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I am an engineer and have been trying to study Chapter 7 of Lee’s Introduction to Smooth Manifolds. I am struggling with the following and would greatly appreciate any help. I do not have an undergraduate education in Mathematics so if my question seems obvious I apologize in advance.

Proposition 7.15 states that every connected component of a Lie group $G$ is diffeomorphic to the identity component, which is the only open connected subgroup of $G$.

This seems to imply that connected components that don’t contain the identity are not open, but how can a non-open component be diffeomorphic to the identity component which is open? Doesn’t a diffeomorphism preserve openness?

I would appreciate if someone can clarify where the flaw is in my reasoning and answer my question.

Many thanks.

Sabiske
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There are two flaws in your attempted reasoning:

  1. The sentence "This seems to imply that connected components that don’t contain the identity are not open" is totally unjustified. I think you are confused by the fact that the identity component $G_0< G$ is the only open subgroup of $G$. But other connected components of $G$ (which all have the form $gG_0$, for some $g\in G$) are not subgroups of $G$ (provided that they are different from $G_0$, i.e. $g\notin G_0$). The reason is quite simple: Each subgroup of $G$ contains $1\in G$ and if some component of $G$ contains $1$, then it is the same component at $G_0$.

  2. As a separate matter, you say "how can a non-open component be diffeomorphic to the identity component which is open? Doesn’t a diffeomorphism preserve openness?" It can easily happen that if $X$ is a topological space and $U\subset X$ is an open subset, then $U$ is homeomorphic to another subset $V\subset X$, where $V$ is not open in $X$. The thing is, such a homeomorphism a priori does not come from a homeomorphism $X\to X$. A simple example of this situation is the topological space $$ X=[0,2]\subset \mathbb R $$ equipped with the subspace topology. Then $U=[0,2]$ is open in $X$ (because it equals the entire $X$) but it is homeomorphic to the subset $V=[0,1]\subset X$, which is not open in $X$.

Unless you already have done so, my suggestion is to read a bit on general (aka point-set) topology before studying Lie groups. There are many textbook recommendations, for instance, here.

Moishe Kohan
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  • Hello, many thanks for taking the time to answer and provide clarification. I appreciate it. I misinterpreted the wording “only open connected subgroup” as the identity component being the only one that is open, i.e., I somehow thought openness was also enjoyed only by the identity component. – Sabiske Jan 05 '25 at 06:07
  • As for the second comment, I gather that my statement about openness being preserved by a diffeomorphism is only correct if the openness in question is relative to the same space over which the diffeomorphism is defined. So in my case, I can take the left-translation $L_g: G \rightarrow G$ which is a diffeomorphism on $G$ and so the openess in $\mathbf{G}$ of the identity component would be preserved under the diffeomorphism.

    Many thanks again.

     – Sabiske Jan 05 '25 at 06:13