sectional curvature determines Riemann curvature tensor
Theorem 1.
The sectional curvature operator Π↦K(Π) completely determines the Riemann curvature tensor
.
In fact, a more general result is true. Recall the Riemann (1,3)-curvature tensor R:TM⊗TM⊗TM→TM satisfies
(x,y,z,t)+(y,z,x,t)+(z,x,y,t) | =0 First Bianchi identity | (1) | ||
(x,y,z,t)+(y,x,z,t) | =0 | (2) | ||
(x,y,z,t)-(z,t,x,y) | =0, | (3) |
where (x,y,z,t):=, and the sectional curvature is defined by
(4) |
Thus Theorem 1 is implied by
Theorem 2.
Let be a real inner product space, with inner product
. Let and be linear maps . Suppose and satisfies
- •
-
•
for all -planes , where are defined by (4) using in of .
Then .
Write
Proof of Theorem 2.
We need to prove, for all ,
From , we get for all . 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
So for all .
Title | sectional curvature determines Riemann curvature tensor |
---|---|
Canonical name | SectionalCurvatureDeterminesRiemannCurvatureTensor |
Date of creation | 2013-03-22 15:55:09 |
Last modified on | 2013-03-22 15:55:09 |
Owner | kerwinhui (11200) |
Last modified by | kerwinhui (11200) |
Numerical id | 10 |
Author | kerwinhui (11200) |
Entry type | Theorem |
Classification | msc 53B21 |
Classification | msc 53B20 |