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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 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.
THE GEOMETRY OF SUBMANIFOLDS
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 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 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 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.
THE GEOMETRY OF SUBMANIFOLDS
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
THE GEOMETRY OF SUBMANIFOLDS
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
_
) + 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." 0fore=0. Therefore, we can state the criterion of stability of a domain in the surface: the domain is stable if its second area variation 62S is positive for all non-trivial variations reducing to zero over the boundary. The domain is unstable if there exist variations such that b2S. Investigations of minimal surface stability in terms of area second variations were carried out by Schwarz in the nineteenth century. He stated the area second variation
formula. To derive that formula, we suppose that a given domain is endowed with lines of a curvature coordinate system (u, v). Then Rodrigues' formula holds: ny
_
- TI r,,
I Trr,
nr
where R, are the principal radii of curvature. We consider variations in the normal direction because it can be shown that more general variations with fixed boundary lead to the same result (see [4)).
On the surface. corresponding to the e-parameter, the vectors tangent to coordinate curves are of the form R
rr = l 1 - R cw) rv. + ew, n. 2
The metric coefficients are
P= / (I
2
-
R1
ew E+e21vu,
G= I I-RZewIG+e2w2. Omitting the infinitesimals of order greater than e'-, we get
E6 -F2 = I_E_G
+Ke2w2+`2Ew2+Gss,2 I
2EG
)
Denote by V1v the Beltrami first differential parameter on the minimal surface F2, i.e. the square of this function gradient:
EG-F'Then we obtain the expression for the second-order area variation of G c F2:
625= J {2Kw2+Vw}dS.
(1)
THE GEOMETRY OF SUBMANIFOLDS
154
Denote by w the area element of spherical image. Then dw/dS = J A1. Let dn2 = e du'- + 2 f du dv + g dv 2 be the spherical image metric with respect to (u, v) coordinates. It is related to the metric ds2 of F'- by dn2 = (q dS2. Let V'w be the first differential parameter of it, in the spherical image metric. We can express it in terms of the first differential parameter of a minimal surface as gw
+
Vw
H', eg _f 22
IK12EG
IK1
Therefore, the expression of second area variation can be rewritten solely in terms of spherical image: b2S = 2
f
w2} dw,
(2)
where is the spherical mapping of a minimal surface. Set G' = P(G). The formula just obtained was found by Schwarz. From this formula it follows that the stability property of a domain in the minimal surface is characterized by its spherical image, i.e. by the domain G'. In some cases the domain G' in a unit sphere can be found easily. If it is sufficiently simple. it is possible to determine the sign of 62S. It turns out that the question of stability can be reduced to the question of the first eigenvalue of the Laplace-Beltrami operator V of the unit sphere metric in G'. Indeed, since ' is zero over the boundary of G', the/p following integral relation holds:
-G'
f
w Vz t dw.
G,
Therefore, we can rewrite the second area variation in the form
b2S=- f (V
(3)
G.
Let us consider the eigenvalue problem in G':
V w4-taw=0 provided that I,x;. = 0. From the general theory of elliptic differential equations it follows that there exists the orthonormal system of eigenfunctions WL corresponding to the eigenvalues 0 < Al < A2 .... Let (p1 be the first eigenfunction in G' corresponding to eigenvalue A,. Suppose al < I. Represent the surface variation in the n. Then form F = r +
62S= -f (-2A, w + 21v)iv dw < 0. U,
Thus, in this case we can point out the variation decreasing the area. It is the domain instability case in a minimal surface.
I»
MINIMAL SUBMANIFOLDS
Suppose \1 > 1. Then \, > I for all i. Any surface variation directed along the normal and vanishing over the boundary we can represent in the form of a linear combination with constant coefficients of eigenfunctions w = E', akcpk. Then V;w = -2 k 1 ak,\kcpk. We have x
r ac
b2S =
- G.J
a,
E{-2akAkcpk +2ak+pk} I: ai'pidw = 2>ak(Ak - 1) > 0. ka1
i=1
k=1
Here we used the orthogonal property of tp1,... , Wok.... . Thus, the following theorem holds.
Theorem (Schwarz) The domain in a minimal surface is stable if the first eigenvalue of the Laplace-Beltrami operator of the domain spherical image Al > I and unstable if Al < 1.
Here Ai are the eigenvalues of the spherical image Laplace-Beltrami operator, i.e. the number for which the equation V;1v = -2A;1v has a non-trivial solution vanishing over the boundary. 7.4 Pogorelov Theorem on the Instability of Large Domains in a Minimal Surface in E3
In the previous section we considered the description of stability or instability properties of a domain in a minimal surface in terms of spherical image. But it is possible to obtain the criterion of instability starting solely from intrinsic properties of the minimal surface metric. In [5] the following theorem has been stated. Theorem (Pogorelov) A simply connected nrinimal.cu{jace is unstable if it contains the geodesic disk of radius p satisfying even one of the following conditions:
(1) Ict(g)I >
P
'tp1 In ip ) for
some < p;
(2) S(p) > 47rp2
Here S(p) is the disk area, l(p) is the length of the circle of radius p, w({) is the integrated curvature over the disk of radius C with the same center. Define the polar coordinate system (u, v) in the considered geodesic disk. It is possible because K < 0. Let ds 2 = dug +g2 dv2 be the metric of the surface with respect to that coordinate system. We consider deformations r = r + ewn such that the function w depends only on the distance from the disk center, i.e. on u. Then the second area variation can be expressed as 2w
62S=
if 00
+1 v )gdudv.
THE GEOMETRY OF SUBMANIFOLDS
156
Suppose that the first condition of the theorem is satisfied. Set ' = I for it < . In the segment { < it < p introduce the new variable a defined by means of equation 2-
du =
1(tt)
where 1(u) =
,
J
gdv.
0
i.e. 1(u) is the length of the geodesic circle of radius u. Let v be a linear function with respect to a over the remaining part of the disk and also equal to I for it = and
to zero for u = u(p). Such a function has the form (u) -
u - u(P)
it(.) - ft(P)
For the function w we have
,+tt(P)
u(t,)
2::
b2S< Jf 2 Kgdudv+ 0 0
J
- tt(0
u(F)
/f tp
=2
+
I
du
1(u).
The function g(u. v) is convex with respect to it because of ,,,/, = -K > 0. Hence 1(u) is also convex downwards. Note that 1(0) = 0. Since the graph of 1(it) lies under
the straight line through the origin and the point (p, 1(p)), then td(p)/p> 1(u) (see Figure 23). Therefore p
1( ) >
In 1(p)
For the second area variation we obtain an estimate: 1(P)
6 2S 2k°kj dS, n,j
i<j
(2)
where dS is the n-dimensional volume (area) element of a submanifold. The field of variation S we took in the form wn.. The field ( defines each sum in (2) uniquely because they are independent of the orthogonal to C basis rotations. We can write the
sum of the first two terms in invariant form as (V(V ), where V is a covariant
162
THE GEOMETRY OF SUBMANIFOLDS
derivative in a normal bundle. Besides, note that Ei<j k f kJ = KK(n.) is the second symmetric function of principal curvatures of F" with respect to np. 7.7 Duschek Formula for Second Volume Variation of Minimal Submanifolds in Riemannian Space
We consider now a more general situation. Let FA be a minimal submanifold in the Riemannian space M"+P. Let V be a normal field of variation on Fk. Through each point x E Fk with V 0 0, draw the geodesics of M"+P in the V-direction. The set of all points of these geodesics form some (k + 1)-dimensional manifold M containing Fk as a hypersurface. Consider the variation of Fk in M. Introduce the coordinate system in M. Take as t the arc-length of a geodesic starting at the point of Fk. Then the metric of M has the form k
ds'`
_
g,
du'du'+dt2.
i.l=l
where (u' ..... u k ) are coordinates in F" and gl depend on face, obtained from Fk by variation, can be represented in ft as
uA, t). The sur-
t = e11'(u'.....Ilk),
where w is some smooth function of class Cl. Varying a we obtain a one-parameter family of surfaces. The surface corresponding toe has metric coefficients of the form 9i1 = g,1 + e21v", w,,, .
In the sequel we set w; We are going to find g = det Ig;;I and its derivatives of first and second order with respect to a for s = 0. Introduce the following columns: till
91i
gi=
a=
Uj
I
Then we can write 2 g = [g1 +6 wla,...gk + e2H'ka) A
_ [gl...g&]+e2 E[gl...a...gk]w;. i=1
Expanding all the determinants [gl ... a ... gk] with respect to the i-th column and summing them over i, we obtain the coefficient of e2: iv; w1g''g = g V, w. Here we use
the standard notation V1w for the square of the gradient off. As gi, depends on t, then g = [g, ...gkj depends on e. We have
dw... 191
atgk]+2egVln'+
MINIMAL SUBMANIFOLDS
163
where dots mean the terms of order higher than e. Note that for e = 0 1 8g;
-
2 8t where L, are the coefficients of second fundamental form Fk C M. If we introduce the column
Ll; L; _
,
Lki
= L;. Differentiating (I) we get the expression of the second then we can write derivative for E = 0 as d 2g __ 2
dE2
1 2 ... gk1 v` + 2g01 w. gi ... agr ... agi ... gk w +kL I ... a'gr at 2 at 8t 19 I
(2)
Introduce in Fk the coordinate system in such a way that at a chosen point .yo would be g; = S;, and coordinate curves would be tangent to the principal directions of the L, du' dui form. If n is the unit vector directed along V then at a chosen point L1; is equal to the normal curvature k; with respect to n. By the very definition, this value is the same either whether we consider Fk in ft or in Mk+P (note that n is tangent to Al). Then
{gi.........]4Lgi...Li...Li...1=4(Liilii_L).z4Jciki. dt
at
(3)
The second term in (2) at xo has the form 492
A
ar-
IV-
Consider the curvature tensor of M at the points of Fk. Denote by Rtgp its components. For convenience, set k + l = v. The metric of M is rewritten in the form
ds2=E j'Igy th' di, where we set x = t, x' = u1 for I
k. The general formula for Riemannian tensor components has the form a2gjj _ a2gk; _ 82g,, ax'8x/ + 8xkav dx'8xp axkaxl
I (82gkp RhJ°
+g1,n(r,,R rhp, - rip, r,JJ), where I'd,,, are Christoffel symbols of the M metric. From this formula it follows that 82gu + g11
2
[(rv;,1)2-rW.1 r;; r,
.
THE GEOMETRY OF SUBMANIFOLDS
164
The Christoffel symbols at x0 are rvi / = I 09-.1 2
ui
0Sul
OS;,
+&J
2 i31
_
- Lii,
r." = 0.
Thus, at x0 10 2 gii + Lz
Rn,;,,
(4)
8 Taking into account (3) and (4), we can rewrite (2) as
Id=g_ ,
_
= 1V
4
2
1: k l + E k? - k ;,,;,, i o
i k2 = 2 >2 k,kj. i<j
i
i<j
Substitute the relations just obtained into (5). To find d2 fg/de2 it is sufficient to find
d2g/de' because dg/de = 0 for e = 0 if Fk is minimal. Thus, we have found the expression for d22 f /de2 at the point xO under the special choice of coordinate system. With respect to an arbitrary coordinate system this expression gets the multiplier f. Integrating over the whole P. we obtain the expression for the second volume variation of minimal submanifold in Riemannian space.
fits=
F,Li 4. Then M is homeomorphic to a sphere.
Also, note that, applying the minimal surfaces theory, Burago and Toponogov in 1973 [51] obtained an interesting result in the theory of geodesics. They stated that in a three-dimensional closed simply connected Riemannian manifold of positive Ricci curvature the length I of any closed geodesic is bounded from below. More precisely, I > 12 a 6/Ri, where Ri is the minimum of Ricci curvature. That paper is related to Toponogov's favorite (by his own admission) problem: prove that if for arty 2-plane cr the sectional curvature K, of a simply connected closed Riemannian manifold satisfies the inequalities 0 < b < K < I then there exists the positive-valued function f (b) such that the length I of any closed geodesic satL /les the inequality l > A b) . In this section and the following section we prove the following result: Theorem (Aminov) Let F2 be a minimal surface homeomorphic to a sphere in a complete simply connected Riemannian manifold R" of curvature from the interval (4 , 1]. Then F2 is an unstable minimal surface.
Note that in [14] Lawson and Simons considered minimal flows in Riemannian space and stated the hypothesis on their instability in a complete simply connected Riemannian space R" of curvature from the interval (4-1. I].
MINIMAL SUBMANIFOLDS
169
One may consider the surface F2 in Riemannian space R" with the boundary defined in the neighborhood of F2. In this case it is necessary to pose the condition of
F2 normal bundle triviality, i.e. of the existence on the surface of the continuous basis normal fields n1,...,n"_2. The following holds: Theorem 2 (Aminov) Let F2 be a minimal surface, homeomorphic to a sphere, in the Riemannian space M" of curvature from the interval (4 1, 1]. Suppose the normal bundle of F2 is trivial. Then F2 is an unstable minimal surface.
For the surface in four-dimensional Riemannian space there hold more strong theorems. But we shall state and prove them in the next section. First we prove Theorem 2. Let M" be a C4 Riemannian space and components of
the position vector of F2 be of class as functions of parameters. As F2 is homeomorphic to S2, it is possible to introduce isothermal coordinates u1, u2 singular at one point (see [15], [18]). The metric of F2 with respect to that coordinate system has the form ds2 = A(dui + dug), where A 0 as 1 if we set z = ul + iu2. Select the variation fields V, and V2 in such a way that they satisfy the following systems simultaneously: (a) '+'u, + i'µ12/2 = 0,
x'", - i'µ12/1 = 0,
(b) Ali/1 + µ2112 = 0, µU/2 - 112j/I = 0.
2.
(1)
Observe that E 3 L({L,j/, + /z2j/2)2 + 011#2 - 112j/1)2] depends on only the plane distribution spanned on V, and V2. From the triviality of the normal bundle on F2 it follows that there exists the field of frames C1, ... -G-2 of normal space N.Y such that Ci are continuous mutually orthogonal unit vector fields on F2. Moreover, if F22 is of the Cm regularity class then Ci can be chosen from the C'"-I class. We can use the Pontriagin method to prove this (see [16] Section 7). In this paper he considered framed manifolds in Euclidean space. By the Nash theorem [17] the Riemannian space M" together with F2 can be regularly and isometrically immersed into some Euclidean space. The method from [16] of constructing the regular framing F2 homotopic to the given continuous framing is applicable to our case. Thus, we can suppose that if F2 is of the C2." regularity class then i E Cl-". Write the decomposition
k = 1..... n - 2.
V k = link =
As I V, = I V21 = iv and V, 1 V2, then K'2=Eair=>a2-,
>a,Ta2T=0
(2)
It is easy to see that the torsion coefficients pjk/i can be expressed in terms of torsion coefficients pjk/i of the 1;i basis as 1t'2 pjk/i = air
8akr
8a- + afr aA,, µm/i,
(3)
THE GEOMETRY OF SUBMANIFOLDS
170
Using (3), the equations of system (1)(b) can be written as
a air
aair aui
raa, r 1\ 0142
+
8a2,
+ an(aI, jro/1 + ago
8142
- 8827 1 8141 /1
0,
(4)
+ atr(a1, lam/2 -agol1) = 0.
Since w2 = Eai,, the equation w,,, + n7112/2 = 0 after multiplying by w gives the form
alr
0a1r 814 1
+aIr
aa2r 814 2
+alra2v{k,o/2 =0,
i.e., taking into account that air aio µro/1 = 0, we conclude that the latter equation can be obtained from (4) if we set j = 1. In an analogous manner we state that (4) produces the equations of system (1)(a) if we set j = 1, 2. Thus, all equations of (1) can be written in the form of (4) if we set j = 1, ... , n - 2. Let ak1 be components of the matrix inverse to IIa;iII Multiplying (4) by ak/ and contracting over j, we get as
as
8141
+ 8142 + a, ILkm/I + a2m Pkn,/2 = 0,
aaik aa2k 5142 - 514i + as,,, Jkm/2 - a2m µkm/l = 0,
(5)
where k = 1,2,...,n- 2. Let U be
a complex vector with components Uk = a1k - ia2k and B the matrix with components Bmk = µmk/1 + iµmk/2. Then we can write (5) as 2
aUk
of
(6)
= Bk,,,
The solution of (6) is the generalized analytical vector function (the definition in tend to zero as [18]). Since (i are regular of class CI-0, the coefficients B4,,, E 1 /IzI2, when z --# oc. Indeed, the torsion coefficients µkm/i have the form {kA-m/; = (nm, nk.i) + IOV+ O1 Yi k Sm,
where [av, I 1 are Christoffel symbols of the M" metric. We can write '
t"
where alas; means the derivative with respect to the arc-length parameter. By virtue of the regularity of k on F2 we have that Ia"I is bounded on F2. Besides, it is evident
that ICI < const. Since v(A- -. 0 as I/IzI2 when z -, oo then µk/; -. 0 as From the conditions I V1 I = I V2I and V, 1 V2 it follows that U2 _ U; = 0. 1/IzI2
Prove that (6) has the solution U defined over the whole z-plane including the infinite
point oo and such that U2 = 0.
MINIMAL SUBMANIFOLDS
171
Since BTU = -B1 for any two solutions X and Y of (6) the functions X2, Y2 and (XY) are holomorphic functions of z. Indeed, for instance, we have 8U2
a_U
I
Un,(Bkm + Bmk) = O.
aZ = 2 Uk In the case of bounded X and Y the holomorphic functions X2, Y2 and (XY) are 2 U aZ
constant complex numbers. If there exist two linearly independent solutions X and Y defined over the whole plane then, taking a linear combination U = al X + a2 Y such
that the equation a X2 + 2a1a2(XY) + a2 Y2 is satisfied, we obtain a non-trivial solution over the whole plane and such that U2 = 0. Show that the above mentioned two solutions X and Y exist. If C is a constant vector and U is a solution of (6) then W = U - C satisfies the inhomogeneous equation 2
=BW+BC.
-
Let B° be the space of continuous vector functions with the usual metric, defined over the whole z-plane and tending to zero when z -' oo. Introduce in that space the linear operator S as
SV=
1
a
f/ BV(t) dE, z - t
E
where E means the complex plane. Since Bj 0 as 1/I:I2 when z oo then, in the same way as in the Pogorelov monograph, ([19], p. 428) for a one-dimensional system, we prove: (1) if V C B° then SV C B°; 2) S is totally continuous. Suppose (6) has the non-trivial solution U tending to zero while z - oc. Since U2 is a bounded holomorphic function, then U2 = 0 and, as a consequence, we suppose below that condition (3) is satisfied and the homogeneous equation has only trivial solutions. Then for an arbitrary C the integral equation W + SW + SC = 0 has the solution W such that W 0 while z - oc. Selecting two linearly independent vectors C, and C2 we get two linearly independent solutions X = W, + C1, Y = W2 + C2 such that X C1, Y -. C2 while z oc. Therefore, there exists the solution U of (6) such that U2 = 0. This fact implies the solvability of (1) in the regularity class. Select the variation fields V, and V2 in such a way that (1) would be satisfied. Below we use the Berger inequalities on curvature tensors (we shall prove them in Section 12): If for any 2-plane o, in the tangent space to M" the sectional curvature K satisfies 6 < K, < I then the components of a Riemannian tensor with respect to the orthogonal coordinate system satisfy
(I - b),
I R,Jk,I < z
j4k
IR+fk1I 5
(1 - b).
i i,j,k0l.
(7)
From this it follows that I T121 < 3 (1 - b). It is easy to show that for a minimal surface F2 Ik12I 5 IKK(nl)I + IKe(n2)I
(8)
THE GEOMETRY OF SUBMANIFOLDS
172
Taking into account (7) and (8), we have
Eb'-S(Vi)=Jsv2 [2K12 -2EIK,.(ni)I+2T,2i-1
K(ei,ni)JdS i.i
i
F_
(
< rtv2L3(I F'
9)
-h)-46JdS 1/4. Therefore, F2 is an unstable minimal surface. Theorem 2 is proved. To prove Theorem I we consider the case of n = 4 separately.
7.10 Minimal Surfaces in Four-Dimensional Riemannian Space
The case of n = 4 is rather distinctive. The instability theorem holds under weaker conditions. It is not necessary to require the completeness of M4 and normal bundle triviality of F2. Theorem (Aminov) Let F2 he a minimal surface homeomorphic to a sphere in an oriented Riemannian space M 4 oJ' curvature froth the interval (,1-t . 11. Then F2 is tatstable.
If there exists the regular unit normals vector field on F2, then the statement on instability is true under the assumption that the curvature of M4 belongs to the closed interval (a . 11. At the same time, in complex projective space P2C with the Fubini-Studi metric of curvature from the interval (,1-t. I) there exists the totally geodesic surface which produces the area minimum among close surfaces. But this surface does not admit the regular field of unit normals. In section 5 of Chapter 6 we considered the Whitney invariant of a surface in fourdimensional Euclidean space. It was defined as a sum of indices of normal vector field singular points. We proved that in the case of a closed oriented surface the Whitney invariant can be evaluated by means of integrated Gauss torsion Kr:
V =dS. JKI F=
In the case of the surface F2 in four-dimensional Riemannian Space M4 a more general formula, stated by Chern, holds: I = 2 V
(t:12 + Tie) dS. F:
This formula is a particular case of the Whitney invariant formula, stated for a closed submanifold FA in Riemannian space M21 (37].
As above, we suppose that F2 is homeomorphic to the sphere. Construct the normal vector field on F2 with only one singularity of index v at = = oc. To do this,
MINIMAL SUBMANIFOLDS
173
take in some neighborhood of z = oc the regular mutually orthogonal normal vector fields SI and S2 which generate a positive orientation in the normal space NT. Set z = IzI e'a. Now define fields i with singularity at z = oc as SI = 1 cos vcp - f2 sin i'
42 = I sin vcp + 6 cos vtp.
,
Since the torsion coefficients it12jj of the (FI.e2) basis are related to the torsion coefficients µ,2/i of the
basis as PI2/i = µ12/i + v aul ,
then {i the singularity index at z = oo is equal to v. Because of it being equal to the Whitney invariant of F2, the fields li have the regular prolongations on the complement to the chosen neighborhood of z = oc. By definition, we have
o = arctan
12
III
,
_ dul
-
U2 1z =
1`
a 0
111
121
Since i are regular about z = oo , then µ12/i -. 0 as 1/1212 while z -+ 00. Therefore, µ,2/i - 0 as 1 /1212. This speed of decrease is not sufficient, in general, for the regular solvability of (6) Section 9, but in the case of v > 0, due to the special choice of {i, we prove that the solution exists. Moreover, IUI 0 while z - 00. Therefore, at z = oc
the variation fields vanish and, as a consequence, the variations will be defined everywhere on F2. Next we shall consider the case of v < 0. The system (6) from Section 9 for F2 C M4 has the form
2a =BI2U2
2aa2=B21U1,
where B12 = µI211 + iµ1212. Besides, it must satisfy U2 + UZ = 0. Set U2 = 1UI. Then
we obtain only one equation
2aa'
=iB12U1.
(1)
In some neighborhood of z = oc we can write iB12 = -A12/2 + iiII2/I = -P12/2 + 1µ12/2 -
V
-.
Suppose that v > 0. Set W = U, (I + Iz1°). Then near z = oc we have
8W out aZ=aZ
(1 + ICI
+v2UI
(Z)e_I
ZY
v
= - 2lµ12/2 + 2µI2/2 - 22(1 +
Izlv), W.
Therefore, W satisfies the equation with a regular coefficient everywhere in the complex plane, where this coefficient tends to zero not slower than 1/1=12 while z -' oc. Then there exists a regular solution W which is bounded in infinity together with the derivative. Then the required solution satisfies
THE GEOMETRY OF SUBMANIFOLDS
174
W
U, =
while z -oo.
If the Whitney invariant v < 0 then we consider the system
2 aU = -BU.
(2)
The point is that the sum of second variations can be transformed to l2S(Vi) =
J1
IVA1212)2 +
" (wu, + IVAA1211)2]/A
F2
+w2[-211.12 -2T12+...]}dS. The latter expression differs from that obtained in Section 8 in the signs before p12/j, ,,I2 and T12. Then the coefficient of U1 on the right of (1) changes to
Ic]2/2 - i,1211 +-;. In this case, set W = UI(1 + Iz]-°). Applying the procedure, analogous to the case of v > 0, we find the solution W. Finally, observe that the contour integral (3) of Section 8, which can be regarded as taken over the circle IzI = const, tends to zero while z -y oc. Theorem 3 is proved. For the surface F2 with zero Whitney invariant we prove a stronger assertion.
Theorem 4 If F2 is homeomorphic to a sphere minimal surface with zero Whitney invariant then it is unstable provided that the curvature of M4 takes its values i n [4 ,
I].
The following holds. Lemma If F'- is homeomorphic to a sphere minimal surface in a Riemannian space M4 of positive curvature and the local imbedding invariant n, n'Z e' e; is sign-preserving on F2 then F2 is an unstable surface. T1 2 =
Here ei and ee are components of mutually orthogonal unit vectors e1 and e2 tangent to F2, ni and n; are components of mutually orthogonal unit normals on F22, R,,m are component of an M4 Riemannian tensor. Since v = 0, there exists on F2 the regular normal basis field (n1,n2). There exist the solutions of (I) and (2). From the two pairs of variations defined by these solutions, select the pair for which the T12 would be included into F_, 62S(VV) with a
negative sign. For that pair E.b2S(Vi) < 0 and, as a consequence, the surface is unstable. The Lemma is proved. Turn now to the proof of Theorem 4. Consider one of the pairs of variations, say that which is defined by system (1). Since the curvature of M4 is from the interval [4, 1], then Eib2S(Vj) < 0 where the equality to zero occurs only in the case when T12 at each point takes its maximum value, i.e. T12 = i (1 - 4) = z Therefore T12 preserves the sign. By the previous lemma, F2 is unstable.
MINIMAL SUBMANIFOLDS
175
By analogous methods, one can prove the instability of a totally geodesic surface homeomorphic to a sphere with zero Whitney invariant in an oriented symmetric space M4 of positive curvature [25]. In this case, from the total geodesity of F2 and symmetry of M4 it follows that T12 = const. Moreover, the condition v = 0 implies T12=0. 7.11 On the Instability of Minimal Surfaces in a Complete Riemannian Space
Here we continue the proof of Theorem l from Section 9 based on the result of the two previous sections. Let M" be a complete simply connected orientable space of curvature from the interval (1, 1]. Under this condition the following theorem holds: Theorem If M" is a complete simply connected orientable Riemannian space of' curvature satisfying inequalities I < K. < 1 then M" is homeomorphic to the sphere S".
The proof of this theorem was begun in the paper by Rauch [43] and was then continued in papers by Berger [21], Toponogov [22], and Klingenberg [23]. It is known - due to Milnor's wonderful result - that on an n-dimensional manifold, which is homeomorphic to a sphere S" for definite values of n, there can be defined differential structures different from the differential structure of the usual sphere. Those spheres are called exotic. There were attempts to prove that in the theorem above M" is diffeomorphic to the usual sphere S". Provided that 0.87 < K < I, Sugimoto and Shiogama [24] proved that M" is diffeomorphic to the usual sphere S", but if a < Ko < 1 it was proved that M" is diffeomorphic to either the usual or the exotic sphere. The exotic sphere E" is the surface obtained from two balls D" and D2 by gluing by means of boundary diffeomorphisms [38]. If one cut out from E" the ball D", however small, then the complement would be diffeomorphic to the Euclidean ball Dl-). As F2 is compact, there exists the point in M" and its sufficiently small neighborhood not intersecting F2. Cut out Di from that neighborhood. Let D. be the complement of D1. In papers [39] and [40] it was proved that the normal bundle of immersed spheref(S2) C E" for n > 5 is trivial. The regular homotopy between f (S2) and the standardly immersed sphere S2 C E3 exists. We can suppose that it is realized in some sufficiently large ball homeomorphic to D". Therefore, if n > 5 then the normal bundle F2 C M" is trivial. In this case we can apply Theorem 2 from Section 9. For n = 4 the instability is stated
by Theorem 3. Thus for any n the surface F2 is unstable. Theorem 1 is proved. In the case of locally conformally flat space, the condition on curvature of the space can be weakened to the condition of curvature positivity. Theorem 5 Let F2 be a minimal homeomorphic to sphere surface in an orientable locally conformally flat space Mn of positive curvature. Suppose the normal bundle F2 is trivial. Then F2 is an unstable minimal surface. If n = 3 then the surface F3 is the minimal hypersurface. In this case the instability
occurs only provided that the curvature of ambient space is positive. Therefore, consider the cases of n > 4.
THE GEOMETRY OF SUBMANIFOLDS
176
Remember the expression for the conformal curvature tensor: Chijk = Rh;jk + n
2 (bhjR;k - bhk R;j + gik Rhj - g,jRhk )
R
+ (n - 1)(n - 2) (bhkgij - 6hj9ik ). As e1,e2, n1 and 112 are mutually orthogonal, by the expression above we find
Therefore, for the case of a surface in locally conformally flat space we have T12 = 0. Then in the integrand in (9) from Section 9 we can replace the number (I - b) with 3 0. Therefore, in this case F_, 6 2S(V,) < 0 which means instability.
7.12 Berger Estimates of Tensor Curvature Components
In previous sections we used Berger inequalities on tensor curvature components. They are useful in many other situations if the curvature K, of the investigated Riemannian manifold satisfies definite conditions over all 2-planes a in the tangent space. Riemannian space is called b-pinched if there exists the constant b such that over all 2-planes the curvature Ko satisfies the inequalities b < Ko < 1.
(1)
For the case of 6-pinched Riemannian space Berger stated the inequalities on Riemannian curvature tensor components R;jA, with respect to some orthonormal basis. They are I Ryk,I
2 (1 - b)
for
i 96 j
IRijkil53(l-b) for (i,j)
k.
(k,!),
(2)
i<j,
k 0,
a2(l - Kij) + 2abRjik + b2(1 - K&) > 0. The discriminant of these two quadratic polynomials with respect to a and b must be non-positive. Hence (Rijk)2 < (Kij - b) (Kik
- 6),
(Rijik)2 < (I - K;1)(I - Kik) As the geometric mean is less than the arithmetic mean, from the above we conclude
RjuI < (K;j+K;k - 2b). Real :5
(2 - Ky - Kk).
Adding the corresponding sides of inequalities just obtained, we get the first inequality under the proof: JR;jkil < 1(I - b), where i # j 36 k. Obtaining the second estimate (3) is more complicated. Consider the inequalities 6 < K(aei + bek, eel + dej)
for any set of basis vectors ei, ej, ek, ei and any real numbers a, b. c, d. Transfer b to the right-hand side and set
F(a,b,c,d,i,k.!,j) = K(ae1+bek,cei+dej) - b.
Let aei + bek, cei + dej be unit and orthogonal to each other. Then a22 + b2 = I , 2 + d2 = 1. We can consider the function Fas a polynomial of degree 4 with respect to a, b, c, d while with respect to each of a, b, c, d it is a polynomial of degree 2. Consider the polynomial C
G(a,b,c,d,i,k,!,j) = 2 [F(a,b,c,d,i,k,1,j) + F(a,-b,c,-d,i,k,1,j)]. It is easy to find that K(ae, + bek, cei + dej) = K,1a2c2 + Kk;b2c2 + K;ja2d22 + Kkjb2d2
- 2Riikiabc2 - 2Rijkjabd2 - 2R;j0a2dc - 2Rkikjb2cd - 2(Riikj + Rijki)abcd.
If we change the signs of b and d, then we obtain an analogous expression which contains the terms in the second line of the above with the sign "+". Hence the function G(a,...) contains terms of the form a2c22, a2d2, b'-c2, b2d2 and abed. Now introduce a polynomial
H(a,b,c,d,i,k,1,j) = G(a,b,c,d,i,k,!,j) +G(-a,b,c.d,i,1,j,k). Observe that the second term differs from the first not only because a is replaced with -a, but also in last the three arguments. That is, the replacement: k --F I -+ j , k has been made. Thus, for the function H we get the expression
THE GEOMETRY OF SUI3MANIFOLDS
178
H=(Kit +K,1)a2C2+(KkJ+K,)b2C2+(K,1+K;a)a2d2+(Kkj+K,k)b2d2 - 2(R,,kj + Rfjk, - R1ik - R;ky)abcd - 26.
We can simplify the coefficients of abcd using the Bianchi identity: R;jk1 + R&1j + R;1jk = 0. We obtain that this coefficient is equal to -GR;jk1. Introduce the notations
A=K;j+Ki1-26, B=K;j+K;k-2b, C=Ku+K1j-26, D=Kkj+Kk,-2b, E = 3R;jk1.
The function H has the form H = Aa2c2 + Ba2d2 + Cb2c2 + Db2d2 - 2Eabcd = (Ac2 + Bd2)a2 - 2Eabcd + (Cc2 + Dd2)b2. Since F > 0, H > 0. If we consider Has a quadratic expression with respect to a and b
then we conclude that the discriminant of this expression must be non-positive. Hence
ACc4 + (AD+BC- E2)c2d2 +BDd4 > 0. From this we get
(AD+BC-E2)22-4ABCD 2 then the choice of n(x) is ambiguous. We would like to have a geometrical object with points which would be 193
194
THE GEOMETRY OF SUBMANIFOLDS
H. GRASSMAN
related invariantly to the points of a given submanifold just like in the case of a spherical image.
To construct it we use not the unit vector field n(x) but the normal spaces N, themselves. The dimension of NT is k. Since N, depends on x, then it is possible to say that N, is a space function on P. although this notion is not commonly used.
Draw k-dimensional space N through the fixed point 0 in E"+k such that N is parallel to N,. N belongs to the set of all k-dimensional subspaces which pass through the fixed point 0 in Euclidean space E"+k. The set above of k-dimensional planes is called a Grassmann manifold. In Klein's book [2] on the history of mathematics in the nineteenth century there are many pages devoted to Grassmann. He did not attend lectures in mathematics but, through his original and distinctive education, he made an essential contribution to the development of mathematical science. His fundamental geometrical work, "The study of extension as a new branch of mathematics", was published twice in his lifetime - in 1844 and in 1862. He was a school teacher in a provincial town, but his ideas featured in University courses and, as Klein wrote, he "exerted an original and strange influence still felt today."
In a preface to the second edition of "The study of extension ..." Grassmann names the mathematicians who influenced him: MBbius with his barycentric calculus
and Belavitis, who in 1835 gave the geometrical algebra of intercepts in complete
GRASSMANN IMAGE OF A SUBMANIFOLD
195
generality. Also, Grassmann was stimulated by one of Gauss's remarks in the preface to a collection of his works (1831). Only the geometrical interpretation of complex numbers leads to the geometrical calculus of intercepts in a plane. Grassmann introduced the concept of the geometrical product of intercepts, invented complex numbers of higher order which have multiplication which may be not commutative and proposed no fewer than 16 types of complex multiplication, one of which - the combinatorial product - he used in the theory of linear extension to the notion of determinant. He also introduced new geometrical objects - multivectors. After the publication of his book in 1845, a paper by Sen-Vennan on the geometrical multiplication of intercepts was published. This paper repeated Grassmann's ideas and he sent two copies of his book to Cauchy with the request that one copy be sent to Sen-Vennan. Surprisingly, O. Cauchy published a paper in Comptes Rendus (1853) with methods based on symbolic quantities. One of these methods was the same, in Grassmann's opinion, as his method published in the book he had sent to Cauchy. The situation was delicate, but although Grassmann did not want to accuse the famous scientist of plagiarism, he sent his claim of priority to the Paris Academy. The commission did not come to any conclusion, but Cauchy published nothing more on this subject. It seems that Grassmann never considered the manifold named after him "in the large" but he studied the objects which are the points of this manifold. It is hard to say now who first gave the name of Grassmann to that manifold. Severi cited the papers of Grassmann in 1909 whereas Shubert did not mention them in 1886. So, the k-dimensional plane in Euclidean space E"+t which passes through the fixed point in En+k is called the point in a Grassmann manifold. The word "plane" has been replaced by another word - "point". One of the mathematical objects the plane - obtained another name - the point. And nothing more. But after this there arose some associations, some questions. For instance, we can also ask: what
"points" are close to each other? How can we measure the distance between points? If we define a neighborhood of these points in such a way that they would satisfy corresponding axioms then we obtain a manifold. At first it seems too abstract and
unusual. Then a question arises: how we can imagine it or model it (in modern language)? Its topological structure is particularly confused and complicated. It is difficult to ascertain what is more abstract - the notion of number, straight line and other objects of so-called elementary geometry or such a modern mathematical notion as the Grassmann manifold. However, the objects of elementary geometry are habitual and have been examined for thousands of years. It seems that the notion of number is a priori habitual to the intellect and nothing to speak about.
Therefore it is necessary to make new mathematical concepts habitual and use them if they are fruitful. From this arises the question: where do we meet more abstraction - in the notion of number and the method of number representation or in the generation of modern mathematical constructions? Endowed with the system of neighborhoods, the set of all k-dimensional planes in the (n + k)-dimensional Euclidean space E"+k that pass through the fixed point 0
1%
THE 41:OM1ETRS OF Si. KMAN8TOLDS
becomes a topological space. Moreover, this topological space forms a manifold. i.e.
for each point there exists a neighborhood which is homeomorphic to the ball in some Euclidean space E' of dimension 1. for which I can be expressed via ri and k. The standard notation for this manifold is GA . &. Denote a point in it by P. We shall study the Grassmann manifold in more detail in the next section. Now turn back to the correspondence x A' as at the beginning of this section. Now we can consider it as a correspondence x -- P. Denote it by 1P. Since the image of 'If is located in a Grassmann manifold. 'I' is naturally called a Grassmwnn nwppnrg (in analogy to spherical) and the image'P(F") is called a Grassinuwm image. Other names such as generalized spherical image or tangential image are used. It seems that this mapping was introduced in the nineteenth century. Such a mapping was widely used in topology, for instance in Pontryagin's papers on characteristic classes [3]. However, from the geometrical point of view the properties of the Grassmann image received scant attention and remain poorly studied still.
S.2 Grassmann Image of a Two-Dimensional Surfaces F2 in E4
Leaving aside for a while the general case, consider the Grassmann image of a two-dimensional surface in E4. Let x' be Cartesian coordinates in El. The twodimensional plane N can be represented by a pair
i1) of mutually orthogonal unit
vectors in this plane. Consider the ordinary bivector p = [ i] with Plucker coordinates p
it/
tit ,
where a and q' are components of 1; and q. There are six Plucker coordinates. P1 '-. p23, p22a, p34. It is easy to check that p'' stay unchanged under a rotation of p13, p'''. the basis k, , in N. Values p'J are related to each other by two relations. First, taking
into account the bivector scalar product formula, we have P"
= Q41111410
1r,
(1)
Also, p12p34
+P I i pa_ + p14p21= o.
Consider P":
We can write the left-hand side of (2) as £ I (q p34 + rt3 p i_ + 1;4p= 3) -Tai (42 p 34 + ' pr' + 4 p23) .
(2)
GRASSMANN IMAGE OF A SUBMANIFOLD
197
The coefficient of 1 can be represented as
Ir
q4
73
2 3 f4 = 0.
it
q3
'T
Analogously, the coefficient of q' is zero, too. Equality (2) can be interpreted geometrically. Consider a bivector q defined by the plane complementary to p. If T, v are mutually orthogonal and orthogonal to N unit vectors such that , i?, T, v form a positively oriented basis in E4 then q = [Tv]. The scalar product of p and q is zero. Indeed, (pq) = ([frl][TV]) = (CT)(r!v)
-
0.
On the other hand, the components q'' of the bivector q can be represented in terms of p'". Namely q'2 = P34,
q13 = _p24
q14 = p23,
q24 = _p1a,
q:3 = pI4,
q34
= Ptz.
(3)
To prove (3), form the matrix of , rl, T, v components: r>z
r
r ill
T1
T2
T3
T4
vl
vz
v3
v4
A=
As A is an orthogonal matrix, the inverse matrix coincides with its transpose: A' = A'. Use the formula for minors of an inverse matrix (see [4], p. 31). Let
/
BIk
1 be a minor of a matrix B consisting of elements of the i-th and j-th rows, k-
th and 1-filth columns. Suppose that il, i2, i3, i4 form a permutation of 1,2,3,4 such that i, < fz, i3 < i4. The same conditions are imposed on k,, k2, k3, k4. Taking into account that the determinant of A is 1, we can write A
i,
iz
k,
k,
_ (-1} i, «i,tk, i k; A
k3 i3
k4 i4
Set, for instance, i, = 1, iz = 2, k, = 1, kz = 2. Then T3
F4
v3
y4
i.e. p12 = q34. Set i, = 1, iz = 3, k, = 1, kz = 2. Then i3 = 2, i4 = 4, k3 = 3, k4 = 4. So we have n3 C3
I-
(_I}4f3I Vr21
74
0 I.
THI (JEOMI I R) 01 St SA14NII'OLDS
198
ti(il RE 2')
Hence. pl' = -e. In an analogous way we get other q". We can now represent the condition (pq) = 0 as
(pNl= E
pugs=)(p12p34+pl pit+p'4p2z)=0.
fr
which coincides with (2).
Consider the six-dimensional space E. Then the endpoint of the vector p with components (p12.....p'4) attached to the origin determines the point in El. Various points in E4 obtained in that way and the coordinates of which satisfy (1) and (2) are located in some algebraic four-dimensional submanifold in E. That submanifold is the Grassmann manifold 624 immersed into E'. In other words. this is its model. That submanifold we also denote by G24Let N be a plane parallel to the normal plane N, of F2. If it,, it, form the basis of
normals then the corresponding bivector p is of the form p = [n1ai2). Since u, = F11 (ill u2) are vector functions of it', u2, p = p(rri. it'). Therefore. the image of a Grassmann mapping * is, in general, some two-dimensional surface r2. That surface r2 is located in the submanifold G_,i (see Figure 29). As an example. consider the Grusrnuriui image of the Clifford trues xi = Cos ri.
X2 = sin o,
N' = cos d.
x4 = sin d.
Unit normals of that surface are
it, = (cosri.sin't.0,0), it, = (0,0,cos,d.sin:3). Hence, the components of the bivector p are of the form p12=0.
p 13 = cosre cos,d, pit = cosh rind,
p21 = sin rt cos d
p=; = sin ri sin .3.
p14 = 0.
GRASSMANN IMAGE OF A SUBMANIFOLD
199
As p12 = p34 = 0, in this case r2 C It is easy to state that r2 is a Clifford torus as well. Study some geometric properties of a Grassmann image of a general surface. Since G2,4 is represented as a submanifold in E6, the ambient space induced Riemannian metric on that submanifold is E4.
do,- = E(dp'1)2.
kj
As t- is a subset in G2,4, the metric arises on t'- as well. Denote it by dp2. State the relation that arises under Grassmann mapping between the metric dp2 and the external geometry of F2. We have dp = [dn1 n2] + [n1 dn2].
This formula has a symbolic sense. It arises from the PlOcker coordinates expression and the way of differentiating determinants. Apply the Weingarten decomposition to differentials of normal vectors dnr, = r'_ L" gll rl +
no du',
where L. are coefficients of the `second fundamental forms of F2, gil are components of inverse metric tensor of P. Using the decomposition rule of a bivector into a sum of bivectors, we get
-L1g''[rln2]du'.
[dnln2] = Analogously we proceed with [n1 dn2j. Thus
dp = - (-L gi'[rl n2] + L2 gil [nl r,J) du'.
To Find the square of dp use the formula for the scalar product of bivectors:
([ab][cdj) = (ac)(bd) - (ad)(be). Note that by virtue of the orthogonality of it, and n2 the relation ([rln2J[nlrk]) = 0 holds. Now write
(Ljg1l[rln2]du,)' = L' Lrlq guIgq",du'duP([rln2][rmn2J)
= L1 L' g" gtm'(rlrn,)du'duP = In the same way 2
(Lfgu'[nIrl]du') = L LMg'ldu'duP.
Thus, we can represent the metric of the Grassmann image r2 as dtt' du'. dp2 = (L;,t Ljl + L2ik L2)gkl j/
THE (d OMLI ItY or S11INIANIFOLUS
2(H)
So. the abstract thing - the Grassmann image I'2 - gives quite real features. On 172 we are able to crawl, to measure distances. angles and so on. Also. we obtain some information on the external geometry of F2. We related the internal geometry of a
Grassmann image to the geometry of F2. We can express the components of clp2 = G ahi char. where G'a - _
(LE
LIra
A
) gAa 2 + L,A L'11 ,
with respect to a special system of coordinates in terms of parameters of the normal curvature indicatrix of F2. Choose uI. ar in F2 and normals in a special way which was described in Section I of Chapter 6. Then at some fixed point r E F2.
Ljj=n+a.
L{2=0.
L;, =n -ca.
L';,=d
L2jI =rii.
Substitution into G,, produces [(LiA)2+(L24)2] _ (c:+(a)2+112+h'.
GPI = A_I
Go =
-I
(L'14 LA +LTAL_A) = 2;3h.
A), =tat-u)-+r12+h-.
A-I
Write the metric of r2 at W (1 ):
dp' = [(u + ca)' + 32 + h' (du' )2+4dhdaa' clan- + (ct
- a)2 +J2 + h2 (daa-')2.
This form of metric has been stated in [7]. Consider the volume element of the Grassmann image dSt =
G12clra' cltr'.
Using the Cartan formula from Section I of Chapter 6, we have ' +' j2 + ca' + h- + ?nra)(ct
GI
((1' + +i=
-2
i-
+ cat + h2 - _ciu) - 432/'2
- ca- - h' )' +4((k21,2 + ri2ra2 )
ti'- +4(n2h' + area' ).
where K is the Gaussian curvature of F2. With respect to the chosen system of coordinates, the volume element of F2 at r,I is dS = dul daa2. So, we have the following generalization of Gauss's theorem: the ratio of the Grassmann image volume element to the volume element of the original is of the form
At dS
K2 +4(n2h' +;i:1(t
()4
GRASSMANN IMAGE OF A SUBMAN[FOLD
201
While this equality has been stated with respect to a special system of coordinates. due to the invariant meaning of both sides of (4). it stays valid with respect to an arbitrary system of coordinates.
From (4) it follows that " > SK I. If F2 is minimal then n = 3 = 0 and dp2 = (K I dS2. Therefore, in this case the Grassmann mapping T is conformal. 8.3 Curvature Tensor of Grassmann Manifold G2 4
Denote by p12. p1z. p1;. p''. p-;, p3 the Cartesian coordinates in E". Standard immersion of G24 into E' is defined with two equations:
F2
(1)
1,
F1
=P12P - -P13P-
+P 14r = 0.
(2)
Gradients of F1 and F2 determine two normals to the submanifold: I
grad Fl = {p12, p1' p14, p' P. p34} = p.
2 grad F, = {p4.
p_4,
P2:4 p14, -PI;, P' } = q.
As (pq) = 0, these normals are mutually orthogonal and unit because P` = q-' = 1. Equation (1) means that 62 4 is located in a five-dimensional sphere Ss: also q is the normal to G34 C S` Let f 1, t2, r , y4 be some curvilinear coordinates about a point in G2 .. A position
vector of the considered submanifold can be represented as a vector function p = p(p1.... , v ). Let drr= = a ! 40 dr i be a metric of G_ 4 induced from ambient ) space Eb with respect to these coordinates. Then a,, . We would like to -0. find the Riemannian tensor components R #,0. To do this. use the Gauss equations
which relate the Riemannian tensor to the second fundamental forms. Let S2,", i dig" dp;, o = 1.2 be the second fundamental forms of G.24 c E° with respect to
normals p and q. Setting a = I corresponds to p. o = 2 corresponds to q. Then sill,
a=6
kV I
P}
0P n { = f7f)r
11
q)
.
Since p and q are normals, these coefficients can be represented as
VO)
SZ,';i=-k q{J.
THE GEOMETRY OF SUBMANIFOLDS
202
In terms of differentials dp and dq, the second fundamental forms of G2,4 c E6 can be represented as
III = -(dp)2,
112
= -(dpdq)
(3)
Up to a sign the form !!I coincides with the submanifold metric. Consider the Gauss equation of immersion of G2,4 into E6:
L(na, a=1
?A - W., TO
a,,, a,m - a,,s ah +
(Lp Lq ) (Lp Lq
ar Or
( ap _q ( Op Oq ).
OYI a?
Or P) Way" Or
Let X = {X" } and Y = { Y" } be mutually orthogonal unit vectors in the tangent space of G2.4. Denote by a a 2-plane spanned on X and Y. Then the curvature of G2.4
with respect to this plane is x(a) = R"chn X° Y.X, Ye
= I + (VxpVxq)(VypDyq) - (OrpVyq)(Vypvxq), where Vx means the covariant derivative in E6 with respect to X. In the last term of the above we have (Vxp Vyq) = (Vyp Vxq). Since G2.4 is in S5 and p is normal to
this sphere, Vxp = -X, Vyp = - Y. Therefore, the curvature of G2,4 can be represented as
K(a) = I + (XVxq)(YVyq) - (XD),q)(YDxq)
(5)
Vector Vxp is unity, i.e. (Vxp)2 = 1. Taking into account that Vxq is formed with the same components as Vxp but taken in a different order, we have (Vxq)2 = 1. In the same way, (Vyq)2 = 1. Vector q is normal to G2.4. Hence (Vxpq) = 0. From this we find (pVxq) = 0. Also note that q is unity. Hence Vxq has decomposition via vectors of tangent space of G2.4. Let X, Y, Z and W be mutually orthogonal unit vectors from that tangent space. Consider decompositions
Vxq =a, X+a2Y+ a3Z+a4W, Vyq = b1X+b2Y+b3Z+b4W.
As Vxq and Vyq are unity, E; a = rib? = 1. As (XVyq) = (YVxq), a2 = b1. So K(a) = 1 + a1 b2 - a;.
Find k(a) variation boundaries. Consider in the plane a pair of vectors with components (a,,a2), (b1,b2). Their lengths are not greater than I. Their vector product modulus Iaib2 - ail is also not greater than 1. Therefore, 0 5 K(a) < 2. Show that
GRASSMANN IMAGE OF A SUBMANIFOLD
203
boundary values are attainable. Take the pair of three-dimensional vectors is = {p'} and v = {v'} of components =P12-p34,
µl =P12+p34, µ2 = p13 +p42,
VI
µ3 = p14 +p23,
V3 = p14 - p23.
V2 = p13 - p4-,
By virtue of (1) and (2), each of them is unity, i.e p22 = 1 and v2 = 1. We can rewrite the forms dp2 and (dpdq) in terms of these vectors: 2dp2 = dµ2 + dv2,
2(dpdq) = dµ2 - dv2.
The normal curvature k of G2,4 with respect to q is
k-_
(dpdq)
- dv2 - dµ2
(dp)2
dv2 + dµ'
From this it follows that the normal curvatures of hypersurface G2.4 C S5 vary in [-1,11 with attainable boundaries. Ask changes its sign, the hypersurface is a saddle. With respect to tangent direction dv = 0 we have k = - I ; with respect to tangent direction dp = 0 we have k = 1. The components of p E G2,4 can be represented via µ' and V. Namely p = (pl' p13 p14 p23 p24 p34)
=2(µl+v1,µ2+v-, µ' t +v -1, 'U3 -v3,
I) -µ +V-1 µ - v. 2
1
Thus, G2.4 is isometric to the direct product of two two-dimensional spheres Si X S
of the same radius I/f. Keeping in mind this correspondence, we shall represent the points of G2.4 as µ x v. Let F, = S1 x vo be a submanifold of p-endpoints while µ
moves along the sphere µ2 = 1 and v is fixed v = vo. Let F2 = µo x SZ be a submanifold corresponding to v-variation while µ is fixed: p = M. Denote by T, and T2 the tangent planes to the respective submanifolds. At every point p = PO x there exist both of these planes. Each vector of the T; plane is principal; also, the normal curvature with respect to directions from T, is - I while from T2 it is 1. Let X and Y be mutually orthogonal in T1. Then Vxq = X,
V}'q = Y.
From (5) we find that k(a) = 2. Note that T, and T2 are perpendicular to each other. Indeed, if X E T, and Y E T2 then X = (dµ 1, dµ2, dµ3, dµ3, -4112,4111),
Y = (dv,, dv2, dv3, -dv3, dv2, -dv, ).
Their scalar product (XY) = 0. Moreover
Vxq=X, Vy=-Y. Therefore, from (5) we see that for the X, Y-plane the curvature of the Grassmann manifold is k(a) = 0.
N4
THh UEOMC1 RY Of S1'HM \Ni1OLD.S
8.4 Curvature of G2 4 with Respect to the Tangent Planes of the Grassmann Image of a Surface
Let F2 be a regular surface in E4. r2 its Grassmann image and p a point in r2 In the
tangent space of G2; consider the plane a tangent to I'= at p. State the relation between the curvature K(cr) and the geometry of F'. Represent I'= in parametric form:
r =}'''(u.r), r+= 1.2.3.4. where it. r were taken out from F2 by Grassmann mapping. Then 11 are tangent to 1`2 but not orthogonal to each other. in general. Use the formula for the G24 curvature tensor in the form equivalent to (4) of the previous section:
+
Jp
W)
CeJt r?,
02P
p
)-(
allrJt,W
atd,,. carrW
"p q
l1r 1c)r'
The curvature A(rr) can be found by
I+ (p
nu W){p ,t
q}-(p
oil
W)`
(1)
Pj,P - l pupr)2.
Consider (dp q) on 1' Find the tangent vectors of t=. They are derivatives of the vector-function p = p(u. r): pff = -G R gA1 [r, F_] -
gc1 [f1
where u1 = it. tr' = 1.
To simplify notations, set L;1 = L' g"1. Bisectors [rr fir] are located in the tangent space of G2 To prove this. consider their scalar products with p = (41 and q = [ri r2] f f (here g is the determinant of the F2 metric tensor). Then
(p[r,£f(} _ ( in,)( ,) -
4.
(r1r,)(r2 ,) - (rlf r)(r_r??) _ 0.
Using Gauss and Weingarten decompositions with respect to F2. find the second derivatives of p: 01P 011,01i
1t
__
-t
1
-Ir11r',-.L. L2 1iti,r
+(_ 1)L'
!
011,
-Lrgr4 +L;U11!1 rI].
,I-1
[,r
f1
f
[E1r
GRASSMANN IMAGE OF A SUBMANIFOLD
205
On the right-hand side of the equality above, the first two terms are linear combinations of [{i r.1. So they are tangent to G2,4. The third is directed along p. The fourth is directed along q because of [r1r2] = vgq. Hence (d 2P q) _ -
ij
f L("11 L2121 dui duj
= 2 {(Li2Li1
vg
- L12Li,)(dul )2 + (Lil
t - Li1L 2) du' due
+ (142"21 - L2'1L42)(du2)2}.
With respect to the special system of coordinates in F2 mentioned above (so that (2) from Section 1 Chapter 6 holds), we obtain
(d2Pq) = -2{-b(a + a)(du' )2 - 2af du1 du2 + b(a - a)(du2)2}.
(2)
Observe that the determinant of the F2 metric tensor matrix (as stated in Section 2) is
P.Pv
-
K2 +4(a2b2 +/32a2)
(3)
du dv + (p,,,q) dv2 with Comparing the expression of (d'-p q) = (p,,,,q)du2 + (2), we find the scalar products (p,,,,,,q) with respect to the chosen system of coordinates:
2af,
(p,, q) = 2b(a + a),
-2b(a - a).
Therefore 4(a2b2 - (b2a2 +a2132))
(4)
Substituting (3) and (4) into (1), we get k(or). The result just obtained is presented as a theorem. Theorem The curvature K(c) of the Grassmann manifold G2,4 with respect to the plane a tangent to the Grassmann image r2 of a surface F2 C E4 has the form:
K(°)
_
K2 + 4a'-b2 K22 + 4(a2#2 + b2a2)
This theorem was proved in [7).
Find the characteristic equation on the principal normal curvature of 1'2 with respect to the normal q. Remember that for this case the principal curvatures are where dp varies in the plane tangent to extremal values of ratio Let p correspond to x. The metric of f'2 with respect to the chosen coordinates has the following form at p: r2.
dp2 = [(a + a)2 + Q2 + b2](dul )2 + 4Qbdul du2 + [(a - a)2 +Q2 + b2](du2)2.
THE GEOMFTRI (N SIIRNIANII.01 DS
_'tx+
Then, the characteristic equation can be written as
[(ri' +d=-a=-h2)'+4(rt`h-+a'rj')] +4). h[n2 +12 -cc2 -h=] +4(h=u'- - h=ri- - a=,3=) = 0. Using the Carton formula on the Gaussian curvature K of T' and the formula on K just obtained we have
!i=+4(nW +,12a2) Then, the characteristic equation can be written as A=+2A
_ct h l;
K2+4(n h=+d (1'-)
+A-I =0.
Let A, be the principal normal curvatures of 1''- with respect to y. Denote by R, and A the symmetric functions of A,. Then
_ Ai + A, 2
-2ahK
R, = A1A: = K - 1.
R' +4(ci-h' + 132cr-) '
(5)
If the principal normal curvatures A, are of the same sign. i.e. the form (d=pq) = -(dpdq) is of fixed sign. then k, = k - I > 0 If A, are of different signs. i.e. the form (dpq) is of alternating signs. then k < 1. By means of (5) we find
h
E= +4a:h: -2obK
Therefore 2ah K
f /it ' - 4N,.
2A. Thus, the Grassmann image r2 of F2 C E4 determines the ratio of Gauss torsion to Gauss curvature; 8.5 An Estimate for the Area of Closed Surfaces in E4 In this section we find an upper bound for the area of it closed surface in terms of the surface enclosing ball radius and the area of the Grassmann image. The following holds. Theorem Suppose that a c loseel on nrahle sw/ac , F2 C E; of class Ca u *h Gaussian curvature K of constant signs. Eider characteristic x and area S is con-
tubred in a ball of radius R. Let the Grasstrrann image area be equal to SZ. Then
S I it follows that F2 is regular if x2u. + x22,,, 54 0. In [10] Polozij stated the following: Theorem In order that functions u(x,y) and v(x,y), which are continuously differentiable solutions of au,,, + bu, - v,. = 0,
d satisfying the Holder condition at zo = xo + iyo, map a one-sheeted neighborhood of zo in the plane z = x + iy onto a one-sheeted neighborhood of vo = f (zo) in the plane w = u + iv, it is necessary and sufficient that the Jacobian uCv,. - v,.u,,. 54 0 at zo.
Zeros of the Jacobian urv,. - v_ru, coincide with zeros of the function u_t + iu,., i.e. with zeros of x2u, + ix2u, in our case. We can also apply, the Levi theorem from [11].
Since a quasi-conformal mapping from the Lavrent'ev theorem is a homeomorphism, by the Polozij theorem x2i, + ix2u, # 0. Therefore, the constructed surface F2 is regular. On 1'22 it is sufficient to require the regularity of C2 Then F2 will be regular of class C2"", a < 1. Theorem 2 is proved. Note that for the hyperbolic case some theorems on the reconstruction of a surface by its Grassmann image have been proved by Kizbickenov [12). Another approach to the problems above has been proposed by Weiner [ 13], [14].
8.9 On Local Projections of Two-Dimensional Surfaces in E4
The local behavior of F2 C E4 essentially depends on the type of Grassmann image.
Consider the projection of F2 c E4 into three-dimensional space E3(r) passing
1H1: C,FA)MLIR\ OF SUBMANII OLDS
2211
through the point x, the normal plane N, and the tangent vector rat x. Let el and e2 be an orthonormal basis in the tangent plane such that e, is a tangent to the u'-curve. Coordinates are consistent with the normal curvature ellipse. Let c'u e4 be normals to
F' parallel to the principal axes. Set r = cosdcl +sinOe2; x' are Cartesian coordinates with respect to the basis e1,--..c4 with the origin Y. Then up to Intinitesimals of 3-d order with respect to 1' (i = 1, 2). a position vector of F2 projection into E'(r) has the form f = r(cos Ox' - sin Ox) -F ea [(ci + (1)(11)2 + (ei -11)(.x)2 + ea
1d(.x' )2 + 2bv' x 2 +3(x2)2 ]
Denote the projection above by P. We suppose, also, that n,13. it. h are not equal to zero simultaneously. Depending on r, the point x in F2 may be either regular or singular. We can find the direction r with respect to which F-(-r) has no singularity at x from the following equation
cth+(1.3sin 20-nhcos20=0.
(1)
Let 0, he the roots. If all coefficients of the equation above are zero then F2(r) is a plane for all T. 2 Further on. we exclude that possibility. Equation (1) has no roots if (u32) + (ah) - (ah) -< 0. i.e. if the Grassmann image is elliptic. In addition. for any r the surface F2 is singular at t (see Figure 30). The structure of F2 (r) is similar to a two-sheeted Riemannian surface. The surface is tangent to z at x. The surface section with a plane through r and it e N, produces a pair of parabolas tangent to r and with branches in distinct half-planes. In some of the sections one of the parabolas degenerates into a half-line which is a line of self-intersections. If (ah) 2 - (ad)2 - (nh) 2 < 0, i.e. if K < I and 0 d, then
FICiLIRE 30
GRASSMANN IMAGE OF A SUBMANIFOLD
221
P2(r) is also singular at v. But in contrast to the previous case. P2(r) is included in some two-sheeted angle (see Figure 31). The section of P with the plane through r
and it E N, is either one point x or the pair of parabolas tangent to r and with branches in the same half-plane. In a particular case one of the parabolas degenerates into the half-line. If 0 = Q, then the projection is a regular parabolic cylinder. In the parabolic case, i.e. when K = 1. generally there exists only one subspace
Ea(rl) such that P has no singularity at x. If r 0 rI then P has the structure as in Figure 32. The section of P with the plane through r and n E N, consists of the parabola and the half-line. For some sections this parabola degenerates into the halfline which is the line of surface self-intersection.
FIGURE 31
FIGURE 32
222
THE GEOMETRY OF SUBMANIFOLDS
8.10 Reconstruction of Two-Dimensional Surfaces in n-Dimensional Euclidean Space by Grassmann Image
Consider the two-dimensional surface F2 in n-dimensional Euclidean space E" for n > 5. Its Grassmann image 1'2 is located in a Grassmann manifold Gn_2," denoted further on by G. As dim G > n for n > 4, the arbitrariness in determination of the two-dimensional surface in G is greater than in E". Hence, an arbitrary surface in G
can not be a Grassmann image of the surface in E". It is possible to state the necessary and sufficient condition for r2 to be the Grassmann image (see [17]). In this section we state the theorem on uniqueness in the general n-dimensional case and then consider in more detail the case of n = 5, where we state the necessary and sufficient condition mentioned above.
Let Ei-4 be an (n - 4)-dimensional Euclidean space passing through 0. Set E= En-4. Denote by G2,4(E) a manifold of all (n - 4)-dimensional subspaces E`2 in E" passing through E. Then G2,4(E) is a Grassmann manifold G2,4 imbedded in G as a submanifold. We say that I'2 is tangent to G2.4(E) if the tangent plane of r2 at x E r2 is in tangent space of G2,4(E) at x. We shall call 1'2 the surface of a general
kind if both F2 is not tangent to G2.4(E) and any tangent to r2 vector is not simultaneously tangent to the pair of G2,4(E) at any point x E 1'2. The following theorem of uniqueness holds:
Theorem (Aminov) A two-dimensional surface in n-dimensional Euclidean space E". n > 5. is determined by its Grassmann image of a general kind uniquely up to homothety and parallel transport.
The proof and also the necessary and sufficient condition mentioned above we present in the case of n = 5. Use a representation of points in the Grassmann image in the form of matrices. Select in E5 an orthonormal basis et,... , e5. Let Eo be a three-dimensional subspace passing through e3, e4, es. Take the normals {,,, v = 3, 4, 5 of F2 in such a way that each & would be of the form 2
= e + E a ,e;,
v=3,4,5.
(1)
The matrix Z = Ila,II uniquely determines E3 - the normal space of F2 - if it is one-to-one projectable onto Ea and vice versa. If UI, u2 are curvilinear coordinates in a regular surface I'22 c G then the components of Z are regular functions of u1, u2. So, Z = Z(ui, U2). Denote by a' the columns of Z. The matrix Z consists of two columns and two rows. By [ ] we denote a determinant. Introduce the quantities f' _ [d",au au,],
i jAk;
i, k = 1, 2.
Set A = f11 f 2 - f2 fi-. From the lemma (see below) we conclude that 0 a general kind.
0 for r,2 of
GRASSMANN IMAGE OF A SUBMANIFOLD
223
Theorem 2 (Aminov) In order that C3-regular surface r2 C G3,5 of general kind is a Grassmann image of C3-regular surface F2 C E5 it is necessary and sufficient that the following equation is satisfied.
a
au,
I
f;u: -f'2.,
A_ a
flu, -f22ui
f2
1
IPq -A.,
f;
art2 ° f2, -f2u, f,2
-
0
(2)
Let r2 be a Grassmann image of some surface F2 of the position vector r = {x1}. Let E3 be a normal space of F2. Suppose that &, = {t j,} are the normals and hence 5
k=1,2.
Exj,,,V,,=0, v=3,4,5;
(3)
j=I
By virtue of decompositions (1), equation (3) can be represented as 2
xy,,, = -
x;,,, a,,.
(4)
;= I
From the equality conditions on the mixed derivatives of x we get three equations on x,, x2: 2
E(x;,,, a',,,,. - x;,,_ a,,,,.) = 0,
v=3,4,5.
(5)
;= I
There exists a function A(u,, u2) such that
x;,,, = of ,
where f =
k, i = 1, 2
(6)
i 0 k, k = 1,2. From the equality conditions on the mixed
derivatives of x, and x2, we obtain a system of two linear equations with respect to the function 'I' = In A: 'P u,f; -'I',,, Al' =fiu: -fz,,, I (7)
wu, f - pu, f12
=fi2-f;,,.
The determinant of the system above is A. By the lemma below, A 4 0. So, we have
4"'. _ ,
1
0
IPU2
-fi,, A
f fl',,, -fiu,
o j2 f,us -fZ, I
(8) .
f,u_ -fiu, fl'- I' The equality conditions on mixed products of produces equation (2). In a sequence of systems (4), (6), (8) each subsequent system is a compatibility condition for the previous one, while (2) is a compatibility condition for the latter system. If (2) is satisfied then 'I' is determined up to an additive constant. Thus, the function A and, as a consequence, the right-hand sides of (6) will be determined up to the constant multiplier. The variation of this constant multiplier means the transformation of F2 into a homothetic one. If the multiplier is fixed then from (6), (4) the function x1 is determined uniquely up to additive constants. The choice of these constants means
224
THE GEOMETRY OF SUBMANIFOLDS
the choice of parallel shift of F2. The surface F2 of position vector r = {x1} is regular because its metric tensor determinant
g = .402 (1 + (a' )2 + (a2)2 + I[aIa2)12)
(9)
is not equal to zero.
Lemma Suppose that r2 is not tangent to G2,4(E). Then 0 = 0 if and only if there exists tangent to r2 a vector such that it is simultaneously tangent to the pair of G2.4(E) For the case of n = 5 the space E is of dimension one. Denote by e the unit vector in E. Suppose that some v = cos a Z,,, + sin a Z,, which is tangent to r2, is tangent to the pair of G2,4(e). Represent the equation on G2,4(e) in terms of the matrix Z. Let e be of the form S
5
j=1
1-3
e = Eb 1e; = > c'
,.,
where {,,, v = 3,4, 5 determine a three-dimensional space E3 which contains e. Using
(1) we get b° = c", b' = E5_3a1c°, v = 3,4,5; i = 1,2. Hence, each of the submanifolds G2.4(e) is determined by the equation µZ = n, where µ = (b3, b4, b5), 17 =(b',b2) are constant vectors. Therefore, if tv,i are curvilinear coordinates in G2,4(e) then the equations UZH., = 0, $ = I,... , 4 are satisfied. Thus, for the vector
W there exists the vector p such that p W = 0. This means that cos a (pa,,. ) + sin a (pa,,) = 0, where () means the scalar product in three-dimensional space. Non-zero /s is orthogonal to the pair of vectors: Ti = cos a ail + sin a au:, i = 1, 2. Since W is tangent to the pair of G2,4(e) then we have a pair of vectors of u type orthogonal to the pair (r', r2). Hence r1 and r2 are parallel to each other. So, r' is in the plane of vectors a;,,, au,, k = i. Hence, the mixed products are
[r'a, a;:]=0, k,i=1,2;k
i.
(10)
From this it follows that rows of determinant
6=
an,
[a
a2. 2
, , an.
2]
'"a" l a"
,
i
au, au, au,
r2
I,]l
a"'a' a'
are proportional, that is 0 = 0. Conversely, suppose that 0 = 0. If for some fixed i the vectors a., and a,,_ are parallel to each other then there exists a number a such that r' = 0. Take as µ any vector in the plane perpendicular to rk, k 0 i. Then the vector W tangent to I'2 satisfies an equation pW = 0, i.e. it is tangent to the pair of G2.4(e) simultaneously. For this case, the lemma is proved. So, suppose that each pair of a',, a, consists of two non-parallel vectors. Then they determine a plane. If four of the vectors a;,, i, k = 1, 2 are in the same plane then there exists the vector p orthogonal to that plane such that (pa,,t) = 0, that is p 0, k = 1, 2. This means that the tangent plane of r2 is in the tangent space of G2.4(e) which contradicts the lemma hypothesis.
GRASSMANN IMAGE OF A SUBMANIFOLD
225
Thus, we suppose that planes of pairs a,,. a,, and a are distinct. From L = 0 it follows that there exists a number a such that the mixed products (10) are zeroes. As the planes of pairs a,F, are distinct then rl and r2 are collinear. There exists a twodimensional plane of vectors p perpendicular to r', i.e. pW = 0. Therefore, W is tangent to the pair of G2,4(e) simultaneously. The lemma is proved. We note that Gorkavy [34]-[36] has proved very interesting theorems on the reconstruction of multidimensional submanifolds by their Grassmann image. 8.11 On Hyperplanar Sections of Two-Dimensional Surfaces In E4 With Elliptic Grassmann Image
Borisenko remarked that a two-dimensional surface F2 C E4 has an elliptic Grassmann image if and only if at each point P E F2 the normal curvature ellipse is not degenerated and P is located in an ellipse bounded domain. This means that for any normal n there exists a system of coordinates x, y such that the second fundamental form with respect to that normal has the form a2 dx 22 - b2 dy2. where a and b are not
equal to zero. As the normal curvature ellipse of the surface with an elliptic Grassmann image is not degenerated at each point, any domain on the surface can not be located in a hyperplane V. Moreover, the following is true. Theorem Let F22 be a C2-regular surface in E4 tt'ith an elliptic Grassmann image. Then the boundary r of any connected compact domain D C F2 can not be located in E3.
Suppose that r is in some hyperplane E3. Introduce Cartesian coordinates X1, x2, x3, x4 in E4 in such a way that the xi, x,, x3 axes are located in E3. Suppose that x4 of P E F2 is some function u(P). Then u = 0 on F. Let us assume that u 0 0. As D is a compact domain and u is continuous, then there exists a point Po in D of maximal or minimal u-value. Introduce about Po in F2 a system of coordinates x,y. Then at P0 the following conditions must be satisfied: ux = 0, u,. = 0, u,,u,.,. > 0. The vector v = (0, 0, 1, 0) is the normal one to F2 at P0 and the second fundamental form with respect to v is 11(v) = u,, dx2 + 2uy dxdy +
dye.
Since F2 has an elliptic Grassmann image, then by a coordinate change 11(v) can be reduced to the form u_Tr dx2 + u,.y dy2, where u,r_ruy1. < 0. This contradicts the extremal property of Po: u,,u,., > 0. Hence, r can not be located in E3. From this theorem it follows that any domain in F2 of a boundary formed by an intersection of F2 and a hyperplane is necessarily non-compact. 8.12 Representation of Points of a General Grassaraan Manifold in Terms of Plticker Coordinates and Matrices
In the previous section the points in the Grassmann manifold G2,4 were represented by Pliicker coordinates. Evidently, the Pucker coordinates can be introduced in any
_.h
THE (DEOME rttl Of SUBUMAMFOLD5
Grassmann manifold GA,, IA. Also. we shall consider the representation of points in terms of matrix elements.
Let NA be the k-dimensional subspace in E"" passing through 0. We shall consider NA taking into account the orientation. Choose in NA an orthonormal frame C),..-CA. With respect to the ambient space basis a .....cl,,,A. each of , obtains some components 4`,. The Plucker coordinates p'f ' of Ni" are the set of values f,
=
?A
p11
Si
..
S)
..
.
f,
. !q
In the other words, the Plucker coordinates of NA are the Plucker coordinates of the
ordinary multivcctor p = .......4J. Order them lexicographically and select the components with distinct strings of i)....,iA. Consider the Euclidean space E" of dimension v = C; (the binomial coefficient). Put the initial point of the vector with components (p) A.. ) at the origin of E''. Then each point p E GA ,, , A corresponds to the unique point in E''. In this standard way the Grassmann manifold GA ,,,A can be immersed into E''. Plucker coordinates of an ordinary hivector are subject to some relations. First 61)=k2l .. The equality above can be represented as (p'
a)=
= I.
=1.
(1)
4
11
Hence, a Grassmann manifold GA,, A is immersed into a unit sphere S" ). Also, we have the set of relations
p" ",p,f 14I=0.
(2)
where [ ] means the cyclic permutation of superscripts, The latter equality can be represented as A
E(-1)'p', f,
I hpM, ,,
1
),
1
1, =0.
(3)
1=1)
where we set ix =J(). The term with a = 0 is a product p" 'kph '/. Denote by .41' a cofactor to e,,4 from the last column of the determinant
SAl
...
SAC
Then for each Plucker coordinate we have Pee
It 11 = At'),
GRASSMANN IMAGE OF A SVBMANIFOLI
229
So. the left-hand side of (3) obtains the form tr++4
r,
.. en
'
cr+11+
...
u
... .. ... ... A ...
4h
L
t4
4
E1
X41+
...
S4111
Each determinant in the latter expression is zero because it contains two identical rows. Thus, (3) is proved. In terms of the Phkker coordinates no GA +A it is possible to introduce the metric clrr=
induced hr the immersion into E":
d4r' = > (dp'+ a )14
Now we describe another way to introduce coordinates in a Grassmann manifold. They will be independent of each other. i.e. the proper coordinates. In this way we shall find a dimension of the Grassmann manifold. Let N, be some k-dimensional fixed subspace in E"lA passing through 0. Define the neighborhood of N,, as a set of all k-dimensional subspaces NA passing through 0 which contain no vectors except zero orthogonal to N,. Let NA be an arbitrary k-dimensional subspace from the neighborhood above. Show that for any e E N, there exists only one f E NA such that a projection off into N coincides with e. Indeed, project all of the vectors from NA into N,4,. If t and 11
have the same image then 4 - q is orthogonal to N, because of the linearity of projection. But NA is from the neighborhood. hence E = n1.
If the projection of N, does not cover N,, then, by virtue of the dimension equality, there exist two vectors in NA with the same projection, which is impossible. Let e,, .... en, k be the orthonormal frame in E"+A, where eI, ... , eA form the basis in N;. In NA take a frame fi, ....ft (non-orthonormal, in general) such that e, is the
projection off into N,4,. Consider the decomposition off with respect to basis of E"+A: +++A
f,=e,+>2a,e,, i=1,...,k. r-A+i
Thus, to each k-dimensional plane NA from the chosen neighborhood of N, there corresponds the matrix Z = IIaaII of k rows and n columns. Conversely, to each
matrix Ila,II there corresponds some frame f,,...,ft and, as a consequence. a kdimensional oriented subspace Nk spanned on that frame, i.e. the point in GA n+A. The correspondence above between points of a Grassmann manifold and matrices defines a homeomorphic mapping of the NQ neighborhood onto a neighborhood of origin in kit-dimensional space EAn, where coordinates are generated by elements of matrices 11u,11. In contrast to PlUcker coordinates, a, are independent. Thus, the dimension of a Grassmann manifold GA n+A is equal to irk.
THE t;LOMFTRt I)} S1 K MANIM)I CIS
22$
8.13 Metric of a Grassmann Image of Arbitrary Submanifold Let F" be a regular submanifold in E" A parametrized by u' , .... ii". Find the metric of the Grassmann image r' of the submanifold, assuming that l"" is parametrized by a' , ... , u" by a Grassmann mapping. Apply the same method as in Section 2.
Consider a differential of an ordinary multivector. Using the Weingarten de-
l
composition. we have dp=(I
A
...i] _
u...
M-
[=;I
ctic'
I ... r,
j
t ]cliff'.
where r, = Or/au'. The square of dp produces the metric of the Grassmann image(dp), -
--r'
r, ...E4]
L;Lm,g"g"
14I
To find the scalar product of ordinary multivectors in the formula above we use (5) Section 1 Chapter 1. where this product is expressed in terms of scalar products of vector components of a multivector, The matrix of these scalar products for rt # ri consists of units in the main diagonal. excluding two places with zeros. Besides, the (ri, '3) position take (r,r,) All the other elements are zero. So, 0
0
1
.
0
0
d 0 0 0
s
...
1
0
...
0
0
0
-
0
...
0
1r,r,)
...
0
...
0
0
...
0
0
.
...
0
0
I
If a 0;1 then the fi-th row and rl-th column consists of zeros. In this case the determinant equals zero. If ct = c3 then the matrices have diagonal form. where (r,r, )
takes the (n, a) position. Thus,
(fr' ... P,.
=(r,r,)=g,%
Tlwre/ore, the metric of the Grassmanni image t" of a suhinani/old has the form R
(clp}= -
LU L`,s g" rlur dit'. t,_1
GRASSMANN IMAGE OF A SUBMANIFOLD
229
Denote by G1; the components of the metric tensor of a Grassmann image. Then k
G11=
Fe, L g.t
The expression above for (dp)2 has been found in [18], [32], [33].
8.14 Angles Between Planes
Let us be given a pair of k-dimensional planes T and E passing through the origin 0. Find the angles that they make with each other. Let u be a non-zero vector in T and v be its orthogonal projection into E. The stationary values of angle 9 between u and v while u rotates in T we shall call the angles between T and E. The direction u1 which corresponds to the stationary value of 9 we shall call the angle direction of T with respect to E. We want to show that if v1 is a projection of ui into E and vi 0 0 then vi is the angle direction of E with respect to T. Let ft,. .. , fk be some basis in T while el, ... , ek is some basis in E. Represent an arbitrary unit vector u E T as k
where E cost a1 = 1.
u = E cos a1 f ,
i=I
i=I
Represent the projection v of u into E as
v = E(ue;)e;. j=I
If 9 is the angle between u and v then L
I.
Cos20 = E(ue;)2 =
(cos ai (f e;)J 2.
(1)
!=1
1=I
We can consider the expression on the right as a quadratic form with respect to cos ai. The matrix of this form consists of k
au = E(.fe1)(fiei)
(2)
1=1
Since the matrix is symmetric, there exist k stationary values of cos2 u and k principal directions ul, ... , uk, which we called the angle directions of T with respect to E. Let
us set f = ui. Then the quadratic form matrix (2) is diagonal, that is k
Df e,)(f e,) = 0, /=I
for
i `i.
(3)
THE GEOMETRY OF SUBMANIFOLDS
230
Let vi be the projection of u; into E. Then k
vi = E(Iel)ei i=I
For i 11-j the vectors vi and vJ are mutually orthogonal. Indeed, by virtue of (3), we have
(vivi) = E(.fei)(fiei) = 0. t-1
Direct et along vi. Then 0 for i 1. Find the angle directions of E with respect to T. Let e be the unit vector in E k
e = E cos Qi ei. i=
Denote by V an angle which e makes with T. Then k
rr
k
k
cos f3 (ef,)=
cos'-
i=1 Li_I
I
cos'- (3J (ei.f')'.
i--I
The expression on the right is a quadratic form with respect to cos(i; moreover, its matrix already has diagonal form. Hence, cos 2 W attains stationary values for e = ei. So, vi are the angle directions of E with respect to T. The plane which passes through ui and vi we shall call the angle plane. It is well defined if vi # 0 and ui 0 vi. Denote by Bi an angle between ui and vi. We shall call it the angle between T and E. So, we can formulate the Wong theorem from [23]. Two k-dimensional planes T and E in E" I k make k angles among which no more than r = min(n, k) can be non-zero.
8.15 Geodesic Unes In a Grassmann Manifold
Let T and E be k-dimensional planes in El" k passing through the origin. Consider them as points p and q in a Grassmann manifold. Find a geodesic passing through these points. Let p and q be represented in terms of ordinary multivectors: p = Jul ... uk],
q = [vI ... vk],
where u, are the angle directions of T with respect to E, vi are the projections of it, into E. Let 0, .... OA be the angles between T and E. Consider a curve -y of parameter t E [0, 1) in a Grassmann manifold p(t) = [al (t) ... ak(1)
where each of ai(t) is a vector function of unit length. Moreover, if ui and vi are distinct then ai(t) is located in the ui, vi-plane, otherwise ai(t) = ui. Let us require that
GRASSMANN IMAGE OF A SUBMANIFOLD
231
a,(0) = it, a,(l) = vv, and ('h) = c-, i.e. the speed of rotation of any a,(t) would be constant. It is evident that either c, = 8, or c, = 9, + 2,rk where k, is integer. Prove that p(t) is a position vector of a geodesic in a Grossmann manifold. We have dt = dp
al ... A ,_I
da, aA . J
dt
(1)
Taking into account the formula for the scalar product of multivectors, we get [171
...
[a[
dt7... aA]
...
+ ... aA
= 0,
for
i
0j.
because the components of the first multivector are orthogonal to a,. If i = j then the corresponding scalar product is {%}-= O. Thus
(5,)2 0
dt
Since I`C'I is constant, the parameter i differs from the arc-length parameter of I by a
+92t. constant multiplier. Namely, if a is the arc length of ry then e = 82 + To show that ry is a geodesic on a Grassmann manifold we verify that the second derivative of p(t) is normal. Differentiating both sides of (1). we get ci-p -it-2
f
l
d -a,
E Lat... 76-2
a*J +
La1
dct,
du,
[it -
Tit
... aLl
(3)
As each of the a,(t) endpoints passes a circle, moreover with constant speed, then d2a,/dt'- = -8, a,. Hence, the first sum on the right-hand side of (3) is -(0? + + 8k')p. Since a Grossmann manifold is located in the unit sphere of center in the origin, then p is one of its normals. Show that each of the muftivectors ..ad from the second sum is orthogonal to the tangent plane of a Grossmann manifold. Every tangent vector of a Grossmann manifold at p(t) can be represented as [0,
b
by=Eful ...ba,,...ctAJ, n=1
where b means differentiation in some direction. For i
(fa1...ba,,...aA] [a
n\, i 0 j we have
da,. tea+..aA])
=0.
because a, from the first multivector is orthogonal to each of the vectors from the second. Note that (a, w) = 0 because a,(t) is in the plane orthogonal to a, for all t. If a = i then a, from the first is orthogonal to each of the vectors from the second.
THE GEOMETRY OF SUBMANIFOLDS
232
So, the second sum on the right-hand side of (3) is perpendicular to the tangent space of a Grassmann manifold. It follows, then, that y is geodesic. Using (2), we can find the distance between planes T and E as an integral of dQ over the shortest geodesic y: I
0
Thus, the square of distance between two planes is equal to the sum of the squares of angles between them.
8.16 Geodesics in a Grassmann Manifold C2,4
Let p be a position vector of submanifold GIA C E6. Normals of this submanifold are mutually orthogonal vectors p and q, where q is a bivector complementary top. If y is a geodesic in G2.4 of a natural parameter s then
p=ap+bq, where a and b are some unknown coefficients. Both sides of the equation above multiply by p. As p and q are mutually orthogonal, we have (d2p
p
','2)
_ d
dp dp ds(pds)-(ds) =-I =a. ''
So,
dip = -p+bq.
(1)
From this it follows that the curvature of -y in E6 is equal to 073Y and b is the normal curvature of G2.4 C SS in the -y-direction. Expand (1) into components and use the expression of components of q in terms of the components of p: dzdpr2l2
u 2p 13
= _p12 + bp-A = d
= _P 13 + b pat =
ds2
d 2ga2
_ _q3a + bp3s
=
-q42 + b pat
ds2
So, the following holds:
d2q=-q+bp. ds2
(2)
GRASSMANN IMAGE OF A SUBMANIFOLD
233
Multiply both sides of (1) by q, taking into account that (q I) = 0:
(q
ds
(q
ds)
- \ds dc)
- (ds ds)
(3)
.
Multiply both sides of (1) by dq/ds, (2) by dp/ds and then add. We obtain d
(ds ds2) + (ds
d sq
)
ds
( d s 1) = 0
because p and q are normals to G2.4. Hence, b = const. So, we come to the conclusion: a normal curvature of G2,4 with respect to q in a geodesic direction stays constant. Find a position vector of a geodesic in explicit form. Set b = k. Then we have
d,pds2 _q
d
-p+kq,
= -q + kp.
Adding and subtracting these equations, we obtain
d2(p q) _ -(I -k)(p+q), d2(p q) dy2
-(1 +k)(p-q)
Since k is constant, the solutions of the latter equations are
p+q=Acos 1-ks+B sin 1-k s,
(4)
p - q=Ccos l +ks+D sin l +ks,
where A, ... , D are constant vectors in six-dimensional space such that A I B, C -L D, JAI _ ... = IDI = vf2-. Also, as all components of q can be expressed in terms of components of p, the components of A, ... , D are not arbitrary. Namely, the three
last components of A and B replicate the first three. For instance, the first and last components of p + q are p12 +p34. Therefore, AI = A6, BI = B6 and so on. The vectors p ± q are of constant length vll While s varies, both of their endpoints trace a circle. Adding, we find p V-1
ks.
Since p2 = q2, then setting s = 0 in above, we get A 1 C. In which case is the geodesic closed? Suppose that there exists a number I such that
1 -kI= 2irm,
1 +kI= 2irn,
where m and n are integers. Then
k= n2-m2
n222+m22*
THE GEOMETRY OF SUBMANIFOLDS
234
From this we find
m2
1-k
i.e. I - k/ I + k is a square of a rational number. Note that IkI < 1 always. The value of k depends on the choice of starting direction of a geodesic. If k is not of the form above then the geodesic is not closed. Project the geodesic -y into spheres Si and S, whose direct product generates the Grassmann manifold G2.4. Denote by y; the results of projections. Consider the three-dimensional position vector p of sphere S? (see Section 3): p = (P12
+P34,P13 +p42,p14 +P13).
Its components coincide with the first three components of p + q. Denote by j and h the three-dimensional vectors consisting of the first three components of A and B respectively. Then, using the representation (4) of p + q, we can find p at points of yl as
µ(s)=AcosvTs+Bsin VI -ks. In an analogous way we find a position vector v(s) of y2
v(.r)=Ccos 1+ks+Dsin l+ks. As A and h are unit mutually orthogonal, the endpoint of p(s) traces a great circle in S2. The same conclusion is valid with respect to y2. So, the projections of geodesic in G2.4 into SZ are the great circles. The points in these circles are related to each other with a parameter s. Let t; be the arc-length parameter of y;. Then we have
dµ=(-A sin 1-ks+Bcos 1-ks) 71-1
I
-kds. dtl
I --k s. Analogously, t2 = 1 -+k s. Hence, the arc-lengths of y; are related to each other linearly: tl = V -R 12 . So, we can give another way to describe a geodesic in G2.4. Take great circles y, in S;2 and set point-to-point correspondence between them such that the ratio of lengths of corresponding arcs is constant. The pairs of corresponding points of y; represent a point in G2,4. Therefore, 11 =
8.17 On the Canonical Form of a Matrix Let Z be a matrix of k columns and n rows. Consider the transformation Z generated by an orthogonal matrix S of order k and an orthogonal matrix Q of order n of the following kind:
2 = QZS.
GRASSMANN IMAGE OF A SUBMANIFOLD
235
Show that by a specific choice of S and Q the matrix Z can be reduced to the simplest form (we shall call it the canonical form) such that 2, = 0 for i 9Q. Let Eo and Eo be mutually perpendicular subspaces in E"+k passing through the fixed point O. Let a, , ... , an and a,,+ I , ... , an+k be orthonormal bases in Ep and Eo
respectively. Then it is possible to map the n-dimensional subspace E" in E"+k to each matrix Z and vice versa (see Section 12). Let e,, . . . , e,, be the angular directions of Eo with respect to E". Let f,, ... f" be the vectors in E" such that a projection of j into Eon coincides with e;. From what was proved above, e; form orthogonal basis in Eo andf form the orthogonal basis in E".
Let f,, ... , fr, r < n be a maximal subset of {f,} with the property of not being located in Eon. Through each pair (e;, ,) i < r draw a two-dimensional plane E,. From what was proved, these planes are mutually orthogonal. In each plane E,2 i = 1, . . . , r orthogonal to ei. We obtain the system of r orselect the vectors thonormal vectors en+i, . , e,+, E E01. If r < k then enlarge this system up to a basis
en+1, ,en+k in E. Write a decompositions of fit. ..,fn as ei+
z,en+I,
e2+
-
-+
Z2en+2,
+
...
er+
Zr
r en+r,
f,, = en.
Matrix Z with respect to this choice of bases has the canonical form
Z=
0
10o
f2
.........
0
...
...
zr
0
...
0
0
0 0
0
We can take the matrix of basis change from a,, ... , a" to et,. .. , en as Q, and the matrix of basis change from a,,1,... , an+k to en+,, ... , en+k as S. 8.18 The Equations of Isometric Immersion of a Grassmann Manifold and Second Fundamental Forms
Our next objective is to find the curvature tensor of a Grassmann manifold. We present two ways to obtain it. The first (which is rather long) exploits equations already found of isometric immersion of a Grassmann manifold into Euclidean space. Therefore, the expression is complete and conceptually simple. In addition, we obtain some lemmas of independent interest. For instance, we state a minimal system
of equations for the standard immersion of a Grassmann manifold into Euclidean
THE (OMETR1 UI' Sti3Ai.tNlt(TLDS
23h
space. The second method which is short and elegant, we shall expound in Section 22. It belongs to Leichtweiss and is based on Cartan's method of external forms. As we already know, the Plucker coordinates p'" in GA,, A satisfy equations 11
pl,
E (p"
;l: pi, III = p.
P
) 2= 1.
(1)
11
But not all of them are independent of another. We are going to find a minimal subsystem of the above equations which is sufficient to represent the immersion. The manifold G4,, , A is imbedded into the Euclidean space E t of dimension A'= C,4, A. Since the dimension of GA ,,,A is equal to kn, to represent the immersion it
is necessary and sufficient to make a system of
lit =Ciit -kit
(2)
independent equations. Consider a point Pi, of Pliicker coordinate p" A = 1. while other coordinates are zero. Represent the immersion with the following system of equations The first of them is (3)
(ii + 1... . n + k). Consider the
Introduce two sets of integers: 11 = (1..... it), equation of the form , Pn
P,,
I i A =0,
(4)
where it..... i A E li . Next, consider the equations f o r superscripts it , .... iA _ I E and it E12:
pnh
ip"
Si
11
(5)
Repeating this process. we obtain n
i..,,
pn
1,
p
I,,
., p .
,r
I
,,,.r:p"hi
Si
A = (),
A ={).
(6)
In the latter equation ii .... , IA 2 E !, and only it, is E 11. The total number Al of equations in (3)-(6) is
A1=&+kC'A I+...+(A'CA +...+CA
(7)
where CI are binomial coefficients. The following formula is known (see [22]. problem 59,b p. 28): A
A
C i,
+
A
CACrI
If ,- {,
(8)
GRASSMANN IMAGE OF A SUBM ANIFOLD
237
where C,',' = C, = 1 is assumed. The right-hand side expression in (7) differs from the kn. Hence, the total number of right-hand side expression in (8) only by equations (3) -(6) is
Af = C44, - kit.
which coincides with (2). Thos, an immersion of GG,,,.A can he represented with
(3)-(6). Each equation from (4)-(6) is determined by a generalized index Denote by F11, the quadratic expressions on the righthand sides of (4)-(6). Then (4)-(6) can be represented as PD =
F,1, = 0. (9)
F,,,,=0. Denote by t,,, the unit normals of a submanifold. To simplify the notations we omit c} in subsequent expressions. The unit normals can be found in terms of gradients as ,, = grad F,,/grad F,,I. Prove that at P these vectors are mutually orthogonal. To do this we represent grad F,, as the vector in E I with coordinates 1/1 ' . Introduce generalized indices
u=(il...f
ii...!A 1 it+13)
t=(i&-, n+2...it+k), I _ (n+ l ...it+k). Let e,,,, be an odd or even number, depending on the c and h sets. Then F,, can be represented as
E(-1)'"'p'v=Pop- +(-1) rIfp"+... Represent the vectors p and grad F,, as
p= {...P"...e...p°... grad F,, _ {... p".. - (-1):- pr... At Pu all of the PlOcker coordinates except p° are zero. Hence, at Pi,
gradF,, _ {...p' ...} _ {...1...}, where the position of 1 corresponds to the position of p. At P11 then IgradF,,I = 1.
For different F the I takes different positions. Therefore, at P the gradients are mutually orthogonal and orthogonal to p. Now find the coefficients of the second fundamental forms with respect to C,, at P0. We have
f 0-.p L'r = 18u,8u, "1
- r7u;
8111)
-8
8u,
grad F,,I .
THE GEOMETRY OF SUBMANIFOLDS
238
The structure of the components of p and grad F, imply
L Lemma
(ap app+(_ly au; aui
ap°apb+apbaps+ap,a.-
au; au;
au1 au;
au; au1)
(10)
If the Plucker superscripts it ... ik contain two or more elements of 11
then at Pa ap;,...;A
au;
=0.
Let us represent p as a multivector [a I.... ak], where a; are En" vector fields which depend on the coordinates u1i ... , ukn in a Grassmann manifold. Let e1, ... , en, en+1,... , e"+k be the orthonormal basis in E"+k We suppose that if the values of parameters uI,... , ukn correspond to P0 then a; = en,;, i = I,. .. , k are the vector fields which correspond to u1i... , ukn. Then for p = [a1,. .. , ak] we have the Plucker coordinate p1 = 1. Further, 5wi
=L[ R_1
.k
The Plucker coordinate °1y of -,° is equal to the sum of the Plucker coordinates of
t,
ak] . It is sufficient to prove that the corresponding Plucker multivectors [a1 coordinates are zero for them. We shall denote corresponding Plucker coordinate by superscripts in brackets. Let be Cartesian components of aR. They are functions
of uI,... ,ukn. We have S;1l
...
elk
as[al...1...ak]41*=L ...
Sk
_ k = 0. Therefore in (11) there are two columns of all zeros except 0(n /au; in the first and aQ; /au; in the second. These columns are proportional. So, the corresponding Plucker coordinate of [al oul° ak] is zero. The lemma is proved. Suppose that i 1 i i2 E 11. Then at Po are C',' = fk =
Consider all equations of (9) for a < 2. Their left-hand sides consist of products of p°pb. One of the generalized indices a, 6 contains two or more elements of 11. Hence,
the second fundamental forms with respect to p are zero by virtue of the lemma. It remains to consider equations of the form pi,...i*_2iili=pn+i...n+kl = 0.
(12)
Superscripts j1 ... A-2 belong to the set of n + 1, ... , n + k. Suppose that among ./l, ... ,jk_2 there are no n + a, n +)3. By permutations of superscripts, represent the
GRASSMANN IMAGE OF A SUBMANIFOLD
239
Plucker coordinate p J' Jk-2'1'2 in such a form that i, and i2 take the a and 0 positions R U
We can represent equation (12) as
respectively:
pn+I...i,...ti2In+,9+I...n+kIpn+1...n+kt = 0,
(13)
where the commutation is taken over i2i n + I,. .. , n +k while superscripts in I I stay immovable. Introduce the generalized superscripts, each of which consists of the k usual superscripts: (I
l+
p= (n+ I ...i, ...i2...n+k), ry = (n+ ...n+a...n+Q...n+k), a, = (n+ ... it ...n+a...n+k), p, _ (n+ 1 ...i2...n+,O...n+k), a2 = (n+ l ...i, ...n+,0...n+k), p2 = (n+ 1 ...n+or ...i2...n+k). In addition, introduce two permutations of (k + 1) elements of the form
... n+a ... n+a n+ 1 n+2 ... i2 ... n+ I n+2 ... n+a ... A2 _ 1n+,0 n+ I n+2 ... n+a ... i2
A
i
=
n+ i n+2
12
... n+k n+/3 ... n+kJ n+/3
,
n+/3 ... n+k
... n+k)-
i2
Let e(A,) be the Kronecker symbol of A,. In terms of generalized superscripts, equation (13) can be represented in a brief form: pPp7 + (-1)`l\')p°'pP' + (_1)t
2)p°2pP2
= 0.
In a, interchange the places of it and n + a, i.e. put i, in the f-th position and n + a in the a-th position. Set a3 = (n + 1... n + a ... i, ... n + k). Then 2K1 Out' .Note au, that the parity of A, and A2 is the same. Indeed, A, is obtained from A2 by two interchanges: first, interchange the places of n + /3 and n + a; then interchange the places of n +,0 and i2. Thus, e(A,) = e(A2). So, PPP" + (-1)E(A') (_p°, pv- + p°2 pot) = 0.
We find the coefficients of the second fundamental form with respect to {P by (10):
Lj=(-1)e(a')[-Cam'a
au, 8uj
'+8p°,-app,
atlj aui
)+( &i l7uj +aanj
r
)J
The vectors Op/&i and Op/&j tangent to the Grassmann manifold at Po are represented in terms of matrices X = Y = IIY°11 of k columns and n rows. The element of X in the a-th column and i,-th row is n+k
aui
THE GEOMETRY OF SIiBMANIFOLDS
240
We have analogous expressions for elements of Y. We can represent coefficients L° as
p= 3,. y;2 + Y"; ll
,9
xa
;i -
k,q < k,
apq = 0,
for p, q > k.
The elements of the k-order matrix A2 are ft>(xfy9 - x7yf) = x,yo - xqy'°. bR4 =1=1
GRASSMANN IMAGE OF A SUBMANIFOLD
241
The numerator in (1) can be represented as
Tr[AIA +A2A;] =
(XPi'vP -xq:P)v` A (2)
R
+
(x r - xa rP)-.
(xglr )`' + 1_4_u.d. P=n
This formula has been cited in [24) without proof. But it was remarked that it was derived with the help of a specific matrix form of a system of equations of geodesics. We shall obtain this formula in another way using the Gauss equation of immersion into a Euclidean space E -v. The curvature tensor of GA +6 can be expressed in terms of coefficients of the second fundamental forms of the immersion. The second fundamental form with respect to normal p coincides with the first fundamental form. Let Lf; be the coefficients of the second fundamental form with respect to 4,,. Let X and I be the tangent vectors to coordinate curves it, and it, respectively. In the previous section we saw that "
,
of,,
Take into account that matrix X has the standard form. Then
it-` n v
rv
for p.qW(jr)e, +
w(j,)e,
I
k
n
k
=1
r=1
=1
bij
k
/ E(W(il)
AW(lj)bilb7;3
1=1 ,=1
+W(ie) AW(,:3)bn,bej +W(n7) Aw(,3)b/ebej
(r,
k
n
W(n)
=1
AW(n))& +bij
,=1 n
w(rn) Aw(l,i) l=1
k
_
AW(hb).
(bilbjkb,bbn;3 + 1,h=1 i 6=1
Thus, at a fixed point the Riemannian tensor components have the form R(i(k)(j 3)(h)(hb) = (bilbjh - bihbjl)bnub,b + 6jblh(bn,6 4 - b06b;37)
Find the sectional curvature of a Grassmann manifold for the 2-plane generated by a = 11X1011 and b = IIYi1I Note that at our point w(i0) = duin and as a consequence 8(in)(j;i) = bijb0;,. Transform the numerator and denominator of the sectional curvature formula separately. Split the numerator into two summands. The first is k
n
E(bi/bjh - bihbjl)bn/36,bXi,,Yj.-iXhYhb =
- XjnYin) E ((xinYin n=I
,
1,1=1
and the second k
L.
bijblh(bn,b;ib - 6,, &tv)XicYj;JXl,Yhb =
(n 1
2
DXinyi;3 - Xi/3Yin)) i=1
It is easy to find the expression for the denominator as n
(8(in)(h)$(j, 00.1) - $(in)(hb)g(h)(j;J)
k
(Xinyji - XjiiYin)
inY%liXhYhb =
i.j=10.8=I
Thus, the curvature of a Grassmann manifold with respect to a 2 -plane of vectors a and b has the form z
k(a, b)
_
v"j=1
(E0=1(XiOYjn
+En.13=1 ( 1(Xinyi/i - Xi3Yk+))2 - XjnYin)) .!c rij=1 Ln,/3=I (XiaYj;3 - Xj1 yin)2
THE GEOMETRY OF SUBMANIFOLDS
252
It also is easy to find that the Ricci tensor of the Grassmann manifold is proportional to the metric tensor: R(,3)(r,) = mg(;;;)(h,), where m = 2 - n - k. So, the Grassmann manifold is the Einstein one. We remark here that the case n = 2, k = 2 is exceptional. On the manifold G2,4 there exists the whole family of invariant metrics d[r2
=
'',,CC
\[(171dd1)2 + (,)I A2)2 + (Y2d 1)2 + (1)2&2)2J
+2)c4(mdd1)(ihdf2) - t]1dy2)lfidS1)1,
where A, it be real constants and A > 1pl.
So. I state that there Ls no another solid, besides five mentioned solids, mhich is contained between equilateral and equiangular polygons equal to each other. Euclid, "Origins
9 Regular Polyhedra in E4 and EN There are rive regular polyhedra in E3. In Euclid's "Origins" the 13th book is devoted to the theory of regular polyhedra. which are often called Platonic solids because in a famous dialog "Thimey" the four of them were described: tetrahedron, icosahedron, octahedron and cube. But there is the background to suppose that the
dodecahedron was unknown to Plato at that time. Morduchaj-Boltowski wrote about this in his comments to Euclid's 13th book. It seems that the dodecahedron was discovered considerably later than the other four regular solids. The science historians (see, for instance, Zmud' "Pythagoras and his school") relate the discovery of the dodecahedron by Hyppas, the Pythagorean, and by the octahedron and the icosahedron by Thiatet. It is possible to construct the regular polyhedra in E4. A polyhedron in E4 is called regular if all its faces of dimension 1,2 and 3 are congruent to each other and the neighborhoods of each vertex are congruent to each other. Their total number is six. Some of them have very complicated structures. These polyhedra are called:
simplex, cube, 16-hedron, 24-hedron, 120-hedron, 600-hedron.
All of them were discovered around 1850 [41 by the Swiss mathematician, physicist, astronomer, and professor of Bern University L. Schlafli (1814-1895). For a long time his paper on regular polyhedra was forgotten and the discovery was attributed to another scientist. Schlafli's mathematical gift was apparent early in his school years. Although he studied in the theological department of Bern University since 1829, he aspired to
study science. From 1837, he taught mathematics and physics in Thun. For 10 years he was a successful teacher, but all his free time was devoted to the fundamental study of high-level mathematics. He also studied botany and was particularly interested in the positions of the leaves on a stem. He tried to relate 253
254
THE GEOMETRY OF SUBMANIFOLDS
LUDWIG SCHLAFLI 1814-1895
this to mathematical law of the Fibonacci series. At this time he met J. Steiner who
worked in Berlin and Steiner recommended the young Schlafli to the German scientists Jacobi and Dirichlet as a translator for their trip to Italy during the winter of 1843-1844. Schlafli was 29 years old and knew the Italian language, and also had brilliant mathematical skills and abilities. He benefited greately from these
journeys with the famous mathematicians. Dirichlet educated him in number theory and Schlafli's later paper on quadratic forms originated from these discussions. Also, he translated two of Steiner's papers and two of Jacobi's papers into Italian and during these journeys Schlafli received many stimulating ideas for his subsequent papers. In 1848 he moved to Bern to work in the department of mathematics in Bern University as an associate professor. He wanted to devote himself to mathematics
but this opportunity proved very expensive because of famine that year. In his memoirs he wrote "...limited by salary of only 400 francs I had to live in relative poverty. However, I went through with this experience with joy. because I had all-helping science". His supernumerary professor assignment in 1853 was not
REGULAR POLYHEDRONS IN E4 AND EN
255
helpful and between 1854 and 1860 he had to earn a salary by performing calculations for the Swiss national insurance office. At the same time (1850) Schldfli discovered the regular polyhedra in E4 and
E". Also, he successfully developed some problems of the Italic geometrical school. He stated the conditions on Riemannian manifold to be of constant curvature: the geodesics in it with respect to an appropriate system of coordinates must be of straight line form. The assertion that an analytic Riemannian space
of dimension n allows the local analytic immersion into Euclidean space of dimension "ice is known as Schlaf is theorem. However, the strict proof of this theorem (rather hypothesis) was given later by Janet and Burstain. SchlAtli also obtained the formula for the volume of a simplex in Lobachevski space.
From 1856 onwards Schlaf i corresponded with the English mathematician A. Cayley. That correspondence lasted, with some breaks, up to 1871 and played an important role for the theory of surfaces of third order, as well as for other branches of mathematics. Schlalli's 70th birthday jubilee was celebrated at Bern University and colleagues from Zurich wrote to him: "This century distinguished due to life of the incomparable Euler, and no other Swiss master of mathematics has such a broad knowledge as you. Now you are known to contemporary mathematicians of all cantons." In 1870 Schldfli received the Steiner award for his geometrical achievements, the French institution presented him with Cauchy's papers, the Norwegians presented him with Abel's papers and the Berlin Academy presented him with Dirichlet's papers. For detailed information on his life see biography by J. J. Burchardt, Ludwig Schldf li, 1814-1895.
9.1 The Four-Dimensional Simplex and Cube
Consider, first, the two simplest polyhedra in E4: the simplex and the cube. Take in some three-dimensional hyperplane E3 a tetrahedron with edge of length a. Through its center draw the straight line I perpendicular to E3. If M is a point
in I then it is equidistant from all vertices of T3. Since the distance from 0 to vertices of T3 is less then a, there exists the point Mo in / such that if M tends in I to infinity then the distance from Mo to the vertices will be equal to a. The set of all points of intercepts joining Mo to every point of T3 forms the four-dimensional simplex.
Consider the four-dimensional cube. Define it as a set of points in E4 whose coordinates satisfy the inequalities: 0 < xi < 1, i = 1, ... , 4. The set of points for which one of the coordinates xi satisfies either r, = 0 or x, = I is called the face. The faces of xi = 0 and .r, = I are called opposite. Denote them by A, and B; respectively. The faces A, and B, have no intersections because they are in parallel hyperplanes. There are four A,-faces and four B,-faces because i takes only four values: 1,2,3,4. Therefore the four-dimensional cube is the 8-hedron. The cube vertex coordinates are formed with 0 and 1. Enumerate all of the vertices. First, consider the vertex with
256
THE GEOMETRY OF SUBMANIFOLDS
all coordinates zero: (0, 0, 0, 0); then all coordinates with one being 1, and so on. We obtain the set of vertices as (0000)
(1000)
(1100)
(1110)
(0100)
(1010)
(1101)
(0010)
(1001)
(1011)
(0001)
(0110)
(0111)
(1111)
(0101) (0011)
Thus, the cube has 16 vertices. In spite of the fact that this is a figure in the manydimensional space, the four-dimensional cube can be represented in a plane as follows. Take four straight lines from one point 01. These straight fines we use in the capacity of coordinate axes in E. Separate off equal intercepts in them. The cube edges are parallel to e1 i e2, e3. So, we obtain eight vertices. Then construct A3, A1, A2
and mark their vertices. Thus we get 15 vertices. The 16th vertex 016 can not be obtained by the procedure above because A, faces do not contain all of the cube vertices, namely (1, 1, 1, 1) does not belong to any of A,. The vertex 016 belongs to one of the B,-faces. To find it, construct a cube in the face x4 = 1. To do this, consider the straight lines passing through 09 in the e4-axis which are parallel to el, e2, e3. All of the B4-vertices are already marked except 016. Drawing the straight lines parallel to corresponding edges, we find 016. The cube A4 of vertices 01, 02, 03. 04, OS, Os, 06 07 is opposite to the cube B4 of vertices O9, Olo, 012, O11, 013, 015, 016. 014 Consider the intersection of faces A, and A. It is defined by the system of equa-
tions xi = 0, x, = 0 and inequalities 0 < xk < 1, 0 < x, < 1, where k and I are subscripts complementary to i and j. Therefore, the intersection of faces Ai and Al is the square. In an analogous way we find that the intersection of Ai and B, faces for i 54j is also a square. Thus, any two faces, except opposite ones, intersect each other by the square (see Figure 33). The intersection of three faces A, n A, n Ak or A; n B; n Bk or Bi n Bk n B,, where all subscripts Q k k are distinct, produces an intercept - the cube edge. How many
edges has the cube? How many vertices is each vertex of four-dimensional cube joined to? From OI there go out four edges. The same number of edges go out from any other vertex. The total number of vertices is 16. Therefore, the total number of edges is 16.4/2 = 32. The total number of two-dimensional faces is equal to the product of the total number of three-dimensional cubes and half the number of faces in each three-dimensional cube, that is 8.6/2 = 24. The boundary of a four-dimensional cube can be represented in another way - by its surface development into three-dimensional space. Consider, first, the development of the surface of the usual cube onto a plane. If we cut it appropriately and develop it onto a plane then we get a cross-like domain. Using the method of edge
gluing, we are able to make the cube surface by an inverse gluing process (see Figure 34).
REGULAR POLYHEDRONS IN E4 AND E" Of$
FIGURE 33
FIGURE 34
044
257
255
THE GFOAIF rRl OF SUHMANIF01 I>S
FI((tRE 31
We shall proceed in an analogous manner to get the three-dimensional surface development of a four-dimensional cube surface. Construct the cube A., in threedimensional space of basis e, . e2. ea. Glue to it three more cubes A,, i = 1.3.3. The AI-cube intersects A. by a two-dimensional face x, = 0. Therefore, the face P1RP:P4 is their common one. Glue A, to A.4 along this face. In an analogous way glue A, to the P, PKPP,-face and A, to the P, P5PSP.,-face. It is possible to pass from the A4-face to the opposite B3-face along any other face,
say B1. In the face chain A4B,B4 each two units intersect each other by a twodimensional face. The B,-face intersects A4 by the square PKPSP,,P7. Glue the cube B, to it. In an analogous way we glue B2 and B; cubes. Note that A, and B, are glued to A4 by the opposite two-dimensional faces. It remains to glue B4 and to show how
to identify free two-dimensional faces. B, and B4 have the common face x, = I. .r; = I which is opposite to the face of gluing B, to 44 because these two-dimensional faces are in parallel planes. Therefore, glue B., to B, by the P,P,aP1, P,2-face. From the development just obtained it is possible to get the four-dimensional cube surface by an ideal identifying process (see figure 35). Now we show how to identify the faces of the cubes A, and B,. If some edge is
common to the faces of two cubes then these faces ought to be identified. For
REGULAR POLYHEDRONS IN F.' AND E'
359
instance. POP-, is the common edge of faces P6PPjPIIPill and P,,P5PI3P1I. Thus, we identify them. Moreover, we see that the edge PIIIPI I is identified with PI.IPI t. Hence, the faces P14PI.P11PI6 and PIQPIIP17PIK ought to be identified, too. In the figure, correspondence between identified faces is marked with arrows.
Thus, all free cube faces are identified to each other. As a result, we obtain the closed three-dimensional manifold - the surface of a four-dimensional cube. From a topological point of view this manifold does not differ from the surface of a fourdimensional ball, i.e. it is homeomorphic to the three-dimensional sphere S3. 9.2 The Analog of an Octahedron: 16-Hedron
Now we proceed with consideration of the next regular polyhedron which is the analog of an octahedron. It is called the 16-hedron. To construct it, take in the space of x1, x+. x3 an octahedron with vertices Q1, Q2, .... Q,,. Suppose the octahedron edges are of length a. Through the octahedron symmetry center 0 draw the straight line parallel to the Y4 axis. Take two points Q7 and QII in this line which are symmetric with respect to the hyperplane x.I = 0. All the distances from Q7 to the octahedron vertices Q; are equal to each other. Since the distance from the octahedron symmetry center 0 to its vertices is a/vl < a, then there exists a point Q, such that Q7Q, is equal to a. The polyhedron just obtained with vertices QI .... , QK is regular. Its three-dimensional face is formed by the set of intercepts joining Q7 or QK to the two-dimensional face of an octahedron. For instance, one such face is QIQ2Q3Q7 - the tetrahedron. The total number of faces through Qi is equal to the number of octahedron faces. i.e. 8. Since every threedimensional face passes through either Q, or QK, the total number of three-dimen-
sional faces is equal to 8 .2 = 16. The number of edges which go out from each vertex of the 16-hedron is equal to the number of octahedron vertices, i.e. to 6. As each edge joins two vertices, the total number of edges is 6.8/2 = 24. Now find the number of two-dimensional faces. We can separate all such faces into three disjoint classes: those passing through Q7, those passing through Qs, those passing through neither Qr nor Q. The number of faces passing through Q7 is equal
to the number of octahedron edges, i.e. to 12. The two-dimensional face passing through neither Q7 nor QK is a two-dimensional octahedron face. There are eight of them. Thus, the total number of two-dimensional faces of a l6-hedron is equal to
12+ 12+8 = 32. For convex polyhedrons in E4 the following analog of the Euler formula is valid. Set PII is the tofu! number of rertices, P, Is the total number of edges, P3 and P3 are the total numbers qt tt o- and three-dimensional faces respectirelr. Then these numbers are related as
Po-PI +P3-P3=0 which is the particular case of the Poincar6 formula. Check it for a 16-hedron:
8-24+32-16=0.
260
THE GEOMETRY OF SUBMANIFOLDS
In an analogous way it is possible to construct the regular polyhedra in n-dimensional Euclidean space E": the simplex, the cube and the 2"-hedron. It happens that for n > 5 there are no other regular polyhedra. Consequently we shall need the concept of director polyhedron. Let A be the vertex of the regular polyhedron M 4. Consider the ends of all edges from A. They are
equidistant from A, symmetric with respect to an axis of symmetry through A and situated in a common three-dimensional plane. The intersection of this plane with M4 is also the regular polyhedron in E3. It is called the director polyhedron. We shall denote it by N3.
9.3 24-Hedron
Take as a director polyhedron the cube N3 in some three-dimensional space E3. Suppose that this E3 is defined by x4 = 0. Let QI,... , QR the cube vertices and a the edge length. Draw the straight line parallel to x4 through the center 0 of the cube N3. Take on this line a point Q9 in such a way that the distance of Q9 to the cube vertices would be equal to a. It is possible to do so because the half-length of the diagonal satisfies avr/2 < a. Draw the three-dimensional space through the twodimensional cube face, say QIQ2Q3Q4, and the point Q9. Construct in this space the octahedron with vertices QI, Q2, Q3, Q4 and the point Q9. Denote by Q1o the vertex which is opposite to Q9. The analogous octahedra we shall construct with respect to other cube faces. Thus, we get six octahedra with the common vertex Q9 and upper faces glued to each other. Let Q10, Q1 1, Q12, Q13, Q14, Q15 be the octahedron vertices
opposite to Q9. All of them are equidistant from Q9 and located in a hyperplane E? parallel to E3. The lower octahedron faces are not glued to each other. Reflect the figure just obtained symmetrically with respect to E, . Let Q16 be the point symmetrical to Q9. Denote by Q17, ... , Q74 the points which are symmetrical to the vertices of the cube N3. Thus, we obtain 24 points which are the vertices of the required polyhedron. In figure 36 we see the schematic picture of this intermediate construction step: we depicted E2 instead of E3 and the square instead of N3. We constructed 12 boundary octahedra. Some of the faces of these octahedra are left
free. Paste the holes with the auxiliary octahedrons. In figure 37 you can see the spatial picture of free octahedron faces. The pyramids QIQ2Q6Q5QI1 and Q6Q2Q3Q7QI5 form the lower parts of an octahedron. They have the common face Q6Q2. The faces QIIQIBQI7 and Q18Q17Q15 belong to unglued octahedron parts which arise after symmetry. Observe that Q1sQ17 is the edge of the cube which was obtained from N3 by symmetry. It is parallel to Q2Q6. loin Q6 to QIS and Q2 to Q17 with edges. The polyhedron QIIQ6Q2QI7QI8QI5 has the edge net of an octahedron. To prove that this is indeed an octahedron, we need to state that the distance between the straight lines Q2Q6 and Q17Q18 is equal to a. The latter is equivalent to the statement that the distance between Q6 and Q1s is equal to a. These points are symmetrical with respect to the plane El. Therefore the distance between them is equal to the distance between E3 and the plane symmetrical to it,
REGULAR POLYHEDRONS IN E4 AND
261
FIGURE 36
4
FIGURE 37
THE GEOMETRY OF SUBMANII.OLDS
262
Q9
FIGURE 38
that is to double the distance from QI I to E3. The distance from Q, I to E3 is equal to the distance from Q9 to E3 (see figure 38). Let h be this latter distance. Then
h=
a---4a---=2a
.
Thus, the distance between Q6 and Q18 is equal to a. The quadrangle Q6QISQI7Q2 is the square and the figure Q6QISQI7Q2QISQII is the octahedron. Since the cube has
12 edges, to paste the holes we need 12 octahedra. Thus, the boundary of a constructed polyhedron consists of 24 octahedrons. The number of vertices is 24. The number of edges is 8.24/2 = 96.
Another way can be given to construct the 24-hedron by pointing out all its vertices and making the convex hull. All the vertices of the 24-hedron can be obtained from the points (±a ± a00) by various coordinate permutations. Separate them into two subsets (a) and (3):
(±a ±a (a) (±a (±a
0
±a
0). 0),
0
0
±a),
0
(0 (/3) (0
(0
±a ±a ±a 0
0),
±a), ±a ±a). 0
From this we see that all vertices are located in the hyperplanes x, = a, x, = -a. Also, the first and last hyperplanes contain six vertices each and they form the (a)set. The hyperplane xI = 0 contains 12 vertices and they form the (i3)-set. Six vertices in the xI = a hyperplane form the boundary octahedron. The (Q)-set makes the vertex set of an icosahedron which forms something like a belt of a 24-hedron. Both of the ways described above produce the same figure. Let 0 be the center of symmetry of a 24-hedron and A be the center of some boundary octahedron, say
Q9Q2Q6Q7Q3QIS. Denote it by T1. The point A is the center of the square
REGULAR POLYHEDRONS IN E' AND E'
263
Q2Q6Q7Q3. Introduce the system of coordinates in E4 in such a way that the x1 axis coincides with OA and the origin coincides with O. Let N be the hyperplane containing T1. The straight line OA is perpendicular to N. Direct the other coordinate axes x2i x3, x4 along the three diagonals of the boundary octahedron T1. Then its
vertices obtain coordinates of the form (a ± a 0 0), (a 0 f a 0), (a 0 0 ± a). Let K be the square in a hyperplane EE' (see figure 38) obtained by projection of the square QIQ4QgQ5 into E2. The boundary octahedron T2, which is formed with Q16 and K, is the image of T, under parallel transport along OA. It can be shown that all of 12 other vertices of the 24-hedron are located in a hyperplane which is parallel to N and passes through O. 9.4 The Symbol and Theorem of Schlifli
Assign to each regular polyhedron P in E" the ordered set of numbers { r 1 , . . ., r,,_ 1). It is called the Schldfli symbol. The numbers for this set are defined inductively. The first number r, is the number of edges of a two-dimensional face P
I) is the Schlaf i symbol for director polyhedron TP. Remember that TP is regular too. Consider the Schlafli symbols for regular polyhedra in E3. They consist of two and { {r2, r 2, . . . ,
numbers {r,,r2}. For the cube we have r, = 2 and the triangle as the director polyhedron (polygon). Hence, the Schlifli symbol of the cube is {4, 3}. This symbol for the tetrahedron is, evidently, {3, 3}. For the octahedron we have r, = 3 and the pentagon as the director polygon. So, the Schlafli symbol is {3, 5}. The dodecahedron is dual to the icosahedron. Its faces are pentagons and the director polygon is a triangle. Therefore, the Schlgfli symbol is {5, 3). The Schli fli symbol for the regular polyhedron in E4 consists of three numbers {ri, r2, r3 }. Evidently, this is 13, 3, 3, } for the tetrahedron, {4, 3, 3} for the cube,
{3,3,4} for the 16-hedron and {3,4,3) for the 24-hedron. For the other regular polyhedra we calculate these symbols later. Let a be the edge length of a regular polyhedron P and r be the radius of the sphere circumscribed around P. Introduce the number p(P) = 4r,. If r"_, } is the Schla"fli symbol for P then the following formula holds
AM = 1 -
2
p(TP) '
(1)
where TP is the director polyhedron for P. Let 0 and O' be the centers of spheres of radii r and r' circumscribed around P and TP respectively. Let A be the vertex of P with AB and AC as the going out edges. Their length is a. The intercept BC is the edge of the director polyhedron. Denote its length by a' (see Figure 39). The points B, A and C are consequent vertices of the two-dimensional face of P. This face has r, sides. Let 0" and r" be the center and radius of the circumscribed circle around this regular polygon (see Figure 40). The
264
THE GEOMETRY OF SUBMANIFOLDS
FIGURE 39
FIGURE 40
angle BO"A is equal to 21r/ri. Therefore, by the cosine theorem a2 = 2r"222ri2 cos. Hence
a=2r"sin-.
(2)
Express a' in terms of a. From the triangle BO"C we see
a'=2r"sin
-. r, 21r
(3)
Therefore, taking into account (2) and (3) we get
a'=2acos-.
(4)
r,
Draw a plane through the points A, B and O. Then draw in it the circle of radius r with the center at O. Let AD be the diameter. Let W be the value of the angle BDA.
RLGULAR POLYHEDRONS IN E' AND E"
265
A
a FIGURE 41
Then the central angle BOA is equal to ABD we find a = 2r sin p. Then
(see Figure 41). From the right triangle
ca -
P(P) =
4r' =
sm- V.
(5)
The point O' is in intercept AO; it is also the projection of B into this intercept. From the right triangle BO'O we find r' = r sin tsp. Hence
r'=acosv.
(6)
Taking into account (4) and (6) we get
a'- 4a2 cos=, p(TP) = 4r,_ = 4u= cos
cos=yo'
(7l
The latter relation together with (5) gives (1).
Since p(P) does not change under a transformation like that, it is possible to represent it as a function of the Schlgfli symbol. So. (I) can be written as
P(rl,...,r,,.l)=1 -
cos' _
P(r,,...,r. I),
This formula allows one to pass from the function of the (n - 1) variable to the function of (n - 2) variables. In addition, we see that p(P) decreases steadily while rl increases. As r> > 3 , then cos=; >t. Write the inequality L COs,-,
d.
From this it follows that p(a .... r 1) > 1/4. The latter inequality allows us to analyze the set of system
IHE(i1:OMEIR1 Of SLIM NIIOLL)S
2t(3
We begin with consideration of the values of 1(P) for regular polygons. The Schlafli symbol consists of one number ri. So,
1(ri) = 4r, = sin- r . In more detailed form: p(3)3.
p(4)p(5)5 Sam
for nyb p(n)=sin'1r
4 From this it follows that only the triangle, the square or the pentagon can be taken as the director polygon for the regular polyhedra in V. For the case of the tetrahedron, the Schlafli symbol is J3.31. Then
p(3.3) = I _ cos=- _ I _ 1/4 3/4 MA)
-2
The Schlafli symbol for the cube is {4.3}. Therefore 1(4.3)=I-cos-=1-1/2
3/4
1(3)
=1
3'
For the case of the octahedron the Schlafli symbol is {3.4} and
p(3.4) = I -
cos-- ii = P(4)
I
- 1 /4 = 1
.
1/2
For the cases of the icosahedron and dodecahedron the Schlittli symbols are 13. 51 and {5.3} respectively. Hence
J)(3.5) = I -
cos' 1(5) cos'
P(5.3) = 1 -
p(3)
5-f 10
3 - f
I
< 4'
6
From the latter inequaht) it follows that the dodecahedron can not be taken as a director polyhedron for the regular polyhedron in E4 Consider the Schlafli symbols of regular polyhedra in E`l. They can be of the forms: (a) 4r1.3,3}. (h) {r,,4.3}, (c) {r1.3,4}. (d) {ri,3.5}. In case (a), rI takes values of 3.4 or 5 only. Indeed, if rI ? 6 then
3/4 1( 3,3)
2/3
9 =1- 5, the only tetrahedron or the octahedron analog can be taken as a director polyhedron. 9.5 The Regular 600-Hedron It was a Diamond! As large as a plover's egg. The light that streamed from it was like the light of the harvest moon. When you looked down into the stone you looked into a
deep yellow that drew your eyes into it so that they saw nothing else. It seemed unfathomable as the heavens themselves. We set it in the sun, and then shut the light out of the room, and it shone awfully out of the depths of its own brightness, with a moony
gleam, in the dark. Wilkie Collins The Moonstone.
Consider a polyhedron of symbol {3, 3, 5}. Its director polyhedron is a {3, 5} polyhedron, i.e. the icosahedron. To represent this polyhedron we use the coordinate method of Coxter and Berger [2]. It is simple to construct this polyhedron using the coordinates of its vertices. Denote by r a positive number satisfying r2 = r+ 1, i.e. T = J 2}'. All vertices of the icosahedron and the dodecahedron can be represented by means of this number. Namely, twelve vertices of the icosahedron are of the form (see section 7):
(0,±T,±l),
(±1,0,±r),
(±r,±1,0).
(1)
They are obtained by even permutation of 0, ±r, ± 1. All twelve vertices of the dodecahedron have coordinates as
(f1,fl,tl), (0,fr-',fr), (fr,0,fr-'),
(fr
±'r' 0).
(2)
Various sections of the 600-hedron by hyperplane represent the polyhedra with the icosahedron and dodecahedron among them. The set of 600-hedron vertices consists of the points (±2,0,0,0),
(0,±2,0,0),
(0,0,±2,O),
(0,0,0,±2),
(t1, fl, fl, tl) and the points obtained by even permutations of coordinates
(± r,±1,±r'',O).
(3)
REGULAR POLYHEDRONS IN L AND E"
269
Set a = ± r, b = ±1, c = ±'r-1, d = 0. Then the latter set of points is of the form (d, b, a, c),
(b, d, c, a),
(a,c,d,b),
(b, a, d, c), (d, a,c,b),
(b,c,a,d),
(d,c,b,a),
(a, d, b, c),
(c, a, b, d),
(c, d, a, b),
(c, b, d, a).
(a, b, c, d),
(4)
The set of the polyhedron vertices consists of 24 vertices of (3) and 96 vertices of (4), i.e. 120 vertices.
The origin is the polyhedron symmetry center. The vertices are located in nine hyperplanes x, = const. Mark in the xi-axis the points that the hyperplanes containing these vertices pass through. Also, subscript the corresponding numbers of vertices in each hyperplane I
I
I
-2 1
12
20
12
I
I
I
0
T'
30
12
1
T
2
20
12
1
In the x, = 2 hyperplane there is only one vertex A (2, 0, 0, 0). Consider the vertices in the x, = T hyperplane. They belong to the set of vertices (4) with a in the first place:
(T,±I,±T-1,0), (T,O,±1,±T-1),
(T,±T ',0,±1).
(5)
In the x I = T hyperplane the variable coordinates are x2, x3, x4. With respect to these coordinates the vertices of (5) can be represented as
T-'(±T,±1,0), r '(O,±T,±1),
r-'(±1,0,±T).
(6)
Comparing with (1), we conclude that (6) represent the vertices of the icosahedron which is obtained from (1) by similitude of the r--' similarity ratio. Join the point A to every vertex of an icosahedron with the edges. Since the lengths of these edges are equal to the icosahedron edge length, all the two-dimensional faces are triangles and three-dimensional face is a tetrahedron. In the hyperplane x, = I there are 8 vertices of set (3):
(1,±1,±1,±1)
(7)
and 12 vertices of set (4) with b in the first place:
(1,±T,0,±T-'), (l,±T-',±T,O),
(1,0.±r',±T).
(8)
Comparing with (2) we conclude that 20 vertices (7)-(8) make the set of dodecahedron vertices. The point A takes a symmetrical position with respect to the icosahedron (5) and dodecahedron (7H8) vertices. The sets (5) and (7)-(8) are invariant under the action of group of the icosahedron. In the hyperplane x, = T-1 there are 12 vertices:
(r',±T,±1,0),
(T-',O,±T,±1),
(r',±1,0,±T),
rHF
2711
OP SLI,)NMANii'()LDS
which coincide with the vertices of the icosahedron (1). Finally, consider the hyperplane xi = 0. It contains 30 vertices: (0.±2,0.0).
(0.0.±2,0),
(0,0,0.±2).
(0.±r.±r '.±l). (0.±1,±r.±r''). (0.±r'',±I,±r).
(9)
The number of these vertices conicides with the number of icosahedron vertices. In the hyperplane v, = 0 consider the icosahedron obtained from (1) by the homothetic transformation of ratio 2r 1. The coordinates of its vertices are
(0,±2.±2r '). (±2r '.O,±2). (*2,±2r '.0)
(10)
Find the midpoints of the edges of this polyhedron. The midpoint of the edge of
vertices (2,2r ',0). (2.-2r 1,0) is (2,0.0). The midpoint of the edge of vertices (2.2r , 0), (0.2.2r ') is (1. r + 1, r 1) = (1. r, r 1). Comparing this with (9). we 1
%ee that these points belong to the set (9). In an analogous way we can state that the other points of (9) are the midpoints of the edges of an icosahedron (10). Therefore. this point set is invariant under the action of the group of the icosahedron Thus, the set of polyhedron vertices is invariant with respect to rotation around the x, -axis which takes the director icosahedron into itself. Therefore, the neighborhood vertices of polyhedra in each separately considered hyperplane .r, = const are congruent to each other.
Note that the polyhedron edges join the iertices located either in the same hyperplane or in neighboring hyperplanes. Mark two points: A(2.0.0.0) and B(I, -1. -1 -1). Draw the hyperplane through the midpoint and perpendicular to the intercept AB. Perform the symmetry transformation with respect to this plane. Let it be the unit normal of the hyperplane. The r = r - 2(rvt)n is a position vector of the symmetric point to the point of position vector r. In coordinate form, taking into account that it = ( . ; . 1 ), this transformation can be expressed as .v,=
(2x, - vi - x' - .ri - .v4) !2.
1 = I ..... 4
Under this transformation the set of vertices (3) passes into itself. For instance. (-2.0, 0.0) passes into (-1.1,1,1). The point symmetric to (u. h, d) has the following coordinates
(cr-h-c h- ct-c
-cc-h-rl
To show that these coordinates can be expressed in terms of it. h, r. d, it is sufficient to state the following equalities-
r+1+r
-r+l+r0 r-l+r.,, r+1
taking into account that I + r = r. Hence, the coordinates in (11) can be expressed via a, h, c, d. For instance. (r. 1, r -1, 0) passes into (0, -r 1. -1. -r). which has the form (d, h, u) and, therefore, belongs to (4).
REGULAR POLYHEDRONS IN E4 AND E"
271
The set of points (4) is invariant under symmetry transformation. Therefore the neighborhoods of A and B are congruent. Conducting the same considerations with respect to B, we conclude that the neighborhood of the vertex A is congruent to the neighborhood of the vertex in the hyperplane x, = r. From this it is easy to state that the neighborhoods of each vertex are congruent to each other. So, the polyhedron is regular.
Each vertex adjoins 20 faces - the tetrahedrons. Since the total number of polyhedron vertices is 120 and the number of tetrahedron vertices is 4, the total number of polyhedron faces is 120.20/4 = 600. 9.6 The Regular 120-Hedron Consider now the polyhedron of the {5, 3, 3} Schlaffli symbol. It is dual to the 600hedron considered above, that is, its vertices coincide with the centers of faces of the
600-hedron. Evidently, the polyhedron dual to the regular one is regular too. Therefore, the polyhedron under consideration has 600 vertices. We shall show that it has 120 faces. The symbol of the director polyhedron is (3, 3). So, it is a tetrahedron. The faces of the 120-hedron are dodecahedrons. Indeed, as {5, 3, 3) is the 120hedron symbol, the number of sides of a two-dimensional face is 5. Among the regular polyhedra in E3 only the dodecahedron has pentagons as faces. This fact can be also proven geometrically. Consider the centers of 600-hedron faces with a common vertex A (2, 0, 0, 0). Each of these faces corresponds to the 2dimensional icosahedron face in the x1 = r hyperplane. Find, for instance, the center of the 3-face of vertices {2,0,0,0}, 17-,,r, 1, 01, Jr, 0, r,1 }, Jr. 0, r, -1 }. We obtain { "r 11, a 2 - y} . In an analogous way we can find the centers of the other three faces. All of them lie in the same hyperplane x, = . Therefore they are vertices of the same face of the 120-hedron. If we draw rays from A (2, 0, 0, 0) through the centers of 3-faces then they will pass through the center of 2-faces of the director icosahedron, i.e. to each 2-face of an icosahedron there corresponds one vertex of a 120-hedron face. Therefore, the faces of a 120-hedron are dodecahedra. Denote by D, the dodecahedron constructed with respect to vertex A, see Figure 42. Two neighboring dodecahedra are glued to each other along the pentagon. The number of dodecahedron faces is 12 and the number of 600-hedron vertices in the hyperplane x, = r is 12. For each of these vertices there is a corresponding dodecahedron D2,..., D1.1. Each of them has a common edge with D1. So, we can imagine
that the dodecahedron DI has 12 dodecahedra glued around it. In the x, = I hyperplane there are 20 vertices of the 600-hedron. T h e r e are 20 corresponding dodecahedra D14, ... , D33 which are glued to the previous layer D2,..., D13. There are 12 vertices of the 600-hedron in the hyperplane x, = r -I. Thus we have 12 more dodecahedra. Finally, 30 dodecahedra correspond to the vertices in the hyperplane x, = 0. The dodecahedra gluing process continues for x1 < 0 also. Thus, the polyhedron consists of nine dodecahedron layers and their total number is I + 12 + 20 + 12 + 30 + 12 + 20 + 12 + I = 120.
272
THE GEOMETRY OF SUBMANIFOLDS
FIGURE 42
Coxeter represented the vertices of a 120-hedron in the following convenient form: these are all points obtained by permutation of coordinates of the points
(±2,±2,0,0),
(±T,±T,±T,±T 2),
(ff,fl,fl,fl), (±T2,±T-',±T-',±T-'), and also the points obtained by even permutations of coordinates of the points ((±T2,±7--2, ±1,O),
(±v7,±7-',±T,±O),
9.7 A Simple Way to Construct the Icosahedron
In this section we shall state a simple and descriptive way to construct the icosahedron. Inscribe into the circle of center 0 two regular pentagons 11, and 112 of vertices A,, . . . , A5 and B1, . . . , B5 respectively in such a way that each B; would be the midpoint of arc A;A141. Let a be the pentagon side length and r the circle radius.
We shall regard these pentagons as located in two coinciding planes. Draw the straight line I perpendicular to the pentagon planes through the circle center 0. Shift these planes along / in such a way that they stay perpendicular to /, the pentagons take symmetrical positions with respect to I and lA;Bjl = a. This is possible because in the starting position IA;Bjl < a and we infinitely increase the distance lA;BjI by the distance increase between planes. Join B ; to A; and A; }, with edges. Then construct two caps for pentagons n, and 112. Mark the point Q, in I such that l Q 1 A j = a. Such a point exists. Indeed, shift Q along I starting from the center of 11,. In the starting position I QA,I = r = 2 si n x/5 As sins > 1 then r < a. If Q tends to infinity along ! then IQAjI increases to infinity. So, there are two required points symmetrically located with respect to the n1-plane. This plane splits the space into two half-spaces. Select Q, in that half-space which contains no 112. Join Q, to Aj with edges. In an
REGULAR POLYHEDRONS IN E4 AND E?'
273
analogous way find Q2 and join it to B; vertices with edges, too. Thus, we obtain the required icosahedron. Denote by C; the vertices of the icosahedron. To get the simple coordinate form of icosahedron vertices we ought to introduce a convenient system of coordinates. Let 0 be the center of symmetry of the icosahedron. Consider the edge C, C2. Draw the
straight line OK perpendicular to C1 C2. Suppose that IOKI = r and the length of edges is equal to 2. Then IKC1 I = 1. Draw a plane through the points 0, C1, C2. This plane contains two more icosahedron vertices C12 and C11 which are symmetrical to
C1 and C2. If we join C, to C12 and C2 to C1 then we obtain the rectangle C1 C2C1I C12. To the edge C, C2 there are adjoined in a symmetrical way the icosahedron faces - the regular triangles C, C2C3 and C, C2C4. The straight line C3C4 is evidently perpendicular to the plane of the rectangle C1 C2C11 C12. Let us now define the system of coordinates. Put the origin into the center of symmetry O. Direct the axes x3 along the straight line OK, the axes x, parallel to C2C1 and the axes x2 along
C4C3. With respect to this system of coordinates we have C, (l, 0, r), C2(-1, 0,,r). Let C3 and C4 be represented as C3 (0, a, b), C4(0, -a, b). Let C10 and G) be the points 0-symmetric to C3 and C4 respectively. They are also the icosahedron vertices. Thus, we know eight of the icosahedron vertices. Let C. C6, C7, C8 be the other four vertices. Suppose that C5 and C6 are joined with an edge. As we saw above, if we draw the plane through 0 and the edge then it contains four icosahe-
dron vertices. These vertices are disposed symmetrically with respect to OK. Therefore, the planes C1 C2C11 CI2 and C5C6C7C8 are perpendicular to each other; also the axis x, is their intersection while the sides of the second rectangle are parallel to the x, and x2 axis. The distance from 0 to the edge C5C6 has to be equal to r and
IMC5I = 1. Hence, we have C5(r,1,0), C6(r, -1,0). The intercept C3Cq is the icosahedron edge. The points C3 and C9 are disposed symmetrically with respect to the plane C5C6C7C8. Therefore C3C9 is perpendicular to this plane. So, the distance from C3 to the plane x1, x2 is equal to 1. Note that the distances from C3 to C1 and C5 are equal to 1. Thus we have two conditions: 12
IC3C1
= I + a2 + (r -
1)2 = 4.
IC3C5I2 = 1+r2+(a- 1)2 =4.
From this we find a = r. Substituting a into the first equation, we obtain the equation with respect to r. The required number is the positive root of the equation r2 = r + 1. After that it is easy to find the coordinates of all icosahedron vertices. We already expressed them above, see (1) Section 5. We can take the centers of icosahedron faces as the vertices of a dodecahedron.
10 Isometric Immersions of Lobachevski Space into Euclidean Space This chapter introduces some background information and provides an individual perspective on work developing from the theory of isometric immersions of the Lobachevski plane into Euclidean space. Section 10.1 gives an account of the author's visit to Nickolina Gora with Professor Efimov and Professor Pozniak and also gives some interesting insights into the work of Professor Efimov and Professor Pozniak and colleagues at Moscow State University. It introduces the conference on Non-Euclidean Geometry held in Kazan in 1976, dedicated to the 150th anniversary of Lobachevski's discovery. Section 10.2 gives an account of the fire in Kazan in 1842. Lobachevski was a professor at the University of Kazan and this section describes his work, achievements and discoveries and the significance of his work in classical mathematics. 10.1 Meeting at Nickolina Gora
The Kharkov-Moscow train arrived in Moscow early in the morning while the meeting with my scientific advisor - Professor Nickolai Vlad mirovich Efimov was scheduled for 5 p.m. Many things depended on that meeting. I hoped to discuss the forthcoming plans and perspectives of my scientific work and I wanted to use my time in Moscow to the best effect. I went to Lenin's library to get copies of scientific
papers not available in Kharkov. At that period I was studying the geometry of vector fields and minimal surfaces and I was interested in the interpretation of mass as a curvature of space. The book by J. Willer and others, The theory of gravitation and gravity collapse, contained a curious formula: in Einstein space the density of mass-energy p is equal to the sum of the scalar curvature of a space-like hypersurface and its second symmetric function of principal curvatures, namely 4ap = R + Ke. 275
276
THE GEOMETRY OF SUBMANIFOLDS
This formula directly relates geometry to physics. In one of my papers I introduced the concept of the curvature source power of a vector field singular point and found
the interpretation of this concept in terms of general relativity as a mass source power. So, I was interested in the reference, where the uncommon hypothesis that the centers of galaxies can be the sources of mass was suggested. I collected a copy of this for further study. At last, the time for the meeting with Professor Efimov arrived and I took the elevator to the 14th floor of the Moscow State University building, to the department of mechanics and mathematical. N. V. Efimov was there and it was good to see him. He had been ill but had recovered and was, as always, friendly, joyful and benevolent. It was late June and the weather was hot and sunny. On the office wall there was a
portrait of Gauss and "the king of mathematicians" gazed at me with a stern, penetrating and slightly mocking look. Efimov was an unusual dean in the history of the mechanics and mathematics department. He was able to pay attention to students, to fascinate them with science and he was respected by professors and students. Consequently he took the position of dean for two terms, i.e. for eight years. He was always attentive to everyone he met regardless of their position. He asked me about my work in science and about the life-style in Kharkov and told me that he was waiting for Edward Genrichovich Pozniak. They planned to go to the cottage at Nickolina Gora to stay in the country and to get some fresh air and he invited me to
go with them. Pozniak arrived and approved the "cottage idea". This was very flattering and unexpected for me, although I knew that academician P. S. Alexandrov had often taken his students out of the city. Sometimes, Pavel Sergeevich came to the student's hostel with a portable record-player and listened to classical music records with them. The rector, Ivan Georgievich Petrovski, also visited students and discussed aspects and perspectives of University-life with them. It was good to be in the company of Efimov again, the famous geometer, Lenin and Lobachevski Prizes laureate. In 1983, the famous French geometer Marcel Berger dedicated the Russian translation of "Geometry" to Prof. Efimov. In a preface he wrote: "In May 1968 Efimov was the guest of Paris VII University and everyone who simply talked to him or discussed mathematical problems was profoundly impressed by him. I would like to express here my deep respect for this extraordinary man." Efimov was not a fisherman, a hunter or a collector. He liked, it seems, only to read books. One day, I went to him for a tutorial consultation and I noticed a book "Dear My Man" by Yu. German on his table; he loved books and once he told that he had reread Tolstoy's War and Peace. I remember also the seminar in geometry when he gave us his impression of Bulgakov's "Master and Margarita" just published in Moscow magazine. "It is something tremendous", he said, and recited the beginning of the first scene with Pontius Pilate. I remember with pleasure and some sadness that hour and a half journey by car to Nickolina Gora with Efimov, "the geometry clan's oldest father" (to use M. Berger's expression), driving with E. Pozniak beside him. The fresh soft air of the evening and the scent of the forests near Moscow generated optimism and the joy of living.
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277
The most recent of my results which I had reported earlier in Moscow were concerned with the instability property of closed minimal surface in a Riemannian space of positive curvature. I hoped that it would provide an approach to the Hopf problem
on the metric on S2 x V. Efimov and Pozniak recommended that I presented a report at the P. K. Rashevski seminar after summer, in the presence of A. Fomenko who studied minimal submanifolds at that time. Then the conversation turned to the importance of geometry. Sometimes geometry was viewed with scepticism although it was accepted that it was an interesting subject. Efimov said: "You know, we. the geometers, studied very deeply the equation w = sin w, which arises in the theory of isometric immersions of Lobachevski plane into Euclidean space. Recently. Sergej Petrovich Novikov told me that this equation occurs in many physical phenomenona. for instance in superconductivity, and it is developing the theory of soliton solutions for it. At the beginning of the century, in geometry the method allowing new solutions to be obtaining by the Backlund transformation was developed, and this method also has applications, So. geometrical approaches justify themselves, maybe not at once but after some time. You must believe in the your work."
At last, the car turned off highway and entered the yard of the dacha which Efimov and his family rented for the summer. A table was placed in the yard which was shaded by trees and we continued the conversation. Efimov told me that it had been proposed to stage a conference in geometry in 1976 in Kazan devoted to the
150th anniversary of Lobachevski s discovery. Efimov said: "Do you know the reason for the obstruction and delay of that discovery? The obstacle was not only the solution of a mathematical problem, but also in overcoming the inertia of secular traditions. It had to be understood that all 'the new' was true although 'the old' was also true. It had to be understood in what sense 'the old' and 'the new' can be true and also mutually exclusive. Before Lobachevski, the geometrical idea was tied to Euclidean space, and Lobachevski opened and released unbounded possibilities of
fruitful generalizations. It was he who got modern geometry off the ground" LQbachevski was regarded as a scientific hero by Efimov. The conversation made an impression on me and I planned to develop the theory of isometric immersion of Lobachevski space into Euclidean spaces.
I remember Efimov's suggestion: "Time is ... nobody knows. It would be very interesting and important to penetrate the secret of time. Only one saying can be expressed definitely - it is inexorable." In 1976, at the All-Union conference in non-Euclidean geometry in Kazan I presented a report: "Lobachevski spaces as a submanifolds in Euclidean spaces" That was my first visit to Kazan and together with other participants in the conference, I visited the monument to Lobachevski. 10.2 Fire in Kazan
In the late summer of 1942, Kazan was hit with a tremendous disaster - fire - and most of the city was destroyed. There was a strong wind. and the fire spread quickly from house to house and the connecting streets up to Kazanka river. The ringing of
278
THE GEOMETRY of SUBMAN1FOLDS
N I LOBACHE% SKI
fire-bells, explosions. people shouting and crying. the unbearable heat the noise of
storm ... People left their houses and tried to save their property. They were in despair. As night fell the dark blue sky was painted with reflections of the fire. The disaster
was exacerbated by the force of the storm and the lire spread to areas nine miles from the city. N. I. Lobache%ski was the rector of the Unnersity and with Musin-Pushkin. the supervisor. organized a plan to protect and sa'e the University from fire. The rector's steel) willpower was pitted against the temf)ing power of the fire. The main University buildings were sa%cd and the students and officials worked hard to protect the library. the anatomy theater, the main University and adjacent building.
The instruments of Astronomic Observatory was carried out into Unicersit} square. The student's library books and equipment from the mineralogy. botany. and zoology departments and other offices were also removed for safety. From the main University library only rarities were carried out while the other books remained. In spite of strong fire around the library building the books remained undamaged. The next morning the fire abated and the people of Kazan were shocked by the
devastation it had casued. About 1300 houses were destroyed together with the University Astronomy Observatory and University residences. including Lobachevski's flat. U. M. Simonov, famous citizen Kazan, was abroad when he learnt of the fire in Kazan. He was a brave and experienced man but tears filled his eyes when he read
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
279
about this disaster in the newspapers. In scientific circles Simonov was more famous than Lobachevski due to his participation, in the capacity of scientist-astronomer, in
the historic Russian round-the-world expedition to the mysterious South Sea in 1819-1821. That expedition of the "Vostok" and "Mirnyj" ships was guided by Bellinsgausen and Lazarev. During the voyage the Antarctic Continent was discovered, in 1820. One of the islands from the Fiji group in the Pacific Ocean was named after Simonov. After that expedition his reputation was enhanced. He corresponded with foreign scientists, regularly visited France, Britain, Austria, and Belgium and became a corresponding member of the Academy of Sciences. Lobachevski did not leave Kazan at that time. He reconstructed the University buildings, gave lectures in various courses, and achieved the discovery of a new geometrical "continent" which remained unrecognized at that time. The expanded account of that discovery, "On origins of geometry", was published in "Kazan University Notices", 1829. Later, he wrote five papers on that topic and
in 1837 his book was translated into French. In 1840, the book Geometric investigations in the theory of parallel lines was published in German. There was also translations into English. That activity played an important role in the international recognition of new ideas. Nevertheless, the respect accorded to Lobachevski was
for different reasons, for other scientific papers and for his term as University rector. His main scientific activity and achievements remained unrecognized by the Academy. However, his papers were carefully studied Gauss and in 1841 he wrote in one of his letters: "I begin to read in Russian with some fluency and benefit. Mr. Knorr send me a short paper in Russian by Lobachevski (Kazan) on parallel lines. Due to this paper and its German translation I am interested to read more papers from this ingenious mathematician." In November, 1842, Gauss proposed that Lobachevski should be awarded the Corresponding Member Diploma of the Konig Gottingen Society. The Diploma
stated that the Society had elected him as a friend and co-member in scientific association, and expressed a hope that he would work collaboratively and share discoveries, notices and thoughts with Society.
In 1843, Lobachevski responded to Gauss and thanked the Konig Gottingen Society for the honor and expressed the desire that all his scientific works be worthy
of the excellent Society. In addition, he wrote: "Please excuse the delay in responding. My health and my personal circumstances have suffered as a result of the fire in Kazan and it was has created many problems for the University." In 1846, Simonov replaced Lobachevski in position of rector, and the government awarded him St. Ann Order first. Lobachevski was rewarded with three Orders much later. But all of the recognition and misunderstandings that Lobachevski suffered in the last years of his life must not undermine the main thing --- the joy of creation that he felt. And time put everything into perspective. It should be remarked that the fate of the Hungarian mathematician Janos Bolyai,
who worked on the same ideas some years later, was much more unhappy. But currently the Hungarian Mathematical Society is named after him.
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THE GEOMETRY OF SUI3MANIFOLDS
103 Eflmov's Theorem on Non-Immersibility of the Lobacbevski Half-Plane into E3
It is well known that a complete Lobachevski plane is not immersible into E3 as a regular surface. This was proved by Hilbert in 1901. Efimov generalized this result in two ways: (I) he proved that a half-plane (i.e. an infinite domain with a complete geodesic line as a side) is not immersible into E3; (2) he stated theorems on nonimmersibility of complete two-dimensional metrics with variable negative Gaussian curvature bounded away from zero: K < -a2 < 0, a = const. In this section we consider Efimov's theorem of the first kind. Theorem I For any a there exists no band in E3 of constant width a located along the complete asymptotic line in a C4-regular surface of constant negative curvature. Suppose that the band from the hypothesis exists. Let -y be a complete asymptotic
line in it. Let y, be a curve in this band of distance a from y. The distance from a point in y, to -y is the length of the shortest line joining the point in y1 to the point in y. Each asymptotic line in the surface belongs to one of two families. In addition, the curve of the first family is distinct from the curve of the second family by the sign of torsion on a surface of Gaussian curvature K; for the first family the torsion is
for the second family the torsion is - vr--K. Let -y be of the second family of asymptotic lines. Emit from 0 E -y the asymptotic of the first family and fix a point A in it such that the length of asymptotic fine arc OA would be smaller than e: OA < C.
Emit from A the asymptotic of the second family. Then it is located in the band. Indeed, suppose that it intersects yI at a point B (see Figure 43). Emit from B an asymptotic of the first family. Suppose that it intersects y at a point C. The arcs OA and OB are opposite sides of an asymptotic quadrangle. Since the net of asymptotic lines
in a surface of constant negative curvature is Chebyshev's one, then
OA = CB < E. However, the length of CB is greater than a because the distance from B to -y is equal to e. So, we come to a contradiction.
FIGURE 43
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
2KI
By means of asymptotic lines, introduce in the surface the coordinates (ir. r) with respect to which the metric has the form d k2 = dl r' + 2 cos w cIu dr + ch=. The Gauss-
Codazzi equations then reduce to a single equation:
' = sin.;:
(1)
The angle w between asymptotic lines must satisfy 0 < w+ < n by means of surface regularity. The first family of asymptotics is represented by u-curves, the second by
r-curves. From what was proved above we can suppose that o, a > 0 in some segment n < it < .I. r = ro = const where a = const. Integrating (1) in r. we get L,, > a for a < ti ,,5 1. r > rp. Let ((i_, .i_) be a segment in (a. d). Let na cr(ir - n) > a(02 - (1) = wu > 0.
w(u, r) - W(ct, r)
Therefore w(u, r) > wy, > 0.
where ' = coast,
(2)
Integrating in it from it up to .3 we get
w,thc>a(d-u)>_a(.1>0 u
where w, = cont. Therefore W(tr, r) < w(3, r) - wl < 7r - wI.
(3)
From (2) and (3) it follows that in a half-band a:! < it < i_, r > rp the inequality sin w > h > 0, where h = const, is satisfied. Integrating both sides of (1) in r. we get
r) -',(ir. rU) =
J
sin'dr > Mr - r0).
Therefore, for any ni there exists r such that w,(u, tip) > in, where a2:5 t t:5 :3±. In this case 0 < w < Tr fails if nt is sufficiently large. The theorem is proved. If we consider a metric ds2 = Edit' + 2Edu dr + G di'2 then the geodesic curvatures
c, of coordinate curves u and r are, respectively, 1
EG - F2
F
E
\T),-I -
EG-F'-Vfir+VGr}
282
THE (GEOMETRY OF SUBMAN[t-OLDS
Therefore, the geodesic curvature of asymptotic lines in a surface of curvature K _ -1 can be expressed in terms of derivatives of net angle j with respect to natural parameters. Namely, NJ = -Vj,,,
n: = W,,.
The integrated geodesic curvature of any asymptotic line are can be estimated from above by ir:
J
h',(iS,I
=
J4idszZ7r.
(4)
We shall use the remark above to prove the next Efimov theorem from [1].
Theorem 2 The Lohaehe rski half-plane does not admit isometric and regular miniersiun into V. Here the regularity assumed is of class Ca while in [21 this condition is lowered to C2. Some lemmas are stated. Lemma I
Let G,, he a simply connected curvilinear polygon of boundary I',, in a
surface of constant negative curvature K= -1. Let r,, consist of smooth arcs Al ...... ,,, where each of the a, is an arc of either an asymptotic or geodesic line. Then the area S(G,,) satisfies the following inequality: S(G,,) N there exists an arc 12 from Bk intersecting 11 at some point B. Consider the asymptotic net quadrangle where k. ni > N (see Figure 45). Fix k while ni -+ oo. As the sequence BI .... , BL.... is divergent in lI and E is complete, the length of the arc BL Bm tends to infinity while ni -' x. However, the length of the arc BABA, is bounded. The arcs B&Bm and B'B,,, are the opposite sides of an asymptotic
quadrangle. By Chebyshev's net property, they must be equal So. we come to a contradiction. Lemma 2 is proved. Let r be a side geodesic line of the half-plane H. Draw in R a complete geodesic line 7 which has a common perpendicular with r. By the property of the Lobachevski plane geodesic lines, the smallest distance between the points of r and , is
THE GFOML1R) (W SUBMANIFOLDS
214
FIGURE. 45
the length of this common perpendicular. So. There exists -- > 0 such that the disk of
radius r and with its center at any point of I is in If. Denote by 1I, a half-plane with , as a side Take an arbitrary closed disk D which is strictly interior to 111. Through the each point of D draw the maximal in III asymptotic of the family (11). In this case all possibilities are exhausted by the following:
(1) the set of all drawn arcs is located in some compact part of the closed half-plane IIi U,)--
(2) among the drawn arcs there exists the arc outgoing to infinity in III U' in both directions; (3) among the drawn arcs there exists the arc outgoing both to infinity in III U -, in one of its directions and to some point A E I in another direction. If the first possibility occurs then there exists a domain G of G type. where n = 2. In this case D C G. The second possibility never occurs because Theorem l implies that 11, would contain an --band along a complete asymptotic line. So. our consideration reduces to the third case. Apply Lemma 2. taking as a one of two rays of the geodesic ; with A as a starting point Denote it by I(,. Moreover, select it in such a way that corresponding to this choice the domain E would include not less than half of the area of disk D. Let S'1 be
a union of the interior of E and all of the points of the ray 1,. Define a mapping ,1': E, - 11 as follows. If x is interior to E then by Lemma 2 it belongs to some set T(B), where B is some point in 11 be some segment [A,y] in the ray 1,. Set 1`(x) = v. If
x E lI then set f (x) = x. The mapping 1 is continuous on Si. Denote by DI the intersection D U E,. As E is closed. D, is a compact set. The mapping f is continuous on D, and the image 1(DI) is a compact set in the ray 11.
From the construction of E it follows that the image of the E interior is not a compact set in 1,. This implies that there exists a point B, E 1i such that D, C T(BI ).
ISOMETRIC IMMERSIONS OF LOBACREVSKI SPACE INTO EUCLIDEAN SPACE
285
FIGURE 46
If T(BI) is a compact set then it is a curvilinear polygon G of G. type, where n = 3,
and in addition D, C G. Suppose that T(B1) is not a compact set. Now consider T(BI) as E, B, as A, /2 from the second family of asymptotic lines as /I. Take an arbitrary point C in 12 and then draw the asymptotic !' of (11) family through C into T(B1). The arc !,, being maximally prolongated in T(BI), must come to rya. This follows from Theorem 1 because if we prolong /1' in the other direction, then across 12 it does not meet either /1 or /2 for the second time and, hence, it goes out to infinity.
Denote by T'(C) a closed part of T(BI) cut off by /' (see Figure 46). Applying Lemma 2 again, we conclude that there exists a point C E 12, such that D, C T'(C). The set T'(C) is a compact curvilinear polygon of type G,,, where n < 4. It contains either a disk D or the part of D, namely D1, of area not less than half the area of D. Thus we come to a contradiction. The theorem is proved. In connection with this theorem the natural problem arises of describing those domains in the Lobachevski plane which can be regularly and isometrically im-
mersed into E3. In Pozniak's paper [3], devoted to the 150th anniversary of Lobachevski geometry, isometrical immersions into E3 of infinite polygons in Lobachevski plane with finite or infinite number of sides but not containing a halfplane were constructed. Here the "infinite polygon" means the intersection of the finite or countable set of closed half-planes with sides - the straight lines in the Lobachevski plane - having no common points. It was proved that any infinite polygon with a finite number of sides can be isometrically immersed into E3. Also, two classes of infinite polygons with infinite numbers of sides were introduced, which can be immersed as well.
Note that the Yugoslavian Mathematician D. Blanusha embedded the Lobachevski plane into E6 while E. Rozendorn immersed it into E5 (see [33], [34]).
THE GEOMETRY OF SUBMANIFOLDS
286
10.4 Local Non-Immersibility of n-Dimensional Lobachevski Space L" into E2 -1
Cartan in papers of 1919-1920 [4] and independently Liber in 1938 [5] proved the following theorem. Theorem The Lobachevski space L" of dimension n can not be immersed even E21-2. locally into (2n - 2)-dimensional Euclidean space
(By the way, it ought be noted that I have never seen [4] and give the citation from [61).
The proof follows from the analysis of the Gauss equations. In the case of space of constant curvature the Riemannian tensor is of the form Rolfld = K(gM g,.* - gnb gih ),
where K means the curvature of space. For the case of L" the curvature K < 0. Consider the solvability problem of the Gauss system P
E(LnLI16 - L.'hLIj,) = K(g.-,g;ra
(I )
n=1
with respect to the p symmetrical tensor a = I , ... , p. The theorem is proved if we show as in [5] that this system has no solutions for
p I.
Evidently, the lemma is true for n = 2. Suppose that La. forms a solution of (1) if n = no and p = po = no - 2. Take any n-dimensional non-zero vector 1 of components 1'1. Set v° = La 1". The vectors va..... vQ are not zero simultaneously because in the opposite case it follows from (1) that 1° ga - g°,1N = 0. The latter implies that 1° = 0 for all a, which is impossible.
Suppose that the system of vectors vrt..... vo contains precisely m which are linearly independent. As we just proved, m > 1. Since po < n - 2, then m < n - 2. By means of the appropriate orthogonal transformation B'", this system can be transformed into a system such that the first p - m vectors will be zero while the other m
are linearly independent. Save the previous notation for the vectors of the new system. So, we obtain the system of vectors with the following properties:
v° =LAP =0 for a= 1,...,po-m v° = L° I' are linearly independent for a > po - m. Consider the (no - m)-dimensional subspace E""-'" in E"° which is orthogonal to each of v°. Let be an arbitrary vector in E"0-". Since v°{° = 0, then contracting (1) with 111f0 we get
E PO
00
Q. t6 - g.6 1.6
(5)
°=Po-m+I
where ub = L' CR. If
96 0 then (5) implies that the metric tensor gb is re-
presented as a sum of m +I dyads. This contradicts Lemma 4 because m + I < no - I + I = no. Hence, we must set 10 CO = 0. Then (5) has the form PO
v°° ub° = K1° s r-Po-m+l
(6)
Let r be an arbitrary vector orthogonal to . Contracting both sides of (6) with T6, we get v° uh Tb = 0.
Since v° are linearly independent, u, Tb = 0, a = po - m + I,. .. ,po. Since T is arbitrary and orthogonal to , ub = This means that
Lxe = A°G,
a =Po - m+ 1,...,m
i.e. f is the eigenvector for each of L. Since 1; is an arbitrary vector from E"°-"', then by Lemma 3 the system (1) is solvable for n = no - m and
p=po-m= (no-m)-2. Setn, =no-m. Since I <m<no-2,2 0 then introduce a linear form .p' _ 11i a, drr. Then .4,' = rA then introduce a linear form pA = V--it a, the. Then A = y7'. The hypothesis of the theorem can be reformulated as follows: let us be given the
linear operators A'.....A" on an n-dimensional vector space I' such that (AIXY) = (XA`)") We can write the condition of external orthogonality as
A'AA"=0. a-I
(2)
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
295
Indeed, consider the value of AAA AA on bivectors [e,e;] generated by the basis vectors e;, ej. We have (A' AA')[e,ei] = [A'e;AAej].
The components of the latter bivector (we mark them with a subscript near the brackets) are [AAe; AAei]kl
_
(AAe;)k I
(AAe,),
(Aae!)k (AAei),
I
L'a
1'Jk
Lu
Lill
Therefore, the system of Gauss equations (1) can really be written as one bivector equation (2). Further, the Lemma hypothesis means that the intersection of kernels of bilinear forms A', ... , A" consists of zero vectors:
nKerAA=0. A
Now show that by means of an orthogonal matrix ([aA.11 we can pass from the system
of operators AA to the system of operators B" of the form
B" =>a. AA. p= 1,...,n, such that B" A Bµ = 0 for each p. Suppose that there exists X E V such that the system of vectors A' (X), ... , A"(X) is linearly independent. Prove it by contradiction. Let X be a fixed vector. Denote by U(X) a space generated by A'(X). A = l .... , n. Let p be a maximum of dimensions of U(X), when X varies in V. Suppose that p < n. If Xo is such a vector that dim U(Xo) = p then without loss of generality we can assume that the subsystem
A'(Xo),...,AP(Xo) is linearly independent while AP" (Xo) = 0, ..., A"(Xo) = 0. If Y is any other vector from V then (2) and the conditions on Xo imply P
EA^(Xo)AA"(Y)=0.
(3)
u=I
If Xo and Y are fixed then we can put the vectors A"(Xo) and A"( Y) we can put into correspondence with the linear forms. The result holds: Lemma of Cartan In order that given linearly independent forms f , ... , fP (the rank matrix of coefficients of these forms is p) satisfy the identity P
A=1
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THE GEOMETRY OF SUBMANIFOLDS
it is necessary and sufficient that each form gyp' is a linear sum of the forms f k with a symmetric matrix of coefficients:
EC,
k
k
k
See the proof in [7] p. 20. So, we have P
A°(Y) = EcnA3(Xo).
(4)
0=1
Thus any vector of the form A° (Y ), a = I,... , p, Y E V, can be expressed in terms
of A1(Xo), 0 = 1, ... p. If we denote by W a subspace generated by A"(Y ), Q = 1, ... , p, Y E V then (4) shows that the dimension of W is p. As p < n, there exists Z E V which is orthogonal to W, i.e. A°(Y, Z) = 0,
a = 1, ... ,p.
But from the properties of A 1, ... , A" it follows that there exist A and N such that AA(N, Z) 0 0,
where, evidently, A > p + 1. If e > 0 is sufficiently small then the vectors A° (cos a Xo + sine N ), I < a < p, generate W. On the other hand Aa (cos a Xo+ sine N) is located beyond W. Indeed, ZA'(coseXo+sin eN) = sin eAA(Z,N) 34 0. Therefore, the subspace generated by
A"(cos e Xo + sine N), µ= 1,...,n is at least (p + 1)-dimensional. This contradicts the definition of p. So, p = n. Let v1,.. . , v,, be a basis in V such that the system
A'(vl),...,A"(v1) is linearly independent. In application to the equation A"(vl) A A"(v') = 0 1i=1
Cartan's lemma implies that there exists a symmetric matrix C(i) = flc(i ) 11 such that
AA(v,) _ Ec(i)NA"(v1),
I < A < n,
1A
where C(1) is a unit matrix. Use the equation A"(vi) A A\(vi) = 0. a=1
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
297
Substituting now the expressions above for AA(v,) and AA(vj), we have
>(c(i )N c(j )"
v2TAavA"(vl) = TABA(vl)
So, every BA(v,) is proportional to BA(vi). Therefore, the bivectors generated by BA(vj)i = 1,...,n are zero: BA(vr) A BA(vj) = 0.
This implies BA A BA = 0.
As was proved above, in this case there exists a one-dimensional form 0A such that e = ftpA (9 V%.
THE GEOMETRY OF SUBMANIFOLDS
298
From (6) we find As:
As = E d BI' _
fag; GPs 0 ips,
where Ild 11 is the matrix inverse to A.
Take ps at the fixed point of an immersed domain as the differentials of new coordinates, i.e. set dria = yoa.
Then each second fundamental form L du' duj is diagonal with respect to a new system of coordinates. Then the theorem is proved. Note that Moore [6] proved the existence of asymptotic coordinates on the immersed domain L" C E2"- 1. 10.6 Lemma on Principle Directions on the Submanifolds of Negative Curvature to be Holonomic
Lemma Let R' be a submanifold of negative curvature in Euclidean space EN. Suppose that at each point in Rthere exist m principal directions. Then they are holonomic.
Denote by T,, a tangent space at x E R"' and by np the normals to RI. Denote by a comma the covariant differentiation in E"'. By definition, the prinicipal direction X E T, is that which satisfies
np.,X'=AX+ayny
(1)
for any np. Denote by X I , .... Xthe field of nr unit principal directions. We say they are holononic if for any X1 there exists a family of hyper-surfaces orthogonal to X,. This hypersurface, then. can be taken as a coordinate surface. To prove the lemma it is necessary and sufficient to prove that for any pair Xk, X; of vectors orthogonal to Xj. the bracket [XkX1] is orthogonal to X. Consider, for instance, [XI, X2] and show that it can be expressed via X, and X2. We have np.r Xj' = Apt X, + > pvp/I nq.
(2)
Let K(XI. Xj) = K, be the curvature of RI with respect to the 2-plane generated by X;, X/. We have K(Xi. Xj) = F, Ap+Apj'
(3)
P
Multiply (2) by Ape and sum over p. Taking into account (3), we obtain Apt np.i X, = KI2 XI + Ape E l4ep/ I ny.
(4)
H
Differentiate equation (4) covariantly in the X2 direction. Thus, we have Ap211p.ij XIX2 + Ap211p.l XI,i XZ = K12 X1.i X4 + ....
(5)
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299
Here and subsequently we denote by dots the linear combinations of X1, X2 and ny. Since np,,, means second covariant derivatives of the vector field np in V. np,, _ ,tp',. So. we can write
n
Xi, _ (11,X2) I X1 -11,X2 r Xi _ (Ap,XX +
,yp.2nq) I
i - np, Y; r X1,
(6)
=Ap2X2,X1, -III, X±IX[+. Substituting (6) into (5). we obtain an equation which we shall use below. Interchanging the roles of XI and X. we get the second equation. Denote by [XI X2}' the j- th component of brackets and set %2 = E ,,(A,,)2. Then the system of equations just obtained can be represented as
K,2XI,X_-X-X',) =EAp2)tpr[XlX2J'+..-. p (7)
K12X2,Xi-X,X;"riEAr, lip, [Y2X,J'+. p
Using system (7). we find that
(Ki,--, - )[XIX2)=(K12-ii)EA, np,[XjX2J, r
+(KI:- )
)EAplnrr[XIX2}'+....
18)
I,
Since RIO' is an integrable submanifold. f X 1 X2J can be expressed in the form of a linear combination of tangent bases. Suppose that [X1 X2[ has a non-zero projection onto XA, k * 1.2. that is [X, X2] = PXA +...,
where p 96 0 and the dots mean the terms without X,1. As XA is a principal direction Ap+np,[X,X2}'=PExMxpr,Z.+...
=pKAXA+ ---
M
Project (8) onto XA. Taking into account (9) and dividing out p y6 0, we get (Ki2 -
i illi ) = K2k(Ki2 -'}i) + KIA(K12 - "r2).
If the curvature of R' With respect to any 2-plane is negative then the right-hand side of the latter equality is positive while the left-hand side is non-positive. Hence, p = 0 and the statement is proved.
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300
10.7 Metric Form with Respect to Curvature Coordinates From the lemma stated in the previous section it follows that in an immersed domain we can introduce an orthogonal system of coordinates uI, ... , u" such that the coordinate curves are tangent to the principal directions. It happens that after scaling the coordinate curves the metric form has a specific form. The following theorem, stated in [8], [9], holds:
Theorem (Aminov) Let a domain in L" be isometrically immersed into E2" Then in this domain we can introduce local coordinates uI..... u such that coordinate curves are tangent to principal directions and the metric obtains the form I
"
sin2 o; d:r.
d52 = r= I
where IT
(I)
sin2o-, = I. r=1
Observe that in E3 the orthogonal systems of coordinates with condition (1) were considered by Gishar and Darboux [10] and in E" by Bianchi [11]. Since at each point of an immersed domain there exist n principal directions, the normal connection is flat. In this case we can choose the normal basis of normal 0. Let vector fields parallel in the normal bundle. With respect to this basis Lp be the coefficients of the second fundamental form with respect to n,,. Since the coordinate curves ui are tangent to the principal directions, Lf = 0 for i 34j. Use, then, the equations of Codazzi for the immersion of L" into E21-I. We have
L°.A-L°ti=0.
(2)
Observe that Lea, = 0 for i 0 k because of the choice of a system of coordinates. Consider (2) in more detail, setting i = j A k: 8Lpa
_ I'LL°,+I';;Lkk =0.
(3)
where rfk are Christoffel symbols. If g;; are metric coefficients then I',k =
2gi; duk'
rk
-
2gkk (JUL
Multiply (3) by Lj° and sum over p. On the left-hand side we obtain
8 duk
E(' " P
)2-
1
agii (Le)2+gjiagii=0. ,i
gii &,,A-
P
(7!!k
(4)
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
101
If we set
h
cot- a,=
() ] /g,,,
_
2
P
then (4) has the form O(ln g,, - In sing a,)
#i
-0
Therefore, by means of a suitable choice of parameter in the it, coordinate curve, we get g,, = sin` a,. Then
E(LF)2 = cos'- a, sin' a,. P
Now we prove (1). Consider the set of g, p = I..... it - 1. where i is fixed, as some vectors of length cos a, sin a, in normal space and project it onto coordinate axes in normal space. Set
p=I.....tt-I; t=1.....n.
(5)
It is evident that the angles y,,n must satisfy n-I
cos3 OW _
(6)
n=1
for any I. Since the curvature of L'" is -1. for any 196 j we have
L! = -g g,, = - stn" a, sin2 a,.
(7)
Using (5) and (7) we obtain n-i
Cosyor Costp,, _ -tang, tan a,.
(8)
P=I
The equalities (6) and (8) express the condition on the n-matrix A to be orthogonal sin al
sin an
COS Sot I COS al
...
cos tat,,-1 cos al
...
Cos.p
cos
cos an cos an
Namely, the product of two distinct columns is zero due to (8) and the square of each column is I due to (6). The orthogonality conditions for rows produce (1).
THE GEOMETRY OF SUBMANIFOLDS
302
10.8 The Fundamental System of Equations for the Immersion of Ln into E2n-' (System "LE")
In consequence we use the notation Hi = sin aj. As ds2 = E" , Hi du; is the metric of the Lobachevski space of curvature -l, the coefficients Hj must satisfy some special system of equations. Let Riikj be the components of the Riemannian tensor of the metric ds2 = E" , H2 du;. It is well known that if dsz = gjjdu; duj is the metric of a Riemannian space of constant curvature Ko then the Riemannian tensor satisfies the relations Rjjk1= Ko (gik gij - gil gik)
For our case we can write these conditions as
Rkyi=0,
i34j;
Ri;ii=Hj2H2,
k3&i.
(1)
In addition, we have (2)
Rewrite (I) in an expanded form while replacing (2) with a system of differential equations. To do this, we use the expressions for a Riemannian tensor with respect to orthogonal coordinates (see (37.4)-(37.6) from Eisenhart's "Riemannian geometry"). In terms of Darboux symbols /3; j = N- 2, i 0j, [3;; = 0, equations(I)can be written as aQ'j+80jj+EQkhOki
auj
auj
k=1
ajki=Q
ji
4 k#i i1 96
kj$j;,
auj
= H,H;, .
Differentiating (2) in uj, we get aHj auj
E
Q%%Hq,
( 3)
q=1.q#J
From the definition of Darboux symbols: aHj = $,; H,, auj
i0j
( 4)
Write the compatibility condition of the system (3)-(4). Differentiating (3) in uj and dividing out Hj # 0, we have a2Hj aujauj
E (10A QiH q
q=1.q#i
+Q. aHg 811i
)
n
(lQi,Qjq H4 + QhQw Hj) - --" Hi + Oilk#i E QtkHk q=1.q#j.i
auj
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
303
In the last term the subscript k can be equal to j. We can write 1l
01r>2/3dH4 = it
>2
/3,rgHH)
v=1.yr!
R#t
Hence r7-H,
= - 1:
d,gdwH, -
du,Oa,
Old''
H, +
(5)
Differentiating (4) in a, implies _q=Hl 011,01t,
= t3r3u if, +
6)
011,
Subtracting (5) from (6). we obtain V3
tad
r3Nd,,, =0.
t1t,,all, +
q-I 4+ !
Thus. we passed from the system (1). (2) to the system with respect to functions If, and 13,,:
(u)
13i,Ht,
r Out 011A
-
(h)
rr
,3xyHy.
q 96 k.
q=I w
{c)
X13 11
u,
= i3 ljA,
(d)
13
Ou14t
+ Ou, +
0.
(7)
,=1
8r3A, _ H,Hh, {e} P++>24s3,t auk 1
where Q k k are distinct from the range 1.... , n. We shall call this system the fundamental ,s wtent of isonnetriak immersion of L" into
E14-1, or in shortened from: the system "LE" (Lobachevsky, Euclid). The former integral of this system is the relation Er,=1 H, = const. Indeed, mul-
tiply (a) by H, and sum over all I. 10 k. Then multiply (b) by Hk and add the expressions obtained in that way. We get
28ak ( !=I
H')
=EJ3k,HkH,->}3tgHtHq =0. 1
q
Due to the choice of initial conditions we can take 1 as the constant in this former integral. Therefore, we can state that (7) is equivalent to the system (1). (2). On the other hand. the fundamental system of equations of isometric immersion of one manifold into another is formed with the equations of Gauss, Codazzi and Ricci. As at each point of the immersed domain there exist tt principal directions. the
304
THE GEOMETRY OF SUBMANIFOLDS
normal connection is flat. Therefore, we can choose the normal basis field parallel in
a normal bundle. In this case the torsion coefficients p,,, , = 0. In addition, with respect to the curvature coordinates the coefficients of the second fundamental forms satisfy L f = 0, i 0 j. Therefore, the Ricci equation is satisfied identically. The Gauss equations are of the form n-1
Rijki = 1(L°&Ljpl - L lLjk).
(8)
P=1
It is known that with respect to the orthogonal coordinate system the Riemannian tensor component is zero if all subscripts Q, k,1 are distinct. On the other hand, the right-hand side of (8) is equal to zero too, because L = 0 for i 36 j. So, if all Q, k, I are distinct then (8) is satisfied. Suppose, now, that two of them, say j and k, coincide while i, j and / are distinct. The right-hand side of (8) stays zero. The equation has the form Rkjii = 0, i.e. it coincides with the second equation of (1). If among Q, k, I there are two pairs of equal subscripts, say i = 1, j = k and i 96 j, then (8) has the form n-1
Rijij = - > Lii4 = sine ai sine aj. p=1
Here we applied (7) from Section 7. Thus, the Gauss equations have the form of system (1). Set'I,,p = cos aj cos Ipjp. Then the expressions from Section 7 for L , can The orthogonal matrix A from Section 7 has the form be written as L, = H1 H1
Hn
A= 4II
I
'In-1
t.-1
...
...
Hn
LI11
LInn
HI
H.
Ln-1
Lnnn-1
11
Hn
HI
We can write the Codazzi equations
6`k-rkL,+r:Lkk =0, i &k as
dip -
=f3ki4Pkp,
k#1.
(9)
Uk
The orthogonality of matrix A implies Ej 1 Hj4bjp = 0. Differentiating the latter relation in u, and using (a), (b) of (7) and (9), we get Qi1Hitlp+IQaHit2p+...+tip(- rl(figHgl
LGg
+HI Oil dip + H2a124bip +
+ Hi
ui
+...
//J1
`0 + -
= 0.
ISOMETRIC IMMERSIONS OF LORACHEYSKI SPACE INTO EUCLIDEAN SPACE
305
Therefore
=0. A
Since for regular immersion H, 94 0, the expression in parentheses above is zero. Thus, fi,,, satisfies the following system: ;1t, 'Pip.
k -A i,
(10)
c
all,
Denote the i-th column of A by a,: H,
oil
u, -
Then (a). (b) of (7) and (10), (11) can be written in the following form OR,
(12)
aq.
+
The compatibility conditions of this system are the equations (c). (d) of (7). Indeed, if k A j 10 k. Then ;3+lr i3!,
8ut8u! Hence.
M= I ,',,,
(tA + A, ,3
Now let i
a,r
02u`
!t
011,0uu
!
r
tt
/. We have
a, - i,!
`d aq +
lII, t!w
!
+
E 13ry aY -
'_ - ` Differentiating
i! J !r! + ;3r r
4
Y.Y#j
;3" uy + 13!r Y Yfr
r
a, + rlrr i3 u,.
E3'y;iry +
=$a, in it,, we get 6Pa, 013#
01110u,
Ou,
U,+l1'lrail u,
Therefore tL
du,J!!,
oil,
+
a, = 0. i,
a)
THE GEOMETRY OF SUBMANIFOLDS
306
by virtue of d) of system (7). Thus, the system (10)-(l 1) together with the Codazzi system has (7) as the compatibility conditions. The system of Gauss-Codazzi-Ricci has (7) as a compatibility condition. Therefore, we shall call system (7) the.fundamental system of isometric immersion of L" into E2"-1. In Section 20 we shall give the matrix form of this system. System (7) serves for the determination of functions H,. If they are found then from (10){11) we find $,p. Then the coefficients of second fundamental forms are defined as LPi = H1 ,p, L,j = 0 i 31j. The initial conditions on d jp can not be taken arbitrarily. Since the system (10)--(l l) contains all the partial derivatives of and moreover the equations are resolved with respect to them, we may take the initial conditions on tip at a single point Po. Show that if the matrix A at Po is orthogonal
then it is orthogonal in all the domain of solvability. Introduce the functions Tjk = (ajak) - 8,k, where (ajak) is a scalar product of aj and ak. If Q, k are distinct then BTjk 0-U,
= Qij(arak) + /3rk(aja,) = f3ijTik + $3,4. Tj,.
Set i = j 34 k. Then BT;k
ou'
- E Qkq (agak) + Aka? _ q
f3Iq Tqk + /3;k TI q#k
Finally, set all subscripts to coincide. Then BT;,
r
_ 8u , = -2 `'f3,q(aiaq) q
2 E l3igTiq q
Thus, we see that Tjk satisfies the homogeneous system of linear differential equations IM BTjk_ Bu,
Bjkr
Tu.,
i+j, k
where B; are the functions of f3,.. If the regular solution of (7) is known then Btj; are definite and regular of the corresponding class. From the theorem of uniqueness for the solution of the system of linear differential equations it follows that if Tjk = 0 at
the initial point then T,. =_ 0. Hence, if A is orthogonal at one point P0 then it is orthogonal in all the domain of solvability of (7), (10)-(l l). 10.9 Gauss and Weingarten Decomposition
A position vector of a submanifold, which corresponds to a given solution of the Gauss-Codazzi-Ricci system, can be found from the system of equations which consists of Gauss decompositions and Weingarten decompositions: k
r,,,,,, = r,j r"k + Ljj no, n,,,, _ -Lijng A r,,,, .
(1)
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307
Remember that torsion coefficients are zero. Consider this system in our case. We have I'u r,,, + L° n,,.
rwu, = I; r,,; +
(2)
kk#i
The Christoffel symbols are
rk=-Hk$ki
ri;=Hji
Let Ti _ as
be unit vectors tangent to coordinate curves. Then (2) can be represented R
8r;
a=-1:OkiTk+IN,n,.
(3)
k=1
Here the summation over a is assumed. For i $ j, we have rau
=rjjru,+rjr.,+E
PUk
k#i.j
But r = 0 for distinct Q, k. In addition
18H; H, Buj '
jTIOHj Hj 0u;
I',j
Therefore, /3;j Tj, i # j. The Weingarten decomposition has the simple form n,,,,, = --Dtt,T. Finally, the system of Gauss and Weingarten decompositions is of the form
Sr;_ 8u;
-
Oki Tk +,k, n,,, k
(4)
uj = A j Ti,
n,
Ti.
The coefficients (3k;, -ti, are known functions of u1.... , up. The compatibility condition for this system gives the system of Gauss-Codazzi-Ricci equations. If we find the solution ri, no of that system then the position vector of the submanifold can be found as r(ui , ... ,
r(Po) + > I
J T;HI du;.
Note that at the initial point Po all the vectors ri(Po), no(Po) are chosen in such a way that they are unit and mutually orthogonal.
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308
10.10 Local Analytic Immersions
Find the arbitrariness in determining the local analytic solution of (7) Section 8. This system belongs to the class of Bourlet-type systems (in Bianchi terminology, see [I I]):
aU, axk = T;k
aUi
UI,...,
Here the system of functions UI, ... , U, forms the required solution. The variable Xk is called main to U; if the system is resolved with respect to All the other variables are called parametric to U,. It is supposed that the right-hand sides contain, maybe, the derivatives aU;/ax, in a parametric variable x, that enters linearly if I > k and if I = k then j > i. We suppose that by virtue of the system the integrability condition r>:u, CPU' x = sk sk is valid with respect to main variables.
If the assumption that'P;k are analytic functions is made the existence theorem holds [2]: in some neighborhood of an initial point (x0,,..., xx) there exists a unique system of analytic functions Ui, ... , U, such that if main variables take the initial values then U, is an arbitrary given analytic function of parametric variables. Apply this theorem to our system. Verify that compatibility conditions are satisfied. System (7) Section 8 is resolved with respect to all derivatives of Hi. The equality of mixed derivatives holds because of (c) and (d). Order the functions /3,;
lexicographically and set /312 = U1, /313 = U2..... The following rule holds: if
(3i;=Uo,/3k,=U1then a B,C'Ck =const. From this follows (3). Denote by 4i the left-hand side of (4) for a = 0. Also, set H = {H;/C2}. From the system we have 1
2 axe
Note that [HHJ = K. Hence,
=(HBH)-(BK).
= 0. Now find the derivative of 4 in x3. Using (6),
we get
28x3
(HFH)+(BT).
Observe that F, and B, are related to each other by found F, = -n;B;/d;. Recalling
the expressions for K and T, it can be found that (HFH) = (KF) = -(BT). So, I = 0. Thus 4 is the first integral. From the theorem just proved follows the next statement. Theorem 2 Each solution of the fundamental system of immersion of L3 into E5 with a hyperplanar Grassmann image of general form is definite over the whole space of parameters uI, u2, u3 and is analytic.
The coordinate change from u; to x; is linear. Therefore, consider the solutions of the system (5)-(6). The right-hand sides of this system are defined for all values of required functions H,, B, and depends on them analytically. However, it is well known that the right-hand sides' regularity does not guarantee the existence of a solution over the whole unbounded domain if the right-hand side is nonlinear with respect to the required solution. But in our case the solution existence is guaranteed by the existence of the first integral (4). Use the following statement proved in [18]. Let f (t, y) be a continuous vector function definite in the closure E of the open (t, y)-set E. Suppose that the Cauchy problem
y, =f(t,Y), Y(to) =Yo, admits the solution over the right maximal interval J. Then either J = [to, oo) or b. J = [to, b), b < oo and (6,y(6)) E 8E or J = [to, 6),6 < oo and [y(t)[ - oo when t
333
THE GLt)61FrRl of SUBAIANIF'ULDS
Apply this statement to each subsystem of (5) and (6). By virtue of the first integral (4) the solution of (5) (6) is bounded. Therefore. for each subsystem only the first possibility is admissible: J = [r(;, x). Analogous statement is true for the left maximal interval. So. the solutions of these systems are defined oNer the whole x2-axis and over the whole x:-axis. We construct the solution of the system (5)--(6) b) means of the solutions of the subsystems. At x±o. xv, we fix the initial conditions t . Ba and find the solution of (5) defined over the whole x_-axis. Then we find the. solutions of (5) at Y _,.x31 as initial conditions. The set of solutions of (6), obtained in this way, depends on the parameter x3, and due to the total integrability of system (5)-(6) it produces the required solution H. B over the whole plane a ^3. v3. The proved theorem does not guarantee that a corresponding immersion of L3 into E` is regular because there the points of H, = 0 and H, < 0 may occur. But for the regular immersion H, > 0 must be true at every point. The existence of singular points of immersion follows from Theorem 3 If the Grassmann image of an immersion of the general form of a complete space L3 into E5 is located in a hyperplane passing through the origin and the first integral of areas equals zero then the immersion has singular points.
Consider system (5). Differentiating the first of these equations in x we get op if =
i).t;
-[KH] - B2H.
(7)
Here we use the condition (HB) = 0 from the hypothesis. Introduce the following
notations: ]q = 1 /Ca - 1/('2, t'_ = I /C2 - 1/(,2. v3 = I /C; - 1/('2. Without loss of generality, assume that C? is the maximal of Cr. Then 1.t > 0, m, < 0. The following two variants are possible: (1) r', > 0. From (7) select the equation on 141:
H+H_i, In this case Him - Hi?I^ > O. (2) vI < 0 From (7) select the equation on H:: t?-H
= -H_(B2 + He=r a
- H= vj ).
Then Hiv3-H12111 > 0.
Consider case (l) because case (2) can be considered in an analogous way.
From the equation 1A= -K. which we represent in coordinate form as -H,HHd it follows that each B, is a monotone function over the v2-axis. Hence, there exists a point x; in the xI-axis such that either over the half-axis (x;. -X.)
the function B; Is greater than some fixed positive or over the half-axis number. Therefore, over this half-axis B2 + H; vi - H? v2 > BB; > 0. where B is a
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
333
positive constant. Since in a half axis we can take a segment of any length however
large with the property that the solution of equation d = -B; j, will have at least two zeros over this segment. then by the Sturm theorem the solution of (8) is zero over the segment. Thus, there exists the point H3 = 0. The submanifold has a singularity at this point. 10.19 isometric Immersion of L3 Into E5 with Hyperplanar Grassmann Image and the Motion of a Rigid Body with Fixed Center of Gravity in the Field of Gravitation
In this section we shall state the relation of the theory of isometric immersion of a Lobachevski space to the known classical problem of mechanics of the motion of a rigid body around a fixed point. If the rigid body is in a central Newton gravitation field then this problem reduces to the solution of the so-called Kirchoff system of six ordinary differential equations. We shall consider this system in the case when the center of gravity of the body coincides with the fixed point:
A 1 = (B - C)(qr - el',) BtY _ (C-A)(pr-F'y>"),
(1)
Crlr = (A - B)(pq -?7') Here A. B and C are the principal moments of inertia of the body, {1.1'. , "} is the unit basis vector of the immovable system of coordinates with respect to the axes of
the body-related moving system of coordinates. { p. q. r} is the vector of instantaneous angular velocity with respect to the moving system of coordinates (see [17D.
Show that if the Grassmann image r3 of a submanifold is located in the hyperplane passing through the origin, i.e. a = 0. then the corresponding system of 12 differential equations (5)-(6) Section 17 contains a subsystem of six equations coinciding with the Kirchoff system (1). This relation between the theory of isometrical immersion and the problem of mechanics shows, on the one hand, the natural, from the physical point of view. character of the many-dimensional fundamental system of immersion of L" into El- 1 and, on the other hand, allows one to apply the methods developed in mechanics to the construction of solutions of an immersion system. Consider the system (5)-(6) from the previous section. Note that d,F, = -n,B,. In
consequence we suppose that it, 0 0. We can always reach this condition by the choice of vector in. Set A, = d/n,. Consider system (6). Write, for instance, the component of OH/8.r3 :
aB3 =
-AIF9 = (Ay - A3)F2F3 + 1
'H3
C:! C3
THE GEOMETRY OF SUBMANIFOLDS
334
Show that e(A2 - A3) _ -n, ICZC3, where e =,';d and A= (md). Indeed, consider the vector product [nd] = [[mc]d] = AC - m(cd).
As (cd) = 0, then [nd] = AC. In a particular case, n2d3 - n3d2 = ACi. Hence, A3 - A2 = ni/eC2C3. So, we can write Al
OF1
8x3
= (A3 - A2)(F2F3 - eH2H3).
In an analogous way we can represent the other two equations of O. Therefore, system (6) Section 17 can be represented as
8H 8x3
= [FH],
Ai 8F' 8x3
_ (Ak - AJ)(FjFk -
"'"z"'
where e = c,c2c,n is the constant number. If e > 0 then these equations coincide up
to notations with the equations of motion of a rigid body having the moment of inertia A, around the center of gravity in the central Newton field of gravitation (1) (see [17]). In this case F is the vector of instantaneous angular velocity, H is the unit basis vector of an immovable axis having components H, with respect to the moving system of coordinates. The time parameter t = x3. So, if there exists the general case immersion of L3 into E5 with a Grassmann image located in a hyperplane E9 passing through the origin then H and B along x3 describe the motion of a rigid body. In [ 171 the value of a is 3g/R, where g is the acceleration of gravity a distance R from the center of gravity. Four of the first integrals (3H4) Section 18 for a = 0 turn into linear combinations of four known first integrals of system (1) (see [17]). In our case they are the first integrals of a wider system in partial derivatives consisting of 12 equations. The moments of inertia A; of a rigid body are positive numbers satisfying the triangle inequality A; + Aj > Ak. In our consideration the values A; = d;/n; satisfy the condition above with some values of C; and m;. In this case e > 0, also. Moreover, for any three positive numbers A; not equal to each other there can be found three distinct numbers C? 34 0 and three numbers mi such that Aj = d;/ni, (md) 4 0 and e > 0. Therefore, any motion of a rigid body with distinct moments of inertia A, around the center of gravity of the body in a central gravity field can be mapped to some isometrical immersion of L3 into E5. The Kirchhoff equations have been considered in the papers of Clebsch, Steklov, Chaplygin and others in connection with the study of rigid body motion inertia in an unbounded ideal liquid. They are intensively studied nowadays because of their importance in mechanics. The quadratures of (1) have been found in papers by Weber and Chaplygin under
the condition that (HB) = 0, i.e. when the integral of areas is zero. Without this condition, the solution in implicit and complicated quadratures was obtained by Cobb. The explicit solutions are known for some particular cases, for example for the Steklov case, see [17]. This solution can be used to construct the immersion system solution, see [16].
ISOMETRIC IMMERSIONS OF LOBACHEVSKI SPACE INTO EUCLIDEAN SPACE
335
10.20 Matrix Fonn of Fundamental Equations of Immersion of L" into E2`1
The system "LE" - the fundamental system (7) Section 8 of immersion of L" into E21-1 - can be represented in a brief form in terms of matrices and vectors. Introduce an n-dimensional vector H and the matrices Aj of order n as follows: 0
...
On
...
0
...
Qi"
...
0
Hl
H=
.
Ai = I -/iii
I,
... -/3i"
1,
H. 0
...
The matrix Aj is skew-symmetric, the i-th column consists of elements Q;j and the i-th row consists of elements -/3; j. The others are zero. Introduce the matrix T of order n as T = IIH;HjIj. This matrix consists of products
H;Hj in position (Q). ). In addition, we use the constant matrices Ei j with I in the (i, j )-position and 0 in all other positions. The fundamental system of immersion of L" into E2n' I can be represented in matrix form as follows: OH 8ai
8A; 8uj
AiH,
- A _ [AjA,J,
(1')
8u,
(1)
18A; E11 8u;
8Aj
I Eij=O, 1