Given a (2,0)-tensor $g$, we can express it as $g_{ij} du^i \otimes du^j$ or $\tilde{g}_{\alpha\beta} dv^\alpha \otimes dv^\beta$ on the overlap of two coordinate patches. If I am not mistaken, we can define the inverse metric as $g^{ij} \frac{\partial}{\partial u^i} \otimes \frac{\partial}{\partial u^j}$, where $g^{ij}$ is the inverse of $g_{ij}$, so that $g^{ik}g_{kj} = \delta^i_j$ and $g_{ik} g^{kj} = \delta^j_i$. I want to derive
$$ g^{ij} = \tilde{g}^{\alpha \beta}\frac{\partial u^i}{\partial v^\alpha}\frac{\partial u^j}{\partial v^\beta} $$
I know that $g_{mn} du^m \otimes du^n = g_{mn} \frac{\partial u^m}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} dv^\beta \otimes dv^\gamma$.
So $\tilde{g}_{\beta \gamma} = g_{mn}\frac{\partial u^m}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma}$ and we can rewrite this in matrix form and use the formula $(ABC)^{-1} = C^{-1}B^{-1}A^{-1}$ to derive the desired formula, but I have done it in an alternative but most probably wrong way.
Applying the inverse, we have $\tilde{g}^{\alpha \beta} \tilde{g}_{\beta \gamma} = \tilde{g}^{\alpha \beta} g_{mn} \frac{\partial u^m}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} = \delta^\alpha_\gamma$. So $ \tilde{g}^{\alpha \beta} g_{mn} g^{nl} \frac{\partial u^m}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} =\tilde{g}^{\alpha \beta} \delta^l_m \frac{\partial u^m}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} = \tilde{g}^{\alpha \beta} \frac{\partial u^l}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} = \delta^\alpha_\gamma g^{nl}$. Replacing $l$ by $\gamma$, $\tilde{g}^{\alpha \beta} \frac{\partial u^\gamma}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} = \delta^\alpha_\gamma g^{n\gamma} = g^{n \alpha}$. We have $g^{n \alpha} = \tilde{g}^{\alpha \beta} \frac{\partial u^\gamma}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma}\neq \tilde{g}^{\gamma \beta} \frac{\partial u^\alpha}{\partial v^\beta} \frac{\partial u^n}{\partial v^\gamma} $. What's wrong here? hmmm…
Best Answer
Just do $$\begin{align} \widetilde{g}^{\alpha\beta}\frac{\partial}{\partial v^\alpha}\otimes \frac{\partial}{\partial v^\beta} &= \widetilde{g}^{\alpha\beta}\left(\frac{\partial u^i}{\partial v^\alpha}\frac{\partial}{\partial u^i}\right)\otimes \left(\frac{\partial u^j}{\partial v^\beta} \frac{\partial}{\partial u^j}\right) \\ &=\left(\widetilde{g}^{\alpha\beta} \frac{\partial u^i}{\partial v^\alpha} \frac{\partial u^j}{\partial v^\beta}\right) \frac{\partial}{\partial u^i}\otimes \frac{\partial}{\partial u^j} ,\end{align}$$so $$g^{ij} = \widetilde{g}^{\alpha\beta} \frac{\partial u^i}{\partial v^\alpha} \frac{\partial u^j}{\partial v^\beta}.$$This transformation law holds for every $2$-contravariant tensors.