# Weyl symmetry and Polyakov action

+ 2 like - 0 dislike
29 views

I have a question in reading Polchinski's string theory volume 1.

p12-p13

Given the Polyakov action $S_P[X,\gamma]= - \frac{1}{4 \pi \alpha'} \int_M d \tau d \sigma (-\gamma)^{1/2} \gamma^{ab} \partial_a X^{\mu} \partial_b X_{\mu}$ (1.2.13),

how to show it has a Weyl invariance

$\gamma'_{ab}(\tau,\sigma) = \exp (2\omega(\tau,\sigma)) \gamma_{ab} (\tau,\sigma)$?

Because both $(-\gamma)^{1/2}$ and $\gamma^{ab}$ give a factor $\exp(2\omega(\tau,\sigma))$, they do not cancel each other

Thank you very much in advance

This post imported from StackExchange Physics at 2014-06-29 15:46 (UCT), posted by SE-user user26143
You might consider splitting this up into three questions and including some more of your thoughts on each like where you're stuck.

This post imported from StackExchange Physics at 2014-06-29 15:46 (UCT), posted by SE-user joshphysics
Do you mean split this into three posts? (since the old post already marked as 1) 2) 3)). I have removed questions 2 and 3 for this moment

This post imported from StackExchange Physics at 2014-06-29 15:46 (UCT), posted by SE-user user26143

+ 4 like - 0 dislike

The transformed action is

\begin{align} \int d\tau d\sigma &\left[-\gamma'(\tau, \sigma)\right]^{1/2} \gamma'^{ab}(\tau, \sigma)\frac{\partial X'^\mu}{\partial \sigma^a}(\tau, \sigma)\frac{\partial X'_\mu}{\partial \sigma^b}(\tau, \sigma) \notag\\ &= \int d\tau d\sigma \left[-[e^{2\omega(\tau, \sigma)}]^2\gamma(\tau, \sigma)\right]^{1/2} e^{-2\omega(\tau, \sigma)}\gamma^{ab}(\tau, \sigma)\frac{\partial X^\mu}{\partial \sigma^a}(\tau, \sigma)\frac{\partial X_\mu}{\partial \sigma^b}(\tau, \sigma) \notag\\ &= \int d\tau d\sigma \left[-\gamma(\tau, \sigma)\right]^{1/2} \gamma^{ab}(\tau, \sigma)\frac{\partial X^\mu}{\partial \sigma^a}(\tau, \sigma)\frac{\partial X_\mu}{\partial \sigma^b}(\tau, \sigma) \end{align} In the first equality, the squared exponential factor inside the square root comes from the identity \begin{align} \det (cA) &= c^n\det A \end{align} for an $n\times n$ matrix $A$. The minus sign in the $e^{-2\omega}$ factor in the transformation of $\gamma^{ab}$ comes from the fact that $\gamma^{ab}$ is the inverse of $\gamma_{ab}$.

This post imported from StackExchange Physics at 2014-06-29 15:46 (UCT), posted by SE-user joshphysics
answered Jul 10, 2013 by (830 points)
Thank you! I forget the inverse!

This post imported from StackExchange Physics at 2014-06-29 15:46 (UCT), posted by SE-user user26143
@user26143 Sure thing.

This post imported from StackExchange Physics at 2014-06-29 15:46 (UCT), posted by SE-user joshphysics

 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$\hbar$ysicsO$\varnothing$erflowThen 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.