# If Spec(A) has a G-fixed point and a dense G-orbit, is Spec(A) a cone?

[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.

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

If you by "cone" mean exactly that $A$ should be isomorpic to $\mathrm{gr}_{\mathfrak m}A$ it seems that the following is counterexample: Let $G=\mathbb G_m$, $A=k[x,y,z]/(x^2+y^3+z^5)$ with $tx=t^{15}x$, $ty=t^{10}y$ and $tz=t^{6}z$ (exponents chosen more or less at random). Then the tangent cone at the origin (the fixed point) has affine algebra $k[x,y,z]/(x^2)$ and hence is not isomorphic to $A$. This is just raising your cusp example one dimension so that it becomes normal.
Addendum: Turning to the modified question let's try for a toric example: $G=\mathbb G_m^2$ so we should look at a monomial subring of $k[x,x^{-1},y,y^{-1}]$ with monoid of monomials saturated (equivalently being the set of integer vectors in a rational cone). Unless I am mistaken $k[xy,x^2y,xy^2,x^3y,xy^3]$ is such a ring (if I have missed some monomial in the saturation it won't matter for my argument). We then have that $x^2y$ and $xy^2$ are elements in $\mathfrak m\setminus\mathfrak m^2$ (this is clear irrespective if I have missed some elements in the saturation) but their product $x^2y\cdot xy^2=(xy)^3\in \mathfrak m^3$ which means that the associated graded is not a domain and hence not isomorphic to $A$. (I think the general condition for being a cone is that the generators of the monoid of monomials should lie on an affine hyperplane which is anyway how I arrived at this example.)
Very nice. Thank you. You probably meant $t\cdot z=t^6z$ so that the ideal is fixed. I feel bad for switching the question on you, but I hope you don't mind if I edit the question to impose the condition that $Spec(A)$ contain a dense open $G$-orbit. –  Anton Geraschenko Mar 31 '11 at 8:26
Wonderful example! The $A_2$, singularity $Spec(k[a,b,c]/(ab=c^3))=Spec(k[xy^2,xy,x^2y])$ seems to work as well, and is slightly less complex. –  Anton Geraschenko Mar 31 '11 at 19:24