## 55.3 Numerical types

Part of the arguments will involve the combinatorics of the following data structures.

Definition 55.3.1. A numerical type $T$ is given by

$n, m_ i, a_{ij}, w_ i, g_ i$

where $n \geq 1$ is an integer and $m_ i$, $a_{ij}$, $w_ i$, $g_ i$ are integers for $1 \leq i, j \leq n$ subject to the following conditions

1. $m_ i > 0$, $w_ i > 0$, $g_ i \geq 0$,

2. the matrix $A = (a_{ij})$ is symmetric and $a_{ij} \geq 0$ for $i \not= j$,

3. there is no proper nonempty subset $I \subset \{ 1, \ldots , n\}$ such that $a_{ij} = 0$ for $i \in I$, $j \not\in I$,

4. for each $i$ we have $\sum _ j a_{ij}m_ j = 0$, and

5. $w_ i | a_{ij}$.

This is obviously a somewhat annoying type of structure to work with, but it is exactly what shows up in special fibres of proper regular models of smooth geometrically connected curves. Of course we only care about these types up to reordering the indices.

Definition 55.3.2. We say two numerical types $n, m_ i, a_{ij}, w_ i, g_ i$ and $n', m'_ i, a'_{ij}, w'_ i, g'_ i$ are equivalent types if there exists a permutation $\sigma$ of $\{ 1, \ldots , n\}$ such that $m_ i = m'_{\sigma (i)}$, $a_{ij} = a'_{\sigma (i)\sigma (j)}$, $w_ i = w'_{\sigma (i)}$, and $g_ i = g'_{\sigma (i)}$.

A numerical type has a genus.

Lemma 55.3.3. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type. Then the expression

$g = 1 + \sum m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii})$

is an integer.

Proof. To prove $g$ is an integer we have to show that $\sum a_{ii}m_ i$ is even. This we can see by computing modulo $2$ as follows

\begin{align*} \sum \nolimits _ i a_{ii} m_ i & \equiv \sum \nolimits _{i,\ m_ i\text{ odd}} a_{ii}m_ i \\ & \equiv \sum \nolimits _{i,\ m_ i\text{ odd}} \sum \nolimits _{j \not= i} a_{ij}m_ j \\ & \equiv \sum \nolimits _{i,\ m_ i\text{ odd}} \sum \nolimits _{j \not= i,\ m_ j\text{ odd}} a_{ij}m_ j \\ & \equiv \sum \nolimits _{i < j,\ m_ i\text{ and }m_ j\text{ odd}} a_{ij}(m_ i + m_ j) \\ & \equiv 0 \end{align*}

where we have used that $a_{ij} = a_{ji}$ and that $\sum _ j a_{ij}m_ j = 0$ for all $i$. $\square$

Definition 55.3.4. We say $n, m_ i, a_{ij}, w_ i, g_ i$ is a numerical type of genus $g$ if $g = 1 + \sum m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii})$ is the integer from Lemma 55.3.3.

We will prove below (Lemma 55.3.14) that the genus is almost always $\geq 0$. But you can have numerical types with negative genus.

Lemma 55.3.5. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type of genus $g$. If $n = 1$, then $a_{11} = 0$ and $g = 1 + m_1w_1(g_1 - 1)$. Moreover, we can classify all such numerical types as follows

1. If $g < 0$, then $g_1 = 0$ and there are finitely many possible numerical types of genus $g$ with $n = 1$ corresponding to factorizations $m_1w_1 = 1 - g$.

2. If $g = 0$, then $m_1 = 1$, $w_1 = 1$, $g_1 = 0$ as in Lemma 55.6.1.

3. If $g = 1$, then we conclude $g_1 = 1$ but $m_1, w_1$ can be arbitrary positive integers; this is case (1) of Lemma 55.6.2.

4. If $g > 1$, then $g_1 > 1$ and there are finitely many possible numerical types of genus $g$ with $n = 1$ corresponding to factorizations $m_1w_1(g_1 - 1) = g - 1$.

Proof. The lemma proves itself. $\square$

Lemma 55.3.6. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type of genus $g$. If $n > 1$, then $a_{ii} < 0$ for all $i$.

Proof. Lemma 55.2.3 applies to the matrix $A$. $\square$

Lemma 55.3.7. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type of genus $g$. Assume $n > 1$. If $i$ is such that the contribution $m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii})$ to the genus $g$ is $< 0$, then $g_ i = 0$ and $a_{ii} = -w_ i$.

Proof. Follows immediately from Lemma 55.3.6 and $w_ i > 0$, $g_ i \geq 0$, and $w_ i | a_{ii}$. $\square$

Definition 55.3.8. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type. We say $i$ is a $(-1)$-index if $g_ i = 0$ and $a_{ii} = -w_ i$.

We can “contract” $(-1)$-indices.

Lemma 55.3.9. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. Assume $n$ is a $(-1)$-index. Then there is a numerical type $T'$ given by $n', m'_ i, a'_{ij}, w'_ i, g'_ i$ with

1. $n' = n - 1$,

2. $m'_ i = m_ i$,

3. $a'_{ij} = a_{ij} - a_{in}a_{jn}/a_{nn}$,

4. $w'_ i = w_ i/2$ if $a_{in}/w_ n$ even and $a_{in}/w_ i$ odd and $w'_ i = w_ i$ else,

5. $g'_ i = \frac{w_ i}{w'_ i}(g_ i - 1) + 1 + \frac{a_{in}^2 - w_ na_{in}}{2w'_ iw_ n}$.

Moreover, we have $g = g'$.

Proof. Observe that $n > 1$ for example by Lemma 55.3.5 and hence $n' \geq 1$. We check conditions (1) – (5) of Definition 55.3.1 for $n', m'_ i, a'_{ij}, w'_ i, g'_ i$.

Condition (1) is immediate.

Condition (2). Symmetry of $A' = (a'_{ij})$ is immediate and since $a_{nn} < 0$ by Lemma 55.3.6 we see that $a'_{ij} \geq a_{ij} \geq 0$ if $i \not= j$.

Condition (3). Suppose that $I \subset \{ 1, \ldots , n - 1\}$ such that $a'_{ii'} = 0$ for $i \in I$ and $i' \in \{ 1, \ldots , n - 1\} \setminus I$. Then we see that for each $i \in I$ and $i' \in I'$ we have $a_{in}a_{i'n} = 0$. Thus either $a_{in} = 0$ for all $i \in I$ and $I \subset \{ 1, \ldots , n\}$ is a contradiction for property (3) for $T$, or $a_{i'n} = 0$ for all $i' \in \{ 1, \ldots , n - 1\} \setminus I$ and $I \cup \{ n\} \subset \{ 1, \ldots , n\}$ is a contradiction for property (3) of $T$. Hence (3) holds for $T'$.

Condition (4). We compute

$\sum \nolimits _{j = 1}^{n - 1} a'_{ij}m_ j = \sum \nolimits _{j = 1}^{n - 1} (a_{ij}m_ j - \frac{a_{in}a_{jn}m_ j}{a_{nn}}) = - a_{in}m_ n - \frac{a_{in}}{a_{nn}}(-a_{nn}m_ n) = 0$

as desired.

Condition (5). We have to show that $w'_ i$ divides $a_{in}a_{jn}/a_{nn}$. This is clear because $a_{nn} = -w_ n$ and $w_ n | a_{jn}$ and $w_ i | a_{in}$.

To show that $g = g'$ we first write

\begin{align*} g & = 1 + \sum \nolimits _{i = 1}^ n m_ i(w_ i(g_ i - 1) - \frac{1}{2}a_{ii}) \\ & = 1 + \sum \nolimits _{i = 1}^{n - 1} m_ i(w_ i(g_ i - 1) - \frac{1}{2}a_{ii}) -\frac{1}{2}m_ nw_ n \\ & = 1 + \sum \nolimits _{i = 1}^{n - 1} m_ i(w_ i(g_ i - 1) - \frac{1}{2}a_{ii} - \frac{1}{2}a_{in}) \end{align*}

Comparing with the expression for $g'$ we see that it suffices if

$w'_ i(g'_ i - 1) - \frac{1}{2}a'_{ii} = w_ i(g_ i - 1) - \frac{1}{2}a_{in} - \frac{1}{2}a_{ii}$

for $i \leq n - 1$. In other words, we have

$g'_ i = \frac{2w_ i(g_ i - 1) - a_{in} - a_{ii} + a'_{ii} + 2w'_ i}{2w'_ i} = \frac{w_ i}{w'_ i}(g_ i - 1) + 1 + \frac{a_{in}^2 - w_ na_{in}}{2w'_ iw_ n}$

It is elementary to check that this is an integer $\geq 0$ if we choose $w'_ i$ as in (4). $\square$

Lemma 55.3.10. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type. Let $e$ be the number of pairs $(i, j)$ with $i < j$ and $a_{ij} > 0$. Then the expression $g_{top} = 1 - n + e$ is $\geq 0$.

Proof. If not, then $e < n - 1$ which means there exists an $i$ such that $a_{ij} = 0$ for all $j \not= i$. This contradicts assumption (3) of Definition 55.3.1. $\square$

Definition 55.3.11. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type $T$. The topological genus of $T$ is the nonnegative integer $g_{top} = 1 - n + e$ from Lemma 55.3.10.

We want to bound the genus by the topological genus. However, this will not always be the case, for example for numerical types with $n = 1$ as in Lemma 55.3.5. But it will be true for minimal numerical types which are defined as follows.

Definition 55.3.12. We say the numerical type $n, m_ i, a_{ij}, w_ i, g_ i$ of genus $g$ is minimal if there does not exist an $i$ with $g_ i = 0$ and $a_{ii} = -w_ i$, in other words, if there does not exist a $(-1)$-index.

We will prove that the genus $g$ of a minimal type with $n > 1$ is greater than or equal to $\max (1, g_{top})$.

Lemma 55.3.13. If $n, m_ i, a_{ij}, w_ i, g_ i$ is a minimal numerical type with $n > 1$, then $g \geq 1$.

Proof. This is true because $g = 1 + \sum \Phi _ i$ with $\Phi _ i = m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii})$ nonnegative by Lemma 55.3.7 and the definition of minimal types. $\square$

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 $M_1$. $\square$

Lemma 55.3.15. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type of genus $g$. Assume $n > 1$. If $i$ is such that the contribution $m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii})$ to the genus $g$ is $0$, then $g_ i = 0$ and $a_{ii} = -2w_ i$.

Proof. Follows immediately from Lemma 55.3.6 and $w_ i > 0$, $g_ i \geq 0$, and $w_ i | a_{ii}$. $\square$

It turns out that the indices satisfying this relation play an important role in the structure of minimal numerical types. Hence we give them a name.

Definition 55.3.16. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a numerical type of genus $g$. We say $i$ is a $(-2)$-index if $g_ i = 0$ and $a_{ii} = -2w_ i$.

Given a minimal numerical type of genus $g$ the $(-2)$-indices are exactly the indices which do not contribute a positive number to the genus in the formula

$g = 1 + \sum m_ i(w_ i(g_ i - 1) - \frac{1}{2} a_{ii})$

Thus it will be somewhat tricky to bound the quantities associated with $(-2)$-indices as we will see later.

Remark 55.3.17. Let $n, m_ i, a_{ij}, w_ i, g_ i$ be a minimal numerical type with $n > 1$. Equality $g = g_{top}$ can hold in Lemma 55.3.14. For example, if $m_ i = w_ i = 1$ and $g_ i = 0$ for all $i$ and $a_{ij} \in \{ 0, 1\}$ for $i < j$.

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