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Let $\pi: V\to W$ be a resolution of singularities, let $E \subset V$ be the exceptional divisor, and let $F$ be a coherent sheaf such that $R^i\pi_*F=0$ for $i>0$.

Can we conclude that $R^i\pi_*F(-E)=0$ for $i>0$?

The idea being to somehow use that$-E$ is nef on $E$.

EDIT: the exceptional fibres of $\pi$ are one-dimensional and $R^{\bullet} \pi_*O_V = O_W$

EDIT 2: we may assume that the support of $F$ is one-dimensional

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Do you have a particular $F$ in mind? –  Karl Schwede Jan 26 '12 at 2:35
    
In particular, if you have the vanishing of the $R^i \pi_* F$ for some Kodaira-type vanishing reason, you may be ok. By the way, is $E$ the reduced exceptional divisor, or something else? –  Karl Schwede Jan 26 '12 at 2:44

2 Answers 2

up vote 7 down vote accepted

1

The statement is still false with the most recently added conditions. The problem is that $-E$ is not necessarily ($\pi$-)nef. I will discuss below what one can do to fix the statement so it will hold. First, here is an example:

Let $W$ be a two dimensional rational double point that's either a $D_n$ or an $E_n$ singularity. The point is to have an exceptional curve in the resolution that intersects at least 3 others.

Let $\pi:V\to W$ be the minimal resolution of $W$ and $E_0,E_1,E_2,E_3$ four exceptional curves with intersection matrix $$\left(\begin{matrix} -2 & 1 & 1 & 1 \\ 1 & -2 & 0 & 0 \\ 1 & 0 & -2 & 0 \\ 1 & 0 & 0 & -2 \end{matrix}\right),$$

and let $E$ be the reduced exceptional divisor. It follows that then $$ E_0\cdot E=E_0\cdot (E_0+E_1+E_2+E_3)=1. $$ In particular, $-E$ is not nef on $E_0$.

Now, let $F=\mathscr O_{E_0}(-1)$. Since $E_0\simeq \mathbb P^1$, it follows that $R^i\pi_*F=0$ for all $i\geq 0$ (even for $i=0$!), but since $F(-E)\simeq \mathscr O_{E_0}(-2)$, $R^1\pi_*F\neq 0$.

2

Here is a statement in the spirit of the question which is true (notice that some assumptions are actually weakened, in particular, it is not needed that the singularities of $W$ be rational):

Claim Let $\pi:V\to W$ be a projective birational morphism over a field of characteristic $0$ such that every fiber of $\pi$ has dimension at most 1 and let $E$ denote the exceptional divisor of $\pi$. Let $\mathscr F$ be a coherent sheaf on $V$ such that $Z=\mathrm{supp}\mathscr F$ is 1-dimensional. Assume that $W$ has $\mathbb Q$-factorial singularities and that $R^1\pi_*\mathscr F=0$. Then there exists an effective exceptional divisor $E'$ such that $R^i\pi_*\mathscr F(-aE')=0$ for all $a>0$ and $i>0$ ($a,i\in \mathbb N$). Furthermore, if $V$ and $W$ are normal, then $\mathrm{supp}E'=\mathrm{supp}E$ in a neighbourhood of $Z$.

Proof First off, notice that since the maximal fiber dimension of $\pi$ is 1, all $R^i\pi_*=0$ for $i>1$. Furthermore, since the claim is local on $W$, we may assume that it is affine.

Next, we claim that one may assume that $Z$ is contained in a fiber of $\pi$. Indeed let $Z=Z_1\cup Z_2$ where $Z_1$ is contained in a union of fibers and $Z_2$ does not have any components contained in a fiber (i.e., $\pi$ is finite on $Z_2$). Since $W$ is affine, it follows that so is $Z_2$. Now let $\mathscr G$ be a coherent sheaf (we will apply this with $\mathscr G=\mathscr F$ and $\mathscr G=\mathscr F(-E)$). Then we have a short exact sequence $$ 0\to \mathscr G\to \mathscr G|_{Z_1}\oplus \mathscr G|_{Z_2}\to \mathscr G|_{Z_1\cap Z_2}\to 0. $$ Since $Z_2$ is affine, it follows that $\pi_*\mathscr G|_{Z_2}\to \pi_*\mathscr G|_{Z_1\cap Z_2}$ is surjective and $R^i\pi_*\left(\mathscr G|_{Z_2}\right)=0$ and $R^i\pi_*\left(\mathscr G|_{Z_1\cap Z_2}\right)=0$ for $i>0$, hence $R^i\pi_*\mathscr G\simeq R^i\pi_*\left(\mathscr G|_{Z_1}\right)$ for $i>0$. This implies that we may assume that $Z=Z_1$ and then, since the statement is local we may assume that $Z$ is mapped to a single point, i.e., it is contained in a fiber.

Finally let $L$ be an arbitrary effective divisor on $V$ such that $\mathrm{supp}L\cap Z\neq\emptyset$ but $Z\not\subseteq \mathrm{supp}L$. (As $\pi$ is projective, such an $L$ exists). Let $H=\pi_*L$ and notice that since $W$ is $\mathbb Q$-factorial, $H$ is $\mathbb Q$-Cartier. Replacing $L$ with an appropriate multiple, we may assume that $H$ is Cartier. Compare $\pi^*H$ and $\widetilde H$ (the strict transform of $H$) and observe that

  1. $\widetilde H=L$ by construction and hence $\widetilde H|_Z\geq 0$ and $\widetilde H|_Z\neq 0$,
  2. $\pi^*H\sim \widetilde H+\sum_i m_iE_i$, where $E_i$ are the irreducible components of the exceptional divisor and $m_i\geq 0$,
  3. $\pi^*H|_Z=0$ since $Z$ is contained in a fiber.

Therefore it follows that $(-\sum m_iE_i)|_Z\geq 0$ and $(-\sum m_iE_i)|_Z\neq 0$. Let $d$ denote the largest common divisor of $m_i$ and let $E'=\sum\frac{m_i}dE_i$. Then we still have that $(-aE')|_Z\geq 0$ and $(-aE')|_Z\neq 0$ for any $a>0$. It follows that then $\mathscr F\subseteq \mathscr F(-aE')$ so we have a short exact sequence $$ 0 \to \mathscr F\to \mathscr F(-aE')\to \mathscr Q\to 0 $$ where $\mathscr Q$ is supported at finitely many points. Then $R^i\pi_*\mathscr Q=0$ for $i>0$ and hence it follows that $0=R^i\pi_*\mathscr F\to R^i\pi_*\mathscr F(-aE')$ is surjective for all $i>0$ and the desired statement follows.

The only thing left to prove is that if $V$ and $W$ are normal, then $\mathrm{supp}E'=\mathrm{supp}E$ in a neighbourhood of $Z$. The above proof actually shows that $-E'$ is $\pi$-nef and then this statement follows from 3.39 of Birational Geometry of Algebraic Varieties by János Kollár and Shigefumi Mori and the choices that ensured that $E'|_Z\neq 0$. $\square$

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Ok, so here's an example when this fails under the assumption that $E$ is the reduced exceptional divisor.

Suppose that $W$ is a smooth variety (one can also do variations with rational singularities) and $V \to W$ is a log resolution of a normal (or even seminormal) subvariety $Z \subseteq W$ which has NON-Du Bois singularities. Further assume that $\pi$ is an isomorphism outside of $Z$ and the the log resolution bit guarantees that $E = \pi^{-1}(Z)_{\text{red}}$ is simple normal crossings.

Set $F = O_V$. Since $W$ is smooth, it has rational singularities and so $R^i \pi_* O_V = 0$ for $i > 0$. Since $Z$ is NOT Du Bois, $R \pi_* O_E \neq O_Z$ in the derived category. Since $Z$ is normal (or at least seminormal), $\pi_* O_E \cong O_Z$. Thus, $R^i \pi_* O_E \neq 0$ for some $i > 0$.

Since we have a long exact sequence

$$0 = R^i \pi_* O_V \to R^i \pi_* O_E \to R^{i+1} \pi_* O_V(-E) \to R^{i+1} \pi_* O_V = 0.$$

It follows that $R^{i+1} \pi_* O_V(-E) \neq 0$.

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I should have mentioned before: the exceptional fibres are one-dimensional and $\pi$ has satisfies $R^{\bullet}\pi_*O_V=O_W$. I think think implies that $R^i\pi_*O_E= 0 i>0$ –  David Steinberg Jan 26 '12 at 18:48
    
I'll have to think about the revised question then, I'm not sure off the top of my head. You really do want the reduced exceptional divisor though? –  Karl Schwede Jan 26 '12 at 21:05
    
Thank you, and yes, the reduced exceptional is what I have in mind. –  David Steinberg Jan 26 '12 at 21:51

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