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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.

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I don't fully follow your remark 3. However, the Hilbert-Mumford stability criterion states that, if $G$ is a reductive group, $V$ a finite dimensional representation, and $x$ a point of $V$ such that $0$ is in the closure of $Gx$, then there is a one parameter multiplicative subgroup $\gamma$ of $G$ such that $0$ is in the closure of $\gamma x$. Does that help? –  David Speyer Apr 5 '11 at 16:33
    
@David: thanks for pointing that out! Perhaps what is confusing about my remark 3 is that $T$ is a maximal torus of the semisimple part $H$ of $G$, not of $G$. The other thing that is confusing is that I have to fix the maximal torus $T\subseteq H$ in order to get the weight space decomposition, which I need in order to define "generic." So even if we can employ Hilbert-Mumford to find a 1-parameter subgroup so that $\gamma x$ has 0 in its closure, it will only tell us that for any maximal torus containing $\gamma$, the weights appearing in $x$ are on one side of a hyperplane. –  Anton Geraschenko Apr 5 '11 at 19:37
    
I'll try to think of a better way to present my Remark 3, and of some way to get some leverage out of the Hilbert-Mumford criterion. If I succeed in either of these, I'll edit the question. –  Anton Geraschenko Apr 5 '11 at 19:38
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If your X is smooth, then by Luna's Slice Theorem it is a finite dimensional representation of G: see Corollaire 2 of archive.numdam.org/article/MSMF_1973__33__81_0.pdf Then the argument with highest weight vectors proves that G is a torus. –  Bart Van Steirteghem Apr 6 '11 at 3:02
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@Bart: Yes, I probably should have said that the smooth case can be handled this way. I'm interested in the singular case. –  Anton Geraschenko Apr 6 '11 at 3:43
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2 Answers

up vote 5 down vote accepted

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.

Proposition: Let $V$ be a representation of a reductive group $G$ and let $X$ be the closure of an orbit of $G$ so that $0\in X$. If $X$ is not contained in a direct sum of 1-dimensional representations, then it contains a positive highest weight vector (with respect to some borel).

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.

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Now this may be a really stupid comment, but since I don't have enough reputation I can only enter this as an "answer" as opposed to "comment". Also, my algebraic-geometry-knowledge is kindergarten level, and there might well be mistakes/clumsiness in it (so please don't rate up unless you have read it fully and feel it is okay and helpful).

Here is what seems to happen when X is a reductive monoid - an irreducible linear algebraic monoid whose (necessarily open dense) group G of units is reductive, and when you think of the action of G on X as given by left multiplication. Suppose X is of the minimal possible dimension. In this case, the theory of Mohan Putcha and Lex Renner says that (as k is algebraically closed) X(k) is the union of the various G(k) e G(k) as e runs over the (finitely many) idempotents in $\bar{T}(k)$, the k-points of the Zariski closure in G of a maximal torus T of G. In this case, the set {g in G : g e = e g e} is (by Putcha's and Renner's theory) a parabolic subgroup P of G, which is easily seen to be proper if e is not central. Their theory also gives that R_u(P) e = {e}, where R_u(P) is the unipotent radical of P. Thus, if P is non-trivial, i.e., if e is not central, then R_u(P) will then be a non-trivial normal unipotent subgroup of your stabilizer {g : g e = e}, forcing e to have a non-reductive stabilizer.

Thus, we need only worry about the case where all the idempotents in $\bar{T}(k)$ are central. In this case X has only finitely many G orbits, namely X(k) is the union of the various G(k) e. Thus, X being normal, the codimension two criterion will force at least one of these e's to have a dimension one stabilizer, which then has to be a dimension one diagonalizable subgroup.

Now consider eX, a reductive monoid with eG as its group of units. ex is an eG-invariant k-point in it, and eX contains a dense open copy of eG. Thus, eX will be a counterexample of dimension smaller than X (we assumed X to be of the minimal possible dimension), once we show that eX is normal. But then eX is the Zariski closure of eG - it is contained in the Zariski closure of eG for continuity reasons, and contains the Zariski closure because the Zariski closure, like eG itself, has to be fixed under multiplication by e. Thus, eX is an orbit closure in the spherical $G \times G$-variety X (recall that e is central), and hence normal (to the best of my knowledge, orbit closures in spherical varieties are normal).

Here the existence of a G(k)-invariant element was crucially used - otherwise there is a chance that eX will be equal to eG, and hence not really a counterexample : this can happen in the case of, say, GL_2 times toric variety.

To repeat, even if you use this (I don't know if the arguments above already cover the monoid case or not), please use with caution as there may be mistakes.

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