32
$\begingroup$

I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism $\phi : G \to \mathbb{Z}$ (without assuming any topological condition on this morphism).

For compact Lie groups, using the exponential map, the answers is no, but in general I don't know.

$\endgroup$
2
  • 4
    $\begingroup$ If G is metrizable then the answer is no - by Dudley's theorem, $\phi$ must be continuous hence is zero everywhere. I don't know what happens in general. $\endgroup$ Nov 15, 2011 at 9:11
  • $\begingroup$ Thanks. However, if I got it well I can use Dudley's theorem has, if G is metric and complete, a morphism from G to Z is continuous? For Dudley's theorem I found this math.univ-lyon1.fr/~melleray/rapport-LeMaitre.pdf , part 3. Otherwise, there is the following (part 3 too), but I didn't understand the proof : math.univ-lyon1.fr/~melleray/Rosendal.pdf $\endgroup$ Nov 15, 2011 at 9:53

3 Answers 3

52
$\begingroup$

The answer is no in general, but this is a rather deep fact.

Theorem: (Nikolov, Segal) If $G$ is any compact Hausdorff topological group, then every finitely generated (abstract) quotient of $G$ is finite.

N. Nikolov and D. Segal, Generators and commutators in finite groups; abstract quotients of compact groups, arXiv, http://arxiv.org/abs/1102.3037

$\endgroup$
3
  • 3
    $\begingroup$ 1+. It's hard to believe that such a result was only proven recently, in 2011. $\endgroup$ Nov 15, 2011 at 11:43
  • 4
    $\begingroup$ this is a 2nd generation proof -- the first one is from 2005 :) $\endgroup$
    – kassabov
    Nov 15, 2011 at 16:43
  • 3
    $\begingroup$ As Alain mentions, the first proof (with target $\mathbf{Z}$) is from R. Alperin in the early 80's; actually the 1982 paper refers to the older [Compact groups acting on trees. Houston J. Math. 6 (1980), no. 4, 439--441.] The proof is subtle but quite elementary, this is considerably simpler than the Nikolov-Segal result. $\endgroup$
    – YCor
    Jun 5, 2012 at 21:48
42
$\begingroup$

Andreas shot first, but I still encourage everybody to have a look at the lemma on p.263 of R. Alperin,Locally compact groups acting on trees and property $T$. Monatsh. Math. 93 (1982), no. 4, 261–265: any homomorphism from a locally compact group to $\mathbb{Z}$, is continuous. This answers Florent's question.

$\endgroup$
1
  • 3
    $\begingroup$ This is a nice lemma. The heart of the matter seems to be in proving that a compact connected group is divisible, which uses a little structure theory. Once you know this then you know that any map $G\to\mathbf{Z}$ factors through the profinite group $G/G_0$, and then by the universal property of profinite completion you need only check there is no nontrivial homomorphism $\widehat{\mathbf{Z}}\to\mathbf{Z}$, which is straightforward. $\endgroup$ Dec 19, 2014 at 11:07
23
$\begingroup$

Sorry for resurrecting such an old question, but I think we can give a much simpler proof here. We'll reduce the problem from $G$ to the Bohr compactification $B\mathbf{Z}$ of $\mathbf{Z}$, then from $B\mathbf{Z}$ to the profinite completion $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ of $\mathbf{Z}$, and then we'll argue directly.

Let $\phi:G\to\mathbf{Z}$ be a homomorphism and fix $x\in G$. The map $\mathbf{Z}\to G$ extending $1\mapsto x$ induces a map $B\mathbf{Z}\to G$ such that $1\mapsto x$, and thus we obtain a map $\phi':B\mathbf{Z}\to \mathbf{Z}$ such that $\phi'(1)=\phi(x)$.

Recall that to construct $B\mathbf{Z}$ one takes the dual of $\mathbf{Z}$, namely $\mathbf{R}/\mathbf{Z}$, strips the topology to get the discrete group $\mathbf{R}_d/\mathbf{Z}\cong\mathbf{R}_d\times\mathbf{Q}/\mathbf{Z}$, then takes the dual again. The result is that $B\mathbf{Z} \cong B\mathbf{R}\times\hat{\mathbf{Z}}$. Since $B\mathbf{R}$ is divisible $\phi'$ must vanish on $B\mathbf{R}$. Since $\prod_{p\neq 2}\mathbf{Z}_p$ is infinitely $2$-divisibile and $\mathbf{Z}_2$ is infinitely $3$-divisible, $\phi'$ vanishes on $\hat{\mathbf{Z}}$. Thus $\phi'$ is identically $0$, so $\phi(x)=\phi'(1)=0$.

$\endgroup$
7
  • $\begingroup$ Very nice. Let me slightly restate your argument: 1) replacing $G$ by the closure of some $x\in\phi^{-1}(\{1\})$, we can suppose that $G$ is abelian with a dense cyclic subgroup. 2) if $G$ is any connected compact abelian group then the result holds because $G$ is divisible (as a projective limit of tori) 3) now supposing $G$ has dense cyclic subgroup, $\phi$ vanishes on its unit connected component because of (2), hence we can suppose $G$ totally disconnected, hence a quotient of the profinite completion $\hat{Z}\simeq\prod_p\mathbf{Z}_p$, and your last argument finishes the job. $\endgroup$
    – YCor
    Jan 22, 2015 at 22:43
  • $\begingroup$ @YCor, I think you mean $x \not\in \phi^{-1}(\{0\})$, right? $\endgroup$
    – LSpice
    Jan 22, 2015 at 23:39
  • 1
    $\begingroup$ You do not need $\phi(x) \not= 0$, as your argument directly shows every homomorphism $\phi \colon G \rightarrow \mathbf Z$ is identically $0$. Pick any $x \in G$ and your argument gives a homomorphism $\phi' \colon B\mathbf Z \rightarrow \mathbf Z$ where $\phi'(1) = \phi(x)$. You show $B\mathbf Z = B\mathbf R \times \widehat{\mathbf Z}$, and you give an argument that any homomorphism $B\mathbf R \rightarrow \mathbf Z$ is $0$ and any homomorphism $\widehat{\mathbf Z} \rightarrow \mathbf Z$ is $0$. Thus $\phi'$ is $0$, so $\phi(x) = \phi'(1) = 0$. Since $x \in G$ was arbitrary we're done. $\endgroup$
    – KConrad
    Jan 23, 2015 at 0:18
  • 7
    $\begingroup$ I'm extremely happy to see an old question resurrected because of a new and informative answer, rather than merely bumped by reformatting or tweaking of existing answers $\endgroup$
    – Yemon Choi
    Jan 23, 2015 at 12:54
  • 2
    $\begingroup$ This is really a proof from the Book. Thanks very much. $\endgroup$
    – Todd Trimble
    Jan 23, 2015 at 14:10

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.