The Stacks project

Proof. If $A$ is as in (1) and $\det (A) = 0$, then there is a nonzero vector $z$ with $Az = 0$. Choose $r$ with $|z_ r|$ maximal. Then

which is a contradiction. To prove (2) apply (1) to the matrix $(a_{ij}m_ j)$ whose determinant is $m_1 \ldots m_ n \det (A)$. $\square$

Proof. After replacing $a_{ij}$ by $a_{ij}m_ j$ we may assume $m_ i = 1$ for all $i$. If $I \subset \{ 1, \ldots , n\} $ such that (1) and (2) are true, then a simple computation shows that $x_ I$ is in the kernel of $A$. Conversely, let $x = (x_1, \ldots , x_ n) \in \mathbf{R}^ n$ be a nonzero vector in the kernel of $A$. We will show by induction on the number of nonzero coordinates of $x$ that $x$ is in the span of the vectors $x_ I$ satisfying (1) and (2). Let $I \subset \{ 1, \ldots , n\} $ be the set of indices $r$ with $|x_ r|$ maximal. For $r \in I$ we have

Thus equality holds everywhere. In particular, we see that $a_{rk} = 0$ if $r \in I$, $k \not\in I$ and equality holds in (55.2.2.1) for $r \in I$. Then we see that we can substract a suitable multiple of $x_ I$ from $x$ to decrease the number of nonzero coordinates. $\square$

First proof. After replacing $a_{ij}$ by $a_{ij}m_ im_ j$ we may assume $m_ i = 1$ for all $i$. Condition (1) means $-a_{ii} = \sum _{j \not= i} a_{ij}$ for all $i$. Recall that $x^ tAx = \sum _{i, j} x_ ia_{ij}x_ j$. Then

This is clearly $\leq 0$. If equality holds, then let $I$ be the set of indices $i$ with $x_ i \not= x_1$. Then $a_{ij} = 0$ for $i \in I$ and $j \not\in I$. Thus $I = \{ 1, \ldots , n\} $ by condition (2) and $x$ is a multiple of $m = (1, \ldots , 1)$. $\square$

Second proof. The matrix $A$ has real eigenvalues by the spectral theorem. We claim all the eigenvalues are $\leq 0$. Namely, since property (1) means $-a_{ii}m_ i = \sum _{j \not= i} a_{ij}m_ j$ for all $i$, we find that the matrix $A' = A - \lambda I$ for $\lambda > 0$ satisfies $|a'_{ii}m_ i| > \sum a'_{ij}m_ j = \sum |a'_{ij}m_ j|$ for all $i$. Hence $A'$ is invertible by Lemma 55.2.1. This implies that the symmetric bilinear form $x^ tAy$ is semi-negative definite, i.e., $x^ tAx \leq 0$ for all $x$. It follows that the kernel of $A$ is equal to the set of vectors $x$ with $x^ tAx = 0$. The description of the kernel in Lemma 55.2.2 gives the final statement of the lemma. $\square$

Proof. Observe that $L \otimes \mathbf{Q} = A \otimes \mathbf{Q} \oplus B \otimes \mathbf{Q}$ because the form is nondegenerate on $A$ (by positivity). We denote $\pi _ B : L \otimes \mathbf{Q} \to B \otimes \mathbf{Q}$ the projection. Observe that $\pi _ B(x) \in B^\# $ for $x \in L$ because the form is integral. This gives an exact sequence

where $Q$ is the cokernel of $L \to B^\# $. Observe that $Q$ is a quotient of $L^\# /L$ as the map $L^\# \to B^\# $ is surjective since it is the $\mathbf{Z}$-linear dual to $B \to L$ which is split as a map of $\mathbf{Z}$-modules. Dividing by $A \oplus B$ we get a short exact sequence

This proves the lemma. $\square$

Proof. Let $x \in L_0$ be an element representing a torsion class in $\mathop{\mathrm{Coker}}(\text{d}^*\text{d})$. Then for some $a > 0$ we can write $ax = \text{d}^*\text{d}(y)$. For any $z \in \mathop{\mathrm{Im}}(\text{d})$, say $z = \text{d}(y')$, we have

Hence $(1/a)\text{d}(y) \in \mathop{\mathrm{Im}}(\text{d})^\# $. We define $\Phi (x) = (1/a)\text{d}(y) \bmod \mathop{\mathrm{Im}}(\text{d})$. We omit the proof that $\Phi $ is well defined, additive, and injective.

To prove $\Phi $ is surjective, let $z \in \mathop{\mathrm{Im}}(\text{d})^\# $. Then $z$ defines a linear map $L_0 \to \mathbf{Z}$ by the rule $x \mapsto \langle z, \text{d}(x)\rangle $. Since the pairing on $L_0$ is unimodular by assumption we can find an $x' \in L_0$ with $\langle x', x \rangle = \langle z, \text{d}(x)\rangle $ for all $x \in L_0$. In particular, we see that $x'$ pairs to zero with $\mathop{\mathrm{Ker}}(\text{d})$. Since $\mathop{\mathrm{Im}}(\text{d}^*\text{d}) \otimes \mathbf{Q}$ is the orthogonal complement of $\mathop{\mathrm{Ker}}(\text{d}) \otimes \mathbf{Q}$ this means that $x'$ defines a torsion class in $\mathop{\mathrm{Coker}}(\text{d}^*\text{d})$. We claim that $\Phi (x') = z$. Namely, write $a x' = \text{d}^*\text{d}(y)$ for some $y \in L_0$ and $a > 0$. For any $x \in L_0$ we get

Hence $z = \Phi (x')$ and the proof is complete. $\square$

Proof. By Lemma 55.2.3 the rank of $A$ is $n - 1$. The composition

has matrix $a_{ij}m_ im_ j$. Since the cokernel of the first and last maps are torsion of order prime to $\ell $ by our restriction on $\ell $ we see that it suffices to prove the lemma for the matrix with entries $a_{ij}m_ im_ j$. Thus we may assume $m = (1, \ldots , 1)$.

Assume $m = (1, \ldots , 1)$. Set $V = \{ 1, \ldots , n\} $ and $E = \{ (i, j) \mid i < j\text{ and }a_{ij} > 0\} $. For $e = (i, j) \in E$ set $a_ e = a_{ij}$. Define maps $s, t : E \to V$ by setting $s(i, j) = i$ and $t(i, j) = j$. Set $\mathbf{Z}(V) = \bigoplus _{i \in V} \mathbf{Z}i$ and $\mathbf{Z}(E) = \bigoplus _{e \in E} \mathbf{Z}e$. We define symmetric positive definite integer valued pairings on $\mathbf{Z}(V)$ and $\mathbf{Z}(E)$ by setting

and all other pairings zero. Consider the maps

and

A computation shows that

in other words, $\text{d}$ and $\text{d}^*$ are adjoint. Next we compute

The coefficient of $i$ in $\text{d}^*\text{d}(i)$ is

because $\sum _ j a_{ij} = 0$ and the coefficient of $j \not= i$ in $\text{d}^*\text{d}(i)$ is $-a_{ij}$. Hence $\mathop{\mathrm{Coker}}(A) = \mathop{\mathrm{Coker}}(\text{d}^*\text{d})$.

Consider the inclusion

The left hand side is an orthogonal direct sum. Clearly $\mathbf{Z}(E)/\mathop{\mathrm{Ker}}(\text{d}^*)$ is torsion free. We claim $\mathbf{Z}(E)/\mathop{\mathrm{Im}}(\text{d})$ is torsion free as well. Namely, say $x = \sum x_ e e \in \mathbf{Z}(E)$ and $a > 1$ are such that $ax = \text{d}y$ for some $y = \sum y_ i i \in \mathbf{Z}(V)$. Then $a x_ e = y_{s(e)} - y_{t(e)}$. By property (2) we conclude that all $y_ i$ have the same congruence class modulo $a$. Hence we can write $y = a y' + (y_1, y_1, \ldots , y_1)$. Since $\text{d}(y_1, y_1, \ldots , y_1) = 0$ we conclude that $x = \text{d}(y')$ which is what we had to show.

Hence we may apply Lemma 55.2.4 to get injective maps

whose cokernels are annihilated by the product of the $a_ e$ (which is prime to $\ell $). Since $\mathop{\mathrm{Ker}}(\text{d}^*)$ is a lattice of rank $1 - n + e$ we see that the proof is complete if we prove that there exists an isomorphism

This is proved in Lemma 55.2.5. $\square$