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