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Euclidean, hyperbolic, discrete, convex, coarse geometry, metric spaces, comparisons in Riemannian geometry, symmetric spaces.

6 votes
1 answer
746 views

Does every ‘curvature’ tensor induce a metric? [duplicate]

So we know that given a Riemannian manifold $(M,g)$ we can calculate its Riemannian curvature $R$. This covariant $4$-tensor field then satisfies some important symmetries \begin{gather*} R_{ijkl} = - …
Matthew Lou's user avatar