Frédéric Wang Subscribe   About Me   Blog Archive   Mathematics   Computer Science

Exercise Curvature

In a 3-dimensional riemannian manifold, the curvature tensor is determined by the Ricci curvature.

Proof:

Let G 1 = ( g 11 g 12 g 13 g 12 g 22 g 23 g 13 g 23 g 33 ) denote the inverse matrix of the riemannian metric, R i , j , k , l the coefficients of the curvature tensor and R i , j those of the Ricci curvature. Using symmetries of the curvature tensor, we have R i , i , k , l = R i , i , k , l = 0 , R i , j , k , k = R i , j , k , k = 0 . Moreover, these symmetries show that it is only determined by the R i , j , k , l for i < j , k < l , ( i , j ) ( k , l ) . An explicit enumeration of these coefficients is:

Since the Ricci curvature is symmetric, the problem is equivalent to showing that the previous coefficient are determined by the R i , j for i j . Using the expression

R i , j = k , m g k , m R k , i , j , m

and all the symmetries, we get a linear system of size 6:

( R 1,1 R 1,2 R 1,3 R 2,2 R 2,3 R 3,3 ) = ( g 22 2 g 23 0 g 33 0 0 g 12 g 13 g 23 0 g 33 0 0 g 12 g 22 g 13 g 23 0 g 11 0 2 g 13 0 0 g 33 0 g 11 g 12 0 g 13 g 23 0 0 0 g 11 2 g 12 g 22 ) A ( R 1,2,1,2 R 1,2,1,3 R 1,2,2,3 R 1,3,1,3 R 1,3,2,3 R 2,3,2,3 )

(note: in the matrix A , the "minus" come from the symmetries where we swap the two first/last coefficients of the curvature tensor. The "2"s come from the sum of two symmetric coefficients of G 1 )

Hence it suffices to show that this system is inversible to prove the expected result. Asking Maxima to compute the determinant of this system, we get the following result:

(%i1) factor(determinant(matrix(
 [-g22, -2*g23,     0, -g33,      0,    0], 
 [ g12,    g13,  -g23,    0,   -g33,    0], 
 [   0,    g12,   g22,  g13,    g23,    0], 
 [-g11,      0, 2*g13,    0,      0, -g33], 
 [   0,   -g11,  -g12,    0,    g13,  g23], 
 [   0,      0,     0, -g11, -2*g12, -g22])));

(%o1) -2*(g11*g22*g33-g12^2*g33-g11*g23^2+2*g12*g13*g23-g13^2*g22)^2

But we recognize the determinant of G 1 , as given by the Sarrus' rule. So det A = 2 ( det G 1 ) 2 0 . □