In Polchinksi's Sec. 10.1, the $bc$-$\beta\gamma$ SCFT is introduced with action

$$S_{BC} = \frac{1}{2\pi} \int d^2z (b \bar \partial c + \beta \bar \partial \gamma)$$

and supercurrent

$$T_F = -\frac12 (\partial \beta)c+ \frac{2\lambda-1}{2}\partial(\beta c)-2b\gamma.$$

I'd like to derive this supercurrent. However, it seems I need to know the variations $\delta b,\delta c,\delta\beta,\delta \gamma$ under the supersymmetry before using Noether's theorem. Are these easy to guess? Is there a faster way to derive $T_F$?

  • 3
    $\begingroup$ A clean derivation is to work with superfields and follow H. Verlinde's PhD thesis procedure, Appendix A, p.25 in: inis.iaea.org/search/search.aspx?orig_q=RN:20024923. (Joe uses a different supercurrent, and I expect you'll find $-b\gamma/2$ instead of $-2b\gamma$ in the last term. I suspect this is related to the fact that Joe's expression doesn't quite lead to the standard PCO, which had puzzled him, see p.125 comment in kitp.ucsb.edu/joep/links/joes-big-book-string/errata. Verlinde's procedure should lead to the expected PCO normalisation.) $\endgroup$ Jun 27, 2019 at 7:32
  • $\begingroup$ @Wakabaloola perhaps I'm not understanding, but does Verlinde derive the $BC$ supercurrent or does he just write it down, in that section you mentioned? $\endgroup$
    – Dwagg
    Jun 27, 2019 at 15:15
  • $\begingroup$ he derives it, he derives both the energy momentum and the supercurrent, both are contained in the superfield expression $\endgroup$ Jun 27, 2019 at 16:42
  • 3
    $\begingroup$ The key concept is that if you want to start from the flat superspace action (which is what you have written) then you need to deform the supercomplex structure in order to read off the supercurrent and energy momentum tensor. If you are not fully accustomed to these concepts I suggest you begin from understanding deformations of complex structure first, see D'Hoker and Phong arxiv.org/abs/1502.03673 for an excellent introduction. (You might find Sec.3.1 in arxiv.org/abs/1209.5461 useful also.) $\endgroup$ Jun 28, 2019 at 8:58

1 Answer 1


The supercurent is fixed, given an energy-momentum tensor $T_{B}$, by the OPEs:

$$ T_F(z)T_F(0)\sim\frac{2c}{3z^3}+\frac{2}{z}T_{B}(0) $$

$$ T_{B}(z)T_F(0)\sim\frac{3}{2z^{2}}T_{F}(0)+\frac{1}{z}\partial T_{F}(0) $$

where $c$ is the central charge. The second OPE state that $T_F$ has conformal weight $3/2$ and the first OPE state that it has odd statistics since there are only odd powers of $z^{-1}$. With this information we have

$$ T_{F}=A(\partial \beta c) + B(\beta\partial c) + C (b\gamma) $$

the constants $A$, $B$ and $C$ are fixed by the first OPE. They depend on the conformal weight of $c$ and $b$.


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