Let $G$ be a reductive algebraic group over an algebraically closed field $K$ of characteristic zero (I am particularly interested in the case $G=GL_n(K))$. Let $H$ be a closed subgroup of $G$. It is known (from e.g. the book of Humphreys on linear algebraic groups) that we have a scheme structure on the space $G/H$. Are there cases in which $H$ is not reductive and $G/H$ is affine? If for example $H=B$ is a Borel subgroup, then $G/H$ is projective. The same holds if we just know that $H$ contains some Borel subgroup. Is there any other choice of nonreductive $H$ for which the quotient space $G/H$ will have a structure of affine variety?

3$\begingroup$ This is Matsushima's Criterion (proved by Richardson in positive characteristic). $\endgroup$– Jason StarrDec 9, 2015 at 17:33

3$\begingroup$ for the lazy: Matsushima says G/H is affine if and only if H is reductive. arxiv.org/abs/math/0506430 $\endgroup$– proDec 9, 2015 at 17:48

4$\begingroup$ Borel proved the result in every characteristic earlier than Richardson (by a slick technique with etale cohomology inspired by his work with HarishChandra over C, not using Haboush's theorem as Richardson does), though Borel only published his argument later than Richardson. Borel's title is something like "Affine homogeneous spaces". In the Introduction he summarizes some of the history around this result. $\endgroup$– nfdc23Dec 9, 2015 at 20:04

1$\begingroup$ Richardson's 1977 paper is found (probably by library use) in the Bulletin of the London Mathematical Society, but another relevant 1977 paper by ClineParshallScott is freely available online: gdz.sub.unigoettingen.de/dms/load/img/?PPN=GDZPPN002314673 $\endgroup$– Jim HumphreysDec 9, 2015 at 21:27

3$\begingroup$ A few more comments: 1) Richardson was told by both Borel and Springer after circulating his draft in 1975 that Borel had already verified the difficult implication in positive characteristic; so Richardson quotes in his published note from Borel's letter to him. 2) The paper by CPS was submitted a little later than Richardson's but gives a wider perspective. 3) Borel's Theorem 1.1(i) shows that $G/H$ is affine iff the identity component $H^\circ$ is reductive. [In his collected papers IV, he also notes (following comments by Knop) that his Theorem 1.1(ii) fails in char 2.] $\endgroup$– Jim HumphreysDec 9, 2015 at 22:51
1 Answer
When $G$ is a nonreductive connected affine algebraic group with $\dim R(G) > 0$, $G$ can be embedded as the unit group of an irreducible affine algebraic monoid $M$. We also assume that $G$ is a nonnilpotent group. Most of time, $M$ has infinitely many minimal idempotents, all lying in the kernel of $M$ (twosided semigrouptheoretic minimal ideal of $M$), $\ker(M)$. Conversely, all idempotents in $\ker(M)$ are minimal.
In this case, the onesided, twosided centralizers of an idempotent $e \in \ker(M)$ in $G$, $Z_G^l(e)$, $Z_G^r(e)$, $Z_G(e)$, all proper connected closed nonnormal subgroups of $G$. Let $H$ be such a subgroup of $G$. Since $R_u(H) < R_u(G)$ (ref: W. Huang, Kernels, regularity, unipotent radicals in a linear algebraic monoids, Forum Math. 23(2011), 803834.), by Grosshans LNM 1673, Theorem 7.1, the homogeneous space $G/H$ is affine.

$\begingroup$ Correction: centralized should be replaced by centralizers. $\endgroup$ May 3, 2018 at 23:47