Reparameterization invariance of strings I'm working through Zweibach's First Course In String Theory.
I'm stuck on a problem (16.1) where I need to show that the coupling Kalb-Ramond field to a worldsheet have the following transformation:
$$\int d\tau' d\sigma' \frac{\partial X^\mu}{\partial \tau'} \frac{\partial X^\nu}{\partial \sigma'} B_{\mu \nu}(X) = {\rm sgn}(\gamma) \int d\tau d\sigma \frac{\partial X^\mu}{\partial \tau} \frac{\partial X^\nu}{\partial \sigma} B_{\mu \nu}(X)$$
where ${\rm sgn}(x)$ is the sign of $x$; and $\gamma = \frac{\partial \tau}{\partial \tau'} \frac{\partial \sigma}{\partial \sigma'} - \frac{\partial\tau}{\partial \sigma '}\frac{\partial \sigma}{\partial \tau '}$; and the reparameterization of the world sheet coordinates go from $(\tau, \sigma)$ to $(\tau'(\tau,\sigma),\sigma'(\tau,\sigma)))$.
I don't have any particular conceptual questions, but I do have a lot of things I've tried. I'm confused about when things are vectors and when they are not in this problem.


*

*gamma is the same as the Jacobian in a change of variables, which is similar  (maybe inverse?) as the square root of the determinant of the metric (of what?).

*$\gamma$ and what I've been calling$\gamma^{-1}$ where  $\gamma^{-1} = \frac{\partial \tau'}{\partial \tau} \frac{\partial \sigma'}{\partial \sigma} - \frac{\partial\tau'}{\partial \sigma }\frac{\partial \sigma'}{\partial \tau }$ are not quite multiplicative inverses of each other unless the cross terms are zero (Or if the gammas were vectors and I took the dot product).

*I can find the total derivatives for $d\tau'$ , $d\sigma'$, and the velocities, and multiply them together, but then I end up with something like $\gamma*\gamma^{-1}*$ (the unprimed coupling), plus a number of extra terms which all go to zero if $\frac{\partial \tau'}{\partial \tau} \frac{\partial \tau'}{\partial \sigma}=0$. This seems reasonable since the vectors along tau and sigma are perpendicular.

*It seems like, algebraically, the only way for sgn(gamma) to appear in an equation would be to have$(\sqrt{{\rm sgn}(\gamma)*\gamma})^2*\gamma^{-1}$ where the sign was installed to ensure that the sign under the square root was positive.
Could someone help me piece these observations together?
 A: You're overthinking this. Note that it's crucial that $B$ is antisymmetric. If it were a symmetric tensor, you wouldn't have a transformation law of that form.
Edit: OK, here is the spiel. I'll write $\sigma^i = (\tau,\sigma)$ and $\sigma'^i = (\tau',\sigma')$ to make the expressions more compact. First note that the integral measure transforms as
$$\int d^2 \sigma' = \int d^2 \sigma \, |\det(\partial \sigma'^i/\partial \sigma^j)|.$$
Next, we consider how the partial derivatives transform. Notice that we can use the antisymmetry to write
$$B_{\mu \nu} \frac{\partial X^{[\mu}}{\partial \tau'} \frac{\partial X^{\nu]}}{\partial \sigma'} = \frac{1}{2}B_{\mu \nu} \frac{\partial X^{\mu}}{\partial \tau} \frac{\partial X^{\nu}}{\partial \sigma} \left[ \frac{\partial \tau}{\partial \tau'}\frac{\partial \sigma}{\partial \sigma'} - \frac{\partial \tau}{\partial \sigma'}\frac{\partial \sigma}{\partial \tau'}\right] = \frac{1}{2}B_{\mu \nu} \frac{\partial X^{\mu}}{\partial \tau} \frac{\partial X^{\nu}}{\partial \sigma} \det(\partial \sigma^i/\partial \sigma'^j).$$
To conclude, you only need to realize that 
$$ |\det(\partial \sigma'^i/\partial \sigma^j)| \det(\partial \sigma^i/\partial \sigma'^j) = \pm 1$$
depending on the sign on the determinant.
