# Profinite extension of a Lie group

Let $H,G,K$ be three topological groups, we say that $G$ is an extension of $K$ by $H$ if the following short sequence $$0\rightarrow H\rightarrow G\rightarrow K\rightarrow 0$$ is exact. (If $H$ is a subgroup of $G$ this is equivalent to $K\cong G/H$)

Now, assume that $H,G,K$ are compact abelian topological groups, therefore, by the structure theorem for compact abelian Lie groups, we have that $K$ is a Lie group if and only if it is isomorphic to $\mathbb{T}^n\times C_k$ where $\mathbb{T}$ is the circle group ($S^1$ or $\mathbb{R}/\mathbb{Z}$), $C_k$ is a finite abelian group and $n\in\mathbb{N}$.

It is well known that every extension of a Lie group by a Lie group is again a Lie group. In particular, every extension of a Lie group by a finite group is a Lie group.

My question is: whether every extension of a compact abelian Lie group by a pro-finite group is a subgroup of $\mathbb{T}^n\times D$ where $D$ is a profinite group for some $n\in\mathbb{N}$.

I am most interested in the case where $H$ is a direct product of finite groups.

Any proof / reference is appreciated.

Edit: I also assume that the groups are Hausdorff. Also every morphism is both topological and algebraic.

• Pontryagin duality translates into an equivalent question about discrete abelian groups. Namely: let $G$ be an abelian group of finite rank (i.e., with a f.g. subgroup such that the quotient is torsion). Is $G$ quotient of $\mathbf{Z}^k\times T$ for some $k$ and some torsion abelian group $T$? A counterexample is $\mathbf{Z}[1/2]$ (or more generally any infinitely generated subgroup of $\mathbf{Q}^n$ for any $n$). Thus its Pontryagin dual (a solenoid) is a counterexample to the question. – YCor Sep 3 '17 at 20:43
• Do you require that G -> K admits a continuous section on a neighbourhood of the identity? – David Roberts Sep 3 '17 at 22:05
• Actually with my previous argument one sees that a discrete abelian group of finite rank is quotient of some $\mathbf{Z^k}\times T$ with $T$ torsion, iff it's isomorphic to some $\mathbf{Z^k}\times T$. Hence for the original question: if a compact abelian group is Lie-by-profinite (with the OP's conventions), then it is isomorphic to a closed subgroup of some $\mathbb{T}^n\times D$ with $D$ profinite iff it's itself isomorphic to some $\mathbb{T}^n\times D$. This is seldom the case (in particular if $G$ is connected, this occurs only if $G$ is a torus). – YCor Sep 3 '17 at 22:34

## 1 Answer

Here is a counterexample.

Take $H=\prod_{n\geq 1}\mathbb{Z}/n\mathbb{Z}$ (or any other profinite group into which $\mathbb{Z}$ injects). Let $\tilde{G}=H\times\mathbb{R}$, and let $G=\tilde{G}/\mathbb{Z}$, where $\mathbb{Z}$ sits inside $\tilde{G}$ diagonally. We get a short exact sequence $$1\to H\to G\to\mathbb{R}/\mathbb{Z}\to 1.$$ Then $G$ is compact, and the connected component of the identity in $G$ is isomorphic to $\mathbb{R}$ as an abstract group (though not as a topological group).

Now suppose $Q$ is a compact subgroup of $\mathbb{T}^n\times D$. The connected component of the identity in $Q$ is a closed subgroup of $\mathbb{T}^n\times \{1_D\}$. An infinite closed subgroup of $\mathbb{T}^n$ must have a non-trivial element of finite order, so cannot be isomorphic to $\mathbb{R}$ as a group. So $Q$ cannot be isomorphic to $G$.

• Hmm.. If I understand your answer then $G$ is not an Hausdorff group is it? (because $\mathbb{Z}$ is not closed in $H$) If not maybe I should edit my question and add the Hausdorff Hypothesis. – Lie groups Sep 3 '17 at 19:45
• It is true that $\mathbb{Z}$ is not closed in $H$, but $\mathbb{Z}$ is closed in $\mathbb{R}$ and also in $\tilde{G}$, so $G$ is Hausdorff. – Julian Rosen Sep 3 '17 at 20:31