# An approach to $\bm{p}$-adic qubits from irreducible representations of
$\bm{SO(3)_{p}}$

Ilaria Svampa  0000-0002-1389-0319<EMAIL_ADDRESS>School of Science
and Technology, University of Camerino, Via Madonna delle Carceri 9, I-62032
Camerino, Italy  INFN–Sezione Perugia, Via A. Pascoli, I-06123 Perugia, Italy
Departament de Física: Grup d’Informació Quàntica, Universitat Autònoma de
Barcelona, ES-08193 Bellaterra (Barcelona), Spain  Stefano Mancini
0000-0002-3797-3987<EMAIL_ADDRESS>School of Science and
Technology, University of Camerino, Via Madonna delle Carceri 9, I-62032
Camerino, Italy  INFN–Sezione Perugia, Via A. Pascoli, I-06123 Perugia, Italy
Andreas Winter  0000-0001-6344-4870<EMAIL_ADDRESS>Departament de
Física: Grup d’Informació Quàntica, Universitat Autònoma de Barcelona,
ES-08193 Bellaterra (Barcelona), Spain  ICREA—Institució Catalana de la
Recerca i Estudis Avançats, Pg. Lluís Companys, 23, ES-08001 Barcelona, Spain

(13 July 2022)

###### Abstract

We introduce the notion of $p$-adic quantum bit ($p$-qubit) in the context of
the $p$-adic quantum mechanics initiated and developed after the seminal paper
of Volovich [Theor. Math. Phys. 71, 574 (1987)]. In this approach, physics
takes place in three-dimensional $p$-adic space rather than Euclidean space.
Based on our prior work describing the $p$-adic special orthogonal group [Di
Martino et al., arXiv:2104.06228 [math.NT] (2021)], we outline a program to
classify its continuous unitary projective representations, which can be
interpreted as a theory of $p$-adic angular momentum. The $p$-adic quantum bit
arises from the irreducible representations of minimal nontrivial dimension
two, of which we construct examples for all primes $p$.

##  I Introduction

A qubit at its most basic is the state space of a two-level quantum system,
i.e., one with two-dimensional complex Hilbert space ${\cal H}$. This
abstracts from all physical characteristics such a system might have, be it an
electron spin, or that of a neutron, the polarization of a photon, a quantum
dot, etc. Its characteristic is that the symmetry group $SU(2)$ acts
transitively on all pure states

${\cal P}({\cal
H})=\\{\rho=|\psi\rangle\\!\langle\psi|\,\mathrel{\mathop{\mathchar
58\relax}}\,\rho\geq 0,\operatorname{Tr}\rho=1,\,\rho\text{ rank-one}\\},$

which geometrically is the surface of the Bloch sphere, ${{\mathbb{R}}}^{3}$,
and the conjugation action on the density matrices is equivalent to the action
of $SO(3)$ on the Bloch sphere. Furthermore, for a spin-$\frac{1}{2}$
particle, this action coincides with the one induced by the spatial rotations,
and as an irreducible representation (irrep) gives rise to all other irreps by
tensor powers $U^{\otimes n}$ and their Clebsch-Gordan decomposition into
irreps. This has been long noticed and has even be taken as a possible
foundation of quantum mechanics Weizsaecker .

However, this link of the most basic quantum system to the geometry and,
indeed, symmetry of physical space Klein  opens up other possibilities. For
example, why should we accept a priori that physical space is described by the
Euclidean ${{\mathbb{R}}}^{3}$? Indeed, the real field is not the only
possible metric completion of the rationals: by Ostrowski’s Theorem Serre ,
there is a countable family of other completions, those of the $p$-adic
numbers ${{\mathbb{Q}}}_{p}$. The fractal geometry of the $p$-adic numbers
made them interesting for attempts at “new” quantum physics with a space-time
that is decidedly non-Euclidean at very short or very large distances, as
seems possible in quantum gravity Vol1st ; volovich10 . In these approaches,
the focus is on the translation symmetry of phase space, which in ordinary
quantum mechanics leads to the framework of the Heisenberg-Weyl group of
position and momentum displacements, parametrized by real numbers, but which
here are displacements by $p$-adic amounts volovich89 .

Here, we consider the rotation symmetry of $\mathbb{Q}^{3}_{p}$, encoded in
the special orthogonal group $SO(3)_{p}$ in dimension three. Recently, we have
characterized the geometry of this group our1st  based on the crucial
observation that, up to equivalence, there is a unique quadratic form on
${{\mathbb{Q}}}_{p}^{3}$ with a compact symmetry group. Motivated by the
analogy with the real case, our program is to find its unitary complex
projective representations, which should constitute a $p$-adic theory of
quantum angular momentum. In particular, the irreps of dimension two are
crucial as $p$-adic analogs of spin-$\frac{1}{2}$ particles, which we
interpret as $p$-adic qubits. In this spirit, we call any pair $({\cal
H},\psi)$, where ${\cal H}$ is a two-dimensional Hilbert space and $\psi$ is a
projective irrep on ${\cal H}$, a qubit. In the present paper, we do not carry
out this entire program, leaving the bulk of it for future work, but contend
ourselves with the construction of the simplest possible two-dimensional
irreps of $SO(3)_{p}$ for every prime $p$. We emphasize that the angular
momentum representations that we are looking for map the $p$-adic group
$SO(3)_{p}$ to groups of linear transformations on complex vector spaces.
Taking this route, moving from $p$-adic numbers to complex numbers, is
different and alternative to the ones that can be considered more
conventional, such as (i) defining $p$-adic Hilbert spaces and working on them
Hilbert1 ; Hilbert2 ; Hilbert3 ; Hilbert4 ; Hilbert5  and (ii) defining
$p$-adic Lie groups Lie .

The structure of the rest of the paper is as follows. In Sec. II, we recall
the basic definitions and facts about the three-dimensional special orthogonal
groups $SO(3)_{p}$, among them the crucial insight that the coefficients of
special orthogonal matrices are $p$-adic integers. In Sec. III, we exploit the
latter property to define a hierarchy of projected groups by reducing the
matrix entries modulo $p^{k}$, $k\geq 1$; each of them is a finite group,
offering an easy way to constructing irreps. In Sec. IV, we turn our
attention, in particular, to the smallest reduction, $SO(3)_{p}\mod p$, for
odd primes $p>2$; we show that each has a homomorphism to some dihedral group
that is onto. This implies, in particular, that they all have two-dimensional
irreps. In the subsequent sections, we explicitly compute the groups
$SO(3)_{3}\mod 3$ (Sec. V) and $SO(3)_{5}\mod 5$ (Sec. VI) and their
representation theory in both cases exhibiting qubits, i.e., two-dimensional
irreps. Finally, in Sec. VII, we do the same for $SO(3)_{2}\mod 2$, which
turns out to be isomorphic to $\SS_{3}$, the permutation group of three
elements. We conclude in Sec. VIII with an outlook to future work and open
questions.

##  II Preliminaries on the compact group $\boldmath{SO(3)_{p}}$

We recall here some basic notions and results drawn from Ref. our1st .

###### Theorem II.1

For every prime $p$, there is a unique quadratic form on
${{\mathbb{Q}}}_{p}^{3}$, up to linear equivalence and scaling, which does not
represent zero,

$Q(\bm{x})=\begin{cases}x_{1}^{2}-vx_{2}^{2}+px_{3}^{2}&\text{if }p>2,\\\
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}&\text{if }p=2,\end{cases}$ (1)

where

$v\coloneqq\begin{cases}-1&\text{if }p\equiv 3\mod 4,\\\ -u&\text{if }p\equiv
1\mod 4,\end{cases}$ (2)

where $u\in\mathbb{U}$ (the group of $p$-adic units, i.e., the group of
invertible elements of $\mathbb{Z}_{p}$) is not a square.

Recalling that every quadratic form $Q$ induces a special orthogonal group as
the set of determinant 1 matrices that preserve $Q$, we have the following
result.

###### Theorem II.2

There is a unique (up to isomorphisms) compact special orthogonal group on
${{\mathbb{Q}}}_{p}^{3}$ for every prime $p$,

$SO(3)_{p}=\\{L\in\mathcal{M}_{3\times
3}({{\mathbb{Q}}}_{p})\,\mathrel{\mathop{\mathchar
58\relax}}\,A=L^{T}AL,\,\det L=1\\}$ (3)

endowed with the common matrix product, where $A$ is the matrix representation
in the canonical basis of ${{\mathbb{Q}}}_{p}^{3}$ of the preserved quadratic
form $Q$,

$A=\begin{cases}\operatorname{diag}(1,-v,p)&\text{if }p>2,\\\
\openone&\text{if }p=2.\end{cases}$ (4)

Since $SO(3)_{p}$ is a topological group, it is natural that the
representations we look for are continuous [with respect to the $p$-adic
metric on the domain and the Lie group metric of the complex $SU(d)$]. All of
the representations we find will turn out to be continuous for trivial
reasons.

Another crucial result proved in Ref. our1st  is the following.

###### Theorem II.3

All special orthogonal matrices have $p$-adic integer entries,

$SO(3)_{p}\subset SL(3,\mathbb{Z}_{p}).$

Next we state a fundamental fact about the geometry of $SO(3)_{p}$.

###### Theorem II.4

The elements of $SO(3)_{p}$ are rotations, i.e., they always have an
eigenvalue $1$, and the corresponding eigenspace is the rotation axis.

We recall the parameterization of matrices of $SO(3)_{p}$.

###### Theorem II.5

A rotation of $SO(3)_{p}$ around
$\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\backslash\\{\bm{0}\\}$ takes the following
matrix form with respect to an orthogonal basis $(\bm{g},\bm{h},\bm{n})$ of
${{\mathbb{Q}}}_{p}^{3}$:

$\mathcal{R}_{\bm{n}}(\sigma)=\begin{pmatrix}\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}&-\frac{2\alpha\sigma}{1+\alpha\sigma^{2}}&0\\\
\frac{2\sigma}{1+\alpha\sigma^{2}}&\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}&0\\\
0&0&1\end{pmatrix}$ (5)

with $\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ and
$\alpha=Q(\bm{h})/Q(\bm{g})$. The set $SO(3)_{p,\bm{n}}$ of rotations around a
given $\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\backslash\\{\bm{0}\\}$ forms an Abelian
subgroup of $SO(3)_{p}$

Last, in parallel to the real orthogonal case, we found that only certain main
angle decompositions of rotations around the reference axes hold for
$SO(3)_{p}$.

###### Theorem II.6

For an odd prime $p>2$, every $M\in SO(3)_{p}$ can be written as any of the
Cardano type compositions,

$\mathcal{R}_{x}\mathcal{R}_{y}\mathcal{R}_{z},\ \
\mathcal{R}_{z}\mathcal{R}_{y}\mathcal{R}_{x},\ \
\mathcal{R}_{z}\mathcal{R}_{x}\mathcal{R}_{y},\ \
\mathcal{R}_{y}\mathcal{R}_{x}\mathcal{R}_{z},$

respectively, with certain rotation parameters
$\sigma,\tau,\omega\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$.

For a prime $p\equiv 1\mod 4$, every $M\in SO(3)_{p}$ can also be written as
any of the Cardano-type compositions,

$\mathcal{R}_{x}\mathcal{R}_{z}\mathcal{R}_{y},\ \
\mathcal{R}_{y}\mathcal{R}_{z}\mathcal{R}_{x},$

respectively, with certain rotation parameters
$\sigma,\tau,\omega\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$.

Note that none of the other possible forms of main angle decomposition,
familiar from the real Euclidean case, hold universally for $SO(3)_{p}$. In
particular, no Euler angle decomposition exists for primes $p>2$, and none of
the Euler and Cardano decompositions exist when $p=2$.

##  III Quotienting the group $\boldmath{SO(3)_{p}}$

A fundamental tool to understand the algebraic aspects of $SO(3)_{p}$ is
quotienting modulo $p^{k}$ the group pigliapochi-msc : this is possible
because the matrix entries are $p$-adic integers (Theorem II.3). These
quotients are finite groups and, thus, much easier to study while generating
$SO(3)_{p}$ as a projective limit. For us, these projections are crucial as
their representations induce representations of the whole $SO(3)_{p}$ and,
thus, allow us to find irreps easily.

We can write a $p$-adic integer $a\in\mathbb{Z}_{p}$ as a sequence
$a=(a_{1},a_{2},a_{3},\ldots)$ with $a_{k}\in\mathbb{Z}/p^{k}$ such that
$a_{k+1}\equiv a_{k}\mod p^{k}$ for all $k>0$, with component-wise addition
and multiplication in the respective rings Serre . This means that the
functions

$\begin{split}\pi_{k}^{\prime}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{Z}_{p}&\rightarrow\mathbb{Z}/p^{k}\\\
a=(a_{i})_{i=1}^{\infty}&\mapsto a_{k}\end{split}$ (6)

are ring homomorphisms,

$\displaystyle\pi_{k}^{\prime}(a+b)$
$\displaystyle=\pi_{k}^{\prime}((a_{i})_{i}+(b_{j})_{j})=\pi_{k}^{\prime}((a_{i}+b_{i})_{i})=a_{k}+b_{k}=\pi_{k}^{\prime}(a)+\pi_{k}^{\prime}(b),$
(7) $\displaystyle\pi_{k}^{\prime}(ab)$
$\displaystyle=\pi_{k}^{\prime}((a_{i})_{i}(b_{j})_{j})=\pi_{k}^{\prime}((a_{i}b_{i})_{i})=a_{k}b_{k}=\pi_{k}^{\prime}(a)\pi_{k}^{\prime}(b).$
(8)

This can be extended to multiplicative groups of matrices over
$\mathbb{Z}_{p}$: the matrix product is defined through sums and products of
elements, for which $\pi_{k}^{\prime}$ are homomorphisms; then the maps

$\pi_{k}\left(M\right)=\pi_{k}\left((m_{ij})_{ij}\right)\mathrel{\mathop{\mathchar
58\relax}}=\left(\pi_{k}^{\prime}\left(m_{ij}\right)\right)_{ij}$ (9)

are group homomorphisms on any group contained in $\mathcal{M}_{n\times
n}(\mathbb{Z}_{p})$ to some other group contained in $\mathcal{M}_{n\times
n}(\mathbb{Z}/p^{k})$.

###### Remark III.1

Note that all the maps $\pi_{k}$ are continuous with respect to the $p$-adic
metric on $SO(3)_{p}$ and the discrete topology on the range, which is a
finite set.

Theorem II.3 means that the projection maps $\pi_{k}$ are well defined on
$SO(3)_{p}$,

$\displaystyle\pi_{k}\mathrel{\mathop{\mathchar 58\relax}}SO(3)_{p}$
$\displaystyle\rightarrow SO(3)_{p}\mod p^{k}\subset
SL(3,{{\mathbb{Z}}}/p^{k}),$ $\displaystyle L=(a_{ij})_{ij}$
$\displaystyle\mapsto(a_{ij}\mod p^{k})_{ij}$ (10)

since all matrix entries of eligible $L$ are $p$-adic integers. This allows us
to build a projective sequence of quotient groups,

$G_{p^{k}}\mathrel{\mathop{\mathchar
58\relax}}=\pi_{k}(SO(3)_{p})=SO(3)_{p}\mod p^{k}.$ (11)

These are finite groups as anticipated, being subgroups of
$SL(3,{{\mathbb{Z}}}/p^{k})$. By $G_{\bm{n},p^{k}}$ we will denote the Abelian
subgroup of $G_{p^{k}}$ of $\pi_{k}$-projected rotations
$\mathcal{R}_{\bm{n}}(\sigma)\in SO(3)_{p}$ around a fixed
$\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\backslash\\{\bm{0}\\}$,

$G_{\bm{n},p^{k}}\mathrel{\mathop{\mathchar
58\relax}}=\pi_{k}(SO(3)_{p,\bm{n}}).$ (12)

The Cardano decompositions (Theorem II.6) hold in $G_{p^{k}}$ too.

It is possible to rewrite the parameterization of the subgroups
$SO(3)_{p,\bm{n}}$ of rotations around a given
$\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\backslash\\{\bm{0}\\}$ (Theorem II.5) only in
terms of $p$-adic integers our1st . To wit, if $\kappa$ denotes the
determinant of the quadratic form preserved by $p$-adic planar rotations, in
the plane orthogonal to the rotation axis, we have

$SO(2)_{p}^{\kappa}=\left\\{\mathcal{R}_{\kappa}(\sigma)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{p}\right\\}\cup\left\\{-\mathcal{R}_{\kappa}\left(\sigma\right)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in p{{\mathbb{Z}}}_{p}\right\\}$

for $p>2$ and $\kappa=-v$ (rotations around the $z$-axis), or $p=2$ and
$\kappa=1,\pm 5$ ($\kappa=1$ corresponds to rotations around $x$, $y$ and $z$
axes), while

$SO(2)_{p}^{\kappa}=\left\\{\mathcal{R}_{\kappa}(\sigma)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{p}\right\\}\cup\left\\{-{\cal
R}_{\kappa}\left(\sigma\right)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{p}\right\\}$

for $p>2$ and $\kappa=p,up$ (rotations around $x$ and $y$ axes), or $p=2$ and
$\kappa=\pm 2,\pm 10$. We also recall that
$\mathcal{R}_{\bm{n}}\left(-\frac{1}{\alpha\sigma}\right)=\mathcal{R}_{\bm{n}}(\infty)\mathcal{R}_{\bm{n}}(\sigma)$,
which allows us to write for every $p>2$,

$\displaystyle G_{z,p^{k}}$ $\displaystyle=\left\\{\mathcal{R}_{z}(\sigma)\mod
p^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{p}\right\\}\cup\left\\{\mathcal{R}_{z}(\infty)\mathcal{R}_{z}(\sigma)\mod
p^{k}\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in
p{{\mathbb{Z}}}_{p}\right\\}$ (13)
$\displaystyle=\left\\{\mathcal{R}_{z}(\sigma)\mod
p^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}/p^{k}\right\\}\cup\left\\{\mathcal{R}_{z}(\infty)\mathcal{R}_{z}(\sigma)\mod
p^{k}\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in
p({{\mathbb{Z}}}/p^{k})\right\\},$

and similarly,

$G_{\bm{n},p^{k}}=\left\\{\mathcal{R}_{\bm{n}}(\sigma)\mod
p^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}/p^{k}\right\\}\cup\left\\{\mathcal{R}_{\bm{n}}(\infty)\mathcal{R}_{\bm{n}}(\sigma)\mod
p^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}/p^{k}\right\\}$ (14)

for the $x$ and $y$ axes.

Moreover, when $p=2$, we have the following for each of the reference axes:

$\displaystyle G_{\bm{n},2^{k}}$
$\displaystyle=\left\\{\mathcal{R}_{\bm{n}}(\sigma)\mod
2^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{2}\right\\}\cup\left\\{\mathcal{R}_{\bm{n}}(\infty)\mathcal{R}_{\bm{n}}(\sigma)\mod
2^{k}\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in
2{{\mathbb{Z}}}_{2}\right\\}$ (15)
$\displaystyle=\left\\{\mathcal{R}_{\bm{n}}(\sigma)\mod
2^{k}\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in
2({{\mathbb{Z}}}/2^{k})\right\\}$
$\displaystyle\phantom{==}\cup\left\\{\mathcal{R}_{\bm{n}}(\infty)\mathcal{R}_{\bm{n}}(\sigma)\mod
2^{k}\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in
2({{\mathbb{Z}}}/2^{k})\right\\}$
$\displaystyle\phantom{==}\cup\left\\{\mathcal{R}_{\bm{n}}(\sigma)\mod
2^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}/2^{k},\,\sigma\equiv 1\mod 2\right\\},$

with the elements in the last set of the union in the plane orthogonal to
$\bm{n}$ written as

$\begin{pmatrix}\frac{-2(\sigma+{\sigma}^{2})}{1+2(\sigma+{\sigma}^{2})}&-\frac{1+2\sigma}{1+2(\sigma+{\sigma}^{2})}\\\
\frac{1+2\sigma}{1+2(\sigma+{\sigma}^{2})}&\frac{-2(\sigma+{\sigma}^{2})}{1+2(\sigma+{\sigma}^{2})}\end{pmatrix}\mod
2^{k}\ ,\ \sigma\in{{\mathbb{Z}}}/2^{k-1}.$ (16)

Eqs. (13), (14) and (15) provide a parameterization of $G_{p^{k}}$ for every
prime $p$ and $k\geq 1$, thanks to the existing Cardano decompositions.

We conclude this section by introducing the groups of solutions modulo
$p^{k}$, $k\geq 1$, of the defining equations of $SO(3)_{p}$,

$\widehat{G}_{p^{k}}=\\{L\in\mathcal{M}_{3\times
3}({{\mathbb{Z}}}/p^{k})\mathrel{\mathop{\mathchar 58\relax}}A\equiv
L^{T}AL\mod p^{k},\,\det L\equiv 1\mod p^{k}\\}.$ (17)

These groups will turn out to be fundamental in the discussion about the
representations of $SO(3)_{p}$ at least for $k=1$ in this paper.

###### Remark III.2

By virtue of Theorem II.3

$G_{p^{k}}\subseteq\widehat{G}_{p^{k}},$ (18)

and a fundamental open question is whether these groups coincide or not, by
varying the values of prime $p>2$ and $k\geq 1$. This question has the
character of Hensel’s Lemma Cassels : do the integer solutions modulo $p$ of
$A=L^{T}AL$ with $\det L=1$ lift to $p$-adic solutions?

This boils down to checking if $\lvert
G_{p^{k}}\rvert=\lvert\widehat{G}_{p^{k}}\rvert$ for some $p$ and $k$.
Focusing on primes $p>2$, this can be done by comparing the order of
$\widehat{G}_{p^{k}}$ with the order of $G_{p^{k}}$ given by one of its
possible Cardano decompositions. One of the main aspects left to be discovered
in this matter is, then, the multiplicity of the Cardano decompositions for
$G_{p^{k}}$ for every prime $p>2,\,k\geq 1$, that is, how many distinct
Cardano decompositions around the same axes exist for each matrix of
$G_{p^{k}}$. Once known that, it will not be hard to find the order of the
subgroups $G_{\bm{n},p^{k}}$ for the $x$, $y$ and $z$ axes from Eqs. (13) and
(14) and, then, deduce $\lvert G_{p^{k}}\rvert$.

##  IV The group $\boldmath{SO(3)_{p}\mod p}$ for $\boldmath{p>2}$

We specialize Eqs. (13) and (14) for odd prime $p>2$ in the case $k=1$: for
the $x$ or $y$ axes,

$G_{\bm{n},p}=\left\\{{\cal R}_{\bm{n}}(\sigma)\mod
p\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}/p\right\\}\cup\left\\{{\cal
R}_{\bm{n}}(\infty){\cal R}_{\bm{n}}\left(\sigma\right)\mod
p\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in{{\mathbb{Z}}}/p\right\\},$
(19)

while for the $z$ axis,

$G_{z,p}=\left\\{{\cal R}_{z}(\sigma)\mod
p,\,\sigma\in{{\mathbb{Z}}}/p\right\\}\cup\left\\{{\cal R}_{z}(\infty)\mod
p\right\\}.$ (20)

By using (5), Eqs. (19) and (20) can be rewritten as follows:

$\displaystyle G_{x,p}$ $\displaystyle=\left\\{\begin{pmatrix}1&0&0\\\
0&1&0\\\ 0&2\xi&1\end{pmatrix}\mod p\ \mathrel{\mathop{\mathchar 58\relax}}\
\xi\in{{\mathbb{Z}}}/p\right\\}\cup\left\\{\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
0&-2\xi&-1\end{pmatrix}\mod p\ \mathrel{\mathop{\mathchar 58\relax}}\
\xi\in{{\mathbb{Z}}}/p\right\\},$ (21) $\displaystyle G_{y,p}$
$\displaystyle=\left\\{\begin{pmatrix}1&0&0\\\ 0&1&0\\\
2\eta&0&1\end{pmatrix}\mod p\ \mathrel{\mathop{\mathchar 58\relax}}\
\eta\in{{\mathbb{Z}}}/p\right\\}\cup\left\\{\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
-2\eta&0&-1\end{pmatrix}\mod p\ \mathrel{\mathop{\mathchar 58\relax}}\
\eta\in{{\mathbb{Z}}}/p\right\\},$ (22) $\displaystyle G_{z,p}$
$\displaystyle=\left\\{\begin{pmatrix}\frac{1+v\zeta^{2}}{1-v\zeta^{2}}&\frac{2v\zeta}{1-v\zeta^{2}}&0\\\
\frac{2\zeta}{1-v\zeta^{2}}&\frac{1+v\zeta^{2}}{1-v\zeta^{2}}&0\\\
0&0&1\end{pmatrix}\mod p\ \mathrel{\mathop{\mathchar 58\relax}}\
\zeta\in{{\mathbb{Z}}}/p\right\\}\cup\left\\{\begin{pmatrix}-1&0&0\\\
0&-1&0\\\ 0&0&1\end{pmatrix}\mod p\right\\}.$ (23)

We will refer to the first set of each of these three unions as “first branch”
and to the second one as “second branch.” They provide a parameterization of
$G_{p}=SO(3)_{p}\mod p$ thanks to the existing Cardano decompositions.

###### Remark IV.1

It is trivial to check that matrices within each of the two branches of
$G_{x,p}$ and $G_{y,p}$ are all distinct by varying the parameter in
${{\mathbb{Z}}}/p$. Moreover a matrix from the first branch cannot be equal to
a matrix of the second branch, for both $G_{x,p}$ and $G_{y,p}$, because
$1\not\equiv-1\mod p$ when $p>2$. Furthermore, by equating two matrices from
the first branch of $G_{z,p}$, we get

$\frac{1+v\zeta^{2}}{1-v\zeta^{2}}\equiv\frac{1+v{\zeta^{\prime}}^{2}}{1-v{\zeta^{\prime}}^{2}}\mod
p,$ (24)

which is equivalent to $\zeta^{2}\equiv{\zeta^{\prime}}^{2}\mod p$. Plugging
this into

$\frac{2\zeta}{1-v\zeta^{2}}\equiv\frac{2\zeta^{\prime}}{1-v{\zeta^{\prime}}^{2}}\mod
p,$ (25)

we obtain $\zeta\equiv\zeta^{\prime}\mod p$. Finally, we equate a matrix from
the first branch of $G_{z,p}$ to the one of the second branch,

$\frac{1+v\zeta^{2}}{1-v\zeta^{2}}\equiv-1\mod p,$ (26)

which is equivalent to $1\equiv-1\mod p$, which is impossible. Therefore, the
matrices in $G_{x,p},\,G_{y,p}$ and $G_{z,p}$ are all distinct by varying the
respective parameters. Then, it follows that

$\lvert G_{x,p}\rvert=\lvert G_{y,p}\rvert=2p,\quad\lvert G_{z,p}\rvert=p+1.$
(27)

As anticipated, a fundamental point in this matter is to discover the
multiplicity of a modulo-$p$-reduced Cardano decomposition for $G_{p}$, for
example that of the kind ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$. A systematic
ambiguity of order 2 in the Cardano representations is based on the relation

${\cal R}_{x}(\infty){\cal R}_{y}(\infty){\cal
R}_{z}(\infty)\equiv\openone\mod p.$ (28)

Moreover, in the calculations below, we are going to use the following
commutation relation

$[{\cal R}_{z}(\infty),{\cal R}_{y}(\eta)]\equiv{\cal R}_{z}(\infty){\cal
R}_{y}(\eta){\cal R}_{z}(\infty)^{-1}{\cal R}_{y}(\eta)^{-1}\equiv{\cal
R}_{y}(-2\eta)\mod p$ (29)

so that

${\cal R}_{z}(\infty){\cal R}_{y}(\eta)\equiv[{\cal R}_{z}(\infty),{\cal
R}_{y}(\eta)]{\cal R}_{y}(\eta){\cal R}_{z}(\infty)\equiv{\cal
R}_{y}(-\eta){\cal R}_{z}(\infty)\mod p.$ (30)

Now, we start analyzing all the possibilities for the branches of the three
rotations involved in the Cardano decomposition of the kind ${\cal R}_{x}{\cal
R}_{y}{\cal R}_{z}$. A triple $ijk$ with $i,j,k\in\\{1,2\\}$ will denote a
Cardano representation of the kind ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$,
where the $x$, $y$ and $z$ rotation modulo $p$ are taken from the $i$th, $j$th
and $k$th branch, respectively. There are $8=2^{3}$ possible Cardano
representations ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$ with respect to the
branches of each of the three involved rotations. Thus, there are 36
possibilities of equating two triple products, i.e., of equating modulo $p$
two Cardano representations depending on their rotation branches.

Suppose that a given $M\in G_{p}$ admits a Cardano representation of the kind
111, i.e., there exist $\xi,\eta,\zeta\in{{\mathbb{Z}}}/p$ such that

$M\equiv{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal R}_{z}(\zeta)\mod p.$ (31)

Then, there exists at least another distinct Cardano representation of $M$
along the same axes: by using Eqs. (28) and (30), we get

$\displaystyle M$ $\displaystyle\equiv{\cal R}_{x}(\xi)\,{\cal
R}_{x}(\infty){\cal R}_{y}(\infty){\cal R}_{z}(\infty)\,{\cal
R}_{y}(\eta){\cal R}_{z}(\zeta)$ $\displaystyle\equiv{\cal R}_{x}(\infty){\cal
R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(-\eta){\cal R}_{z}(\infty){\cal
R}_{z}(\zeta)\mod p.$ (32)

The latter is a Cardano representation of $M$ of the kind either 221 if
$\zeta\not\equiv 0\mod p$ or 222 if $\zeta\equiv 0\mod p$. The reverse is also
true: if $M\in G_{p}$ admits a Cardano representation 221 with non-zero
parameter for the $z$-rotation (or a Cardano representation 222), then it
admits a Cardano representation 111 too.

Similarly, if for a given $M\in G_{p}$ there exist
$\xi,\eta,\zeta\in{{\mathbb{Z}}}/p$ such that $M$ has Cardano representation
of the kind 121,

$M\equiv{\cal R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)\mod p,$ (33)

then also

$\displaystyle M$ $\displaystyle\equiv{\cal R}_{x}(\xi)\,{\cal
R}_{x}(\infty){\cal R}_{y}(\infty){\cal R}_{z}(\infty)\,{\cal
R}_{y}(\infty){\cal R}_{y}(\eta){\cal R}_{z}(\zeta)$ $\displaystyle\equiv{\cal
R}_{x}(\infty){\cal R}_{x}(\xi){\cal R}_{y}(\infty)^{2}{\cal
R}_{z}(\infty){\cal R}_{y}(\eta){\cal R}_{z}(\zeta)$ $\displaystyle\equiv{\cal
R}_{x}(\infty){\cal R}_{x}(\xi){\cal R}_{y}(-\eta){\cal R}_{z}(\infty){\cal
R}_{z}(\zeta)\mod p,$ (34)

which is a Cardano representation of $M$ of the kind either 211 if
$\zeta\not\equiv 0\mod p$ or 212 if $\zeta\equiv 0\mod p$.

Moreover, if for a given $M\in G_{p}$ there exist
$\xi,\eta\in{{\mathbb{Z}}}/p$ such that $M$ has Cardano representation 112,

$M\equiv{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal R}_{z}(\infty)\mod p,$ (35)

then also

$\displaystyle M$ $\displaystyle\equiv{\cal R}_{x}(\xi)\,{\cal
R}_{x}(\infty){\cal R}_{y}(\infty){\cal R}_{z}(\infty)\,{\cal
R}_{y}(\eta){\cal R}_{z}(\infty)$ $\displaystyle\equiv{\cal
R}_{x}(\infty){\cal R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(-\eta){\cal
R}_{z}(0)\mod p,$ (36)

that is, a Cardano representation 221 of $M$.

Finally, if for a given $M\in G_{p}$ there exist $\xi,\eta\in{{\mathbb{Z}}}/p$
such that $M$ has Cardano representation 122,

$M\equiv{\cal R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(\eta){\cal
R}_{z}(\infty)\mod p,$ (37)

then also

$\displaystyle M$ $\displaystyle\equiv{\cal R}_{x}(\xi)\,{\cal
R}_{x}(\infty){\cal R}_{y}(\infty){\cal R}_{z}(\infty)\,{\cal
R}_{y}(\infty){\cal R}_{y}(\eta){\cal R}_{z}(\infty)$
$\displaystyle\equiv{\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}(\infty)^{2}{\cal R}_{z}(\infty){\cal R}_{y}(\eta){\cal R}_{z}(\infty)$
$\displaystyle\equiv{\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}(-\eta){\cal R}_{z}(0)\mod p,$ (38)

that is, a Cardano representation 211 of $M$.

We have analyzed each of the eight initial triples $ijk$ since the reverse of
each of the above reasonings is also valid. This provides six different
modular congruences of Cardano representations with respect to certain triples
of branches. We deduce the following.

###### Remark IV.2

Given a Cardano representation of $M$ of the kind ${\cal R}_{x}{\cal
R}_{y}{\cal R}_{z}$ of parameters $\xi,\eta,\zeta$, respectively, then $M$
admits at least another distinct Cardano representation of the kind ${\cal
R}_{x}{\cal R}_{y}{\cal R}_{z}$ with parameters
$\xi^{\prime},\eta^{\prime},\zeta^{\prime}$, respectively: that obtained by
changing the branches of the $x$\- and $y$-rotation and

  * •

by changing the branch of the $z$-rotation if its parameter $\zeta$ is
$0\in{{\mathbb{Z}}}/p$ or $\infty$ and

  * •

by fixing the branch of the $z$-rotation if
$\zeta\in{{\mathbb{Z}}}/p,\,\zeta\not\equiv 0\mod p$,

and with parameters

$\xi^{\prime}\equiv\xi,\ \ \ \eta^{\prime}\equiv-\eta,\ \ \
\zeta^{\prime}\equiv-\frac{1}{\alpha\zeta}\mod p.$ (39)

Now, we show that a given $M\in G_{p}$ cannot admit two distinct Cardano
representations of the kind ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$ with
respect to the same three branches. Here we need to compare each of the eight
branch triples $ijk$ with itself. The thesis is trivial to check in the four
cases $ij2$ where the $z$-rotation is ${\cal R}_{z}(\infty)$. We now focus on
the case 111 where the three rotations of the Cardano representations all come
from the first branch since the calculations for the remaining three cases
$ij1$ are very similar.

Suppose that a given $M\in G_{p}$ can be represented as

$M\equiv{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal R}_{z}(\zeta)\equiv{\cal
R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal
R}_{z}(\zeta^{\prime})\mod p$ (40)

for certain
$\xi,\xi^{\prime},\,\eta,\eta^{\prime},\,\zeta,\zeta^{\prime}\in{{\mathbb{Z}}}/p$.
It means that

$\displaystyle M$
$\displaystyle\equiv\begin{pmatrix}\frac{1+v\zeta^{2}}{1-v\zeta^{2}}&\frac{2v\zeta}{1-v\zeta^{2}}&0\\\
\frac{2\zeta}{1-v\zeta^{2}}&\frac{1+v\zeta^{2}}{1-v\zeta^{2}}&0\\\
\frac{2}{1-v\zeta^{2}}\big{(}\eta(1+v\zeta^{2})+2\xi\zeta\big{)}&\frac{2}{1-v\zeta^{2}}\big{(}\xi(1+v\zeta^{2})+2v\eta\zeta\big{)}&1\end{pmatrix}$
$\displaystyle\equiv\begin{pmatrix}\frac{1+v{\zeta^{\prime}}^{2}}{1-v{\zeta^{\prime}}^{2}}&\frac{2v\zeta^{\prime}}{1-v{\zeta^{\prime}}^{2}}&0\\\
\frac{2\zeta^{\prime}}{1-v{\zeta^{\prime}}^{2}}&\frac{1+v{\zeta^{\prime}}^{2}}{1-v{\zeta^{\prime}}^{2}}&0\\\
\frac{2}{1-v{\zeta^{\prime}}^{2}}\big{(}\eta^{\prime}(1+v{\zeta^{\prime}}^{2})+2\xi^{\prime}\zeta^{\prime}\big{)}&\frac{2}{1-v{\zeta^{\prime}}^{2}}\big{(}\xi^{\prime}(1+v{\zeta^{\prime}}^{2})+2v\eta^{\prime}\zeta^{\prime}\big{)}&1\end{pmatrix}\mod
p.$

By equating the entries of the $2\times 2$ upper left minors, we get Eqs. (24)
and (25), giving $\zeta\equiv\zeta^{\prime}\mod p$, i.e., ${\cal
R}_{z}(\zeta)\equiv{\cal R}_{z}(\zeta^{\prime})\mod p$. Then the following are
equivalent:

$\displaystyle{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)\equiv{\cal R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal
R}_{z}(\zeta^{\prime})\mod p,$ (41) $\displaystyle{\cal R}_{x}(\xi){\cal
R}_{y}(\eta)\equiv{\cal R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime})\mod
p,$ (42) $\displaystyle\begin{pmatrix}1&0&0\\\ 0&1&0\\\
2\eta&2\xi&1\end{pmatrix}\equiv\begin{pmatrix}1&0&0\\\ 0&1&0\\\
2\eta^{\prime}&2\xi^{\prime}&1\end{pmatrix}\mod p,$ (43)
$\displaystyle\xi^{\prime}\equiv\xi\mod
p\quad\text{and}\quad\eta^{\prime}\equiv\eta\mod p,$ (44) $\displaystyle{\cal
R}_{x}(\xi^{\prime})\equiv{\cal R}_{x}(\xi)\mod p\quad\text{and}\quad{\cal
R}_{y}(\eta^{\prime})\equiv{\cal R}_{y}(\eta)\mod p.$ (45)

Thus each $M\in G_{p}$ has a unique Cardano representation of the kind ${\cal
R}_{x}{\cal R}_{y}{\cal R}_{z}$ with respect to a certain triple of rotation
branches.

There are $36-6-8=22$ possibilities left of equating two triples $ijk$,
representing Cardano decompositions ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$
with respect to those triples of branches. For all of them, equating modulo
$p$ the first or last matrix entry of the two Cardano products leads to
$1\equiv-1\mod p$, which is impossible. It means that each $M\in G_{p}$ does
not have Cardano representations of the kind ${\cal R}_{x}{\cal R}_{y}{\cal
R}_{z}$ with respect to different triples of branches, except from the cases
of Remark IV.2.

We have, thus, proved the following.

###### Theorem IV.3

Every $M\in G_{p}$, for every odd prime $p$, has exactly two distinct Cardano
representations of the kind ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$. They are
one of the following six (depending on $M$) according to Remark IV.2, for
certain $\xi,\eta,\zeta\in{{\mathbb{Z}}}/p$:

$\begin{array}[]{ll}111\leftrightarrow 221&{\cal R}_{x}(\xi){\cal
R}_{y}(\eta){\cal R}_{z}(\zeta)\equiv{\cal R}_{x}(\infty){\cal
R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(-\eta){\cal
R}_{z}\left(\frac{1}{v\zeta}\right)\mod p\ \text{if}\ \zeta\not\equiv 0\mod
p,\\\ 111\leftrightarrow 222&{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(0)\equiv{\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal R}_{y}(\infty){\cal
R}_{y}(-\eta){\cal R}_{z}(\infty)\mod p,\\\ 121\leftrightarrow 211&{\cal
R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)\equiv{\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}(-\eta){\cal R}_{z}\left(\frac{1}{v\zeta}\right)\mod p\ \text{if}\
\zeta\not\equiv 0\mod p,\\\ 121\leftrightarrow 212&{\cal R}_{x}(\xi){\cal
R}_{y}(\infty){\cal R}_{y}(\eta){\cal R}_{z}(0)\equiv{\cal R}_{x}(\infty){\cal
R}_{x}(\xi){\cal R}_{y}(-\eta){\cal R}_{z}(\infty)\mod p,\\\
112\leftrightarrow 221&{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\infty)\equiv{\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}(\infty){\cal R}_{y}(-\eta){\cal R}_{z}(0)\mod p,\\\ 122\leftrightarrow
211&{\cal R}_{x}(\xi){\cal R}_{y}(\infty){\cal R}_{y}(\eta){\cal
R}_{z}(\infty)\equiv{\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}(-\eta){\cal R}_{z}(0)\mod p.\end{array}$

###### Corollary IV.4

Given a prime $p>2$, the order of the group $SO(3)_{p}\mod p=G_{p}$ is given
by

$\lvert G_{p}\rvert=\frac{1}{2}\lvert G_{x,p}\rvert\lvert G_{y,p}\rvert\lvert
G_{z,p}\rvert=2p^{2}(p+1).$ (46)

###### Proof.

The rotation groups $G_{x,p}$, $G_{y,p}$, and $G_{z,p}$ are subgroups of
$G_{p}$; hence, each product ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$ such that
${\cal R}_{x}\in G_{x,p}$, ${\cal R}_{y}\in G_{y,p}$, and ${\cal R}_{z}\in
G_{z,p}$ is an element of $G_{p}$. By the the Cardano decomposition ${\cal
R}_{x}{\cal R}_{y}{\cal R}_{z}$, each element of $G_{p}$ has a decomposition
into a product of these three, and Theorem IV.3, more precisely, states that
each element has exactly two such decompositions. Therefore, $\lvert
G_{p}\rvert=\frac{1}{2}\lvert G_{x,p}\rvert\lvert G_{y,p}\rvert\lvert
G_{z,p}\rvert$.  $\sqcap$$\sqcup$

###  IV.1 Equivalence with $\widehat{G}_{p}$

Now we consider the groups of Eq. (17) for $k=1$: we want to find a
parameterization of these groups $\widehat{G}_{p}$ too. Being
$\pi_{1}(A)=\operatorname{diag}(1,-v,0)$ and setting
$L=\left(a_{ij}\right)_{i,j=1}^{3}\in SO(3)_{p}$, the defining conditions of
$SO(3)_{p}$ reduced modulo $p$ translate in the following system of equations
for $\widehat{G}_{p}$:

$\begin{cases}a_{11}^{2}-va_{21}^{2}\equiv 1\mod p,\\\
a_{12}^{2}-va_{22}^{2}\equiv-v\mod p,\\\ a_{13}^{2}-va_{23}^{2}\equiv 0\mod
p,\\\ a_{11}a_{12}-va_{21}a_{22}\equiv 0\mod p,\\\
a_{11}a_{13}-va_{21}a_{23}\equiv 0\mod p,\\\ a_{12}a_{13}-va_{22}a_{23}\equiv
0\mod p,\\\
a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12}\equiv
1\mod p.\end{cases}$ (47)

The elements $a_{31},a_{32},a_{33}$ appear only in the last equation of (47),
the one deriving from the determinant: this means that we could at first try
to find the possible solutions for the other two rows, looking at the other
equations.

If $a_{23}\not\equiv 0\mod p$, the third equation of (47) would give $v\equiv
a_{13}^{2}a_{23}^{-2}\mod p$, which is in contradiction to the fact that $v$
is not a square modulo $p$. Hence it must be $a_{13},a_{23}\equiv 0\mod p$,
and from (47), we have

$\begin{cases}a_{11}^{2}-va_{21}^{2}\equiv 1\mod p,\\\
a_{12}^{2}-va_{22}^{2}\equiv-v\mod p,\\\ a_{13},a_{23}\equiv 0\mod p,\\\
a_{11}a_{12}-va_{21}a_{22}\equiv 0\mod p,\\\
a_{33}(a_{11}a_{22}-a_{21}a_{12})\equiv 1\mod p.\end{cases}$ (48)

With $a_{33}(a_{11}a_{22}-a_{21}a_{12})$ being non-zero and, thus, with its
two factors, we can write $a_{33}\equiv(a_{11}a_{22}-a_{21}a_{12})^{-1}$.
Actually, we can say more about these values: let us multiply the first two
equations of (48) member by member, obtaining

$\displaystyle-v$ $\displaystyle\equiv
a_{11}^{2}a_{12}^{2}-va_{11}^{2}a_{22}^{2}-va_{12}^{2}a_{21}^{2}+v^{2}a_{21}^{2}a_{22}^{2}$
$\displaystyle\equiv
a_{11}^{2}a_{12}^{2}-va_{11}^{2}a_{22}^{2}-va_{12}^{2}a_{21}^{2}+v^{2}a_{21}^{2}a_{22}^{2}+2v{a_{11}}{a_{12}}{a_{21}}{a_{22}}-2v{a_{11}}{a_{12}}{a_{21}}{a_{22}}$
$\displaystyle\equiv\left({a_{11}}{a_{12}}-v{a_{21}}{a_{22}}\right)^{2}-v\left({a_{11}}{a_{22}}-{a_{12}}{a_{21}}\right)^{2}$
$\displaystyle\equiv-v\left({a_{11}}{a_{22}}-{a_{12}}{a_{21}}\right)^{2}\mod
p.$

We can, thus, deduce

$a_{11}a_{22}-a_{21}a_{12}\equiv a_{33}\equiv\pm 1.$ (49)

Now, the whole remaining study concerns the submatrix

$\begin{pmatrix}a_{11}&a_{12}\\\ a_{21}&a_{22}\end{pmatrix}$ (50)

with

$\displaystyle a_{11}^{2}-va_{21}^{2}\equiv 1\mod p,$ (51) $\displaystyle
a_{12}^{2}-va_{22}^{2}\equiv-v\mod p,$ (52)
$\displaystyle{a_{11}}{a_{12}}-v{a_{21}}{a_{22}}\equiv 0\mod p,$ (53)
$\displaystyle{a_{11}}{a_{22}}-{a_{12}}{a_{21}}\equiv\pm_{1}1\mod p.$ (54)

Here, $a_{11}$ and $a_{21}$ cannot be both divisible by $p$; otherwise, Eq.
(51) would be impossible: suppose that $a_{11}\not\equiv 0\mod p$. Equation
(53) provides $a_{12}\equiv va_{11}^{-1}a_{21}a_{22}\mod p$. We plug this into
Eq. (52) and use Eq. (51),

$\displaystyle-v$ $\displaystyle\equiv
v^{2}a_{11}^{-2}a_{21}^{2}a_{22}^{2}-va_{22}^{2}\equiv-
va_{22}^{2}\left(1-va_{11}^{-2}a_{21}^{2}\right)$ $\displaystyle\equiv-
va_{22}^{2}\big{(}1+a_{11}^{-2}(1-a_{11}^{2})\big{)}\equiv-
va_{11}^{-2}a_{22}^{2}\mod p,$

which implies that

$a_{22}\equiv\pm_{2}a_{11}.$ (55)

Now, from Eqs. (51) and (52), we get $a_{12}^{2}\equiv v(a_{22}^{2}-1)\equiv
v(a_{11}^{2}-1)\equiv v^{2}a_{21}^{2}\mod p$, and so

$a_{12}\equiv\pm_{3}va_{21}\mod p.$ (56)

The choice is now only about $a_{11},a_{21}$ and three signs. However, we show
that these three signs $\pm_{1},\pm_{2},\pm_{3}$ are actually the same.
Equations (55) and (56) make Eq. (53) become
$va_{11}a_{21}(\pm_{3}1\mp_{2}1)\equiv 0\mod
p\Leftrightarrow(\pm_{3}1\mp_{2}1)a_{21}\equiv 0\mod p$. If $a_{21}\not\equiv
0\mod p$, then $\pm_{2}=\pm_{3}$; if $a_{21}\equiv 0\mod p$, then the choice
of $\pm_{3}$ is meaningless. Hence, we can write $\pm_{2}1=\pm_{3}1$, in
general. Now, Eqs. (51) and (54) give
$\pm_{1}1\equiv\pm_{2}\left(a_{11}^{2}-va_{21}^{2}\right)\equiv\pm_{2}1$;
therefore $\pm_{1}=\pm_{2}=\pm_{3}$.

Equation (53) becomes an identity, while Eqs. (52) and (54) become equivalent
to Eq. (51). Very similar steps in the hypothesis that $a_{21}\not\equiv 0\mod
p$ provide the same results, which we collect in the following equations:

$\left\\{\begin{aligned} &a_{11}^{2}-va_{21}^{2}\equiv 1\mod p,\\\
&{a_{22}}\equiv\pm_{1}{a_{11}}\mod p,\\\ &{a_{12}}\equiv\pm_{1}v{a_{21}}\mod
p,\\\ &{a_{13}}\equiv{a_{23}}\equiv 0\mod p,\\\ &{a_{33}}\equiv\pm 1\mod
p.\end{aligned}\right.$ (57)

In addition, $\pm=\pm_{1}$: $a_{33}^{-1}\equiv
a_{11}a_{22}-a_{12}a_{21}\Leftrightarrow\pm
1\equiv\pm_{1}(a_{11}^{2}-va_{21}^{2})\equiv\pm_{1}1\mod p$. We conclude that
matrix solutions of Eq. (47) are of the form

$\begin{pmatrix}{a_{11}}&\pm v{a_{21}}&0\\\ {a_{21}}&\pm{a_{11}}&0\\\
{a_{31}}&{a_{32}}&\pm 1\end{pmatrix}\in{\cal M}_{3\times 3}({{\mathbb{Z}}}/p)$
(58)

with $a_{11}^{2}-va_{21}^{2}\equiv 1\mod p$.

All of the above can be summarized in the following result.

###### Theorem IV.5

The matrices of $\widehat{G}_{p}$ for any prime $p>2$ are uniquely defined by
$a,b,c,d\in\mathbb{Z}/p$ (the first two solutions of $a^{2}-vb^{2}\equiv 1\mod
p$ and the others chosen freely) and by a sign $s=\pm 1$,

$M(a,b,c,d,s)=\begin{pmatrix}a&svb&0\\\ b&sa&0\\\ c&d&s\end{pmatrix}.$ (59)

###### Corollary IV.6

Given a prime $p>2$, the order of the matrix group $\widehat{G}_{p}$ is

$\lvert\widehat{G}_{p}\rvert=2p^{2}(p+1).$ (60)

###### Proof.

For a fixed odd prime $p$, the number of solutions of the equation
$a^{2}-vb^{2}\equiv 1\mod p$ is $p+1$ (Appendix A). Then, the total number of
combinations of $a,b,c,d,s$ is $(p+1)\cdot p\cdot p\cdot 2$ such that every
choice represents a distinct matrix. Then the number of distinct matrices is
$2p^{2}(p+1)$.  $\sqcap$$\sqcup$

###### Remark IV.7

Proving that $G_{p}$ coincides with $\widehat{G}_{p}$ is equivalent to proving
that $\lvert G_{p}\rvert=\lvert\widehat{G}_{p}\rvert$ (Remark III.2). This is
true for every prime $p>2$, by comparing Eqs. (46) and (60). This implies that
Eq. (59) is an equivalent parameterization for $G_{p}$.

The study of $G_{p}\mathrel{\mathop{\mathchar 58\relax}}=SO(3)_{p}\mod p$
regarding its order and parameterization turns out to be fundamental to build
a first list of representations of $SO(3)_{p}$. Examining the commutators and
the abelianization of a group is enough to exhaust the list of one-dimensional
representations simon : this will be helpful to understand how many other
representations are present and their form. It is for this reason that here we
anticipate the structure of the commutators between matrices of $G_{p}$ for
every prime $p>2$ by exploiting Remark IV.7. We calculate
$[M,N]=MNM^{-1}N^{-1}$ with $M,\,N\in G_{p}$ written through the
parameterization (59) as

$M=\begin{pmatrix}a_{11}&tva_{21}&0\\\ a_{21}&ta_{11}&0\\\
a_{31}&a_{32}&t\end{pmatrix},\ \ N=\begin{pmatrix}b_{11}&rvb_{21}&0\\\
b_{21}&rb_{11}&0\\\ b_{31}&b_{32}&r\end{pmatrix}.$

We get

$[M,N]=\begin{pmatrix}c_{11}&vc_{21}&0\\\ c_{21}&c_{11}&0\\\
c_{31}&c_{32}&1\end{pmatrix},$ (61)

where

$\displaystyle
c_{11}=(b_{11}^{2}-tvb_{21}^{2})(a_{11}^{2}-rva_{21}^{2})+(t+r-tr-1)va_{11}a_{21}b_{21}b_{11},$
(62) $\displaystyle
c_{21}=(t-1)(a_{11}^{2}-rva_{21}^{2})b_{21}b_{11}-(r-1)(b_{11}^{2}-tvb_{21}^{2})a_{11}a_{21}.$

###  IV.2 From $p$-adic rotations to dihedral groups

We introduce a homomorphism, which will turn out to be useful to our aim of
finding two-dimensional irreducible representations of $SO(3)_{p}$.

###### Proposition IV.8

The map

$\displaystyle K_{p}^{\prime}\mathrel{\mathop{\mathchar 58\relax}}$
$\displaystyle G_{p}\rightarrow GL((\mathbb{Z}/p)^{2}),$ (63) $\displaystyle
M\mapsto M_{33},$

where $M_{33}$ is the the $2\times 2$ upper-left minor of $M$, is a group
homomorphism.

###### Proof.

According to the parameterization (59) and Remark IV.7, the map
$K_{p}^{\prime}$ acts as follows:

$\begin{pmatrix}a&svb&0\\\ b&sa&0\\\
c&d&s\end{pmatrix}\mapsto\begin{pmatrix}a&svb\\\ b&sa\end{pmatrix}.$

The image $\mathfrak{I}(K_{p}^{\prime})$ is parameterized by
$(a,b)\in({{\mathbb{Z}}}/p)^{2}$, which are solutions of $a^{2}-vb^{2}\equiv
1\mod p$ and $s\in\\{\pm 1\\}$.

  * •

$\mathfrak{I}(K_{p}^{\prime})\subset GL({{\mathbb{Z}}}/p)$,

$\det\begin{pmatrix}a&svb\\\ b&sa\end{pmatrix}=s(a^{2}-vb^{2})\equiv\pm 1\neq
0.$

  * •

$K_{p}^{\prime}$ is a group homomorphism,

$K_{p}^{\prime}\begin{pmatrix}a&svb&0\\\ b&sa&0\\\
c&d&s\end{pmatrix}K_{p}^{\prime}\begin{pmatrix}a^{\prime}&s^{\prime}vb^{\prime}&0\\\
b^{\prime}&s^{\prime}a^{\prime}&0\\\
c^{\prime}&d^{\prime}&s^{\prime}\end{pmatrix}=\begin{pmatrix}a&svb\\\
b&sa\end{pmatrix}\begin{pmatrix}a^{\prime}&s^{\prime}vb^{\prime}\\\
b^{\prime}&s^{\prime}a^{\prime}\end{pmatrix}=\begin{pmatrix}aa^{\prime}+vsbb^{\prime}&vs^{\prime}(ab^{\prime}+sa^{\prime}b)\\\
a^{\prime}b+sab^{\prime}&s^{\prime}(saa^{\prime}+vbb^{\prime})\end{pmatrix},$

which equals

$\displaystyle K_{p}^{\prime}\left(\begin{pmatrix}a&svb&0\\\ b&sa&0\\\
c&d&s\end{pmatrix}\begin{pmatrix}a^{\prime}&s^{\prime}vb^{\prime}&0\\\
b^{\prime}&s^{\prime}a^{\prime}&0\\\
c^{\prime}&d^{\prime}&s^{\prime}\end{pmatrix}\right)=K_{p}^{\prime}\begin{pmatrix}aa^{\prime}+svbb^{\prime}&vs^{\prime}(ab^{\prime}+sa^{\prime}b)&0\\\
a^{\prime}b+sab^{\prime}&s^{\prime}(saa^{\prime}+vbb^{\prime})&0\\\
\dots&\dots&ss^{\prime}\end{pmatrix}.$

$\sqcap$$\sqcup$

A point of interest lies in the behaviour of $a$ and $b$ as solutions of
$a^{2}-vb^{2}\equiv 1\mod p$ while composing matrices in
$\mathfrak{I}(K_{p}^{\prime})$, pigliapochi-msc . In
$\mathfrak{I}(K_{3}^{\prime})$ there are only the matrices

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix},\quad\begin{pmatrix}0&1\\\
1&0\end{pmatrix},$ (64)

up to sign changes on the components, and it is trivial that products of
matrices of this form give back matrices of this form. However, which
mathematical property guarantees that a product

$\begin{pmatrix}a&svb\\\
b&sa\end{pmatrix}\begin{pmatrix}a^{\prime}&s^{\prime}vb^{\prime}\\\
b^{\prime}&s^{\prime}a^{\prime}\end{pmatrix}=\begin{pmatrix}aa^{\prime}+vsbb^{\prime}&vs^{\prime}(ab^{\prime}+sa^{\prime}b)\\\
a^{\prime}b+sab^{\prime}&s^{\prime}(saa^{\prime}+vbb^{\prime})\end{pmatrix}$
(65)

still gives a matrix coherent with the definition in
$\mathfrak{I}(K_{p}^{\prime})$? Can we find some deeper meaning about it?

As proved in Appendix A, solutions of $a^{2}-vb^{2}\equiv 1$ in $\mathbb{Z}/p$
form a cyclic group: the key to prove it lies in looking at the pair $(a,b)$
as a single “complex” number $a+ib$, $i^{2}=v$, and introducing products
between pairs as products between complex numbers. As a consequence, the
“product” of two solutions gives another solution according to the product of
matrices. As a matter of fact, another way to represent the complex product is
by writing the components in a $2\times 2$ matrix of the form

$\begin{pmatrix}a&vb\\\ b&a\end{pmatrix}$ (66)

and applying the common matrix product.

This is of great interest for us, as these are exactly half of the matrices in
$\mathfrak{I}(K_{p}^{\prime})$, the ones with $s=1$. The other half, the
matrices with $s=-1$, are obtained simply by multiplying these by the (Pauli)
matrix $Z$,

$\begin{pmatrix}a&-vb\\\ b&-a\end{pmatrix}=\begin{pmatrix}a&vb\\\
b&a\end{pmatrix}Z,\qquad Z=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$ (67)

Note that $Z\in\mathfrak{I}(K_{p}^{\prime})$ because ${\cal
R}_{x}(\infty)=\operatorname{diag}(1,-1,-1)_{{{\mathbb{Z}}}_{p}}\in SO(3)_{p}$
for every prime $p$.

This means that the group $\mathfrak{I}(K_{p}^{\prime})$ is generated by two
matrices,

$\begin{array}[]{cc}C=\begin{pmatrix}a_{0}&-b_{0}\\\
b_{0}&a_{0}\end{pmatrix},&Z=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}\end{array},$ (68)

where $C$ is the matrix obtained by $(a_{0},b_{0})$, a generator of the cyclic
group of the solutions $(a,b)$.

Finally we observe that

  * •

$C^{p+1}=\openone$ because of the isomorphism with the solutions subgroup of
$\mathbb{Z}/p[i]^{*}$,

  * •

$Z^{2}=\openone$, and

  * •

$ZCZ=C^{-1}$, as can be checked readily.

This gives us a complete presentation for our group,

$\mathfrak{I}(K_{p}^{\prime})=\langle C,Z\mid
C^{p+1}=Z^{2}=\openone,ZCZ=C^{-1}\rangle,$ (69)

which is exactly the presentation for the dihedral group $D_{p+1}$: in other
words, the group $\mathfrak{I}(K_{p}^{\prime})$ is isomorphic to the group of
symmetries of a $(p+1)$-gon. Denoting this isomorphism with $\varphi$, we have

$\varphi\mathrel{\mathop{\mathchar 58\relax}}\mathfrak{I}(K_{p}^{\prime})\
\stackrel{{\scriptstyle\sim}}{{\rightarrow}}\ D_{p+1},\ C\mapsto a,\,Z\mapsto
x,$ (70)

where $a$ and $x$ are the generators (rotation and reflection, respectively)
of $D_{p+1}$.

This will allow us to exploit the known representation theory of dihedral
groups for the $p$-adic group $SO(3)_{p}$, some basic facts about which are
reported in Appendix B. While we will not discuss this in detail, Appendix B.1
shows that $\mathfrak{I}(K_{p}^{\prime})$ has always four one-dimensional
representations. Furthermore, Appendix B.2 shows that there always exists a
two-dimensional irrep of $\mathfrak{I}(K_{p}^{\prime})$, and for $p>3$, there
are more than one 2-irreps, leading to different $p$-adic qubits.

###### Remark IV.9

We summarize the preceding discussion by recording that

$J_{p}^{(i)}\mathrel{\mathop{\mathchar 58\relax}}=\rho_{i}\circ\varphi\circ
K_{p}^{\prime}\circ\pi_{1},$ (71)

where $\rho_{i}$’s are the $2$-irreps of $D_{p+1}$, are $2$-irreps of
$SO(3)_{p}$, providing examples of $p$-adic qubits for every $p>2$.

##  V Representations for $SO(3)_{3}\mod 3$

In the present section, we exploit the previous results for the case $p=3$ to
study some irreducible representations of $SO(3)_{3}$. From these will emerge
one irrep of degree 2, which we will propose as the foundation for the
$3$-adic qubit.

###  V.1 Abelianization

As we anticipated, a starting point in the study of the irreducible
representations of $SO(3)_{3}$ is through those of its projection $G_{3}$,
which is a finite group. First, we find its abelianization to understand how
many one-dimensional representations $G_{3}$ has, which benefits the analysis
of the degrees of its remaining irreps.

We specify Eq. (59) for $p=3$: by Remark IV.7, the matrices of $G_{3}$ are
parametrized by

$M(a,b,c,d,s)=\begin{pmatrix}a&-sb&0\\\ b&sa&0\\\ c&d&s\end{pmatrix}$ (72)

with $(a,b)\in\\{(\pm 1,0),\,(0,\pm 1)\\}$. Now we consider Eqs. (61) and
(LABEL:eq:commG35entries) for $p=3$: $c_{21}\equiv 0\mod 3$ because
$a_{11}a_{21}\equiv b_{11}b_{21}\equiv 0\mod 3$. Moreover, $c_{11}\equiv\pm
1\mod 3$, so a commutator of $G_{3}$ is of the form

$[M,N]=\begin{pmatrix}\pm 1&0&0\\\ 0&\pm 1&0\\\ c_{31}&c_{32}&1\end{pmatrix}.$

The set of commutators of this form is closed under multiplication, meaning
that

$[G_{3},G_{3}]=\left\\{\begin{pmatrix}c_{11}&0&0\\\ 0&c_{11}&0\\\
c_{31}&c_{32}&1\end{pmatrix},\ c_{11}\in\\{\pm
1\\},\,c_{31},c_{32}\in{{\mathbb{Z}}}/3\right\\}.$ (73)

By comparing this with Eq. (72), we can state the following.

###### Proposition V.1

$M(a,b,c,d,s)\in G_{3}$ belongs to the commutator subgroup $[G_{3},G_{3}]$ if
and only if $s=1$ and $(a,b)=(\pm 1,0)$.

The order of $[G_{3},G_{3}]$ is $18$, so the abelianization
$\text{Ab}(G_{3})\mathrel{\mathop{\mathchar 58\relax}}=G_{3}/[G_{3},G_{3}]$ is
of order $72/18=4$.

###### Proposition V.2

$\text{Ab}(G_{3})$ is isomorphic to $\mathbb{Z}/2\times\mathbb{Z}/2$: a class
of matrices is characterized by $s$ and $t\mathrel{\mathop{\mathchar
58\relax}}=a^{2}-b^{2}$ using the form in (72). The following equations are
possible representatives for the classes $\text{Ab}(s,t)$:

$\begin{array}[]{cc}\text{Ab}(1,1)\rightsquigarrow\begin{pmatrix}1&0&0\\\
0&1&0\\\
0&0&1\end{pmatrix},&\text{Ab}(-1,1)\rightsquigarrow\begin{pmatrix}1&0&0\\\
0&-1&0\\\ 0&0&-1\end{pmatrix},\\\
\text{Ab}(1,-1)\rightsquigarrow\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
0&0&1\end{pmatrix},&\text{Ab}(-1,-1)\rightsquigarrow\begin{pmatrix}0&1&0\\\
1&0&0\\\ 0&0&-1\end{pmatrix}.\end{array}$

###### Proof.

According to Eqs. (72) and (73) we have

$M(a,b,c,d,s)[G_{3},G_{3}]=\begin{pmatrix}a&-sb&0\\\ b&sa&0\\\
c&d&s\end{pmatrix}\begin{pmatrix}c_{11}&0&0\\\ 0&c_{11}&0\\\
c_{31}&c_{32}&1\end{pmatrix}=\begin{pmatrix}ac_{11}&-sc_{11}b&0\\\
c_{11}b&sac_{11}&0\\\ c_{11}c+sc_{31}&c_{11}d+sc_{32}&s\end{pmatrix}$

with $c_{11},s\in\\{\pm 1\\},\,(a,b)\in\\{(\pm 1,0),(0,\pm 1)\\}$.

With fixed $a,b$ and $s$, we have
$M(a,b,c,d,s)[G_{3},G_{3}]=M(a,b,c^{\prime},d^{\prime},s)[G_{3},G_{3}]$ for
every $c,c^{\prime},d,d^{\prime}\in{{\mathbb{Z}}}/3$. Moreover, for every
$c,c^{\prime},\,d,d^{\prime}\in{{\mathbb{Z}}}/3$,

$M(1,0,c,d,1)[G_{3},G_{3}]=M(-1,0,c^{\prime},d^{\prime},1)[G_{3},G_{3}]=\left\\{\begin{pmatrix}c_{11}&0&0\\\
0&c_{11}&0\\\ c_{31}^{\prime}&c_{32}^{\prime}&1\end{pmatrix},\ c_{11}\in\\{\pm
1\\},\,c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/3\right\\}.$

A representative of this class is $\openone$, and the class is located by
$s=1$ and $t=a^{2}-b^{2}=1$.

Similarly, for every $c,c^{\prime},\,d,d^{\prime}\in{{\mathbb{Z}}}/3$,

$\displaystyle
M(1,0,c,d,-1)[G_{3},G_{3}]=M(-1,0,c^{\prime},d^{\prime},-1)[G_{3},G_{3}]=\left\\{\begin{pmatrix}c_{11}&0&0\\\
0&-c_{11}&0\\\ c_{31}^{\prime}&c_{32}^{\prime}&-1\end{pmatrix},\
c_{11}\in\\{\pm
1\\},\,c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/3\right\\},$
$\displaystyle
M(0,1,c,d,1)[G_{3},G_{3}]=M(0,-1,c^{\prime},d^{\prime},1)[G_{3},G_{3}]=\left\\{\begin{pmatrix}0&-c_{11}&0\\\
c_{11}&0&0\\\ c_{31}^{\prime}&c_{32}^{\prime}&1\end{pmatrix},\ c_{11}\in\\{\pm
1\\},\,c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/3\right\\},$
$\displaystyle
M(0,1,c,d,-1)[G_{3},G_{3}]=M(0,-1,c^{\prime},d^{\prime},-1)[G_{3},G_{3}]=\left\\{\begin{pmatrix}0&c_{11}&0\\\
c_{11}&0&0\\\ c_{31}^{\prime}&c_{32}^{\prime}&1\end{pmatrix},\ c_{11}\in\\{\pm
1\\},\,c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/3\right\\}.$

$\sqcap$$\sqcup$

As a consequence of Proposition V.2, we have the following further result.

###### Proposition V.3

$G_{3}$ cannot have more than four orthogonal one-dimensional irreducible
representations $U\mathrel{\mathop{\mathchar
58\relax}}G_{3}\rightarrow\mathbb{C}^{\ast}$: a possible choice for the four
1-irreps is as follows:

  * •

$\det\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto 1$,

  * •

$s\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto s$,

  * •

$t\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto a^{2}-b^{2}$, and

  * •

$st\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto s(a^{2}-b^{2})$.

###### Proof.

A $1$-irrep for a group is always the trivial representation mapping any
element of the group to $1$. This can be seen as the determinant on matrices
of $G_{3}$.

The irreducible character $\chi_{U}$ of a 1-irrep $U$ is $U$ itself: in fact,
the trace of a $1\times 1$ matrix is equal to the only value it contains.
Since the product in ${{\mathbb{C}}}$ is commutative, the $1$-irreps of
$G_{3}$ have to be invariant on abelianized classes. As consequences, we can
actually build the character table for $\text{Ab}(G_{3})$ in order to find the
1-irreps of $G_{3}$, and the two characterizing quantities $s$ and $t$ in
Proposition V.2 can be good candidates as characters. Indeed, we show that
$\det,\,s,\,t,\,st$, as defined in the Proposition, are the four $1$-irreps of
this group.

Now, $\det,\,s,\,t,\,st$ are homomorphisms from $G_{3}$ to
${{\mathbb{C}}}^{\ast}$ because

$M(a,b,c,d,s)M(a^{\prime},b^{\prime},c^{\prime},d^{\prime},s^{\prime})=\begin{pmatrix}aa^{\prime}-sbb^{\prime}&\dots&0\\\
sab^{\prime}+a^{\prime}b&\dots&0\\\ \dots&\dots&ss^{\prime}\end{pmatrix},$

so $s(MM^{\prime})=ss^{\prime}=s(M)s(M^{\prime})$,
$t(MM^{\prime})=(aa^{\prime}-sbb^{\prime})^{2}-(sab^{\prime}+a^{\prime}b)^{2}=(a^{2}-b^{2})({a^{\prime}}^{2}-{b^{\prime}}^{2})=t(M)t(M^{\prime})$,
and similarly for $st$.

Furthermore, in accordance with general Peter-Weyl theory simon ,
$\det,\,s,\,t,\,st$ are orthogonal representations. This can be seen by
looking at the character table of $\text{Ab}(G_{3})$.

$\begin{array}[]{ l c c c c
}\hline\cr\hline\cr\text{Ab}(G_{3})\simeq\\{(s,t)\\}&\quad(1,1)&\quad(1,-1)&\quad(-1,1)&\quad(-1,-1)\\\
\det&1&1&1&1\\\ s&1&1&-1&-1\\\ t&1&-1&1&-1\\\ st&1&-1&-1&1\\\
\hline\cr\hline\cr\end{array}$

The rows are all different, and they correspond to the different values the
$1$-irreps associate with each element of $G_{3}$.

We have shown four 1-irreps of $G_{3}$: any other 1-irrep would be equivalent
to these because of Proposition V.2.  $\sqcap$$\sqcup$

###  V.2 Degree of representations

The Peter-Weyl theorem simon  states that the sum of the squares of the
degrees $d_{\alpha}$ of the irreducible representations of $G$ located by
$\alpha$ is equal to the order of $G$.

The order of $G_{3}$ is $72$ by Corollary IV.4, and there are $4$ irreducible
representations of $G_{3}$ of degree $1$ by Proposition V.3. Then,

$\sum_{\alpha\mid d_{\alpha}>1}d_{\alpha}^{2}=68.$ (74)

Unfortunately, this is not sufficient to deduce the nature of higher-degree
irreps: we need another proposition (proved in Appendix C).

###### Proposition V.4

$G_{3}$ is partitioned in nine conjugacy classes. Their orders and lists are
in Appendix C.

As a consequence of the fact that the number of conjugacy classes of a group
equals the number of its irreducible representations simon , there are nine
non equivalent irreps of $G_{3}$, among which five have degree greater than 1.
Suppose, then, to sort the degrees of these irreps from the smaller to the
greater as $d_{1},\ldots,d_{9}$: we can state that

$\displaystyle d_{1}=d_{2}=d_{3}=d_{4}=1,$ (75) $\displaystyle
d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}+d_{9}^{2}=68.$ (76)

It is known that $d_{\alpha}$ divides $\lvert G\rvert$ when $\alpha$ locates
an irreducible representation of $G$. This is enough to deduce all of the
$d_{\alpha}$:

###### Proposition V.5

The only solution for (76) for integers $d_{\alpha}$ divisors of $72$ greater
than $1$ is the following up to permutations:

$d_{5}=2,\ d_{6}=d_{7}=d_{8}=d_{9}=4.$ (77)

The proof of this proposition is in Appendix D. Looking at $d_{5}$, we can now
state the following.

###### Proposition V.6

There exists a two-dimensional irrep $V$ of $G_{3}$, that is,

$K_{3}\mathrel{\mathop{\mathchar 58\relax}}G_{3}\rightarrow GL(V),\quad
V\simeq{{\mathbb{C}}}^{2},$

such that for any subspace $Y\subset V$,

$(K_{3}(G_{3}))Y=Y\ \text{implies}\ Y=\\{0_{V}\\}\text{ or }Y=V.$

###  V.3 A $3$-adic qubit

Now it is not hard to find an explicit form of a 2-irrep for $SO(3)_{3}$,
pigliapochi-msc . An important consequence of the way $G_{3}=SO(3)_{3}\mod
3=\pi_{1}(SO(3)_{3})$ was constructed according to Sec. III is as follows.

###### Proposition V.7

There exists a 2-irrep $J_{3}$ of $SO(3)_{3}$,

$J_{3}\mathrel{\mathop{\mathchar 58\relax}}=K_{3}\circ\pi_{1}.$

###### Proof.

$J_{3}$ is a homomorphism because it is composition of homomorphisms. $J_{3}$
is irreducible since $K_{3}$ is so: if $Y\subset V$, then the following are
equivalent:

$\displaystyle(J_{3}(SO(3)_{3}))Y=Y,$ (78)
$\displaystyle(K_{3}(\pi_{1}(SO(3)_{3})))Y=Y,$ (79)
$\displaystyle(K_{3}(G_{3}))Y=Y.$ (80)

The latter is equivalent to $Y=\\{0_{V}\\}$ or $Y=V$.  $\sqcap$$\sqcup$

The first ingredient to find an explicit form for a 2-irrep $K_{3}$ is the
homomorphism $K_{3}^{\prime}$ of Eq. (63). Furthermore, one considers the
embedding of the integers modulo $3$ into the complex numbers

$\imath\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Z}}}/3\hookrightarrow{{\mathbb{C}}},\
0_{{{\mathbb{Z}}}/3}\mapsto 0_{{\mathbb{C}}},\,\pm
1_{{{\mathbb{Z}}}/3}\mapsto\pm 1_{{\mathbb{C}}}.$ (81)

This can be extended to a map between groups of matrices on these fields,
respectively, which we will always call $\imath$ with abuse of notation. Note
that the solutions $(a,b)\in\\{(\pm 1,0),(0,\pm
1)\\}\subset({{\mathbb{Z}}}/3)^{2}$ of $a^{2}+b^{2}\equiv 1\mod 3$ are such
that their embedding $\\{(\pm 1,0),(0,\pm 1)\\}\subset{{\mathbb{C}}}^{2}$ is
solution of $a^{2}+b^{2}=1$ in ${{\mathbb{C}}}$.

###### Proposition V.8

$\imath\mathrel{\mathop{\mathchar
58\relax}}\mathfrak{I}(K_{3}^{\prime})\rightarrow GL({{\mathbb{C}}}^{2})$ is a
group homomorphism.

###### Proof.

  * •

$\mathfrak{I}(\imath)\subset GL({{\mathbb{C}}}^{2})$,

$\det\begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}_{{{\mathbb{C}}}}=s(a^{2}+b^{2})=s\neq 0,$

where $(a,b)\in\\{(\pm 1,0),(0,\pm 1)\\}\subset{{\mathbb{C}}}^{2}$.

  * •

$\\{0,\pm 1\\}\subset{{\mathbb{C}}}$ is not a subfield since it is not closed
under the addition operation. However, if we look at the entries of
$\begin{pmatrix}aa^{\prime}-sbb^{\prime}&-s^{\prime}(ab^{\prime}+sa^{\prime}b)\\\
a^{\prime}b+sab^{\prime}&s^{\prime}(saa^{\prime}-bb^{\prime})\end{pmatrix}$,
when these operations are to be performed both in ${{\mathbb{Z}}}/3$ and in
${{\mathbb{C}}}$, they never present sums between $\pm 1$ and $\pm 1$ because
$(a,b),(a^{\prime},b^{\prime})\in\\{(\pm 1,0),(0,\pm 1)\\}$. It follows that

$\displaystyle\imath\begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}_{{{\mathbb{Z}}}/3}\imath\begin{pmatrix}a^{\prime}&-s^{\prime}b^{\prime}\\\
b^{\prime}&s^{\prime}a^{\prime}\end{pmatrix}_{{{\mathbb{Z}}}/3}=\begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}_{{{\mathbb{C}}}}\begin{pmatrix}a^{\prime}&-s^{\prime}b^{\prime}\\\
b^{\prime}&s^{\prime}a^{\prime}\end{pmatrix}_{{{\mathbb{C}}}}=\begin{pmatrix}aa^{\prime}-sbb^{\prime}&-s^{\prime}(ab^{\prime}+sa^{\prime}b)\\\
a^{\prime}b+sab^{\prime}&s^{\prime}(saa^{\prime}-bb^{\prime})\end{pmatrix}_{{{\mathbb{C}}}}$
$\displaystyle=\imath\begin{pmatrix}aa^{\prime}-sbb^{\prime}&-s^{\prime}(ab^{\prime}+sa^{\prime}b)\\\
a^{\prime}b+sab^{\prime}&s^{\prime}(saa^{\prime}-bb^{\prime})\end{pmatrix}_{{{\mathbb{Z}}}/3}=\imath\left(\begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}_{{{\mathbb{Z}}}/3}\begin{pmatrix}a^{\prime}&-s^{\prime}b^{\prime}\\\
b^{\prime}&s^{\prime}a^{\prime}\end{pmatrix}_{{{\mathbb{Z}}}/3}\right).$

$\sqcap$$\sqcup$

Now we are able to state the following.

###### Proposition V.9

The function

$K_{3}\mathrel{\mathop{\mathchar 58\relax}}G_{3}\rightarrow
GL({{\mathbb{C}}}^{2}),\ \ \ \ K_{3}\mathrel{\mathop{\mathchar
58\relax}}=\imath\circ K_{3}^{\prime}$ (82)

is the unitary $2$-irrep of $G_{3}$ up to isomorphisms.

###### Proof.

As said, parameterization (72) is fundamental to understand the action of
$G_{3}$, that is,

$\begin{pmatrix}a&-sb&0\\\ b&sa&0\\\
c&d&s\end{pmatrix}_{{{\mathbb{Z}}}/3}\stackrel{{\scriptstyle
K_{3}^{\prime}}}{{\mapsto}}\ \begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}_{{{\mathbb{Z}}}/3}\stackrel{{\scriptstyle\imath}}{{\mapsto}}\
\begin{pmatrix}a&-sb\\\ b&sa\end{pmatrix}_{{\mathbb{C}}}.$

The image $\mathfrak{I}(K_{3})$ of $K_{3}$ is parameterized by
$(a,b)\in\\{(\pm 1,0),(0,\pm 1)\\}\subset{{\mathbb{C}}}^{2},s\in\\{\pm
1\\}\subset{{\mathbb{C}}}$.

  * •

$\mathfrak{I}(K_{3})\subset GL({{\mathbb{C}}}^{2})$ as noticed in Proposition
V.8.

  * •

$K_{3}$ is a group homomorphism, being the composition of two group
homomorphisms $K_{3}^{\prime}$ and $\imath$.

  * •

$K_{3}$ is irreducible:
if $Y$ is a subrepresentation such that $Y\neq\\{0\\},{{\mathbb{C}}}^{2}$,
then $\text{dim}(Y)=1$, i.e., $Y={{\mathbb{C}}}y$ for any $y\in Y$. By
definition of subrepresentation, $K_{3}(G_{3})Y\subset Y$ means that
$K_{3}(M)y\in Y$ for every $M\in G_{3}$ and $y\in Y$: it must be
$K_{3}(M)y=\lambda y$ for some $\lambda\in{{\mathbb{C}}}$ depending on $y$ and
$M$. In other words, $y$ is a common eigenvector for every action of the
group. This must be true, in particular, for

$K_{3}\big{(}M(1,0,c,d,1)\big{)}=\begin{pmatrix}1&0\\\
0&1\end{pmatrix}_{{\mathbb{C}}}\ \ \ \text{of eigenvectors}\ \ \
\begin{pmatrix}1\\\ 0\end{pmatrix},\ \begin{pmatrix}0\\\ 1\end{pmatrix}$

and for

$K_{3}\big{(}M(0,1,c,d,1)\big{)}=\begin{pmatrix}0&-1\\\
1&0\end{pmatrix}_{{{\mathbb{C}}}}\ \ \ \text{of eigenvectors}\ \ \
\begin{pmatrix}i\\\ 1\end{pmatrix},\ \begin{pmatrix}-i\\\ 1\end{pmatrix}.$

These two group actions have no common eigenvectors. This is a contradiction:
the representation $K_{3}$ is irreducible.

  * •

$K_{3}$ is unitary,
namely, $\mathfrak{I}(K_{3})$ is composed by unitary linear transformations on
${{\mathbb{C}}}^{2}$,

$\displaystyle\langle\begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}\begin{pmatrix}x_{1}\\\
x_{2}\end{pmatrix},\begin{pmatrix}a&-sb\\\
b&sa\end{pmatrix}\begin{pmatrix}y_{1}\\\
y_{2}\end{pmatrix}\rangle=\langle\begin{pmatrix}ax_{1}-sbx_{2}\\\
bx_{1}+sax_{2}\end{pmatrix},\begin{pmatrix}ay_{1}-sby_{2}\\\
by_{1}+say_{2}\end{pmatrix}\rangle$
$\displaystyle=(ax_{1}-sbx_{2})(ay_{1}-sby_{2})+(bx_{1}+sax_{2})(by_{1}+say_{2})$
$\displaystyle=(a^{2}+b^{2})(x_{1}y_{1}+x_{2}y_{2})=x_{1}y_{1}+x_{2}y_{2}=\langle\begin{pmatrix}x_{1}\\\
x_{2}\end{pmatrix},\begin{pmatrix}y_{1}\\\ y_{2}\end{pmatrix}\rangle$

for every $\begin{pmatrix}x_{1}\\\ x_{2}\end{pmatrix},\begin{pmatrix}y_{1}\\\
y_{2}\end{pmatrix}\in{{\mathbb{C}}}^{2}$ according to its Euclidean scalar
product.

$\sqcap$$\sqcup$

We have, then, an explicit form for $J_{3}$ too: it is a function that makes a
$2\times 2$ complex matrix out of the first components of the $p$-adic
sequences in the upper-left $2\times 2$ part of a matrix in $SO(3)_{3}$, that
is,

$\displaystyle J_{3}\mathrel{\mathop{\mathchar
58\relax}}=K_{3}\circ\pi_{1}\mathrel{\mathop{\mathchar
58\relax}}SO(3)_{3}\rightarrow GL({{\mathbb{C}}}^{2})$ (83)
$\displaystyle\begin{pmatrix}(a_{111},a_{112},\ldots)&(a_{121},a_{122},\ldots)&(a_{131},a_{132},\ldots)\\\
(a_{211},a_{212},\ldots)&(a_{221},a_{222},\ldots)&(a_{231},a_{232},\ldots)\\\
(a_{311},a_{312},\ldots)&(a_{321},a_{322},\ldots)&(a_{331},a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{3}}\stackrel{{\scriptstyle
J_{3}}}{{\longmapsto}}\begin{pmatrix}a_{111}&a_{121}\\\
a_{211}&a_{221}\end{pmatrix}_{{\mathbb{C}}}.$

We, thus, have the following final result.

###### Proposition V.10

The pair $({{\mathbb{C}}}^{2},J_{3})$ with $J_{3}$ defined as in (83) is a
3-adic qubit, that is, a continuous unitary irreducible linear (and then
projective) representation of degree $2$ on ${{\mathbb{C}}}$ of the $3$-adic
special orthogonal group $SO(3)_{3}$ defined in (3).

The isomorphism $\varphi\mathrel{\mathop{\mathchar
58\relax}}\mathfrak{I}(K_{3}^{\prime})\rightarrow D_{4}$ of Eq. (70) gives us
an equivalent way to derive the $2$-irrep $J_{3}$ of $SO(3)_{3}$, applying the
representation theory of dihedral groups of even degree to
$\mathfrak{I}(K_{3}^{\prime})$. However, first, Appendix B.1 confirms the four
one-dimensional representations we found in Proposition V.3 for the case
$p=3$. Then, Appendix B.2 shows that for $p=3$ there is a unique 2-irrep of
$\mathfrak{I}(K_{3}^{\prime})$,

$C_{{{\mathbb{Z}}}/3}\mapsto\begin{pmatrix}0&-1\\\
1&0\end{pmatrix}_{{\mathbb{C}}},\ \ \ \
Z_{{{\mathbb{Z}}}/3}\mapsto\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}_{{\mathbb{C}}}.$ (84)

This coincides with $\imath$ of Proposition V.8 if we choose
$(a_{0},b_{0})=(0,1)$, leading to the same qubit representation $J_{3}$ of Eq.
(83), which, then, can be written as $J_{3}=\rho\circ\varphi\circ
K_{3}^{\prime}\circ\pi_{1}$ too, where $\rho$ denotes the $2$-irrep of
$D_{4}$.

##  VI Representations for $SO(3)_{5}\mod 5$

We have $\widehat{G}_{5}=G_{5}$ according to Remark IV.7. We choose $u=2$ as
non-square invertible $5$-adic integer. In this way, the parameterization (59)
for the matrices in $G_{5}$ becomes

$M(a,b,c,d,s)=\begin{pmatrix}a&-2sb&0\\\ b&sa&0\\\ c&d&s\end{pmatrix},$ (85)

where $(a,b)\in\\{(\pm 1,0),\,(\pm 2,1),\,(\pm 2,-1)\\}$ are the solutions of
$a^{2}+2b^{2}\equiv 1\mod 5$.

###  VI.1 Abelianization

We specify Eqs. (61) and (LABEL:eq:commG35entries) for $p=5$: it turns out
that $(c_{11},c_{21})\in\\{(1,0),\,(2,\pm 1)\\}$, so a commutator of $G_{5}$
is of one of the following forms:

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ c_{31}&c_{32}&1\end{pmatrix},\ \
\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ c_{31}&c_{32}&1\end{pmatrix},\ \
\begin{pmatrix}2&2&0\\\ -1&2&0\\\ c_{31}&c_{32}&1\end{pmatrix}.$ (86)

The set of commutators of these forms is closed under multiplication, meaning
that

$[G_{5},G_{5}]=\left\\{\begin{pmatrix}c_{11}&-2c_{21}&0\\\ c_{21}&c_{11}&0\\\
c_{31}&c_{32}&1\end{pmatrix},\ (c_{11},c_{21})\in\\{(1,0),(2,\pm
1)\\},\,c_{31},c_{32}\in{{\mathbb{Z}}}/5\right\\}.$ (87)

By comparing this with Eq. (85), we can state the following.

###### Proposition VI.1

$M(a,b,c,d,s)\in G_{5}$ belongs to the commutator subgroup $[G_{5},G_{5}]$ if
and only if $s=1$ and $(a,b)\in\\{(1,0),(2,\pm 1)\\}$.

The order of $[G_{5},G_{5}]$ is $75$, so the abelianization
$\text{Ab}(G_{5})\mathrel{\mathop{\mathchar 58\relax}}=G_{5}/[G_{5},G_{5}]$ is
of order $300/75=4$.

###### Proposition VI.2

$\text{Ab}(G_{5})$ is isomorphic to $\mathbb{Z}/2\times\mathbb{Z}/2$: a class
of matrices is characterized by $s$ and $t\mathrel{\mathop{\mathchar
58\relax}}=\text{sign}(a)$ (taking the representatives of the elements in
${{\mathbb{Z}}}/5$ as $-2,-1,0,1,2$) using the form in (85). The following are
possible representatives for the classes $\text{Ab}(s,t)$:

$\begin{array}[]{cc}\text{Ab}(1,1)\rightsquigarrow\begin{pmatrix}1&0&0\\\
0&1&0\\\
0&0&1\end{pmatrix},&\text{Ab}(-1,1)\rightsquigarrow\begin{pmatrix}1&0&0\\\
0&-1&0\\\ 0&0&-1\end{pmatrix},\\\
\text{Ab}(1,-1)\rightsquigarrow\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
0&0&1\end{pmatrix},&\text{Ab}(-1,-1)\rightsquigarrow\begin{pmatrix}-1&0&0\\\
0&1&0\\\ 0&0&-1\end{pmatrix}.\end{array}$

###### Proof.

According to Eqs. (85) and (87), we have

$M(a,b,c,d,s)[G_{5},G_{5}]=\begin{pmatrix}a&-2sb&0\\\ b&sa&0\\\
c&d&s\end{pmatrix}\begin{pmatrix}c_{11}&-2c_{21}&0\\\ c_{21}&c_{11}&0\\\
c_{31}&c_{32}&1\end{pmatrix}=\begin{pmatrix}ac_{11}-2sbc_{21}&-2ac_{21}-2sc_{11}b&0\\\
c_{11}b+sac_{21}&-2bc_{21}+sac_{11}&0\\\
c_{11}c+c_{21}d+sc_{31}&c_{11}d-2c_{21}c+sc_{32}&s\end{pmatrix}$

with $(c_{11},c_{21})\in\\{(1,0),(2,\pm 1)\\},s\in\\{\pm
1\\},\,(a,b)\in\\{(\pm 1,0),(\pm 2,1),(\pm 2,-1)\\}$.

With fixed $a,b$ and $s$, we have
$M(a,b,c,d,s)[G_{5},G_{5}]=M(a,b,c^{\prime},d^{\prime},s)[G_{5},G_{5}]$ for
every $c,c^{\prime},d,d^{\prime}\in{{\mathbb{Z}}}/5$. Moreover, for every
$c,c^{\prime},c^{\prime\prime},\,d,d^{\prime},d^{\prime\prime}\in{{\mathbb{Z}}}/5$,

$\displaystyle M$
$\displaystyle(1,0,c,d,1)[G_{5},G_{5}]=M(2,1,c^{\prime},d^{\prime},1)[G_{5},G_{5}]=M(2,-1,c^{\prime\prime},d^{\prime\prime},1)[G_{5},G_{5}]$
$\displaystyle=\left\\{\begin{pmatrix}c_{11}&-2c_{21}&0\\\ c_{21}&c_{11}&0\\\
c_{31}^{\prime}&c_{32}^{\prime}&1\end{pmatrix},\
(c_{11},c_{21})\in\\{(1,0),(2,\pm
1)\\},c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/5\right\\}.$

A representative of this class is $\openone$, and the class is located by
$s=1,\,t=\text{sign}(a)=1$.

Similarly, for every
$c,c^{\prime},c^{\prime\prime},\,d,d^{\prime},d^{\prime\prime}\in{{\mathbb{Z}}}/5$,

$\displaystyle M$
$\displaystyle(-1,0,c,d,1)[G_{5},G_{5}]=M(-2,1,c^{\prime},d^{\prime},1)[G_{5},G_{5}]=M(-2,-1,c^{\prime\prime},d^{\prime\prime},1)[G_{5},G_{5}]$
$\displaystyle=\left\\{\begin{pmatrix}-c_{11}&2c_{21}&0\\\
-c_{21}&-c_{11}&0\\\ c_{31}^{\prime}&c_{32}^{\prime}&1\end{pmatrix},\
(c_{11},c_{21})\in\\{(1,0),(2,\pm
1)\\},c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/5\right\\},$
$\displaystyle M$
$\displaystyle(1,0,c,d,-1)[G_{5},G_{5}]=M(2,1,c^{\prime},d^{\prime},-1)[G_{5},G_{5}]=M(2,-1,c^{\prime\prime},d^{\prime\prime},-1)[G_{5},G_{5}]$
$\displaystyle=\left\\{\begin{pmatrix}c_{11}&-2c_{21}&0\\\
-c_{21}&-c_{11}&0\\\ c_{31}^{\prime}&c_{32}^{\prime}&-1\end{pmatrix},\
(c_{11},c_{21})\in\\{(1,0),(2,\pm
1)\\},c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/5\right\\},$
$\displaystyle M$
$\displaystyle(-1,0,c,d,-1)[G_{5},G_{5}]=M(-2,1,c^{\prime},d^{\prime},-1)[G_{5},G_{5}]=M(-2,-1,c^{\prime\prime},d^{\prime\prime},-1)[G_{5},G_{5}]$
$\displaystyle=\left\\{\begin{pmatrix}-c_{11}&2c_{21}&0\\\ c_{21}&c_{11}&0\\\
c_{31}^{\prime}&c_{32}^{\prime}&-1\end{pmatrix},\
(c_{11},c_{21})\in\\{(1,0),(2,\pm
1)\\},c_{31}^{\prime},c_{32}^{\prime}\in{{\mathbb{Z}}}/5\right\\}.$

$\sqcap$$\sqcup$

As a consequence of Proposition VI.2, we have the following further result.

###### Proposition VI.3

$G_{5}$ has four orthogonal one-dimensional irreducible representations
$U\mathrel{\mathop{\mathchar 58\relax}}G_{5}\rightarrow\mathbb{C}^{\ast}$: a
possible choice for the four 1-irreps is as follows:

  * •

$\det\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto 1$,

  * •

$s\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto s$,

  * •

$t\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto\text{sign}(a)$
(taking the representatives of the elements in ${{\mathbb{Z}}}/5$ as
$-2,-1,0,1,2$), and

  * •

$st\mathrel{\mathop{\mathchar 58\relax}}M(a,b,c,d,s)\mapsto
s\cdot\text{sign}(a)$.

###### Proof.

As for the case $p=3$, the $1$-irreps of $G_{5}$ correspond to the irreducible
characters of $\text{Ab}(G_{5})$. The character table of
$\text{Ab}(G_{5})\simeq{{\mathbb{Z}}}/2\times{{\mathbb{Z}}}/2$ is as follows.

$\begin{array}[]{ l c c c c }\hline\cr\hline\cr
Ab(G_{5})\simeq\\{Ab(s,t)\\}&\quad Ab(1,1)&\quad Ab(1,-1)&\quad Ab(-1,1)&\quad
Ab(-1,-1)\\\ \det&1&1&1&1\\\ s&1&1&-1&-1\\\ t&1&-1&1&-1\\\ st&1&-1&-1&1\\\
\hline\cr\hline\cr\end{array}$

The bijective correspondence between the $1$-irreps of $G_{5}$ and the irreps
of $\text{Ab}(G_{5})$ is as follows: if
$\widetilde{\rho}\mathrel{\mathop{\mathchar
58\relax}}Ab(G_{5})\rightarrow{{\mathbb{C}}}^{\ast}$ is a $1$-irrep of
$\text{Ab}(G_{5})$, then $\rho(M)\mathrel{\mathop{\mathchar
58\relax}}=\widetilde{\rho}(M[G_{5},G_{5}])$ is a $1$-irrep of $G_{5}$ (and
vice versa). This concludes the proof.  $\sqcap$$\sqcup$

###  VI.2 Degree of representations

The Peter-Weyl theorem simon  for $G_{5}$ states that
$\sum_{\alpha}d_{\alpha}^{2}=\lvert G_{5}\rvert=300$, where $d_{\alpha}$ are
the degrees of the irreducible representations of $G_{5}$ located by $\alpha$.
Then,

$\sum_{\alpha\mid d_{\alpha}>1}d_{\alpha}^{2}=296$ (88)

according to Proposition VI.3.

This is not enough to deduce the nature of higher-degree irreps, but another
proposition (proved in Appendix E) comes to our help.

###### Proposition VI.4

$G_{5}$ is partitioned in $14$ conjugacy classes. Their orders and lists are
in Appendix E.

It means that there are $14$ non-equivalent irreps of $G_{5}$, among which
four have degree 1. Sorting the degrees of these irreps from the smaller to
the greater as $d_{1},\ldots,d_{14}$, we have

$\displaystyle d_{1}=$ $\displaystyle d_{2}=d_{3}=d_{4}=1,$ (89)
$\displaystyle\sum_{\alpha=5}^{14}d_{\alpha}^{2}=296.$ (90)

Solving Eq. (90) for $d_{\alpha}$ divisors of $\lvert G_{5}\rvert$ greater
than $1$ provides $18$ possible solutions, among which four present no degree
equal to $2$, which we are interested relatively to $5$-adic qubits. However,
there exists a stronger result due to It$\hat{\text{o}}$ ito , namely,
$d_{\alpha}$ divides the index in $G_{5}$ of any of its (maximal) Abelian
normal subgroups. In order to exploit it, we prove the following.

###### Proposition VI.5

$A\mathrel{\mathop{\mathchar
58\relax}}=\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/5\big{\\}}$ (91)

is the maximal normal Abelian subgroup of $G_{5}$.

###### Proof.

Normal subgroups are unions of conjugacy classes. The elements in each
$C_{i}$, $i=6,\dots,14$, $C_{i}$ conjugacy classes of $G_{5}$ in Appendix E,
do not commute, in general. These classes, then, are not of interest for our
purposes here. On the other hand, the matrices in
$\bigcup_{i=1}^{5}C_{i}=\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/5\big{\\}}$ commute. This set is closed under
matrix multiplication and contains the identity matrix.  $\sqcap$$\sqcup$

The index of $A$ in $G_{5}$ is $\lvert G_{5}\mathrel{\mathop{\mathchar
58\relax}}A\rvert=\lvert G_{5}\rvert/\lvert A\rvert=300/25=12$. It follows by
It$\hat{\text{o}}$’s theorem that $d_{\alpha}\in\\{2,3,4,6,12\\}$ for every
$\alpha=5,\dots,14$.

###### Proposition VI.6

Equation (90) has three solutions for integers $d_{\alpha}\in\\{2,3,4,6,12\\}$
up to permutations.

$\begin{array}[]{lccccccccccc}\hline\cr\hline\cr&\quad d_{5}&\quad d_{6}&\quad
d_{7}&\quad d_{8}&\quad d_{9}&\quad d_{10}&\quad d_{11}&\quad d_{12}&\quad
d_{13}&\quad d_{14}\\\ \textup{First}&2&2&3&3&3&3&6&6&6&\quad 12\\\
\textup{Second}&2&2&6&6&6&6&6&6&6&\quad 6\\\
\textup{Third}&2&4&4&4&4&4&4&4&6&\quad 12\\\ \hline\cr\hline\cr\end{array}$

The proof of this proposition is in Appendix F.

In order to further restrict this set of solutions, we recall Eq. (70): here
$\mathfrak{I}(K_{5}^{\prime})$ is isomorphic to the dihedral group $D_{6}$.
Appendix (B.1) confirms the four one-dimensional representations we found in
(VI.3). Moreover, Appendix (B.2) shows that $\mathfrak{I}(K_{5}^{\prime})$ has
$\frac{p-1}{2}=2$ irreps of degree $2$; hence, $G_{5}$ has at least these two
2-irreps. This reduces the possibilities of Proposition VI.6 to the following
solutions.

$\begin{array}[]{lccccccccccc}\hline\cr\hline\cr&\quad d_{5}&\quad d_{6}&\quad
d_{7}&\quad d_{8}&\quad d_{9}&\quad d_{10}&\quad d_{11}&\quad d_{12}&\quad
d_{13}&\quad d_{14}\\\ \textup{First}&2&2&3&3&3&3&6&6&6&\quad 12\\\
\textup{Second}&2&2&6&6&6&6&6&6&6&\quad 6\\\ \hline\cr\hline\cr\end{array}$

In any case, we conclude that $G_{5}$ leads exactly to two qubit
representations: those coming from the two unitary $2$-irreps of the dihedral
group $D_{6}$.

###  VI.3 $5$-Adic qubits

Similarly to Proposition V.7, we have the following.

###### Proposition VI.7

There exist two unitary 2-irreps $J_{5}^{(1)},\,J_{5}^{(2)}$ of $SO(3)_{5}$,

$J_{5}^{(i)}\mathrel{\mathop{\mathchar 58\relax}}=\rho_{i}\circ\varphi\circ
K_{5}^{\prime}\circ\pi_{1},$ (92)

where $\varphi$ represents isomorphism (70) and $\rho_{i},\ i=1,2$ are the two
$2$-irreps of $D_{6}$ (Appendix B.2),

$\displaystyle\rho_{1}\,\mathrel{\mathop{\mathchar 58\relax}}\quad
a\mapsto\begin{pmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\
\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix},\quad
x\mapsto\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix},$
$\displaystyle\rho_{2}\,\mathrel{\mathop{\mathchar 58\relax}}\quad
a\mapsto\begin{pmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\
\frac{\sqrt{3}}{2}&-\frac{1}{2}\end{pmatrix},\quad
x\mapsto\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

###### Proof.

$J_{5}^{(i)}$ for $i=1,2$ are group homomorphisms from $SO(3)_{5}$ to
$GL({{\mathbb{C}}}^{2})$ because they are compositions of group homomorphisms.
They are irreducible since the $\rho_{i}$ are so: if
$Y\subset{{\mathbb{C}}}^{2}$, then the following are equivalent:

$\displaystyle\left(J_{5}^{(i)}(SO(3)_{5})\right)Y=Y,$ (93)
$\displaystyle\rho_{i}\big{(}(\varphi\circ
K_{5}^{\prime}\circ\pi_{1})(SO(3)_{5})\big{)}Y=Y,$ (94)
$\displaystyle\rho_{i}(D_{6})Y=Y.$ (95)

The latter is equivalent to $Y=\\{0\\}$ or $Y={{\mathbb{C}}}^{2}$.
$\sqcap$$\sqcup$

To write them explicitly, it is enough to tell the action of the generators
$C,Z$ of $\mathfrak{I}(K_{5}^{\prime})$. Choosing $(a_{0},b_{0})=(-2,1)$ as
the generator of the solutions $(a,b)$ to $a^{2}+2b^{2}\equiv 1\mod 5$ and
writing $5$-adic numbers as sequences of integer numbers modulo $5^{k}$, we
get

$\displaystyle J_{5}^{(1)}\mathrel{\mathop{\mathchar 58\relax}}$
$\displaystyle\begin{pmatrix}(-2,a_{112},\ldots)&(-2,a_{122},\ldots)&(0,a_{132},\ldots)\\\
(1,a_{212},\ldots)&(-2,a_{222},\ldots)&(0,a_{232},\ldots)\\\
(a_{311},a_{312},\ldots)&(a_{321},a_{322},\ldots)&(1,a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{5}}\stackrel{{\scriptstyle
K_{5}^{\prime}\circ\pi_{1}}}{{\longmapsto}}C=\begin{pmatrix}-2&-2\\\
1&-2\end{pmatrix}_{{{\mathbb{Z}}}/5}\stackrel{{\scriptstyle\rho_{1}\circ\varphi}}{{\longmapsto}}\begin{pmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\
\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}_{{\mathbb{C}}}$ (96)
$\displaystyle\begin{pmatrix}(1,a_{112},\ldots)&(0,a_{122},\ldots)&(0,a_{132},\ldots)\\\
(0,a_{212},\ldots)&(-1,a_{222},\ldots)&(0,a_{232},\ldots)\\\
(a_{311},a_{312},\ldots)&(a_{321},a_{322},\ldots)&(-1,a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{5}}\stackrel{{\scriptstyle
K_{5}^{\prime}\circ\pi_{1}}}{{\longmapsto}}Z=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}_{{{\mathbb{Z}}}/5}\stackrel{{\scriptstyle\rho_{1}\circ\varphi}}{{\longmapsto}}\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}_{{\mathbb{C}}}$

and

$\displaystyle J_{5}^{(2)}\mathrel{\mathop{\mathchar 58\relax}}$
$\displaystyle\begin{pmatrix}(-2,a_{112},\ldots)&(-2,a_{122},\ldots)&(0,a_{132},\ldots)\\\
(1,a_{212},\ldots)&(-2,a_{222},\ldots)&(0,a_{232},\ldots)\\\
(a_{311},a_{312},\ldots)&(a_{321},a_{322},\ldots)&(1,a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{5}}\stackrel{{\scriptstyle
K_{5}^{\prime}\circ\pi_{1}}}{{\longmapsto}}C=\begin{pmatrix}-2&-2\\\
1&-2\end{pmatrix}_{{{\mathbb{Z}}}/5}\stackrel{{\scriptstyle\rho_{2}\circ\varphi}}{{\longmapsto}}\begin{pmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\
\frac{\sqrt{3}}{2}&-\frac{1}{2}\end{pmatrix}_{{\mathbb{C}}}$ (97)
$\displaystyle\begin{pmatrix}(1,a_{112},\ldots)&(0,a_{122},\ldots)&(0,a_{132},\ldots)\\\
(0,a_{212},\ldots)&(-1,a_{222},\ldots)&(0,a_{232},\ldots)\\\
(a_{311},a_{312},\ldots)&(a_{321},a_{322},\ldots)&(-1,a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{5}}\stackrel{{\scriptstyle
K_{5}^{\prime}\circ\pi_{1}}}{{\longmapsto}}Z=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}_{{{\mathbb{Z}}}/5}\stackrel{{\scriptstyle\rho_{2}\circ\varphi}}{{\longmapsto}}\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}_{{\mathbb{C}}}.$

The matrix $\begin{pmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\
\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}$ is of order $6$ as $C$, and
$\rho_{1}$ is a faithful representation of
$D_{6}\simeq\mathfrak{I}(K_{5}^{\prime})$. On the other hand,
$\begin{pmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\
\frac{\sqrt{3}}{2}&-\frac{1}{2}\end{pmatrix}$ is of order $3$, so $\rho_{2}$
is not injective.

We, thus, arrive to the following final result.

###### Proposition VI.8

The pairs $({{\mathbb{C}}}^{2},J_{5}^{(1)})$,
$({{\mathbb{C}}}^{2},J_{5}^{(2)})$, with $J_{5}^{(i)}$ defined as in (96) and
(97), are 5-adic qubits, i.e., continuous unitary irreducible linear (and then
projective) representations of degree $2$ on ${{\mathbb{C}}}$ of the $5$-adic
special orthogonal group $SO(3)_{5}$ defined in (3).

Note that, in contrast to the case $p=3$,
$\imath^{\prime}\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Z}}}/5\hookrightarrow{{\mathbb{C}}},\
0_{{{\mathbb{Z}}}/5}\mapsto 0_{{\mathbb{C}}},\,\pm
1_{{{\mathbb{Z}}}/5}\mapsto\pm 1_{{\mathbb{C}}},\,\pm
2_{{{\mathbb{Z}}}/5}\mapsto\pm 2_{{\mathbb{C}}}$ does not lead to a group
homomorphism $\imath^{\prime}\circ K_{5}^{\prime}$; the four solutions
$(a,b)\in\\{(\pm 2,1),(\pm 2,-1)\\}\subset({{\mathbb{Z}}}/5)^{2}$ of
$a^{2}+2b^{2}\equiv 1\mod 5$ are not embedded by $\imath^{\prime}$ to
solutions of $a^{2}+2b^{2}=1$ in ${{\mathbb{C}}}$.

##  VII Representations of $SO(3)_{2}\mod 2$

We exploit Eqs. (15) and (16) for $k=1$. Since ${\cal
R}_{\bm{n}}(\infty)\equiv\openone\mod 2$, we just have

$G_{\bm{n},2}=\left\\{{\cal R}_{\bm{n}}\left(\sigma\right)\mod
2\,\mathrel{\mathop{\mathchar 58\relax}}\,\sigma\in{{\mathbb{Z}}}/2\right\\}.$
(98)

The elements of $G_{\bm{n},2}$ written on the plane orthogonal to $\bm{n}$ are

$\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\ \ \ \text{and}\ \ \
\begin{pmatrix}0&1\\\ 1&0\end{pmatrix},$ (99)

which are the identity and the exchange of the two reference axes orthogonal
to $\bm{n}$. This holds for every $\bm{n}$ of the canonical basis of
${{\mathbb{Q}}}_{2}^{3}$, for which the groups $G_{\bm{n},2}$ provide the
matrices

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix},\ \
\begin{pmatrix}1&0&0\\\ 0&0&1\\\ 0&1&0\end{pmatrix},\ \
\begin{pmatrix}0&0&1\\\ 0&1&0\\\ 1&0&0\end{pmatrix},\ \
\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&1\end{pmatrix}.$ (100)

These exchanges generate the group of permutations of the three reference axes
of ${{\mathbb{Q}}}_{2}^{3}$. We conclude that there exists an isomorphism
$\phi$ between $SO(3)_{2}\mod 2$ and the group $\mathbb{S}_{3}$ of
permutations of $3$ elements,

$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \phi\mathrel{\mathop{\mathchar
58\relax}}SO(3)_{2}\mod 2\ \stackrel{{\scriptstyle\sim}}{{\rightarrow}}\
\mathbb{S}_{3},$ (101) $\displaystyle\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&1\end{pmatrix}\mapsto(12),\quad\begin{pmatrix}1&0&0\\\ 0&0&1\\\
0&1&0\end{pmatrix}\mapsto(23),$

where $(12),\ (23)$ are the exchanges that generate $\mathbb{S}_{3}$.

This isomorphism allows us to apply the representation theory of
$\mathbb{S}_{3}$ to $SO(3)_{2}\mod 2$. In particular, we know that
$\mathbb{S}_{3}$ has a unique $2$-irrep, which we propose as the foundation
for the $2$-adic qubit.

In detail repsSL , $\sum_{\alpha}d_{\alpha}^{2}=\lvert\mathbb{S}_{3}\rvert=6$,
where $d_{\alpha}$ are the degrees of all the irreducible representations of
$\mathbb{S}_{3}$ located by $\alpha$. This means that no irreducible
representation of degree higher or equal than $3$ is allowed for
$\mathbb{S}_{3}$. The commutator subgroup $[\mathbb{S}_{3},\mathbb{S}_{3}]$ is
the group of even permutations, so
$\lvert\mathbb{S}_{3}/[\mathbb{S}_{3},\mathbb{S}_{3}]\rvert=2$ is the number
of $1$-irreps of $\mathbb{S}_{3}$. Hence, $\mathbb{S}_{3}$ has two $1$-irreps
and one $2$-irrep.

The complex 1-irreps for $\mathbb{S}_{3}$ are as follows:

  * •

the trivial representation

$\operatorname{TRIV}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{S}_{3}\rightarrow{{\mathbb{C}}}^{\ast},\ \ \pi\mapsto 1,$

  * •

the sign representation

$\operatorname{SIGN}\mathrel{\mathop{\mathchar
58\relax}}\mathbb{S}_{3}\rightarrow{{\mathbb{C}}}^{\ast},\ \ \
\pi\mapsto\text{sign}(\pi)=\left\\{\begin{aligned} &+1,\ \text{if}\ \pi\
\text{even},\\\ &-1,\ \text{if}\ \pi\ \text{odd}.\end{aligned}\right.$

A complex representation of degree $3$ of $\mathbb{S}_{3}$ is

$\tau\mathrel{\mathop{\mathchar 58\relax}}\mathbb{S}_{3}\rightarrow
GL({{\mathbb{C}}}^{3}),\ \ \tau(\pi)(e_{i})=e_{\pi(i)},$

which permutes the basis elements $\\{e_{i}\\}_{i=1}^{3}$ of
${{\mathbb{C}}}^{3}$. This representation is reducible, in fact there exists
the subrepresentation $\tau^{\prime}$ on

${{\mathbb{C}}}^{2}\simeq\text{span}\\{e_{1}-e_{2},e_{2}-e_{3}\\}=\\{x^{1}e_{1}+x^{2}e_{2}+x^{3}e_{3}\in{{\mathbb{C}}}^{3}\,\mathrel{\mathop{\mathchar
58\relax}}\,x^{1}+x^{2}+x^{3}=0\\}.$ (102)

This has the following matrix form for the generators $(12)$ and $(23)$ of
$\mathbb{S}_{3}$ with respect to the basis $\\{e_{1}-e_{2},e_{2}-e_{3}\\}$:

$(12)\longmapsto\begin{pmatrix}-1&1\\\ 0&1\end{pmatrix},\ \ \ \ \
(23)\longmapsto\begin{pmatrix}1&0\\\ 1&-1\end{pmatrix}.$

$\tau^{\prime}$ can be easily shown to be the unitary irreducible
representation of degree $2$ of $\mathbb{S}_{3}$. In fact, as in the proof of
irreducibility of Proposition V.9, the actions of $(12)$ and $(23)$ have no
common eigenvectors.

###### Proposition VII.1

There exists a unitary 2-irrep $J_{2}$ of $SO(3)_{2}$,

$J_{2}\mathrel{\mathop{\mathchar
58\relax}}=\tau^{\prime}\circ\phi\circ\pi_{1}.$ (103)

###### Proof.

$J_{2}$ is a homomorphism because it is composition of homomorphisms. $J_{2}$
is irreducible since $\tau^{\prime}$ is so: if $Y\subset{{\mathbb{C}}}^{2}$,
then the following are equivalent:

$\displaystyle(J_{2}(SO(3)_{2})Y=Y,$ (104)
$\displaystyle\tau^{\prime}\big{(}(\phi\circ\pi_{1})(SO(3)_{2})\big{)}Y=Y,$
(105) $\displaystyle\tau^{\prime}(\mathbb{S}_{3})Y=Y.$ (106)

The latter is equivalent to $Y=\\{0\\}\ \text{or}\ Y={{\mathbb{C}}}^{2}$.
$\sqcap$$\sqcup$

To write $J_{2}$ explicitly, it is enough to tell the action of the generators
$\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&1\end{pmatrix},\ \begin{pmatrix}1&0&0\\\
0&0&1\\\ 0&1&0\end{pmatrix}$ of $SO(3)_{2}\mod 2$,

$\displaystyle J_{2}\mathrel{\mathop{\mathchar 58\relax}}$
$\displaystyle\begin{pmatrix}(0,a_{112},\ldots)&(1,a_{122},\ldots)&(0,a_{132},\ldots)\\\
(1,a_{212},\ldots)&(0,a_{222},\ldots)&(0,a_{232},\ldots)\\\
(0,a_{312},\ldots)&(0,a_{322},\ldots)&(1,a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{2}}\stackrel{{\scriptstyle\phi\circ\pi_{1}}}{{\longmapsto}}(12)\in\mathbb{S}_{3}\stackrel{{\scriptstyle\tau^{\prime}}}{{\longmapsto}}\begin{pmatrix}-1&1\\\
0&1\end{pmatrix}_{{\mathbb{C}}}$ (107)
$\displaystyle\begin{pmatrix}(1,a_{112},\ldots)&(0,a_{122},\ldots)&(0,a_{132},\ldots)\\\
(0,a_{212},\ldots)&(0,a_{222},\ldots)&(1,a_{232},\ldots)\\\
(0,a_{312},\ldots)&(1,a_{322},\ldots)&(0,a_{332},\ldots)\end{pmatrix}_{{{\mathbb{Z}}}_{2}}\stackrel{{\scriptstyle\phi\circ\pi_{1}}}{{\longmapsto}}(23)\in\mathbb{S}_{3}\stackrel{{\scriptstyle\tau^{\prime}}}{{\longmapsto}}\begin{pmatrix}1&0\\\
1&-1\end{pmatrix}_{{\mathbb{C}}}.$

We, thus, arrive to the following final result.

###### Proposition VII.2

The pair $({{\mathbb{C}}}^{2},J_{2})$ with $J_{2}$ defined as in (107) is a
2-adic qubit, i.e., a continuous unitary irreducible linear (and then
projective) representation of degree $2$ on ${{\mathbb{C}}}$ of the $2$-adic
special orthogonal group $SO(3)_{2}$ defined in (3).

##  VIII Discussion

We have examined the $p$-adic special orthogonal group $SO(3)_{p}$ in three
dimensions, outlining some of its geometric and algebraic properties with the
help of modular arithmetics and representation theory. Our aim was to show,
for each $p$, at least one two-dimensional continuous unitary projective
irrep, thus showing the existence of $p$-adic qubits, in the sense of the
construction of a spin-$\frac{1}{2}$ particle through an angular momentum
representation.

As a matter of fact, for $p\geq 5$, we found that there is more than one such
$p$-adic qubit, and it remains an important open question to classify all two-
dimensional projective irreps for every $p$, not to mention the higher-
dimensional ones.

A special role is played by the one-dimensional irreps, which act by
multiplication on the other irreps and which themselves form an Abelian group,
isomorphic to the abelianization of $SO(3)_{p}$. It remains largely unknown,
except that we know for odd $p>2$ that it contains the Klein group
${{\mathbb{Z}}}/2\times{{\mathbb{Z}}}/2$.

In any case, the present work has the potential to open new lines of
investigations, as follows:

  * •

$p$-adic spin: while there are works treating representations of $SO(3)_{p}$
as a characterizing property of particles, such as Ref. varvir2010 , they do
not work to distinguish the behaviour of the group for different values of
$p$.

  * •

$p$-adic quantum computation: the qubit is the simplest object in this
subject, and computational gates can be formulated in this framework with the
potential of leading to new eventually more efficient algorithms.

  * •

$p$-adic quantum information: stochastic processes over the field of $p$-adic
numbers VVZ  can be formalized for qubit, thus enlarging the perspectives of
quantum communication.

## Acknowledgments

The authors acknowledge fruitful discussions with Sara Di Martino and Michele
Pigliapochi in the early stages of the present project. A.W. acknowledges the
financial support by the Spanish MINECO (Project Nos. FIS2016-86681-P and
PID2019-107609GB-I00/AEI/10.13039/501100011033), with the support of FEDER
funds, and the Generalitat de Catalunya (Project No. 2017-SGR-1127).

## Author Declarations

### Conflict of Interest

The authors have no conflicts to disclose.

## Data availability

Data sharing is not applicable to this article as no new data were created or
analyzed in this study.

##  Appendix A Cyclic structure of solutions of the equation
$a^{2}-vb^{2}\equiv 1$ in ${{\mathbb{Z}}}/p$, $p>2$ prime

Recall that $v$ is not a square modulo $p$ for every prime $p>2$. Thus, we can
construct the field extension of the Galois field
${{\mathbb{F}}}_{p}={{\mathbb{Z}}}/p$ galois , with $i$ being a fixed root of
$x^{2}\equiv v$. This is denoted as ${{\mathbb{F}}}_{p}[i]$ and can be seen as
${{\mathbb{F}}}_{p}+i{{\mathbb{F}}}_{p}$. The Galois group
$\text{Aut}({{\mathbb{F}}}_{p}[i]/{{\mathbb{F}}}_{p})=\\{\text{automorphisms}\,\psi\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{F}}}_{p}[i]\rightarrow{{\mathbb{F}}}_{p}[i]\,\mathrel{\mathop{\mathchar
58\relax}}\,\psi(x)=x\ \forall x\in{{\mathbb{F}}}_{p}\\}$ contains only two
elements: the identity and $i\mapsto-i$.

Next, ${{\mathbb{F}}}_{p}[i]^{\ast}$ with the multiplicative operation is a
cyclic group (since ${{\mathbb{F}}}_{p}[i]$ is a finite field) of order
$p^{2}-1$. This means there exists an element $\alpha+i\beta,\
\alpha,\beta\in{{\mathbb{F}}}_{p}^{\ast}$ of order $p^{2}-1$ which generates
the whole group:

${{\mathbb{F}}}_{p}[i]^{\ast}=\langle\alpha+i\beta\rangle=\left\\{(\alpha+i\beta)^{k}\,\mathrel{\mathop{\mathchar
58\relax}}\,k=0,\dots,p^{2}-2\right\\}.$ (108)

Our aim is to find all the solutions of $1\equiv
a^{2}-vb^{2}\equiv(a+ib)(a-ib)\mod p$. We have

$a+ib\equiv(\alpha+i\beta)^{m},\quad a-ib\equiv(\alpha-i\beta)^{m}$ (109)

for some $m=0,\dots,p^{2}-2$ because $i\mapsto-i$ is the only nontrivial
automorphism of the field ${{\mathbb{F}}}_{p}[i]\supset{{\mathbb{F}}}_{p}$.
Hence, our equation becomes $(\alpha^{2}-v\beta^{2})^{m}\equiv 1\mod p$, from
which we see that $m$ is a multiple of the order of $\alpha^{2}-v\beta^{2}$.

###### Proposition A.1

If $\alpha+i\beta$ is generator of ${{\mathbb{F}}}_{p}[i]^{\ast}$, then
$\alpha^{2}-v\beta^{2}$ is generator of ${{\mathbb{F}}}_{p}^{\ast}$.

###### Proof.

First, we want to show that the map

$f\mathrel{\mathop{\mathchar
58\relax}}\mathbb{F}_{p}[i]\rightarrow\mathbb{F}_{p}[i],\ x\mapsto x^{p}$
(110)

corresponds to the non trivial automorphism $i\mapsto-i$.
$f\in\text{Aut}(\mathbb{F}_{p}[i]/\mathbb{F}_{p})$ because of the following
conditions:

  * •

$f(x+y)\equiv(x+y)^{p}\equiv\sum_{j=0}^{p}{p\choose j}x^{j}y^{p-j}\equiv
x^{p}+y^{p}\equiv f(x)+f(y),\ \forall x,y\in\mathbb{F}_{p}[i]$,
since the binomial coefficients ${p\choose j}$ for $j=1,\dots,p-1$ are
multiples of $p$.

  * •

$f(xy)\equiv\underbrace{xy\dots xy}_{p\,\text{times}}\equiv x^{p}y^{p}\equiv
f(x)f(y),\ \forall x,y\in\mathbb{F}_{p}[i]$.

  * •

$f(x)\equiv 0\Leftrightarrow x\equiv 0$.

  * •

For every $x\in\mathbb{F}_{p}^{\ast}$, we have $f(x)\equiv x^{p}\equiv
x^{p-1}x\equiv x$ since $\mathbb{F}_{p}^{\ast}$ is a multiplicative cyclic
group of order $p-1$.

Furthermore, $f\neq id$; otherwise $x^{p}\equiv x$ for every
$x\in\mathbb{F}_{p}[i]$ would imply $x^{p-1}\equiv 1$ for every
$x\in\mathbb{F}_{p}[i]^{\ast}$, which leads to a contradiction with the fact
that the order of $\mathbb{F}_{p}[i]^{\ast}$ is $p^{2}-1$.
It follows that $a-ib\equiv f(a+ib)\equiv(a+ib)^{p},\forall
a,b\in\mathbb{F}_{p}$.

If $\alpha+i\beta$ is generator of $\mathbb{F}_{p}[i]^{\ast}$, it is an
element of order $p^{2}-1$ and
$1\equiv(\alpha+i\beta)^{p^{2}-1}\equiv\big{[}(\alpha+i\beta)^{(p+1)}\big{]}^{(p-1)}$.
Therefore
$\alpha^{2}-v\beta^{2}\equiv(\alpha+i\beta)(\alpha-i\beta)\equiv(\alpha+i\beta)^{p+1}$
is a generator of $\mathbb{F}_{p}^{\ast}$.  $\sqcap$$\sqcup$

Since $\alpha^{2}-v\beta^{2}\in\mathbb{F}_{p}^{\ast}$ is of order $p-1$,
together with $(\alpha^{2}-v\beta^{2})^{m}\equiv 1$, we can write $m=(p-1)n$
where $n=0,\dots,p$, and

$a+ib\equiv\left((\alpha+i\beta)^{p-1}\right)^{n}\mathrel{\mathop{\mathchar
58\relax}}\equiv(a_{0}+ib_{0})^{n}.$ (111)

We have
$a_{0}^{2}-vb_{0}^{2}\equiv(\alpha+i\beta)^{p-1}(\alpha-i\beta)^{p-1}\equiv(\alpha^{2}-v\beta^{2})^{p-1}\equiv
1$. The pairs $(a,b)$ such that $a+ib\equiv(a_{0}+ib_{0})^{n}$ are all
distinct by varying $n$ because
$(a_{0}+ib_{0})^{n}\equiv(a_{0}+ib_{0})^{n^{\prime}}\Leftrightarrow
1\equiv(a_{0}+ib_{0})^{n-n^{\prime}}\equiv(\alpha+i\beta)^{(p-1)(n-n^{\prime})}\Leftrightarrow(n-n^{\prime})(p-1)=M(p^{2}-1)\Leftrightarrow
n-n^{\prime}=M(p+1)$, which is impossible for every $M\in{{\mathbb{Z}}}_{>0}$.
Furthermore, $(a_{0}+ib_{0})^{n}\equiv(\alpha+i\beta)^{(p-1)n}\equiv
1\Leftrightarrow(p-1)n=q(p^{2}-1)\Leftrightarrow n=q(p+1)$ for
$q\in{{\mathbb{Z}}}$.

This means that the solutions $(a,b)\in{{\mathbb{F}}}_{p}^{2}$ of
$a^{2}-vb^{2}\equiv 1\mod p$ form a cyclic group of order $p+1$, endowed with
the product

$(a,b)\cdot(c,d)\mathrel{\mathop{\mathchar 58\relax}}=(ac+vbd,\,ad+bc).$ (112)

This is because the pairs $(a,b)$ are regarded as a single number $a+ib$ and
$(a+ib)(c+id)=ac+iad+ibc+i^{2}bd=(ac+vbd)+i(ad+bd)$. Another equivalence is

$a+ib\leftrightarrow\begin{pmatrix}a&vb\\\ b&a\end{pmatrix}$ (113)

that gives again

$(a,b)\cdot(c,d)=\begin{pmatrix}a&vb\\\ b&a\end{pmatrix}\begin{pmatrix}c&vd\\\
d&c\end{pmatrix}=\begin{pmatrix}ac+vbd&v(ad+bc)\\\ ad+bc&ac+vbd\end{pmatrix}.$
(114)

##  Appendix B Representations of dihedral groups of even degree

Consider the dihedral group $D_{n}$, where $n$ is even repsSL ,

$D_{n}\coloneqq\langle a,x\mid a^{n}=x^{2}=e,\,xax=a^{-1}\rangle.$ (115)

This group has $(n+6)/2$ conjugacy classes: the identity element, the element
$a^{n/2}$, $(n-2)/2$ other conjugacy classes in $\langle a\rangle$, and two
conjugacy classes outside $\langle a\rangle$, with representatives $x$ and
$ax$. The representations we are going to show are all unitary.

###  B.1 The four one-dimensional representations

The commutator subgroup is $\langle a^{2}\rangle$, which has index four, and
the quotient group (the abelianization) is a Klein four-group. There are,
thus, four one-dimensional representations:

  * •

The trivial representation, sending all elements to the $1\times 1$ matrix
$(1)$.

  * •

The representation sending all elements in $\langle a\rangle$ to $(1)$ and all
the others to $(-1)$.

  * •

The representation sending all elements in $\langle a^{2},x\rangle$ to $(1)$
and all the others to $(-1)$.

  * •

The representation sending all elements in $\langle a^{2},ax\rangle$ to $(1)$
and all the others to $(-1)$.

###  B.2 Two-dimensional representations

There are $(n-2)/2$ irreducible two-dimensional representations. The $k$th
representation into real orthogonal matrices is as follows:

$a\mapsto\begin{pmatrix}\cos\frac{2\pi k}{n}&-\sin\frac{2\pi k}{n}\\\
\sin\frac{2\pi k}{n}&\cos\frac{2\pi k}{n}\end{pmatrix},\quad
x\mapsto\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$ (116)

$k$ ranges between $1$ and $(n-2)/2$, because $k$ and $n-k$ form equivalent
complex irreducible representations. Furthermore, for $k=0$ and $k=n/2$, the
representation is not irreducible and can be broken down into one-dimensional
representations.

##  Appendix C Conjugacy classes of $G_{3}$

In Proposition V.4, we need to know the number of conjugacy classes of

$G_{3}=\pi_{1}(SO(3)_{3})=SO(3)_{3}\mod 3.$

$G_{3}$ is partitioned in the following nine conjugacy classes [using the form
in (72)]:

  * •

one class of order 1, composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&1\end{pmatrix}.\end{array}$

  * •

Two classes of order 4,

$\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(\pm 1,0)\\}\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&-1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
1&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
-1&0&1\end{pmatrix},\end{array}$
$\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(1,\pm 1),(-1,\pm 1)\\}\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
1&-1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
-1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&1&0\\\
-1&-1&1\end{pmatrix}.\end{array}$

  * •

Two classes of order 6,

$\big{\\{}M(a,b,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(a,b)\in\\{(0,\pm 1)\\},(c,d)\in\\{(0,0),(\pm 1,\pm
b)\\}\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
1&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
-1&-1&-1\end{pmatrix},&\\\ \\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
0&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
1&-1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
-1&1&-1\end{pmatrix},&\end{array}$
$\big{\\{}M(1,0,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,0),(\pm
1,0)\\}\big{\\}}\cup\big{\\{}M(-1,0,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,0),(0,\pm 1)\\}\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
0&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
-1&0&-1\end{pmatrix},&\\\ \\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
0&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
0&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
0&-1&-1\end{pmatrix}.&\end{array}$

  * •

One class of order 9,

$\big{\\{}M(-1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/3\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
0&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
0&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
0&-1&1\end{pmatrix},&\\\ \\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
1&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
1&-1&1\end{pmatrix},&\\\ \\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
-1&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
-1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\
-1&-1&1\end{pmatrix}.&\end{array}$

  * •

Two classes of order 12,

$\big{\\{}M(0,\pm 1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(\pm 1,0),(\pm 1,\mp b)\\}\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
0&-1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
1&-1&-1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
-1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ 1&0&0\\\
-1&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
0&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
0&-1&-1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
1&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
-1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ -1&0&0\\\
-1&-1&-1\end{pmatrix},\end{array}$ $\big{\\{}M(1,0,c,\pm
1,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c\in{{\mathbb{Z}}}/3\big{\\}}\cup\big{\\{}M(-1,0,\pm
1,d,-1)\,\mathrel{\mathop{\mathchar 58\relax}}\,d\in{{\mathbb{Z}}}/3\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
0&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
0&-1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
1&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
1&-1&-1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
-1&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}1&0&0\\\ 0&-1&0\\\
-1&-1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
1&1&-1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
1&-1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
-1&0&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
-1&1&-1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}-1&0&0\\\ 0&1&0\\\
-1&-1&-1\end{pmatrix}.\end{array}$

  * •

One class of order 18,

$\big{\\{}M(0,\pm 1,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/3\big{\\}}$

composed of

$\begin{array}[]{llll}\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
0&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
0&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
0&-1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
1&0&1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
1&-1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
-1&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
-1&1&1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}0&1&0\\\ -1&0&0\\\
-1&-1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
0&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
0&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
0&-1&1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
1&0&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
1&-1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
-1&0&1\end{pmatrix},\\\ \\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
-1&1&1\end{pmatrix},&\\!\\!\\!\begin{pmatrix}0&-1&0\\\ 1&0&0\\\
-1&-1&1\end{pmatrix}.&&\end{array}$

##  Appendix D Proof of Proposition V.5

We need to solve $\sum_{\alpha\mid d_{\alpha}>1}d_{\alpha}^{2}=68$ for
integers $d_{\alpha}$ divisors of $72$ greater than $1$ up to permutation,
$\lvert\\{\alpha\mid d_{\alpha}>1\\}\rvert=5$. Without loss of generality, we
can suppose the $d_{\alpha}$ in non-decreasing order. We will study case by
case, starting from the largest value, $d_{9}$, with decreasing values.

$d_{9}$ cannot be larger than 8, because its square would exceed 68, which is
the sum of positive numbers.

Case $d_{9}=8$: this would mean that
$d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}=68-64=4$, but that is not possible as
$d_{\alpha}>1$ for any $\alpha=5,6,7,8$.

Case $d_{9}=6$: assuming this, we have
$d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}=32$. What can we say about $d_{8}$?

  * •

$d_{8}=4$: then, $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}=16$.

    * –

$d_{7}=3$: $d_{5}^{2}+d_{6}^{2}=7$, with only the contradictory chance
$d_{5}=d_{6}=2$.

    * –

$d_{7}=2$: then, $d_{5}=d_{6}=2$, but $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}$ results
as 12 and not 16.

  * •

$d_{8}=3$: $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}=23$.

    * –

$d_{7}=3$: $d_{5}^{2}+d_{6}^{2}=14$.

      * *

$d_{6}=3$: $d_{5}^{2}=5$, impossible.

      * *

$d_{6}=2$: it would be $d_{5}^{2}+d_{6}^{2}=8=14$, absurd.

    * –

$d_{7}=2$: then, $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}=12$, while it should be 23.

  * •

$d_{8}=2$: in this case, $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}=16$ instead
of 32.

Case $d_{9}=4$: consequently, $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}=52$.

  * •

$d_{8}=4$: $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}=36$.

    * –

$d_{7}=4$: $d_{5}^{2}+d_{6}^{2}=20$.

      * *

$d_{6}=4$: then, $d_{5}^{2}=4\Rightarrow d_{5}=2$, solution.

      * *

$d_{6}=3$: then, $d_{5}^{2}=11$, impossible.

      * *

$d_{6}=2$: then, $d_{5}^{2}+d_{6}^{2}=8<16$.

    * –

$d_{7}=3$: then, $d_{5}^{2}+d_{6}^{2}=27$. The odd sum forces
$d_{6}=3\Rightarrow d_{5}^{2}=18$, impossible.

    * –

$d_{7}=2$: then, $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}=12<36$.

  * •

$d_{8}=3$: $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}$ should be 43, but
$d_{5}^{2}+d_{6}^{2}+d_{7}^{2}\leq 27$.

  * •

$d_{8}=2$: then, $d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}=16<52$.

Case $d_{9}=3$: if the larger value is 3,
$d_{5}^{2}+d_{6}^{2}+d_{7}^{2}+d_{8}^{2}+d_{9}^{2}$ can be at most $5\cdot
9=45<68$.

Case $d_{9}=2$: it is ruled out for the same reason as $d_{9}=3$.

The only solution is $d_{5}=2,\ d_{6}=d_{7}=d_{8}=d_{9}=4$.

##  Appendix E Conjugacy classes of $G_{5}$

In Proposition VI.4, we need to know the number of conjugacy classes of

$G_{5}=\pi_{1}(SO(3)_{5})=SO(3)_{5}\mod 5.$

$G_{5}$ is partitioned in the following $14$ conjugacy classes [referring to
the form in (85)]:

  * •

one class of order $1$, say $C_{1}$, composed of

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}.$

  * •

Four classes of order $6$,

$C_{2}=\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(1,\pm 1),(-1,\pm 1),(\pm 2,0)\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 1&1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 1&-1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -1&1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -1&-1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 2&0&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -2&0&1\end{pmatrix}$,

$C_{3}=\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(\pm 1,0),(2,\pm 2),(-2,\pm 2)\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 1&0&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -1&0&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 2&2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 2&-2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -2&2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -2&-2&1\end{pmatrix}$,

$C_{4}=\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 2),(2,\pm 1),(-2,\pm 1)\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&-2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 2&1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 2&-1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -2&1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -2&-1&1\end{pmatrix}$,

$C_{5}=\big{\\{}M(1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(1,\pm 2),(-1,\pm 2)\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&-1&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 1&2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 1&-2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -1&2&1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ -1&-2&1\end{pmatrix}$.

  * •

Two classes of order $15$,

$C_{6}=\big{\\{}M(a,b,c,sbc,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(a,b)\in\\{(1,0),(2,\pm 1),c\in{{\mathbb{Z}}}/5\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&0&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 2&0&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -1&0&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 1&0&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -2&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 0&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -2&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -1&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 0&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 2&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 1&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -2&-2&-1\end{pmatrix}$,

$C_{7}=\big{\\{}M(-1,0,0,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,d\in{{\mathbb{Z}}}/5\big{\\}}\cup\big{\\{}M(a,b,c,abc,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(a,b)\in\\{(-2,\pm 1),c\in{{\mathbb{Z}}}/5\\}\big{\\}}$

composed of
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&0&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 0&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 0&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -1&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 2&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 1&-2&-1\end{pmatrix}$
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 0&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 1&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -2&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -1&-2&-1\end{pmatrix}$.

  * •

One class of order $25$,

$C_{8}=\big{\\{}M(-1,0,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/5\big{\\}}$

composed of
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 0&0&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -2&0&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 1&0&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -1&0&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 2&0&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 0&-2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -2&-2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 1&-2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -1&-2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 2&-2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 0&1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -2&1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 1&1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -1&1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 2&1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 0&-1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -2&-1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 1&-1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -1&-1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 2&-1&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 0&2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -2&2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 1&2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ -1&2&1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&-1&0\\\ 2&2&1\end{pmatrix}$.

  * •

Four classes of order $30$,

$\displaystyle C_{9}=$ $\displaystyle\big{\\{}M(-1,0,\pm
2,d,-1)\,\mathrel{\mathop{\mathchar 58\relax}}\,d\in{{\mathbb{Z}}}/5\big{\\}}$
$\displaystyle\cup\big{\\{}M(-2,1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0\pm 2),(\pm 1,0),(\pm 1,\pm 1),(\pm 2,\mp 1),(\pm
2,\mp 2)\\}\big{\\}}$
$\displaystyle\cup\big{\\{}M(-2,-1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 2),(\pm 1,0),(\pm 1,\mp 1),(\pm 2,\pm 1),(\pm
2,\pm 2)\\}\big{\\}}$

composed of
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -2&0&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 2&0&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -2&2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 2&2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -2&1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 2&1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -2&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 1&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 0&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 1&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 0&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -2&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -1&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 2&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -1&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -1&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 0&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 0&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 2&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 1&0&-1\end{pmatrix}$,

$\displaystyle C_{10}=$ $\displaystyle\big{\\{}M(1,0,c,\pm
1,-1)\,\mathrel{\mathop{\mathchar 58\relax}}\,c\in{{\mathbb{Z}}}/5\big{\\}}$
$\displaystyle\cup\big{\\{}M(2,1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0\pm 2),(\pm 1,\pm 1),(\pm 1,\pm 2),(\pm 2,0),(\pm
2,\pm 1)\\}\big{\\}}$
$\displaystyle\cup\big{\\{}M(2,-1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 2),(\pm 1,\mp 1),(\pm 1,\mp 2),(\pm 2,0),(\pm
2,\mp 1)\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 2&1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -1&1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 1&1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -2&1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&-1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 1&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 2&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 2&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 0&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 0&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -2&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 1&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -1&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -2&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 0&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -2&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 0&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -1&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 2&0&-1\end{pmatrix}$,

$\displaystyle C_{11}=$ $\displaystyle\big{\\{}M(-1,0,\pm
1,d,-1)\,\mathrel{\mathop{\mathchar 58\relax}}\,d\in{{\mathbb{Z}}}/5\big{\\}}$
$\displaystyle\cup\big{\\{}M(-2,1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(\pm 1,\mp 1)(\pm 1,\pm 2),(\pm 2,0),(\pm
2,\pm 2)\\}\big{\\}}$
$\displaystyle\cup\big{\\{}M(-2,-1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(\pm 1,\pm 1),(\pm 1,\mp 2),(\pm 2,0),(\pm
2,\mp 2)\\}\big{\\}}$

composed of
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 1&0&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -1&0&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 1&2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -1&2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 1&1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -1&1&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ 1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-1&0&0\\\ 0&1&0\\\ -1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -1&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 0&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 2&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 1&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ -2&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 0&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ 1&2&0\\\ 2&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 1&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -1&-1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 0&1&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -2&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -1&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 2&0&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ -2&2&-1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ -1&2&0\\\ 0&-1&-1\end{pmatrix}$,

$\displaystyle C_{12}=$
$\displaystyle\big{\\{}M(1,0,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 2),(\pm 1,\pm 2),(\pm 1,\mp 2),(\pm 2,\pm
2),(\pm 2,\mp 2)\\}\big{\\}}$
$\displaystyle\cup\big{\\{}M(2,1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(\pm 1,0)(\pm 1,\mp 2),(\pm 2,\mp 1),(\pm
2,\pm 2)\\}\big{\\}}$
$\displaystyle\cup\big{\\{}M(2,-1,c,d,-1)\,\mathrel{\mathop{\mathchar
58\relax}}\,(c,d)\in\\{(0,\pm 1),(\pm 1,0),(\pm 1,\pm 2),(\pm 2,\pm 1),(\pm
2,\mp 2)\\}\big{\\}}$

composed of
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&-2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 0&2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 2&2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -1&2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ 1&2&-1\end{pmatrix}$,
$\begin{pmatrix}1&0&0\\\ 0&-1&0\\\ -2&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 2&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 1&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 0&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -1&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -2&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ 0&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ 1&-2&0\\\ -1&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -2&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 2&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 0&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -1&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 1&2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 2&1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -1&-2&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ -2&-1&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 1&0&-1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ -1&-2&0\\\ 0&1&-1\end{pmatrix}$.

  * •

Two classes of order $50$,

$C_{13}=\big{\\{}M(-2,\pm 1,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/5\big{\\}}$

composed of
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 0&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 0&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 1&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 1&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 2&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 2&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -2&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -2&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -1&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -1&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -2&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 2&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -1&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -2&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 0&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -1&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 1&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 0&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 2&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 1&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 1&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -1&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 2&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 0&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -2&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 1&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -1&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 2&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 0&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -2&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -1&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 1&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 0&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 2&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 1&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -2&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 2&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -1&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -2&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 0&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 2&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -2&-1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -2&-2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ -1&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ -1&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 0&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 0&1&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 1&2&1\end{pmatrix}$,
$\begin{pmatrix}-2&2&0\\\ -1&-2&0\\\ 1&0&1\end{pmatrix}$,
$\begin{pmatrix}-2&-2&0\\\ 1&-2&0\\\ 2&-2&1\end{pmatrix}$,

$C_{14}=\big{\\{}M(2,\pm 1,c,d,1)\,\mathrel{\mathop{\mathchar
58\relax}}\,c,d\in{{\mathbb{Z}}}/5\big{\\}}$

composed of
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 0&0&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 0&0&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -1&1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -1&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -2&2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -2&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 2&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 2&2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 1&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 1&1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 2&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -2&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 1&0&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 2&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 0&1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 1&2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -1&2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 0&1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -2&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -1&0&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -1&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 1&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -2&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 0&2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 2&0&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -1&1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 1&1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -2&0&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 0&2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 2&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 1&2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -1&2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 0&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -2&1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -1&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 2&0&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -2&0&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 1&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 2&1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 0&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -2&1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 2&1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 2&2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 1&0&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 1&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ 0&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ 0&-1&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -1&-2&1\end{pmatrix}$,
$\begin{pmatrix}2&-2&0\\\ 1&2&0\\\ -1&0&1\end{pmatrix}$,
$\begin{pmatrix}2&2&0\\\ -1&2&0\\\ -2&2&1\end{pmatrix}$.

##  Appendix F Proof of Proposition VI.6

To solve $\sum_{\alpha=5}^{14}d_{\alpha}^{2}=296$ for
$d_{\alpha}\in\\{2,3,4,6,12\\}$ up to permutations, we proceed analogously to
the Proof of Proposition V.5 (Appendix D).

Case $d_{14}=12$: this gives
$\sum_{\alpha=5}^{13}d_{\alpha}^{2}=296-12^{2}=152$.

  * •

$d_{13}=12\mathrel{\mathop{\mathchar 58\relax}}$
$\sum_{\alpha=5}^{12}d_{\alpha}^{2}=152-12^{2}=8<8\cdot
2^{2}=\min\left(\sum_{\alpha=5}^{12}d_{\alpha}^{2}\right)$, impossible.

  * •

$d_{13}=6$: $\sum_{\alpha=5}^{12}d_{\alpha}^{2}=152-6^{2}=116$.

    * –

$d_{12}=6$: $\sum_{\alpha=5}^{11}d_{\alpha}^{2}=116-6^{2}=80$.

      * *

$d_{11}=6$: $\sum_{\alpha=5}^{10}d_{\alpha}^{2}=80-6^{2}=44$.

        * ·

$d_{10}=6$: $\sum_{\alpha=5}^{9}d_{\alpha}^{2}=44-6^{2}=8<5\cdot 2^{2}$
impossible.

        * ·

$d_{10}=4$: $\sum_{\alpha=5}^{9}d_{\alpha}^{2}=44-4^{2}=28$.

          1. 1.

$d_{9}=4$: $\sum_{\alpha=5}^{8}d_{\alpha}^{2}=28-4^{2}=12<4\cdot 2^{2}$
impossible.

          2. 2.

$d_{9}=3$: $\sum_{\alpha=5}^{8}d_{\alpha}^{2}=28-3^{2}=19$.

            1. (a)

$d_{8}=3$: $\sum_{\alpha=5}^{7}d_{\alpha}^{2}=19-3^{2}=10<3\cdot 2^{2}$,
impossible.

            2. (b)

$d_{8}=2$: this forces $\sum_{\alpha=5}^{8}d_{\alpha}^{2}=4\cdot 2^{2}=16=19$,
absurd.

          3. 3.