# Is there a compact, connected, totally path-disconnected topological group?

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?

• (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). – YCor Mar 8 '20 at 8:14
• I delete my answer because it is incomplete, and there is a complete answer already... – Bugs Bunny Mar 8 '20 at 11:46

(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.