I was reading the paper A Characterization of Rational Singularities by Professor Kovács.

The main theorem is stated as follows:

THEOREM 1. Let $\phi: Y \rightarrow X$ be a morphism of varieties over $\mathbb{C}$, and let $\rho:\mathcal{O}_X \to R \phi_* \mathcal{O}_Y$ be the associated natural morphism. Assume that $Y$ has rational singularities and there exists a morphism (in the derived category of $\mathcal{O}_X$-modules) $\rho^{\prime}: R \phi_* \mathcal{O}_Y \rightarrow \mathcal{O}_X$ such that $\rho^{\prime} \circ \rho$ is a quasi-isomorphism of $\mathcal{O}_X$ with itself. Then $X$ has only rational singularities.

The proof goes as follows, first, we take a resolution on both $X$ and $Y$ (denote it $\pi:\tilde{X}\to X$ and $\sigma:\tilde{Y}\to Y$) which makes the diagram commute (i.e. there exist a morphism $\psi: \tilde{Y}\to \tilde{X}$ with $\phi\circ \sigma = \pi \circ \psi$) thus it will induce the following diagram in the derived category:

$$\require{AMScd} \begin{CD} \mathcal{O}_X @>{\rho}>> R\phi_*\mathcal{O}_Y;\\ @VVV @VVV \\ R\pi_*\mathcal{O}_{\tilde{X}} @>{\gamma}>> R\phi_*R\sigma_*\mathcal{O}_{\tilde{Y}}; \end{CD}$$

where $\alpha:\mathcal{O}_X\to R\pi_*\mathcal{O}_{\tilde{X}}$ and $\beta:R\phi_*\mathcal{O}_Y\to R\phi_*R\sigma_*\mathcal{O}_{\tilde{Y}}$, by the assumption of the main theorem, we have $\left(\rho^{\prime} \circ \beta^{-1} \circ \gamma\right) \circ \alpha$ is a quasi-isomorphism of $\mathcal{O}_X$ with itself, then the paper state that thus we may assume $\phi$ is the resolution of singularity.

And finally we can apply the standard result to prove $X$ is Cohen-Macaulay and the direct image of canonical sheaf is canonical sheaf.(using some vanishing theorem and quasi isomorphism conditions)

The question is why the quasi isomorphism $\left(\rho^{\prime} \circ \beta^{-1} \circ \gamma\right) \circ \alpha$ implies $\phi$ be a resolution?

By the definition of resolution, $Y$ must be nonsingular, the assumption in the theorem assumes only $Y$ being rational. The quasi-isomorphism condition only implies something on the cohomology level, why does it imply smoothness?


1 Answer 1


I think he means that instead of the data $(\phi \colon Y \to X, \rho)$ you can consider the data $(\pi \colon \tilde{X} \to X, \rho' \circ \beta^{-1} \circ \gamma)$. The new data satisfy the same assumptions as the old data do, but this time $\pi$ is a resolution.


Your Answer

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

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