# Does the maximal compact subgroup always act transitively on a compact homogeneous space?

Let $$G$$ be a Lie group, $$H$$ a closed subgroup, and $$G/H$$ compact. Under what conditions do we have that $$G/H \cong K/(K\cap H)$$ where $$K$$ is a maximal compact subgroup of $$G$$? Obviously the result is trivial if $$G$$ is compact.

If $$G=\mathbb{R}$$ and $$H=\mathbb{Z}$$ this is not true. I guess that is because $$\mathbb{Z}$$ is not the real points of a linear algebraic group. As long as $$G$$ and $$H$$ are real linear algebraic groups and $$G/H$$ is compact do we have $$G/H \cong K/(K\cap H)$$ ?

EDIT:

I poked around more on the internet and it looks like my example of $$\mathbb{Z}$$ in $$\mathbb{R}$$ generalizes to a whole class of cocompact but not Zariski closed lattices in nilpotent lie groups and these things called (compact) nilmanifolds which are all iterated torus bundles that can be realized as a nilpotent Lie group mod a cocompact lattice (e.g. Heisenberg group mod integer Heisenberg group). Nilpotent Lie groups are not compact so the maximal compact has strictly smaller dimension and thus cannot act transitively on a compact nilmanifold (since it is just $$G$$ mod a lattice and thus has same dimension as $$G$$).

I think a more complicated but vaguely similar story holds when generalizing to all solvable lie groups and (compact) solvmanifolds. So lots of solvable Lie groups have non Zariski closed cocompact subgroups and the resulting homogeneous spaces are not acted on transitively by the maximal compact subgroup.

Anyway that's a bunch of no go results...now let's talk sufficient conditions.

Thanks to Ycor for the argument that if $$H$$ is Zariski closed cocompact then $$K$$ still acts transitively on $$G/H$$.

There was claim in this question

https://math.stackexchange.com/questions/2043479/homogeneous-space-of-sl3-r/2043623#2043623

that if a compact simply connected manifold is homogeneous for a Lie group $$G$$ then it is also homogeneous for the maximal compact subgroup of $$G$$. I'm a bit surprised by this fact. Can anyone provide an argument or reference for this fact?

• You're asking when $K$ acts transitively on $G/H$, and (for every group $G$ and subgroups $K,H$) this holds iff $KH=G$.
– YCor
Commented Nov 19, 2021 at 5:40
• So yes, I thinks $KH$ holds when $G$ is real algebraic and $H$ is Zariski-closed cocompact. For $G=KAN$ and $H$ then contains a conjugate of $AN$.
– YCor
Commented Nov 19, 2021 at 5:44
• The result about simply connected homogeneous spaces quoted in your (currently) last paragraph is Montgomery’s Theorem (1950). Commented Nov 19, 2021 at 16:13
• Oh wow this article is absolute little 3 page gem exactly what I wanted you could honestly just post it as an answer and it would be great. Here's a link that worked better for me ams.org/journals/proc/1950-001-04/S0002-9939-1950-0037311-6/… Commented Nov 19, 2021 at 16:32
• Anyway since any Zariski closed subgroup has finitely many connected components Corollary 2 subsumes the answer from YCor. In particular Corollary 2 states that as long as $H$ has finitely many connected components then $G/H$ is acted on transitively by a maximal compact subgroup. Commented Nov 19, 2021 at 16:36

The “surprising result” about simply connected homogeneous spaces in your (currently) last paragraph is Montgomery's Theorem (1950): generally (in your notation) if $$G/H$$ is compact and $$H$$ closed connected, then $$K$$ is transitive on $$G/H$$. The theorem is also discussed in Samelson (1952, p. 17).
Yes it is true that if $$G$$ and $$H$$ are the real points of a linear algebraic group and $$G/H$$ is compact then the maximal compact of $$G$$ acts transitively on $$G/H$$. We know from
that $$G$$ and $$H$$ have finitely many connected components. So we can apply Corollary 2 of the Montgomery paper to conclude that the maximal compact subgroup of $$G$$ acts transitively on $$G/H$$.