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Consider the Polyakov action

$$S[x^{\mu},\gamma_{ab}]=-\frac{1}{4\pi\alpha'}\int_M d\tau\,d\sigma\,\sqrt{-\gamma}\gamma^{ab}\partial_ax^{\mu}\partial_ax_{\mu}$$ Consider the case of a closed string. According to Polchinski's book on string theory (page 15) the following is invariant under Weyl transformations, just as the Polyakov action. $$\chi=\frac{1}{4\pi}\int_Md\tau\,d\sigma\,\sqrt{-\gamma}\,R$$ where $R$ is the Ricci tensor. I want to prove this statement. In order to do this I Weyl transform the metric $$\gamma_{ab}\to{}e^{2\omega}\gamma_{ab}$$ under such a transformation we clearly have $$\sqrt{-\gamma}\to\sqrt{-e^{4\omega}\gamma}$$ it is also straighforward (but quite lengthy) to probe that the Ricci tensor transforms as $$R\to{}e^{-2\omega}(R-2\nabla^2\omega)$$ with both of these ingredients we get $$\chi\to\frac{1}{4\pi}\int_Md\tau\,d\sigma\,\sqrt{-\gamma}\,(R-2\nabla^2\omega)$$. In order to prove that $\chi$ is invariant under Weyl transformations, I should prove that $$-\frac{1}{2\pi}\int_Md\tau\,d\sigma\,\sqrt{-\gamma}\,\nabla^2\omega$$ is zero. Using Stokes' theorem I can write $$-\frac{1}{2\pi}\int_Md\tau\,d\sigma\,\sqrt{-\gamma}\,\nabla^2\omega=-\frac{1}{2\pi}\int_{\partial{}M}ds\,n^a\partial_a\omega.$$ I don't know how to follow. Any ideas?

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  • $\begingroup$ What you are trying to prove is true for a world-sheet without boundary. For a world-sheet with boundary, there is an additional surface term (exercise 1.3 in Polchinski). $\endgroup$ – Evan Rule Sep 24 '15 at 14:00
  • $\begingroup$ I know that, but why is it true? $\endgroup$ – Yossarian Sep 24 '15 at 18:04
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    $\begingroup$ The integral over the boundary of a manifold without boundary is zero. $\endgroup$ – Evan Rule Sep 24 '15 at 18:38
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Your last equation can be rewritten as follows:

$$ -\frac{1}{2\pi}\int_M d\tau d\sigma\sqrt{-\gamma}\ \nabla^2\omega = -\frac{1}{2\pi}\int_M d\tau d\sigma \ \partial_{a}\left(\sqrt{-\gamma} \ \nabla^a \omega\right) \ , $$ since $\sqrt{-\gamma}\nabla_a f^a = \partial_{a}(\sqrt{-\gamma}f^a)$ for any tensor $f_a$. Then you can apply Stokes for a world-sheet without boundary and kill the total derivative.

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