## 55.2 Linear algebra

A couple of lemmas we will use later on.

Lemma 55.2.1. Let $A = (a_{ij})$ be a complex $n \times n$ matrix.

1. If $|a_{ii}| > \sum _{j \not= i} |a_{ij}|$ for each $i$, then $\det (A)$ is nonzero.

2. If there exists a real vector $m = (m_1, \ldots , m_ n)$ with $m_ i > 0$ such that $|a_{ii} m_ i| > \sum _{j \not= i} |a_{ij}m_ j|$ for each $i$, then $\det (A)$ is nonzero.

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

$|a_{rr} z_ r| = |\sum \nolimits _{k \not= r} a_{rk}z_ k| \leq \sum \nolimits _{k \not= r} |a_{rk}||z_ k| \leq |z_ r| \sum \nolimits _{k \not= r} |a_{rk}| < |a_{rr}||z_ r|$

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

Lemma 55.2.2. Let $A = (a_{ij})$ be a real $n \times n$ matrix with $a_{ij} \geq 0$ for $i \not= j$. Let $m = (m_1, \ldots , m_ n)$ be a real vector with $m_ i > 0$. For $I \subset \{ 1, \ldots , n\}$ let $x_ I \in \mathbf{R}^ n$ be the vector whose $i$th coordinate is $m_ i$ if $i \in I$ and $0$ otherwise. If

55.2.2.1
\begin{equation} \label{models-equation-ineq} -a_{ii}m_ i \geq \sum \nolimits _{j \not= i} a_{ij}m_ j \end{equation}

for each $i$, then $\mathop{\mathrm{Ker}}(A)$ is the vector space spanned by the vectors $x_ I$ such that

1. $a_{ij} = 0$ for $i \in I$, $j \not\in I$, and

2. equality holds in (55.2.2.1) for $i \in I$.

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

$|a_{rr} x_ r| = |\sum \nolimits _{k \not= r} a_{rk}x_ k| \leq \sum \nolimits _{k \not= r} a_{rk}|x_ k| \leq |x_ r| \sum \nolimits _{k \not= r} a_{rk} \leq |a_{rr}||x_ r|$

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. $\square$

Lemma 55.2.3. Let $A = (a_{ij})$ be a symmetric real $n \times n$ matrix with $a_{ij} \geq 0$ for $i \not= j$. Let $m = (m_1, \ldots , m_ n)$ be a real vector with $m_ i > 0$. Assume

1. $Am = 0$,

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

Then $x^ t A x \leq 0$ with equality if and only if $x = qm$ for some $q \in \mathbf{R}$.

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

\begin{align*} \sum \nolimits _{i \not= j} -a_{ij}(x_ j - x_ i)^2 & = \sum \nolimits _{i \not= j} -a_{ij}x_ j^2 + 2a_{ij}x_ ix_ i - a_{ij}x_ i^2 \\ & = \sum \nolimits _ j a_{jj} x_ j^2 + \sum \nolimits _{i \not= j} 2a_{ij}x_ ix_ i + \sum \nolimits _ j a_{jj} x_ i^2 \\ & = 2x^ tAx \end{align*}

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. $\square$

Lemma 55.2.4. Let $L$ be a finite free $\mathbf{Z}$-module endowed with an integral symmetric bilinear positive definite form $\langle \ ,\ \rangle : L \times L \to \mathbf{Z}$. Let $A \subset L$ be a submodule with $L/A$ torsion free. Set $B = \{ b \in L \mid \langle a, b\rangle = 0,\ \forall a \in A\}$. Then we have injective maps

$A^\# /A \leftarrow L/(A \oplus B) \rightarrow B^\# /B$

whose cokernels are quotients of $L^\# /L$. Here $A^\# = \{ a' \in A \otimes \mathbf{Q} \mid \langle a, a'\rangle \in \mathbf{Z},\ \forall a \in A\}$ and similarly for $B$ and $L$.

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

$0 \to A \to L \xrightarrow {\pi _ B} B^\# \to Q \to 0$

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

$0 \to L/(A \oplus B) \to B^\# /B \to Q \to 0$

This proves the lemma. $\square$

Lemma 55.2.5. Let $L_0$, $L_1$ be a finite free $\mathbf{Z}$-modules endowed with integral symmetric bilinear positive definite forms $\langle \ ,\ \rangle : L_ i \times L_ i \to \mathbf{Z}$. Let $\text{d} : L_0 \to L_1$ and $\text{d}^* : L_1 \to L_0$ be adjoint. If $\langle \ ,\ \rangle$ on $L_0$ is unimodular, then there is an isomorphism

$\Phi : \mathop{\mathrm{Coker}}(\text{d}^*\text{d})_{torsion} \longrightarrow \mathop{\mathrm{Im}}(\text{d})^\# /\mathop{\mathrm{Im}}(\text{d})$

with notation as in Lemma 55.2.4.

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

$\langle (1/a)\text{d}(y), z \rangle = \langle (1/a)\text{d}(y), \text{d}(y') \rangle = \langle x, y' \rangle \in \mathbf{Z}$

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

$\langle z, \text{d}(x)\rangle = \langle x', x \rangle = \langle (1/a)\text{d}^*\text{d}(y), x \rangle = \langle (1/a)\text{d}(y),\text{d}(x) \rangle$

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

Lemma 55.2.6. Let $A = (a_{ij})$ be a symmetric $n \times n$ integer matrix with $a_{ij} \geq 0$ for $i \not= j$. Let $m = (m_1, \ldots , m_ n)$ be an integer vector with $m_ i > 0$. Assume

1. $Am = 0$,

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

Let $e$ be the number of pairs $(i, j)$ with $i < j$ and $a_{ij} > 0$. Then for $\ell$ a prime number coprime with all $a_{ij}$ and $m_ i$ we have

$\dim _{\mathbf{F}_\ell }(\mathop{\mathrm{Coker}}(A)[\ell ]) \leq 1 - n + e$

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

$\mathbf{Z}^{\oplus n} \xrightarrow {\text{diag}(m_1, \ldots , m_ n)} \mathbf{Z}^{\oplus n} \xrightarrow {(a_{ij})} \mathbf{Z}^{\oplus n} \xrightarrow {\text{diag}(m_1, \ldots , m_ n)} \mathbf{Z}^{\oplus n}$

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

$\langle i, i \rangle = 1\text{ for }i \in V, \quad \langle e, e \rangle = a_ e\text{ for }e \in E$

and all other pairings zero. Consider the maps

$\text{d} : \mathbf{Z}(V) \to \mathbf{Z}(E), \quad i \longmapsto \sum \nolimits _{e \in E,\ s(e) = i} e - \sum \nolimits _{e \in E,\ t(e) = i} e$

and

$\text{d}^*(e) = a_ e(s(e) - t(e))$

A computation shows that

$\langle d(x), y\rangle = \langle x, \text{d}^*(y) \rangle$

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

\begin{align*} \text{d}^*\text{d}(i) & = \text{d}^*( \sum \nolimits _{e \in E,\ s(e) = i} e - \sum \nolimits _{e \in E,\ t(e) = i} e) \\ & = \sum \nolimits _{e \in E,\ s(e) = i} a_ e(s(e) - t(e)) - \sum \nolimits _{e \in E,\ t(e) = i} a_ e(s(e) - t(e)) \end{align*}

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

$\sum \nolimits _{e \in E,\ s(e) = i} a_ e + \sum \nolimits _{e \in E,\ t(e) = i} a_ e = - a_{ii}$

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

$\mathop{\mathrm{Im}}(\text{d}) \oplus \mathop{\mathrm{Ker}}(\text{d}^*) \subset \mathbf{Z}(E)$

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

$\mathop{\mathrm{Im}}(\text{d})^\# /\mathop{\mathrm{Im}}(\text{d}) \leftarrow \mathbf{Z}(E)/(\mathop{\mathrm{Im}}(\text{d}) \oplus \mathop{\mathrm{Ker}}(\text{d}^*)) \rightarrow \mathop{\mathrm{Ker}}(\text{d}^*)^\# /\mathop{\mathrm{Ker}}(\text{d}^*)$

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

$\Phi : M_{torsion} \longrightarrow \mathop{\mathrm{Im}}(\text{d})^\# /\mathop{\mathrm{Im}}(\text{d})$

This is proved in Lemma 55.2.5. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).