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Apr 19, 2017 at 12:22 answer added ChanaG timeline score: 2
Jan 12, 2017 at 10:14 answer added user05811 timeline score: 3
Jan 11, 2017 at 20:05 comment added YCor I think that for $G$ compact $Sub(G)/\sim$ is always totally disconnected (this is trivial if $G$ is profinite but otherwise $Sub(G)$ need not be totally disconnected, e.g., for $G=SO(3)$).
Jan 11, 2017 at 15:44 comment added YCor For the bare definition, the topology makes sense on the set of all closed subsets: when the compact group is metrizable and endowed with a compatible metric, it is just given by the Hausdorff distance between nonempty closed subsets. This is a compact set, and the set of closed subgroup is a compact subsets therein. In the totally disconnected case, it's naturally a profinite set.
Jan 11, 2017 at 15:42 comment added YCor There are many references, usually called "Chabauty topology" (for an arbitrary locally compact group); actually Chabauty introduced it in the 50s to compactify the space of lattices in a Euclidean space. This topology appeared at many places, including Bourbaki. A number of papers appeared in Russian journals in the 80s about spaces of subgroups. In the last 10 years it became again fashionable.
Jan 11, 2017 at 15:30 history asked PrimeRibeyeDeal CC BY-SA 3.0