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