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Aug 12, 2012 at 0:17 comment added user22479 @Jesko: If $G$ is an affine group scheme over a ring $R$ and $M$ is any $R$-module then a "representation" of $G$ on $M$ is an assignment to each $R$-algebra $S$ of an $S$-linear action of $G(S)$ on $M_S$ functorially in $S$. By a Yoneda argument, this data quantified over "all" $S$ is encoded in a single such action with $S=R[G]$. This applies to $G$-actions on affine $R$-schemes via the coordinate ring, so $M=R[G]$ is a representation of $G$ and thus of any closed $R$-subgroup of $G$. For $G=GL_1$, a representation on $M$ is a $Z$-grading! The proof early in SGA3 is a clever computation.
Jul 27, 2012 at 12:22 history bounty ended Jesko Hüttenhain
Jul 23, 2012 at 12:22 comment added Jesko Hüttenhain @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?
Jul 21, 2012 at 22:37 comment added user22479 @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).
Jul 20, 2012 at 11:27 history bounty started Jesko Hüttenhain
Jul 19, 2012 at 5:51 history edited Jesko Hüttenhain CC BY-SA 3.0
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Jul 18, 2012 at 15:16 comment added Marc Palm 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.
Jul 18, 2012 at 12:45 comment added Jesko Hüttenhain At the risk of sounding dumb, how does the rational canonical form relate to my question?
Jul 18, 2012 at 11:31 history edited Jesko Hüttenhain CC BY-SA 3.0
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Jul 18, 2012 at 9:16 comment added Marc Palm 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.
Jul 18, 2012 at 9:10 history asked Jesko Hüttenhain CC BY-SA 3.0