6

Given a topological space $(X,\tau)$, a connected component is a maximal connected subset of $X$ with respect to $\tau$ (which exists). Now, consider the group $G$ of homeomorphisms $\varphi:X\to X$; it naturally induces a graph on $X$ where an edge $(x,y)$ is given by a $\varphi\in G$ with $\varphi(x)=y$. Under what assumptions about the topology can we conclude that the connected components of this graph are precisely the connected components in the topological sense?

Considering the two extremes, we see that when $\tau$ is the trivial topology $\{X,\varnothing\}$, $G=S(X)$; on the other hand, when $\tau$ is the discrete topology and $X$ is bigger than a singleton, we also have $G=S(X)$, but now it is plainly impossible for the two senses of connectedness to coincide, so I think this condition may have something to do with the coarseness of the topology, but I have no clue of where to obtain any more information starting from here.

RobPratt
  • 50,938

2 Answers2

4

The property of connectivity of your graph can be described with standard terminology from topology and dynamics: $X$ is topologically homogeneous if the homeomorphism group $\text{Homeo}(G)$ acts transitively on $X$, meaning (as you say) that for all $x,y \in X$ there exists $\phi \in \text{Homeo}(X)$ such that $\phi(x)=y$.

One necessary condition for $X$ to be topologically homogeneous is that all of the components of $X$ are in the same homeomorphism class. That condition is not sufficient for $X$ to be homogeneous, though, because the space $\{0\} \cup \{1/2,1/3,1/4,1/5,\ldots\}$ is a counterexample.

Another necessary condition of a different type is that the "connectedness quotient" of your space, namely the quotient space defined by the decomposition of $X$ into its components, be topologically homogeneous. An example of this is the Cantor set, which is topologically homogeneous, and so the Cantor set is an example of a disconnected topologically homogeneous space (and a more exotic example than a space with the discrete topology).

I doubt that those two conditions, taken together, are sufficient for $X$ to be topologically homogeneous, although I don't have a counterexample in mind.

So, let's add a condition to $X$, requiring that the topology of its connectedness quotient is discrete --- equivalently, every connected component of $X$ is a clopen set (both closed and open). In this case, topological homogeneity of $X$ is equivalent to the statement that all components are in the same homeomorphism class, and that each member of that class is itself topologically homogeneous. As an effect, for the class of spaces whose connectedness quotient is discrete, your problem is reduced to the connected case.

So, what are the connected topologically homogeneous spaces? There is a large and important class of examples, namely all connected manifolds, as explained here on this site.

There are other scattered examples lying around the mathematical landscape. One example I particularly like is the universal R-tree which is homeomorphic to the asymptotic cone of the hyperbolic plane as shown by Bestvina, and as shown at that link is also homeomorphic to the asymptotic cone of any Gromov hyperbolic group as shown by Erschler. The Menger cube (a.k.a. the Menger sponge) is also connected and topologically homogeneous; see this MathOverflow answer.

Lee Mosher
  • 135,265
0

A trivial condition would be that there is only one connected component in both notions. This would mean that the space is homogeneous with respect to $Homeo(X)$ and also a connected space. I think that cases where the topology comes from a connected topological group will all be such examples.

Keen-ameteur
  • 8,404
  • I don't see immediately how this would be a sufficient condition -- given $x,y$ in the same component, how can we derive a homeomorphism sending $x$ to $y$ out of thin air? And this is certainly not a necessary condition, since we can have a space where two components have different cardinality and each component has homeomorphisms sending points to points. – Akira Satou Sep 15 '24 at 15:31
  • 2
    Although I agree that the wording is confusing, I believe all that @Keen-ameteur is intending to say is that in the connected case, the property of homogeneity is (trivially) equivalent to connectivity of your graph. – Lee Mosher Sep 15 '24 at 16:13
  • @LeeMosher That is indeed what I tried to convey. – Keen-ameteur Sep 15 '24 at 19:28