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Let $D=\sum d_iD_i$ be an exceptional divisor on a smooth projective surface $X$. i.e., the intersection matrix $(D_i.D_j)$ is negative definite.

I have 2 stupid questions.

  1. Fix an ample divisor $A$ on $X$. Then why do there exist $r_i\in \mathbb{Q}$ s.t. $(\sum r_iD_i).D_j=-(A.D_j)$ for every $j$? (Why negative definiteness of $(D_i.D_j)$ implies this?)

  2. If $\sum r_iD_i$ is effective and $(\sum r_iD_i).D_j=0$ for all $j$, then $r_i>0$ for all $i$

Thanks.

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I don't know the answers, but are you familiar with the standard reference on the topic by M. Artin? mathoverflow.net/questions/123375/… –  roy smith May 18 '13 at 15:24
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1) is elementary linear algebra: the $r_i$ are the unknowns in a $mxm$ linear system (where $m$ is the number of the $D_i$) and the linear system has a unique solution because the matrix $(D_iD_j)$ has maximal rank. I don't understand 2): taking $r_i=0$ for every $i$ seems to give a contradiction to the claim. –  rita May 19 '13 at 15:14
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1 Answer

As Rita points out, 1) is trivial, because the matrix $(D_i\cdot D_j)$ is invertible. Also, as she points out, the second statement is not quite true. Well, actually, it is completely false. Again by the fact that the matrix $(D_i\cdot D_j)$ is invertible, if $(\sum r_iD_i)\cdot D_j=0$ for all $j$, then necessarily $r_i=0$ for all $i$. On the other hand, this is easily fixable.

In fact, one can make a somewhat stronger statement:

Claim If the $r_i$ are arbitrary and $(\sum r_iD_i)\cdot D_j\leq 0$ for all $j$, then $r_i\geq 0$, that is, $\sum r_iD_i$ is effective and if furthermore $\cup D_i$ is connected and there exists a $j$ such that $(\sum r_iD_i)\cdot D_j\neq 0$, then $r_i>0$ for all $i$.

Proof Let $\sum r_iD_i=A-B$ with $A,B$ effective and suppose $B\neq 0$. Then $B^2<0 \leq A\cdot B$, so $$ 0 < A\cdot B - B^2= \left(\sum r_iD_i\right)\cdot B = \sum_{r_j<0} (-r_j)\left(\sum r_iD_i\right)\cdot D_j \leq 0, $$ which is a contradiction and hence $B=0$, that is, $\sum r_iD_i$ is effective.

Next we want to prove that if there exists a $k$ such that $r_k=0$, then $(\sum r_iD_i)\cdot D_j=0$ for all $j$. Let $k$ be such that $r_k=0$. Then $$ \left(\sum r_iD_i\right)\cdot D_k = \sum_{i: D_i\cdot D_k>0} r_i (D_i\cdot D_k) \geq 0. $$

Now if there exists a $j$ among these such that $r_j\neq 0$, then $\left(\sum r_iD_i\right)\cdot D_k>0$, otherwise $r_j=0$ for all $j$ such that $D_j\cap D_k\neq \emptyset$. Now repeat this step with one of these $j$'s. Since $\cup D_i$ is connected, this proves the claim. $\square$

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