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## Units

Yesterday I wrote down a definition of an action of a group $G$ on a category $C$ that was slightly incorrect because I neglected to write down any conditions involving units. With the definition I gave it is possible for $F(e)$ to fail to be an automorphism (it might instead be a nontrivial idempotent endofunctor $C \to C$).

The condition we need on units is first that we should have a unit isomorphism $\varepsilon : \text{id}_C \cong F(e)$

and second that this unit isomorphism should be compatible with the isomorphisms $\eta(g, h) : F(g) F(h) \cong F(gh)$ in the sense that the composites $\displaystyle F(g) \cong \text{id}_C F(g) \xrightarrow{\varepsilon F(g)} F(e) F(g) \xrightarrow{\eta(e, g)} F(g)$

and $\displaystyle F(g) \cong F(g) \text{id}_C \xrightarrow{F(g) \varepsilon} F(g) F(e) \xrightarrow{\eta(g, e)} F(g)$

should both be the identity. If we use the unit isomorphism to replace $F(e)$ with $\text{id}_C$ (which changes the $\eta(g, h)$), this is just the condition that $\eta(g, e)$ and $\eta(e, g)$ should both be the identity.

Similarly, in our definition of an equivalence $\alpha$ between two group actions $(F_1, \eta_1, \varepsilon_1), (F_2, \eta_2, \varepsilon_2)$, $\alpha$ needs to respect these unit isomorphisms in the sense that $\displaystyle \left( \text{id}_C \xrightarrow{\varepsilon_1} F_1(e) \xrightarrow{\alpha(e)} F_2(e) \right) = \left( \text{id}_C \xrightarrow{\varepsilon_2} F_2(e) \right).$

Again, if we use the unit isomorphisms to replace $F_1(e), F_2(e)$ with $\text{id}_C$ on the nose, this is just the condition that $\alpha(e)$ should be the identity.

Fortunately, in the special case we considered in the previous post, where $\pi_0(\text{Aut}(C))$ vanishes (and perhaps in general), this produces the same classification of group actions as before, so nothing has gone too badly wrong. Details below the fold.

Concrete details

Recall that in the special case of group actions $F$ where each $F(g)$ is the identity $\text{id}_C$, the only information left is in the natural isomorphisms $\eta(g, h) \in Z(C)^{\times}$. Yesterday we only imposed the usual 2-cocycle condition on these, and saw that equivalences of group actions corresponded to 2-coboundaries. Today we want the additional unit conditions above. This produces a variant of 2-cocycles and 2-coboundaries which we’ll call unital. Explicitly, for 2-cocycles this means $\eta(e, h) = \eta(g, e) = e \in Z(C)^{\times}$, and for 2-coboundaries it means we look at 2-coboundaries of the form $\frac{\alpha(g) \alpha(h)}{\alpha(gh)}$ where $\alpha(1) = e \in Z(C)^{\times}$.

We want to show that these give rise to the same cohomology group $H^2(BG, Z(C)^{\times})$, and we’ll do this very explicitly, by showing that any 2-cocycle is cohomologous to a unital 2-cocycle, and that two unital 2-cocycles are 2-cohomologous iff they are unitally 2-cohomologous. First, using the 2-cocycle condition $\displaystyle \eta(g, hk) \eta(h, k) = \eta(gh, k) \eta(g, h)$

and substituting first $g = e$ and then $k = e$, we get $\displaystyle \eta(e, hk) \eta(h, k) = \eta(h, k) \eta(e, h)$

and $\displaystyle \eta(g, h) \eta(h, e) = \eta(gh, e) \eta(g, h)$

from which it follows that $\eta(e, -)$ and $\eta(-, e)$ are both constant, and hence must both be equal to $\eta(e, e)$. Now we just need to modify this 2-cocycle by the 2-coboundary $\frac{\alpha(g) \alpha(h)}{\alpha(gh)}$ where $\alpha(e) = \eta(e, e)^{-1}$ and $\alpha(g) = e$ for $g \neq e$; we get a new 2-cocycle $\eta'$ satisfying $\displaystyle \eta'(g, h) = \eta(g, h) \frac{\alpha(g) \alpha(h)}{\alpha(gh)}$

and we see that $\eta'(e, h) = \eta'(g, e) = e$ as desired. Hence every 2-cocycle is cohomologous to a unital 2-cocycle.

Next, suppose $\eta_1, \eta_2$ are two unital 2-cocycles which are cohomologous via a 2-coboundary $d \alpha = \frac{\alpha(g) \alpha(h)}{\alpha(gh)}$, so that $\displaystyle \frac{\eta_1(g, h)}{\eta_2(g, h)} = \frac{\alpha(g) \alpha(h)}{\alpha(gh)}$.

Then setting $g = e$ gives $e = \frac{\alpha(e) \alpha(h)}{\alpha(h)} = \alpha(e)$

and hence $d \alpha$ must be a unital 2-coboundary.

### 4 Responses

1. on January 29, 2020 at 8:02 am | Reply Johannes Hahn

So with that in mind, a group action on C is the same as a monodial functor $G \to EndoFun(C)$, right? (Where $G$ is the monodial category with $G$ as objects, only identities as morphisms and multiplication as tensor product)

• on March 2, 2020 at 12:17 pm | Reply Qiaochu Yuan

Yes, that’s right, sorry for taking so long to reply!

2. […] from this data agree. We’ll also want unit isomorphisms satisfying the same compatibility as before. This is just spelling out the definition of a 2-functor from the category of separable extensions […]