# 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.

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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 '09 at 22:24
Yes, I'm really thinking about algebraic groups. –  Anton Geraschenko Nov 3 '09 at 0:28

## 1 Answer

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.

I've read that Kapranov paper recently, I'll see if I can find something more useful to say.

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I must have misunderstood the argument. The line "This means that each generic point x∈C lies in the closure of some orbit H⋅y not coinciding with H⋅x" is really confusing me now. –  Anton Geraschenko Nov 3 '09 at 0:31