# Quotient variety and subgroups

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

• 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. Oct 4, 2021 at 17:54
• Oct 4, 2021 at 18:00
• Unfortunately I don't have any hypothesis on normality or smoothness in the case I'm interested :( Oct 4, 2021 at 20:42
• 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. Oct 6, 2021 at 14:20

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.

• This is a great counterexample ! Thank you! If I can ask you: what was the reason to think of this exact situation? Oct 6, 2021 at 14:26
• 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.
– afh
Oct 6, 2021 at 17:01