Let $X=Spec(A)$ be a reduced normal affine scheme over an algebraically closed field $k$ of characteristic $0$, with an action of a connected reductive group $G$. Suppose

- $x\in X$ is a $G$-invariant $k$-point,
- $X$ contains a dense open stabilizer-free $G$-orbit (i.e. a dense open copy of $G$), and
- the stabilizers of all points of $X$ are reductive.
Must $G$ be a torus (and therefore $X$ a toric variety)?

## Thoughts so far

I'll include the ideas I've had so far in the hope that somebody might see a way to use them to answer the question.

**Remark 1:** Choosing a set of generators $f_1,\dots, f_r\in \mathfrak m_x$ for $A$ as an algebra, there is a finite-dimensional $G$-invariant vector space $V^\ast\subseteq \mathfrak m_x$ that contains them. The surjection $Sym^\ast(V^\ast)\to A$ induces $G$-equivariant a closed immersion $X=Spec(A)\to Spec(Sym^\ast(V^\ast))\cong V$ sending $x$ to the origin. So we may assume $X$ is a $G$-invariant closed subvariety of a faithful representation $V$ and that $x$ is the origin.

To fix our ideas, choose a decomposition $G=T_0\cdot H$, where $T_0$ is a central torus and $H$ is semisimple. Choose a borel $B\subseteq H$ and a maximal torus $T\subseteq B$, and let $V = \bigoplus_{\mu\in T^\vee}V_\mu$ be the weight decomposition of $V$ as a representation of $T$. To answer the question affirmatively, it suffices to show that $V$ is trivial as a representation of $H$. That is, that only the weight $\mu=0$ appears in the decomposition.

**Remark 2:** If $X$ contains a maximal highest weight vector $v$, then $v$ is stabilized by the unipotent radical of $B$. Since points of $X$ have reductive stabilizers, $v$ must be stabilized by all of $H$. Since $v$ was a *maximal* highest weight vector in $V$, all of $V$ must be stabilized by $H$.

Peter McNamara came up with a trick for cooking up a maximal highest weight vector in the case where $X\subseteq V$ is a cone. However, Torsten Ekedahl showed that we cannot hope that $X$ will be a cone in general.

**Remark 3:** Another possible way to get an affirmative answer is to find a vector $v\in X$ which is generic (i.e. has non-trivial component in each weight space, or at least in enough weight spaces) such that the torus orbit $T\cdot v$ is not closed. If such a $v$ exists, then there is a non-trivial 1-parameter subgroup $\gamma:\mathbb G_m\to T$ such that $\gamma(t)\cdot v$ has a limit. This means that all (or enough of) the weights of $V$ lie on one side of the hyperplane in $T^\vee$ determined by $\gamma$. Since the character of $V$ is symmetric under the action of the Weyl group of $H$, all the weights must be zero.

This is sort of a combination of my two previous questions, both of which have gotten wonderful answers:

If a representation has enough reductive stabilizers, is it a direct sum of characters?

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

In particular, they helped me pin down this question, which is what I think I'm really after.