Is every Hausdorff, locally compact group that does not contain any non-trivial compact group, finitely dimensional?

## 1 Answer

Yes, it's even a Lie group whose unit component is a semidirect product $R\rtimes S^n$, where $R$ is a simply connected solvable Lie group and $S$ is the universal covering of $\mathrm{SL}_2(\mathbf{R})$.

Indeed, by van Dantzig, every locally compact group $G$ has an open subgroup $U$ such that $U/U^\circ$ is compact. By the solution to Hilbert's fifth problem, $U$ has a compact normal subgroup $W$ such that $U/W$ is Lie (necessarily with finitely many components). Hence, in the current setting, $W=1$, so $U$ is Lie and $U^\circ=G^\circ$ is open in $G$.

Every connected Lie group with maximal compact subgroup $K$ is homeomorphic to $\mathbf{R}^d\times K$ for some $d$ (Iwasawa). Hence, $G^0$ is homeomorphic to $\mathbf{R}^d$ for some $d$. Being simply connected, it is semidirect product $R\rtimes T$ with $R$ its radical, necessarily simply connected, and $T$ a semisimple Levi factor, simply connected and hence $T$ is a direct product of simple simply connected Lie group. The only possibility for such simple factor to be contractible is indeed the 3-dimensional $S$.

Added: here's a characterization. The contractible Lie groups are characterized above.

**Proposition.** A locally compact group $G$ has no nontrivial compact subgroup iff $G$ is Lie (i.e., $G^0$ is Lie and open) and the discrete quotient $G/G^0$ is torsion-free.

Proof: clearly, these are necessary conditions. Conversely, suppose that $G$ has no nontrivial compact subgroup. By the above, $G^0$ is open and is a contractible Lie group. We claim that $G/G^0$ is torsion-free (note that this is not so immediate since the surjection $G\to G/G^0$ may not be split). Otherwise, let $F$ be a nontrivial finite subgroup of $G/G^0$, and let $H$ be its inverse image in $G$: this is an open subgroup of $G$, which is Lie, virtually connected but not connected. A result of Mostow (valid in any virtually connected Lie group $H$) is that $H^0K=H$ for some compact subgroup $K$ of $H$. So $K=\{1\}$. Hence $H^0=H$, thus $H$ is connected, contradiction.