Examples of non parallelizable homogeneous spaces $G/H$ where $H$ is discrete

Question

Let $G$ be a Lie group and $H$ a closed Lie subgroup, so that $G/H$ is a homogeneous space. From this question we know that even though all Lie groups are parallelizable, their quotients aren't necessarily. I'm wondering about the case where $H$ is a discrete subgroup (i.e. of dimension $0$).

Are there examples of homogeneous spaces $G/H$ with $H$ discrete that are not parallelizable?

I know about three dimensional lens spaces, which still are parallelizable since orientable $3$-manifolds are parallelizable, and which are homogeneous spaces that arise from the quotient of discrete subgroups of $S^3$. But I can't think of any examples for my question.

Any references to more about this subject would also be appreciated. Many thanks.

Answer 1

In the case where $H$ is discrete, there exist no such non-parallelizable examples due to the following argument.

For $H$ discrete, the map $\pi: G\to G/H$ is a covering map. Let $F: G\to FG$ be a frame on $G$. We can choose $F$ to be left invariant, i.e. $T_{g'}\phi_g F(g')=F(gg')$ for all $g,g'\in G$, where $\phi_g$ denotes the left action of $G$ on itself. Let $\pi_{G/H}:F(G/H)\to G/H$ and $\pi_{G}:FG\to G$ denote the frame bundle projections for $G/H$ and $G$ respectively.

Because $\pi: G\to G/H$ is a local diffeomorphism it induces a map $\pi_*: FG\to F(G/H)$, so we have a map $\pi_*\circ F=F': G\to F(G/H)$. The map $F'$ has the property that $\pi_{G/H}\circ F'=\pi_{G/H}\circ \pi_*\circ F=\pi\circ \pi_G\circ F=\pi$. If we can show that $F'(g)=F'(g')$ for all $g,g'$ with $\pi(g)=\pi(g')$ then we automatically get a map $\tilde{F}: G/H\to F(G/H)$ such that $\tilde{F}\circ \pi=F'$.

One can view $\pi_*: FG\to F(G/H)$ as the quotient $FG\to (FG)/H$ with group action given by $T\phi$ which means that $\pi_* B=\pi_*B'$ if $T\phi_h(B)=B'$ for some $h\in H$. This implies that $F'(g)=\pi_*(F(g))=\pi_*F(hg)=F'(hg)$ for each $h\in H$. From this we conclude that such an $\tilde{F}$ exists.

We then see that $\pi_{G/H}\circ \tilde{F}\circ \pi=\pi_{G/H}\circ F'=\pi$ and hence $\pi_{G/H}\circ \tilde{F}=1_{G/H}$. Since $\pi_{G/H}: F(G/H)\to G/H$ has a section, $G/H$ is parallelizable.