# Slice regular functions as covering maps and global $\star$-roots

Amedeo Altavilla  Altavilla Amedeo: Dipartimento di Matematica, Università
degli Studi di Bari “Aldo Moro”, via Edoardo Orabona, 4, 70125, Bari, Italy.
<EMAIL_ADDRESS>and  Samuele Mongodi  Mongodi Samuele: Politecnico
di Milano, Piazza Leonardo da Vinci, 32, 20133, Milano, Italy.
<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to prove that a large class of quaternionic slice
regular functions result to be (ramified) covering maps. By means of the
topological implications of this fact and by providing further topological
structures, we are able to give suitable natural conditions for the existence
of $k$-th $\star$-roots of a slice regular function. Moreover, we are also
able to compute all the solutions which, quite surprisingly, in the most
general case, are in number of $k^{2}$. The last part is devoted to compute
the monodromy and to present a technique to compute all the $k^{2}$ roots
starting from one of them.

###### Key words and phrases:

Slice-regular functions, $k$-th roots, covering maps, monodromy

###### 2010 Mathematics Subject Classification:

Primary 30G35, 30C25; secondary 58K10, 32A10

Partially supported by GNSAGA of INdAM and by the INdAM project “Teoria delle
funzioni ipercomplesse e applicazioni”.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries on quaternionic slice regular functions
    1. 2.1 Two slightly different approaches
  3. 3 Differential and covering properties of slice regular functions
  4. 4 Powers in ${\mathbb{C}}\otimes{\mathbb{H}}$ as covering maps
  5. 5 Global $\star$-roots of a slice function
  6. 6 Covering automorphisms and monodromy
    1. 6.1 Permutations of $k$-th roots

##  1\. Introduction

The present work aims at studying slice regular functions of a quaternionic
variable as covering maps and the existence and nature of global $k$-th
$\star$-roots of slice functions.

Geometric function theory is the study of geometric and topological properties
of analytic functions of a complex variable; one could very well say that one
of the fundamental (although rather simple) results that originate and
motivate such a study is the local nature of branched coverings of holomorphic
functions of one variable.

This characteristic is even more striking when we move from open planar
domains to Riemann surfaces: holomorphic maps between Riemann surfaces are
locally branched coverings. In other terms, we can always find local
coordinates such that a holomorphic map between Riemann surfaces is locally
written as $z\mapsto z^{k}$, with $k\in\mathbb{N}$.

The global counterpart of these statements is obtained, for instance, in the
case when the function has some finiteness properties: as an example,
polynomial functions are branched coverings of ${\mathbb{C}}$ over
${\mathbb{C}}$, whose branching set is related to the zeros of their
derivative. The monodromy around branching points can be quite complicated and
it is related to the Galois group of the corresponding extension of rings of
rational functions, as opposed to the rather simple behaviour of the local
model $z\mapsto z^{k}$. However, if we are interested in the problem of
lifting holomorphic functions via such coverings, even the local model
presents interesting phenomena: if $p$ is a holomorphic polynomial, the
Riemann surface $w^{k}=p(z)$ possesses a number of geometric features that are
relevant in the general study of Riemann surfaces.

This geometric description of the local behaviour of holomorphic maps in one
dimension brings together several ingredients: Rouché theorem, winding
numbers, logarithmic indicator, local invertibility, conformality, nature of
zeroes.

In the setting of slice regular functions of a quaternionic variable, all
these results are, to some degree, true, however the difficulties which are
inherent in the quaternionic theory prevent us from merging them in a
comprehensive geometric description of slice regular functions as mappings
from ${\mathbb{H}}$ to ${\mathbb{H}}$.

We are referring to the unavailability of the usual (quaternionic) product and
composition operations, which do not leave the set of slice regular functions
invariant.

The case of the product is quite representative of the challenges posed by the
quaternionic setting: given two slice regular functions
$f,g:{\mathbb{H}}\to{\mathbb{H}}$ their product $q\mapsto f(q)g(q)$ (where, on
the right, we consider the product operation in the algebra of the
quaternions) is not a slice regular function; this problem is overcome by
considering a suitable product, called $\star$-product (see Definition 2.3).
However, the value of $f\star g$ at a quaternion $q$ is not, apart from some
particular cases, the product of the values of $f$ and $g$ at $q$, nor can be
obtained from these two values alone.

This results in a cumbersome way of dealing with powers and exponentials.

Quite recently, a number of papers addressed the problem of finding the
analogues of a logarithm or a $k$-th root for slice regular functions, see
[AltavillaPAMS, altavillaLOG, GPV, GPV2].

Our purpose, in the present work, is to study the existence and nature of th
$k$-th $\star$-roots of a slice regular function, i.e. the solutions of
$g^{\star k}=f$, from the point of view of covering maps, thus obtaining
global results and allowing a study of the monodromy of the solution of the
functional equation $g^{\star k}=f$.

The main difference with other previous attempts at this task lays in the
techniques we employ, which stem from merging our two different, but related,
interpretations of slice regularity, developed in our respective previous
works in the field.

On the one hand, any slice regular function can be interpreted, via a suitable
complex analytic representation, as a holomorphic curve in
${\mathbb{C}}^{4}\equiv{\mathbb{C}}\otimes{\mathbb{H}}$, so we can apply all
the techniques of classical complex analysis and relate geometric properties
of such curve to characteristic of the slice regular function (see
[Mongodi:HolSR, Mongodi:AssAlg]). In this setting, the $\star$-product emerges
naturally as the operation induced on such curves by the algebra product of
${\mathbb{C}}\otimes{\mathbb{H}}$.

On the other hand, the space of slice regular functions can be given the
structure of a rank $4$ module over the set of multiplicative commutators,
i.e. the set of slice preserving functions $\mathcal{S}_{\mathbb{R}}$ (see
[AltavillaAMPA, AltavillaLAA]); this construction depends on the choice of a
basis $(1,\ I,\ J,\ K)$ of ${\mathbb{H}}$ as a real vector space, so that,
given $f_{0},\ f_{1},\ f_{2},\ f_{3}$ slice preserving, the map

$(f_{0},\ f_{1},\ f_{2},\ f_{3})\mapsto f_{0}+f_{1}I+f_{2}J+f_{3}K$

is a bijection. In this case, the $\star$-product is recovered by observing
that the slice preserving functions are such that $f\star g=g\star f=fg=gf$
and this gives us a way to extend the product from ${\mathbb{H}}$ to
$(\mathcal{S}_{\mathbb{R}})^{4}$.

As different in spirit as they seem, these two viewpoints are, in fact, two
sides of the same coin; while the complex analytic approach is useful in
giving clear and general proofs based on known techniques in complex analysis
and geometry, the algebraic approach closely relates the peculiar
characteristics of quaternions to the properties of slice regular functions,
particularly when the operations of the algebra structure are involved, making
it easier to understand the computational side and allowing to produce several
explicit examples.

They are therefore both useful in separating those behaviours which come
seamlessly from the theory of one complex variable from the phenomena that
properly arise from the unique properties of ${\mathbb{H}}$, thus revealing
the true rôle of the quaternions.

We believe that the combination of these two approaches could be useful in
dealing with other similar problems and, particularly, in exploring the
geometric implications of such results, venturing in the scarcely explored
realm of quaternionic Riemann surfaces.

The content is organized as follows.

Section 2 contains a review of the basic material needed for our purposes,
following the notations and strategies introduced in [GP:AltAlg,
Mongodi:HolSR]. In particular, we introduce the formalism of stem functions
and of slice preserving functions and their relations with the
$\star$-product.

In the last part of the section, namely in Subsection 2.1, we recall the
general ideas leading the authors to develop the two aforementioned
interpretations of slice regularity, showing how these are linked and part of
their potential.

In Section 3 we show that, under suitable natural hypotheses, a slice regular
function that is also a finite map is, in fact, a covering map (see Theorem
3.4).

To obtain this result we present a couple of technical lemmas characterizing
geometrically, in terms of tangent vectors, the non-invertibility of the real
differential of a slice regular function (see Lemma 3.1 and 3.2). Of course,
part of the content of these results was already known in the literature
(maybe with different notations), but the proofs we present here are new
(especially in giving geometric insights on the involved objects), and the
techniques will be exploited in the mentioned main result.

As a consequence, we obtain a new proof (based on covering properties of slice
regular functions), of the fact that an injective slice regular function has
real differential that is invertible in the whole domain (see [GSS:Twistor,
Altavilla:RealDiff, Ghiloni2020]). To offer the right perspective on Theorem
3.4, we underline that from the content of [Ghiloni2020], it can be already
inferred that slice regular functions are locally covering maps, while our new
techniques give a global result.

The rest of the paper is devoted to study $k$-th $\star$-roots.

We begin, in Section 4, with the more abstract case arising from the complex
analytic interpretation of slice regularity. In particular, having transferred
the notion of $\star$-product to curves in ${\mathbb{C}}^{4}$, we coherently
define the analog of the $k$-th $\star$-power, denote by $\sigma_{k}$. In
Lemma 4.1 we give an explicit expression of the differential of $\sigma_{k}$,
describing its zero set completely.

This is related to a result given in the last section of [AltavillaAMPA],
where the authors give algebraic conditions in order to have that the $k$-th
$\star$-power of a slice regular function is slice preserving.

Passing to a more abstract interpretation we are then able to prove that,
under suitable hypotheses, $\sigma_{k}$ is generically $k^{2}$-to-$1$ and that
it is, in fact, a covering map (see Proposition 4.3 and Theorem 4.4). In the
last part of this section we show how the hypotheses of the previously
mentioned results translate in the language of slice regular functions.

Section 5 contains the main outcomes of this paper as mentioned in the
abstract. Building on the results of the previous section, we are able to
prove that, under suitable natural hypotheses, any slice regular function
defined on a domain without real points admits $k^{2}$ $k$-th $\star$-roots
(see Theorem 5.3).

This result was quite unexpected, but is not so surprising once one realizes
that the space of quaternions minus the real line
${\mathbb{H}}\setminus{\mathbb{R}}$ is biholomorphic to a half complex plane
times the Riemann sphere ${\mathbb{C}}^{+}\times\mathbb{CP}^{1}$. In a certain
sense these two basic complex spaces produces two independent monodromies,
each one counting $k$ sheets. If we impose the condition for the domain to
intersect the real axis, then many of these solutions are not anymore well
defined and the amount of survivors is the more expected number $k$ (see
Theorem 5.4).

In the last part of this section we provide two explicit examples showing, in
a heuristic way, the different behaviours between the case $k$ odd and $k$
even. In both examples, a key role is played by the so-called slice polynomial
functions introduced in [AltavillaMATHZ].

The last section is devoted to compute the monodromy in detail, explaining, in
particular, the $k^{2}$ factor discussed before. We find that, for a fixed
$k$, it is possible to define two different actions of the set of $k$-th
rooths of unity in ${\mathbb{C}}$ on the set of $k$-th $\star$-roots of a
given slice regular function. In terms of the algebraic representation of
slice regular functions, these two actions are related to two different
representations of complex numbers (as a field extension of ${\mathbb{R}}$ and
as a subset of $2\times 2$ real matrices).

We explore in detail these two actions, highlighting their different nature in
the two cases, $k$ odd and $k$ even; this allows us to obtain a general
existence result for $k$-th roots of slice functions (see Theorems 6.8 and
6.10).

We notice that this double monodromy, producing the $k^{2}$ factor, is
coherent with the results obtained for the logarithm, where, under consistent
natural hypotheses, a given slice regular function has $\infty^{2}$
$\star$-logarithms (see [altavillaLOG, Theorem 1.2] and [GPV2, Section 6]).

##  2\. Preliminaries on quaternionic slice regular functions

We refer to [GSS:RegFunc] for a general introduction to the slice regularity.

Let ${\mathbb{H}}$ be the algebra of quaternions and $\\{1,i,j,k\\}$ its
standard basis satisfying the usual multiplicative rules. Then, any quaternion
$q\in{\mathbb{H}}$ can be written as $q=q_{0}+q_{1}i+q_{2}j+q_{3}k$, with
$q_{0},q_{1},q_{2},q_{3}\in{\mathbb{R}}$. We endow ${\mathbb{H}}$ with the
standard involution ${\mathbb{H}}\ni q=q_{0}+q_{1}i+q_{2}j+q_{3}k\mapsto
q^{c}=q_{0}-(q_{1}i+q_{2}j+q_{3}k)\in{\mathbb{H}}$. The euclidean norm of $q$
can therefore be computed as $\|q\|=\sqrt{qq^{c}}$. Moreover, thanks to such a
conjugation, it is possible to define the scalar and vector part of any
$q\in{\mathbb{H}}$ as

$q_{0}=\frac{q+q^{c}}{2}\;,\qquad q_{v}=\frac{q-q^{c}}{2}\;,$

respectively. Therefore, any $q\in{\mathbb{H}}$ can be also written as
$q=q_{0}+q_{v}$. Then, we will identify ${\mathbb{R}}\subset{\mathbb{H}}$ with
the set $\\{q\in{\mathbb{H}}\ :\ q_{v}=0\\}$ and ${\mathbb{R}}^{3}$ with the
set of purely imaginary quaternions $\\{q\in{\mathbb{H}}\ :\ q_{0}=0\\}$. With
this representation, the product of two quaternions $q=q_{0}+q_{v}$ and
$p=p_{0}+p_{v}$ can be written as

$qp=q_{0}p_{0}-\langle q_{v},p_{v}\rangle+q_{0}p_{v}+p_{0}q_{v}+q_{v}\wedge
p_{v}\;,$

where $\langle\cdot,\cdot\rangle$ and $\wedge$ denote the standard euclidean
and cross product in ${\mathbb{R}}^{3}$.

The set of imaginary units in ${\mathbb{H}}$ is diffeomorphic to a $2$-sphere
$S^{2}$ and will be denoted as follows

${\mathbb{S}}=\\{I\in{\mathbb{H}}\ :\ I^{2}=-1\\}=\\{q\in{\mathbb{H}}\ :\
q_{0}=0,\ \|q_{v}\|=1\\}\;.$

With this notation, any $q=q_{0}+q_{v}\in{\mathbb{H}}\setminus{\mathbb{R}}$
can be written as $q=\alpha+I\beta$, where $\alpha=q_{0}$, $\beta=\|q_{v}\|$
and $I=q_{v}/\|q_{v}\|\in{\mathbb{S}}$ and hence
${\mathbb{H}}=\cup_{I\in{\mathbb{S}}}{\mathbb{C}}_{I}$, where
${\mathbb{C}}_{I}={\rm Span}_{\mathbb{R}}\\{1,I\\}$. Clearly, for any
$\alpha,\beta\in{\mathbb{R}}$ and any $I\in{\mathbb{S}}$, $\alpha+I\beta$ is a
well defined quaternion. Therefore, we can define the map
$\pi:{\mathbb{C}}\times{\mathbb{S}}\to{\mathbb{H}}$ by

(1) $\pi(z,I)={\sf Re}(z)+I\mathsf{Im}(z)\;.$

We will denote the imaginary unit in ${\mathbb{C}}$ by $\imath$. Notice that,
for any $\alpha+\imath\beta\in{\mathbb{C}}$,
$\pi(\\{z\\}\times{\mathbb{S}})=\\{\alpha+I\beta\ :\ I\in{\mathbb{S}}\\}\simeq
S^{2}$.

In view of the introduction of slice regularity, we turn our attention to the
real tensor product ${\mathbb{C}}\otimes{\mathbb{H}}$. The imaginary unit of
this complexification will be denoted by $\sqrt{-1}$. We will write the
elements in ${\mathbb{C}}\otimes{\mathbb{H}}$ both as $z\otimes q$, with
$z\in{\mathbb{C}}$ and $q\in{\mathbb{H}}$, or as $v_{0}+\sqrt{-1}v_{1}$, with
$v_{0},v_{1}\in{\mathbb{H}}$. We extend the previously defined quaternionic
conjugation in the following way: if
$v_{0}+\sqrt{-1}v_{1}\in{\mathbb{C}}\otimes{\mathbb{H}}$, then
$(v_{0}+\sqrt{-1}v_{1})^{c}:=v_{0}^{c}+\sqrt{-1}v_{1}^{c}$. Given
$v=v_{0}+\sqrt{-1}v_{1},w=w_{0}+\sqrt{-1}w_{1}\in{\mathbb{C}}\otimes{\mathbb{H}}$,
their product is defined as

$vw=v_{0}w_{0}-v_{1}w_{1}+\sqrt{-1}(v_{0}w_{1}+v_{1}w_{0})\;.$

The map defined in Formula (1) induces a new map (denoted with the same
symbol) $\pi:{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}\to{\mathbb{H}}$
by requiring the linearity in the first component and that

$\pi(z\otimes q,I)=\pi(z,I)q\;.$

In other words, if $w=w_{0}+\sqrt{-1}w_{1}\in\mathbb{C}\otimes\mathbb{H}$,
with $w_{1},w_{2}\in\mathbb{H}$, and $I\in{\mathbb{S}}$, then

$\pi(w,I)=w_{0}+Iw_{1}\;.$

We recall that the complex structure defined by left multiplication by
$\sqrt{-1}$ gives a structure of complex affine space to
${\mathbb{C}}\otimes{\mathbb{H}}$ for which it results to be biholomorphic to
${\mathbb{C}}^{4}$. Having chosen the standard basis $(1,i,j,k)$, if
$p=p_{0}+p_{1}i+p_{2}j+p_{3}k,q=q_{0}+q_{1}i+q_{2}j+q_{3}k\in{\mathbb{H}}$,
then such a biholomorphism
$\phi:{\mathbb{C}}\otimes{\mathbb{H}}\to{\mathbb{C}}^{4}$ easily reads as
follows

(2) $\phi(p+\sqrt{-1}q)\mapsto(p_{0}+\imath q_{0},p_{1}+\imath
q_{1},p_{2}+\imath q_{2},p_{3}+\imath q_{3})\;.$

The complex conjugation extends to a linear map, still denoted in the same
way, such that $\overline{z\otimes q}=\overline{z}\otimes q$, or, if
$w=w_{0}+\sqrt{-1}w_{1}$, then $\overline{w}=w_{0}-\sqrt{-1}w_{1}$. For any
$w\in{\mathbb{C}}\otimes{\mathbb{H}}$, we have that
$(\overline{w})^{c}=\overline{(w^{c})}$.

We have now all the tools needed to introduce slice regularity with the
approach of Ghiloni and Perotti [GP:AltAlg].

######  Definition 2.1.

Let ${\mathcal{U}}\subset{\mathbb{C}}$ be an open domain and
$U=\pi({\mathcal{U}}\times{\mathbb{S}})$. A function
$F:{\mathcal{U}}\to{\mathbb{C}}\otimes{\mathbb{H}}$ such that
$F(\overline{z})=\overline{F(z)}$ is said to be a _stem function_. A function
$f:U\to{\mathbb{H}}$ is called _slice_ if there exists a stem function
$F:{\mathcal{U}}\to{\mathbb{C}}\otimes{\mathbb{H}}$ such that
$f(\pi(z,I))=\pi(F(z),I)$. We will write $f=\mathcal{I}(F)$. Moreover, if $F$
is a holomorphic function, then $f$ is said to be _slice regular_.

For convenience, we make here the following assumption.

######  Assumption 2.2.

In what follows ${\mathcal{U}}\subset{\mathbb{C}}$ will be an open domain and
$U=\pi({\mathcal{U}}\times{\mathbb{S}})$. Moreover, we will fix a slice
function $f=\mathcal{I}(F):U\to{\mathbb{H}}$ and we will write $F=1\otimes
F_{0}+\sqrt{-1}\otimes F_{1}$ where $F_{0},\
F_{1}:{\mathcal{U}}\to{\mathbb{H}}$. In other words if
$\alpha+\imath\beta\in{\mathcal{U}}$, $F=F_{0}+\sqrt{-1}F_{1}$ and
$f=\mathcal{I}(F)$, then
$f(\alpha+I\beta)=F_{0}(\alpha+\imath\beta)+IF_{1}(\alpha+\imath\beta)$.

As the pointwise product of two slice functions does not preserve sliceness, a
different, yet natural, product may be introduced.

######  Definition 2.3.

Let $f=\mathcal{I}(F),g=\mathcal{I}(G)$ be two slice functions defined on $U$.
We define their _slice product_ or $\star$-_product_ as

$f\star g:=\mathcal{I}(FG):U\to{\mathbb{H}}\;.$

Hence if $F=F_{0}+\sqrt{-1}F_{1}$ and $G=G_{0}+\sqrt{-1}G_{1}$, then $f\star
g$ is the slice function induced by
$F_{0}G_{0}-F_{1}G_{1}+\sqrt{-1}(F_{0}G_{1}+F_{1}G_{0})$.

A special subset of slice regular functions is given in the following
definition.

######  Definition 2.4.

A slice function $f=\mathcal{I}(F):U\to{\mathbb{H}}$ is said to be _slice
preserving_ if, for any $I\in{\mathbb{S}}$,
$f(U\cap{\mathbb{C}}_{I})\subset{\mathbb{C}}_{I}$. Equivalently, this happens
if and only if $F$ takes values in
${\mathbb{C}}\otimes{\mathbb{R}}\subset{\mathbb{C}}\otimes{\mathbb{H}}$.

A slice preserving function $f=\mathcal{I}(F_{0}+\sqrt{-1}F_{1})$ is therefore
a slice function such that $F_{0}$ and $F_{1}$ are real valued. However, this
does not imply that $f$ is real valued. Notice that if $f$ is a slice
preserving function and $g$ is any slice function, then $f\star g=g\star
f=fg$.

We conclude this section by defining the slice conjugate and symmetrized of a
slice function.

######  Definition 2.5.

Let $f=\mathcal{I}(F_{0}+\sqrt{-1}F_{1}):U\to{\mathbb{H}}$ be a slice
function. We define its _slice conjugate_ as the slice function
$f^{c}:=\mathcal{I}(F^{c}):U\to{\mathbb{H}}$ and its _symmetrized function_ or
_symmetrization_ as $f^{\mathsf{s}}=f\star f^{c}=\mathcal{I}(FF^{c})$.

The symmetrized of any slice function is a slice preserving function.
Moreover, if $f$ is slice preserving, then $f^{c}=f$ and so
$f^{\mathsf{s}}=f^{\star 2}=f^{2}$.

From the general theory [GSS:RegFunc, GP:AltAlg], we know that if $f(q)=0$,
then $f^{\mathsf{s}}(q)=0$. In particular, if
$q=\alpha+I\beta\in{\mathbb{H}}\setminus{\mathbb{R}}$, then
$f^{\mathsf{s}}(\alpha+J\beta)=0$, for any $J\in{\mathbb{S}}$. Moreover, if
$f^{\mathsf{s}}(\alpha+I\beta)=0$, then there exists $J\in{\mathbb{S}}$, such
that $f(\alpha+J\beta)=0$.

###  2.1. Two slightly different approaches

The introduction of slice regularity by means of
${\mathbb{C}}\otimes{\mathbb{H}}$ and all the structures presented above
allows us to exploit complex analysis and geometry in order to obtain our
results.

First of all the identification between ${\mathbb{C}}\otimes{\mathbb{H}}$ and
${\mathbb{C}}^{4}$ was already noticed in [GP:AltAlg, Remark 3(2)] and used in
[AltavillaCVEE, Theorem 3.4] to prove a result about a particular family of
slice regular functions. Later, this approach was widely exploited in
[Mongodi:HolSR, Mongodi:AssAlg] to produce alternative proofs of known and new
results on slice regularity, highlighting their holomorphic nature.

On the other hand, such approach is an underlying element of a series of paper
by the first author and de Fabritiis [AltavillaAMPA, AltavillaPAMS,
AltavillaLAA, altavillaLOG]. In these papers many algebraic properties of
slice regular functions are proved and existence and uniqueness results for
the quaternionic exponential and logarithm are given. The basic result and
idea behind these researches is a result due to Colombo, Gonzalez-Cervantes
and Sabadini [CGCS, Proposition 3.12] and Ghiloni, Moretti and Perotti [GMP,
Lemma 6.11] stating that any slice regular function $f:U\to{\mathbb{H}}$ can
be uniquely written as a sum

$f=f_{0}+f_{1}i+f_{2}j+f_{3}k,$

where $f_{0},f_{1},f_{2},f_{3}$ are slice preserving regular functions defined
on the same domain. Thanks to this equality, it is possible to define the
“real” and “vector” part of $f=f_{0}+f_{v}$ as

$f_{0}=\frac{f+f^{c}}{2}\;,\qquad f_{v}=\frac{f-f^{c}}{2}\;,$

and, with these observation, it is possible to represent the $\star$-product
of two slice regular functions $f=f_{0}+f_{v},g=g_{0}+g_{v}$ as

$f\star g=f_{0}g_{0}-\langle
f_{v},g_{v}\rangle_{\star}+f_{0}g_{v}+g_{0}f_{v}+f_{v}\operatorname{\wedge\mspace{-9.5mu}_{\star}\
}g_{v}\;,$

where $\langle\cdot,\cdot\cdot\rangle_{\star}$ and
$\operatorname{\wedge\mspace{-9.5mu}_{\star}\ }$ are algebraic operators
acting, formally, as the usual euclidean and cross product (see
[AltavillaAMPA]). In particular, given $f=f_{0}+f_{1}i+f_{2}j+f_{3}k$, we have
that $f^{c}=f_{0}-(f_{1}i+f_{2}j+f_{3}k)=f_{0}-f_{v}$, and
$f^{\mathsf{s}}=f_{0}^{2}+f_{1}^{2}+f_{2}^{2}+f_{3}^{2}$.

It is clear that this interpretation of slice regular functions is suggested
and underlined by the previously described complex one. However, if the
complex approach will be effective to produce general proofs, the second one
will be exploited to produce and describe explicit examples.

##  3\. Differential and covering properties of slice regular functions

In this section, we recover some known fact about the real differential of
slice regular function by looking at it at the level of
${\mathbb{C}}\otimes{\mathbb{H}}$. This approach is not only interesting by
itself, but will be exploited later in this section when we will discuss
covering properties.

With a slightly different language and notation, in [Mongodi:HolSR] the second
author defined, for $q\in{\mathbb{H}}$, the sets

$\displaystyle Z_{q}$ $\displaystyle=\\{w\in{\mathbb{C}}\otimes{\mathbb{H}}\
:\ \pi(w,I)=q\textrm{ for some }I\in{\mathbb{S}}\\}\;,$
$\displaystyle\mathfrak{Z}_{q}$
$\displaystyle=\\{(w,I)\in{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}\
:\ \pi(w,I)=q\\}\;.$

######  Remark 3.1.

We note that, if $w=w_{0}+\sqrt{-1}w_{1}\in{\mathbb{C}}\otimes{\mathbb{H}}$ is
a real point in the complexification of ${\mathbb{H}}$, i.e. if $w_{1}=0$,
then $w\in Z_{q}$ if and only if $w=q$.

As in [Mongodi:AssAlg], we also consider the map
$\mathscr{F}:{\mathcal{U}}\times{\mathbb{S}}\to{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}$
given by $\mathscr{F}(z,I)=(F(z),I)$, obtaining the equality
$f\circ\pi=\pi\circ\mathscr{F}$, expressed by the following commutative
diagram.

$\begindc{\commdiag}[3]\obj(0,180)[A]{{\mathcal{U}}\times{\mathbb{S}}}\obj(280,180)[B]{{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}}\obj(0,0)[C]{U}\obj(280,0)[D]{{\mathbb{H}}}\mor{A}{B}{\mathscr{F}}[\atleft,\solidarrow]\mor{A}{C}{\pi}[\atright,\solidarrow]\mor{C}{D}{f}[\atright,\solidarrow]\mor{B}{D}{\pi}[\atleft,\solidarrow]\enddc$

Moreover, for any $(w,I)\in{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}$
we have that $\ker D\pi_{(w,I)}=T_{\mathscr{F}(z,I)}\mathfrak{Z}_{f(q)}$. It
is known [Mongodi:HolSR] that $Z_{q}$ is a complex quadratic cone in
${\mathbb{C}}\otimes{\mathbb{H}}$, with one singular point (the vertex), which
is the only point with real coordinates, and that $\mathfrak{Z}_{q}$ is a
complex submanifold of ${\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}$,
once ${\mathbb{S}}$ is given the appropriate complex structure (which is
actually the natural one on ${\mathbb{CP}}^{1}$). As $q$ varies in
${\mathbb{H}}$, the manifolds $\mathfrak{Z}_{q}$ give a foliation of
${\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}$.

######  Remark 3.2.

Notice that the definition of slice regularity can be interpreted in terms of
(pseudo-)holomorphic curves as follows. Let $f:U\to\mathbb{H}$ be a slice
regular function and $p=\alpha+I\beta\in U\setminus\mathbb{R}$. Locally, the
tangent space $T_{p}U$ can be split into a sum
$T_{p}U=T_{\alpha+\imath\beta}\mathbb{C}\oplus T_{I}\mathbb{S}$. Hence, if
$v\in T_{\alpha+\imath\beta}\mathbb{C}$, then $df(\imath v)=L_{I}(df(v))$,
where $L_{I}$ denotes the left multiplication by $I\in{\mathbb{S}}$.

In the following lemma we give a geometric interpretation for the degeneration
of the real differential of a slice regular function.

######  Lemma 3.1.

Suppose that $q=\pi(z,I)\in U$ is such that $\mathsf{Im}(z)\neq 0$,
$F_{1}(z)\neq 0$ and that the (real) differential $Df_{q}$ is singular. Then
$F({\mathcal{U}})$ is tangent to $Z_{f(q)}$ in $F(z)$.

###### Proof.

If we set ${\mathcal{U}}^{+}=\\{z\in{\mathcal{U}}\ :\ \mathsf{Im}(z)>0\\}$,
then the restriction of $\pi$ to ${\mathcal{U}}^{+}\times{\mathbb{S}}$ is a
diffeomorphism between the latter and $U\setminus{\mathbb{R}}$. Hence,
$Df_{q}$ is singular if and only if $D(\pi\circ\mathscr{F})_{(z,I)}$ is
singular, i.e. $Df_{q}[v]=0$ for some $v\in T_{q}U$ if and only if there
exists $v^{\prime}\in T_{(z,I)}({\mathcal{U}}\times{\mathbb{S}})$ such that

$D\pi_{\mathscr{F}(z,I)}\circ D\mathscr{F}_{(z,I)}[v^{\prime}]=0\;.$

This happens if and only if $D\mathscr{F}_{(z,I)}[v^{\prime}]\in\ker
D\pi_{\mathscr{F}(z,I)}$, i.e. by definition if and only if
$D\mathscr{F}_{(z,I)}[v^{\prime}]\in T_{\mathscr{F}(z,I)}\mathfrak{Z}_{f(q)}$.

We now show that the former analysis implies that $F({\mathcal{U}})$ is
tangent to $Z_{f(q)}$. Consider the projection map
$p_{1}:{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}\to{\mathbb{C}}\otimes{\mathbb{H}}$,
so that $p_{1}\circ\mathscr{F}(z,I)=F(z)$, and let
$Z_{f(q)}^{*}\subset{\mathbb{C}}\otimes{\mathbb{H}}$ denote the set $Z_{f(q)}$
minus its real points, i.e. $Z_{f(q)}^{*}=Z_{f(q)}\setminus\\{f(q)\\}$ (see
Remark 3.1). The map $p_{1}$ induces a biholomorphism between $Z^{*}_{f(q)}$
and $\mathfrak{Z}^{*}_{f(q)}=p_{1}^{-1}(Z^{*}_{f(q)})$ (see the proof of
[Mongodi:HolSR]*Theorem 3.3). Therefore, as $F_{1}(z)\neq 0$,

$0\neq Dp_{1}(D\mathscr{F}_{(z,I)}[v^{\prime}])\in
T_{p_{1}(\mathscr{F}(z,I))}Z_{f(q)}=T_{F(z)}Z_{f(q)}$

so, as $p_{1}\circ\mathscr{F}(z,I)=F(z)$,
$D(p_{1}\circ\mathscr{F})_{(z,I)}[v^{\prime}]=DF_{z}[v^{\prime\prime}]$, where
$v^{\prime\prime}$ is the component of $v^{\prime}$ along $T_{z}{\mathbb{C}}$;
therefore $F({\mathcal{U}})$ is tangent to $Z_{f(q)}$ in $F(z)$. ∎

######  Remark 3.3.

We consider the case when $F({\mathcal{U}})\subseteq Z_{f(q)}$ as a tangency
case.

From this point of view, we obtain another proof of a known result (see
[GSS:Twistor]*Proposition 3.3 and [Ghiloni2020]) regarding the invertibility
of the real differential of a slice regular function. Before stating it, we
set up the following bits of notations. For $I\in{\mathbb{S}}$, we identify
$T_{I}{\mathbb{S}}\subset T_{I}{\mathbb{H}}\cong{\mathbb{H}}$ with

$A_{I}=\\{p\in{\mathbb{H}}\ :\ Ip+pI=0\\}$

(see [Mongodi:AssAlg]*Lemma 3.1). Moreover, given
$w\in{\mathbb{C}}\otimes{\mathbb{H}}$, $w=1\otimes a+\sqrt{-1}\otimes b$, we
identify $T_{w}({\mathbb{C}}\otimes{\mathbb{H}})$ with
$T_{a}{\mathbb{H}}\oplus T_{b}{\mathbb{H}}$.

######  Lemma 3.2.

Suppose that $q=\pi(z,I)$ with $\mathsf{Im}(z)\neq 0$. The differential
$Df_{q}$ vanishes if and only if one of the following three cases occurs:

  1. (1)

$F_{1}(z)=0$;

  2. (2)

$DF_{z}=0$;

  3. (3)

$\pi(DF_{z},I)F_{1}(z)^{-1}\in A_{I}$.

######  Remark 3.4.

Notice that the three cases of the previous Lemma, correspond to $(1)$ the
spherical derivative of $f$ being zero (see [GSS:RegFunc, Definition 1.18]);
$(2)$ the slice derivative being zero (see again [GSS:RegFunc, Definition
1.7]); $(3)$ the product of the slice derivative with the inverse of the
spherical derivative belonging to the orthogonal complement of ${\rm
Span}\\{1,q\\}$ (see [Altavilla:RealDiff, Proposition 29]). Indeed, for this
last claim, notice that ${\rm Span}\\{1,q\\}^{\bot}=\\{p\in{\mathbb{H}}\ :\
qp+pq=0\\}$, and if $q=\alpha+I\beta$, the equality $qp+pq=0$ is equivalent to
$Ip+pI=0$, which define $A_{I}$.

###### Proof.

Consider $D\mathscr{F}_{(z,I)}:T_{z}{\mathbb{C}}\oplus T_{I}{\mathbb{S}}\to
T_{a}{\mathbb{H}}\oplus T_{b}{\mathbb{H}}\oplus T_{I}{\mathbb{S}}$, where
$a=F_{0}(z)$, $b=F_{1}(z)$; we have

$D\mathscr{F}=\begin{pmatrix}DF_{0}&0\\\ DF_{1}&0\\\ 0&id_{I}\end{pmatrix}$

where $DF_{0}:T_{z}{\mathbb{C}}\to T_{a}{\mathbb{H}}$,
$DF_{1}:T_{z}{\mathbb{C}}\to T_{b}{\mathbb{H}}$ are the real differentials of
$F_{0}$ and $F_{1}$ and $id_{I}$ is the identity map on $T_{I}{\mathbb{S}}$.

Likewise, the differential $D\pi_{(w,I)}:T_{a}{\mathbb{H}}\oplus
T_{b}{\mathbb{H}}\oplus T_{I}{\mathbb{S}}\to T_{\pi(w,I)}{\mathbb{H}}$ is
given by

$D\pi_{(w,I)}=\begin{pmatrix}id_{a}&L_{s}&R^{\prime}_{b}\end{pmatrix}$

where $id_{a}$ is the identity map on $T_{a}{\mathbb{H}}$,
$L_{I}:{\mathbb{H}}\to{\mathbb{H}}$ is the left multiplication by $I$, i.e.
$L_{I}(q)=Iq$, and $R^{\prime}_{b}:A_{s}\to{\mathbb{H}}$ is the right
multiplication by $b$, restricted to the subspace $A_{I}$, i.e.
$R^{\prime}_{b}(h)=hb$.

Given $v^{\prime}\in T_{z}{\mathbb{C}}\oplus T_{I}{\mathbb{S}}$, we write it
$v^{\prime}=v^{\prime}_{1}+v^{\prime}_{2}$, with $v^{\prime}_{1}\in
T_{z}{\mathbb{C}}$ and $v^{\prime}_{2}\in T_{I}{\mathbb{S}}$, then

$D(\pi\circ\mathscr{F})_{z,I}[v^{\prime}]=(DF_{0}+L_{I}DF_{1})_{z}[v^{\prime}_{1}]+R_{F_{1}(z)}^{\prime}[v^{\prime}_{2}]\;.$

Therefore, the rank of $Df$ drops in the following three cases:

  1. (1)

there exists $v^{\prime}_{2}\neq 0$ such that $v^{\prime}_{2}\in\ker
R^{\prime}_{F_{1}(z)}$,

  2. (2)

there exists $v^{\prime}_{1}\neq 0$ such that
$v^{\prime}_{1}\in\ker(DF_{0}+L_{I}DF_{1})_{z}$,

  3. (3)

there exist $v^{\prime}_{1},v^{\prime}_{2}\neq 0$ such that
$(DF_{0}+L_{I}F_{1})_{z}[v^{\prime}_{1}]=R_{F_{1}(z)}^{\prime}[v^{\prime}_{2}]$.

In the first case, as the map $R_{b}$ is always invertible, unless $b=0$, we
conclude that $F_{1}(z)=0$. By Remark 3.2, we have that
$(DF_{0}+L_{I}DF_{1})[\imath
v^{\prime}_{1}]=L_{I}(DF_{0}+L_{I}DF_{1})[v^{\prime}_{1}]=0$, so $DF_{z}=0$.

In the third case, we are saying that there exists $v^{\prime}_{1}$ such that

$R_{F_{1}(z)}^{-1}(DF_{0}+L_{I}DF_{1})_{z}[v^{\prime}_{1}]\in A_{I}\;.$

We note that, as the right multiplication operator commutes with the left one,
we have that

$R_{F_{1}(z)}^{-1}(DF_{0}+L_{I}DF_{1})_{z}[\imath
v^{\prime}_{1}]=R_{F_{1}(z)}^{-1}L_{I}(DF_{0}+L_{u}DF_{1})_{z}[v^{\prime}_{1}]=L_{I}R_{F_{1}(z)}^{-1}(DF_{0}+L_{u}DF_{1})_{z}[v^{\prime}_{1}]$

and the subspace $A_{I}$ is stable under $L_{I}$, therefore, as both the
spaces have real dimension $2$, the third case happens if and only if
$R_{F_{1}(z)}^{-1}(DF_{0}+L_{I}DF_{1})$ is an isomorphism between
$T_{z}{\mathbb{C}}$ and $A_{I}=T_{I}{\mathbb{S}}$.

By the definition of $A_{I}$, this happens if and only if

$L_{I}R_{F_{1}(z)}^{-1}(DF_{0}+L_{I}DF_{1})+R_{I}R_{F_{1}(z)}^{-1}(DF_{0}+L_{I}DF_{1})=0\;,$

i.e. if and only if $\pi(DF_{z},I)F_{1}(z)^{-1}\in A_{I}$. ∎

We remark that cases (2) and (3) are both contained in the geometric statement
of Lemma 3.1, i.e. in both cases $F(z)$ is a point of tangency between
$F({\mathcal{U}})$ and $Z_{f(q)}$ (and a regular point for the latter).

Let us now consider the holomorphic function
$\Phi_{q}:{\mathbb{C}}\otimes{\mathbb{H}}\to{\mathbb{C}}$, where
$q=q_{0}+q_{1}i+q_{2}j+q_{3}k$ is a quaternion, given by

$\Phi_{q}(z_{0}+z_{1}i+z_{2}j+z_{3}k)=\sum_{h=0}^{3}(z_{h}-q_{h})^{2}\;,$

where $z_{0}+z_{1}i+z_{2}j+z_{3}k=\phi^{-1}(z_{0},z_{1},z_{2},z_{3})$. The
function $\Phi_{q}$ is such that its zero set equals $Z_{q}$ when embedded in
${\mathbb{C}}^{4}$ (see [Mongodi:HolSR, Corollary 3.4]). Notice that, if
$F:{\mathcal{U}}\to{\mathbb{C}}\otimes{\mathbb{H}}$ is a map inducing a slice
function $f=f_{0}+f_{1}i+f_{2}j+f_{3}k$, and
$z=\alpha+\imath\beta\in{\mathcal{U}}$, then $\Phi_{q}(F(z))=0$ if and only if
$(F(z)-q)(F^{c}(z)-q^{c})=0$, if and only if
$(f-q)^{\mathsf{s}}=(f_{0}-q_{0})^{2}+(f_{1}-q_{1})^{2}+(f_{2}-q_{2})^{2}+(f_{3}-q_{3})^{2}=0$.

As we will see in the next corollary, properties of $\Phi_{q}$ relate with the
rank of slice regular functions.

######  Corollary 3.3.

Given $q=\pi(z,I)\in{\mathbb{H}}$, the real differential $Df_{q}$ is singular
if and only if $z$ is a zero of multiplicity greater than $1$ for the
holomorphic function $\Phi_{f(q)}\circ F:{\mathcal{U}}\to{\mathbb{C}}$.

###### Proof.

Since $Z_{f(q)}=\Phi_{f(q)}^{-1}(0)$ (see [Mongodi:HolSR, Corollary 3.4]), as
long as $F_{1}(z)\neq 0$ the tangent space to $Z_{f(q)}$ at $F(z)$ is given by
the equation $D(\Phi_{f(q)})_{F(z)}=0$; therefore,

$0=(\Phi_{f(q)}\circ F)^{\prime}(z)=D(\Phi_{f(q)})_{F(z)}DF_{z}$

if and only if $DF_{z}[v^{\prime}_{1}]$ is contained in $T_{F(z)}Z_{f(q)}$ for
all $v^{\prime}_{1}\in T_{z}{\mathbb{C}}$, i.e if and only if
$F({\mathcal{U}})$ is tangent to $Z_{f(q)}$ at $F(z)$.

If $F_{1}(z)=0$, then $F(z)$ has all real components, hence it is the vertex
of the cone $Z_{f(q)}$, which is the only singular point of $\Phi_{f(q)}$,
which is a zero of multiplicity $2$, so the conclusion is trivial in this
case. ∎

######  Remark 3.5.

This last corollary is a complex analyitic interpretation of a result obtained
independently in the quaternionic setting stating, essentially, that a slice
regular function $f$ is singular at a point $q_{0}$ if and only if,
$f(q)=f(q_{0})+(q-q_{0})\star(q-\tilde{q_{0}})g(q)$ for some
$\tilde{q_{0}}\in{\mathbb{S}}_{q_{0}}$ and some slice regular function $g$
(see [GSS:Twistor, Proposition 3.6] and [Altavilla:RealDiff, Theorem 30].

We conclude this section by showing that a large class of slice regular
functions are global covering maps outside their (real) singular locus. The
fact that slice regular functions are local covering can be excerpt already
from the content of [GSS:Twistor, Altavilla:RealDiff, Ghiloni2020]. Before
stating the theorem, let us define the following sets related to singular
points and values of a slice regular function. We denote by $C_{0}(f)$ the set
of critical points of $f$, i.e.

$C_{0}(f)=\\{p\in U\,:\,Df_{q}\text{ is singular}\\}\;;$

we consider its image under $f$, i.e. the set of critical values

$C(f)=f(C_{0}(f))\;,$

and its saturation under $f$, i.e. the singular locus of $f$,

$S(f)=f^{-1}(C(f))\;.$

######  Theorem 3.4.

Suppose $f$ is a finite map. Then, the restriction

$f|_{U\setminus S(f)}:U\setminus S(f)\to f(U\setminus
S(f))\subseteq{\mathbb{H}}\setminus C(f)$

is a covering map.

###### Proof.

Take a point $p\in f(U\setminus S(f))$ and a relatively compact neigborhood
$V$ of $p$ in $f(U\setminus S(f))$ such that $V\Subset{\mathbb{H}}\setminus
C(f)$. We consider the family of holomorphic maps

$\\{\Phi_{p^{\prime}}\circ F:{\mathcal{U}}\to{\mathbb{C}}\\}_{p^{\prime}\in
V}\;$

If $p^{\prime}\to p$, then $\Phi_{p^{\prime}}\circ F$ converges uniformly on
compact sets to $\Phi_{p}\circ F$.

Suppose that $f^{-1}(p)=\\{q_{1},\ldots,q_{n}\\}$ and write
$q_{m}=\pi(z_{m},I_{m})$. As $q_{1},\ldots,q_{n}\in U\setminus S(f)$, the
differential of $f$ is non singular around such points, so $f$ is locally
injective and, by Corollary 3.3, $\Phi_{p}\circ F$ has a zero of multiplicity
$1$ in each of these points.

Therefore, by Hurwitz’s theorem, $\Phi_{p^{\prime}}\circ F$ has $n$ zeroes
(counted with multiplicities) for $p^{\prime}\in V$ (up to shrinking $V$, if
necessary). However, again by Corollary 3.3, as long as
$p^{\prime}\in{\mathbb{H}}\setminus C(f)$, the zeroes of
$\Phi_{p^{\prime}}\circ F$ have multiplicity $1$, so each point $p^{\prime}\in
V$ has exactly $n$ preimages.

This shows that $f^{-1}(V)$ is the disjoint union of $n$ open sets
$V_{1},\ldots,V_{n}$. The differential of $f$ is non singular on every $V_{m}$
and $f|_{V_{m}}$ is injective, so $f$ is a diffeomorphism between $V_{m}$ and
$V$, for $m=1,\ldots,n$. This implies that $f$ is a covering map from
$U\setminus S(f)$ to $f(U\setminus S(f))\subseteq{\mathbb{H}}\setminus C(f)$.
∎

Along the same lines of proof, we obtain the following result, which already
appeared, proved with different methods, in [GSS:RegFunc, Altavilla:RealDiff,
Ghiloni2020]. The proof we are going to present differs from those in the
cited papers for being more topological, rather than analytical.

######  Corollary 3.5.

If $f:U\to{\mathbb{H}}$ is slice regular and injective, its real differential
is invertible at all points.

###### Proof.

Suppose that the differential is not invertible at $q=\pi(z_{0},I)$, then
either $F_{1}(z_{0})=0$ and then $f$ is constant on the sphere
$\pi(\\{z_{0}\\}\times{\mathbb{S}})$, or $\Phi_{f(q)}\circ F$ has a zero of
multiplicity greater than $1$ in $z_{0}$.

If $\Phi_{f(q)}\circ F$ vanishes identically, then the function $f$ is
constant on some $2$-dimensional surface in $U$ (see [Mongodi:HolSR, Remark
3.4] and [Ghiloni2020, Proposition 5.2]). If $\Phi_{f(q)}\circ F$ does not
vanish identically, by Hurwitz’s theorem, $\Phi_{q^{\prime}}\circ F$ has more
than one zero (counting multiplicities) for all $q^{\prime}$ close enough to
$f(q)$. However, suppose that there exists a neighborhood $V$ of $f(q)$ such
that $\Phi_{q^{\prime}}\circ F$ has a zero of multiplicity greater than one:
this means that $\mathscr{F}({\mathcal{U}}\times{\mathbb{S}})$ is tangent to
$\mathfrak{Z}_{q^{\prime}}$ for all $q^{\prime}\in V$.

Without loss of generality, we can suppose that the tangency points are such
that $F_{1}(z)\neq 0$. On ${\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}$,
we consider the smooth subbundle $E$ of the tangent bundle given by

${\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}\ni p\mapsto
E_{p}=T_{p}\mathfrak{Z}_{\pi(p)}\subseteq
T_{p}({\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}})\;.$

The integral manifolds of this subbundle are obviously the manifolds
$\mathfrak{Z}_{\pi(p)}$; on the other side,
$V^{\prime}=\mathscr{F}^{-1}(\pi^{-1}(V))$ is an open set in
${\mathcal{U}}\times{\mathbb{S}}$ and, for every $(z,I)\in V^{\prime}$,
$T_{\mathscr{F}(z,I)}\mathscr{F}(V^{\prime})\subseteq E_{\mathscr{F}(z,I)}$.
Therefore $\mathscr{F}(V^{\prime})$ is contained in an integral manifold of
the distribution $p\mapsto E_{p}$, so in $\mathfrak{Z}_{f(q)}$. This implies
that $f$ is not injective, as $f_{|{\pi(V^{\prime})}}\equiv f(q)$.

$\begindc{\commdiag}[3]\obj(0,180)[A]{{\mathcal{U}}\times{\mathbb{S}}}\obj(-105,185)[G]{V^{\prime}\subset}\obj(280,180)[B]{{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}}\obj(0,0)[C]{U}\obj(280,0)[D]{{\mathbb{H}}}\obj(340,0)[G]{\supset
V}\obj(280,360)[E]{E\subset
T({\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}})}\obj(560,180)[F]{{\mathbb{C}}\times{\mathbb{S}}}\mor{A}{B}{\mathscr{F}}[\atleft,\solidarrow]\mor{A}{C}{\pi}[\atright,\solidarrow]\mor{C}{D}{f}[\atright,\solidarrow]\mor{B}{D}{\pi}[\atleft,\solidarrow]\mor{B}{E}{}\mor{B}{F}{(\Phi_{\cdot},id)}\enddc$

∎

##  4\. Powers in ${\mathbb{C}}\otimes{\mathbb{H}}$ as covering maps

Having proved Theorem 3.4 it is natural to study the case of $k$-th
$\star$-powers, aiming to a result giving necessary and sufficient conditions
in order to have the existence of global $k$-th $\star$-roots of a given slice
function. Notice, however, that the $k$-th $\star$-power corresponds to the
pointwise $k$-th power only on slice preserving functions. Therefore, to
better analyse its features it is more convenient to move our attention to its
stem function, living in ${\mathbb{C}}\otimes{\mathbb{H}}$, where such
operator corresponds to the pointwise $k$-th power.

Let, hence,
$\sigma_{k}:{\mathbb{C}}\otimes{\mathbb{H}}\to{\mathbb{C}}\otimes{\mathbb{H}}$
be the function that raises an element of ${\mathbb{C}}\otimes{\mathbb{H}}$ to
its $k$-th power. We recall the identification given in Formula (2). In
particular, we denote by $e_{m}$, $m=0,1,2,3$, the elements of the standard
basis $(1\otimes 1,\ 1\otimes i,\ 1\otimes j,\ 1\otimes k)$, respectively. The
corresponding coordinates will be denoted as $z_{0},\ z_{1},\ z_{2},\ z_{3}$.
From now on, whenever working on ${\mathbb{C}}^{4}$ we will keep in mind such
an identification.

We therefore denote with the same symbol $\sigma_{k}$ the corresponding map
defined on ${\mathbb{C}}^{4}$ with values on ${\mathbb{C}}^{4}$. We have

$\sigma_{k}(z_{0},z_{1},z_{2},z_{3})=(p^{k}_{0}(z_{0},\underline{z}^{2}),z_{1}p^{k-1}_{1}(z_{0},\underline{z}^{2}),z_{2}p^{k-1}_{1}(z_{0},\underline{z}^{2}),z_{3}p^{k-1}_{1}(z_{0},\underline{z}^{2}))$

where $\underline{z}=z_{1}e_{1}+z_{2}e_{2}+z_{3}e_{3}$,
$\underline{z}^{2}=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}$ and $p_{0}^{k},p_{1}^{k-1}$
are such that

$(x+Iy)^{k}=p^{k}_{0}(x,y^{2})+Iyp^{k-1}_{1}(x,y^{2})\;.$

######  Remark 4.1.

Let $f=\mathcal{I}(F):U\to{\mathbb{H}}$ be a slice regular function such that
$f=f_{0}+f_{v}=f_{0}+f_{1}i+f_{2}j+f_{3}k$, with $f_{0},f_{1},f_{2},f_{3}$
slice preserving functions defined on $U$. With this representation, the
function $\sigma_{k}(F)$ corresponds to the $k$-th $\star$-power

$f^{\star
k}=p_{0}^{k}(f_{0},f_{v}^{\mathsf{s}})+p_{1}^{k-1}(f_{0},f_{v}^{\mathsf{s}})(f_{1}i+f_{2}j+f_{3}k)=p_{0}^{k}(f_{0},f_{v}^{\mathsf{s}})+p_{1}^{k-1}(f_{0},f_{v}^{\mathsf{s}})f_{v}\;.$

This representation was used in [AltavillaAMPA, Section 7] to study algebraic
properties of the $k$-th $\star$-power of a slice regular function.

For the convenience of what follows, let us define the following function

$N(z):=(z_{0}^{2}+z_{1}^{2}+z_{2}^{2}+z_{3}^{2})^{\frac{1}{2}}\;.$

Then, we have the following equalities:

$p_{0}^{k}(z_{0},\underline{z}^{2})=N(z)^{k}T_{k}(t)\;,\qquad
p_{1}^{k-1}(z_{0},\underline{z}^{2})=N(z)^{k-1}U_{k-1}(t)\;,$

where $t=z_{0}/N(z)$ and $T_{k}$ and $U_{k-1}$ denote the Chebyshev
polynomials of first kind and degree $k$ and of second kind and degree $k-1$,
respectively. We now give an explicit formula for the differential of
$\sigma_{k}$.

######  Lemma 4.1.

$D\sigma_{k}(z)$ is a function of $z_{0}^{2}$ and $\underline{z}^{2}$. More
precisely,

$D\sigma_{k}(z)=k^{2}N(z)^{2}[p_{1}^{k-1}(z_{0},\underline{z}^{2})]^{2}$

###### Proof.

Assume first that $N(z)\neq 0$. Before starting with the explicit
computations, we recall some well known formula. First of all the Pell
equation states that, for any $n\in\mathbb{N}$,

$T_{n}^{2}(x)+(1-x^{2})U_{n-1}^{2}(x)=1\;.$

Moreover, the derivation rules of the Chebyshev polynomials read as follows:

$\frac{dT_{n}(x)}{dx}=nU_{n-1}(x),\qquad\frac{dU_{n-1}}{dx}=\frac{xU_{n}(x)-(n+1)T_{n+1}(x)}{1-x^{2}}\;.$

Furthermore, recalling that $t=z_{0}/N(z)$, we have that

$\frac{\partial t}{\partial
z_{0}}=\frac{1}{N(z)}(1-t^{2}),\qquad\frac{\partial t}{\partial
z_{n}}=-t\frac{z_{i}}{N(z)^{2}}\;,$

for $n=1,2,3$.

Now, as

$\sigma_{k}(z_{0},z_{1},z_{2},z_{3})=(p^{k}_{0}(z_{0},\underline{z}^{2}),z_{1}p^{k-1}_{1}(z_{0},\underline{z}^{2}),z_{2}p^{k-1}_{1}(z_{0},\underline{z}^{2}),z_{3}p^{k-1}_{1}(z_{0},\underline{z}^{2}))$

and

$p_{0}^{k}(z_{0},\underline{z}^{2})=N(z)^{k}T_{k}(t),\qquad
p_{1}^{k-1}(z_{0},\underline{z}^{2})=N(z)^{k-1}U_{k-1}(t),$

thanks to the previous formulas we have

$\frac{\partial p^{k}_{0}(z_{0},\underline{z}^{2})}{\partial
z_{0}}=\frac{\partial\left(N(z)^{k}T_{k}(t)\right)}{\partial
z_{0}}=kN(z)^{k-1}[tT_{k}(t)+(1-t^{2})U_{k-1}(t)]=:A\;,$

for $n=1,2,3$,

$\frac{\partial p^{k}_{0}(z_{0},\underline{z}^{2})}{\partial
z_{n}}=\frac{\partial\left(N(z)^{k}T_{k}(t)\right)}{\partial
z_{n}}=kN(z)^{k-2}[T_{k}(t)-tU_{k-1}(t)]z_{n}=:Bz_{n}\;.$

Passing now to the other components, for $n,m=1,2,3$, we get

$\frac{\partial z_{m}p^{k-1}_{1}(z_{0},\underline{z}^{2})}{\partial
z_{0}}=\frac{\partial\left(z_{m}N(z)^{k-1}U_{k-1}(t)\right)}{\partial
z_{0}}=kN(z)^{k-2}[tU_{k-1}(t)-T_{k}(t)]z_{m}=:-Bz_{m}\;,$

and

$\displaystyle\frac{\partial
z_{m}p^{k-1}_{1}(z_{0},\underline{z}^{2})}{\partial z_{n}}$
$\displaystyle=\frac{\partial\left(z_{m}N(z)^{k-1}U_{k-1}(t)\right)}{\partial
z_{n}}$
$\displaystyle=\delta_{nm}N(z)^{k-1}U_{k-1}(t)+N(z)^{k-3}[(k-1)U_{k-1}(t)+\frac{t}{1-t^{2}}(kT_{k}(t)-tU_{k-1}(t))]z_{n}z_{m}$
$\displaystyle=:\delta_{nm}H+Cz_{n}z_{m}\;,$

where $\delta_{nm}$ denotes the Kronecker delta.

Thanks to these computations the Jacobian of $\sigma_{k}$ is equal to

$\displaystyle D\sigma_{k}(z)$
$\displaystyle=\det\begin{pmatrix}A&Bz_{1}&Bz_{2}&Bz_{3}\\\
-Bz_{1}&H+Cz_{1}^{2}&Cz_{1}z_{2}&Cz_{1}z_{3}\\\
-Bz_{2}&Cz_{1}z_{2}&H+Cz_{2}^{2}&Cz_{2}z_{3}\\\
-Bz_{3}&Cz_{1}z_{3}&Cz_{2}z_{3}&H+Cz_{3}^{2}\end{pmatrix}$
$\displaystyle=H^{2}[AH+(AC+B^{2})(z_{1}^{2}+z_{2}^{2}+z_{3}^{2})]$
$\displaystyle=H^{2}[AH+(AC+B^{2})N(z)^{2}(1-t^{2})]\;.$

We now analyze the term $(AC+B^{2})N(z)^{2}(1-t^{2})$. This can be written as
follows

$\displaystyle(AC+B^{2})N(z)^{2}(1-t^{2})=$ $\displaystyle
N(z)^{2}(1-t^{2})\\{kN(z)^{k-1}[tT_{k}(t)+(1-t^{2})U_{k-1}(t)]\times$
$\displaystyle\times
N(z)^{k-3}[(k-1)U_{k-1}(t)+\frac{t}{1-t^{2}}(kT_{k}(t)-tU_{k-1}(t))]+$
$\displaystyle+kN(z)^{k-2}[T_{k}(t)-tU_{k-1}(t)]^{2}$ $\displaystyle=$
$\displaystyle
kN(z)^{2(k-1)}(1-t^{2})\\{T_{k}(t)U_{k-1}(t)[t(k-1)-t\frac{t^{2}}{1-t^{2}}+tk-2tk]+$
$\displaystyle+T_{k}^{2}(t)[k\frac{t^{2}}{1-t^{2}}+k]+U_{k-1}^{2}(t[(1-t^{2})(k-1)-t^{2}+kt^{2}])\\}$
$\displaystyle=$ $\displaystyle
kN(z)^{2(k-1)}\\{kT_{k}^{2}(t)+(1-t^{2})(k-1)U_{k-1}^{2}(t)-tT_{k}(t)U_{k-1}(t)\\}\;.$

Therefore we have

$\displaystyle D\sigma_{k}(z)=$ $\displaystyle H^{2}[AH+(AD-
BC)N(z)^{2}(1-t^{2})]$ $\displaystyle=$
$\displaystyle[N(z)^{k-1}U_{k-1}(t)]^{2}\\{kN(z)^{k-1}[tT_{k}(t)+(1-t^{2})U_{k-1}(t)]\times$
$\displaystyle\times N(z)^{k-1}U_{k-1}(t)+$
$\displaystyle+kN(z)^{2(k-1)}\\{kT_{k}^{2}(t)+(1-t^{2})(k-1)U_{k-1}^{2}(t)-tT_{k}(t)U_{k-1}(t)\\}\\}$
$\displaystyle=$ $\displaystyle k^{2}N(z)^{2(k-1)}[N(z)^{k-1}U_{k-1}(t)]^{2}$
$\displaystyle=$ $\displaystyle
k^{2}(z_{0}^{2}+z_{1}^{2}+z_{2}^{2}+z_{3}^{2})[p_{1}^{k-1}(z_{0},\underline{z}^{2})]^{2}\;,$

where in the third equality we used the Pell equation. Hence we have proved
the equality whenever $N(z)\neq 0$. Now, as the resulting polynomial is of the
correct degree, this equality extends also to the set of points where
$N(z)=0$. ∎

We now want to analyze the zero set of $D\sigma_{k}$. We set

$Q^{k}(t)=\sum_{h=0}^{[k/2]}(-1)^{h}{k\choose
2h+1}t^{k-1-2h}=\mathsf{Im}((t+\imath)^{k})$

and we notice that
$Q^{k}(t)=p_{1}^{k-1}(t,1)=(t^{2}+1)^{k-1}U_{k-1}(\frac{t}{1+t^{2}})$.
Therefore $Q^{k}(t)=0$ if and only if $U_{k-1}(t)=0$. Now, the zeroes of the
Chebyshev polynomials are well known (in particular, $U_{k-1}(t)=0$ if and
only if $t=\cos(\frac{n}{k}\pi)$ with $n=1,\dots,k-1$), however to be self-
contained, we present here a proof of the fact that such roots are all simple.
We will later make use of techniques involved in the proof.

######  Lemma 4.2.

$Q^{k}(t)$ has all real and simple roots.

###### Proof.

By definition, for $t\in\mathbb{R}$

$Q^{k}(t)=\frac{(t+\imath)^{k}-(t-\imath)^{k}}{2\imath}$

so, $Q^{k}(t)=0$ if and only if

$\left(\frac{t-\imath}{t+\imath}\right)^{k}=1$

i.e. if and only if $(t-\imath)/(t+\imath)$ is a $k$-th root of unity. Now,
the map

$\mathfrak{C}(\zeta)=\frac{\zeta-\imath}{\zeta+\imath}$

is the Cayley transform which is a biholomorphism between the upper half-plane
and the unit disc; in particular, it maps the real line bijectively on the
unit circle minus the point $\\{1\\}$, a set which contains $k-1$ of the
$k$-th roots of unity. For $k$ odd, $Q^{k}(t)$ is a polynomial in $t^{2}$, for
$k$ even, $Q^{k}(t)/t$ is a polynomial in $t^{2}$; therefore, for $k$ odd,
$Q^{k}(t)$ has $(k-1)/2$ positive roots and $(k-1)/2$ negative roots, for $k$
even, $Q^{k}(t)$ has a zero root, $(k-2)/2$ positive roots and $(k-2)/2$
negative roots.

Therefore, $Q^{k}(t)$ has $k-1=\deg Q^{k}(t)$ real roots, which are then all
simple roots, and $[(k-1)/2]$ of them are positive roots. ∎

We now pass to prove that, under suitable hypotheses on the domain and range,
$\sigma_{k}$ is a covering map.

Let us consider the set of imaginary units in
${\mathbb{C}}\otimes{\mathbb{H}}$

$\mathcal{S}=\\{s=(0,\underline{z})\in{\mathbb{C}}^{4}\ :\
\underline{z}^{2}=1\\}\simeq\\{p+\sqrt{-1}q\in{\mathbb{C}}\otimes{\mathbb{H}}\
:\ (p+\sqrt{-1}q)^{2}=-1\\}.$

######  Remark 4.2.

To clarify the definition of $\mathcal{S}$, if
$p=\alpha_{1}i+p_{2}j+p_{3}k,q=q_{1}+q_{2}j+q_{3}k\in{\mathbb{H}}$, then
$(p+\sqrt{-1}q)^{2}=p^{2}-q^{2}+\sqrt{-1}(pq+qp)=-\|p\|^{2}+\|q\|^{2}+2\sqrt{-1}\langle
p,q\rangle$. Therefore, $p+\sqrt{-1}q\in\mathcal{S}$ if and only if

$\begin{cases}-\|p\|^{2}+\|q\|^{2}=1\\\ \langle p,q\rangle=0.\end{cases}$

Notice that, if $\mathsf{Im}(z)=0$ and $p\in{\mathbb{S}}$ (which corresponds
to $q=0$), then $z\otimes p\in\mathcal{S}$, while if $z\otimes
p\in\mathcal{S}$, then $p\in{\mathbb{S}}$ does not hold in general. In fact,
the element $i+j+\sqrt{-1}(\frac{1}{3}i-\frac{1}{3}j+\frac{\sqrt{7}}{3}k)$
belongs to $\mathcal{S}$, but in general
$\pi(i+j+\sqrt{-1}(\frac{1}{3}i-\frac{1}{3}j+\frac{\sqrt{7}}{3}k),I)\notin{\mathbb{S}}$:
if $I=i$, then
$(i+j+i(\frac{1}{3}i-\frac{1}{3}j+\frac{\sqrt{7}}{3}k))^{2}\neq-1$.

In analogy with the map $\pi:{\mathbb{C}}\times{\mathbb{S}}\to{\mathbb{H}}$,
we define

(3) $\rho:{\mathbb{C}}^{2}\times\mathcal{S}\to\mathbb{C}^{4}$

given by

$\rho((u_{0},u_{1}),s)=u_{0}e_{0}+u_{1}s\;.$

The restriction of $\rho$ to
$W^{\prime}=({\mathbb{C}}^{2}\setminus\\{u_{1}=0\\})\times\mathcal{S}$ is a
$2$-to-$1$ cover of its image $\Omega^{\prime}=\rho(W^{\prime})$, given by

$\Omega^{\prime}=\\{(z_{0},\underline{z})\in{\mathbb{C}}^{4}\ :\
\underline{z}^{2}\neq 0\\}\;.$

Given $(z_{0},\underline{z})\in\Omega^{\prime}$ we have that

$\rho^{-1}(z_{0},\underline{z})=\left\\{\left((z_{0},\sqrt{\underline{z}^{2}}),\frac{\underline{z}}{\sqrt{\underline{z}^{2}}}\right),\,\left((z_{0},-\sqrt{\underline{z}^{2}}),-\frac{\underline{z}}{\sqrt{\underline{z}^{2}}}\right)\right\\}.$

Moreover, if $(z_{0},\underline{z})\in{\mathbb{C}}^{4}$ and
$(w_{0},\underline{w})=\sigma_{k}((z_{0},\underline{z}))$, then

$\begin{cases}w_{0}=p_{0}^{k}(z_{0},\underline{z}^{2})\\\
\underline{w}=p_{1}^{k-1}(z_{0},\underline{z}^{2})\underline{z}\end{cases}$

therefore, if we put

$\Omega^{\prime}_{k}=\\{(z_{0},\underline{z})\in{\mathbb{C}}^{4}\ :\
\underline{z}^{2}\neq 0,\ p_{1}^{k-1}(z_{0},\underline{z}^{2})\neq 0\\}$

we have that $\sigma_{k}(\Omega^{\prime}_{k})\subseteq\Omega^{\prime}$.

We consider
$W_{k}^{\prime}=\rho^{-1}(\Omega^{\prime}_{k})\subset{\mathbb{C}}^{2}\times\mathcal{S}$
and the map $\mathfrak{s}_{k}:W^{\prime}_{k}\to W^{\prime}$ such that
$\rho\circ\mathfrak{s}_{k}=\sigma_{k}\circ\rho$ given by

$\mathfrak{s}_{k}((u_{0},u_{1}),s)=((p_{0}^{k}(u_{0},u_{1}^{2}),u_{1}p_{1}^{k}(u_{0},u_{1}^{2})),s)\;.$

$\begindc{\commdiag}[3]\obj(0,180)[A]{W_{k}^{\prime}}\obj(280,180)[B]{W^{\prime}}\obj(0,0)[C]{\Omega_{k}^{\prime}}\obj(280,0)[D]{\Omega^{\prime}}\mor{A}{B}{\mathfrak{s}_{k}}[\atleft,\solidarrow]\mor{A}{C}{\rho}[\atright,\solidarrow]\mor{C}{D}{\sigma_{k}}[\atright,\solidarrow]\mor{B}{D}{\rho}[\atleft,\solidarrow]\enddc$

Finally, let

$\Omega=\\{(z_{0},\underline{z})\in\Omega^{\prime}\ :\
z_{0}^{2}+\underline{z}^{2}\neq 0,\ z_{0}\neq 0\\}$

and set $\Omega_{k}=\Omega^{\prime}_{k}\cap\Omega$.

######  Remark 4.3.

Let $f:U\to{\mathbb{H}}$ be a slice regular function, then with the notation
introduced in [AltavillaAMPA], we have that, for any $I\in{\mathbb{S}}$,

  * •

$F(z)\in\Omega^{\prime}$ if and only if $f_{v}^{\mathsf{s}}(\pi(z,I))\neq 0$;

  * •

$F(z)\in\Omega_{k}^{\prime}$ if and only if $f_{v}^{\mathsf{s}}(\pi(z,I))\neq
0$ and $p_{1}^{k-1}(f_{0}(\pi(z,I)),f_{v}^{\mathsf{s}}(\pi(z,I)))\neq 0$ (see
[AltavillaAMPA, Section 7]);

  * •

$F(z)\in\Omega$ if and only if $f^{\mathsf{s}}(\pi(z,I))\neq 0$ and
$f_{0}(\pi(z,I))\neq 0$.

######  Proposition 4.3.

$\sigma_{k}$ is $k^{2}$-to-$1$ from $\Omega_{k}$ to $\Omega$.

###### Proof.

We consider $W=\rho^{-1}(\Omega)$ and $W_{k}=\rho^{-1}(\Omega_{k})$; if we
show that the map $\mathfrak{s}_{k}$ is $k^{2}$-to-$1$ from $W_{k}$ to $W$,
the thesis will follow, as $\mathfrak{s}_{k}$ acts as the identity on
$\mathcal{S}$, so the preimages of a point through $\mathfrak{s}_{k}$ are sent
to different values by $\rho$.

Given $((v_{0},v_{1}),s^{\prime})\in W$, the equation
$\mathfrak{s}_{k}((u_{0},u_{1}),s)=((v_{0},v_{1}),s^{\prime})$ is satisfied if
and only if $s=s^{\prime}$ and

$v_{0}=p_{0}^{k}(u_{0},u_{1}^{2})\qquad
v_{1}=u_{1}p_{1}^{k-1}(u_{0},u_{1}^{2})\;.$

These two are polynomial equations in the variables $(u_{0},u_{1})$ of (total)
degree $k$, so, by the affine Bezout theorem, the system they form has at most
$k^{2}$ solutions.

We set $t=u_{0}/u_{1}$, as in $W_{k}$ we have that $u_{1}\neq 0$, then

$p_{0}^{k}(t,1)=\frac{(t+\imath)^{k}+(t-\imath)^{k}}{2}\qquad
p_{1}^{k-1}(t,1)=\frac{(t+\imath)^{k}-(t-\imath)^{k}}{2\imath}$

so, as $v_{1}=u_{1}p_{1}^{k-1}(u_{0},u_{1})=u_{1}^{k}p^{k-1}_{1}(t,1)\neq 0$
in $W_{k}$,

$\lambda=\frac{v_{0}}{v_{1}}=\imath\frac{(t+\imath)^{k}+(t-\imath)^{k}}{(t+\imath)^{k}-(t-\imath)^{k}}=\imath\frac{1+\mathfrak{C}^{k}(t)}{1-\mathfrak{C}^{k}(t)}$

where $\mathfrak{C}$ is the previously introduced Cayley transform, which is
an automorphism of the Riemann sphere, sending $\infty$ to $1$. We have

(4)
$\mathfrak{C}^{k}(t)=\frac{\lambda-\imath}{\lambda+\imath}=\mathfrak{C}(\lambda)$

and we note that $\mathfrak{C}(\lambda)=0,\infty$ if and only if
$\lambda=\pm\imath$ if and only if $v_{0}=\pm\imath v_{1}$ if and only if
$v_{0}^{2}+\underline{v}^{2}=0$ which does not happen in $W$.

Therefore, for $((v_{0},v_{1}),s)\in W$, we have $k$ distinct solutions
$t_{1},\ldots,t_{k}$ to equation (4). If we fix $m\in\\{1,\ldots,k\\}$, then
$u_{0}=t_{m}u_{1}$ and

$v_{1}=u_{1}^{k}p_{1}^{k-1}(t_{m},1)\;.$

Hence, we obtain

$u_{1}^{k}=\frac{v_{1}}{p_{1}^{k-1}(t_{m},1)}$

which has $k$ different solutions $\omega_{m,1},\ldots,\omega_{m,k}$ for each
$m\in\\{1,\ldots,h\\}$. This gives a total of $k^{2}$ solutions of the form
$(t_{m}\omega_{m,n},\omega_{m,n})$ for $m,n\in\\{1,\ldots,k\\}$. ∎

We are now able to state our result.

######  Theorem 4.4.

The function $\sigma_{k}:\Omega_{k}\to\Omega$ is a covering map of degree
$k^{2}$.

###### Proof.

By Lemma 4.1, $\sigma_{k}$ is a local biholomorphism from $\Omega_{k}$ to
$\Omega$ and by Proposition 4.3 it has a constant number of preimages.
Therefore it is a covering map. ∎

In the last part of this section we give a brief description of the action of
$\sigma_{k}$ outside $\Omega_{k}$. We define
$X_{k}={\mathbb{C}}^{4}\setminus\Omega_{k}$ and
$X={\mathbb{C}}^{4}\setminus\Omega$. For $\alpha\in{\mathbb{C}}$ we set

$V_{\alpha}=\\{(z_{0},\underline{z})\in{\mathbb{C}}^{4}\ :\
z_{0}^{2}=\alpha\underline{z}^{2}\\}$

and

$V_{\infty}=\\{(z_{0},\underline{z})\in{\mathbb{C}}^{4}\ :\
\underline{z}^{2}=0\\}\;.$

Notice that if a slice function $f=\mathcal{I}(F)$ is such that $F$ takes
values in $V_{-1}$, then $f^{\mathsf{s}}\equiv 0$, i.e. $f$ is identically
zero or a zero divisor with respect to the $\star$-product (see [AltavillaLAA,
Section 2.4]).

Collecting everything, if $R_{k}$ denotes the set of positive roots of
$Q^{k}(t)$, then

$X_{k}=V_{-1}\cup V_{0}\cup V_{\infty}\cup\bigcup_{r\in
R_{k}}V_{r^{2}}\quad\textrm{and}\quad X=V_{-1}\cup V_{0}\cup V_{\infty}\;.$

We now describe the action of $\sigma_{k}$ where it is not a covering map.

######  Proposition 4.5.

Let $(z_{0},\underline{z})\in{\mathbb{C}}^{4}$. We have the following
relations:

  1. (1)

if $(z_{0},\underline{z})\in V_{-1}$, then $\sigma_{k}(z_{0},\underline{z})\in
V_{-1}$;

  2. (2)

if $(z_{0},\underline{z})\in V_{0}$ and $k$ is odd, then
$\sigma_{k}(z_{0},\underline{z})\in V_{0}$;

  3. (3)

if $(z_{0},\underline{z})\in V_{0}$ and $k$ is even, then
$\sigma_{k}(z_{0},\underline{z})\in V_{\infty}$;

  4. (4)

if $(z_{0},\underline{z})\in V_{\infty}$, then
$\sigma_{k}(z_{0},\underline{z})\in V_{\infty}$;

  5. (5)

if $(z_{0},\underline{z})\in V_{r^{2}}$, with $r\in R_{k}$, then
$\underline{\sigma_{k}(z_{0},\underline{z})}=0$ and hence
$\sigma_{k}(z_{0},\underline{z})\in V_{\infty}$.

###### Proof.

For point $(1)$ it sufficient to notice that if
$z_{0}^{2}+\underline{z}^{2}=0$, then
$\sigma_{2}(z_{0},\underline{z})=(z_{0}^{2}-\underline{z}^{2},2z_{0}\underline{z})$
and
$(z_{0}^{2}-\underline{z}^{2})^{2}+4z_{0}^{2}\underline{z}^{2}=(z_{0}^{2}+\underline{z}^{2})^{2}=0$.

For points $(2)$ and $(3)$ it is sufficient to recall symmetry properties for
Chebyshev polynomials:

$T_{n}(-x)=(-1)^{n}T_{n}(x)\;,\qquad U_{n}(-x)=(-1)^{n}U_{n}(x)\;.$

We pass now to point $(4)$: in this case, as $\underline{z}^{2}=0$, then

$\sigma_{k}(z_{0},\underline{z})=(z_{0}^{k}T_{k}(1),\underline{z}z^{k-1}U_{k-1}(1))=(z_{0}^{k},\underline{z}z^{k-1}U_{k-1}(1))\;,$

and hence $\underline{\sigma_{k}(z_{0},\underline{z})}^{2}=0$.

The last point $(5)$ is trivial as $R_{k}$ is the set of positive roots of
$Q^{k}(t)=(t^{2}+1)^{k-1}U_{k-1}(\frac{t}{1+t^{2}})$. ∎

######  Remark 4.4.

Notice that, if $(z_{0},\underline{z})\in V_{-1}$, then

$p_{0}^{k}(z_{0},\underline{z}^{2})=z_{0}^{k}\qquad
p_{1}^{k-1}(z_{0},\underline{z}^{2})=z_{0}^{k-1}\;.$

Therefore, $\sigma_{k}(z_{0},\underline{z})=z_{0}^{k-1}(z_{0},\underline{z})$
is a $k$-to-$1$ covering of $V_{-1}$ on itself. The same happens with
$V_{\infty}$.

In the next corollary we interpret Proposition 4.5 in terms of slice regular
functions.

######  Corollary 4.6.

Let $f=f_{0}+f_{1}i+f_{2}j+f_{3}k=f_{0}+f_{v}:U\to{\mathbb{H}}$ be a slice
function. Then we have the following relations:

  1. (1)

if $f^{\mathsf{s}}\equiv 0$, then $(f^{\star k})^{\mathsf{s}}\equiv 0$;

  2. (2)

if $f_{0}\equiv 0$ and $k$ is odd, then $(f^{\star k})_{0}\equiv 0$;

  3. (3)

if $f_{0}\equiv 0$ and $k$ is even, then $(f^{\star k})_{v}^{\mathsf{s}}\equiv
0$;

  4. (4)

if $f_{v}^{\mathsf{s}}\equiv 0$, then $(f^{\star k})_{v}^{\mathsf{s}}\equiv
0$;

  5. (5)

if $f_{0}^{2}=r^{2}f_{v}^{\mathsf{s}}$, with $r\in R_{k}$, then $(f^{\star
k})_{v}\equiv 0$ and hence $(f^{\star k})_{v}^{\mathsf{s}}\equiv 0$.

The proof of this last corollary is a direct application of Proposition 4.5
taking into account Remark 4.1, but indeed many of these points are silly
statements. However, notice that point $(5)$ confirms [AltavillaAMPA,
Proposition 7.4].

##  5\. Global $\star$-roots of a slice function

In this section we apply the results obtained at the end of the last section
for the $k$-th power in ${\mathbb{C}}\otimes{\mathbb{H}}$, to obtain results
for the $k$-th $\star$-root of a slice function and at the end we also provide
a couple of explicit examples. We recall Assumption 2.2 and we add the
following.

######  Assumption 5.1.

From now on the set ${\mathcal{U}}\subset{\mathbb{C}}$ will always be a simply
connected open domain.

This last assumption is consistent with the one used in [altavillaLOG] and in
[GPV2] where the authors call the resulting set $U$ basic domain.

Let $f:U\to{\mathbb{H}}$ be a slice function. We denote by
$F:\mathcal{U}\to{\mathbb{C}}\otimes{\mathbb{H}}$ its stem function. As in the
previous sections, with a small abuse of notation, we will consider elements
in ${\mathbb{C}}\otimes{\mathbb{H}}$ or in ${\mathbb{C}}^{4}$ as the same. So,
we will denote with the same symbol the function $F$ when its range is
${\mathbb{C}}^{4}$ instead of ${\mathbb{C}}\otimes{\mathbb{H}}$ (we recall
that this identification is explicitly given in Formula (2)).

######  Proposition 5.2.

If $F(\mathcal{U})\subseteq\Omega$, then there exist $k^{2}$ functions
$F_{1},\ldots,F_{k^{2}}:\mathcal{U}\to{\mathbb{C}}\otimes{\mathbb{H}}$ such
that $\sigma_{k}\circ F_{m}=F$ for $m=1,\ldots,k^{2}\;.$

###### Proof.

This is a direct consequence of Theorem 4.4 and the lifting property of
covering maps. ∎

We notice that, by the very definition of $\sigma_{k}$, if $F$ and $G$ are
stem functions defining two slice functions $f$ and $g$, respectively, then

$F=\sigma_{k}\circ G\Leftrightarrow f=(g)^{\star k}\;.$

######  Remark 5.1.

As $F$ is a stem function, then $F(\overline{z})=\overline{F(z)}$. Moreover,
by the definition of $\sigma$, we have
$\sigma_{k}(\overline{z})=\overline{\sigma_{k}(z)}$. It follows that, if
$\sigma_{k}\circ G=F$, then

$\sigma_{k}(G(\overline{z}))=\overline{\sigma_{k}(G(z))}=\sigma_{k}(\overline{G(z)})\;.$

Therefore, if $G$ is a solution of the equation $\sigma_{k}(G)=F$ on the
domain $\mathcal{U}$, then the function $\widehat{G}(z)=\overline{G(\bar{z})}$
is another solution on the same domain.

Thanks to the previous remark we are able to prove the following unexpected
result.

######  Theorem 5.3.

If $F(\mathcal{U})\subset\Omega$ and $\mathcal{U}\cap{\mathbb{R}}=\emptyset$,
then there exist $k^{2}$ slice functions $f_{m}:U\to{\mathbb{H}}$,
$m=1,\dots,k^{2}$, such that

$(f_{m})^{\star k}=f\;.$

###### Proof.

From Proposition 5.2 we know that there are $k^{2}$ functions
$G_{1},\ldots,G_{k^{2}}:\mathcal{U}\to{\mathbb{C}}\otimes{\mathbb{H}}$ such
that $\sigma_{k}\circ G_{m}=F$. Now, from Remark 5.1 for any $m=1,\dots,k^{2}$
there exists $n=1,\dots,k^{2}$, such that $\widehat{G}_{m}(z)=G_{n}(z)$, for
all $z\in\mathcal{U}$. Moreover, such $n$ is unique from Theorem 4.4.
Therefore, we can define a function
$\tau:\\{1,\dots,k^{2}\\}\to\\{1,\dots,k^{2}\\}$ as $\tau(m)=n$, where $n$ is
such that $\widehat{G}_{m}(z)=G_{n}(z)$ for all $z\in\mathcal{U}$. Hence, we
are able to define the following stem functions
$F_{\mu}:\mathcal{U}\to{\mathbb{C}}\otimes{\mathbb{H}}$ as

$F_{\mu}(z)=\begin{cases}G_{\mu}(z),&\mbox{if }z\in\mathcal{U}^{+}\;,\\\
G_{\tau(\mu)}(z),&\mbox{if }z\in\mathcal{U}^{-}.\end{cases}$

Then, the induced stem functions $f_{1},\dots,f_{k^{2}}$ are such that
$(f_{\mu})^{\star k}=f$ for any $\mu=1,\dots,k^{2}$. ∎

The case in which the domain of $f$ has real points follows the general
expectation. However, imposing such a condition on the domain implies greater
effort in the proof.

######  Theorem 5.4.

If $F(\mathcal{U})\subseteq\Omega$ and
$\mathcal{U}\cap{\mathbb{R}}\neq\emptyset$, then there exist $k$ slice
functions $f_{m}:U\to{\mathbb{H}}$, $m=1,\dots,k$, such that

$(f_{m})^{\star k}=f\;.$

###### Proof.

Let ${\mathbb{R}}^{4}$ denote the set of real points in ${\mathbb{C}}^{4}$,
corresponding to the set $\\{p+\sqrt{-1}\cdot 0\ :\
p\in{\mathbb{H}}\\}\subset{\mathbb{C}}\otimes{\mathbb{H}}$, then, on it the
projection
$\pi:{\mathbb{C}}\otimes{\mathbb{H}}\times{\mathbb{S}}\to{\mathbb{H}}$
restricts to the map $h:{\mathbb{R}}^{4}\to{\mathbb{H}}$ given by
$h(x_{0},x_{1},x_{2},x_{3})=x_{0}+x_{1}i+x_{2}j+x_{3}k$. We notice that
$\sigma_{k}({\mathbb{R}}^{4})\subseteq{\mathbb{R}}^{4}$ for all $k$ and

(5) $h\circ({\sigma_{k}}_{|{\mathbb{R}}^{4}})=h^{k}\;,$

where in the right hand side we mean the $k$-th power in the algebra
${\mathbb{H}}$.

As we know from Theorem 3.4, the map $g(q)=q^{k}$ is a covering map from
${\mathbb{H}}\setminus S(g)$ to ${\mathbb{H}}\setminus C(g)$. An easy
computation shows that ${\mathbb{H}}\setminus
C(g)=h({\mathbb{R}}^{4}\cap\Omega)$ and ${\mathbb{H}}\setminus
S(g)=h({\mathbb{R}}^{4}\cap\Omega_{k})$, where $\Omega$ and $\Omega_{k}$ are
the sets defined in the previous section.

Let now $x^{0}$ be a point in $\mathcal{U}\cap{\mathbb{R}}$, then
$F(x^{0})\in{\mathbb{R}}^{4}$. Consider $q_{0}=h(F(x^{0}))$. Then, by
hypothesis, $q_{0}\in{\mathbb{H}}\setminus C(g)$, so there exist $k$ preimages
of $q_{0}$ via $g$ in ${\mathbb{H}}\setminus S(g)$. Denoting them by
$q_{1},\ldots,q_{k}$, by (5), we have $\sigma_{k}(h^{-1}(q_{m}))=F(x^{0})$.

We consider the lift $F_{m}$ of $F$ through $\sigma_{k}$ such that
$F_{m}(x^{0})=h^{-1}(q_{m})$. We will show that $F_{m}$ is a stem function.

Consider $G(z)=\overline{F_{m}(\bar{z})}$. As
$\sigma_{k}(\bar{z})=\overline{\sigma_{k}(z)}$, we have

$\sigma_{k}(G(z))=\sigma_{k}(\overline{F_{m}(\bar{z})})=\overline{\sigma_{k}(F_{m}(\bar{z}))}\;.$

Now,

$\sigma_{k}(F_{m}(\bar{z}))=F(\bar{z})=\overline{F(z)}$

because $F$ is a stem function; so

$\sigma_{k}(G(z))=F(z)$

hence $G(z)$ is one of the $k^{2}$ lifts of $F$ via $\sigma_{k}$. However,
$G(x^{0})=h^{-1}(q_{m})$, so $G$ must coincide with $F_{m}$. So

$F_{m}(\bar{z})=\overline{F_{m}(z)}$

which means that $F_{m}$ is a stem function. Let $f_{m}$ be the corresponding
slice function, then, by construction $(f_{m})^{\star k}=f$. ∎

As an obvious corollary, we get the corresponding statement for slice regular
functions, as the map $\sigma_{k}$ is holomorphic.

######  Corollary 5.5.

A slice regular function $f$ satisfying the hypothesis of the previous theorem
has $k$ slice regular $k$-th $\star$-roots.

In the hypothesis of the previous theorem, given $F_{m}$ as above and given
$\xi\in{\mathbb{C}}$ a primitive $k$-th rooth of unity, then
$F_{ak+m}=\xi^{a}F_{m}$ is again a lift of $F$: $\sigma_{k}$ is
$k$-homogeneous, so

$\sigma_{k}(\xi^{a}F_{m}(z))=\xi^{ak}\sigma_{k}(F_{m}(z))=\sigma_{k}(F_{m}(z))=F(z)\;.$

If $F_{ak+m}(z)=F_{bk+n}(z)$, then

$\xi^{a}F_{m}(z)=\xi^{b}F_{n}(z)$

again, for $z=x^{0}$, we obtain that $F_{m}(x^{0}),\
F_{n}(x^{0})\in{\mathbb{R}}^{4}$, therefore $\xi^{a-b}\in{\mathbb{R}}$. This
is possible if $a=b$ or if $2(a-b)=k$ (and $k$ is even); in the first case, it
follows that also $n=m$. So, for $k$ odd, the $F_{ak+m}$ are all the $k^{2}$
lifts given by Proposition 5.2.

Moreover,

$\overline{F_{ak+m}(z)}=\overline{\xi^{a}F_{m}(z)}=\xi^{-a}\overline{F_{m}(z)}=\xi^{-a}F_{m}(\overline{z})=\xi^{-2a}F_{m}(\overline{z})\;.$

Again, if $k$ is odd, the only possibility is $a=0$, so that we have $k$
families of $k$ functions each such that $G(\bar{z})=\xi^{c}\overline{G(z)}$
and each family has a different value of $c\in\\{0,\ldots,k-1\\}$.

We now want to show a couple of explicit examples suggesting the general form
of such $k$-th $\star$-roots. For the convenience of what follows we need to
introduce a special slice regular function.

######  Definition 5.6.

Let $U\subset{\mathbb{H}}$ be an open domain such that
$U\cap{\mathbb{R}}=\emptyset$. We define the slice regular function
$\mathcal{J}:U\to{\mathbb{H}}$ as $J(q)=\frac{q_{v}}{\|q_{v}\|}$.

The function ${\mathcal{J}}$ is slice preserving and it is induced by the
function

$J(z)=\begin{cases}\sqrt{-1},&z\in{\mathcal{U}}^{+},\\\
-\sqrt{-1},&z\in{\mathcal{U}}^{-},\end{cases}$

corresponding to the locally constant complex curve

$z\mapsto\begin{cases}(\imath,0,0,0),&z\in{\mathcal{U}}^{+},\\\
(-\imath,0,0,0),&z\in{\mathcal{U}}^{-}.\end{cases}$

From the function ${\mathcal{J}}$ it is possible to construct idempotent
functions and zero-divisors for the $\star$-product, the prototypes of which
are the functions $\ell_{+},\ell_{-}:U\to{\mathbb{H}}$ defined as

$\ell_{+}(q)=\frac{1-{\mathcal{J}}(q)i}{2}\;,\qquad\ell_{-}(q)=\frac{1+{\mathcal{J}}(q)i}{2}\;.$

Thanks to the so called _Peirce decomposition_ , any slice regular function
$f$ defined on a domain without real points $U$ can be written as
$f=f_{+}\star\ell_{+}+f_{-}\star\ell_{-}$, where
$f_{+},f_{-}:U\to{\mathbb{H}}$ are slice regular functions and not zero
divisors. The case in which both $f_{+}$ and $f_{-}$ are regular polynomials
is studied in some details in [AltavillaMATHZ] and corresponds to the family
of slice polynomial functions.

######  Example 5.1.

Let ${\mathcal{U}}$ be a simply connected domain in ${\mathbb{C}}$ and
consider the regular polynomial $f:U\to{\mathbb{H}}$ defined as
$f(q)=q^{3}+3q^{2}i-3q-i=(q+i)^{\star 3}$. This function preserves the slice
${\mathbb{C}}_{i}$, meaning that
$f(U\cap{\mathbb{C}}_{i})\subset{\mathbb{C}}_{i}$. It corresponds to the
complex curve $F:{\mathcal{U}}\to{\mathbb{C}}^{4}$ given by
$F(z)=(z^{3}-3z,3z^{2}-1,0,0)$. A trivial $3$-rd $\star$-root of $f$ is given
by $g:U\to{\mathbb{H}}$ defined as $g_{0}(q)=q+i$ corresponding to the complex
curve $G_{0}:{\mathcal{U}}\to{\mathbb{H}}$ defined as $G_{0}(z)=(z,1,0,0)$.
Moreover, if
$\eta_{1}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,\eta_{2}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$
are the two non-trivial cubic root of the unity in ${\mathbb{C}}_{i}$, then
$g_{1}=g_{0}\eta_{1}$ and $g_{2}=g_{0}\eta_{2}$ are other cubic $\star$-roots
of $f$, corresponding to the complex curves
$G_{1}(z)=(-z/2-\sqrt{3}/2,q\sqrt{3}/2-1/2,0,0)$ and
$G_{2}(z)=(-z/2+\sqrt{3}/2,-q\sqrt{3}/2-1/2,0,0)$, respectively. If
$U\cap{\mathbb{R}}=\emptyset$ it is possible to compute the remaining roots as
follows

$\begin{matrix}g_{0}^{\prime}(q)=\left(-\frac{1}{2}+\mathcal{J}(q)\frac{\sqrt{3}}{2}\right)g_{0}(q)\;,&&g_{0}^{\prime\prime}(q)=\left(-\frac{1}{2}-\mathcal{J}(q)\frac{\sqrt{3}}{2}\right)g_{0}(q)\;,\\\
g_{1}^{\prime}(q)=\left(-\frac{1}{2}+\mathcal{J}(q)\frac{\sqrt{3}}{2}\right)g_{1}(q)\;,&&g_{1}^{\prime\prime}(q)=\left(-\frac{1}{2}-\mathcal{J}(q)\frac{\sqrt{3}}{2}\right)g_{1}(q)\;,\\\
g_{2}^{\prime}(q)=\left(-\frac{1}{2}+\mathcal{J}(q)\frac{\sqrt{3}}{2}\right)g_{2}(q)\;,&&g_{2}^{\prime\prime}(q)=\left(-\frac{1}{2}-\mathcal{J}(q)\frac{\sqrt{3}}{2}\right)g_{2}(q)\;,\end{matrix}$

corresponding to the complex curves

$\begin{matrix}G_{0}^{\prime}(z)=\begin{cases}\eta_{1}G_{0}(z)&z\in{\mathcal{U}}^{+}\\\
\eta_{2}G_{0}(z)&z\in{\mathcal{U}}^{-}\end{cases}\;,&&G_{0}^{\prime\prime}(z)=\begin{cases}\eta_{2}G_{0}(z)&z\in{\mathcal{U}}^{+}\\\
\eta_{1}G_{0}(z)&z\in{\mathcal{U}}^{-}\end{cases}\;,\\\
G_{1}^{\prime}(z)=\begin{cases}\eta_{1}G_{1}(z)&z\in{\mathcal{U}}^{+}\\\
\eta_{2}G_{1}(z)&z\in{\mathcal{U}}^{-}\end{cases}\;,&&G_{1}^{\prime\prime}(z)=\begin{cases}\eta_{2}G_{1}(z)&z\in{\mathcal{U}}^{+}\\\
\eta_{1}G_{1}(z)&z\in{\mathcal{U}}^{-}\end{cases}\;,\\\
G_{2}^{\prime}(z)=\begin{cases}\eta_{1}G_{2}(z)&z\in{\mathcal{U}}^{+}\\\
\eta_{2}G_{2}(z)&z\in{\mathcal{U}}^{-}\end{cases}\;,&&G_{2}^{\prime\prime}(z)=\begin{cases}\eta_{2}G_{2}(z)&z\in{\mathcal{U}}^{+}\\\
\eta_{1}G_{2}(z)&z\in{\mathcal{U}}^{-}\end{cases}\;,\end{matrix}$

respectively. Notice that all these functions are slice polynomial functions.
For instance we can write the first one as

$g_{0}^{\prime}(q)=g_{0}(q)\eta_{1}\star\ell_{+}+g_{0}(q)\eta_{2}\star\ell_{-}\;.$

The case $k$ even is more involved. For the moment we only give an example of
what happens for $k=2$. In the next section we will try to explain this
phenomenon from an algebraic point of view.

######  Example 5.2.

Let us consider the function
$F:{\mathbb{C}}\to{\mathbb{C}}^{4}(\simeq{\mathbb{C}}\otimes{\mathbb{H}})$
defined by $F(z)=(z^{2},-z,-z,1)$. This function intersects $V_{\infty}$ for
$z=\pm\imath/\sqrt{2}$ (it corresponds to the slice regular function
$f(q)=(q-i)\star(q-j)=q^{2}-q(i+j)+k$. We have four solutions of the equation
$\sigma_{2}(G)=F$ around $z=\imath$, namely

$G_{1}(z)=\frac{\sqrt{2}}{2}(\imath,\imath z,\imath z,-\imath)$
$G_{2}(z)=\frac{\sqrt{2}}{2}(-\imath,-\imath z,-\imath z,\imath)$

and, fixing a determination of the square root of $2z^{2}+1$ on some open set,

$G_{3}(z)=\left(\frac{\sqrt{4z^{2}+2}}{2},\frac{-z}{\sqrt{4z^{2}+2}},\frac{-z}{\sqrt{4z^{2}+2}},\frac{1}{\sqrt{4z^{2}+2}}\right)$
$G_{4}(z)=\left(-\frac{\sqrt{4z^{2}+2}}{2},\frac{z}{\sqrt{4z^{2}+2}},\frac{z}{\sqrt{4z^{2}+2}},-\frac{1}{\sqrt{4z^{2}+2}}\right)\;.$

These lifts “collide” when $z$ approaches $\pm\imath$: if we pick the square
root such that $\sqrt{-1/2}=\imath/\sqrt{2}$, then
$G_{1}(\imath)=G_{3}(\imath)$ and $G_{2}(\imath)=G_{4}(\imath)$. Moreover, for
$z=\pm\imath/\sqrt{2}$, $f(\pm\imath/\sqrt{2})\in\Omega^{\prime}$, where we do
not have $4$ square roots, but only $2$.

Notice that since $G_{m}(\overline{z})=\overline{G_{m}(z)}$, for $m=3,4$, then
$G_{3}$ and $G_{4}$ correspond to stem functions but they are not defined on
${\mathbb{C}}$: we can extend $G_{3}$ and $G_{4}$ to any simply connected open
domain $\mathcal{V}$ which does not contain $\pm\imath/\sqrt{2}$. If
$V=\pi(\mathcal{V})$, then the slice regular functions
$g_{3}:V\to{\mathbb{H}}$ and $g_{4}:V\to{\mathbb{H}}$ induced by $G_{3}$ and
$G_{4}$ can be written as

$g_{3}(q)=\sqrt{q^{2}+\frac{1}{2}}-\frac{q}{\sqrt{4q^{2}+2}}(i+j)+\frac{k}{\sqrt{4q^{2}+2}},\quad
g_{4}(q)=-g_{3}(q),$

respectively, where the existence of the square root of $4q^{2}+2$ is
implicitly implied by our assumptions.

The phenomenon of reduced solutions (from $4$ to $2$) is not due to the fact
that some preimages collide when $z$ approaches $\pm\imath/\sqrt{2}$ (this is
what happens at the points where $D\sigma_{k}$ is not a local diffeomorphism),
but rather to the fact that some preimages go to infinity, “disappearing”.
This is a manifestation of the fact that $\sigma_{k}$ is not proper from
${\mathbb{C}}^{4}\to{\mathbb{C}}^{4}$, but from
${\mathbb{C}}^{4}\setminus\Omega^{\prime}_{k}$ to
${\mathbb{C}}^{4}\setminus\Omega^{\prime}$.

Lastly, as in the previous example, notice that
$G_{1}(\bar{z})=\overline{G_{2}(z)}$, hence, if $\mathcal{V}$ has no real
points, then it is possible to construct the other two square roots of $f$ as
explained in the proof of Theorem 5.3. In particular, the resulting slice
regular functions are slice polynomial functions:

$g_{1}(q)=-R(q)\star\ell_{+}+R(q)\star\ell_{-}\;,\quad g_{2}(q)=-g_{1}(q)\;,$

where $R(q)=q(1+k)-i+j$.

##  6\. Covering automorphisms and monodromy

In this section we reinterpret the results given in the previous one in terms
of covering automorphisms, obtaining more refined statements and the monodromy
of the maps $\mathfrak{s}_{k}$ and $\sigma_{k}$. This will allow us to give a
more precise description of what happen in explicit cases.

Let us begin by considering the following commutative diagram

$\begindc{\commdiag}[3]\obj(0,180)[A]{W_{k}}\obj(280,180)[B]{W}\obj(0,0)[C]{\Omega_{k}}\obj(280,0)[D]{\Omega}\mor{A}{B}{\mathfrak{s}_{k}}[\atleft,\solidarrow]\mor{A}{C}{\rho}[\atright,\solidarrow]\mor{C}{D}{\sigma_{k}}[\atright,\solidarrow]\mor{B}{D}{\rho}[\atleft,\solidarrow]\enddc$

where $\rho$ is the map defined in Formula (3) and each arrow is a covering
map: more precisely, $\rho$ is a $2$-to-$1$ covering map and $\sigma_{k}$,
$\mathfrak{s}_{k}$ are $k^{2}$-to-$1$ covering maps.

Given a covering map $\pi:X\to Y$, we will denote by
$\operatorname{Aut}_{\pi}$ the group of automorphisms $f$ of $X$ such that
$\pi\circ f=\pi$.

$\begindc{\commdiag}[3]\obj(0,180)[A]{X}\obj(280,180)[B]{X}\obj(140,0)[C]{Y}\mor{A}{B}{f}[\atleft,\solidarrow]\mor{A}{C}{\pi}[\atright,\solidarrow]\mor{B}{C}{\pi}[\atleft,\solidarrow]\enddc$

######  Remark 6.1.

Let
$\Gamma:{\mathbb{C}}^{2}\times\mathcal{S}\to{\mathbb{C}}^{2}\times\mathcal{S}$
be given by $\Gamma((u_{0},u_{1}),s)=((u_{0},-u_{1}),-s)$. Then

$\operatorname{Aut}_{\rho}=\\{1,\ \Gamma\\}\;,$

where we denoted by $1$ the identity automorphism.

We define two actions of ${\mathbb{C}}$ on ${\mathbb{C}}^{2}$, as the
multiplication by the following two matrices

$z\mapsto z\mathrm{Id}\qquad z\mapsto A_{z}$

where $z\in{\mathbb{C}}$, $\mathrm{Id}$ is the $2\times 2$ identity matrix and

$A_{z}=\begin{pmatrix}{\sf Re}z&-\mathsf{Im}z\\\ \mathsf{Im}z&{\sf
Re}z\end{pmatrix}\,.$

We denote by $U_{k}$ the group of complex $k$-th roots of unity. From now on
we will only consider maps from ${\mathbb{C}}^{2}\times\mathcal{S}$ to itself
of the form
$((u_{0},u_{1}),s)\mapsto((g_{0}(u_{0},u_{1}),g_{1}(u_{0},u_{1})),s)$, so we
will consistently omit the component relative to $\mathcal{S}$.

######  Remark 6.2.

Let $f:U\to{\mathbb{H}}$ be a slice regular function. Then if
$f=f_{0}+f_{1}i+f_{2}j+f_{3}k=f_{0}+f_{v}$, we have that the corresponding
complex curve is given by
$F(z)=(f_{0}(z),f_{1}(z),f_{2}(z),f_{3}(z))=(f_{0},\underline{f_{v}})$. Assume
that $F\in\Omega$, then its lifts via $\rho$ is given by

$\left((f_{0},\sqrt{f_{v}^{\mathsf{s}}}),\frac{f_{v}}{\sqrt{f_{v}^{\mathsf{s}}}}\right)\in{\mathbb{C}}^{2}\times\mathcal{S}\;.$

In the next proposition we compute the monodromy of $\mathfrak{s}_{k}$ in the
case in which $k$ is odd.

######  Proposition 6.1.

Let $k$ be odd. Then

$\operatorname{Aut}_{\mathfrak{s}_{k}}=\\{\xi A_{\eta}\ :\ \xi,\eta\in
U_{k}\\}\;,$

which is isomorphic to the group $\mathbb{Z}_{k}\times\mathbb{Z}_{k}$.

###### Proof.

The map $(\xi,\eta)\mapsto\xi A_{\eta}$ is a group homomorphism from
$U_{k}\times U_{k}$ to $M_{2,2}({\mathbb{C}})$, indeed,

$(\xi
A_{\eta})(\xi^{\prime}A_{\eta^{\prime}})=(\xi\xi^{\prime})A_{\eta\eta^{\prime}}\;.$

Now, its kernel is given by the condition $\xi A_{\eta}=\mathrm{Id}$, which
implies (taking the determinant)

$\xi^{2}|\eta|^{2}=1$

i.e. $\xi^{2}=1$, i.e. $\xi=\pm 1$, but as $k$ is odd, then $\xi=1$
(otherwise, if $\xi=-1$, then $1=\xi^{k}=(-1)^{k}=-1$). Moreover, taking the
trace, we obtain $2{\sf Re}\eta=2$, i.e. ${\sf Re}\eta=1$, which implies
$\eta=1$. Hence the kernel is trivial and the map is injective.

Finally, as $\mathfrak{s}_{k}$ is $k$-homogeneous in the ${\mathbb{C}}^{2}$
component, it is obvious that
$\xi\mathrm{Id}\in\operatorname{Aut}_{\mathfrak{s}_{k}}$ for all $\xi\in
U_{k}$; on the other hand,

$\mathfrak{s}_{k}((u_{0},u_{1}),s)=((p_{0}^{k}(u_{0},u_{1}^{2}),u_{1}p_{1}^{k-1}(u_{0},u_{1}^{2})),s)$

and by definition
$(x+\sqrt{-1}y)^{k}=p_{0}^{k}(x,y^{2})+\sqrt{-1}yp_{1}^{k-1}(x,y^{2})$, so,
from the fact that

$\displaystyle(({\sf
Re}(\eta)x-\mathsf{Im}(\eta)y)+\sqrt{-1}(\mathsf{Im}(\eta)x+{\sf
Re}(\eta)y))^{k}$ $\displaystyle=(({\sf
Re}(\eta)+\sqrt{-1}\mathsf{Im}(\eta))(x+\sqrt{-1}y))^{k}$
$\displaystyle=((x+\sqrt{-1}y))^{k}$

for $\eta\in U_{k}$, it follows that
$A_{\eta}\in\operatorname{Aut}_{\mathfrak{s}_{k}}$.

Given that $U_{k}\times U_{k}$ is generated by $(\xi,1)$ and $(1,\eta)$, we
obtain that $(\xi,\eta)\to\xi A_{\eta}$ is an injective homomorphism from
$U_{k}\times U_{k}$ to $\operatorname{Aut}_{\mathfrak{s}_{k}}$. However, the
latter has at most $k^{2}$ elements (as $\mathfrak{s}_{k}$ has degree
$k^{2}$), so the map is an isomorphism. ∎

We now give an abstract example generalizing Example 5.1. Before providing it,
we recall from [AltavillaPAMS, Proposition 2.10] that two slice regular
functions $f=f_{0}+f_{v},g=g_{0}+g_{v}$ commute with respect to the
$\star$-product if and only if there exist two slice preserving functions
$\alpha$ and $\beta$ not both identically zero such that $\alpha f_{v}+\beta
g_{v}\equiv 0$.

######  Example 6.1.

Let ${\mathcal{U}}$ be a simply connected domain such that
${\mathcal{U}}\cap{\mathbb{R}}=\emptyset$ and $f:U\to{\mathbb{H}}$ be a slice
regular function such that $f=f_{0}+f_{v}$. Then clearly $f$ is a cubic
$\star$-root of $f^{\star 3}$. By following the procedure of Example 5.1 and
the presentation of $Aut_{\mathfrak{s}_{k}}$ given in Proposition 6.1 we
compute the other $8$ roots as follows.

$\begin{array}[]{rrr}f\;,&f\star\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\frac{f_{v}}{\sqrt{f_{v}^{s}}}\right)\;,&f\star\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}\frac{f_{v}}{\sqrt{f_{v}^{s}}}\right)\;,\\\
\left(-\frac{1}{2}+{\mathcal{J}}\frac{\sqrt{3}}{2}\right)f\;,&\left(-\frac{1}{2}+{\mathcal{J}}\frac{\sqrt{3}}{2}\right)f\star\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\frac{f_{v}}{\sqrt{f_{v}^{s}}}\right)\;,&\left(-\frac{1}{2}+{\mathcal{J}}\frac{\sqrt{3}}{2}\right)f\star\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}\frac{f_{v}}{\sqrt{f_{v}^{s}}}\right)\;,\\\
\left(-\frac{1}{2}-{\mathcal{J}}\frac{\sqrt{3}}{2}\right)f\;,&\left(-\frac{1}{2}-{\mathcal{J}}\frac{\sqrt{3}}{2}\right)f\star\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\frac{f_{v}}{\sqrt{f_{v}^{s}}}\right)\;,&\left(-\frac{1}{2}-{\mathcal{J}}\frac{\sqrt{3}}{2}\right)f\star\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}\frac{f_{v}}{\sqrt{f_{v}^{s}}}\right)\;,\\\
\end{array}$

where ${\mathcal{J}}={\mathcal{J}}(q)$ is the slice regular function given in
Definition 5.6. Notice that, thanks to the previous consideration on the
commutativity of the $\star$-product, the factors in all the functions
presented above commute. All these functions can be computed by letting the
group $Aut_{\mathfrak{s}_{k}}$ presented in Proposition 6.1 act on the element

$\left((f_{0},\sqrt{f_{v}^{\mathsf{s}}}),\frac{f_{v}}{\sqrt{f_{v}^{\mathsf{s}}}}\right)\in{\mathbb{C}}^{2}\times\mathcal{S}\;.$

In detail, using the notation of Proposition 6.1, the action of $\xi$
corresponds to the (left) multiplication by a complex number and the action of
$A_{\eta}$ corresponds to the right $\star$-multiplication.

We now turn our attention to the case when $k$ is even which, as we saw in
Example 5.2, is more subtle.

######  Remark 6.3.

If $k$ is even, when computing the kernel of $(\xi,\eta)\mapsto\xi A_{\eta}$,
we obtain

$\\{(1,1),(-1,-1)\\}$

so the map is not injective and its image is just an index $2$ subgroup of
$\operatorname{Aut}_{\mathfrak{s}_{k}}$.

We notice that if $\lambda,\mu\in U_{2k}$ are primitive $2k$-th roots of $1$,
then $\lambda^{k}=\mu^{k}=-1$, so

$(\lambda A_{\mu})^{k}=\lambda^{k}A_{\mu^{k}}=-1(-\mathrm{Id})=\mathrm{Id}\;.$

Thanks to this observation we are able to compute the monodromy of
$\mathfrak{s}_{k}$ when $k$ is even.

######  Proposition 6.2.

Let $k$ be even and $\lambda,\mu\in U_{2k}$ be primitive roots. If we set
$S=\lambda A_{\mu}$, then

$\operatorname{Aut}_{\mathfrak{s}_{k}}=\\{\xi A_{\eta}S^{\delta}\ :\
(\xi,\eta,\delta)\in U_{k}\times U_{k}\times\\{0,1\\}\\}\;,$

which is isomorphic ${\mathbb{Z}}_{k}\times{\mathbb{Z}}_{k}$.

###### Proof.

The image of the map $(\xi,\eta,\delta)\mapsto\xi A_{\eta}S^{\delta}$ is
clearly contained in $\operatorname{Aut}_{\mathfrak{s}_{k}}$.

This map from $U_{k}\times U_{k}\times{\mathbb{Z}}_{2}$ to
$\operatorname{Aut}_{\mathfrak{s}_{2k}}$ is not a group homomorphism, however,
we can factor it through the map from $U_{2k}\times
U_{2k}\to\operatorname{Aut}_{\mathfrak{s}_{2k}}$, by sending (injectively)
$(\xi,\eta,\delta)$ to $(\xi\lambda^{\delta},\eta\mu^{\delta})$, so the former
is a $2$-to-$1$ map.

Again, by cardinality, we conclude that

$(\xi,\eta,\delta)\mapsto\xi A_{\eta}S^{\delta}$

is surjective from $U_{k}\times U_{k}\times{\mathbb{Z}}_{2}$ onto
$\operatorname{Aut}_{\mathfrak{s}_{k}}$.

If now $\xi\in U_{k},\ \mu\in U_{2k}$ are primitive roots of unity (of orders
$k$ and $2k$ respectively) such that $\mu^{2}=\xi$, then
$\operatorname{Aut}_{\mathfrak{s}_{k}}$ is generated by

$\xi,A_{\xi},\ \mu A_{\mu}\;.$

Moreover, $\xi^{k}=A_{\xi}^{k}=1$, $(\mu A_{\mu})^{2}=\xi A_{\xi}$, so the
group $\operatorname{Aut}_{\mathfrak{s}_{k}}$ can be presented as

$\langle a,b,c\ |\ a^{k},\ b^{k},\ c^{2}(ab)^{-1},\ [a,b],\ [a,c],\
[b,c]\rangle$

which is isomorphic ${\mathbb{Z}}_{k}\times{\mathbb{Z}}_{k}$ as in the case
$k$ odd. ∎

######  Example 6.2.

Going back to Example 5.2, we notice that $U_{2}={\mathbb{Z}}_{2}$ and, in
fact, $g_{1}=-g_{2}$; moreover, the primitive roots of $1$ of order $4$ are
$\pm\iota$, so that

$g_{3}(q)=\mathcal{J}(q)g_{1}(q)\star\mathcal{J}(g_{1}(q))$

and, accordingly, $g_{4}=-g_{3}$.

From the theory of covering maps, as the group of automorphism of
$\mathfrak{s}_{k}$ acts transitively on the fibers of $\mathfrak{s}_{k}$, we
obtain the following result.

######  Corollary 6.3.

The covering map $\mathfrak{s}_{k}$ is normal (or regular, or Galois).

Knowing the automorphisms of $\mathfrak{s}_{k}$, we now pass to analyze
$\sigma_{k}$.

Given $w\in\Omega$, the fiber $\rho^{-1}(w)$ is composed of two points $v$ and
$\Gamma(v)$ in $W$, where $\Gamma$ is the function defined in Remark 6.1;
$\rho$ is then a bijection from $\mathfrak{s}_{k}^{-1}(v)$ to
$\sigma_{k}^{-1}(w)$ and from $\mathfrak{s}_{k}^{-1}(\Gamma(v))$ to the same
set. As $\mathfrak{s}_{k}$ is normal, the elements of
$\operatorname{Aut}_{\mathfrak{s}_{k}}$ are uniquely defined by their action
on $\mathfrak{s}_{k}^{-1}(v)$, so that each of them corresponds to a different
element of the group of permutation of $\mathfrak{s}^{-1}(v)$ denoted by
$\mathrm{Perm}(\mathfrak{s}_{k}^{-1}(v))$, which is isomorphic, via $\rho$, to
$\mathrm{Perm}(\sigma_{k}^{-1}(w))$.

Given a neighborhood $U$ of $w$ in $\Omega$, which is uniformly covered by
$\sigma_{k}$, we can extend each deck transformation of $\sigma_{k}^{-1}(w)$
to an automorphism of $\sigma_{k}^{-1}(U)$ (which is diffeomorphic to the
disjoint union of $k^{2}$ copies of $U$). We can suppose (up to shrinking)
that $U$ is also uniformly covered by $\rho$ and that, in turn, each connected
component of $\rho^{-1}(U)$ is uniformly covered by $\mathfrak{s}_{k}$.

The restriction of the automorphisms in
$\operatorname{Aut}_{\mathfrak{s}_{k}}$ to the connected components of
$\mathfrak{s}_{k}^{-1}(\rho^{-1}(U))$ correspond, via $\rho$, to the extension
of the deck transformations of $\sigma_{k}^{-1}(w)$ to $\sigma_{k}^{-1}(U)$.

######  Proposition 6.4.

The group $\operatorname{Aut}_{\sigma_{k}}$ is isomorphic to
${\mathbb{Z}}_{k}$. In particular, $\sigma_{k}$ is never regular.

###### Proof.

By the discussion above, given $F\in\operatorname{Aut}_{\sigma_{k}}$, there is
$G\in\operatorname{Aut}_{\mathfrak{s}_{k}}$ such that

$F\circ\rho=\rho\circ G\;.$

Therefore, $\rho\circ G=\rho\circ G\circ\Gamma$ and, as $G$ is an
automorphism, this happens if and only if $G\circ\Gamma=\Gamma\circ G$. Given
$(u,s)\in{\mathbb{C}}^{2}\times\mathcal{S}$, $G(u,s)=(\xi A_{\eta}u,s)$, so we
want that

$(\xi QA_{\eta}u,-s)=(\xi A_{\eta}Qu,-s)$

for all $(u,s)$, where

$Q=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}\,.$

In conclusion, we need that $A_{\eta}Q=QA_{\eta}$. Writing $\eta=a+\imath b$,
we have

$\begin{pmatrix}a&-b\\\ -b&-a\end{pmatrix}=\begin{pmatrix}a&b\\\
b&-a\end{pmatrix}$

which holds if and only if $b=-b$, i.e. $b=0$.

For $k$ odd, this implies $\eta=1$, for $k$ even, we obtain $\eta=\pm 1$. In
both cases, we obtain that $G$ is of the form

$G(u,s)=G(\xi u,s)$

so the corresponding $F\in\operatorname{Aut}_{\sigma_{k}}$ is of the form
$F(z)=\xi z$, with $z\in\Omega_{k}\subset{\mathbb{C}}^{4}$ and $\xi\in U_{k}$,
which obviously form, under composition, the group ${\mathbb{Z}}_{k}$.

Given that $\sigma_{k}$ is a covering map of degree $k^{2}$, this implies that
it is not a regular covering. ∎

We already exploited the existence of the automorphisms $z\mapsto\xi z$ in the
discussion after Theorem 5.4, to obtain, in the odd case, all the roots from
the intrinsic ones. However, if our slice function corresponds to a stem
function defined on a simply connected open domain
$\mathcal{U}\subseteq{\mathbb{C}}$, we completely describe all the $k^{2}$
roots. In the last part of this paper we provide algebraic techniques to
compute all the $k$-th $\star$-roots of a given slice functions, starting from
one of them.

###  6.1. Permutations of $k$-th roots

Suppose now that we are given a slice function $f:U\to{\mathbb{H}}$ and its
stem function
$F:\mathcal{U}\to{\mathbb{C}}^{4}\simeq{\mathbb{C}}\otimes{\mathbb{H}}$, with
$\mathcal{U}$ simply connected and $F(\mathcal{U})\subseteq\Omega$.

By Proposition 5.2, the set

$\mathcal{G}=\\{G:\mathcal{U}\to\Omega_{k}\ :\ \sigma_{k}\circ G=F\\}$

contains $k^{2}$ elements. By the properties of covering automorphisms we have
that $\operatorname{Aut}_{\sigma_{k}}$ acts on $\mathcal{G}$ by
(post-)composition, partitioning it in $k$ orbits with $k$ elements each.
Explicitly, given $\xi\in U_{k}$, the map $G\mapsto\xi G$ is a permutation of
$\mathcal{G}$ without fixed points (unless $\xi=1$).

######  Proposition 6.5.

Let $G\in\mathcal{G}$ and fix $H:\mathcal{U}\to W_{k}$ such that $\rho\circ
H=G$. Then for each $\tau\in\operatorname{Aut}_{\mathfrak{s}_{k}}$ the
function

$G_{\tau}=\rho\circ\tau\circ H$

belongs to $\mathcal{G}$.

###### Proof.

We have that $\sigma_{k}\circ G_{\tau}=\sigma_{k}\circ\rho\circ\tau\circ H$.
We know that $\sigma_{k}\circ\rho=\rho\circ\mathfrak{s}_{k}$ and
$\mathfrak{s}_{k}\circ\tau=\mathfrak{s}_{k}$, so

$\sigma_{k}\circ G_{\tau}=\rho\circ\mathfrak{s}_{k}\circ\tau\circ
H=\rho\circ\mathfrak{s}_{k}\circ H=\sigma_{k}\circ\rho\circ H=\sigma_{k}\circ
G\;.$

Therefore, $G_{\tau}\in\mathcal{G}$. ∎

With the construction from this last Proposition, we recover all the $k^{2}$
functions in $\mathcal{G}$ from one of them. However, the previous proposition
does not define an action of $\operatorname{Aut}_{\mathfrak{s}_{k}}$ on
$\mathcal{G}$, as $G_{\tau}$ depends on how we choose $H$. In order to obtain
a result in this direction, we now reinterpret some computations made in the
previous section with the language of cover automorphisms (see Remark 5.1).

######  Definition 6.6.

Given $G:\mathcal{U}\to{\mathbb{C}}^{4}$, we define
$TG:\mathcal{U}\to{\mathbb{C}}^{4}$ as $TG(z)=\overline{G(\bar{z})}$.

The previous definition reinterprets in the language of curves in
${\mathbb{C}}^{4}$ what previously was defined in the context of
${\mathbb{C}}\otimes{\mathbb{H}}$ (see Remark 5.1).

######  Lemma 6.7.

We have that

  1. (1)

$TTG=G$;

  2. (2)

if $\xi\in U_{k}$, $T(\xi G)=\bar{\xi}TG$;

  3. (3)

if $G\in\mathcal{G}$, then $TG\in\mathcal{G}$;

  4. (4)

the subgroup of $\mathrm{Perm}\mathcal{G}$ generated by
$\operatorname{Aut}_{\sigma_{k}}$ and $T$ is isomorphic to the dihedral group.

###### Proof.

The first and second points are obvious.

Given $G\in\mathcal{G}$, $\sigma_{k}(G(z))=F(z)$ so $T(\sigma_{k}\circ
G)=TF=F$ (because $F$ is the stem function of a slice function). Moreover,
$\sigma_{k}$ has polynomial components with real coefficients, so

$\sigma_{k}TG(z)=\sigma_{k}(\overline{G(\bar{z})})=\overline{\sigma_{k}(G(\bar{z}))}=T\sigma_{k}\circ
G(z)=F(z)\;,$

which implies $TG\in\mathcal{G}$.

$\operatorname{Aut}_{\sigma_{k}}\cong U_{k}\cong{\mathbb{Z}}_{k}$ acts on
$\mathcal{G}$ by scalar multiplication. Given $\xi$ a primitive $k$-th rooth
of unity, we have that the group generated by $\xi$ and $T$ has the following
presentation

$\langle\xi,\ T\ |\ \xi^{k},T^{2},\xi T\xi=T\rangle$

(once we notice that $\bar{\xi}=1/\xi$), which is the standard presentation of
the dihedral group. ∎

The last item of the previous Lemma gives an action of the dihedral group on
$\mathcal{G}$.

In the next theorem we show, in the case $k$ odd, how to recover all the stem
functions in $\mathcal{G}$ starting from one of them.

######  Theorem 6.8.

If $k$ is odd and $F$ is a fixed point for $T$ (i.e. if it is a stem
function), then there exist $k$ functions $G:\mathcal{U}\to{\mathbb{C}}^{4}$
such that

$\sigma_{k}\circ G=F\qquad TG=G\;.$

###### Proof.

If $k$ is odd, there is always at least one element of $\mathcal{G}$ such that
$TG=G$, because $T$ is an involution. If we consider the elements $\tau$ of
$\operatorname{Aut}_{\mathfrak{s}_{k}}$ induced by the matrices $A_{\eta}$,
$\eta\in U_{k}$, then the functions $G_{\tau}$ obtained thanks to Proposition
6.5 are all fixed points of $T$: $G_{\tau}=\rho\circ\tau\circ H$ and
$H(z)=(u(z),s(z))$, so $\tau\circ H(z)=(A_{\eta}u(z),s(z))$, which means that

$TG_{\tau}=\overline{\rho(A_{\eta}u(\bar{z}),s(\bar{z}))}=\rho(A_{\eta}\overline{u(z)},\overline{s(\bar{z})})$

because $A_{\eta}$ is a real matrix, and, as $TG=G$,
$\overline{u(\bar{z})}=u(z)$ and $\overline{s(\bar{z})}=s(z)$.

Therefore, $k$ elements of $\mathcal{G}$ are fixed points for $T$. ∎

Notice that, in the previous theorem, we are not assuming here that
$\mathcal{U}$ contains any real points.

As in the previous section, the result for $k$ even needs some additional
effort: let $F^{\prime}:\mathcal{U}\to{\mathbb{C}}^{2}\times\mathcal{S}$ be
such that $\rho\circ F^{\prime}=F$; define

$\mathcal{G}^{\prime}=\\{G^{\prime}:\mathcal{U}\to{\mathbb{C}}^{2}\times\mathcal{S}\
:\ \mathfrak{s}_{k}\circ G^{\prime}=F^{\prime}\\}$

and define $TG^{\prime}(z)=\overline{G^{\prime}(\bar{z})}$, where, if
$z=((u_{0},u_{1}),s),then$
$\overline{z}=\overline{((u_{0},u_{1}),s)}=((\bar{u}_{0},\bar{u}_{1}),\bar{s})$.

If $F$ is a stem function, then $TF^{\prime}=F^{\prime}$.

######  Lemma 6.9.

We have that

  1. (1)

$TTG^{\prime}=G^{\prime}$

  2. (2)

if $G^{\prime}\in\mathcal{G}^{\prime}$, then
$TG^{\prime}\in\mathcal{G}^{\prime}$

  3. (3)

for each $G^{\prime}\in\mathcal{G^{\prime}}$, there exists
$\tau_{G^{\prime}}\in\operatorname{Aut}_{\mathfrak{s}_{k}}$ such that
$TG^{\prime}=\tau_{G^{\prime}}\circ G^{\prime}$

  4. (4)

$\tau_{G^{\prime}}$ is always of the form $(u,s)\mapsto(\xi u,s)$.

###### Proof.

The first and the second points are analogous to the ones in Lemma 6.7.

The third point follows from the uniqueness of the lifts through a covering
map.

For the fourth, we note that, if $\tau(u,s)=(A_{\eta}u,s)$, then $T(\tau\circ
G^{\prime})=\tau\circ TG^{\prime}$. By the first point,
$TTG^{\prime}=G^{\prime}$, so, if $TG^{\prime}=\tau\circ G^{\prime}$, we have
that

$TTG^{\prime}=T(\tau\circ G^{\prime})=\tau\circ TG^{\prime}=\tau\circ\tau\circ
G^{\prime}$

i.e. $\tau\circ\tau$ is the identity, which implies $A_{\eta}=\pm I$ (so, it
is of the form $(u,s)\mapsto(\xi u,s)$. ∎

######  Remark 6.4.

Given $\eta\in U_{k}$, let
$\tau_{\eta}\in\operatorname{Aut}_{\mathfrak{s}_{k}}$ be of the form
$\tau_{\eta}(u,s)=(\eta u,s)$, then

$T(\tau_{\eta}\circ
G^{\prime})=\tau_{\bar{\eta}}TG^{\prime}=\tau_{\bar{\eta}}\circ\tau_{G^{\prime}}\circ
G^{\prime}=\tau_{\bar{\eta}^{2}}\circ\tau_{G^{\prime}}\circ\tau_{\eta}\circ
G^{\prime}\;.$

Therefore, if $G^{\prime\prime}=\tau_{\eta}\circ G^{\prime}$, then

$\tau_{G^{\prime\prime}}=\tau_{\bar{\eta}^{2}}\circ\tau_{G^{\prime}}\;.$

On the other hand, if $\tau(u,z)=(A_{\eta}u,s)$ and
$G^{\prime\prime}=\tau\circ G^{\prime}$, then
$\tau_{G^{\prime\prime}}=\tau_{G^{\prime}}$, as shown in the end of the proof
of Lemma 6.9.

Therefore, given $G^{\prime}\in\mathcal{G}^{\prime}$,
$\tau_{G^{\prime}}(u,s)=(\xi u,s)$ with $\xi\in U_{k}$; now, we can always
find $\tau$ in $\operatorname{Aut}_{\mathfrak{s}_{k}}$ such that
$\tau\circ\tau=\tau_{\bar{\xi}}$, so that, setting $G^{\prime\prime}=\tau\circ
G^{\prime}$, we have

$\tau_{G^{\prime\prime}}=1\;.$

So, we have produced an element of $\mathcal{G}^{\prime}$ such that
$TG^{\prime\prime}=G^{\prime\prime}$; we also know that, for all $\eta\in
U_{k}$, given $\tau(u,s)=(A_{\eta}u,s)$, $\tau\circ G^{\prime\prime}$ is again
a fixed point for $T$.

By composing with $\rho$ and considering also the previous Theorem, we have
proved the following

######  Theorem 6.10.

If $F$ is a fixed point for $T$ (i.e. if it is a stem function), then there
exist $k$ functions $G:\mathcal{U}\to{\mathbb{C}}^{4}$ such that

$\sigma_{k}\circ G=F\qquad TG=G\;.$

## References