3
$\begingroup$

Is there an infinite $T_2$-space $X$ with $X\cong \text{Aut}(X)$? (Here, $\text{Aut}(X)$ is the set of automorphisms $\varphi:X\to X$ and it carries the topology inherited from the product topology on $X^X$.)

What's the answer if we endow $\text{Aut}(X)$ with the compact-open topology instead of the product topology?

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ Is there a particular reason you want to use the product topology (as opposed to something like the compact-open topology)? $\endgroup$
    – Eric Wofsey
    Commented Jun 18, 2015 at 8:19
  • 3
    $\begingroup$ And is there a particular reason why you would want to want such an homeomorphism in the first place? What's the motivation for this question? $\endgroup$
    – Johannes Hahn
    Commented Jun 18, 2015 at 8:21
  • $\begingroup$ I thought that's the most "natural" topology on $\text{Aut}(X)$... But I will include your suggestion in the question $\endgroup$
    – Dominic van der Zypen
    Commented Jun 18, 2015 at 8:31
  • 1
    $\begingroup$ The compact open topology on $\text{Aut}(X)$ only seems natural for compact spaces; the compact-open topology only behaves the way it is supposed to for locally compact spaces $X$. Furthermore, there are locally compact spaces where the compact-open topology on $\text{Aut}(X)$ is generally not a topological group since the operation $f\mapsto f^{-1}$ is generally not continuous mathoverflow.net/q/58690/22277. However, for compact spaces $X$, the space $\text{Aut}(X)$ however is a topological group. $\endgroup$
    – Joseph Van Name
    Commented Jun 18, 2015 at 21:15
  • $\begingroup$ The most natural structure to put on sets of continuous functions from a non-locally compact space $X$ to a space $Y$ is not a topological structure but a convergence structure. Of course, if $X$ is a uniform space, then the uniformity of uniform continuity on $X^{Y}$ is a fairly natural uniformity to put on such function spaces, but this uniformity induces the compact open topology whenever $X$ is a compact space. $\endgroup$
    – Joseph Van Name
    Commented Jun 18, 2015 at 21:19

1 Answer 1

Reset to default
6
$\begingroup$

In this paper Jan van Mill has constructed a separable metrizable Boolean topological group $X$ such that each homeomorphism $h:X\to X$ is equal to the translation $X\to X$, $x\mapsto x+h(0)$. This implies that the homomorphism group $Aut(X)$ of $X$ (endowed with any topology intermediate between the topology of pointwise convergence and the topology of uniform convergence) is topologically isomorphic (and hence homeomorphic) to $X$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .