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



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> 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?
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[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 $C$.



If a representation has enough reductive stabilizers, is it a direct sum of characters?
The answer is yes. We know G is reductive, take B=TU a Borel.


Decompose V into a direct sum of weight spaces for T: $V=\oplus V\_\lambda$. I claim that C contains a non-zero wieght vector. Proof: First lets consider T=Gm. If v=Σivi in C with t in Gm acting on vi by multiplication by ti, consider $\lim\_{t\to \infty}t^{-N}(t\cdot v)$ (t-N is scalar multiplication, dot is action). Since C is closed under action, scalar multiplication and taking limits, this is a weight vector in C for appropriate N. (this argument produces a weight vector of maximal possible weight in span(Gmv).) A general T is a product of Gm's. We pick a vector in C, run the argument for each factor Gm in succession, and end up with a weight vector for T.


If v in Vλ and u in U, then uv-v is in $\oplus V\_\mu$ where only wieghts μ greater than λ appear in the sum. Let v be a weight vector in C of maximal possible weight . Applying u to v and running the torus argument again, we see that for this maximalilty assumption to hold, v must be annihilated by all of U. Since stabaliser of v is reductive, containing U, it must contain derived group of G.


Let $\alpha^\vee$ be a simple coroot and λ be a weight appearing in V. Suppose $\langle\alpha^\vee,\lambda\rangle$>0. Since C spans V, there exists c in C with non-zero projection onto Vλ. Run the Gm argument with the vector c and the subgroup $\alpha^\vee(\mathbb{G}\_m)$. Now run the torus argument on this output, we get a weight vector in C of weight μ where $\langle\alpha^\vee,\mu\rangle>0$.


Now under this assumption, the paragraph with the unipotent element shows that there is a weight vector v in V, of weight ν greater than μ with derived group in stabaliser. Also $\langle\alpha^\vee,\nu\rangle>0$. This is a contradiction. Simlilaraly $\langle\alpha^\vee,\lambda\rangle$<0 is a contradiction. So for every weight λ and every simple coroot $\alpha^\vee$, we have $\langle\alpha^\vee,\lambda\rangle$=0, which forces the representation to be a direct sum of characters.


