Timeline for How to think about parabolic induction.

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Dec 10, 2012 at 17:55	comment	added	Sam Gunningham		As mentioned above, $G/P$ and $G/Q$ are not even bi-holomorphic (let alone $G$-equivariantly so) even when $P$ and $Q$ have the same Levi. However, for various notions of ``function'' there is $G$-equivariant isomorphism $Funct(G/P) \cong Funct(G/Q)$. This is just a special case of independence of parabolic, as both of these are the parabolic induction of the trivial module on $L$. For example, the derived category of $D$-modules on $G/P$ and $G/Q$ are equivalent as $G$-categories.
Dec 18, 2009 at 10:22	comment	added	Kevin McGerty		That's true for the GL(3) example as you say, but not equivariantly -- the isomorphism uses the outer automorphism of GL(n) (the transpose map). Also, I suspect that's it, in terms of isomorphisms, even without demanding equivariance.
Dec 18, 2009 at 5:41	comment	added	Ben Webster♦		But there's no G-equivariant isomorphisms between G/P's where the parabolic aren't conjugate so something more subtle is happening.
Dec 18, 2009 at 3:46	history	answered	S Kitchen	CC BY-SA 2.5