11
$\begingroup$

A $k$-marked groups is a pair $(G,S)$ where $G$ is a group and $S$ is an ordered set of $k$ generators of $G$. Each such pair can be identified with a normal subgroup of the free group $F_k$ of rank $k$. Hence, the space of $k$-marked groups $\mathcal{M}_k$ can be identified with the set of normal subgroups of $F_k$ which has a natural topology (inherited from the set of all subsets of $F_k$) making it compact and totally disconnected.

My question is, can $\mathcal{M}_k$ be homeomorphic to $\mathcal{M}_\ell$ when $k\neq \ell$?

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ I think it's an open question (for $k\ge 2$ of course). All we know about these topologies consists in zooming at some given (very specific) places and describe what's in there. $\endgroup$
    – YCor
    Sep 4, 2013 at 5:14
  • $\begingroup$ I do not think this problem has been discussed before. It may not be transcendentally hard. For example, does $Out(F_n)$ act on ${\mathcal M}_n$ by homeomorphisms? Perhaps it does not act on ${\mathcal M}_2$ if $n$ is big enough. That would imply ${\mathcal M}_2\ne {\mathcal M}_n$. $\endgroup$
    – user6976
    Sep 5, 2013 at 9:47
  • 2
    $\begingroup$ Well, I think the debate here demonstrates that it's a nice question! $\endgroup$
    – HJRW
    Sep 6, 2013 at 12:16
  • 1
    $\begingroup$ @Mark: I didn't say the problem is hopeless, but the approach :) On the other hand, the approach by embedding $n$-generated groups into 2-generated groups is more convincing. A first thing would be to show that every point in $\mathcal{M}_n$ has a clopen neighborhood homeomorphic to a clopen subset in $\mathcal{M}_2$, but you really need a special kind of embedding (both injectivity and surjectivity are issues). $\endgroup$
    – YCor
    Sep 6, 2013 at 12:37
  • 1
    $\begingroup$ Here is another approach. $\mathcal M_k$ is a profinite space and so it is determined up to homeomorphism by its boolean algebra of clopen sets. This Boolean algebra is the universal boolean algebra of the join semilattice (with identity) of finitely normally generated subgroups of $F_k$. Said differently, it is the Zariski spectrum of the monoid algebra of this semilattice over the two-element field. It follows that $\mathcal M_k$ is homeomorphic to $\mathcal M_n$ iff the corresponding commutative rings are isomorphic. $\endgroup$
    – Benjamin Steinberg
    Sep 6, 2013 at 14:38

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.