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?



Are Gromov-Witten invariants birational invariants?
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*](http://dx.doi.org/10.1112/S0010437X17007667), Compos. Math. **154**, No. 3, 595-620 (2018). [ZBL1420.14124](https://zbmath.org/?q=an:1420.14124). 


