# Dualizing sheaf under birational morphism

Let $$f:X\to Y$$ be a birational morphism of irreducible projective varieties over $$\mathbb{C}$$ (so $$f$$ is defined everywhere but invertible only on a open dense subset). If $$X$$ and $$Y$$ are smooth, then there is a natural isomorphism between the vector spaces $$\Gamma(X,\omega_X)$$ and $$\Gamma(Y,\omega_Y)$$ where $$\omega_X$$ and $$\omega_Y$$ are the dualizing sheaves of $$X$$ and $$Y$$ (as for example shown in Hartshorne's book). Now assume that only $$X$$ is smooth and $$Y$$ is not necessarily smooth (but say the singular locus has codimension at least two if necessary). Do we still have such a natural isomorphism?

This should not be true in general. Note that your condition on $$f:X\to Y$$ implies that it's actually a proper birational map from a smooth variety and thus a resolution of singularities of $$Y$$. If $$Y$$ is Cohen-Macaulay and has rational singularities, one has that $$f_*\Omega_Y \to \Omega_X$$ must be isomorphism of sheaves (actually, of dualizing complexes) which would imply the result you ask for on global sections. But there are certainly non-rational singularities.

• Thanks! So just to be sure: $Y$ having rational singularities would be a sufficient condition for my statement being true but in general there might be counter-examples, right? Is there a good source where I can read about details?
– Hans
Dec 13, 2019 at 11:14
• One slight adjustment - I should have required that $Y$ be Cohen-Macaulay as well. But yes, $Y$ Cohen-Macaulay and rational singularities implies the statement. The original proof of the statement in this post can be found in R. Elkik, Singularites rationnelles et deformations, Inventiones math. 47, (1978) if you already know that Cohen-Macaulay means that the dualizing complex is quasi-isomorphic to a sheaf placed in the correct degree (I'm sure this is covered in StacksProject, for instance). If you have trouble finding this, let me know and I can add the details directly to the post. Dec 13, 2019 at 20:31