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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.

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    $\begingroup$ Are you assuming the group field is $\mathbf{C}$? or $\mathbf{R}$? otherwise I'm not sure the question makes sense. Also have in mind that a real algebraic group is not exactly the same as its set of real points. $\endgroup$
    – YCor
    Commented Dec 10, 2021 at 7:49
  • $\begingroup$ Also, "direct product" is a group-theoretic notion, or more generally in the presence of laws to mean that the operations are taken coordinate-wise. For instance for groups, the semidirect product is a non-direct group operation on the Cartesian product. For topological spaces, I think "Cartesian product" is better suited. $\endgroup$
    – YCor
    Commented Dec 10, 2021 at 9:55

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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 $7$ !

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    $\begingroup$ So (just to make sure I'm understanding you correctly) in the $ n=2 $ case this would be the sub variety of $ \mathbb{C}^3 $ defined by $ x^2+y^2+z^2=1 $? Also what would happen if we took $ G= O_{n+1}(\mathbb{C}) $ and $ H=O_n(\mathbb{C}) \times O_1(\mathbb{C}) $ ? I think we get the normal bundle over real projective space $ P_\mathbb{R}^n $. Is that bundle also nontrivial? $\endgroup$
    – Ian Gershon Teixeira
    Commented Dec 10, 2021 at 16:20
  • $\begingroup$ yes exactly, you'd get the tangent bundle to $P^n_{\mathbb R}$ which is also non-trivial unless $n\in \{1,3,7\}$ (since the sphere is its double cover). $\endgroup$
    – Nicolast
    Commented Dec 10, 2021 at 20:27
  • $\begingroup$ Is it true in general that as a real smooth manifold $ G_{\mathbb{C}}/H_{\mathbb{C}} $ is always the tangent bundle of $ G_{\mathbb{R}}/H_{\mathbb{R}} $ or is there something special about this case with the sphere? $\endgroup$
    – Ian Gershon Teixeira
    Commented Dec 29, 2021 at 13:54
  • $\begingroup$ I think it is always true, at least when $G$ and $H$ are reductive. $\endgroup$
    – Nicolast
    Commented Jan 1, 2022 at 13:18
  • $\begingroup$ Hmm but what about when $ G=SL_2 $ and H is trivial? $ SL_2(\mathbb{C}) $ is homotopy equivalent to $ SU_2 $ but the tangent bundle of $ SL_2(\mathbb{R}) $ is homotopy equivalent to $ SO_2(\mathbb{R}) $ so they can't be diffeomorphic. Maybe it's true as long as the real points of $ G $ and $ H $ are both compact? $\endgroup$
    – Ian Gershon Teixeira
    Commented Jan 1, 2022 at 21:26

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