# Homogeneous Manifold Deformation Retracts Onto Compact Submanifold

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?

• 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: 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 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

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."

• This looks like a great counterexample. Could you give an example of matrices in $SL(2,\mathbb{R})$ that generate such a $\Gamma$? – Ian 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 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 math.stackexchange.com/questions/3123098/… 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 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 mathoverflow.net/questions/43726/… – Ian Teixeira Nov 14 '19 at 16:51