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Let us consider a functional determinant $$\det G^{-1}(x,y;g_{\mu\nu})$$ where the operator $G^{-1}(x,y;g_{\mu\nu})$ reads $$G^{-1}(x,y;g_{\mu\nu})=\delta^{(4)}(x-y)\sqrt{-g(y)}\left(g^{\mu\nu}(y)\nabla^{(y)}_\mu\nabla^{(y)}_\nu+m^2\right).$$ Such an operator appears in the one-loop effective action for a scalar field in curved spacetime (in the metric signature (+,-,-,-)). My question is how to do the following derivative $$\frac{\delta \log\det G^{-1}{(x,y;g_{\mu\nu})}}{\delta g^{\mu\nu}}?\tag{1}$$ I would think that such a calculation may show up in some textbooks on QFT in curved spacetime. So any recommendation on references is also greatly appreciated.

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Let me explain more why I found that doing the above derivative is difficult. In a scalar field theory in flat spacetime, we also have a term $$\log\det G^{-1}(x,y;\varphi)$$ with $$G^{-1}(x,y;\varphi)=\delta^{(4)}(x-y)\left(\eta^{\mu\nu}\partial^{(y)}_\mu\partial^{(y)}_\nu+V''(\varphi)\right)$$ in the effective action in presence of a scalar background $\varphi$. But now the derivative can be calculated as follows. $$\frac{\delta \log\det G^{-1}(x,y;\varphi)}{\delta\varphi}=\frac{\delta {\rm Tr}\log G^{-1}(x,y;\varphi)}{\delta\varphi}\\ =\int d^4 x d^4y\frac{\partial G^{-1}(x,y;\varphi)}{\partial\varphi}G(y,x;\varphi)=V'''(\varphi(x))G(x,x;\varphi)$$ where $G$ is the Green's function of the operator $G^{-1}$.

However, for the derivative (1), there are subtleties. First $$\frac{\partial G^{-1}(x,y;g_{\mu\nu})}{\partial g^{\mu\nu}}$$ is a differential operator. If one follows the above procedure, one should find $$\frac{\delta \log\det G^{-1}{(x,y;g_{\mu\nu})}}{\delta g^{\mu\nu}}\supset \int d^4x \left[\nabla^{(x)}_\mu\nabla^{(x)}_\nu G(x,y;g_{\rho\sigma})|_{y=x}\right]$$ which is divergent (recall $G$ is the Green's function and you have $\delta^{(4)}(0)$ in the above expression).

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  • $\begingroup$ The question has a functional determinant tag but from the little I know it doesn't seems like it have been tried to be solved as such, is that because the space is curved? do you know the answer when the space is flat? just interested in the problem and trying to learn myself $\endgroup$ – Dabed Mar 15 at 20:07
  • $\begingroup$ There is always the hard way, in which you vary $g_{\mu\nu} \to g_{\mu\nu} + \delta g_{\mu\nu}$, and determine what is the first order correction on $ \delta g_{\mu\nu}$. $\endgroup$ – JGBM Mar 16 at 7:58
  • $\begingroup$ @DanielD. As you can see from the example of flat spacetime when doing such a derivative, it is not necessary to carry out the computation of the functional determinant. But I do guess that the question in the post may be answered using zeta functions. But I don't know. That's why I asked this question... $\endgroup$ – Wein Eld Mar 16 at 12:31
  • $\begingroup$ Sorry I see I had really not really understood your question, now on the contrary I hope this can be answered without zeta function or path integrals $\endgroup$ – Dabed Mar 16 at 13:15
  • $\begingroup$ @WeinEld The determinant is ill-defined. This is clear in the simplest case, namely flat spacetime with $m=0$, because then the eigenvalues of $G^{-1}$ are clearly unbounded. Sometimes we can get a well-defined answer from an ill-defined starting point, just because our calculation might happen to be consistent with a regulator that happens to be removable in the final answer. QFT textbooks are filled with examples of that phenomenon, but this is apparently not one of those cases. Would you be content with some intuition about why it isn't? (I'm not saying that I have that intuition yet.) $\endgroup$ – Chiral Anomaly Mar 31 at 4:23

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