The Mac Lane strictness theorem states that any monoidal category is monoidally equivalent to a strict monoidal category (see here section 2.8).

Q1: Is it true that any fusion category is monoidally equivalent to a strict fusion category?
Q2: Is the Grothendieck ring of a fusion category invariant by monoidal equivalence?

Application: If Q1 and Q2 admit positive answers then a fusion ring is categorifiable into a fusion category iff it is categorifiable into a strict fusion category, which seems much easier to check, indeed too easy, that's why I have serious doubts and would like to understand where is the failure.

  • 1
    $\begingroup$ If strict fusion category simply means a fusion category which is strict monoidal as a monoidal category, then I'm having trouble seeing why the answer to both questions isn't a trivial 'yes'. Could you amplify on why "too easy"? $\endgroup$
    – Todd Trimble
    Jan 18, 2016 at 13:45
  • $\begingroup$ @ToddTrimble: because for a strict monoidal category we don't need the pentagon axiom, so we don't need to know if the pentagon equation admits a solution (which is very hard a priori). $\endgroup$ Jan 18, 2016 at 13:53
  • 2
    $\begingroup$ I must be very confused then. The pentagon condition is after all satisfied in a strict monoidal category. But I should probably bow out of the discussion since I'm not an expert on fusion categories and so am ill-placed to discuss where the real issue presumably lies. $\endgroup$
    – Todd Trimble
    Jan 18, 2016 at 14:05
  • 1
    $\begingroup$ I can't quite tell what you're asking. If you're asking whether a fusion ring is the (Grothendieck ring) of a rig obtained by decategorifying a rigid strict monoidal category with coproducts over which the monoidal product distributes, even that doesn't seem trivially achievable (much less the added conditions you'd need for an actual strict fusion category). I just don't see how you'd manufacture such a structured category out of thin air, as it were. $\endgroup$
    – Todd Trimble
    Jan 18, 2016 at 15:25
  • 3
    $\begingroup$ I think the confusion here is that one might think that for a strict category the 6j symbols are trivial and one could build a skeletal fusion category with trivial associater. The point is that this will in general not fulfill all the axioms. It is in general pretty difficult to check if a fusion ring has a categorification. To me it seems to be completely magical if a categorification exists, because the system of equations looks pretty overdetermined. That's why people strife to find a common source from which all fusion categories come from, conformal field theory is a big candidate. $\endgroup$ Jan 18, 2016 at 23:18

1 Answer 1


@Todd Trimble is correct; a strict fusion category is just a fusion category that is strict as a monoidal category. (Being strict or being fusion is just a property of a monoidal category that can either hold or fail.) In particular, the pentagon axiom holds, where the associators are identities on the nose. Moreover, there is a version of MacLane's theorem for pivotal fusion categories, see Section 2 of arXiv:0503167v2.

As someone who studies planar algebras, at one time I was very confused about this, and I think I know where the confusion lies. Let's focus on the fusion category $\operatorname{Vec}(G,\omega)$ where $\omega\in Z^3(G,\mathbb{C}^\times)$ is a cohomologically nontrivial 3-cocycle. The discussion below comes from a conversation I had with Scott Morrison which cleared up my confusion.

We will give two realizations of the fusion category $\operatorname{Vec}(G,\omega)$. The first has only one representative for each isomorphism class of simple objects. There are exactly $|G|$ many 1-dimensional simple objects denoted $V_g$ for $g\in G$, with fusion rules given by $V_g \otimes V_h = V_{gh}$, and associator given by $$ (V_g\otimes V_h) \otimes V_k \xrightarrow{\omega_{g,h,k}} V_g\otimes (V_h\otimes V_k). $$ The associators are easily seen to satisfy the pentagon axiom because of the 3-cocycle condition.

Now there's also a planar presentation of $\operatorname{Vec}(G,\omega)$. Usually, when we talk about a planar presentation, we're working with strict categories, because drawing strings next to each other in the plane is associative on the nose. (One can get around this by bending strings closer or further from other strings, or by drawing coupons for associators, but let's not do this here.)

The category $\operatorname{Vec}(G,\omega)$ has a planar presentation via oriented strands labelled by group elements (considering an oriented label the same as the reverse orientation with the inverse label), trivalent vertices

enter image description here

which correspond to a choice of isomorphism $g\otimes h \to gh$ for each $g,h \in G$, and relations

enter image description here

The 3-cocycle condition ensures the pentagon axiom holds for the trivalent vertices. (This is an important exercise to do!)

Since $\operatorname{Vec}(G,\omega)$ is pointed, there is only a one-dimensional space of maps $g\otimes h \to gh$, so choosing different isomorphisms for the trivalent vertices is equivalent to multiplying each trivalent vertex by a scalar $\mu(g,h)\in \mathbb{C}^\times$. This results in changing the third relation above by a coboundary: $$ \omega'(g,h,k) = \omega(g,h,k) \mu(g,hk)\mu(h,k)\mu(g,h)^{-1}\mu(gh,k)^{-1}. $$

Notice that in the second description, there's not just one simple object in each isomorphism class. For example, $g\otimes h \neq gh$. One of these objects corresponds to two oriented strands labelled g and h, and the other corresponds to one oriented strand labelled gh. However, they are isomorphic. The associator is hidden in the trivalent vertices. If one forms a skeletal category that is equivalent to the second category by choosing one representative of each isomorphism class of simple objects, one would see a non-trivial associator reemerge in the pentagon axiom.

The moral is as follows. Your category can be strict, or it can have one simple object for each isomorphism class, but in general, it can't have both of these properties at the same time.

As for your application, even trying to categorify a fusion ring into a strict fusion category is extremely difficult. You can think of this as parallel to constructing non-trivial planar algebras, which is highly non-trivial...

  • 7
    $\begingroup$ Thanks for the detailed answer, Dave. The moral at the end reminds me of a common confusion about the process of strictifying a monoidal category. It's not that you strictify by quotienting the collection of objects; rather you add a whole bunch of new objects and isomorphisms between them, where the new objects are formal tensor products. I call this the method of cliques (reminiscent of cliques in graph theory); see e.g. here for a description: ncatlab.org/nlab/show/clique $\endgroup$
    – Todd Trimble
    Jan 18, 2016 at 15:43
  • $\begingroup$ The difficulty for proving the existence of a categorification by the skeletal way is that we have to prove that the pentagon equation (i.e. a huge polynomial system of degree 3) admits a solution. What is the difficulty by the strict way, what kind of equations we have to deal with? $\endgroup$ Jan 19, 2016 at 1:33

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.