So I don't have to worry about formalities, in the following let $\mathscr{C}$ be a sufficiently nice category--at least nice enough so that the following definition makes sense. I believe concrete and locally small is enough, but I really am just thinking of tame examples like $\mathbf{Grp},\mathbf{Ring},R$-$\mathbf{Mod}$,$\mathbf{Top}$, etc.

Definition: Let us say that a homgeneity pair in $\mathscr{C}$ consists of a pair $(X,G)$ where $X$ is an object of $\mathscr{C}$ and $G$ is a subgroup of $\text{Aut}_\mathscr{C}(X)$ such that the usual action $G\curvearrowright X$ is transitive. Let us say that an object $X$ of $\mathscr{C}$ has a homogeneity principle if there exists a subgroup $G$ of $\text{Aut}_\mathscr{C}(X)$ such that $(X,G)$ is a homogeneity pair.

I have two questions. The first is simple:

  • What are some examples of categories $\mathscr{C}$ for which we have a full description of the elements with a homogeneity principle?

EDIT: As Thomas Andrews points out below, there is obviously a big boo-boo with the below. Non-trivial groups can never have a homogeneity principle, because the identity element is uniquely different from other elements. The below shows that the only f.g. groups such that any two non-zero points look alike are the elementary abelian $p$-groups. All vector spaces have the same issue--everything looks the same, except the zero point. I will try to think about how to remedy this, to make it coherent.

For example, in the category $f.g.\mathbf{Grp}$ of finitely generated groups, I claim that the only objects with a homogeneity principle are the elementary abelian $p$-groups. First note that necessarily any such $G$ must be equal to any of its characteristic subgroups, and so in particular, $G=Z(G)$. Now, let us then note that any such $G$ cannot have a free part. Indeed, suppose that $G\cong \mathbb{Z}^r\times A$ where $A$ is some finite abelian group. Note that there is no element of $\text{Aut}(G)$ sending $(1,0,\ldots,0,0)$ to $(2,0,\ldots,0,0)$ since $2x=(1,0,\ldots,0,0)$ has no solution with $x\in G$. Thus, we see that $G$ is necessarily finite abelian. Note necessarily every element of $G$ has the same order. Immediately from this we see that if

$$G\cong (\mathbb{Z}/p_1^{e_1}\mathbb{Z})\times\cdots\times(\mathbb{Z}/\mathbb{Z}p_m^{e_m})$$

then $p_1^{e_1}=\cdots=p_m^{e_m}=p$. Thus, $G$ is necessarily an elementary abelian $p$-group. Conversely, note that if $G$ is an elementary abelian $p$-group then $\text{Aut}_{f.g.\mathbf{Grp}}(G)=\text{Aut}_{\mathbf{Vec}_{\mathbb{F}_p}}(G)$ from where the fact that $G$ has a homogeneity principle as an element of $f.g.\mathbf{Grp}$ follows from the following trivial observation:

Observation: For any field $k$, every object of $\mathbf{Vec}_k$ has a homogeneity principle.

Examples of objects with homogeneity principle in $\mathbf{Top}$ and $\mathbf{Man}_\infty$ are topological groups and Lie groups respectively. Note though that, in general, group objects of a category do not possess a homogeneity principle. For example, in $\mathbf{Grp}$ the group objects are just abelian groups, and so can't have a homogeneity principle in general (by the above). The problem for $\mathbf{Grp}$ is that the map $G\to G\times G:x\mapsto (g,x)$ is a morphism if and only if $g=e$.

So, can we classify all the elements of $\mathbf{Top}$ or $\mathbf{Man}_\infty$ with a homogeneity principle? What about for other common categories?

The second question is almost certainly too complicated to answer. For any category $\mathscr{C}$ as above, we can form a new category $\mathscr{D}$, the category of homogenous pairs. The objects of $\mathscr{D}$ consist of homogeneity pairs $(X,G)$ and the morphisms $(X,G)\to (Y,H)$ consist of pairs $(f,\varphi)$ where $f:X\to Y$ is a $\mathscr{C}$-morphism, and $\varphi:G\to H$ is a group map such that for all $x\in X$ and $g\in G$ we have that $f(gx)=\varphi(g)f(x)$ (i.e. $f$ and $\varphi$ intertwine the actions of $(X,G)$ and $(Y,H)$).

For example, if we let $\mathscr{D}$ be the category of homogeneity pairs for $\mathbf{Man}_\infty$, then the category $\mathbf{Lie}$ of Lie groups is naturally identifiable with a subcategory of $\mathscr{D}$. Indeed, we can identify $\mathbf{Lie}$ with the category of pairs $(G,G)$ and $(f,f)$ where $f$ is a Lie group map.

So the second question is:

  • Are there any categories $\mathscr{C}$ for which we can find get a handle on the category of homogeneity pairs (e.g. find a more familiar category it is equivalent to)?

EDIT: Also, if anyone has any tag suggestions, that would be greatly appreciated!

  • $\begingroup$ That only works if the objects in your category are sets and maps are functions, which makes the definition non-categorical. $\endgroup$ – Thomas Andrews Aug 6 '13 at 20:50
  • $\begingroup$ @ThomasAndrews That's correct, I didn't mean this to be categorical per se (as I said in the beginning, this really only makes sense in concrete locally small categories) I just wanted a framework to define the concept for all the common categories we know and love ($\mathbf{Grp},\mathbf{Top}$, etc.) and this just seemed like convenient phrasing. $\endgroup$ – Alex Youcis Aug 6 '13 at 21:13
  • $\begingroup$ Also, why define in terms of subgroups? If a subgroup of $\mathrm{Aut}(X)$ acts transitively, then $\mathrm{Aut}(X)$ acts transitively. $\endgroup$ – Thomas Andrews Aug 6 '13 at 21:41
  • $\begingroup$ @ThomasAndrews That's a good point. This was all coming from thinking of Lie groups where, technically, you're focused on the actual Lie group as a subgroup of the automorphism group. $\endgroup$ – Alex Youcis Aug 6 '13 at 21:47
  • $\begingroup$ Oh, and as sets, the automorphism of a group can't act transitively, because they can't send the identity to anything other than the identity. Same with vector spaces and $0$. You need some other definition of transitivity if you want to deal with groups and vector spaces. $\endgroup$ – Thomas Andrews Aug 6 '13 at 21:47

A definition which might be a useful extension is to say pick a set of objects, $\mathcal Y$ and say an object $X$ is $\mathcal Y$-homogeneous if, for any $Y\in\mathcal Y$ the action of $\mathrm{Aut}(X)$ on the set of monomorphisms $Y\to X$ is transitive.

This definition has the advantage that is it categorical, at least for locally small categories.

It's not clear how we can get your idea with $\mathbf{Ab}$.

In $\mathbf{Vec}_k$ your notion of transitivity is actually $k$-transitivity - that is, $\mathcal Y=\{k\}$.

If the category is $\mathbf{Top}$, and we define $\mathcal Y=\{1\}$ then we can say $X$ is $1$-homogeneous in the normal sense. If $\mathcal Y=\{\mathbf n\}$ consists of the space of $n$ discrete points, then a space is $\mathbf n$-homogenous if $\mathrm{Aut}(X)$ is $n$-transitive. (A while back, I asked on this site for examples of spaces that are $1$-homogenous but not $2$-homogenous.)

  • $\begingroup$ This also covers the geometric case, I believe. This implies that in the case of, say $\mathbf{Top}$, $\text{Homeo}(X)$ acts transtiively on the open sets of $X$, by considering inclusions. Right? $\endgroup$ – Alex Youcis Aug 6 '13 at 22:22
  • $\begingroup$ Actually, it implies it acts transitively on the points of $X$ since there is a space $1$. $\mathbf{Top}$ is a "well-pointed category." en.wikipedia.org/wiki/Well-pointed_category $\endgroup$ – Thomas Andrews Aug 6 '13 at 22:26
  • $\begingroup$ What do you mean? @AlexYoucis Nothing in this definition says you need to "get from" $1\to A$ to $A\to A$. For distinct $Y_1,Y_2$, there is no way that $Aut(X)$ can send a monomorphism $Y_1\to X$ to a mono $Y_2\to X$. It only has to be transitive for each fixed $Y$. $\endgroup$ – Thomas Andrews Aug 6 '13 at 22:30
  • $\begingroup$ And no, $\mathrm{Homeo}(X)$ doesn't act transitively on the open sets - it can't sent a disconnected open set to a connected open set. $\endgroup$ – Thomas Andrews Aug 6 '13 at 22:32
  • $\begingroup$ Hmmm, but this definition is much stronger than transitivity in topological spaces. So maybe you need to pick specific base examples $Y$. $\endgroup$ – Thomas Andrews Aug 6 '13 at 22:35

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.