce the notations: ai = [ri ti], pi = [Ti ri;], vi j = [r,-rj]. Using (7)-(8), write the derivatives of normal basis vectors of r3 with respect to the u2 coordinate. In these derivatives expressions in t;] and [ri t2] occur with i 96 2. We represent them with respect to basis (4) as follows. Set tpij = gyp; - tpj. Applying (3), we find j = coscpji6 +sin cpjir,;,

'ibj = -sin ,pjiti +cosWjirh

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

325

Therefore, we can represent, for instance, j via 2 and t]2 and then substitute into [r2Ei}. As a result we obtain the following expression for the derivatives of normals to P3 with respect to the u2 coordinate: aa;

atp,

H2Hi h,

aa2_

E,3,2 (cos co,, Jai + sin (P,2i µi) + P2

au2 -

r=1.3

M2,

P2

&2

(9)

avn 0142

&'13 aU2

=

-l3k2Vik + h

2 ( cos(pa, J

j+s in(p2jpli),

= fl 12V23 + 1332V12.

Here i,k 34 2. Now differentiate both sides of (6) in u2:

as au2

aaj

=

8U2

aai

k + aj

au2

abk

+ au2

E jA Vii

+ bk ei k

avi j

au2

-

( 10 )

Multiply scalarly both sides by v,,,:

+&2Ersk +bkfijk( O 2 Vrs) = 0.

ai{

Taking three distinct vectors of v,,, and using (9), we obtain the system:

ab, au2

-

H2 H3 h3

a: - 1312b2 = 0 ,

+ 1312b1 +03A = 0, 03+HH,a1 h1

0b2

(I)

-f332b2=0.

This system gives the expressions for the first derivatives of coefficients bi. Multiplying scalarly (10) by µr, we obtain aj

aai aN2

ll

Pr) + bkEijk

(,ii P) - (Ybr2 = 0.

In more detailed form the latter gives a' au2 - a242

sin VJ2 + bkh2 sin

j = 0, j = 1, 3,

a2"+1: aiQj2 sin cPJ2+ah2=0.

THE GEOMETRY OF SIIRMANIFOLDS

326

These equations do not contain the derivatives of a, and h,. Finally, we obtain the third system after multiplication of (10) by A,: 111

t?u:

t1tt

+{

1=1.3,

hAh±) cos p2, = 0,

k-762.1.

(III) + E a,, J,, cos w^+, = 0. ,=13

By differentiation in u1 and u3 from (6) we find new equations, which, however, can be found by an appropriate change of subscripts from the already obtained equations. Analyze the systems (I)-(ITI). Consider, first, the equations of (II). Use the Codazzi equation for the immersion of L; into E5 (see (3) Section 11): Vik.

h

sin w,1 dar

!

Substituting this expression into the first two equations of (11) and dividing out by

sin,.,. we obtain a,.12,I' -t12.1,2h,+hA/i I,, =0. With the help of this equation. replace the expression -a242 + hAh2 in first two equations of (III) and. also, use (8) Section 7. Then

0lna, _0InI,=0, 1=1,3. ()it

tht2

Therefore, there exists a function C,(ul. u) such that (t, = h,C,011. a:) With the help of the equations obtained after differentiation of (6) in ul and a3 we can see that a,/h, for j = 2, 3 does not depend on it, and a,/h, for i = I,2 does not depend on tt3. From this it follows that C, depend only on tt,. Later we shall show that as a matter of fact C, are constants. From the third equation of (111) and (7) Section 8 we obtain

By changing subscripts 2 -- 3 -- I - 2 we get two more equations. Set 1"A, = C )A, Note that TA, = -TA. Then these three equations can he represented in a common form: h;

i)C, i)tt,

= H,{ TA,HA +

where subscripts i, /. k form a cyclic permutation of 1.2.3. Consider the third equation of (11). For the derivative equation: hY

= falht sin y711 +

3h3 sin vji,

use the Codazzi

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

327

Substitute the latter expression into the third equation of (II) and replace sin1pjj by (1) Section 11. Then we get T21 H3 + T32HI + ahZ = 0.

(12)

Performing the cyclic permutation of 1,2.3 we get T32H1 + T13H2 + ah3 = 0,

(13)

T13H2+T21H3+ah1 =0.

(14)

Subtracting (13) from (12) and adding (14), we get 2T21 H3 + a(hi + h2 - h;) = 0. As h' + h2 + h2 = 2, T21 = -aH3. In general, we have Tip = aerykHk. i.e. the required relation (2) holds. Substitute these expressions into (11). Then we get h2 8u2

= H2(T12H1 + T32H3) = 0.

Therefore, Cj = const. Theorem I is proved. Consider now the equations of (I). By means of (5) and (6) we get b1 = C3023 - C21332,

b3 = C2012 - C1021

The permutation of subscripts also gives b2 = C11331 - C3013-

Replace b, in the first equation of (II) with the expressions just obtained. Then we find C3

(2L '- H2H3 + 13121313) - C2

- 012131 CI = 0.

Using (c) and (e) of (7) Section 8, we find

Ciauj1332=0. This equation means that 032 is a constant with respect to the direction C = {C,}. From the other two equations of (I) we obtain that (313,1312 are also constant with respect to the direction C. The coefficients Hi are constant with respect to C, too. Consider, for instance, the derivative of HI. Using (a) and (b) of (7) Section 8, we get Ci!8Hj o-uj

H2 T21 + H3 T31 = 0.

So, the following statement holds: if the Grassmann image I'3 of an immersion of three-dimensional Lobachevski space L3 into E5 is hyperplanar then on the immersed domain there exists a constant (with respect to the coordinate system uj) direction such that all Hi and (33k are constant along this direction.

From this it follows that the submanifold admits the motion by itself along the curves tangent to this field of directions. See [16] for a more detailed proof.

328

THE GEOMETRY OF SUBMANIFOLDS

10.17 The Fundamental System of Immersion of L3 into ES with Hyperplanar Grassmann Image

We separate the general (in other words, not particular) case, when C,? 0 C? and all C; 9k 0. In this case we say that the immersion is of general form. Theorem In the general case the system of immersion of L3 into E5 with hyperplanar Grassmann image is totally integrable and its solution is definite with given constants.

To simplify the explanation we consider the case of a = 0, when r3 is located in a hyperplane E9 passing through the origin. Without this assumption but with difficult calculations the proof was given in [16].

Besides the special direction C = {Ci} there exists one more special direction r = { 1 /Ci}. Taking into account n = 3, a = 0 and (2) Section 16, write one of the equations of system (7) Section 8 aQu

aQki

-, + auk + QjiQjk = HiHk

(1)

Make the substitution f31k = Ck,iki/Ci. We get Ck aQki

T, aU, Use the relations Qij,3jk =

aQki

C,

+ auk + C' Qi jQjk = HiHk

= Ck ° 1C,. Therefore. (I) has the form

I aQki

I

aQki

I aQki

H,Hk

C, Ou, + Ck ank + U, 0uj

(2)

Ck

On the right-hand side we have the derivatives of,3k1 in the r-direction. Introduce in the submanifold the coordinate system xi by setting ui = Cix1 + c, x2 + mix3, where

mi are constants such that the Jacobian 0 of coordinate change is not zero. Introduce the vectors m = {mi} and d = [CT]. Then d will have the components dk = sijk(C? - Cj2)/CiC1. The Jacobian 0 = (md). These coordinates xi are chosen in such a way that Hi and fiij depend only on x2, x3 and the derivatives of Qki in x2 are given by equations (2), i.e. = H;H,t/Ck. Write the system of immersion (7) Section 8 with respect to curvature coordinates. Set bk = eijk(Cj,3i j - C,$,,). Let us take as new unknown functions the components of three-dimensional vectors H = {Hi} and B = {Bi}, where Bi = bi /CjCk. We shall rewrite the required system in vector notation. Introduce the following vectors K = {Ki}, where Ki = HjHkdi /CjCk, and M = {Bi/C?}. With the help of the first two equations of (1) and (2) we obtain OH

OB

_ -K,

(3)

X2

where [ ] means the vector product of three-dimensional vectors. Let F be a vector of components Fi = -(f3kjmk - Qjk mj)Eijk. Set n = {n,} = [mC],

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

329

(m,C,-,lkCL )/ C,Ct. The components F, can be expressed in terms of B, from d,F, + nB, = 0. From the first two equations of (7) Section 8 we find that

p,

E= [FHI.

To find the derivative of B in x3, we use the third equation from (7) Section 8, by taking the derivatives = (p, v + n, oiz +d, ) of 44 and .3j,. Note that B and H do not depend on xl, and the derivative of B in r2 can be found by (3). Introduce one more vector T = {T, = n,H,Ht/C,CL }. We can make the following system:

H=[FH].

B=-[BF]+T.

(4)

So. the system of immersion of L3 into E5 with a hyperplanar Grassmann image for n = 0 has in the general case the form of a system of four differential equations with two independent variables x_, x3;

c?H=[BH] ,

N =[FH] .

-K

(5 )

,

=-[BF ]+ T.

(

6)

Now we pass to the proof of the theorem. Prove, first, that

B, - T. = [FBI.

(7)

This is a sufficiently simple problem. Consider, for instance, the first component of the relation. By virtue of (6) we have t

8x3

H2 H3

_ [FB]i + C,C3

where (J, is the first component of the vector product. Using (5), we get

8F,

t7

I

0-V2 = L-+C, atr,

_)?13332)

=

H_H;ni C_C3

Subtracting the latter equation from the previous. we get 'fit, - A-= [FB] 1, Analogous relations hold for other components. By means of (7) it is easy to obtain the equality of mixed derivatives of H.

__ _ __ tlt210x3

_

8r30x= [[FB]H] + [[BH]FJ + [[HFJB] = 0.

Further, by means of (5)-(6). we obtain 02B

d=B

60x36x2

8x28x3

=-K,,,-T,.-[KF] -[TB] =Q.

Taking into account (5)-(6). by direct calculation we find that Q = 0. Hence, the equality of the mixed derivatives of H holds. The theorem is proved.

THE GEOMETRY OF SUBMANIFOLDS

330

Since the system (5)-(6) is resolved with respect to derivatives of H and B in x2 and

x3, then we can give these vector components arbitrary values only at the initial point. If we take x2 = 0, x3 = 0 as the initial point, then we see that the arbitrariness of the immersion system solution depends on five constants (we set E, H12 = 1). Since the initial condition is analytic and the right-hand side of the system is analytic, then the solution is analytic, too. The corresponding immersion of L3 into E5 in the domain, where this immersion is of class C3, is also analytic.

10.18 Seven First Integrals

If the reader was patient enough to get over the long analytic calculations in the previous section then he or she will be rewarded with seven first integrals. Immersions with a hyperplanar Grassmann image are outstanding due to the property of the fundamental system of immersion to have a large number of first integrals.

Theorem I (Aminov)

If a domain in L3 is immersed into E5 with a hyperplanar

Grassmann image of general form then the fundamental system of immersion has seven

first integrals: three of them are linear with respect to Hi and /3,j:

Cjj-Ciu31r=aHAe;

(1)

and four of them are quadratic:

H2 = const, 3

(HB) = const,

(2)

3

H,C2-BrIC2C2k =const, iOjy6 k#i,

(3)

r=1

r=1

(H1 + aBr}

/-+

B2 = const.

(4)

Here, as above, H = {HI}, B = {Bi = eij* (C'r'c.-cc's")}. In analogy to the first integral in the problem of the rigid body moving, we shall call the first integral (H8) = const by the integral of areas.

From this theorem it follows that the fundamental system of immersion is integrable in quadratures. The existence of first integrals of type (1) has been stated in Section 16. The first integral H2 = coast is the first integral of the original system (7) Section 8. Prove that (HB) = const. We have axe

(HB) = (BHB) - (HK) _ -(HK), (HB) = (FHB) + (H, - [BF) + T) = (HT).

3

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

331

It happens that K and T are orthogonal to H. Indeed

(HK)=HIH2H3(dC)=0, CI CZC3

(HT) =

HIH2H3 CI CZC3

(nC) = 0.

The first integral (,3) can be found in a purely geometrical way. Come back to the expression of the normal vector a to the hyperplane E3 which contains 13 (see Section 16). As a is constant, it has a constant length. The coefficients of decomposition have been found above: a, = h;C;,

b; = B,C,Ck.

We have

a2=I:a?+Eb; =E C?(l -H?)+> B,C'Ck =const. From this follows (3). Denote by 4i the left-hand side of (4) for a = 0. Also, set H = {H;/C2}. From the system we have 1

2 axe

Note that [HHJ = K. Hence,

=(HBH)-(BK).

= 0. Now find the derivative of 4 in x3. Using (6),

we get

28x3

(HFH)+(BT).

Observe that F, and B, are related to each other by found F, = -n;B;/d;. Recalling

the expressions for K and T, it can be found that (HFH) = (KF) = -(BT). So, I = 0. Thus 4 is the first integral. From the theorem just proved follows the next statement. Theorem 2 Each solution of the fundamental system of immersion of L3 into E5 with a hyperplanar Grassmann image of general form is definite over the whole space of parameters uI, u2, u3 and is analytic.

The coordinate change from u; to x; is linear. Therefore, consider the solutions of the system (5)-(6). The right-hand sides of this system are defined for all values of required functions H,, B, and depends on them analytically. However, it is well known that the right-hand sides' regularity does not guarantee the existence of a solution over the whole unbounded domain if the right-hand side is nonlinear with respect to the required solution. But in our case the solution existence is guaranteed by the existence of the first integral (4). Use the following statement proved in [18]. Let f (t, y) be a continuous vector function definite in the closure E of the open (t, y)-set E. Suppose that the Cauchy problem

y, =f(t,Y), Y(to) =Yo, admits the solution over the right maximal interval J. Then either J = [to, oo) or b. J = [to, b), b < oo and (6,y(6)) E 8E or J = [to, 6),6 < oo and [y(t)[ - oo when t

333

THE GLt)61FrRl of SUBAIANIF'ULDS

Apply this statement to each subsystem of (5) and (6). By virtue of the first integral (4) the solution of (5) (6) is bounded. Therefore. for each subsystem only the first possibility is admissible: J = [r(;, x). Analogous statement is true for the left maximal interval. So. the solutions of these systems are defined oNer the whole x2-axis and over the whole x:-axis. We construct the solution of the system (5)--(6) b) means of the solutions of the subsystems. At x±o. xv, we fix the initial conditions t . Ba and find the solution of (5) defined over the whole x_-axis. Then we find the. solutions of (5) at Y _,.x31 as initial conditions. The set of solutions of (6), obtained in this way, depends on the parameter x3, and due to the total integrability of system (5)-(6) it produces the required solution H. B over the whole plane a ^3. v3. The proved theorem does not guarantee that a corresponding immersion of L3 into E` is regular because there the points of H, = 0 and H, < 0 may occur. But for the regular immersion H, > 0 must be true at every point. The existence of singular points of immersion follows from Theorem 3 If the Grassmann image of an immersion of the general form of a complete space L3 into E5 is located in a hyperplane passing through the origin and the first integral of areas equals zero then the immersion has singular points.

Consider system (5). Differentiating the first of these equations in x we get op if =

i).t;

-[KH] - B2H.

(7)

Here we use the condition (HB) = 0 from the hypothesis. Introduce the following

notations: ]q = 1 /Ca - 1/('2, t'_ = I /C2 - 1/(,2. v3 = I /C; - 1/('2. Without loss of generality, assume that C? is the maximal of Cr. Then 1.t > 0, m, < 0. The following two variants are possible: (1) r', > 0. From (7) select the equation on 141:

H+H_i, In this case Him - Hi?I^ > O. (2) vI < 0 From (7) select the equation on H:: t?-H

= -H_(B2 + He=r a

- H= vj ).

Then Hiv3-H12111 > 0.

Consider case (l) because case (2) can be considered in an analogous way.

From the equation 1A= -K. which we represent in coordinate form as -H,HHd it follows that each B, is a monotone function over the v2-axis. Hence, there exists a point x; in the xI-axis such that either over the half-axis (x;. -X.)

the function B; Is greater than some fixed positive or over the half-axis number. Therefore, over this half-axis B2 + H; vi - H? v2 > BB; > 0. where B is a

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

333

positive constant. Since in a half axis we can take a segment of any length however

large with the property that the solution of equation d = -B; j, will have at least two zeros over this segment. then by the Sturm theorem the solution of (8) is zero over the segment. Thus, there exists the point H3 = 0. The submanifold has a singularity at this point. 10.19 isometric Immersion of L3 Into E5 with Hyperplanar Grassmann Image and the Motion of a Rigid Body with Fixed Center of Gravity in the Field of Gravitation

In this section we shall state the relation of the theory of isometric immersion of a Lobachevski space to the known classical problem of mechanics of the motion of a rigid body around a fixed point. If the rigid body is in a central Newton gravitation field then this problem reduces to the solution of the so-called Kirchoff system of six ordinary differential equations. We shall consider this system in the case when the center of gravity of the body coincides with the fixed point:

A 1 = (B - C)(qr - el',) BtY _ (C-A)(pr-F'y>"),

(1)

Crlr = (A - B)(pq -?7') Here A. B and C are the principal moments of inertia of the body, {1.1'. , "} is the unit basis vector of the immovable system of coordinates with respect to the axes of

the body-related moving system of coordinates. { p. q. r} is the vector of instantaneous angular velocity with respect to the moving system of coordinates (see [17D.

Show that if the Grassmann image r3 of a submanifold is located in the hyperplane passing through the origin, i.e. a = 0. then the corresponding system of 12 differential equations (5)-(6) Section 17 contains a subsystem of six equations coinciding with the Kirchoff system (1). This relation between the theory of isometrical immersion and the problem of mechanics shows, on the one hand, the natural, from the physical point of view. character of the many-dimensional fundamental system of immersion of L" into El- 1 and, on the other hand, allows one to apply the methods developed in mechanics to the construction of solutions of an immersion system. Consider the system (5)-(6) from the previous section. Note that d,F, = -n,B,. In

consequence we suppose that it, 0 0. We can always reach this condition by the choice of vector in. Set A, = d/n,. Consider system (6). Write, for instance, the component of OH/8.r3 :

aB3 =

-AIF9 = (Ay - A3)F2F3 + 1

'H3

C:! C3

THE GEOMETRY OF SUBMANIFOLDS

334

Show that e(A2 - A3) _ -n, ICZC3, where e =,';d and A= (md). Indeed, consider the vector product [nd] = [[mc]d] = AC - m(cd).

As (cd) = 0, then [nd] = AC. In a particular case, n2d3 - n3d2 = ACi. Hence, A3 - A2 = ni/eC2C3. So, we can write Al

OF1

8x3

= (A3 - A2)(F2F3 - eH2H3).

In an analogous way we can represent the other two equations of O. Therefore, system (6) Section 17 can be represented as

8H 8x3

= [FH],

Ai 8F' 8x3

_ (Ak - AJ)(FjFk -

"'"z"'

where e = c,c2c,n is the constant number. If e > 0 then these equations coincide up

to notations with the equations of motion of a rigid body having the moment of inertia A, around the center of gravity in the central Newton field of gravitation (1) (see [17]). In this case F is the vector of instantaneous angular velocity, H is the unit basis vector of an immovable axis having components H, with respect to the moving system of coordinates. The time parameter t = x3. So, if there exists the general case immersion of L3 into E5 with a Grassmann image located in a hyperplane E9 passing through the origin then H and B along x3 describe the motion of a rigid body. In [ 171 the value of a is 3g/R, where g is the acceleration of gravity a distance R from the center of gravity. Four of the first integrals (3H4) Section 18 for a = 0 turn into linear combinations of four known first integrals of system (1) (see [17]). In our case they are the first integrals of a wider system in partial derivatives consisting of 12 equations. The moments of inertia A; of a rigid body are positive numbers satisfying the triangle inequality A; + Aj > Ak. In our consideration the values A; = d;/n; satisfy the condition above with some values of C; and m;. In this case e > 0, also. Moreover, for any three positive numbers A; not equal to each other there can be found three distinct numbers C? 34 0 and three numbers mi such that Aj = d;/ni, (md) 4 0 and e > 0. Therefore, any motion of a rigid body with distinct moments of inertia A, around the center of gravity of the body in a central gravity field can be mapped to some isometrical immersion of L3 into E5. The Kirchhoff equations have been considered in the papers of Clebsch, Steklov, Chaplygin and others in connection with the study of rigid body motion inertia in an unbounded ideal liquid. They are intensively studied nowadays because of their importance in mechanics. The quadratures of (1) have been found in papers by Weber and Chaplygin under

the condition that (HB) = 0, i.e. when the integral of areas is zero. Without this condition, the solution in implicit and complicated quadratures was obtained by Cobb. The explicit solutions are known for some particular cases, for example for the Steklov case, see [17]. This solution can be used to construct the immersion system solution, see [16].

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

335

10.20 Matrix Fonn of Fundamental Equations of Immersion of L" into E2`1

The system "LE" - the fundamental system (7) Section 8 of immersion of L" into E21-1 - can be represented in a brief form in terms of matrices and vectors. Introduce an n-dimensional vector H and the matrices Aj of order n as follows: 0

...

On

...

0

...

Qi"

...

0

Hl

H=

.

Ai = I -/iii

I,

... -/3i"

1,

H. 0

...

The matrix Aj is skew-symmetric, the i-th column consists of elements Q;j and the i-th row consists of elements -/3; j. The others are zero. Introduce the matrix T of order n as T = IIH;HjIj. This matrix consists of products

H;Hj in position (Q). ). In addition, we use the constant matrices Ei j with I in the (i, j )-position and 0 in all other positions. The fundamental system of immersion of L" into E2n' I can be represented in matrix form as follows: OH 8ai

8A; 8uj

AiH,

- A _ [AjA,J,

(1')

8u,

(1)

18A; E11 8u;

8Aj

I Eij=O, 1
2

"

(I )

Here, as usual, [AjAj] = AjA, - A;Aj. It is easy to see that (a) and (b) of system (7) Section 8 can be represented as = A;H, while (c) and (d) are represented as

&jj

8Aj _ [AjAi}.

8Ai

(2)

Direct verification shows that equation (e) has the form Ejj 8Ai

8Aj + EA2+T 8uj 2

Eij=O, I
/

Observe that the multiplication from the right and from the left by Ejj selects the element of the i-th row and j-th column from the matrix in parentheses above and puts it in the i-th column and j-th row. The other elements become zero. Use matrix A from Section 7. Let A' be the transpose. Then the system (12) Section 8: 8a; Out

= Okiak,

aal Our = '-

aigaq, q

k#i

THE GEOMETRY OF SUBMANIFOLDS

336

can be represented in matrix form as

= A;A'. Multiplying from the left by

A" = A we get A A; =

au;

A,

i = 1,...,n.

(3)

If we take A; in this form then equations (2) are fulfilled automatically. Indeed, we have 0A;

8A;

_ 8A' 8A

au;

au;

8u; au;

8A' 8A au, au;

(4)

On the other hand, observing that A is an inverse matrix to A', we get

[A;A;Jau; =AA A A-au;- A au; AA au; =OA. A(2A'A-A'8A'\ -aA'A(aA'A-A, aAl au; l au; au;) 8u; l 8u, au,) aA aA

aA' aA

=au;au;-au;au;' Comparing with (4) we conclude that (2) is fulfilled. The system (I") can be represented in terms of a matrix A. In matrix A? the i-th row and i-th column both consist of zeros except for the diagonal elements. Thus,

E;;A?E;; = 0. In an analogous way we find that E;;AJ E;1 = 0. Note that;am,P _ -A?. Therefore E11 "' &A E11 = 0. Then, the matrix T is constructed with elements of

the first row of A. It is easy to see that T= A'E,1A. Substitute the expression of A; from (3) into (1"). Taking into account the observation above, we obtain E,,

suz

8u?

+ E a Q A auQ + A' E AE;; = 0, i < j.

(5)

These equations are necessary consequences of (1). However, in order that a solution A of the latter system generates the isometric immersion it is necessary, in addition, A consists of zeros except for the i-th row and i-th column. that the matrix A; _

From (5) we note that if A is a solution and C is an orthogonal matrix which commutes with E then CA is also a solution. Geometrically the latter operation means the same rotation of the normal basis at all points. The matrix form of a differential system of immersion of L" into EN with flat normal connection has been obtained in [19]. 10.21 On the Spaces of Variable Metric Signature

Physical theories usually consider pseudo-Riemannian spaces with constant metric signature. The metric tensor matrix g& has three positive and one negative eigenvalues. This corresponds to the separation of all coordinates into three space-like coordinates and one time-like coordinate.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

CD

C

i

2

3

337

It

FIGURE 51

In theories of the Kaluza-Klein type the number of time-like coordinates stays equal to I while the number of space-like coordinates increases by compactified ones. In papers of Vilenkin [27], Sakharov and others the hypothesis on the possibility of physical states with different metric signature is discussed. Let n be a dimension of the space-time continuum M and or be a number of time-

like coordinates in a given domain of M. The signs of metric time-like principal values are supposed negative while the space-like metric principal values are supposed positive. In the usual theory a = I and the signs of principal values are - + ++. In [28] the domain in M with a = 1 is denoted by U - "Universe". The domain with a = 0, i.e. a purely space domain without time, Sakharov denoted by P - the capital letter of the name of the Greek philosopher Parmenides who reflected on a universe without motion. Parmenides was the successor of Zenon - the author of well-known paradoxes. In the capacity of visual illustration to the arguments above, consider the twodimensional surfaces based on one-dimensional rings A and B. The surfaces I and 2 describe some physical states. They differ in their topology. In addition, different one-dimensional sections of the same surface 2 have different one-dimensional to-

pology. The physical states on surfaces 3, 4, 5 differ from I, 2 in the signature structure. Their different one-dimensional sections have different signature structures (see Figure 51). To the author's mind is the hypothesis that the differences in signature structure of the space-time continuum are as natural as the differences in topological structure. The principal idea in this hypothesis is its relation to the causality principal. If the number of time-like coordinates a > I then the light cone divides the space of ray directions at a fixed point x only into two domains - space-like and time-like - but not into three domains as for a = 1. If a > I then the intersection of a light cone with the unit sphere Sn-1 of tangent space at point x is, from the topological viewpoint, the product of spheres Sn-o-1 x So-1 which is connected. Therefore, if a > I then locally the past is not separated from the future. However, the property of global order can be stated, it seems, if we suppose that all time coordinates except one are compactified.

THE GEOMETRY OF SUBMANIFOLDS

338

Together with ideas of Kaluza-Klein type theories we come to the theory of spaces, where some space-like and some time-like coordinates are compactified. In [28] the author considered the transition from P to U and also the relation of the hypothesis to the so-called "anthropological principle". In the next section we construct a model of the electromagnetic field in a Rie-

mannian space (in Lobachevski space) and give the condition, when one of the coordinates is compactified. In accordance with the theory presented above, these considerations may have physical sense. 10.22 Gauge Field Model

In a previous section we stated the relations between the theory of isometric immersions of Lobachevski space L; into E5 and the theory of rigid body rotation around the fixed point in a central Newton gravity field. This relation points out the importance of fundamental equations of immersion of L" into Euclidean space from the physical viewpoint. In this section we shall find the analogy between these equations and Maxwell equations and construct a model of a gauge field. Consider the isometric immersion of L4 into E7. In this case the normal connection is flat automatically. The set of matrices F,,,, _ [A,,A J we shall call "the electromagnetic tension" tensor. The classical electromagnetic tension tensor has six essential components. The matrix "electromagnetic tension" tensor introduced above consists of six skew-symmetric matrices of order four with components being quadratic expressions with respect to 12 quantities lij. From (I') of system (1) Section 20 follows the first group of Maxwell equations: aFjj aFjk aFk;

auk+au;+auj

-0.

The second group of classical Maxwell equations consists of the following four equations (see [29], p. 94): 4

aF;j_4rr

where j = { jj} is the four-dimensional current density vector which characterizes the

distribution and motion of electric charge. We obtain the analog of this group of equations as a consequences of (1") of system (1) Section 20. Let Q, is be distinct subscripts. Differentiate the left-hand side of (I") in u,,. Denote by D the matrix Az + T. Then _ a2Aj 8D

Ei1(y

5A,,

aujaull

+

su,,

0.

Since , Erj = 0,

E,j r

E;j02A" E;j = 0, auj

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

339

for distinct Q, p, then, using (I'), the equations just obtained can be transformed to

E;j{BFw -BFµj+OD1E;j=0, 8u;

8uj

.'c j.

8u,, J

This system gives the analog to the second group of Maxwell equations. Thus, the derivatives of the components of matrix D play the role of current density. In gauge field theory the topological charge has been introduced (see [30]):

N=

1

I > SF V, dul ... du4i

I W4

where w4 is a volume of unit four-dimensional sphere, 'F,,,, = is the dual tensor. Under some restrictions on the speed of the gauge field decreasing at infinity the number N is integer. The integrand is called the topological charge density. In an analogous way it is possible to introduce the topological charge density in our theory, too. But this is zero identically. Probably it is so because of the flat normal connection of immersion L4 into E5. Prove this assumption. The dual tensor has the components 'F12 = F34, 'FI3 = -F24, 'F14 = F23, 'F24 = F14, 'F24 = -F13, 'F34 = F12. Hence, the topological charge density has the form E,,,, T,F,,,,'FN = 2T, (F12F34 - F13F24 + F14F23) Set 0ij = /iik/3jk +,0i1,0j1, where Q, k, I are fixed and distinct. Then the matrices F,,,,

have the form

F12 =

F13 =

F14 =

0

-021 0

/312023

012024

-021013

-021014

*

*

0

0

*

*

0

0

0

013032

-031

/313034

*

0

031/312

0

,

I

F34 =

F24 =

0

0

-034041

/343/331

0

0

-/334/342

043032

* *

* *

0

-043

*

0

0

-024041

0

/34202I

*

0

024043

-042

0

042,023

*

0

,

* *

*

0

-/331/314

0

0

*

0

*

* *

0

/314/342

014043

-041

0

-023031

032021

0

*

0

0

040312

*

0

-032

023034

* *

0

0

/34I/313

*

*

0

*

*

0

0

*

*

-032024 0

F23 =

These matrices are skew-symmetric, so we presented only the upper diagonal elements. Set 7#k, = ,13; j,3jk,3k1,31j. Evidently, 7;jkI = 7jkli. Then T, (F12F34 - F13F24 + FI4F23) = 71234 - 71243 - 72134 + 72143 + 71234 - ^12134 - 71243 + 72143 - 71324 + 71342 - 71324 + 73124 + 73124 - 73142

+ 71342 - 73142 + 71423 - 71432 + 71423 - 74123 + 74132 - 74123 + 74132 = 0.

THE GEOMETRY OF Si1BMANIFOLDS

340

Here we use, for instance, the equality Biz;a =14I2%. Therefore, the density and, as a consequence. the topological charge are zero. The Field F,,,, is called self-dual if F,,,. = "F,,,,. Since T, F,,,,F,,,. < 0. then from the equality, proved above. F T, F,,,.'F,,,, = 0. follows the statement: the field F,,,. is selfdual if F,,,. = 0. Construct now the analogs of electric and magnetic fields. Separate one of the coordinates , say 04. Introduce the vectors of matrix components: E as the "electric tension" vector. H as the "magnetic tension" vector. Set E= (FI4.FF4,Fza).

H=

Define the scalar product (ah) of matrix component vectors it = {a,} and h = {h, } by (ah) = T, F, o,h,. It is easy to see that the equality to :ero o/'tire topological charge density implies (EH) = 0. The fields E and H are constructed over the domain of the positive definite metric. that is, over a P-type domain in the previous section's notations. It was possible to set u = it. where i is a purely imaginary number -- the time analog. We say that the metric ds-' _ "_1 H,- dn; of an immersed domain L" into E'" I is stationary it all the coefficients H, do not depend on it,,. In this case the coordinate hypersurfaces it,, = const. which consist of lines of curvature, are totally geodesic in U', i.e. they are (n - 1)-dimensional Lobachevski spaces L" 1.

Theorem I (Anrinor) For the innner.uort of L4 into F7 with it slationari metro the "electric tension " E - 0. the "niagnetle tension- H (h)('.% not depend on 14 and IN of sonic is conipaetified. The intniersion of L4 is represented as the of dirnc nsuuurl srthnreuiilold F'3 C E s and the circle S' of radius as it

points in P. Since `om = 0..14, = 0. From the expressions above for F,,,, it follows that E = 0. It is easy to see that .44 = 0. Show that the Darhoux symbols .3,R do not depend on 14. The following equation hold: i4

U.i,i 7)114

f

4

#

i),i,.i

i).14,

0111

oil,

0. Therefore. the vector field H does not depend on ui. So. this case corresponds to the constant magnetic field. Since .14, = 0. ;1

Lemma The fundamental system of immersion of L" into F'-" I with stationary metric has the first integral: -i

const.

Equation (e) (7) Section 8 for k = n and .1,,, = 0 has the form

+ V' .l,i I

e11,

11-1 px1

=H.14-

1=1.. .. .

I?-)

.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

341

Multiply the both sides of this equation by 2p," and take into account the equalities

b , Hi/3t = 0, We obtain "-t

a

(32 +

aui

'gyp" -

Hn = 0.

p=1,p$i

The expression in parentheses does not depend on i and coincides with the expression of the first integral. The lemma is proved.

The existence of this first integral leads to the "compactification" of the last coordinate a". We mean it in such a way that under the isometric immersion of L" into E21-1 with a stationary metric the curve u" turns into the circle. If the point moves along this curve then at some time it will take the initial position. Let r(ul, ... , un) be a parametric representation of the immersed domain of V into

E". Write one of the Gauss decompositions: 0

We also have the expressions for Christoffel symbols and i

I'nn

-- - H"3" H

,

in

i rin - 0,

= HiHSin

coefficients:

a Lo' -

It

I

Hence, the next Gauss and Weingarten decompositions have the form: HiQin

-6na

r,,,,. = H H.

Differentiating

H H.

in un and using the decompositions above, we get n-I

n-1

Hn

ru u4u, _ 1=1

Mn, no au" n=1

t.r2 b.

For the derivatives of matrix A columns from Section 7 there hold relations (12). In a particular case,

aa aun

n-1

-

4=1

= 0. Hence = 0. From the orthogonality of matrix A it follows that 1 - H;. Set w2 = ,nall I - H. By the lemma, w is a constant. So, we can write that -w2r,,,. Integrating this equation, we find Since

0,

r = a cos win + b sin wun + p(ut,... sun-01

where a, b and p are some vector functions of uI, ... , u"_ I . The plane of vectors a

and b stays fixed. Indeed, differentiating r and using Gauss decomposition r",", = Hj,3jr",/Hn we get a,,, =

alnH; a, sui

b,,, =

alnH, b. 811i,

342

THE GEOMETRY OF SUBMANIFOLDS

Therefore, a = b = H,,bo, where ao, bo are constant vectors. Since r. = HR does not depend on IaoI = Ibol = and ao, bo are mutually orthogonal. Therefore, each u,, line is an arc of a circle of radius The planes containing these circles are parallel to each other. The vectors r,,, and r,,,, are mutually orthogonal and, hence,

(pb) = 0. With the origin shift we can achieve that p will be orthogonal to ao, bo at each point. The vector function p represents some (n - 1)-dimensional submanifold F"T' in some EI-3. This submanifold is a locus of S' centers. Thus, the position vector of an immersed domain of L" is represented as a sum of the position vector p and the position vector of the circle S1. The theorem is proved. The coordinate hypersurface u = const is the (n - 1)-dimensional Lobachevski space L"-'. The immersion of domain of L" into E21-' induces the immersion of

some domain of L' into E2i-2. This follows from the expression of a position vector.

Consider the question of the immersion "in the large" of the whole space L" into E2n-' with stationary metric. The following theorem holds: Theorem 2 There is no C2 regular isometric immersion of the whole space L" into E21-1 with stationary metric.

Consider the first integral found above. The expression for

can be represented as

"-1 (Jq)2The

left-hand side is grad H,,12 in V. Therefore, H,, level surfaces are geodesically parallel to each other. Observe that in regular immersion of L" the function H. can not be constant on V. Indeed, if H = const then (3; = 0. The equation

0(i. 8u;

+ 2 PpiPpn = Hifn P

implies that 0. But in a regular immersion all of Hj 96 0. Consider the variation of H. along the geodesic line orthogonal to level surfaces. Let s be the arc-length parameter on this geodesic and the increasing s direction coincides with

the H,, increasing direction. Since H,2 + c $ 0, then integrating the equation dfl = H" + c, we get H,, = (e'+'0 - ce '"1°)/2. Therefore, H,, is unbounded when s -+ 00. This contradicts the condition

I. The theorem is proved.

10.23 Immersions and Continuous Bendings of a Horodisk in the Lobachevski Plane in E4 Let el, e2, e3, e4 be a coordinate basis in E4. Construct the immersion of a horodisk in Lobachevski space into E4 different to the standard immersion into E3 as a universal

covering of the pseudosphere. We represent the position vector of the surface as (e, cos u + e2 sin u) + p(v),

r(u, V) = V

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

343

where p = p(v) is an arbitrary curve in the plane (e3,e4) and v is related to the arclength parameter by

s= J

(1)

dv.

that F2 of the position vector above has the

We denote the curve p = p(v) by -y.

metric of the Lobachevski plane. Indeed,

r = v (-ei sin u + e2 cos u),

rr = - 4 (el cos u + e2 sin u) + pr.

Therefore

r2 =

0,

r2

I

Consider the derivative of v in s. By virtue of (1), we have ds dv

v2

On the other hand, pr = ps -4. Hence, pr =

lps12 ()2=. Therefore

The metric on F2 has the form of a Lobachevski plane metric in the Poincafe interpretation: ds,

- due + dv2 v2

From (1) it follows that v > 1. The curve in the Lobachevski plane with v = I is a horocircle. Hence, the domain v > l is the horodisk. The obtained horodisk immersion is induced one by cylindrical immersion of E3 into E3 with 'y as an element. In this case the surface F2 in E3 is a pseudosphere. The curve ry in the (e3i e4)-plane can be taken arbitrary. For instance, if y is a circle then the surface will be like a thinning tube twisted around this circle. If 7 varies in the (e3, e4)-plane then F2 also varies. When ry varies continuously, we obtain a continuous bending of the horodisk immersion. It is easy to verify that the Gauss torsion of the surface is zero.

10.24 The Generalization of the Banchi Transformation

In the theory of surfaces of constant negative curvature in E3 there are known Bianchi and Bgcklund transformations which take these surfaces into the surfaces of

constant negative curvature. By means of these transformations it is possible to construct new classes of immersions starting from some already known immersions.

344

THE GEOMETRY OF SUBMANIFOLDS

In this section we state the generalized Bianchi transformation of n-dimensional

Lobachevski space L" domains immersed into E2". Let x = x(y1,...,y") be a position vector of immersed L", where y; forms a semi-geodesic system of coordinates in L" such that the metric has the form ds 2 = e 'j'- (dy, + ... + dy'._,) + dy..

(1)

Coordinate hypersurfaces yn = const form in L" the family of geodesically parallel horospheres with geodesics yn as the orthogonal trajectories. At any point in L" of the position vector x consider the unit tangent vector to y"-geodesic at this point with

opposite sign. The set of endpoints of those vectors forms a new submanifold (perhaps with singularities). We say that this submanifold is obtained from a given submanifold by Bianchi transformation. Thus, the Bianchi transformation maps the submanifold V" of position vector t to L" of position vector x in the following way: ax fr =x-ay-.

(2)

n

Theorem (Aminov) The Bianchi transformation takes an n-dimensional submanifold of constant negative curvature L" C E2n-I into the submanifold of the same constant negative curvature. To prove this, we find the metric of V". Let q be Christoffel symbols of ds2 with respect toys,. .. , yn. Let n i , ... , nn_ i be the orthogonal frame of normals to L" with zero torsion coefficients. This is possible because the normal connection of L" is flat.

Let L°. be the coefficients of the second fundamental forms with respect to that frame. Then

Xy, = xy, - xy.y, = x,, -14, xy, - L, no. Write the Christoffel symbols:

g/i+agl"-agrn rn-gU I OiYn ay; ay/ We see that rn, = 0 if i O j. Besides, we have

rj=1, j96 n; I",=0.

(3)

Therefore

Xyj=xy,-x,, -L°,no=-L.0rno, i
Lnn0

no

By means of (4), we have

dz = > Xy, dyr = -nc (t L .-j dy/J + xy, dyn. f=1

/=1

(4)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

Taking into account the orthogonality of n, and

we obtain the metric on V": 2

n / n

o-I

345

(5)

j-I

Show that there exist the functions y, , ... , yn-, of parameters y, , ... , y,, such that n

dyo = ef^

(6)

L"/ d v/. /=I

For this it is necessary and sufficient that the second fundamental forms' coefficients satisfy the equations \\

1

49Yj

Show that these equations are satisfied due to the Codazzi equations on the original submanifold. Since the torsion coefficients are zero, we have (8)

L d.r - LR;.i = 0. In expanded form:

-

8L"

'J

aY/

+(I' -Vni)L°i/ =0.

(9)

Multiply (9) by ey". If i, j 0 n then I' = r;;; = 1. In this case (7) is satisfied. Suppose that one of indices, say i = n while j # n. As rn,, = 0, we get 8ey^L °J

aYn

a y/

"/

nJ

-0.

Thus, in this case (7) is fulfilled, also. Taking into account (5) and (6), the metric has the form

Therefore, V" is the submanifold of constant negative curvature -1. From (4) it follows that (zJ.,x,,) = 0, where i= 1,... , n; j = 1. This means that the tangent plane to the horosphere yn = const is the normal plane to V" and the normal plane to L" is tangent to V". The vector x,.. is the common tangent vector of L" and V". The field of unit vectors xy,/ey'", j = 1, ... , n - I is parallel in the normal bundle of V".

10.25 Classical and Generalized Backlund Transformations

Consider, first, the classical Backlund transformation of a two-dimensional surface

F2 in E3 of constant negative curvature K = -1. We obtain a new surface F2 as follows. Take the straight line intercept denoted by ry of the constant length cos a.

THE GEOMETRY OF SUBMANIFOLDS

346

Apply the initial point of the intercept y to each point x E F2. The set of all endpoints of y forms a surface F2. Moreover, impose the following conditions: (1) the intercept -1 is tangent to F2 and E simultaneously; (2) the angle between normals n and n to F2 and P2 respectively is constant and equal to or + a/2. The following theorem holds:

Theorem I The surface obtained by the Bdcklund transformation from the surface of constant Gaussian curvature K = -1 is also the surface of constant Gaussian

curvature K = - I. We show that the conditions (1), (2) allow us to determine the direction of y with some arbitrariness. Namely, it is sufficient to give the direction of y at one point. In the construction of the Backlund transformation we use the system of coordinates based on fines of curvature. With respect to curvature coordinates the metric of the surface of Gaussian curvature K = -1 has the form ds2 = H, dud + H2 du;, where H, = cos w, H2 = sinw and the function w satisfies the equation 02w

02w

au -W2 =sin wcosw.

(1)

Let r(u,,u2) be a position vector of the surface. Let jP be the angle which -y makes with the first principal direction. Then we can represent the position vector of the surface P2 obtained from F2 with a Backlund transformation as r = r + CH; cos cp + H2 sin cpl cos or.

(2)

Set rj = r,,; /Hj. Consider the mutually orthogonal unit vectors a, b in the tangent plane. Then

a = r,

+r2 sin V,

b = -r, sin
Find the vectors tangent to coordinate curves in F2:

cos +

rej = r,,, +

cos a.

sin jP + b

From the Gauss decomposition we get 871

&d

8u, = 0u2

72 +

Or;,

H, '

8u,

_

o'ti

-F r, 2

Taking into account that with respect to our system of coordinates L = H, H2 = -L22, we get

k = r,,,

+cosv(bI - +i 1 +nH2 cosjp).

(5)

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

347

Analogous calculations lead to the expression

=r,,+cosor (,b(°--- + J -nH1 8142 ()U 1

(6)

Now use properties (1), (2) of the Backlund transformation. Since the vector ry = a cos o is tangent to both F2 and F22, then a is orthogonal to it and n. Therefore, n is a linear combination of n and b. Using (2), we can write

n = -sin an+cosab.

(7)

Expand the conditions (F,,h) = 0. Taking into account (5) and (6), we get the system of differential equations on cp as a function of u1, u2: a"Iv

_ cos w sin cp - sin o sin w cos cp COsa

8141 +8142

8142

(8)

sin w cos V -sin o cos w sin W

8cp

8ul

COs a

A remarkable and unexpected property of this system is the fact that its compatibility condition coincides with (1). Indeed, differentiating the first equations of (8) in u2 and the second in ul and subtracting the second from the first, we get by virtue of (8) the following: 82w

82w

au; - au;

cos w cos 5' - sin o sin w sin wP

8w

81p

+ 8u1) _ sin w sin gp - sin a cosw cos p 80 da cos or

cos a

(8112

814 + 8142 1

= - sin w cos w.

Find the metric on F2. Set A 1 = cos w sin p + sin or sin w cos gyp,

A2 = sin w cos W + sin or cos w sin W.

Then (5) and (6) by virtue of (8) can be represented as F, = r,,, + bA 1 + n cos a sin w cos (p, F,,. = r,, - bA2 - n cos a cos w sin cp.

So, the metric coefficients of F2 are E = cost w + 2(r,,, b)A I +A + cost a sing w cost p = cost

,

F = (r,,, b) A 1 - (r,,, b) A2 - A I A2 - cost a sin w Cos w sin p cos ip = 0,

G = sine V.

Hence, the metric on P2 has the form fist = cost cP du2 + sin 2 V du,. From this it follows that F2 is regular at those points where ry is not tangent to the principal

THE GEOMETRY OF SUBMANIFOLDS

348

directions. The Gaussian curvature k of this metric can be expressed by the simple formula K

_

i w_ O2

1

(

cosw sinw

(9)

OU2

By means of (8) we find

a

I

cos -

8u2

[(cosw cos p - sin a s i n 141

- (sinw sin w - sin a cosw cosw)

(

iiU)

(10)

+ a )J = sinw cosw.

Using (9) and (10), we obtain the statement of the theorem K = -1. State one more property: the Bdcklund transformation takes lines of curvature into lines of curvature in the new surface. To prove this, verify that (r,,,,,2n) = 0. We have fu,ui = ru, u_ + bu, A I + bA I,,, + n,,, cos a sinw cos w + n cos a

L(sinw cos V).

Multiplying scalarly by n = - sin an + cos a b, we obtain sin a A I (b,,,n) - sin a cos a

2

(sinw cosw)

+cosa(r,,,,,,b) +cosaA1u, + (bn,,,) cos2a sin w cosw.

Take into account that b,,,n = -(bn,,,) _ -cosw cosw. Then

sinw+cosw cosw a

(r,,,,,,b) = -1'12H1 sinw+F12H2 cosw = sinw After simple transformations we find (F,,,,,,n) =coswcosw Icosal

8 +0J +A2

that\\\(F,,,,,,n)

= 0. The statement is proved.

From the second equation of (8) it follows

Find the geodesic curvature of vector field a streamlines, when a defines the Backlund transformation. If some curve in the surface makes the angle w with the first coordinate curve then its geodesic curvature is 1

_ dw+cosw+sinw' ds

Ps

p,

(I1)

P2

and are geodesic curvatures of coordinate curves (see Blaschke: "Differential geometry", p. 211 ). The geodesic curvatures of coordinate curves are

where

1

PI

do cosw 0U2' 1

EG

1

P2

(mo)w, _

fEG

1

i

sinw 8ui

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

349

Since the curve is a streamline which makes the angle cp with coordinate curve uI,

_ cos cp

dui ds

duZ

_ sin V

cosw' ds sinw The derivative of V with respect to the arc-length parameter in the streamline is dV

_ ap cos cp

sin p

8ui cosw+81,2 sin w'

ds

1, 2 by means of (8). Then

Replace

_

1

pg

cos cp cos w sin p + sin a sin w cos p cosw cos a sin w cos

sin cp

+ sin a COS w sin tp

+sinw +

61)

8u2 J

cosa

cos cosw 8u2

8uI

sin cp 8p sin(w - cp) sin(w + cp) +--=tang sinw cosw sinw 8uI

Therefore, if the length of the intercept joining corresponding points in F2 and F2 is equal to I then cos a = I and, as a consequence, tan a = 0. In this case the geodesic curvature of streamlines is zero. Thus we obtain the geodesic flow in the surface. To characterize this flow, find the geodesic curvature of orthogonal trajectories. To do this, replace V in (11) with V + n/2. Also, on the orthogonal trajectory: du1

_

sin cp

duZ

_ cos(p

cosw' ds sinw Denote the curvature of orthogonal trajectory by 1 /pp. We have IFS

1

p

dip

sin V

ds

PI

cosV)

+

cos V

-

wd

6L;

cos a

8142

sin w cos p + sin a cosw sin Vp

+ sin w + son

coswsinV +sinasinwcosW cosw

P2

cos or

cos a

-

c8uI v

sin w olp - cosw 8142

+ tan a sin V COs W(cot w - tan w).

If cos a = I then the geodesic curvature of an orthogonal trajectory is constant and equals -1, i.e. the orthogonal trajectory is the horocircle. Thus, our geodesic flow consists of a geodesic orthogonal to the horocircle. Therefore, in this case the Backlund transformation coincides with the Bianchi transformation. In papers by Tenenblat and Terng [24], [25] the n-dimensional generalization of the

Backlund transformation was obtained, based on two properties of the classical transformation mentioned above. The linear congruence of two submanifolds Fand Fof dimension n in E2i-I is the diffeomorphism 1: F -. F such that for any P E F the line joining P to P = l(P) is the common tangent of F and F. The linear congruence is called pseudospherical if

THE GEOMETRY OF SUBMANIFOLDS

350

(1) the length iPPJ = p is constant independent of P; (2) each of the (n - 1) angles between normal planes Np and Np is equal to the same number 0 independent of P; (3) the normal plane distributions N and N are of flat normal connection; (4) the mapping r: N - N given by orthogonal projection commutes with normal connection.

As a matter of fact, in the proof of the next theorem the conditions (3) and (4) mean that normal bases consisting of angular directions of normal planes Np and Np have zero torsion coefficients on both surfaces. In [25] the following theorem was proved.

Theorem 2 (Tenenblat. Terng) Suppose that there exists the pseudospherical congruence 1: F F of n-dimensional submanifolds in E2"- I of distance p between corresponding points and 0 between corresponding normal planes. Then both of the submanifolds F and F are of constant sectional curvature -sin 2 0/p'-.

If the initial submanifold is of constant negative curvature and the intercept direction is given by means of some totally integrable system then (4) will be satisfied automatically. It is natural to pose the question of the relations between the Bianchi and Backlund transformation. Masaltscv proved that the Bianchi transformation coincides with the Backlund transformation if 0 = 7r/2 [31].

10.26 The Lost World?

"The sun was just above the western sky-line and the evening was a particularly bright and clear one, so that the whole extent of the plateau was visible beneath me." A. Conan Doyle, The Lost World.

In late September-early October of 1990 in a picturesque place on the Black Sea shore, not far from the Abrau-Durso village in a sports camp of Rostov University, the geometry conference dedicated to the 80th anniversary of the birth of N. V. Efimov was held. Here I present a simple account of the conference. The road comes out in a loop from Novorossijsk to mountains, first, and then descends into a small valley, open to the Black Sea. The two steep mountains by the sea-shore stand out like sentry-towers, guarding the sides of the valley. In this valley the conference took place in a hall next to the sea, a perfect location for the meeting which attracted participants from Leningrad and Moscow, Yerevan and Kharkov, Rostov, Odessa, Tashkent, Khabarovsk, Tartu and many other cities. It was a week spent out of touch with the outside world, and no information - either printed or

verbal - came into that "Lost World". It was a week of Geometry, scientific conversations, walks along the valley and the sea shore, with a violin concert on one of the closing evenings. This part of the world is beautiful. To feel the grandeur and beauty of nature you do not need to go to faraway lands. The calm atmosphere of

ISOMETRIC IMMERSIONS OF LO13ACHEVSKI SPACE INTO EUCLIDEAN SPACE

351

N. V EFIMOV

the valley and the renouncement of the outside world assisted the rapprochement of

viewpoints, clarified approaches to solutions of problems, and provided the opportunity to get acquainted with new advances in geometry. At the opening of the conference, a letter from Academician A. D. Alexandrov was read to the participants. The following is an abbreviated version: "Dear comradesl You cannot imagine how sorry I was not to be able to come to the Conference, because I did not feel well. My regret is particularly strong as the Conference was devoted to the memory of Nickolai Vladimirovich Efmov, an outstanding geometrician. remarkable man and close friend with whom I was happy to be in contact with for almost 40 years. I take this opportunity to say some words about him. using an old phrase, for the edification of his descendants. For my part. the major feature of N. V. Efimov's character was, to put it in a nutshell, his complete honesty which manifested itself in all situations. In mathematics, Nickolai Vladimirovich obtained world-class results in three scientific directions: he discovered and studied the unbendability properties of surfaces with planar points; his works in functional analysis are largely devoted to the problems of approximations (Efimov-Stechkin theorems); and he studied negative curved surfaces, where the main result has been achieved - the Efimov theorem of

352

THE GEOMETRY OF SUBMANIFOLDS

the non-existence of complete regular surfaces of less than some negative constant curvature. This theorem seems to be the best and most concrete result in the theory of surfaces "in the large", which was obtained in a final form by a single person. The problem was stubborn. It was studied by S. N. Bernstein. for instance, who once said that he had solved the problem for the case of surfaces which can be represented as z = z(x, y), but a week later he confessed that he was wrong. Nickolai Vladimirovich published several books, in particular The Highest geometry which was a classic work for that time. In his usual honest manner, he performed his professor's duties and, later, the dean's duties for several years. Here, he also followed his usual rule: if you do things then you must direct all your efforts to doing them well. N. V. was a very gifted man, in particular, he was very fond of poetry and knew it well. He was a Russian intellectual with a typical Russian character without any indication of nationalism; he was modest and benevolent. Let the Efimov theorem and Etimov's spirit serve as examples for us all and let the participants of the Conference follow them. I wish you great success and the happiness which science can grant." A. D. Alexandrov

The results of that conference which relate to the subject of this book were pre-

sented in about 50 reports. Some of them considered problems related to the Grassmann image. Borisenko [32) proposed the uniqueness theorem for isometric submanifolds with the same Grassmann image: Theorem (Borisenko) Let Ff be a I-dimensional C3-regular submanijold in Euclidean space Eti`P. If at some point there exists the normal with respect to which the rank of the second fundamental form is equal to l then every submaniftdd which is isometric to F1 and having the sane Grassinann image coincides with Ft up to a parallel shift. For the case of a hypersurface the analogous theorem has been proved by Abe and Erbaher. Fomenko studied the infinitesimally small deformations preserving Grassmann image of two-dimensional surfaces in four-dimensional Riemannian space W. Under that deformation the surface F passes into the surface F, in such a way that the normal planes Nt and N. at corresponding points x and xc are parallel to each other in the Levi-Civita sense. Fomenko called the Grassmann image preserving deformations G-deformations. His report contained the following theorem: Theorem (Fontenko) A closed two-dimensional orientable C" (A > 3, 0 < a < 1) surface of genus p > 0 with the center x circle as the normal curvature ellipse at every point x admits

(1) 10 linearly independent G-deformations fl 'p = 0: (2) two linearly independent G-deformations iif p = I ; (3) no, except identical, G-deformations if p > 1.

ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE

353

The isometric immersions were discussed in Pozniak and Shikin's report. Using

the small parameter method, they proved a series of "in the large" theorems. Namely, they proved the possibility of the regular realization of arbitrary compact domains, horodisks and expanding zones of negatively curved manifolds in threedimensional Euclidean space E3. Earlier, Pozniak, making use of the small parameter method, constructed the analytic immersion "in the large" of compact domains in arbitrary analytic manifolds without conjugate points into E3. Verner and Sovertkov. continuing the study of complete regular negatively curved surfaces in E3, showed the way to construct the complete saddle surface homeomorphic to the torus with one point deleted. The conference was organized by the geometry Chair of Rostov-na-Donu University, headed by Professor S. Klimentov. The young participants (there were many of them, especially from Rostov-na-Donu) presented interesting reports, which shows that the younger generation knows about the classic achievements and has taken up complicated problems. Shapovalova stated the following theorem: if the surface F2 C E4 has a projection into some three-dimensional space E3 C E4 of the form of a closed convex surface of Gaussian curvature K > 0 then the surface F2 ahnits the regular bendings in E4. The generalization of the Efimov theorem on the non-immersibility of negatively curved metrics was proposed by Perel'man. Applying the theory of generalized analytic functions, Bickchantaev studied Gdeformations of two-dimensional surfaces in V. It is not our aim to present all the results. Many of them, even interesting ones, stayed out of the presentation. The mountains and the sea put us in the mood for poetry, born of day-dreams. Staying at the sea-shore reflecting with melancholy on the unrealized dreams of youth and watching the sea and its steady changing wave profiles, you think: Where are the triangles, perpendiculars, prisms and other geometrical figures? Does that ever changeable element belong to the geometrical realm? Is geometry the key to the

description of nature? The world of three dimensions, but what a freedom and variety there is in these motions. What kind of powerful forces raised the mass of mountains? How weak we are with our theories, if we do not know with sufficient exactness, what is going on underfoot or thousands of miles away. The conclusion that our knowledge is not exact follows from the fact that in spite of extravagant theories and supercomplicated generalizations, in spite of the ability of mankind to look into the depth of the Universe, we are never able to predict the very moment of an earthquake. The mighty forces deep in the earth test the tensile strength of the external shell. At some moment, unexpectedly, either as a consequence of depth

processes or as a consequence of reaching the critical limit point of the tensile strength of the shell or. perhaps, due to other external causes, these mighty forces find an exit, bringing death and destruction. The processes in the depths of the Earth are complicated and, it seems, beyond geometrical description. However, the geometrical component of the Earth's behavior is presented, at least, in the fact that as a solid, or almost a solid, it is rotating around its axes. That everlasting motion, having a different influence on the external firm shell and the internal shell, together with

354

THE GEOMETRY OF SUBMANIFOLDS

graotty and the convective motion of liquid magma. leads to the strains, deformations, shifts, breaks. and continental drifts. The geometrical mentality played its role

in the hypothesis of Alfred Wegner on a united motherland - PanGaia. In his universal interest, the famous geometer Gauss, the founder of differential geometry, did not avoid the global properties of the Earth, He studied the magnetic field of our planet. Is geometry the key to the description of nature?

In the last days of the conference Nature imperiously called us. Some of the participants explored the neighboring forests in their free time, others. L. Verner for instance, went so far away along the sea-shore to wild solitary places that we worried about their return, but fortunately there were no accidents. The most experienced walkers. Borisenko and Sovertkov. made their way over the mountains up to ten

miles away to a dolphin pool to meet possible "sapient" brothers and then said farewell to the dolphins. As for me. together with Sabitov I climbed the mountain beside the valley to view

the seascape and neighboring "lost world" from this height. The mountain was covered with forest and low bushes and there were wild-plum trees (kizil) with ripened red fruits. Sometimes we saw the sites of extinguished fires marking where modem "homo sapiens" has been. The steep and dangerous cliff fell away to the seashore, so that a stone. thrown from the cliff, fell for a long time before we heard the water splash. From this height. we could see the road ribbon into the outside world During the climb we discussed the problem of immersion of the complete Lo-

bachevski plane into E;. In 1989, Sabitov published the paper On isometrical immersions of the Lobachevski plane into E4" in Siberian Mathematical Journal (see rel:) and the paper contains the proof of the following theorem: Theorem (Sabitor ) Tlrc Lohcre lret;ski pltrne c can he analytic surface at cktxr C"

itutrrcrsccl iota E4

it% a

It seems difficult to obtain the immersion of the whole Lobachevski plane into E't as a surface of classical regularity class C; or C'=. However, just as mountaineers, who having conquered one peak prepare themselves for another, more complicated ascent, so mathematicians, passing on their experience and knowledge. gradually achieve their objectives and ambitions. They can he likened to an aquanaut who is lowered into deep ocean hollows. As E. Pozniak says. Mathematics is an abyss. Has everybody made up his mind to look into it?

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1. Lumiste Yu., Differential geometry of submanifolds. Itogi Nauki i Tehniki. Algebra, Topology, Geometry. 13 (1975) 273-340 VINITI (In Russian).

2. Pozniak E., Sokolov D., Isometric immersion of Riemannian space into Euclidean space. Itogi Nauki i Tehniki. Algebra, Topology, Geometry. 15 (1977) 173211 VINITI (In Russian).

3. Aminov Yu., The problems of immersions: geometric and topologic aspects. Itogi Nauki i Tehniki. Problems of Geometry. 13 (1982) 113-156 VINITI (In Russian). 4. Chen B.-Y., Geometry of submanifolds and its applications. Sci. Univ. Tokyo (1981).

5. Burago Yu., Zalgaller V., Geometric inequalities. Nauka. Leningrad (1980) (In Russian). 6. Pogorelov A., Extrinsic geometry of convex surfaces. Translations of Math. Monographs Amer. Math. Soc., 35 Providence (1973). Chapter 1

1. Schouten J. A., Struik D. J., Einfarung in die neuren Methoden der Differential geomeirie. P. Nardoff. Groningen - Batavia (1935). 2. Pontryagin L., Ordinary differential equations. Nauka. Moscow (1975) (In Russian). 3. Aminov Yu., Differential geometry and topology of curves. Nauka. Moscow (1987)

(In Russian). 4. Aminov Yu., On closed curves - the trigonometrical polynomials of arc-length parameter. Investigations in geometry "in the large" and mathematical analysis. Trudy Inst. Math. Sibirskoe otdelenie AN SSSR. 9 (1987) 25-34 (In Russian). 355

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5. Gorkaviy V., One integral inequality for closed curve in Euclidean space. C. R. Acad. Sci. Paris. t. 321 Serie T (1995) 1587-1591. Chapter 3 1. Eisenhart L., A Treatise on the differential geometry of curves and surfaces. Cinn and Company, Boston (1909), reprinted by Dover, New York (1960). 2. Aminov Yu., The condition to be holonomic. Matem. Zametki., 41 (1987) N 4 543-548 (In Russian). 3. Riemann B., Gesammelte Werke. Springer, Berlin (1919).

4. Christoffel E. B., Uber die Transformation der homogenen Differentialausdruke zweiten Grades. Journal fur die reine and angew. Math (Crell) 70 (1869) 46-70. 5. Milnor J. On the immersion of n-manifolds in (n + 1)-space. Comm. Math. Hely. 30 (1956) N 4 275-184. 6. Pogorelov A. V., The Minkowski multidimensional problem. Scripta. Washington (1978).

7. Alexandrov A. D., To the theory of mixed volumes of convex bodies. Mat. Sb. 2 (1937) 917-970 (In Russian). 8. Alexandrov A. D. II. New inequalities between mixed volumes and applications, Math. Sbornik, 2 N 6 (1937) 1205-1235 (In Russian). 9. Alexandrov A. D. III Expansion of two Minkowski theorems on convex poly-

hedrons to arbitrary convex bodies, Math. Sbornik. 3 N 2 (1938) 27-44 (In Russian). 10. Alexandrov A. D., Mixed discriminants and mixed volumes. Math. Sbornik, 3 N 2 (1938) 227-249 (in Russian). 11. Volkov Yu., Stability of Minkowski problem solution If Vestnik LGU, N 1, ser.: Mathematics, Mechanics, Astronomy, (1963) 33-43 (In Russian). 12. Diskant V., Stability of Minkowski equation solution // Sibirski Math. Journal 14 N 3 (1973) 669-673 (In Russian).

13. Diskant V., Stability of Minkowski equation solution for the area of convex body surface. Ukrainski Geom. Sbornik, 25 (1982) 43-51 (In Russian). 14. Diskant V., Refined analogue of generalized isometric inequality. Ukrainski Geom. Sbornik, 31 (1988) 56-59 (In Russian).

15. Efimov N., Qualitative question on deformations of surfaces "in the small". Trudy Steklov Math. Institution Acad. Sci. USSR 30 (1949) 1-128 (In Russian). 16. Alexandrov A. D., On some class of closed surfaces. Math. Sbornik, 4 (46) (1937) 69-77 (In Russian). 17. Nirenberg L., Rigidity of a class of closed surfaces. Nonlinear problems. Univ. Of Wisconsin Press (1963) 177-193.

18. Sabitov I., Investigations on rigidity of analytic surfaces of revolution with pranar point at the pole. Vestnik MGU, ser. I. Math. and Mech., 5 (1986) 29-36 (In Russian).

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19. Aminov Yu., n-dimensional analogues of Bernstein integral formula. Math. Sbornik, 75 N 3 (1968) 375-399 (In Russian). 20. Aminov Yu., On diameter and volume estimates of submanifold in Euclidean space. Ukrainski Geom. Sbornik, 18 (1975) 3-15 (In Russian). 21. Aminov Yu., On spherical image of the surface with curvature close to zero. Ukrainski Geom. Sbornik, 9 (1970) 10-13 (In Russian). 22. Masaltsev L., On the first eigenvalue of Laplacian on compact orientable hypersurface in Euclidean space. Ukrainski Geom. Sbornik, 19 (1976) 104-108 (In Russian).

23. Aminov Yu., On rigidity of closed trigonometry type hypersurfaces. Math. Sbornik, 181 N 12 (1990) 1710-1720 (In Russian).

24. Pogorelov A., Regularity of convex hypersurfaces with regular metric. Sov. Math. Dokl. 224 N 1 (1975) 39-42 (In Russian). 25. Milka A., Shortest lines in convex surfaces. Sov. Math. Dokl. 248 N 1 (1979) 34-36 (In Russian). 26. Milka A., Convex hypersurfaces in pseudo-Euclidean space. Soy. Math. Dokl. 284 N 6 (1985) 1314-1316 (In Russian). 27. Gantmacher F., The theory of matrices. Nauka, Moscow (1967) (In Russian). 28. Smyth B., Xavier F., Efimov's theorem in dimension greater than two. Invent. math. 90 (1987) 443-450. 29. Okayasu 0., 0(2) x 0(2)-invariant hypersurfaces with constant negative scalar curvature in E4. Proc. of AMS, 107 (1989) 1045-1050.

Chapter 4 1. Chern S. S., Kuiper N., Some theorems on the isometric embedding of compact Riemann manifolds in Euclidean space. Ann. Math. 56 (1952) 422-430. 2. Otsuki T.. On existence of solutions of quadratic equations and its geometrical application. Proc. Japan Acad., 29 (1953) 99-100. 3. Shefel S., On two classes of k-dimensional surfaces in n-dimensional Euclidean space. Sibirski Math. Journ., 10 N 2 (1969) 459-466 (In Russian). 4. Shefel S., Surfaces in Euclidean space. Math., Analysis and related problems in mathematics. Novosibirsk, (1978) 297-318 (In Russian). 5. Glazyrin V., Riemannian curvature of multidimensional saddle surfaces. Sibirski Math. Journ., 18 N 3 (1977) 541-548 (In Russian). 6. Glazyrin V., Topological and metric properties of k-saddle surfaces. Soy. Math. Dokl. 233 N 6 (1977) 1028-1031 (In Russian).

7. Glazyrin V., Topological and metric structure of k-saddle surfaces. Sibirski Math. Journ., 19 N 3 (1978) 555-565 (In Russian).

8. Borisenko A., On compact surfaces of negative external curvature in Riemannian space. Ukrainski Geom. Sbornik, 19 (1976) 9-11 (In Russian). 9. Borisenko A., On complete parabolic surfaces. Ukrainski Geom. Sbornik, 28 (1985) 8-19 (In Russian).

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10. Borisenko A., Complete n-dimensional surfaces of nonpositive extrinsic curvature in Riemannian space. Math. Sbornik, 104 (1977) 559-576 (In Russian).

11. Borisenko A., On characteristic Pontrjagin classes of compact surface in spherical space. Ukrainski Geom. Sbornik. 23 (1980) 24-30 (In Russian).

12. Lisitsa V., On surfaces with principal directions field in spaces of constant curvature. Ukrainski Geom. Sbornik, 23 (1980) 74-83 (In Russian). 13. Toponogov V., Shefel S. Geometry of immersed surfaces. Math. Encyclopedia, 4 (In Russian). 14. Borisenko A., Yampolsky A., Riemannian geometry of bundles. Uspehi Math. Nauk. 46 (1991) (In Russian).

Chapter 5

I. Ricci G., Sulle superficie geodetiche in una variety qualunque et in particolare nella varieta a tre dimensioni. Rend. dei Lincei., ser. 5, 12' (1903) 409-420. 2. Frankel T. T., Manifold with positive curvature. Pacific J. Math. 11 (1961) 165-174.

3. Borisenko A., On extremal properties of compact parabolic surfaces in Ricmannian space. Math. Sbornik. 133 (1987) N 1 112-126 (In Russian). 4. Borisenko A., On cylindricity of manydimensional surfaces in Lobachevski space. Ukrainski Geom. Sbornik, 33 (1990) 18-27 (In Russian). 5. Lisitsa V., Immersion of n-dimensional cylindrical metric as a cylindrical surface in Lobachevski space. Ukrainski Geom. Sbornik, 34 (1991) 66-69 (In Russian).

6. Sidorov L., Some properties of surfaces of negative extrinsic curvature in Lobachevski space. Math. Zametki, 4 N 2 (1968) 165-169 (In Russian). 7. Il'ina L., Kogan A., Theory of surfaces in Lobachevski space in Cayley-Klein model and infinitesimal bendings. Vestnik LGU., ser. Math., 19 N 2 (1972) 21(In Russian) 8. Volkov Yu., Vladimirova S., Isometric immersions of Euclidean plane in the Lobachevski space. Math. Zametki, 10 (1971) 327-332 (In Russian). 9. Jorgens K., Ober die Losungen der Differentialgleichung rt - s2 = 1. Math. Ann. 127 (1954) 130-134.

10. Kotlyar A., On supremum of external curvatures for some classes of many dimensional surfaces in n-dimensional Lobachevski space. Izvestia VUZov, ser. Mathematics, N 8 (1993) 9-18 (In Russian). Chapter 6

1. Aminov Yu., Torsion of 2-dimensional surfaces in Euclidean space. Ukrainski Geom. Sbornik, 17 (1975) 3-14 (In Russian). 2. Artin E., Zur Isotopie zweidimensionaler Flachen im R4. Abh. Math. Sem. Univ. Hamburg, 4 (1925) 174-177.

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3. FBssler W., Ober die normaltorsion von Flachen im vierdimensionalen euklidischen Raum. Comm. Math. Helv.. 33 N 2 (1959) 89-108. 4. Kadomtsev S., Investigation of some properties of 2-dimensional surface normal torsion in 4-dimensional space. Itogi Nauki i Techniki. Problems in Geometry, 7 (1975) 267-278 M.: VINITI (In Russian). 5. Kadomtsev S.. Investigation of uniqueness of 2-dimensional surfaces in Euclidean spaces. Itogi Nauki i Techniki. Problems in Geometry, 8 (1977) 243-256 M.: VINITI (In Russian).

6. Pozniak E., Isometric immersion of 2-dimensional Riemannian metrics into Euclidean spaces. Uspehi Math. Nauk, 28 N 4 (1973) 47-76 (In Russian). 7. Fomenko V., Some properties of 2-dimensional surfaces with zero normal torsion in E4. Math. Sbornik, 106 N 4 (1978) 589-603. 8. Fomenko V., Zubkov A. Some properties of 2-dimensional surfaces related to the torsion. Ukrainski Geom. Sbornik, 33 (1990) 45-52 (In Russian). 9. Chen B. Y., Differential geometry of submanifolds with planar sections. Ann. Mat. Pura App!., 130 (1982) 59-66. 10. Chen B. Y., Classification of surfaces with planar normal sections. J. Geometry, 20 (1983) 122-127.

11. Chen B. Y., Li S. Y. Some new characterizations of Veronese surface and standard tori. Revista, Union Mat. Argentina, 29 (1984) 291-309. 12. Chen B. Y., Li S. I., Classification of surfaces with pointwise planar normal sections and its application to Fomenkov's conjecture. J. Geometry, 26 (1986) 21-34. 13. Rozendorn E., On complete surfaces of negative curvature K < -I in Euclidean spaces E3 and E4. Math. Sbornik, 117 N 2 (1962) 147-160 (In Russian).

14. Rozendorn E., To the question on immersion of 2-dimensional Riemannian metric into 4-dimensional Euclidean space. Vestnik MGU, ser. Math. and Mech., N 2 (1979) 47-50 (In Russian). 15. Aminov Yu., On surfaces in E4 with constant sign Gaussian torsion. Ukrainski Geom. Sbornik, 31 (1988) 3-14 (In Russian). 16. Hu S.-T., Homotopy theory. New York: accademic Press (1959).

17. Aminov Yu., Surfaces in E4 with a Gaussian curvature coinciding with a Gaussiantorsion up to the sign. Mat. Zametki, 56 (1994) 1211-1245 (in Russian). 18. Masaltsev L., On helicoidal surfaces in Euclidean spaces Em. Ukrainski Geom. Sbornik, 26 (1975) 100-103 (In Russian). 19. Lisitsa V., On n-dimensional helicoidal surfaces in Euclidean space E. Mat. Zametki. 41 (1987) 549-556 (In Russian). Chapter 7

1. Dao Ch.-T., Fomenko A., Minimal surfaces and Plateau problem. M. Nauka (1987) (In Russian). 2. Backenbach., Minimal surfaces in Euclidean n-space. Amer. J. Math. 55 (1933) 458-468. 3. Shabat B., Introduction into complex analysis. M. Nauka (1969) (In Russian).

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4. Aminov Yu., Minimal surfaces. Kharkov. KhGU publish house (1978) (in Russian). 5. Pogorelov A., On stability of minimal surfaces. Sov. Math. Dokl., 260 N 2 (1981) 293-295 (In Russian).

6. do Carmo M., Peng C. K., The stable minimal surfaces in R3 are plane. Bull. Amer. Math. Soc., 1 N 6 (1979) 903-906. 7. Blaschke W., Differential geometry. M., ONTI (1935) 8-14. 8. Duschek A., Zur geometrischen Variationsrechnung. Math. Zeit., 40 (1936) 279291.

9. Schoen R., Yau S. T., Existence of incompressible minimal surfaces and the topology of three dimensional manifolds with non-negative scalar curvature. Ann. Math., 110 N I (1979) 127-142. 10. Micallef M. F., Stable minimal surfaces in Euclidean space. J. Dif. Geom., 19 (1984) 57-84. 11. Micallef M. F., On the topology of positively curved manifolds. The Australian Nat. University. Research Report Centre for Math. Analysis (1986). 12. Moore J. D., On stability of minimal spheres and a two-dimensional version of Synge's theorem. Arch. Math., 44 N 3 (1985) 278-281. 13. Micaleff M. F., Moore J. D., Minimal two-spheres and the topology of manifolds with positive curvature in totally isotropic two-planes. Ann. Math., 127 (1988) 199-227.

14. Lawson H. B., Simons J., On stable currents and their application to global problems in real and complex geometry. Ann. Math., 98 (1973) 427-450. 15. Reshetnjak Yu., Isometric coordinate system in manifolds of bunded curvature. Sibirski Math. Journal, 1 (1960) (In Russian). 16. Pontryagin L., Smooth manifolds and applications in homotopy theory. Trudy Steklov Math. Inst., 45 (1955) (In Russian). 17. Nash J. F. The embedding problem for Riemannian manifold. Ann. Math., 63 (1956) 20-63. 18. Vekua I., Generalized analytic functions. M. Physmatgiz (1959).

19. Pogorelov A., Extrinsic geometry of convex surfaces. M. Nauka (1969) (in Russian). 20. Berger M., Les varietes riemanniennes 1 /4-pinces. Ann. Scuola Norm. Sup. Pisa., 14 (1960) 161-170. 21. Berger M., Sur quelques varietes riemanniennes suffisamment pinces. Bull. Soc. Math. France, 88 (1960) 57-71.

22. Toponogov V., Relation between curvature and topological structure of Riemann space of even dimension. Sov. Math. Dokl., 133 N 5 (1960) 1031-1033 (In Russian).

23. Klingenberg W., Uber Riemannsche Mannigfaltigkeiten mit positiver Krummung. Comm. Math. Helv., 35 (1961) 35-54. 24. Sugimoto M., Shiohama K., On the Differentiable Pinching Problem. Math. Ann., 195 N 1 (1971) 1-16. 25. Aminov Yu., On unstability of minimal surface in n-dimensional Riemannian space of positive curvature. Math. Sbornik, 100 N 3 (1976) 400-419 (In Russian).

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26. Aminov Yu., To the problem of minimal surface stability in Riemannian space of positive curvature. Sov. Math. Dokl., 224 N 4 (1975) 745-747 (In Russian). 27. Calabi E., Minimal immersion of surfaces in Euclidean spheres. J. D5 Geom., I N 2 (1967) 111-127. 28. Do Carmo M., Stability of minimal submanifolds. Lect. Not. Math. 838 (1981) 129-139.

29. Almgren F. J., Some interior regularity theorems for minimal surfaces and extension of Bernstein's theorem. Ann. Math., 85 (1966) 277-292. 30. Hopf H., Ober Flachen mit einer Relation zwischen den Hauptkrummungen. Math. Nach., 4 (1950-51) 232-249. 31. Kon-Fossen S., Some problems in differential geometry in the large. M. GIFML (1959).

32. Alexandrov A. D., Surfaces uniqueness theorems in the large. Vestnik LGU, ser. Math., 11 (1956) 5-17 12 (1957) 15-44 13 (1958) 14-26 27-34 (In Russian). 33. Hisang W. Y., Teng Z. H., Yu W. C., New examples of constant mean curvature immersions of (2k - 1)-spheres into Euclidean 2k-space. Ann. Math., 117 (1983) 609-625. 34. Wente H. C., Counterexample to a Conjecture of H. Hopf. Pacific J. Math., 121 (1968) 193-243. 35. Walter R., Explicit examples to the H-problem of Heinz Hopf. Geom. Dedicata, 23 (1987) 187-213. 36. Lawson H. B., Complete minimal surfaces in S3. Ann. Math., 92 (1970) 335-374. 37. Chern S. S., On the curvature Integra in a Riemannian Manifold. Ann. Math., 46 (1945) 674-684. 38. Milnor J., Differential topology. Lectures on Modern Mathematics. New York: Willey, 11 (1964) 165-183. 39. Smale S., A classification of immersions of the two-sphere. Trans. Amer. Math. Soc., 90 (1958) 281-190. 40. Kervaire M., Sur le fibre normal a une sphere immergee dans un espace euclidien. Comm. Math. Helv., 33 (1959) 121-131. 41. Aminov Yu., Vector field curvature sources. Math. Sbornik, 80 N 2 (1969) 210224 (In Russian).

42. Aminov Yu., Intrinsic Ricci condition analogue for minimal submanifold in Riemannian space. Ukrainski Geom. Sbornik, 17 (1975) 15-22 (In Russian). 43. Rauch H. E., A contribution to differential geometry in the large. Ann. Math., 54 (1951) 38-55.

44. Lumiste Yu., On geometric structure of complex-analytic surface VZ in V. Sov. Math. Dokl., 114 N 2 (1957) 259 (In Russian). 45. Lumiste Yu., To the theory of 2-dimensional minimal surfaces. Ucheni je Zapiski Tartu Univ., 102 (1961) 3-15 129 (1962).

46. Goroch V., On hypersurface with minimal submanifold. Ukrainski Geom. Sbornik, 24 (1981) 18-26 (In Russian). 47. Goroch V., On area minimum of surface with given boundary contour in pseudo Riemannian space. Ukrainski Geom. Sbornik, 30 (1987) 18-22 (In Russian).

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48. Goroch V., On 2-dimensional minimal surfaces in pseudo Euclidean space. Ukrainski Geom. Sbornik, 31 (1988) 36-47 (In Russian). 49. Goroch V., On stability of minimal surface in pseudo Euclidean space. Ukrainski Geom. Sbornik, 33 (1990) 41-46 (In Russian).

50. Borisenko A., Complete 1-dimensional surfaces with nonpositive extrinsic curvature in Riemannian space. Mat. Sbornik. 104 (1997) 559-576 (In Russian). 51. Burago Yu., Toponogov V., On 3-dimensional Riemannian spaces of curvature bounded from above. Mat. Zametki, 13 N 6 (1973) 881-887.

Chapter 8 1. Efimov N., Singularity rise on the surface of negative curvature. Math. Sbornik, 64 N 2 (1964) 286-320 (In Russian). 2. Klein F., Lectures on development of mathematics in the nineteenth century. I M. Nauka (1989). 3. Pontryagin L., Some topological invariants of closed Riemannian manifolds. Izvestia Acad. Sci. USSR, ser. Math., 13 N 2 (1949) (In Russian). 4. Gantmacher F., Matrix theory. M. Nauka, (1988). 5. Obata M., The Gauss map immersions of Riem. manifolds in space of const. curvature. J. Diff. Geom., 2 N 2 (1968) 217-223. 6. Leichtweiss K., Zur Riemannschen Geometric in Grassmanschen Mannigfultigkeiten. Math. Zeit., 76 N 4 (1961) 334-366. 7. Aminov Yu., On Grassman image of 2-dimensional surface in 4-dimensional Euclidean space. Ukrainski Geom. Sbornik, 23 (1980) 3-16 (In Russian).

8. Aminov Yu., Determination of the surface in E4 by prescribed Grassmann image. Math. Sbornik, 117 N 2 (1982) 147-160 (In Russian). 9. Lavrent'ev M., Fundamental theorem in the theory of quasiconformal mappings of flat domains. Izvestia Acad Sci. USSR, ser. Math., 12 (1948) 513-554 (In Russian). 10. Polozij G., Domain preserve theorem for some differential equations and applications. Math. Sbornik, 32 (1963) 485-492 (In Russian) . 11. Hilbert D.. Courant R., Methods of Mathematical Physics. M.-L. 1 (1951). 12. Kizbikenov K., 2-dimensional surface in 4-dimensional Euclidean space with prescribed Grassman image. LGPI, L. (1983), Dep. Manuscript, VINITI, 12.83, N 6568-83 DEP. 13. Weiner J. L., The Gauss map for surfaces in 4-space. Math. Ann., 269 N 4 (1984) 541-560. 14. Weiner J. L., The Gauss map for surfaces. The affine case. Part I. Trans. AMS.,

293 N 2 (1986) 431-446. Partil. The Euclidean case. Trans. AMS., 293 N 2 (1986) 447-466.

15. Hoffman D., Osserman R., The geometry of the generalized Gauss map. Memoirs AMS., 28 N 236 (1) (1980) 1-105. 16. Hoffman D., Osserman R. The Gauss map of surfaces in R". J. Diff. Geom., 18 N 1 (1983) 733-754.

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17. Aminov Yu., Reconstruction of 2-dimensional surface in n-dimensional Euclidean space by Grassmann image. Math. Zametki, 36 N 2 (1984) 223-228 (In Russian).

18. Muto Y.. The Gauss map of submanifold in a Euclidean space. J. Math. Soc. Japan, 30 N I (1978) 85-100. 19. Muto Y., Submanifolds of a Euclidean space with homothetic Gauss map. J. Math. Soc. Japan, 32 N 3 (1980) 531-555. 20. Muto Y., Deformation of a submanifold in Euclidean space with fixed Gauss image. I. Geom. Dedic., 11 N1 (1981) 1-18. II. Ibid., 12 N3 (1982) 321-335. 21. Muto Y., Deformability of a submanifolds in a Euclid. space whose image by the Gauss map fixed. Proc. A MS., 76 N 1 (1979) 140-144. 22. Yaglom A., Yaglom 1., Nonelementary problems in elementary exposition. M. GITL (1954) 23-24. 23. Wong Y. C., Differential geometry of Grassmann manifolds. Proc. Nat. Acad. Sci. USA, 57 (1967) 589-594.

24. Wong Y. C., Sectional curvatures of Grassmann manifolds. Proc. Nat. Acad. Sci. USA, 60 (1968) 75-79.

25. Borisenko A., Nikolaevsky Yu., On surfaces of maximal Grassmann image curvature. Math. Zametki, 48 N 3 (1990) 12-13 (In Russian). 26. Nikolaevsky Yu., On surfaces of Grassmann image curvature nonless then 1. Ukrainski Geom. Sbornik, 33 (1990) 77-91 (In Russian). 27. Borisenko A., Nikolaevsky Yu., Classification of points of 3-dimensional surfaces by Grassmann image. Ukrainski Geom. Sbornik, 32 (1989) 11-27 (In Russian). 28. Borisenko A., Affine classification of points of manydimensional surfaces. Sibirski Math. Journ., 31 N 3 (1990) 17-29. 29. Borisenko A., Nickolaevsky Yu., Grassmann manifolds and Grassmann image of submanifolds. Uspehi Math. Nauk, 46 N 2 (1991) 41-83 (In Russian). 30. Masaltsev L., Minimal surfaces in R5 with constant curvature of Gauss image. Math. Zametki, 35 N 6 (1984) 927-932 (In Russian). 31. Courant R., Equations with partial derivatives. M. Mir (1964).

32. Leichtweiss K., Uber eine Art von Krummungsinvatianten beliebiger Untermannigfaltigkeiten des n-dimensionalen euklidischen Raumes. Abh. Math. Sent. Univ. Hamburg, 26 (1963/64) 155-190. 33. Berezina L., Moving frame of nt-dimensional surface in n-dimensional space of constant curvature. Izvestia VUZov, Mathematics, 5 (1964) 8-11 (In Russian). 34. Gorkavy V., On reconstruction of a 3-dimensional submanifold of a 5-dimensional Euclidean space from a degenerate 2-dimensional Grassmann image. Mat. Physics, Analysis, Geometry, 2 (1995) 25-41 (In Russian). 35. Gorkavy V., On reconstraction of a submanifold in Euclidean space from degenerate in a line Grassmann image. Mat.Zametki 59 (1996) 681-691 (In Russian).

36. Gorkavy V., The theorem of reduction in the problem of reconstruction of submanifolds in Euclidean space from given Grassmann image. Mat. Physics, Analysis,Geometry. (1997) 4 v. 3 309-333 (In Russian).

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37. Savel'ev V., On the theory of curvature for Grassmann image of submanifold in

Euclidean space. Math. Physics. Analysis. Geometry. 1 (1994) 520-528 (In Russian).

38. Sheffel S. Z., Surfaces in Euclidean space. Matematicheskii analiz, i smeshnie voprosi matematiki, Novosibirsk, Nauka, (1978) 297-318. 39. Sheffel G. S., Groups of transformations of Euclidean space. Sihir. Math. J. 26 (1985) 187-215.

Chapter 9

1. Coxeter H. S. M., Introduction to Geometry. Moscow: Nauka (1966) (In Russian).

2. Berger M., Geometry. Moscow: Mir (1984) (In Russian). 3. Gordevskii D. Z., Leibin A. S., Popular introduction to manydintensional geometrv. Kharkov State University, Kharkov (1964) (In Russian). 4. Schlafli L., Gesammelte mathematische Ahhandlungen. Basel. 1 (1950). 5. Stringham W. I., Regular figures in n-dimensional space. Amer Journal of Math. 111 (1880) 1-14.

Chapter 10

1. Efimov N., Nonembeddability of Lobachevski halfplane. Vestnik Moskovskogo Univ., ser. Math. and Mech., 2 (1975) 83-86 (In Russian). 2. Vorobjova L., Impossibility of C2 isometric immersion of Lobachevski halfplane into E3. Vestnik Moskovskogo Univ., ser. Math. and Mech., 5 (1976) 42-46 (In Russian). 3. Pozniak E., Isometric immersion into E3 of some noncompact domains in Lobachevski plane. Math. Sbornik, 102 N 1 (1977) 3-12 (In Russian) .

4. Cartan E., Sur Ies varietes de curbure constante d'une espace euclidien ou noneuclidien. Bull. Soc. Math. France, 47 (1919) 125-160; 48 (1920) 132-208.

5. Liber A., On classes of Riemannian spaces of constant negative curvature. Uchenie Zapiski Saratov Univ., ser. Phys.-Math., 1 (1938) (In Russian).

6. Moore J. D., Isometric immersions of space forms in space forms. Pacific J. Math. 40 N 1 (1972) 157-167. 7. Cartan E., Riemannaian geometry in orthogonal frame. M., MGU (1960) (In Russian). 8. Aminov Yu., On immersion of n-dimensional Lobachevski space domains into (2n - 1)-dimensional Euclidean space. Sov. Math. Dokl., 236 N 3 (1977) 521-524 (In Russian). 9. Aminov Yu., Isometric immersions of n-dimensional Lobachevski space domains into (2n - 1) Euclidean space. Math. Sbornik, 111 N 3 (1980) 402-403 (In Russian).

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10. Darboux G.. Lecons sur les svstems orthogonaux et les coordones curvilignes. Paris: Gauthier-Villars (1910). 11. Bianchi L.. Lezioni di geonietria dijferentiale. Bologna: Zanichelli, 3rd ed. 2 part 2 (1924).

12. Bourlet M. C., Sur les equations aux derivees partielles simultannes qui contiennent plusieurs fontions in connues. Ann. Ecole Norm. Sup., 8 ser 3 (1981) 3-63.

13. Brill A., Bemerkung fiber pseudospherische Mannigfaltigkeiten von drei Dimensionen. Math. Ann. B.26, Heft 2 (1986) 300-303.

14. Schur F., Uber die Deformation der Raume constanten Riemann'schen Krfimmungsmaasses. Math. Ann. B.27, Heft 2 (1986) 163-176. 15. Aminov Yu., Manydimensional analogue of "sine-Gordon" equation and rigid body motion. Sov. Math. Dokl., 264 N 5 (1982) 1113-1116 (In Russian). 16. Aminov Yu., Isometric immersions of 3-dimensional Lobachevski space domains into 5-dimensional Euclidean space and rigid body motion. Math. Sbornik. 122 N 1 (1983) 12-30 (In Russian). 17. Arhangel'ski Yu., Analytic dynamics of rigid body. M. Nauka (1977) (In Russian).

18. Hartman Ph., Ordinary differential equations. M. Mir (1970). 19. Aminov Yu., Isometric immersions with flat connection of n-dimensional Lobachevski space domains into Euclidean spaces. Gauge fields model. Math. Sbornik, 137 N 3 (1988) 275-298 (In Russian). 20. Aminov Yu., On immersions with n fields of principal directions of n-dimensional Lobachevski space into 2n-dimensional Euclidean space. Ukrainski Geom. Sbornik, 28 (1985) 3-8 (In Russian). 21. Aminov Yu., On functionally degenerated immersions of Lobachevski plane into E4. Ukrainski Geom. Sbornik, 33 (1990) 8-18 (In Russian).

22. Aminov Yu. Bianchi transformation of manydimensional Lobachevski space domain. Ukrainski Geom. Sbornik, 21 (1978) 3-5 (In Russian). 23. Borisenko A., On isometric immersion of pseudo Riemannian space of constant curvature. Ukrainski Geom. Sbornik, 19 (1976) 11-18 (In Russian). 24. Tenenblat K., Terng C. L., A higher dimensiona generalization of the sineGordon equation and its Backlund transformation. Bull. Amer. Math. Soc. (New ser.), I N 3 (1979) 589-593. 25. Tenenblat K., Terng C. L., Backlund theorem for n-dimensional submanifolds of RZ"-'. Ann. Math., 111 (1980) 477-490. 26. Terng C. L., A higher dimension generalization of the sine-Gordon equations and its solution theory. Ann. Math., 111 (1980) 491-510. 27. Vilenkin A., Phys. Rev. D., 27 (1983) 2848. 28. Sakharov A., Cosmological passes with metric signature change. J. E.xp. Theor. Phys., 87 N 2 (1984) 375-383 (In Russian). 29. Landau L., Lifshits E., The Classical Theory of Fields, fourth edition Pergamon, Oxford (1975).

30. Prasad M.. Instantons and monopoles in Yang-Mills fields gauge theory. In: Geometrical ideas in physics. M. Mir, (1983).

Subject Index

F First curvature 3 flat normal connection 22

A

Angle Direction 222 asymptotic line 45 - direction

Frankel formulas 4

321

Fubini-Study metric

191

B

Berstein theorem

38 40

C Cartan formula 113 Christoffel symbol 24 codimention 62 conformal curvature tensor

G Gauss-Kronecker curvature 35 Gauss torsion 113 geodesic curvature vector 20 125

D Darboux equation generalized 60 degree of normal mapping 52 divergent form of mean curvature 32 dyad 288 E Elementary curve

1

- submanifold 19 Euler formula generalized 4.1 Euler-Poincare formula 259 exotic spheres L25 explicit representation 36

H Half-tangent line 2 hinge deformation 24

I Indicatrix of normal curvature 2Q isometric immersion

J Jorgen's theorem

91

192

L

left and right helixes

Q

line of curvature 40 23 367

SUBJECT INDEX

368

M

R

Mean curvature 35

Rectifiable curve 2 regular curve 2 - submanifold 20 representation of curve with constant curvatures 1$ Rodrige's formula 41

Milnor lemma 52 mixed representation

26

N

Natural parameter 3 normal curvature 31 -vector 69 -space

21

0 Operator of the second fundamental form 22 ordinary multivector 6

S

scalar product of two multivectors second area variation 1611 - fundamental form 93 small parameter L35 smooth curve 2 space 6-pinched 126

T Tangent line 2

-space P

Parametric representation L 20. pluriharmonic conditions 146 Poincare theorem 190 position vector L9 principal curvature 34 - direction 80 pseudospherical congruence 342

21

three-dimensional torus 22 Toponogov's problem favorite trigonometric polynomial L3 V

Vector function 2 Veronese submanifold Volkov formula 198

2

151

1.68

2

Author Index

Alexandrov A. D. 351 Allendoerfer C. B. 51 Almgren F. J., Jr. 38.185 Aminov Yu. 168, 215. 222. 31 Artin E. 116

Cayley A. 255 Chaplygin S. A. 334 Chebyshev P. L. 312 Chem S. S. 83 83 172 249, 290 Clebsch R. F. A. 334 Collins W. 26$

B

do Carmo M. UL 1.52 Coxeter EL S. M. 272

A

Biicklund A. V. 34L 345 Beckenbach E. F. L43 Beltrami E. 181 Bernstein S.

157. 165. 291. 352

Berger M. 1668. l.2.L 175. 176. 275 Bianchi L. 44,3 345 B1anuga D. 285 Blaschke W. 291 Bonnet 0. 42 Borisenko A. A. 22, 225.248, 352.354 Borisov Yu. F. 9 Bourbaki N. 290 Bourlet M. C. 342 Brill A. 3.16

D Darboux G. 136 Dirichlet L. 254 Doyle A. C. 3M1 Duschek A. 1.62. 165 E

Efimov N. V. 58.67 Einstein A. 251 Eisenhart L. 42. 143 Euclid, 253

Burago Yu. D. 168 Burchardt J. J. 255 Burstain C. L. 255

F Fleming W. H. 39 Fomenko A. T. 277 Fomenko V. T. 352 Frankel T. T. 102

C Calabi E. 151. 192 Cartan E. 51. 113. 290. 295

G

Cauchy 0.

125

Gauss K. 3147 De-Giorgi 38 369

25

12; 75 2, 351

AUTHOR INDEX

370

Gishar 3411 Giusti E. 34

Mordochaj-Boltovskoj D. D. Muto Y. 248 Moore J. D. 168. 293.228

Glazyrin V. 82 Gorkavij V. 225 Gulliver R. D. 1$Q

N

Nash J. F. 162. 1RQ Nikolaevsky Yu. A. 248 Novikov P. S. 291 Noviov S. P. 277

U

Hadamard J. 291 Hilbert D. 134 28Q Hildebrandt S. L811 Hopf H. 190 Hyppas 253

0 Ossermann R. 1.84 Otsuki T. 81. 83

I Il'ina L.

253

P Parmcnid 332

1116

J Jacobi K. G. J. 254 Janet M. L. R. 255

Peng C. K. 152 Perelman G.Ya. 1153 Platon 253 Pliicker J. 227 Pogorelov A. V. 24 155, 121

K

Kagan V. F. 291 Kazdan J. L. 181 Kaul13. L80 Kirchhoff G. R. 334 Kizbickenov K. O. 214 Klien F. 47 144 Klimentov S. 351 Klingenberg W. 168. 125

Polozij G.

212

Pozniak E. H.

3344 216. 285.353

R

Rashevski P. K. 277 Rauch H.,

L68

Riemann G. F. B. 47 AS Rozendorn E. R. 137. 285

Kobb M.G. 334 Kogan A. 106 Kolmogorov A. A. Kuiper N. 81

S 221

L

Lagrange J. L. 159 Laplace P. S. 152. 181 Lavrcnt'ev M. A. 218, 214 Leichtwiess K. 240 248 Levi-Civita T. 332 Liber A. 51. 286 M

Masaltsev L. A. 334 Maxwell J. C. 338 Maummary S. xv Micallef M. F. L6$ Milnor J. 52, 115 Monten P. 221

Sabitov L II. 354 Sakharov A. D. 331 Savel'ev V. M. 248 Schlklli L. 2,2 263 Schmidt O. 291 Schoen R. LL. 180. 181 Schwarz H. A. 1533.. 155. L51 L58 Sheffel S. Z. xv, $5. 81 Shapovalova L. N. 353 Shikin E. V. 353 Shiogama K. L25 Shur A. 316 Sidorov L. 106 Simonov L 222 Simons 38. 165 Sintsov D. 221 Smyth B. 61 Sovertkov P. L 353. 354

AUTHOR INDEX

371

Steklov V. A.

334

Sugimoto M.

175

Weierstrass K. Th. W. Weiner J. 219 Weingarten J. 306

51 350

Wente H. C. 190 Whitney H. 124. 172 Willer J. 275 Wong Y. C. 240. 242

T Thomas J. M. Tenenblat K.

Terng C -L 350 Toponogov V. A. 89, 168, 175 V

Vekua 1. N. 219 Verner A. L. 353,354 Vilenkin A. 337 Vladimirova S 109 Volkov Yu. A. 109

W Walter R. 190 Warner F. W. 181 Weber H. 334 Wegner A. 353

X Xavier F.

143

67

Y

Yampolsky A. 79 Yau S -T. 168, 179. 180, 181 Z

Zulanke R. xv Zmud' L. Ya. 253

Zenon 337

THE GEOMETRY OF SUBMANIFOLDS Yu. Aminov. Indit utr, t()r Lies Irmprvature PIl\.ir ,inrl I n uu-ering, Kharkov, Ukraine Ihis volunnC pro%icles a comprehensive presental loll of the geometry of suhmanifolds and expands ( lassie II results in the Iheory of curves and stir far es. The geometrv of submanifolds begins Irnm the idea of the extrinsic geometry of a surface and the theory studies the position and prodxertins of a submanifold in ambient space, in lot al and global aspe Is. I he volume ,Ibo highlights the (ontributiuns made by great geumeters, past and present, to the geometrv it .uhm.mifolds and the developing areas ul tppli( at ion

About the author I'rulcssr.r Yu .\nnnv is a leading resean h fellow at the Institute Ior Low Eemperalure I'hysr( s and Engineering of the Ukrainian Academy of Sciences in Kharkov, Ukraine. I It, works in classical differential geometryand has published extensively in this area of mathematics.

Titles of related interest I htlerrnhal Yn. \nunnv

lrc and topology of Climes

The Get imetry of Vector Fields

Yu. Aminov

Linear Algchra and Geometry A.I. Koslrikin and Yu.l. Manin tensor and Vector Analysis: Cevrnetry, ,bh' hanks and Physf( edited by A. T. Fomenku, O.V. Manturov and V.V. Trnfimuv

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