# When does an affine subset of an orbit have affine preimage under the orbit map?

I have an algebraic group $$G$$ acting on an affine variety $$X$$, the orbit $$O(m)$$ of an element $$m \in X$$, and an affine curve $$C$$ contained in the Zariski closure $$\overline{O(m)}$$ of $$O(m)$$, such that $$m \in C$$.

If we define $$C^\prime = C \cap O(m)$$ then it's not hard to see that $$C^\prime$$ is open in $$C$$. And therefore $$C^\prime$$ is again affine (follows for example from an exercise in Hartshorne).

Let $$p: G \to O(m)$$ defined by $$p(g) = g \cdot m$$ be the orbit map. I want to prove that there is an affine subset $$G^\prime$$ of $$G$$ such that $$p(G^\prime)$$ is dense in $$C^\prime$$.

Is the preimage of $$X^\prime$$ affine? If so why? If not, how do I construct such a $$G^\prime$$?

For reference: The statement that such an $$G^\prime$$ exists can be found in the article A characterization of orbit closure and applications (MR0944153) on the first page of the proof of Theorem 1.2. Clearly he has more assumptions on $$X$$ and $$G$$, but I do not see anything else that could be useful here.

I was able to figure out an elementary answer with some help: It seems like a good hint is to look at abstract varieties or even easier use that every quasi-affine variety is an abstract variety.

The preimage is presumably not affine, but it is a locally closed subset of $$G$$ since $$C$$ is locally closed. Therefore $$p^{-1}(C)$$ is a quasi-affine variety and as such it has an open cover $$U_i$$ where each $$U_i$$ has the structure of an affine varieties such that for each intersection $$U_i \cap U_j$$ the induced variety structures are isomorphic (see for example Hartshorne - Algebraic Geometry Section I.8 page 58 where this is stated without proof). So let

$$p^{-1}(C) = U_1 \cup \dots \cup U_\alpha$$.

Since $$p$$ is surjective we have

$$C = p(U_1) \cup \dots \cup p(U_\alpha)$$.

Since $$C$$ is irreducible it can't be the union of two proper closed subsets and therefore some $$p(U_i)$$ has to be dense which finishes the proof.

• That is much cleaner and more elementary. It is much better to not kill a small bird with the giant gun of GIT. – schemer Oct 29 at 22:50

I do not know if the pre-image of $$C^{'}$$ (I assume that you mean $$C^{'}$$ not $$X^{'}$$) is affine. However, here is an idea for how to construct such a $$G^{'}$$. For notational purposes I will relabel $$m$$ as $$x$$ so that people don't confuse $$G_{m}$$ with $$\mathbb{G}_{m}$$. When you said that $$x$$ was an element of $$X$$, I assumed that you meant that it is a closed point of $$X$$.

If $$G_{x}$$ is the stabilizer of the point $$x \in X$$, then $$G_{x}$$ acts on $$G$$. Let us assume that $$G$$ is quasi-projective. If we denote the maximal ideal of a closed point $$y$$ by $$\mathfrak{m}_{y}$$, then let $$\mathcal{L}$$ be an invertible sheaf on $$O(x)$$ such that $$H^{0}(O(x), \mathcal{L})$$ contains a non-constant global section $$s_{1}$$ such that $$s_{1} \notin \mathfrak{m}_{x} \mathcal{L}_{x}$$. Since $$O(x)$$ is a sub-variety of an affine variety such an $$\mathcal{L}$$ must exist.

The pull-back $$p^{\ast}(\mathcal{L})$$ is a $$G_{x}$$ linearized invertible sheaf $$\mathcal{M}$$ with a non-constant global section $$s$$ such that $$s \notin \mathfrak{m}_{e_{G}} \mathcal{M}$$. Let $$U$$ be the points $$g \in G$$ such that $$s_{g} \notin \mathfrak{m}_{g} \mathcal{M}_{g}$$. The neighborhood $$U$$ is an open, affine $$G_{x}$$-stable sub-scheme of $$G$$. Assume that $$U \cong \operatorname{Spec}(A)$$.

Since $$U$$ is $$G_{x}$$ stable, there is a map $$p_{1}: U \to U//G_{x}$$ (here $$U//G_{x}$$ denotes $$\operatorname{Spec}(A^{G_{x}})$$). By the universal property of categorical quotients there is a map $$f: U//G_{x} \to p(U)$$ such that $$f \circ p_{1} = p$$. Since quotients are unique, $$U//G_{x} \cong p(U)$$. The map $$p_{1}$$ is affine, so since $$p(U) \cap C^{'}$$ is non-empty as $$x \in p(U) \cap C^{'}$$, it is an open sub-scheme of $$C^{'}$$. Therefore, $$U$$ should be the affine set you seek.

• I suppose that all my points should be closed (my variety is $mod^d_A(k)$ and an element of it can be viewed as an alpha-tuple (alpha is the dimension of the f.d. $k$-algebra $A$) of dxd matrices. This should be closed. But as I was thinking about it in the sense of varieties as in the first section of Hartshorne I do not realy think about open and closed points. I see that you assume that G is quasi-projective and I am not sure about that in my case ($G=GL_d(k)$). I can´t say if what you are doing for the rest of the proof is correct or not because I do not know anything about sheafes. – Bobior Oct 23 at 21:30