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According to arXiv:1507.08553v1, the superconformal index, defined by

$$I(\beta_j) = \mbox{Tr}_{\mathcal{H}}(-1)^F e^{-\gamma\{Q,Q^\dagger\}}e^{-\sum_{j}\beta_j t_j}$$

is independent of the parameter $\gamma$.

(Here, $F$ is the fermion number, $Q$ is the supercharge, and $t_j$'s are generators of the Cartan subalgebra of the superconformal and flavor symmetry algebra that commute with Q).

Is this a standard result? How is it obvious?

EDIT: I think this makes sense physically for states with $\{Q, Q^\dagger\} > 0$, as they come in boson/fermion pairs due to supersymmetry. So I would expect only the $\gamma = 0$ term to contribute to the trace.

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  • $\begingroup$ Hint: compute the derivative w.r.t. $\gamma$. $\endgroup$
    – Hans Moleman
    Feb 4, 2016 at 15:33
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    $\begingroup$ That would bring down a $-\{Q,Q^\dagger\}$ on the right of the $(-1)^F$ but the anticommutator is bosonic so it should commute with $F$ (I know that $F$ anticommutes with $Q$ and $Q^\dagger$). $\endgroup$
    – leastaction
    Feb 4, 2016 at 15:40
  • $\begingroup$ I think this makes sense physically for states with $\{Q, Q^\dagger\} > 0$, as they come in boson/fermion pairs due to supersymmetry. So I would expect only the $\gamma = 0$ term to contribute to the trace (comment added to the question). $\endgroup$
    – leastaction
    Feb 4, 2016 at 15:53

2 Answers 2

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I cannot comment yet so I have to write an answer but I really want to address on of your comments. If you read the original paper by Maldacena et. al. you will see that the superconformal index can only receive contributions from states annihilated by $Q^{i\alpha}, \bar{Q}_{i}^{\dot{\alpha}},S_{j\alpha }, \bar{S}_{\dot{\alpha}}^j$ since any other contribution actually cancels outs.

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By construction, the superconformal index only receives contributions from states that satisfy $\{Q,Q^\dagger\}=0$: the index counts states that preserve some of the supersymmetry. Hence, the coefficient in front of the anticommutator is arbitrary and the index does not depend on it.

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  • $\begingroup$ Thanks for your answer Frederic. If I forget about $\gamma$, this just looks like the Witten index of ordinary supersymmetric quantum mechanics. If the superconformal index doesn't depend on $\gamma$, why does the definition include it? I read that the $\beta$ term can be thought of as regulating the index (much like the Witten index is sometimes written with an $e^{-\beta\mathcal{H}}$). $\endgroup$
    – leastaction
    Mar 18, 2016 at 0:53
  • $\begingroup$ Secondly, your response made sense to me when I saw it first. But now that I think about it, I am not sure if I understand your first statement: why does it only get contribution from states which satisfy $\{Q, Q^\dagger\} = 0$ by construction? $\endgroup$
    – leastaction
    Mar 18, 2016 at 0:53
  • $\begingroup$ Oh that's because states with E >0 occur in pairs... $\endgroup$
    – leastaction
    Mar 18, 2016 at 2:08

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