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If I'm right, then please provide a reference because I was not able to find any online.

Let $G$ be a Lie group with identity $e$ and dimension $n$. Let $G^0$ be the identity component of $G$, and let $\langle S \rangle$ be the subgroup generated by a subset $S$, which need not be a subgroup or submanifold, of $G$.

Because $G$ is a smooth manifold, $G$ is locally Euclidean of dimension $n$. This is defined as that every $g \in G$ has a neighbourhood $V$ that is homeomorphic (and thus diffeomorphic, if I remember correctly) to an open subset of $\mathbb R^n$. However, this equivalently means that every $g \in G$ has a neighbourhood $U$ that is homeomorphic (and thus diffeomorphic, if I remember correctly) to $\mathbb R^n$ itself. For $g=e$, obtain any such $U$, and call it $U_e$.

Questions: Because $\mathbb R^n$ is a Lie group, each $U$ for each $g \in G$, including the $U_e$ for $e$, is (PLEASE read the sentence to its completion before commenting) a Lie group too but not necessarily an embedded Lie subgroup or even a Lie subgroup (PLEASE also see the Note below) .

  1. Specifically for the neighbourhood $U_e$ of $e$ that is homeomorphic to $\mathbb R^n$ and not for arbitrary neighbourhood $U$ of $g$ that is homeomorphic to $\mathbb R^n$ (and certainly not for arbitrary neighbourhood $V$ of $g$ that is homeomorphic to an open subset of $\mathbb R^n$), is $U_e$ an embedded Lie subgroup of $G$?

  2. If not, then does $e$ have an open neighbourhood that is both an embedded Lie subgroup and homeomorphic to $\mathbb R^n$? (This could be $G^0$, but I wasn't able to find anything hinting at this. Also, I don't know how to prove this, if true.)

Note: When I say $U$ is a Lie group but not necessarily a Lie subgroup or an embedded Lie subgroup, I mean this in the same way that $[0,1)$ can be a group that is a subset of $\mathbb R$ (as an additive group) but not a subgroup of $\mathbb R$. I recall this happens when $[0,1)$ is isomorphic to the circle group. Equivalently, this happens when the group operation on $[0,1)$ is, if I remember correctly, $f: [0,1) \times [0,1) \to [0,1)$, $f(a,b)=a+b$ if $a+b < 1$ and $f(a,b)=a+b-1$ if $a+b \geq 1$. Essentially, the group operations of $\mathbb R$ (as an additive group) restricted to $[0,1)$ are not the same as the operations of $[0,1)$ that make $[0,1)$ a group that is a isomorphic to the circle group and of course do not make $[0,1)$ a group.

Some unfortunate background: It can be shown that $G^0$ is both closed and open. Thus, because $U_e$ is connected, we have, by this, that $U_e \subseteq G^0$. Therefore, by definition of $\langle U_e \rangle$ as the intersection of all subgroups of $G$ that contain $U_e$, $U_e \subseteq \langle U_e \rangle \subseteq G^0 \subseteq G$. Initially, I unfortunately incorrectly deduced that $U_e \supseteq G^0$ instead of $U_e \subseteq G^0$. Thus, I was supposed to ask about how each $U_e$ is actually equal to $G^0$ itself. In case I was wrong about $U_e = G^0$ but still right about $U_e \supseteq G^0$, my next bet was that at least $U_e$ was an embedded Lie subgroup. Now that I know I am wrong about $U_e \supseteq G^0$ or at least the way I deduced $U_e \supseteq G^0$, my bet is either that $U_e$ is still an embedded Lie subgroup or that, at least, $e$ has an open neighbourhood that is both an embedded Lie subgroup and homeomorphic to $\mathbb R^n$.

Some context: In relation to my other question In what sense is differential of exponential map an identity map?, I am trying to prove that '$\exp_{\{*, Z_e\}}$' is '$\gamma$', so please take note of this to avoid any circular arguments. I potentially run into a problem if I cannot find an open neighbourhood of $e$ that is both an embedded Lie subgroup and homeomorphic to $\mathbb R^n$.

So far what I've done is show $\langle U_e \rangle = G^0$: Because $U_e$ is homeomorphic to $\mathbb R^n$ and because $\mathbb R^n$ is connected, $U_e$ is connected too. It can be shown that $\langle H \rangle$ is connected and open for any subset $H$ of $G$ that is connected and open. However, it can also be shown that $G^0$ is the only connected open subgroup of $G$. Therefore, $\langle U_e \rangle = G^0$.

Thanks in advance!

By the way, there seems to be a deleted post:

Here is a proof that $U_e=G^0$ if $U_e$ is a subgroup.

Since $U_e$ is a subgroup, $G$ is a disjoint union of cosets $gU_e$ for $g\in G$, which are all homeomorphic to $\mathbb{R}^n$ hence are open. If $U_e$ is a proper subset of $G^0$, then $U^e$ and $G^0-U_e$ are both open and hence form a separation of $G^0$, which is impossible. Thus we must have had that $U^e=G^0$.

Initially the post did not contain 'if $U_e$ is a subgroup'. I asked why $U_e$ was a subgroup, and then the deleted post had a comment that went something like 'It is not necessarily a subgroup, but if it is, then it is equal to $G^0$...For most Lie groups, no subset homeomorphic to $\mathbb R^n$ is an embedded Lie subgroup'.

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Ekhin Taylor R. Wilson
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    "Because $R^$ is a Lie group, each $$ for each $\in $ is a Lie group too" is simply false (assuming that the group structure is restricted from $G$). – Moishe Kohan Nov 08 '19 at 16:54
  • @MoisheKohan I know. See the 'Note' please. – Ekhin Taylor R. Wilson Nov 08 '19 at 16:55
  • @MoisheKohan Additionally my statement is 'Because $\mathbb R^n$ is a Lie group, each $U$ for each $g \in G$, including the $U_e$ for $e$, is a Lie group too but not necessarily an embedded Lie subgroup or even a Lie subgroup.' rather than 'Because $\mathbb R^n$ is a Lie group, each $U$ for each $g \in G$, including the $U_e$ for $e$, is a Lie group too', so the 'but...' phrase should have hinted at what I meant. – Ekhin Taylor R. Wilson Nov 08 '19 at 16:56
  • @MoisheKohan I mean, what else is the text saying if I assumed $U$ is a Lie group with group structure restricted from $G$? If that were true, then $U$ is by definition an embedded Lie subgroup. – Ekhin Taylor R. Wilson Nov 08 '19 at 16:59
  • @MoisheKohan Anyway, I think you (might) have answered question 1. What about question 2? – Ekhin Taylor R. Wilson Nov 08 '19 at 17:17
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    The circle group $S^1 \subset \mathbb C$ is a Lie group. Its identity is $1$, which has many different open neighborhoods diffeomorphic to $\mathbb R^1$. Not one of those neighborhoods is an embedded Lie subgroup of $S^1$. In fact, every open neighborhood of $1$ generates the entire Lie group $S^1$. Does this answer your question? – Lee Mosher Nov 08 '19 at 17:44
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    Or, does the following answer your question? In every connected Lie group $G$, every neighborhood of the identity element generates the whole Lie group, so not one of those neighborhoods is an embedded Lie subgroup of $G$, except for $G$ itself of course. – Lee Mosher Nov 08 '19 at 17:47
  • Saying that a subset of a topological group is a topological group does not quite make sense unless you specify a group operation. I suggest in the future you say "is homeomorphic to a topological group", in order to avoid the confusion. As for the question that $G^0$ is a subgroup of $G$, it is a duplicate of this question: https://math.stackexchange.com/questions/139979/component-of-identity?rq=1. – Moishe Kohan Nov 08 '19 at 18:43
  • @MoisheKohan Thanks for suggesting. When did I ask if $G^0$ is a subgroup? – Ekhin Taylor R. Wilson Nov 12 '19 at 03:55
  • @LeeMosher I wasn't asking about only connected Lie groups, but it seems you want me to think about how to extend your answer to disconnected Lie groups. This is what I understand your answer is: For Number 1, 'no'. For Number 2, 'No. However, yes if $G^0$ is homeomorphic to $\mathbb R^n$. Additionally, $G^0$ is not always homeomorphic to $\mathbb R^n$ such as when $G=S^1$'. Do I understand you correctly? – Ekhin Taylor R. Wilson Nov 12 '19 at 04:02
  • @LeeMosher Can you answer about the exponential map here also please? This question is intended actually for that question. I'm trying to prove differential of $\exp$ is $\gamma$ itself (assuming that is true, which is what that question is asking). – Ekhin Taylor R. Wilson Nov 12 '19 at 04:03
  • @MoisheKohan Thanks for suggesting, but when did I ask if $G^0$ is a subgroup? I already knew that. Also, can you answer about the exponential map here also please? This question is intended actually for that question. I'm trying to prove differential of $\exp$ is $\gamma$ itself (assuming that is true, which is what that question is asking). – Ekhin Taylor R. Wilson Nov 12 '19 at 04:04
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    Your understanding of my comments is correct, although I would word the answer to 2 more informatively: the answer is no in the generality asked; but for given $G$ and a given neighborhood $U$ of $e$, the answer is yes if and only if $U = G^0$ and $G^0$ is homeomorphic to $\mathbb R^n$. – Lee Mosher Nov 12 '19 at 17:13
  • @LeeMosher Oh yes, I should have been clearer and have said 'yes if $G^0$ is homeomorphic to $\mathbb R^n$, in which case we take $U_e=G^0$'. Of course, you would still add the converse. Understood (I think). You can answer now. (I also hope you can please answer my main question about the exponential map.) – Ekhin Taylor R. Wilson Nov 13 '19 at 08:08

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I'll rewrite my comments above as an answer.

To start, for any Lie group $G$, the component $G^0$ of $e$ is indeed an embedded Lie subgroup. Furthermore, for any neighborhood $U_e$ of $e$ that is homeomorphic to $\mathbb R^n$, the subgroup of $G$ that is generated by $U_e$ is $G^0$.

It follows that the only possible neighborhood of $e$ that is homeomorphic to $\mathbb R^n$ and is an embedded Lie subgroup is $G^0$ itself. This answers question 2: "no" if $G^0$ is not homeomorphic to $\mathbb R^n$; and "yes" if $G^0$ is homeomorphic to $\mathbb R^n$, the only such neighborhood being $G^0$ itself.

As a consequence, for examples where no such neighborhood exists, pick $G$ so that $G^0$ is not homeomorphic to $\mathbb R^n$, for example the circle group $G=S^1$, which answers question 1.

Lee Mosher
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