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?


1 Answer 1


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.

  • $\begingroup$ By definition, GW invariants are only invariant under complex deformations. Why are they biregular invariants? $\endgroup$ 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. $\endgroup$ 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 $X$. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ Nov 13, 2019 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.