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Let $X$ and $Y$ be two smooth complex projective varieties. Then they are also symplectic manifolds. We know Gromov-Witten (GW) invariants are symplectic invariants. That means if there exists a a bijective symplectomorphism $f: X \to Y$, then GW invariants of $X$ and $Y$ are the same.

Now suppose $f: X \to Y$ is an algebraic isomorphism, or more generally, a birational map, then $f$ may not be a symplectomorphism. In this case, are GW invariants of $X$ and $Y$ different in general?

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On a complex projective variety, Gromov-Witten invariants can be interpreted as virtual counts of curves, so they are biregular invariants.

However, they are not birational invariant in general. The behaviour of Gromov-Witten invariants under an arbitrary birational modification is in fact rather subtle.

For more details and examples you can have a look at Section 1.4 of the paper

D. Abramovich, J. Wise Birational invariance in logarithmic Gromov-Witten theory, Compos. Math. 154, No. 3, 595-620 (2018). ZBL1420.14124.

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  • $\begingroup$ By definition, GW invariants are only invariant under complex deformations. Why are they biregular invariants? $\endgroup$
    – Yuhang Chen
    Commented Nov 13, 2019 at 13:27
  • $\begingroup$ Biregular maps (= algebraic isomorphisms) do not change virtual curve counting, because biregularly equivalent varieties contain the same curves. $\endgroup$
    – Francesco Polizzi
    Commented Nov 13, 2019 at 13:33
  • $\begingroup$ 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 $X$. $\endgroup$
    – Yuhang Chen
    Commented Nov 13, 2019 at 13:42
  • $\begingroup$ 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. $\endgroup$
    – Francesco Polizzi
    Commented Nov 13, 2019 at 13:52
  • $\begingroup$ 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. $\endgroup$
    – Yuhang Chen
    Commented Nov 13, 2019 at 14:05

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