291: Coordinates Matrix of Inverse Riemannian Metric Is Inverse of Coordinates Matrix of Riemannian Metric

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A description/proof of that coordinates matrix of inverse Riemannian metric is inverse of coordinates matrix of Riemannian metric

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Topics

About: Riemannian manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of Riemannian metric.
• The reader knows a definition of inverse Riemannian metric.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold, any Riemann metric on the manifold, and any chart on the manifold, the coordinates matrix of the inverse Riemann metric is the inverse of the coordinates matrix of the Riemann metric.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any topological space, $T$, any Riemann metric, $g$, on $T$, and any chart on $T$, the coordinates matrix, ${g^{-1}}^{ij}$, of the inverse Riemann metric is the inverse of the coordinates matrix, $g^{ij}$, of $g$, which is ${g^{-1}}^{ij}{g}_{jk}={\delta }_k^i$.

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2: Proof

$g(V,\bullet )$ is a covectors field where $V$ is any vectors field. $g^{-1}(\mathrm{\Omega },g(V,\bullet ))=\mathrm{\Omega }(b^{-1}(g(V,\bullet )))=\mathrm{\Omega }(V)$ where $\mathrm{\Omega }$ is any covectors field, because $b(V)(\bullet )=g(V,\bullet )$ and $b^{-1}(b(V)(\bullet ))=V=b^{-1}(g(V,\bullet ))$. Let us put $\mathrm{\Omega }={ϵ}^i$ and $V=e_j$ there where $e_j$ is the j-th member of the standard basis on the chart open set and ${ϵ}^i$ is the i-th member of the dual basis. $g^{-1}({ϵ}^i,g(e_j,\bullet ))={ϵ}^i(e_j)={\delta }_j^i$. $g(e_j,\bullet )={\sigma }_{j-k}{ϵ}_k$ where ${\sigma }_{j-k}$s are some functions on the chart open set. Let us put $\bullet =e_l$ there. $g(e_j,e_l)=g_{jl}={\sigma }_{j-k}{ϵ}_k(e_l)={\sigma }_{j-k}{\delta }_l^k={\sigma }_{j-l}$, so, $g(e_j,\bullet )=g_{jk}{ϵ}^k$. $g^{-1}({ϵ}^i,g_{jk}{ϵ}^k)={g^{-1}}^{ik}{g}_{jk}={g^{-1}}^{ik}{g}_{kj}={\delta }_j^i$.

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References

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