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May 31, 2010 at 3:44 vote accept Eric O. Korman
May 30, 2010 at 12:46 comment added BCnrd @Scott, Angelo: my intended comment about with $G/G_x \rightarrow X$ being an (injective) immersion is correct, without any extra hypotheses on $X$ (a paracompact Hausdorff analytic manifold, say). In this way, orbits always have a natural submanifold structure, but of course generally not embedded.
May 30, 2010 at 10:26 comment added Minhyong Kim It's not the same situation as irrational winding, but thinking about orbits of analytic reductive group actions was the motivation for this comment I made a while ago: mathoverflow.net/questions/12/…
May 30, 2010 at 7:23 comment added S. Carnahan Okay, Angelo's irrational winding example drives the point home. Thanks for clearing that up.
May 30, 2010 at 7:03 answer added Angelo timeline score: 5
May 30, 2010 at 6:03 comment added Victor Protsak Note also that the main matrix of interest is semisimple, so that its orbit is closed in Zariski topology and bad stuff doesn't happen.
May 30, 2010 at 4:56 comment added BCnrd @Angelo: Whoops, I meant to say "immersion" rather than "embedding". It may be that one still needs some mild hypotheses for this case to be true (my copy of Bourbaki is far away), but my recollection is that orbits are immersed submanifolds.
May 30, 2010 at 4:34 comment added Angelo To Brian: I don't think this is true; the standard examples of actions of $\mathbb R$, or $\mathbb C$, on a torus with a dense orbit are analytic actions. However, it is certainly true for algebraic group actions on algebraic manifold (over $\mathbb C$, but from this case it is easy to deduce the real case).
May 30, 2010 at 4:16 comment added BCnrd @Scott: strictly speaking, it doesn't make sense to say "tangent space at each point of an orbit" until one first addresses the issue of submanifold structure at that point for the orbit. More generally, if an analytic Lie group $G$ acts on the left on an analytic manifold $X$ then for $x \in X$ with stablizer $G_x$ the issue is whether or not $G/G_x \rightarrow X$ is an embedding. This must be proved somewhere in Chapter 3 of Bourbaki Lie Groups & Lie Algebras. As usual, Google Books seems to be missing the interesting relevant pages.
May 30, 2010 at 2:07 comment added S. Carnahan I'm having some difficulty seeing the obstruction to being an embedded submanifold. Isn't the tangent space at each point of an orbit naturally isomorphic to Lie($GL_n$)/Lie(centralizer)?
May 30, 2010 at 0:45 history asked Eric O. Korman CC BY-SA 2.5