The Stacks project

Proof. By Lemma 55.7.1 it suffices to show $m_ i|a_{ii}| \leq 768g$ for all $i$. Let $J \subset \{ 1, \ldots , n\} $ be the set of non-$(-2)$-indices as in Lemma 55.7.2. Observe that $J$ is nonempty as $g \geq 2$. Also $m_ j|a_{jj}| \leq 6g$ for $j \in J$ by the lemma.

Suppose we have $j \in J$ and a sequence $i_1, \ldots , i_7$ of $(-2)$-indices such that $a_{ji_1}$ and $a_{i_1i_2}$, $a_{i_2i_3}$, $a_{i_3i_4}$, $a_{i_4i_5}$, $a_{i_5i_6}$, and $a_{i_6i_7}$ are nonzero. Then we see from Lemma 55.7.1 that $m_{i_1}w_{i_1} \leq 6g$ and $m_{i_1}a_{ji_1} \leq 6g$. Because $i_1$ is a $(-2)$-index, we have $a_{i_1i_1} = -2w_{i_1}$ and we conclude that $m_{i_1}|a_{i_1i_1}| \leq 12g$. Repeating the argument we conclude that $m_{i_2}w_{i_2} \leq 12g$ and $m_{i_2}a_{i_1i_2} \leq 12g$. Then $m_{i_2}|a_{i_2i_2}| \leq 24g$ and so on. Eventually we conclude that $m_{i_ k}|a_{i_ ki_ k}| \leq 2^ k(6g) \leq 768g$ for $k = 1, \ldots , 7$.

Let $I \subset \{ 1, \ldots , n\} \setminus J$ be a maximal connected subset. In other words, there does not exist a nonempty proper subset $I' \subset I$ such that $a_{i'i} = 0$ for $i' \in I'$ and $i \in I \setminus I'$ and $I$ is maximal with this property. In particular, since a numerical type is connected by definition, we see that there exists a $j \in J$ and $i \in I$ with $a_{ij} > 0$. Looking at the classification of such $I$ in Proposition 55.5.17 and using the result of the previous paragraph, we see that $w_ i|a_{ii}| \leq 768g$ for all $i \in I$ unless $I$ is as described in Lemma 55.5.8 or Lemma 55.5.9. Thus we may assume the nonvanishing of $a_{ii'}$, $i, i' \in I$ has either the shape

(which has 3 subcases as detailed in Lemma 55.5.8) or the shape

We will prove the bound holds for the first subcase of Lemma 55.5.8 and leave the other cases to reader (the argument is almost exactly the same in those cases).

After renumbering we may assume $I = \{ 1, \ldots , t\} \subset \{ 1, \ldots , n\} $ and there is an integer $w$ such that

The equalities $a_{ii}m_ i + \sum _{j \not= i} a_{ij}m_ j = 0$ imply that we have

Equality holds in $2m_ i \geq m_{i - 1} + m_{i + 1}$ if and only if $i$ does not “meet” any indices besides $i - 1$ and $i + 1$. And if $i$ does meet another index, then this index is in $J$ (by maximality of $I$). In particular, the map $\{ 1, \ldots , t\} \to \mathbf{Z}$, $i \mapsto m_ i$ is concave.

Let $m = \max (m_ i, i \in \{ 1, \ldots , t\} )$. Then $m_ i|a_{ii}| \leq 2mw$ for $i \leq t$ and our goal is to show that $2mw \leq 768g$. Let $s$, resp. $s'$ in $\{ 1, \ldots , t\} $ be the smallest, resp. biggest index with $m_ s = m = m_{s'}$. By concavity we see that $m_ i = m$ for $s \leq i \leq s'$. If $s > 1$, then we do not have equality in $2m_ s \geq m_{s - 1} + m_{s + 1}$ and we see that $s$ meets an index from $J$. In this case $2mw \leq 12g$ by the result of the second paragraph of the proof. Similarly, if $s' < t$, then $s'$ meets an index from $J$ and we get $2mw \leq 12g$ as well. But if $s = 1$ and $s' = t$, then we conclude that $a_{ij} = 0$ for all $j \in J$ and $i \in \{ 2, \ldots , t - 1\} $. But as we've seen that there must be a pair $(i, j) \in I \times J$ with $a_{ij} > 0$, we conclude that this happens either with $i = 1$ or with $i = t$ and we conclude $2mw \leq 12g$ in the same manner as before (as $m_1 = m = m_ t$ in this case). $\square$