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Let $G$ be a group. Let $H$ be a subgroup of $R_u(G)$. Then $G/H\rightarrow G/R_u(G)$ is a $R_u(G)/H$ fibration. It is well known that $R_u(G)/H=\mathbb{A}^n$. Is $G/H$ a quasi-affine variety?

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  • $\begingroup$ What is your definition of an $\mathbb{A}^n$-fibration? For a separated, finitely presented morphism, the property of being affine can be checked after flat, surjective base change. This can be proved using Serre's criterion for affineness, among other methods, $\endgroup$
    – Jason Starr
    Commented Feb 14, 2018 at 18:53
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    $\begingroup$ For $G$ equal to $\textbf{SL}_2$, and for $H$ equal to $R_u(G)$, the quotient $G/H$ equals $\mathbb{A}^2\setminus\{(0,0)\}$. This is not affine. Did you intend to ask whether $G/H$ is quasi-affine? $\endgroup$
    – Jason Starr
    Commented Feb 15, 2018 at 10:29
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    $\begingroup$ I thought $R_u(G)$ was the unipotent radical of $G$. For $G=\mathrm{SL}_2$ it is trivial. $\endgroup$
    – Laurent Moret-Bailly
    Commented Feb 15, 2018 at 14:00
  • $\begingroup$ @LaurentMoret-Bailly You are correct. I was using the unipotent radical of a Borel subgroup. $\endgroup$
    – Jason Starr
    Commented Feb 15, 2018 at 15:27

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$G/H$ is affine. More generally, assume $G$ is an affine algebraic group over a field $k$ (the affine condition is needed, although the OP does not mention it) and $H<N\lhd G$ be $k$-subgroups (with $N$ normal). Then $G/H$ is affine if and only if $N/H$ is.
The "only if" part being obvious, assume $N/H$ affine. It suffices to prove that $G/H\twoheadrightarrow G/N$ is an affine morphism (because $G/N$ itself is affine). This property is local for the fppf topology, so it suffices to prove that the first projection $$G\times_{G/N}G/H \longrightarrow G$$ is affine. But this is a "trivial $N/H$-fibration" via the isomorphism $$\begin{array}{rcl} G\times_{G/N}G/H & \xrightarrow{\sim} & G\times(N/H)\\ (g,g'H) & \longmapsto & (g,g^{-1}g'H) \end{array} $$ valid, in fact, without assumption on $H$.

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Yes, $G/H$ is quasi-affine, because $H$ is unipotent (more precisely because $X^\ast(H)$ is trivial). This is Exercise 5.5.9.(a) in Springer: Linear algebraic groups. A sketch of the proof is as follows:

First consider a the proof of quasi-projectiveness for general $H$, which follows from the construction of $G/H$, as described in Springer's book 5.5.3-5.5.5: One chooses a representation $V$ of $G$ such that there exists a one dimensional subvectorspace $L \subset V$ such that $H$ (and ${\rm Lie}(H)$) is the stabiliser of $L$ inside $G$ (and ${\rm Lie}(G)$, respectively). Then $G/H$ is isomorphic to the orbit of $L$ in $\mathbb{P}(V)$ under the $G$-action, which is locally closed and thus quasi-projective.

If $X^\ast(H)$ is trivial, then $H$ fixes every point of $L$; thus it is the stabiliser of any fixed $x \in L \setminus \{0\}$. By the same arguments as above (more precisely 5.5.4/5 in Springer's book), $G/H$ is isomorphic to the orbit of $x$ in $V$ under the $G$-action, which is again locally closed and thus quasi-affine.

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