## Do subgroups respect the orbit-closure relation?

Suppose G is a Lie group (or algebraic group) acting on a manifold (or scheme) X, and H⊆G is a subgroup. Let x,y∈X be points such that x is in the closure of the orbit H⋅y (but not in H⋅y itself). Then obviously x is in the closure of G⋅y, but can it happen that x is actually in the orbit G⋅y (not just in the closure)?

Background: 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 Chow quotients of Grassmannian I, 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⊆G is a torus and X is a smooth projective variety.

-
 I guess you mean H is a closed subgroup? Otherwise, think of G=the reals, H=the rationals, X=the reals, x=0, y=pi. – Andrew Critch Nov 2 2009 at 22:24 Yes, I'm really thinking about algebraic groups. – Anton Geraschenko♦ Nov 3 2009 at 0:28