PlanetMath: sectional curvature determines Riemann curvature tensor

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sectional curvature determines Riemann curvature tensor

(Theorem)

Theorem 1 The sectional curvature operator $\Pi\mapsto K(\Pi)$ completely determines the Riemann curvature tensor.

In fact, a more general result is true. Recall the Riemann -curvature tensor $R\colon TM\otimes TM\otimes TM\to TM$ satisfies

$\displaystyle (x,y,z,t)+(y,z,x,t)+(z,x,y,t)$	$\displaystyle =0$ First Bianchi identity	(1)
$\displaystyle (x,y,z,t)+(y,x,z,t)$	$\displaystyle =0$	(2)
$\displaystyle (x,y,z,t)-(z,t,x,y)$	$\displaystyle =0,$	(3)

where

, and the sectional curvature is defined by

$\displaystyle K(\Pi={\mathrm{span}}\{x,y\})=\frac{g(R(x,y,x),y)}{g(x,x)g(y,y)-g(x,y)^2}.$

(4)

Thus Theorem 1 is implied by

Theorem 2 Let be a real inner product space, with inner product $\langle-,-\rangle$ . Let and be linear maps $V^{\otimes 3}\to V$ . Suppose and satisfies

Equations (1),(2),(3), and
$K(\sigma)=K'(\sigma)$ for all -planes $\sigma$ , where are defined by (4) using $\langle -,-\rangle$ in place of .

Then .

Write

$\displaystyle (x,y,z,t)$	$\displaystyle :=\langle R(x,y,z),t\rangle$
$\displaystyle (x,y,z,t)'$	$\displaystyle :=\langle R'(x,y,z),t\rangle.$

Proof. [Proof of Theorem 2] We need to prove, for all $x,y,z,t\in V$ ,

$\displaystyle (x,y,z,t)=(x,y,z,t)'.$

From , we get for all $x,y\in V$ . The first step is to use polarization identity to change this quadratic form (in ) into its associated symmetric bilinear form. Expand and use (3), we get

$\displaystyle (x,y,x,y)+2(x,y,z,y)+(z,y,z,y)=(x,y,x,y)'+2(x,y,z,y)'+(z,y,z,y)'.$

for all $x,y,z\in V$ .

Unfortunately, the form is not symmetric in and , so we need to work harder. Expand , we get

$\displaystyle (x,y,z,t)+(x,t,z,y)=(x,y,z,t)'+(x,t,z,y)'.$

Now use (2) and (3), we get

$\displaystyle (x,y,z,t)-(x,y,z,t)'$	$\displaystyle =(x,t,z,y)'-(x,t,z,y)$
	$\displaystyle =(z,y,x,t)'-(z,y,x,t)$
	$\displaystyle =(y,z,x,t)-(y,z,x,t)'.$

is invariant under cyclic permutation of

. But the cyclic sum is zero by (1). So

$\displaystyle (x,y,z,t)=(x,y,z,t)'\quad\forall x,y,z,t\in V$

as desired. $\qedsymbol$

"sectional curvature determines Riemann curvature tensor" is owned by kerwinhui.

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Cross-references: sum, cyclic, cyclic permutation, invariant, symmetric, expand, symmetric bilinear form, quadratic form, polarization identity, equations, linear maps, inner product, inner product space, real, tensor, Riemann curvature tensor, operator, sectional curvature

This is version 7 of sectional curvature determines Riemann curvature tensor, born on 2006-05-22, modified 2007-05-17.
Object id is 7924, canonical name is SectionalCurvatureDeterminesRiemannCurvatureTensor.
Accessed 1474 times total.

Classification:

AMS MSC:	53B20 (Differential geometry :: Local differential geometry :: Local Riemannian geometry)
	53B21 (Differential geometry :: Local differential geometry :: Methods of Riemannian geometry)

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