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Hello i am trying to prove the following proposition :

Let $G$ be a connected Lie group, and $U\subset G$ a neighborhood of the identity element. Also, let $U^k = \{g_1 . g_2 . \dots g_k : g_i \in U\}$ be the set of k - fold products of elements of U.

Then, $G=\cup_{k=1} ^ \inf U^k$.

I read somewhere that, this result follows immediately from connectedness of $G$ but it is not so obvious to me.

Thank you for your time!

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    $\begingroup$ $G$ is the union of its $1$-parameter subgroups. $\endgroup$ – Alex Degtyarev Jun 5 '14 at 12:51
  • $\begingroup$ This is a standard result in Lie group theory (and in fact valid for a much larger class of topological groups) You find this in many textbooks on Lie groups as eg in Hilgert-Neeb $\endgroup$ – Stefan Waldmann Jun 5 '14 at 13:02
  • $\begingroup$ The union above, say $\tilde U$, is open. Define an equivalence relation on $G$ by declaring $g\sim h$ if $gh^{-1}\in \tilde U$. Show that equivalence classes are open. The rsult is in Frank Warner's book. $\endgroup$ – Claudio Gorodski Jun 5 '14 at 13:03
  • $\begingroup$ Thank you all. Stefan it is my first contact with the subject. Your answer was really helpfull. $\endgroup$ – Mark Jun 5 '14 at 13:22
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This is true for any connected topological group $G$.

Let $V = \bigcup_{k=1}^\infty U^k$. It is clear that $V$ is open. On the other hand, let $g \in \bar{V}$. Since $gU^{-1}$ is a neighborhood of $g$, it must intersect $V$. Let $h \in V \cap gU^{-1}$. Then

  • $h = gu^{-1}$ for some $u \in U$, and
  • $h = u_1 \ldots u_k$ for some $k \in \mathbb{N}$ and $u_1, \ldots, u_k \in U$.

Hence $g = u_1 \ldots u_k u \in U^{k+1} \subset V$, proving $V$ is closed. Since $G$ is connected, and $V$ is both open and closed, we have $V = G$.

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  • $\begingroup$ I might be wrong, but don't you need that connected components are closed and open for your argument? This needs not to be true for general topological spaces. You're done if $G$ is, say, locally path-connected (which is the case for a Lie group). $\endgroup$ – Stefan Waldmann Jun 5 '14 at 13:05
  • $\begingroup$ @Stefan: if $G$ is connected, it only has one connected component, which is open and closed. $\endgroup$ – Alberto García-Raboso Jun 5 '14 at 13:10
  • $\begingroup$ Sorry, my fault. I was thinking of a statement that the above construction yields the connected component of the identity (if G is not necessarily connected). Just forgot my stupidity... $\endgroup$ – Stefan Waldmann Jun 5 '14 at 13:14
  • $\begingroup$ @Stefan: you made question whether I was being stupid! $\endgroup$ – Alberto García-Raboso Jun 5 '14 at 13:21
  • $\begingroup$ Thank you Alberto! everything is clean and clear now!:) $\endgroup$ – Mark Jun 5 '14 at 13:23

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