# Fermion version of Gauss-Milgram sum?

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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?

This post imported from StackExchange Physics at 2015-06-28 18:32 (UTC), posted by SE-user Yingfei Gu

asked Jun 23, 2015
edited Jun 28, 2015

Nice question. What do you mean by fermionic version though?

Hi Ryan, I just consulted Meng Cheng about a precise statement of fermionic topological order: there is a fermion in the particle content, which braid trivially with everyone else. And obviously this particle makes the tensor category not modular. So we need a new formalism to work with.

And a fermionic version of this formula means an equation whose left hand consists of data like topological spins and quantum dimensions of each anyon, and right hand consists of e.g. total quantum dimension and chiral central charge. So that we can read out the chiral central charge up to some integer from the formula, just like Gauss-Milgram.

I don't know how correct it is field-theoretically to include the electron as a quasiparticle. I would believe some formula like this exists for 3d spin TQFTs with the modularity constraint $Z(S^2) = \mathbb{C}$.

The previous Gauss-Milgram formula is derived via modular symmetry. So field theoretically, we need to use restricted modular group, which preserve spin structure.

BTW, could you remind me why $Z(S^2)=\mathbb{C}$ is related to modularity?

$Z(S^2)$ is the space of point operators in the theory. If there are non-trivial point operators, then they can be inserted with their inverse on a non-trivial Wilson line and used to "open it up". Such a Wilson line will have trivial braiding with everyone else because you can use this trick to unlink it from anything. I don't believe $Z(S^2) = \mathbb{C}$ implies modularity, but it approximates it. Anyway I think you should look up the Landsberg-Schaar relation and see the formula for the partition function of the BF theory in my paper http://arxiv.org/abs/1308.2926 . It seems like your formula above is best understood via these formulas and the 4d anomaly theory of Chern-Simons.

## 1 Answer

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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 computer the chiral central charge of the fermionic topological orders.

answered Aug 8, 2015 by (3,485 points)

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