general relativity - Variation of the metric with respect to the metric

For a variation of the metric $g^{\mu\nu}$ with respect to $g^{\alpha\beta}$ you might expect the result (at least I did):

$$\begin{equation} \frac{\delta g^{\mu\nu}}{\delta g^{\alpha\beta}}= \delta^\mu_\alpha\delta^\nu_\beta. \end{equation}$$

but then to preserve the fact that $g^{\mu\nu}$ is symmetric under interchange of $\mu$ and $\nu$ we should probably symmetrise the right hand side like this:

$$\begin{equation}\frac{\delta g^{\mu\nu}}{\delta g^{\alpha\beta}}= \delta^\mu_\alpha\delta^\nu_\beta + \delta^\mu_\beta\delta^\nu_\alpha.\end{equation}$$

Is this reasonable/correct? If not, why not?

It seems that I can derive some weird results if this is right (or maybe I'm just making other mistakes).

Answer

Since the metric $g_{\mu\nu}=g_{\nu\mu}$ is symmetric, we must demand that

$$\tag{1} \delta g_{\mu\nu}~=~\delta g_{\nu\mu}~=~\frac{1}{2}\left(\delta g_{\mu\nu}+\delta g_{\nu\mu}\right)~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta g_{\alpha\beta},$$

and therefore

$$\tag{2} \frac{\delta g_{\mu\nu}}{\delta g_{\alpha\beta}} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right).$$

The price we pay to treat the matrix entries $g_{\alpha\beta}$ as $n^2$ independent variables (as opposed to $\frac{n(n+1)}{2}$ symmetric elements) is that there appears a half in the off-diagonal variations.

Another check of the formalism is that the RHS and LHS of eq. (2) should be idempotents because of the chain rule. For further motivation, see e.g. this Phys.SE post.