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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!

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  • $\begingroup$ 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
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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.)

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  • $\begingroup$ 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

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