Exercise Curvature
In a 3-dimensional riemannian manifold, the curvature tensor is determined by the Ricci curvature.
Proof:
Let denote the inverse matrix of the riemannian metric, the coefficients of the curvature tensor and those of the Ricci curvature. Using symmetries of the curvature tensor, we have , . Moreover, these symmetries show that it is only determined by the for . An explicit enumeration of these coefficients is:
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- :
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Since the Ricci curvature is symmetric, the problem is equivalent to showing that the previous coefficient are determined by the for . Using the expression
and all the symmetries, we get a linear system of size 6:
(note: in the matrix , 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 )
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 , as given by the Sarrus' rule. So . □