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

Question

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 $C$.

Answer 1

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=G_m. If v=Σ_iv_i in C with t in G_m acting on v_i 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(G_mv).) A general T is a product of G_m's. We pick a vector in C, run the argument for each factor G_m in succession, and end up with a weight vector for T.

If v in V_λ 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 V_λ. Run the G_m 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.

Answer 2

This is meant to be a slightly cleaned up version of Peter's answer above. If I'm not mistaken, it is not necessary to assume the characteristic of $k$ is zero. However, I don't know the theory of reductive groups in positive characteristic—even in characteristic 0, I've only dealt with semi-simple groups—so I may be making wrong assumptions about how the representation theory works.

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Let $B\subseteq G$ be a borel with torus $T$ and unipotent radical $U$. Let $V=\bigoplus V_\lambda$ be the decomposition of $V$ into weight spaces with respect to $T$.

Lemma: Suppose $v_\lambda\in V_\lambda$, $v=\sum v_\lambda$ is a point of $C$, and $\mu\in T^\vee$ is extremal among weights in the support of $v$ (i.e. weights for which $v_\mu\neq 0$). Then $C$ contains $v_\mu \def\l{\langle}\def\r{\rangle}$.

Proof: Since $\mu$ is extremal among weights in the support of $v$, there is a 1-parameter subgroup $\alpha\in \hom(\mathbb G_m,T)\cong \hom(T^\vee,\mathbb Z)$ such that $\l\alpha,\mu\r > \l\alpha,\lambda\r$ for any $\lambda\neq \mu$ in the support of $v$. Consider the map $f\colon\mathbb A^1\to V$ given by $f(t)= \sum t^{\l\alpha,\mu\r-\l\alpha,\lambda\r}v_\lambda$. Away from $t=0$, we have that $f(t)=t^{\l\alpha,\mu\r}\alpha(t^{-1})\cdot v$ is in $C$. Since $C$ is closed, we get that $f(0)=v_\mu$ is in $C$. $\square$

Now suppose $\mu$ is an extremal highest (with respect to $B$) weight appearing in the decomposition of $V$. Since $C$ spans $V$, there is a point of $C$ with a component in $V_\mu$, so by the lemma, there is a non-zero vector $v\in V_\mu\cap C$. Since $\mu$ is a highest weight, we have that $U$ stabilizes $v$. Since the stabilizer of $v$ is reductive, it must contain the derived subgroup of $G$, so the irreducible representation with highest weight $\mu$ is a character, so $\mu$ pairs to zero with every coroot. It follows that every weight in the decomposition of $V$ pairs to zero with every coroot, so $V$ is a direct sum of characters.