Do you have an example of an infinite Hausdorff nonabelian topological group $(G,\mathcal T)$ such that for any nontrivial group topology $\mathcal S$ on $G$ with $\mathcal S\subseteq \mathcal T$ we have $\mathcal S = \mathcal T$?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ I assume, which is a standard, that only Hausdorff spaces are considered in the context of topological groups. Then arbitrary compact group would do. Mentioning groups in this context does not add anything to the topological situation when only examples are needed. $\endgroup$– Włodzimierz HolsztyńskiCommented Feb 7, 2015 at 17:31
-
$\begingroup$ @WłodzimierzHolsztyński: In some contexts only Hausdorff group topologies are considered. But Hausdorffness is not part of definition. By these comments (click link) , an abelian group cannot have Haussdorff atom. So group structure matter. btw, I do not have access to this paper and just read the abstract. $\endgroup$– Minimus HeximusCommented Feb 7, 2015 at 17:44
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
Any compact group with no nontrivial normal closed subgroups has this property, since there can be no coarser Hausdorff topology and in any coarser non-Hausdorff topology the closure of the identity would be a normal closed subgroup in the original topology. For instance, this includes all (centerless) compact simple Lie groups.
answered Feb 7, 2015 at 15:40
-
$\begingroup$ Do you have an example of a centerless compact simple Lie group, assuming that I just know a definition of Lie groups? $\endgroup$ Commented Feb 16, 2015 at 0:22
-
$\begingroup$ There may be a problem with Any in your answer. In fact the topology must also be nontrivial. So there must be at least three distinct group topologies on the group. $\endgroup$ Commented Feb 16, 2015 at 1:15
-
1$\begingroup$ The smallest example is $SO(3)$. $\endgroup$ Commented Feb 16, 2015 at 2:31