# Quaternionic Regularity via Analytic Functional Calculus

Florian-Horia Vasilescu
Department of Mathematics, University of Lille,
59655 Villeneuve d’Ascq, France
e-mail<EMAIL_ADDRESS>
###### Abstract

Denoting by ${\mathbb{M}}$ the complexification of the quaternionic algebra
${\mathbb{H}}$, we characterize the family of those ${\mathbb{M}}$-valued
functions, defined on subsets of ${\mathbb{H}}$, whose values are actually
quaternions, using an intrinsic approach. In particular, we show that the
slice quaternionic regularity can be characterized via a Cauchy type transform
acting on the space of analytic ${\mathbb{M}}$-valued stem functions.

Keywords: quaternionic valued functions; analytic functional calculus; slice
regularity

Mathematics Subject Classification 2010: 47A10; 30G35; 30A05; 47A60

##  1 Introduction

The quaternions form a unital non commutative division algebra, with numerous
applications in mathematics and physics. In mathematics, one of the most
important investigation in the quaternionic context has been to find a
convenient manner to express the ”analyticity“ of functions depending on
quaternions. Among the pioneer contributions in this direction one should
mention the works [9] and [5]. Concerning the connections with physics, the
foundations of the quaternion quantum mechanics can be found in the work [4].

In 2007, a concept of slice regularity for functions of one quaternionic
variable was introduced in [6], leading to a large development sythesized in
[2] (see also [8], [7], [1] etc.).

Unlike in [6], in the paper [12] the regularity of a quaternionic-valued
function was investigated via the analytic functional calculus acting on
quaternions. This was done by considering the algebra of quaternions as a real
subalgebra of the complex algebra of $2\times 2$ matrices with complex
entries. This well known matrix representation allowed us to view the
quaternions as linear operators on a complex space, and thus commuting with
the complex numbers. Moreover, each quaternion was regarded as a normal
operator, having a spectrum which was used to define various compatible
functional calculi, including the analytic one. Specifically, a ”quaternionic
regular function“ was obtained by a pointwise construction of the analytic
functional calculus with stem functions on a conjugate symmetric open set $U$
in the complex plane, applied to quaternions whose spectra was in $U$, via the
matrix version of Cauchy’s formula, with no need of slice derivatives. In
fact, all of the ”regular functions“, regarded as quaternionic Cauchy
transforms of stem functions, happened to have some unexpected multiplicative
properties, and several properties of slice regular functions (see [2]) was
recaptured.

In the present paper we treat similar problems to those from the first part of
[12], in the context of the abstract Hamilton’s algebra, embedding the real
$C^{*}$-algebra of quaternions into its complexification, organized as a
complex $C^{*}$-algebra. Using elements of spectral theory, we succed to
considerably simplify the proof the main corresponding results from [12].
Nevertheless, the second part of [12], dealing with real and quaternionic
operators, is not covered by the present text. Unlike in [12], the actual
arguments are not only much simpler but the framework is intrisic, that is, it
does not depend of any representation of Hamilton’s algebra.

Let us briefly describe the contents of this work. The next chapter is
dedicated to some preliminaries, including Hamilton’s algebra of quaternions,
denoted by ${\mathbb{H}}$, and its complexification ${\mathbb{M}}$, endowed
with a unique $C^{*}$-algebra structure. The class of ${\mathbb{M}}$-valued
slice regular functions is also mentioned.

In the third chapter, the spectrum of a quaternion as an element of the
algebra ${\mathbb{M}}$ is introduced (see Remark 1), and the slice regularity
of an associated ${\mathbb{M}}$-valued Cauchy kernel is proved. As the
spectrum of a quaternion consists of at most two points, the structure of the
associated spectral projections is also exhibited, for later use.

The representation of a quaternion in terms of a resolution of identity, as a
particular case of the concept of scalar operator (see [3], Part III), allows
us the define a functional calculus with arbitrary ${\mathbb{M}}$-valued
functions for each quaternion, naturally extended to sets of quaternions in a
pointwise manner (see Definition 4). In addition, the function given by
formula (7) is ${\mathbb{H}}$-valued if and only if it is constructed by using
a stem function, via Theorem 2, leading to a general functional calculus with
arbitrary functions (see Theorem 3).

A quaternionic Cauchy transform is introduced in the fifth chapter, which can
be defined for all analytic ${\mathbb{M}}$-valued functions, but especially of
interest when working with analytic stem functions.

In the last chapter the analytic functional calculus for quaternionic
functions is obtained (see Theorem 5), as a particular case of the general
functional calculus given by Theorem 3. We also recapture one of the main
results from [12], showing that the slice regularity of a quaterninic function
is equivalent with its property of being the Cauchy transform of a stem
function (Theorem 6). The main ingredients of this result are the
representation formula (14) and a similar result from [2] (Lemma 4.3.8).

##  2 Preliminaries

###  2.1 Hamilton’s Algebra

For the sake of completeness, and to fix the notation and terminology, we
start this discussion with some well known facts. Abstract Hamilton’s algebra
${\mathbb{H}}$ is the four-dimensional ${\mathbb{R}}$-algebra with unit $1$,
generated by the ”imaginary units“ $\\{\bf{j,k,l}\\}$, which satisfy

${\bf jk=-kj=l,\,kl=-lk=j,\,lj=-jl=k,\,jj=kk=ll}=-1.$

We may assume that ${\mathbb{H}}\supset{\mathbb{R}}$ identifying every number
$x\in{\mathbb{R}}$ with the element $x1\in{\mathbb{H}}$.

The algebra ${\mathbb{H}}$ has a natural multiplicative norm given by

$\|{\bf x}\|=\sqrt{x_{0}^{2}+x_{1}^{2}+x_{2}^{2}+x_{0}^{2}},\,\,{\bf
x}=x_{0}+x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l},\,\,x_{0},x_{1},x_{2},x_{3}\in{\mathbb{R}},$

and a natural involution

${\mathbb{H}}\ni{\bf x}=x_{0}+x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l}\mapsto{\bf x}^{*}=x_{0}-x_{1}{\bf j}-x_{2}{\bf k}-x_{3}{\bf
l}\in{\mathbb{H}}.$

Every element ${\bf x}\in{\mathbb{H}}\setminus\\{0\\}$ is invertible, and
${\bf x}^{-1}=\|{\bf x}\|^{-2}{\bf x}^{*}$.

For an arbitrary quaternion ${\bf x}=x_{0}+x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l},\,\,x_{0},x_{1},x_{2},x_{3}\in{\mathbb{R}}$, we set $\Re{\bf x}=x_{0}=({\bf
x}+{\bf x}^{*})/2$, and $\Im{\bf x}=x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l}=({\bf x}-{\bf x}^{*})/2$, that is, the real and the imaginary part of ${\bf
x}$, respectively.

Using one possible equivalent definition, a real $C^{*}$-algebra is a real
Banach $*$-algebra $A$ satisfying the $C^{*}$-identity $\|a^{*}a\|=\|a\|^{2}$
for all $a\in A$, also having the property $\|a^{*}a\|\leq\|a^{*}a+b^{*}b\|$
for all $a,b\in A$ (see for instance [11]). It is clear that the algebra
${\mathbb{H}}$ is a real $C^{*}$-algebra.

###  2.2 The Associated Complex $C^{*}$-Algebra

The ”imaginary units“ $\bf j,k,l$ of the algebra ${\mathbb{H}}$ will be
considered independent of the imaginary unit $i$ of the complex plane
${\mathbb{C}}$. Specifically, we construct the complexification
${\mathbb{C}}\otimes_{\mathbb{R}}{\mathbb{H}}$ of the ${\mathbb{R}}$-vector
space ${\mathbb{H}}$ (see also [7]), which will be identified with the direct
sum ${\mathbb{M}}={\mathbb{H}}+i{\mathbb{H}}$, with the the natural
multiplicative structure given by

$({\bf x}_{1}+i{\bf x}_{2})({\bf y}_{1}+i{\bf y}_{2})=({\bf x}_{1}{\bf
y}_{1}-{\bf x}_{2}{\bf x}_{2})+i({\bf x}_{1}{\bf y}_{2}+{\bf x}_{2}{\bf
y}_{1}),\,\,{\bf x}_{1},{\bf x}_{2},{\bf y}_{1},{\bf y}_{2}\in{\mathbb{H}},$

Of course, the algebra ${\mathbb{M}}$ contains the complex field
${\mathbb{C}}$, that is, every complex number $z=s+it,\,s,t\in{\mathbb{R}}$ is
identified with the element $s1+it1\in{\mathbb{M}}$. In this way,
${\mathbb{M}}$ becomes an associative complex algebra, with unit $1$ and
involution $({\bf x}_{1}+i{\bf x}_{2})^{*}={\bf x}_{1}^{*}-i{\bf x}_{2}^{*}$,
where ${\bf x}_{1},{\bf x}_{2}\in{\mathbb{H}}$ are arbitrary, making
${\mathbb{M}}$ an involutive algebra. Moreover, in the algebra ${\mathbb{M}}$,
the elements of ${\mathbb{H}}$ commute with all complex numbers.

In the algebra ${\mathbb{M}}$ there also exists a natural conjugation given by
$\bar{\bf a}={\bf b}-i{\bf c}$, where ${\bf a}={\bf b}+i{\bf c}$ is arbitrary
in ${\mathbb{M}}$, with ${\bf b},{\bf c}\in{\mathbb{H}}$ (see also [7]). Note
that $\overline{\bf a+b}=\bar{\bf a}+\bar{\bf b}$, and $\overline{\bf
ab}=\bar{\bf a}\bar{\bf b}$, in particular $\overline{r\bf a}=r\bar{\bf a}$
for all ${\bf a},{\bf b}\in{\mathbb{M}}$, and $r\in{\mathbb{R}}$. Moreover,
$\bar{{\bf a}}={\bf a}$ if and only if ${\bf a}\in{\mathbb{H}}$, which is a
useful characterization of the elements of ${\mathbb{H}}$ among those of
${\mathbb{M}}$.

Using some results from [10] (see also [11]), the algebra ${\mathbb{M}}$ may
be endowed with a unique $C^{*}$-algebra structure, containing the algebra
${\mathbb{H}}$ as a real $C^{*}$-algebra, via the natural enbedding.

In our particular case, we can apply a more direct procedure, via a standard
matricial representation (as in [12]). Namely, we have the following.

###### Theorem 1

The complex algebra ${\mathbb{M}}$ has a unique $C^{*}$-algebra structure such
that ${\mathbb{H}}$ is a real $C^{*}$-subalgebra of ${\mathbb{M}}$.

Sketch of proof. There exists a $*$-isomorphism between the involutive algebra
${\mathbb{M}}$ and the algebra ${\mathbb{M}}_{2}$ consisting of all $2\times
2$-matrices with complex entries, which is a complex $C^{*}$-algebra. Then the
norm of ${\mathbb{M}}_{2}$ induces a norm on ${\mathbb{M}}$, making it a
$C^{*}$-subalgebra. We omit the details.

###  2.3 Slice Regular Functions

There exists a large literature dedicated to a concept of ”slice regularity“,
which is a form of holomorphy in the context of quaternions (see for instance
[2] and the works quoted within).

For ${\mathbb{M}}$-valued functions defined on subsets of ${\mathbb{H}}$, the
concept of slice regularity (see [2]) is defined as follows.

Let ${\mathbb{S}}=\\{\mathfrak{s}=x_{1}{\bf j}+x_{2}{\bf k}+x_{3}{\bf
l};x_{1},x_{2},x_{3}\in{\mathbb{R}},x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\\}$, that
is, the unit sphere of purely imaginary quaternions. It is clear that
$\mathfrak{s}^{*}=-\mathfrak{s}$, and so
$\mathfrak{s}^{2}=-1,\,\mathfrak{s}^{-1}=-\mathfrak{s}$, and
$\|\mathfrak{s}\|=1$ for all $\mathfrak{s}\in{\mathbb{S}}$.

Let also $\Omega\in{\mathbb{H}}$ be an open set, and let
$F:\Omega\mapsto{\mathbb{M}}$ be a differentiable function. In the spirit of
[2], we say that $F$ is right slice regular on $\Omega$ if for all
$\mathfrak{s}\in{\mathbb{S}}$,

$\bar{\partial}_{\mathfrak{s}}F(x+y\mathfrak{s}):=\frac{1}{2}\left(\frac{\partial}{\partial
x}+R_{\mathfrak{s}}\frac{\partial}{\partial y}\right)F(x+y\mathfrak{s})=0,$

on the set $\Omega\cap({\mathbb{R}}+{\mathbb{R}}\mathfrak{s})$, where
$R_{\mathfrak{s}}$ is the right multiplication of the elements of
${\mathbb{M}}$ by $\mathfrak{s}$.

Note that, unlike in [6], we use the right slice regularity rather than the
left one because of a reason to be later seen. Nevertheless, a left slice
regularity can also be defined via the left multiplication of the elements of
${\mathbb{M}}$ by elements from ${\mathbb{S}}$. In what follows, the right
slice regularity will be simply called slice regularity.

We are particularly interested by slice regularity of ${\mathbb{H}}$-valued
functions, but the concept is valid for ${\mathbb{M}}$-valued functions and
plays an important role in our discussion.

###### Example 1

(1) The convergent series of the form $\sum_{k\geq 0}a_{k}{\bf q}^{k}$, on
quaternionic balls $\\{{\bf q}\in{\mathbb{H}};\|{\bf q}\|<r\\}$, with $r>0$
and $a_{k}\in{\mathbb{H}}$ for all $k\geq 0$, are ${\mathbb{H}}$-valued slice
regular on their domain of definition. In fact, if actually
$a_{k}\in{\mathbb{M}}$, such functions are ${\mathbb{M}}$-valued slice regular
on their domain of definition.

(2) An important example of slice regular ${\mathbb{M}}$-valued function will
be further given by Example 2.

##  3 Spectrum of a Quaternion

In the complex algebra ${\mathbb{M}}$ we have a natural concept of spectrum,
which can be easily described in the case of quaternions. In fact, this
spectrum coincides with that one introduced in [12] (see also [2]).

###### Remark 1

(1) As each quaternion commutes in ${\mathbb{M}}$ with every complex number,
we have the identities

$(\lambda-{\bf x}^{*})(\lambda-{\bf x})=(\lambda-{\bf x})(\lambda-{\bf
x}^{*})=\lambda^{2}-\lambda({\bf x}+{\bf x}^{*})+\|{\bf
x}\|^{2}\in{\mathbb{C}},$

for all $\lambda\in{\mathbb{C}}$ and ${\bf x}\in{\mathbb{H}}$. Therefore, the
element $\lambda-{\bf x}\in{\mathbb{M}}$ is invertible if and only if the
complex number $\lambda^{2}-2\lambda\Re{\bf x}+\|{\bf x}\|^{2}$ is nonnull,
and in that case

$(\lambda-{\bf x})^{-1}=\frac{1}{\lambda^{2}-2\lambda\Re{\bf x}+\|{\bf
x}\|^{2}}(\lambda-{\bf x}^{*}).$

Hence, the element $\lambda-{\bf x}\in{\mathbb{M}}$ is not invertible if and
only if $\lambda=\Re{\bf x}\pm i\|\Im{\bf x}\|$. In this way, the spectrum of
a quaternion ${\bf x}\in{\mathbb{H}}$ is given by the equality $\sigma({\bf
x})=\\{s_{\pm}(\bf x)\\}$, where $s_{\pm}(\bf x)=\Re{\bf x}\pm i\|\Im{\bf
x}\|$ are the eigenvalues of $\bf x$.

(2) As usually, the resolvent set $\rho({\bf x})$ of a quaternion ${\bf
x}\in{\mathbb{H}}$ is the set ${\mathbb{C}}\setminus\sigma({\bf x})$, while
the function

$\rho({\bf x})\ni\lambda\mapsto(\lambda-{\bf x})^{-1}\in{\mathbb{M}}$

is the resolvent $($function$)$ of ${\bf x}$, which is a ${\mathbb{M}}$-valued
analytic function on $\rho({\bf x})$.

(3) Note that two quaternions ${\bf x},{\bf y}\in{\mathbb{H}}$ have the same
spectrum if and only if $\Re{\bf x}=\Re{\bf y}$ and $\|\Im{\bf x}\|=\|\Im{\bf
y}\|$.

(4) As before, let ${\mathbb{S}}$ be the unit sphere of purely imaginary
quaternions. It is clear that every quaternion ${\bf
q}\in{\mathbb{H}}\setminus{\mathbb{R}}$ can be written as ${\bf
q}=x+y\mathfrak{s}$, where $x,y$ are real numbers, with $x=\Re{\bf q}$,
$y=\pm\|\Im{\bf q}\|$, and $\mathfrak{s}=\pm\Im{\bf q}/\|\Im{\bf
q}\|\in{\mathbb{S}}$. Anyway, we always have $\sigma({\bf q})=\\{x\pm iy\\}$,
because $\Im{\bf q}=y\mathfrak{s}$. Note that, for fixed real numbers $x,y$,
the spectrum of $\bf q$ does not depend on $\mathfrak{s}$. Thus, for every
$\lambda=u+iv\in{\mathbb{C}}$ with $u,v\in{\mathbb{R}}$, we have
$\sigma(u+v\mathfrak{s})=\\{\lambda,\bar{\lambda}\\}$.

(5) Fixing an element $\mathfrak{s}\in{\mathbb{S}}$, we define an isometric
${\mathbb{R}}$-linear map from the complex plane ${\mathbb{C}}$ into the
algebra ${\mathbb{H}}$, say $\tau_{\mathfrak{s}}$, given by
$\tau_{\mathfrak{s}}(u+iv)=u+v\mathfrak{s},\,u,v\in{\mathbb{R}}$. For every
subset $A\in{\mathbb{C}}$, we put

$A_{\mathfrak{s}}=\\{x+y\mathfrak{s};x,y\in{\mathbb{R}},x+iy\in
A\\}=\tau_{\mathfrak{s}}(A).$ (1)

Note that, if $A$ is open in ${\mathbb{C}}$, then $A_{\mathfrak{s}}$ is open
in the ${\mathbb{R}}$-vector space ${\mathbb{C}}_{\mathfrak{s}}$.

###### Definition 1

The ${\mathbb{M}}$-valued Cauchy kernel on the open set
$\Omega\subset{\mathbb{H}}$ is given by

$\rho({\bf q})\times\Omega\ni(\zeta,{\bf q})\mapsto(\zeta-{\bf
q})^{-1}\in{\mathbb{M}}.$ (2)

###### Example 2

The ${\mathbb{M}}$-valued Cauchy kernel on the open set
$\Omega\subset{\mathbb{H}}$ is slice regular. Specifically, choosing an
arbitrary relatively open set
$V\subset\Omega\cap({\mathbb{R}}+{\mathbb{R}}\mathfrak{s})$, and fixing
$\zeta\in\cap_{{\bf q}\in V}\rho({\bf q})$, we can write for ${\bf q}\in V$
the equalities

$\frac{\partial}{\partial x}(\zeta-x-y\mathfrak{s})^{-1}=(\zeta-
x-y\mathfrak{s})^{-2},$ $R_{\mathfrak{s}}\frac{\partial}{\partial y}(\zeta-
x-y\mathfrak{s})^{-1}=-(\zeta-x-y\mathfrak{s})^{-2},$

because $\zeta$, $\mathfrak{s}$ and $(\zeta-x-y\mathfrak{s})^{-1}$ commute in
${\mathbb{M}}$. Therefore,

$\bar{\partial}_{\mathfrak{s}}((\zeta-q)^{-1})=\bar{\partial}_{\mathfrak{s}}((\zeta-
x-y\mathfrak{s})^{-1})=0,$

implying the assertion.

###### Remark 2

(1) The discussion about the spectrum of a quaternion can be enlarged, keeping
the same background. Specifically, we may regard an element $\bf
q\in{\mathbb{H}}$ as a left multiplication operator on the $C^{*}$-algebra
${\mathbb{M}}$, denoted by $L_{\bf q}$, and given by $L_{\bf q}{\bf a}={\bf
qa}$ for all ${\bf a}\in{\mathbb{M}}$. It is easily seen that $\sigma(L_{\bf
q})=\sigma(\bf q)$. In this context, we may find the eigenvectors of $L_{\bf
q}$, which would be of interest in what follows. Therefore, we should look for
solutions of the equation ${\bf q}\nu=s\nu$ in the algebra ${\mathbb{M}}$,
with $s\in\sigma(\bf q)$. Writing ${\bf q}=q_{0}+\Im{\bf q}$ with
$q_{0}\in{\mathbb{R}}$, $s_{\pm}=q_{0}\pm i\|\Im{\bf q}\|$, and $\nu={\bf
x}+i{\bf y}$ with ${\bf x},{\bf y}\in{\mathbb{H}}$, we obtain the equivalent
equations

$(\Im{\bf q}){\bf x}=\mp\|\Im{\bf q})\|{\bf y},\,\,(\Im{\bf q}){\bf
y}=\pm\|\Im{\bf q})\|{\bf x},$

leading to the solutions

$\nu_{\pm}({\bf q})=\left(1\mp i\frac{\Im\bf q}{\|{\Im\bf q}\|}\right){\bf x}$

of the equation ${\bf q}\nu_{\pm}=s_{\pm}\nu_{\pm}$, where ${\bf
x}\in{\mathbb{H}}$ is arbitrary, provided ${\Im\bf q}\neq 0$.

When ${\Im\bf q}=0$, the solutions are given by $\nu=q_{0}{\bf a}$, with ${\bf
a}\in{\mathbb{M}}$ arbitrary.

(2) Every quaternion $\mathfrak{s}\in\mathbb{S}$ may be associated with two
elements $\iota_{\pm}(\mathfrak{s})=(1\mp i\mathfrak{s})/2$ in ${\mathbb{M}}$,
which are commuting idempotents such that
$\iota_{+}(\mathfrak{s})+\iota_{-}(\mathfrak{s})=1$ and
$\iota_{+}(\mathfrak{s})\iota_{-}(\mathfrak{s})=0$. For this reason, setting
${\mathbb{M}}^{\mathfrak{s}}_{\pm}=\iota_{\pm}(\mathfrak{s}){\mathbb{H}}$, we
have a direct sum decomposition
${\mathbb{M}}={\mathbb{M}}^{\mathfrak{s}}_{+}+{\mathbb{M}}^{\mathfrak{s}}_{-}$.
Indeed, it is clear that
${\mathbb{M}}^{\mathfrak{s}}_{+}\cap{\mathbb{M}}^{\mathfrak{s}}_{-}=\\{0\\}$.
In addition, if ${\bf a}={\bf u}+i{\bf v}$, with $\bf u,v\in{\mathbb{H}}$, the
equation $\iota_{+}(\mathfrak{s}){\bf x}+\iota_{-}(\mathfrak{s}){\bf y}=\bf a$
has the solution ${\bf x}={\bf u+\mathfrak{s}v},\,{\bf y}={\bf
u-\mathfrak{s}v}\in{\mathbb{H}}$, because $\mathfrak{s}^{-1}=-\mathfrak{s}$.
Therefore, we also have,
${\mathbb{M}}={\mathbb{M}}^{\mathfrak{s}}_{+}+{\mathbb{M}}^{\mathfrak{s}}_{-}$.

In particular, if ${\bf q}\in{\mathbb{H}}$ and ${\Im\bf q}\neq 0$, setting
$\mathfrak{s}_{\tilde{\bf q}}=\tilde{\bf q}\|\tilde{\bf q}\|^{-1}$, where
$\tilde{\bf q}=\Im\bf q$, the elements $\iota_{\pm}(\mathfrak{s}_{\tilde{\bf
q}})$ are idempotents, as above. Moreover,

${\bf q}\iota_{\pm}(\mathfrak{s}_{\tilde{\bf q}}){\bf h}=s_{\pm}({\bf
q})\iota_{\pm}(\mathfrak{s}_{\tilde{\bf q}}){\bf h},\,\,{\bf
h}\in{\mathbb{H}}.$ (3)

The next result provides explicit formulas of the spectral projections (see
[3], Part I, Section VII.1) associated to the operator $L_{\bf q},\,{\bf
q}\in{\mathbb{H}}.$ Of course, this is not trivial only if ${\bf
q}\in{\mathbb{H}}\setminus{\mathbb{R}}$ because if ${\bf q}\in{\mathbb{R}}$,
its spectrum a real singleton, and the the only spectral projection is the
identity.

###### Lemma 1

Let ${\bf q}\in{\mathbb{H}}\setminus{\mathbb{R}}$ be fixed. The spectral
projections associated to $s_{\pm}({\bf q})$ are given by

$P_{\pm}({\bf q}){\bf a}=\iota_{\pm}(\mathfrak{s}_{\tilde{\bf q}}){\bf
a},\,\,{\bf a}\in{\mathbb{M}}.$

Moreover, $P_{+}({\bf q})P_{-}({\bf q})=P_{-}({\bf q})P_{+}({\bf q})=0$, and
$P_{+}({\bf q})+P_{-}({\bf q})$ is the identity on ${\mathbb{M}}$.

When ${\bf q}\in{\mathbb{R}}$, the corresponding spectral projection is the
identity on ${\mathbb{M}}$.

Proof. Let us fix a quaternion $\bf q$ with ${\Im\bf q}\neq 0$. Next, write
the general formulas for its spectral projections. Setting
$s_{\pm}=s_{\pm}({\bf q})$, the points $s_{+},s_{-}$ are distinct and not
real. We fix an $r>0$ sufficiently small such that, setting
$D_{\pm}:=\\{\zeta\in\rho({\bf q});|\zeta-s_{\pm}|\leq r\\}$, we have
$D_{\pm}\subset\rho({\bf q})$ and $D_{+}\cap D_{-}=\emptyset$. Then we have

$P_{\pm}({\bf q})=\frac{1}{2\pi i}\int_{\Gamma_{\pm}}(\zeta-L_{\bf
q})^{-1}d\zeta$

where $\Gamma_{\pm}$ is the boundary of $D_{\pm}$.

Using the equality $L_{\bf q}\nu_{\pm}({\bf q})=s_{\pm}({\bf q})\nu_{\pm}({\bf
q})$ (see Remark 2), for every $\zeta\in\rho({\bf q})$ and ${\bf
h}\in{\mathbb{H}}$, we have

$(\zeta-L_{\bf q})^{-1}(1\mp i\mathfrak{s}_{\tilde{\bf q}}){\bf h}=(\zeta-
s_{\pm})^{-1}(1\mp i\mathfrak{s}_{\tilde{\bf q}}){\bf h},$

by Remark 2. Therefore,

$P_{+}({\bf q})(1\mp i\mathfrak{s}_{\tilde{\bf q}}){\bf h}=\frac{1}{2\pi
i}\int_{\Gamma_{+}}(\zeta-s_{\pm})^{-1}(1\mp i\mathfrak{s}_{\tilde{\bf
q}}){\bf h}d\zeta,$

and

$P_{-}({\bf q})(1\mp i\mathfrak{s}_{\tilde{\bf q}}){\bf h}=\frac{1}{2\pi
i}\int_{\Gamma_{-}}(\zeta-s_{\pm})^{-1}(1\mp i\mathfrak{s}_{\tilde{\bf
q}}){\bf h}d\zeta.$

Using Cauchy’s formula, we deduce that

$P_{+}({\bf q})(1-i\mathfrak{s}_{\tilde{\bf q}}){\bf
h}=(1-i\mathfrak{s}_{\tilde{\bf q}}){\bf h},\,P_{+}({\bf
q})(1+i\mathfrak{s}_{\tilde{\bf q}}){\bf h}=0,$

and

$P_{-}({\bf q})(1-i\mathfrak{s}_{\tilde{\bf q}}){\bf h}=0,\,P_{-}({\bf
q})(1+i\mathfrak{s}_{\tilde{\bf q}}){\bf h}=(1+i\mathfrak{s}_{\tilde{\bf
q}}){\bf h},$

for all $\bf h\in{\mathbb{H}}$.

Fixing an arbitrary element ${\bf a}={\bf u}+i{\bf v}\in{\mathbb{M}}$, writing

${\bf a}=\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})({\bf
u}+\mathfrak{s}_{\tilde{\bf q}}{\bf v})+\iota_{-}(\mathfrak{s}_{\tilde{\bf
q}})({\bf u}-\mathfrak{s}_{\tilde{\bf q}}{\bf v})$

with ${\bf u}\pm\mathfrak{s}_{\tilde{\bf q}}{\bf v}\in{\mathbb{H}}$ (see
Remark 2(2)), and noticing that $\iota_{\pm}(\mathfrak{s}_{\tilde{\bf
q}})\mathfrak{s}_{\tilde{\bf q}}=\pm i\iota_{\pm}(\mathfrak{s}_{\tilde{\bf
q}})$, as $P_{\pm}({\bf q})$ are ${\mathbb{C}}$-linear, we obtain

$P_{\pm}({\bf q}){\bf a}=\iota_{\pm}(\mathfrak{s}_{\tilde{\bf q}}){\bf
a},\,\,{\bf a}\in{\mathbb{M}},$ (4)

which are precisely the formulas from the statement.

The properties $P_{+}({\bf q})P_{-}({\bf q})=P_{-}({\bf q})P_{+}({\bf q})=0$,
and $P_{+}({\bf q})+P_{-}({\bf q})$ is the identity on ${\mathbb{M}}$ are
direct consequences of the analytic functional calculus associated to a fixed
element ${\bf q}\in{\mathbb{H}}$ in the algebra ${\mathbb{M}}$.

By a slight abuse of terminology, the projections $P_{\pm}({\bf q})$ will be
also called the spectral projections of $\bf q$. In fact, as formula (4)
shows, they depend only on the imaginary part of ${\bf q}$.

##  4 A General Functional Calculus

In this section, starting from some spaces of ${\mathbb{M}}$-valued functions,
defined on subsets of the complex plane, we construct and characterize
functional calculi, taking values in the quaternionic algebra ${\mathbb{H}}$.

###### Remark 3

Regarding the $C^{*}$-algebra ${\mathbb{M}}$ as a (complex) Banach space, and
denoting by $\mathcal{B}({\mathbb{M}})$ the Banach space of all linear
operators acting on ${\mathbb{M}}$, the operator $L_{\bf q},\,\bf
q\in{\mathbb{H}}$, (see Remark 2(1)) is a (very) particular case of a scalar
type operator, as defined in [3], Part III, XV.4.1. Its resolution of the
identity consists of four projections $\\{0,P_{\pm}({\bf q}),$I$\\}$,
including the null operator $0$ and the identity I on ${\mathbb{M}}$, where
$P_{\pm}({\bf q})$ are the spectral projections of $L_{\bf q}$, and its
integral representation is given by

$L_{\bf q}=s_{+}({\bf q})P_{+}({\bf q})+s_{-}({\bf q})P_{-}({\bf
q})\in\mathcal{B}({\mathbb{M}}),$

via formulas (3) and (4). For every function $f:\sigma({\bf
q})\mapsto{\mathbb{C}}$ we may define the operator

$f(L_{\bf q})=f(s_{+}({\bf q}))P_{+}({\bf q})+f(s_{-}({\bf q}))P_{-}({\bf
q})\in\mathcal{B}({\mathbb{M}}).$

which provides a functional calculus with arbitrary functions on the spectrum.
More generally, we may extend this formula to functions of the form
$F:\sigma({\bf q})\mapsto{\mathbb{M}}$, putting

$F(L_{\bf q})=F(s_{+}({\bf q}))P_{+}({\bf q})+F(s_{-}({\bf q}))P_{-}({\bf
q}),$ (5)

and keeping this order, which is a ”left functional calculus“, not
multiplicative, in general. It is this idea which leads us to try to define
${\mathbb{H}}$-valued functions on subsets of ${\mathbb{H}}$ via some
${\mathbb{M}}$-valued functions, defined on subsets of ${\mathbb{C}}$.

###### Definition 2

(1) A subset $S\subset{\mathbb{C}}$ is said to be conjugate symmetric if
$\zeta\in S$ if and only if $\bar{\zeta}\in S$.

(2) A subset $A\subset{\mathbb{H}}$ is said to be spectrally saturated (see
[12]) if whenever $\sigma({\bf h})=\sigma({\bf q})$ for some ${\bf
h}\in{\mathbb{H}}$ and ${\bf q}\in A$, we also have ${\bf h}\in A$.

For an arbitrary $A\subset{\mathbb{H}}$, we put $\mathfrak{S}(A)=\cup_{{\bf
q}\in A}\sigma({\bf q})$. We also put $S_{\mathbb{H}}=\\{{\bf
q}\in{\mathbb{H}};\sigma({\bf q})\subset S\\}$ for an arbitrary subset
$S\subset{\mathbb{C}}$.

###### Remark 4

(1) If $A\subset{\mathbb{H}}$ is spectrally saturated, then
$S=\mathfrak{S}(A)$ is conjugate symmetric, and conversely, if
$S\subset{\mathbb{C}}$ is conjugate symmetric, then $S_{\mathbb{H}}$ is
spectrally saturated, which can be easily seen. Moreover, the assignment
$S\mapsto S_{\mathbb{H}}$ is injective. Indeed, if $\lambda=u+iv\in
S,\,u,v\in{\mathbb{R}}$, then $\lambda\in\sigma(u+v\mathfrak{s})$ for a fixed
$\mathfrak{s}\in{\mathbb{S}}$. If $S_{\mathbb{H}}=T_{\mathbb{H}}$ for some
$T\subset{\mathbb{C}}$, we must have ${\bf q}=u+v\mathfrak{s}\in
T_{\mathbb{H}}$. Therefore $\sigma(u+v\mathfrak{s})\subset T$, implying
$\lambda\in T$, and so $S\subset T$. Clearly, we also have $T\subset S$.

Similarly, the assignment $A\mapsto\mathfrak{S}(A)$ is injective and
$A=S_{\mathbb{H}}$ if and only if $S=\mathfrak{S}(A)$. These two assertions
are left to the reader.

(2) If $\Omega\subset{\mathbb{H}}$ is an open spectraly saturated set, then
$\mathfrak{S}(\Omega)\subset{\mathbb{C}}$ is open. To see that, let
$\lambda_{0}=u_{0}+iv_{0}\in\mathfrak{S}(\Omega)$ be fixed, with
$u_{0},v_{0}\in{\mathbb{R}}$, and let ${\bf q}_{0}=u_{0}+v_{0}\mathfrak{s}$,
where $\mathfrak{s}\in{\mathbb{S}}$ is also fixed. Because $\Omega$ is
spectrally saturated, we must have ${\bf q}_{0}\in\Omega$. Because the set
$\Omega\cap{\mathbb{C}}_{\mathfrak{s}}$ is relatively open, there is a
positive number $r$ such that the open set

$\\{{\bf q}=u+v\mathfrak{s};u,v\in{\mathbb{R}},|{\bf q}-{\bf q}_{0}|<r\\}$

is in $\Omega\cap{\mathbb{C}}_{\mathfrak{s}}$, where ${\bf
q}=u+v\mathfrak{s}$. Therefore, the set of the points $\lambda=u+iv$,
satisfying $|\lambda-\lambda_{0}|<r$ is in $\mathfrak{S}(\Omega)$, implying
that it is open.

Conversely, if $U\subset{\mathbb{C}}$ is open and conjugate symmetric, the set
$U_{\mathbb{H}}$ is also open via the upper semi-continuity of the spectrum
(see [3], Part I, Lemma VII.6.3.).

An important particular case is when
$U={\mathbb{D}}_{r}:=\\{\zeta\in{\mathbb{C}};|\zeta|<r\\}$, for some $r>0$.
Indeed, if $|\kappa|<r$ and $\theta$ has the property $\sigma({\bf
q})=\sigma({\bf h})$, from the equality $\\{\Re(\kappa)\pm
i|\Im(\kappa)|\\}=\\{\Re({\bf h})\pm i|\Im({\bf h})|\\}$ it follows that
$|{\bf h}|<r$.

(3) A subset $\Omega\subset{\mathbb{H}}$ is said to be axially symmetric if
for every ${\bf q}_{0}=u_{0}+v_{0}\mathfrak{s}_{0}\in\Omega$ with
$u_{0},v_{0}\in{\mathbb{R}}$ and $\mathfrak{s}_{0}\in{\mathbb{S}}$, we also
have ${\bf q}=u_{0}+v_{0}\mathfrak{s}\in\Omega$ for all
$\mathfrak{s}\in{\mathbb{S}}$. This concept is similar to the corresponding
one in [2], Definition 2.2.17. In fact, we have the following.

###### Lemma 2

A subset $\Omega\subset{\mathbb{H}}$ is axially symmetric if and only if it is
spectrally saturated.

Proof. Let $\Omega\subset{\mathbb{H}}$ be axially symmetric and ${\bf
q}=u+v\mathfrak{s}\in\Omega$. Therefore, $\sigma({\bf q})=\\{u\pm iv\\}$. Let
also ${\bf h}\in{\mathbb{H}}$ be such that $\sigma({\bf h})=\sigma({\bf q})$.
Hence ${\bf h}=u+w\mathfrak{s}^{\prime}$. Changing if necessaery, the sign of
$\mathfrak{s}^{\prime}$, we may assume, with no loss of generality, that
$w=v$. Then ${\bf h}\in\Omega$, showing that $\Omega$ is spectrally saturated.

Conversely, assuming that $\Omega$ is spectrally saturated, and fixing an
element ${\bf q}_{0}=u_{0}+v_{0}\mathfrak{s}_{0}\in\Omega$, then each element
of the form ${\bf q}=u_{0}+v_{0}\mathfrak{s}$ has the same spectrum as ${\bf
q}_{0}$, and thus it must belong to $\Omega$. Consequently, $\Omega$ is
axially symmetric.

Nevertheless, we continue to use the expression ”spectrally saturated set“ to
designate an ”axially symmetric set“, because the former is more compatible
with our spectral approach.

As noticed above, the algebra ${\mathbb{M}}$ is endowed with a conjugation
given by $\bar{\bf a}={\bf b}-i{\bf c}$, when ${\bf a}={\bf b}+i{\bf c}$, with
${\bf b},{\bf c}\in{\mathbb{H}}$. Note also that, because ${\mathbb{C}}$ is a
subalgebra of ${\mathbb{M}}$, the conjugation of ${\mathbb{M}}$ restricted to
${\mathbb{C}}$ is precisely the usual complex conjugation.

The next definition has an old origin, going back to [5] (see for instance
[8]).

###### Definition 3

Let $U\subset{\mathbb{C}}$ be conjugate symmetric, and let
$F:U\mapsto{\mathbb{M}}$. We say that $F$ is a stem function if
$F(\bar{\lambda})=\overline{F(\lambda)}$ for all $\lambda\in U$.

For an arbitrary conjugate symmetric subset $U\subset{\mathbb{C}}$, we put

$\mathcal{S}(U,{\mathbb{M}})=\\{F:U\mapsto{\mathbb{M}};F(\bar{\zeta})=\overline{F(\zeta)},\zeta\in
U\\},$ (6)

that is, the ${\mathbb{R}}$-vector space of all ${\mathbb{M}}$-valued stem
functions on $U$. Replacing ${\mathbb{M}}$ by ${\mathbb{C}}$, we denote by
$\mathcal{S}(U)$ the real algebra of all ${\mathbb{C}}$-valued stem functions,
which is an ${\mathbb{R}}$-subalgebra in $\mathcal{S}(U,{\mathbb{M}})$. In
addition, the space $\mathcal{S}(U,{\mathbb{M}})$ is a
$\mathcal{S}(U)$-bimodule.

###### Definition 4

Let $U\subset{\mathbb{C}}$ be conjugate symmetric. For every
$F:U\mapsto{\mathbb{M}}$ and all ${\bf q}\in U_{\mathbb{H}}$ we define a
function $F_{\mathbb{H}}:U_{\mathbb{H}}\mapsto{\mathbb{M}}$, via the
assignment

$U_{\mathbb{H}}\ni{\bf q}\mapsto F_{\mathbb{H}}({\bf q})=F(s_{+}({\bf
q}))\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})+F(s_{-}({\bf
q}))\iota_{-}(\mathfrak{s}_{\tilde{\bf q}})\in{\mathbb{M}},$ (7)

where $\tilde{\bf q}=\Im\bf q,\,\mathfrak{s}_{\tilde{\bf q}}=\tilde{\bf
q}\|\tilde{\bf q}\|^{-1}$, and $\iota_{\pm}(\mathfrak{s}_{\tilde{\bf
q}})=2^{-1}(1\mp i\mathfrak{s}_{\tilde{\bf q}})$.

Formula (7) is strongly related to formula (5) because the spectral
projections $P_{\pm}({\bf q})$ are the left multiplications defined by
$2^{-1}(1\mp i\mathfrak{s}_{\tilde{\bf q}})$ respectively, via formula (4).

The next result is an intrinsic version of Theorem 1 from [12], with a much
shorter proof.

###### Theorem 2

Let $U\subset{\mathbb{C}}$ be a conjugate symmetric subset, and let
$F:U\mapsto{\mathbb{M}}$. The element $F_{\mathbb{H}}({\bf q})$ is a
quaternion for all ${\bf q}\in U_{\mathbb{H}}$ if and only if $F$ is a stem
function.

Proof. We first assume that $F_{\mathbb{H}}({\bf q})$ is a quaternion for all
${\bf q}\in U_{\mathbb{H}}$. We fix a point $\zeta\in U$, supposing that
$\Im\zeta>0$. Then we choose a quaternion ${\bf q}\in U_{\mathbb{H}}$ with
$\sigma({\bf q})=\\{\zeta,\bar{\zeta}\\}$. Therefore, $s_{+}({\bf q})=\zeta$
and $s_{-}({\bf q})=\bar{\zeta}$. We write
$F(\zeta)=F_{+1}+iF_{+2},F(\bar{\zeta})=F_{-1}+iF_{-2}$, with
$F_{+1},F_{+2},F_{-1},F_{-2}\in{\mathbb{H}}$. According to (7), we infer that

$2F({\bf q})=F_{+1}+F_{+2}\mathfrak{s}_{\tilde{\bf
q}}+F_{-1}-F_{-2}\mathfrak{s}_{\tilde{\bf
q}}+i(-F_{+1}\mathfrak{s}_{\tilde{\bf
q}}+F_{+2}+F_{-2}+F_{-1}\mathfrak{s}_{\tilde{\bf q}}),$

so

$2\overline{F({\bf q})}=F_{+1}+F_{+2}\mathfrak{s}_{\tilde{\bf
q}}+F_{-1}-F_{-2}\mathfrak{s}_{\tilde{\bf q}}+i(F_{+1}\mathfrak{s}_{\tilde{\bf
q}}-F_{+2}-F_{-2}-F_{-1}\mathfrak{s}_{\tilde{\bf q}}).$

Because $F({\bf q})=\overline{F({\bf q})}$, we must have

$-F_{+1}\mathfrak{s}_{\tilde{\bf
q}}+F_{+2}+F_{-2}+F_{-1}\mathfrak{s}_{\tilde{\bf
q}}=F_{+1}\mathfrak{s}_{\tilde{\bf
q}}-F_{+2}-F_{-2}-F_{-1}\mathfrak{s}_{\tilde{\bf q}},$

which is equivalent to

$F_{+2}+F_{-2}=(F_{+1}-F_{-1})\mathfrak{s}_{\tilde{\bf q}}$

Assuming $F_{+1}\neq F_{-1}$, we deduce that

$(F_{+1}-F_{-1})^{-1}(F_{+2}+F_{-2})=\mathfrak{s}_{\tilde{\bf q}}.$

This equality is impossible because the left hand side depends only on $\zeta$
and $\bar{\zeta}$ while the right hand side has infinitely many distinct
values, when replacing $\bf q$ by another element from the set $\\{{\bf
h}\in{\mathbb{H}};s_{+}({\bf h})=\zeta\\}\subset U_{\mathbb{H}}$. Therefore,
the equality $F({\bf q})=\overline{F({\bf q})}$ implies the equalities
$F_{+1}=F_{-1}$ and $F_{+2}=-F_{-2}$, meaning that
$\overline{F(\zeta)}=F(\bar{\zeta})$.

If $\Im\zeta=0$, so $\zeta=x_{0}\in{\mathbb{R}}$, taking ${\bf q}=x_{0}$, we
have $\sigma({\bf q})=\\{x_{0}\\}$, and $F({\bf q})=F(x_{0})$ is a quaternion.

If $\Im\zeta<0$, applying the above argument to $\bar{\zeta}$ we obtain
$\overline{F(\bar{\zeta})}=F({\zeta})$. Consequently, $F$ is a stem function
on $U$.

Conversely, if $\overline{F(\zeta)}=F(\bar{\zeta})$ for all $\zeta\in U$,
choosing a ${\bf q}\in{\mathbb{H}}$ with $\Im{\bf q}\neq 0$, and fixing
$\zeta=s_{+}({\bf q})$, we obtain from (7) the equality

$2F_{\mathbb{H}}({\bf
q})=F(\zeta)+F(\bar{\zeta})-i(F(\zeta)-F(\bar{\zeta})\mathfrak{s}_{\tilde{\bf
q}}.$

Therefore,

$2\overline{F_{\mathbb{H}}({\bf
q})}=F(\bar{\zeta})+F(\zeta)+i(F(\bar{\zeta})-F(\zeta))\mathfrak{s}_{\tilde{\bf
q}},$

showing that $F({\bf q})\in{\mathbb{H}}$ for all ${\bf q}\in U_{\mathbb{H}}$,
because the case ${\bf q}=x_{0}\in{\mathbb{R}}$ is evident.

###### Corollary 1

Let $U\subset{\mathbb{C}}$ be a conjugate symmetric subset, and let
$f:U\mapsto{\mathbb{C}}$. The element $f_{\mathbb{H}}({\bf q})$ is a
quaternion for all ${\bf q}\in U_{\mathbb{H}}$ if and only if $f$ is a stem
function.

###### Remark 5

Let $U\subset{\mathbb{C}}$ be a conjugate symmetric set and let
$F\in\mathcal{S}(U,{\mathbb{M}})$ be arbitrary. We can easily describe the
zeros of $F_{\mathbb{H}}$. Indeed, if $F_{\mathbb{H}}({\bf q})=F(s_{+}({\bf
q}))\iota_{+}({\tilde{\bf q}})+F(s_{-}({\bf q}))\iota_{-}({\tilde{\bf q}})=0$,
we must have $F(s_{+}({\bf q}))\iota_{+}({\tilde{\bf q}})=0$ and $F(s_{-}({\bf
q}))\iota_{-}({\tilde{\bf q}})=0$, via a direct manipulation with the
idempotents $\iota_{\pm}({\tilde{\bf q}})$. In other words, we must have
$F(s_{\pm}({\bf q}))=\pm iF(s_{\pm}({\bf q}))\mathfrak{s}_{\tilde{\bf q}}$.
Choosing another quaternion ${\bf h}$ with $s_{+}({\bf q})\in\sigma({\bf h})$
and $\tilde{{\bf q}}\neq\tilde{{\bf h}}$, we obtain $F(s_{+}({\bf
q}))(\mathfrak{s}_{\tilde{\bf q}}-\mathfrak{s}_{\tilde{\bf h}})=0$. Therefore,
$F(s_{+}({\bf q}))=0$ because $\mathfrak{s}_{\tilde{\bf
q}}-\mathfrak{s}_{\tilde{\bf h}}$ is invertible. Similarly, $F(s_{-}({\bf
q}))=0$. Conqequently, setting $\mathcal{Z}(F):=\\{\lambda\in
U;F(\lambda)=0\\}$, and $\mathcal{Z}(F_{\mathbb{H}}):=\\{{\bf q}\in
U_{\mathbb{H}};F_{\mathbb{H}}({\bf q})=0\\}$, we must have

$\mathcal{Z}(F_{\mathbb{H}})=\\{{\bf q}\in U_{\mathbb{H}};\sigma({\bf
q})\subset\mathcal{Z}(F)\\}.$

For every subset $\Omega\subset{\mathbb{H}}$, we denote by
$\mathcal{F}(\Omega,{\mathbb{H}})$ the set of all ${\mathbb{H}}$-valued
functions on $\Omega$.

The next result offers an ${\mathbb{H}}$-valued general functional calculus
with arbitrary stem functions.

###### Theorem 3

Let $\Omega\subset{\mathbb{H}}$ be a spectrally saturated set, and let
$U=\mathfrak{S}(\Omega)$. The map

${\mathcal{S}}(U,{\mathbb{M}})\ni F\mapsto
F_{\mathbb{H}}\in\mathcal{F}(\Omega,{\mathbb{H}})$

is ${\mathbb{R}}$-linear, injective, and has the property
$(Ff)_{\mathbb{H}}=F_{\mathbb{H}}f_{\mathbb{H}}$ for all
$F\in{\mathcal{S}}(U,{\mathbb{M}})$ and $f\in{\mathcal{S}}(U)$. Moreover, the
restricted map

${\mathcal{S}}(U)\ni f\mapsto
f_{\mathbb{H}}\in\mathcal{F}(\Omega,{\mathbb{H}})$

is unital and multiplicative.

Proof. The map $F\mapsto F_{\mathbb{H}}$ is clearly ${\mathbb{R}}$-linear. The
injectivity of this map follows from Remark 5. Note also that

$F_{\mathbb{H}}({\bf q})f_{\mathbb{H}}({\bf q})=(F(s_{+}({\bf
q}))\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})+F(s_{-}({\bf
q}))\iota_{-}(\mathfrak{s}_{\tilde{\bf q}}))\times$ $(f(s_{+}({\bf
q}))\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})+f(s_{-}({\bf
q}))\iota_{-}(\mathfrak{s}_{\tilde{\bf q}})=$ $(Ff)(s_{+}({\bf
q}))\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})+(Ff)(s_{-}({\bf
q}))\iota_{-}(\mathfrak{s}_{\tilde{\bf q}})=(Ff)_{\mathbb{H}}({\bf q}),$

because $f$ is complex valued, and by the properties of the idempotents
$\iota_{\pm}(\mathfrak{s}_{\tilde{\bf q}})$ In particular, this computation
shows that if $f,g\in{\mathcal{S}}(U)$, we have
$(fg)_{\mathbb{H}}=f_{\mathbb{H}}g_{\mathbb{H}}=g_{\mathbb{H}}f_{\mathbb{H}}$,
so the map $f\mapsto f_{\mathbb{H}}$ is multiplicative. It is also clearly
unital.

##  5 The Quaternionic Cauchy Transform

Using the ${\mathbb{M}}$-valued Cauchy kernel, we may define a concept of
Cauchy transform, whose main properties will be discussed in this section.

The frequent use of versions of the Cauchy formula is simplified by adopting
the following definition. Let $U\subset{\mathbb{C}}$ be open. An open subset
$\Delta\subset U$ will be called a Cauchy domain (in $U$) if
$\Delta\subset\bar{\Delta}\subset U$ and the boundary $\partial\Delta$ of
$\Delta$ consists of a finite family of closed curves, piecewise smooth,
positively oriented. Note that a Cauchy domain is bounded but not necessarily
connected.

For a given open set $U\subset{\mathbb{C}}$, we denote by
$\mathcal{O}(U,{\mathbb{M}})$ the complex algebra of all ${\mathbb{M}}$-valued
analytic functions on $U$.

If $U\subset{\mathbb{C}}$ is open and conjugate symmetric, let
$\mathcal{O}_{s}(U,{\mathbb{M}})$ be the real subalgebra of
$\mathcal{O}(U,{\mathbb{M}})$ consisting of all stem functions from
$\mathcal{O}(U,{\mathbb{M}})$.

Because ${\mathbb{C}}\subset{\mathbb{M}}$, we have
$\mathcal{O}(U)\subset\mathcal{O}(U,{\mathbb{M}})$, where $\mathcal{O}(U)$ is
the complex algebra of all complex-valued analytic functions on the open set
$U$. Similarly, when $U\subset{\mathbb{C}}$ is open and conjugate symmetric,
$\mathcal{O}_{s}(U)\subset\mathcal{O}_{s}(U,{\mathbb{M}})$, where
$\mathcal{O}_{s}(U)$ is the real subalgebra consisting of all functions $f$
from $\mathcal{O}(U)$ which are stem functions.

As un example, if $\Delta\subset{\mathbb{C}}$ is an open disk centered at $0$,
each function $F\in\mathcal{O}_{s}(\Delta,{\mathbb{M}})$ can be represented as
a convergent series $F(\zeta)=\sum_{k\geq 0}a_{k}\zeta^{k},\,\zeta\in\Delta$,
with $a_{k}\in{\mathbb{H}}$ for all $k\geq 0$.

###### Definition 5

Let $U\subset{\mathbb{C}}$ be a conjugate symmetric open set, and let
$F\in\mathcal{O}(U,{\mathbb{M}})$. For every ${\bf q}\in U_{\mathbb{H}}$ we
set

$C[F]({\bf q})=\frac{1}{2\pi i}\int_{\Gamma}F(\zeta)(\zeta-{\bf
q})^{-1}d\zeta,$ (8)

where $\Gamma$ is the boundary of a Cauchy domain in $U$ containing the
spectrum $\sigma({\bf q})$. The function
$C[F]:U_{\mathbb{H}}\mapsto{\mathbb{M}}$ is called the (quaternionic) Cauchy
transform of the function $F\in\mathcal{O}(U,{\mathbb{M}})$. Clearly, the
function $C[F]$ does not depend on the choice of $\Gamma$ because the function
$U\setminus\sigma({\bf q})\ni\zeta\mapsto F(\zeta)(\zeta-{\bf
q})^{-1}\in{\mathbb{M}}$ is analytic.

We put

$\mathcal{R}(U_{\mathbb{H}},{\mathbb{M}})=\\{C[F];F\in\mathcal{O}(U,{\mathbb{M}})\\}.$
(9)

###### Proposition 1

Let $U\subset{\mathbb{C}}$ be open and conjugate symmetric, and let
$F\in\mathcal{O}(U,{\mathbb{M}})$. Then function
$C[F]\in\mathcal{R}(U_{\mathbb{H}},{\mathbb{M}})$ is slice regular on
$U_{\mathbb{H}}$.

Proof. Let $F\in\mathcal{O}(U,{\mathbb{M}})$, let ${\bf q}\in U_{\mathbb{H}}$
and let $\Delta\ni\sigma({\bf q})$ be a conjugate symmetric Cauchy domain in
$U$, whose boundary is denoted by $\Gamma$. We use the representation of
$C[F]({\bf q})$ given by (8). Because we have

$\bar{\partial}_{\mathfrak{s}}((\zeta-{\bf
q})^{-1})=\bar{\partial}_{\mathfrak{s}}((\zeta-x-y\mathfrak{s})^{-1})=0$

for ${\bf
q}=x+y\mathfrak{s}\in\Delta_{\mathbb{H}}\cap({\mathbb{R}}+{\mathbb{R}}\mathfrak{s})$,
via Example 2, we infer that

$\bar{\partial}_{\mathfrak{s}}(C[F]({\bf q}))=\frac{1}{2\pi
i}\int_{\Gamma}F(\zeta)\bar{\partial}_{\mathfrak{s}}((\zeta-{\bf
q})^{-1})d\zeta=0,$

which implies the assertion.

###### Remark 6

(1) Because the function $F$ does not necessarily commute with the left
multiplication by $\mathfrak{s}$, the choice of the right multiplication in
the slice regularity is necessary to get the stated property of $C[F]$.

(2) Let $r>0$ and let $U\supset\\{\zeta\in{\mathbb{C}};|\zeta|\leq r\\}$ be a
conjugate symmetric open set. Then for every $F\in\mathcal{O}(U,{\mathbb{M}})$
one has

$C[F]({\bf q})=\sum_{n\geq 0}\frac{F^{(n)}(0)}{n!}{\bf q}^{n},\,\,{\bf q}\in
U_{\mathbb{H}},\,\,\|\bf q\|<r,$

where the series is absolutely convergent. Of course, using the convergent
series $(\zeta-{\bf q})^{-1}=\sum_{n\geq 0}\zeta^{-n-1}{\bf q}^{n}$ in
$\\{\zeta;|\zeta|=r\\}$, the assertion follows easily, via formula (8).
Moreover, by Proposition 1, the function $C[F]$ is a slice regular
${\mathbb{M}}$-valued function in $U_{\mathbb{H}}$. Nevertheless, we are
particularly interested in slice regular ${\mathbb{H}}$-valued functions.

###### Theorem 4

Let $U\subset{\mathbb{C}}$ be a conjugate symmetric open set and let
$F\in\mathcal{O}(U,{\mathbb{M}})$. The Cauchy transform $C[F]$ is
${\mathbb{H}}$-valued if and only if $F\in\mathcal{O}_{s}(U,{\mathbb{M}})$.

Proof.  We first fix a ${\bf q}\in U\setminus{\mathbb{R}}$. If $\sigma({\bf
q})=\\{s_{+},s_{-}\\}$, the points $s_{+},s_{-}$ are distinct and not real. We
then fix an $r>0$ sufficiently small such that, setting $D_{\pm}:=\\{\zeta\in
U;|\zeta-s_{\pm}|\leq r\\}$, we have $D_{\pm}\subset U$ and $D_{+}\cap
D_{-}=\emptyset$. Then

$C[F]({\bf q})=\frac{1}{2\pi i}\int_{\Gamma_{+}}F(\zeta)(\zeta-{\bf
q})^{-1}d\zeta+\frac{1}{2\pi i}\int_{\Gamma_{-}}F(\zeta)(\zeta-{\bf
q})^{-1}d\zeta,$

where $\Gamma_{\pm}$ is the boundary of $D_{\pm}$. We may write
$F(\zeta)=\sum_{k\geq 0}(\zeta-s_{+})^{k}a_{k}$ with $\zeta\in D_{+}$,
$a_{k}\in{\mathbb{M}}$ for all $k\geq 0$, as a uniformly convergent series.
Similarly, $F(\zeta)=\sum_{k\geq 0}(\zeta-s_{-})^{k}b_{k}$ with $\zeta\in
D_{-}$, $b_{k}\in{\mathbb{M}}$ for all $k\geq 0$, as a uniformly convergent
series.

Note that

$\frac{1}{2\pi i}\int_{\Gamma_{+}}F(\zeta)(\zeta-{\bf
q})^{-1}d\zeta=\sum_{k\geq 0}\left(a_{k}\frac{1}{2\pi
i}\int_{\Gamma_{+}}(\zeta-s_{+})^{k}(\zeta-{\bf
q})^{-1}d\zeta\right)=a_{0}\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})$

because we have

$\frac{1}{2\pi i}\int_{\Gamma_{+}}(\zeta-s_{+})^{k}(\zeta-{\bf
q})^{-1}d\zeta=({\bf q}-s_{+})^{k}\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})$

by the analytic functional calculus of ${\bf q}$ (see also Lemma 1), which is
equal to $\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})$ when $k=0$, and it is equal
to $0$ when $k\geq 1$, via the equality ${\bf
q}\iota_{+}(\mathfrak{s}_{\tilde{\bf
q}})=s_{+}\iota_{+}(\mathfrak{s}_{\tilde{\bf q}})$

Similarly

$\frac{1}{2\pi i}\int_{\Gamma_{-}}F(\zeta)(\zeta-q)^{-1}d\zeta=\sum_{k\geq
0}\left(b_{k}\frac{1}{2\pi i}\int_{\Gamma_{-}}(\zeta-s_{-})^{k}(\zeta-{\bf
q})^{-1}d\zeta\right)=b_{0}\iota_{-}(\mathfrak{s}_{\tilde{\bf q}})$

because, as above, we have

$\frac{1}{2\pi i}\int_{\Gamma_{-}}(\zeta-s_{-})^{k}(\zeta-{\bf
q})^{-1})d\zeta=({\bf q}-s_{-})^{k}\iota_{-}(\mathfrak{s}_{\tilde{\bf q}}),$

which is equal $\iota_{-}(\mathfrak{s}_{\tilde{\bf q}})$ when $k=0$, and it is
equal to $0$ when $k\geq 1$. Consequently,

$C[F]({\bf q})=F(s_{+})\iota_{+}(\mathfrak{s}_{\tilde{\bf
q}})+F(s_{-})\iota_{-}(\mathfrak{s}_{\tilde{\bf q}}),$

and the right hand side of this equality coincides with the expression from
formula (5).

Assume now that $\sigma({\bf q})=\\{s\\}$, where
$s:=s_{+}=s_{-}\in{\mathbb{R}}$. We fix an $r>0$ such that the set
$D:=\\{\zeta\in U;|\zeta-s|\leq r\\}\subset U$, whose boundary is denoted by
$\Gamma$. Then we have

$C[F]({\bf q})=\frac{1}{2\pi i}\int_{\Gamma}F(\zeta)(\zeta-{\bf
q})^{-1}d\zeta=F(s),$

via the usual analytic functional calculus.

In all of these situations, the element $C[F]({\bf q}))$ is equal to the right
hand side of formula (7). Therefore, we must have $C[F]({\bf
q})\in{\mathbb{H}}$ if and only if $F(s_{+})=\overline{F(s_{-})}$, via Theorem
2. Consequently, $C[F]({\bf q})\in{\mathbb{H}}$ for all ${\bf q}\in
U_{\mathbb{H}}$ if and only if $F:U\mapsto{\mathbb{M}}$ is a stem function.

###### Remark 7

(1) It follows from the proof of the previous theorem that the element
$C[F]({\bf q}))$, given by formula (8), coincides with the element
$F_{\mathbb{H}}({\bf q}))$ given by (7). To unify the notation, from now on
this element will be denoted by $F_{\mathbb{H}}({\bf q})$, whenever $F$ is a
stem function, analytic or not.

(2) An important particular case is when let $f:U\mapsto{\mathbb{C}}$ is an
analytic function, where $U\subset{\mathbb{C}}$ is a conjugate symmetric open
set. In this case we may also consider the (quaternionic) Cauchy transform of
$f$ given by

$C[f]({\bf q})=\frac{1}{2\pi i}\int_{\Gamma}f(\zeta)(\zeta-{\bf
q})^{-1}d\zeta,$ (10)

where $\Gamma$ is the boundary of a Cauchy domain in $U$ containing the
spectrum $\sigma({\bf q})$. According to Theorem 5, we have $C[f]({\bf
q})\in{\mathbb{H}}$ if and only if $f(s_{+}({\bf q}))=\overline{f(s_{-}({\bf
q}))}$ for all ${\bf q}\in U_{\mathbb{H}}$. In ohter words, $C[f]({\bf
q})\in{\mathbb{H}}$ for all ${\bf q}\in{\mathbb{H}}$ if and only if $f$ is a
stem function, that is $f\in\mathcal{O}_{s}(U)$. Of course, in this case we
may (and shall) also use the notation $C[f]=f_{\mathbb{H}}$.

##  6 Analytic Functional Calculus in Quaternionic Framework

Let $\Omega\subset{\mathbb{H}}$ be a spectrally saturated open set, and let
$U=\mathfrak{S}(\Omega)\subset{\mathbb{C}}$ (which is also open by Remark
4(2)). We set

$\mathcal{R}(\Omega)=\\{f_{\mathbb{H}};f\in\mathcal{O}_{s}(U)\\},$

which is an ${\mathbb{R}}$-algebra, and

$\mathcal{R}(\Omega,{\mathbb{H}})=\\{F_{\mathbb{H}};F\in\mathcal{O}_{s}(U,{\mathbb{M}})\\},$

which, according to the next theorem, is a right $\mathcal{R}(\Omega)$-module.

In fact, these ${\mathbb{R}}$-linear spaces have some important properties, as
already noticed in a version of the next theorem (see Theorem 2 in [12]).

###### Theorem 5

Let $\Omega\subset{\mathbb{H}}$ be a spectrally saturated open set, and let
$U=\mathfrak{S}(\Omega)$. The space $\mathcal{R}(\Omega)$ is a unital
commutative ${\mathbb{R}}$-algebra, the space
$\mathcal{R}(\Omega,{\mathbb{H}})$ is a right $\mathcal{R}(\Omega)$-module,
the map

${\mathcal{O}}_{s}(U,{\mathbb{M}})\ni F\mapsto
F_{\mathbb{H}}\in\mathcal{R}(\Omega,{\mathbb{H}})$

is a right module isomorphism, and its restriction

${\mathcal{O}}_{s}(U)\ni f\mapsto f_{\mathbb{H}}\in\mathcal{R}(\Omega)$

is an ${\mathbb{R}}$-algebra isomorphism.

Moreover, for every polynomial
$P(\zeta)=\sum_{n=0}^{m}a_{n}\zeta^{n},\,\zeta\in{\mathbb{C}}$, with
$a_{n}\in{\mathbb{H}}$ for all $n=0,1,\ldots,m$, we have
$P_{\mathbb{H}}(q)=\sum_{n=0}^{m}a_{n}q^{n}\in{\mathbb{H}}$ for all
$q\in{\mathbb{H}}$.

Proof.  Thanks to Theorem 4, this statement is a particular case of Theorem 3.
Indeed, the ${\mathbb{R}}$-linear maps

${\mathcal{O}}_{s}(U,{\mathbb{M}})\ni F\mapsto
F_{\mathbb{H}}\in\mathcal{R}(\Omega,{\mathbb{H}}),\,{\mathcal{O}}_{s}(U)\ni
f\mapsto f_{\mathbb{H}}\in\mathcal{R}(\Omega),\,\,$

are restrictions of the maps

${\mathcal{S}}(U,{\mathbb{M}})\ni F\mapsto
F_{\mathbb{H}}\in\mathcal{F}(\Omega,{\mathbb{H}}),\,\,{\mathcal{S}}(U)\ni
f\mapsto f_{\mathbb{H}}\in\mathcal{F}(\Omega,{\mathbb{H}}),$

respectively. Moreover, they are ${\mathbb{R}}$-isomorphisms, the latter being
actually unital and multiplicative. Note that, in particular, for every
polynomial $P(\zeta)=\sum_{n=0}^{m}a_{n}\zeta^{n}$ with $a_{n}\in{\mathbb{H}}$
for all $n=0,1,\ldots,m$, we have $P_{\mathbb{H}}({\bf
q})=\sum_{n=0}^{m}a_{n}{\bf q}^{n}\in{\mathbb{H}}$ for all ${\bf
q}\in{\mathbb{H}}$.

###### Remark 8

For every function $F\in\mathcal{O}_{s}(U,{\mathbb{M}})$, the derivatives
$F^{(n)}$ also belong to $\mathcal{O}_{s}(U,{\mathbb{M}})$, where
$U\subset{\mathbb{C}}$ is a conjugate symmetric open set.

Now fixing $F\in\mathcal{O}_{s}(U,{\mathbb{M}})$, we may define its extended
derivatives with respect to the quaternionic variable via the formula

$F^{(n)}_{\mathbb{H}}({\bf q})=\frac{1}{2\pi
i}\int_{\Gamma}F^{(n)}(\zeta)(\zeta-{\bf q})^{-1}d\zeta,$ (11)

for the boundary $\Gamma$ of a Cauchy domain $\Delta\subset U$, $n\geq 0$ an
arbitrary integer, and $\sigma({\bf q})\subset\Delta$.

In particular, if $\Delta$ is a disk centered at zero and
$F\in\mathcal{O}_{s}(\Delta,{\mathbb{M}})$, so we have a representation of $F$
as a convergent series $\sum_{k\geq 0}a_{k}\zeta^{k}$ with coefficients in
${\mathbb{H}}$, then (11) gives the equality $F^{\prime}_{\mathbb{H}}({\bf
q})=\sum_{k\geq 1}ka_{k}{\bf q}^{k-1}$, which looks like a (formal) derivative
of the function $F_{\mathbb{H}}({\bf q})=\sum_{k\geq 0}a_{k}{\bf q}^{k}$.

The functions from the space $\mathcal{R}(\Omega,{\mathbb{H}})$ admit a series
development around any real point of their domain of definition. In this
sense, we have the following.

###### Proposition 2

Let $U\subset{\mathbb{C}}$ be conjugate symmetric, let
$F\in\mathcal{O}_{s}(U,{\mathbb{M}})$, let $s_{0}\in{\mathbb{R}}\cap U$, and
let $r>0$ be such that $D_{0}=\\{\zeta\in U;|\zeta-
s_{0}|<r\\}\subset\bar{D_{0}}\subset U$. Then we have

$F_{\mathbb{H}}({\bf q})=\sum_{n\geq 0}\frac{F^{(n)}(s_{0})}{n!}({\bf
q}-s_{0})^{n},\,\,\sigma({\bf q})\subset D_{0}.$

Proof Fixing ${\bf q}=x+y\mathfrak{s}$ with $x,y\in{\mathbb{R}}$ and
$\mathfrak{s}\in{\mathbb{S}}$, such that $\sigma({\bf q})=\\{x\pm iy\\}\subset
D_{0}$, we must have

$\|{\bf q}-s_{0}\|^{2}=|x-s_{0}|^{2}+y^{2}=|x\pm iy-s_{0}|^{2}<r^{2},$

implying the convergence of the series $\sum_{n\geq 0}r^{-n-1}({\bf
q}-s_{0})^{n}.$ Therefore,

$F_{\mathbb{H}}({\bf q})=\frac{1}{2\pi i}\int_{\Gamma_{0}}F(\zeta)(\zeta-{\bf
q})^{-1}d\zeta=\frac{1}{2\pi i}\int_{\Gamma_{0}}F(\zeta)\sum_{n\geq
0}\frac{({\bf q}-s_{0})^{n}}{(\zeta-s_{0})^{n+1}}d\zeta=$ $\sum_{n\geq
0}\frac{F^{(n)}(s_{0})}{n!}({\bf q}-s_{0})^{n},$

where $\Gamma_{0}$ is the boundary of $D_{0}$.

###### Remark 9

As already noticed in the framework of [12], Theorem 5 suggests a definition
for ${\mathbb{H}}$-valued ”analytic functions“ as elements of the set
$\mathcal{R}(\Omega,{\mathbb{H}})$, where $\Omega$ is a spectrally saturated
open subset of ${\mathbb{H}}$. Because the expression ”analytic function“ is
quite improper in this context, the elements of
$\mathcal{R}(\Omega,{\mathbb{H}})$ will be called Q-regular functions on
$\Omega$. In fact, the functions from $\mathcal{R}(\Omega,{\mathbb{H}})$ are
Cauchy transforms of the stem functions from
$\mathcal{O}_{s}(U,{\mathbb{M}})$, with $U=\mathfrak{S}(\Omega)$. Moreover, as
Proposition 2 seems to suggest, the functions from
$\mathcal{R}(\Omega,{\mathbb{H}})$ are ”real analytic“ rather than ”analytic“.

Except for Theorem 5, many other properties of $Q$-regular functions can be
obtained directly from the definition, by recapturing the corresponding
results from [12]. We omit the details.

###### Remark 10

Let $U\subset{\mathbb{C}}$ be conjugate symmetric, let $x,y\in{\mathbb{R}}$
with $y\neq 0$ and $z_{\pm}=x\pm iy\in U$, let
$F\in\mathcal{O}_{s}(U,{\mathbb{M}})$, and let $\mathfrak{s}\in{\mathbb{S}}$

Assuming $y>0$, we consider the quaternions ${\bf q}_{\pm}=x\pm y\mathfrak{s}$
for which for which $s_{+}({\bf q}_{\pm})=x+iy,\,s_{-}({\bf q}_{\pm})=x-iy$.

As we have $\tilde{\bf q}_{\pm}=\pm y\mathfrak{s}$, then $s_{\tilde{\bf
q}_{\pm}}=\pm\mathfrak{s}$, and $\iota_{\pm}(s_{\tilde{\bf q}_{+}})=(1\mp
i\mathfrak{s})/2,\iota_{\pm}(s_{\tilde{\bf q}_{-}})=(1\pm i\mathfrak{s})/2.$
Therefore,

$2F_{\mathbb{H}}({\bf
q}_{+})=F(z_{+})(1-i\mathfrak{s})+F(z_{-})(1+i\mathfrak{s}),$
$2F_{\mathbb{H}}({\bf
q}_{-})=F(z_{+})((1+i\mathfrak{s})+F(z_{-})(1-i\mathfrak{s}).$

From these equations we deduce that

$2F(z_{+})=F_{\mathbb{H}}({\bf q}_{+})(1-i\mathfrak{s})+F_{\mathbb{H}}({\bf
q}_{-})(1+i\mathfrak{s}),$ (12) $2F(z_{-})=F_{\mathbb{H}}({\bf
q}_{+})(1+i\mathfrak{s})+F_{\mathbb{H}}({\bf q}_{-})(1-i\mathfrak{s}).$ (13)

If $y<0$, for the quaternions ${\bf q}_{\pm}=x\pm y\mathfrak{s}$ we have
$s_{+}({\bf q}_{\pm})=x-iy,\,s_{-}({\bf q}_{\pm})=x+iy$. Moreover, as
$\tilde{\bf q}_{\pm}=\pm y\mathfrak{s}$, then $s_{\tilde{\bf
q}_{\pm}}=\mp\mathfrak{s}$, and $\iota_{\pm}(s_{\tilde{\bf q}_{+}})=(1\pm
i\mathfrak{s})/2,\break\iota_{\pm}(s_{\tilde{\bf q}_{-}})=(1\mp
i\mathfrak{s})/2.$ Therefore

$2F_{\mathbb{H}}({\bf
q}_{+})=F(z_{-})(1+i\mathfrak{s})+F(z_{+})(1-i\mathfrak{s}),$
$2F_{\mathbb{H}}({\bf
q}_{-})=F(z_{-})((1-i\mathfrak{s})+F(z_{+})(1+i\mathfrak{s}).$

These formulas lead again to equations (12) and (13). Consequently, we have
the following.

###### Proposition 3

Let $U\subset{\mathbb{C}}$ be conjugate symmetric, let $x,y\in{\mathbb{R}}$
with $x\pm iy\in U$, let $\mathfrak{s}\in{\mathbb{S}}$, and let
$F\in\mathcal{O}_{s}(U,{\mathbb{M}})$. Then we have the formulas

$F(x\pm iy)=F_{\mathbb{H}}(x\pm y\mathfrak{s})\left(\frac{1\mp
i\mathfrak{s}}{2}\right)+F_{\mathbb{H}}(x\mp y\mathfrak{s})\left(\frac{1\pm
i\mathfrak{s}}{2}\right).$ (14)

As the proof has been previously done, we only note that equality (14) also
holds for $y=0$.

###### Lemma 3

Let $U\subset{\mathbb{H}}$ be a conjugate symmetric open set, let
$\mathfrak{s}\in{\mathbb{S}}$ be fixed, and let
$\Psi:U_{\mathfrak{s}}\mapsto{\mathbb{H}}$ be such that
$\bar{\partial}_{\pm\mathfrak{s}}\Psi=0$. Then there exists a function
$\Phi\in\mathcal{R}(U_{\mathbb{H}},{\mathbb{H}})$ with
$\Psi=\Phi|U_{\mathfrak{s}}$, where
$U_{\mathfrak{s}}=\\{x+y\mathfrak{s},x+iy\in U\\}$.

Proof. For arbitrary points $z_{\pm}=x\pm iy\in U$ with $x,y(\neq
0)\in{\mathbb{R}}$, as in Remark 10, we consider the quaternions ${\bf
q}_{\pm}=x\pm y\mathfrak{s}$, so $s_{+}({\bf q}_{\pm})=x+i|y|$, and $s_{(}{\bf
q}_{\pm})=x-i|y|$. Inspired by formula (14), we set

$2F(z_{+})=\Psi({\bf q}_{+})(1-i\mathfrak{s})+\Psi({\bf
q}_{-})(1+i\mathfrak{s}),$ $2F(z_{-})=\Psi({\bf
q}_{+})(1+i\mathfrak{s})+\Psi({\bf q}_{-})(1-i\mathfrak{s}).$

Then we have

$2\frac{\partial F(z_{+})}{\partial x}=\frac{\partial\Psi({\bf
q}_{+})}{\partial x}(1-i\mathfrak{s})+\frac{\partial\Psi({\bf
q}_{-})}{\partial x}(1+i\mathfrak{s}),$

and

$2i\frac{\partial F(z_{+})}{\partial y}=\frac{\partial\Psi({\bf
q}_{+})}{\partial y}\mathfrak{s}(1-i\mathfrak{s})+\frac{\partial\Psi({\bf
q}_{-})}{\partial y}(-\mathfrak{s})(1+i\mathfrak{s}),$

because $i(1-i\mathfrak{s})=\mathfrak{s}(1-i\mathfrak{s})$ and
$i(1+i\mathfrak{s})=-\mathfrak{s}(1+i\mathfrak{s})$.

Therefore,

$\frac{\partial F(z_{+})}{\partial x}+i\frac{\partial F(z_{+})}{\partial
y}=(\bar{\partial}_{\mathfrak{s}}\Psi({\bf
q}_{+}))(1-i\mathfrak{s})+(\bar{\partial}_{\mathfrak{-s}}\Psi({\bf
q}_{-}))(1+i\mathfrak{s})=0,$

showing that the function $z_{+}\mapsto F(z_{+})$ is analytic in $U$.

Because $\overline{F(z_{-})}=\overline{F(\overline{z_{+}})}=F(z_{+})$, and
when $y=0$ we have $\overline{F(z_{-})}=F(z_{+})=F(x)$, we have constructed a
function $F\in\mathcal{O}_{s}(U,{\mathbb{M}})$. Hence, taking
$\Phi=F_{\mathbb{H}}$, we have
$\Phi\in\mathcal{R}(U_{\mathbb{H}},{\mathbb{H}})$ with
$\Psi=\Phi|U_{\mathfrak{s}}$, via Remark 10.

###### Theorem 6

Let $\Omega\subset{\mathbb{H}}$ be a spectrally saturated open set, and let
$\Phi:\Omega\mapsto{\mathbb{H}}$. The following conditions are equivalent:

$(i)$ $\Phi$ is a slice regular function;

$(ii)$ $\Phi\in\mathcal{R}(\Omega,{\mathbb{H}})$, that is, $\Phi$ is
$Q$-regular.

Proof. If $\Phi\in\mathcal{R}(\Omega,{\mathbb{H}})$, then $\Phi$ is slice
regular, by Lemma 1, so $(ii)\Rightarrow(i)$.

Conversely, let $\Phi$ be slice regular in $\Omega$. Fixing a
$\mathfrak{s}\in{\mathbb{S}}$, we have
$\bar{\partial}_{\pm\mathfrak{s}}\Phi_{\mathfrak{s}}=0$, where
$\Phi_{\mathfrak{s}}=\Phi|U_{\mathfrak{s}}$. It follows from Lemma 3 that
there exists $\Psi\in\mathcal{R}(U_{\mathbb{H}},{\mathbb{H}})$ with
$\Psi_{\mathfrak{s}}=\Phi_{\mathfrak{s}}$. This implies that $\Phi=\Psi$,
because both $\Phi,\Psi$ are uniquely determined by
$\Phi_{\mathfrak{s}},\Psi_{\mathfrak{s}}$, respectively, the former by (the
right hand version of) Lemma 4.3.8 in [2], and the latter by Remark 5.
Consequently, we also have $(i)\Rightarrow(ii)$.

Final Remark Theorem 6 allows us to obtain the properties of what we called
$Q$-regular functions via those of the slice regular functions, as in [2].
Clearly, there is no coincidence that a result like Proposition 2 looks like
Corollary 4.2.3 from [2]. Nevertheless, as mentioned before, they can also be
obtained directly, with the our techniques (see also [12]).

## References

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  * [2] F. Colombo, I. Sabadini and D. C. Struppa: Noncommutative Functional Calculus, Theory and Applications of Slice Hyperholomorphic Functions: Progress in Mathematics, Vol. 28 Birkhäuser/Springer Basel AG, Basel, 2011.
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ON THE NILRADICAL OF A LEIBNIZ ALGEBRA

DAVID A. TOWERS

Department of Mathematics and Statistics

Lancaster University

Lancaster LA1 4YF

England

<EMAIL_ADDRESS>

###### Abstract

The purpose of this short note is to correct an error which appears in the
literature concerning Leibniz algebras $L$: namely, that $N(L/I)=N(L)/I$ where
$N(L)$ is the nilradical of $L$ and $I$ is the Leibniz kernel.

Mathematics Subject Classification 2000: 17B05, 17B20, 17B30, 17B50.

Key Words and Phrases: Leibniz algebras, nilradical, radical, Frattini ideal,
Leibniz kernel.

##  1 Introduction

An algebra $L$ over a field $F$ is called a Leibniz algebra if, for every
$x,y,z\in L$, we have

$[x,[y,z]]=[[x,y],z]-[[x,z],y]$

In other word,s the right multiplication operator $R_{x}:L\rightarrow
L:y\mapsto[y,x]$ is a derivation of $L$. As a result, such algebras are
sometimes called right Leibniz algebras, and there is a corresponding notion
of left Leibniz algebra. Clearly, the opposite of a right Leibniz algebra is a
left Leibniz algebra so, for our purposes, it does not matter which is used.
Every Lie algebra is a Leibniz algebra and every Leibniz algebra satisfying
$[x,x]=0$ for every element is a Lie algebra.

Put $I=span\\{x^{2}:x\in L\\}$. Then $I$ is an ideal of $L$ and $L/I$ is a Lie
algebra called the liesation of $L$. We define the following series:

$L^{1}=L,L^{k+1}=[L^{k},L]\hbox{ and
}L^{(1)}=L,L^{(k+1)}=[L^{(k)},L^{(k)}]\hbox{ for all }k=2,3,\ldots$

Then $L$ is nilpotent (resp. solvable) if $L^{n}=0$ (resp. $L^{(n)}=0$) for
some $n\in{\mathbb{N}}$. The nilradical, $N(L)$, (resp. radical, $R(L)$) is
the largest nilpotent (resp. solvable) ideal of $L$.

In [10] it is claimed that $R(L/I)=R(L)/I$ and $N(L/I)=N(L)/I$. However,
whereas the former is clearly true, the latter is false in general. The claim
appears in the proof of [10, Proposition 4], which has two corollaries.
Although this paper appears only to have been published on arxiv it has been
quite widely cited. Moreover, the results following from this assertion have
been quoted in [8, Proposition 4.4], [12, Propositions 3.3, 3.4 and
Corollaries 3.5,3.6] and referenced to [10], and the same incorrect assertion
is used to prove two results in a recent book ([3, Proposition 2.1 and 2.2]).
The results which Gorbatsevich uses this assertion to prove are true, as are
[3, Proposition 2.1 and 2.2]. The purpose of this short note is to give a
correct characterisation of $N(L/I)$ and to provide or reference proofs for
the results in which the incorrect assertion is used.

Throughout, $L$ will denote a finite-dimensional (right) Leibniz algebra over
a field $F$. The Frattini ideal of $L$ is the largest ideal of $L$ contained
in every maximal subalgebra of $L$. We will denote algebra direct sums by
$\oplus$ and vector space direct sums by $\dot{+}$.

##  2 The nilradical

The literature concerning Leibniz algebras is quite diverse and a number of
results have been duplicated, so the survey articles [9], [10] and the new
book [3] are useful. The nilradical of a Leibniz algebra is a well-defined
object.

###### Theorem 2.1

The sum of two nilpotent ideals of a Leibniz algebra is nilpotent.

Proof. See [9, Theorem 5.14] or [5, Lemma 1.5]. $\Box$

###### Corollary 2.2

Any Leibniz algebra has a maximal nilpotent ideal containing all nilpotent
ideals of $L$.

Proof. A valid proof of this can also be found as [7, Corollary 4]. Note that
the proof given in [10, Proposition 1] and [3, Proposition 2.1] is incorrect.
$\Box$

Next, we show that $N(L/I)\neq N(L)/I$ in general.

###### Example 2.1

Let $L=Fx+Fx^{2}$ where $[x^{2},x]=x^{2}$ is the two-dimensional solvable
cyclic Leibniz algebra, we have $I=Fx^{2}=N(L)$ and $L/I$ is the nilradical of
$L/I$.

The fact that $N(L)=I$ is not significant in the above example as the
following class of algebras shows.

###### Example 2.2

Let $L=Fx_{1}+\ldots+Fx_{r}+Fx_{r+1}+\ldots+Fx_{n}+Fy$ where $[x_{i},y]=x_{i}$
for $r+1\leq i\leq n$ and all other products are zero. Then
$I=Fx_{r+1}+\ldots+Fx_{n}$, $N(L)=Fx_{1}+\ldots+Fx_{n}$ and $N(L/I)=L/I$.

Nor is the fact that $N(L/I)=L/I$ significant in the above examples as can be
seen by taking the direct sum of them with a simple Lie algebra.

Next we look for more information on $N(L/I)$. First note the following.

###### Lemma 2.3

If $I\subseteq\phi(L)$ then $N(L/I)=N(L)/I$.

Proof. Clearly $N(L)/I\subseteq N(L/I)=K/I,$ say. But $K$ is nilpotent, by [4,
Theorem 5.5]. $\Box$

###### Lemma 2.4

If $I\not\subseteq\phi(L)$ then there is a subalgebra $B$ of $L$ such that
$L=I+B$ and $I\cap B\subseteq\phi(B)$.

Proof. This is [13, Lemma 7.1]. $\Box$

Then, using the same notation as in Lemma 2.4, we have the following.

###### Theorem 2.5

The nilradical of $L/I$ is $N(L/I)=(I+N(B))/I$ and this is the same as
$N(L)/I$ if and only if $R_{n}|_{I}$ is nilpotent for all $n\in N(B)$.

Proof. We have

$N\left(\frac{L}{I}\right)\cong N\left(\frac{B}{I\cap
B}\right)=\frac{N(B)}{I\cap B}=\frac{N(B)}{I\cap N(B)}\cong\frac{I+N(B)}{I}.$

But $(I+N(B)/I\subseteq N(L/I)$, so equality results.

Now, $I+N(B)=N(L)$ if and only if $N(B)$ acts nilpotently on the right on $I$.
$\Box$

##  3 Some results where $N(L/I)=N(L)/I$ were used

First we have the following analogue of a well-known result for Lie algebras.
The only references to this in the literature of which we are aware are [10,
Proposition 4] and [3, Proposition 2.2]. However, the proof in each case is
incorrect, though the result is true.

###### Proposition 3.1

Let $L$ be a Leibniz algebra over a field of characteristic zero, $R$ be its
radical and $N$ its nilradical. Then $[L,R]\subseteq N$.

Proof. Since $N(L/\phi(L))=N(L)/\phi(L)$ we can assume that $L$ is
$\phi$-free. Then $L=$Asoc$(L)\dot{+}V$, where $V=S\oplus Z(V)$ and $S$ is a
semisimple Lie algebra, by [6, Corollary 2.9]. Also $R=$Asoc$(L)+Z(V)$ and
$N=$Asoc$(L)$, by [6, Theorem 2.4], from which the result is clear. $\Box$

The above result has the following corollary which appears in several places
in the literature. It occurs as [10, Corollaries 5 and 6] and [3, Corollaries
2.2 and 2.3] but the proofs are incorrect. It appears with correct proofs as
[9, Corollary 6.8], [11, Corollary 3], [2, Theorem 4], [1, Theorem 2] and [4,
Theorem 2.6].

###### Corollary 3.2

([10, Corollary 5]) With same notation as in Proposition 3.1, $[R,R]\subseteq
N$; in particular, $[R,R]$ is nilpotent; in fact, $L$ is solvable if and only
if $[L,L]$ is nilpotent.

## References

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