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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?

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    $\begingroup$ (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 at 8:14
  • $\begingroup$ I delete my answer because it is incomplete, and there is a complete answer already... $\endgroup$ – Bugs Bunny Mar 8 at 11:46
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(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.

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