Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
add comment

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.

share|improve this answer
    
Thank you, Francesco. But I don't know how to use Chern classes and Picard groups. –  emmy Jun 19 '12 at 15:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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