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For Bosonic topological order, a very useful formula was proved to be true:

$\sum_a d_a^2 \theta_a=\mathcal{D} \exp(\frac{c_-}{8}2\pi i) $

(for more detail: $d_a$ is the quantum dimension of anyon labeled by a, and $\theta_a$ is the topological spin.D is the total quantum dimension, $\mathcal{D}^2=\sum_a d_a^2$. And $c_-$ is the chiral central charge. If we assume bulk boundary correspondence, $c_-$ can be defined as $c_-=c_L-c_R$, the chiral combination of the central charge of boundary CFT. Alternatively, the chiral central charge is also well defined without referring to CFT, that is via the thermal Hall effect when we have an edge termination.)

So my question is straightforward: what's the fermionic version of this formula?

I also post this question in physics stackexchange: https://physics.stackexchange.com/questions/190902/fermion-version-of-gauss-milgram-sum

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  • $\begingroup$ Good question. I suspect there exists a nice answer, but that answer is not yet known. $\endgroup$
    – Kevin Walker
    Jun 24, 2015 at 0:34
  • $\begingroup$ @KevinWalker Hi Kevin, one related, maybe simpler question is about fermionic Short Range Entangled state: the above Gauss-Milgram formula suggest Bosonic SRE has $c_-=8n$; although there is no fermionic Gauss-Milgram, is there a statement/proof about $c_-$ for fermionic SRE? $\endgroup$
    – Yingfei Gu
    Jul 3, 2015 at 1:27

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We just posted a paper http://arxiv.org/abs/1507.04673 addressing this issue. For fermion topological orders, the fermionic version of this formula is $\Theta=\sum_a d_a^2 \theta_a=0$. See eq. 14 of the paper. So we cannot use eq. 14 to compute the chiral central charge of the fermionic topological orders. We have to use the bosonic extension of the fermionic topological orders to compute the chiral central charge of the fermionic topological orders.

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  • $\begingroup$ And $\Theta=0$ seems to suggest $S^\dagger TS=0$ for fermionic TO, so I am curious about the physical meaning of it. I was naively guessing there should be some restricted modularity for the fermionic TO, which preserve the spin structure of the torus, e.g. a subgroup of $\mathbb{SL}_2(\mathbb{Z})$ maybe generated by $T^2$ an S. Does your result $S^\dagger TS=0$ something along this line? $\endgroup$
    – Yingfei Gu
    Jul 21, 2015 at 2:52
  • $\begingroup$ Following the above comment, the fermionic version of Gauss-Milgram in my mind was something derived from the restricted modularity(no matter what it is). Now you suggest to consider an embedding to bosonic TO, and I saw your paper with T. Lan conjectures the $c_-$ mod 1/2 is independent of embedding, I am very interested and excited about this point. Is there any hint or heuristic argument for that? $\endgroup$
    – Yingfei Gu
    Jul 21, 2015 at 2:58
  • $\begingroup$ There are two sets: $S^{TC},T^{TC}$ ($N$ by $N$ matrix) and $S,T$ ($N/2$ by $N/2$ matrix). The ground state degeneracy is $N/2$. In our paper, we mainly used $S^{TC},T^{TC}$ . $\endgroup$
    – Xiao-Gang Wen
    Jul 22, 2015 at 0:52
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There is no fermionic analogue of the Gauss-Milgram formula. It applies to modular topological quantum field theories, while phases built out of fermions are not modular. A simple example, showing where the Gauss-Milgram sum fails, is the Laughlin state, as described here (see text below equation 30).

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    $\begingroup$ There is a way to generalize Gauss-Milgram sum to Abelian fermionic topological order, which was discussed in arxiv.org/abs/1310.5708 (also arxiv.org/abs/hep-th/0505235). For the general case, although there is still a bulk-edge correspondence, the precise relation is not known. $\endgroup$
    – Meng Cheng
    Jun 23, 2015 at 16:20

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