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

+ 2 like - 0 dislike
180 views

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?

+ 4 like - 0 dislike

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 (540 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverflo$\varnothing$Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.