# Vanishing associated to a resolution of singularities

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

• Do you have a particular $F$ in mind? Commented Jan 26, 2012 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? Commented Jan 26, 2012 at 2:44

## 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 $E'|_Z\neq 0$. $\square$

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

• 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 $R^i\pi_*O_E= 0 i>0$ Commented Jan 26, 2012 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? Commented Jan 26, 2012 at 21:05
• Thank you, and yes, the reduced exceptional is what I have in mind. Commented Jan 26, 2012 at 21:51