3
$\begingroup$

Let $G$ be a connected Lie group. Then by a theorem of Cartan there is a diffeomorphism $$ G \cong K \times \mathbb{R}^n $$ where $K$ is a maximal compact subgroup of $G$. Now, let $M$ be a homogeneous manifold. In other words, there exists a Lie group $G$ acting transitively on $M$. Is it true that $M$ deformation retracts onto a compact submanifold? Slightly stronger, is it true that there is a diffeomorphism $$ M \cong K \times \mathbb{R}^n $$ where $K$ is a compact submanifold of $M$?

$\endgroup$
2
  • $\begingroup$ You can find information about the topology of homogeneous spaces of Lie groups in A. Borel, Les bouts des espaces homogènes des groupes de Lie, Ann. Math. 58(3), 1953, 443-457, behind JSTOR's rapacious paywall here. $\endgroup$ – YCor Nov 13 '19 at 20:15
  • $\begingroup$ Crossposted from MathSE: math.stackexchange.com/questions/3433385 $\endgroup$ – YCor Nov 13 '19 at 23:01
3
$\begingroup$

Mostow-Karpelevich theorem says that if $G/G^\prime$ is a homogeneous space where $G$, $G^\prime$ are Lie groups with finitely many connected components, and maximal compact subgroups $K\supset K^\prime$, respectively, then $G/G^\prime$ is a vector bundle over $K/K^\prime$. In fact, it is a homogeneous $K$-vector bundle. A more precise statement is in [Mostow, G. D., Covariant fiberings of Klein spaces II, Amer. J. Math. 84 (1962), 466–474]. There are examples where the bundle is nontrivial.

Another result in this direction addresses the situation when $G$ is solvable, and $G^\prime$ is any closed subgroup. Then $G/G^\prime$ is a vector bundle over a compact of the form $H/K^\prime$ where $H$ is solvable Lie group and $H^\prime$ is a closed subgroup. See [Auslander, L. and Tolimieri, R., Splitting theorems and the structure of solvmanifolds, Ann. of Math. (2) 92 (1970), 164–173].

In order for the quotient to be a vector bundle some assumptions are clearly necessary. For example consider a nonabelian free discrete subgroup of $SL(2,\mathbb R)$, and lift it to a nonabelian free discrete subgroup $F$ of the universal cover $\widetilde{SL}(2,\mathbb R)$. The latter is diffeomorphic to $\mathbb R^3$, so the quotient $\widetilde{SL}(2,\mathbb R)/F$ is homotopy equivalent to a wedge of circles, and in particular it is not a vector bundle over a manifold.

$\endgroup$

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.