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Is it true that a connected, simply connected, nilpotent $n$-dimensional Lie group $G$ is homeomorphic to $\mathbb R^n$?

EDIT: Maybe a possible argument is the following: Since $G$ is simply connected, $G$ cannot contain any non-trivial maximal compact subgroups. By a theorem associated to Iwasawa and Malcev, all maximal compact subgroups are conjugate and thus have the same dimension. By a theorem by Hochschild(?), $G/K$ is diffeomorphic to $\mathbb R^n$, where $K$ is a(ny) maximal compact subgroup. But for $G$ simply connected, connected, nilpotent, $K$ must be trivial, whence $G$ itself is diffeomorphic to $\mathbb R^n$.

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Your argument does work, but it is orders of magnitude more complex that the statement you want to prove, because of its dependencies :) –  Mariano Suárez-Alvarez Jul 7 '12 at 7:56
    
Thanks, I was hesitating to ask you directly for an opinion. –  Earthliŋ Jul 7 '12 at 8:23

2 Answers 2

up vote 3 down vote accepted

If a Lie group $G$ is nilpotent, its Lie algebra $\mathfrak g$ is nilpontent, and there is an ideal $\mathfrak h\subseteq\mathfrak g$ of codimension $1$. It follows that there is a normal subgroup $H\subseteq G$ whose Lie algebra is $\mathfrak h$, this subgroup is closed because $G$ is simply connected, and we have a short exact sequence of Lie groups $$1\to H\to G\to\mathbb R\to 0$$ This extension of groups is split, so that $G$ is in fact a semidirect product of $\mathbb R$ and $H$; to exhibit a splitting, let $\mathfrak a$ be a subspace of $\mathfrak g$ complementary to $\mathfrak h$: the $1$-dimensional closed Lie subgroup of $G$ tangent to it maps isomorphically to the $\mathbb R$ here, so its inverse splits the exact sequence. Since $H$ is nilpotent and of smaller dimension than $G$, by induction we can assume that $H$ is diffeomorphic to $\mathbb R^n$ for some $n$ and then $G\cong\mathbb R\rtimes H$ is diffeomorphic to $\mathbb R^{n+1}$.

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How do we know that $H$ is simply connected (to be able to apply the induction hypothesis)? –  Ted Jul 7 '12 at 20:44
    
Use, for example, the long exact sequence for homotopy groups for the fibration $G\to\mathbb R$ with fiber $H$. –  Mariano Suárez-Alvarez Jul 7 '12 at 21:02
    
Why is $\mathfrak h$ of codimension 1? This may be elementary, but can you suggest a reference? –  Earthliŋ Jul 9 '12 at 2:46
    
A nilpotent Lie algebra $\mathfrak g$ always has a codimension-$1$ ideal. Take any subspace containing the derived subalgebra $[\mathfrak g,\mathfrak g]$ and of codimension $1$. –  Mariano Suárez-Alvarez Jul 9 '12 at 2:58

$S^1$ is abelian, hence nilpotent, but it is not homeomorphic to $\mathbb{R}$.

Edit: With the additional assumption that the Lie group $G$ is simply connected, yes. The exponential map defines a local homeomorphism from $\mathbb{R}^n$ to $G$ (every point $x \in \mathbb{R}^n$ has a neighborhood $U$ such that $\exp$ restricted to $U$ is a homeomorphism onto its image).

But if 2 connected and simply-connected spaces are locally homeomorphic, then they are homeomorphic. (There need to be some assumptions on the spaces to make this true but surely they are fulfilled for $\mathbb{R}^n$ and a smooth manifold $G$.)

Edit 2: This still isn't quite right. See comments below. For one thing, we haven't used the nilpotency assumption. If $G$ is nilpotent, then $\exp$ is surjective, but we may need more to conclude local homeomorphism $\implies$ homeomorphism.

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True. Make that a connected, simply connected nilpotent Lie group. –  Earthliŋ Jul 7 '12 at 3:37
    
Are you sure about your last claim? For example, isn't the inverse of stereographic projection a local homeomorphism from $\mathbb{R}^2$ to $S^2$? –  Jason DeVito Jul 7 '12 at 3:55
    
@Jason You're right. The map needs to be surjective. –  Ted Jul 7 '12 at 3:56
    
Great, never heard of the argument in your last paragraph. There must be some assumptions on $G$ other than $G$ being smooth to deduce 2-connectedness, though. Can you give a reference? –  Earthliŋ Jul 7 '12 at 3:57
    
It's not quite right. See edits. –  Ted Jul 7 '12 at 3:59

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