Let $X \subset Y$ be an inclusion of compact complex (possibly Kähler) manifolds. I'm wondering if "Riemann-Roch without Denominators" [1, Thm 15.3] holds in that situation. The statement is
For a vector bundle $E$ on $X$ one has $$c(f_* E) = 1 + f_*(P(N,E)), \label{eq:1}\tag{1}$$ where $N$ is the normal bundle of $X$ in $Y$ and $P(N,E)$ is a specific computable polynomial in the Chern classes of $E$ and $N$ (see[1]).
I didn't find any reference to this in the setting of complex manifolds, though I found a version of Grothendieck-Riemann-Roch[2].
I had the impression that one can basically copy the proof from Fulton's book, but I'm not sure if there are any subtle problems I'm not seeing. So my strategy would be:
- Define the Chern classes for a coherent sheaf $\mathcal F \in \operatorname{Coh}(Y)$ by taking a locally free resolution of $C^\infty_Y$-sheaves $$E^* \to \mathcal F \otimes C^\infty_Y \to 0.$$ This is necessary because on a non-projective complex manifold, a coherent sheaf does not necessarily have a holomorphic locally free resolution (cf. [3]). This defines Chern classes $c_k(\mathcal F) \in H^{2k}(Y, \mathbb Z)$, so equality \eqref{eq:1} is supposed to hold in the singular cohomology ring $H^*(Y,\mathbb Z)$.
- Deformation to the normal cone of $X$ should work exactly as in [1, Chapter 5], where one basically takes the blowup of $Y \times \mathbb P^1$ at $X \times \{\infty\}$.
- Basically follow the proof from Fulton's book [1, p.287-288], replacing $E$ by $E \otimes C^\infty_Y$ at the appropriate places.
Is that correct?
[1] Fulton, Intersection Theory.
[2] O'Brian, Toledo, Tong (1981) Grothendieck-Riemann-Roch for Complex Manifolds https://www.ams.org/journals/bull/1981-05-02/S0273-0979-1981-14939-9/S0273-0979-1981-14939-9.pdf
[3] Voisin (2004) Recent Progresses in Kähler and Complex Algebraic Geometry, https://www.cmls.polytechnique.fr/perso/voisin/Articlesweb/C07Voisi.pdf
