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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?

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    $\begingroup$ Skip disclaimer? $\endgroup$
    – user1504
    Commented Aug 5, 2013 at 21:54
  • $\begingroup$ Yep. I often say "I have a stupid question" something like that, and joshphysics suggested that I don't have to use such "disclaimer", although i always feel uncomfortable for encountering so many problems during reading Polchinski >_< (now removed) physics.stackexchange.com/q/71928 $\endgroup$
    – user26143
    Commented Aug 5, 2013 at 22:07
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    $\begingroup$ I guess the next step is to remove the "remove disclaimer" at the beginning of your questions. You can do it! People who judge you for asking so many questions can...well you get the point. $\endgroup$ Commented Aug 5, 2013 at 23:05

2 Answers 2

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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 $\int_M X^*B = \int_M X^*(d\Lambda) = \int_M d (X^*\Lambda) = \int_{\partial M} X^*\Lambda = 0$.

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  • $\begingroup$ Does your "*" mean taking Hodge star on $B$? How does $X$ correspond to $\epsilon^{ab} \partial_a X^{\mu} \partial_b X^{\nu}$? $\endgroup$
    – user26143
    Commented Aug 6, 2013 at 0:14
  • $\begingroup$ $*$ is the notation for pullback. $X^\mu$ is the pullback $X^*z^\mu$ of the $\mu$-th coordinate function $z^\mu$ on the target spacetime. $\partial_a X^\mu$ is the $a$-the component of the exterior derivative of $X^u$. $\endgroup$
    – user1504
    Commented Aug 6, 2013 at 0:17
  • $\begingroup$ How to get the expression of $X^*$? is that $\epsilon^{ab} \partial_a X^{\mu} \partial_b X^{\nu}$ (I can get $B=B_{\mu \nu} dx^{\mu} \wedge dx^{\nu}$, but not other part between non-differential-form and differential-form of $S_B$)? And in your last equation, should $\int_M X^* (dB)$ be $\int_M X^* (d\Lambda)$? $\endgroup$
    – user26143
    Commented Aug 6, 2013 at 1:01
  • $\begingroup$ Yep, that was typo. For the rest, you're going to have to spend a little time learning about forms. It will be time well spent, but it's not going to happen here in the comments. $\endgroup$
    – user1504
    Commented Aug 6, 2013 at 1:55
  • $\begingroup$ I took a course on differential geometry, although i didn't done well. I think I need at least a reference for the techniques of pullback $X$ in the content of world sheet..(seems Nakahara's book is not directly useful here) $\endgroup$
    – user26143
    Commented Aug 6, 2013 at 2:12
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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.

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