most recent 30 from http://mathoverflow.net 2013-05-24T07:36:07Z http://mathoverflow.net/feeds/question/3849 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3849/do-subgroups-respect-the-orbit-closure-relation Anton Geraschenko 2009-11-02T21:10:32Z 2009-11-02T21:46:11Z

<p>Suppose G is a Lie group (or algebraic group) acting on a manifold (or scheme) X, and H&sube;G is a subgroup. Let x,y&isin;X be points such that x is in the closure of the orbit H&sdot;y (but not in H&sdot;y itself). Then obviously x is in the closure of G&sdot;y, but can it happen that x is actually in the orbit G&sdot;y (not just in the closure)?</p> <p><strong>Background:</strong> I got stuck on this point when trying to understand the very last line of the proof of Theorem 0.3.1 of Kapranov's <a href="http://arxiv.org/abs/alg-geom/9210002" rel="nofollow">Chow quotients of Grassmannian I</a>, which states that every irreducible component of an algebraic cycle corresponding to a point in the Chow quotient X//G is the closure of a single G-orbit. In this case, H&sube;G is a torus and X is a smooth projective variety.</p>

http://mathoverflow.net/questions/3849/do-subgroups-respect-the-orbit-closure-relation/3853#3853 David Speyer 2009-11-02T21:46:11Z 2009-11-02T21:46:11Z

<p>Sure, that can happen. G = PGL_2, H is the torus, X is the Riemann sphere, x is the north pole, y is some point other than the two poles.</p> <p>I've read that Kapranov paper recently, I'll see if I can find something more useful to say.</p>