Question

I seem to recall that there is a straightforward subfactor construction that yields fusion categories given by G-graded vector spaces and representations of G, for finite groups G. Is there an analogous construction for 2-groups?

Some background: A 2-group is a monoidal groupoid, for which the isomorphism classes of objects form a group. Sinh showed that up to monoidal equivalence, these are classified by a group G (isom. classes of objects), a G-module H (automorphisms of identity), and an element of H³(G,H). In the context of this discussion, we can limit our attention to G finite, H=C^x. One notable feature is that when the action of G on H is trivial, the three-cocycle twists the associator in the G-graded vector space category.

I'm mostly curious about how to tell when two elements of H³(G,H) yield Morita-equivalent fusion categories, and am wondering if subfactors or planar algebras make it easy to detect this.

Answer 1

This is a standard construction in Subfactor theory see the intro of http://arxiv.org/abs/0811.1084v2 for details. The construction goes back a long long way (if I remember correctly both Vaughan Jones and Adrian Ocneanu's theses were related to this question, but I could be wrong there).

From a category theory perspective recall that a subfactor (N < M) is a unitary tensor category C (the N-N bimodules) together with a Frobenius algebra object A in C (M as an N-N bimodule with conditional expectation as trace). In this case the tensor category C is the twisted category of G-graded vector spaces (where you use the 3-cocycle to change the associator), and the algebra object is a twisted version of the group algebra (or maybe just the group algebra? I'm getting confused, shouldn't group algebras be twisted by 2-cocycles?).

Answer 2

Any spherical fusion category leads to a 3-manifold invariant by the Turaev-Viro construction — this was explained in an old arXiv paper by Barrett and Westbury arXiv:hep-th/9311155. The invariant is the same as the Reshetikhin-Turaev invariant of the doubled category, so if two of these categories are Morita equivalent, they yield the same 3-manifold invariant. If the category is made from your finite group $G$ together with your cohomology class in $H^3(G,\mathbb{C}^*)$, then the corresponding invariant was defined by Dijkgraaf and Witten; they interpreted it as Chern-Simons field theory with gauge group a finite group. If $\omega$ is the cohomology class, then the invariant is the sum over all homotopy classes of maps $f:M \to B_G$ of $\langle f^*(\omega),[M]\rangle$. Here $B_G$ is the classifying space of $G$ and $[M]$ is the fundamental class of $M$. If you find a 3-manifold to distinguish two of these 2-groups, then they are not Morita equivalent.

For example, let $G = C_3$ and let $\omega \in H^3(G,\mathbb{C}^*)$ be non-trivial. If you let $M$ be the lens space $L(3,1)$, then actually $B_G$ is an infinite-dimensional lens space that contains $M$. If $f$ is the inclusion map, then $f^*(\omega)$ cannot be trivial in this case. The Dijkgraaf-Witten invariant of $M$ is the sum of the three roots of unity, which vanishes. On the other hand, if $\omega$ is trivial, then the Dijkgraaf-Witten invariant is 3. So, no Morita equivalence for these two choices of $\omega$.

Despite the homotopy-theoretic language, these computations are generally tractable when $G$ is not too complicated.

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Also, I don't know a lot about the subfactor end of this, but I would suppose that $(G,\omega)$ does give you a subfactor. I'm not sure how much that by itself says about Morita equivalence though.