# Zeta functions for table algebras and fusion rings with irrational-valued
characters

Angelica Babei∗ and Allen Herman† Department of Mathematics $\&$ Statistics,
McMaster University, Hamilton Hall, 1280 Main Street West, Hamilton, ON, L8S
4K1, Canada<EMAIL_ADDRESS>Department of Mathematics and Statistics,
University of Regina, Regina, Saskatchewan S4S 0A2, Canada
<EMAIL_ADDRESS>
###### Abstract.

We calculate ideal zeta functions for certain orders of rank $3$ defined by
standard integral table algebras and integral fusion rings that have
irrational-valued irreducible characters. The calculations are obtained from
explicit calculations of zeta integrals.

###### Key words and phrases:

Zeta functions, orders, association schemes, integral adjacency algebras,
table algebras, fusion rings

###### 1991 Mathematics Subject Classification:

Primary: 11S45. Secondary: 11R54, 05E30.

∗ Department of Mathematics $\&$ Statistics, McMaster University, Hamilton
Hall, 1280 Main Street West, Hamilton, ON, L8S 4K1, Canada

† Department of Mathematics and Statistics, University of Regina, Regina,
Saskatchewan S4S 0A2, Canada

† The work of the second author is supported by an NSERC Discovery Grant.

##  1\. Introduction

Let $\Lambda$ be an order in a semisimple $\mathbb{Q}$-algebra $A$, and let
$M$ be a $\Lambda$-lattice. In [20], Solomon defined a zeta function

$\zeta_{\Lambda}(M;s)=\sum_{n\geq 1}a_{n}n^{-s}$

where $a_{n}$ is the number of $\Lambda$-sublattices of $M$ of index $n$ for
all $n\in\mathbb{Z}^{+}$. In the case $M=\Lambda$,
$\zeta_{\Lambda}(s):=\zeta_{\Lambda}(\Lambda;s)$ is called the zeta function
of $\Lambda$, and is the generating function for the sequence $\\{a_{n}\\}$
that counts the number of ideals of index $n$.

In [20], Solomon gave a calculation of $\zeta_{\Lambda}(s)$ in the case where
$\Lambda=\mathbb{Z}G$ is an integral group ring of a finite cyclic group of
prime order $p$. Work of Hironaka ([16], [15]) and Takegahara ([21]) in the
1980s produced the calculations of $\zeta_{\mathbb{Z}G}(s)$ where $G$ is any
abelian of order $pq$, where $p$ and $q$ are not-necessarily distinct primes.
The same elementary approach was used by Hanaki and Hirasaka for integral
adjacency rings of association schemes that have prime order or rank $2$ [11],
and by Hirasaka and Oh for quotient polynomial rings of the form
$\mathbb{Z}[x]/(x-k)(x-a)(x-b)$ for $k,a,b\in\mathbb{Z}$ [14]. In [2], the
authors applied a formula involving local zeta integrals due to Bushnell and
Reiner in [6] to give explicit calculations of $\zeta_{\mathbb{Z}B}(s)$, where
$\mathbb{Z}B$ was the integral adjacency ring of certain small association
schemes with rational character tables.

In this paper, we improve the methods of [2] to include cases where
$\mathbb{C}\mathbf{B}$ has an irrational character table with respect to the
$\mathbb{Z}$-basis $\mathbf{B}$. This allows us to compute zeta functions of
$\Lambda=\mathbb{Z}\mathbf{B}$ in more situations. We illustrate these
techniques by calculating $p$-local zeta functions for the integral adjacency
rings of association schemes of rank $3$ that correspond to doubly regular
tournaments or conference graphs whose order $n$ has $p$-valuation $1$ or $3$
at an odd prime $p$, and the global zeta functions of all categorifiable
integral fusion rings of rank $3$.

##  2\. Orders defined by table algebras and fusion rings

An integral table algebra $(A,\mathbf{B})$ is a finite-dimensional algebra $A$
with (skew-linear) involution $*$ over $\mathbb{C}$ whose defining basis
$\mathbf{B}=\\{1=b_{0},b_{1},\dots,b_{d}\\}$ contains $1$, is $*$-closed,
admits non-negative integer structure constants $\\{\lambda_{ijk}:0\leq
i,j,k\leq d\\}$ satisfying

$b_{i}b_{j}=\sum_{k=0}^{d}\lambda_{ijk}b_{k},$

and satisfies the pseudo-inverse condition: for all $b_{i}\in B$, $b_{i}^{*}$
is the unique $b_{j}\in\mathbf{B}$ such that $\lambda_{ij0}>0$, and
furthermore if we denote the index of this $b_{j}$ by $i^{*}$, then
$\lambda_{ii^{*}0}=\lambda_{i^{*}i0}$.

Examples of integral table algebras include the group algebras of finite
groups $G$ (whose skew-linear involution is given by $g^{*}=g^{-1}$ for all
$g\in G$), the adjacency algebras (a.k.a. Bose-Mesner algebras) of association
schemes (for which $\mathbf{B}$ can be taken to be the standard basis of
adjacency of adjacency matrices of the association scheme and the involution
is the conjugate transpose), and the complexification of an integral fusion
ring (which is precisely a commutative integral table algebra with the
additional property $\lambda_{ii^{*}0}=\lambda_{i^{*}i0}=1$.) When the algebra
is commutative, the fact that the basis elements $b_{i}\in\mathbf{B}$ are
represented by non-negative integer matrices in the regular representation
implies that the largest eigenvalue $k_{i}>0$ of each matrix will be obtained
from a common Perron-Frobenius eigenvector, and thus the map
$\delta(b_{i})=k_{i}$ extends to a linear character of $A$. This unique
irreducible character that takes positive values on $\mathbf{B}$ is called the
degree map of the table algebra, it corresponds to the augmentation map for
group rings, the valency map for association schemes, and the Perron-Frobenius
eigenvalue for fusion rings. If we re-scale the basis so that
$\lambda_{ii^{*}0}=k_{i}$ for all $i$, we call the new basis $\mathbf{B}$
standard, and if we re-scale the basis so that $\lambda_{ii^{*}0}=1$ for all
$i$, we call the new basis $\mathbf{B}$ transitional. (This terminology comes
from Blau’s survey, see [3].) In our group algebra and adjacency algebra
situations, the defining basis is standard, and for fusion rings the defining
basis is transitional. Note that re-scaling can affect the integrality of
structure constants, which is necessary for $\mathbb{Z}\mathbf{B}$ to be an
order in $\mathbb{Q}\mathbf{B}$.

When $\mathbf{B}$ is the standard basis of a table algebra,
$n=\sum_{i=0}^{d}\delta(b_{i})$ is called the order of the table algebra, and
the linear map extending $\rho(b_{i})=\begin{cases}n&i=0\\\ 0&i\neq
0\end{cases}$ is called the standard feasible trace of $\mathbb{C}\mathbf{B}$.
The standard feasible trace induces a bilinear form $\langle
a,b\rangle=\rho(ab^{*})$, $a,b\in\mathbb{C}\mathbf{B}$, which leads to a
formula for the centrally primitive idempotents of the table algebra (see [12,
§5], [13, §5], [7, Proposition (9.17)], and [1, Theorem 3.6]). For the
standard basis of a table algebra, this formula says the centrally primitive
idempotent of $\mathbb{C}\mathbf{B}$ associated to an irreducible character
$\chi$ of $\mathbb{C}\mathbf{B}$ is

(1)
$e_{\chi}=\frac{m_{\chi}}{n}\sum_{i=0}^{d}\frac{\chi(b_{i}^{*})}{\delta(b_{i})}b_{i}$

where the numbers $m_{\chi}$ are the positive real numbers that occur in the
expression of the standard feasible trace as a linear combination of the
irreducible characters of $\mathbb{C}\mathbf{B}$; i.e. $\rho=\sum_{\chi\in
Irr(\mathbb{C}\mathbf{B})}m_{\chi}\chi$. The positive real number $m_{\chi}$
is called the multiplicity of $\chi$, and it follows from uniqueness of the
Perron-Frobenius eigenvector that the degree character $\delta$ has
multiplicity $m_{\delta}=1$. When $\mathbf{B}$ is a transitional basis, the
re-scalings $b_{i}\mapsto\delta(b_{i})b_{i}:=B_{i}$ and
$B_{i}\mapsto\frac{1}{\sqrt{\delta(B_{i})}}B_{i}:=b_{i}$ allow us to change
back-and-forth from a standard basis, so in the case when
$\mathbb{Z}\mathbf{B}$ is a fusion ring the above formula translates to

(2) $e_{\chi}=\frac{m_{\chi}}{n}\sum_{i=0}^{d}\chi(b_{i}^{*})b_{i}.$

##  3\. Zeta functions of Orders

Let $\Lambda$ be an order in a commutative semisimple $\mathbb{Q}$-algebra
$A$. In this case $A$ has a unique maximal order

$\Lambda_{0}=\oplus_{\chi}\mathbb{Z}(\chi)\tilde{e}_{\chi},$

where $\chi$ runs over a set of representatives of the Galois conjugacy
classes of irreducible characters of $\mathbb{C}\mathbf{B}$, and the primitive
idempotents $\tilde{e}_{\chi}$ of $\mathbb{Q}\mathbf{B}$ are the sums of the
primitive idempotents $e_{\psi}$ of $\mathbb{C}\mathbf{B}$ where $\psi$ runs
over the Galois congujates of $\chi$, and $\mathbb{Z}(\chi)$ are the rings of
integers of the number fields $\mathbb{Q}(\chi)$. The Solomon zeta function of
$\Lambda_{0}$ will be the product of the Dedekind zeta functions for the
number fields $\mathbb{Q}(\chi)$. Recall that when $R$ is the ring of integers
of a number field, its Dedekind zeta function is

$\zeta_{R}(s)=\prod_{\mathcal{P}}(1-[R:\mathcal{P}]^{-s})^{-1},$

where $\mathcal{P}$ runs over the nonzero prime ideals of $R$.

In most of the cases we consider in this paper, our integral table algebra
$\mathbb{Z}\mathbf{B}$ is not equal to the maximal order of
$\mathbb{Q}\mathbf{B}$. In [20], Solomon proved the Euler product identity
$\zeta_{\mathbb{Z}\mathbf{B}}(s)=\prod_{p}\zeta_{\mathbb{Z}_{p}\mathbf{B}}(s)$,
where $p$ runs over the rational primes, and $\mathbb{Z}_{p}\mathbf{B}$
denotes the $p$-adic completion $\mathbb{Z}_{p}\otimes\mathbb{Z}\mathbf{B}$.
Furthermore, he showed each factor $\zeta_{\mathbb{Z}_{p}\mathbf{B}}(s)$ of
this Euler product is equal to $\zeta_{\Lambda_{0,p}}(s)$ times the value of a
polynomial $\delta_{p}(x)$ evaluated at $p^{-s}$, where $\Lambda_{0,p}$ is a
maximal order of $\mathbb{Q}_{p}\mathbf{B}$ containing
$\mathbb{Z}_{p}\mathbf{B}$. In all but finitely many cases, Solomon showed
$\mathbb{Z}_{p}\mathbf{B}$ is equal to $\Lambda_{0,p}$, the exceptions being
the primes $p$ that divide either the discriminant of $\mathbb{Z}\mathbf{B}$
or the least positive integer $f$ with
$f\Lambda_{0}\subset\mathbb{Z}\mathbf{B}$.

This reduces the calculation of $\zeta_{\mathbb{Z}\mathbf{B}}(s)$ to the
calculation of the zeta functions of the local orders
$\mathbb{Z}_{p}\mathbf{B}$ for finitely many primes $p$. To calculate these,
we will use the approach Bushnell and Reiner introduced in [6] for calculating
local zeta integrals. First, Bushnell and Reiner showed that when
$\Lambda_{p}$ is a $\mathbb{Z}_{p}$-order in a maximal order $\Lambda_{0,p}$
of a $\mathbb{Q}_{p}$-algebra $A_{p}$,

$\zeta_{\Lambda_{p}}(s)=\sum_{M\in\mathscr{G}}Z_{\Lambda_{p}}(\Lambda_{p},M;s),$

where $\mathscr{G}$ is a set of representatives up to isomorphism for all full
$\Lambda_{p}$-lattices in $A_{p}$. For each $M\in\mathscr{G}$ the genus zeta
function $Z_{\Lambda_{p}}(\Lambda_{p},M;s)$ computes the contribution from all
$\Lambda_{p}$-sublattices isomorphic to $M$. In the special case of the
regular module of $\Lambda_{p}$, Bushnell and Reiner’s integral formula for
the gunus zeta function reduces to

(3)
$Z_{\Lambda_{p}}(\Lambda_{p},M;s)=\mu(\operatorname{Aut}M)^{-1}(\Lambda_{p}:M)^{-s}\int_{A_{p}^{\times}}\Phi_{\\{M:\Lambda_{p}\\}}(x)\left\|x\right\|^{s}d^{\times}x,$

where the index is
$\displaystyle(\Lambda_{p}:M)=\frac{(\Lambda_{p}:(\Lambda_{p}\cap
N))}{(M:(\Lambda_{p}\cap M))}$, $\Phi_{\\{M:\Lambda_{p}\\}}$ is the
characteristic function in $A$ of the lattice

$\\{M:\Lambda_{p}\\}=\\{x\in A\,|\,Mx\subseteq\Lambda_{p}\\},$

the norm is $\left\|x\right\|=(Nx:N)$ where $N$ is any full
$\mathbb{Z}_{p}$-lattice in $A_{p}$, and $d^{\times}x$ denotes a
multiplicative Haar measure on $A_{p}^{\times}$, which is normalized so that
integrating over $\Lambda_{0,p}^{\times}$ gives $1$. The first factor in their
integral formula is the inverse of
$\mu(\operatorname{Aut}M)=\mu(\\{M:M\\}^{\times})$ in this same measure. When
the $\Lambda_{p}$-lattice $M$ is a $\mathbb{Z}_{p}$-order,
$\mu(\operatorname{Aut}M)=\mu(M^{\times})$.

The first challenge in applying Bushnell and Reiner’s integral formula is to
determine a full set of representatives for the full
$\mathbb{Z}_{p}\mathbf{B}$-lattices in $A_{p}$. In [2], all of the cases under
consideration were orders in split semisimple $\mathbb{Q}_{p}$-algebras, which
made this task a bit easier. The next Lemma generalizes [2, Lemma 3.2 and
3.4], it will enable us to do this in the non-split situation.

######  Lemma 3.1.

Let $A$ be a commutative semisimple $\mathbb{Q}_{p}$-algebra. Let
$\mathscr{E}=\\{e_{0},\dots,e_{d}\\}$ be the set of primitive idempotents of
$A$, and suppose $A\simeq K_{0}e_{0}\oplus\dots\oplus K_{d}e_{d}$ for $p$-adic
number fields $K_{0},\dots,K_{d}$. For each $i=0,\dots,d$ let $R_{i}$ be the
ring of integers of $K_{i}$ and let $\pi_{i}R_{i}$ be the maximal ideal of
$R_{i}$.

Let $\Lambda$ be a full $\mathbb{Z}_{p}$-order in the maximal order
$\Lambda_{0}=R_{0}e_{0}\oplus\dots\oplus R_{d}e_{d}$ of $A$.

Let $N$ be any $\Lambda$-lattice in $A$. Then the following hold.

(i) $N$ has a block upper triangular basis as a $\mathbb{Z}_{p}$-module; i.e.
there exists a $\mathbb{Z}_{p}$-basis

$\\{v_{0,1},v_{0,2},\dots,v_{0,r_{0}},v_{1,1},\dots,v_{d,r_{d}}\\}$

of $\Lambda_{0}$ and powers $a_{i,j_{i}}\geq 0$ for $i=0,\dots,d$,
$j_{i}=1,\dots,r_{i}$ such that $\\{v_{i,1}e_{i},\dots,v_{i,r_{i}}e_{i}\\}$ is
a $\mathbb{Z}_{p}$-basis of $R_{i}e_{i}$ for $i=0,\dots,d$;
$\\{\pi_{i}^{a_{i,j_{i}}}v_{i,j_{i}}:i=0,\dots,d,j_{i}=1,\dots,r_{i}\\}$ is a
$\mathbb{Z}_{p}$-basis of $N$; and $v_{i,j_{i}}e_{k}=0$ if $k<i$, for
$j_{i}=1,\dots,r_{i}$, $i=1,\dots,d$, and $k=0,\dots,d-1$.

(ii) There exists a $\Lambda$-lattice $M$ in $A$ such that $M\simeq N$ as
$\Lambda$-lattices, $1\in M$, and $M\subseteq\Lambda_{0}$.

(iii) Suppose $M$ and $N$ are an isomorphic pair of $\Lambda$-lattices in
$\Lambda_{0}$ that both contain $1$. Then any $\Lambda$-lattice isomorphism
from $M$ to $N$ is realized as multiplication by a unit $u\in
N\cap\Lambda_{0}^{\times}$ for which $u^{-1}\in M\cap\Lambda_{0}^{\times}$.

###### Proof.

(i). Let $U$ be a $\mathbb{Z}_{p}$-basis of $N$. We choose a subset
$W_{0}=\\{w_{0,1},\dots,w_{0,r_{0}}\\}$ of $U$ so that
$\\{w_{0,1}e_{0},\dots,w_{0,r_{0}}e_{0}\\}$ is a $\mathbb{Z}_{p}$-basis of
$Ne_{0}$. Since $R_{0}e_{0}$ is a local ring with maximal ideal generated by
$\pi_{0}$, we can find powers $\pi_{0}^{a_{0,j_{0}}}$ with
$a_{0,j_{0}}\in\mathbb{Z}$ for $j_{0}=1,\dots,r_{0}$ such that
$\pi_{0}^{-a_{0,j_{0}}}w_{0,j_{0}}e_{0}$ is a unit of $R_{0}e_{0}$ for
$j_{0}=1,\dots,r_{0}$. Define $v_{0,j_{0}}\in\Lambda_{0}$ to be
$\pi_{0}^{-a_{0,j_{0}}}w_{0,j_{0}}e_{0}+w_{0,j_{0}}(1-e_{0})$ for
$j_{0}=1,\dots,r_{0}$. It is easy to see that
$V_{0}=\\{v_{0,j_{0}}e_{0}:j_{0}=1,\dots,r_{0}\\}$ is a $\mathbb{Z}_{p}$-basis
of $R_{0}e_{0}$, and
$\\{\pi_{0}^{a_{0,j_{0}}}v_{0,j_{0}}e_{0}:j_{0}=1,\dots,r_{0}\\}=\\{w_{0,j_{0}}e_{0}:j_{0}=1,\dots,r_{0}\\}$
is a $\mathbb{Z}_{p}$-basis of $Ne_{0}$.

For $u\in U-W_{0}$, we have that $ue_{0}=w_{u}e_{0}$, where $w_{u}$ is a
specific $\mathbb{Z}_{p}$-linear combination of
$\\{w_{0,j_{0}}:j_{0}=1,\dots,r_{0}\\}$. Let $U_{1}=\\{u-w_{u}:u\in
U-W_{0}\\}$. Then $W_{0}\cup U_{1}$ is a $\mathbb{Z}_{p}$-basis of $N$ for
which $U_{1}e_{0}=0$. Repeat this process with the idempotents
$e_{1},\dots,e_{d}$ in succession, to produce $\mathbb{Z}_{p}$-independent
sets $W_{0},W_{1},\dots,W_{d}$ whose union is a block upper triangular basis
of $N$ with respect to the idempotents $e_{0},\dots,e_{d}$.

(ii). Let $N$ be a full $\Lambda$-lattice in $A$. Since each $K_{i}$ is a
$p$-local field, we can multiply by an appropriate power of $p$ to arrange
that $N\subseteq\Lambda_{0}$ but $p^{-1}N\not\subseteq\Lambda_{0}$. If $N$
contains a unit $u$ of $\Lambda_{0}$, then $Nu^{-1}$ would be a
$\Lambda$-lattice satisfying $1\in Nu^{-1}\subseteq\Lambda_{0}$. So suppose
$N$ does not contain a unit. Then there is a primitive idempotent $e_{i}$ for
which $Ne_{i}\subseteq\pi_{i}R_{i}e_{i}$. Suppose $a_{i}\geq 1$ is the largest
power of $\pi_{i}$ such that $Ne_{i}\subsetneq\pi_{i}^{a_{i}}R_{i}e_{i}$. Let
$\phi_{i}:N\rightarrow\Lambda_{0}$ be the map
$\phi_{i}(n)=n(1-e_{i})+\pi_{i}^{-a_{i}}ne_{i}$ for all $n\in N$. It is easy
to see that $\phi_{i}$ is $\Lambda_{0}$-linear and injective, so $\phi_{i}(N)$
is isomorphic to $N$ as a $\Lambda$-lattice and the number of primitive
idempotents $e_{j}$ for which $\phi_{i}(N)e_{j}\subseteq\pi_{j}R_{j}e_{j}$ is
one fewer than the number for $N$. By repeatedly applying the maps $\phi_{i}$
we will eventually arrive at a $\Lambda$-lattice $M$ that is isomorphic to $N$
and will contain an element $u$ for which $ue_{i}\in
R_{i}e_{i}-\pi_{i}R_{i}e_{i}$ for $i=0,\dots,d$. So this element $u$ of $M$ is
a unit of $\Lambda_{0}$, and (ii) follows.

(iii). Let $k$ be a positive integer for which $p^{k}M\subseteq\Lambda$, and
let $\phi:M\rightarrow N$ be a $\Lambda$-lattice isomorphism. Then for all
$m\in M$

$\phi(m)=p^{-k}p^{k}\phi(m)=p^{-k}\phi(p^{k}m)=p^{-k}(p^{k}m)\phi(1)=m\phi(1).$

Let $u=\phi(1)$. We have that $u\in\Lambda_{0}$ because
$\phi(M)=N\subseteq\Lambda_{0}$. Since $1\in N$, there is a $v\in M$ for which
$1=\phi(v)=vu$. Therefore, $u\in N\cap\Lambda_{0}^{\times}$ and $v=u^{-1}\in
M\cap\Lambda_{0}^{\times}$. ∎

Once we have a set of representatives $\\{M\\}$ of the genera of
$\mathbb{Z}_{p}\mathbf{B}$-lattices in $A$, the next challenge is determining
the genus zeta functions
$Z_{\mathbb{Z}_{p}\mathbf{B}}(\mathbb{Z}_{p}\mathbf{B},M;s)$. The following
lemma is helpful in determining these integrals.

######  Lemma 3.2.

Let $K$ be a $p$-adic number field with ring of integers $R$, whose maximal
ideal is $\pi R$, such that the residue class field $R/\pi R$ is isomorphic to
the finite field with $q$ elements $\mathbb{F}_{q}$, where $q=p^{f}$. Let
$d^{\times}x$ denote a multiplicative Haar measure on $K^{\times}$ such that
$\int_{R}d^{\times}x=1$ and $\left\|x\right\|=(R:Rx)^{-1}$. Then the following
hold:

  1. (1)

For any measurable subset $S\subseteq R$ and $i\geq 0$,
$\int_{\pi^{i}S}\left\|x\right\|^{s}d^{\times}x=q^{-is}\int_{S}\left\|x\right\|^{s}d^{\times}x$;

  2. (2)

$\int_{\pi^{i}R-\\{0\\}}\left\|x\right\|^{s}d^{\times}x=\frac{q^{-is}}{1-q^{-s}}$;

  3. (3)

For $i\geq 1$, denote the $i$-th higher unit group by $U^{(i)}=1+\pi^{i}R$.
Then $\int_{U^{(i)}}\left\|x\right\|^{s}d^{\times}x=\frac{1}{q^{i-1}(q-1)}$.
Moreover, for any $u\in(R/\pi R)^{\times}$, $a_{k}\in R/\pi R$ and $i\geq 1$,
consider the subset $S=u+\sum_{k=1}^{i}a_{k}\pi^{k}+\pi^{i}R$. Then
$\int_{S}\left\|x\right\|^{s}d^{\times}x=\frac{1}{q^{i-1}(q-1)}$.

###### Proof.

  1. (1)

Write $S$ as a disjoint union $S=\dot{\bigcup}_{j}\pi^{j}S_{j}$, where each
$S_{j}\subseteq R^{\times}$ and $j\geq 0$. Since $\pi^{i}(\pi^{j}S_{j})$ and
$\pi^{j}S_{j}$ are multiplicative translates,
$\int_{\pi^{i}(\pi^{j}S_{j})}d^{\times}x=\int_{\pi^{j}S_{j}}d^{\times}x$.
Moreover, $(R:R\pi^{j})=q^{j}$, so

$\displaystyle\int_{\pi^{i}S}\left\|x\right\|^{s}d^{\times}x$
$\displaystyle=\sum_{j}\int_{\pi^{i+j}S_{j}}\left\|x\right\|^{s}d^{\times}x=\sum_{j}\int_{\pi^{i+j}S_{j}}q^{-(i+j)s}d^{\times}x=q^{-is}\sum_{j}\int_{\pi^{i}(\pi^{j}S_{j})}q^{-js}d^{\times}x$
$\displaystyle=q^{-is}\sum_{j}\int_{\pi^{j}S_{j}}(R:R\pi^{j})^{-s}d^{\times}x=q^{-is}\int_{S}\left\|x\right\|^{s}d^{\times}x.$

  2. (2)

The claim follows from (1), since we have the disjoint union
$R-\\{0\\}=\dot{\bigcup}_{j=0}^{\infty}\pi^{j}R^{\times}$ and

$\int_{R-\\{0\\}}\left\|x\right\|^{s}d^{\times}x=\sum_{j=0}^{\infty}\int_{\pi^{j}R^{\times}}\left\|x\right\|^{s}d^{\times}x=\sum_{j=0}^{\infty}q^{-js}=\frac{1}{1-q^{-s}}.$

  3. (3)

Since $(R:Rx)=1$ for any $x\in U^{(1)}$, this follows from
$\int_{U^{(i)}}\left\|x\right\|^{s}d^{\times}x=\mu(U^{(i)})=\frac{1}{q^{i-1}(q-1)}$
by [18, Proposition 3.10]. The second claim is due to
$S=(u+\sum_{k=1}^{i}a_{k}\pi^{k})U^{(i)}$ with
$u+\sum_{k=1}^{i}a_{k}\pi^{k}\in R^{\times}$.

∎

We are now ready to investigate zeta functions for integral table algebras
with irrational character table. The smallest standard integral table algebra
examples have rank $3$, and occur in both the asymmetric and symmetric
situations.

######  Example 3.3.

The standard basis $\mathbf{B}=\\{1,b,b^{*}\\}$ of the adjacency algebra of an
asymmetric table algebra of rank $3$ satisfies $bb^{*}=(2u+1)1+ub+ub^{*}$ and
$b^{2}=ub+(u+1)b^{*}$ for some $u\geq 0$. (These table algebras are realized
as the adjacency algebras of association schemes corresponding to doubly
regular tournaments of order $n=4u+3$.) The minimal polynomial of both $b$ and
$b^{*}$ is $\mu(x)=(x-(2u+1))(x^{2}+x+(u+1))$, and it follows that
$\mathbb{Q}\mathbf{B}=\mathbb{Q}\oplus\mathbb{Q}\left[\frac{-1+\sqrt{-n}}{2}\right]$
for $n=4u+3$. Since the second component does not split over $\mathbb{Q}$, the
idempotents of $\mathbb{Q}\mathbf{B}$ will be

$e_{0}=\frac{1}{n}(1+b+b^{*})\mbox{ and
}e_{1}=\frac{n-1}{n}1-\frac{1}{n}(b+b^{*}).$

A $\mathbb{Z}$-basis for the maximal order $\Lambda_{0}$ of
$\mathbb{Q}\mathbf{B}$ is $\\{e_{0},e_{1},be_{1}\\}$, and when we write $1$,
$b$, and $b^{*}$ in terms of this basis we get

$1=e_{0}+e_{1},b=(2u+1)e_{0}+be_{1}\mbox{ and
}b^{*}=(2u+1)e_{0}-e_{1}-be_{1}.$

Reducing the basis $\\{1,b,b^{*}\\}$ of $\mathbb{Z}\mathbf{B}$ gives

$\mathbb{Z}\mathbf{B}=\langle ne_{0},e_{0}+e_{1},(2u+1)e_{0}+be_{1}\rangle.$

(Here we denote $\langle v_{1},v_{2},v_{3}\rangle$ for the
$\mathbb{Z}$-lattice spanned by $\\{v_{1},v_{2},v_{3}\\}$, we will also use
the same notation later for locally integral $\mathbb{Z}_{p}$-lattices.) This
tells us $[\Lambda_{0}:\mathbb{Z}\mathbf{B}]=n$, and so to compute the zeta
function of $\mathbb{Z}\mathbf{B}$ we need to consider its completions at the
primes $p$ dividing $n$.

######  Example 3.4.

Let $\mathbf{B}=\\{1,b_{1},b_{2}\\}$ be the standard basis of a symmetric
table algebra of rank $3$ that has an irrational character table. In this
situation there exists a $u\geq 0$ such that
$b_{1}^{2}=(2u)1+(u-1)b_{1}+ub_{2}$ and $b_{1}b_{2}=ub_{1}+ub_{2}=b_{2}b_{1}$,
and the minimal polynomial of $b_{1}$ and $b_{2}$ has the irreducible
factorization $\mu(x)=(x-2u)(x^{2}+x-u)$. (These table algebras are realized
by the adjacency algebras of association schemes corresponding to conference
graphs of order $n=4u+1$ when $n$ is not a perfect square. Payley graphs are
one such example.) For these table algebras,
$\mathbb{Q}\mathbf{B}\simeq\mathbb{Q}\oplus\mathbb{Q}[\frac{-1+\sqrt{n}}{2}]$
and the maximal order $\Lambda_{0}$ of $\mathbb{Z}\mathbf{B}$ is $\langle
e_{0},e_{1},b_{1}e_{1}\rangle$. Writing the elements of $\mathbf{B}$ in terms
of this basis, we have

$1=e_{0}+e_{1},b_{1}=(2u)e_{0}+b_{1}e_{1},\mbox{ and
}b_{2}=(2u)e_{0}-e_{1}-b_{1}e_{1}.$

Reducing the basis $\\{1,b_{1},b_{2}\\}$ gives $\mathbb{Z}\mathbf{B}=\langle
e_{1}+e_{0},b_{1}e_{1}+2ue_{0},ne_{0}\rangle$, so
$[\Lambda_{0}:\mathbb{Z}\mathbf{B}]=n$. Again for the zeta function
calculation this tells us we need to consider completions at the primes $p$
that divide $n$.

In both examples, we are interested in the $p$-local zeta function of
$\mathbb{Z}_{p}\mathbf{B}$, for odd primes $p$ dividing $n$. Let
$n=p^{v_{p}(n)}v$, where $\gcd(p,v)=1$. Then we have three cases to consider:
$v_{p}(n)$ is odd, $v_{p}(n)$ is even and $v$ is a nonsquare modulo $p$, and
$v_{p}(n)$ is even and $v$ is a square modulo $p$. Since we focus on table
algebras of rank $3$ with irrational-valued characters, there must be some odd
prime $p$ dividing $n$ such that $v_{p}(n)$ is odd, so we start with this
case, where one of the simple components of $\mathbb{Q}_{p}\mathbf{B}$ will be
a ramified quadratic extension.

######  Theorem 3.5.

Let $\mathbf{B}$ be the standard basis of a rank $3$ table algebra with
irrational character table; i.e. one of the above two cases depending on
whether the order $n\equiv\pm 1\mod 4$.

Let $p$ be a prime such that $p$ divides $n$ but $p^{2}$ does not divide $n$.

Then
$\zeta_{\mathbb{Z}_{p}\mathbf{B}}(s)=(p^{1-2s}-p^{-s}+1)\zeta_{\mathbb{Z}_{p}}(s)^{2}$.

###### Proof.

We have that $p$ is an odd prime with $n=pv$ for an integer $v$ with
$(p,v)=1$. Then the completion at $p$ gives
$A:=\mathbb{Q}_{p}\mathbf{B}\cong\mathbb{Q}_{p}\oplus K$ where
$K=\mathbb{Q}_{p}(\pi)$ is a ramified quadratic extension of $\mathbb{Q}_{p}$
with ring of integers $R$ and uniformizer $\pi$ satisfying $(\pi R)^{2}=pR$,
so the residue class field $R/\pi R\cong\mathbb{F}_{p}$. Since
$\frac{1}{2}\in\mathbb{Z}_{p}$, we have that $R=\mathbb{Z}_{p}[\sqrt{\pm n}]$,
the sign depending on the congruence of $n$ mod $4$. The uniformizer of $R$
can be taken to be $\pi=\sqrt{\pm n}$ since $\pi^{2}=\pm pv$ will always have
$p$-valuation $1$ in $\mathbb{Z}_{p}$.

The maximal order of $\mathbb{Q}_{p}\mathbf{B}$ is
$\Lambda^{\prime}_{p}\cong\mathbb{Z}_{p}\oplus R$, where
$R=\mathbb{Z}_{p}[\pi]$. Let $e_{0}$ and $e_{1}$ be the primitive idempotents
of $\mathbb{Q}_{p}\mathbf{B}$, with
$\mathbb{Q}_{p}\mathbf{B}=\mathbb{Q}_{p}e_{0}\oplus Ke_{1}$. In
$\mathbb{Q}_{p}\mathbf{B}e_{1}$, the element $e_{1}+2b_{1}e_{1}$ can be
identified with $\pi e_{1}=\sqrt{\pm n}e_{1}$. Then $\Lambda^{\prime}_{p}$ has
$\mathbb{Z}_{p}$-basis $\\{e_{0},e_{1},\pi e_{1}\\}$, and with respect to this
basis, $\Lambda_{p}\coloneqq\mathbb{Z}_{p}\mathbf{B}=\langle
e_{0}+e_{1},ne_{0},\pi e_{1}\rangle$. Therefore, we can represent
$\Lambda_{p}$ as

$\Lambda_{p}=\\{(\alpha,\beta)\in\mathbb{Z}_{p}\oplus
R:\alpha\pmod{p}\equiv\beta\pmod{\pi}\\}.$

From this point the calculation of the zeta function $\zeta_{\Lambda_{p}}$
follows an identical argument as the calculation of the zeta function of the
group ring for the cyclic group of order $p$ outlined in [6, Section 3.4].
Since $(\Lambda_{p}^{\prime}:\Lambda_{p})=p$, there are only two genera of
$\Lambda_{p}$-lattices, with representatives $\Lambda^{\prime}_{p}$ and
$\Lambda_{p}$. Therefore, we need two genus zeta functions
$Z(\Lambda_{p}^{\prime},\Lambda_{p};s)$ and $Z(\Lambda_{p},\Lambda_{p};s)$.

We first calculate

$Z(\Lambda_{p}^{\prime},\Lambda_{p};s)=\mu({\Lambda^{\prime}_{p}}^{\times})^{-1}(\Lambda_{p}:\Lambda^{\prime}_{p})^{-s}\int_{A^{\times}}\Phi_{\\{\Lambda^{\prime}_{p}:\Lambda_{p}\\}}(x)\left\|x\right\|_{A}^{s}d^{\times}x.$

Note that $\\{\Lambda_{p}^{\prime}:\Lambda_{p}\\}=p\mathbb{Z}_{p}\oplus\pi R$,
so by Lemma 3.2,

$Z(\Lambda_{p}^{\prime},\Lambda_{p};s)=p^{s}\int_{p\mathbb{Z}_{p}-\\{0\\}}\left\|a\right\|_{\mathbb{Q}_{p}}^{s}d^{\times}a\int_{\pi
R-\\{0\\}}\left\|b\right\|_{K}^{s}d^{\times}b=p^{-s}(1-p^{-s})^{-2}.$

On the other hand, $\\{\Lambda_{p}:\Lambda_{p}\\}=\Lambda_{p}$, and
$A^{\times}\cap\Lambda_{p}$ equals the disjoint union between
$\Lambda_{p}^{\times}\dot{\cup}(A^{\times}\cap(p\mathbb{Z}_{p}\oplus\pi R))$,
the latter being the domain of integration in the previous genus zeta
function. Therefore,

$\begin{array}[]{rcl}Z(\Lambda_{p},\Lambda_{p};s)&=&\mu(\Lambda_{p}^{\times})^{-1}(\Lambda_{p}:\Lambda_{p})^{-s}\left(\int_{\Lambda_{p}^{\times}}\left\|x\right\|_{A}^{s}d^{\times}x+p^{-2s}(1-p^{-s})^{-2}\right)\\\
&=&1+(p-1)p^{-2s}(1-p^{-s})^{-2}.\end{array}$

Since the local Dedekind zeta function is
$\zeta_{\mathbb{Z}_{p}}(s)=(1-p^{-s})^{-1}$, we get

$\zeta_{\Lambda_{p}}(s)=(p^{1-2s}-p^{-s}+1)\zeta_{\mathbb{Z}_{p}}(s)^{2}.$

∎

It is interesting to see that
$\zeta_{\Lambda_{p}}(s)=\zeta_{\mathbb{Z}_{p}C_{p}}(s)$, the (local) zeta
function of the group ring for the cyclic $p$-group $C_{p}$. This was also the
case for general association schemes of order $p$ in [11].

When we consider the extension of the previous result to the case where $n$ is
divisible by a higher powers of $p$, we need to first find representatives for
the isomorphism classes of $\mathbb{Z}_{p}\mathbf{B}$-lattices between
$\mathbb{Z}_{p}\mathbf{B}$ and the maximal order of
$\mathbb{Q}_{p}\mathbf{B}$.

######  Lemma 3.6.

Let $\mathbf{B}$ be the defining basis of a rank $3$ table algebra of order
$n$ with irrational character table. From Examples 3.3 and 3.4, we know that
$n>1$ is odd, $n$ is not a square, and its character table is determined by
the congruence of $n$ mod $4$.

Let $p$ be an odd prime for which $v_{p}(n)=2m+1$ is odd. Let $v$ be the
integer coprime to $p$ for which $\pm n=p^{2m+1}v$ and $n\equiv\pm 1\mod 4$.

  1. (1)

$\mathbb{Q}_{p}\mathbf{B}\simeq\mathbb{Q}_{p}\oplus\mathbb{Q}_{p}(\sqrt{\pm
n})$, and the maximal order $\Lambda_{0}$ of $\mathbb{Q}_{p}\mathbf{B}$ is
isomorphic to $\mathbb{Z}_{p}\oplus\mathbb{Z}_{p}[\pi]$, where $\pi^{2}=pv$.

  2. (2)

Let $\Lambda=\mathbb{Z}_{p}\mathbf{B}$ and let $\Lambda_{0}$ be the maximal
$\mathbb{Z}_{p}$-order of $\mathbb{Q}_{p}\mathbf{B}$. The $\Lambda$-lattices
satisfying $\Lambda\subseteq M\Lambda_{0}$ are $\mathbb{Z}_{p}$-lattices
$M(r,i,j)=\langle e_{0}+e_{1},re_{0}+p^{i}\pi,p^{j}e_{0}\rangle$, where $0\leq
i\leq m$, $0\leq j\leq 2m+1$, and $0\leq r<p^{2m+1}$ satisfy the conditions
$m+i+1\geq j$ and $m-i\geq j-k$.

  3. (3)

Two of the $\Lambda$-lattices $M(r,i,j)$ and $M(s,i^{\prime},j^{\prime})$ with
$(r,i,j)\neq(s,i^{\prime},j^{\prime})$ in the previous list will be isomorphic
as $\Lambda$-lattices if and only if $i=i^{\prime}$, $j=j^{\prime}$, and one
of the following holds:

$\bullet$ $r,s\not\equiv 0\mod p$,

$\bullet$ $s=0$ and $1\leq 2i+1\leq v_{p}(r)<j$, or

$\bullet$ $1\leq v_{p}(s)=2i+1\leq v_{p}(r)<j$.

###### Proof.

  1. (1)

Since $p$ is odd and
$\mathbb{Q}\mathbf{B}\simeq\mathbb{Q}\oplus\mathbb{Q}(\frac{-1+\sqrt{\pm
n}}{2})$, when we complete the second component with respect to the $p$-adic
valuation its ring of integers will be equal to $\mathbb{Z}_{p}[\sqrt{\pm
n}]$, the sign agreeing with $n\equiv\pm 1\mod 4$. Since $v_{p}(n)=2m+1$,
$\mathbb{Q}_{p}(\sqrt{\pm n})$ will be a ramified quadratic extension of
$\mathbb{Q}_{p}$. Writing $\pm n=p^{2m+1}v$ with $v$ a positive or negative
integer coprime to $p$, we have that the maximal order $\Lambda_{0}$ in
$\mathbb{Q}_{p}\mathbf{B}$ is $\mathbb{Z}_{p}\oplus\mathbb{Z}_{p}[\pi]$, where
$\pi^{2}=pv$.

  2. (2)

Let $e_{0}$ and $e_{1}$ be the primitive idempotents of $\Lambda_{0}$. We fix
the ordered $\mathbb{Z}_{p}$-basis $\\{\pi e_{1},e_{1},e_{0}\\}$ of
$\Lambda_{0}$. Then $\Lambda$ is the $\mathbb{Z}_{p}$-lattice $\langle
p^{m}\pi e_{1},e_{1}+e_{0},p^{2m+1}e_{1}\rangle$. This is in Hermite normal
form with respect to our fixed basis of $\Lambda_{0}$, which we represent with
the matrix $\left(\begin{smallmatrix}p^{m}&0&0\\\ 0&1&1\\\
0&0&p^{2m+1}\end{smallmatrix}\right).$

The $\Lambda$-lattices $M$ such that $\Lambda\subseteq M\subseteq\Lambda_{0}$
will correspond to the matrix forms given by
$\left(\begin{smallmatrix}p^{i}&0&r\\\ 0&1&1\\\
0&0&p^{j}\end{smallmatrix}\right),$ where $0\leq i\leq m$ and $0\leq j\leq
2m+1$, and $0\leq r<p^{2m+1}$, that satisfy the arithmetic conditions imposed
by the requirements $\Lambda\subseteq M$ and $\Lambda M\subseteq M$.

Let $M(r,i,j)$ be the $\mathbb{Z}_{p}$-lattice whose Hermite normal form is
determined by the parameters $r$, $i$, and $j$. If $r=0$, it is clear that
$\Lambda\subseteq M(0,i,j)$. Suppose $r\neq 0$, and write $r=p^{k}r_{0}$ with
$(r_{0},p)=1$ and $0\leq k<j$. It is easy to see that $\Lambda\subseteq
M(r,i,j)$ will hold if and only if $p^{m}\pi e_{1}\in M$, in which case

$\begin{array}[]{rcl}p^{m}\pi e_{1}&=&p^{m-i}(p^{i}\pi
e_{1}+r_{0}p^{k}e_{0})-p^{m-i+k-j}r_{0}(p^{j}e_{0}),\end{array}$

for $p^{m-i+k-j}r_{0}\in\mathbb{Z}_{p}$, so it must be the case that $m+k\geq
i+j$.

Suppose that either $r=0$, or $r=p^{k}r_{0}$ with $(r_{0},p)=1$, $0\leq k<j$,
and $m+k\geq i+j$. To show $\Lambda M(r,i,j)\subseteq M(r,i,j)$, it suffices
to show $M(r,i,j)$ is closed under multiplication by the Hermite normal form
basis elements of $\Lambda$. Since it is clear that $\Lambda$ contains
$p^{2m+1}\Lambda_{0}$ and $\Lambda\subseteq M(r,i,j)$, we only need to check
that $M(r,i,j)$ is closed under multiplication by $p^{m}\pi e_{1}$, and this
comes down to $p^{m}\pi e_{1}(re_{0}+p^{i}\pi e_{1})=p^{m+i+1}ve_{1}\in
M(r,i,j)$. So there needs to be a $\mathbb{Z}_{p}$-solution to

$p^{m+i+1}ve_{1}=a(e_{0}+e_{1})+b(p^{k}r_{0}e_{0}+p^{i}\pi
e_{1})+c(p^{j}e_{0})=(a+bp^{k}r_{0}+cp^{j})e_{0}+(a+bp^{i}\pi)e_{1}.$

Solving this we get $a=p^{m+i+1}v$, $b=0$, and $c=p^{m+i+1-j}v$. This is a
$\mathbb{Z}_{p}$-solution if and only if $m+i+1\geq j$. Note that this
condition does not depend on whether or not $r=0$. So $M(r,i,j)$ is a
$\Lambda$-lattice with $\Lambda\subseteq M(r,i,j)\subseteq\Lambda_{0}$ when
$m+i+1\geq j$, or when $r\neq 0$, $v_{p}(r)=k$, and $m+k\geq i+j$.

  3. (3)

Now we consider the $\Lambda$-lattice isomorphism classes of the $M(r,i,j)$
satisfying the above conditions using condition (iii) of Lemma 3.1. In order
for two of these lattices $M=M(s,i^{\prime},j^{\prime})$ and $N=M(r,i,j)$ to
be isomorphic, there must be a unit $u\in N\cap\Lambda_{0}$ for which
$u^{-1}\in M\cap\Lambda_{0}$ and $uM=N$. Since multiplication by a unit of
$\Lambda_{0}$ preserves the index of $M$ in $\Lambda_{0}$, we must have that
$M$ and $N$ share the same $i$ and $j$ parameters. Since $j=0$ forces $r=s=0$,
when $M\neq N$ we must have that $j>0$. For
$u=\alpha(e_{0}+e_{1})+\beta(re_{0}+p^{i}\pi e_{1})+\gamma(p^{j}e_{0})\in N$
to be a unit of $\Lambda_{0}$, we must have that $\alpha\not\equiv 0\mod p$
and $(\alpha+\beta r+\gamma p^{j})\not\equiv 0\mod p$. Since scaling by units
of $\mathbb{Z}_{p}$ preserves $M$ and $N$, we can assume $\alpha=1$, and our
condition for $u$ to be a unit becomes $(1+\beta r)\not\equiv 0\mod p$.

Next, we determine the conditions on $\beta$ and $\gamma$ in order for
$u^{-1}\in M=M(s,i,j)$. We have that $u^{-1}=(1+\beta r+\gamma
p^{j})^{-1}e_{0}+(1+\beta p^{i}\pi)^{-1}e_{1}$, so the inverse of its second
component is $(1+\beta p^{i}\pi)^{-1}=1-\beta
p^{i}\pi+\sum_{\ell>1}(-1)^{\ell}\beta^{\ell}(p^{i}\pi)^{\ell}$. Therefore,
for $u^{-1}\in M$ we must be able to find
$\delta,\epsilon,\phi\in\mathbb{Z}_{p}$ for which

$u^{-1}=\delta(e_{0}+e_{1})+\epsilon(se_{0}+p^{i}\pi
e_{1})+\phi(p^{j}e_{0})=(\delta+\epsilon s+\phi p^{j})e_{0}+(\delta+\epsilon
p^{i}\pi)e_{1}.$

Now,

$\begin{array}[]{rcl}(\delta+\epsilon p^{i}\pi)&=&(1+\beta p^{i}\pi)^{-1}\\\
&=&1-\beta p^{i}\pi+\beta^{2}p^{2i}\pi^{2}-\beta^{3}p^{3i}\pi^{3}+\dots\\\
&=&(1+\beta^{2}p^{2i}(pv)+\beta^{4}p^{4i}(pv)^{2}+\dots)-\beta
p^{i}\pi(1+\beta^{2}p^{2i}(pv)+\beta^{4}p^{4i}(pv)^{2}+\dots),\end{array}$

so $(1+\beta p^{i}\pi)^{-1}=\delta-\beta\delta p^{i}\pi$ with
$\delta=\sum_{\ell\geq 0}\beta^{2\ell}p^{2i\ell}(pv)^{\ell}=(1-\beta
p^{2i+1}v)^{-1}\in\mathbb{Z}_{p}$. We also have $\epsilon=-\beta\delta$. We
must also have $(\delta-\beta\delta s+\phi p^{j})=(1+\beta r+\gamma
p^{j})^{-1}$. Let $\alpha=(1+\beta r)^{-1}$. Then

$(1+\beta r+\gamma p^{j})^{-1}=\alpha(1-\gamma\alpha
p^{j}+\gamma^{2}\alpha^{2}p^{2j}-\gamma^{3}\alpha^{3}p^{3j}+\dots),$

so we must have $(\delta-\beta\delta s)\equiv\alpha\mod p^{j}$, which is
equivalent to $(1-\beta^{2}p^{2i+1}v)\equiv(1-\beta s)(1+\beta r)\mod p^{j}$.
This reduces to $\beta^{2}(rs-p^{2i+1}v)\equiv\beta(r-s)\mod p^{j}$. Once we
have found a $\beta\in\mathbb{Z}_{p}$ satisfying this congruence the existence
of a suitable $\psi\in\mathbb{Z}_{p}$ for which
$u^{-1}=(e_{0}+e_{1})-\beta\delta(se_{0}+p^{i}\pi e_{1})+\phi p^{j}e_{0}$ will
be automatic.

Finally, consider the condition $uM=N$. Since $u(p^{j}e_{0})=(1+\beta
r)p^{j}e_{0}$, we have that $uM=\langle e_{0}+e_{1},u(se_{0}+p^{i}\pi
e_{1}),p^{j}e_{0}\rangle$, and so showing $uM=N$ reduces to showing
$u(se_{0}+p^{i}\pi e_{1})\in N$. Now,

$\begin{array}[]{rcl}u(se_{0}+p^{i}\pi)&=&(se_{0}+p^{i}\pi e_{1})((1+\beta
r+\gamma p^{j})e_{0}+(1+\beta p^{i}\pi)e_{1}\\\ &=&(s+\beta sr+s\gamma
p^{j})e_{0}+(\beta p^{2i+1}v+p^{i}\pi)e_{1}\\\ &=&\beta
p^{2i+1}(e_{0}+e_{1})+(re_{0}+p^{i}\pi)+(s-r+\beta rs-\beta p^{2i+1}v+s\gamma
p^{j})e_{0},\end{array}$

so this will lie in $N$ if and only if $\beta$ is a solution to
$\beta(sr-p^{2i+1}v)\equiv r-s\mod p^{j}$. Note that this condition implies
the condition needed for $u^{-1}\in M$.

The existence of a solution to $\beta(sr-p^{2i+1}v)\equiv(r-s)\mod p^{j}$
depends on $j,i,r$, and $s$. If $s=0$ and $r\neq 0$ with $v_{p}(r)=k$, then
the congruence has a solution if and only if $j>k\geq 2i+1\geq 1$. When
$r,s\neq 0$, we can assume $j>v_{p}(r)=k\geq v_{p}(s)=\ell\geq 0$. In the case
when $2i+1\geq j$, the congruence reduces to $\beta(rs)\equiv(r-s)\mod p^{j}$,
and this will have a solution only in the case $k=\ell=0$. In the case
$2i+1<j$, a solution exists if and only if $k=\ell=0$ or $1\leq\ell=2i+1\leq
k<j$.

∎

We can now compute the $p$-local zeta function of a rank $3$ integral table
algebra when the order $n$ has $v_{p}(n)=3$ for an odd prime $p$.

######  Theorem 3.7.

Let $\mathbf{B}$ be the standard basis of a rank $3$ table algebra order
$n=p^{3}v$, $(p,v)=1$, that has an irrational character table.

Then
$\zeta_{\mathbb{Z}_{p}\mathbf{B}}(s)=(p^{4-8s}-p^{3-7s}+p^{3-6s}+p^{3-5s}-p^{2-5s}+p^{2-3s}-p^{1-3s}+p^{1-2s}-p^{-s}+1)\zeta_{\mathbb{Z}_{p}}(s)^{2}$.

###### Proof.

Let $\Lambda=\mathbb{Z}_{p}\mathbf{B}$. The maximal order of
$\mathbb{Q}_{p}\mathbf{B}$ is $\Lambda_{0}\cong R\oplus\mathbb{Z}_{p}$, where
$R=\mathbb{Z}_{p}[\pi]$ with residue class field $R/\pi R\cong\mathbb{F}_{p}$.
We denote the $i$-th higher unit group in $R$ by $U_{R}^{(i)}$ and the $j$-th
higher unit group in $\mathbb{Z}_{p}$ by $U_{\mathbb{Z}_{p}}^{(j)}$.

Since we are in the $m=1$ case of Lemma 3.6, a complete list of
representatives of the isomorphism classes of $\Lambda$-lattices $M(r,i,j)$
that lie strictly between $\Lambda=\mathbb{Z}_{p}\mathbf{B}=M(0,1,3)$ and
$\Lambda_{0}=M(0,0,0)$ is: $M(0,1,2)$, $M(0,1,1)$, $M(0,0,2)$, $M(0,1,0)$,
$M(0,0,1)$, and $M(1,0,1)$. The indices of $\Lambda$ in these
$\Lambda$-lattices are: $1$ for $\Lambda$, $p$ for $M(0,1,2)$, $p^{2}$ for
$M(0,1,1)$ and $M(0,0,2)$, $p^{3}$ for $M(0,1,0)$, $M(0,0,1)$, and $M(1,0,1)$,
and $p^{4}$ for $\Lambda_{0}$.

To compute $\mu(\operatorname{Aut}M)^{-1}$ for each of the lattices above, we
note all of them except $M(0,0,2)$ and $M(1,0,1)$ are orders. In these two
cases, we calculate that $\\{M(0,0,2):M(0,0,2)\\}=M(0,1,2)$ and
$\\{M(1,0,1):M(1,0,1)\\}=M(0,1,1)$. We can compute the indices of the groups
of units of their automorphism groups in $\Lambda_{0}^{\times}$ by tracking
the unit groups of their simple components $Me_{0}$ and $Me_{1}$ and
congruence conditions that exist between these components. Doing so gives the
following indices of the automorphism groups: $p^{3}(p-1)$ for $\Lambda$,
$p^{2}(p-1)$ for $M(0,1,2)$ and $M(0,0,2)$, $p(p-1)$ for $M(0,1,1)$ and
$M(1,0,1)$, $p$ for $M(0,1,0)$, $(p-1)$ for $M(0,0,1)$, and $1$ for
$\Lambda_{0}$.

For each $M(r,i,j)$ found above, we compute the lattices
$\\{M(r,i,j):\Lambda\\}$ in Table 1,

Table 1. Complementary lattices $M(r,i,j)$ | $\\{M(r,i,j):\Lambda\\}$
---|---
$M(0,0,0)$ | $\langle p^{2}\pi e_{1},p^{3}e_{1},p^{3}e_{0}\rangle$
$M(0,1,0)$ | $\langle p\pi e_{1},p^{3}e_{1},p^{3}e_{0}\rangle$
$M(0,0,1)$ | $\langle p^{2}\pi e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle$
$M(1,0,1)$ | $\langle p\pi e_{1}+p^{2}ve_{1}+p^{2}ve_{0},p^{3}e_{1},p^{3}e_{0}\rangle$
$M(0,1,1)$ | $\langle p\pi e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle$
$M(0,0,2)$ | $\langle p^{2}\pi e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle$
$M(0,1,2)$ | $\langle p\pi e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle$
$M(0,1,3)$ | $\langle p\pi e_{1},e_{1}+e_{0},p^{3}e_{0}\rangle=\mathbb{Z}_{p}\mathbf{B}$

Next, we calculate the genus zeta function for each $\Lambda$-lattice in the
table. In the first case $Z(M(0,0,0),\Lambda;s)$, we notice that $\langle
p^{2}\pi e_{1},p^{3}e_{1},p^{3}e_{0}\rangle\cong\pi^{5}R\oplus
p^{3}\mathbb{Z}_{p}$, so by Lemma 3.2, we have

(4) $\int_{A^{\times}\cap(\pi^{5}R\oplus
p^{3}\mathbb{Z}_{p})}\left\|x\right\|_{A}^{s}d^{\times}x=\int_{\pi^{5}R-\\{0\\}}\left\|a\right\|_{\mathbb{Q}_{p}[\pi]}^{s}d^{\times}a\int_{p^{3}\mathbb{Z}_{p}-\\{0\\}}\left\|b\right\|_{\mathbb{Q}_{p}}^{s}d^{\times}b=p^{-8s}\zeta^{2},$

where $\zeta=(1-p^{-s})^{-1}$. Therefore,

(5) $Z(M(0,0,0),\Lambda;s)=p^{-4s}\zeta^{2}.$

For the remaining lattices, we examine a general element, and record in tables
the possible disjoint subsets such an element could belong to, and the value
of the integral corresponding to each subset as it follows from Lemma 3.2. To
illustrate this, consider $M(0,1,0)$. An element of
$\\{M(0,1,0):\Lambda\\}=\langle p\pi
e_{1},p^{3}e_{1},p^{3}e_{0}\rangle\subseteq R\oplus\mathbb{Z}_{p}$ is of the
form $(\pi^{3}(a_{0}+\pi^{2}R),p^{3}\mathbb{Z}_{p})$, where
$a_{0}\in\mathbb{F}_{p}$. We have two choices: either $a_{0}=0$ or
$a\in\mathbb{F}_{p}^{\times}$. When $a_{0}=0$, we get the subset
$\pi^{5}R\oplus p^{3}\mathbb{Z}_{p}$, whose integral we calculated in Equation
4. When $a_{0}=u\in\mathbb{F}_{p}^{\times}$, we obtain
$(\pi^{3}(u+\pi^{2}R),p^{3}\mathbb{Z}_{p})=u\pi^{3}U_{R}^{(2)}\oplus
p^{3}\mathbb{Z}_{p}$, which is a multiplicative translate of
$\pi^{3}U_{R}^{(2)}\oplus p^{3}\mathbb{Z}_{p}$ by
$(u,1)\in\Lambda_{0}^{\times}$, and therefore by Lemma 3.2, gives the integral
$\frac{p^{-6s}}{p(p-1)}\zeta$. Since there are $(p-1)$ choices for
$u\in\mathbb{F}_{p}^{\times}$, the value of the integral over the disjoint
union
$\displaystyle\dot{\bigcup}_{u\in\mathbb{F}_{p}^{\times}}u\pi^{3}U_{R}^{(2)}\oplus
p^{3}\mathbb{Z}_{p}$ is $\frac{p^{-6s}}{p}\zeta$. This data is summarized in
Table 2.

Table 2. Integrals for $M(0,1,0)$ $a_{0}$ | Subset | Value of the integral
---|---|---
$0$ | $A^{\times}\cap(\pi^{5}R\oplus p^{3}\mathbb{Z}_{p})$ | Equation 4
$u\in\mathbb{F}_{p}^{\times}$ |  $(p-1)$ multiplicative translates of $\pi^{3}U_{R}^{(2)}\oplus p^{3}\mathbb{Z}_{p}$ | $\frac{p^{-6s}}{p}\zeta$

Adding the two integrals, we get

(6) $\displaystyle\int_{A^{\times}\cap\langle p\pi
e_{1},p^{3}e_{1},p^{3}e_{0}\rangle}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=p^{-8s}\zeta^{2}+\frac{p^{-6s}}{p}\zeta=\frac{1}{p}(p^{1-8s}-p^{-7s}+p^{-6s})\zeta^{2},$

and

(7) $Z(M(0,1,0),\Lambda;s)=p\cdot
p^{3s}\frac{1}{p}(p^{1-8s}-p^{-7s}+p^{-6s})\zeta^{2}=(p^{1-5s}-p^{-4s}+p^{-3s})\zeta^{2}.$

For $M(0,0,1)$, an element of $\langle p^{2}\pi
e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle$ belongs to
$(\pi^{4}(a_{0}v^{-2}+\pi R),p^{2}(a_{0}+p\mathbb{Z}_{p}))$, with
$a_{0}\in\mathbb{F}_{p}$. The disjoint subsets and corresponding integrals are
recorded in Table 3.

Table 3. Integrals for $M(0,0,1)$ $a_{0}$ | Subset | Value of the integral
---|---|---
$0$ | $A^{\times}\cap(\pi^{5}R\oplus p^{3}\mathbb{Z}_{p})$ | Equation 4
$u\in\mathbb{F}_{p}^{\times}$ |  $(p-1)$ multiplicative translates of $\pi^{4}U_{R}^{(1)}\oplus p^{2}U_{\mathbb{Z}_{p}}^{(1)}$ | $\frac{p^{-6s}}{p-1}$

Adding the integrals, we obtain

(8) $\displaystyle\int_{A^{\times}\cap\langle p^{2}\pi
e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=p^{-8s}\zeta^{2}+\frac{p^{-6s}}{p-1}=\frac{1}{p-1}(p^{1-8s}-2p^{-7s}+p^{-6s})\zeta^{2}.$

and

(9)
$Z(M(0,0,1),\Lambda;s)=p^{3s}(p-1)\frac{1}{p-1}(p^{1-8s}-2p^{-7s}+p^{-6s})\zeta^{2}=(p^{1-5s}-2p^{-4s}+p^{-3s})\zeta^{2}.$

In the case of $M(1,0,1)$, an element of $\langle p\pi
e_{1}+p^{2}ve_{1}+p^{2}ve_{0},p^{3}e_{1},p^{3}e_{0}\rangle$ belongs to
$(\pi^{3}(a_{0}v^{-1}+a_{0}v^{-1}\pi+\pi^{2}R),p^{2}(a_{0}v+p\mathbb{Z}_{p}))$,
$a_{0}\in\mathbb{F}_{p}$. This case is recorded in Table 4.

Table 4. Integrals for $M(1,0,1)$ $a_{0}$ | Subset | Value of the integral
---|---|---
$0$ | $A^{\times}\cap(\pi^{5}R\oplus p^{3}\mathbb{Z}_{p})$ | Equation 4
$u\in\mathbb{F}_{p}^{\times}$ |  $(p-1)$ multiplicative translates of $\pi^{3}U_{R}^{(2)}\oplus p^{2}U_{\mathbb{Z}_{p}}^{(1)}$ | $\frac{p^{-5s}}{p(p-1)}$

This gives

$\displaystyle\int_{A^{\times}\cap\langle p\pi
e_{1}+p^{2}ve_{1}+p^{2}ve_{0},p^{3}e_{1},p^{3}e_{0}\rangle}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=\frac{1}{p(p-1)}(p^{2-8s}-p^{1-8s}+p^{-7s}-2p^{-6s}+p^{-5s})\zeta^{2},$

and

(10)
$Z(M(1,0,1),\Lambda;s)=(p^{2-5s}-p^{1-5s}+p^{-4s}-2p^{-3s}+p^{-2s})\zeta^{2}.$

.

For $M(0,1,1)$, an element of $\langle p\pi
e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle$ belongs to
$(\pi^{3}(a_{0}v^{-1}+a_{1}v^{-2}\pi+\pi^{2}R),p^{2}(a_{1}+p\mathbb{Z}_{p}))$,
$a_{0},a_{1}\in\mathbb{F}_{p}$. The choices for $a_{0},a_{1}$ are recorded in
Table 5.

Table 5. Integrals for $M(0,1,1)$ $(a_{0},a_{1})$ | Subset | Value of the integral
---|---|---
$(0,0)$ or $(u_{1},0)$ | $A^{\times}\cap\langle p\pi e_{1},p^{3}e_{1},p^{3}e_{0}\rangle$ | Equation 6
$(0,u_{2})$ |  $(p-1)$ translates of $\pi^{4}U_{R}^{(1)}\oplus p^{2}U_{\mathbb{Z}_{p}}^{(1)}$ | $\frac{p^{-6s}}{p-1}$
$(u_{1},u_{2})\in(\mathbb{F}_{p}^{\times})^{2}$ |  $(p-1)^{2}$ translates of $\pi^{3}U_{R}^{(2)}\oplus p^{2}U_{\mathbb{Z}_{p}}^{(1)}$ | $\frac{p^{-5s}}{p}$

Adding up the integrals, we obtain

(11) $\displaystyle\int_{A^{\times}\cap\langle p\pi
e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=\frac{1}{p(p-1)}(p^{2-8s}-2p^{1-7s}+p^{-6s}+p^{1-5s}-p^{-5s})\zeta^{2},$

and

(12)
$Z(M(0,1,1),\Lambda;s)=(p^{2-6s}-2p^{1-5s}+p^{-4s}+p^{1-3s}-p^{-3s})\zeta^{2}.$

.

For $M(0,0,2)$, an element of $\langle p^{2}\pi
e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle$ belongs to
$(\pi^{2}(a_{0}v^{-1}+a_{1}v^{-2}\pi^{2}+\pi^{3}R),p(a_{0}+a_{1}p+p^{2}\mathbb{Z}_{p}))$,
$a_{0},a_{1}\in\mathbb{F}_{p}$. We have the following choices, outlined in
Table 6.

Table 6. Integrals for $M(0,0,2)$ $(a_{0},a_{1})$ | Subset | Value of the integral
---|---|---
$(0,0)$ or $(0,u_{2})$ | $A^{\times}\cap\langle p^{2}\pi e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle$ | Equation 8
$(u_{1},0)$ or $(u_{1},u_{2})$ |  $p(p-1)$ translates of $\pi^{2}U_{R}^{(3)}\oplus pU_{\mathbb{Z}_{p}}^{(2)}$ | $\frac{p^{-3s}}{p^{2}(p-1)}$

Adding them up, we obtain

$\displaystyle\int_{A^{\times}\cap\langle p^{2}\pi
e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=\frac{1}{p^{2}(p-1)}(p^{3-8s}-2p^{2-7s}+p^{2-6s}+p^{-5s}-2p^{-4s}+p^{-3s})\zeta^{2},$

and

(13)
$Z(M(0,0,2),\Lambda;s)=(p^{3-6s}-2p^{2-5s}+p^{2-4s}+p^{-3s}-2p^{-2s}+p^{-s})\zeta^{2}.$

.

For $M(0,1,2)$, an element of $\langle p\pi
e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle$ belongs to
$(\pi^{2}(a_{0}v^{-1}+a_{1}v^{-1}\pi+a_{2}v^{-2}\pi^{2}+\pi^{3}R),p(a_{0}+a_{2}p+p^{2}\mathbb{Z}_{p}))$,
with $a_{0},a_{1},a_{2}\in\mathbb{F}_{p}$. However, based on previously
computed integrals, we only need to consider the choice of $a_{0}$, as
described in Table 7.

Table 7. Integrals for $M(0,1,2)$ $a_{0}$ | Subset | Value of the integral
---|---|---
$0$ | $A^{\times}\cap\langle p\pi e_{1},p^{2}(e_{1}+e_{0}),p^{3}e_{0}\rangle$ | Equation 11
$u\in\mathbb{F}_{p}^{\times}$ |  $p^{2}(p-1)$ multiplicative translates of $\pi^{2}U_{R}^{(3)}\oplus pU_{\mathbb{Z}_{p}}^{(2)}$ | $\frac{p^{-3s}}{p(p-1)}$

Adding them up, we obtain

(14) $\displaystyle\int_{A^{\times}\cap\langle p\pi
e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=\frac{1}{p(p-1)}(p^{2-8s}-2p^{1-7s}+p^{-6s}+p^{1-5s}-2p^{-4s}+p^{-3s})\zeta^{2},$

and

(15)
$Z(M(0,1,2),\Lambda;s)=(p^{3-7s}-2p^{2-6s}+p^{1-5s}+p^{2-4s}-2p^{1-3s}+p^{1-2s})\zeta^{2}.$

.

Finally, for $M(0,1,3)$, an element of $\Lambda=\langle p\pi
e_{1},e_{1}+e_{0},p^{3}e_{0}\rangle$ belongs to
$(a_{0}+a_{1}v^{-1}\pi^{2}+a_{2}v^{-1}\pi^{3}+a_{3}v^{-2}\pi^{4}+\pi^{5}R),a_{0}+a_{1}p+a_{3}p^{2}+p^{3}\mathbb{Z}_{p}))$,
with $a_{0},a_{1},a_{2},a_{3}\in\mathbb{F}_{p}$. Again, we need only consider
the choice of $a_{0}$, recorded in Table 8.

Table 8. Integrals for $M(0,1,3)$ $a_{0}$ | Subset | Value of the integral
---|---|---
$0$ | $A^{\times}\cap\langle p^{2}\pi e_{1},p(e_{1}+e_{0}),p^{3}e_{0}\rangle$ | Equation 14
$u\in\mathbb{F}_{p}^{\times}$ | $\Lambda^{\times}$ | $\frac{1}{p^{3}(p-1)}$

Adding up these integrals, we get

$\displaystyle\int_{A^{\times}\cap\Lambda}\left\|x\right\|_{A}^{s}d^{\times}x$
$\displaystyle=\frac{1}{p^{3}(p-1)}(p^{4-8s}-2p^{3-7s}+p^{2-6s}+p^{3-5s}-2p^{2-4s}+p^{2-3s}+p^{-2s}-2p^{-s}+1)\zeta^{2},$

and

(16)
$Z(M(0,1,3),\Lambda;s)=(p^{4-8s}-2p^{3-7s}+p^{2-6s}+p^{3-5s}-2p^{2-4s}+p^{2-3s}+p^{-2s}-2p^{-s}+1)\zeta^{2}.$

.

Finally, adding Equations 5, 7, 9, 10, 12, 13, 15, and 16, we obtain

(17)
$\zeta(\Lambda;s)=(p^{4-8s}-p^{3-7s}+p^{3-6s}+p^{3-5s}-p^{2-5s}+p^{2-3s}-p^{1-3s}+p^{1-2s}-p^{-s}+1)\zeta^{2}.$

∎

The two cases left to examine are for $v_{p}(n)$ even. If $v$ is a square
modulo $p$, then the algebra
$\mathbb{Q}_{p}\mathbf{B}\cong\mathbb{Q}_{p}\oplus\mathbb{Q}_{p}\oplus\mathbb{Q}_{p}$
splits, and the zeta function of $\mathbb{Z}_{p}\mathbf{B}$ can be examined
using the techniques of [2].

When $v$ is not a square modulo $p$,
$\mathbb{Q}_{p}\mathbf{B}\cong\mathbb{Q}_{p}\oplus K$, where $K$ is an
unramified quadratic extension of $\mathbb{Q}_{p}$. If we let $e_{0}$ and
$e_{1}$ be the primitive idempotents of $\mathbb{Q}_{p}\mathbf{B}$, then the
maximal order of $\mathbb{Q}_{p}\mathbf{B}$ is
$\Lambda_{0}=\langle\sqrt{v}e_{1},e_{1},e_{0}\rangle$, and we can represent
the Hermite normal form of $\mathbb{Z}_{p}\mathbf{B}$ with respect to this
basis by $\left(\begin{smallmatrix}p^{m}&0&0\\\ 0&1&1\\\
0&0&p^{2m}\end{smallmatrix}\right)$. Therefore, the calculation of the local
$p$-zeta function of $\mathbb{Z}_{p}\mathbf{B}$ can be obtained using the same
ideas as Theorem 3.5, Lemma 3.6 and Theorem 3.7.

##  4\. Zeta functions of Small Fusion Rings

The most widely studied examples of fusion rings are those that arise as the
Grothendieck rings of monoidal tensor categories; these are called
categorifiable fusion rings (see [10]). In these representation categories
there are finitely many irreducible representations that play the role of
basis elements. The direct sum of representations is naturally a
representation of the algebra, and the presence of a compatible co-algebra
structure implies that the tensor product of representations also gives a
representation. So the nonnegative integer span of the irreducibles is closed
under the tensor product, and using similarity of representations as an
equality, the integer span of the irreducibles produces a fusion ring with the
irreducible representations as the defining basis. We will consider the zeta
functions for the following categorifiable fusion rings of ranks $2$ and $3$:

(Rank 2:)

the Fibonacci fusion ring: $\mathbf{B}=\\{1,b\\}$, $b^{2}=1+b$; and

the group $C_{2}$;

(Rank 3)

(a) the Ising fusion ring: $\mathbf{B}=\\{1,b,d\\},b^{2}=1,bd=d,d^{2}=1+b$;

(b) $Rep(S_{3})$: $\mathbf{B}=\\{1,b,d\\},b^{2}=1,bd=d,d^{2}=1+b+d$;

(c) $PSU(5)$ level $2$: $\mathbf{B}=\\{1,b,d\\},b^{2}=1+d,bd=b+d,d^{2}=1+b+d$;

(d) the $E_{6}$-subfactor fusion ring:
$\mathbf{B}=\\{1,b,d\\},b^{2}=1,bd=d,d^{2}=1+b+2d$; and

(e) the cyclic group $C_{3}$.

The Fibonacci fusion ring is the only categorifiable one of dimension $2$ that
is not a group. The five types of $3$-dimensional fusion categories
corresponding to pivotal fusion categories were recently classified by Ostrik
in [19] \- the pivotal assumption on these is believed to be unnecessary. For
dimension $4$ and higher complete classification remains open, but a
considerable amount of recent work on special cases has appeared - see [17],
[8], [9], [4], and [5]. A list of small multiplicity-free (i.e. commutative)
fusion rings is under development at
https://anyonwiki.github.io/pages/Lists/losmffr.html [22].

######  Example 4.1.

The Fibonacci fusion ring. The integral basis of the Fibonacci fusion ring is
$\mathbf{B}=\\{1,b\\}$, $b^{2}=1+b$. The largest eigenvalue of $b$ is
$\delta(b)=\frac{1+\sqrt{5}}{2}$, the golden ratio. Since
$\mathbb{Z}\mathbf{B}=\mathbb{Z}[b]\simeq\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$,
and this is the ring of integers of
$\mathbb{Q}\mathbf{B}\simeq\mathbb{Q}(\sqrt{5})$, we have that
$\zeta_{\mathbb{Z}\mathbf{B}}(s)$ is equal to the Dedekind zeta function for
the ring $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$.

######  Example 4.2.

3a. The Ising fusion ring. For the Ising fusion ring, the standard basis will
be $\mathbf{B}^{s}=\\{1,b,\sqrt{2}d\\}$. The standard integral table algebra
$\mathbb{Q}\mathbf{B}^{s}$ agrees with the adjacency algebra of the
association scheme of order $4$ corresponding to the square. The character
table of this standard integral table algebra (with columns corresponding to
elements of $\mathbf{B}^{s}$, rows to the irreducible characters whose
multiplicities appear on the right) is

$P=\begin{bmatrix}1&1&2\\\ 1&1&-2\\\ 1&-1&0\end{bmatrix}\begin{array}[]{l}1\\\
1\\\ 2\end{array}.$

Since the basis element $d\in\mathbf{B}$ has a quadratic irrational Perron-
Frobenius eigenvalue $\sqrt{2}$, the degree character $\delta$ of $\mathbf{B}$
has a Galois conjugate $\delta^{\tau}$, and the third irreducible character
$\phi$ will be rational-valued. So we have that the primitive idempotents of
$\mathbb{C}\mathbf{B}$ are

$e_{\delta}=\frac{1}{4}(1+b+\sqrt{2}d),e_{\delta^{\tau}}=\frac{1}{4}(1+b-\sqrt{2}d),\mbox{
and }e_{\phi}=\frac{1}{2}(1-b).$

The primitive idempotents of $\mathbb{Q}\mathbf{B}$ are $e_{0}=e_{\phi}$ and
$e_{1}=e_{\delta}+e_{\delta^{\tau}}=\frac{1}{2}(1+b)$. Note that $de_{0}=0$.
We have $\mathbb{Q}\mathbf{B}\simeq\mathbb{Q}\oplus\mathbb{Q}(\sqrt{2})$, and
the maximal order is $\Lambda_{0}=\langle de_{1}=d,e_{1},e_{0}\rangle$. When
we write $\mathbf{B}$ in terms of this basis, we get $1=e_{0}+e_{1}$,
$b=e_{1}-e_{0}$, and $d=de_{1}$, which gives $\mathbb{Z}\mathbf{B}=\langle
de_{1},e_{1}+e_{0},2e_{0}\rangle$. So $[\Lambda_{0}:\mathbb{Z}\mathbf{B}]=2$
and we only need to calculate the local zeta function at $p=2$.

When we complete at $p=2$, the maximal order is
$\Lambda_{0,2}\simeq\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}[\sqrt{2}]$ and
$\mathbb{Z}_{2}\mathbf{B}=\\{xde_{1}+ye_{1}+(y+2z)e_{0}:x,y,z\in\mathbb{Z}_{2}\\}.$
Therefore, we can represent $\mathbb{Z}_{2}\mathbf{B}$ as

$\mathbb{Z}_{2}\mathbf{B}\cong\\{(a,b)\in\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}[\sqrt{2}]:b\pmod{2}\equiv
a\pmod{\sqrt{2}}\\}.$

Since $\sqrt{2}$ is a uniformizer in the local ring $\mathbb{Z}_{2}[\sqrt{2}]$
with residue class field $\mathbb{F}_{2}$, the same local calculations as in
[6, Section 3.4] and Theorem 3.5 apply, and

$\zeta_{\mathbb{Z}_{2}\mathbf{B}}(s)=(2^{1-2s}-2^{-s}+1)\zeta_{\mathbb{Z}_{2}}(s)^{2}.$

Therefore,

$\zeta_{\mathbb{Z}\mathbf{B}}(s)=(2^{1-2s}-2^{-s}+1)\zeta_{\mathbb{Z}}(s)\zeta_{\mathbb{Z}[\sqrt{2}]}(s).$

3b. $Rep(S_{3})$: This time the standard basis will be
$\mathbf{B}^{s}=\\{1,b,2d\\}$. This the standard basis of an integral table
algebra of order $6$ whose character table is

$P=\begin{bmatrix}1&1&4\\\ 1&1&-2\\\ 1&-1&0\end{bmatrix}\begin{array}[]{l}1\\\
2\\\ 3\end{array}.$

This is the character table of the imprimitive association scheme
corresponding to $K_{2}^{(3)}$ and its multipartite complement $K_{3\times
2}$; it is dual to the conjugacy class scheme of $S_{3}$. The primitive
idempotents of $\mathbb{Q}\mathbf{B}$ are:

$e_{\delta}=\frac{1}{6}(1+b+2d),e_{\psi}=\frac{1}{3}(1+b-d),\mbox{ and
}e_{\phi}=\frac{1}{2}(1-b),$

so $b=e_{\delta}+e_{\psi}-e_{\phi}$ and $d=2e_{\delta}-e_{\psi}$. Simplifying
the basis $\\{1,b,d\\}$, it follows that $\mathbb{Z}\mathbf{B}=\mathbb{Z}\cdot
1+2\mathbb{Z}(e_{\delta}+e_{\psi})+3\mathbb{Z}e_{\psi}$, so it has nontrivial
completions at the primes $p=2$ and $p=3$. The $2$\- and $3$\- local zeta
functions of this order match those of of
$\mathbb{Z}_{p}K_{p}\oplus\mathbb{Z}_{p}$, $p=2,3$, where $K_{p}$ is the
complete graph association scheme on $p$ vertices. The zeta function of
$\mathbb{Z}_{p}K_{p}$ has been worked out for all primes $p$ in [11] and [2].
From the formula in [2] we get

$\zeta_{\mathbb{Z}\mathbf{B}}(s)=(2^{1-2s}-2^{-s}+1)(3^{1-2s}-3^{-s}+1)\zeta_{\mathbb{Z}}(s)^{3}.$

3c. $PSU(5)$ level $2$: Note that $b=d^{2}-d-1$. This implies
$\mathbb{Z}\mathbf{B}=\mathbb{Z}[d]$, and $d$ has minimal polynomial
$x^{3}-2x^{2}-x+1$. The roots of this polynomial are the Galois conjugates of
$\alpha=-\zeta_{7}-\zeta_{7}^{2}-\zeta_{7}^{5}-\zeta_{7}^{6}$. As this element
of the ring of integers of the totally real cubic subfield
$\mathbb{Q}(\zeta_{7})^{+}$ of $\mathbb{Q}(\zeta_{7})$ has norm $1$,
$\mathbb{Z}[\alpha]$ is equal to the ring of integers of
$\mathbb{Q}(\zeta_{7})^{+}$. Therefore, $\zeta_{\mathbb{Z}\mathbf{B}}(s)$ will
be equal to the Dedekind zeta function of $\mathbb{Z}[\alpha]$.

3d. The rank $3$ $E_{6}$ subfactor fusion ring: This time $b=d^{2}-2d-1$, so
$\mathbb{Z}\mathbf{B}=\mathbb{Z}[d]$. Since $\delta(b)=1$, we have
$\delta(d)^{2}=2+2\delta(d)$, which means $\delta(d)=1+\sqrt{3}$. The minimal
polynomial of $d$ is $x^{3}-2x^{2}-2x$, so
$\mathbb{Q}\mathbf{B}\simeq\mathbb{Q}\oplus\mathbb{Q}[1+\sqrt{3}]$. The
primitive idempotents of $\mathbb{Q}\mathbf{B}$ are $e_{0}=\frac{1}{2}(1-b)$
and $e_{1}=\frac{1}{2}(1+b)$, and the maximal order is $\Lambda=\langle
de_{1}=d,e_{1},e_{0}\rangle$. Since $1=e_{0}+e_{1}$, $b=e_{1}-e_{0}$, and
$d=de_{0}$, we get $\mathbb{Z}\mathbf{B}=\langle
de_{1},e_{1}+e_{0},2e_{0}\rangle$. So we only need the local zeta function at
$p=2$. The completion at $p=2$ of the maximal order of
$\mathbb{Q}_{2}\mathbf{B}$ is
$\Lambda_{0}=\mathbb{Z}_{2}e_{0}\oplus\mathbb{Z}_{2}\mathbf{B}e_{1}\simeq\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}[1+\sqrt{3}]$,
and $\pi=1+\sqrt{3}$ is a uniformizer of $\mathbb{Z}_{2}[1+\sqrt{3}]$ of norm
$-2$. Then $\mathbb{Z}_{2}\mathbf{B}$ corresponds to
$\\{(\alpha,x+y\pi)\in\Lambda_{0}:a\mod 2\equiv x+y\pi\mod\pi\\}$, and this
condition is equivalent to $x\equiv\alpha\mod 2$, $y\in\mathbb{Z}_{2}$. Again
we have just two genera of $\mathbb{Z}_{2}\mathbf{B}$-lattices, which are
represented by $\Lambda_{0}$ and $\mathbb{Z}_{2}\mathbf{B}$. The calculation
of the local zeta function in this case matches that of the Ising fusion ring
in case 3a, so in the end we will get:

$\zeta_{\mathbb{Z}\mathbf{B}}(s)=(2^{1-2s}-2^{-s}+1)\zeta_{\mathbb{Z}}(s)\zeta_{\mathbb{Z}[\sqrt{3}]}(s).$

3e. The cyclic group $C_{3}$: Solomon computed $\zeta_{\mathbb{Z}C_{p}}(s)$
for $p$ prime in [20], the result matches that of the asymmetric rank $3$
table algebra discussed in the previous section, in the case where the order
is $3$:

$\zeta_{\mathbb{Z}C_{3}}(s)=(3^{1-2s}-3^{-s}+1)\zeta_{\mathbb{Z}}(s)\zeta_{\mathbb{Z}[\sqrt{-3}]}(s).$

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