Skip to main content
8 events
when toggle format what by license comment
Mar 9, 2019 at 14:24 vote accept rori
Mar 9, 2019 at 13:48 history edited YCor
edited tags
Mar 9, 2019 at 13:45 comment added YCor Your reduction to the connected case is not correct. The connected case is easy (find a discontinuous automorphism of $Z(G)^\circ$ that is identity on $Z(G)^\circ\cap [G,G]$. In general, one needs to extend such an automorphism and this requires some argument and I don't think you're giving one (you need to construct an automorphism).
Mar 9, 2019 at 13:44 answer added YCor timeline score: 2
Mar 9, 2019 at 12:40 review Close votes
Mar 9, 2019 at 13:50
Mar 9, 2019 at 12:30 comment added YCor As regards your last question, yes: in a compact abelian Lie group, any countable subset is contained in a countable direct summand (for the abstract group structure).
Mar 9, 2019 at 11:45 review First posts
Mar 9, 2019 at 12:36
Mar 9, 2019 at 11:40 history asked rori CC BY-SA 4.0