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S Oct 8, 2013 at 14:56 history bounty ended CommunityBot
S Oct 8, 2013 at 14:56 history notice removed CommunityBot
Oct 5, 2013 at 8:43 answer added Rami timeline score: -1
Oct 3, 2013 at 21:05 comment added Jesko Hüttenhain @Rami: You are completely correct, that's what I mean. If it's easy to see that $M.v$ is a variety, please answer with a proof =).
Oct 3, 2013 at 16:09 comment added Rami 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?
S Sep 30, 2013 at 13:10 history bounty started Jesko Hüttenhain
S Sep 30, 2013 at 13:10 history notice added Jesko Hüttenhain Draw attention
Sep 28, 2013 at 13:27 comment added Jesko Hüttenhain @AndyB: Yea, that's a really rare case unfortunately.
Sep 27, 2013 at 21:53 comment added Andy B 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 $Spec( \mathbb{C}[V]/\ker(f))$.
Sep 27, 2013 at 21:18 comment added Jesko Hüttenhain @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.
Sep 27, 2013 at 16:47 comment added Benjamin Steinberg Lex Renner and Mohan Putcha are the experts in algebraic monoids. You might ask them.
Sep 27, 2013 at 13:01 history asked Jesko Hüttenhain CC BY-SA 3.0