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

  • 1
    $\begingroup$ @NickS it's the topological dimension. So a discrete group has dimension 0, is not a counterexample. $\endgroup$ – YCor Mar 8 at 6:46

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.