Metric variation of Laplace-Beltrami
I am trying to vary the laplace-Beltrami operator with respect to the metric. Using the following two rules \begin{align} \frac{\delta g^{\alpha \beta}}{\delta g^{\mu \nu}} &=\frac{1}{2} \left[\delta^\alpha_\mu \delta^\beta_\nu + \delta^\alpha_\nu \delta^\beta_\mu \right]\ \frac{\delta g_{\alpha \beta}}{\delta g^{\mu \nu}} &=- \frac{1}{2} \left[g_{\mu \alpha} g_{\nu \beta} + g_{\mu \beta} g_{\nu \alpha} \right], \end{align} I have come across the following conundrum when taking the metric variation of the Laplace-Beltrami operator $\square_x = g^{\mu \nu} \partial^x_\mu \partial^x_\nu$
\begin{align} \frac{\delta \square}{\delta g^{\mu \nu}} &= \frac{\delta}{\delta g^{\mu \nu}} \left[g_{\alpha \beta} \partial^\alpha \partial^\beta \right]=-\frac{1}{2} \left[g_{\mu \alpha} g_{\nu \beta} + g_{\mu \beta} g_{\nu \alpha} \right] \partial^\alpha \partial^\beta \ &=-\partial_\mu \partial_\nu\ &=\frac{\delta}{\delta g^{\mu \nu}} \left[g^{\alpha \beta} \partial_\alpha \partial_\beta \right] = + \frac{1}{2} \left[\delta^\alpha_\mu \delta^\beta_\nu + \delta^\alpha_\nu \delta^\beta_\mu \right] \partial_\alpha \partial_\beta\ &=+ \partial_\mu \partial_\nu. 818
\end{align} Where did I make a mistake ? I assume it has something to do with the gradient being a covariant object, but I am yet to spot the missing minus sign - where does it come into play ? Thanks in advance for your help.
Edit: How I derived the formulas for the metric variation: Starting from $g_{\mu \alpha} g^{\alpha \nu} = \delta^\nu_\mu$, I find that \begin{align} \delta g_{\mu \alpha} g^{\alpha \nu} + g_{\mu \alpha} \delta g^{\alpha \nu} &=0 \ \Rightarrow \quad g_{\lambda \nu} g^{\alpha \nu} \delta g_{\mu \alpha} = \delta g_{\mu \lambda} &= -g_{\lambda \nu} g_{\mu \alpha} \delta g^{\alpha \nu} \ \Leftrightarrow \quad \frac{\delta g_{\mu \lambda}}{\delta g^{\alpha \nu}} &= - g_{\lambda\nu} g_{\mu \alpha}, \end{align} which I subsequently symmetrized in the <PERSON> indices. Then I used that $g^{\mu \nu} g_{\mu \nu}=d$ to derive that \begin{align} \delta(g^{\mu \nu} g_{\mu \nu}) &=0 = g^{\mu \nu } \delta g_{\mu \nu} + \delta g^{\mu \nu} g_{\mu \nu} \ \Rightarrow \quad g^{\mu \nu} \delta g_{\mu \nu} &=- \delta g^{\mu \nu} g_{\mu \nu}, \end{align} such that the other rule follows analogously with a minus sign and raised indices. Is this correct ? 818