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Sep 22, 2014 at 3:55 vote accept user1832
Apr 1, 2010 at 5:51 comment added Marty I don't know the real case as well as I should. But relevant aspects are described well in a paper by Casselman, Hecht, and Milicic,"Bruhat filtrations and Whittaker vectors for real groups" in the smooth setting. You can find this paper easily online. In the unitary setting, you're right that B has measure zero, so it doesn't cut out any proper subspace or quotient of the Hilbert space $L^2(G/B, \chi)$ of the unitary principal series representation.
Apr 1, 2010 at 3:26 comment added user1832 Thanks, Marty. I think your explanation works perfectly in p-adic setting, but in real setting, if we consider smooth Frechet (or Banach) representation, the irreducible subspaces means closed ones, so it seems still possible that $V(BwB)$ is all of $V$. Another issue in unitary case, I think, is that $B$ has Haar mesaure zero, so the function supported on it is zero in L^2.
Mar 30, 2010 at 19:41 history edited Marty CC BY-SA 2.5
fixed typo.
Mar 30, 2010 at 15:52 history answered Marty CC BY-SA 2.5