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# Geometry of the $p$-adic special orthogonal group $SO(3)_{p}$ 111The present
work grew out of a BSc thesis Ilaria:tesina  and two MSc theses Michele:tesi ;
Ilaria:tesi  at the University of Camerino, co-supervised at the Autonomous
University of Barcelona.

Sara Di Martino  Física Teòrica: Grup d’Informació Quàntica, Departament de
Física, Universitat Autònoma de Barcelona, ES-08193 Bellaterra (Barcelona),
Spain  Stefano Mancini  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  Michele Pigliapochi  School
of Science and Technology, University of Camerino, Via Madonna delle Carceri
9, I-62032 Camerino, Italy  Ilaria Svampa  School of Science and Technology,
University of Camerino, Via Madonna delle Carceri 9, I-62032 Camerino, Italy
Física Teòrica: Grup d’Informació Quàntica, Departament de Física, Universitat
Autònoma de Barcelona, ES-08193 Bellaterra (Barcelona), Spain  Andreas Winter
Física Teòrica: Grup d’Informació Quàntica, Departament de Física, Universitat
Autònoma de Barcelona, ES-08193 Bellaterra (Barcelona), Spain  ICREA —
Institució Catalana de la Recerca i Estudis Avançats, Pg. Lluís Companys, 23,
ES-08010 Barcelona, Spain

###### Abstract

We derive explicitly the structural properties of the $p$-adic special
orthogonal groups in dimension three, for all primes $p$, and, along the way,
the two-dimensional case. In particular, starting from the unique definite
quadratic form in three dimensions (up to linear equivalence and rescaling),
we show that every element of $SO(3)_{p}$ is a rotation around an axis. An
important part of the analysis is the classification of all definite forms in
two dimensions, yielding a description of the rotation subgroups around any
fixed axis, which all turn out to be abelian and parametrised naturally by the
projective line.

Furthermore, we find that for odd primes $p$, the entire group $SO(3)_{p}$
admits a representation in terms of Cardano (aka nautical) angles of rotations
around the reference axes, in close analogy to the real orthogonal case.
However, this works only for certain orderings of the product of rotations
around the coordinate axes, depending on the prime; furthermore, there is no
general Euler angle decomposition. For $p=2$, no Euler or Cardano
decomposition exists.

$p$-adic numbers, special orthogonal group, nautical angles

##  1 Introduction

Large parts of classical geometry can be interpreted, following Klein’s
programme Klein , as statements about the orthogonal and special orthogonal
groups in dimensions $2$, $3$, and generally $n$. It is thus no surprise that
the special orthogonal groups $SO(n)_{{{\mathbb{R}}}}$ are among the most
studied and most well-known groups in mathematics. The cases $n=2$ and $n=3$
stand out particularly, both for their structural properties and the
possibility to visualise the action of the group on Euclidean space. In
particular, $SO(2)_{{{\mathbb{R}}}}$ is the group of planar rotations, i.e.,
isomorphic to the commutative group of adding angles mod $2\pi$; and
$SO(3)_{{{\mathbb{R}}}}$ consists entirely of rotations around different axis,
admitting essentially unique decompositions into Euler and Cardano angles.

Given the importance of $p$-adic numbers in number theory, it is natural that
orthogonal groups should have been studied also over the fields
${{\mathbb{Q}}}_{p}$, which a priori are a multitude of symmetry groups, one
for each nontrivial quadratic form. Just as in the real case, a fundamental
property of the quadratic form is whether it is definite (i.e., only has the
trivial zero) or indefinite (i.e., represents zero in nontrivial ways), which
distinguishes compact symmetry groups in the latter and non-compact groups in
the former case. Unlike the real case, definite quadratic forms over
${{\mathbb{Q}}}_{p}$ exist only in dimensions two, three and four. As we shall
see here, in dimensions two and three the symmetry groups, denoted by
$SO(2)_{p}$ and $SO(3)_{p}$, are largely governed by structures familiar from
Euclidean geometry, reinterpreted $p$-adically.

The structure of the paper is as follows. In Section 2, we review and re-
derive the unique quadratic forms on dimensions three and four that do no
represent $0$ nontrivially (‘‘definite’’ forms), up to linear and rescaling
equivalence, for all primes $p$, introducing some useful notation for the rest
of the paper. This material, albeit well-known, is included for the sake of a
self-contained exposition. This classification allows us to define the
$p$-adic special orthogonal groups $SO(3)_{p}$ and $SO(4)_{p}$ in a unique
way, though we will not consider the latter afterwards in the present work.

Then, in Section 3, we begin our new contributions, by first making several
basic observations about $SO(3)_{p}$, most importantly that it is compact and
profinite, and that every of its elements is a rotation around some axis in
$p$-adic three-space ${{\mathbb{Q}}}_{p}^{3}$. This then motivates the
investigation of the rotations around a fixed given axis, which are special
orthogonal transformations of the plane orthogonal to that axis, in Sections 4
and 5. They are naturally the orthogonal symmetry groups of the definite form
restricted to a plane, and there are three ($p$ odd), resp. seven ($p=2$)
equivalence classes of those. We denote them $SO(2)_{p}^{\kappa}$, and they
all turn out to be abelian groups. Furthermore, we derive a parametrisation of
each of these rotation groups by the $p$-adic projective line
$P^{1}({{\mathbb{Q}}}_{p})$, allowing us to identify the groups
$SO(2)_{p}^{\kappa}$ with certain abelian subgroups of
$PGL(2,{{\mathbb{Q}}}_{p})$, illuminating in particular the composition law.
Finally, in Section 6, we show that every element of $SO(3)_{p}$ has a
decomposition into Cardano principal angles, for odd primes $p$: there exist
orderings of the coordinate axes such that every special orthogonal group
element is a product of rotations around the axes in that order (in exactly
two different ways). For $p=2$, however, no fixed ordering of the product can
recover the entire group $SO(3)_{2}$, and we exhibit an example of a special
orthogonal matrix that cannot be written in any of the twelve possible Euler
or Cardano ways.

##  2 Quadratic forms over $p$-adic numbers

In this section we review, and in some cases, re-derive the elementary
properties we shall need about the quadratic forms over ${{\mathbb{Q}}}_{p}$,
according to dimension ($n=3$ and $n=4$) as well as type of prime ($p\equiv
1\mod 4$, $p\equiv 3\mod 4$ and $p=2$). Comprehensive treatments of this
material can be found in the books of Cassels Cassels  and Serre Serre .

A quadratic form for our purposes is a homogeneous function on the
$n$-dimensional ${{\mathbb{Q}}}_{p}$-vector space $V$ that can be written as

$Q(\bm{x})=\sum_{ij}a_{ij}x_{i}x_{j}=x^{\top}Ax,$

where $\bm{x}=\sum_{i}x_{i}\bm{e}_{i}\in V$ is a vector,
$(\bm{e}_{1},\ldots,\bm{e}_{n})$ is a basis of $V$, and $A$ is an $n\times
n$-matrix. Throughout we will assume that $A$ is nondegenerate, i.e., has rank
$n$. Equivalently, we could speak about symmetric bilinear forms
$b(\bm{x},\bm{y})$ on $V$ such that $Q(\bm{x})=b(\bm{x},\bm{x})$, since we
recover the bilinear form via
$b(\bm{x},\bm{y})=\frac{1}{2}\bigl{(}Q(\bm{x}+\bm{y})-Q(\bm{x})-Q(\bm{y})\bigr{)}$.
In either case, the orthogonal group is defined as the set of linear maps on
$V$ that are symmetries of the quadratic/bilinear form

$\begin{split}O(Q)&=\\{L\in\text{End}(V)\mathrel{\mathop{\mathchar
58\relax}}Q(L\bm{x})=Q(\bm{x})\ \forall\bm{x}\in V\\}\\\
&=\\{L\in\text{End}(V)\mathrel{\mathop{\mathchar
58\relax}}b(L\bm{x},L\bm{y})=b(\bm{x},\bm{y})\ \forall\bm{x},\bm{y}\in V\\}\\\
&\simeq\\{L\in M_{n\times n}({{\mathbb{Q}}}_{p})\mathrel{\mathop{\mathchar
58\relax}}L^{\top}AL=A\\},\end{split}$

the latter under the identification of $V$ with ${{\mathbb{Q}}}_{p}^{n}$ via
the basis $(\bm{e}_{i})$,
$\bm{x}=\sum_{i}x_{i}\bm{e}_{i}\leftrightarrow(x_{1},\ldots,x_{n})$, which
turns the linear maps on $V$ into $n\times n$-matrices. The subset of $O(Q)$
consisting of matrices $L$ with unit determinant, $\det L=1$, is the special
orthogonal group, denoted $SO(Q)$.

We are interested in the abstract group structure of $O(Q)$ and $SO(Q)$, which
do not change when going to an equivalent form. First, $Q^{\prime}$ is
_similar_ to $Q$, $Q\sim Q^{\prime}$, if there exists an invertible linear map
$S$ such that $Q^{\prime}(\bm{x})=Q(S\bm{x})$ for all $\bm{x}\in V$, meaning
for the matrix representation $A^{\prime}$ of $Q^{\prime}$ that
$A^{\prime}=S^{\top}AS$. In that case, $O(Q^{\prime})\simeq O(Q)$ and
$SO(Q^{\prime})\simeq SO(Q)$, the isomorphism being $O(Q)\ni L\mapsto
S^{-1}LS\in O(Q^{\prime})$. Furthermore, $Q^{\prime}$ is a _scaling_ of $Q$ if
$Q^{\prime}=tQ$ with $t\in{{\mathbb{Q}}}_{p}^{*}$; in this case, clearly
$O(Q^{\prime})=O(Q)$ and $SO(Q^{\prime})=SO(Q)$. Hence, our first task is to
classify the quadratic forms up to similarity and scaling, which we sum up
into _equivalence_. Indeed, up to similarity, we can write every quadratic
form with a diagonal matrix $A$, i.e.,

$Q(\bm{x})=\sum_{j}a_{j}x_{j}^{2},$

with $\bm{x}=(x_{1},x_{2},\dots,x_{n})\in{{\mathbb{Q}}}_{p}^{n}$, and
$a_{j}\in{{\mathbb{Q}}}_{p}^{*}$. Multiplying $x_{j}$ by $\lambda_{j}^{-1}\neq
0$ will change $a_{j}$ to $\lambda_{j}^{2}a_{j}$, hence to classify the
quadratic forms up to coordinate changes $GL_{n}({{\mathbb{Q}}}_{p})$, we only
need to consider
$a_{j}\in{{\mathbb{Q}}}_{p}^{*}/({{\mathbb{Q}}}_{p}^{*})^{2}$. We thus have to
understand the structure of the group
$K={{\mathbb{Q}}}_{p}^{*}/({{\mathbb{Q}}}_{p}^{*})^{2}$. The following
descriptions of $K$ are well-known Cassels :

If $\mathbf{p\neq 2}$, then $K=\langle u,p\rangle=\\{1,u,p,up\\}$, with $u$ a
unit in ${{\mathbb{Z}}}_{p}$ that is not a square. Clearly
$K\simeq({{\mathbb{Z}}}/2{{\mathbb{Z}}})^{2}$ is the Klein group. For $p\equiv
3\mod 4$, we may choose $u=-1$.

If $\mathbf{p=2}$, then $K=\langle-1,2,5\rangle=\\{1,-1,2,-2,5,-5,10,-10\\}$.
In this case $K\simeq({{\mathbb{Z}}}/2{{\mathbb{Z}}})^{3}$.

Given the structure of $K$, it remains to find the invariants of quadratic
forms. It can be proven Cassels  that apart from the rank of the form, there
are only two more invariants: the discriminant $d(Q)=\Pi_{j}a_{j}=\det A$, as
well as $\varepsilon(Q)=\Pi_{j<k}(a_{j},a_{k})$, where $(a,b)$ is the Hilbert
symbol defined as

$(a,b)\mathrel{\mathop{\mathchar 58\relax}}=\begin{cases}\phantom{-}1&\text{
iff }z^{2}-ax^{2}-by^{2}=0\text{ admits nontrivial solutions,}\\\ -1&\text{
otherwise.}\end{cases}$

Now we have all the information we need in order to classify quadratic forms
on ${{\mathbb{Q}}}_{p}^{n}$ for all $n$.

######  Theorem 1.

Two quadratic forms over ${{\mathbb{Q}}}_{p}$ are similar if and only if they
have same rank $n$, same determinant $d$ and same Hasse invariant
$\varepsilon$.  $\sqcap$$\sqcup$

######  Theorem 2.

The quadratic form $Q$ on ${{\mathbb{Q}}}_{p}^{n}$ represents $0$ nontrivially
if and only if

  * •

$n=2$ and $d\simeq-1$ in $K$;

  * •

$n=3$ and $\varepsilon=(-1,-d)$;

  * •

$n=4$ and either $d\not\simeq 1$ or $d\simeq 1$ and $\varepsilon=(-1,-1)$;

  * •

$n\geq 5$.  $\sqcap$$\sqcup$

We record explicitly the quadratic forms for $n=3$ up to equivalence, separate
by odd and even primes $p$.

Prime $\mathbf{p}$ odd: There are exactly two inequivalent forms on
${{\mathbb{Q}}}_{p}^{3}$,

$\displaystyle Q_{+}(\bm{x})$ $\displaystyle=x_{1}^{2}-vx_{2}^{2}+px_{3}^{2},$
$\displaystyle Q_{0}(\bm{x})$ $\displaystyle=x_{1}^{2}+x_{2}^{2}+x_{3}^{2},$

where

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

is a particular choice of a non-square in ${{\mathbb{Q}}}_{p}$.

Prime $\mathbf{p=2}$:  There are exactly two inequivalent forms on
${{\mathbb{Q}}}_{2}^{3}$,

$\displaystyle Q_{+}(\bm{x})$ $\displaystyle=x_{1}^{2}+x_{2}^{2}+x_{3}^{2},$
$\displaystyle Q_{0}(\bm{x})$ $\displaystyle=x_{1}^{2}+x_{2}^{2}-x_{3}^{2}.$

Real Euclidean case: Also here, there are exactly two inequivalent forms on
${{\mathbb{R}}}^{3}$,

$\displaystyle Q_{+}^{{\mathbb{R}}}(\bm{x})$
$\displaystyle=x_{1}^{2}+x_{2}^{2}+x_{3}^{2},$ $\displaystyle
Q_{0}^{{\mathbb{R}}}(\bm{x})$ $\displaystyle=x_{1}^{2}+x_{2}^{2}-x_{3}^{2}.$

In all cases, the real and the $p$-adic ones, one form ($Q_{+}$) is definite,
in the sense that it does not represent zero nontrivially (i.e.,
$Q_{+}(\bm{x})=0$ iff $\bm{x}=0$), while the other ($Q_{0}$) is indefinite, in
the sense that it has isotropic vectors (i.e., $Q_{0}(\bm{x})=0$ for some
$\bm{x}\neq 0$). The symmetry group $SO(Q_{0})$ is always non-compact, as is
well-known in the real case and easy to see in general in the $p$-adic case,
since it has a hyperbolic component GrossReeder . On the other hand, in the
real Euclidean case, $SO(Q_{+})$ is just the real $SO(3)_{{{\mathbb{R}}}}$,
which is a compact Lie group, and as we would like to preserve the compactness
in the $p$-adic case, we define $SO(3)_{p}=SO(Q_{+})$ for all primes $p$.
According to the above classification, this is a unique and well-defined group
for every prime $p$, which will indeed turn out to be compact.

For $n=4$, we similarly focus only on the definite forms. It turns out that
again, for ${{\mathbb{R}}}$ and all ${{\mathbb{Q}}}_{p}$, there is exactly one
up to equivalence

$Q_{+}^{(4)}(\bm{x})=\begin{cases}x_{1}^{2}-vx_{2}^{2}+px_{3}^{2}-vpx_{4}^{2}&\text{
if }p\text{ odd},\\\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}&\text{ if
}p=2,\\\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}&\text{ in the real
case}.\end{cases}$

All other inequivalent forms represent zero nontrivially (and hence their
associated orthogonal groups are non-compact GrossReeder ), a fact that we
will exploit later. This means, that also for $n=4$, both
${{\mathbb{Q}}}_{p}^{4}$ and ${{\mathbb{R}}}^{4}$ have essentially unique
groups $SO(4)_{p}$ and $SO(4)_{{{\mathbb{R}}}}$. On the other hand, for $n\geq
5$, it is known that for all primes $p$, there are no definite quadratic forms
on ${{\mathbb{Q}}}_{p}^{n}$ Cassels ; Serre , unlike the real Euclidean case
${{\mathbb{R}}}^{n}$, where of course the sum of the squares of the standard
coordinates is the unique definite form up to equivalence, and where we thus
have an unambiguous compact Lie group $SO(n)_{{{\mathbb{R}}}}$. Because of the
lack of definite quadratic forms for $n\geq 5$, we refrain from speaking of
$SO(n)_{p}$ without qualification.

Even in the real Euclidean case, $n=3$ is distinguished by several geometric
peculiarities. We highlight two of them, see (Eulerangle, , Ch. 4).

######  Theorem 3.

Elements of $SO(3)_{{{\mathbb{R}}}}$ are rotations about axes in
${{\mathbb{R}}}^{3}$, meaning that they always have eigenvalue $1$, for which
the corresponding eigenspace is the rotation axis.  $\sqcap$$\sqcup$

######  Theorem 4.

With $R_{x}(\theta)$, $R_{y}(\eta)$, $R_{z}(\phi)$ denoting the rotations
around the reference axes of ${{\mathbb{R}}}^{3}$ by a given angle $\theta$,
$\eta$ and $\phi$, every $R\in SO(3)$ can be written as a product of three
such rotations in any of the following forms

$\begin{array}[]{cccccc}R_{x}R_{y}R_{z},&R_{y}R_{z}R_{x},&R_{z}R_{x}R_{y},&R_{x}R_{z}R_{y},&R_{z}R_{y}R_{x},&R_{y}R_{x}R_{z},\\\
R_{x}R_{y}R_{x},&R_{x}R_{z}R_{x},&R_{y}R_{x}R_{y},&R_{y}R_{z}R_{y},&R_{z}R_{x}R_{z},&R_{z}R_{y}R_{z}.\end{array}$

These angles are called _Cardano_ or _Tait–Bryan angles_ , and also nautical
angles, when applied to a form from the first row, and (proper) _Euler angles_
when applied to a form from the second row.

In both cases, the choice of the angles is unique (up to isolated points)
modulo $2\pi$ radians for $\theta$ and $\phi$, and with the range of $\eta$
covering $\pi$ radians.  $\sqcap$$\sqcup$

This theorem implies that three successive rotations relative to coordinate
axes generate every rotation in ${{\mathbb{R}}}^{3}$. Hence,
$SO(3)_{{{\mathbb{R}}}}$ is generated by the rotations around the three
reference axes of ${{\mathbb{R}}}^{3}$; in the case of the Euler
decomposition, actually only two reference axes. Each of these rotation groups
is an infinite cyclic Lie group, so $SO(3)_{{{\mathbb{R}}}}$ is generated by
two cyclic subgroups linked by a non-commutative relation.

We have left out $n=2$ until now, which may seem strange coming from the real
Euclidean case, where it is the same as for all other $n$: there is a unique
definite quadratic form on ${{\mathbb{R}}}^{2}$. Also on
${{\mathbb{Q}}}_{p}^{2}$ there are definite quadratic forms, but now they are
not unique up to equivalence. We will come back to them in detail below.

##  3 Basic observations about $SO(3)_{p}$

We start by deriving a few basic facts about the $p$-adic special orthogonal
groups in dimension $3$ from the definition. It is evidently a group under the
usual matrix multiplication. In fact, it is a topological group, with the
operations of multiplication and inverse being continuous with respect to the
$p$-adic metric. Explicitly, the topology on $SO(3)_{p}$ is the one generated
by the open balls with respect to the $p$-adic norm
$||L||_{p}=||(\ell_{ij})_{ij}||_{p}\coloneqq\operatorname{max}_{i,j=1,2,3}|\ell_{ij}|_{p}$,
where $|\,\cdot\,|_{p}$ denotes the $p$-adic absolute value on
$\mathbb{Q}_{p}$.

######  Theorem 5.

For every prime $p$, the group $SO(3)_{p}$ is compact. As a matter of fact, it
is a closed subset (with respect to the $p$-adic metric) of $M_{3\times
3}({{\mathbb{Z}}}_{p})$, the set of matrices with $p$-adic integer entries,
and so $SO(3)_{p}\subset SL(3,{{\mathbb{Z}}}_{p})$.

###### Proof.

Let $L=(\ell_{ij})_{ij}\in SO(3)_{p}$, and write
$\ell_{ij}=p^{\nu_{ij}}u_{ij}$, where
$\nu_{ij}=\nu_{p}(\ell_{ij})\in{{\mathbb{Z}}}\cup\\{+\infty\\}$ is the
$p$-adic valuation of $\ell_{ij}$, and $u_{ij}\in{{\mathbb{U}}}_{p}$ is a unit
in ${{\mathbb{Z}}}_{p}$. We need to show that $\nu_{ij}\geq 0$ for all
$i,j\in\\{1,2,3\\}$.

When $p$ is odd, the defining condition $A=L^{\top}AL$ of $SO(3)_{p}$ implies
the following three relations

$\begin{split}\ell_{11}^{2}-v\ell^{2}_{21}+p\ell_{31}^{2}&=1,\\\
\ell_{12}^{2}-v\ell^{2}_{22}+p\ell_{32}^{2}&=-v,\\\
\ell_{13}^{2}-v\ell^{2}_{23}+p\ell_{33}^{2}&=p.\end{split}$ (2)

The first one is equivalent to

$p^{2\nu_{11}}u_{11}^{2}-vp^{2\nu_{21}}u_{21}^{2}+p^{1+2\nu_{31}}u_{31}^{2}=1.$

Let us assume by contradiction that $\min\\{\nu_{11},\nu_{21},\nu_{31}\\}<0$.
If $\min\\{2\nu_{11},\,2\nu_{21},1+2\nu_{31}\\}=2\nu_{11}<0$ (similarly for
$\nu_{21}$), we multiply both sides of the equation by
$p^{2\lvert\nu_{11}\rvert}$ to obtain

$u_{11}^{2}-vp^{2(\nu_{21}-\nu_{11})}u_{21}^{2}+p^{1+2(\nu_{31}-\nu_{11})}u_{31}^{2}=p^{2\lvert\nu_{11}\rvert},$

where every term is a $p$-adic integer. Hence, for every
$k=1,\dots,2\lvert\nu_{11}\rvert$,

$u_{11}^{2}-vp^{2(\nu_{21}-\nu_{11})}u_{21}^{2}+p^{1+2(\nu_{31}-\nu_{11})}u_{31}^{2}\equiv
0\mod p^{k},$

and in particular $u_{11}^{2}-vp^{2(\nu_{21}-\nu_{11})}u_{21}^{2}\equiv 0\mod
p$. The quadratic form $x^{2}-vy^{2}$ does not represent $0$ in
${{\mathbb{Q}}}_{p}$: if $x,y\in{{\mathbb{Z}}}_{p}$ as in our case, the only
solution to $x^{2}-vy^{2}\equiv 0\mod p$ is
$(0,0)\in({{\mathbb{Z}}}/p{{\mathbb{Z}}})^{2}$. This gives a contradiction,
since $u_{11}\in{{\mathbb{U}}}_{p}$, in particular $u_{11}^{2}\not\equiv 0\mod
p$.

If instead $\min\\{2\nu_{11},\,2\nu_{21},1+2\nu_{31}\\}=1+2\nu_{31}<0$, we
rewrite the equation in terms of $p$-adic integers as

$p^{2(\nu_{11}-\nu_{31})-1}u_{11}^{2}-vp^{2(\nu_{21}-\nu_{31})-1}u_{21}^{2}+u_{31}^{2}=p^{2\lvert\nu_{31}\rvert-1}.$

Reducing it modulo $p$ we get $u_{31}^{2}\equiv 0\mod p$, which is in
contradiction with the hypothesis that $u_{31}\in{{\mathbb{U}}}_{p}$.

The same can be done for the second and third equations of Eq. (2), since
$-v,\,p\in{{\mathbb{Z}}}_{p}$.

When $p=2$, the defining condition $L^{\top}L=I$ implies

$2^{2\nu_{1i}}u_{1i}^{2}+2^{2\nu_{2i}}u_{2i}^{2}+2^{2\nu_{3i}}u_{3i}^{2}=1,\text{
for }i=1,2,3.$

Again assuming by contradiction
$\min\\{\nu_{1i},\nu_{2i},\nu_{3i}\\}=\nu_{1i}<0$ (the latter without loss of
generality, by symmetry), we rewrite the last equation in terms of $2$-adic
integers as

$u_{1i}^{2}+2^{2(\nu_{2i}-\nu_{1i})}u_{2i}^{2}+2^{2(\nu_{3i}-\nu_{1i})}u_{3i}^{2}=2^{2\lvert\nu_{1i}\rvert}.$

As a consequence it must hold

$u_{1i}^{2}+2^{2(\nu_{2i}-\nu_{1i})}u_{2i}^{2}+2^{2(\nu_{3i}-\nu_{1i})}u_{3i}^{2}\equiv
0\mod 2^{k},\ k=1,\dots,2\lvert\nu_{1i}\rvert.$

The quadratic form $x^{2}+y^{2}+z^{2}$ does not represent $0$ in
${{\mathbb{Q}}}_{2}$: as a matter of fact, as $x,y,z$ are in
${{\mathbb{Z}}}_{2}$, it does not so in ${{\mathbb{Z}}}/4{{\mathbb{Z}}}$,
unless all three variables are $\equiv 0\mod 4$; but this is again in
contradiction to $u_{1i}\in{{\mathbb{U}}}_{p}$.

We have found $SO(3)_{p}\subset M_{3\times 3}({{\mathbb{Z}}}_{p})$, and the
defining condition $\det L=1$ implies $SO(3)_{p}\subset
SL(3,{{\mathbb{Z}}}_{p})$.  $\sqcap$$\sqcup$

######  Theorem 6.

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

###### Proof.

Let $L\in SO(3)_{p}$, if $\lambda$ is an eigenvalue of $L$, then also
$\lambda^{-1}$ is. In fact if $x$ is an eigenvector for $\lambda$, we have
$Ax=L^{\top}ALx=\lambda L^{\top}Ax\Rightarrow\lambda^{-1}Ax=L^{\top}Ax$. In
other words, $\lambda^{-1}$ is an eigenvalue of $L^{\top}$ with eigenvector
$Ax$. On the other hand, $L$ and $L^{\top}$ share the characteristic
polynomial and hence the eigenvalues.

Suppose now that none of the eigenvalues $\lambda_{i}$ of $L$, for $i=1,2,3$,
are equal to $1$. Then, $\lambda_{i}=\lambda_{i}^{-1}$ for all $i$
(equivalently $\lambda_{i}^{2}=1$, thus $\lambda_{i}=\pm 1$), otherwise we can
suppose, for example, $\lambda_{1}\neq\lambda_{1}^{-1}=\lambda_{2}$ without
loss of generality, and hence $1=\det
L=\lambda_{1}\lambda_{2}\lambda_{3}=\lambda_{3}$, which is a contradiction.

Summing up, the only case which allows the condition proved before, excluding
the presence of $1$ among the eigenvalues, is
$\lambda_{1}=\lambda_{2}=\lambda_{3}=-1$, which is in contradiction with the
condition $\det L=1$. Thus, $L\in SO(3)_{p}$ has always $1$ as eigenvalue.
$\sqcap$$\sqcup$

Theorem 5 means that we have well-defined projection maps

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

since all matrix entries of eligible $L$ are $p$-adic integers. The images
$SO(3)_{p}\mod p^{k}=\pi_{k}\bigl{(}SO(3)_{p}\bigr{)}$ are all finite groups,
forming a projective system under the (commuting) projection maps of taking a
number in ${{\mathbb{Z}}}/p^{k^{\prime}}{{\mathbb{Z}}}$ modulo $p^{k}$
($k<k^{\prime}$), which by slight abuse of notation we denote by $\pi_{k}$,
too. Just as the $p$-adic integers ${{\mathbb{Z}}}_{p}$ are the inverse (aka
projective) limit of the rings ${{\mathbb{Z}}}/p^{k}{{\mathbb{Z}}}$ connected
by the modulo $p^{k}$ projections, we conclude that $SO(3)_{p}$ is the inverse
limit of the finite groups $SO(3)_{p}\mod p^{k}$ connected by the projections
$\pi_{k}$. This makes $SO(3)_{p}$ a so-called _profinite_ group.

Theorem 6 means that we can describe arbitrary element $L\in SO(3)_{p}$ by
first picking an axis of rotation, ${{\mathbb{Q}}}_{p}\mathbf{n}$, with a
nonzero vector $\mathbf{n}\in{{\mathbb{Q}}}_{p}^{3}$, construct the two-
dimensional subspace $V=\mathbf{n}^{\perp}=\\{x\mathrel{\mathop{\mathchar
58\relax}}b(\mathbf{n},x)=0\\}\subset{{\mathbb{Q}}}_{p}^{3}$ and consider the
quadratic form $Q_{+|V}$, which is necessarily definite. Thus, $L$ can be
written as $L=L_{V}+b(\mathbf{n},\cdot)\frac{1}{Q_{+}(\mathbf{n})}\mathbf{n}$,
with a two-dimensional special orthogonal transformation $L_{V}\in
SO(Q_{+|V})$. In the next sections, we shall consider what forms can appear as
restrictions $Q_{+|V}$, and which are the different appearances of
$SO(2)_{p}$.

######  Lemma 7.

For any two orthogonal bases $\mathcal{B}=(\bm{v}_{1},\bm{v}_{2},\bm{v}_{3})$
and $\mathcal{C}=(\bm{w}_{1},\bm{w}_{2},\bm{w}_{3})$ of
${{\mathbb{Q}}}_{p}^{3}$, there exists an orthogonal transformation
$M\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Q}}}_{p}^{3}\to{{\mathbb{Q}}}_{p}^{3}$ such that
$M\bm{v}_{i}=\bm{w}_{i}$ for all $i=1,2,3$ if and only if
$Q_{+}(\bm{v}_{i})=Q_{+}(\bm{w}_{i})$.

In that case, either $M$ is special or the orthogonal transformation
$M^{\prime}$ sending $\mathcal{B}$ to
$\mathcal{C}^{\prime}=(\bm{w}_{1},-\bm{w}_{2},\bm{w}_{3})$ is special.

###### Proof.

There exists a unique linear transformation $M\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Q}}}_{p}^{3}\to{{\mathbb{Q}}}_{p}^{3}$ sending
$\mathcal{B}$ to $\mathcal{C}$. $M$ is an orthogonal transformation if and
only if it preserves the bilinear form on ${{\mathbb{Q}}}_{p}^{3}$ associated
to $Q_{+}$. It is enough to show
$b(\bm{v}_{i},\bm{v}_{j})=b(\bm{w}_{i},\bm{w}_{j})$ for all
$i,j\in\\{1,2,3\\}$. For $i\neq j$ this holds as we assume orthogonal bases.
For $i=j$ it amounts to $Q_{+}(\bm{v}_{i})=Q_{+}(\bm{w}_{i})$ for every
$i\in\\{1,2,3\\}$.

$\mathcal{C}^{\prime}$ is an orthogonal basis like $\mathcal{C}$, and the
orthogonal transformation $M$ exists if and only if the orthogonal
$M^{\prime}$ exists. An orthogonal transformation has determinant $\pm 1$, and
as $\det M^{\prime}=-\det M$, exactly one of the two will have determinant
$1$.  $\sqcap$$\sqcup$

In contrast to ${{\mathbb{Q}}}_{p}^{3}$, every vector in ${{\mathbb{R}}}^{3}$
can be normalized to $1$, as $Q_{+}(\bm{v})\simeq 1$ modulo squares for all
$\bm{v}\in{{\mathbb{R}}}^{3}\setminus\bm{0}$ (indeed
${{\mathbb{R}}}^{\ast}/({{\mathbb{R}}}^{\ast})^{2}\simeq\\{\pm 1\\}$ and
$Q_{+}^{{\mathbb{R}}}(\bm{v})>0$ for every
$\bm{v}\in{{\mathbb{R}}}^{3}\setminus\bm{0}$). As a consequence, every
orthogonal basis of ${{\mathbb{R}}}^{3}$ can be made orthonormal, and there
always exists a proper or improper rotation mapping any orthonormal basis to
any other orthonormal basis. On the other hand, in ${{\mathbb{Q}}}_{p}^{3}$ we
can only consider orthogonal bases due to the existence of different kinds of
vectors, depending on the value in
$K={{\mathbb{Q}}}_{p}^{\ast}/({{\mathbb{Q}}}_{p}^{\ast})^{2}$ that the
quadratic form $Q_{+}$ takes on the vectors of
${{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$. In this case the condition of
preservation of the quadratic form becomes relevant.

######  Proposition 8.

Given $\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$, for any non-zero
vectors $\bm{v},\,\bm{w}\in\bm{n}^{\perp}\subset{{\mathbb{Q}}}_{p}^{3}$ there
exists a rotation ${\cal R}_{\bm{n}}\in SO(3)_{p}$ such that ${\cal
R}_{\bm{n}}\bm{v}=\bm{w}$ if and only if $Q_{+}(\bm{v})=Q_{+}(\bm{w})$.

###### Proof.

The direct implication is trivial by definition of $SO(3)_{p}$.

Conversely, for the orthogonal pair $\bm{v},\bm{n}$ there exists a unique
vector $\bm{v}^{\prime}\in{{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$, up to
nonzero scalar multiples, which completes $\bm{v},\bm{n}$ to an orthogonal
basis $\mathcal{B}=(\bm{v},\bm{v}^{\prime},\bm{n})$. Similarly, we can
complete the pair $\bm{w},\bm{n}$ to an orthogonal basis
$\mathcal{C}=(\bm{w},\bm{w}^{\prime},\bm{n})$.

Now we prove that $Q_{+}(\bm{v}^{\prime})$ and $Q_{+}(\bm{w}^{\prime})$ are in
the same class in $K$. Indeed, $Q_{+}$ has diagonal matrix representation on
any orthogonal basis. We write the matrix representation of $Q_{+}$ on the
basis $\mathcal{B}$ as $M=\text{diag}(q_{1},q_{2},q_{3})$ for every
$\bm{v}^{\prime}\in\\{\bm{v},\bm{n}\\}^{\perp}$ and on $\mathcal{C}$ as
$M^{\prime}=\text{diag}(q_{1}^{\prime},q_{2}^{\prime},q_{3}^{\prime})$ for
every $\bm{w}^{\prime}\in\\{\bm{w},\bm{n}\\}^{\perp}$. The hypothesis
$Q_{+}(\bm{v})=Q_{+}(\bm{w})$ translates into $q_{1}^{\prime}=q_{1}$, and
$q_{3}^{\prime}=q_{3}$, the latter since $\mathcal{B}$ and $\mathcal{C}$ share
the third vector $\bm{n}$. The determinant is an invariant of $Q_{+}$ in $K$,
$\det M\simeq\det M^{\prime}$: we have
$q_{1}q_{2}q_{3}=q_{1}q_{2}^{\prime}q_{3}\,\lambda^{2}$ for some
$\lambda\in{{\mathbb{Q}}}_{p}^{\ast}$, which gives
$q_{2}=q_{2}^{\prime}\,\lambda^{2}$, i.e.,
$Q_{+}(\bm{v}^{\prime})=Q_{+}(\bm{w}^{\prime})\lambda^{2}=Q_{+}(\lambda\bm{w}^{\prime})$.

Therefore, there exists $\tilde{\bm{w}}^{\prime}=\lambda\bm{w}^{\prime}$ such
that $\tilde{\mathcal{C}}=(\bm{w},\tilde{\bm{w}}^{\prime},\bm{n})$ is an
orthogonal basis for ${{\mathbb{Q}}}_{p}^{3}$ and
$Q_{+}(\bm{v}^{\prime})=Q_{+}(\tilde{\bm{w}}^{\prime})$. To this basis we
apply Lemma 7: there exists a special orthogonal transformation ${\cal
R}_{\bm{n}}\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Q}}}_{p}^{3}\to{{\mathbb{Q}}}_{p}^{3}$ such that ${\cal
R}_{\bm{n}}\bm{v}=\bm{w}$, ${\cal
R}_{\bm{n}}\bm{v}^{\prime}=\tilde{\bm{w}}^{\prime}$ and ${\cal
R}_{\bm{n}}\bm{n}=\bm{n}$. In fact, either the orthogonal transformation
${\cal R}_{\bm{n}}$ is already special, or we choose
$\tilde{\bm{w}}^{\prime}=-\lambda\bm{w}^{\prime}$ rather than
$\tilde{\bm{w}}^{\prime}=\lambda\bm{w}^{\prime}$ in the basis
$\tilde{\mathcal{C}}$.  $\sqcap$$\sqcup$

We remark that there exists a transformation in $SO(3)_{p}$ rotating the line
${{\mathbb{Q}}}_{p}\bm{v}$ to the line ${{\mathbb{Q}}}_{p}\bm{w}$ if and only
if $Q_{+}(\bm{v})\simeq Q_{+}(\bm{w})$. This is trivially satisfied in
${{\mathbb{R}}}^{3}$, where any line can be rotated to any other line.
However, the necessary condition of preservation of $Q_{+}$ implies that there
does not always exist a matrix in $SO(3)_{p}$ rotating a direction in
${{\mathbb{Q}}}_{p}^{3}$ to another one. For example, there is no element of
$SO(3)_{p}$ for $p$ odd, rotating the $x$-axis of ${{\mathbb{Q}}}_{p}^{3}$ to
the $z$-axis: $Q_{+}(\mu\bm{e}_{1})=\mu^{2}\simeq 1$ and
$Q_{+}(\lambda\bm{e}_{3})=\lambda^{2}p\simeq p$ belong to different classes in
$K$ for every $\mu,\lambda\in{{\mathbb{Q}}}_{p}^{\ast}$.

The reverse implication of the previous proposition implies the following.

######  Proposition 9.

Given a direction $\bm{n}\in{{\mathbb{Q}}}_{p}^{3}$, the action of the
subgroup of $SO(3)_{p}$ of matrix rotations around $\bm{n}$ is transitive on
the equivalence classes
$Q_{+}^{-1}(q)=\\{\bm{v}\in{{\mathbb{Q}}}_{p}^{3}\,\mathrel{\mathop{\mathchar
58\relax}}\,Q_{+}(\bm{v})=q\\}$.

More generally, for any two vectors
$\bm{v},\bm{w}\in{{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$ with
$Q_{+}(\bm{v})=Q_{+}(\bm{w})$, there exists an $L\in SO(3)_{p}$ with
$L\bm{v}=\bm{w}$. This is achieved by choosing
$\bm{n}\in\\{\bm{v},\bm{w}\\}^{\perp}$.  $\sqcap$$\sqcup$

As noted above, there are different classes of vectors in
$\mathbb{Q}_{p}^{3}$, i.e.,
$\\{\bm{v}\in\mathbb{Q}_{p}^{3}\backslash\bm{0}\,\mathrel{\mathop{\mathchar
58\relax}}\,Q_{+}(\bm{v})\simeq k\\}$ for
$k\in\mathbb{Q}_{p}^{\ast}/(\mathbb{Q}_{p}^{\ast})^{2}$. This is one of the
fundamental obstacles in the study of $SO(3)_{p}$, which marks a first
departure from the real case.

##  4 Planar case: the three (seven) incarnations of $SO(2)_{p}$

By virtue of Theorem 6, the elements of $SO(3)_{p}$ are transformations
occurring in the plane $V$ orthogonal to the rotation axis $\bm{n}$ in
${{\mathbb{Q}}}_{p}^{3}$. Thus we move to classify the definite quadratic
forms on ${{\mathbb{Q}}}_{p}^{2}$ and analyze the corresponding symmetry
groups. We will derive a parameterization for the rotations in these groups
and for $SO(3)_{p}$, and show that the restrictions of the rotations of
$SO(3)_{p}$ to the plane orthogonal to their rotation axis realize all the
classes of rotation groups of the $p$-adic plane. Indeed, we will show that
the restrictions $Q_{+|V}$ of the three-dimensional definite form $Q_{+}$,
where $V\subset{{\mathbb{Q}}}_{p}^{3}$ varies over all two-dimensional
subspaces, exhaust all two-dimensional definite quadratic forms, up to
equivalence (similarity and scaling).

Beginning with odd primes $p$, the invariants of the two-dimensional quadratic
forms $Q(x,y)=ax^{2}+by^{2}$, $x,y\in{{\mathbb{Q}}}_{p}$, characterized by
pairs $(a,b)\in K^{2}$, give the following $8$ distinct equivalence classes

$d$ | $\varepsilon$ | Representative $(a,b)$ |  $(a,b)$ of equivalent forms
---|---|---|---
$1$ | $1$ | $(1,1)$ |  $(u,u)$ and $(p,p)$, $(up,up)$ for $p\equiv 1\mod 4$
$1$ | $-1$ |  $(p,p)$ for $p\equiv 3\mod 4$ |  $(up,up)$ for $p\equiv 3\mod 4$
$u$ | $1$ | $(1,u)$ |  $(p,up)$ for $p\equiv 3\mod 4$
$u$ | $-1$ |  $(p,up)$ for $p\equiv 1\mod 4$ |
$p$ | $1$ | $(1,p)$ |
$p$ | $-1$ | $(u,up)$ |
$up$ | $1$ | $(1,up)$ |
$up$ | $-1$ | $(u,p)$ |

Considering scaling, we can always make $a=1$, without loss of generality.
Thus, there are four different equivalence classes of quadratic forms on
${{\mathbb{Q}}}_{p}^{2}$ up to equivalence for every odd prime $p$

$x^{2}+y^{2},\quad x^{2}+uy^{2},\quad x^{2}+py^{2},\quad ux^{2}+py^{2}$ (3)

(it will become clear in a moment why we choose the last form as written, and
not as $x^{2}+upy^{2}$ or $x^{2}+p/uy^{2}$).

To determine which of them represents zero, we need to check whether
$d\simeq-1$ in $K$. First, $-1\in({{\mathbb{Q}}}_{p}^{\ast})^{2}$ for $p\equiv
1\mod 4$, so $-1\simeq 1\in K$; on the other hand, $-1\simeq u\in K$ for
$p\equiv 3\mod 4$ because $-1\notin({{\mathbb{Q}}}_{p}^{\ast})^{2}$. Now,
$x^{2}+y^{2}$ has $d=1$: it represents $0$ for $p\equiv 1\mod 4$, but not for
$p\equiv 3\mod 4$. $x^{2}+uy^{2}$ has $d=u$: it represents $0$ for $p\equiv
3\mod 4$, but not for $p\equiv 1\mod 4$. The last two forms of (3) do not
represent $0$.

Proceeding similarly for $p=2$, we get the following $16$ classes of quadratic
forms on ${{\mathbb{Q}}}_{2}^{2}$

$d$ | $\varepsilon$ | Representative $(a,b)$ |  $(a,b)$ of equivalent forms
---|---|---|---
$1$ | $1$ | $(1,1)$ | $(2,2),\,(5,5),\,(10,10)$
$1$ | $-1$ | $(-1,-1)$ | $(-2,-2),\,(-5,-5),\,(-10,-10)$
$-1$ | $1$ | $(1,-1)$ | $(2,-2),\,(5,-5),\,(10,-10)$
$-1$ | $-1$ | N/A† | N/A
$2$ | $1$ | $(1,2)$ | $(-5,-10)$
$2$ | $-1$ | $(-1,-2)$ | $(5,10)$
$-2$ | $1$ | $(1,-2)$ | $(-1,2)$
$-2$ | $-1$ | $(5,-10)$ | $(-5,10)$
$5$ | $1$ | $(1,5)$ | $(-2,-10)$
$5$ | $-1$ | $(-1,-5)$ | $(2,10)$
$-5$ | $1$ | $(1,-5)$ | $(-1,5)$
$-5$ | $-1$ | $(2,-10)$ | $(-2,10)$
$10$ | $1$ | $(-2,-5)$ | $(1,-10)$
$10$ | $-1$ | $(2,5)$ | $(-1,-10)$
$-10$ | $1$ | $(1,-10)$ | $(-1,10)$
$-10$ | $-1$ | $(2,-5)$ | $(-2,5)$

† No $2$-adic quadratic form in dimension two can have determinant $-1$
together with Hasse invariant $-1$.

They reduce to the following $8$ classes up to scaling

$\displaystyle x^{2}+y^{2},$ $\displaystyle x^{2}-y^{2},$ $\displaystyle
x^{2}+2y^{2},$ $\displaystyle x^{2}-2y^{2},$ $\displaystyle x^{2}+5y^{2},$
$\displaystyle x^{2}-5y^{2},$ $\displaystyle x^{2}+10y^{2},$ $\displaystyle
x^{2}-10y^{2}.$

Only the second one represents $0$. In summary, we get the following
classification.

######  Proposition 10.

The quadratic forms on ${{\mathbb{Q}}}_{p}^{2}$ that do not represent $0$,
are, up to equivalence

  * •

the following $3$ for every odd prime $p$

$Q_{-v}(\bm{x})=x^{2}-vy^{2},\quad Q_{p}(\bm{x})=x^{2}+py^{2},\quad
Q_{up}(\bm{x})=ux^{2}+py^{2},$

where $v$ is as in Eq. (1);

  * •

the following $7$ for $p=2$

$\displaystyle Q_{1}(\bm{x})$ $\displaystyle=x^{2}+y^{2},\quad$ $\displaystyle
Q_{\pm 2}(\bm{x})$ $\displaystyle=x^{2}\pm 2y^{2},$ $\displaystyle Q_{\pm
5}(\bm{x})$ $\displaystyle=x^{2}\pm 5y^{2},\quad$ $\displaystyle Q_{\pm
10}(\bm{x})$ $\displaystyle=x^{2}\pm 10y^{2}.$

The subscript of these quadratic forms denotes their determinant.
$\sqcap$$\sqcup$

When $p\equiv 1\mod 4$, the three definite forms are the restrictions of the
definite three-dimensional one to the planes $xy$, $xz$ and $yz$. This is not
true for $p\equiv 3\mod 4$, where $Q_{+}(\bm{x})$ reduces to $Q_{-v}(\bm{x})$
for $z=0$ and to the same $Q_{p}(\bm{x})$ for both $x=0$ and $y=0$. Similarly,
the restriction of the definite three-dimensional form for $p=2$ to any
reference plane $xy$, $xz$ and $yz$ gives only $Q_{1}(\bm{x})$ among the seven
possible forms in dimension two.

In contrast to the unique special orthogonal group $SO(2)_{{\mathbb{R}}}$ on
the real plane, there are three (compact) special orthogonal groups on
${{\mathbb{Q}}}_{p}^{2}$ for odd prime $p$, and seven for $p=2$, up to
isomorphisms, according to the last proposition. We call them
$SO(2)_{p}^{\kappa}$ with $\kappa$ denoting the determinant of the preserved
quadratic form: these groups are

$SO(2)_{p}^{\kappa}=\\{L\in{\cal M}_{2\times
2}({{\mathbb{Q}}}_{p})\,\mathrel{\mathop{\mathchar
58\relax}}\,A_{\kappa}=L^{\top}A_{\kappa}L,\ \det L=1\\},$ (4)

endowed with the common matrix product. $A_{\kappa}$ denotes the matrix
representation of the quadratic form $Q_{\kappa}$ in the canonical basis

$A_{-v}=\begin{pmatrix}1&0\\\ 0&-v\end{pmatrix},\quad
A_{p}=\begin{pmatrix}1&0\\\ 0&p\end{pmatrix},\quad
A_{up}=\begin{pmatrix}u&0\\\ 0&p\end{pmatrix},$ $A_{1}=\begin{pmatrix}1&0\\\
0&1\end{pmatrix},\quad A_{\pm 2}=\begin{pmatrix}1&0\\\ 0&\pm
2\end{pmatrix},\quad A_{\pm 5}=\begin{pmatrix}1&0\\\ 0&\pm
5\end{pmatrix},\quad A_{\pm 10}=\begin{pmatrix}1&0\\\ 0&\pm 10\end{pmatrix}.$

We now proceed to the parametrisation of the two-dimensional rotation groups.
Consider the transformation of a vector in ${{\mathbb{Q}}}_{p}^{2}$ by some
${\cal R}_{\kappa}\in SO(2)_{p}^{\kappa}$

$\begin{pmatrix}x^{\prime}\\\ y^{\prime}\end{pmatrix}=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\begin{pmatrix}x\\\ y\end{pmatrix}.$

The orthogonality condition for ${\cal R}_{\kappa}$ yields the system of
equations

$\left\\{\begin{aligned} &a^{2}+\alpha_{\kappa}c^{2}=1,\\\
&b^{2}+\alpha_{\kappa}d^{2}=\alpha_{\kappa},\\\
&ab+\alpha_{\kappa}cd=0,\end{aligned}\right.$ (5)

where $\alpha_{\kappa}=-v,p,\frac{p}{u},1,\pm 2,\pm 5,\pm 10$, respectively
for the quadratic forms with $\kappa=-v,p,up,\,p\neq 2$ and $\kappa=1,\pm
2,\pm 5,\pm 10,\,p=2$. We parametrize the solutions of this system of three
equations in four $p$-adic unknowns through

$\sigma=\pm\frac{c}{1+a}\in{{\mathbb{Q}}}_{p}$ (6)

for $a\in{{\mathbb{Z}}}_{p}\backslash\\{-1\\}$, and treat the case $a=-1$
separately. The first equation of the system (5) gives

$\alpha_{\kappa}=\frac{1-a^{2}}{c^{2}}$

when $c\neq 0$. Manipulating Eq. (6), we get
$\frac{1+a}{c^{2}}=\frac{1}{(1+a)\sigma^{2}}$, $\sigma\neq 0$, so

$\alpha_{\kappa}=\frac{1-a}{(1+a)\sigma^{2}}\Rightarrow(1+\alpha_{\kappa}\sigma^{2})a=1-\alpha_{\kappa}\sigma^{2}.$
(7)

The parameter $-\alpha_{\kappa}$ is not a square for any $\kappa$ of the
considered quadratic forms (because none of them represents zero). This means
that $1+\alpha_{\kappa}\sigma^{2}\neq 0$ for all $\alpha_{\kappa}$ and
$\sigma\in{{\mathbb{Q}}}_{p}$. Now Eq. (7) gives

$a=\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}.$

The first equation of (5) now provides

$c^{2}=\frac{1-a^{2}}{\alpha_{\kappa}}=\frac{4\sigma^{2}}{\left(1+\alpha_{\kappa}\sigma^{2}\right)^{2}}\Rightarrow
c=\pm_{c}\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}.$

From the third equation of (5), when $a\neq 0$, we get

$b^{2}=\alpha_{\kappa}\left(\frac{1}{a^{2}}-1\right)d^{2}$ (8)

which plugged into the second one, gives $d^{2}=a^{2}$, hence $d=\pm_{d}a$.

Now the second equation of (5) gives

$b^{2}=\alpha_{\kappa}^{2}c^{2}\Rightarrow
b=\pm_{b}\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}}.$

We arrive at

${\cal R}_{\kappa}(\sigma)=\begin{pmatrix}a(\sigma)&b(\sigma)\\\
c(\sigma)&d(\sigma)\end{pmatrix}=\begin{pmatrix}\phantom{\pm_{c}}\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}&\pm_{b}\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}}\\\
\pm_{c}\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}&\pm_{d}\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}\end{pmatrix},$
(9)

where the signs $\pm_{b},\pm_{c},\pm_{d}$ are a priori unrelated. The
determinant of the matrix (9) is

$\pm_{d}\left(\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}\right)^{2}-\pm_{b}\pm_{c}\alpha_{\kappa}\left(\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}\right)^{2}=\frac{\pm_{d}1-2\alpha_{\kappa}\sigma^{2}(\pm_{d}1+\pm_{b}\pm_{c}2)\pm_{d}\alpha_{\kappa}^{2}\sigma^{4}}{1+2\alpha_{\kappa}\sigma^{2}+\alpha_{\kappa}^{2}\sigma^{4}},$

and it must be $1$. This happens when $\pm_{d}1=1$ and
$\pm_{d}1+\pm_{b}\pm_{c}2=-1$, i.e., $\pm_{c}=\mp_{b}$. Therefore,

${\cal
R}_{\kappa}(\sigma)=\begin{pmatrix}\phantom{\pm}\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}&\mp\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}}\\\
\pm\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}&\phantom{\pm}\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}\end{pmatrix},$
(10)

with linked signs.

In the derivation, at Eq. (8), we had to assume $a\neq 0$. When $a=0$, we have
$b^{2}=\alpha_{\kappa}$, $c^{2}=1/\alpha_{\kappa}$, $d=0$, which provides a
solution when $\alpha_{\kappa}$ is a square. In this case, $a=0$ corresponds
to $\sigma=\pm c$, included in Eq. (10).

The case $c=0$ left out in the system of equations gives ${\cal
R}_{\kappa}=\pm I$, using that the determinant is $1$. Thus, $c=0$ is
equivalent to $\sigma=0$ when $a\neq-1$ in Eq. (6), leading to ${\cal
R}_{\kappa}(0)=I$. This leaves one solution of Eq. (10) unaccounted for, which
is ${\cal R}_{\kappa}=-I$, reached from the system of equations with $a=-1$,
for which $\sigma$ is not well-defined in ${{\mathbb{Q}}}_{p}$. But this sign
change from the identity matrix can be included in Eq. (10) by considering
$\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$.

######  Remark 11.

Note that $\lim\limits_{n\to\infty}p^{n}=0$ in ${{\mathbb{Q}}}_{p}$ with
respect to the $p$-adic norm, since $|p^{n}|_{p}$ converges to $0$ by
increasing $n$. Then, $\infty$ on the $p$-adic field is formally introduced as
the limit $\lim\limits_{n\to\infty}p^{-n}$ (the inverse of $0$). Indeed the
matrix entries in Eq. (10) are such that

$\displaystyle\lim_{n\to\infty}\frac{1-\alpha_{\kappa}p^{-2n}}{1+\alpha_{\kappa}p^{-2n}}=\lim_{n\to\infty}\frac{p^{-2n}(p^{2n}-\alpha_{\kappa})}{p^{-2n}(p^{2n}+\alpha_{\kappa})}=-1,$
$\displaystyle\lim_{n\to\infty}\frac{2p^{-n}}{1+\alpha_{\kappa}p^{-2n}}=\lim_{n\to\infty}\frac{p^{-2n}(2p^{n})}{p^{-2n}(p^{2n}+\alpha_{\kappa})}=0$

yielding ${\cal R}_{\kappa}(\infty)=-I$.

As a parameter $\sigma$ has a sign that is linked to its inverse (as we are
going to see in a moment): it can swap between $+$ and $-$ variations of $b$
and $c$ even when we fix one choice of $\pm_{c}$. Hence we can keep only one
sign in Eq. (10) without loss of generality. We have thus proved the
following.

######  Theorem 12.

A rotation of $SO(2)_{p}^{\kappa}$ takes the following matrix form in the
canonical basis of ${{\mathbb{Q}}}_{p}^{2}$

${\cal
R}_{\kappa}(\sigma)=\begin{pmatrix}\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}&-\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}}\\\
\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}&\phantom{-}\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}\end{pmatrix},$
(11)

with $\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$,
$\alpha_{\kappa}\in\\{-v,p,\frac{p}{u}\\}$ and $\alpha_{k}\in\\{1,\pm 2,\pm
5,\pm 10\\}$ respectively for $\kappa=-v,p,up$ ($p$ odd) and $\kappa=1,\pm
2,\pm 5,\pm 10$ ($p=2$).

In addition, this parametrization is one-to-one, i.e., it results in different
special orthogonal transformations for different $\sigma$.  $\sqcap$$\sqcup$

######  Remark 13.

For every $\kappa$ of the definite rank-$2$ quadratic forms

${\cal R}_{\kappa}\left(-\frac{1}{\alpha_{\kappa}\sigma}\right)=-{\cal
R}_{\kappa}\left(\sigma\right),\ \
\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\},$ (12)

because if we replace $\sigma\mapsto-\frac{1}{\alpha_{\kappa}\sigma}$ in the
matrices (11), then the matrix elements change as follows

$\displaystyle\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}$
$\displaystyle\mapsto\frac{1-\alpha_{\kappa}(\alpha_{\kappa}\sigma)^{-2}}{1+\alpha_{\kappa}(\alpha_{\kappa}\sigma)^{-2}}=\frac{\alpha_{\kappa}\sigma^{2}-1}{\alpha_{\kappa}\sigma^{2}+1}=-\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}},$
$\displaystyle-\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}}$
$\displaystyle\mapsto-\frac{2\alpha_{\kappa}(-\alpha_{\kappa}\sigma)^{-1}}{1+\alpha_{\kappa}(\alpha_{\kappa}\sigma)^{-2}}=\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}},$
$\displaystyle\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}$
$\displaystyle\mapsto\frac{2(-\alpha_{\kappa}\sigma)^{-1}}{1+\alpha_{\kappa}(\alpha\sigma)^{-2}}=-\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}.$

We need $\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ in order for this
transformation to be well-defined for $\sigma=0$, too, which is guaranteed by

${\cal R}_{\kappa}\left(\infty\right)=-{\cal R}_{\kappa}\left(0\right)=-I.$

######  Remark 14.

As a corollary of the parameterization (11), we can see directly that the
matrix entries
$\frac{1-\alpha_{\kappa}\sigma^{2}}{1+\alpha_{\kappa}\sigma^{2}}$,
$\frac{2\sigma}{1+\alpha_{\kappa}\sigma^{2}}$ and
$-\frac{2\alpha_{\kappa}\sigma}{1+\alpha_{\kappa}\sigma^{2}}$ of the rotations
are $p$-adic integers for every $\alpha_{\kappa}$, every
$\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ and every prime $p$, in
accordance with Theorem 5. This can be easily checked for parameters
$\sigma\in{{\mathbb{Z}}}_{p}$:

  * •

$1+\alpha_{\kappa}\sigma^{2}\not\equiv 0\mod p$, for every
$\kappa,\,\sigma\in{{\mathbb{Z}}}_{p},\,p\neq 2$, for $\kappa=\pm 2,\pm
10,\,\sigma\in{{\mathbb{Z}}}_{2}$, and for $\kappa=1,\pm 5,\,\sigma\in
2{{\mathbb{Z}}}_{2}$. In these cases
$(1+\alpha_{\kappa}\sigma^{2})^{-1}\in{{\mathbb{U}}}$, which multiplied with
$1-\alpha_{\kappa}\sigma^{2},\,2\sigma,\,-2\alpha_{\kappa}\sigma\in{{\mathbb{Z}}}_{p}$
give $p$-adic integer matrix entries;

  * •

If $\sigma\in{{\mathbb{Z}}}_{2}$, $\sigma\equiv 1\mod 2$, then
$1+\sigma^{2},\,1\pm 5\sigma^{2}\equiv 0\mod 2$: they are invertible in
${{\mathbb{Q}}}_{2}$ but not in ${{\mathbb{Z}}}_{2}$. Here the calculus modulo
$2$ is not enough to verify that the associated matrix entries are $2$-adic
integers. But writing
$\sigma=1+2\sigma^{\prime},\,\sigma^{\prime}\in{{\mathbb{Z}}}_{2}$ we get

$\displaystyle\frac{1-\sigma^{2}}{1+\sigma^{2}}=\frac{-4(\sigma^{\prime}+{\sigma^{\prime}}^{2})}{2\big{(}1+2(\sigma^{\prime}+{\sigma^{\prime}}^{2})\big{)}}=\frac{-2(\sigma^{\prime}+{\sigma^{\prime}}^{2})}{1+2(\sigma^{\prime}+{\sigma^{\prime}}^{2})},$
$\displaystyle\frac{2\sigma}{1+\sigma^{2}}=\frac{2(1+2\sigma^{\prime})}{2\big{(}1+2(\sigma^{\prime}+{\sigma^{\prime}}^{2})\big{)}}=\frac{1+2\sigma^{\prime}}{1+2(\sigma^{\prime}+{\sigma^{\prime}}^{2})}.$

$1+2(\sigma^{\prime}+{\sigma^{\prime}}^{2})\not\equiv 0\mod 2$ is invertible
in ${{\mathbb{Z}}}_{2}$.

We have shown that the entries of the matrices of $SO(2)_{p}^{\kappa}$ are
$p$-adic integers when $\sigma\in{{\mathbb{Z}}}_{p}$, for every $\kappa$ and
prime $p$. But then we can distinguish two branches for
$\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$: either
$\sigma\in{{\mathbb{Z}}}_{p}$ or $\sigma^{-1}\in p{{\mathbb{Z}}}_{p}$
(including $\infty$ formally when $\sigma=0$). Since we want to exploit Eq.
(12) for a non integer parameter, either $\sigma\in{{\mathbb{Z}}}_{p}$ or:

  * •

$\sigma=-\frac{1}{\alpha_{\kappa}\tau},\ \tau\in p{{\mathbb{Z}}}_{p}$, for
$p\neq 2,\kappa=-v$ and for $p=2,\kappa=1,\pm 5$,

  * •

$\sigma=-\frac{1}{\alpha_{\kappa}\tau},\ \tau\in{{\mathbb{Z}}}_{p}$, for
$p\neq 2,\kappa=p,up$ and for $p=2,\kappa=\pm 2,\pm 10$.

In each of the cases of the first point

$\displaystyle SO(2)_{p}^{\kappa}$ $\displaystyle=\left\\{{\cal
R}_{\kappa}(\sigma)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}\right\\}$
$\displaystyle=\left\\{{\cal R}_{\kappa}(\sigma)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{p}\right\\}\cup\left\\{{\cal
R}_{\kappa}\left(-\frac{1}{\alpha_{\kappa}\tau}\right)\,\mathrel{\mathop{\mathchar
58\relax}}\,\tau\in p{{\mathbb{Z}}}_{p}\right\\}$ $\displaystyle=\left\\{{\cal
R}_{\kappa}(\sigma)\,\mathrel{\mathop{\mathchar
58\relax}}\,\sigma\in{{\mathbb{Z}}}_{p}\right\\}\cup\left\\{-{\cal
R}_{\kappa}\left(\tau\right)\,\mathrel{\mathop{\mathchar 58\relax}}\,\tau\in
p{{\mathbb{Z}}}_{p}\right\\}.$

The second set includes ${\cal R}_{\kappa}(\infty)$ for $\tau=0$.

Similarly for all the cases in the second point

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

In this way, the parameterisation of the groups $SO(2)_{p}^{\kappa}$ is in
terms of $p$-adic integers only. The second branch for the parameter $\sigma$
gives the same matrix entries of the first one up to a sign: these entries are
$p$-adic integers as shown for the first branch.

This is useful when projecting $SO(3)_{p}$ modulo $p^{k}$, to be able to
enumerate the elements of its subgroups of rotations around the reference axes
Ilaria:tesi .

Interlude: Euclidean geometrical interpretation. The introduction of a
parameter $\sigma$ leading to
$a=\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}$ in Eq. (11) is inspired by
Euclidean geometry. Indeed, recalling the _tangent half-angle formulae_

$\cos\theta=\frac{1-\tan^{2}(\theta/2)}{1+\tan^{2}(\theta/2)},\quad\sin\theta=\frac{2\tan(\theta/2)}{1+\tan^{2}(\theta/2)},$

and letting

$\alpha_{\kappa}=1,\quad\sigma=\tan\left(\frac{\theta}{2}\right)\in{{\mathbb{R}}},$
(13)

Eq. (11) takes the form

${\cal R}(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\
\sin\theta&\cos\theta\end{pmatrix},$

which is the matrix for a rotation by an angle $\theta$ in the real plane. The
special case $-I$ is given by $\theta=\pi$, for which $\tan(\theta/2)$
diverges. It is as if we have been treating rotations in
${{\mathbb{Q}}}_{p}^{2}$ in terms of the tangent of the rotation angle rather
than the angle itself. It is worth noticing that the usage of tangent of
angles (in place of angles) in trigonometry dates back to ancient Babylon: the
Babylonian table, _Plimpton 322_ (written about 1800 BC), may be interpreted
as a first trigonometric table, where Pythagorean triples were written as
ratios between the sides of right triangles Cowen .

This geometrical interpretation motivates our notation ${\cal
R}_{\kappa}(\sigma)$ for a rotation of parameter $\sigma$.

Composition laws of rotations. The composition of two rotations of
$SO(2)_{p}^{\kappa}$, for any fixed $\kappa$, is given by the product of their
matrix forms (11). It turns out to take the very simple form

${\cal R}_{\kappa}(\sigma){\cal R}_{\kappa}(\tau)={\cal
R}_{\kappa}\left(\frac{\sigma+\tau}{1-\alpha_{\kappa}\sigma\tau}\right),$ (14)

for every $\sigma,\tau\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$. For the
infinitely many values of $\sigma,\tau\in{{\mathbb{Q}}}_{p}$ such that
$1-\alpha_{\kappa}\sigma\tau=0$, the right-hand-side is ${\cal
R}_{\kappa}(\infty)=-I$, according to Remark 11. If
$\tau\in{{\mathbb{Q}}}_{p}$ while $\sigma$ is $\infty$ (or vice versa), the
argument on the right-hand side is

$\lim_{n\to\infty}\frac{p^{-n}+\tau}{1-\alpha_{\kappa}p^{-n}\tau}=\lim_{n\to\infty}\frac{p^{-n}(1+p^{n}\tau)}{p^{-n}(p^{n}-\alpha_{\kappa}\tau)}=-\frac{1}{\alpha_{\kappa}\tau},$

so Eq. (14) becomes $-{\cal R}_{\kappa}(\tau)={\cal
R}_{\kappa}\left(-\frac{1}{\alpha_{\kappa}\tau}\right)$ in agreement with Eq.
(12). If both $\sigma$ and $\tau$ are $\infty$, the above composition gives
correctly the identity, since the rotation parameter on the right is

$\lim_{n,m\to\infty}\frac{p^{-n}+p^{-m}}{1-\alpha_{\kappa}p^{-n-m}}=\lim_{n,m\to\infty}\frac{p^{-n-m}(p^{m}+p^{n})}{p^{-n-m}(p^{n+m}-\alpha_{\kappa})}=0.$

We recall that

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

which is equivalent to the arguments of Eq. (14) if we use Eq. (13).

######  Proposition 15.

$SO(2)_{p}^{\kappa}$ defined in Eq. (4) is an abelian group, for every
$\kappa$ and prime $p$.

###### Proof.

This is straightforward to prove via Eq. (14), and it is in analogy with
$SO(2)_{{{\mathbb{R}}}}$, which is isomorphic to the group of addition or real
angles $\theta$ modulo $2\pi$, but reinterpreted through the half-angle
tangent. Furthermore, ${\cal R}_{\kappa}(\sigma)^{-1}={\cal
R}_{\kappa}(-\sigma)$ for all $\sigma\in{{\mathbb{Q}}}_{p}$ and ${\cal
R}_{\kappa}(\infty)^{-1}={\cal R}_{\kappa}(\infty)$.  $\sqcap$$\sqcup$

Rotations parametrized by the $p$-adic projective line. The form of the
parametrization (11) of two-dimensional rotations, and their composition law
(14), and in particular the necessity to add a point at infinity to the
$p$-adic affine line, suggest a description in terms of projective geometry.

As usual, we identify the projective line $P^{1}({{\mathbb{Q}}}_{p})$ with the
equivalence classes $[s\mathrel{\mathop{\mathchar
58\relax}}t]={{\mathbb{Q}}}_{p}^{*}(s,t)\subset{{\mathbb{Q}}}_{p}^{2}\setminus(0,0)$
of nonzero vectors under multiplication by (nonzero) scalars. Representatives
are $[s\mathrel{\mathop{\mathchar 58\relax}}1]$ and
$\infty\equiv[1\mathrel{\mathop{\mathchar 58\relax}}0]$, allowing us to think
of $P^{1}({{\mathbb{Q}}}_{p})={{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ as the
$p$-adic line closed with a point at infinity. The parameter in Eq. (11) is
$\sigma=s/t$, which is well-defined on each equivalence class. We can write
the rotations as bona fide functions of the points
$[s\mathrel{\mathop{\mathchar 58\relax}}t]$ of $P^{1}({{\mathbb{Q}}}_{p})$

${\cal R}_{\kappa}([s\mathrel{\mathop{\mathchar
58\relax}}t])=\begin{pmatrix}\frac{t^{2}-\alpha_{\kappa}s^{2}}{t^{2}+\alpha_{\kappa}s^{2}}&-\frac{2\alpha_{\kappa}st}{t^{2}+\alpha_{\kappa}s^{2}}\\\
\frac{2st}{t^{2}+\alpha_{\kappa}s^{2}}&\phantom{-}\frac{t^{2}-\alpha_{\kappa}s^{2}}{t^{2}+\alpha_{\kappa}s^{2}}\end{pmatrix},$
(15)

where $\alpha_{\kappa}\in\\{-v,p,\frac{p}{u}\\}$ and $\alpha_{k}\in\\{1,\pm
2,\pm 5,\pm 10\\}$ respectively for $\kappa=-v,p,up$ ($p$ odd) and
$\kappa=1,\pm 2,\pm 5,\pm 10$ ($p=2$). Furthermore, the uniqueness of the
parametrization (15) as a function of $[s\mathrel{\mathop{\mathchar
58\relax}}t]\in P^{1}({{\mathbb{Q}}}_{p})$ is inherited from Theorem 12.

Looking at $\sigma$ as a ratio reveals the underlying structure, too, of the
composition law (14), which in projective line coordinates becomes

${\cal R}_{\kappa}([s\mathrel{\mathop{\mathchar 58\relax}}t]){\cal
R}_{\kappa}([u\mathrel{\mathop{\mathchar 58\relax}}v])={\cal
R}_{\kappa}\left([sv+tu\mathrel{\mathop{\mathchar
58\relax}}tv-\alpha_{\kappa}su]\right).$ (16)

Note that for each fixed $[s\mathrel{\mathop{\mathchar 58\relax}}t]$, this is
a projective linear transformation of $[u\mathrel{\mathop{\mathchar
58\relax}}v]$ (and vice versa). Indeed, the mapping

$\displaystyle T_{\kappa}\mathrel{\mathop{\mathchar
58\relax}}P^{1}({{\mathbb{Q}}}_{p})$ $\displaystyle\longrightarrow
PGL(2,{{\mathbb{Q}}}_{p})$ $\displaystyle[s\mathrel{\mathop{\mathchar
58\relax}}t]$ $\displaystyle\longmapsto
T_{\kappa}([s\mathrel{\mathop{\mathchar
58\relax}}t])\mathrel{\mathop{\mathchar
58\relax}}={{\mathbb{Q}}}_{p}^{*}\begin{pmatrix}t&-\alpha_{\kappa}s\\\
s&t\end{pmatrix}$

defines abelian subgroups ${\cal T}_{\kappa}\mathrel{\mathop{\mathchar
58\relax}}=T_{\kappa}(P^{1}({{\mathbb{Q}}}_{p}))\subset
PGL(2,{{\mathbb{Q}}}_{p})$ (the invertible $2\times 2$-matrices over
${{\mathbb{Q}}}_{p}$ modulo scalars ${{\mathbb{Q}}}_{p}^{*}$), each a rational
image of the projective line. $T_{\kappa}$ translates the composition law Eq.
(16) into a simple matrix multiplication

$T_{\kappa}([s\mathrel{\mathop{\mathchar
58\relax}}t])T_{\kappa}([u\mathrel{\mathop{\mathchar
58\relax}}v])=T_{\kappa}([sv+tu\mathrel{\mathop{\mathchar
58\relax}}tv-\alpha_{\kappa}su]),$

This means that we can identify $SO(2)_{p}^{\kappa}$ with the subgroup ${\cal
T}_{\kappa}\subset PGL(2,{{\mathbb{Q}}}_{p})$, $T_{\kappa}$ providing the
isomorphism. If, by abuse of notation, we refer to $T_{\kappa}$ as the map
from $P^{1}({{\mathbb{Q}}}_{p})$ to its image ${\cal T}_{\kappa}$, the
isomorphic map from ${\cal T}_{\kappa}$ to $SO(2)_{p}^{\kappa}$ is explicitly
given by first applying $T_{\kappa}^{-1}\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Q}}}_{p}^{*}\begin{pmatrix}t&-\alpha_{\kappa}s\\\
s&t\end{pmatrix}\mapsto[s,t]$ and then parameterizing an element of
$SO(2)_{p}^{\kappa}$ as in Eq. (15).

######  Remark 16.

Consider the quotient topologies on $P^{1}(\mathbb{Q}_{p})$ and
$PGL(2,{{\mathbb{Q}}}_{p})$, where $\mathbb{Q}_{p}^{2}\backslash(0,0)$ and
$GL(2,\mathbb{Q}_{p})$ are naturally endowed with their $p$-adic topology, and
consider the topology of the $T_{\kappa}$ as subspaces of
$PGL(2,{{\mathbb{Q}}}_{p})$. The map $T_{\kappa}$, from
$P^{1}({{\mathbb{Q}}}_{p})$ to its image ${\cal T}_{\kappa}$, is clearly a
homeomorphism, for every $\kappa$: it is continuous, as well as its inverse
$T_{\kappa}^{-1}$, as it can be directly seen from their explicit expressions
above. Now, let $SO(2)_{p}^{\kappa}$ be supplied with its natural $p$-adic
topology. The parameterization Eq. (15) is continuous in $s$ and $t$, as its
components are well defined rational functions. Hence it is continuous in
$[s\mathrel{\mathop{\mathchar 58\relax}}t]$, since the parameterization does
not depend on the representatives of the equivalence classes
$[s\mathrel{\mathop{\mathchar 58\relax}}t]$. The inverse map, providing
$[s\mathrel{\mathop{\mathchar 58\relax}}t]$ from such a parameterized matrix
${\cal R}_{\kappa}([s\mathrel{\mathop{\mathchar 58\relax}}t])$, is given by

$\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\coloneqq\begin{pmatrix}\frac{t^{2}-\alpha_{\kappa}s^{2}}{t^{2}+\alpha_{\kappa}s^{2}}&-\frac{2\alpha_{\kappa}st}{t^{2}+\alpha_{\kappa}s^{2}}\\\
\frac{2st}{t^{2}+\alpha_{\kappa}s^{2}}&\phantom{-}\frac{t^{2}-\alpha_{\kappa}s^{2}}{t^{2}+\alpha_{\kappa}s^{2}}\end{pmatrix}\mapsto[c\mathrel{\mathop{\mathchar
58\relax}}1+a]=\left[s\frac{2t}{t^{2}+\alpha_{\kappa}s^{2}}\mathrel{\mathop{\mathchar
58\relax}}t\frac{2t}{t^{2}+\alpha_{\kappa}s^{2}}\right]=[s\mathrel{\mathop{\mathchar
58\relax}}t]$

whenever $a\neq-1$, otherwise

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix}=\begin{pmatrix}-1&0\\\
0&-1\end{pmatrix}\mapsto[1\mathrel{\mathop{\mathchar 58\relax}}0].$

This inverse map is again continuous, thus Eq. (15) provides a homeomorphism
between $P^{1}(\mathbb{Q}_{p})$ and $SO(2)_{p}^{\kappa}$, for every $\kappa$.
It immediately follows that the parameterization (11) is a homeomorphism. We
have also shown that the isomorphic map between ${\cal T}_{\kappa}$ and
$SO(2)_{p}^{\kappa}$ is a homeomorphism, for every $\kappa$.

##  5 Parametrization of rotations in $SO(3)_{p}$

Returning to $p$-adic three-space ${{\mathbb{Q}}}_{p}^{3}$, we found that
every element of $SO(3)_{p}$ has a fixed axis of rotation. So, now we consider
a rotation around a general direction
$\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$. Let
$\bm{g},\bm{h}\in{{\mathbb{Q}}}_{p}^{3}$ be vectors spanning
$V=\bm{n}^{\perp}$. Restricting our attention to $V$ means reducing a vector
$\bm{s}\in{{\mathbb{Q}}}_{p}^{3}$, with coordinates $(s_{1},s_{2},s_{3})$ with
respect to the basis $(\bm{g},\bm{h},\bm{n})$, to a vector
$\bm{s}_{\left|V\right.}$ with coordinates $(s_{1},s_{2},0)$. If
$\bm{g}=(g_{1},g_{2},g_{3})$, $\bm{h}=(h_{1},h_{2},h_{3})$ and
$\bm{n}=(n_{1},n_{2},n_{3})$ with respect to the canonical basis
$(\bm{e}_{1},\,\bm{e}_{2},\,\bm{e}_{3})$ of ${{\mathbb{Q}}}_{p}^{3}$, then the
canonical coordinates of $\bm{s}_{|V}$ are given by

$\begin{pmatrix}w_{1}\\\ w_{2}\\\
w_{3}\end{pmatrix}=\begin{pmatrix}g_{1}&h_{1}&n_{1}\\\ g_{2}&h_{2}&n_{2}\\\
g_{3}&h_{3}&n_{3}\end{pmatrix}\begin{pmatrix}s_{1}\\\ s_{2}\\\ 0\end{pmatrix}$

or in short

$\bm{w}=M\bm{s}_{\left|V\right.}.$

The quadratic form preserved by $SO(3)_{p}$ is
$Q_{+}(\bm{x})=\sum_{i=1}^{3}a_{i}x_{i}^{2}$ in the canonical basis where,
$(a_{i})=(1,-v,p)$ for $p\neq 2$ and $(a_{i})=(1,1,1)$ for $p=2$. Its
restriction to $V$ is

$\begin{split}Q_{+|V}(\bm{w})&=\sum_{i=1}^{3}a_{i}w_{i}^{2}=\sum_{i=1}^{3}a_{i}\left(g_{i}s_{1}+h_{i}s_{2}\right)^{2}\\\
&=\left(\sum_{i=1}^{3}a_{i}g_{i}^{2}\right)s_{1}^{2}+2\left(\sum_{i=1}^{3}a_{i}g_{i}h_{i}\right)s_{1}s_{2}+\left(\sum_{i=1}^{3}a_{i}h_{i}^{2}\right)s_{2}^{2}\\\
&=Q_{+}(\bm{g})s_{1}^{2}+2b(\bm{g},\bm{h})s_{1}s_{2}+Q_{+}(\bm{h})s_{2}^{2}.\end{split}$

If $\bm{g}\perp\bm{h}$, then

$Q_{+|V}(\bm{w})=gs_{1}^{2}+hs_{2}^{2}$

where $g=Q_{+}(\bm{q})$, $h=Q_{+}(\bm{h})$.

We consider $\bm{t}\in{{\mathbb{Q}}}_{p}^{3}$ with coordinates
$(t_{1},t_{2},t_{3})$ with respect to $(\bm{g},\bm{h},\bm{n})$ such that

$\begin{pmatrix}t_{1}\\\ t_{2}\end{pmatrix}=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}\begin{pmatrix}s_{1}\\\ s_{2}\end{pmatrix}$

and we want to find this matrix rotating $\bm{s}_{\left|V\right.}$ to
$\bm{t}_{\left|V\right.}$. If $\bm{u}\mathrel{\mathop{\mathchar
58\relax}}=M\bm{t}_{\left|V\right.}$, we must have

$Q_{+|V}(\bm{u})=Q_{+|V}(\bm{w}).$

This implies

$\left\\{\begin{aligned} ga^{2}+hc^{2}&=g,\\\ gb^{2}+hd^{2}&=h,\\\
gab+hcd&=0,\end{aligned}\right.$ (17)

which generalizes Eq. (5) by setting

$\alpha\mathrel{\mathop{\mathchar
58\relax}}=\frac{h}{g}\in{{\mathbb{Q}}}_{p}^{\ast}.$

Now we proceed as in the previous section, and introduce

$\sigma=\pm\frac{c}{1+a}\in{{\mathbb{Q}}}_{p},$

leaving out the case $a\neq-1$ for the moment. This leads to

$(1+\alpha\sigma^{2})a=1-\alpha\sigma^{2},$

and in order to extract $a$ we need the following results.

######  Proposition 17.

For any orthogonal basis $(\bm{g},\bm{h},\bm{n})$ of ${{\mathbb{Q}}}_{p}^{3}$,

$Q_{+}(\bm{g})Q_{+}(\bm{h})Q_{+}(\bm{n})=\begin{cases}-vpt^{2}&\text{ for
}p\text{ odd},\\\ t^{2}&\text{ for }p=2,\end{cases}$

for some $t\in{{\mathbb{Q}}}_{p}^{\ast}$.

###### Proof.

We recall that the determinant is an invariant of quadratic forms in
$K={{\mathbb{Q}}}_{p}^{\ast}/({{\mathbb{Q}}}_{p}^{\ast})^{2}$ under change of
basis. Given bases $\mathcal{B}$ and $\mathcal{C}$ of
${{\mathbb{Q}}}_{p}^{3}$, there exists a unique linear transformation
$L\mathrel{\mathop{\mathchar
58\relax}}{{\mathbb{Q}}}_{p}^{3}\rightarrow{{\mathbb{Q}}}_{p}^{3}$ mapping
$\mathcal{B}$ to $\mathcal{C}$. If a quadratic form over ${{\mathbb{Q}}}_{p}$
has matrix representation $M$ with respect to $\mathcal{B}$, it has a matrix
representation $M^{\prime}=LML^{\top}$ with respect to $\mathcal{C}$. Now,
$\det LML^{\top}=(\det L)^{2}(\det M)$ implies $\det M\simeq\det M^{\prime}$
in $K$. Our form $Q_{+}$ is represented in the canonical basis
$\mathcal{B}=(\bm{e}_{1},\bm{e}_{2},\bm{e}_{3})$ of ${{\mathbb{Q}}}_{p}^{3}$
by $A=\text{diag}(1,-v,p)$ for $p\neq 2$ and by $A=I$ for $p=2$. We have $\det
A=Q_{+}(\bm{e}_{1})Q_{+}(\bm{e}_{2})Q_{+}(\bm{e}_{3})$, which is equal to
$-vp$ when $p\neq 2$ and to $1$ when $p=2$. On the other hand, $Q_{+}$ has
diagonal matrix form with respect to any orthogonal basis
$\mathcal{C}=(\bm{g},\bm{h},\bm{n})$ of ${{\mathbb{Q}}}_{p}^{3}$, where its
determinant is $Q_{+}(\bm{g})Q_{+}(\bm{h})Q_{+}(\bm{n})$. Hence,
$Q_{+}(\bm{g})Q_{+}(\bm{h})Q_{+}(\bm{n})\simeq-vp$ in $K$ for $p\neq 2$, and
$Q_{+}(\bm{g})Q_{+}(\bm{h})Q_{+}(\bm{n})\simeq 1$ for $p=2$.  $\sqcap$$\sqcup$

######  Corollary 18.

$-\alpha\in{{\mathbb{Q}}}_{p}^{\ast}$ is never a square.

###### Proof.

From Proposition 17, we have

$Q_{+}(\bm{g})Q_{+}(\bm{h})\simeq\begin{cases}-vpQ_{+}(\bm{n})&\text{ if
}p\text{ odd},\\\ Q_{+}(\bm{n})&\text{ if }p=2,\end{cases}$

which gives

$-\alpha=-\frac{Q_{+}(\bm{h})}{Q_{+}(\bm{g})}\simeq-
Q_{+}(\bm{g})Q_{+}(\bm{h})\simeq\left\\{\begin{aligned} vpQ_{+}(\bm{n}),&\
p\neq 2,\\\ -Q_{+}(\bm{n}),&\ p=2.\end{aligned}\right.$

Accordingly, $-\alpha$ is a square if and only if $Q_{+}^{(4)}(\bm{n},s)=0$
for some $s\in{{\mathbb{Q}}}_{p}^{\ast}$. However $Q_{+}^{(4)}$ does not
represent $0$.  $\sqcap$$\sqcup$

This implies $1+\alpha\sigma^{2}\neq 0$ for all
$\alpha\in{{\mathbb{Q}}}_{p}^{\ast},\sigma\in{{\mathbb{Q}}}_{p}$, and we get

$a=\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}},$

and, therefore,

${\cal
R}_{\bm{n}}(\sigma)_{|\bm{n}^{\perp}}=\begin{pmatrix}a(\sigma)&b(\sigma)\\\
c(\sigma)&a(\sigma)\end{pmatrix}=\begin{pmatrix}\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}&-\frac{2\alpha\sigma}{1+\alpha\sigma^{2}}\\\
\frac{2\sigma}{1+\alpha\sigma^{2}}&\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}\end{pmatrix}.$
(18)

We need to consider $\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ as in Remark
11 in order to include the solution to Eq. (17) with $a=-1$, which is

${\cal R}_{\bm{n}}(\infty)_{|\bm{n}^{\perp}}=-I.$

We extend the result of Eq. (18) to the whole of ${{\mathbb{Q}}}_{p}^{3}$:

$\begin{pmatrix}t_{1}\\\ t_{2}\\\
t_{3}\end{pmatrix}=\begin{pmatrix}a(\sigma)&b(\sigma)&0\\\
c(\sigma)&a(\sigma)&0\\\ 0&0&1\end{pmatrix}\begin{pmatrix}s_{1}\\\ s_{2}\\\
s_{3}\end{pmatrix}$

choosing $t_{3}=s_{3}$, or in short

$\bm{t}={\cal R}_{\bm{n}}(\sigma)\bm{s}.$ (19)

This concludes the proof of the following theorem.

######  Theorem 19.

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

${\cal
R}_{\bm{n}}(\sigma)=\begin{pmatrix}\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}&-\frac{2\alpha\sigma}{1+\alpha\sigma^{2}}&0\\\
\frac{2\sigma}{1+\alpha\sigma^{2}}&\frac{1-\alpha\sigma^{2}}{1+\alpha\sigma^{2}}&0\\\
0&0&1\end{pmatrix}$ (20)

with $\sigma\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ and
$\alpha=Q_{+}(\bm{h})/Q_{+}(\bm{g})$, for every prime $p$.  $\sqcap$$\sqcup$

Eq. (19) can be written with respect to the canonical basis
$(\bm{e}_{1},\bm{e}_{2},\,\bm{e}_{3})$ as

$\bm{u}=M{\cal R}_{\bm{n}}(\sigma)M^{-1}\bm{w},$

so that $M{\cal R}_{\bm{n}}(\sigma)M^{-1}$ is the rotation matrix on
${{\mathbb{Q}}}_{p}^{3}$ around $\bm{n}$ with respect to
$(\bm{e}_{1},\bm{e}_{2},\,\bm{e}_{3})$.

######  Remark 20.

Rotations around the reference axes of ${{\mathbb{Q}}}_{p}^{3}$ are particular
cases of Eq. (20), in which we choose the canonical basis as
$(\bm{g},\bm{h},\bm{n})$ in some order. When $p\neq 2$, if
$(\bm{g},\bm{h},\bm{n})$ equals

  * •

$(\bm{e}_{1},\bm{e}_{2},\bm{e}_{3})$, then $g=1,\ h=-v\Rightarrow\alpha=-v$,
which determine a rotation around the third reference axis, ${\cal
R}_{z}(\sigma)$, characterized by

$\begin{pmatrix}\frac{1+v\sigma^{2}}{1-v\sigma^{2}}&\frac{2v\sigma}{1-v\sigma^{2}}\\\
\frac{2\sigma}{1-v\sigma^{2}}&\frac{1+v\sigma^{2}}{1-v\sigma^{2}}\end{pmatrix};$

  * •

$(\bm{e}_{1},\bm{e}_{3},\bm{e}_{2})$, then $g=1,\ h=p\Rightarrow\alpha=p$,
which determine a rotation around the second reference axis of the canonical
basis, ${\cal R}_{y}(\sigma)$, characterized by

$\begin{pmatrix}\frac{1-p\sigma^{2}}{1+p\sigma^{2}}&-\frac{2p\sigma}{1+p\sigma^{2}}\\\
\frac{2\sigma}{1+p\sigma^{2}}&\frac{1-p\sigma^{2}}{1+p\sigma^{2}}\end{pmatrix};$

  * •

$(\bm{e}_{2},\bm{e}_{3},\bm{e}_{1})$, then $g=-v,\
h=p\Rightarrow\alpha=-\frac{p}{v}$, which determine a rotation around the
first reference axis, ${\cal R}_{x}(\sigma)$, characterized by

$\begin{pmatrix}\frac{1+\frac{p}{v}\sigma^{2}}{1-\frac{p}{v}\sigma^{2}}&\frac{2\frac{p}{v}\sigma}{1-\frac{p}{v}\sigma^{2}}\\\
\frac{2\sigma}{1-\frac{p}{v}\sigma^{2}}&\frac{1+\frac{p}{v}\sigma^{2}}{1-\frac{p}{v}\sigma^{2}}\end{pmatrix}.$

These matrices are associated respectively to the conservation of the rank-2
quadratic forms $Q_{-v},\,Q_{p}$ and $-vy^{2}+pz^{2}$. The latter is $Q_{up}$
if $p\equiv 1\mod 4$, but it is again $Q_{p}$ for $p\equiv 3\mod 4$, in which
case ${\cal R}_{x}$ and ${\cal R}_{y}$ are characterized by the same $2\times
2$ submatrix, only differently positioned inside the $3\times 3$ matrix on the
canonical basis.

When $p=2$ we have $Q_{+}(\bm{e}_{i})=1,\ i=1,2,3$, which gives $\alpha=1$
whenever $(\bm{g},\bm{h},\bm{n})$ equals any permutation of the canonical
basis. Then, a $2$-adic rotation around any reference axis of
${{\mathbb{Q}}}_{2}^{3}$ is characterized by

$\begin{pmatrix}\frac{1-\sigma^{2}}{1+\sigma^{2}}&-\frac{2\sigma}{1+\sigma^{2}}\\\
\frac{2\sigma}{1+\sigma^{2}}&\frac{1-\sigma^{2}}{1+\sigma^{2}}\end{pmatrix}$

associated to the conservation of the rank-2 quadratic form $Q_{1}$.

Eq. (12) translates in three dimensions into

${\cal R}_{\bm{n}}\left(-\frac{1}{\alpha\sigma}\right)={\cal
R}_{\bm{n}}\left(\infty\right){\cal R}_{\bm{n}}\left(\sigma\right).$ (21)

The important question we have to face now is, which of the three (seven)
equivalence classes of definite quadratic forms in dimension two for odd $p$
($p=2$), according to Proposition 10, arise from restricting $Q_{+}$ to two-
dimensional subspaces $V=\bm{n}^{\perp}$ of ${{\mathbb{Q}}}_{p}^{3}$. It will
turn out that all of them are realized. Since the transformations ${\cal
R}_{\bm{n}}\in SO(3)_{p}$ are rotations around an axis
$\bm{n}\in{{\mathbb{Q}}}_{p}^{3}$ preserving the rank-$3$ quadratic form
$Q_{+}$, this is equivalent to asking if the restrictions $({\cal
R}_{\bm{n}})_{|\bm{n}^{\perp}}$, by varying the classes of $\bm{n}$, cover all
the classes of special orthogonal symmetries of the $p$-adic plane preserving
the definite rank-$2$ forms (recall that Theorem 6 identifies all elements of
$SO(3)_{p}$ as such rotations around axes).

Consider a basis change between orthogonal bases of ${{\mathbb{Q}}}_{p}^{3}$
of the kind
$(\bm{g},\bm{h},\bm{n})\mapsto(\lambda\bm{g},\mu\bm{h},\nu\bm{n})$, with
$\lambda,\mu,\nu\in{{\mathbb{Q}}}_{p}^{\ast}$. The parameter $\alpha$ in the
parameterization (20) is affected by this transformation as

$\alpha=\frac{Q_{+}(\bm{h})}{Q_{+}(\bm{g})}\mapsto\frac{Q_{+}(\mu\bm{h})}{Q_{+}(\lambda\bm{g})}=\left(\frac{\mu}{\lambda}\right)^{2}\frac{Q_{+}(\bm{h})}{Q_{+}(\bm{g})}.$

Hence, $\alpha$ is well-defined in
$K={{\mathbb{Q}}}_{p}^{\ast}/({{\mathbb{Q}}}_{p}^{\ast})^{2}$; there are at
most $4$ classes of rotations in $SO(3)_{p}$ for $p\neq 2$ and $8$ classes for
$p=2$.

In Corollary 18 we found a bijective relation modulo squares between the value
of the form $Q_{+}$ on a rotation axis $\bm{n}$ and $\alpha$

$\alpha\simeq\begin{cases}-vpQ_{+}(\bm{n})&\text{ for }p\text{ odd},\\\
Q_{+}(\bm{n})&\text{ for }p=2.\end{cases}$ (22)

This allows us to determine the class of a rotation realized on a plane via
the class $Q_{+}(\bm{n})$ of its orthogonal axis.

As a consequence of the fact that $Q_{+}^{(4)}$ does not represent $0$,
Corollary 18 showed that $\alpha\not\simeq-1$. Hence there are at most $3$ (or
$7$) classes of rotations for odd $p$ (for $p=2$, respectively), depending on
the equivalence class modulo squares of $\alpha$:

$\displaystyle\alpha\in\\{-v,p,up\\},\ p>2,$ $\displaystyle\alpha\in\\{1,\pm
2,\pm 5,\pm 10\\},\ p=2.$

Focusing on the language of quadratic forms, there are left at most $3$ (or
$7$) equivalence classes up to scaling of restrictions of $Q_{+}$ to a plane,
identified by the equivalence class modulo squares of $\alpha$.

We want to see if there exist classes of $\bm{n}$ in ${{\mathbb{Q}}}_{p}^{3}$
realizing all these classes of $\alpha$, uniquely associated to the definite
rank-$2$ forms through their determinant. We make use of Eq. (22).

When $p$ is odd, $Q_{+}(\bm{n})\simeq-vp\alpha\Leftrightarrow
n_{1}^{2}-vn_{2}^{2}+pn_{3}^{2}+vp\alpha s^{2}=0$ for some
$s\in{{\mathbb{Q}}}_{p}^{\ast}$, $\bm{n}$ of canonical coordinates
$(n_{1},n_{2},n_{3})$. The quadratic form
$n_{1}^{2}-vn_{2}^{2}+pn_{3}^{2}+vp\alpha s^{2}$ represents $0$ in a
nontrivial way for every $\alpha\in\\{-v,p,up\\}$, for its determinant is
$d\simeq-\alpha\not\simeq 1$ (Theorem 2). If $n_{1}=n_{2}=n_{3}=0$, then
$vp\alpha s^{2}=0,\,s\in{{\mathbb{Q}}}_{p}^{\ast}$ is impossible. It means
that $Q_{+}(\bm{n})\simeq-vp\alpha$ admits nontrivial solutions $\bm{n}$ for
every $\alpha$.

Similarly, when $p=2$, $Q_{+}(\bm{n})\simeq\alpha\Leftrightarrow
n_{1}^{2}+n_{2}^{2}+n_{3}^{2}-\alpha s^{2}=0$ for some
$s\in{{\mathbb{Q}}}_{p}^{\ast}$. The form $x^{2}+y^{2}+z^{2}-\alpha s^{2}$
represents $0$ for every $\alpha\in\\{1,\pm 2,\pm 5,\pm 10\\}$, because its
determinant is $d\simeq-\alpha\not\simeq 1$. In summary, we have proved the
following.

######  Proposition 21.

For every prime $p$ and every two-dimensional definite quadratic form
$Q_{\kappa}$, there exists a rotation axis determined by
$\bm{n}\in{{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$ such that
$Q_{+|\bm{n}^{\perp}}\sim Q_{\kappa}$.  $\sqcap$$\sqcup$

In fact, when $p$ is odd,

  * •

$Q_{+}(\bm{e}_{3})=p\ \Leftrightarrow\ \alpha\simeq-v$, as it can be seen from
Remark 20 too. This is associated to the conservation of $Q_{-v}$;

  * •

$Q_{+}(\bm{e}_{2})=-v\ \Leftrightarrow\ \alpha\simeq p$, associated to
$Q_{p}$;

  * •

$Q_{+}(\bm{e}_{1})=1\ \Leftrightarrow\ \alpha\simeq-\frac{p}{v}$. This is
associated to $Q_{up}$ for $p\equiv 1\mod 4$ and $Q_{p}$ for $p\equiv 3\mod
4$.

On the other hand, when $p=2$, then $Q_{+}(\bm{e}_{i})=1$ for all $i=1,2,3$,
thus $\alpha=1$, associated to the conservation of $Q_{1}$.

This means that when $p\equiv 1\mod 4$ the canonical basis is enough to
realize all the classes of rank-$2$ forms; however it leaves out the class
associated to $\alpha\simeq up$ for $p\equiv 3\mod 4$, and it realizes only
one class ($\alpha\simeq 1$) out of the seven ones for $p=2$. These remaining
classes are covered by non-reference axes. Indeed, $Q_{+}$ maintains some
symmetry on $x,y$ for $p\equiv 3\mod 4$, contrary to $p\equiv 1\mod 4$, and
the symmetry is total for $p=2$.

The axes needed to realize all the classes of definite rank-$2$ forms when
$p\equiv 3\mod 4$ do not form an orthogonal basis. Moreover, in the case of
$p=2$, we need seven axes, so we cannot have a basis of
${{\mathbb{Q}}}_{2}^{3}$ to realize all classes of rank-$2$ forms.

Furthermore, the map $\bm{n}\mapsto Q_{+|\bm{n}^{\perp}}$ from vectors
$\bm{n}$ in ${{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$ to definite quadratic
forms in dimension two induces bijective correspondence between the
equivalence classes of $Q_{+}(\bm{n})$ in $K$ and the rank-two definite forms
$Q_{\kappa}$ up to equivalence. In fact, if
$\bm{n},\bm{n}^{\prime}\in{{\mathbb{Q}}}_{p}^{3}\setminus\bm{0}$ are such that
$Q_{+}(\bm{n})\not\simeq Q_{+}(\bm{n}^{\prime}),$ then
$Q_{+|\bm{n}^{\perp}}\not\simeq Q_{+|{\bm{n}^{\prime}}^{\perp}}$. This is
because, as in the proof of Proposition 17, the determinant is an invariant of
quadratic forms under basis changes, and $\det Q_{+}\simeq
Q_{+}(\bm{g})Q_{+}(\bm{h})Q_{+}(\bm{n})\simeq
Q_{+}(\bm{g}^{\prime})Q_{+}(\bm{h}^{\prime})Q_{+}(\bm{n}^{\prime})$,
respectively on the orthogonal bases $(\bm{g},\bm{h},\bm{n})$ and
$(\bm{g}^{\prime},\bm{h}^{\prime},\bm{n}^{\prime})$. As a consequence, if
$Q_{+}(\bm{n})\not\simeq Q_{+}(\bm{n}^{\prime})$, then
$Q_{+}(\bm{g})Q_{+}(\bm{h})\not\simeq
Q_{+}(\bm{g}^{\prime})Q_{+}(\bm{h}^{\prime})$, where
$Q_{+}(\bm{g})Q_{+}(\bm{h})\simeq\det Q_{+|\bm{n}^{\perp}}$,
$Q_{+}(\bm{g}^{\prime})Q_{+}(\bm{h}^{\prime})\simeq\det
Q_{+|{\bm{n}^{\prime}}^{\perp}}$. Two quadratic forms $Q_{+|\bm{n}^{\perp}}$
and $Q_{+|{\bm{n}^{\prime}}^{\perp}}$ with different determinants modulo
squares cannot be equivalent (Theorem 1).

It is possible to choose representatives in the three (resp. seven) classes of
definite rank-$2$ forms for odd $p$ (reps. $p=2$) in such a way that if
$\bm{n},\bm{n}^{\prime}\in{{\mathbb{Q}}}_{p}^{3}$ satisfy $Q_{+}(\bm{n})\simeq
Q_{+}(\bm{n}^{\prime})$, then $Q_{+|\bm{n}^{\perp}}\simeq
Q_{+|{\bm{n}^{\prime}}^{\perp}}$ by a linear transformation (scaling is not
involved). This can be shown in an abstract way, without resorting to the
explicit classification of $p$-adic quadratic forms. The statement is trivial
if $\bm{n}=\lambda\bm{n}^{\prime}$ for some
$\lambda\in{{\mathbb{Q}}}_{p}^{\ast}$. Otherwise, consider a vector
$\bm{v}\in{{\mathbb{Q}}}_{p}^{3}$ orthogonal to $\bm{n},\bm{n}^{\prime}$. By
Proposition 8, there exists some ${\cal R}_{\bm{v}}(\sigma)\in SO(3)_{p}$ such
that ${\cal R}_{\bm{v}}(\sigma)\bm{n}=\lambda\bm{n}^{\prime}$ with
$\lambda\in{{\mathbb{Q}}}_{p}^{\ast}$. Then, ${\cal R}_{\bm{v}}(\sigma)$
transforms the plane $\bm{n}^{\perp}$ to the plane
${\bm{n}^{\prime}}^{\perp}$. This is a linear map, which necessarily
implements the equivalence $Q_{+|\bm{n}^{\perp}}\simeq
Q_{+|{\bm{n}^{\prime}}^{\perp}}$

##  6 $p$-adic Cardano and Euler decompositions

The previous development of $SO(3)_{p}$ showed a sufficiently close analogy to
the real Euclidean $SO(3)_{{{\mathbb{R}}}}$, in that the three-dimensional
space has an essentially unique definite quadratic form, and all special
orthogonal transformations are actually rotations. The rotation groups around
fixed axes themselves turned out to be rather more complex: there are three
(for odd $p$) and seven (for $p=2$) types of rotation axes with associated
different rotation groups.

Now we ask if it is possible to express any element of $SO(3)_{p}$ as a
composition of rotations around the reference axes of
${{\mathbb{Q}}}_{p}^{3}$, as it happens in ${{\mathbb{R}}}^{3}$ according to
Theorem 4. In the following we will revert to calling the axes $x$, $y$ and
$z$, to make them typographically more distinguishable.

######  Theorem 22 ($p$-adic Cardano decomposition).

For every odd prime $p$, any $M\in SO(3)_{p}$ can be decomposed into

$M={\cal R}_{z}(\zeta){\cal R}_{y}(\eta){\cal R}_{x}(\xi),$

for some parameters $\xi,\eta,\zeta\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$.

###### Proof.

Let $M\bm{e}_{1}$ be a vector with canonical coordinates
$(m_{1},m_{2},m_{3})$, such that $Q_{+}(M\bm{e}_{1})=Q_{+}(\bm{e}_{1})=1$
since $M\in SO(3)_{p}$ preserves $Q_{+}$.

We need to find a composition of rotations around $\bm{e}_{2}$ and
$\bm{e}_{3}$ transforming $M\bm{e}_{1}$ back to $\bm{e}_{1}$.

First, we show that there exists a $z$-rotation ${\cal R}_{z}(\zeta)\in
SO(3)_{p}$ such that

${\cal R}_{z}(\zeta)^{-1}M\bm{e}_{1}\in\text{span}(\bm{e}_{1},\bm{e}_{3}).$
(23)

This vector would be ${\cal
R}_{z}(\zeta)^{-1}M\bm{e}_{1}=m_{1}^{\prime}\bm{e}_{1}+m_{3}\bm{e}_{3}$ for
some $m_{1}^{\prime}\in{{\mathbb{Q}}}_{p}$, where the third component of
$M\bm{e}_{1}$ is left unchanged by a rotation around $\bm{e}_{3}$. As noted in
Proposition 8, necessary and sufficient condition for the existence of such a
${\cal R}_{z}(\zeta)^{-1}$ is

$Q_{+}(m_{1}^{\prime}\bm{e}_{1}+m_{3}\bm{e}_{3})=Q_{+}(M\bm{e}_{1})\Leftrightarrow{m_{1}^{\prime}}^{2}+pm_{3}^{2}=1.$

$Q_{+}(M\bm{e}_{1})=m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=1$ implies that
$m_{1},m_{2},m_{3}\in{{\mathbb{Z}}}_{p}$ (cf. the compactness proof for
$SO(3)_{p}$ in Theorem 5). Therefore, we resort to Hensel’s Lemma to show that
$f(m_{1}^{\prime})={m_{1}^{\prime}}^{2}-1+pm_{3}^{2}$ admits roots
${m_{1}^{\prime}}\in{{\mathbb{Z}}}_{p}$.
$f(m_{1}^{\prime})\equiv{m_{1}^{\prime}}^{2}-1\mod p$ has zeros
$m_{1}^{\prime}\equiv\pm 1\mod p$, in which the derivative
$f^{\prime}(m_{1}^{\prime})=2m_{1}^{\prime}\not\equiv 0\mod p$. Then, Hensel’s
Lemma allows us to (uniquely) lift each of these solutions to a solution of
the same equation $\mod p^{k}$, converging to a $p$-adic solution
$m_{1}^{\prime}$. It means that ${\cal R}_{z}(\zeta)^{-1}$ as in Eq. (23)
exists.

Next, there exists a $y$-rotation ${\cal R}_{y}(\eta)^{-1}\in SO(3)_{p}$ such
that

${\cal R}_{y}(\eta)^{-1}{\cal R}_{z}(\zeta)^{-1}M\bm{e}_{1}=\bm{e}_{1},$ (24)

because $Q_{+}\big{(}{\cal
R}_{z}(\zeta)^{-1}M\bm{e}_{1}\big{)}=Q_{+}(\bm{e}_{1})$.

Eq. (24) means that ${\cal R}_{y}(\eta)^{-1}{\cal R}_{z}(\zeta)^{-1}M\in
SO(3)_{p}$ has eigenvector $\bm{e}_{1}$ with corresponding eigenvalue $1$,
i.e., ${\cal R}_{y}(\eta)^{-1}{\cal
R}_{z}(\zeta)^{-1}M\mathrel{\mathop{\mathchar 58\relax}}={\cal R}_{x}(\xi)$ is
a rotation around the $x$-axis.  $\sqcap$$\sqcup$

######  Corollary 23.

For every odd prime $p$, any $M\in SO(3)_{p}$ can be decomposed into

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

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

###### Proof.

To prove the existence of a decomposition of the kind ${\cal R}_{z}{\cal
R}_{x}{\cal R}_{y}$, it is enough to repeat the steps of the proof of Theorem
22. In particular, given
$M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, there exists
${\cal R}_{z}(\sigma)\in SO(3)_{p}$ such that

${\cal R}_{z}(\sigma)^{-1}M\bm{e}_{2}\in\text{span}(\bm{e}_{2},\bm{e}_{3}).$

In fact, there exists $m_{2}^{\prime}\in{{\mathbb{Z}}}_{p}$ such that ${\cal
R}_{z}(\sigma)^{-1}M\bm{e}_{2}=m_{2}^{\prime}\bm{e}_{2}+m_{3}\bm{e}_{3}$ and

$Q_{+}(m_{2}^{\prime}\bm{e}_{2}+m_{3}\bm{e}_{3})=-v{m_{2}^{\prime}}^{2}+pm_{3}^{2}=-v=Q_{+}(\bm{e}_{2}).$

To show this, we apply Hensel’s Lemma to
$f(m_{2}^{\prime})=v{m_{2}^{\prime}}^{2}-pm_{3}^{2}-v$:
$f(m_{2}^{\prime})\equiv 0\mod p\Leftrightarrow v{m_{2}^{\prime}}^{2}\equiv
v\Leftrightarrow m_{2}^{\prime}\equiv\pm 1\mod p$, and
$f^{\prime}(m_{2}^{\prime})=2m_{2}^{\prime}\not\equiv 0\mod p$ in these
solutions, which then are lifted to $p$-adic solutions.

The existence of the decomposition with respect to the $x$, $y$ and $z$ axes
follows straightforwardly from Theorem 22: if any $M^{-1}\in SO(3)_{p}$ can be
written as $M^{-1}={\cal R}_{z}(\zeta){\cal R}_{y}(\eta){\cal R}_{x}(\xi)$ for
certain parameters $\xi,\eta,\zeta$, then any $M\in SO(3)_{p}$ can be
decomposed into $M={\cal R}_{x}(\xi)^{-1}{\cal R}_{y}(\eta)^{-1}{\cal
R}_{z}(\zeta)^{-1}\mathrel{\mathop{\mathchar 58\relax}}={\cal
R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal R}_{z}(\zeta^{\prime})$,
where $\xi^{\prime}=-\xi$ if $\xi\in{{\mathbb{Q}}}_{p}$ or $\xi=\xi^{\prime}$
are infinite, and similarly for $\eta^{\prime},\zeta^{\prime}$.

One similarly proves the existence of the nautical decomposition ${\cal
R}_{y}{\cal R}_{x}{\cal R}_{z}$ for $SO(3)_{p}$ from the one as ${\cal
R}_{z}{\cal R}_{x}{\cal R}_{y}$.  $\sqcap$$\sqcup$

######  Theorem 24 ($p$-adic Cardano decomposition, $p\equiv 1$ mod 4).

For every odd prime $p\equiv 1\mod 4$, any $M\in SO(3)_{p}$ can be decomposed
into

$M={\cal R}_{x}(\zeta){\cal R}_{z}(\eta){\cal R}_{y}(\xi),$

for some parameters $\xi,\eta,\zeta\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$.

By applying this to $M^{-1}$, we get a decomposition $M={\cal
R}_{y}(\zeta){\cal R}_{z}(\eta){\cal R}_{x}(\xi)$.

###### Proof.

The strategy is similar to the previous Theorem 22. We look at
$M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, which has
$m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=Q_{+}(M\bm{e}_{2})=Q_{+}(\bm{e}_{2})=-v$. We
first look for a $\zeta$ with ${\cal
R}_{x}(\zeta)^{-1}M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{2}^{\prime}\bm{e}_{2}$, which
is guaranteed if $-v{m_{2}^{\prime}}^{2}=-vm_{2}^{2}+pm_{3}^{2}$, because in
this way
$Q_{+}(m_{1}\bm{e}_{1}+m_{2}^{\prime}\bm{e}_{2})=Q_{+}(M\bm{e}_{2})=Q_{+}(\bm{e}_{2})$,
and we can invoke Proposition 8. We prove that we can find $m_{2}^{\prime}$
with ${m_{2}^{\prime}}^{2}=m_{2}^{2}-\frac{p}{v}m_{3}^{2}$. Indeed, note first
that $m_{2}\not\equiv 0\mod p$, since otherwise $-v=Q_{+}(M\bm{e}_{2})\equiv
m_{1}^{2}\mod p$, which is impossible since $-v$ is a non-square in
${{\mathbb{Z}}}_{p}$ and hence modulo $p$ (it is here that we use $p\equiv
1\mod 4$, since then $-1$ is a square in ${{\mathbb{Q}}}_{p}$, thus $v$ and
$-v$ are both non-squares). Thus, we have solutions $m_{2}^{\prime}\equiv\pm
m_{2}\mod p$, which are both nonzero. Since the derivative $2m_{2}^{\prime}$
is then nonzero, as well, we can invoke Hensel’s Lemma to obtain a solution of
the equation in ${{\mathbb{Z}}}_{p}$.

Now we proceed as before: since now ${\cal
R}_{x}(\zeta)^{-1}M\bm{e}_{2}\perp\bm{e}_{3}$, we can find $\eta$ such that
${\cal R}_{z}(\eta)^{-1}{\cal R}_{x}(\zeta)^{-1}M\bm{e}_{2}=\bm{e}_{2}$ (once
more invoking Proposition 8). But this means that ${\cal
R}_{z}(\eta)^{-1}{\cal R}_{x}(\zeta)^{-1}M$ is a rotation around the $y$-axis,
i.e., ${\cal R}_{z}(\eta)^{-1}{\cal R}_{x}(\zeta)^{-1}M={\cal R}_{y}(\xi)$ for
some $\xi$, and we are done.  $\sqcap$$\sqcup$

######  Remark 25.

Suppose a general Cardano or Euler type decomposition

$SO(3)_{p}\ni M={\cal R}_{\bm{n}_{1}}(\sigma){\cal R}_{\bm{n}_{2}}(\tau){\cal
R}_{\bm{n}_{3}}(\omega),$

where $\bm{n}_{1}\perp\bm{n}_{2}$ and $\bm{n}_{2}\perp\bm{n}_{3}$ (in the
Cardano decomposition, $\bm{n}_{1}\perp\bm{n}_{3}$, while in the Euler one
$\bm{n}_{1}=\bm{n}_{3}$). We have

$\begin{split}M={\cal R}_{\bm{n}_{1}}(\sigma){\cal R}_{\bm{n}_{2}}(\tau){\cal
R}_{\bm{n}_{3}}(\omega)&\Leftrightarrow{\cal R}_{\bm{n}_{2}}(\tau)^{-1}{\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M={\cal R}_{\bm{n}_{3}}(\omega)\\\
&\Leftrightarrow{\cal R}_{\bm{n}_{2}}(\tau)^{-1}{\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}=\bm{n}_{3}.\end{split}$

Now, ${\cal R}_{\bm{n}_{2}}(\tau)^{-1}$ preserves the component of the vector
${\cal R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}$ along $\bm{n}_{2}$, which must
be $0$ to get $\bm{n}_{3}$ as a result, therefore ${\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}\perp\bm{n}_{2}$.

This shows that ${\cal R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}\perp\bm{n}_{2}$
is a necessary condition for the existence of a decomposition $M={\cal
R}_{\bm{n}_{1}}(\sigma){\cal R}_{\bm{n}_{2}}(\tau){\cal
R}_{\bm{n}_{3}}(\omega)$. Since orthogonal transformations preserve $Q_{+}$,
we also have $Q_{+}\big{(}{\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}\big{)}=Q_{+}(\bm{n}_{3})$.

Conversely, these conditions are also sufficient. To be precise, assume that
there exists a vector $\bm{v}\perp\bm{n}_{2}$ with
$\bm{n}_{1}^{\top}\bm{v}=\bm{n}_{1}^{\top}M\bm{n}_{3}$ and
$Q_{+}(\bm{v})=Q_{+}(M\bm{n}_{3})=Q_{+}(\bm{n}_{3})$. Then, by Proposition 8
there exists $\sigma$ such that ${\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}=\bm{v}\perp\bm{n}_{2}$. Note that
after the choice of $\sigma$, it follows that $\tau$ and $\omega$ are
determined uniquely.

######  Remark 26.

The Cardano decompositions for $SO(3)_{p}$ of the kind ${\cal R}_{x}{\cal
R}_{z}{\cal R}_{y}$ and ${\cal R}_{y}{\cal R}_{z}{\cal R}_{x}$ do not exist in
general for odd primes $p\equiv 3\mod 4$. In fact, one can construct
counterexamples of matrices $M\in SO(3)_{p}$ for which there does not exist
${\cal R}_{\bm{n}_{1}}(\sigma)\in SO(3)_{p}$ such that ${\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}\perp\bm{n}_{2}$ according to Remark
25, by exploiting $Q_{+}\big{(}{\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}\big{)}=Q_{+}(\bm{n}_{3})$.

As an example, let us show the existence of special orthogonal transformations
that cannot be written in the Cardano product form ${\cal R}_{x}{\cal
R}_{z}{\cal R}_{y}$ (the form ${\cal R}_{y}{\cal R}_{z}{\cal R}_{x}$ follows a
closely similar reasoning). According to the recipe, we consider
$M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, which has
$m_{1}^{2}+m_{2}^{2}+pm_{3}^{2}=Q_{+}(M\bm{e}_{2})=Q_{+}(\bm{e}_{2})=1$. To
get a corresponding Cardano decomposition, we would need $\sigma$ with ${\cal
R}_{x}(\sigma)^{-1}M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{2}^{\prime}\bm{e}_{2}$ such
that ${m_{2}^{\prime}}^{2}=m_{2}^{2}+pm_{3}^{2}$, but this is impossible if
$m_{2}=0$, since $p$ is not a square in ${{\mathbb{Q}}}_{p}$. It remains to
show that this case can occur, namely there exists an $M\in SO(3)_{p}$ with
$M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{3}\bm{e}_{3}$. By Proposition 9, it is
sufficient to find solutions of $m_{1}^{2}+pm_{3}^{2}=1$, which indeed has two
roots for $m_{1}$, as we can see by first solving it modulo $p$, resulting in
$m_{1}\equiv\pm 1\mod p$, and then using Hensel’s Lemma.

######  Remark 27.

For odd primes $p$, none of the six possible Euler decompositions for
$SO(3)_{p}$ exist in general. We indicate how to construct counterexamples,
following the same strategy as in the previous remark, based on Remark 25.

(i & ii) Nonexistence of XYX and YXY: Let us focus on the decomposition form
${\cal R}_{x}{\cal R}_{y}{\cal R}_{x}$ (the form ${\cal R}_{y}{\cal
R}_{x}{\cal R}_{y}$ follows a closely similar reasoning). According to the
recipe, we consider
$M\bm{e}_{1}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, which has
$m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=Q_{+}(M\bm{e}_{1})=Q_{+}(\bm{e}_{1})=1$. To
get a corresponding Euler decomposition, we would need $\xi$ with ${\cal
R}_{x}(\xi)^{-1}M\bm{e}_{1}=m_{1}\bm{e}_{1}+m_{3}^{\prime}\bm{e}_{3}$ such
that $p{m_{3}^{\prime}}^{2}=-vm_{2}^{2}+pm_{3}^{2}$, but this is impossible if
$m_{2}\not\equiv 0\mod p$. It remains to show that this case can occur, namely
that there exists an $M\in SO(3)_{p}$ with $m_{2}\not\equiv 0\mod p$. By
Hensel’s Lemma, this boils down to finding solutions to
$m_{1}^{2}-vm_{2}^{2}\equiv 1\mod p$ in integers modulo $p$, such that
$m_{2}\not\equiv 0\mod p$, which always exist. (Indeed, for $p\equiv 3\mod 4$,
when $-v=1$, we can choose $m_{2}=1$.)

(iii) Nonexistence of XZX: According to the recipe, we consider
$M\bm{e}_{1}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, which has
$m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=Q_{+}(M\bm{e}_{1})=Q_{+}(\bm{e}_{1})=1$. To
get a corresponding Euler decomposition, we would need $\xi$ with ${\cal
R}_{x}(\xi)^{-1}M\bm{e}_{1}=m_{1}\bm{e}_{1}+m_{2}^{\prime}\bm{e}_{2}$, such
that $-v{m_{2}^{\prime}}^{2}=-vm_{2}^{2}+pm_{3}^{2}$, but this is impossible
if $m_{2}=0$. This case can occur: there exists an $M\in SO(3)_{p}$ with
$m_{2}=0$, because $m_{1}^{2}+pm_{3}^{2}=1$ always admits solutions according
to Hensel’s Lemma.

(iv) Nonexistence of YZY: Similarly, consider
$M\bm{e}_{2}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, which has
$m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=Q_{+}(M\bm{e}_{2})=Q_{+}(\bm{e}_{2})=-v$. To
get a corresponding Euler decomposition, we would need $\eta$ with ${\cal
R}_{y}(\eta)^{-1}M\bm{e}_{2}=m_{1}^{\prime}\bm{e}_{1}+m_{2}\bm{e}_{2}$, such
that ${m_{1}^{\prime}}^{2}=m_{1}^{2}+pm_{3}^{2}$, but this is impossible if
$m_{1}=0$. However, there exists an $M\in SO(3)_{p}$ with $m_{1}=0$, because
$-vm_{2}^{2}+pm_{3}^{2}=-v$ always admits solutions, as before using Hensel’s
Lemma.

(v & vi) Nonexistence of ZXZ and ZYZ: Let us focus on ${\cal R}_{z}{\cal
R}_{y}{\cal R}_{z}$ (the form ${\cal R}_{z}{\cal R}_{x}{\cal R}_{z}$ follows a
closely similar reasoning), and consider
$M\bm{e}_{3}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$, which has
$m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=Q_{+}(M\bm{e}_{3})=Q_{+}(\bm{e}_{3})=p$. To
get a corresponding Euler decomposition, we would need $\zeta$ with ${\cal
R}_{z}(\zeta)^{-1}M\bm{e}_{3}=m_{1}^{\prime}\bm{e}_{1}+m_{3}\bm{e}_{3}$, such
that ${m_{1}^{\prime}}^{2}=m_{1}^{2}-vm_{2}^{2}$. This ends in a contradiction
if the right hand side happens to be a non-square in ${{\mathbb{Q}}}_{p}$.
Indeed, we will show that $m_{1}^{2}-vm_{2}^{2}=vp^{2}$ can occur. To see
this, consider first the condition $m_{1}^{2}-vm_{2}^{2}+pm_{3}^{2}=p$, from
which we get $m_{1}^{2}-vm_{2}^{2}\equiv 0\mod p$, and since $v$ is a non-
square modulo $p$, this means that $m_{1}\equiv m_{2}\equiv 0\mod p$, i.e., we
can write $m_{1}=px$, $m_{2}=py$ with $x,y\in{{\mathbb{Z}}}_{p}$, and our
condition after cancellation of $p$ factors becomes
$px^{2}-vpy^{2}+m_{3}^{2}=1$. First solving modulo $p$ and then using Hensel’s
Lemma shows that this always has solutions for $m_{3}$, regardless of the
integers $x$ and $y$: $m_{3}=\pm\sqrt{1-px^{2}+vpy^{2}}\in{{\mathbb{Q}}}_{p}$.
This means we can choose $m_{1}$ and $m_{2}$ freely as multiples of $p$, and
always satisfy the quadratic form constraint. On the other hand, we claim that
there are solutions of $m_{1}^{2}-vm_{2}^{2}=vp^{2}$, which upon substituting
$x$ and $y$ becomes $x^{2}-vy^{2}=v$. Indeed, the quadratic form on the left
hand side does not represent zero nontrivially, but $x^{2}-vy^{2}-vz^{2}$
does. So, we can take any solution of $x^{2}-vy^{2}-vz^{2}=0$, which
necessarily must have $z\neq 0$. By homogeneity, we can assume $z=1$, which
gives the desired solution.

######  Remark 28.

When $p=2$, none of the Cardano and Euler decompositions exist for all of
$SO(3)_{2}$. The counterexample is a single matrix

$M=\begin{pmatrix}-2&-2&\sqrt{-7}\\\
\frac{1}{2}(\sqrt{-7}+1)&\frac{1}{2}(\sqrt{-7}-1)&2\\\
\frac{1}{2}(\sqrt{-7}-1)&\frac{1}{2}(\sqrt{-7}+1)&2\end{pmatrix}.$

Note that $-7$ is a square in ${{\mathbb{Q}}}_{2}$, and indeed
$\sqrt{-7}=1+2^{2}+2^{4}+2^{5}+2^{7}+\ldots$, so that
$\frac{1}{2}(\sqrt{-7}+1)=1+2+2^{3}+2^{4}+2^{6}+\ldots$ is odd, and
$\frac{1}{2}(\sqrt{-7}-1)=2+2^{3}+2^{4}+2^{6}+\ldots$ is even in
${{\mathbb{Z}}}_{2}$.

As a matter of fact, $M$ does not have a decomposition of any of the six
Cardano and six Euler forms shown in Theorem 4. Let us show that it cannot be
written as a product ${\cal R}_{x}{\cal R}_{y}{\cal R}_{z}$. Namely, by Remark
25 it is enough to show that there does not exist a $\xi$ such that ${\cal
R}_{x}(\xi)^{-1}M\bm{e}_{3}\perp\bm{e}_{2}$. With
$M\bm{e}_{3}=m_{1}\bm{e}_{1}+m_{2}\bm{e}_{2}+m_{3}\bm{e}_{3}$ such that
$m_{1}^{2}+m_{2}^{2}+m_{3}^{2}=1$, we would need a vector
$m_{1}\bm{e}_{1}+m_{3}^{\prime}\bm{e}_{3}$ with quadratic form evaluating to
$1$, i.e., ${m_{3}^{\prime}}^{2}=m_{2}^{2}+m_{3}^{2}=8$, which however is not
a square in ${{\mathbb{Q}}}_{2}$; contradiction.

Likewise, it cannot be decomposed in the form ${\cal R}_{y}{\cal R}_{x}{\cal
R}_{z}$, because then we would need ${\cal
R}_{y}(\eta)^{-1}M\bm{e}_{3}\perp\bm{e}_{1}$, which amounts to a vector
$m_{2}\bm{e}_{2}+m_{3}^{\prime}\bm{e}_{3}$ with quadratic form evaluating to
$1$, i.e., ${m_{3}^{\prime}}^{2}=m_{1}^{2}+m_{3}^{2}=-3$, which however is not
a square in ${{\mathbb{Q}}}_{2}$, as it is $\equiv 5\mod 8$; contradiction.

Any other decomposition results in a contradiction of the same type, because,
as can be checked, every column of $M$ has the property that, while the
squares of its elements sum to $1$, the sum of any two squares is not a square
in ${{\mathbb{Q}}}_{2}$. To prove this, it is enough to consider all occurring
cases and check that the sum in question is not a square modulo $8$ or $16$.

Although these results are in contrast to the Euclidean case, $SO(3)_{p}$ is
still generated by its subgroups $G_{x},\,G_{y},\,G_{z}$ of rotations around
the reference axes, at least for all odd primes $p$. Now we will find the
multiplicity of the above Cardano representations of $SO(3)_{p}$ for odd $p$.

######  Proposition 29.

If $M\in SO(3)_{p}$ and $M={\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)$, then there exists at least another distinct Cardano
representation of $M$ along the same axes

$M={\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}\left(\frac{1}{\alpha\eta}\right){\cal R}_{z}(\infty){\cal
R}_{z}(\zeta).$

###### Proof.

A systematic ambiguity of order $2$ in the products ${\cal R}_{x}(\xi){\cal
R}_{y}(\eta){\cal R}_{z}(\zeta)$ is based on the relation

${\cal R}_{x}(\infty){\cal R}_{y}(\infty){\cal R}_{z}(\infty)=I.$

By using it with Eq. (21) we get

$\displaystyle M$ $\displaystyle={\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)$ $\displaystyle={\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{x}(\infty){\cal R}_{y}(\infty){\cal R}_{z}(\infty){\cal R}_{z}(\zeta)$
$\displaystyle={\cal R}_{x}(\infty){\cal R}_{x}(\xi)\big{(}{\cal
R}_{y}(\eta)+{\cal R}_{x}(\infty)\left[{\cal R}_{y}(\eta),{\cal
R}_{x}(\infty)\right]\big{)}{\cal R}_{y}(\infty)\,{\cal R}_{z}(\infty){\cal
R}_{z}(\zeta)$ $\displaystyle={\cal R}_{x}(\infty){\cal
R}_{x}(\xi)\begin{pmatrix}-e(\eta)&0&f(\eta)\\\ 0&1&0\\\
g(\eta)&0&-e(\eta)\end{pmatrix}{\cal R}_{z}(\infty){\cal R}_{z}(\zeta),$

where the product of ${\cal R}_{x}(\infty)$ with the commutator of matrices
$\left[{\cal R}_{y}(\eta),{\cal R}_{x}(\infty)\right]$ provides the
transformation $\eta\mapsto-\eta$ (an infinite parameter remains the same) on
the parameter of the $y$-rotation (sign change of the off-diagonal elements),
which together with ${\cal R}_{y}(\infty)$ globally gives
$\eta\mapsto\frac{1}{\alpha\eta}$ (sign change of the diagonal entries).
$\sqcap$$\sqcup$

We give two elementary results in order to prove that the Cardano
representation of $SO(3)_{p}$ along the axes $x$, $y$ and $z$ is exactly
twofold for odd $p$.

######  Proposition 30.

For odd $p$, and all ${\cal R}_{x}(\xi)$, ${\cal R}_{x}(\xi^{\prime})$, ${\cal
R}_{y}(\eta)$, ${\cal R}_{y}(\eta^{\prime})\in SO(3)_{p}$,

${\cal R}_{x}(\xi){\cal R}_{y}(\eta)={\cal R}_{x}(\xi^{\prime}){\cal
R}_{y}(\eta^{\prime})\Leftrightarrow{\cal R}_{x}(\xi)={\cal
R}_{x}(\xi^{\prime}),\,{\cal R}_{y}(\eta)={\cal R}_{y}(\eta^{\prime}).$

###### Proof.

This is immediate to prove by equating two matrices of the kind

${\cal R}_{x}(\xi){\cal R}_{y}(\eta)=\begin{pmatrix}1&0&0\\\
0&a(\xi)&b(\xi)\\\
0&c(\xi)&a(\xi)\end{pmatrix}\begin{pmatrix}e(\eta)&0&f(\eta)\\\ 0&1&0\\\
g(\eta)&0&e(\eta)\end{pmatrix}=\begin{pmatrix}e(\eta)&0&f(\eta)\\\
b(\xi)g(\eta)&a(\xi)&b(\xi)e(\eta)\\\
a(\xi)g(\eta)&c(\xi)&a(\xi)e(\eta)\end{pmatrix}$

thanks to the fact that $e(\eta),\,a(\xi)\neq 0$ for every parameter.
$\sqcap$$\sqcup$

######  Corollary 31.

Let ${\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal R}_{\bm{n}}(\sigma)={\cal
R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal
R}_{\bm{n}}(\sigma^{\prime})$, $p>2$. If $\big{(}{\cal R}_{x}(\xi),\,{\cal
R}_{y}(\eta),\,{\cal R}_{\bm{n}}(\sigma)\big{)}\neq\big{(}{\cal
R}_{x}(\xi^{\prime}),\,{\cal R}_{y}(\eta^{\prime}),\,{\cal
R}_{\bm{n}}(\sigma^{\prime})\big{)}$ then ${\cal R}_{\bm{n}}(\sigma)\neq{\cal
R}_{\bm{n}}(\sigma^{\prime})$.

###### Proof.

We prove the contrapositive: if ${\cal R}_{\bm{n}}(\sigma)={\cal
R}_{\bm{n}}(\sigma^{\prime})$, then

${\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal R}_{\bm{n}}(\sigma)={\cal
R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal
R}_{\bm{n}}(\sigma^{\prime})\Rightarrow{\cal R}_{x}(\xi){\cal
R}_{y}(\eta)={\cal R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}).$

By Proposition 30, this is equivalent to $\big{(}{\cal R}_{x}(\xi),\,{\cal
R}_{y}(\eta),\,{\cal R}_{\bm{n}}(\sigma)\big{)}=\big{(}{\cal
R}_{x}(\xi^{\prime}),\,{\cal R}_{y}(\eta^{\prime}),\,{\cal
R}_{\bm{n}}(\sigma^{\prime})\big{)}$.  $\sqcap$$\sqcup$

######  Theorem 32.

Every $M\in SO(3)_{p}$, for odd prime $p$, has exactly two distinct Cardano
decompositions with respect to the $x$, $y$ and $z$ axes

$M={\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal R}_{z}(\zeta)$

for some $\xi,\eta,\zeta\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$, and

$M={\cal R}_{x}(\infty){\cal R}_{x}(\xi){\cal
R}_{y}\left(\frac{1}{\alpha\eta}\right){\cal R}_{z}(\infty){\cal
R}_{z}(\zeta).$

###### Proof.

We look for nontrivial solutions of ${\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)={\cal R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal
R}_{z}(\zeta^{\prime})$. Using the parameterization (20), we have

$\begin{split}{\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)&=\begin{pmatrix}1&0&0\\\ 0&a(\xi)&b(\xi)\\\
0&c(\xi)&a(\xi)\end{pmatrix}\begin{pmatrix}e(\eta)&0&f(\eta)\\\ 0&1&0\\\
g(\eta)&0&e(\eta)\end{pmatrix}\begin{pmatrix}l(\zeta)&m(\zeta)&0\\\
n(\zeta)&l(\zeta)&0\\\ 0&0&1\end{pmatrix}\\\
&=\begin{pmatrix}e(\eta)l(\zeta)&e(\eta)m(\zeta)&f(\eta)\\\
a(\xi)n(\zeta)+b(\xi)g(\eta)l(\zeta)&a(\xi)l(\zeta)+b(\xi)g(\eta)m(\zeta)&b(\xi)e(\eta)\\\
c(\xi)n(\zeta)+a(\xi)g(\eta)l(\zeta)&c(\xi)l(\zeta)+a(\xi)g(\eta)m(\zeta)&a(\xi)e(\eta)\end{pmatrix}.\end{split}$

We equate two matrices of this kind, whose entries we call $m_{ij}$ and
$m_{ij}^{\prime}$ respectively ($i,j=1,2,3$). Thus,

$m_{13}=m_{13}^{\prime}\Leftrightarrow f(\eta)=f(\eta^{\prime}),$

equivalent to
$-\frac{2p\eta}{1+p\eta^{2}}=-\frac{2p\eta^{\prime}}{1+p{\eta^{\prime}}^{2}}\Leftrightarrow
2p(\eta-\eta^{\prime})(1-p\eta\eta^{\prime})=0\Leftrightarrow\eta=\eta^{\prime}$
or $\eta^{\prime}=1/(p\eta)$. Furthermore,
$f(\eta)=f(\eta^{\prime})\Leftrightarrow g(\eta)=g(\eta^{\prime})$, since
$g(\eta)=\frac{2\eta}{1+p\eta^{2}}$.

If $\eta=\eta^{\prime}\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$, then ${\cal
R}_{y}(\eta)={\cal R}_{y}(\eta^{\prime})$, otherwise
$\eta^{\prime}=1/(p\eta)\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$ and

${\cal R}_{y}(\eta)=\begin{pmatrix}-e(\eta^{\prime})&0&f(\eta^{\prime})\\\
0&1&0\\\ g(\eta^{\prime})&0&-e(\eta^{\prime})\end{pmatrix}.$

As a consequence, from $m_{11}=m_{11}^{\prime}$, $m_{12}=m_{12}^{\prime}$,
$m_{23}=m_{23}^{\prime}$, $m_{33}=m_{33}^{\prime}$ we deduce $l(\zeta)=\pm
l(\zeta^{\prime})$, $m(\zeta)=\pm m(\zeta^{\prime})$, $a(\xi)=\pm
a(\xi^{\prime})$, $b(\xi)=\pm b(\xi^{\prime})$, where the $+$ sign is always
related to $\eta=\eta^{\prime}$ and the $-$ sign to $\eta\eta^{\prime}=1/p$.
This is because $e(\eta)\neq 0$ for every
$\eta\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$, as
$e(\eta)=\frac{1-p\eta^{2}}{1+p\eta^{2}}=0\Leftrightarrow\eta^{2}=1/p$, but
$1/p$ is not a square; $e(\infty)=-1$.

Then, $m_{22}=m_{22}^{\prime}$ is always satisfied, whereas

$\displaystyle m_{21}=m_{21}^{\prime}$ $\displaystyle\Leftrightarrow
a(\xi)n(\zeta)+b(\xi)g(\eta)l(\zeta)=a(\xi^{\prime})n(\zeta^{\prime}+b(\xi^{\prime})g(\eta^{\prime})l(\zeta^{\prime})$
$\displaystyle\Leftrightarrow a(\xi)n(\zeta)=\pm a(\xi)n(\zeta^{\prime})\
\Leftrightarrow n(\zeta)=\pm n(\zeta^{\prime});$ $\displaystyle m_{31}=m_{32}$
$\displaystyle\Leftrightarrow
c(\xi)n(\zeta)+a(\xi)g(\eta)l(\zeta)=c(\xi^{\prime})n(\zeta^{\prime})+a(\xi^{\prime})g(\eta^{\prime})l(\zeta^{\prime})$
$\displaystyle\Leftrightarrow c(\xi)n(\zeta)=\pm
c(\xi^{\prime})n(\zeta)\Leftrightarrow c(\xi)=\pm c(\xi^{\prime}).$

Note that $a(\xi)\neq 0$ for every $\xi\in{{\mathbb{Q}}}_{p}\cup\\{\infty\\}$,
because
$a(\xi)=\frac{1+\frac{p}{v}\xi^{2}}{1-\frac{p}{v}\xi^{2}}=0\Leftrightarrow\xi^{2}=-v/p$
but $-v/p$ is not a square; $a(\infty)=-1$.

In the last step it might be
$n(\zeta)=\frac{2\zeta}{1-v\zeta^{2}}=0\Leftrightarrow\zeta=0$, for which the
condition $m(\zeta)=\pm m(\zeta^{\prime})=0$ gives ${\cal R}_{z}(\zeta)={\cal
R}_{z}(\zeta^{\prime})=I$. This implies ${\cal R}_{x}(\xi)={\cal
R}_{x}(\xi^{\prime}),\,{\cal R}_{y}(\eta)={\cal R}_{y}(\eta^{\prime})$ by
Corollary 31, in particular $c(\xi)=c(\xi^{\prime})$, in agreement with what
we deduced above. Then, $m_{32}=m_{32}^{\prime}$ is satisfied.

We have found that ${\cal R}_{x}(\xi){\cal R}_{y}(\eta){\cal
R}_{z}(\zeta)={\cal R}_{x}(\xi^{\prime}){\cal R}_{y}(\eta^{\prime}){\cal
R}_{z}(\zeta^{\prime})$ if and only if either

$\bigl{(}{\cal R}_{x}(\xi),{\cal R}_{y}(\eta),{\cal
R}_{z}(\zeta)\bigr{)}=\bigl{(}{\cal R}_{x}(\xi^{\prime}),{\cal
R}_{y}(\eta^{\prime}),{\cal R}_{z}(\zeta^{\prime})\bigr{)}$

or

$\left\\{\begin{aligned} &{\cal R}_{x}(\xi)={\cal R}_{x}(\infty){\cal
R}_{x}(\xi^{\prime}),\\\ &{\cal
R}_{y}(\eta)=\begin{pmatrix}-e(\eta^{\prime})&0&f(\eta^{\prime})\\\ 0&1&0\\\
g(\eta^{\prime})&0&-e(\eta^{\prime})\end{pmatrix},\\\ &{\cal
R}_{z}(\zeta)={\cal R}_{z}(\infty){\cal
R}_{z}(\zeta^{\prime}),\end{aligned}\right.$

which concludes the proof.  $\sqcap$$\sqcup$

######  Remark 33.

The above results of the duplicity of the Cardano decomposition can be
understood qualitatively, and for any $p$, from Remark 25.

Indeed, assume that $SO(3)_{p}\ni M={\cal R}_{\bm{n}_{1}}(\sigma){\cal
R}_{\bm{n}_{2}}(\tau){\cal R}_{\bm{n}_{3}}(\omega)$, where
$\bm{n}_{1}\perp\bm{n}_{2}\perp\bm{n}_{3}$. We have seen that this is possible
if and only if ${\cal R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}\perp\bm{n}_{2}$,
and chosen ${\cal R}_{\bm{n}_{1}}(\sigma)$, the other two rotations are
uniquely determined. There are at most two solutions, since there is only one
free parameter in conditions for $\bm{v}={\cal
R}_{\bm{n}_{1}}(\sigma)^{-1}M\bm{n}_{3}$, and they boil down to a single
quadratic equation. On the other hand, with ${\cal R}_{\bm{n}_{1}}(\sigma)$,
another solution of this condition is always ${\cal
R}_{\bm{n}_{1}}(\infty){\cal R}_{\bm{n}_{1}}(\sigma)$.

##  7 Discussion

We have seen that not only is there a well-motivated $p$-adic special
orthogonal group $SO(3)_{p}$, but that it shares many geometric features with
its real counterpart $SO(3)_{{{\mathbb{R}}}}$, although there are also crucial
differences, owing to the specific number theory of ${{\mathbb{Z}}}_{p}$ for
different primes $p$, in particular distinguishing odd $p$ from $p=2$, and
within odd primes between $p\equiv 1\mod 4$ and $p\equiv 3\mod 4$. These
differences manifested themselves most blatantly in the discussion of Cardano
and Euler angle decompositions of general special orthogonal transformations
in terms of rotations around the reference axes: they are not available in
general for $p=2$, but certain (all) Cardano decompositions hold for odd
primes $p\equiv 3\mod 4$ ($p\equiv 1\mod 4$). We leave open the possibility of
modified principal angle decompositions beyond the use of the reference axes.

Note that we have treated the groups essentially as algebraic groups, and in
future development, it may pay off to follow the analytic approach of $p$-adic
Lie groups and Lie algebras Serre:Lie , which allows for a local description
in terms of infinitesimal generators.

An open question that we have left, might be answerable in this way, which is
that after the cyclicity of the two-dimensional rotation groups
$SO(2)_{p}^{\kappa}$. Since the groups are definitely profinite, the correct
question seems to be whether the groups $SO(2)_{p}^{\kappa}$ are procyclic. To
prove this, we consider their projections modulo $p^{k}$,
$\pi_{k}\bigl{(}SO(2)_{p}^{\kappa}\bigr{)}\subset
SL(2,{{\mathbb{Z}}}/p^{k}{{\mathbb{Z}}})$, and should show that for every $p$
and $\kappa$, and sufficiently large $k$ these are cyclic, cf. Ilaria:tesi .

Finally, we want to discuss our motivation for considering $p$-adic spatial
rotations in the first place: it grows out of Volovich’s $p$-adic quantum
theory, in which Euclidean space is replaced by $p$-adic space to define the
underlying phase space Volovich , see also V+V ; Varadarajan ; Vuvuzela . The
idea is to realise quantum systems as unitary representations of the symmetry
group of the $p$-adic space, according to Noether’s theorem Weyl , and this
has been realised for the displacement operations in position and momentum,
resulting in a $p$-adic Heisenberg–Weyl algebra of position and momentum
operators. What has not been done yet is to develop a quantum theory of
$p$-adic angular momentum. By the same philosophy, this is identical to the
classification of all the projective unitary irreducible representations of
$SO(3)_{p}$. We get infinitely many such representations from reducing the
group modulo $p^{k}$, noting that $SO(3)_{p}\mod p^{k}$ are finite groups, for
which the irreps can be found by standard tools FultonHarris . Indeed, in
Michele:tesi ; Ilaria:tesi , this has been done for reduction modulo $p$ and
$p^{2}$, for certain odd primes. An open question is whether all irreps of
$SO(3)_{p}$ arise in this way. We suspect that this problem cannot be solved
using standard $p$-adic techniques based on Hensel’s Lemma, but may require
non-standard $p$-adic analysis and Gretel’s Lemma Grimm .

## Acknowledgments

This project originated with the research visits of IS and MP with the Grup
d’Informació Quàntica at UAB, Barcelona, in the context of their degree and
MSc projects. MP and IS were co-funded by the Erasmus+ programme of the
European Union. AW acknowledges support by the EU (STREP ‘‘RAQUEL’’), the ERC
(AdG ‘‘IRQUAT’’), the Spanish MINECO (grants FIS2013-40627-P and
FIS2016-86681-P) with the support of FEDER funds, as well as by the
Generalitat de Catalunya CIRIT, projects 2014-SGR-966 and 2017-SGR-112.

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