Lemma 55.3.14. If $n, m_ i, a_{ij}, w_ i, g_ i$ is a minimal numerical type with $n > 1$, then $g \geq g_{top}$.

Proof. The reader who is only interested in the case of numerical types associated to proper regular models can skip this proof as we will reprove this in the geometric situation later. We can write

$g_{top} = 1 - n + \frac{1}{2}\sum \nolimits _{a_{ij} > 0} 1 = 1 + \sum \nolimits _ i (-1 + \frac{1}{2}\sum \nolimits _{j \not= i,\ a_{ij} > 0} 1)$

On the other hand, we have

\begin{align*} g & = 1 + \sum m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii}) \\ & = 1 + \sum m_ iw_ ig_ i - \sum m_ iw_ i + \frac{1}{2} \sum \nolimits _{i \not= j} a_{ij}m_ j \\ & = 1 + \sum \nolimits _ i m_ iw_ i(-1 + g_ i + \frac{1}{2} \sum \nolimits _{j \not= i} \frac{a_{ij}}{w_ i}) \end{align*}

The first equality is the definition, the second equality uses that $\sum a_{ij}m_ j = 0$, and the last equality uses that uses $a_{ij} = a_{ji}$ and switching order of summation. Comparing with the formula for $g_{top}$ we conclude that the lemma holds if

$\Psi _ i = m_ iw_ i(-1 + g_ i + \frac{1}{2} \sum \nolimits _{j \not= i} \frac{a_{ij}}{w_ i}) - (-1 + \frac{1}{2}\sum \nolimits _{j \not= i,\ a_{ij} > 0} 1)$

is $\geq 0$ for each $i$. However, this may not be the case. Let us analyze for which indices we can have $\Psi _ i < 0$. First, observe that

$(-1 + g_ i + \frac{1}{2}\sum \nolimits _{j \not= i} \frac{a_{ij}}{w_ i}) \geq (-1 + \frac{1}{2}\sum \nolimits _{j \not= i,\ a_{ij} > 0} 1)$

because $a_{ij}/w_ i$ is a nonnegative integer. Since $m_ iw_ i$ is a positive integer we conclude that $\Psi _ i \geq 0$ as soon as either $m_ iw_ i = 1$ or the left hand side of the inequality is $\geq 0$ which happens if $g_ i > 0$, or $a_{ij} > 0$ for at least two indices $j$, or if there is a $j$ with $a_{ij} > w_ i$. Thus

$P = \{ i : \Psi _ i < 0\}$

is the set of indices $i$ such that $m_ iw_ i > 1$, $g_ i = 0$, $a_{ij} > 0$ for a unique $j$, and $a_{ij} = w_ i$ for this $j$. Moreover

$i \in P \Rightarrow \Psi _ i = \frac{1}{2}(-m_ iw_ i + 1)$

The strategy of proof is to show that given $i \in P$ we can borrow a bit from $\Psi _ j$ where $j$ is the neighbour of $i$, i.e., $a_{ij} > 0$. However, this won't quite work because $j$ may be an index with $\Psi _ j = 0$.

Consider the set

$Z = \{ j : g_ j = 0\text{ and } j\text{ has exactly two neighbours }i, k\text{ with } a_{ij} = w_ j = a_{jk}\}$

For $j \in Z$ we have $\Psi _ j = 0$. We will consider sequences $M = (i, j_1, \ldots , j_ s)$ where $s \geq 0$, $i \in P$, $j_1, \ldots , j_ s \in Z$, and $a_{ij_1} > 0, a_{j_1j_2} > 0, \ldots , a_{j_{s - 1}j_ s} > 0$. If our numerical type consists of two indices which are in $P$ or more generally if our numerical type consists of two indices which are in $P$ and all other indices in $Z$, then $g_{top} = 0$ and we win by Lemma 55.3.13. We may and do discard these cases.

Let $M = (i, j_1, \ldots , j_ s)$ be a maximal sequence and let $k$ be the second neighbour of $j_ s$. (If $s = 0$, then $k$ is the unique neighbour of $i$.) By maximality $k \not\in Z$ and by what we just said $k \not\in P$. Observe that $w_ i = a_{ij_1} = w_{j_1} = a_{j_1j_2} = \ldots = w_{j_ s} = a_{j_ sk}$. Looking at the definition of a numerical type we see that

\begin{align*} m_ ia_{ii} + m_{j_1}w_ i & = 0,\\ m_ iw_ i + m_{j_1}a_{j_1j_1} + m_{j_2}w_ i & = 0,\\ \ldots & \ldots \\ m_{j_{s - 1}}w_ i + m_{j_ s}a_{j_ sj_ s} + m_ kw_ i & = 0 \end{align*}

The first equality implies $m_{j_1} \geq 2m_ i$ because the numerical type is minimal. Then the second equality implies $m_{j_2} \geq 3m_ i$, and so on. In any case, we conclude that $m_ k \geq 2m_ i$ (including when $s = 0$).

Let $k$ be an index such that we have a $t > 0$ and pairwise distinct maximal sequences $M_1, \ldots , M_ t$ as above, with $M_ b = (i_ b, j_{b, 1}, \ldots , j_{b, s_ b})$ such that $k$ is a neighbour of $j_{b, s_ b}$ for $b = 1, \ldots , t$. We will show that $\Phi _ j + \sum _{b = 1, \ldots , t} \Phi _{i_ b} \geq 0$. This will finish the proof of the lemma by what we said above. Let $M$ be the union of the indices occurring in $M_ b$, $b = 1, \ldots , t$. We write

$\Psi _ k = -\sum \nolimits _{b = 1, \ldots , t} \Psi _{i_ b} + \Psi _ k'$

where

\begin{align*} \Psi _ k' & = m_ kw_ k\left(-1 + g_ k + \frac{1}{2} \sum \nolimits _{b = 1, \ldots t} (\frac{a_{kj_{b, s_ b}}}{w_ k} - \frac{m_{i_ b}w_{i_ b}}{m_ kw_ k}) + \frac{1}{2} \sum \nolimits _{l \not= k,\ l \not\in M} \frac{a_{kl}}{w_ k} \right) \\ & -\left( -1 + \frac{1}{2}\sum \nolimits _{l \not= k,\ l \not\in M,\ a_{kl} > 0} 1 \right) \end{align*}

Assume $\Psi _ k' < 0$ to get a contradiction. If the set $\{ l : l \not= k,\ l \not\in M,\ a_{kl} > 0\}$ is empty, then $\{ 1, \ldots , n\} = M \cup \{ k\}$ and $g_{top} = 0$ because $e = n - 1$ in this case and the result holds by Lemma 55.3.13. Thus we may assume there is at least one such $l$ which contributes $(1/2)a_{kl}/w_ k \geq 1/2$ to the sum inside the first brackets. For each $b = 1, \ldots , t$ we have

$\frac{a_{kj_{b, s_ b}}}{w_ k} - \frac{m_{i_ b}w_{i_ b}}{m_ kw_ k} = \frac{w_{i_ b}}{w_ k}(1 - \frac{m_{i_ b}}{m_ k})$

This expression is $\geq \frac{1}{2}$ because $m_ k \geq 2m_{i_ b}$ by the previous paragraph and is $\geq 1$ if $w_ k < w_{i_ b}$. It follows that $\Psi _ k' < 0$ implies $g_ k = 0$. If $t \geq 2$ or $t = 1$ and $w_ k < w_{i_1}$, then $\Psi _ k' \geq 0$ (here we use the existence of an $l$ as shown above) which is a contradiction too. Thus $t = 1$ and $w_ k = w_{i_1}$. If there at least two nonzero terms in the sum over $l$ or if there is one such $k$ and $a_{kl} > w_ k$, then $\Psi _ k' \geq 0$ as well. The final possibility is that $t = 1$ and there is one $l$ with $a_{kl} = w_ k$. This is disallowed as this would mean $k \in Z$ contradicting the maximality of $M_1$. $\square$

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