13
$\begingroup$

There exist homogeneous spaces such as the pseudo-arc, which are compact, connected, and totally path-disconnected. Is there a nontrivial, Hausdorff topological group with the same properties, i.e. that is compact, connected, and totally path-disconnected? What about a metrizable example?

$\endgroup$
2
  • 1
    $\begingroup$ (a) Here's the proof that such $G\neq0$ abelian doesn't exist. Path-free implies $Hom(\mathbf{R},G)=0$. Connected means $Hom(G,$finite$)=0$. So, $A$ being the Pontryagin dual: $Hom(A,\mathbf{R})=0$ (i.e. $A$ is torsion) and $Hom($finite$,A)=0$. So $A$ torsion and torsion-free, hence $A=0$. (b) There's no point in elaborating about Hausdorff, since $G$ every path from $G/\overline{\{1_G\})}$ lifts to a path in $G$. (c) The singleton is both connected and totally disconnected, so the question should ask $G\neq 1$ Hausdorff (or that $G$ doesn't carry the indiscrete topology). $\endgroup$
    – YCor
    Mar 8, 2020 at 8:14
  • $\begingroup$ I delete my answer because it is incomplete, and there is a complete answer already... $\endgroup$
    – Bugs Bunny
    Mar 8, 2020 at 11:46

1 Answer 1

18
$\begingroup$

(I'm assuming the groups to be Hausdorff to avoid the discussion degenerate into idle banter.)

The answer is yes: $\{1\}$ is such a group.

The answer to the intended question (which is probably whether there's a nontrivial such group) is no.

Andrew M. Gleason. Arcs in locally compact groups. Proc. Nat. Acad. Sci. U.S.A. 36 (1950), 663-667. Link

1st line of MR review: The author gives an outline of the proof of the following theorem: Every locally compact group which is not totally disconnected contains an arc.


Edit: the above is for arbitrary locally compact groups, but for compact groups it's significantly easier. Indeed, if $G$ is a compact connected group, it follows from the Peter-Weyl theorem and the basic structure of compact connected Lie groups that there exists a group $H=A\times\prod_{i\in I}S_i$, where $A$ is a compact connected abelian group, and each $S_i$ is a simple, simply connected compact Lie group, and a surjective homomorphism $H\to G$ with totally disconnected kernel (this is for instance in Bourbaki, Lie, Chap 9, appendix). If $G\neq 1$, then $H\neq 1$, and then $\mathrm{Hom}(\mathbf{R},H)\neq\{1\}$ (since either $I$ is non-empty, or $A\neq 1$, and the abelian case is settled by Pontryagin duality as I mentioned in a comment. The composition map $\mathrm{Hom}(\mathbf{R},H)\to \mathrm{Hom}(\mathbf{R},G)$ being injective (because $\mathrm{Ker}(H\to G)$ is totally disconnected), one deduces $\mathrm{Hom}(\mathbf{R},G)\neq\{1\}$.

(Note: Peter-Weyl was established around 1925, and Pontryagin duality in the early 1930's; the basic structure of compact Lie groups was known before these dates; I'm not sure of an early reference for the structural result on compact connected groups but it follows easily so I guess was known to people working on Hilbert's 5th problem in the late 1940's).


Edit 2: one of the results in Hilbert's 5th problem is that for every connected locally compact group $G$, every neighborhood of $1$ contains a compact normal subgroup $W$ such that $G/W$ is Lie. Also it was proved by Iwasawa around 1950 that every connected Lie group $G$ has a maximal compact subgroup $K$, and that such $K$ is connected.

One this is granted, one reduces from the compact case to the locally compact case as follows: let $G$ be a nontrivial connected locally compact group. Let $W$ be a compact normal subgroup such that $G/W$ is Lie. Let $K/W$ be a maximal compact subgroup of $G/W$. I claim that $K$ is connected. Granting the claim and the compact case, we're done if $K\neq 1$. Otherwise $K=1$ and hence $W=1$, so $G$ is Lie and this case is fine.

If $K$ were not connected, $K$ would have the nontrivial profinite quotient $K/K^\circ$, and hence a nontrivial finite quotient, say with kernel $K'$. Hence there exists a symmetric neighborhood $N$ of $1$ in $G$ such that $NK'\cap K=K$. Let $W'$ be a compact normal subgroup of $G$ contained in $N$, such that $G/W'$ is Lie Since $K$ is maximal compact, we have $W'\subset K$, and $K/W'$ is a maximal compact subgroup of $G/W'$, but it is not connected, contradicting Iwasawa's result.

$\endgroup$

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.