Let $G=\mathrm{Gl}_n(\mathbb C)$ and $V$ an irreducible $G$-module on which $G$ acts polynomially. In other words, the algebraic group action of $G$ on the affine space $V$ extends to an algebraic action of the monoid $M=\mathbb C^{n\times n}$ on $V$. Now, let $v\in V$ and consider the subset $M.v\subseteq V$. I was told that $M.v$ is a variety, but noone could tell me a reference. I would be very happy if you could provide that. I am also curious whether it is an affine variety, or at least quasi-projective.
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ Lex Renner and Mohan Putcha are the experts in algebraic monoids. You might ask them. $\endgroup$– Benjamin SteinbergCommented Sep 27, 2013 at 16:47
-
$\begingroup$ @BenjaminSteinberg: I had originally thought that this would translate to a general question about algebraic monoids, but apparently it is rather special to the case of the general linear group (and the matrix monoid). Of course, I will probably follow your advice if this question gets no answers. $\endgroup$– Jesko HüttenhainCommented Sep 27, 2013 at 21:18
-
1$\begingroup$ If it is closed it's easy to identify it as a scheme: Let $f : \mathbb{C}[V] \to \mathbb{C}[G]$ be defined by taking the $i$-th coordinate of $V$ to the $i$-th entry of $g.v$ (where $g$ is a matrix of variables). Your matrix orbit wants to be $Spec( \mathbb{C}[V]/\ker(f))$. $\endgroup$– Andy BCommented Sep 27, 2013 at 21:53
-
$\begingroup$ @AndyB: Yea, that's a really rare case unfortunately. $\endgroup$– Jesko HüttenhainCommented Sep 28, 2013 at 13:27
-
$\begingroup$ Am I understand correctly that: 1.$M$ is the monoid of $n \times n$ matrices. 2. $M.v=\{mv|m\in M\}$? If yes, it's look to easy. What did I miss? $\endgroup$– RamiCommented Oct 3, 2013 at 16:09
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
8
As explained in the comments, the following Lemma is wrong:
Let $f:W \to V$ be an homogeneous algebraic map (i.e. $f(\alpha v)=\alpha^k f(v)$ for some $k$). Then the image of $f$ is closed sub-variety of $V$.
Proof: let $\bar f: \mathbb P(W) \to \mathbb P(V)$ be the corresponding map of protective spaces. Since $\mathbb P(W) $ is complete, the image of $\bar f$ is a closed sub-variety of $\mathbb P(V)$. So it is given be a collection of homogeneous polynomials on $V$. The same polynomials are the one that defines the image of $f$.
The rest is probably correct now (I had to correct it too), but probably useless:
The map $m \mapsto mv$ is homogeneous map (since the representation $V$ is irreducible).
-
$\begingroup$ The problem is that $\bar f$ is usually not well-defined. There could be matrices $m\in M\setminus\{0\}$ such that $mv=0$. $\endgroup$ Commented Oct 5, 2013 at 13:29
-
$\begingroup$ Does it really matter? The "deprojectivization" is a cone anyway and so it does include zero. What I mean is that you can just replace $f$ by $\tilde{f} \colon \mathrm{End}(\mathbb{C}^n)/\mathrm{Ker}(f) \to V$. $\endgroup$ Commented Oct 5, 2013 at 14:31
-
$\begingroup$ What is $\mathrm{Ker}(f)$? Do you mean the fiber over $0$? It does not have to be a sub-vectorspace. You might look at $\mathbb P(M\setminus f^{-1}(0))$, but that is probably no longer projective, so the argument will not work. $\endgroup$ Commented Oct 5, 2013 at 15:26
-
1$\begingroup$ Unfortunately, the "lemma" is false. Take the map $(x,y) \to (x^2, xy)$. The image is the union of $(0,0)$ and the set where $x$ is nonzero, which is not even a variety. $\endgroup$ Commented Oct 5, 2013 at 19:20
-
$\begingroup$ I'm confused. I thought we are dealing with linear representation, are we not? $\endgroup$ Commented Oct 6, 2013 at 0:00