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Nov 12, 2013 at 20:50 answer added André Henriques timeline score: 4
Mar 4, 2010 at 23:39 comment added Allen Knutson Thanks. I'd feel terrible if anyone misattributed this question to some other guy Bertrand Kostant.
Mar 3, 2010 at 16:58 comment added Mariano Suárez-Álvarez Bert == Bertram I guess... Most people on the site are not that familiar with him :P
Feb 11, 2010 at 2:08 answer added David E Speyer timeline score: 7
Feb 10, 2010 at 16:51 comment added David E Speyer A basic test case: does the Bott-Samelson over SL_3/B have the same topology as (P^1)^3? I tried to work this out last night, but couldn't. (Of course, they have the same betti numbers.)
Feb 10, 2010 at 0:45 comment added Allen Knutson An interesting idea. Is there a way to degenerate the pair $(({\mathbb P}^1)^R, $ample line bundle$)$ to (Bott-Samelson, nef line bundle)? I am more used to embedded degenerations, in which the line bundle necessarily stays ample.
Feb 9, 2010 at 20:59 comment added VA. Perhaps because of the Bott-Samelson resolution $X\to G/B$, which is an iterated $\mathbb P^1$-fibration?
Feb 9, 2010 at 16:34 comment added Allen Knutson You're right, $F_0$ degenerates to $F_2$, in Hirzebruch-surface notation. (It even does so $T^1$-equivariantly.) But those manifolds, unlike mine, are diffeomorphic. Question edited to reflect that.
Feb 9, 2010 at 16:32 history edited Allen Knutson CC BY-SA 2.5
added 37 characters in body
Feb 9, 2010 at 15:51 comment added David E Speyer "Since the two varieties are both smooth, there won't be a flat family over an irreducible base in which one is a general fiber, one the special." Are you sure about this? I thought there was a flat family whose general fibers were $P^1 \times P^1$ and whose special fiber was $\Sigma_2$. Namely, set $E:=\mathrm{Ext}_{P^1}(\mathcal{O}(2), \mathcal{O}(0))$. Form the universal rank two vector bundle on $E \times P^1$; this is $\mathcal{O}(2) \oplus \mathcal{O}(0)$ over the origin of $E$ and $\mathcal{O}(1) \oplus \mathcal{O}(1)$ everywhere else. Projectivize to get the claimed example.
Feb 9, 2010 at 15:11 comment added David E Speyer Hartshorne connectivity does not hold when you keep track of the multigrading. See arxiv.org/abs/math/0201271 . I don't know about any of the more interesting questions you pose.
Feb 9, 2010 at 14:57 history asked Allen Knutson CC BY-SA 2.5