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Feb 29, 2016 at 14:54 vote accept YHBKJ
Feb 20, 2016 at 9:40 answer added Honglu timeline score: 4
Oct 20, 2015 at 0:50 comment added Allen Knutson I'm guessing the natural equivariance is w.r.t. $T^n$, though obviously bigger groups act.
Oct 18, 2015 at 13:37 comment added YHBKJ @JasonStarr I'm talking about GW invariants taking values in cohomology. There may still be some non-trivial GW invariants, and when the rank of the bundle is not small compared to the dimension of the case, I think the computaton of $QH$ will still be involved as there are potentially many non-trivial GW invariants.
Oct 18, 2015 at 13:31 comment added Jason Starr Regarding Allen's question: are you asking about Gromov-Witten invariants taking values in cohomology / Chow theory, or are you asking about equivariant Gromov-Witten invariants for equivariant cohomology / equivariant Gromov-Witten theory. When $n\gg m$, of course the "usual" Gromov-Witten invariants are zero, while the equivariant theory can be nonzero. If you are asking about equivariant invariants, what is your choice of group action?
Oct 18, 2015 at 7:51 comment added pro the question made me think of this paper, but I have no idea whether it's relevant (thought I'd share anyway) projecteuclid.org/euclid.nmj/1415383898
Oct 18, 2015 at 3:26 history edited YHBKJ CC BY-SA 3.0
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Oct 18, 2015 at 3:10 history edited YHBKJ CC BY-SA 3.0
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Oct 18, 2015 at 3:06 comment added YHBKJ @AllenKnutson Sure.
Oct 17, 2015 at 20:24 comment added Allen Knutson Naive question: since these bundles have no sections, won't any holomorphic curves all land in the base $\mathbb P^m$?
Oct 17, 2015 at 19:10 history asked YHBKJ CC BY-SA 3.0