# Polchinski, Making a good gauge choice

I was reading Polchinski book and there he tries to make a gauge choice to fix the redundacies in the Polyakov action.

In order to make this choice, he fixes $$\tau=X^{+}\tag{1.3.8a}$$ and then he says $f=\gamma_{\sigma\sigma}\left(-\det\gamma\right)^{-1/2}$ transforms as $$f'd\sigma'=fd\sigma \tag{1.3.9},$$ under reparameterizations of $\sigma$ where $\tau$ is left fixed. I would like to show this last equation, but I couldn't. Can you help me?

• Hint: Write down the transformation law for a metric tensor $\gamma_{ab}$ under a coordinate transformation $(\tau,\sigma)\to (\tau,\sigma^{\prime})$. Commented Feb 21, 2017 at 21:20
• I make that , I got this transformation under the supposition $\sigma'\left(\sigma\right)$. But I would like to know if It's unnecessary this supposition.
– 7919
Commented Feb 21, 2017 at 21:33
• $\gamma'_{\sigma'\sigma'}=\frac{\partial\sigma^{a}}{\partial\sigma'}\frac{\partial\sigma^{b}}{\partial\sigma'}\gamma_{ab}$ . In order to eliminate terms as $\gamma_{\tau\sigma}$ or $\gamma_{\tau\tau}$ . I used the assumption $\sigma'(\sigma)$
– 7919
Commented Feb 21, 2017 at 21:45
• $\uparrow$ You should. Commented Feb 22, 2017 at 14:14

Note that the determinant of the metric behave as a $$()_{\sigma\sigma}$$ (two lower $$\sigma$$ indices) under $$\sigma$$-reparametrizations with $$\tau$$ held fixed.
$$\det(\gamma_{ab})=\gamma_{\sigma\sigma}\gamma_{\tau\tau}-\gamma_{\sigma\tau}\gamma_{\sigma\tau}$$ You can see that from the counting how many $$\sigma$$ indices there are in each term of the determinant. The square root of it, $$\sqrt{\det(\gamma_{ab})}$$ will behave as $$()_{\sigma}$$, just one index. Finally, taking the inverse this will behave as $$()^{\sigma}$$, just one upper index. Now, $$f=\gamma_{\sigma\sigma}(\det(\gamma_{ab}))^{-1/2}$$ will behave as $$()_{\sigma}$$ since $$2-1=1$$. Knowing that $$d\sigma$$ behave as $$()^{\sigma}$$ under $$\sigma$$-reparametrization you get:
$$fd\sigma=f'd\sigma'$$