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Mar 3, 2010 at 16:54 comment added David E Speyer I fixed some sign errors, which had no effect on the answer.
Mar 3, 2010 at 16:54 history edited David E Speyer CC BY-SA 2.5
edited body; edited body; edited body
Feb 11, 2010 at 2:41 comment added Allen Knutson While I haven't fully checked your calculation, I will note that the Bott-Samelson for a Schubert divisor in $Flags({\mathbb C}^3)$ is $F_1$, not homemorphic to $F_0$, so it seems very plausible.
Feb 11, 2010 at 2:08 history answered David E Speyer CC BY-SA 2.5