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Jul 4, 2011 at 15:12 comment added Allen Knutson The orbit through $x\in X$ is the image of the composite map $G \to G \times \{x\} \to G \times X \to X$. Images of algebraic maps are constructible sets. Constructible sets (are nasty but) have an open set on which they are locally closed. So $G\cdot x$ has an open dense set on which it's locally closed. And $G\cdot x$ is homogeneous, so it's locally closed everywhere.
Jun 30, 2011 at 3:06 comment added Dr Shello Hmmmm.......why not?
May 5, 2010 at 14:51 history answered Allen Knutson CC BY-SA 2.5