In my lecture notes on tensor categories it says:

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


  • 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?

  • 2
    $\begingroup$ 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. $\endgroup$ Jul 27, 2020 at 10:08

1 Answer 1


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, and in Example 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.

  • 1
    $\begingroup$ 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) $\endgroup$
    – Javi
    Jul 27, 2020 at 13:05
  • $\begingroup$ Is it clear that they aren't monoidall equivalent ? (in characteristic $\neq 2$, say, or $=0$ if it's any easier) $\endgroup$ Jul 27, 2020 at 13:54
  • $\begingroup$ @Daniel Plácido: Why does the triangle diagram commute? Or do you adjust the left/right unit constraint accordingly? $\endgroup$
    – M.C.
    Sep 19, 2020 at 14:26
  • 1
    $\begingroup$ @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 $(-1)^iz\cdot x\in X_i$. $\endgroup$ Feb 2, 2021 at 13:35
  • 1
    $\begingroup$ Where $\mathbb 1$ is the graded vector space with $V_0 = \mathbb K$ and zero in other degrees. $\endgroup$ Feb 2, 2021 at 13:37

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