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 $C$.

  • $\begingroup$ Your proof for the case $C = V$ ia quite nice. $\endgroup$ – Angelo Mar 25 '11 at 7:00
  • $\begingroup$ I'm confused by the formulation of the question. If the answer is yes (as expected), then the group must be just a torus if the representation is faithful (?) The wording is also too elaborate, since "linearly reductive" = "reductive" in characteristic 0. $\endgroup$ – Jim Humphreys Mar 25 '11 at 10:46
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    $\begingroup$ Side remark: For a closed subgroup $H$ 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 11:07
  • $\begingroup$ @Jim: You're right that if the answer is "yes" then $G$ is a torus if the representation is faithful. I suppose I should have mentioned this, but I don't see how to actually use this fact to modify the phrasing of the problem. Thanks for the references. I only knew the result in the case when $G$ is linearly reductive, 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
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    $\begingroup$ I wrote my answer before alg. closed. char 0 condition was dropped. If k is not alg.closed, what do you mean by a point of V? If you only mean k-point, then G=units of division algebra acts on division algebra with reductive stabalisers. $\endgroup$ – Peter McNamara Mar 25 '11 at 18:13

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.

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    $\begingroup$ I don't understand how you're using that C spans V. I think this argument shows that if λ is any weight whose weight space intersects C, then there is a larger weight which is the weight of a 1-dimensional representation. So the only weight spaces that can intersect C have weights corresponding to characters. But it can happen that C does not intersect every weight space of V. In other words, C need not be contained in the span of the weight spaces it intersects. $\endgroup$ – Anton Geraschenko Mar 25 '11 at 20:01
  • $\begingroup$ @Peter: I'm also confused by this aspect of your argument, but even more confused about how you can reduce the proof at the outset to rank 1. Aside from this, it makes sense to think through some concrete situations involving $G$ of rank 1, say acting irreducibly on $V$ or with two irreducible summands. Even this case isn't so obvious, with many possible choices of a maximal torus (all conjugate) and resulting weight space decompositions. $\endgroup$ – Jim Humphreys Mar 25 '11 at 20:34
  • $\begingroup$ I think I've fixed the issue now (and added some more details). $\endgroup$ – Peter McNamara Mar 25 '11 at 23:15
  • $\begingroup$ Beautiful! This trick for peeling off extremal weight vectors while remaining in $C$ is excellent. Btw, the example that was frustrating me most was the case where $C$ is the closure of the $\mathbb G_m^2$-orbit $(s,st,st^2,\dots, st^{n-1})$ in $\mathbb A^n$. In this case, $C$ intersects only the extremal weight spaces. $\endgroup$ – Anton Geraschenko Mar 26 '11 at 2:31

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.

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.

  • $\begingroup$ Now that the argument is making more sense to me I've deleted my attempted negative version. $\endgroup$ – Jim Humphreys Mar 27 '11 at 20:58

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