Use Index Notation properly when indices are already used in identifying which bases is the matrix metric calculated with respect to

Question

I am wondering how to apply the usual linear algebra to the rather unfamiliar case of 'matrices' with indices in special relativity or even general relativity. In particular, consider$$f=\sqrt{-\det\gamma_{\alpha\beta}}$$ where $\gamma_{\alpha\beta}$ represents a metric so that it has inverse $\gamma^{\beta\gamma}$. My understanding is that the $\alpha$ and $\beta$ here are not real indices, they are just there to let you know that this matrix we calculate determinant from, is the metric with both indices down. I would like to calculate $\frac{\partial f}{\gamma^{ab}}$ so the first step is$$\frac{\partial f}{\partial \gamma^{ab}}=-\frac{1}{2f}\frac{\partial (\det \gamma_{\alpha\beta})}{\partial \gamma^{ab}}$$ Proceed with Jacobi's formula for determinant:\begin{align}\frac{\partial\text{det}\gamma_{\alpha\beta}}{\partial \gamma^{ab}}&=\text{Tr}(\det\gamma_{\alpha\beta}\gamma^{-1}_{\alpha\beta}\frac{\partial\gamma_{\alpha\beta}}{\partial\gamma^{ab}})\end{align}

Here is the reason I fail to proceed. To calculate the quantity $\frac{\gamma_{\alpha\beta}}{\gamma^{ab}}$, one needs to treat $\alpha$ and $\beta$ as real indices and in particular they cannot be arbitrary $\alpha $ and $\beta$ if we want the derivative to be non-zero, as we will use $\gamma_{ab}\gamma^{bc}=\delta_a~^c$ to calculate this quantity. It would be extremely helpful if someone could write out the steps after this as an example for me to follow.

Answer 1

Let $\gamma\equiv -\det \gamma_{\alpha\beta}$ where $\alpha\beta$ is used to indicate the covariant metric tensor. We are interested in $$\frac{\partial\sqrt{-\gamma}}{\partial\gamma^{ab}}=\frac{1}{2\sqrt{-\gamma}}\frac{\partial\gamma}{\partial \gamma^{ab}}$$ and we are left to obtain a form of $\partial \det \gamma_{\alpha\beta}/\partial \gamma^{ab}$. Referring to Jacobi's formula, \begin{align} \frac{\partial \det \gamma_{\alpha\beta}}{\partial\gamma^{ab}}&=\gamma~\text{Tr}((\gamma_{\alpha\beta})^{-1}\cdot \frac{\partial\gamma_{\alpha\beta}}{\partial\gamma^{ab}})\\&=\gamma (\gamma^{\alpha\beta})(-1)\gamma_{\beta a}\gamma_{b\gamma}\\&=-\gamma\gamma_{ab} \end{align} and in particular when we let $a=\alpha, b=\beta$ we get $-\gamma \gamma_{\alpha\beta}$ so $$\frac{\partial \sqrt{-\gamma}}{\partial \gamma^{\alpha\beta}}=-\frac{1}{2}\sqrt{-\gamma}\gamma_{\alpha\beta}$$