Probably this is a very easy question. Let $f:X\rightarrow S$ be a resolution of a projective surface such that $$K_X = f^{*}K_S+\sum_ia_iE_i$$ with $a_i>0$. By Grauert-Mumford theorem the intersection matrix of the $E_i$'s is negative definite. I found the following claim: there exists an $E_j$ such that $$E_j\cdot(\sum_ia_iE_i)<0.$$ Why is this true?
1 Answer
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You can argue also in the following way (let us do the case of two componets $E_1$, $E_2$ for simplicity of notation. The general case will be clear):
the intersection matrix $$ I = \left(\begin{array}{cc} E_1^2 & E_1E_2 \\ E_1E_2 & E_2^2 \end{array}\right) $$ is negative definite. In particular if you take the vector $a=(a_1,a_2)$ you get $$a\cdot I\cdot a^{t} = a_1^2E_1^2+2a_1a_2E_1E_2+a_2^2E_2^2 <0.$$ On the other hand $$a_1^2E_1^2+2a_1a_2E_1E_2+a_2^2E_2^2 = a_1E_1(a_1E_1+a_2E_2)+a_2E_2(a_1E_1+a_2E_2)<0.$$ Since $a_1,a_2>0$ the last inequality yields either $E_1(a_1E_1+a_2E_2)<0$ or $E_2(a_1E_1+a_2E_2)<0$.
