# Weyl symmetry and Polyakov action

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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
asked Jul 10, 2013
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

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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 (835 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

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