I was reading Kiritsis' notes (http://arxiv.org/abs/hep-th/9709062), at page 105/106 (equation 10.1), where he has a covariant action of the superstring including the gravitino. I have problems showing that this is invariant under the given supersymmetry. The action is $$S=\int \sqrt{g}\left(g^{ab}\partial_a X^\mu \partial_b X_\mu+\frac{i}{2}\psi^\mu\gamma^a\partial_a\psi_\mu+\frac{i}{2}(\chi_a\gamma^b\gamma^a\psi^\mu)\left(\partial_bX^\mu-\frac{i}{4}\chi_b\psi^\mu\right)\right)$$

and the supersymmetry transformation is supposed to be

$$\delta g_{ab}=i\epsilon(\gamma_a\chi_b+\gamma_b\chi_a)\\ \delta\chi_a=2\nabla_a\epsilon\\ \delta X^\mu=i\epsilon\psi^\mu\\ \delta\psi^\mu=\gamma^a\left(\partial_aX^\mu-\frac{i}{2}\chi_a\psi^\mu\right)\epsilon$$

But when doing the variation I fail. For instance, trying to locate all terms that only have one $X$ and one $\psi$, I do not get that these sum to zero. However, I am not sure I do it correctly. If I understand it correctly (which is not explained well in the paper) the gamma matrices satisfy $\{\gamma^a,\gamma^b\}=-2g^{ab}$ since the metric is not gauge fixed yet. But under the supersymmetry variation, the metric also changes. Does this mean that we need a variation $\delta \gamma=\ldots$? If so what is this variation of these Dirac matrices? (I tried something like $\gamma\sim \epsilon (\chi^a)^T$ but did not make it work with the anti commutation relation)

If someone has some other reference about the covariant action of the superstring and BRST quantization, let me know, many books seem to only deal with the gauge fixed version.

  • $\begingroup$ If you have access to Peter van Nieuwenhuizen's lecture notes on string theory, there is a nice discussion at about p 130. Note with spinors, you should work with vielbeins and not the metric. The SUSY variation of a vielbein is proportional to the gravitino \chi. Also, the gamma matrices in curved space are then defined by contracting a vielbein with a flat space gamma matrix. Other potential subtleties: In PvN's notes, he introduces an auxiliary field F. I believe you will need a Fierz identity to cancel everything involving the fermions. $\endgroup$ Commented Jan 27, 2016 at 21:00
  • $\begingroup$ Thank you I, yes, I had a feeling that this should be formulated using vielbeins, which is why I thought it was surprising that Kiritsis just wrote everything using the metric. Where can I find these lecture notes? $\endgroup$ Commented Jan 27, 2016 at 21:09
  • $\begingroup$ I would just ask him. His email can be found on this webpage: insti.physics.sunysb.edu/itp/www/people. I have it but don't feel comfortable sending it to you without his permission. $\endgroup$ Commented Jan 28, 2016 at 15:14
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    $\begingroup$ is there still interest in a more complete answer to this question? $\endgroup$ Commented Oct 30, 2023 at 11:23
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    $\begingroup$ @Wakabaloola sure, any intuition on doing this quicker/easier is helpful. $\endgroup$
    – bolbteppa
    Commented Dec 9, 2023 at 13:07

1 Answer 1


I would guess that you forgot to vary $\sqrt{g}$. Furthermore, there is a chance that you fail when you Fierz the fermionic terms. If you can write down the Fierz identity that you used, i can help you with the explicit calculations.


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