# Monoidal category: Inequivalent associators

Context

In my lecture notes on tensor categories it says:

"For a given category $$C$$ and a given tensor product $$\otimes$$, inequivalent associators can exist."

Questions

• What notion of equivalence is (usually) meant here?

• Is it simply the equality of two natural transformations?
That is, does the proposition read 'There exist monoidal categories $$(C, \otimes, I, a_1, l_1, r_1)$$ and $$(C, \otimes, I, a_2, l_2, r_2)$$ with associators $$a_2 \neq a_1$$'?

• What are examples of such categories with inequivalent associators?

• My guess would be that it means the resulting monoidal categories are not monoidally equivalent. Indeed this is possible, I'll write a full answer if I have time later. Jul 27, 2020 at 10:08

Take the category $$Vect^\mathbb{Z}_\mathbb{K}$$ of $$\mathbb{Z}$$ graded $$\mathbb{K}$$-vector spaces with the graded tensor product: $$(V\otimes W)_n=\bigoplus_{i+j = n}(V_i\otimes W_{j}).$$

There's the usual associator: $$(a\otimes b)\otimes c \in(U\otimes V)\otimes W\mapsto a\otimes (b\otimes c)\in U\otimes (V\otimes W).$$

Another associator takes the grading into account: $$(a\otimes b)\otimes c \in(U\otimes V)\otimes W \mapsto (-1)^{i+k} a\otimes (b\otimes c)\in U\otimes (V\otimes W),$$ where $$i$$ and $$k$$ are the gradings of $$a$$ and $$c$$, respectively. The index $$j$$ of $$b$$ was omitted for the pentagon axiom to work.

The monoidal categories defined by these associators aren't monoidally equivalent. In fact, a function $$a:\mathbb{Z}^3\to\mathbb{K}^*$$ defines an associator for $$Vect^\mathbb{Z}_\mathbb{K}$$ iff $$a(r,s,t)a(r,st,v)a(s,t,v)a(r,s,tv)^{-1}a(rs,t,v)^{-1} = 1$$ for all $$r,s,t,v\in\mathbb{Z}$$. This is the same as saying that $$a$$ is a nontrivial 3-cocycle of $$\mathbb{Z}$$ with coefficients in $$\mathbb{K}^*$$.

For more details you can see Example 1.7 of these lectures notes.

edit: An example very similar to this one is worked out in detail at Kerodon: the monoidal structure of a 3-cocycle is defined in Example 2.1.3.3, and in Example 2.1.6.8 it's proven that 3-chains $$a,a'$$ define equivalent monoidal structures if and only if they are cohomologous. This is also revisited at Example 2.1.15.

• I was waiting for your answer and I liked it. The associator that takes grading into account can be thought as the symbols ( and ) having odd degree, so aplying the Koszul sign rule conceptually explains why the degree of $b$ is ommitted (although in the end it is needed for the axioms to work)
– Javi
Jul 27, 2020 at 13:05
• Is it clear that they aren't monoidall equivalent ? (in characteristic $\neq 2$, say, or $=0$ if it's any easier) Jul 27, 2020 at 13:54
• @Daniel Plácido: Why does the triangle diagram commute? Or do you adjust the left/right unit constraint accordingly?
– M.C.
Sep 19, 2020 at 14:26
• @M.C. I believe the triangles commute if we adjust accordingly, e.g. $\rho_X:X\otimes \mathbb 1\to X$ maps $x\otimes z\in X_i\otimes \mathbb 1$ to $(-1)^iz\cdot x\in X_i$. Feb 2, 2021 at 13:35
• Where $\mathbb 1$ is the graded vector space with $V_0 = \mathbb K$ and zero in other degrees. Feb 2, 2021 at 13:37