Let $G$ be an affine algebraic group (let's say over $\mathbb{C}$). If necessary one can assume $G$ to be reductive. Imagine one has $X$ over which $G$ acts freely: moreover, we have a locally closed subvariety $Y$ such that $X=G \cdot Y$.

Moreover, one has a subgroup $H \subseteq G$ such that $H$ stabilizes $Y$. We know also that given any two points $y_1,y_2 \in Y$ and $g \in G$ such that $g y_1=y_2$ we have $g \in H$.

As everything acts freely one can make up two quotient varieties $Y/H$ and $X/G$ with the natural morphism $$f:Y/H \to X/G .$$

The hypotheses imply that $f$ is bijective. Is it also an isomorphism? I'm not assuming anything on normality or smoothness

  • 3
    $\begingroup$ Suppose that $X$ is normal. Then $X/G$ is also normal. Also, $f$ is quasi-finite and birational (because the extension of function fields is separable, and hence of degree one, since we are in char. $0$). Hence by Zariski's main theorem $f$ is an open immersion and hence an isomorphism. $\endgroup$ Oct 4, 2021 at 17:54
  • $\begingroup$ See also mathoverflow.net/questions/264204/… $\endgroup$ Oct 4, 2021 at 18:00
  • $\begingroup$ Unfortunately I don't have any hypothesis on normality or smoothness in the case I'm interested :( $\endgroup$ Oct 4, 2021 at 20:42
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    $\begingroup$ The issue is (I believe this is what afh's answer illustrates) is that the scheme-theoretic intersection of a $G$-orbit in $X$ with $Y$ might be non reduced. $\endgroup$ Oct 6, 2021 at 14:20

1 Answer 1


I don't think that this is true without extra hypotheses like normality of $X$ (if $X$ is normal then it follows as explained by Damian Rossler in the comments). Here is a possible example.

We will take $G = GL_2$ and $H=1$ (the latter is just the trivial group). Let $C$ be the cuspidal curve $C = Spec(k[x,y]/(y^2-x^3))$.

Let's take $X = GL_2 \times C$, equipped with the natural action of $GL_2$ by right multiplication on the first factor (this is just the trivial $GL_2$-bundle over $C$).

Inside $GL_2$ we have the additive group $\mathbb{G}_a$ of strictly upper triangular matrices. So we have a closed subvariety $Z = \mathbb{G}_a \times C$ inside $X$. We can write $Z = Spec(k[x,y,t](y^2-x^3))$. Inside $Z$ we can define $Y$ to be the vanishing locus of the polynomial $xt-y$. So we get a closed subvariety $Y \hookrightarrow X$ given by

$$ Y = Spec(k[x,y,t]/(y^2-x^3, xt-y))$$

Notice that $X/GL_2 =C$ and the projection $Y \to C$ onto the second component is the map $Y/H \to X/GL_2$ in this case.

$Y$ is actually isomorphic to $\mathbb{A}^1$, and the projection $Y \to C$ exhibits $Y$ as the normalization of the cuspidal curve $C$. In particular $Y \to C$ is not an isomorphism.

Note that the projection $Y \to C$ is bijective on points, so $Y$ contains exactly one point in each fiber of the $GL_2$-bundle $X \to C$. This implies that $GL_2 \cdot Y = X$ and also that there are no two closed points of $Y$ that belong to the same $GL_2$-orbit. In other words, the hypotheses are satisfied for $Y \hookrightarrow X$ (with $H = 1$) and the map $Y/H \to X/G$ is not an isomorphism.

One can probably do variations of this to get examples with $H$ nontrivial.

  • $\begingroup$ This is a great counterexample ! Thank you! If I can ask you: what was the reason to think of this exact situation? $\endgroup$ Oct 6, 2021 at 14:26
  • $\begingroup$ Hi Tommaso. The normalization of a cusp is an example I knew where you have a bijection of varieties that is not an isomorphism. I tried to reverse engineer the situation so that the final morphism f was that map. $\endgroup$
    – afh
    Oct 6, 2021 at 17:01

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