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Consider algebraic actions of unipotent groups $G$ on affine spaces $X=\mathbb{C}^n$. I am looking for a condition that would guarantee that the quotient $X/G$ exists and is also an affine space. For instance:

Suppose $X$ is itself isomorphic to a unipotent group and $G$ acts via compositions of left translations with group automorphisms. Suppose the action is such that every point of $X$ has trivial stabilizer. Is it true that $X/G$ is isomorphic to affine space?

UPDATE: It turns out, by replacing $X/G$ with $G\backslash X\rtimes A/A$ where $A$ is the image of $G$ in the group of automorphisms of $X$, the question is reduced to the corresponding question for double cosets, which is again a special case of the original question. So here is an equivalent question:

Suppose a unipotent group $G$ contains unipotent subgroups $G_1, G_2$ such that $G_1\cap x G_2 x^{-1} = \{e\}$ for all $x\in G$. Is the double coset space $G_1 \backslash G / G_2$ isomorphic to affine space?

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  • $\begingroup$ "The action doesn't have fixed points" has two possible meanings: no global fixed point, vs free action. What do you mean? $\endgroup$
    – YCor
    Commented Jun 8, 2020 at 13:07
  • $\begingroup$ @YCor Sorry, I rephrased the question. I meant free action. I would also like to understand the non-free case, but I don't know what kind of property to expect. $\endgroup$
    – Anton Mellit
    Commented Jun 8, 2020 at 15:39
  • $\begingroup$ If it is indeed isomorphic, then I think I know a formula for an isomorphism. You may assume that your unipotent groups $G$ is embedded in ${\rm GL}(V)$. Then its Lie algebra $\frak g$ is embedded in ${\frak gl}(V)$, and the exponential map $\exp\colon {\frak g}\to G$ is a (polynomial) isomorphism of varieties. $\endgroup$
    – Mikhail Borovoi
    Commented Jun 9, 2020 at 18:25
  • $\begingroup$ Let $U\subset V$ be a complement in $\frak g$ to ${\frak g}_1\oplus{\frak g}_2$. In other words, ${\frak g}= {\frak g}_1\oplus U\oplus{\frak g}_2$. I expect that for a suitable choice of $U$ (or even for any choice of $U$), the map $$e\colon \,{\frak g}= {\frak g}_1\oplus U\oplus{\frak g}_2\,\to\, G,\quad g_1+u+g_2\,\mapsto\, \exp(g_1)\cdot\exp(u)\cdot\exp(g_2)$$ will be an isomorphism of varieties. This will answer your question. $\endgroup$
    – Mikhail Borovoi
    Commented Jun 9, 2020 at 18:49
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    $\begingroup$ @Mikhail Borovoi, it would be great if it works. Note that for arbitrary $U$ it doesn't work even for trivial $\mathfrak g_2$ and one-dimensional $\mathfrak g_1$. Unless $\mathfrak g_1$ is in the center. $\endgroup$
    – Anton Mellit
    Commented Jun 10, 2020 at 1:03

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The answer is "no". In this paper: Winkelmann, J. On free holomorphic ℂ-actions on ℂn and homogeneous stein manifolds. Math. Ann. 286, 593–612 (1990) a free affine linear action of $G_a\times G_a$ on $\mathbb{C}^6$ is given in such a way that the quotient is not an affine variety. So for $X=G_a^6$, $G=G_a^2$, we obtain a counter-example.

By the way, in op. cit. this action is reduced to a triangular algebraic action of $G_a$ on $\mathbb{C}^5$. There is also an example there of a free triangular algebraic action of $G_a$ on $\mathbb{C}^4$ with non-Hausdorff quotient. It is not hard to check that the class of triangular algebraic actions coincides with the class of actions of $G_1$ on $G/G_2$ where $G_1, G_2$ range over unipotent subgroups of unipotent groups $G$.

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