3.5 Properties of the intersection pairing related to fibers

Here $X$ is a smooth, projective surface. The goal is to study the intersection pairing on the components of the fiber of a morphism $f:X\to Y$ in the case when the fiber is a divisor. Curiously, we don’t really use the Hodge Index Theorem.

Lemma 3.33

Suppose $V$ is a $\mathbb {Q}$-vector space with a symmetric bilinear form $(x,y)\mapsto x\cdot y$, and suppose $\{ e_i\} _{i\in I}$ generated $V$ and with $e_i\cdot e_j\geq 0$ for all $i\neq j$. Suppose further that there exists a $z = \sum a_i e_i$ with $a_i {\gt} 0$ such that $z\cdot e_i = 0$ for all $i$.

Then $x\cdot x \leq 0$ for all $x\in V$ and $W := \{ x\in V: x\cdot x = 0\} $ is a liner subspace. Moreover, if the $\{ e_i\} _{i\in I}$ form a basis, then the dimension of $W$ is the number of connected components of the graph whose vertices are the elements of $I$ and whose edge $\{ i,j\} $ correspond to pairs $\{ i,j\} $ with $e_i\cdot e_j {\gt} 0$.

Proof ▼

We can replace $e_i$ with $a_i e_i$ and then assume that $z = \sum e_i$. Then pick $x\in V$. We have $x = \sum c_i e_i$ with $c_i\in \mathbb {Q}$. Then we have

\begin{align*} x\cdot x & = \sum _i c_i^2 e_i\cdot e_i + 2 \sum _{i{\lt}j} c_i c_j e_i\cdot e_j\\ & \leq \sum _i c_i^2 e_i\cdot e_i + \sum _{i{\lt}j} (c_i^2 + c_j^2) e_i\cdot e_j\\ & = \sum _{i,j} c_i^2 (e_i\cdot e_j) = \sum _i c_i^2 z\cdot e_i = 0. \end{align*}

Now, suppose the $e_i$ are linearly independent. Then the first inequality is strict if and only if $2 c_i c_j = c_i^2 + c_j^2$ for all $i\neq j$, and this holds if and only if $e_i\cdot e_j\neq 0 \Rightarrow c_i = c_j$. That implies the last two statements.

Corollary 3.34

Suppose $f: X\to C$ is a morphism from $X$ to a nonsingular curve $C$, and let $D = [f^{-1}(y)]$ be the fiber of $D$. Write $D = \sum a_i [D_i]$ where the $D_i$ are prime divisors and with the $a_i{\gt}0$. Then, for every divisor of the form $D' = \sum a_i D_i$ with $a_i\in \mathbb {Z}$, we have $(D')^2 \leq 0$. Moreover, if $f$ has connected fibers then $(D')^2 = 0$ if and only if there exists $a\in \mathbb {Q}$ with $D' = aD$.

Proof ▼

Since $C$ a curve, $f$ is actually flat. So $f^{!} = f^*$ and $f^*[y] = D$. If we pick another point $y^{\dagger }\in C$ and set $D^{\dagger } = f^{*}y^{\dagger }$, then, since $[y]$ and $[y^{\dagger }]$ are numerically equivalent, so are $D$ and $D^{\dagger }$. So, $[D][D_i] = [D^{\dagger }][D_i] = 0$ for all $i$. On the other hand, $D_i\cdot D_j \geq 0$ for all $i,j$ because the $D_i$ are distinct. The result then follows from the Lemma 3.33. If we pick another point $y^{\dagger }\in C$ and set $D^{\dagger } = f^{*}[y^{\dagger }]$, then, since $y$ and $y^{\dagger }$ are numerically equivalent, so are $D$ and $D^{\dagger }$. Therefore, that $D_i\cdot D = D_i\cdot D^{\dagger } = 0$ for all $i$.

So, in Lemma 3.33, take $V$ to be the free vector space generated by the divisors $[D_i]$ and take $z = D$.

Remark 3.35

If $f:X\to C$ is a morphism as in Corollary 3.34, then, by Stein factorization, we can factor $f$ as $X\xrightarrow {g} C' \xrightarrow {h} C$ where $g$ has connected fibers and $h$ is finite.

Corollary 3.36

Suppose $f:X\to Y$ is a birational morphism from a smooth surface $X$ to a normal surface $Y$. Let $y\in Y$ be a closed point and suppose that $[f^{-1} y] = \sum _{i=1}^r E_i$, where the $E_i$ are pairwise distinct prime divisors on $X$. Then the intersection matrix $E_i\cdot E_j$ is negative definite.

Proof ▼

Pick a very ample divisor $H$ on $X$. Using Bertini [ 4 , Theorem II.8.18 and III.7.9.1 ] , we can find an integral curve $C\in |H|$, which is distinct from all the divisors $E_i$. Let $\bar C = f(C)$. Then $\bar C$ is necessarily a curve in $Y$ containing the point $y$. Let $U$ be a Zariski open neighborhood of $y\in Y$. Then we can find a nonzero $g\in \Gamma (U,\mathcal{O}_Y)$ such that $\bar C\cap U = V(g)$. Regarding $g$ as an element of the function field $K(Y)$, we see that $\operatorname{\mathrm{div}}f^*(g) = \sum _{i=1}^r m_i E_i + D$ with $C\cap f^{-1} U = D\cap f^{-1} U$. Since $D|_{f^{-1}(U)}$ is effective and $E_i\subseteq f^{-1}(U)$, we get that $E_i\cdot D = E_i\cdot C = E_i\cdot H\neq 0$ for all $i$. But then

\begin{align} E_j \cdot (\sum m_i E_i + D) & = E_j\cdot \operatorname{\mathrm{div}}f^*(g) = 0\text{ for all $j$},\\ D\cdot (\sum m_i E_i + D) & = 0. \end{align}

Now, apply Lemma 3.33 to the free vector space on the $E_i$ and $D$, with $z = \sum m_i E_i + D$. We get the result.