TITLE: Applications of the inequality $e^x\geqslant 1+x$
QUESTION [6 upvotes]: I am interested in proofs of famous theorems or inequalities which can be proved by the inequality
$$
e^x\geqslant 1+x.
$$
For example, the divergence of harmonic series can be proved by assigning $x=\frac{1}{k}$.
$e^{\pi}>\pi^e$ can be proved by assigning $x=\frac{\pi}{e}-1$.
The inequality of the arithmetic average of n-th degree can be proved by assigning $x=\frac{a_k}{(a_1*a_2*\cdots*a_n)^{\frac{1}{n}}} -1$.
The divergence of $\frac{e^t}{t^n}$ can be proved by assigning $x=\frac{t}{n+1}$.

I want to know other possible proofs.

REPLY [1 votes]: A standard use is that
$\prod (1+a_k)
\le \prod e^{a_k}
=  e^{\sum a_k}
$
so that,
if $\sum a_k$ converges then
$\prod (1+a_k)$
also converges.


text:
\begin{document}
\maketitle

\begin{abstract}

Recovery of the sparsity pattern (or support) of a sparse vector from a small number of noisy linear samples is a common problem that arises in signal processing and statistics. In the high dimensional setting, it is known that recovery with a vanishing fraction of errors is impossible if the sampling rate and per-sample signal-to-noise ratio (SNR) are finite constants independent of the length of the vector. In this paper, it is shown that recovery with an arbitrarily small but constant fraction of errors is, however, possible, and that in some cases a computationally simple thresholding estimator is near-optimal. Upper bounds on the sampling rate needed to attain a desired fraction of errors are given in terms of the SNR and various key parameters of the unknown vector for two different estimators. The tightness of the bounds in a scaling sense, as a function of the SNR and the fraction of errors, is established by comparison with existing necessary bounds. Near optimality is shown for a wide variety of practically motivated signal models.
\end{abstract}

\begin{keywords}
compressed sensing, information-theoretic bounds, random matrices, random projections, regression, sparse approximation, sparsity, subset selection.
\end{keywords}

\input{sections/introduction}

\input{sections/formulation}

\input{sections/upper-bounds}

\input{sections/rate-sharing}

\input{sections/scalings-LB}

\input{sections/examples}

\input{sections/optimality}

\appendices

\input{sections/proof-MLnew}
\input{sections/proof-MCnew}
\input{sections/proof-rate-sharing}

\input{sections/scalings-proofs-UB}

\section*{Acknowledgment} We would like to thank Martin Wainwright for helpful discussions and pointers. This work was supported in part by ARO MURI No. W911NF-06-1-0076.

\bibliographystyle{ieeetran}

\bibliography{appox-sampling-upper-bounds.bbl}

\end{document}

config: arxiv


TITLE: Why is it accepted that Gödel uses a self-referential statement in his incompleteness theorems?
QUESTION [2 upvotes]: (I have only read second-hand about Gödel's incompleteness theorems, so I hope that there is nothing I have missed by doing that which would render my questions easily clarified.)
Gödel's first incompleteness theorem says that any consistent mathematical system is incomplete.
But the whole proof seems to depend solely on using a self-referential statement $G$ where
$G = $"G cannot be proven".
(I hope it is correct/ okay to put it like that)
It seems intuitive, however, that this statement is nonsensical.
$1.$ But then why is he allowed to use a nonsensical statement for proof?
To maybe explain at least a bit why I think it is nonsensical (if someone should not have that intuition strongly enough):
A. If we look at it as a definition of $G$, then it makes no sense to use $G$ itself in the definition of it.
This intuition could be partly the same as B.ii., but also $G$ doesn't add anything new to $G$. The self-referential aspect seems to make $G$ non-substantial.
So, $G$ seems to be without "real content".
B. Let $G_1= "G_2$ cannot be proven".
i. If $G_1$ and$ G_2 $are different, taken to be the same, this makes no sense (reminds of equivocation).
ii. If $G_1=G_2(=G)$, then we get an infinite regress.
Substitute $G_1 $for $G_2$, since it is the same thing, and we get ""$G_2$ cannot be proven" cannot be proven""= $G_1$. And do it again and we get " $G_2$ cannot be proven" cannot be proven" cannot be proven" =$ G_1.$ And so on infinitely.
In other words: "$G$ cannot be proven" can be infinitely substituted for $ G$ in "$G$ cannot be proven".
And maybe infinite regress (as a consequence) alone isn't enough to say of something that doesn't make any sense or should be discarded (we have Münchhausen Trilemma), but since $G$ intuitively also seems nonsensical (see A.), may it not be enough to discard it (/self-referential statements)?

Also, if we do not use $G$ (or self-referential statements in general?), we would have no proof for any of what Gödel's incompleteness theorems show, is this correct? Or: if we discard $G$, then there is still the possibility that we can have consistent and complete and decidable systems?

Someone has said that there are other proofs for it using diagonalization. But if this is true, (since I also have intuitive issues with diagonalization arguments, though somewhat less), apart from these, there are none?

REPLY [11 votes]: No. A standard Gödel sentence $G$ (for a given theory $T$) is NOT identical to the sentence "$G$ cannot be proven) (in $T$)".
But $G$ is so constructed that it is true if and only if $G$  cannot be proven in $T$.
Being materially equivalent and being identical are not the same. The difference is crucial here.
You don't need to read Gödel himself to understand that. There are many good expositions out there. For example, you could try the first three short chapters of my Gödel Without (Too Many) Tears, downloadable from https://www.logicmatters.net/igt


TITLE: infinite primes $p\equiv1\pmod n$ without cyclotomic polynomial
QUESTION [2 upvotes]: Without cyclotomic polynomial, is there an elementary proof of  the following: for each integer $n>1$, there are infinitely many primes $p$ such that $p\equiv1\pmod n$ ?
please don't refer to Dirichlet's theorem on arithmetic progressions, or analytic number theory. Thanks

REPLY [2 votes]: An elementary proof of this result is outlined in problem 36 on p. 108 of An Introduction to the Theory of Numbers by Niven, Zuckerman and Montgomery. That problem references the article by I. Niven and B. Powell "Primes in certain arithmetic progressions," Amer. Math. Monthly, 83 (1976), 467-469, as simplified by R.W. Johnson.


TITLE: What function could describe this GIF animation?
QUESTION [13 upvotes]: I found this image on Beautiful Mathematical GIFs Will Mesmerize You and this GIF really caught my attention. From what I see, it's a 2D circle morphing into the 3D sphere. What function could describe this GIF animation? All comments are appreciated.
z < 0 -> red
z > 0 -> green
z = 0 -> blue

REPLY [3 votes]: The source code shows it: in spherical coordinates, $\phi=t_w\cos(\theta)$ where $t_w$ changes periodically over time between $0$ and $1$ (as a third degree polynomial of the cosine). The "sphere" also rotates uniformly and is displayed with perspective.


TITLE: Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant.
QUESTION [2 upvotes]: Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant.
I want to prove $f$ is bounded. Then by using Liouville theorem, $f$ is constant. But I can't proceed. Can anyone give some hint?

REPLY [6 votes]: Note : If $f$ is entire, then, in particular it is bounded on closed balls around the origin. Hence, if $D = \{z : |z| \leq 1\}$, then $f$ is bounded on $D$ by some $M > 0$.
Hence, if $|z| \geq 1$, then $1/z \in D$, and hence
$$
|f(z)| = |f(1/z)| \leq M
$$
And hence, $|f(z)| \leq M$ for all $z\in \mathbb{C}$. Hence, Louiville applies.


TITLE: Continuity of the supremum of a continuous function over a closed interval
QUESTION [3 upvotes]: Proposition: let $F:R\times [0,1]\rightarrow R$, be a countinuous function. If $g(x):= Sup\{F(x,t):t\in [0,1]\}$, then $g(x)$ is continuous at all points.
My Idea:
Assume $g$ is not continuous at $x_0$. Then there exists $\epsilon >0$ such that there exists a sequence $\{y_n\}$ which is converges to $x_0$ and $|g(y_n)-g(x_0)|>\epsilon$ and $y_n\ne x_0$ for all $n$ .
$g(y_n)=F(y_n,t_n)$ for some $t_n\in [0,1]$ as $[0,1]$ is compact and $F$ is continuous. Thus we have the sequence $\{(y_n,t_n)\}$. Now since the sequence lies in a closed interval by the Bolzano–Weierstrass theorem there is a sub-sequence $\{(y_m,t_m)\}$ which converges. It is easy to note $\{(y_m,t_m)\}$ converges to $(x_0,t_0)$ where $t_0$ is some real number between 0 and 1 .
Hence $F(y_m,t_m)>g(x_0)+\epsilon$ or $F(y_m,t_m)<g(x_0)-\epsilon$
as
$g(y_m)>g(x_0)+\epsilon$ or $g(y_m)<g(x_0)-\epsilon$, which implies
$F(x_0,t_0)\ge g(x_0)+\epsilon$ or $F(x_0,t_0)\le g(x_0)-\epsilon$
If it were only for the first inequality we would be done, alas it is not the case.
I would appreciate if someone helped me prove/disprove the proposition.

REPLY [1 votes]: We show $g(x)$ is continuous by the $\epsilon- \delta$ definition using uniform continuity of $F(x)$.
To show $g(x)$ is continuous at $x_0$, given $\epsilon>0$ we note that $F(x)$ is uniformly continuous in the interval $[x_0-1\ ,\ x_0+1]\times [0,1]$.
Next given ${\epsilon}$ by uniform continuity, $\exists\ \delta>0$ such that $|F(\textbf{x})-F(\textbf{y})|<\epsilon \ \ \ \forall \ \textbf{x,y}\in[x_0-1\ ,\ x_0+1]\times [0,1]\ \ \ \&\ \ \ ||\textbf{x-y}||<\delta$
Now given $x\in \ (x_0-min(\delta,1),\ x_0+min(\delta,1))$

Case 1: $g(x)> g(x_0)+\epsilon$ : let $F(x,t)=g(x)$ we know such a $t$ exists as $F$ is continuous. Note $|F(x,t)-F(x_0,t)|<\epsilon$ or $F(x_0,t)>g(x)-\epsilon>g(x_0)$  which is a contradiction as $g(x_0)\ge F(x_0,a)\ \forall a\in[0,1]$
Case 2: $g(x)< g(x_0)-\epsilon$ : let $F(x_0,t)=g(x_0)$ we know such a $t$ exists as $F$ is continuous. Note $|F(x,t)-F(x_0,t)|<\epsilon$ or $F(x,t)>g(x_0)-\epsilon>g(x)$  which is a contradiction as $g(x)\ge F(x,a)\ \forall a\in[0,1]$

what remains is $|g(x)-g(x_0)|\le \epsilon$
Hence $g(x)$ is continuous everywhere!!


TITLE: What is the analog of a uniform matroid when the ground set is not finite?
QUESTION [1 upvotes]: If $E$ is a finite set and $k \in \mathbb{N}$, then the uniform matroid of rank $k$ is defined as the matroid generated by taking the collection of $k$ element subsets of $E$ as a basis. Is there an analog of this notion when $E$ is not finite?
Currently reading this paper, other references or examples would be great.

REPLY [1 votes]: An uniform matroid is a matroid such that: $\forall I\in \mathcal{I}$, $\forall x\in I$, $\forall y\in E\backslash I$, $I-x+y\in \mathcal{I}$.
Here some slides of a talk about the implications of the existance of infinite uniform matroids, http://settheory.mathtalks.org/wp-content/uploads/2014/03/Geschke.pdf


TITLE: In $\epsilon$-$\delta$ continuity over an interval, $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$. Since both $c$ and $x$ vary, what's the difference?
QUESTION [0 upvotes]: If a function $f$ is continuous at every $c$ point in $[a, b]$ then we have for every $\epsilon$ we can find a $\delta$ such that whenever $|x-c|<\delta$, we have $|f(x)-f(c)|<\epsilon$.

Now, in this definition, $c$ varies in the interval $[a, b]$, and also $x$ varies in $[a, b]$, then what is the difference between $x$ and $c$?

REPLY [2 votes]: Working with some function $f:[a,b]\to\mathbb R$ and using quantors we have:$$f\text{ is continuous at }c\iff\forall \epsilon>0\;\exists \delta>0\;\forall x\in[a,b]\;\left[|x-c|\leq\delta\implies |f(x)-f(c)|<\epsilon\right]$$Note that here $c$ is a free variable and $x$ is a bound variable (i.e. connected with a quantor).
We are dealing here with a statement about $f$ and $c$.

Further:$$f\text{ is continuous }\iff\forall c\in[a,b]\;\left[f\text{ is continuous at }c\right]\iff$$$$\forall c\in[a,b]\;\forall \epsilon>0\;\exists \delta>0\;\forall x\in[a,b]\;\left[|x-c|\leq\delta\implies |f(x)-f(c)|<\epsilon\right]$$
Note that here $c$ and $x$ are both bound variables and that the meaning of the statement on RHS does not change if the variables $c$ and $x$ are switched.
We are dealing here with a statement about $f$.


text:
\begin{document}
\title[A generalization of Roberts' counterexample]{A note on "a generalization of Roberts' counterexample to the fourteenth problem of Hilbert by S.\ Kuroda"}
\author{Mikiya Tanaka}
\address{Graduate School of Science and Technology, Kwansei Gakuin University,
2-1 Gakuen, Sanda, Hyogo 669-1337, Japan}
\email{mtanaka@kwansei.ac.jp}
\keywords{locally nilpotent derivation, the fourteenth problem of Hilbert}
\subjclass[2000]{Primary: 14R20; secondary: 13A50, 13N15}
\maketitle

\begin{abstract}
In \cite{Kuroda}, Kuroda generalized Roberts' counterexample \cite{Roberts} to the fourteenth problem of Hilbert.
The counterexample is given as the kernel of a locally nilpotent derivation on a polynomial ring.
We replace his construction of the invariant elements by a more straightforward construction and give a more precise form of invariant elements.
\end{abstract}

\section{Introduction}

Let $k$ be a field of characteristic zero and let $B$ be a $k$-algebra.
We denote by ${\rm LND}_k(B)$ the set of $k$-derivations of $B$.
In \cite{Kuroda}, Kuroda proved the following result.

\begin{thm}\label{thm1.1}
Let $B=k[x_1,\dots,x_n,y_1,\dots,y_n,y_{n+1}]$ be a polynomial $k$-algebra and define $\delta\in {\rm LND}_k(B)$ by $\delta(x_i)=0$
and $\delta(y_i)={x_i}^{2}$ for all $1\leq i\leq n$, and $\delta(y_{n+1})=x_1\cdots x_n$.
Suppose that $n\geq 4$.
Then $A:=\Ker \delta$ is not finitely generated over $k$.
\end{thm}

In order to prove this theorem, he made use of the following lemma.

\begin{lem}\label{lem1.2}
With the notations and assumptions in the above theorem, there exists a positive integer $\alpha$ such that the $k$-subalgebra $A$ contains elements of the form
$$
{x_1}^\alpha{y_{n+1}}^\ell+(\text{terms of lower degree in }y_{n+1})
$$
for each $\ell \geq 1$.
\end{lem}

In this paper, we prove that we can take $\alpha=1$.
Namely, we prove the following.

\begin{thm}\label{thm1}
With the notations and assumptions in the above theorem, the $k$-subalgebra $A$ contains elements of the form
$$
x_1{y_{n+1}}^\ell+(\text{terms of lower degree in }y_{n+1})
$$
for each $\ell \geq 1$.
\end{thm}

\section{Proof of Theorem \ref{thm1}}

In a subsequent proof, we use the following result.

\begin{lem}
Let $B=k[x_2,\dots,x_n,y_2,\dots,y_n]$ and define $\delta \in {\rm LND}_k(B)$ by $\delta(x_i)=0$ and $\delta(y_i)={x_i}^2$ for each $i$.
Then $\Ker \delta$ is a $k$-algebra generated by $x_2,\dots,x_n$ and ${x_i}^2y_j-{x_j}^2y_i\ (2\leq i,j\leq n, i\neq j)$.
\end{lem}

We can prove this lemma by the same argument in \cite[Theorem 1.2]{Kojima-Miyanishi}.

Now, for each monomial $m={x_1}^{a_1}\cdots{x_n}^{a_n}{y_1}^{b_1}\cdots{y_{n+1}}^{b_{n+1}}$, define
$$
\tau(m)=\left[\frac{a_2}{2}\right]+\cdots+\left[\frac{a_n}{2}\right]-(b_1+\cdots+b_n),
$$
where we write $[a]=\max\{n\in \Z\mid n\leq a\}$ for any $a\in \R$.
Let $f_{1,n+1}=x_1y_{n+1}-x_2\cdots x_ny_1$ and let $f_{i,j}={x_i}^2y_j-{x_j}^2y_i$ for each pair $(i,j)$ with $1\leq i,j\leq n$ and $i\neq j$.
It is easy to see that all of $f_{1,n+1}$, $x_i$ and $f_{i,j}$ belong to $A$.
Let $A'$ be the $k$-subalgebra generated by $f_{1,n+1}$, $x_i\ (1\leq \i \leq n)$ and $f_{i,j}\ (1\leq i,j \leq n, i\neq j)$.
Since $A$ is factorially closed in $B$, i.e., $a=b_1b_2$ with $b_1,b_2\in B$ implies $b_1,b_2\in A$, it suffices to show that there exists $f\in A'$ such that ${f_{1,n+1}}^\ell -f$ is of the form
\begin{align*}
&{x_1}^\ell {y_{n+1}}^\ell +{x_1}^{\ell-1}(\text{terms of lower degree in }y_{n+1})\\
&={x_1}^{\ell-1}\left(x_1{y_{n+1}}^{\ell}+(\text{terms of lower degree in }y_{n+1})\right).
\end{align*}
We have
\begin{align*}
{f_{1,n+1}}^\ell=&{x_1}^\ell{y_{n+1}}^\ell-\ell {x_1}^{\ell-1}{y_{n+1}}^{\ell-1}x_2\cdots x_n y_1\\
&+\binom{\ell}{2}{x_1}^{\ell-2}{y_{n+1}}^{\ell-2}{x_2}^2\cdots{x_n}^2{y_1}^2+\cdots+(-1)^\ell {x_2}^\ell \cdots {x_n}^\ell {y_1}^\ell
\end{align*}
and we construct $f\in A'$ which, when subtracted from ${f_{1,n+1}}^\ell$, cancels the terms in ${f_{1,n+1}}^\ell$ of degree $<\ell-1$ in $x_1$ and produces only the terms of degree $\geq \ell-1$ in $x_1$.
Namely, as the element ${f_{1,n+1}}^\ell-f$, we construct an element in $A'$ of the form
$$
{x_1}^\ell {y_{n+1}}^\ell + g_{l-1}{y_{n+1}}^{\ell-1}+g_{\ell-2}{y_{n+1}}^{\ell-2}+\cdots+g_0,
$$
where $g_i\in k[x_1,\dots,x_n,y_1,\dots,y_n]$ and ${x_1}^{\ell-1}$ divides every $g_i$.

By the descending induction on $r$, we suppose that we obtain an element in $A'$ of the form
$$
G_{r}={x_1}^\ell {y_{n+1}}^\ell + g_{l-1}{y_{n+1}}^{\ell-1}+g_{\ell-2}{y_{n+1}}^{\ell-2}+\cdots+g_r{y_{n+1}}^r+\cdots+g_0
$$
with $g_i\in k[x_1,\dots,x_n,y_1,\dots,y_n]$ and $g_{\ell-1},\dots,g_{r+1}$ divisible by ${x_1}^{\ell-1}$.
We show that $G_r$ is modified by an element of $A'$ so that a new $g_r$ is divisible by ${x_1}^{\ell-1}$ without changing the terms $g_{\ell-1},\dots,g_{r+1}$.
Furthermore, we suppose the following conditions are satisfied.
\begin{enumerate}
\item[(1)]For $0\leq i \leq \ell-1$, if we write $g_i=\sum_j {x_1}^j {y_1}^{q_{i,j}} h_{i,j}$ with $h_{i,j}\in k[x_2,\dots,x_n,y_2,\dots,y_n]$, then $i+j+2q_{i,j}=2\ell$.
\item[(2)]We have $h_{i,0}=\cdots=h_{i,i-1}=0$ for $0\leq i\leq \ell-1$, i.e., for each ${x_1}^j {y_1}^{q_{i,j}} h_{i,j}$ appearing in $g_i$, we have $j\geq i$.
\item[(3)]For each monomial $m={x_1}^j{x_2}^{a_2}\cdots{x_n}^{a_n}{y_1}^{b_1}\cdots{y_n}^{b_n}{y_{n+1}}^i$\\
in ${y_{n+1}}^i{x_1}^j {y_1}^{q_{i,j}} h_{i,j}$, we have
\begin{enumerate}
\item[(i)]$2\tau(m)\geq \ell-j-3$ and $a_2,\dots,a_n$ are all odd integers if $j \equiv \ell-1 \pmod{2}$,
\item[(ii)]$2\tau(m)\geq \ell-j$ and $a_2,\dots,a_n$ are all even integers if $j\equiv \ell \pmod{2}$.
\end{enumerate}
\end{enumerate}
In order to improve the term $g_r$ in such a way that $h_{r,0}=\cdots=h_{r,\ell-2}=0$, we suppose by a double induction that $h_{r,0}=\cdots=h_{r,p-1}=0$ and $h_{r,p}\neq 0$ with $r\leq p \leq \ell-2$.
With this hypothesis taken into account, we denote the polynomial $G_r$ by $G_{r,p}$.
The beginning polynomial for induction is $G_{\ell-2,\ell-2}={f_{1,n+1}}^\ell$, for which $g_i=(-1)^i \binom{\ell}{i}{x_1}^i(x_2\cdots x_n y_1)^{\ell-i}$, $h_{i,i}=(-1)^i \binom{\ell}{i}(x_2\cdots x_n)^{\ell-i}$, $h_{i,j}=0\ (i\neq j)$ and $q_{i,i}=\ell-i$ for $0\leq i,j\leq \ell-1$.
One can check easily that the above conditions are satisfied for $G_{\ell-2,\ell-2}={f_{1,n+1}}^\ell$

We explain the process of improving $g_r$.
Since $g_{r+1}$ is divisible by ${x_1}^{\ell-1}$ and
\begin{align*}
&\delta({x_1}^p{y_1}^{q_{r,p}}h_{r,p}{y_{n+1}}^r)={x_1}^p{y_1}^{q_{r,p}}{y_{n+1}}^r\delta(h_{r,p})\\
&+(\text{terms of degree }>p\text{ in }x_1)+(\text{terms of degree }<r\text{ in }y_{n+1}),
\end{align*}
we have
\begin{align*}
&0=\delta(G_{r,p})={x_1}^p{y_1}^{q_{r,p}}{y_{n+1}}^r\delta(h_{r,p})\\
&+(\text{terms of degree }>p\text{ in }x_1)+(\text{terms of degree }\neq r\text{ in }y_{n+1})
\end{align*}
and hence $\delta(h_{r,p})=0$.
Lemma 2.1 implies that $h_{r,p}$ is a sum of polynomials of the form
$$
c{x_2}^{d_2}\cdots{x_n}^{d_n}\prod_{i,j\in \{2,\dots,n\}}{f_{i,j}}^{t_{i,j}}
$$
with $c\in k$ and non-negative integers $d_i$, $t_{i,j}$.
Note that all of $d_2,\dots,d_n$ are odd integers (resp.\ even integers) if $p\equiv \ell-1 \pmod{2}$ (resp.\ if $p\equiv \ell\pmod{2}$).
In fact, since the contributions of the $f_{i,j}$ to the exponent $d_2,\dots,d_n$ are even, the remark follows from the conditions (i) and (ii) of (3).
Now we choose any one of the above polynomials and let $H=\prod_{i,j\in \{2,\dots,n\}}{f_{i,j}}^{t_{i,j}}$.
Then, for each monomial $m$ in ${y_{n+1}}^r{x_1}^p{y_1}^{q_{r,p}}{x_2}^{d_2}\cdots{x_n}^{d_n}H$, we have in view of (i) and (ii) of (3),
\begin{align*}
2\tau({y_1}^{q_{r,p}}{x_2}^{d_2}\cdots{x_n}^{d_n})&=2\tau(m)\\
&\geq
\begin{cases}
\ell-p-3 & \text{if } p\equiv \ell-1 \pmod{2}\\
\ell-p & \text{if } p\equiv \ell \pmod{2}
\end{cases},
\end{align*}
where multiplying ${y_1}^{q_{r,p}}{x_2}^{d_2}\cdots{x_n}^{d_n}$ by any ${x_i}^2y_j\ (i\neq 1)$, $y_{n+1}$ or $x_1$ does not change the value of $\tau$.
Note that $r\leq p\leq \ell-2$ and that if $p\equiv \ell-1 \pmod{2}$, then $p\leq \ell-3$ and hence $\ell-p-3\geq 0$.
Thus we have $\tau({y_1}^{q_{r,p}}{x_2}^{d_2}\cdots{x_n}^{d_n})\geq 0$ and there exists an element $F\in A'$ of the form
\begin{align*}
F&=c{x_1}^{p-r}{f_{1,n+1}}^r{f_{2,1}}^{q_2}\cdots{f_{n,1}}^{q_n}{x_2}^{d_2-2q_2}\cdots{x_n}^{d_n-2q_n}\\
&=c{x_1}^p{y_1}^{q_{r,p}}{y_{n+1}}^r{x_2}^{d_2}\cdots{x_n}^{d_n}\\
&+(\text{terms of degree }>p\text{ in }x_1)+(\text{terms of degree }<r\text{ in }y_{n+1}),
\end{align*}
where $q_2+\cdots+q_n=q_{r,p}$.
We can prove that $G_{r,p}-FH$ satisfies the same conditions as $G_{r,p}$ does except for the condition $h_{r,p}\neq 0$ but the number of nonzero terms in $h_{r,p}$ gets smaller.
We prove this below.
By repeating this process finitely many times, we obtain a new $G_r$ satisfying the condition $h_{r,p}=0$.
Further, continuing this process finitely many times, we obtain a modified $G_r$ satisfying the condition $h_{r,0}=\cdots=h_{r,\ell-2}=0$, i.e., $g_r$ is divisible by ${x_1}^{\ell-1}$.
Hence by induction on $r$, we completes a proof.

Now we show that $G_{r,p}-FH$ satisfies the same conditions as $G_{r,p}$ does but the number of nonzero monomial terms in $h_{r,p}$ becomes less.
We have only to show that each monomial in $F$ satisfies the conditions (1)-(3) since none of $y_{n+1}$, $x_1$ and $y_1$ appears in $H$ and multiplication of any monomial in $H$ to a monomial does not chage the value of $\tau$.
Each nonzero monomial $m_F$ in $F$ is of the form
\begin{align*}
{x_1}^{p-r}(x_1y_{n+1})^{r_1}(x_2\cdots x_n y_1)^{r_2}{x_2}^{d_2-2q_2}\cdots{x_n}^{d_n-2q_n}
\prod_{i=2}^n ({x_i}^2 y_1)^{\alpha_i}({x_1}^2y_i)^{\beta_i}
\end{align*}
with $r_1+r_2=r$ and $\alpha_i+\beta_i=q_i$ for $i=2,\dots,n$.
We choose one $m_F$ and let $w$, $z_1$, and $z_{n+1}$ be the exponents of $x_1$, $y_1$, and $y_{n+1}$ in $m_F$ respectively.
Then we have
\begin{align*}
w&=p-r+r_1+2\beta_2+\cdots+2\beta_n=p-r_2+2\beta_2+\cdots+2\beta_n,\\
z_1&=r_2+\alpha_2+\cdots+\alpha_n \text{\quad and\quad}z_{n+1}=r_1.
\end{align*}
First we prove $m_F$ satisfies the conditions (1) and (2).
Indeed, we have
\begin{align*}
&z_{n+1}+w+2z_1=p+r_1+r_2+2(\alpha_2+\beta_2)+\cdots+2(\alpha_n+\beta_n)\\
&=p+r+2q_2+\cdots+2q_n=p+r+2q_{r,p}=2\ell
\end{align*}
and
\begin{align*}
&w-z_{n+1}=p-(r_1+r_2)+2(\beta_2+\cdots+\beta_n)\\
&=p-r+2(\beta_2+\cdots+\beta_n)
\geq p-r\geq 0.
\end{align*}

In order to prove that $m_F$ satisfies the condition (3), we consider four cases
\begin{enumerate}
\item[(a)]$p\equiv \ell-1 \pmod{2}$ and $r_2=2u+1$
\item[(b)]$p\equiv \ell-1 \pmod{2}$ and $r_2=2u$
\item[(c)]$p\equiv \ell \pmod{2}$ and $r_2=2u+1$
\item[(d)]$p\equiv \ell \pmod{2}$ and $r_2=2u$,
\end{enumerate}
where $u$ is an integer.
We only consider the case (a).
The remaining cases can be treated in a similar fashion.
Then we have
$$w\equiv \ell-1-2u-1+2\beta_2+\cdots+2\beta_n\equiv \ell \pmod{2}.$$
The exponent of each $x_i\ (i\neq 1)$ in $m_F$ is equal to $2u+1+d_i-2q_i+2\alpha_i$.
Since each $d_i$ is an odd integer by the condition (i) of (3), it is an even integer.
In addition, we have
\begin{align*}
2\tau(m_F)=&2((n-1)u+(n-1)+\tau({x_2}^{d_2-2q_2}\cdots{x_n}^{d_n-2q_n})\\
&-(2u+1)-(\beta_2+\cdots+\beta_n))\\
=&2u(n-3)+2(n-2)+2\tau({x_2}^{d_2-2q_2}\cdots{x_n}^{d_n-2q_n})\\
&-2(\beta_2+\cdots+\beta_n)\\
=&(r_2-1)(n-3)+2(n-2)+2\tau({y_1}^{q_{r,p}}{x_2}^{d_2}\cdots{x_n}^{d_n})\\
&-2(\beta_2+\cdots+\beta_n)\\
\geq &r_2-1+2\cdot 2+\ell-p-3-2(\beta_2+\cdots+\beta_n)\\
=&\ell-(p-r_2+2\beta_2+\cdots+2\beta_n)=\ell-w,
\end{align*}
where the term $(n-1)$ in the first equality is due to the condition that all the $d_i$ and $r_2$ are odd integers and we use the condition $n\geq 4$ to show the inequality.
Thus the condition (3) holds for $m_F$.
This induction completes a proof of Theorem \ref{thm1}.

\section{Application to module derivations}

In this section, we give application of Theorem \ref{thm1} to locally nilpotent module derivations.
First, we recall the following definition (see \cite{Tanaka}).

\begin{defn}{\em
Let $\delta\in {\rm LND}_k(B)$ and let $M$ be a $B$-module with a $k$-linear endomorphism $\delta_M:M\to M$.
A pair $(M,\delta_M)$ is called a $(B,\delta)$-module (a $\delta$-module, for short) if the following two conditions are satisfied.
\begin{enumerate}
\item[(1)]For any $b\in B$ and $m\in M$, $\delta_M(bm)=\delta(b)m+b\delta_M(m)$.
\item[(2)]For each $m\in M$, there exists a positive integer $N$ such that ${\delta_M}^n(m)=0$ if $n\geq N$.
\end{enumerate}
Let $A=\Ker \delta$.
Then $\delta_M$ is an $A$-module endomorphism.
Whenever we consider $\delta$-modules, the derivation $\delta$ on $B$ is fixed once for all.
We call $\delta_M$ a {\em module derivation} (resp.\ {\em locally nilpotent module derivation}) on $M$ if it satisfies the condition (1) (resp.\ both conditions (1) and (2)).
}\end{defn}

If there is no fear of confusion, we simply say that $M$ is a $\delta$-module instead of saying that $(M,\delta_M)$ is a $\delta$-module.
If $M$ is a $\delta$-module, then $M_0:=\Ker \delta_M=\{m\in M \mid \delta_M(m)=0\}$ is an $A$-module.
We retain below the notations $A,M_0$ for this specific purposes.
For the basic properties of $\delta$-modules, we refer the readers to \cite{Tanaka}.

We consider the following problem.

\begin{problem}\label{problem}
Let $B$ be an affine $k$-domain with a locally nilpotent derivation $\delta$ and let $M$ be a finitely generated $B$-module with $\delta$-module structure.
Is $M_0$ a finitely generated $A$-module?
\end{problem}

We have positive answers to Problem \ref{problem} if one of the following conditions is satisfied (see \cite{Tanaka2}).
\begin{enumerate}
\item[(i)]$M$ is torsion-free as a $B$-module and $A$ is a noetherian domain.
\item[(ii)]$M$ is torsion-free as a $B$-module and $\dim B\leq 3$.
\item[(iii)]$M_0$ is a free $A$-module.
\item[(iv)]The $B$-module $BM_0$ generated by $M_0$ is a free $B$-module with a basis $\{e_1,\dots,e_n\}$ such that $e_i\in M_0$.
\item[(v)]$B=A[y]$ is a polynomial ring over a noetherian domain $A$, $a:=\delta(y)$ is a nonzero element of $A$ and $a$ has no torsion in $M$.
\end{enumerate}
In \cite{Tanaka}, there is an easy counterexample to Problem \ref{problem} in the case where $M$ has torsion as a $B$-module.
In addition, there are counterexamples in the free case by making use of the counterexamples to the fourteenth problem of Hilbert given by Roberts \cite{Roberts}, Kojima-Miyanishi \cite{Kojima-Miyanishi}, Freudenburg \cite{Freudenburg} and Daigle-Freudenburg \cite{Daigle-Freudenburg}.
In such examples, we take $B$ to be a polynomial ring and $M$ to be the differential module $\Omega_{B/k}$.
We can give $\Omega_{B/k}$ a natural module derivation as follows.

\begin{lem}
Let $B$ be a $C$-algebra and let $\delta$ be a locally nilpotent $C$-derivation of $B$.
Then the differential module $M:=\Omega_{B/k}$ is a $\delta$-module if we define $\delta_M$ by $\delta_M(db)=d\delta(b)$ for $b\in B$.
\end{lem}

We can prove this lemma easily (see \cite{Tanaka2}).
Theorem \ref{thm1} gives a new counterexample to Problem \ref{problem}.
Namely, we have the following assertions.

\begin{thm}\label{thm}
With the notations and assumptions in Theorem \ref{thm1.1}, let $M=\Omega_{B/k}$ be the differential module with natural $\delta$-module structure.
Namely, $M$ is a free $B$-module
$$
M=\bigoplus_{i=1}^n Bdx_i\oplus \bigoplus_{i=1}^n Bdy_i
$$
with a free basis $\{dx_1,\dots,dx_n,dy_1,\dots,dy_{n+1}\}$ and a module derivation defined by
\begin{align*}
&\delta(dx_i)=0,\quad \delta(dy_i)=2x_idx_i\ (1\leq i\leq n)\quad \text{and}\\
&\delta(dy_{n+1})=\sum_{i=1}^n x_1\cdots \stackrel{\vee}{x_i} \cdots x_n dx_i.
\end{align*}
Then $M_0$ is not a finitely generated $A$-module.
\end{thm}

We can prove this in a fashion similar to \cite[Theorem6.2]{Tanaka2} by making use of Theorem \ref{thm1}.
The fact that we can take $\alpha=1$ in Lemma \ref{lem1.2} plays an important role in the proof of Theorem \ref{thm}.


TITLE: Value of c so that $c(2-|x|-|y|)$ is a probability distribution function(see picture)
QUESTION [0 upvotes]: Hint: Use the formula of volume of pyaramid.
My approach: I know that the integral of a pdf from $-\infty to +\infty$ gives you $1$. I tried taking the double integral, but got stuck in as how to evaluate the absolute value and use the volume of pyramid.

REPLY [0 votes]: If you graph $f(x,y) = c(2-|x|-|y|)$ over the region $|x|+|y| \le 2$ (i.e. the square with vertices $(2,0), (0,2), (-2,0), (0,-2)$), then you get a pyramid shape.
The base of the pyramid is the square with vertices $(2,0), (0,2), (-2,0), (0,-2)$. The height of the pyramid is $f(0,0)$. The formula for the volume of a pyramid is $$\text{Volume} = \dfrac{1}{3}(\text{Base Area}) \cdot (\text{height}).$$
So, first find the base area and the height. Then, use the above formula to find the volume of the pyramid (in terms of $c$). Finally, figure out what value of $c$ makes that volume $1$.


TITLE: Determine the number of solutions for $x^2 = c$ such that $c \in \mathbb{F}$
QUESTION [1 upvotes]: I am working on the problem from number system class taught by the professor who shares his knowledge about the relation of abstract algebra and number systems.  Here is the problem:
Assume $c$ and $\mathbb{F}$ are arbitrary.  For $c \in \mathbb{F}$, how many distinct solutions can the equation $x^2 = c$ have?  Be specific about what properties of $c$ lead to what number of solution and justify your reasoning using properties a field must have.
Thoughts for the Problem
I believe that there is a solution when $c = 0$ for any field.  If $c = 1$, and we have $\mathbb{F}_2$, then we obtain exactly one solution.  BUT if $\mathbb{F}$ is any arbitrary field and $c = 1$, then we have two solutions.  This situation could occur the same way for any arbitrary field and $c \geq 2$, being an integer.  The thing is: what if $c$ is either trascendental or irrational (like roots number)?  Is it possible for the equation to have solutions like this?  I am having a hard time, approaching this problem.

REPLY [2 votes]: Equivalently, we are talking about the solutions over $x \in \mathbb{F}$ to $x^2 - c= 0$, where $c\in \mathbb{F}$.
In any case, one of two things happens: either there is no solution, or there is a solution $d$ such that $d^2 = c$.  In such a case, we may write
$$
x^2 - c = (x+d)(x-d)
$$
So that $x^2 = 0$ if and only if $x = d$ or $x = -d$.  If (and only if) $d = -d$, our solution is unique.  Otherwise, there are two solutions.
Note that
$d = -d \iff 2d = 0$.  So, if $c$ has a unique square root, then either $c = 0$ or there is an element $d \in \mathbb{F}$ for which $2d = 0$.
I'm not sure what you mean with the statement "what if $c$ is transcendental or irrational". What about it?


TITLE: Abrikosov's Vortex Lattice (Beta Parameter)
QUESTION [1 upvotes]: In order to find the correct vortex lattice configuration (i.e. ground state) in Ginzburg-Landau theory (or the Abelian Higgs Model), it is standard practice to minimize the beta parameter:
$\beta=\frac{\langle |\phi|^{4}\rangle}{\langle |\phi|^{2}\rangle}$.
What is the difference between minimizing $\beta$ and minimizing $\langle |\phi|^{4}\rangle$? Why are they not equivalent?

REPLY [0 votes]: The answer can be found in Abrikosov's paper: Sov. Phys. JETP 5 1174 (1957). The free energy for the vortex lattice is
$F=H_{0}^{2}-\frac{(H_{c}+H_{0})(H_{c}-H_{0})}{\beta(2\chi^{2}-1)}$, where $H_{0}$ is the external magnetic field, $H_{c}\equiv\chi$ is the critical field above which the vortex lattice is unstable and the ground state is the normal vacuum, $\beta_{A}$ is the Abrikosov parameter that determines the lattice structure and $\chi$ is the only free parameter in the theory. The minimum value $\beta$ can take is $1.1596$ for a hexagonal lattice. This assertion is valid for $\chi>\frac{1}{\sqrt{2}}$. Therefore, minimizing $\beta$ gives the correct ground state, it is not equivalent to minimizing $\langle |\phi|^{4}\rangle$.


text:
\begin{document}

\title[$(p,q)$-harmonic morphisms]{Complex-valued $(p,q)$-harmonic morphisms\\ from Riemannian manifolds}

\dedicatory{* The corresponding author}

\author{Elsa Ghandour}
\address{Mathematics, Faculty of Science\\
	University of Lund\\
	Box 118, Lund 221 00\\
	Sweden}
\email{Elsa.Ghandour@math.lu.se}

\author{Sigmundur Gudmundsson*}
\address{Mathematics, Faculty of Science\\
	University of Lund\\
	Box 118, Lund 221 00\\
	Sweden}
\email{Sigmundur.Gudmundsson@math.lu.se}

\begin{abstract}
We introduce the natural notion of $(p,q)$-harmonic morphisms between Riemannian manifolds.  This unifies several theories that have been studied during the last decades.  We then study the special case when the maps involved are complex-valued.  For these we find a  characterisation and provide new non-trivial examples in important cases.
\end{abstract}

\subjclass[2010]{53C43, 58E20}

\keywords{harmonic morphisms}

\maketitle

\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 $\cn$ to harmonic functions in the $3$-dimensional Euclidean space $\rn^3$.  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 $T^{\cn}M$ be the complexification of the tangent bundle $TM$ of $M$. We extend the metric $g$ to a complex-bilinear form on $T^{\cn}M$.  Then the gradient $\nabla z$ of a complex-valued function $z:(M,g)\to\cn$ is a section of $T^{\cn}M$.  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 $z,w:(M,g)\to\cn$ 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 $z=x+i\,y:(M,g)\to\cn$ 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 $z,w:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ 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 $f:U\subset N\to\rn$, defined on an open subset $U$ such that $\phi^{-1}(U)$ is not empty, the composition $f\circ\phi:\phi^{-1}(U)\subset M\to\rn$ 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 $z:(M,g)\to\cn$ is a smooth complex-valued function on a Riemannian manifold and that $f:U\to\cn$ is differentiable and defined on an open subset $U$ of $\cn$ containing the image $z(M)$ of $z$.  Further let $\phi:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ be a complex-valued function on a Riemannian manifold and $F,G:U\to\cn$ be differentiable functions defined on an open subset $U$ of $\cn$ 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 $z:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ is a $(1,1)$-harmonic morphism if and only if, for any harmonic $f:U\to\cn$ defined on an open subset $U$ of $\cn$ 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 $z:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ is a $(2,1)$-harmonic morphism if and only if, for any harmonic $f:U\to\cn$ defined on an open subset $U$ of $\cn$ 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 $\phi:M\to\cn$ 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 $\rn^4$ be the standard $4$-dimensional Euclidean space and $U$ be the open subset given by 	$$U=\{(x_1,x_2,x_3,x_4)\in\rn^4|\ x_1^2+x_2^3+x_3^2>0\}.$$  Then $z:U\to\cn$ 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 $z:M\to\cn$ be a $(2,1)$-harmonic morphism. Further let $f:U\to\cn$ be a holomorphic function defined on an open subset of $\cn$ such that $z(M)\subset U$.  Then the composition $f\circ z:M\to\cn$ 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 $\phi:N\to\cn$ be a $(2,1)$-harmonic morphism.  Then the composition $\phi\circ f:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ 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 $z:(M,g)\to\cn$ is a $(2,2)$-harmonic morphism if and only if, for any $2$-harmonic $f:U\to\cn$ defined on an open subset $U$ of $\cn$ 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, $\phi:M\to\cn$ be a $(2,1)$-harmonic morphism and $\psi:N\to\cn$ be a $(1,1)$-harmonic morphism. Then the sum $\Phi=\phi\oplus\psi:M\times N\to\cn$, 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 $\rn^n$. Example \ref{example-first} shows that such objects can easily be constructed for any dimension $n\ge 4$.

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

\begin{example}\label{example-first}
Let $\rn^n$ be the standard $n$-dimensional Euclidean space of dimension $n\ge 4$ and $U$ be the open subset given by 	$$U=\{x\in\rn^n|\ x_1^2+x_2^3+x_3^2>0\}.$$
Then the complex-valued function $\phi:U\to\cn$ 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 $\rn^4$ with $U=\{(x_1,x_2,x_3,x_4)\in\rn^4|\ x_2^2+x_3^2> 0\}$ and define the function $\phi:U\to\cn$ 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 $\phi:\rn^4\setminus\{0\}\to\cn$ 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 $r\in\rn^+$, the well-known local $(1,1)$-harmonic morphism $\phi_r:U\subset\rn^3\to\cn$, 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 $\rn^4$.
\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 $a,b,c,d\in\cn$ with $a^2+b^2+c^2+d^2=0$, let $\phi:\rn^4\setminus\{0\}\to\cn$ 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 $a,b,c,d\in\cn$, define the complex-valued function $\phi:U\subset\rn^4\to\cn$ 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 $\phi:\rn^4\setminus\{0\}\to\cn$ 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 $\phi:M\to\cn$, $\psi:N\to\cn$ be two horizontally conformal functions.  Let $U$ be an open subset of $\cn^2$ such that $\phi(M)\times \psi(N)\subset U$
and $f:U\to\cn$ be a holomorphic function.  Then the composition
$\Phi:M\times N\to\cn$ 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 $\B_M$ and $\B_N$ be local orthonormal frames for the tangent bundles
$TM$ and $TN$, respectively.  Then
\begin{eqnarray*}
\kappa(\Phi,\Phi)&=&\sum_{X\in\B_M}\bigl(\frac{\partial f}{\partial \phi}\cdot X(\phi)\bigr)^2+
\sum_{Y\in\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 $\rn^m$ and $\rn^n$, respectively. Let $\phi:U\to\cn$ be a $(2,1)$-harmonic morphism and $\psi:V\to\cn$ be a $(1,1)$-harmonic morphism. Let $W$ be an open subset of $\cn^2$ such that $\phi(U)\times \psi(V)\subset W$ and
$f:W\to\cn$ be a holomorphic function.  Then the composition $\Phi:U\times V\to\cn$ 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 $\phi:\rn^4\setminus\{0\}\to\cn$ 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 $\psi:\rn^4\to\cn$ satisfying
$$\psi(x)=\log(x_5+ix_6)\cdot\sin(x_7+ix_8)$$ is a $(1,1)$ harmonic morphism.
Calculations confirm that $\Phi:U\subset\rn^8\to\cn$ 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 $\phi:\rn^6\setminus\{0\}\to\cn$ 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 $\phi^*=\phi\circ i_3:\rn^6\setminus\{0\}\to\cn$ 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 $\rn^6$.  They are proper if and only if the stated $p$ is $3$ and not proper otherwise.

\begin{table}[ht]
\begin{center}
\caption{$\phi:U\subset\rn^6\setminus\{0\}\to\cn$.}
\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{$\phi:U\subset\rn^6\setminus\{0\}\to\cn$.}
\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 $c^p_{jk}:U\to\cn$ 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 $c^p_{jk}:U\to\cn$ 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 $z:(M,g)\to\cn$ is a $(p,q)$-harmonic morphism if and only if, for any $q$-harmonic function $f:U\to\cn$ defined on an open subset $U$ of $\cn$ 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 $p\in\zn^+$.  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{$\phi:U\subset\rn^8\setminus\{0\}\to\cn$.}
		\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{$\phi:U\subset\rn^8\setminus\{0\}\to\cn$.}
		\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 $\phi:\rn^8\to\cn$ 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 $\phi^*=\phi\circ i_4:\rn^8\setminus\{0\}\to\cn$ is a proper $(4,1)$-harmonic morphism.
\end{example}

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

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

\begin{theorem}\label{theorem-inversion}
Let $i_p:\rn^{2p}\setminus\{0\}\to\rn^{2p}\setminus\{0\}$ be the inversion about the unit sphere $S^{2p-1}$ in $\rn^{2p}$ 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 $n\in\zn^+$ let the map  $\phi:\rn^{p}\setminus\{0\}\to\rn^{p}\setminus\{0\}$ 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 $p\in\zn^+$ be a positive integer and $i_p=(F_1,F_2,\dots ,F_{2p}):\rn^{2p}\setminus\{0\}\to\rn^{2p}\setminus\{0\}$ be the inversion about the unit sphere $S^{2p-1}$ in $\rn^{2p}$.  Then $z:\rn^{2p}\setminus\{0\}\to\cn$ with
$$z=a_1F_1+a_2F_2\cdots+a_{2p}F_{2p}$$ is a complex-valued $(p,p)$-harmonic morphism for any element $a\in\cn^{2p}$.
\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 $\rn^{3}$, not even locally.  For this we have the following.

\begin{conjecture}
Let $p\ge 2$ and $\phi:U\to\cn$ be a complex-valued $(p,1)$-harmonic morphism defined locally on the standard Euclidean space $\rn^{2p-1}$.  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.


TITLE: Existence of smooth solutions to the heat equation with variable coefficients
QUESTION [5 upvotes]: Consider $$\partial_{t} u - a  \partial_{x}^{2}u = f, $$ in $[0,1]\times [0,T]$,  where $a, f : \mathbb{R} \times \mathbb{R}^+ \to \mathbb{R}$ are smooth. Additionally, the initial data
$$u(x,0) = u_{0} $$ is smooth and we further endow $u$ with boundary conditions $$u(0,t) = u(1,t).  $$

Question: How can you prove the existence of a smooth solution to this problem?
I have read somewhere that this follows from the theory of linear parabolic equations, but I don't know which theory exactly. I did see a proof in Evans for the homogeneous heat equation in $\mathbb{R}$ but that's about it.

REPLY [4 votes]: One way to proceed would be to explicitly construct the solution to this equation (which is relatively simple since it's linear). To satisfy the periodicity condition on the interval $[0,1]$ one needs to expand the solution in a Fourier series. Following through with this procedure, one finds that the function given by
$$u(x,t)=\int_0^1 dy ~u_0(y)\Theta(x-y,t)+\int_0^1 dy\int_0^{t}dt' f(y,t')\Theta(x-y,t-t')~~,~~ t>0$$
satisfies the equation and the initial conditions. Here $\Theta(x,t)$ is a kernel with the properties
$$1.~~\Theta(x,t)\in C^\infty([0,1]\times(0,T])$$
$$2.~~~~~~~~~~~~~~~~~~\int_0^\infty \Theta(x,t)dx=1$$
Can you show properties 1. and 2. and replicate the proof of Evans?

HINT:
$$\Theta(x,t)=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}e^{-4\pi^2 a n^2 t }e^{i2\pi n x}$$


TITLE: Monotone Convergence Theorem for nonnegative functions (not quite a decreasing sequence)
QUESTION [3 upvotes]: This question suggests that the MCT for functions is to be applied, but I can't see how this could be done.

Assume $g_n: X \to \bar{\mathbb{R}}$ is a sequence of nonnegative measurable functions satisyfing $$\int g_nd\mu < \frac{1}{n^2}$$
  for each $n\geq1$. Using the Monotone Convergence Theorem for nonnegative functions, or otherwise, prove that $$\sum_{N=1}^{\infty}g_n(x)<+\infty$$ $\mu$-almost everywhere.

It seems to me that $g_n$ is similar to a decreasing sequence, although not necessarily for all $x$; and that as $n$ tends to infinity, the integral must tend to $0$, so somehow the 'limit' of $g_n$ must be equal to $0$ $\mu$-almost everywhere. But since $g_{n+1}$ is not necessarily less than $g$ for all $x$, I can't see that the pointwise limit even exists.

REPLY [3 votes]: You need to know (or to prove) that $\sum \frac 1 {n^2}$ is finite, with sum $S$ (you don't need to know what the sum is - that is a much harder problem than just knowing the sum is finite). Now if $f_M = \sum_1^M g_n$, the $f_M$ form a monotonically increasing sequence (why?) and the integral of $f_M$ is bounded by S, so the integral of the limit (that is, the integral of the infinite sum of $g_n$) exists and is finite. So the pointwise sum of $g_n(x)$ may only be infinite on a set of measure zero.


TITLE: Discontinuity of a screw dislocation
QUESTION [1 upvotes]: Consider a screw dislocation,

The displacement field is given by
$$u_{z}=\frac{b\cdot\theta}{2\pi}=\frac{b}{2\pi}\cdot\tan^{-1}(y/x)=\frac{b}{2\pi}\cdot\arctan(y/x).$$
And the strain field is derived by
$$\varepsilon_{yz}=\varepsilon_{zy}=\frac{b}{4\pi}\cdot\frac{x}{x^{2}+y^{2}}=\frac{b}{4\pi}\cdot\frac{\cos\theta}{r}$$
My question is that the displacement seems discontinuous across the dislocation surface ($y=0$), so why it is still differentiable on the surface?

REPLY [2 votes]: I think that it is better to leave the displacement in polar coordinates, that is
$$u_z = \frac{b}{2\pi} \theta\, .$$
If you want to express it in cartesian coordinates you should use the function atan2 instead of atan.
Also, in polar coordinates you have
$$\epsilon_{z\theta} = \frac{b}{4\pi r}\, .$$
We could read this as material points rotating in the same direction. This can be visualized in the following image.

The effect looks different because some of the points are moving upward and the others downward, but all of them are rotating and skewing in the same direction.