Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a question of terminology. I want to talk about the category whose...

  • ...objects are pairs $(G,M)$, where $G$ is a group and $M$ is a $G$-set.
  • ...morphisms $(G,M)\rightarrow (G',M')$ are pairs $(f_G,f_M)$, where $f_G:G\rightarrow G'$ is a group homomorphism, and $f_M:M\rightarrow M'$ is a set map such that $$ f_M(g\cdot m) = f_G(g)\cdot f_M(m)$$ for all $g\in G$ and $m\in M$.

I've been calling these decorated groups (and their morphisms), since they've been arising in connection with decorated local systems. However, I'd prefer a more standard name, hopefully one which evokes the correct idea before explanation.

share|improve this question
The category of group-sets? –  Noah Snyder May 12 '11 at 16:44
If you replace group by semi-group your objects are what some call dynamical systems and your morphisms are usually called semi--conjugacies. To some "the category of invertible dynamical systems" would evoke the right idea, but "the category of actions" as proposed by Angelo sounds much better to me. –  jvp May 12 '11 at 17:05
Most of the time people tend to fix a specific group $G$ and then refer to "the category of $G$-sets" (with obvious morphisms). What is wanted here seems to come up rarely in practice: letting the group as well as the set vary arbitrarily. I'd be content just to refer to the category as before, but with no specific $G$ fixed. I also have no objection to "the category of group actions" (whereas "the category of actions" is really too vague). Until mathematicians elect a dictator to decide such things, we'll probably all do our own thing anyway. –  Jim Humphreys May 12 '11 at 17:21
There are other similar categories--for instance, the category with pairs (ring R, R-module) as objects, in which a morphism $(R, M) \to (R', M')$ is a ring morphism $R \to R'$, together with an $R$-module morphism $M \to M'$. It would be nice to have a similar naming conventions for all of these. Perhaps these categories turn out to be less useful than one might suppose (e.g., the category I just described is not abelian), and so they don't make it into very many published works, hence don't have well-known names. –  Charles Staats May 12 '11 at 19:54
Such a pair $(G,M)$ gives a groupoid $M//G$ with object set $M$, and one morphism from $x$ to $y$ for each $g\in G$ with $gx=y$. Every morphism $(G,M)\to (G',M')$ gives a functor $M//G\to M'//G'$. Most things that you might want to do can be made to work for general groupoids, and that extra generality is often useful, as well as sidestepping the problem of terminology. –  Neil Strickland May 12 '11 at 22:35
show 3 more comments

2 Answers

up vote 13 down vote accepted

How about "actions"? "The category of actions" sounds good to me.

share|improve this answer
Hmm, I like it, though I might call it the 'category of group actions' for clarity. –  Greg Muller May 12 '11 at 16:51
In my opinion, the best term would be the category of group actions. –  TaQ May 12 '11 at 17:31
Actually, Vezzosi and I used the terminology "actions" and "morphisms of actions" (in the context group schemes) in <arxiv.org/abs/math/9912155>;. "Group actions" is probably better. –  Angelo May 12 '11 at 17:58
I agree with "the category of group actions". It's natural (in the non-technical sense of the word), and inventing new terminology rather than just putting old terminology together into a phrase just makes things complicated. –  Michael Hardy May 12 '11 at 23:31
add comment

Angelo's answer above is clearly the correct one. But, somewhat tongue-in-cheek, I would also like to recommend "The category of trivialized groupoids". As I'm sure you're very aware, there's a pretty good analogy

vector bundles : trivialized vector bundles :: groupoids : group actions.

share|improve this answer
"action groupoids" would also work. –  S. Carnahan May 13 '11 at 7:33
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.