# Is a quotient of a solvable group always affine?

After several useful comments, I understood that I meant two questions rather than one. Sorry for messing it up, and feel free to answer any of them or vote for closure!

1. If $G$ is a connected, solvable affine algebraic group (over $\mathbb C$) and $H$ is a closed subgroup, is $G/H$ an affine variety? It is th. 4.3, p. 184 of Hochschild's Basic theory of algebraic groups and Lie algebras, thanks to Francois Ziegler's comment.

2. If $G$ is a connected, unipotent affine algebraic group and $H$ is a connected closed subgroup, is $G/H$ an affine space? As $G=\mathbb A^n$ and $H=\mathbb A^k$ as varieties, intuitively, it is a question whether $\mathbb A^n / \mathbb A^k = \mathbb A^{n-k}$. It is true if $H$ is normal, because $G/H$ is connected, unipotent (consists of unipotent elements), and also affine by Ziegler's comment.

For a normal $H$, the second question can also be proven directly: the orbits of a unipotent group $H$ on an affine variety $G$ are closed, therefore the affine quotient $\mathrm{Spec} \; \mathbb C[G]^H$ (where, a priori, the ring is not finitely generated) is geometrical and coincides with $G/H$.

Old version of the question:

If I am not wrong, any flag variety $G/P$ (where $G$ is semi-simple, $P$ is parabolic) is covered by the orbits of a Borel subgroup $B$, each of which is isomorphic to $\mathbb A^k$ for some $k$.

So I wonder whether every quotient of a connected solvable affine algebraic group by a connected but not necessarily normal subgroup is an affine space?

If not, maybe it is an affine or quasi-affine, variety, or what are the conditions that it is? My motivation is why any spherical variety (that is, a $G$-variety with an open $B$-orbit) is covered by open $B$-stable affine subsets.

• why did this get migrated? for a start, there needs to be a assumption that the initial solvable group is affine algebraic. – Peter McNamara Mar 8 '17 at 1:46
• @YCor, why, $G$ is semi-simple, not solvable. – evgeny Mar 8 '17 at 5:30
• I edited the title as evgeny seems to mean affine in the sense affine space, not affine variety. (In the latter sense, quotients of an irreducible solvable algebraic group over an algebraically closed field by an algebraic subgroup are always affine: Hochschild (1981, p. 184).) – Francois Ziegler Mar 8 '17 at 6:34
• It would be useful to clarify this mess. A torus is not isomorphic to the affine space, so I don"t know how you get this interpretation, and I don't know how you define "quasi-affine" then (isomorphic to open subset of an affine space?). Also I don't see any meaning to the question for non Zariski-closed subgroups. – YCor Mar 8 '17 at 12:53
• @FrancoisZiegler, is it better this way? Maybe, you could write it as an answer, so that I can accept it? Thank you for all the help! – evgeny Mar 8 '17 at 16:52