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May 7, 2021 at 0:55 comment added Robert Bryant @SeanEberhard: Ah, you are right. Thanks for the explanation!
May 6, 2021 at 22:17 vote accept aglearner
May 6, 2021 at 21:40 answer added Sean Eberhard timeline score: 4
May 6, 2021 at 21:10 comment added Sean Eberhard @RobertBryant The problem is that the non-compact skew line still doesn't work, because its intersection with a nbd of 1 is dense there
May 6, 2021 at 20:31 comment added Robert Bryant I'm not sure why the OP is modifying the question. In the original question, the OP did not ask that the Lie subgroup $G\subset \mathrm{SO}(n)$ be compact, i.e., closed, just that it exist, and the 'skew-line' that Sean writes down is obviously part of a (non-closed) Lie subgroup. Did the OP mean to require that $G$ be a closed Lie subgroup?
May 6, 2021 at 16:29 history edited aglearner CC BY-SA 4.0
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May 6, 2021 at 16:24 comment added aglearner Dear Sean, thanks! This is indeed a counter-example, but it am not afraid of it. I'll modify the question so that there is still a meaning, which is good enough for me
May 6, 2021 at 15:42 comment added Sean Eberhard What if (a trite example) $X = \{(x, \sqrt{2}x) : x \in [-\epsilon, \epsilon]\} \subset \mathbf{T}^2$ (and the torus is embedded in $\mathrm{SO}(n)$)?
May 6, 2021 at 13:22 history edited YCor CC BY-SA 4.0
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May 6, 2021 at 13:20 history asked aglearner CC BY-SA 4.0