Lemma 55.3.9 (0C77)—The Stacks project

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

$n' = n - 1$,
$m'_ i = m_ i$,
$a'_{ij} = a_{ij} - a_{in}a_{jn}/a_{nn}$,
$w'_ i = w_ i/2$ if $a_{in}/w_ n$ even and $a_{in}/w_ i$ odd and $w'_ i = w_ i$ else,
$g'_ i = \frac{w_ i}{w'_ i}(g_ i - 1) + 1 + \frac{a_{in}^2 - w_ na_{in}}{2w'_ iw_ n}$.

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). $\square$

The Stacks project

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