# Is the inverse of the metric tensor such that $g_{ij}g^{jk}=\delta_{i}^k$?

The Minkowski metric tensor have the relation $$\eta_{ij} \eta^{jk}=\delta_i {^k}$$. That is the inverse of the Minkowski matrix is the matrix itself.
Analogously, is it true that $$g_{ij} g^{jk}=\delta_i {^k}$$, where $$g_{ij}$$ is the metric tensor in a curved space? If yes how to prove this? I came up with the confusion while finding the chrischoffel symbol I came up with a equation $$2 \Gamma^{\gamma}_{\alpha j} g_{i \gamma}=g_{ij,\alpha}+g_{\alpha i,j}-g_{j\alpha,i} .$$ To eliminate $$g_{i\gamma}$$ I have to find the inverse of the metric tensor in tensor notation. Can anyone suggest how would I solve this problem?

The matrix $$(g^{-1})^{\mu\nu}$$ of components of the inverse metric tensor field$$^1$$ is not necessarily equal to the matrix $$g_{\mu\nu}$$ of components of the metric tensor field, if that's what you're asking.

However, there is a standard notational shorthand convention to write $$(g^{-1})^{\mu\nu}$$ as $$g^{\mu\nu}$$ because the upper indices already indicate that we're talking about the inverse metric.

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$$^1$$ The metric tensor field is a covariant symmetric $$(0,2)$$ tensor field, while the inverse metric tensor field is a contravariant symmetric $$(2,0)$$ tensor field.

• 1. $g^{j m}$ in the equation $\Gamma$: $\Gamma^k_{li}=\frac{1}{2}g^{jm}(\frac{\partial g_{ij}}{\partial q^l}+\frac{\partial g_{lj}}{\partial q^i}-\frac{\partial g_{il}}{\partial q^j})$ of Christoffel symbol is the component of the inverse of the matrix representing the metric tensor. Is it right? 2. But it is it a covariant metric tensor too? Mar 23, 2019 at 14:14
• 1. Yes. 2. No, it is instead contravariant. Mar 23, 2019 at 17:03
• What does (0, 2) tensor field and (2, 0) tensor field mean? Jan 26, 2021 at 13:54
• Hi @HelloGoodbye: Plainly speaking, it counts the number of upper and lower indices of the tensor components. Jan 26, 2021 at 13:57
• @HelloGoodbye: E.g. $g_{\mu\nu}=2\eta_{\mu\nu}$. Feb 17, 2021 at 23:15

Even in flat spacetime, the metric does not have to have the form $$\operatorname{diag}(1,-1,-1,-1)$$. For example, if you want to do physics in SI units rather than natural relativistic units where $$c=1$$, then you want something like $$g=\operatorname{diag}(c,-1,-1,-1)$$. This is not the same as its own inverse.

• And if you use anything else than Cartesian coordinates, the diagonal may look different, or the off-diagonal elements may even be non-zero if the grid axes are not orthogonal. I don't know why you would use such a coordinate system, especially in flat spacetime; I'm just saying that you can, and if you do it will cause the metric tensor to look different. Jan 26, 2021 at 14:00

I'm surprised this hasn't been mentioned before, but$$g_{ij}g^{jk}=g_{i\mu}g^{\mu k}-g_{i0}g^{0k}=\delta_i^k-g_{i0}g^{0k}.$$