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In the paper On classification of modular tensor categories by Rowell, Stong and Wang, they list the Ising modular category $I$ as having 3 objects $1$, $\sigma$ and $\psi$, with fusion rules $\sigma^2 = 1 + \psi$, $\sigma \psi = \psi \sigma = \sigma$ and $\psi^2 = 1$. There are in fact 16 modular tensor categories with these fusion rules (see DGNO, App B). What distinguishes $I$ from the rest is that in $I$, if I understand correctly, the dimension of $\sigma$ equals $\sqrt{2}$ (as opposed to $-\sqrt{2}$) and that the twist $\theta_\sigma = e^\frac{\pi i}{8}$ (as opposed to some other eighth root of $-1$).

RSW say that that `$I$ can be obtained as a quantum group category as the complex conjugate of $E8$ at level = 2'.

Is this just a coincidence? Or is there some kind of straight-up $E8$ symmetry in the underlying statistical mechanical Ising model, at zero magnetic field?

I'm aware of the Zamolodchikov $E8$ symmetry that apparently arises when one perturbs the zero-field Ising model to have a magnetic field (see this article, and these previous math overflow posts A, B). Perhaps it is related? But I'm talking about an $E8$ symmetry that somehow shows up already in the zero-field case (i.e. in the conformal field theory, since the perturbed theory is not a conformal field theory).

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Bruce Bartlett
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  • All the 16 modular tensor category with Ising fusion rules can be made unitary (so $d_\sigma=\sqrt{2}$) and the difference is only the twist. This MTC can be represented as conjugate of $(E_8)_2$, or the conjugate of $SO(15)_1$, so I doubt it has anything to do with the $E_8$ symmetry of Zamolodchikov. – Meng Cheng Jul 01 '15 at 01:24
  • Your first statement isn't right; it's easy to get confused in this business. See DGNO App B. There are two spherical structures for each Ising fusion category. In one spherical structure, $\dim \sigma = \sqrt{2}$, in the other, $\dim \sigma = -\sqrt{2}$. "Making it unitary" has the effect of selecting a preferred spherical structure (the one with positive dimensions). There are only 8 different twists, all of them an eighth root of -1. That is how I arrived at my statement, that the MTC's could be distguished by the sign of dim$\sigma$ and $\theta_\sigma$. – Bruce Bartlett Jul 01 '15 at 09:43
  • When you say that the Ising MTC can be represented as the conjugate of $SO(15)_1$, how do you compute that? – Bruce Bartlett Jul 01 '15 at 09:45
  • First of all every modular tensor coming from a conformal field theory comes from many different conformal field theories. The 8 unitary Ising MTCs are realized for example by Virasoro $c= 1/2$ minimal model, $\mathrm{Spin}(2n+1)1$ for $n=1,\ldots,7$. The $n=7$ case is apparently also realized by $E{8,2}$. This CFT has a $E_8$ symmetry. – Marcel Bischoff Jul 04 '15 at 03:44
  • See here for the non uniqueness of the realization by a CFT http://mathoverflow.net/q/207325/10718 – Marcel Bischoff Jul 04 '15 at 03:53
  • Yes, I am aware of the nonuniqueness of realizing an MTC by a CFT. Nevertheless, the fact remains. On the one hand, we have the Ising statistical mechanics model. On the other hand, we have the Ising conformal field theory. I know of three "natural" ways to formulate the CFT: as a minimal model, as the free fermion, or as a coset model, as in the big yellow book, namely $(\hat{E_8})_1 \oplus (\hat{E_8})_1/(\hat{E_8})_2$. The question remains: is there thus an $E8$ symmetry in the stat-mech model? – Bruce Bartlett Jul 06 '15 at 12:36
  • So on the CFT side Virasoro $c=1/2$ cannot have any symmetries. If it had one you could take the orbifold, but Virasoro is minimal. – Marcel Bischoff Jul 15 '15 at 09:09

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