In a general relativity context, I am required to compute this quantity: $\frac{\partial g^{\mu\nu}}{\partial g_{\alpha\beta}}$ where $g_{\mu\nu}$ is the metric tensor and $g^{\mu\nu}$ its inverse.
How can I calculate this derivative? I thought to use the rule for differentiating the inverse funcion (thinking $g^{\mu\nu}=(g_{\mu\nu})^{-1}$), but I don't know how to do the computation.
Moreover i found an article where the relation was $\frac{\partial g^{\mu\nu}}{\partial g_{\alpha\beta}}=-g^{\mu\alpha}g^{\nu\beta}$ (but the proof was not provided) is this true? how can I do the computation?
Thanks for help!
Starting with the definition of the inverse metric tensor- i.e. the inverse of the metric, we have
$$g g^{-1} = I$$
But we could also write the inverse metric as a matrix-valued function of the metric:
$$ g F(g) = I $$
$$ \sum_k g_{ik} F_{kj}(g) = \delta_{ij} $$
Since $g$ is full rank and square, this is a linear system in $F$ and so a unique solution exists for $F$. So long as $g$ remains a metric tensor- which via its positive definiteness implies full rank- so does the solution for $F$. This justifies the approach of differentiating both sides of the system and solving for the required differential. With some minor abuse of notation-
$$ \sum_k d g_{ik}\cdot F_{kj}(g) + g_{ik} \cdot dF_{kj}(g) = 0 \forall i,j$$
More compactly:
$$ dg \cdot F + g \cdot dF = 0 $$
Which can be rearranged and simplified with $F = g^{-1}$:
$$ dg^{-1} = -g^{-1} \cdot dg \cdot g^{-1}$$
Then, for a derivative with respect to a particular element $g_{kl}$, $dg$ is going to be a matrix of all zeroes except for a 1 in row $k$ and column $l$, but also in row $l$ and column $k$ if $k\neq l$ as $g_{kl}=g_{lk}$ must always hold. We can write this "structure matrix" as $S = e_k \otimes e_l + e_l \otimes e_k - \delta_{kl} e_k \otimes e_k$ (where $e_k$ is the kth row of the identity matrix) in order to unify the case where we differentiate with respect to a diagonal element with the case of an off-diagonal element. Putting it all together:
$$ \frac{dg^{-1}}{d g_{kl}} = - g^{-1} S g^{-1} $$ $$ = \left\{ -g^{ik} g^{lj}-g^{il}g^{kj}+\delta_{kl}g^{ik} g^{kj}\right\}_{ij}$$
This agrees with the formula you found on diagonal entries- and that is true for general matrices*- but it fails for the off-diagonal entries because of the restriction to symmetric matrices.
I've intentionally avoided any of the trappings of Riemannian geometry or relativity in general because it isn't necessary here and I think the calculus is instructive, but it is also possible to derive this from the property that the covariant derivative of the inverse metric tensor is identically 0, which can be derived from the covariant derivative of the metric tensor being identically 0. $$\nabla (g g^{-1}) = g \nabla g^{-1} + (\nabla g) g^{-1} = g \nabla g^{-1} = 0 \implies \nabla g^{-1}= 0 $$
At which point you can expand $\nabla g^{ij}$ in terms of Christoffel symbols and an ordinary derivative and do the same matrix algebra as above.