Let $ G $ be the real points of a linear algebraic group and $ G' $ a Zariski closed subgroup. **Then is $ G/G' $ a Cartesian product
$$
(K/K') \times F
$$
where $ F $ is contractible?** Here $ K,K' $ are maximal compacts of $ G,G' $.


Some relevant information:
Let $ G,G' $ be Lie groups with finitely many connected components (for example the real points of linear algebraic groups). They can be expressed as cartesian products
$$
G= K \times E\, , \, G'=K' \times E'
$$
where $ K,K' $ are maximal compacts and $ E,E' $ are contractible. According to [Mostow, G. D., Covariant fiberings of Klein spaces II, Amer. J. Math. 84 (1962), 466–474] $ G/G' $ is of the form
$$
G/G' \cong K \times\_{K'} F
$$
where $ F \times E' \cong E $ and the $ \times\_{K'} $ means taking the Cartesian product but then identifying
$$
 (k\_0,f\_0) \sim (k\_0k,kf\_0k^{-1}) 
$$
for all $ k \in K' $. We can essentially summarize this by saying that $ G/G' $ is a $ K' $ homogeneous vector bundle over $ K/K' $.


Follow-up: Great answer. Exactly what I wanted. $ SO\_3(\mathbb{C})/SO\_2(\mathbb{C}) $ is the normal bundle over the 2 sphere (so 4 dimensional) and is nontrivial. Upon further reflection I found a less beautiful but more minimal counterexample. $ SE\_2(\mathbb{R}) $ is a linear algebraic group. There is a Zariski closed subgroup given by taking $ x\_{1,1}x\_{2,2}=1 $ and $ x\_{1,3}=0 $ and the quotient is the Moebius strip.



Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors?
The answer is no.


At least when $G$ and $H$ are semisimple, the quotient $G/H$ is diffeomorphic to the normal bundle of $K\_G/K\_H$ inside $G/H$ (where $K\_G$ and $K\_H$ denote respectively maximal compact subgroups of $G/H$), but this normal bundle might not be trivial.


For a concrete example take $G= SO(n+1,\mathbb C)$ and $H= SO(n,\mathbb C)$ (seen as real algebraic groups). Then $G/H$ is the smooth complex affine quadric of dimension $n$ and $K\_G/K\_H = SO(n+1,\mathbb R)/SO(\mathbb R)$ is the real sphere of dimension $n$ inside it. In particular, it is a real form of $G/H$, so its normal bundle is isomorphic to the tangent bundle of the sphere, which is trivial if and only if $n= 1,3$ or $7$ !


