11
$\begingroup$

It is well known that for a finite group $G$, the associator of the fusion category of $G$-graded $k$-vector spaces is given by an element of $H^3(G,k^*)$, up to equivalence of categories. ($k^*$ is the multiplicative group of units in $k$.)

A crucial step when showing this is the fact that in $G$-graded vector spaces, all simple objects are invertible, and therefore the tensor product of two simples is simple again. Denote the simple object generating the $g$-graded subcategory, $g \in G$, by $k_g$. Then $k_{g_1} \otimes (k_{g_2} \otimes k_{g_3}) = k_{g_1 g_2 g_3}$ is again simple, and thus is automorphisms are given by an element of $k^*$, for each triple of group elements. Two categories of $G$-graded vector spaces may be monoidally equivalent via a monoidal functor $(F, F^2, F^0)$, but if so, the coherence morphism $F^2_{g,h}\colon Fk_g \otimes Fk_h \to F(k_g \otimes k_h)$ is again given by a number in $k^*$ for every tuple $(g,h)$ of group elements, and it's in fact the coboundary for two representatives of the same cohomology class.

We might want to generalise this result and start with a based fusion ring, corresponding to the Grothendieck ring of our future fusion category, with the simples as chosen basis. For the previous example, the fusion ring is $\mathbb{C}[G]$ (the group ring), and the chosen basis is $G$ (as a subset of the group ring).

Of course, most fusion categories don't have all simples invertible. This means, e.g. that the associators and the monoidal functor coherences live in entirely different spaces, and it's not so obvious how to repeat the cohomology construction.

Is it still possible to classify associators as elements of some cohomology? How about other data, such as pivotal structures and braidings? Ideally, the cohomology theory could just be formulated given the based fusion ring.

Additionally, what's the relation to already known homotopical and (co)homological data typically associated to fusion categories? I.e. is such a cohomology related to the cohomology of the Brauer-Picard groupoid? In which way is its "tangent cohomology" the Davydov-Yetter cohomology (as discussed here?

$\endgroup$
9
$\begingroup$

There is no such cohomology theory known (in particular, this is not related to Davydov-Yetter cohomology which is about deformations and vanishes for finite groups). In my mind this is a very important open problem in the field which could have some major applications to classification of Izumi categories. One can speculate about what the other small cohomology groups should be:

  • If A is the fusion ring of some C, then $H^1(A)$ should give the tensor natural isomorphisms of the identity functor.
  • If A is the fusion ring of some C, then $H^2(A)$ should give the "gauge automorphisms" of C (i.e. the tensor autoequivalences whose underlying functor is the identity modulo natural isos).
  • $H^3(A)$ should classify possible C with fusion ring A up to tensor equivalences whose underlying functor is the identity.

I think one can even write down some kind of definitions that make the above work, but the problem is to understand in what sense it's a "cohomology theory." For example does one ever get spectral sequences or any other sort of "algebraic topology" computational tools?

It's possible that fusion rings are not the right input. In particular, you might be better off thinking of "Ocneanu cells" as an $H^3$ not for fusion rings but for certain kinds of graphs related to module tensor categories over a fixed braided tensor category. But again I don't know any meaningful sense in which these form a "cohomology theory."

I spent some time discussing trying to build such a cohomology theory with Chris Schommer-Pries and Chris Douglas, but we never made any substantial progress.

$\endgroup$
  • $\begingroup$ Thank you very much! Is there any reference to these "certain kinds of graphs"? I think I've never seen anything like this. $\endgroup$ – Manuel Bärenz Nov 13 '17 at 11:24
  • $\begingroup$ My point about Davydov-Yetter was just this: In the finite group case, and a field like $\mathbb{R}$ or $\mathbb{C}$, $H^3(G, k^*)$ is a 0-dimensional manifold. It classifies (up to outer automorphisms) fusion categories of $G$-graded vector spaces. Since the tangent space of $k^*$ is $k$, and Davydov-Yetter classifies deformations of the associator, I found it reasonable to identify $H^3(G,k)$ (for a given $\operatorname{Vec}_G^\omega$) with the tangent space of $H^3(G,k^*)$ (at $\omega$). Now I was hoping that this viewpoint could be generalised somehow. $\endgroup$ – Manuel Bärenz Nov 13 '17 at 11:27
  • $\begingroup$ One detail that puzzles me in what sense $H^3(A)$ would even be a group. It would mean in particular that every fusion ring has a "trivial" fusion category realising it (which can't be true in general, I think). So $H^3(A)$ seems less like a group but more like affine spaces? $\endgroup$ – Manuel Bärenz Apr 27 '18 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.