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I'm studying how a string coupled to a Kalb-Ramond 2-form $B_{\mu \nu}$ is affected by a gauge transformation of the K-R field, $\delta B_{\mu \nu} = \partial_{\mu} C_{\nu} - \partial_{\nu} C_{\mu}$ from David Tong's notes, chapter 7, pages 190-191. I cannot work out the last step in the following: $$ S_{B} = \frac{1}{4 \pi \alpha'} \int_{\mathcal{M}} d\sigma d\tau \epsilon^{\alpha \beta}\partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} B_{\mu \nu} \rightarrow S_{B} + \frac{1}{2 \pi \alpha'} \int_{\mathcal{M}} d\tau d\sigma \epsilon^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} \partial_{\mu} C_{\nu}$$ $$ = S_{B} + \frac{1}{2 \pi \alpha'}\int_{\mathcal{M}} d\tau d\sigma \epsilon^{\alpha \beta} \partial_{\alpha} (\partial_{\beta}X^{\nu}C_{\nu}). $$ Here, $\alpha, \beta $ run over $D$-brane coordinates $\sigma, \tau$ and $\mu,\nu$ run over spacetime. I have tried integrating by parts and am not sure how to proceed.

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  • $\begingroup$ Hint: Try work backwards from the last expression. $\endgroup$
    – Qmechanic
    Aug 28 '20 at 12:39
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Just use the fact that $$ \varepsilon^{\alpha\beta}\partial_{\alpha}\partial_{\beta}=0 $$ and the chain rule $$ \partial_{\alpha}=\partial_{\alpha}x^{\mu}\partial_{\mu}\,. $$

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  • $\begingroup$ Got it, thank you. I was hesitant to use $\partial_{\alpha} x^{\mu} = \delta^{\mu}_{\alpha}$ because I wasn’t sure that two of the spacetime directions would coincide with the world sheet directions. $\endgroup$
    – saad
    Sep 10 '20 at 16:54
  • $\begingroup$ $\partial_{\alpha}x^{\mu}\neq \delta_{\alpha}^{\mu}$ in general. Note that depending in the Dp-brane dimension this might not even agree in terms of number of indices. It is also not reparametrization invariant. In fact you can use it for $p+1$ of your $x$'s to fix reparametrization invariance. I did not use it here $\endgroup$
    – Nogueira
    Sep 10 '20 at 17:01
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    $\begingroup$ Right, got it. I misinterpreted the chain rule to assume the presence of a Kronecker delta. The chain rule is true irrespective of what gauge you use. $\endgroup$
    – saad
    Sep 10 '20 at 17:02
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    $\begingroup$ Just used that $(\partial_{\alpha}x^{\mu})\partial_{\mu} C=\partial_{\alpha}C$. $\endgroup$
    – Nogueira
    Sep 10 '20 at 17:03

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