In this article, the following claims are made:
- The Yetter cohomology of $G$-graded $k$-vector spaces, for $G$ a finite group and $k$ algebraically closed, is given by group cohomology of $G$ with values in (the unit group of) $k$. (Example 7.2)
- The Yetter cohomology of a multi-fusion category vanishes. (Theorem 2.27)
It seems to me that these two statements are in direct contradiction. Clearly for finite $G$, the category of $G$-graded vector spaces is multifusion, since it is in particular fusion. But clearly, group cohomology doesn't vanish.
What's the resolution to this paradox?
Qiaochu Yuan has brought me to the solution of the paradox, I believe.
Consider the category of finite dimensional $G$-graded $k$-vector spaces. It is a $k$-linear, semisimple category with finitely many simple objects, which correspond to the elements of the group. Call them $k_g, g \in G$. Their tensor product is then $k_{g_1} \otimes k_{g_2} = k_{g_1 g_2}$.
To choose an associator for the whole category, it suffices to define it on the simple objects. This means we have isomorphisms $\alpha_{g_1,g_2,g_3}\colon (k_{g_1} \otimes k_{g_2}) \otimes k_{g_3} \to k_{g_1} \otimes (k_{g_2} \otimes k_{g_3})$ satisfying the pentagon axiom. Since $k_{g_1 g_2 g_3}$ is a simple object, these correspond to invertible numbers in $k$ which we will call $\omega(g_1,g_2,g_3)$, and the pentagon axiom turns out to be the multiplicative cocycle condition:
$$ \omega(g_1, g_2, g_3) \omega(g_1, g_2 g_3, g_4) \omega(g_2, g_3, g_4) = \omega(g_1 g_2, g_3, g_4) \omega(g_1, g_2, g_3 g_4) $$
Furthermore, a monoidal equivalence between two copies of the same category with different associators is given by (an outer automorphism of $G$ and) a coboundary between the two cocycles, so indeed any category of $G$-graded vector spaces is given by a class in $H^3(G,k^*)$ (divided by outer automorphisms (since inner automorphisms don't touch cohomology)).
Now let us deform the associator. Replace $k$ by $k[\varepsilon]/\varepsilon^2$. An associator is now given by $\omega(g_1, g_2, g_3) + \varepsilon \omega'(g_1, g_2, g_3)$, where $\omega$ satisfies the cocycle condition as before, and $\omega'$ satisfies the additive cocycle condition:
$$ \omega'(g_1, g_2, g_3) + \omega'(g_1, g_2 g_3, g_4) + \omega'(g_2, g_3, g_4) = \omega'(g_1 g_2, g_3, g_4) + \omega'(g_1, g_2, g_3 g_4) $$
It is, in a sense, the derivative of the multiplicative cocycle condition.
Up to equivalence of categories, we have in particular $[\omega] \in H^3(G, k^*)$ and $[\omega'] \in H^3(G, k)$, in the latter considering the underlying additive group of $k$. $H^3(G, k)$ classifies in which directions we can "continuously deform" the associator.
Now for $k$ having characteristic 0 (or in general, as Qiaochu says, if the characteristic of $k$ doesn't divide the order of the group), $H^3(G,k) = 0$, which means that we cannot deform the associator at all. In particular, there are only finitely many different possible associators. This is a special case of a statement called "Ocneanu rigidity".