A question for the generalization of gauge transformation with two antisymmetric indices I have a question about the generalization of gauge transformation with two antisymmetric indices.
Starting from Eq. (3.7.6) in Polchinski's string theory book p. 108.
$$S_{\sigma} = \frac{1}{4 \pi \alpha'} \int_M d^2 \sigma g^{1/2} \left[ \left( g^{ab} G_{\mu \nu}(X) + i \epsilon^{ab} B_{\mu \nu} (X)  \right)  \partial_a X^{\mu} \partial_b X^{\nu} + \alpha' R 
\Phi(X) \right] \tag{3.7.6} $$
where $B_{\mu \nu}(X)$ is the antisymmetric tensor.
It is said variation
$$ \delta B_{\mu \nu} (X) = \partial_{\mu} \zeta_{\nu}(X) - \partial_{\nu} \zeta_{\mu}(X) \tag{3.7.7} $$
can add a total derivative to the Lagrangian density.
I tried to do some integration by part for $\partial_{\mu}$, $\partial_{\nu}$, $\partial_a$, and/or $\partial_b$ in (3.7.7), miserably I didn't get a total derivative. 
My question is, how to see Eq. (3.7.7) gives a total derivative?
 A: The answer to this question is best understood in terms of differential forms.
Thanks to the antisymmetry of $B_{\mu\nu}$ and $\epsilon^{ab}$, the component $S_B = k \int_M \epsilon^{ab} B_{\mu\nu} \partial_a X^u \partial_b X^\nu$ of the action which contains $B$ can be written $S_B = k\int_M X^*B$, where $X: \mbox{M} \to \mbox{Spacetime}$ is the string's worldsheet and $B = B_{\mu\nu} dx^\mu \wedge dx^\nu$ is a 2-form on the target spacetime.  
If $B = d\Lambda$ and the boundary $\partial M = \emptyset$, then $\int_M X^*B = \int_M X^*(d\Lambda) = \int_M d (X^*\Lambda) = \int_{\partial M} X^*\Lambda = 0$.
A: The variation of the Lagrangian density is : 
$$\delta \mathbb L= i\epsilon^{ab}  \partial_a X^\mu \partial_b X^\nu(\partial_{\mu} \zeta_{\nu} - \partial_{\nu} \zeta_{\mu}) \tag{1}$$
The chain rule for partial derivatives gives : 
$$\partial_a \zeta_{\nu} = \partial_a X^\mu ~~\partial_{\mu} \zeta_{\nu}\tag{2}$$
and : 
$$\partial_b \zeta_{\mu} = \partial_b X^\nu ~~\partial_{\nu} \zeta_{\mu}\tag{3}$$
Using $(2),(3)$ in $(1)$, we get : 
$$\delta \mathbb L= i\epsilon^{ab}(\partial_b X^\nu~\partial_a \zeta_{\nu} -\partial_a X^\mu~\partial_b \zeta_{\mu})\tag{4}$$
$\epsilon^{ab}$ and the second term of the above equation are antisymmetric in $a,b$, so we could write : 
$$\delta \mathbb L= 2i\epsilon^{ab}\partial_b X^\nu~\partial_a \zeta_{\nu}\tag{5}$$
With an integration by parts, we get : 
$$\delta \mathbb L= 2i\epsilon^{ab}(\partial_b (X^\nu~\partial_a \zeta_{\nu}) -  X^\nu~\partial_a \partial_b \zeta_{\nu})\tag{6}$$
The expression $\epsilon^{ab}\partial_a \partial_b \zeta_{\nu}$ vanishes, because of the antisymmetry of $\epsilon^{ab}$.
So, finally, we get : 
$$\delta \mathbb L= \partial_b (2i\epsilon^{ab}X^\nu~\partial_a \zeta_{\nu}) \tag{7}$$
So, the variation of the Lagrangian density is a total derivative.
