text:
Cellular Chair' by Danish designer, Mathias Bengtsson, is a genius blend of art, design and biotechnology. Created by Bengtsson by means of a specific computer program invented by the artist himself, which is based on the evolution of cellular bone tissues, 'Cellular Chair' is not only organic in form but in structure as well: rather than composed, the chair has been "grown" obeying the growth logic of a living organism. It is produced in a limited edition of 8 unique pieces, each with it's own DNA so to speak. Made of resin and silver, 'Cellular Chair' was exhibited at the Victoria and Albert Museum in the exhibition 'Industrial Revolution 2.0.' in September 2011.
Always seeking to take his thinking in new directions, Bengtsson breaks down established boundaries between design, art, craft and technology. As a result of several years of research, Bengtsson has invented a computer program in which a 'digital seed' grows in a virtual world and simulates natural growth, imitating some of nature's rules and methods. Combining high technology and traditional craftsmanship, the result is a strikingly beautiful piece where art, crafts and design dialogue in the most exquisite manner.
Mathias Bengtsson has always described himself as a designer, but his works are closer to fine art than to traditional industrial design. Working with diverse industrial materials and processes, Bengtsson pushes forward the sculptural, technical, and philosophical possibilities of three-dimensional design. Accordingly, the majority of his works are unique or created as editions, requiring both high-tech production facilities and labour-intensive processes. Museums on both sides of the Atlantic have been quick to acquire Bengtsson's works, and his pieces have entered the permanent collections of institutions including the Museum of Modern Art, New York, the Contemporary Art Museum Houston, the Montreal Museum of Fine Arts, and Manchester Art Gallery. Almost uniquely for a designer around 40, his work has been shown extensively both in fine art and design contexts, at venues such as the Design Museum, London and the Rohsska Museum, Gotheberg, and in public galleries such as The Lowry and Northern Gallery for Contemporary Art.
Bengtsson's work sprang to prominence even before his graduation from the Royal College of Art, when he exhibited his 'Slice' chairs at Galleria Post Design in Milan, in 1998. Such auspicious beginnings have sometimes distracted from the variety of his subsequent oeuvre, which have seen him move into uncharted territory in each year since. The designer has reamined best known for his works which adapt the archetypal forms of seating and domestic furniture into new territories: the easy chair; the chaise longue; the bench; the dining table. Each of these object-types offers a different point of departure, which Bengtsson uses to expand the language of furniture design. In his hands, the cubic volumes of a lounge chair provide opportunity to bring to mind human forms, those of the automotive industry, and those of the landscape or sea. We are invited to engage with his objects through different frames of reference and on different levels beyond their functional use. The designer has said, "I don't count myself as a product designer. Furniture is just a platform or medium to present my ideas and a framework to 'build' from. What is crucial is that furniture has a set of known rules and conventions, and a set of minimum requirements." All of Bengtsson's work is able to be used as furniture, and indeed intentionally recalls the "organic shapes of the Danish design tradition" in his words. However, this tradition, and indeed the need to sit or lie down are simply an armature for his ideas. These include arguments about the possibilities of design; about the technologies – digital and analogue – which determine our present day means of production; and about the imagined relation between the body, the spaces it occupies, and its environment.
Bengtsson's determining logic is then, not that form follows function, nor of function following form, but of form surpassing our imaginative boundaries. His aim is to give life to what we could not previously have been thought possible. Bengtsson begins with what he calls "the promise of technology" – the desire to push the limits of what is imaginable, and the will to create objects unlike any that have existed before. In the words of one critic, Bengtsson's works are a "quest to redefine the boundaries of design". This has involved treating materials in ways which make them defy and surpass their capabilities or characteristics. Bengtsson's palette of materials has expanded radically in recent years to encompass a bewildering diversity of industrially manufactured materials – from aluminium to cardboard, and from tinted acrylic to carbon fibre. But there is a common denominator: each of Bengtsson's works exceeds what we thought design could be, and what the material in question could achieve.
For Bengtsson, it is crucial that the unfamiliarity of his works is countered by a familiar frame of reference. They are, or resemble domestic objects which are made strange rather than artworks made functional. Rather than creating autonomous sculptures, Bengtsson prefers the challenge of creating a "tension" between achieving functionality and distilling ideas about the big issues of design. Indeed the question of functionality in furniture determines why Bengtsson continues to define himself as a designer: it is simply that his objects serve different functions to what he calls "padded furniture". As he notes, "the functions of providing optimal comfort and convenience have long been solved – the study of ergonomics was completed in the 1960s," he argues. "My furniture is about challenging your senses: that is its function."
Bengtsson's solutions to the challenge as outlined often result in him creating seemingly organic forms which 'nature' could not have possibly generated. He sees his work as a branch of research and experimentation: that some of his findings may find future wider applicability is an open possibility; but in the main, Bengtsson's activities should be seen as pure rather than applied research. "My area of investigation is to explore what could be the ideas of the future" he has remarked, but a popular misreading of Bengtsson's works is that they are intentionally futuristic in appearance. Such an assessment is faulty on two counts. Firstly, the designer's works begin with, and derive from a love of materials and their possibilities, rather than from an attempt to evoke an image of the future. It is crucial to the designer that "my designs are materials and process-based, not 'image-based' – I am a fanatic about materials, technologies, and the processes dependent on them. I prefer the shapes of my works to derive from materials and processes, rather than solely from a pre-existing concept or image unrelated to them." As outlined, his process is not that of function following form: each object comes out of an investigation into the gap between what materials are capable of and what it is unimaginable that they should achieve.
Secondly, Bengtsson's works are almost all based upon achieving what he calls a "tension" between the artisanal and the precision-engineered. "I have never seen anything made solely on a computer which has that 'extra something' which you have never seen before," he notes. Those unfamiliar with Bengtsson's works often begin by observing that he utilizes high-performance materials and high-tech processes to achieve organic forms. The genesis of each object is far more complicated than any first impression could convey. The designer's first 'slice' chair provides a perfect example of the complexity of Bengtsson's process, and how each part of the process contributes to the finished product. He has said that "I always start work with a pen and paper, or brush: drawing provides the shortest distance between an idea in your imagination and a form. The problems are then how you 'translate' an analogue idea into digital data, and how you can produce a digital design by hand." Having conceived how the very first sketches for the 'slice' chair, the designer then created a wax model which was literally 'sliced' with a knife. Each slice was scanned, then redrawn and amended digitally. Only at a very late stage was computer-controlling cutting deployed; and after this, the chair was constructed by hand.
By definition, then, "every slice chair is different, is an individual" rather than being a clinically perfect, computer-generated object. Rather, Bengtsson's works are artefacts which yoke the work of "machines and the human hand" so that they feel "somewhere between both." Technologies, for the designer, are seldom a completely self-contained means of production. They are more often a way of complicating our relationship to objects, and our relationship to the modern world. Many of the machine processes the designer has employed originated in highly capitalized and high-tech industries, such as motor racing and aviation.
The associations of these industries and their artefacts are unavoidable, but Bengtsson's use of materials are often functional in the most obvious senses. His carbon fibre works, for example, being made of "high-performance materials", are the lightest chairs ever made. They are not only hollow but created from slender woven strips of filament able to support a human being with the minimum materials necessary. A proposal for a carbon fibre bridge linking two islands has seen Bengtsson operate on an urban scale, marrying his ideas to those of civic engineering to suggest entirely new solutions to age-old problems. Bengtsson's bridge, yet to be realized, could be the first introduction of entirely new materials into the vocabulary of bridge-building for nearly a century. If such a process sounds singularity programmatic on the other hand, it is worth remembering that Bengtsson has also drawn ideas from the most unlikely of places. The process of laser-cutting enabled his 'slice' works to be fabricated, but his imagination was sparked to create them by visceral ideas. He has noted that a generation ago, some executed criminals (elsewhere in the world) were frozen in nitrogen before being sliced into sections. Encountering the human body as cross-sections made it "unrecognizable" to him, both estranged and fascinating. This alerted him the designer to how other objects could be both made comprehensible and unfamiliar at the same time. The designer's work has exploited a disjuncture between our imagined archetypes of the forms of furniture and the range of new metaphors he has brought to the field, and the sculptural adventures he has taken his materials upon. Even at first impressions, his 'slice' chairs offer up to our imagination the craggy forms of a coastal landscape more than those of a domestic object, as though they had been eroded by the elements over decades. In more recent works still under development, computer-carved forms resemble the quicksilver zip of mercury underwater – a microscopic element of the landscape rather than a section of it. By comparison, the basic cylindrical form of a chaise longue offers up a voluptuous form whose proportions echo those of both a panoramic landscape and a recumbent female body. Both the acrylic and carbon fibre chaise longues conjure Henry Moore-like hills and hollows, evoking both the macrocosmic scale of the land and the familiar contours of the human form.
His unorthodox methods, and insistence on combining craft skills with mechanical means help the designer's works accrue several levels of meaning. And he continues to push forward boundaries of what might be possible in three-dimensional design in ways we could scarcely expect.
Resin, Silver


text:
\section{Introduction}\label{section-introduction}

The history of {\it harmonic morphisms} can be traced back to the pioneering work \cite{Jac} of Jacobi from 1848.  He studies complex-valued functions pulling back harmonic functions in the complex plane ${\mathbb C}$ to harmonic functions in the $3$-dimensional Euclidean space ${\mathbb R}^3$.  The notion is then generalised to the Riemannian setting in the late 1970s, independently by Fuglede and Ishihara, see \cite{Fug} and \cite{Ish}.  This has led to a lively development that can be followed both in \cite{Bai-Woo-book} and at the regularly updated on-line bibliography \cite{Gud-bib}.

Loubeau and Ou study {\it biharmonic morphisms} between Riemannian manifolds, see \cite{Lou-Ou-1} and \cite{Lou-Ou-2}.  These are maps pulling back biharmonic functions to biharmonic functions.  In his work \cite{Mae}, Maeta introduces the notion of {\it triharmonic morphisms}.  These are mappings pulling back triharmonic functions to triharmonic functions.

Recently, Ghandour and Ou introduce the notion of {\it generalised harmonic morphisms} between Riemannian manifolds, see \cite{Gha-thesis} and \cite{Gha-Ou-1}.  These are maps pulling back harmonic functions to biharmonic functions.  They also find a characterisation for these non-linear objects.
\medskip

In this work we unify the above notions by defining the {\it $(p,q)$-harmonic morphisms}.  These are maps between Riemannian manifolds pulling back $q$-harmonic functions to $p$-harmonic  functions.  Just as the harmonic morphisms and their above mentioned variants, they are solutions to an over-determined system of {\it non-linear} partial differential equations.  This means that they have no general existence theory.  For this reason it is interesting to develop methods for constructing solutions in particular cases.

In this paper we focus our attention on complex-valued $(p,q)$-harmonic morphisms from Riemannian manifolds. The aim is to extend the known characterisation to this case and to manufacture new non-trivial examples to this non-linear problem.  The explicit examples presented here involve rather demanding computations.  They were all tested, by the computer algebra systems Maple and Mathematica, independently.

\section{Preliminaries}\label{section-preliminaries}

Let $(M,g)$ be an $m$-dimensional  Riemannian manifold and $T^{{\mathbb C}}M$ be the complexification of the tangent bundle $TM$ of $M$. We extend the metric $g$ to a complex-bilinear form on $T^{{\mathbb C}}M$.  Then the gradient $\nabla z$ of a complex-valued function $z:(M,g)\to{\mathbb C}$ is a section of $T^{{\mathbb C}}M$.  In this situation, the well-known complex linear {\it Laplace-Beltrami operator} (alt. {\it tension field}) $\tau$ on $(M,g)$ acts locally on $z$ as follows
$$
\tau(z)=\Div (\nabla z)=\sum_{i,j=1}^m\frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x_j}
\left(g^{ij}\, \sqrt{|g|}\, \frac{\partial z}{\partial x_i}\right).
$$
For two complex-valued functions $z,w:(M,g)\to{\mathbb C}$ we have the following well-known relation
\begin{equation}\label{equation-basic}
\tau(z\cdot w)=\tau(z)\cdot w +2\cdot\kappa(z,w)+z\cdot\tau(w),
\end{equation}
where the complex bilinear {\it conformality operator} $\kappa$ is given by $\kappa(z,w)=g(\nabla z,\nabla w)$.  Locally this satisfies
$$\kappa(z,w)=\sum_{i,j=1}^mg^{ij}\cdot\frac{\partial z}{\partial x_i}\frac{\partial w}{\partial x_j}.$$
For the naming of the operator $\kappa$, we have the following.
\begin{remark}
Note that for a complex-valued function $z=x+i\,y:(M,g)\to{\mathbb C}$ we have
$$\kappa(z,z)=(|\nabla x|^2-|\nabla y|^2)+2i\, g(\nabla x,\nabla y)=0$$
if and only if the two gradients $\nabla x$ and $\nabla y$ are orthogonal and of the same length at every point $p\in M$ i.e. $z$ is {\it horizontally conformal}, see \cite{Bai-Woo-book}.
\end{remark}

As a direct consequence of the complex linearity and bi-linearity of the operators $\tau$ and $\kappa$, respectively, we have the following.

\begin{lemma}\label{lemma-conjugate}
Let $(M,g)$ be a Riemannian manifold and $z,w:(M,g)\to{\mathbb C}$ be two complex-valued functions.  Then the tension field $\tau$ and the conformality operator $\kappa$ satisfy
\begin{equation}
\overline{\tau(z)}=\tau(\bar z)\ \ \text{and}\ \ \overline{\kappa(z,w)}=\kappa(\bar z,\bar w).
\end{equation}
\end{lemma}

We are now ready to define the complex-valued proper $p$-harmonic functions, the main objects of our study.

\begin{definition}\label{definition-proper-r-harmonic}
For a positive integer $p$, the iterated Laplace-Beltrami operator $\tau^p$ is given by
$$\tau^{0} (z)=z\ \ \text{and}\ \ \tau^p (z)=\tau(\tau^{(p-1)}(z)).$$	We say that a complex-valued function $z:(M,g)\to{\mathbb C}$ is
\begin{enumerate}
\item[(a)] {\it $p$-harmonic} if $\tau^p (z)=0$, and
\item[(b)] {\it proper $p$-harmonic} if $\tau^p (z)=0$ and $\tau^{(p-1)} (z)$ does not vanish identically.
	\end{enumerate}
\end{definition}

We now introduce the natural notion of a $(p,q)$-harmonic morphism.  For $(p,q)=(1,1)$ this is the classical case of harmonic morphisms introduced by Fuglede and Ishihara, in \cite{Fug} and \cite{Ish}, independently.

\begin{definition}\label{definition-pq}
A map $\phi:(M,g)\to (N,h)$ between Riemannian manifolds is said to be a {\it $(p,q)$-harmonic morphism} if, for any {\it $q$-harmonic} function $f:U\subset N\to{\mathbb R}$, defined on an open subset $U$ such that $\phi^{-1}(U)$ is not empty, the composition $f\circ\phi:\phi^{-1}(U)\subset M\to{\mathbb R}$ is  {\it $p$-harmonic}.
\end{definition}

As an immediate consequence of Definition \ref{definition-pq} we have the following natural composition law.

\begin{lemma}\label{lemma-composition}
Let $\phi:(M,g)\to (\bar N,\bar h)$ be a $(p,r)$-harmonic morphism between Riemannian manifolds.  If $\psi:(\bar N,\bar h)\to (N,h)$ is an $(r,q)$-harmonic morphism then  the composition $\psi\circ\phi:(M,g)\to (N,h)$ is a $(p,q)$-harmonic morphism.
\end{lemma}

Another useful consequence of Definition \ref{definition-pq} is the following.

\begin{lemma}\label{lemma-higher-lower}
Let $\phi:(M,g)\to (N,h)$ be a $(p,q)$-harmonic morphism between Riemannian manifolds.  Then $\phi$ is a $(p,q^\prime)$-harmonic morphism for $q>q^\prime$ and is a $(p^\prime,q)$-harmonic morphism for $p^\prime >p$.
\end{lemma}

In our Theorem \ref{theorem-(p,q)}, we show that a complex-valued  $(p,q)$-harmonic morphism is a $p$-harmonic function and for this situation we define the following.

\begin{definition}A complex-valued $(p,q)$-harmonic morphism  $z:(M,g)\to\mathbb{C}$ from a Riemannian manifold is said to be {\it proper} if it is proper as a $p$-harmonic function i.e. $\tau^p(z)=0$ and $\tau^{p-1}(z)\neq 0$.
\end{definition}

\section{Complex-valued $(2,q)$-Harmonic Morphisms}

Throughout this work we assume that $z:(M,g)\to{\mathbb C}$ is a smooth complex-valued function on a Riemannian manifold and that $f:U\to{\mathbb C}$ is differentiable and defined on an open subset $U$ of ${\mathbb C}$ containing the image $z(M)$ of $z$.  Further let $\phi:(M,g)\to{\mathbb C}$ be the composition $\phi=f\circ z$.  For this situation we have the following result that later will be employed several times.  This is easily obtained by using the chain rule.

\begin{lemma}\label{lemma-fundamental}
Let $z:(M,g)\to{\mathbb C}$ be a complex-valued function on a Riemannian manifold and $F,G:U\to{\mathbb C}$ be differentiable functions defined on an open subset $U$ of ${\mathbb C}$ containing the image $z(M)$ of $z$. Then the tension field $\tau$ and the conformality operator $\kappa$ satisfy
\begin{eqnarray*}
\tau(F(z,\bar z))&=&\frac{\partial F}{\partial z}\cdot\tau(z)+\frac{\partial F}{\partial \bar z}\cdot\tau(\bar z)\\
& &+\frac{\partial^2 F}{\partial z^2}\cdot\kappa(z,z)+
2\,\frac{\partial^2 F}{\partial z\partial\bar z}\cdot\kappa(z,\bar z)+\frac{\partial^2 F}{\partial \bar z^2}\cdot\kappa(\bar z,\bar z).
\end{eqnarray*}
and
\begin{equation*}
\kappa(F(z,\bar z),G(z,\bar z))=\frac{\partial F}{\partial z}\cdot\kappa(z,G(z,\bar z))+\frac{\partial F}{\partial \bar z}\cdot\kappa (\bar z,G(z,\bar z)).
\end{equation*}
\end{lemma}

As a direct consequence of Lemma \ref{lemma-fundamental}, we now see that the tension field $\tau(\phi)$ of the composition $\phi=f\circ z$ is given by
\begin{equation}\label{equation-tau1}
\begin{aligned}
\tau(\phi)&=\frac{\partial f}{\partial z}\cdot\tau(z)+\frac{\partial f}{\partial \bar z}\cdot\tau(\bar z) \\
&\qquad +\frac{\partial^2 f}{\partial z^2}\cdot\kappa(z,z)+
2\,\frac{\partial^2 f}{\partial z\partial\bar z}\cdot\kappa(z,\bar z)+\frac{\partial^2 f}{\partial \bar z^2}\cdot\kappa(\bar z,\bar z).
\end{aligned}
\end{equation}

For the completeness of our exposition we now state the following.  This  recovers the classical result of Fuglede and Ishihara in our special case of complex-valued functions.

\begin{theorem}\label{theorem-(1,1)}
A complex-valued function $z:(M,g)\to{\mathbb C}$ from a Riemannian manifold is a $(1,1)$-harmonic morphism if and only if $$\kappa(z,z)=0\ \ \text{and}\ \ \tau(z)=0.$$
\end{theorem}

\begin{proof}
The function $z:(M,g)\to{\mathbb C}$ is a $(1,1)$-harmonic morphism if and only if, for any harmonic $f:U\to{\mathbb C}$ defined on an open subset $U$ of ${\mathbb C}$ containing the image $z(M)$ of $z$, the tension field $\tau(\phi)$ of the composition $\phi=f\circ z$ vanishes. Since the function $f$ is assumed to be harmonic we have
$$\tau(f)=\frac{\partial^2 f}{\partial z\partial\bar z}=0.$$
It now follows from Lemma \ref{lemma-conjugate} and equation (\ref{equation-tau1}) that $\tau(\phi)=0$ is equivalent to $$\kappa(z,z)=0\ \ \text{and}\ \ \tau(z)=0.$$
\end{proof}

\begin{proposition}\label{proposition-constant}
Let $z:(M,g)\to{\mathbb C}$ be a complex-valued $(1,q)$-harmonic morphism from a Riemannian manifold. If $1<q$ then the function $z$ is constant.
\end{proposition}

\begin{proof}
The condition $1<q$ implies from (3.1) that both $\kappa(z,z)=0$ and $\kappa(z,\bar z)=0$ or equivalently that the function $z$ is constant.
\end{proof}

The next result is our fundamental tool for analysing the case of $(2,q)$.

\begin{lemma}\label{lemma-tau2}
Let $z:(M,g)\to\mathbb{C}$ be a complex-valued function from a Riemannian manifold and $f:U\to\mathbb{C}$ be defined on an open subset $U$ of $\mathbb{C}$ containing the image $z(M).$ Then the $2$-tension field $\tau^2(\phi)$ of the composition $\phi=f\circ z$ satisfies
\begin{eqnarray*}
	& &\tau^2(\phi)\\
	&=&\tau^2(z)\cdot\frac{\partial f}{\partial z}+\tau^2(\bar z)\cdot\frac{\partial f}{\partial \bar z}\\
	& &+\bigl[\tau(z)^2+2\cdot\kappa(z,\tau(z))+\tau(\kappa(z,z))\bigr]\cdot\frac{\partial^2 f}{\partial z^2}\\
	& &+2\cdot \bigl[\tau(z)\tau(\bar z)+\kappa(z,\tau(\bar z))+\kappa(\bar z,\tau(z))+\tau(\kappa(z,\bar z))\bigr]\cdot\frac{\partial^2 f}{\partial z\partial\bar z}\\
	& &+\bigl[\tau(\bar z)^2+2\cdot\kappa(\bar z,\tau(\bar z))+\tau(\kappa(\bar z,\bar z))\bigr]\cdot\frac{\partial^2 f}{\partial\bar z^2}\\
	& &+2\cdot
	\bigl[\kappa(z,z)\tau(z)+\kappa(z,\kappa(z,z))\bigr]\cdot\frac{\partial^3 f}{\partial z^3}\\
	& &+2\cdot
	\bigl[2\cdot\kappa(z,\bar z)\tau(z)+\kappa(z,z)\tau(\bar z)\\
	& &\quad\quad  +\kappa(\bar z,\kappa(z,z))+2\cdot\kappa(z,\kappa(z,\bar z))\bigr]\cdot\frac{\partial^3 f}{\partial z^2\partial\bar z}\\
	& &+2\cdot\bigl[2\cdot\kappa(z,\bar z)\tau(\bar z) +\kappa(\bar z,\bar z)\tau(z)\\
	& &\quad\quad+\kappa(z,\kappa(\bar z,\bar z))+2\cdot\kappa(\bar z,\kappa(z,\bar z))
	\bigr]\cdot\frac{\partial^3 f}{\partial z\partial\bar z^2}\\
	& &+2\cdot
	\bigl[\kappa(\bar z,\bar z)\tau(\bar z)+\kappa(\bar z,\kappa(\bar z,\bar z))\bigr]\cdot\frac{\partial^3 f}{\partial\bar z^3}\\
	& &+\kappa(z,z)^2\cdot\frac{\partial^4 f}{\partial z^4}+4\cdot
	\kappa(z,z)\kappa(z,\bar z)\cdot\frac{\partial^4 f}{\partial z^3\partial\bar z}\\
	& &+2\cdot\bigl[\kappa(z,z)\kappa(\bar z,\bar z)+2\cdot\kappa(z,\bar z)^2\bigr]\cdot\frac{\partial^4 f}{\partial z^2\partial\bar z^2}\\
	& &+4\cdot
	\kappa(\bar z,\bar z)\kappa(z,\bar z)\cdot\frac{\partial^4 f}{\partial z\partial\bar z^3}+\kappa(\bar z,\bar z)^2\cdot\frac{\partial^4 f}{\partial \bar z^4}.\\
\end{eqnarray*}

\end{lemma}

\begin{proof}
Utilising the two basic equations (\ref{equation-basic}) and (\ref{equation-tau1}) we see that the $2$-tension field $\tau^2(\phi)$ of the composition $\phi=f\circ z$ satisfies
\begin{eqnarray*}
& &\tau^2(\phi)\\
&=&\tau(\frac{\partial f}{\partial z})\cdot\tau(z)+2\cdot\kappa(\frac{\partial f}{\partial z},\tau(z))+\frac{\partial f}{\partial z}\cdot\tau^2(z)\\
& &\quad +\tau(\frac{\partial f}{\partial \bar z})\cdot\tau(\bar z)+2\cdot\kappa(\frac{\partial f}{\partial \bar z},\tau(\bar z))+\frac{\partial f}{\partial \bar z}\cdot\tau^2(\bar z)\\
& &\quad +\tau(\frac{\partial^2 f}{\partial z^2})\cdot\kappa(z,z)
+2\cdot\kappa(\frac{\partial^2 f}{\partial z^2},\kappa(z,z))
+\frac{\partial^2 f}{\partial z^2}\cdot\tau(\kappa(z,z))\\
& &\quad +2\cdot\tau(\frac{\partial^2 f}{\partial z\partial\bar z})\cdot\kappa(z,\bar z)
+4\cdot\kappa(\frac{\partial^2 f}{\partial z\partial\bar z},\kappa(z,\bar z))
+2\cdot\frac{\partial^2 f}{\partial z\partial\bar z}\cdot\tau(\kappa(z,\bar z))\\
& &\quad +\tau(\frac{\partial^2 f}{\partial \bar z^2})\cdot\kappa(\bar z,\bar z)
+2\cdot\kappa(\frac{\partial^2 f}{\partial \bar z^2},\kappa(\bar z,\bar z))
+\frac{\partial^2 f}{\partial \bar z^2}\cdot\tau(\kappa(\bar z,\bar z)).
\end{eqnarray*}
By applying Lemma 3.1 and reordering the terms we then obtain the stated result.
\end{proof}

For later use, we now reformulate Lemma \ref{lemma-tau2} and thereby show that the $2$-tension field $\tau^2(\phi)$ of $\phi$ can be presented in terms of the different partial derivatives of $f$ with coefficients determined by the functions $z$,$\bar z$ and their various tension fields.

\begin{lemma}\label{lemma-tau2-expansion}
	Let $z:(M,g)\to\mathbb{C}$ be a complex-valued function from a Riemannian manifold and $f:U\to\mathbb{C}$ be defined on an open subset $U$ of $\mathbb{C}$ containing the image $z(M).$ Then the $2$-tension field $\tau^2(\phi)$ of the composition $\phi=f\circ z$ satisfies
	\begin{eqnarray*}
		& &\tau^2(\phi)\\
		&=&\tau^2(z)\cdot\frac{\partial f}{\partial z}+\tau^2(\bar z)\cdot\frac{\partial f}{\partial \bar z}\\
		& &+\bigl[\tfrac 12\tau^2(z^2)-z\, \tau^2(z)\bigr]\cdot\frac{\partial^2 f}{\partial z^2}\\
		& &+\bigl[\tau^2(z\bar{z})-\bar{z}\tau^2(z)-z\tau^2(\bar{z})\bigr]\cdot\frac{\partial^2 f}{\partial z\partial\bar z}\\
		& &+\bigl[\tfrac{1}{2}\tau^2(\bar{z}^2)-\bar{z}\tau^2(\bar{z})\bigr]\cdot\frac{\partial^2 f}{\partial\bar z^2}\\
		& &+
		\bigl[\tfrac{1}{6}\tau^2(z^3)-\tfrac{1}{2}z\tau^2(z^2)+\tfrac{1}{2}z^2\tau^2(z)\bigr]\cdot\frac{\partial^3 f}{\partial z^3}\\
		& &+
		\bigl[\frac 12\tau^2(z^2\bar z)-\frac 12\bar z\tau^2(z^2)+z\bar z\tau^2(z)- z\tau^2(z\bar z)+\frac 12z^2\tau^2(\bar z)\bigr]\cdot\frac{\partial^3 f}{\partial z^2\partial\bar z}\\
		& &+\bigl[\frac 12\tau^2(z\bar z^2)-\frac 12 z\tau^2(\bar z^2)+z\bar z\tau^2(\bar z)- \bar z\tau^2(z\bar z)+\frac 12 \bar z^2\tau^2(z)
		\bigr]\cdot\frac{\partial^3 f}{\partial z\partial\bar z^2}\\
		& &+
		\bigl[\tfrac{1}{6}\tau^2(\bar z^3)-\tfrac{1}{2}\bar z\tau^2(\bar z^2)+\tfrac{1}{2}\bar z^2\tau^2(\bar z)\bigr]\cdot\frac{\partial^3 f}{\partial\bar z^3}\\
		& &+\bigl[\tfrac{1}{24}\tau^2(z^4)-\tfrac{1}{6}z\tau^2(z^3)+\tfrac{1}{4}z^2\tau^2(z^2)-\tfrac{1}{6}z^3\tau^2(z)\bigr]\cdot\frac{\partial^4 f}{\partial z^4}+\\
		& &+ \bigl[\tfrac{1}{6}\tau^2(z^3\bar{z})-\tfrac{1}{6}\bar{z}\tau^2(z^3)+\tfrac{1}{2}z\bar z\tau^2(z^2)-\tfrac{1}{2}z\tau^2(z^2\bar{z})+\tfrac{1}{2}z^2\tau^2(z\bar{z})\bigr.\\
		& &\bigl. \quad \quad -\tfrac{1}{6}z^3\tau^2(\bar{z})-\tfrac{1}{2}z^2\bar{z}\tau^2(z)\bigr]\cdot\frac{\partial^4 f}{\partial z^3\partial\bar z}\\
		& &+\bigl[\tfrac{1}{4}\tau^2(z^2\bar{z}^2)+\tfrac{1}{4}\bar z^2\tau^2(z^2)+\tfrac{1}{4}z^2\tau^2(\bar z^2)-\tfrac{1}{2}\bar{z}\tau^2(z^2\bar z)-\tfrac{1}{2}z\bar{z}^2\tau^2(z)+z\bar{z}\tau^2(z\bar{z})\bigr.\\
		& &\quad\quad \bigl.-\tfrac{1}{2}z^2\bar{z}\tau^2(\bar{z})-\tfrac{1}{2}z\tau^2(\bar{z}^2z)\bigr]\cdot\frac{\partial^4 f}{\partial z^2\partial\bar z^2}\\
		& &+ \bigl[\tfrac{1}{6}\tau^2(z\bar{z}^3)-\tfrac{1}{6}z\tau^2(\bar z^3)+\tfrac{1}{2} z\bar z\tau^2(\bar z^2)-\tfrac{1}{2}\bar z\tau^2(\bar z^2 z)+\tfrac{1}{2}\bar z^2\tau^2(z\bar{z})\bigr.\\
		& &\bigl. \quad \quad -\tfrac{1}{6}z^3\tau^2(z)-\tfrac{1}{2}\bar z^2 z\tau^2(\bar z)\bigr]\cdot\frac{\partial^4 f}{\partial z\partial\bar z^3}\\
		& & +\bigl[\tfrac{1}{24}\tau^2(\bar z^4)-\tfrac{1}{6}\bar z\tau^2(\bar z^3)+\tfrac{1}{4}\bar z^2\tau^2(\bar z^2)-\tfrac{1}{6}\bar z^3\tau^2(\bar z)\bigr]\cdot\frac{\partial^4 f}{\partial \bar z^4}.\\
	\end{eqnarray*}

\end{lemma}

\begin{proof}
The statement follows directly by inserting the following identities, and their conjugates, into the formula given in Lemma \ref{lemma-tau2}.  For this see Lemma \ref{lemma-conjugate}.
\begin{eqnarray*}\label{1}
\quad \tau(z)^2+2\cdot\kappa(z,\tau(z))+\tau(\kappa(z,z))
=\frac 12\tau^2(z^2)-z\, \tau^2(z),
\end{eqnarray*}
\begin{eqnarray*}\label{2}
& &\quad\tau(z)\tau(\bar z)+\kappa(z,\tau(\bar z))+\kappa(\bar z,\tau(z))+\tau(\kappa(z,\bar z))\\
& &=\frac 12\bigl(\tau^2(z\bar z)-\tau^2(z)\bar z-z\tau^2(\bar z)\bigr),
\end{eqnarray*}
\begin{eqnarray*}\label{3}
2\,\tau(z)\kappa(z,z)+2\kappa(z,\kappa(z,z))=\tfrac{1}{6}\tau^2(z^3)-\tfrac{1}{2}z\tau^2(z^2)+\tfrac{1}{2}z^2\tau^2(z),
\end{eqnarray*}
\begin{eqnarray*}\label{4}
& &2\,\kappa(z,\bar z)\tau(z)+\kappa(\bar z,\kappa(z,z))+\tau(\bar z)\kappa(z,z)+2\,\kappa(z,\kappa(z,\bar z))	\\
&=&\frac 14\tau^2(z^2\bar z)-\frac 14\bar z\tau^2(z^2)+\frac 12z\bar z\tau^2(z)-\frac 12z\tau^2(z\bar z)+\frac 14z^2\tau^2(\bar z),\\
\end{eqnarray*}
\begin{eqnarray*}\label{5}
& &2\kappa(z,z)\kappa(\bar{z},\bar{z})+4\kappa(z,\bar{z})^2\\
&=&\tfrac{1}{4}\tau^2(z^2\bar{z}^2)+\tfrac{1}{4}\bar z^2\tau^2(z^2)+\tfrac{1}{4}z^2\tau^2(\bar z^2)
-\tfrac{1}{2}\bar{z}\tau^2(z^2\bar z)\\
& &-\tfrac{1}{2}z\bar{z}^2\tau^2(z)+z\bar{z}\tau^2(z\bar{z})
-\tfrac{1}{2}z^2\bar{z}\tau^2(\bar{z})
-\tfrac{1}{2}z\tau^2(\bar{z}^2z),
\end{eqnarray*}
\begin{eqnarray*}\label{6}
4\,\kappa(z,z)\kappa(z,\bar{z})
&=&\tfrac{1}{6}\tau^2(z^3\bar{z})-\tfrac{1}{6}\bar{z}\tau^2(z^3)+\tfrac{1}{2}z\bar z\tau^2(z^2)-\tfrac{1}{2}z\tau^2(z^2\bar{z})\\
& &+\tfrac{1}{2}z^2\tau^2(z\bar{z})-\tfrac{1}{6}z^3\tau^2(\bar{z})-\tfrac{1}{2}z^2\bar{z}\tau^2(z),
\end{eqnarray*}
\begin{eqnarray*}\label{7}
\kappa(z,z)^2=\tfrac{1}{24}\tau^2(z^4)-\tfrac{1}{6}z\tau^2(z^3)+\tfrac{1}{4}z^2\tau^2(z^2)-\tfrac{1}{6}z^3\tau^2(z).
\end{eqnarray*}

\end{proof}

In their paper \cite{Gha-Ou-1}, the authors introduce the notion of {\it generalised harmonic morphisms} between Riemannian manifolds.  These are exactly the $(2,1)$-harmonic morphisms in the sense of our Definition \ref{definition-pq}.  They give a characterisation of these objects between Riemannian manifolds.  In general this is rather complicated, see  Theorem 2.2 of \cite{Gha-Ou-1}.  In our context, of complex-valued functions, it is the following.

\begin{theorem}\label{theorem-(2,1)}
A complex-valued function $z:(M,g)\to{\mathbb C}$ from a Riemannian manifold is a $(2,1)$-harmonic morphism if and only if
$$\kappa(z,z)=0,\ \ \tau^2(z)=0\ \ \text{and}\ \ \tau^2(z^2)=0.$$
\end{theorem}

\begin{proof}
The function $z:(M,g)\to{\mathbb C}$ is a $(2,1)$-harmonic morphism if and only if, for any harmonic $f:U\to{\mathbb C}$ defined on an open subset $U$ of ${\mathbb C}$ containing the image $z(M)$ of $z$, the $2$-tension field $\tau^2(\phi)$ of the composition $\phi=f\circ z$ vanishes.  It follows immediately from Lemma \ref{lemma-fundamental} that
$$\kappa(z,z)=\kappa(\bar z,\bar z)=0.$$
Since the test function $f$ is assumed to be harmonic we also have
$$\tau(f)=\frac{\partial^2 f}{\partial z\partial\bar z}=0.$$
This means that the formulae for the $2$-tension field $\tau^2(\phi)$, presented in Lemmas \ref{lemma-fundamental} and \ref{lemma-tau2-expansion}, simplify considerably.  The statement is then a direct consequence of the latter.
\end{proof}

\begin{remark}
	In the case when the Riemannian manifold $(M,g)$ is a surface, i.e. of dimension $2$, then the horizontal conformality of $\phi:M\to{\mathbb C}$ and the Cauchy-Riemann equations imply harmonicity.  That means that in this case no proper $(2,1)$-harmonic morphisms do exist.
\end{remark}

In their paper \cite{Gha-Ou-1} the authors construct the following first known proper $(2,1)$-harmonic morphism.  This was basically the only known example before this current study.

\begin{example}\label{example-basic}
	Let ${\mathbb R}^4$ be the standard $4$-dimensional Euclidean space and $U$ be the open subset given by 	$$U=\{(x_1,x_2,x_3,x_4)\in{\mathbb R}^4|\ x_1^2+x_2^3+x_3^2>0\}.$$  Then $z:U\to{\mathbb C}$ satisfying $z(x)= \sqrt{x_1^2+x_2^2+x_3^2}+i\, x_4$ is a {proper} $(2,1)$-harmonic morphism.
\end{example}

Furthermore they introduce several interesting general methods for constructing solutions to our non-linear $(2,1)$-problem from Euclidean spaces. The following result is a direct consequence of Corollary 3.1. of \cite{Gha-Ou-1}.

\begin{proposition}\label{proposition-holomorphic-composition}
Let $(M,g)$ be a Riemannian manifold and $z:M\to{\mathbb C}$ be a $(2,1)$-harmonic morphism. Further let $f:U\to{\mathbb C}$ be a holomorphic function defined on an open subset of ${\mathbb C}$ such that $z(M)\subset U$.  Then the composition $f\circ z:M\to{\mathbb C}$ is a $(2,1)$-harmonic morphism.
\end{proposition}

\begin{proof}
	It is a classical result that any such holomorphic function $f$ is a $(1,1)$-harmonic morphism. The statement then is a direct concequence of our Lemma \ref{lemma-composition}.
\end{proof}

The following is a special case of Lemma \ref{lemma-composition}, noted already in \cite{Gha-Ou-1}.

\begin{proposition}\label{proposition-known-composition-2}
	Let $(M,g)$ and $(N,h)$ be Riemannian manifolds, $f:(M,g)\to (N,h)$ be a $(2,2)$-harmonic morphism and $\phi:N\to{\mathbb C}$ be a $(2,1)$-harmonic morphism.  Then the composition $\phi\circ f:(M,g)\to{\mathbb C}$ is a $(2,1)$-harmonic morphism.
\end{proposition}

\begin{remark}\label{remark-proper}
	The reader should note that the word {\it "proper"} does not appear in Proposition \ref{proposition-known-composition-2}.  As we will see later, there is a good reason for this.
\end{remark}

From the above calculations of the $2$-tension field $\tau^2(\phi)$ we now have the following result in the case when $(p,q)=(2,2)$.  This should be compared with Theorem 4.1 of \cite{Lou-Ou-1} and Theorem 3.3 of \cite{Lou-Ou-2}.

\begin{theorem}\label{theorem-(2,2)}
A complex-valued function $z:(M,g)\to{\mathbb C}$ from a Riemannian manifold is a $(2,2)$-harmonic morphism if and only if
$$\kappa(z,z)=0,\ \ \tau^2(z)=0,\ \ \tau^2(z^2)=0,$$ $$\tau^2(z\bar z)=0,\ \ \tau^2(z^2\bar z)=0.$$
\end{theorem}

\begin{proof}
The function $z:(M,g)\to{\mathbb C}$ is a $(2,2)$-harmonic morphism if and only if, for any $2$-harmonic $f:U\to{\mathbb C}$ defined on an open subset $U$ of ${\mathbb C}$ containing the image $z(M)$ of $z$, the $2$-tension field $\tau^2(\phi)$ of the composition $\phi=f\circ z$ vanishes. It follows directly from Lemma \ref{lemma-fundamental} that
$$\kappa(z,z)=\kappa(\bar z,\bar z)=0.$$
Since the function $f$ is assumed to be $2$-harmonic we also have
$$\tau^2(f)=\frac{\partial^4 f}{\partial z^2\partial\bar z^2}=0.$$
This means that the formulae for the $2$-tension field $\tau^2(\phi)$, presented in Lemmas \ref{lemma-fundamental} and \ref{lemma-tau2-expansion}, simplify considerably.  The statement is then an immediate consequence of the latter.
\end{proof}

The next statement follows immediately from Proposition 3.2. of \cite{Gha-Ou-1}.

\begin{proposition}\label{proposition-sum}
	Let $(M,g)$, $(N,h)$ be Riemannian manifolds, $\phi:M\to{\mathbb C}$ be a $(2,1)$-harmonic morphism and $\psi:N\to{\mathbb C}$ be a $(1,1)$-harmonic morphism. Then the sum $\Phi=\phi\oplus\psi:M\times N\to{\mathbb C}$, with  $$\Phi:(x,y)\mapsto \phi(x)+\psi(y),$$ is a $(2,1)$-harmonic morphism on the Riemannian product $M\times N$.
\end{proposition}

\section{New $(2,1)$-harmonic morphisms}

In this section we present several new proper complex-valued $(2,1)$-harm\-onic morphisms locally defined on Euclidean ${\mathbb R}^n$. Example \ref{example-first} shows that such objects can easily be constructed for any dimension $n\ge 4$.

\begin{definition}
For a positive integer $p\in{\mathbb Z}^+$ we denote by $i_p$ the {\it inversion} $i_p:{\mathbb R}^{2p}\setminus\{0\}\to{\mathbb R}^{2p}\setminus\{0\}$ of the unit sphere $S^{2p-1}$ in ${\mathbb R}^{2p}$ satisfying $$i_p(x)=\frac x{|x|^2}.$$
Let $\phi:U\to{\mathbb C}$ be a function defined locally on an open subset $U$ of ${\mathbb R}^{2p}\setminus\{0\}$. Then by its {\it dual function} $\phi^*$ we shall mean the composition $\phi^*=\phi\circ i_p:U\to{\mathbb C}$.
\end{definition}

\begin{example}\label{example-first}
Let ${\mathbb R}^n$ be the standard $n$-dimensional Euclidean space of dimension $n\ge 4$ and $U$ be the open subset given by 	$$U=\{x\in{\mathbb R}^n|\ x_1^2+x_2^3+x_3^2>0\}.$$
Then the complex-valued function $\phi:U\to{\mathbb C}$ defined by
$$\phi(x)= \sqrt{x_1^2+x_2^2+x_3^2}+\sum_{k=4}^n b_k\cdot x_k$$ is a proper $(2,1)$-harmonic morphism if and only if the complex coefficients satisfy the relation
$$1+b_4^2+\cdots +b_n^2=0.$$
The same applies to the dual function $\phi^*=\phi\circ i_{p}$ in the case when $n=2p$.  It should be noted that, when $n=4$, the function $\phi$ reduces to Example 2 in \cite{Gha-Ou-1}.
\end{example}

\begin{example}\label{example-second}
Let $U$ be the open subset of the standard Euclidean space ${\mathbb R}^4$ with $U=\{(x_1,x_2,x_3,x_4)\in{\mathbb R}^4|\ x_2^2+x_3^2> 0\}$ and define the function $\phi:U\to{\mathbb C}$ by
$$\phi(x)=\frac{ x_2(1-|x|^2)+2\,x_1x_3}{x_2^2+x_3^2}+i\cdot\frac{x_3(1-|x|^2)-2\,x_1x_2}{x_2^2+x_3^2}.$$
Then $\phi$ is a proper $(2,1)$-harmonic morphism.  Furthermore, its dual function $\phi^*=\phi\circ i_2$ is the proper $(2,1)$-harmonic morphism given by
$$\phi^*(x)=4\,x_1\cdot\frac{x_3-i\,x_2}{x_2^2+x_3^2}-\phi(x).$$
\end{example}

\begin{example}\label{example-third}
The complex-valued function $\phi:{\mathbb R}^4\setminus\{0\}\to{\mathbb C}$ satisfying
$$\phi(x)=\log{\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}}+i\cdot\arccos\Bigl({\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}}}\Bigr)$$ is a proper $(2,1)$-harmonic morphism. Its dual function $\phi^*=\phi\circ i_2$ is the proper $(2,1)$-harmonic morphism satisfying
$$\phi^*(x)=-\log{\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}}+i\cdot\arccos\Bigl({\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}}}\Bigr).$$
\end{example}

\begin{example}\label{example-fourth}
For a positive $r\in{\mathbb R}^+$, the well-known local $(1,1)$-harmonic morphism $\phi_r:U\subset{\mathbb R}^3\to{\mathbb C}$, often called the {\it outer-disc example}, is given by
$$\phi_r(x)=\frac{-(x_3+ir)+\sqrt{x_1^2+x_2^2+x_3^2-r^2+2ir\cdot x_3}}{x_1-ix_2}.$$
Then the dual map $\phi_r^*=\phi_r\circ i_2$ satisfies
$$\phi_r^*(x)=\frac{\sqrt{x_1^2+x_2^2+x_3^2+2ir\cdot x_3\cdot|x|^2-r^2\cdot|x|^4}-(x_3+ir\cdot|x|^2)}{x_1-ix_2}.$$
This is a proper $(2,1)$-harmonic morphism on ${\mathbb R}^4$.
\end{example}

In the above Examples \ref{example-first}-\ref{example-fourth} we have seen that the  constructed complex-valued $(2,1)$-harmonic morphisms $\phi$ and its dual $\phi^*$ are both proper.  The next three examples show that this is not true in general, see Remark \ref{remark-proper}.

\begin{example}
For complex numbers $a,b,c,d\in{\mathbb C}$ with $a^2+b^2+c^2+d^2=0$, let $\phi:{\mathbb R}^4\setminus\{0\}\to{\mathbb C}$ be the proper $(2,1)$-harmonic morphism
$$\phi(x)= \frac{a\cdot x_1+b\cdot x_2+c\cdot x_3+d\cdot x_4}{x_1^2+x_2^2+x_3^2+x_4^2}.$$
Then its dual function
$\phi^*=\phi\circ i_2$ is the globally defined $(1,1)$-harmonic morphism satisfying
$$\phi^*:(x_1,x_2,x_3,x_4)\mapsto a\cdot x_1+b\cdot x_2+c\cdot x_3+d\cdot x_4.$$
By Lemma 2.5, this is a $(2,1)$-harmonic morphism, but it is not proper.
\end{example}

\begin{example}
For elements $a,b,c,d\in{\mathbb C}$, define the complex-valued function $\phi:U\subset{\mathbb R}^4\to{\mathbb C}$ by
\begin{eqnarray*}
\phi(x)&=&\frac{a\cdot(x_1^2+x_2^2+x_3^2+x_4^2)+b\cdot(x_3+ix_4)}{x_1+ix_2}\\
& &\quad +\frac{c\cdot(x_1^2+x_2^2+x_3^2+x_4^2)+d\cdot(x_1+ix_2)}{x_3+ix_4}.
\end{eqnarray*}
Then $\phi$ is a proper $(2,1)$-harmonic morphism and its dual  $\phi^*=\phi\circ i_2$ is the holomorphic function
$$\phi^*(x)=\frac{
d\cdot(x_1+i\,x_2)^2+c\cdot(x_1+i\,x_2)+
b\cdot(x_3+i\,x_4)^2+a\cdot(x_3+i\,x_4)}
{(x_1+i\,x_2)\cdot(x_3+i\,x_4)}.
$$  This is clearly a $(2,1)$-harmonic morphism which is not proper.
\end{example}

\begin{example}
Define the complex-valued function $\phi:{\mathbb R}^4\setminus\{0\}\to{\mathbb C}$ by
$$\phi(x)=\cos(\frac{x_1+i\,x_2}{x_1^2+x_2^2+x_3^2+x_4^2})+i\cdot\sin(\frac{x_3+i\,x_4}{x_1^2+x_2^2+x_3^2+x_4^2}).$$
Then $\phi$ is a proper $(2,1)$-harmonic morphism and its dual satisfying
$$\phi^*(x)=\phi\circ i_2(x)=\cos(x_1+i\,x_2)+i\cdot\sin(x_3+i\,x_4)$$
is holomorphic and hence a $(2,1)$-harmonic morphism, but not proper as it is a $(1,1)$-harmonic morphism.
\end{example}

\section{A Generalised Construction Method}

The main purpose of this section is to prove Theorem \ref{theorem-general-2} which is a wide generalisation of Proposition \ref{proposition-sum}.

\begin{lemma}\label{lemma-hor-conf}
Let $(M,g)$, $(N,h)$ be Riemannian manifolds and $\phi:M\to{\mathbb C}$, $\psi:N\to{\mathbb C}$ be two horizontally conformal functions.  Let $U$ be an open subset of ${\mathbb C}^2$ such that $\phi(M)\times \psi(N)\subset U$
and $f:U\to{\mathbb C}$ be a holomorphic function.  Then the composition
$\Phi:M\times N\to{\mathbb C}$ with $\Phi(x,y)=f(\phi(x),\psi(y))$ is horizontally conformal on the Riemannian product space $M\times N$.
\end{lemma}

\begin{proof}
Let $\mathcal B_M$ and $\mathcal B_N$ be local orthonormal frames for the tangent bundles
$TM$ and $TN$, respectively.  Then
\begin{eqnarray*}
\kappa(\Phi,\Phi)&=&\sum_{X\in\mathcal B_M}\bigl(\frac{\partial f}{\partial \phi}\cdot X(\phi)\bigr)^2+
\sum_{Y\in\mathcal B_N}\bigl(\frac{\partial f}{\partial \psi}\cdot Y(\psi)\bigr)^2\\
&=&\bigl(\frac{\partial f}{\partial \phi}\bigr)^2\cdot\kappa(\phi,\phi)+
\bigl(\frac{\partial f}{\partial \psi}\bigr)^2\cdot\kappa(\psi,\psi)\\
&=&0.
\end{eqnarray*}
\end{proof}

\begin{theorem}\label{theorem-general-2}
Let $U$, $V$ be open subsets of ${\mathbb R}^m$ and ${\mathbb R}^n$, respectively. Let $\phi:U\to{\mathbb C}$ be a $(2,1)$-harmonic morphism and $\psi:V\to{\mathbb C}$ be a $(1,1)$-harmonic morphism. Let $W$ be an open subset of ${\mathbb C}^2$ such that $\phi(U)\times \psi(V)\subset W$ and
$f:W\to{\mathbb C}$ be a holomorphic function.  Then the composition $\Phi:U\times V\to{\mathbb C}$ with $\Phi(x,y)=f(\phi(x),\psi(y))$ is a $(2,1)$-harmonic morphism.
\end{theorem}

\begin{proof}
It follows from Lemma \ref{lemma-hor-conf} that $\Phi$ is horizontally conformal i.e. $\kappa(\Phi,\Phi)=0$.  For the tension field $\tau(\Phi)$ of $\Phi$ we have
\begin{eqnarray*}
\tau(\Phi )&=&\sum_{k=1}^m\frac{\partial^2}{\partial x_k^2}(f(\phi(x),\psi(y)))+\sum_{r=1}^n\frac{\partial^2}{\partial y_r^2}(f(\phi(x),\psi(y)))\\	&=&\sum_{k=1}^m\frac{\partial}{\partial x_k}\bigl(\frac{\partial f}{\partial \phi}\cdot
\frac{\partial \phi}{\partial x_k}\bigr)+
\sum_{r=1}^n\frac{\partial}{\partial y_r}\bigl(\frac{\partial f}{\partial \psi}\cdot
\frac{\partial \psi}{\partial y_r}\bigr)\\
&=&\sum_{k=1}^m\bigl(\frac{\partial^2 f}{\partial \phi^2}\cdot\bigl(\frac{\partial \phi}{\partial x_k}\bigr)^2+\frac{\partial f}{\partial \phi}\cdot\frac{\partial^2\phi}{\partial x_k^2}\bigr)+
\sum_{r=1}^n\bigl(\frac{\partial^2 f}{\partial \psi^2}\cdot\bigl(\frac{\partial \psi}{\partial y_r}\bigr)^2+\frac{\partial f}{\partial \psi}\cdot\frac{\partial^2\psi}{\partial y_r^2}\bigr)\\
&=&\frac{\partial^2 f}{\partial \phi^2}\cdot\kappa(\phi,\phi)+\frac{\partial f}{\partial \phi}\cdot\tau(\phi)+\frac{\partial^2 f}{\partial \psi^2}\cdot\kappa(\psi,\psi)+\frac{\partial f}{\partial \psi}\cdot\tau(\psi)\\
&=&\frac{\partial f}{\partial \phi}\cdot\tau(\phi).
\end{eqnarray*}
With this at hand, we can now calculate  the $2$-tension field $\tau^2(\Phi)$ of $\Phi$ as follows.
\begin{eqnarray*}
\tau^2(\Phi)&=&\tau(\frac{\partial f}{\partial \phi}\cdot\tau(\phi))\\
&=&\tau(\frac{\partial f}{\partial \phi})\cdot\tau(\phi)+2\cdot \kappa(\frac{\partial f}{\partial \phi},\tau(\phi))+\frac{\partial f}{\partial \phi}\cdot\tau^2(\phi)\\
&=&\tau(\phi)\cdot\Bigl(\sum_{k=1}^m\frac{\partial^2}{\partial x_k^2}\bigl(\frac{\partial f}{\partial \phi}\bigr)+\sum_{r=1}^n\frac{\partial^2}{\partial y_r^2}\bigl(\frac{\partial f}{\partial \phi}\bigr)\Bigr)\\
& &\quad+2\cdot\Bigl(\sum_{k=1}^m\frac{\partial}{\partial x_k}\bigl(\frac{\partial f}{\partial \phi}\bigr)\cdot\frac{\partial}{\partial x_k}(\tau(\phi))+\sum_{r=1}^n\frac{\partial}{\partial y_r}\bigl(\frac{\partial f}{\partial \phi}\bigr)\cdot\frac{\partial}{\partial y_r}(\tau(\phi))\Bigr)\\
&=&\tau(\phi)\cdot\Bigl(
\sum_{k=1}^m\frac{\partial}{\partial x_k}\bigl(\frac{\partial^2f}{\partial \phi^2}\cdot\frac{\partial \phi}{\partial x_k}\bigr)+\sum_{r=1}^n\frac{\partial}{\partial y_r}\bigl(\frac{\partial^2f}{\partial \psi\partial \phi}\cdot\frac{\partial \psi}{\partial y_r}\bigr)
\Bigr)\\
& &\quad +2\cdot\sum_{k=1}^m\frac{\partial^2f}{\partial \phi^2}\cdot\frac{\partial \phi}{\partial x_k}\cdot\frac{\partial}{\partial x_k}(\tau(\phi))\\
&=&\tau(\phi)\cdot
\sum_{k=1}^m\Bigl(\frac{\partial^3f}{\partial \phi^3}\cdot\bigl(\frac{\partial \phi}{\partial x_k}\bigr)^2+\frac{\partial^2f}{\partial \phi^2}\cdot\frac{\partial^2\phi}{\partial x_k^2}\Bigr)\\
& &\quad +\tau(\phi)\cdot\sum_{r=1}^n\Bigl(
\frac{\partial ^3f}{\partial \psi^2\partial \phi}\cdot\bigl(\frac{\partial \psi}{\partial y_r}\bigr)^2+\frac{\partial^2f}{\partial \psi\partial \phi}\cdot\frac{\partial^2 \psi}{\partial y_k^2}
\Bigr)\\
& &\quad\quad +2\cdot\frac{\partial^2 f}{\partial \phi^2}\cdot\kappa(\phi,\tau(\phi))\\
&=&
\tau(\phi)\cdot\Bigl(\frac{\partial^3 f}{\partial \phi^3}\cdot\kappa(\phi,\phi)+\frac{\partial ^3f}{\partial \psi^2\partial \phi}\cdot\kappa(\psi,\psi)+\frac{\partial^2f}{\partial \psi\partial \phi}\cdot\tau(\psi)\Bigr)\\
& &\quad +\frac{\partial ^2f}{\partial \phi^2}\cdot(2\cdot\kappa(\phi,\tau(\phi))+\tau(\phi)^2)\\
&=&0.
\end{eqnarray*}
For the tension field $\tau(\Phi^2)$ of $\Phi^2$ we have
$$\tau(\Phi^2)=2\cdot\Phi\cdot\tau(\Phi)+2\cdot\kappa(\Phi,\Phi)=2\cdot\Phi\cdot\tau(\Phi).$$
Hence the bi-tension field $\tau^2(\Phi^2)$ of $\Phi^2$ satisfies
\begin{eqnarray*}
\tau^2(\Phi^2)
&=&2\cdot\tau(\Phi\cdot\tau(\Phi))\\
&=&2\cdot(\tau(\Phi)^2+2\cdot\kappa(\Phi,\tau(\Phi))+\Phi\cdot\tau^2(\Phi))\\
&=&2\cdot\Bigl(\frac{\partial f}{\partial \phi}\Bigr)^2\cdot\tau(\phi)^2+4\cdot\kappa(\Phi,\frac{\partial \Phi}{\partial \phi}\cdot\tau(\phi))\\
&=&2\cdot\Bigl(\frac{\partial f}{\partial \phi}\Bigr)^2\cdot\tau(\phi)^2
+4\cdot\sum_{k=1}^m\frac{\partial\Phi}{\partial x_k}\cdot\frac{\partial}{\partial x_k}\bigl(\frac{\partial f}{\partial \phi}\cdot\tau(\phi)\Bigr)\\
& &\qquad\qquad +4\cdot\sum_{r=1}^n\frac{\partial\Phi}{\partial y_r}\cdot\frac{\partial}{\partial y_r}\bigl(\frac{\partial f}{\partial \phi}\cdot\tau(\phi)\Bigr)\\
&=&2\cdot\Bigl(\frac{\partial f}{\partial \phi}\Bigr)^2\cdot\tau(\phi)^2\\
& &\quad +\,4\cdot\sum_{k=1}^m\frac{\partial \phi}{\partial x_k}\cdot\frac{\partial f}{\partial \phi}\cdot\Bigl(\frac{\partial^2 f}{\partial \phi^2}\cdot\frac{\partial \phi}{\partial x_k}\cdot\tau(\phi)+\frac{\partial f}{\partial \phi}\cdot\frac{\partial}{\partial x_k}(\tau(\phi))\Bigr)\\
& &\quad\quad +\,4\cdot\sum_{r=1}^n\frac{\partial \psi}{\partial y_r}\cdot\frac{\partial f}{\partial \psi}\cdot\Bigl(\frac{\partial^2 f}{\partial \psi\partial \phi}\cdot\frac{\partial \psi}{\partial y_r}\cdot\tau(\phi)+\frac{\partial f}{\partial \phi}\cdot\frac{\partial}{\partial y_r}(\tau(\phi))\Bigr)\\
&=&2\cdot\Bigl(\frac{\partial f}{\partial \phi}\Bigr)^2\cdot\tau(\phi)^2
+4\cdot\frac{\partial f}{\partial \phi}\cdot\frac{\partial^2 f}{\partial \phi^2}\cdot\tau(\phi)\cdot\kappa(\phi,\phi)\\
& &\quad +4\cdot\Bigl(\frac{\partial f}{\partial \phi}\Bigr)^2\cdot\kappa(\phi,\tau(\phi))+4\cdot\frac{\partial f}{\partial \psi}\cdot\frac{\partial^2 f}{\partial \psi\partial \phi}\cdot\tau(\phi)\cdot\kappa(\psi,\psi)\\
&=&2\cdot\Bigl(\frac{\partial f}{\partial \phi}\Bigr)^2 (\tau(\phi)^2+2\cdot\kappa(\phi,\tau(\phi)))\\
&=&0.
\end{eqnarray*}
The conclusion follows from Theorem 3.6.
\end{proof}

\begin{example}
We have already seen that the complex-valued function $\phi:{\mathbb R}^4\setminus\{0\}\to{\mathbb C}$ satisfying
$$\phi(x)=\log{\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}}+i\cdot\arccos\Bigl({\frac{x_1}{\sqrt{x_1^2+x_2^2+x_3^2+x_4^2}}}\Bigr)$$ is a proper $(2,1)$-harmonic morphism.  It is clear that the holomorphic function $\psi:{\mathbb R}^4\to{\mathbb C}$ satisfying
$$\psi(x)=\log(x_5+ix_6)\cdot\sin(x_7+ix_8)$$ is a $(1,1)$ harmonic morphism.
Calculations confirm that $\Phi:U\subset{\mathbb R}^8\to{\mathbb C}$ given by $\Phi=f(\phi,\psi)=\phi\cdot\psi$ is a proper $(2,1)$-harmonic morphism for any holomorphic function f.
\end{example}

\section{Complex-valued $(3,q)$-Harmonic Morphisms}

In this section we present a formula for the $3$-tension field $\tau^3(\phi)$, of the composition $\phi=f\circ z$. It turns out that, just as in the case of $(2,q)$, horizontal conformality, i.e. $\kappa(z,z)=0$, is a necessary condition.  Elementary but rather tedious calculations provide the following useful result.

\begin{lemma}\label{lemma-tau3-horizontally-conformal}
	Let $z:(M,g)\to\mathbb{C}$ be a horizontally conformal complex-valued function from a Riemannian manifold and $f:U\to\mathbb{C}$ be defined on an open subset $U$ of $\mathbb{C}$ containing the image $z(M).$ Then the $3$-tension field $\tau^3(\phi)$ of the composition $\phi=f\circ z$ satisfies
	\begin{eqnarray*}
		& &\tau^3(\phi)\\
		&=&\tau^3(z)\cdot\frac{\partial f}{\partial z}+\tau^3(\bar{z})\cdot\frac{\partial f}{\partial\bar{z}}\\
		& & +\left[\tfrac{1}{2}\tau^3(z^2)-z\tau^3(z)\right]\cdot\frac{\partial^2f}{\partial z^2}\\
		& &
		+\left[\tau^3(z\bar{z})-\bar{z}\tau^3(z)-z\tau^3(\bar{z})\right]\cdot\frac{\partial^2f}{\partial z\partial\bar{z}}\\
		& & +\left[\tfrac{1}{2}\tau^3(\bar{z}^2)-\bar{z}\tau^3(\bar{z})\right]\cdot\frac{\partial^2f}{\partial \bar{z}^2}\\
		& &
		+\left[\tfrac{1}{6}\tau^3(z^3)-\tfrac{1}{2}z\tau^3(z^2)+\tfrac{1}{2}z^2\tau^3(z)\right]\cdot\frac{\partial ^3f}{\partial z^3}\\
		& &
		+\left[\tfrac{1}{2}\tau^3(z^2\bar{z})-\tfrac{1}{2}\bar{z}\tau^3(z^2)-z\tau^3(z\bar{z})+z\bar{z}\tau^3(z)+\tfrac{1}{2}z^2\tau^3(\bar{z})\right]\cdot\frac{\partial ^3f}{\partial{z^2}\partial\bar{z}}\\
		& &
		+\left[\tfrac{1}{2}\tau^3(z\bar{z}^2)-\bar{z}\tau^3(z\bar{z})+\tfrac{1}{2}\bar{z}^2\tau^3(z)-\tfrac{1}{2}z\tau^3(\bar{z}^2)+z\bar{z}\tau^3(\bar{z})\right]\cdot\frac{\partial ^3f}{\partial{z}\partial\bar{z}^2}\\
		& &
		+\left[\tfrac{1}{6}\tau^3(\bar{z}^3)-\tfrac{1}{2}\bar{z}\tau^3(\bar{z}^2)+\tfrac{1}{2}\bar{z}^2\tau^3(\bar{z})\right]\cdot\frac{\partial ^3f}{\partial \bar{z}^3}\\
		& &
		+\left[\tfrac{1}{6}\tau^3(z^3\bar{z})-\tfrac{1}{6}\bar{z}\tau^3(z^3)-\tfrac{1}{2}z\tau^3(z^2\bar{z})+\tfrac{1}{2}z\bar{z}\tau^3(z^2)+\tfrac{1}{2}z^2\tau^3(z\bar{z})-\tfrac{1}{2}z^2\bar{z}\tau^3(z)\right.\\
		& & \quad\quad\left.-\tfrac{1}{6}z^3\tau^3(\bar{z})\right]\cdot\frac{\partial ^4f}{\partial z^3\partial \bar{z}}\\
		& &
		+\left[\tfrac{1}{4}\tau^3(z^2\bar{z}^2)-\tfrac{1}{2}\bar{z}\tau^3(z^2\bar{z})+\tfrac{1}{4}\bar{z}^2\tau^3(z^2)-\tfrac{1}{2}z\tau^3(\bar{z}^2z)+z\bar{z}\tau^3(z\bar{z})-\tfrac{1}{2}z\bar{z}^2\tau^3(z)\right.\\
		& &
		\quad\quad\left. +\tfrac{1}{4}z^2\tau^3(\bar{z}^2)-\tfrac{1}{2}z^2\bar{z}\tau^3(\bar{z})\right]\cdot\frac{\partial ^4f}{\partial z^2\partial\bar{z}^2}\\
		& &
		+\left[\tfrac{1}{6}\tau^3(z\bar{z}^3)-\tfrac{1}{2}\bar{z}\tau^3(\bar{z}^2z)+\tfrac{1}{2}\bar{z}^2\tau^3(z\bar{z})-\tfrac{1}{6}\bar{z}^3\tau^3(z)-\tfrac{1}{6}z\tau^3(\bar{z}^3)+\tfrac{1}{2}z\bar{z}\tau^3(\bar{z}^2)\right.\\
		& &
		\quad\quad\left.-\tfrac{1}{2}z\bar{z}^2\tau^3(\bar{z})\right]\cdot\frac{\partial ^4f}{\partial z\partial \bar{z}^3}\\
		& &
		+\left[\tfrac{1}{12}\tau^3(z^3\bar{z}^2)-\tfrac{1}{6}\bar{z}\tau^3(z^3\bar{z})+\tfrac{1}{12}\bar{z}^2\tau^3(z^3)-\tfrac{1}{4}z\tau^3(z^2\bar{z}^2)+\tfrac{1}{2}z\bar{z}\tau^3(z^2\bar{z})-\tfrac{1}{4}z\bar{z}^2\tau^3(z^2)\right.\\
		& &\left.\quad +\tfrac{1}{4}z^2\tau^3(z\bar{z}^2)-\tfrac{1}{2}z^2\bar{z}\tau^3(z\bar{z})+\tfrac{1}{4}z^2\bar{z}^2\tau^3(z)-\tfrac{1}{12}z^3\tau^3(\bar{z}^2)+\tfrac{1}{6}z^3\bar{z}\tau^3(\bar{z})\right]\cdot\frac{\partial ^5f}{\partial z^3\partial\bar{z}^2}\\
		& &
		+\left[\tfrac{1}{12}\tau^3(z^2\bar{z}^3)-\tfrac{1}{4}\bar{z}\tau^3(z^2\bar{z}^2)+\tfrac{1}{4}\bar{z}^2\tau^3(z^2\bar{z})-\tfrac{1}{12}\bar{z}^3\tau^3(z^2)-\tfrac{1}{6}z\tau^3(z\bar{z}^3)+\tfrac{1}{2}z\bar{z}\tau^3(z\bar{z}^2)\right.\\
		& &\left.\quad -\tfrac{1}{2}z\bar{z}^2\tau^3(z\bar{z})+\tfrac{1}{6}z\bar{z}^3\tau^3(z)+\tfrac{1}{12}z^2\tau^3(\bar{z}^3)-\tfrac{1}{4}z^2\bar{z}\tau^3(\bar{z}^2)+\tfrac{1}{4}z^2\bar{z}^2\tau^3(\bar{z})\right]\cdot\frac{\partial ^5f}{\partial z^2\partial\bar{z}^3}\\
		& &
		+8\,\kappa(z,\bar{z})^3\cdot\frac{\partial^6f}{\partial z^3\partial\bar{z}^3}\\
	\end{eqnarray*}
\end{lemma}

\begin{theorem}\label{theorem-(3,1)}
A complex-valued function $z:(M,g)\to\mathbb{C}$ from a Riemannian manifold is a $(3,1)$-harmonic morphism if and only if
$$
\kappa(z,z)=0,
$$
$$\tau^3(z)=0,\ \ \tau^3(z^2)=0,\ \ \tau^3(z^3)=0,$$
\end{theorem}

\begin{proof}
The method used here is exactly the same as that we have employed in the proof of Theorem \ref{theorem-(2,1)} employing the fact that $f$ is harmonic i.e.
$$\tau(f)=\frac{\partial^2 f}{\partial z\partial \bar z}=0.$$
\end{proof}

\begin{example}
Let $\phi:{\mathbb R}^6\setminus\{0\}\to{\mathbb C}$ be the $(1,1)$-harmonic morphism given by
$$\phi(x)=(x_1+ix_2)(x_3+ix_4)+\sin (x_5+ix_6).$$
Then calculations show that its dual map $\phi^*=\phi\circ i_3:{\mathbb R}^6\setminus\{0\}\to{\mathbb C}$ is a proper $(3,1)$-harmonic morphism.
\end{example}

In the following Tables 1 and 2 we give several new examples of $(3,1)$-harmonic morphisms defined on the appropriate open subsets $U$ of ${\mathbb R}^6$.  They are proper if and only if the stated $p$ is $3$ and not proper otherwise.

\begin{table}[ht]
\begin{center}
\caption{$\phi:U\subset{\mathbb R}^6\setminus\{0\}\to{\mathbb C}$.}
\label{tab:table1}
\begin{tabular}{|c|c|}
\hline
$\phi(x)$ & $(p,q)$ \\
\hline\hline
$\phi_{11}(x)=(x_1+ix_2)+(x_3+ix_4)+(x_5+ix_6)$ & $(1,1)$ \\
$\phi_{12}(x)=\sqrt{x_1^2+x_2^2+x_3^2}+ix_4+ (x_5+ix_6)$ & $(2,1)$ \\
$\phi_{13}(x)=\sqrt{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}+ix_6$ & $(3,1)$ \\
\hline
$\phi^*_{11}(x)=\phi_{11}\circ i_3(x)$ & $(3,1)$ \\
$\phi^*_{12}(x)=\phi_{12}\circ i_3(x)$ & $(3,1)$ \\
$\phi^*_{13}(x)=\phi_{13}\circ i_3(x)$ & $(3,1)$ \\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{table}[ht]
\begin{center}
\caption{$\phi:U\subset{\mathbb R}^6\setminus\{0\}\to{\mathbb C}$.}
\label{tab:table2}
\begin{tabular}{|c|c|}
\hline
$\phi(x)$ & $(p,q)$ \\
\hline\hline
$\phi_{21}(x)=\log\sqrt{x_1^2+x_2^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+x_2^2}})$
& $(1,1)$ \\
$\phi_{22}(x)=\log\sqrt{x_1^2+\cdots +x_4^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+\cdots +x_4^2}})$
& $(2,1)$ \\
$\phi_{23}(x)=\log\sqrt{x_1^2+\cdots +x_6^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+\cdots +x_6^2}})$
& $(3,1)$ \\
\hline
$\phi^*_{21}(x)=\phi_{21}\circ i_3(x)$ & $(3,1)$ \\
$\phi^*_{22}(x)=\phi_{22}\circ i_3(x)$ & $(3,1)$ \\
$\phi^*_{23}(x)=\phi_{23}\circ i_3(x)$ & $(3,1)$ \\
\hline
\end{tabular}
\end{center}
\end{table}

\begin{theorem}\label{theorem-(3,2)}
	A complex-valued function $z:(M,g)\to\mathbb{C}$ from a Riemannian manifold is a $(3,2)$-harmonic morphism if and only if
	$$
	\kappa(z,z)=0,
	$$
	$$\tau^3(z)=0,\ \ \tau^3(z^2)=0,\ \ \tau^3(z^3)=0,$$
	$$\tau^3(z\bar{z})=0,\ \ \tau^3(z^2\bar{z})=0,\ \ \tau^3(z^3\bar{z})=0.$$
\end{theorem}

\begin{proof}
The statement follows easily from the fact that $\kappa(z,z)=\kappa(\bar z,\bar z)=0$, Lemma \ref{lemma-tau3-horizontally-conformal} and
$$
\tau^2(f)=\frac{\partial^4f}{\partial z^2\partial \bar{z}^2}=0,
$$
since $f$ is assumed to be a general 2-harmonic function.
\end{proof}

Our next result gives a characterisation in the complex-valued $(3,3)$-case.  This recovers a particular case of Corollary 6.2 of the interesting work \cite{Mae} of Maeta.

\begin{theorem}\label{theorem-(3,3)}
	A complex-valued function $z:(M,g)\to\mathbb{C}$ from a Riemannian manifold is a $(3,3)$-harmonic morphism if and only if
	$$
	\kappa(z,z)=0,
	$$
	$$\tau^3(z)=0,\ \ \tau^3(z^2)=0,\ \  \tau^3(z^3)=0,
	$$
	$$
	\tau^3(z\bar{z})=0,\ \ \tau^3(z^2\bar{z})=0,\ \  \tau^3(z^3\bar{z})=0,
	$$
	$$
	\tau^3(z^2\bar{z}^2)=0,\ \  \tau^3(z^3\bar{z}^2)=0.
	$$
\end{theorem}

\begin{proof}
Here we use exactly the same method as above, utilising Lemma \ref{lemma-tau3-horizontally-conformal} and the fact that in this case we have $$\tau^3(f)=\frac{\partial^6f}{\partial z^3\partial\bar z^3}=0.$$
\end{proof}

\section{Complex-valued $(p,q)$-harmonic morphisms}

In this section we investigate the $p$-tension field $\tau^p(\phi)$ of the composition $\phi=f\circ z$ and derive several consequences from  the condition $\tau^p(\phi)=0$ i.e. of $\phi$ being $p$-harmonic.  It turns out that $\tau^p(\phi)$ takes the following form
$$\tau^p(\phi)=\displaystyle{
	\sum\limits_{\substack{ 1\leq j+k\leq 2p}}
	c_{jk}^p\cdot\frac{\partial^{j+k}f}{\partial z^j\partial\bar{z}^{k}}},$$
where the coefficients $c^p_{jk}:U\to{\mathbb C}$ are differentiable functions involving various tension fields and conformality operators of the functions  $z$ and $\bar z$ and independent of $f$.

We have already presented the tension fields $\tau(\phi)$, $\tau^2(\phi)$ and $\tau^3(\phi)$ of $\phi$. When calculating the $4$-tension field $\tau^4(\phi)$ a clear pattern comes to light.  These calculations are far too extensive to be presented here.  For the $p$-tension field $\tau^p(\phi)$ we have the following  result.

\begin{lemma}\label{lemma-pq}
Let $z:(M,g)\to\mathbb{C}$ be a complex-valued function from a Riemannian manifold and $f:U\to\mathbb{C}$ be defined on an open subset $U$ of $\mathbb{C}$ containing the image $z(M).$ Then for $p\geq 2$ the $p$-tension field $\tau^p(\phi)$ of the composition $\phi=f\circ z$ is of the form
$$\tau^p(\phi)=\displaystyle{
\sum\limits_{\substack{1\leq j+k\leq 2p}}
c_{jk}^p\cdot\frac{\partial^{j+k}f}{\partial z^j\partial\bar{z}^{k}}}.$$
The coefficients $c^p_{jk}:U\to{\mathbb C}$ are differentiable functions, independent of $f$, involving various tension fields and conformality operators of the functions $z$ and $\bar z$. Furthermore, they are symmetric with respect to complex conjugation i.e. $c_{jk}^p=\bar c_{kj}^p$ and
$$c_{10}=\tau^p(z),\ \ c_{01}=\tau^p(\bar z),\ \ c_{2p,0}=\kappa(z,z)^p,\ \  c_{0,2p}=\kappa(\bar z,\bar z)^p.$$
\end{lemma}

This leads to the following general result which should be compared with Theorems \ref{theorem-(1,1)}, \ref{theorem-(2,1)}, \ref{theorem-(2,2)}, \ref{theorem-(3,1)}, \ref{theorem-(3,2)} and \ref{theorem-(3,3)} above.

\begin{theorem}\label{theorem-(p,q)}
A complex-valued function $z:(M,g)\to\mathbb{C}$ from a Riemannian manifold is a $(p,q)$-harmonic morphism if and only if
$$\kappa(z,z)=0,$$
$$\tau^p(z)=0,\ \ \tau^p(z^2)=0,\ \ \cdots\ \  ,\tau^p(z^p)=0,$$
$$\tau^p(z\bar{z})=0,\ \ \tau^p(z^2\bar{z})=0,\ \  \cdots\ \ ,\tau^p(z^p\bar{z})=0,$$
$$\tau^p(z^2\bar{z}^2)=0,\ \ \tau^p(z^3\bar{z}^2)=0,\ \ \cdots\ \ ,\tau^p(z^p\bar{z}^2)=0,$$
$$\vdots$$
$$
\tau^p(z^{q-1}\bar{z}^{q-1})=0,\ \ \tau^p(z^{q}\bar{z}^{q-1})=0,\ \ \cdots\ \ , \tau^p(z^p\bar{z}^{q-1})=0.$$
\end{theorem}

\begin{proof}
The function $z:(M,g)\to{\mathbb C}$ is a $(p,q)$-harmonic morphism if and only if, for any $q$-harmonic function $f:U\to{\mathbb C}$ defined on an open subset $U$ of ${\mathbb C}$ containing the image $z(M)$ of $z$, the $p$-tension field $\tau^p(\phi)$ of the composition $\phi=f\circ z$ vanishes. Since the function $f$ is assumed to be $q$-harmonic we know that
$$\tau^q(f)=\frac{\partial^{2q}f}{\partial z^q\partial\bar z^q}=0.$$
According to Lemma \ref{lemma-pq} we also have
$$\tau^p(z)=\tau^p(\bar z)=0\ \ \text{and}\ \
\kappa(z,z)=\kappa(\bar z,\bar z)=0.$$  If we now plug these indentities into the expression for $\tau^p(\phi)$ this simplifies considerably to
$$\tau^p(\phi)=\displaystyle{
	\sum\limits_{\substack{0\le j,k\le p\\ 2\leq j+k\leq 2p}}
	c_{jk}^p\cdot\frac{\partial^{j+k}f}{\partial z^j\partial\bar{z}^{k}}},$$
where
$c_{pp}^p=2^p\cdot\kappa(z,\bar z)^p$. Hard work then shows that the remaining coefficients satisfy
$$
c_{jk}^p=\sum\limits_{\substack{0\leq r\leq j\\ 0\leq s\leq k}}(-1)^{j-r+k-s}\frac{1}{j!}\frac{1}{k!}\binom jr\binom ks
z^{j-r}\bar{z}^{k-s}\tau^p(z^{r}\bar{z}^{s}).
$$
The rest follows by the same method as applied in the proof of Theorem \ref{theorem-(2,2)}.
\end{proof}

\begin{remark}
In the paper \cite{Mae}, Maeta presents his interesting Conjecture 7.6.  In our language his statement is: {\it "A $(p,p)$-harmonic morphism is characterized as a special horizontally weakly conformal $2p$-harmonic map."}

In our Theorem \ref{theorem-(p,q)} we study the special case of complex-valued $(p,p)$-harmonic morphisms.  We obtain a characterisation of these  objects and show that they are both horizontally conformal and $2p$-harmonic, as Maeta suggests. But additionally, they must satisfy several rather non-trivial conditions.  They can therefore rightly be said to be "special horizontally weakly conformal $2p$-harmonic maps".
\end{remark}

We conclude this section by presenting further examples. They suggest that one should be able to produce $(p,1)$-harmonic morphisms for any positive integer $p\in{\mathbb Z}^+$.  The question marks '?' in Table \ref{tab:table4} tell us that the calculations needed, in those cases, were too heavy for the tools available to us.

\begin{table}[ht]
	\begin{center}
		\caption{$\phi:U\subset{\mathbb R}^8\setminus\{0\}\to{\mathbb C}$.}
		\label{tab:table3}
		\begin{tabular}{|c|c|}
			\hline
			$\phi(x)$ & $(p,q)$ \\
			\hline\hline
			$\phi_{31}(x)=(x_1+ix_2)+(x_3+ix_4)+(x_5+ix_6)+(x_7+ix_8)$ & $(1,1)$ \\

			$\phi_{32}(x)=\sqrt{x_1^2+x_2^2+x_3^2}+ix_4+ (x_5+ix_6)+(x_7+ix_8)$ & $(2,1)$ \\

			$\phi_{33}(x)=\sqrt{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}+ix_6+(x_7+ix_8)$ & $(3,1)$ \\
			$\phi_{34}(x)=\sqrt{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+x_6^2+x_7^2}+ix_8)$ & $(4,1)$ \\
			\hline
			$\phi^*_{31}(x)=\phi_{31}\circ i_4(x)$ & $(4,1)$ \\
			$\phi^*_{32}(x)=\phi_{32}\circ i_4(x)$ & $(4,1)$ \\
			$\phi^*_{33}(x)=\phi_{33}\circ i_4(x)$ & $(4,1)$ \\
			$\phi^*_{34}(x)=\phi_{34}\circ i_4(x)$ & $(4,1)$ \\
			\hline
		\end{tabular}
	\end{center}
\end{table}

\begin{table}[ht]
	\begin{center}
\caption{$\phi:U\subset{\mathbb R}^8\setminus\{0\}\to{\mathbb C}$.}
		\label{tab:table4}
		\begin{tabular}{|c|c|}
\hline
$\phi(x)$ & $(p,q)$ \\
\hline\hline
$\phi_{41}(x)=\log\sqrt{x_1^2+x_2^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+x_2^2}})$ & $(1,1)$ \\
$\phi_{42}(x)=\log\sqrt{x_1^2+\cdots +x_4^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+\cdots +x_4^2}})$ & $(2,1)$ \\
$\phi_{43}(x)=\log\sqrt{x_1^2+\cdots +x_6^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+\cdots +x_6^2}})$ & $(3,1)$ \\
$\phi_{44}(x)=\log\sqrt{x_1^2+\cdots +x_8^2}+i\arccos (\frac {x_1} {\sqrt{x_1^2+\cdots +x_8^2}})$ & $(4,1)$ \\
\hline
$\phi^*_{41}(x)=\phi_{41}\circ i_4(x)$ & $(4,1)$ \\
$\phi^*_{42}(x)=\phi_{42}\circ i_4(x)$ &  ? \\
$\phi^*_{43}(x)=\phi_{43}\circ i_4(x)$ &  ? \\
$\phi^*_{44}(x)=\phi_{44}\circ i_4(x)$ &  ? \\
			\hline
		\end{tabular}
	\end{center}
\end{table}

\begin{remark}
	In the process of obtaining Lemma \ref{lemma-pq}, it is easily seen that every $(p,q)$-harmonic morphism is constant in the cases when $p<q$.  This is due to the fact that in these cases we have
$$c_{2p,0}^p=\kappa(z,z)=0\ \ \text{and}\ \ c_{pp}^p=\kappa(z,\bar z)=0.$$
	The reader should compare this with Proposition  \ref{proposition-constant}.
\end{remark}

\begin{example}
	Let $\phi:{\mathbb R}^8\to{\mathbb C}$ be the holomorphic $(1,1)$-harmonic morphism defined by
	$$\phi(x)=(x_1+ix_2+x_3+ix_4)+\sin(x_5+ix_6+x_7+ix_8).$$
	Then its dual map $\phi^*=\phi\circ i_4:{\mathbb R}^8\setminus\{0\}\to{\mathbb C}$ is a proper $(4,1)$-harmonic morphism.
\end{example}

\section{The Inversion about the unit sphere $S^{2p-1}$ in ${\mathbb R}^{2p}$}

In this section we investigate the inversion $i_p:{\mathbb R}^{2p}\setminus
\{0\}\to{\mathbb R}^{2p}\setminus\{0\}$ about the unit sphere $S^{2p-1}$ in ${\mathbb R}^{2p}$.

\begin{theorem}\label{theorem-inversion}
Let $i_p:{\mathbb R}^{2p}\setminus\{0\}\to{\mathbb R}^{2p}\setminus\{0\}$ be the inversion about the unit sphere $S^{2p-1}$ in ${\mathbb R}^{2p}$ given by
$$i_p=(F_1,\dots ,F_{2p}):x\mapsto\frac {(x_1,\dots ,x_{2p})}{|x|^2}.$$  Then the map $i_p$ is horizontally conformal and $p$-harmonic.
\end{theorem}

\begin{proof}
The fact that $i_p$ is conformal is classic, but we prove it here for the reader's convenience.  For $1\le j,k\le 2p$ the conformality operator $\kappa$ satisfies
	\begin{eqnarray*}
		\kappa (F_j,F_k)
		&=&\sum_{s=1}^{2p}\frac{\partial F_j}{\partial x_s}\cdot\frac{\partial F_k}{\partial x_s}\\
		&=&\sum_{s=1}^{2p}
		\frac{(\delta_{js}|x|^2-2\,x_jx_s)}{|x|^4}\cdot
		\frac{(\delta_{ks}|x|^2-2\,x_kx_s)}{|x|^4}\\
		&=&\sum_{s=1}^{2p}
		\frac{(\delta_{js}\delta_{ks}|x|^4-2\,\delta_{js}x_kx_s|x|^2-2\,\delta_{ks}x_jx_s|x|^2+4x_jx_kx_s^2)}{|x|^8}\\
		&=&\frac{\delta_{jk}|x|^4-2|x|^2x_jx_k-2|x|^2x_kx_j+4x_jx_k|x|^2}{|x|^8}\\
		&=&\frac{\delta_{jk}}{|x|^4}.
	\end{eqnarray*}

	The fact that the map $i_p$ is proper $p$-harmonic is a direct consequence of the following repeated application of Lemma \ref{lemma-Laplacian}.
	$$\tau( i_p)=\frac{2(2-2p)}{|x|^2}\cdot i_p,$$
	$$\tau^2( i_p)=\frac{2(2-2p)4(4-2p)}{|x|^4}\cdot i_p$$
	$$\vdots$$

	$$\tau^p( i_p)=\frac{2(2-2p)4(4-2p)\cdots 2p(2p-2p)}{|x|^{2p}}\cdot i_p=0.$$
\end{proof}

\begin{lemma}\label{lemma-Laplacian}
	For a positive integer $n\in{\mathbb Z}^+$ let the map  $\phi:{\mathbb R}^{p}\setminus\{0\}\to{\mathbb R}^{p}\setminus\{0\}$ be given by
	$$\phi=(\phi_1,\dots ,\phi_p):x\mapsto\frac {(x_1,\dots ,x_p)}{|x|^n}.$$
	Then the tension field $\tau(\phi)$ of $\phi$ satisfies
	$$\tau(\phi)=\frac {n(n-p)}{|x|^{n+2}}\cdot\phi.$$
\end{lemma}

\begin{proof}
First we notice that
$$\frac{\partial}{\partial x_j}|x|^n=n\, x_j|x|^{n-2}.$$
Applying this several times we then get
$$\frac{\partial\phi_k}{\partial x_j}=\frac{\delta_{jk}|x|^n-nx_kx_j\, |x|^{n-2}}{|x|^{2n}}$$
and
\begin{eqnarray*}
\frac{\partial^2\phi_k}{\partial x_j^2}
&=&\frac 1{|x|^{2n+2}}\Bigl(
\frac{\delta_{jk}nx_j|x|^{n-2}-\delta_{jk}nx_j|x|^{n-2}}{|x|^{2n+2}}\\
& &
\qquad\qquad  -\frac{nx_k\,|x|^{n-2}+n(n-2)x_kx_j^2|x|^{n-4}}{|x|^{2n+2}}\\
& &\qquad\qquad\qquad -\frac{2n(\delta_{jk}|x|^n-nx_kx_j\, |x|^{n-2}|x|^{2n})x_j}{|x|^{2n+2}}\Bigr).
\end{eqnarray*}
This means that for the tension field $\tau(\phi_k)$ we yield
\begin{eqnarray*}
& &\tau(\phi_k)\\
&=&\frac{-npx_k|x|^{3n-2}-n(n-2)x_k|x|^{3n-2}-2nx_k|x|^{3n-2}+2n^2x_k|x|^{3n-2}}{|x|^{4n}}\\
&=&(2n^2-2n-n(n-2)-np)\frac {x_k}{|x|^{n+2}}\\
&=&n(n-p)\frac{x_k}{|x|^{n+2}}.
\end{eqnarray*}
\end{proof}

\section{Two Conjectures}

We conclude this work with two conjectures that have come to our minds while working on this project.

\begin{conjecture}\label{conjecture-1}
Let $p\in{\mathbb Z}^+$ be a positive integer and $i_p=(F_1,F_2,\dots ,F_{2p}):{\mathbb R}^{2p}\setminus\{0\}\to{\mathbb R}^{2p}\setminus\{0\}$ be the inversion about the unit sphere $S^{2p-1}$ in ${\mathbb R}^{2p}$.  Then $z:{\mathbb R}^{2p}\setminus\{0\}\to{\mathbb C}$ with
$$z=a_1F_1+a_2F_2\cdots+a_{2p}F_{2p}$$ is a complex-valued $(p,p)$-harmonic morphism for any element $a\in{\mathbb C}^{2p}$.
\end{conjecture}

Our rather extensive computer calculations show that this Conjecture \ref{conjecture-1} is true in the cases when $p=1,2,3,4$, but the statement seems to be difficult to prove in general.
\medskip

No proper $(2,1)$-harmonic morphism is known to exist from the three dimensional Euclidean spaces ${\mathbb R}^{3}$, not even locally.  For this we have the following.

\begin{conjecture}
Let $p\ge 2$ and $\phi:U\to{\mathbb C}$ be a complex-valued $(p,1)$-harmonic morphism defined locally on the standard Euclidean space ${\mathbb R}^{2p-1}$.  Then $\phi$ is a $(1,1)$-harmonic morphism i.e. $\tau(\phi)=0$.
\end{conjecture}

\section{Acknowledgements}

The authors would like to thank the referee for several useful remarks that have improved the presentation. The first author would like to thank the Department of Mathematics at Lund University for its great hospitality during her time there as a postdoc.

redpajama_set_name: RedPajamaArXiv


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