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What is the rank of the matrix of a reimannienne metric ?

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- Thread starter math6
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- #1

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What is the rank of the matrix of a reimannienne metric ?

- #2

quasar987

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It is of full rank: if the manifold is of dimension n, then the rank is n.

This is because a riemannian metric g, when evaluated at a point p of the manifold M, is a positive definite symmetric bilinear form g_p(.,.)on TpM. In particular, its matrix can be diagonalized: there is a basis e_1,...,e_n of T_pM such that g(e_i,e_j) = 0 if i and j are different. Then the rank is equal to the number of nonzero diagonal element. Suppose the ith diagonal element is 0. Then g_p(e_i,e_i) = 0, violating the positive definiteness of g_p.

More generally, a bilinear form has maximal rank iff it is nondegenerate.

- #3

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Therefore the rank of the fundamental metric tensor Finslérienne has rank n-1?

- #4

quasar987

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I don't know enough about Finslérienne géometry to answer that, sorry.

- #5

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the fundamental tensor g is also a symmetric bilinear form defined poitive USUALLY so it will rank n, I find no reason why it is of rank n-1, is a property called strong convexity

- #6

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the fact that g is positive definite on TM \ {0} can not give a solution to the fact that the rank is (n-1)?

- #7

quasar987

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For a bilinear form B:V x V --> R, positive definiteness on V-0 or on V is the same thing since positive definiteness is a property that concerns vectors in V-0.

- #8

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oki thnx in all case i will try to found the result :)

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