# Weyl symmetry and Polyakov action

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

• You might consider splitting this up into three questions and including some more of your thoughts on each like where you're stuck. – joshphysics Jul 10 '13 at 3:01
• 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 – user26143 Jul 10 '13 at 3:02
• @user26143: Yes, that's what he means. – Abhimanyu Pallavi Sudhir Sep 1 '13 at 2:58

\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 $\gamma_{ab}$.