Timeline for Are Gromov-Witten invariants birational invariants?
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17 events
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| Nov 14, 2019 at 0:14 | comment | added | Yuhang Chen | @LevSoukhanov Thanks for the confirmation. | |
| Nov 13, 2019 at 19:56 | comment | added | Lev Soukhanov | Virtual fundamental classes are an intrinsic notion - they can be defined as Euler classes of obstruction bundle, and obstruction bundle is intrinsic. | |
| Nov 13, 2019 at 17:19 | vote | accept | Yuhang Chen | ||
| Nov 13, 2019 at 17:18 | comment | added | Yuhang Chen | @Qfwfq The construction was carried out in two papers: "The intrinsic normal cone" by Behrend and Fantechi, and "Gromov-Witten invariants in algebraic geometry" by Behrend. I haven't read the details yet, but everything looks intrinsic. Thanks for the discussion. This cleared my doubt. | |
| Nov 13, 2019 at 15:16 | comment | added | Qfwfq | "I was thinking of GW invariants via algebraic geometry" - I'm no expert and I don't know how the GW virtual fundamental class is actually constructed. When you construct it do you make some choices (such as a polarization of $X$ etc) during the journey? If not, they're "biregular invariants" just by definition... | |
| Nov 13, 2019 at 15:09 | comment | added | Yuhang Chen | @FrancescoPolizzi Now I have one more doubt: are GW invariants (defined in algebraic geometry) always symplectic invariants? | |
| Nov 13, 2019 at 14:59 | comment | added | Yuhang Chen | @Qfwfq Suppose we use the symplectic definition of GW invariants. Then they are symplectic invariants. Considering algebraic isomorphisms may not preserve symplectic structures, we should suspect GW invariants are not algebraic invariants, right? | |
| Nov 13, 2019 at 14:53 | comment | added | Yuhang Chen | @Qfwfq I was thinking of GW invariants via algebraic geometry. I am not sure whether they are intrinsic via the virtual normal cone construction using obstruction theory. | |
| Nov 13, 2019 at 14:24 | comment | added | Qfwfq | @Yuhang Chen: ...or you're using a symplectic definition of those invariants, and the way you attach a symplectic form to an abstract smooth projective variety is via the Fubini-Studi form of an embedding so it's not a priori intrinsic to the holomorphic/algebraic structure? | |
| Nov 13, 2019 at 14:19 | comment | added | Qfwfq | @Yuhang Chen: if you start with two isomorphic mathematical objects and perform the same intrinsic construction to them, you'll get the same result won't you? (where "intrinsic" and "same" have to be interpreted in the usual mathematical ways of course). Like: do two isomorphic groups have the same derived group? Yes (in the obvious sense). | |
| Nov 13, 2019 at 14:05 | comment | added | Yuhang Chen | I guess my concern is mainly about the virtual fundamental class. I haven't looked into the details of the definition of the virtual fundamental class. So I am not sure if different algebraic descriptions (i.e., different polynomial equations) of a projective variety will affect anything. | |
| Nov 13, 2019 at 13:52 | comment | added | Francesco Polizzi | I am not sure that I understand your concern. At the end of of the story, you are simply (virtually) counting how many curves there are that intersect $n$ chosen submanifolds of $X$. This number is clearly invariant under any biregular map. The counting is only virtual, as there can be non-integer contributions given by the stabilizers at the orbifold points of the moduli stack of stable maps, but again these contributions are biregularly invariant. | |
| Nov 13, 2019 at 13:42 | comment | added | Yuhang Chen | But I doubt whether isomorphic varieties $X$ and $Y$ must have same "virtual counting of curves". It's not obvious by looking at the definition of GW invariants of $X$, i.e., the integration of cohomology classes over the virtual fundamental class of the moduli stack of stable maps into $X$. | |
| Nov 13, 2019 at 13:33 | comment | added | Francesco Polizzi | Biregular maps (= algebraic isomorphisms) do not change virtual curve counting, because biregularly equivalent varieties contain the same curves. | |
| Nov 13, 2019 at 13:27 | comment | added | Yuhang Chen | By definition, GW invariants are only invariant under complex deformations. Why are they biregular invariants? | |
| Nov 13, 2019 at 13:12 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Nov 13, 2019 at 13:06 | history | answered | Francesco Polizzi | CC BY-SA 4.0 |