Let $G$ be any group. Is there a topological space $(X,\tau)$ such that the automorphism group $\textrm{Aut}(X,\tau)$ is isomorphic to $G$?
$\begingroup$
$\endgroup$
5
asked Nov 25, 2014 at 7:53
-
2$\begingroup$ At least for finite groups it is true. $\endgroup$– DuneNov 25, 2014 at 8:13
-
3$\begingroup$ According to the thread mathoverflow.net/questions/37356/… on MathOverflow, the answer is yes. $\endgroup$– Jeremy RickardNov 25, 2014 at 8:36
-
$\begingroup$ @JeremyRickard I don't understand; the topological automorphisms of a graph is a big group, much bigger than the group of graph theoretic automorphisms. $\endgroup$– user98602Nov 25, 2014 at 18:02
-
1$\begingroup$ @MikeMiller The thread I pointed to contains a reference to a construction for topological spaces as well as one for graphs: de Groot, J. (1959), Groups represented by homeomorphism groups, Mathematische Annalen 138 $\endgroup$– Jeremy RickardNov 25, 2014 at 19:29
-
$\begingroup$ @JeremyRickard Ah, thanks. Sorry, don't know how I missed that. $\endgroup$– user98602Nov 25, 2014 at 19:30
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
As pointed out in the comments, this has been answered by Tony Huynh on MathOverflow. In
de Groot, J. ($1959$), Groups represented by homeomorphism groups, Mathematische Annalen $138$
the author shows that:
"for every group $G$ one can find a complete, connected, locally connected metric space $M$ of any positive dimension such that $G \cong A(M)$"
where $A(M)$ denotes the autohomeomorphism group of $M$.
answered Jan 26, 2015 at 2:53