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=Cx. 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.

  • $\begingroup$ Do you already have a notion of "representation of a 2-group" in mind? Are you thinking of representations on vector spaces or on some suitable sort of category? Or is that all open to interpretation? $\endgroup$ Nov 17 '09 at 19:25
  • $\begingroup$ I'm mostly looking for something that works. Since vector spaces don't detect extensions by BH, I guess it's categories or something with similar complexity. $\endgroup$
    – S. Carnahan
    Nov 18 '09 at 1:07

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?).

  • $\begingroup$ Group algebras are indeed twisted by 2-cocycles, but I believe this is a version of the group algebra that lives in C, so the associativity gets twisted by the associator. $\endgroup$
    – S. Carnahan
    Nov 18 '09 at 17:00
  • $\begingroup$ Yeah, that makes sense. There's probably some additional twisting one could do by a 2-cocycle, but the basic version of the group algebra as an object in the twisted category of G-graded vector spaces needs an H^3 twist. $\endgroup$ Nov 18 '09 at 18:38
  • $\begingroup$ The sum over all group elements is a Frobenius algebra object if and only if the 3-cocycle is trivial... otherwise one has to go to the diagonal subfactor, which is reducible. Namely, take $V$ to be the direct sum object over all group elements, then $V\otimes \bar V$ has canonically the structure of a Frobenius algebra. $\endgroup$ Oct 20 '14 at 0:40

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.

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.

  • $\begingroup$ I'm afraid cyclic groups are deceptively easy - one can find the Morita invariance classes with just a quadratic form calculation, with no visible topology. This also applies to a certain class of cocycles of abelian groups (Mason, Ng, math.QA/0002246). BG tends to get rather complicated at about the same level of complexity where these techniques fail. $\endgroup$
    – S. Carnahan
    Nov 18 '09 at 4:36
  • $\begingroup$ Fine, then. Let $G$ be the binary icosahedral group and let $M$ be the Poincare homology sphere. (Or let $G$ be any spherical 3-manifold group and let $M$ be the corresponding 3-manifold.) Then my example construction still works. I admit that more work is needed to find the 3-manifolds systematically, but I bet that you can at least make many examples. $\endgroup$ Nov 18 '09 at 5:01
  • $\begingroup$ For that matter, although I don't know the value of the pullback calculation, you do at least have good control of the maps from a lens space to the classifying space of any finite group, since the fundamental group is cyclic. $\endgroup$ Nov 18 '09 at 5:39
  • $\begingroup$ Although this is not quite what I was looking for, it is quite interesting. Thanks for answering. $\endgroup$
    – S. Carnahan
    Nov 18 '09 at 16:57

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