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Let $G$ be 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 $G$.

Consider the exponential map $\exp: T_eG \to G$. Let $U$ be an open neighbourhood of $e$ in $G$ ($U$ need not be homeomorphic to an open subset of $\mathbb R^n$). Then $T_eU$ is isomorphic to $T_eG$ under $i_{\{*,e\}}: T_eU \to T_eG$ where $i: U \to G$ is the inclusion map.

Questions:

  1. Is it true that $\exp(i_{\{*,e\}}(T_eU)) \subseteq U$? (Equivalently, $\exp(T_eU) \subseteq U$, once you identify $T_eU$ with $i_{\{*,e\}}(T_eU)$.)

  2. If no, then what if $U$ is homeomorphic to an open neighbourhood of $\mathbb R^n$?

  3. If still no, then what if $U$ is homeomorphic to $\mathbb R^n$? (I think that if $U$ is homeomorphic to $\mathbb R^n$, then $U$ is diffeomorphic to $\mathbb R^n$, so there's no need to inquire further for the case that $U$ is diffeomorphic to $\mathbb R^n$)

All I know so far is that

  1. If $U$ is connected (such as when $U$ is homeomorphic to $\mathbb R^n$), then $U \subseteq G^0$, by this and $\langle U \rangle = G^0$.

  2. If $H$ is an open subgroup of $G$, then $H \supseteq G^0$

  3. Image of $\exp$ is a subset of $G^0$.

  4. Considering $T_eG$ as a manifold (diffeomorphic and isomorphic, possibly Lie group isomorphic, to $\mathbb R^n$), $i_{\{*,e\}}(T_eU)$ is an open subset of $T_eG$ that is diffeomorphic to (and I guess isomorphic and even Lie group isomorphic to) $\mathbb R^n$, even if $U$ is not homeomorphic to $\mathbb R^n$.

  5. For every $V$ open in $T_eG$ (such as $V = i_{\{*,e\}}(T_eU)$, I think), $\langle \exp(V) \rangle = G^0$

Note: Please try not to use anything like $\exp$ is a local diffeomorphism at $Z_e$ because that's what I'm trying to prove (here: Differential of exponential map at identity)

Thanks in advance!

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    $\begingroup$ It seems you are getting horribly mixed up in thinking formally about the definitions. If you stopped and wrote down what all this actually means in some simple examples the answer would be obvious. $\endgroup$ – Eric Wofsey Nov 8 at 17:36
  • $\begingroup$ @EricWofsey 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). $\endgroup$ – Ekhin Taylor R. Wilson Nov 12 at 4:00
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First of all, there is absolutely no reason to talk about $T_eU$. Since $i_{*,e}$ is surjective, $\exp(i_{\{*,e\}}(T_eU))$ is just the same as $\exp(T_eG)$. So your question is, must the image of the exponential map be contained in an arbitrary open neighborhood of $e$?

The answer is then obviously no, since the intersection of all open neighborhoods of $e$ is just $\{e\}$ itself, and the image of the exponential map is more than just $\{e\}$ (assuming $n>0$). The same holds if you restrict to $U$ that are homeomorphic to $\mathbb{R}^n$, since $e$ has arbitrarily small neighborhoods that are homeomorphic to $\mathbb{R}^n$.

For a very simple explicit example, if $G=\mathbb{R}$, then the exponential map is surjective, so any open interval around $0$ is an open neighborhood homeomorphic to $\mathbb{R}$ that does not contain the image of the exponential map.

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