Suppose that $(M\_{1},\omega\_{1})$ and $(M\_{2},\omega\_{2})$ are compact symplectic $4$-manifolds, that are (oriented) diffeomorphic. Is it true that the Todd genus ($\frac{1}{12} (c\_{1}^{2} + c\_{2})(M\_{i})$) are equal?


I know a reference that the answer is yes for algebraic surfaces (since the Todd genus equals $\chi(\mathcal{O}\_{S}) = 1 - \frac{1}{2}b\_{1}(S) + P\_{1}(S) $ and the plurigenera are known to be an (oriented) smooth invariant by Seiberg-Witten theory). 



Todd genus of symplectic $4$-manifolds a smooth invariant?
In dimension 4, the Todd genus does not depend on the choice of a symplectic structure or even on an almost complex structure. If $M$ is an almost complex 4-manifold, then $\langle c\_1(M)^2, [M]\rangle = 2\chi(M) + 3\sigma(M)$ (see [here](http://web.stanford.edu/~lstarkst/Notes3), p. 9), and $\langle c\_2(M), [M]\rangle = \chi(M)$; here $\chi(M)$ is the Euler characteristic of $M$ and $\sigma(M)$ is its signature.


Thus any orientation-preserving diffeomorphism preserves the Todd genus, and the orientation-preserving assumption is necessary, because the signature depends on the choice of orientation.


This is also mentioned at the end of §2 of [this paper](https://arxiv.org/pdf/1211.2640.pdf) by Łukasz Bąk.


