The Stacks project

Proof. For every index $j$ we have $m_ j a_{jj} + \sum _{i \not= j} m_ ia_{ij} = 0$. Thus if we have an upper bound on $|a_{jj}|$ and $m_ j$, then we also get an upper bound on the nonzero (and hence positive) $a_{ij}$ as well as $m_ i$. Recalling that $w_ i$ divides $a_{ij}$, the reader easily sees the lemma is correct. $\square$

Proof. Recall that $g = 1 + \sum m_ j(w_ j(g_ j - 1) - \frac{1}{2} a_{jj})$. For $j \in J$ the contribution $m_ j(w_ j(g_ j - 1) - \frac{1}{2} a_{jj})$ to the genus $g$ is $> 0$ and hence $\geq 1/2$. This uses Lemma 55.3.7, Definition 55.3.8, Definition 55.3.12, Lemma 55.3.15, and Definition 55.3.16; we will use these results without further mention in the following. Thus $J$ has at most $2(g - 1)$ elements. This proves (1).

Recall that $-a_{ii} > 0$ for all $i$ by Lemma 55.3.6. Hence for $j \in J$ the contribution $m_ j(w_ j(g_ j - 1) - \frac{1}{2} a_{jj})$ to the genus $g$ is $> m_ jw_ j(g_ j - 1)$. Thus

This indeed implies $g_ j < g$ which proves (2).

For $j \in J$ if $g_ j > 0$, then the contribution $m_ j(w_ j(g_ j - 1) - \frac{1}{2} a_{jj})$ to the genus $g$ is $\geq -\frac{1}{2}m_ ja_{jj}$ and we immediately conclude that $m_ j|a_{jj}| \leq 2(g - 1)$. Otherwise $a_{jj} = -kw_ j$ for some integer $k \geq 3$ (because $j \in J$) and we get

Plugging this back into $a_{jj} = -km_ jw_ j$ we obtain

This proves (3).

Part (4) follows from Lemma 55.7.1 and (3). $\square$

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). $\square$

Proof. Say $T$ is given by $n, m_ i, a_{ij}, w_ i, g_ i$. If $T$ is not minimal, then there exists a $(-1)$-index. After replacing $T$ by an equivalent type we may assume $n$ is a $(-1)$-index. Applying Lemma 55.4.4 we find $\mathop{\mathrm{Pic}}\nolimits (T) \subset \mathop{\mathrm{Pic}}\nolimits (T')$ where $T'$ is a numerical type of genus $g$ (Lemma 55.3.9) with $n - 1$ indices. Thus we conclude by induction on $n$ provided we prove the lemma for minimal numerical types.

Assume that $T$ is a minimal numerical type of genus $\geq 2$. Observe that $g_{top} \leq g$ by Lemma 55.3.14. If $A = (a_{ij})$ then since $\mathop{\mathrm{Pic}}\nolimits (T) \subset \mathop{\mathrm{Coker}}(A)$ by Lemma 55.4.3. Thus it suffices to prove the lemma for $\mathop{\mathrm{Coker}}(A)$. By Lemma 55.7.3 we see that $m_ i|a_{ij}| \leq 768g$ for all $i, j$. Hence the result by Lemma 55.2.6. $\square$