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Let $k$ be an algebraically closed field of characteristic 0. Let $G$ be a connected linear algebraic group over $k$. Let $H\subset G$ be an algebraic $k$-subgroup. Let $Y$ be an algebraic variety over $k$ and let $y_0\in Y(k)$ be a $k$-point. Let $$\theta\colon G\times_k Y\to Y$$ be an action of $G$ on $Y$. I need a criterion to check whether $Y$ is isomorphic (as a pointed $G$-variety) to $G/H$.

Consider the induced action of $G(k)$ on $Y(k)$ and assume that (1) $G(k)$ acts on $Y(k)$ transitively, and (2) the stabilizer of $y_0$ in $G(k)$ is $H(k)$.

Question 1. Does if follow that $Y$ is canonically isomorphic, as a pointed $G$-variety, to $G/H$ ?

By definition (see Section 5.5 of Springer's book "Linear Algebraic Groups, 2nd ed.) a quotient of $G$ by $H$ is a $G$-variety $X$ with a $k$-point $x_0$ having stabilizer $H$ with the following universal property:

(Q) For any $G$-variety $Z$ over $k$ with a $k$-point $z_0$ having stabilizer containing $H$, there exists a unique $G$-morphism $\nu\colon X\to Z$ sending $x_0$ to $z_0$.

It follows immediately from the universal property that if $(X', x'_0)$ is another such $G$-variety, then there exists a unique $G$-isomorphism $X\to X'$ sending $x_0$ to $x'_0$.

If $G$ is a linear algebraic group over $k$ and $H$ any $k$-subgroup, then there exists a quotient of $G$ by $H$, see Theorem 5.5.5 of Springer's book.

This quotient $X$ is denoted by $G/H$. If $g\in G(k)$, then the $k$-point $g\cdot x_0$ is denoted $gH$.

I am trying to answer my Question 1. Let $X=G/H$. Using the universal property, we obtain a $G$-morphism $$\nu\colon X\to Y$$ taking $x_0$ to $y_0$, and one can check easily that $\nu$ is bijective of $k$-points. Note that both $X$ and $Y$ are nonsingular and, by assumption, ${\rm char}\,k=0$.

Question 2. Does it follow that $\nu$ an isomorphism of varieties?

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    $\begingroup$ Once you know flatness of the morphism $\theta_{y_0}:G\to Y$, $g\mapsto \theta(g,y_0)$, you can use faithfully flat descent to construct the isomorphism with $G/H$. By generic flatness, there is a maximal, dense, open subscheme $U$ of $Y$ over which $\theta_{y_0}$ is flat. The closed complement is a proper, $G$-invariant, closed subset of $Y$. Since $Y$ is $G$-homogeneous, this proper subset is the empty set, i.e., $\theta_{y_0}$ is flat. $\endgroup$
    – Jason Starr
    Commented Aug 6, 2017 at 12:15
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    $\begingroup$ . . . The fiber product $G\times_Y G$ is isomorphic to $H\times G$ via the conjugation morphism. So now you can construct the isomorphism between the datum $(\text{pr}_1,\text{pr}_2:G\times_Y G \to G)$ and the datum $(m,\text{pr}_2:H\times G \to G)$ and apply faithfully flat descent for morphisms to get isomorphisms of $Y$ and $G/H$. $\endgroup$
    – Jason Starr
    Commented Aug 6, 2017 at 12:18
  • $\begingroup$ @JasonStarr: Thank you for prompt comment! However, is the answer to Question 1 "yes" and to Question 2 "yes" as well? $\endgroup$
    – Mikhail Borovoi
    Commented Aug 6, 2017 at 12:20
  • $\begingroup$ @JasonStarr: Could you please write an answer? $\endgroup$
    – Mikhail Borovoi
    Commented Aug 6, 2017 at 12:21
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    $\begingroup$ This lemma also seems to be the second part of Springer's Theorem 5.5.5 that you quote ("In fact, if $X$ and $x$ are as in 5.5.4 then $(X, x)$ is such a quotient"). $\endgroup$
    – LSpice
    Commented Aug 6, 2017 at 12:44

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