Are irregular points of an action necessarily in the closure of a larger orbit? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:16:34Z http://mathoverflow.net/feeds/question/265 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/265/are-irregular-points-of-an-action-necessarily-in-the-closure-of-a-larger-orbit Are irregular points of an action necessarily in the closure of a larger orbit? Anton Geraschenko 2009-10-10T21:28:21Z 2009-10-11T22:49:14Z <p>Suppose G is an affine algebraic group acting linearly on a vector space V. A point v&isin;V is <em>stable</em> if the orbit Gv is closed and v is <em>regular</em> (the dimension of the stabilizer of v is locally constant, or equivalently, <a href="http://mathoverflow.net/questions/193/when-is-fiber-dimension-upper-semi-continuous/202#202" rel="nofollow">locally minimum</a>). I would really like to say this is equivalent to the orbit Gv being closed and <em>not</em> being in the closure of another orbit. </p> <p>Since the orbit of a regular point has locally maximum dimension, it can't be in the closure of another orbit. But is the converse true? If a point is not in the closure of an orbit larger than its own, is it regular?</p> <p>The answer is no ... we have to throw in some hypotheses. If you consider the action of <b>G</b><sub>a</sub> on <b>A</b><sup>2</sup> given by t(x,y)=(x,tx+y), then all the orbits are closed, but points of the form (0,y) are irregular. So let's throw in the hypothesis that G <em>is linearly reductive</em>. I feel like we might also want to insist that v&ne;0, but I'm not sure about that.</p> <p>Linearly reductive seems like a strange hypothesis, so feel free to modify it. I was thinking that you could somehow show that if v is not in the closure of a larger-dimensional orbit, then span(Gv) would be an invariant subspace with no complement, but I haven't been able to get this argument to work.</p> http://mathoverflow.net/questions/265/are-irregular-points-of-an-action-necessarily-in-the-closure-of-a-larger-orbit/266#266 Answer by Ben Webster for Are irregular points of an action necessarily in the closure of a larger orbit? Ben Webster 2009-10-10T21:39:32Z 2009-10-10T21:39:32Z <p>Linearly reductive is a very good hypothesis; it tells you that the closed orbits of your group are the points of an affine variety (they are Spec of invariant polynomials). Maybe you can use some kind of semi-continuity of fiber dimension?</p> http://mathoverflow.net/questions/265/are-irregular-points-of-an-action-necessarily-in-the-closure-of-a-larger-orbit/325#325 Answer by Jarod Alper for Are irregular points of an action necessarily in the closure of a larger orbit? Jarod Alper 2009-10-11T22:49:14Z 2009-10-11T22:49:14Z <p>Anton, it seems like you answered your own question but, yes, your desired result is in fact true. In fact, it's true in the more general setting of good moduli spaces. See Prop 9.2 on the website version of my paper "Good moduli spaces for Artin stacks."</p>