In this post Quantum String action , the user does the following manipulation :

$$\delta (H_{mnp})H^{mnp} =\partial_m\delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}] $$

I can't quite understand how he does that, I get where the derivative comes from but how does it appear three combinations of $H^{mnp}$?


1 Answer 1


If $H_{mnp}=\partial_m B_{np}+\partial_n B_{pm}+\partial_p B_{mn}$, then

$$\delta(H_{mnp})=\partial_m(\delta B_{np})+\partial_n(\delta B_{pm})+\partial_p(\delta B_{mn})$$


\begin{align} \delta (H_{mnp})H^{mnp} &=\left[\partial_m(\delta B_{np})+\partial_n(\delta B_{pm})+\partial_p(\delta B_{mn})\right]H^{mnp}\\ &=\partial_m(\delta B_{np})H^{mnp}+\partial_n(\delta B_{pm})H^{mnp}+\partial_p(\delta B_{mn})H^{mnp}\\ &=\partial_m(\delta B_{np})H^{mnp}+\partial_m(\delta B_{np})H^{pmn}+\partial_m(\delta B_{np})H^{npm}\\ &=\partial_m(\delta B_{np})\left[H^{mnp}+H^{pmn}+H^{npm}\right]. \end{align}

So it's just a matter of renaming indices in the last two terms.

  • $\begingroup$ Thank you very much for your help! $\endgroup$
    – RKerr
    Dec 9, 2020 at 17:49

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