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.
Best Answer
Vera Serganova showed me the following (affirmative) answer. I'll use the setup from Remark 1. Note that Remark 2 shows that $X$ cannot contain a positive highest weight vector. So the following result does the job.
Proof. We have that $X=\overline{G\cdot v}$, and we are given that $v$ is not contained in a sum of 1-dimensional representations (otherwise $X$ would be as well). Since $0\in \overline{G\cdot v}$, the Hilbert-Mumford criterion (Proposition 2.4 of GIT) tells us that there is a 1-parameter subgroup $\gamma$ so that $\gamma(t)\cdot v$ contains $0$ in its closure. Let $V=\bigoplus_{i\in \mathbb Z}V_i$ be the weight space decomposition of $V$ with respect to $\gamma$. Then $v=v_p+v_{p+1}+\cdots$ with $v_i\in V_i$, $v_p\neq 0$, and $p>0$.
Let $T$ be a maximal torus of $G$ which contains $\gamma$, and let $B\subseteq G$ be a borel containing $T$ so that $\gamma$ pairs non-negatively with all the positive roots. Since $V$ is finite-dimensional, we can tweak $\gamma$ so that it pairs positively with all the positive roots.
If $v$ is a highest weight vector, we are done. Otherwise there is some positive root $\alpha$ so that $e_\alpha\in \mathfrak g_\alpha$ does not annihilate $v$. Let $\exp(t\cdot e_\alpha)\cdot v = \sum_{i\ge p} f_i(t)$, where $f_i(t)\in k[t]\otimes V_i$, and let $m_i=\deg(f_i)$. Since $e_\alpha\cdot V_i\subseteq V_{i+\langle \gamma,\alpha\rangle}$ and $\langle \gamma,\alpha\rangle>0$, we have that $m_p=0$. Since $e_\alpha\cdot v\neq 0$, we have that $m_i>0$ for some $i>p$.
Let $a/b\in \mathbb Q$ be the positive rational number so that $m_i\cdot b \le i\cdot a$ for all $i$, with $m_j\cdot b = i\cdot a$ for some $j>p$. Define the morphism $g: \mathbb A^1 \to V$ by the formula $g(t) = \sum_{i\ge p} t^{i\cdot a}f(t^{-b})$. The condition that $m_i\cdot b \le i\cdot a$ implies this is a well-defined morphism, and the condition that $m_j\cdot b = j\cdot a$ for some $j>p$ implies that $g(0)\neq 0$. But for $t\neq 0$, we have that $g(t) = \gamma(t^a)\exp(t^{-b}e_\alpha)\cdot v\in X$, so since $X$ is closed, we have that $g(0)\in X$.
Note that $g(0)$ is non-zero, it lies in $\bigoplus_{i>p} V_i$, and it is not contained in a sum of 1-dimensional representations since it is in the image of $e_\alpha$ (which annihilates 1-dimensional representations). Since $V$ is finite dimensional, this procedure can only be repeated a finite number of times, so we get a highest weight vector after a finite number of iterations.