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Let G be a connected Lie group. Then by 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 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?

Ian Gershon Teixeira
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  • As a general rule: Do not ask the same question simultaneously here and at Mathoverflow. – Moishe Kohan Nov 13 '19 at 20:22
  • Crossposted to MathOF: https://mathoverflow.net/questions/345905/ – YCor Nov 13 '19 at 23:02
  • Oh sorry! I’m a newbie— in the future I’ll just post to one place. For this particular question do you think I should have posted to stack exchange or overflow? – Ian Gershon Teixeira Nov 14 '19 at 12:31
  • @Ianteixeira: It is up to you, but the best choice is to first post at MSE, then wait a week and if you did not get a (satisfactory) answer, then post at MO. – Moishe Kohan Nov 14 '19 at 17:07

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Consider the following example. Let $G=SL(2, {\mathbb R})$ and $\Gamma< G$ a discrete subgroup which is free of infinite rank. Form the quotient manifold $M=G/\Gamma$. Then $G$ acts on $M$ via left multiplication: The action is smooth and transitive, thus, $M$ is homogeneous. But $M$ has non-finitely generated fundamental group, hence, cannot be homotopy-equivalent to a compact manifold.

Of course, the answer is different for Riemannian homogeneous manifolds, i.e. for Riemannian manifolds $M$ admitting transitive isometric Lie group actions $G\times M\to M$. Such an action has compact point-stabilizer $H< G$ and, thus, $M$ is homeomorphic to $G/H$. Taking a maximal compact subgroup $K< G$ containing $H$, we see that $G/H$ is homotopy-equivalent to $K/H$, which is a compact manifold.

Edit. Here is a semi-explicit construction. Start with countably many round circles $C_n, n\in {\mathbb Z}- \{0\}$, in the complex plane ${\mathbb C}$, whose centers lie on the x-axis and which bound pairwise disjoint open disks. For instance, take centers which are even integers and unit radii. For each circle $C=C(a,r)$ in the complex plane define the inversion $J_C$ in this circle by the formula: $$ J_C(z)= \frac{r^2}{\bar{z} +a} -a. $$ Now, for each $n$ let $g_n$ denote the composition of the inversions $$ g_n=J_{C_n}\circ J_{C_{-n}}. $$ These will be linear-fractional transformations of the extended complex plane preserving the upper half-plane: $$ g_n(z)= \frac{a_n z+b_n}{c_n z+ d_n}, a_nd_n -b_n c_n=1, a_n, b_n, c_n, d_n\in {\mathbb R}. $$ I leave it to you to compute the coefficients in terms of the centers and the radii.

It is a standard fact that the transformations $g_n$ freely generate a discrete subgroup of $PSL(2, {\mathbb R})$. My favorite reference for this staff is Beardon's book "The Geometry of Discrete Groups."

Moishe Kohan
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  • This looks like a great counterexample. Could you give an example of matrices in $ SL(2,\mathbb{R}) $ that generate such a $ \Gamma $? – Ian Gershon Teixeira Nov 13 '19 at 17:11
  • @Ianteixeira: I prefer not to write matrices but I can if you really want me to. A better example is to take a compact hyperbolic surface $S$ and let $\Gamma$ be, say, the commutator subgroup of $\pi_1(S)$. The group $\pi_1(S)$ embeds as a discrete subgroup in $SL(2,R)$ and so does $\Gamma$. – Moishe Kohan Nov 13 '19 at 17:14
  • If it’s not too much trouble I would love it if you could give generators for such a $ \Gamma $. – Ian Gershon Teixeira Nov 14 '19 at 12:33
  • I found a construction of $ \Gamma $. Apparently almost any two matrices $ a,b \in SL(2,\mathbb{Z}) $ generate a free group on two elements. Once you have such $ a,b $ then this answer https://math.stackexchange.com/questions/3123098/show-that-the-free-group-on-three-generators-is-a-subgroup-of-the-free-group-on claims that the subgroup of $ SL(2,\mathbb{Z}) $ generated by the countable list of matrices $ { a^nb^maba^{-1}b^{-1}a^{-n}b^{-m} } $ for all $ n,m \in \mathbb{N} $ generates a free group on countably many generators which is a subgroup of the discrete subgroup $ SL(2,\mathbb{Z}) $ – Ian Gershon Teixeira Nov 14 '19 at 16:43
  • An explicit example of $ a,b \in SL(2,\mathbb{Z}) $ such that $ F_2 \cong <a,b> $ are the matrices $ a= \begin{pmatrix} 1 & 2 \ 0 &1 \end{pmatrix} $ and $ b= \begin{pmatrix} 1 & 0 \ 2 &1 \end{pmatrix} $. These in particular generate a free subgroup of finite index in $ SL(2,\mathbb{Z}) $. I found the claim that they generate a copy of $ F_2 $ in this answer here https://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz – Ian Gershon Teixeira Nov 14 '19 at 16:51
  • @Ianteixeira: Yes, this would work as well. – Moishe Kohan Nov 14 '19 at 17:06
  • I know this is an old question but I was rereading it. Why do we know that $ G/H $ deformation retracts onto $ K/H $ for the Riemannian homogeneous case? – Ian Gershon Teixeira Dec 09 '21 at 04:41