# String coupled to Kalb-Ramond field under gauge transformation

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.

• Hint: Try work backwards from the last expression. Aug 28 '20 at 12:39

## 1 Answer

Just use the fact that $$\varepsilon^{\alpha\beta}\partial_{\alpha}\partial_{\beta}=0$$ and the chain rule $$\partial_{\alpha}=\partial_{\alpha}x^{\mu}\partial_{\mu}\,.$$

• 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.
– saad
Sep 10 '20 at 16:54
• $\partial_{\alpha}x^{\mu}\neq \delta_{\alpha}^{\mu}$ 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 Sep 10 '20 at 17:01
• 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.
– saad
Sep 10 '20 at 17:02
• Just used that $(\partial_{\alpha}x^{\mu})\partial_{\mu} C=\partial_{\alpha}C$. Sep 10 '20 at 17:03