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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 24, 2011 at 23:03 comment added Ryan Budney @algori: there are of course notions of manifold that do not demand 2nd countability. That was the point of my comment.
May 24, 2011 at 21:53 comment added algori Ryan -- a Lie group is a smooth manifold, so it is second countable.
May 24, 2011 at 21:48 comment added Ryan Budney If you don't demand Lie groups are 2nd countable, any abstract group with the discrete topology is a $0$-dimensional Lie group. I suspect this is what Claudio and Qiaochu are talking about. If you demand 2nd countability, these are not examples.
May 24, 2011 at 21:46 history edited algori CC BY-SA 3.0
fixed the end of the argument
May 24, 2011 at 21:42 comment added algori Claudio -- I assume $G$ is a Lie group, so $G$ with discrete topology would not do. But you are right that in the end the argument uses connectedness. However, this is easily fixed.
May 24, 2011 at 21:32 comment added Qiaochu Yuan Yes, I really think you want to talk about "unique connected Lie group structure" or else you can give any Lie group the discrete topology (making it a $0$-dimensional Lie group).
May 24, 2011 at 21:28 answer added Jim Humphreys timeline score: 2
May 24, 2011 at 21:21 comment added Claudio Gorodski In 3., it seems you are assuming $G$ is compact and connected. For instance, you could have the discrete topology in $G$.
May 24, 2011 at 20:37 comment added algori woops! Thanks, Ben, will correct this.
May 24, 2011 at 20:35 comment added Ben Webster Um, $T^n$ has $2^n$ elements of order 2. Not that that changes your point.
May 24, 2011 at 20:34 history edited algori CC BY-SA 3.0
corrected a mistake
May 24, 2011 at 20:28 history asked algori CC BY-SA 3.0