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Consider a (closed) Riemann surface and let $G(x,y)$ be the Green function of the Laplace-Beltrami operator. We can informally identify $G$ with the two-point correlation function for the Gaussian random field:

$$G(x,y)=\left<\phi(x)\phi(y)\right>=\frac{1}{Z} \int \mathcal{D}\phi\;\phi(x)\phi(y) \exp\left(-\frac{1}{2}\int \left| \nabla\phi(z)\right|^2 dV_g(z)\right).$$ This is easy to vary with respect to the metric $g_{\mu\nu}$:

$$\delta G(x,y)=-\frac{1}{2}\int dV_g(z)\;\delta g^{ij}(z)\left<\phi(x)\phi(y) \nabla_i\phi(z)\nabla_j\phi(z)\right>_c,$$ where $\left<\right>_c$ is the connected correlation function (the disconnected diagram $x\leftrightarrow y,z\leftrightarrow z$ is cancelled by the variation of the partition function). We can use Wick's theorem to compute this, getting

$$\boxed{\frac{\delta G(x,y)}{\delta g^{ij}(z)}=-\nabla_{(i} G(x,z)\nabla_{j)} G(z,y)}$$

(derivatives w.r.t. $z$). This formula looks wickedly similar to the Hadamard variation formula for the variation of the boundary of the domain in flat space. Yet I haven't been able to find any mentions of this in the mathematical literature.

Furthermore, if we define the regularized Green's function (a.k.a. the Robin function) by $$G^R(x)=\lim_{y\to x} \left(G(x,y)-\frac{1}{2\pi}\ln d(x,y)\right),$$ where $d(x,y)$ is the local geodesic distance, then I'm conjecturing the following variational formula: $$\frac{\delta G^R(x)}{\delta g^{ij}(z)}=-\nabla_{(i} G(x,z)\nabla_{j)} G(x,z)-\frac{1}{4\pi}\nabla_i \nabla_j G(x,z).$$ The second term is motivated by the well-known formula for conformal variations (where it is $\frac{1}{4\pi}\delta_x(z)$) and seems to be necessary to cancel the second order pole in this variation. Edit: this guess turned out to be wrong, see my answer below.

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It seems that the naive derivation from the path integral only picks up the term coming from quasiconformal variations. Combining the known result for the quasiconformal variation with the much more well known conformal variation, I arrived at the expression

$$\frac{\delta G(x,y)}{\delta g^{\mu\nu}(z)}=(-\nabla_{(\mu}G(z,x)\nabla_{\nu)}G(z,y)+\frac{1}{2}g_{\mu\nu}(z)\nabla_\rho G(z,x)\nabla^\rho G(z,y))+\frac{1}{2V}(G(x,z)+G(z,y))g_{\mu\nu}(z).$$

The traceless part in the brackets represents the quasiconformal variation (which in complex conformal coordinates reduces to just $8\partial_z G(z,x)\partial_z G(z,y)$, as found in the literature). The last "trace" term is the conformal variation, which is not probed by the path integral because it only shows up at finite volume.

I believe this formula to be correct based on the independence of conformal and quasiconformal variations. The resulting tensor is also divergence-free away from $x$ and $y$, as required by general covariance. The more difficult part is proving the following conjecture for the Robin function:

$$\frac{\delta G^{R}\left(x\right)}{\delta g^{\mu\nu}\left(z\right)}=-\nabla_{(\mu}G\left(z,x\right)\nabla_{\nu)}G^R\left(z\right)+\frac{1}{2}g_{\mu\nu}(z)\nabla_\rho G(z,x)\nabla^\rho G^R(z)+\frac{1}{V}G\left(z,x\right)g_{\mu\nu}\left(z\right)-\frac{1}{4\pi}\nabla_{\mu}\nabla_{\nu}G\left(x,z\right)-\frac{\delta(x,z)}{4\pi}g_{\mu\nu}(z)+\frac{1}{8\pi V}g_{\mu\nu}(z)+\frac{1}{2V}f_{\mu\nu},$$ where $f_{\mu\nu}$ is an unknown symmetric traceless tensor defined by $$\nabla^\mu f_{\mu\nu}=\nabla_\nu G^R,$$ as required by the divergence-free condition on the metric variation of $G^R$. This result is based on the exact formula $$G^R(x)=\frac{1}{4\pi}\int G\left(x,y\right)R\left(y\right)\mathrm{d}V(y)+\frac{1}{V}\zeta^{R}\left(1\right)-c,$$ where $\zeta^{R}(s)=\zeta(s)-\frac{V}{4\pi}\frac{1}{s-1}$ and $\zeta$ is the spectral zeta function of the positive Laplace-Beltrami operator, $c=\frac{1}{2\pi}\left(\gamma-\ln2\right)$ (reference below).

Steiner, Jean, A geometrical mass and its extremal properties for metrics on $S^2$, Duke Math. J. 129, No. 1, 63-86 (2005). ZBL1144.53055.

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  • $\begingroup$ Does the bracket in the indices mean anti-symmetrization? Or symmetrization? To me it seems that it must be symmetrization in order to get the reduction to $8 \partial_z G(z,x) \partial_z G(z,y)$ in complex coordinates. $\endgroup$
    – desos
    Sep 3, 2020 at 14:22
  • $\begingroup$ @desos yes, the whole tensor has to be symmetric. I guess I mixed up the convention when I first wrote this post. $\endgroup$
    – level1807
    Sep 3, 2020 at 14:23
  • $\begingroup$ Where do the additional terms come from in the first equation? You explained the origin of the first term but it's not clear to me where the rest come from. I'm interested in looking at a variation $\frac{\delta G(x,y)}{\delta g^{zz}}$ in complex coordinates, and especially the limit $y \to x$ of this variation. I tried computing it another way and it looked like it is not well-defined on the diagonal, but according to your formula it seems to be well-defined. $\endgroup$
    – desos
    Sep 3, 2020 at 14:28
  • $\begingroup$ @desos to prove the formula for the variation of G, it's essentially enough to assume that it's true and check that its divergence satisfies the Ward identity $\nabla_\mu \delta G/\delta g^{\mu\nu}=\frac12 \delta_x \nabla_\nu G_y+\frac12 \delta_y \nabla_\nu G_x$, and that the trace part (conformal variation) is correct, which is pretty trivial. Also G is known to be modular invariant, so this formula remains true on higher genus surfaces too. $\endgroup$
    – level1807
    Sep 4, 2020 at 19:04
  • $\begingroup$ @desos And yes, the variation has a second order pole on the diagonal $x=y$, so in the sense of a normal function it's not well defined there. But it really needs to be treated as a distribution that's always integrated against some smooth tensor field, and in that sense it's completely well defined because such integrals will always converge. $\endgroup$
    – level1807
    Sep 4, 2020 at 19:05

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