## Line bundles and rational singularities

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Hi, I have some problem to understand the proof of lemma 3.2 of this article: http://www.ams.org/journals/jams/2001-14-03/S0894-0347-01-00368-X/.

The lemma states the following: Let `$X$` be a variety and `$f: Y \rightarrow X$` a resolution of singularities. Assume that `$X$` has rational singularities. Then a line bundle `$L$` on `$Y$` is the pullback `$f^*M$` of some linebundle `$M$` on `$X$` if and only if the restriction of $L$ to each formal fibre of $f$ is trivial. Moreover, when this holds, `$M=f_*L$`.

For the proof of the "if" part, suppose that the restriction of `$L$` to each formal fibre is trivial. The teorem on formal functions shows that the completions of the stalks of the sheaves `$R^if_* \mathcal{O}_Y$` and `$R^if_*L$` at any point `$x \in X$` are isomorphic for each $i$. Since $X$ has rational singularities, `$R^if_*L=0$` for all `$i>0$` and `$M=f_*L$` is a linebundle on $X$.

Since `$f^*M$` is torsion free, the natural adjunction map `$\eta: f^*f_*L \rightarrow L$` is injective, so there is a short exat sequence `$$ 0 \rightarrow f^*f_*L \stackrel{\eta}{\rightarrow} L \rightarrow Q \rightarrow 0.$$` By tha projection formula and the fact that $X$ has rational singularities, `$R^if_*(f^*M)=M \otimes R^if_* \mathcal{O}_Y=0$` for all `$i>0$`. The fact that `$\eta$` is the unit of adjunction implies that `$f_* \eta$` has a left inverse, and in particular is surjective. Applying `$f_*$` to the exact triple we conclude that `$f_*Q=0$`, and, by the theorem on formal functions `$f_*(Q \otimes L^{-1})=0$`, in particular `$Q \otimes L^{-1}$` has no nonzero global sections. Tensoring the exact triple with `$L^{-1}$` gives a contradiction, unless $Q=0$. Hence $\eta$ is an isomorphism and we are done.

I did not understand this last step. Tensoring the short exact sequence with `$L^{-1}$` and then taking global sections, we get `$ \Gamma(Y, f^*M \otimes L^{-1}) \cong \Gamma(Y, \mathcal{O}_Y)$`, because the last term is zero. How can I deduce from this that `$f^*M \otimes L^{-1} \simeq \mathcal{O}_Y$` and then `$f^*M \cong L$`? Where is the contradiction? Thank you

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## 2 Answers

For any line bundle $F$, giving a map of sheaves $\mathcal O_X \to F$ is equivalent to giving a global section of $F$.

In your case, take $F=Q \otimes L^{-1}$ with the map given by your exact sequence tensorized by $L ^{-1}$. As $F$ has no non-zero global section, the aforesaid map is trivial, so $\eta \otimes Id_{L^{-1}}$ is an isomorphism, and therefore so is $\eta$ too.

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Since $X$ has rational singularities, it is normal. Then the singular locus of $X$ has codimension at most $2$ and $f \colon X \to Y$ is an isomorphism in codimension $1$.

This imples that the support of $Q$ has codimension at least $2$, hence $c_1(Q)=0$.

Therefore your exact sequences gives `$c_1(f^*M)=c_1(L)$`, that is `$c_1(f^*M \otimes L^{-1})=0$`, that is $f^*M \otimes L^{-1} \in \textrm{Pic}^0(Y)$.

Now $H^0(Y, f^*M \otimes L^{-1})=H^0(Y, \mathcal{O}_Y)=\mathbb{C}$ implies that $f^*M \otimes L^{-1}$ is a line bundle of degree $0$ with a non-zero global section, hence it must be isomorphic to $\mathcal{O}_Y$.

This implies $Q=0$: in fact, we obtain $$\textrm{Hom}(f^*M, L)=H^0(Y, f^*M^{-1} \otimes L)=H^0(Y, \mathcal{O}_Y)=\mathbb{C},$$ so $\eta\colon f^*M \longrightarrow L$ is necessarily an isomorphism.

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Thank you, Francesco. But I don't know how to use Chern classes and Picard groups. – emmy Jun 19 at 15:35
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