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
 A: 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 $\gamma_{ab}$.
