[Edited to include a dense orbit]

Let $X=Spec(A)$ be a normal affine scheme over an algebraically closed field $k$, with an action of a linearly reductive group $G$. Suppose $x\in X$ is a $G$-invariant $k$-point and that $X$ contains a dense open $G$-orbit. Note that this implies that every $G$-orbit contains $x$ in its closure (the good quotient $Spec(A^G)$ is $Spec(k)$). Must $X$ by $G$-equivariantly isomorphic to a cone inside a representation of $G$?

The natural cone in the picture is the *tangent cone* of $X$ at $x$, $\def\m{\mathfrak m}Spec(gr_\m A)$, where $\m$ is the maximal ideal corresponding to $x$ and $gr_\m A = A/\m\oplus \m/\m^2\oplus \m^2/\m^3\oplus \cdots$. This tangent cone is a closed cone inside the tangent space at $x$, $Spec(Sym^*(\m/\m^2))$, which is a representation of $G$.

Since $G$ is linearly reductive, there is a $G$-equivariant isomorphism *of vector spaces* $A\cong gr_\m A$. The question is whether this can be made into an isomorphism *of rings*.

Associated graded doesn't have a universal property, which suggests that it should be hard (or impossible) to construct such an isomorphism, but I can't think of a counterexample.

**Remark 1:** The normality assumption is necessary. Otherwise consider the action of $\mathbb G_m$ on the cuspidal cubic $Spec(k[x^2,x^3])$ given by $t\cdot x^n=t^nx^n$. The tangent space at the fixed point is 2-dimensional, so if the cuspidal cubic were isomorphic to a cone, it would be a cone it $\mathbb A^2$, but all 1-dimensional cones in $\mathbb A^2$ are unions of lines.

**Remark 2:** Since $X$ contains a dense orbit, a natural place to look for ideas for a proof or counterexample is in literature on spherical varieties. I haven't been able to understand very much of it yet. If $X$ is an affine normal spherical variety with a $G$-invariant $k$-point, must it be a cone? In my situation, $X$ actually contains a dense open copy of $G$.

**Remark 3:** I should probably also impose the condition that $X$ is reduced, though I don't see why that should make much of a difference.

`$G$`

(if connected) is just an algebraic torus, which may make it more efficient to treat this case separately (?) $\endgroup$ – Jim Humphreys Mar 31 '11 at 17:29