Assume that the spectrum of some supersymmetric theory is discrete, then the Witten index is expected to be independent of temperature given by $T = 1/\beta$. However, it is well-known (see this) that when we choose periodic boundary conditions for both fermions and bosons, the partition function/generating functional ($\mathcal{Z}$) is simply the Witten Index ($W$) i.e $\mathcal{Z}_{pbc} = W$. If the theory is now put at finite temperature by choosing anti-periodic fermions along the time direction, why is $\mathcal{Z}$ no longer equal to $W$? This means that somehow $W$ depends on $\beta$, which cannot be true (also since it is a regulator-dependent term in the definition of $W$). Can someone point out the cause of this discrepancy? Also, in this regard, a quick remark on the "twisted" boundary conditions for several of these calculations would be appreciated.


1 Answer 1


You're using different boundary conditions around the thermal circle, so there's no reason to expect the results to be related.

When you use periodic boundary conditions around the thermal circle, it's equivalent to inserting a factor $(-1)^F$ in the trace

$$Z = {\rm Tr\ } (-1)^F e^{-\beta H}$$

and because of supersymmetry, only the groundstates can contribute to this, so the result is independent of $\beta$. On the other hand, clearly all states will contribute to the anti-periodic partition function

$$Z = {\rm Tr\ } e^{-\beta H}$$

since the operator $e^{-\beta H}$ is manifestly positive.

  • $\begingroup$ Thanks! What if I take the $\beta \to \infty $ limit with the apbc boundary conditions? Will I force myself into Z_pbc? Also, I was curious to understand why one should compute the index conveniently for twisted bc's and not for any other ? For ex : E. Witten, Nucl. Phys. B202 (1982) 253 $\endgroup$
    – R.G.J
    Commented Sep 12, 2018 at 22:21
  • $\begingroup$ In the limit $\beta \to \infty$, the second partition function just computes the ground state degeneracy, while the first partition function computes the difference between the fermion parity even and fermion parity odd ground states. $\endgroup$ Commented Sep 12, 2018 at 22:43
  • $\begingroup$ It's very hard to compute the second partition function. Maybe I don't understand what you're asking about? $\endgroup$ Commented Sep 12, 2018 at 22:43

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