3
$\begingroup$

If $H \lhd G$, then $G$ acts on $H$ by conjugation. I need to talk about this action but in a situation where $H$ is not (necessarily) normal. When $H \leq G$, there is a "partial action" of $G$ on $H$ by $G$-conjugation, and clearly it has very nice properties, for example $h^g h^{g'} = h^{gg'}$ and $(hk)^g = h^g k^g$ when these expressions are defined. To my amazement (at how bad I am at searching), I did not find a discussion of this anywhere. I'm sure it is very standard, and I'm looking for some pointers for a good formal setting.

Question. Is there a good basic reference for the "partial conjugation action" of a group on a subgroup, in some formalism? Is there a standard way to talk about this object?

I know many ways to talk about partial actions, and I can specialize them to my case, so in theory I can solve this question in many ways myself. However, it seems like such a natural example of a partial action that the fact I am not finding this discussed anywhere suggests to me that I may be missing something, so a reference or situation where this appears would be nice.

A more serious problem that tripped me up is "generation". In many of the settings of partial actions, it is difficult to state that a partial group action is finitely-generated. If $G = \langle S \rangle$, then the partial actions given by partial conjugation by $s \in S$ on a subgroup $H \leq G$ obviously "generate" the partial action of $G$ in some sense. But I don't really know how to say this in a good way, especially I run into issues with domains, when trying to state write down the axioms, and I don't want to reinvent the wheel.

More specifically, it is natural enough to say that a group action $G \curvearrowright H$ is finitely generated when $G$ is, so...

Question. Is there a standard way to say a "partial group action" is "finitely-generated", so that in the case above of a finitely-generated group $G$ partially acting on its subgroup $H$ by conjugation, the partial action of $G$ would indeed be finitely-generated?

I tried to look at some existing formalisms for partial actions. One thing you can do is form the action groupoid of $G$ acting on itself by conjugation, so take the action groupoid with $\Gamma = G$ acting group, $\Omega = G$ the set where it acts, then take the subgroupoid for the restriction $H \subset \Omega$. Unfortunately, then it is a bit awkward (I think) to discuss individual partial bijections that may or may not end up being part of such an action (they should of course be "partial automorphisms", but the list of axioms does not seem to be suggested by the subgroupoid definition). It is also not obvious to me how to state finite generation correctly. Groupoidification drops the group, and after taking the subgroupoid corresponding to $H$, it might not even be determined up to isomorphism, so when you say that a groupoid of partial automorphisms on $H$ is finitely-generated, there is no $G$ this could possibly refer to. Furthermore, a "finitely-generated groupoid" seems to usually have a finite number of objects, so this doesn't look correct.

I then thought of pseudogroups, but all pseudogroup references I found deal with pseudogroups of homeomorphisms, and discuss mostly orthogonal issues. By any definition of a pseudogroup I could find, an action by partial automorphisms is not really a pseudogroup (unless I introduce some topological structure that I'm not going to use). Furthermore, I did not find a discussion of finite generation that tells me what I should do with domains.

There is also literally the notion of a partial action of a group. I thought of this last because I had never actually seen this before (people I know only talk about groupoids), but this was maybe the most promising formalism. I guess I would like to discuss the representations of groups by the "inverse semigroup of partial automorphisms of a group", but I don't know the jargon, and at least based on a brief look I did not find a notion of finite generation with the correct properties.

$\endgroup$
7
  • $\begingroup$ There's a good formalism of partial action of a group $G$ on a set $X$ due to Exel (each group element acts as a partial bijection of $X$, that is, a bijection between two given subsets of $X$, with natural axioms), and in particular, if $G$ acts on a set $M$ and $X$ is an arbitrary subset of $M$, then $G$ naturally partially acts on $X$ by restriction. Orbits of partial actions are well-defined (here these are just intersections of orbits on $M$ with $X$). $\endgroup$ – YCor Feb 22 at 9:23
  • $\begingroup$ About finite generation: for a genuine action, what would you mean by "finitely generated"? $\endgroup$ – YCor Feb 22 at 9:24
  • $\begingroup$ "it is natural enough to say that a group action $G \curvearrowright H$ is finitely generated when $G$ is" I mean this to be the definition. It's just a property of the group, and I'm ok with just saying that $G$ is finitely-generated, my point was just that that's already non-trivial e.g. in the groupoid approach. $\endgroup$ – Ville Salo Feb 22 at 9:27
  • $\begingroup$ So you are saying, I can just say $G$ partially acts on itself by conjugation, and we take the partial subaction on $H$, and then I just say $G$ is finitely-generated. Maybe that is indeed the good setting. But I would like to say that "$G$ partially acts on $H$", and only have the conjugation partial action as a special case; the issue is, what is the big group where $G$ acts? Anyway, maybe I just need to think about this more... $\endgroup$ – Ville Salo Feb 22 at 9:32
  • $\begingroup$ You should look for inverse semigroup instead of pseudogroups. You'll see that there is lots of literature on partial group actions and inverse semigroups starting with the work of Exel @YCor mentioned. A good early introduction is the IJAC paper of Lawson and Kellendonk which also discusses discusses globalizing partial actions. They don't speak of your particular partial action so far as I remember. $\endgroup$ – Benjamin Steinberg Feb 22 at 12:44
5
$\begingroup$

This is not a real answer but things are getting too long for a comment. If $X$ is a set, let $I_X$ be the symmetric inverse monoid, the monoid of all partial bijections of $X$. It is naturally ordered by the restriction relation and this partial order is compatible with multiplication and preserved by inversion. It can also be defined by $f\leq g$ if $f=ge$ for some idempotent $e$ and this definition works in any inverse semigroup. Inverse semigroup is the abstract algebraic structure that encodes pseudogroups of transformations.

A partial group action is intuitively what you get when a group $G$ acts on a set $X$ and you have a subset $Y$ and view each $g\in G$ as giving a partial bijection from $Y$ to itself by defining $gy$ only when $gy\in Y$. In fact, it is shown in the Kellendonk-Lawson article I mentioned in the comments that every partial action arises this way. However, if the partial action preserves some extra structure on $Y$, there is no reason that the globalized set $X$ on which $G$ acts satisfies this extra structure.

The formal definition (which was given by Exel in a different but equivalent way) is a partial action of $G$ on a set $X$ is a dual prehomomorphism $\theta\colon G\to I_X$. What does this mean? It means $\theta(1)=1$, $\theta(g)\theta(h)\leq \theta(gh)$ and $\theta(g^{-1}) = \theta(g)^{-1}$. The middle axioms says that first acting by $g$ and then by $h$ is a restriction of the action of $gh$. This makes sense because maybe $hy$ is not in $Y$ but $ghy$ is.

Because $\theta$ is not a genuine homomorphism, there is no reason that $\theta$ should be determined by what it does to generators of $G$ and in fact it isn't.

There is a way to replace dual prehomomorphisms by inverse semigroup homomorphism using the so-called Birget-Rhodes expansion of a group $G$ (rediscovered by Exel). It is the inverse monoid $M(G)$ whose element consist of pairs $(A,g)$ where $A$ is a finite subset of $G$ containing $1,g$ and you multiply by the rule $(A,g)(B,h) = (A\cup gB,gh)$. There is a natural homomorphism $\pi\colon M(G)\to G$ that projects to the second coordinate and it has the property the the preimage of each element of $g$ has a maximum element in the natural partial order (namely $(\{1,g\},g)$) and so $M(G)$ is what is called an $F$-inverse monoid. There is a dual prehomomorphism $\theta\colon G\to M(G)$ given by $\theta(g) =(\{1,g\},g)$ and it is universal in the sense that any dual prehomomorphism from $G$ into an inverse monoid factors through this one. In particular, partial actions of $G$ on a set $X$ correspond to genuine inverse semigroup actions of $M(G)$ on $X$ via a homomorphism $M(G)\to I_X$.

Now the issue is that if a proper subset $S$ generates $G$, it is rarely the case that $S$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups. So you have some partial actions that are not determined on generators.

When a partial action of a free group on a set is determined by generators is understood, but I don't know about the general case. If you have any $F$-inverse cover of $G$, that is an inverse monoid $M$ with a surjective homomorphism $\pi\colon M\to G$ such that each fiber has a maximum element, then you can define a dual prehomomorphism $\theta\colon G\to M$ by $\theta(g)$ is the max element of its fiber. Then any action of $M$ on a set $X$ can be turned into a partial action of $G$.

Now if $S$ is a set, the free inverse monoid $FIM(S)$ is an $F$-inverse cover of the free group $FG(S)$ under the natural projection. The maximum element of each fiber is the element represented by the reduced word in $S$ and its generators mapping to the group element. So there is a dual prehomomorphism $\theta\colon FG(S)\to FIM(S)$ and using the universal property of $FIM(S)$, it is easy to see that any collection of partial bijections of a set $X$ indexed by elements of $S$ give a partial action of $FG(S)$ in which the action is determined by what happens to $S$. Now somebody I think wrote up in some old papers axiomatically which actions of $FG(S)$ come about this way, but I forget where. If you had any partial action of $FG(S)$ generated by $S$ in any reasonable sense, it will have to factor through this one. It is in some sense, that I forget how to make precise, the universal generator preserving dual prehomomorphism. But I don't know for other groups how to do this.

$\endgroup$
17
  • 1
    $\begingroup$ The multiplication formula in the Birget-Rhodes construction has a typo, $h$ should be $gh$. $\endgroup$ – Ville Salo Feb 22 at 13:18
  • $\begingroup$ I understood some of this, it suggests to me that the problems are real if trying to generalize. Actually, I did not even realize that indeed knowing the partial action of generators does not determine the entire action, I had simpler problems in mind. I think I will abandon the attempt to generalize for now. $\endgroup$ – Ville Salo Feb 22 at 13:34
  • $\begingroup$ The issues I had in mind were more about the formula $\theta(g)\theta(h) \leq \theta(gh)$; I see why you need this, but the issue is that these domain restrictions seem to necessarily lead to having the same element with various domains, and finite generation would require you to get the domains right... which is not really what I care about because I only care about the action. Probably "maximum element" is about exactly this. $\endgroup$ – Ville Salo Feb 22 at 13:40
  • $\begingroup$ The Kellendonk-Lawson paper is here worldscientific.com/doi/10.1142/S0218196704001657 $\endgroup$ – Benjamin Steinberg Feb 22 at 14:12
  • $\begingroup$ If you try to generated an inverse monoid by the the partial bijections coming from your group, then you a quotient inverse monoid of that $F$-inverse cover which may no longer be $F$-inverse, that is, each element of your inverse monoid is a restriction of group elements but it may be a restriction of more than one of them. You will certainly get many elements which are restrictions of a given group element but the case that elements are restrictions of more than one element is in some sense worse. $\endgroup$ – Benjamin Steinberg Feb 22 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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