MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Fix some field $k$. I am not very prudent about the field - although I'd prefer to assume as little as possible, you may assume as much as you want, the case of primary interest being $k=\mathbb{C}$.

Let $G\subseteq\mathrm{Gl}_n(k)$ be a closed, reductive subgroup and $X\subseteq\mathbb{A}_k^m$ an affine $G$-variety. Assume that both $G$ and $X$ are cones, i.e. cut out by homogeneous equations. Note that $\mathrm{Gl}_n(k)$ is the open affine subset of $\mathbb{A}^{n\times n}_k$ where the determinant (a homogeneous polynomial) does not vanish. This yields an action of $k^\times$ on both $G$ and $X$. Under this action, assume that the action map $$\begin{align*}\alpha_x:G&\longrightarrow X \\ g&\longmapsto g.x\end{align*} $$ is a morphism of $k^\times$-varieties for each $x\in X$, i.e. $\lambda g.x=g.\lambda x$ for $g\in G$ and $\lambda\in k^\times$.


I have a point $x\in X$ such that $H:=G_x$ is reductive. Let $U:= G.x$ be the orbit of $x$. Now in this very friendly setting, I have some questions about the closure $Z:=\overline U$.

  1. How does the coordinate ring $k[Z]$ of $Z$ look like? It is known that $U$ itself is affine, and its coordinate ring can be described as the $H$-invariants of $k[G]$. However, what about $k[Z]$?

  2. The orbit $U$ is smooth, but $Z$ isn't (in general). Since $Z$ is the union of orbits, however, the singular locus of $Z$ should also be a union of orbits. Is there some nice way to characterize $\mathrm{Sing}(Z)$?

share|cite|improve this question
The similarity classes in $M_{n \times n}(k)$ (conjugation action by $Gl(n,k)$) are well understood by the rational canonical form, but you probably know this? If not, probably you want to have a look at Laumon "Drinfeld modules .. " Part 1 Chapter 4 Section 3. – Marc Palm Jul 18 '12 at 9:16
At the risk of sounding dumb, how does the rational canonical form relate to my question? – Jesko Hüttenhain Jul 18 '12 at 12:45
The closure of the conjugacy classes are describable by this theory. This is at least the canonical example with which I would start with. The closure of an orbit contains a finite number of smaller dimensional orbits. Only the diagonazable elements (in some algebraic extension) have closed orbits. Unipotent matrices have never closed orbits and so forth. – Marc Palm Jul 18 '12 at 15:16
@Jesko: Rather than speak in terms of cones and homogeneous polynomials, it seems more intrinsic (albeit equivalent) to say that we are given the action of an abstract connected reductive $G$ on a smooth affine $X$, choosing a central ${\rm{GL}}_1$ in $G$, and contemplating the singular locus of the closure $Z$ of the orbit of a point with (connected?) reductive stabilizer as well as the graded ring $k[Z]$ (grading defined by the ${\rm{GL}}_1$-action on $Z$). I know this doesn't constitute any genuine progress, so feel free to ignore it (but from this viewpoint it looks kind of hopeless). – user22479 Jul 21 '12 at 22:37
@quasi-coherent: Actually, I thought about formulating the question in a similar manner, and my problem was the following: Why does the action of our central $\mathrm{GL}_1$ on $Z$ define a grading? I can call an $f\in k[Z]$ "homogeneous of degree $d$" if $f(\lambda z)=\lambda^d f(z)$ for all $z\in Z$ and $\lambda\in\mathrm{GL}_1$, but how do I know that this yields a decomposition into direct summands? – Jesko Hüttenhain Jul 23 '12 at 12:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.