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6 votes
1 answer
766 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} = - ...
2 votes
1 answer
137 views

Mean Gaussian curvature from non-unit vector

Pg.248 of "Textbook in Tensor Calculus and Differential Geometry" by Prasun Nayak. Let us suppose that $\lambda_{h|}^i$ is not a unit vector and therefore, the mean curvature $M_h$ in this ...
1 vote
0 answers
922 views

On a Riemannian manifold, calculate the metric from the distance [closed]

Given a Riemannian manifold $(M,g)$ is it possible to calculate the distance between two points on this manifold. Is it possible the inverse? That means: given a formula of the distance, for example: ...
10 votes
2 answers
1k views

Does the hyperdeterminant calculate a quantity akin to the volume of a parallelepiped?

If $M$ is an $n \times n$ matrix, $|\det(M)|$ is the volume of the $n$-dimensional parallelepiped spanned by the column vectors of $M$.                  ...