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Trouble proving $\frac{1}{4\pi}\int_Md\tau\,d\sigma\,\sqrt{-\gamma}\,R$ is Weyl invariant

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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?

asked Sep 23, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]

1 Answer

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The quantity is only Weyl invariant on closed manifolds (with \(\partial M = 0\)). When the manifold has a boundary, the Weyl invariant quantity (Euler characteristic) is

\[\chi = \frac{1}{4\pi} \int_M d^2 \sigma \sqrt{-\gamma } R [ \gamma ] + \frac{1}{2\pi} \int_{\partial M} ds k\]

Here, $k$ is the extrinsic curvature of the boundary given by

\[k = \pm n_b t^a \nabla_a t^b\]

where $t^a$ is the unit vector tangent to the boundary with $t^2 = \mp 1$, $n^a$ is the outward pointing unit normal ($n^2 = 1$ and $n \cdot t = 1$). The upper sign is for a Lorentzian world-sheet and lower one for a Euclidean one). Now, under a Weyl transformation

\[ds' \to e^{\omega} ds,~~~n'_a = e^\omega n_a,~~~ t'^a = e^{-\omega} t^a \implies ds' k' = ds k + ds n^b \nabla_b \omega \]

Combining this with the result you derived, we find that $\chi$ is Weyl invariant. 

answered Oct 3, 2015 by prahar21 (535 points) [ no revision ]

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