• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

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

+ 2 like - 0 dislike

Consider the Polyakov action

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.
where $R$ is the Ricci tensor. I want to prove this statement. In order to do this I Weyl transform the metric
under such a transformation we clearly have
it is also straighforward (but quite lengthy) to probe that the Ricci tensor transforms as
with both of these ingredients we get
In order to prove that $\chi$ is invariant under Weyl transformations, I should prove that
is zero. Using Stokes' theorem I can write
I don't know how to follow. Any ideas?

asked Sep 23, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ revision history ]

1 Answer

+ 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 prahar21 (545 points) [ no revision ]

Your answer

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):
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:
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).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights