A Basis that Reduces to Zero as many Curvature Components as Possible

Abh. Math. Sem. Univ. Hamburg 61 (1991), 243-248

A Basis that Reduces to Zero as many Curvature C o m p o n e n t s as Possible By R. KLINGER

1

Introduction

Some calculations in Riemannian geometry would become much simpler if a basis in the tangent space could be chosen such that as many curvature components as possible vanish. In 1956 S.S. CHERN ([2]) simplified the discussion of the Gauss-Bonnet integrand by constructing a suitable basis in the tangent space TmM of an n-dimensional Riemannian manifold M. Thus at least 3 ( n - 2) independent curvature components of the Riemannian tensor R~bcd could be reduced to zero. The object of this paper is to improve on CHERN'S achievement and to show that i) for n > 4 it is possible to increase this number (for 3 < n < 4 the CHERN result is optimal) ii) moreover, it is not possible to choose a better basis for a general curvature tensor, where even more curvature components vanish.

2

The Construction of the Basis

In an arbitrary orthonormal coordinate system we rotate (action of SO(n)) the tangent vectors of the coordinate lines in TraM so that as many curvature components Rabcd as possible vanish. Since SO(n)is an (2)-parameter group, at best (2) independent components could be reduced to zero. That there actually exists such a basis will be shown in the proof of the theorem below. Let 9 V :=

T,,M

and V A2 be the space of bivectors

9 (el,...,en) be an orthonormal basis in V 9 g be a metric in V, and gA2 be the metric in V A2 with

gA2(ea A eb, ec A ed) := det ( g(ea, ec) g(ea, ed) ) g(eb, ec) g(eb, ed) "

244

R. Klinger

The curvature tensor R induces the curvature transformation VA2

e. A eb

g~

,

,

V A2

RabrSerA es

(r < s).

Because o f R~bcd = gA2(R(e. A eb), ec A ed) the sectional curvature o f the plane <e.,eb> (linear span o f ea and eb) is

k(<ea, eb>) : = k(ea, eb) = gA2(R(ea A eb),ea A eb) = Rabab =" kab. The case n = 2 is an exception. In the following we will only consider the case n>3. The following lemma shows how three o r t h o n o r m a l vectors (ea, eb, ec) are to be chosen so that certain c o m p o n e n t s o f the curvature tensor vanish. L e m m a ([1]). Let {ea, eb, e~} be a subset o f an orthonormal basis. k(e., cos cr eb + sin e ec) be stationary for ct = O. Then Rob,c = O.

ea

Let

~ eb

A plane rotates a r o u n d eo; k(ea, cos o~eb q- sin aec) is stationary for ~ = 0.

Proof. Set f(~)

:=

k(ea, COS ~ eb + sin ~ ec)

=

g^2(R(ea A (cosaeb + sinaec)),e~ A (cosaeb + sinaec))

=

COS2 0~ Rabab"-[-sin 2 ~ Racac + 2 sin c~cos ct Rabac.

F r o m this follows

df

0 = ~ 1 ~ = 0 = 2Rabac.

[]

Theorem. Let Rabcd with a < c, b < d, a < b, c < d, be the independent compo-

nents o f the curvature tensor. I f n > 3, there is an orthonormal basis (el .... , en) o f the tangent space TraM where at least the following (~) (independent) components vanish: Rtil)

for

2 < i < j <_ n ,

R122j

for

3 <_ j <_ n,

R1323.

In this basis the (symmetric) matrix of the curvature transformation has the following zeros above the diagonal:

A Basis that Reduces to Zero as many Curvature Components as Possible 12 12 13 14

13

*'-,~ *"~

14

15

. . . l ( n - 1 ) ln

23

24

"'"

2n

0

0

'"

0

0

[0

0

"'"

O[

0

"'"

0

0

*",,,,O

"'"

0

0

..

!

i

245

...(n-1)n

15

l ( n - 1) In 23

*

24

*

2n

*

( n - 1)n

*

-Rlilj for 2 <_ i < j < n are the components in the triangle, R122j for 3 < j <_ n those in the rectangle. Proof a) In the following step-by-step construction o f (~t . . . . . ~n) we i) normalize all the vectors ~k for k = 1. . . . . n ii) orient (~1. . . . . 3,) so that (~i . . . . . 3,) and (el . . . . . e,) have the same orientation.

1.Step: We determine a plane El2 := < e l , e 2 > so that the sectional curvature k(E12)

becomes maximal.

2.Step: We consider all the planes which are spanned by a vector x E El2 and a vector y _1_E12, and choose x -- ~1 and y = ~3 so that k(~1,~3) Then we choose ~2 in E12 so that

becomes maximal.

e2 -J- el.

~2

The first three vectors in a Chern-basis

246

R. Klinger

In the i-th step we choose ~i+1 for 3 < i < n complement o f < ~ 1 , . . . , ~ i > so that

2 from the orthogonal

k(~l, ~i+1) becomes maximal. Continuing in this way, after n - 2

steps, we get the desired o r t h o n o r m a l basis

(el . . . . . en).

N o w we apply the lemma: F r o m the choice o f ~1 and e2 follows (zero entries in row one): k(el, c o s ~ e2 7t- sin ~ ~j)) is stationary for ~ = 0 =>

RlZlj = 0 for 3 _< j _< n,

k(-~2, cos a ~1 + sin ~ ~j)) is stationary for a = 0 ::=>

RI22j = 0 for 3 < j _< n.

F r o m the choice o f ~3 follows (zero entries in row two)" k(~l,cos c~3 + sin a~j)) is stationary for ~ -- 0 R131j = 0 for 4 < j < n, k(~3,cos c~1 + sin ~ 2 ) ) is stationary for ~ = 0 =:~

R3132 = 0 .

F r o m the choice o f ~i for 4 < i < n - 1 follows (zero entries in the rows three to n - 2): k(~l, cos ~ ~ + sin c~~j)) is stationary for ~ = 0

=r

Rlilj = 0 for 4 < i < j < n.

b) Since the components that vanish in this basis have two equal indices, they are not involved in any nontrivial Bianchi identities and hence are independent. []

Definition. (el . . . . . en) is a Chern-basis .r i) (el . . . . . en) is an o r t h o n o r m a l basis in T,,M ii) the following curvature c o m p o n e n t s vanish:

Rlilj for 2 < i < j < n , R122j for 3 < j < n, R1323 9

Remark.

i) For a space o f constant curvature any o r t h o n o r m a l basis in the tangent space is a Chern-basis. ii) In general a Chern-basis is uniquely determined up to a finite group o f reflections. Corollary. g has signature (n+, n_), n+ > 2, n_ > 2. Then it is possible to reduce to zero at least (n2+)+ ( 2 ) curvature components.

A Basis that Reduces to Zero as many Curvature Components as Possible

247

Proposition. In general it is not possible to find a basis where more than (2)

curvature components vanish. Proof. Let N - n2(n2 - 1) 12 The curvature tensors R = (Rabcd) are represented by points of RN. We consider the SO(n)-action in ]Ru given by the tensor transformation of R induced by the action of SO(n) in the tangent space. From the theorem above we know that on each SO(n)-orbit in p N there is a point with at least (2) zeros. If there were a point with (2) § 1 zeros on each SO(n)-orbit, this would mean a contradiction: Let W be the subset of all the points of RN with at least (2) + 1 zeros. (W is a finite union of subspaces of IRN). Then dim W = N' := N -

(

Since R u would be the union R N = [,.J SO(n)-orbit through w wEW

and since the orbits are differentiable manifolds of dimension < (2) we get dimRN = dim{ [,.J S O ( n ) - o r b i t t h r o u g h w } < N ' + ( ~ ) = N - 1 wEW

[]

which is a contradiction.

Remark.

is the number of independent scalar invariants which can be formed from the components of the metric tensor and its derivatives up to order two ([3], [4]). Thus the remaining N - (2) non-zero components of a generic curvature tensor R are local coordinates for the scalar invariants.

References [1] R.L. BISHOPand S.G. GOLDBERG,Some Implications of the Generalized GaussBonnet Theorem, Trans. Amer. Math. Soc. 112, 508-535. [2] S.S. CHERN, On the Curvature and Characteristic Classes of a Riemannian Manifold, Abh. Math. Sem. Univ. Hamburg 20, 117-126. [3] C.N. HASKINS,On the Invariants of Quadratic Differential Forms, Trans. Amer. Math. Soc. 3, 71-91.

248

R. Klinger

[4] R. KLINGER, Differentialinvarianten und Symmetrieklassen von Tensoren, Dissertation ETH Ziirich, 1990.

Eingegangen am: 20.03.1991

Author's address: Schweiz.

Ruedi Klinger, K~iferholzstrasse 254, CH-8046 Ziirich,