Talk:Identity component

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On properties, the paragraph talking about the identity component being open in some cases, what is that talking about exactly? It is a component, so by definition it is automatically open and closed. 88.14.189.237 (talk) 16:36, 3 January 2017 (UTC)Reply

Indeed, there is no reason for a non-continuous automorphism to fix the identity component. Consider ${\displaystyle G=(\mathbb {R} ,+)\times (\mathbb {Q} ,+)}$, then ${\displaystyle G^{0}=(\mathbb {R} ,+)\times \{0\}}$. ${\displaystyle G}$ can be viewed as a ${\displaystyle \mathbb {Q} }$-vector space (so an automorphism of ${\displaystyle G}$ as a group is just a an automorphism of ${\displaystyle G}$ as a ${\displaystyle \mathbb {Q} }$-vector space), and ${\displaystyle G^{0}}$ is a subspace of codimension 1. Must every automorphism of a ${\displaystyle \mathbb {Q} }$-vector space fix a certain subspace of codimension 1? Of course not! 129.104.241.17 (talk) 20:33, 5 February 2025 (UTC)Reply

The identity component of a topological group is closed, but perhaps the identity path component is not. I think that a solenoid may be an example? It is connected, but its path components are not closed?

This should be mentioned in the entry if the assertion is true and we can find some references. 129.104.241.17 (talk) 20:38, 5 February 2025 (UTC)Reply

Note that the page inverse limit mentions the p-adic solenoid. I believe that this space is connected, while the path-connected component of ${\displaystyle 0}$ is ${\displaystyle \mathbb {R} =\mathbb {Z} +[0,1)}$ which is dense dans the whole space, just like ${\displaystyle \mathbb {Z} }$ is dense in ${\displaystyle \mathbb {Z} _{p}}$. 129.104.241.17 (talk) 22:19, 5 February 2025 (UTC)Reply