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Consider a scalar masless field (in 2d for concreteness) in a curved space with standard action $$S=\frac{1}{4\pi}\int d^2x \sqrt{g}g^{ab}\nabla_a\phi\nabla_b \phi$$ There is an elegant way to derive equation for the propagator from the functional integral. To this end one writes $$0=\int \mathcal{D}\phi \frac{\delta}{\delta \phi(x)}\left(e^{-S[\phi]}\phi(y)\right)=\int \mathcal{D}\phi e^{-S[\phi]}\left(-\frac{\delta S[\phi]}{\delta \phi(x)}\phi(y)+\delta(x-y)\right)$$ and hence $$\left\langle \frac{\delta S[\phi]}{\delta \phi(x)} \phi(y)\right\rangle=\delta(x-y)$$ Computing functional derivative of our action one gives $$\frac{\delta S[\phi]}{\delta \phi(x)}=-\frac{\sqrt{g}}{2\pi}\Delta \phi(x)$$ and hence the standard equation for the Green function $$\sqrt{g}\Delta_x \left\langle \phi(x)\phi(y)\right\rangle=-2\pi\delta(x-y)$$ This gives the correct equation for the curved plane. However, it can not be correct on a closed surface (such as 2d sphere) because integrating the left part gives zero while integral of the right side is $-2\pi$. Correct equation for the Green function on the closed surface includes constant term proportional to area in the rhs.

The question is what in the above derivation via functional integral fails in the case of a closed surface?

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  • $\begingroup$ Why will integration of left part give zero? You have not usual derivative, but rather covariant. $\endgroup$ – Nikita Apr 3 at 10:46
  • $\begingroup$ @Nikita $\Delta=\frac{1}{\sqrt{g}}\partial_a(\sqrt{g}g^{ab}\partial_b)$ $\endgroup$ – Weather Report Apr 3 at 10:53

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