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I was introduced to geometry of curved space in General Relativity but now I am more interested in learning more about it in general.The tensors which describe a curved 3D space can be represented in the form of a matrix A.The eleements of the main diagonal of $A$ give the distance conversion of each dimension and $A[i,j] , j \ne i$ give the skew angle between corresponding dimensions.Suppose we get the inverse of A $A^{-1}$ .Which is the relationship between the curved space described by $A$ and $A^{-1}$?

Cerise
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    Welcome to MSE! <> To a mathematician, a metric isn't a matrix or matrix-valued function, but a smooth quadratic form field on some manifold $M$. Such a thing is represented by a matrix-valued function with respect to a local frame, but transforms differently from a linear transformation under change of frame. It's not immediately clear to me that this notion of inversion is well-defined (independent of frame/choice of coordinates). <> That aside, there is little (as in, no) geometric relationship between the respective curvatures, even if we work in a single coordinate system. – Andrew D. Hwang Aug 26 '23 at 16:02
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    @AndrewD.Hwang sorry.As I have said I come from a physics background so my terminology may be a bit off. – Cerise Aug 26 '23 at 19:21
  • No need to apologize; just pointing out a potential source of confusion with the use of matrix in this context, and the possibly-surprising fact that the question might not strictly make sense depending how we interpret it mathematically. :) – Andrew D. Hwang Aug 26 '23 at 20:44

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As evidence for the sweeping comment "There is little (as in no) geometric relationship between the respective curvatures, even if we work in a single coordinate system": Fix a smooth, positive function $\varphi$ of one variable, and consider the associated Clairaut metric $$ g_{\varphi} = \frac{1}{\varphi(u)}\, du^{2} + \varphi(u)\, dv^{2} $$ in the $(u, v)$-plane. The proposed "inverse" metric is $$ g_{\varphi}^{-1} = \varphi(u)\, du^{2} + \frac{1}{\varphi(u)}\, dv^{2}. $$ The Gaussian curvature of $g$ has strikingly simple formula $$ K_{g} = -\tfrac{1}{2}\varphi''(u). $$ Consequently, the inverse metric has Gaussian curvature $$ K_{g^{-1}} = -\biggl[\frac{1}{2\varphi}\biggr]''(u) = \frac{\varphi(u)\varphi''(u) - \varphi'(u)^{2}}{2\varphi(u)^{2}}. $$ For example, letting $k$ be a constant and taking $\varphi(u) = 1 + ku^{2}$ gives a metric of constant curvature $-k$, but the inverse metric need not even have curvature of constant sign.

Andrew D. Hwang
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