Suppose $G\to GL(V)$ is a linear representation of an irreducible algebraic group over a field $k$.

Suppose $C\subseteq V$ is a $G$-invariant closed cone that spans $V$, and that the stabilizer of any point of $C$ is linearly reductive. Must $V$ be a direct sum of 1-dimensional representations?

[Edit] I mostly care about the case where $k$ is algebraically closed and of characteristic zero. In this case, all instances of "linearly reductive" may be changed to "reductive."

[Edit] **Remark 0:** If they answer is "yes," then the image of $G$ in $GL(V)$ is a torus.

**Remark 1:** In my situation, $C$ is actually the closure of a $G$-orbit, so I'm happy to assume $C$ contains a dense open copy of $G$.

**Remark 2:** I can prove this result in the case $C=V$ as follows.

Taking $v=0$, we see that $G$ is linearly reductive. A subgroup $H\subseteq G$ is linearly reductive if and only if $G/H$ is affine (I can't recall the name associated to this result). It follows that for every $v\in V$, the orbit $G\cdot v \cong G/Stab(v)$ is affine. Every non-zero orbit is either a $k^\times$-torsor or a $\mu_n$-torsor (for some $n$) over its image in $\def\P{\mathbb P}\P(V)$, so the image of each orbit in $\P(V)$ is also affine (this uses that $k^\times$ and $\mu_n$ are linearly reductive). Choose a non-zero $v\in V$ so that the image $Z$ of $G\cdot v$ in $\P(V)$ is minimal dimensional. Minimality of dimension implies that $Z$ is closed. A closed subscheme of $\P(V)$ is affine if and only if it is finite. Since $G$ is irreducible, we have that $Z$ is a single point, so $G$ fixes the 1-dimensional subspace $k\cdot v$. $G$ is linearly reductive, so you can find a complement to $k\cdot v$ and repeat this argument until you've decomposed all of $V$ into 1-dimensional $G$-invariant subspaces.

In general, this proof produces $\dim(C)$ many $G$-invariant 1-dimensional spaces (contained in $C$), but the span of these subspaces need not contain $C$.

`$H$`

of a reductive group`$G$`

in any characteristic, it's true that`$G/H$`

is affine iff`$H$`

is reductive. The history is complicated, but in char 0 the hard direction goes back to Matsushima and in char`$p$`

Mumford's Conjecture (proved by Haboush) is needed. See R.W. Richardson,Bull. London Math. Soc.9 (1977). $\endgroup$ – Jim Humphreys Mar 25 '11 at 11:07linearlyreductive, which is why I left that word there. I suppose I'm simultaneously broadcasting my general preference not to assume characteristic 0 and my lack of understanding of (non-linearly) reductive groups. I'll edit the question. $\endgroup$ – Anton Geraschenko Mar 25 '11 at 17:39