Lemma 55.7.2. Fix $g \geq 2$. For every minimal numerical type $n, m_ i, a_{ij}, w_ i, g_ i$ of genus $g$ with $n > 1$ we have
the set $J \subset \{ 1, \ldots , n\} $ of non-$(-2)$-indices has at most $2g - 2$ elements,
for $j \in J$ we have $g_ j < g$,
for $j \in J$ we have $m_ j|a_{jj}| \leq 6g - 6$, and
for $j \in J$ and $i \in \{ 1, \ldots , n\} $ we have $m_ ia_{ij} \leq 6g - 6$.
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
\[ g - 1 > m_ jw_ j(g_ j - 1) \Rightarrow g_ j < (g - 1)/m_ jw_ j + 1 \]
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
\[ m_ jw_ j(-1 + \frac{k}{2}) \leq g - 1 \Rightarrow m_ jw_ j \leq \frac{2(g - 1)}{k - 2} \]
Plugging this back into $a_{jj} = -km_ jw_ j$ we obtain
\[ m_ j|a_{jj}| \leq 2(g - 1) \frac{k}{k - 2} \leq 6(g - 1) \]
This proves (3).
Part (4) follows from Lemma 55.7.1 and (3).
$\square$
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