7
$\begingroup$

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 $\mathbb R^n$.

$\endgroup$
  • $\begingroup$ Your argument does work, but it is orders of magnitude more complex that the statement you want to prove, because of its dependencies :) $\endgroup$ – Mariano Suárez-Álvarez Jul 7 '12 at 7:56
5
$\begingroup$

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 $\mathbb R^{n+1}$.

$\endgroup$
  • $\begingroup$ How do we know that $H$ is simply connected (to be able to apply the induction hypothesis)? $\endgroup$ – Ted Jul 7 '12 at 20:44
  • $\begingroup$ Use, for example, the long exact sequence for homotopy groups for the fibration $G\to\mathbb R$ with fiber $H$. $\endgroup$ – Mariano Suárez-Álvarez Jul 7 '12 at 21:02
  • $\begingroup$ Why is $\mathfrak h$ of codimension 1? This may be elementary, but can you suggest a reference? $\endgroup$ – Earthliŋ Jul 9 '12 at 2:46
  • $\begingroup$ 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 $1$. $\endgroup$ – Mariano Suárez-Álvarez Jul 9 '12 at 2:58
3
$\begingroup$

$S^1$ 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.

$\endgroup$
  • $\begingroup$ True. Make that a connected, simply connected nilpotent Lie group. $\endgroup$ – Earthliŋ Jul 7 '12 at 3:37
  • $\begingroup$ Are you sure about your last claim? For example, isn't the inverse of stereographic projection a local homeomorphism from $\mathbb{R}^2$ to $S^2$? $\endgroup$ – Jason DeVito Jul 7 '12 at 3:55
  • $\begingroup$ @Jason You're right. The map needs to be surjective. $\endgroup$ – Ted Jul 7 '12 at 3:56
  • $\begingroup$ 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? $\endgroup$ – Earthliŋ Jul 7 '12 at 3:57
  • 1
    $\begingroup$ I'm pretty sure even surjectivity is not enough, but I don't have a counter example at my fingertips. Wikipedia says that if $X$ and $Y$ are locally compact (which we have here), then a proper surjective local homeo is a covering, but we are not guaranteed to have a proper map in this case. $\endgroup$ – Jason DeVito Jul 7 '12 at 3:59
0
$\begingroup$

Let $G$ be a simply connected nilpotent Lie group with Lie algebra $\mathfrak g$. The nilpotency of $G$ implies that the Baker-Campbell-Hausdorff formula has only finitely many terms, so it converges everywhere on $\mathfrak g$. So it can be used to define a group law $\times$ on the Lie algebra $\mathfrak g$ such that $(\mathfrak g,\times)$ is a Lie group whose Lie algebra is isomorphic to $\mathfrak g$. Since two simply connected Lie groups with the same Lie algebra are isomorphic, $G$ is isomorphic to $(\mathfrak g,\times)$. In particular, $G$ is homeomorphic to $\mathfrak g$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.