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\title{Intersection Pairing}
\author{Kapil Hari Paranjape}
\begin{document}
\maketitle
We will study curves and line bundles on a smooth projective surface $S$
over a perfect field $k$.

\section{Negativity of Intersection Forms on Fibre Components}
\begin{enumerate}
    \item Suppose that $V$ is a $\bbQ$-vector space with a symmetric
        bi-linear form (indicated by ($\cdot$)). Given $e_i\in V$ a
        finite collection of elements so that $e_i\cdot e_j \geq 0$ for
        $i\neq j$. Suppose $z=\sum_i e_i$ is such that $z\cdot e_i\leq 0$.
    \item Putting $c=\sum_i c_i e_i$, we see that
        \begin{multline*}
            c\cdot c = \sum_i c_i^2 e_i\cdot e_i +
                2 \sum_{i<j} 2 c_i c_j e_i\cdot e_j \\
                \leq \sum_i c_i^2 e_i\cdot e_i +
                \sum_{i<j} (c_i^2 + c_j^2) e_i\cdot e_j \\
                = \sum_i c_i^2 e_i\cdot e_i +
                \sum_{i<j} c_i^2 e_i\cdot e_j \\
                \sum_{i>j} c_i^2 e_i\cdot e_j \\
                = \sum_i \sum_j c_i^2 e_i\cdot e_j =
                \sum_i c_i^2 z\cdot e_i \leq  0
        \end{multline*}
        It follows that $c\dot c\leq 0$ with equality only if $2c_i
        c_j=c_i^2 + c_j^2$ (equivalently, $c_i=c_j$) whenever $e_i\cdot
        e_j > 0$ \emph{and} $z\cdot e_i=0$ for all $i$ such that
        $c_i\neq 0$.
    \item Replacing $e_i$ by a positive multiple of $e_i$, we see that
        the same result holds with $z$ replaced by a positive linear
        combination of the $e_i$'s.
    \item We now apply this to the case where $e_i$ are the classes of
        components of the fibre over a (smooth) point $b$ of a morphism
        $f:S\to C$ where $S$ is a smooth surface and $C$ is a curve.
        Since the fibre divisor $z$ is a positive linear combination of
        an effective divisor supported on the point. On the other hand,
        any divisor supported on $b$ is linearly equivalent to a divisor
        whose support excludes $b$. Hence $z\cdot e_i=0$ and the above
        calculations apply to show that \emph{the intersection pairing on
        the subspace generated by the fibre components of a map from a
        smooth surface to a curve is negative semi-definite}.
    \item If we moreover assume that the fibre is connected then for all
        $i$ there is a $j$ so that $e_i\cdot e_j\neq 0$. In this case,
        the pairing is negative definite.
    \item Suppose that there is a dominant morphism $f:S\to S'$ for a
        surface $S'$ so that the image $f(E_i)=p$ of the curves $E_i$ in
        $S$ is (the same) point $p$ for all $i$. Let $D_i$ be curves in
        $S$ so that $E_i\cdot D_i>0$. Let $g$ be a regular function near
        $p$ on $S'$ that vanishes on the images of $D_i$. We write the
        divisor of zeros of $g$ on $S$ as $D''=\sum_i a_i E_i + D$ where $D$
        consists of all components other than $E_i$; we put $Z=\sum_i
        a_i E_i$. The divisor $D''$
        is linearly equivalent to the divisor $D'$ of poles of $g$ on
        $S$ and the image of $D'$ in $S'$ avoids $E_i$. It follows that
        $E_i\cdot D''=E_i\cdot D'=0$. Since $E_i\cdot D>0$, it follows
        that $Z\cdot E_i < 0$. The above linear algebra then implies
        that \emph{the intersection pairing on the subspace generated by
        the components of the fibres of a map from a smooth surface
        to another surfaces is negative definite}.
\end{enumerate}
\end{document}