text:
\begin{document}
\title[Nondegeneracy of harmonic maps]{Nondegeneracy of harmonic maps from $\mathbb R^2$ to $\mathbb S^2$}
\author[G. Chen]{Guoyuan Chen}
\address{\noindent
School of Data Sciences, Zhejiang University of Finance \& Economics, Hangzhou 310018, Zhejiang, P. R. China}
\email{gychen@zufe.edu.cn}

\author[Y. Liu]{Yong Liu}
\address{\noindent School of Mathematics and Physics, North China Electric Power University, Beijing, China}
\email{liuyong@ncepu.edu.cn}

\author[J. Wei]{Juncheng Wei}
\address{\noindent
Department of Mathematics,
University of British Columbia, Vancouver, B.C., Canada, V6T 1Z2}
\email{jcwei@math.ubc.ca}

\begin{abstract}
We prove that all harmonic maps from $\mathbb R^2$ to $\mathbb S^2$ with finite energy are nondegenerate. That is, for any harmonic map $u$ from $\mathbb R^2$ to $\mathbb S^2$ of degree $m$ (in $\mathbb Z$), all bounded kernel maps of the linearized operator $L_u$ at $u$ are generated by these harmonic maps near $u$ and hence the real dimension of bounded kernel space of $L_u$ is $4|m|+2$.
\end{abstract}
\maketitle
\section{Introduction}

In this paper, we consider the harmonic maps given by
\begin{eqnarray}\label{e:harmonic-map}
\Delta u + |\nabla u|^2u=0, \ \ u:\mathbb R^2\to \mathbb S^2,
\end{eqnarray}
where $\mathbb S^2=\{u=(u_1,u_2,u_3):\,|u|=1\}$ is the unit sphere in $\mathbb R^3$, $\Delta $ is the Laplacian in $\mathbb R^2$ and
$$|\nabla u|^2=\sum_{j=1}^2 \sum_{i=1}^3 \left(\frac{\partial u_i}{\partial x_j}\right)^2.$$
Harmonic maps are critical points of the following associated energy
\begin{eqnarray}\label{e:energy1}
\mathcal E(u)=\int_{\mathbb R^2}|\nabla u|^2dx.
\end{eqnarray}
Harmonic maps also can be defined between general Riemannian manifolds. We refer the interested readers to  \cite{eells1978report}, \cite{eells1988another}, \cite{schoen1997lectures}, \cite{lin2008analysis}, \cite{helein2008harmonic} and the references therein for more results in this direction.

One important case is the harmonic maps between Riemann surfaces.
It is well known that oriented two-dimensional Riemannian manifold has a natural complex structure. Hence any holomorphic or anti-holomorphic map between oriented surfaces is a harmonic map. Conversely, harmonic map is not necessarily holomorphic or anti-holomorphic. There is a topological obstruction. For example, if a harmonic map $u$ from a surface $M$ to $\mathbb S^2$ satisfies certain degree conditions (for example, $|{\rm deg}(u)|$ is larger than the genus of $M$), then $u$ is holomorphic or anti-holomorphic. For more related results, see for example \cite{eells1964harmonic}, \cite{eells1978report}, \cite{eells1988another}, \cite{jost2006harmonic}, \cite{helein2008harmonic}, and the references therein.

A map $u$ from $\mathbb R^2$ to $\mathbb S^2$ is harmonic if and only if $u$ is holomorphic or anti-holomorphic. In particular, if we choose local coordinates of $\mathbb S^2$ as stereographic projections $\mathcal S$ and $\mathcal S'$ (see (\ref{e:stereo1}), (\ref{e:stereo2}) below) and consider $\mathbb R^2$ as the complex plane, then a map $u$ from $\mathbb R^2$ to $\mathbb S^2$ with finite topological degree is harmonic if and only if there is a (irreducible) rational complex valued function $f=q/p$ in $\mathbb C$ such that
$u=\mathcal S(f)$. Here $q$ and $p$ are complex polynomials with algebraic order $l$ and $n$ respectively. Some basic computations imply that ${\rm deg}(u)=\max\{l,n\}$ (see Section \ref{e:classification} below). We should mention that the classifications of harmonic maps from $\mathbb C$ to Lie groups or other symmetric spaces were also investigated, see, for example, \cite{uhlenbeck1989harmonic}, \cite{eells1983harmonic}, \cite{guest1997harmonic}, and the references therein.

It is clear that changing the coefficients (complex numbers) of $f$ continuously yields a family of harmonic maps.
Therefore, it generates kernel maps for the linearized operators $L_u$
$$ L_u [v]:= \Delta v +|\nabla u |^2 v + 2 (\nabla u \cdot \nabla v) u $$
of (\ref{e:harmonic-map}) at some fixed harmonic maps $u$. Let us consider a simple example. Assume $u(z)=\mathcal S\left(\frac{z^2+1}{z}\right)$. The degree of $u$ is $2$. Let $$u(z;a_0,a_1,a_2;b_0,b_1,b_2)=\mathcal S\left(\frac{(1+a_0)z^2+a_1z+1+a_2}{b_0z^2+(1+b_1)z+b_2}\right)$$
with some small complex numbers $a_0,a_1,a_2;b_0,b_1,b_2$. Then $u(z;a_0,a_1,a_2;b_0,b_1,b_2)$ are close to $u(z)$. Then this family of harmonic maps generates $10$ linearly independent bounded kernels of $L_u$.

Our main theorem asserts the nondegeneracy of all harmonic maps. We prove that
\begin{theorem}\label{t:general}
Let $u$ be a harmonic map from $\mathbb R^2$ to $\mathbb S^2$ of degree $m\in \mathbb Z$. Then all the bounded maps in the kernel of $L_u$ are generated by harmonic maps close to $u$. In particular, the real dimension of the bounded kernel space of $L_u$ is $4|m|+2$.
\end{theorem}

\begin{remark}
(1) We should point out that the degree $m$ of harmonic maps can be negative in Theorem \ref{t:general}. It corresponds to anti-holomorphic functions. See Section \ref{e:classification} below. Moreover, we emphasize that the harmonic maps we discussed here are general, especially, we do not restrict the harmonic maps to be \emph{corotational}. For nondegeneracy within corotational classes, see \cite{GKT1} and \cite{GKT2}.

(2) All bounded kernels of $L_u$ can be written down explicitly. See the proof of Proposition \ref{p:cr}. As an example, we will give the explicit formula of kernels for $m$-corotational case (Corollary \ref{t:main}).
\end{remark}

\begin{remark}
(1) For $m=1$, the nondegeneracy of $1$-corotational harmonic maps was obtained by \cite{davila2017singularity}. For case of half-harmonic map from $\mathbb R$ to $\mathbb S^1$ of degree $1$, the nondegeneracy was verified by \cite{sire2017nondegeneracy}. In \cite{chanillo2005asymptotic}, the authors proved the nondegeneracy of standard bubbles of degree $1$ for $\Delta u=2u_x\times u_y$ where $u$ is a map from $\mathbb R^2$ to $\mathbb R^3$.

(3) Since the set of harmonic maps from $\mathbb R^2$ to $\mathbb S^2$ with finite topological degree  corresponds to the space of complex rational functions by stereographic projection, it introduces a natural algebraic structure. Similar phenomena were found in Toda system, see \cite{wei2011nondegeneracy} \cite{lin2012classification}.

(2) Theorem \ref{t:general} asserts that the kernel space of $L_u$ is generated by changing the $2m+2$ complex coefficients of corresponding rational functions. Such kind of result was obtained for half wave maps from $\mathbb R$ to $\mathbb S^2$ in a recent paper \cite{lenzmann2017energy}. To be more precise, traveling solitary waves from $\mathbb R$ to $\mathbb S^2$ with degree $m$ can be generated by finite Blaschke products with degree $m$ which are holomorphic or anti-holomorphic functions. These finite Blaschke products depend on $2m+1$ real parameters. The authors proved that the null space of linearized operators is generated by differentiation these parameters and $x$-, $y$-rotations. An important ingredient in the proof is the classification and nondegeneracy of half-harmonic maps from $\mathbb R$ to $\mathbb S^1$. We should point out that the proof of  nondegeneracy in \cite{lenzmann2017energy} is done through analyzing the spectrum of linearized operator.

Our proof relies on the fact that all harmonic maps from $\mathbb R^2$ to $\mathbb S^2$ are minimizers (locally stable critical points) in its degree class (see (\ref{e:energy-first-order}) and (\ref{e:minus-degree}) below). That is, $u$ is a harmonic map with degree $m\ge 0$ (resp. $m<0$) if and only if $u$ satisfies the first order equation (\ref{e:JH}) (resp. (\ref{e:anti-JH})). Let $L_{1,u}$ be the linearized (first order) operator corresponding to (\ref{e:energy-first-order}) at $u$. Then we prove that for $m\ge 0$ the bounded kernel space of $L_u$ is the same as that of $L_{1,u}$ (see Proposition \ref{l:Lu-L1u} below). Then we verify that all
maps in the kernel of $L_{1,u}$ satisfies Cauchy-Riemann equations after stereographic projection (see Proposition \ref{p:cr} below). For $m<0$, the similar results correspond to anti-holomorphic and anti-version of Cauchy-Riemann equations.
\end{remark}

As a special case, we consider $m$-corotational harmonic maps. That is, $U_m(z)=\mathcal S(z^m)$, $m\in \mathbb N$, in $\mathbb C$.
In polar coordinates $(r,\theta)$ in $\mathbb R^2$,
the $m$-corotational harmonic maps can be written as
\begin{eqnarray}\label{e:standard}
U_m(r,\theta)=\left(
              \begin{array}{c}
                \cos m\theta\sin Q_m(r) \\
                \sin m\theta\sin Q_m(r) \\
                \cos Q_m(r) \\
              \end{array}
            \right)=\left(
                      \begin{array}{c}
                        \dfrac{2r^m\cos m\theta}{1+r^{2m}} \\
                        \dfrac{2r^m\sin m\theta}{1+r^{2m}} \\
                        \dfrac{r^{2m}-1}{r^{2m}+1} \\
                      \end{array}
                    \right),
\end{eqnarray}
where
$Q_m(r)=\pi-2\arctan (r^m).$
Theorem \ref{t:general} implies the explicit bounded kernel of the linearized operator at the standard $m$-corotational harmonic map $U_m$.
For simplicity of notations, we set
\begin{equation*}
E_1=\left(
             \begin{array}{c}
               -\sin m\theta \\
               \cos m\theta  \\
               0\\
             \end{array}
           \right)=\left(
                     \begin{array}{c}
                       ie^{im\theta} \\
                       0 \\
                     \end{array}
                   \right)
           ,\quad E_2=\left(
                                       \begin{array}{c}
                                         \cos m\theta\cos Q_m \\
                                         \sin m\theta\cos Q_m \\
                                         -\sin Q_m \\
                                       \end{array}
                                     \right)=\left(
                                               \begin{array}{c}
                                                 e^{im\theta}\cos Q_m \\
                                                 -\sin Q_m \\
                                               \end{array}
                                             \right)
                                     .
\end{equation*}
Then we have the following corollary.
\begin{cor}\label{t:main}
Assume that $m\ge 1$. The standard $m$-corotational harmonic map $U_m$ is nondegenerate. That is, all bounded maps in the kernel of $L_{U_m}$ are linear combinations of $\{E_{kj}, \tilde E_{\nu j}, \tilde E_{mj}\}$ ($k=0,1,\cdots,m$, $\nu=1,\cdots,m-1$ and $j=1,2$). Here
\begin{eqnarray}\label{e:ek1}
E_{k1}=\frac{r^{m-k}}{1+r^{2m}}(-\cos k\theta E_1 +\sin k\theta E_2),\quad
 E_{k2}=\frac{r^{m-k}}{1+r^{2m}}(\sin k\theta E_1 +\cos k\theta E_2),
\end{eqnarray}
where $k=0,1,\cdots, m$, and
\begin{eqnarray}\label{e:ek3}
\tilde E_{\nu1}=\frac{r^{m+\nu}}{1+r^{2m}}(-\cos \nu\theta E_1 +\sin \nu\theta E_2),
\quad
\tilde  E_{\nu2}=\frac{r^{m+\nu}}{1+r^{2m}}(\sin \nu\theta E_1 +\cos \nu\theta E_2),
\end{eqnarray}
where $\nu=1,\cdots, m-1$, and
\begin{eqnarray}\label{e:ek5}
\tilde E_{m1}=\left(
                \begin{array}{c}
                  0 \\
                  \dfrac{r^{2m}-1}{r^{2m}+1} \\
                  -\dfrac{2r^m\sin m\theta}{r^{2m}+1} \\
                \end{array}
              \right)
,
\quad
\tilde  E_{m2}=\left(
                \begin{array}{c}
                  \dfrac{r^{2m}-1}{r^{2m}+1} \\
                  0\\
                  -\dfrac{2r^m\cos m\theta}{r^{2m}+1} \\
                \end{array}
              \right),
\end{eqnarray}
In particular, $\dim {\rm ker}L_{U_m}=4m+2$.
\end{cor}

\begin{remark}
(1) For $m\le -1$, $U_m(z)=\mathcal S(\bar z^m)$. The similar results can be verified by reversing the rotation direction.

(2) Nondegeneracy for $m$-corotational harmonic map $U_m$ can also be proved by ODE method. See Appendix below.
\end{remark}

This paper is organized as follows. In Section 2, we recall some basic classifications results and compute the degree of general harmonic maps in terms of algebraic order of rational functions. Section 3 is devoted to the proof of Theorem \ref{t:general}. In Section 4, we give an explicit formula of the kernel maps in $m$-corotational case. In Appendix, we illustrate some details of computations, and prove Corollary \ref{t:main} by ODE methods.

\subsection*{Acknowledgement}
G. Chen is partially supported by Zhejiang Provincial Science Foundation of China (No. LY18A010023) and Zhejiang University of Finance and Economics.
J. Wei is partially supported by NSERC of Canada. Part of this work was finished while the first two authors were visiting the University of British Columbia in 2017.

\section{Preliminaries}\label{e:classification}

In this section, we first recall the classification facts for harmonic maps from $\mathbb R^2$ to $\mathbb S^2$ as well as compute their degrees.

It is well known that $\mathbb S^2$ is a complex manifold. For our application, we give a system of local charts by the stereographic projection. Let $z=(x,y)\in \mathbb R^2=\mathbb C$ and $s=(s_1,s_2,s_3)\in \mathbb S^2$. Define
\begin{eqnarray}\label{e:stereo1}
\mathcal S:\mathbb C\to \mathbb S^2\setminus\{N\}  \quad \mbox{by } s_1=\frac{2x}{1+|z|^2},\,s_2=\frac{2y}{1+|z|^2}\mbox{ and }s_{3}=\frac{|z|^2-1}{|z|^2+ 1}.
\end{eqnarray}
and
\begin{eqnarray}\label{e:stereo2}
\mathcal S':\mathbb C\to \mathbb S^2\setminus\{S\}  \quad \mbox{by } s_1=\frac{2x}{1+|z|^2},\,s_2=-\frac{2y}{1+|z|^2}\mbox{ and }s_{3}=\frac{1-|z|^2}{1+|z|^2},
\end{eqnarray}
Here $S=(0,0,-1)$ and $N=(0,0,1)$ denote the south and north pole respectively. Alternatively, in complex variable form,
\begin{eqnarray}\label{e:steregraphic-projection}
\mathcal S=\frac{1}{1+|z|^2}\left(
               \begin{array}{c}
                 2 z \\
                 |z|^2-1 \\
               \end{array}
             \right),\quad\mathcal S'=\frac{1}{1+|z|^2}\left(
               \begin{array}{c}
                 2\bar z \\
                 1-|z|^2 \\
               \end{array}
             \right).
\end{eqnarray}
Therefore, the transition function between these two charts is
$\frac{1}{z}$. It is holomorphic in $\mathbb C\setminus \{0\}$.

Topological degree of a $C^1$ map $u$ from $\mathbb R^2$ to $\mathbb S^2$ can be defined by the de Rham approach
\begin{eqnarray}\label{e:de-rham-degree}
{\rm deg}(u)=\frac{1}{4\pi}\int_{\mathbb R^2}u\cdot(u_y\times u_{x})=\frac{1}{4\pi}\int_{\mathbb R^2} u_{x}\cdot (u\times u_y).
\end{eqnarray}
It is well known that (\ref{e:de-rham-degree}) is equivalent to the Brouwer's degree for all $C^1$ maps. See, for example, \cite[Chapter III]{outerelo2009mapping}.

Since $u\perp u_{x}$ and $u\perp u_{y}$,
it holds that
$|u\times u_{y}|=|u||u_{y}|$. Then
we have
\begin{eqnarray}
\mathcal E(u)=\frac{1}{2}\int_{\mathbb R^2}(|u_{x}|^2+|u\times u_{y}|^2)dxdy.\notag
\end{eqnarray}
Rewrite
\begin{eqnarray}\label{e:energy-first-order}
\mathcal E(u)&=&\frac{1}{2}\int_{\mathbb R^2}|u_{x}-u\times u_{y}|^2+\int_{\mathbb R^2} u_{x}\cdot (u\times u_y)\notag\\
&=&\frac{1}{2}\int_{\mathbb R^2}|u_{x}-u\times u_{y}|^2+4\pi {\rm deg}(u).
\end{eqnarray}
If
\begin{eqnarray}\label{e:JH}
u_{x}=u\times u_{y},
\end{eqnarray}
then $\mathcal E(u)=4\pi {\rm deg}(u)$. Hence ${\rm deg}(u)\ge 0$. Since Brouwer's degree is invariant by homotopy deformations, we find that $u$ is a minimizer in its degree class.

On the other hand, we also have that
\begin{eqnarray}\label{e:minus-degree}
\mathcal E(u)&=&\frac{1}{2}\int_{\mathbb R^2}(|u\times u_{x}|^2+|u_{y}|^2)dxdy\notag\\
&=&\frac{1}{2}\int_{\mathbb R^2}|u\times u_{x}-u_{y}|^2dxdy+\int_{\mathbb R^2}(u\times u_{x})\cdot u_{y}dxdy\notag\\
&=&\frac{1}{2}\int_{\mathbb R^2}|u\times u_{x}-u_{y}|^2dxdy-4\pi {\rm deg}(u).
\end{eqnarray}
Hence if $u$ satisfies
\begin{eqnarray}\label{e:anti-JH}
u_{y}=u\times u_{x},
\end{eqnarray}
then $\mathcal E(u)=-4\pi  {\rm deg}(u)$. It follows that ${\rm deg}(u)\le0$. Similarly by the invariance of deformations of Brouwer's degree, we have that $u$ is a local minimizer in its degree class.

Before discussing the nondegeneracy problem, we first recall the classification result.
\begin{prop}\label{p:harmonic-holomorphic}
A map $u$ from $\mathbb R^2$ to $\mathbb S^2$ is harmonic if and only if $u$ is holomorphic or anti-holomorphic.
\end{prop}

\begin{remark}
For the relation between harmonic maps and holomorphic (or anti-holomorphic) maps from general surfaces to $\mathbb S^2$, we refer the reader to \cite[11.6]{eells1978report}, \cite[Section 2.2]{helein2008harmonic} and the references therein.
\end{remark}

Note that if $u$ satisfies (\ref{e:JH}) or (\ref{e:anti-JH}), then $u$ is a critical point of $\mathcal E$. That means that $u$ is a solution of (\ref{e:harmonic-map}). Conversely, Proposition \ref{p:harmonic-holomorphic}, Lemma \ref{l:first-cr} and Remark \ref{r:anti-cr} below tell us that if $u$ is a solution of (\ref{e:harmonic-map}), then $u$ satisfies (\ref{e:minus-degree}) or (\ref{e:anti-JH}). Hence we find that all harmonic maps from $\mathbb R^2$ to $\mathbb S^2$ are stable.

By Proposition \ref{p:harmonic-holomorphic}, in local coordinates, the harmonic maps are holomorphic or anti-holomorphic functions.
\begin{cor}
A map $u$ from $\mathbb R^2$ to $\mathbb S^2$ is harmonic with ${\rm deg}(u)\ge 0$ (resp. ${\rm deg}(u)< 0$) if and only if $u=\mathcal S(p/q)$ where $p$ and $q$ are complex polynomials of $z$ (resp. $\bar z$). That is, $u$ is a harmonic map if and only if $u$ is a stereographic projection of a rational function of $z$ (resp. $\bar z$). We call (irreducible) rational function $p/q$ the generate function function of harmonic map $u$. Moreover, all harmonic maps from $\mathbb R^2$ to $\mathbb S^2$ are stable.
\end{cor}
\begin{remark}
In what follows, we shall focus on the nonnegative degree harmonic maps. Harmonic maps of negative degree correspond to anti-holomorphic functions. The proof of nondegeneracy is similar.
\end{remark}

We now in the position to compute the degree of harmonic maps. The degree of rational functions was computed in \cite{segal1979topology} (see also \cite{cohen1991topology}). For our utilities, we give a basic computation. Let us investigate some examples first.
\begin{example}
Some direct computations verify that
\begin{itemize}
  \item[(1)] ${\rm deg}(\mathcal S( z))={\rm deg}(\mathcal S(\frac{1}{z}))=1$.
${\rm deg}(\mathcal S(z^2))={\rm deg}(\mathcal S(\frac{1}{z^2}))=2$.
${\rm deg}(\mathcal S\circ z^m)={\rm deg}(\mathcal S(\frac{1}{z^m}))=m$ with $m\ge 0$.
  \item[(2)] ${\rm deg}(\mathcal S(z+\frac{1}{z}))=2$.
  \item[(3)] ${\rm deg}(\mathcal S(\bar z))={\rm deg}(\mathcal S(\frac{1}{\bar z}))=-1$. And ${\rm deg}(\mathcal S(\bar z^m))={\rm deg}(\mathcal S(\frac{1}{\bar z^m}))=-m$ with $m>0$.
\end{itemize}
\end{example}

In general, we have the following result.
\begin{prop}\label{p:degree}
Assume that $u=\mathcal S (g+\dfrac{p}{q})$ where $g$, $p(\ne 0)$ and $q$ are three polynomial complex functions with order $s$, $l$ and $t$ respectively, $l<t$, and $p/q$ is irreducible. Then
\begin{eqnarray}\label{e:degree-f-g}
{\rm deg}(u)=s+t.
\end{eqnarray}
\end{prop}
\begin{proof}
Let $f(z)=g(z)+\dfrac{p(z)}{q(z)}$. Then $f(z)=0$ yields that
\begin{eqnarray}\label{e:polynomial}
g(z)q(z)+p(z)=0.
\end{eqnarray}
Without loss of generality, we may assume that there are $s+t$ different roots, $a_1,\cdots, a_s,a_{s+1},\cdots, a_{s+t}$, for (\ref{e:polynomial}).

Assume that $q(z)=0$ has roots $b_1,\cdots, b_t$. It is clear that $b_1,\cdots, b_t$ can not be roots of (\ref{e:polynomial}). Now we prove that in $\Omega:=\mathbb C\setminus \{b_1,\cdots, b_t\}$, ${\rm deg}(f)=s+t$. Indeed, since $\mathcal S\circ f$ maps $b_1,\cdots, b_t$ to $N$,
\begin{eqnarray}
{\rm deg}(f,\Omega, 0)=\sum_{j=1}^{s+t}{\rm deg}(f,B_{\varepsilon}(a_j), 0),\notag
\end{eqnarray}
where $\varepsilon>0$ is a sufficient small constant. Recall that
\begin{eqnarray}
{\rm deg}(f,B_{\varepsilon}(a_j), 0)=1,\quad \forall \varepsilon>0 \mbox{ sufficiently small},\notag
\end{eqnarray}
and
\begin{eqnarray}
{\rm deg}(\mathcal S,B_{\delta}(a_j), S)=1.\notag
\end{eqnarray}
Hence by formula for degree of composition maps, we obtain (\ref{e:degree-f-g}). This completes the proof.
\end{proof}

\begin{remark}
(1) In general, a rational function $f(z)$ in $\mathbb C$ can be represented by
\begin{eqnarray}
f(z)=\sum_{l=0}^s a_{s-l}z^l+\sum_{k=1}^K\sum_{j=1}^{J_k}\frac{a_{k,j}}{(z-z_k)^j},\quad \mbox{for }z\in \mathbb C\setminus\{z_1,\cdots,z_K\}.\notag
\end{eqnarray}
where $a_s\ne0$ and $a_{k,J_k}\ne 0$ ($k=1,\cdots,K$). In what follows, we call $J_k$ the singular order of $z_k$.
Hence the degree of harmonic map $u=\mathcal S(f)$ is
\begin{eqnarray}
{\rm deg}(u)=s+\sum_{k=1}^K J_k.\notag
\end{eqnarray}

(2) If we write (irreducible) rational functions as $f=p/q$ where $p$ and $q$ are two polynomial complex functions with order $l$ and $n$, then
\begin{eqnarray}
{\rm deg} ({\mathcal S } (f))=\max \{l,n\}. \notag
\end{eqnarray}

\end{remark}

We now consider the relation between harmonic maps $u(z)=\mathcal S(f)(z)$ and $w(z)=\mathcal S(1/f)(z)$ where $f(z)=f_1(z)+if_2(z)$ is a rational function. That is,
\[
u_{1}=\frac{2f_1}{|f|^2+1},\quad
u_{2}=\frac{2f_2}{|f|^2+1},\quad
u_{3}=\frac{|f|^2-1}{|f|^2+1}.
\]
Let $\tilde u=Q_{1,\alpha}(u)$ be a rotation of $u$ with respect to $u_1$-axis. Hence,
\[
\tilde u_{1}=\frac{2f_1}{|f|^2+1}
\]
\[
\tilde u_{2}=\frac{2}{|f|^2+1}(f_2\cos \alpha-(|f|^2-1)\sin
\alpha)
\]
\[
\tilde u_{3}=\frac{2}{|f|^2+1}(f_2\sin \alpha+(|f|^2-1)\cos
\alpha)\]

Some direct computations yield that
\begin{eqnarray}\label{e:u1-rotation}
\mathcal S^{-1}(Q_{1,\alpha}(u))=\frac{u_1+i\tilde u_2}{1-\tilde u_3}=-i\cot\frac{\alpha}{2}+\frac{1}{\sin^2\frac{\alpha}{2}(f-i\cot\frac{\alpha}{2})}.
\end{eqnarray}
In particular, let $\alpha=\pi$, then (\ref{e:u1-rotation}) implies
\begin{eqnarray}
\frac{u_1+i\tilde u_2}{1-\tilde u_3}=-i\cot\frac{\alpha}{2}+\frac{1}{\sin^2\frac{\alpha}{2}(f-i\cot\frac{\alpha}{2})}=\frac{1}{f}.\notag
\end{eqnarray}
That means that the difference between $u(z)=\mathcal S(f)$ and $w(z)=\mathcal S(\frac{1}{f})$ is a rotation.

Moreover, note that
the tangent map of $Q_{1,\alpha}$ is given by
\begin{eqnarray}
v:=\frac{d}{d\alpha}Q_{1,\alpha}(u)|_{\alpha=0}=\left(
                                            \begin{array}{c}
                                              0 \\
                                              -u_3 \\
                                              u_2 \\
                                            \end{array}
                                          \right).\notag
\end{eqnarray}

\section{Proof of Theorem \ref{t:general}}\label{s:proof}
In this section, we shall prove the nondegeneracy of harmonic maps.

Let us compute the associated linearized operators at harmonic maps first.
The associated linearized operators of (\ref{e:harmonic-map}) and (\ref{e:JH}) at $u$ are given by
\begin{eqnarray}
L_u[v]=\Delta v+|\nabla u|^2v+2(\nabla u\cdot\nabla v)u,\notag
\end{eqnarray}
and,
\begin{eqnarray}
L_{1,u}[v]= v_x-v\times u_y-u\times v_y,\notag
\end{eqnarray}
respectively, where $v$ is a smooth map from $\mathbb C$ to $T\mathbb S^2$. Let $C^2_b(\mathbb C,T\mathbb S^2)$ be the space of maps $v$ from $\mathbb C$ to $T\mathbb S^2$ satisfying $\sup_{z\in \mathbb C}(|v(z)|+|\partial v(z)|+|\partial^2 v(z)|)$. Define
\begin{eqnarray}
{\rm ker} L_u:=\{v\in C^2_b(\mathbb C,T\mathbb S^2)\,|\, L_u[v]=0\}\,\, \mbox{and}\,\,{\rm ker} L_{1,u}:=\{v\in C^2_b(\mathbb C,T\mathbb S^2)\,|\, L_{1,u}[v]=0\}.\notag
\end{eqnarray}

\begin{prop}\label{l:Lu-L1u}
Assume that $u$ is a harmonic map with degree $m\ge 0$. It holds that
\begin{eqnarray}\label{e:kernel-1-2}
{\rm ker}L_u={\rm ker}L_{1,u}.
\end{eqnarray}
\end{prop}
\begin{proof}
When $m=0$, (\ref{e:kernel-1-2}) holds. Hence we only consider the case $m>0$.

Since ${\rm deg}(u)>0$ and $\mathcal S^{-1}(u)$ is holomorphic, we find that $|u_x|=|u_y|\ne 0$ almost everywhere in $\mathbb C$.

1. ``${\rm ker}L_{1,u}\Rightarrow {\rm ker}L_{u}$''. From
$v_x-v\times u_y-u\times v_y=0$,
we have that
\begin{eqnarray}
v_y+v\times u_x+u\times v_x=0,\notag
\end{eqnarray}
\begin{eqnarray}
v_{xx}-v_x\times u_y-v\times u_{xy}-u_x\times v_y-u\times v_{xy}=0,\notag
\end{eqnarray}
and
\begin{eqnarray}
v_{yy}+v_y\times u_x+v\times u_{xy}+u_y\times v_x+u\times v_{xy}=0.\notag
\end{eqnarray}
Hence we obtain that
\begin{eqnarray}\label{e:vxxyy}
v_{xx}+v_{yy}-v_x\times u_y+v_y\times u_x-u_x\times v_y+u_y\times v_x=0.
\end{eqnarray}
Since
\begin{eqnarray}
&-(v_x\times u_y)\cdot u=u_x\cdot v_x,\quad (v_y\times u_x)\cdot u=u_y\cdot v_y,\notag\\
&-(u_x\times v_y)\cdot u=u_y\cdot v_y,\quad (u_y\times v_x)\cdot u=u_x\cdot v_x,\notag
\end{eqnarray}
it holds that
\begin{eqnarray}
(v_{xx}+v_{yy})\cdot u+2\nabla u\cdot \nabla v=0.\notag
\end{eqnarray}
Similarly, \eqref{e:vxxyy}$\cdot u_x$ and \eqref{e:vxxyy}$\cdot u_y$  yield that
\begin{eqnarray}
(v_{xx}+v_{yy})\cdot u_x+|\nabla u|^2v\cdot u_x=0,\quad (v_{xx}+v_{yy})\cdot u_y+|\nabla u|^2v\cdot u_y=0.\notag
\end{eqnarray}
Therefore, we obtain that
\begin{eqnarray}
\Delta v+|\nabla u|^2v+2(\nabla u\cdot\nabla v)u=0.\notag
\end{eqnarray}

2. ``${\rm ker}L_{u}\Rightarrow {\rm ker}L_{1,u}$''. We shall prove this by Liouville theorem.
Note that
\begin{eqnarray}\label{e:reduce}
&&(v_x-v\times u_y-u\times v_y)\cdot u=0,\notag\\
&&(v_x-v\times u_y-u\times v_y)\cdot u_x=u_x\cdot v_x-u_y\cdot v_y,\\
&&(v_x-v\times u_y-u\times v_y)\cdot u_y=u_x\cdot v_y+u_y\cdot v_x.\notag
\end{eqnarray}

\emph{Claim: It holds that
\begin{eqnarray}\label{e:uvy}
u_x\cdot v_x-u_y\cdot v_y=0,
\end{eqnarray}
\begin{eqnarray}\label{e:uv-x}
u_x\cdot v_y+u_y\cdot v_x=0.
\end{eqnarray}}

Indeed, we compute $\Delta (u_x\cdot v_x-u_y\cdot v_y)$. By some direct computations, we get that
\begin{eqnarray}
\partial_x(u_x\cdot v_x-u_y\cdot v_y)=u_{xx}\cdot v_x+u_x\cdot v_{xx}-u_{xy}\cdot v_y-u_y\cdot v_{xy},\notag
\end{eqnarray}
\begin{eqnarray}
\partial_x^2(u_x\cdot v_x-u_y\cdot v_y)&=&u_{xxx}\cdot v_x+2u_{xx}\cdot v_{xx}+u_x\cdot v_{xxx}\notag\\
&&-u_{xxy}\cdot v_y-2u_{xy}\cdot v_{xy}-u_{y}\cdot v_{xxy}.\notag
\end{eqnarray}
Similarly, it holds that
\begin{eqnarray}
\partial_y(u_x\cdot v_x-u_y\cdot v_y)=u_{xy}\cdot v_x+u_x\cdot v_{xy}-u_{yy}\cdot v_y-u_y\cdot v_{yy},\notag
\end{eqnarray}
\begin{eqnarray}
\partial_y^2(u_x\cdot v_x-u_y\cdot v_y)&=&u_{xyy}\cdot v_x+2u_{xy}\cdot v_{xy}+u_x\cdot v_{xyy}\notag\\
&&-u_{yyy}\cdot v_y-2u_{yy}\cdot v_{yy}-u_{y}\cdot v_{yyy}.\notag
\end{eqnarray}
Hence we have that
\begin{eqnarray}
\Delta (u_x\cdot v_x-u_y\cdot v_y)&=&(u_{xxx}+u_{xyy})\cdot v_x+2(u_{xx}\cdot v_{xx}-u_{yy}\cdot v_{yy})\notag\\
&&+u_x\cdot (v_{xxx}+v_{xyy})-(u_{xxy}+u_{yyy})\cdot v_y-u_{y}\cdot (v_{xxy}+v_{yyy})\notag\\
&=&\partial_x(\Delta u)\cdot v_x+2((\Delta u)\cdot v_{xx}-u_{yy}\cdot (\Delta v))\notag\\
&&+u_x\cdot \partial_x(\Delta v)-\partial_y((\Delta u)\cdot v_y)-u_{y}\cdot \partial_y(\Delta v),\notag\\
&=&\partial_x((\Delta u)\cdot v_x)-\partial_y (u_{y}\cdot (\Delta v))+(\Delta u)\cdot v_{xx}+u_x\cdot \partial_x(\Delta v)\notag\\
&&-u_{yy}\cdot (\Delta v)-\partial_y(\Delta u)\cdot v_y.\notag
\end{eqnarray}
Further, compute
\begin{eqnarray}
\partial_x((\Delta u)\cdot v_x)=-\partial_x(|\nabla u|^2u\cdot v_x),\notag
\end{eqnarray}
\begin{eqnarray}
-\partial_y((\Delta u)\cdot v_y)=\partial_y(|\nabla u|^2 u\cdot v_y),\notag
\end{eqnarray}
\begin{eqnarray}
&&(\Delta u)\cdot v_{xx}+u_x\cdot \partial_x(\Delta v)\notag\\
&=&-|\nabla u|^2 u\cdot v_{xx}+\partial_x[u_x\cdot (\Delta v)]-u_{xx}\cdot (\Delta v)\notag\\
&=&-|\nabla u|^2 u\cdot v_{xx}-\partial_x[u_x\cdot (|\nabla u|^2v+2(\nabla u\cdot \nabla v)u)]
+u_{xx}\cdot [|\nabla u|^2v+2(\nabla u\cdot \nabla v)u]\notag\\
&=&-|\nabla u|^2 u\cdot v_{xx}-\partial_x[|\nabla u|^2v \cdot u_x]+|\nabla u|^2v\cdot u_{xx}+2(\nabla u\cdot \nabla v)u\cdot u_{xx},\notag
\end{eqnarray}
and
\begin{eqnarray}
&&-u_{yy}\cdot (\Delta v)-\partial_y(\Delta u)\cdot v_y\notag\\
&=&u_{yy}\cdot (|\nabla u|^2v+2(\nabla u\cdot \nabla v)u)-\partial_y[(\Delta u)\cdot v_y]+(\Delta u)\cdot v_{yy}\notag\\
&=&|\nabla u|^2v\cdot u_{yy}+2(\nabla u\cdot \nabla v)u\cdot u_{yy}+\partial_y[|\nabla u|^2u\cdot v_y]-|\nabla u|^2u\cdot v_{yy}.\notag
\end{eqnarray}
Moreover, note that
\begin{eqnarray}
\partial_x(|\nabla u|^2u\cdot v_x)+\partial_x[|\nabla u|^2v \cdot u_x]=0,\notag
\end{eqnarray}
\begin{eqnarray}
\partial_y(|\nabla u|^2 u\cdot v_y)+\partial_y[|\nabla u|^2u\cdot v_y]=0,\notag
\end{eqnarray}
\begin{eqnarray}
-|\nabla u|^2 u\cdot (v_{xx}+v_{yy})=|\nabla u|^2 u\cdot [|\nabla u|^2v+2(\nabla u\cdot \nabla v)u]=2|\nabla u|^2 (\nabla u\cdot \nabla v),\notag
\end{eqnarray}
\begin{eqnarray}
|\nabla u|^2v\cdot [u_{xx}+u_{yy}]=-|\nabla u|^2v\cdot [|\nabla u|^2u]=0,\notag
\end{eqnarray}
\begin{eqnarray}
2(\nabla u\cdot \nabla v)u\cdot [u_{xx}+u_{yy}]=-2|\nabla u|^2(\nabla u\cdot \nabla v).\notag
\end{eqnarray}
Therefore, we have that
\begin{eqnarray}
\Delta (u_x\cdot v_x-u_y\cdot v_y)=0.\notag
\end{eqnarray}
Hence $u_x\cdot v_x-u_y\cdot v_y$ is a harmonic function. Since $v\in C^{1}(\mathbb C, T\mathbb S^2)$ and $\mathcal E(u)<\infty$, the claim \eqref{e:uvy} holds.

Similar computations yield
\begin{eqnarray}
\Delta (u_x\cdot v_y+u_y\cdot v_x)=0.\notag
\end{eqnarray}
Hence we have (\ref{e:uv-x}).

Summarizing \eqref{e:reduce} and the claim, we obtain that
\begin{eqnarray}
v_x-v\times u_y-u\times v_y=0.\notag
\end{eqnarray}
This completes the proof.
\end{proof}

We now compute the tangent map $D\mathcal S^{-1}$. Let $u=(u_1,u_2,u_3)\in\mathbb S^2$. Then
\begin{eqnarray}
\mathcal S^{-1}(u)=\left(\frac{u_1}{1-u_3},\frac{u_2}{1-u_3}\right)=\frac{u_1+iu_2}{1-u_3}.\notag
\end{eqnarray}
Let $u$ be a harmonic map with degree $m\ge 0$. Let $u(t)$ ($t\in [0,\epsilon)$) be a family of harmonic maps with $u'(0)=v$. Hence $v$ is a map from $\mathbb C$ to $T\mathbb S^2$ with $u\cdot v=0$.
Direct calculations yield that
\begin{eqnarray}
D\mathcal S^{-1}_{u}(v)=\left.\frac{d}{dt}\mathcal S^{-1}(u(t))\right|_{t=0}
=\left(\frac{u_1v_3-u_3v_1+v_1}{(1-u_3)^2}, \frac{u_2v_3-u_3v_2+v_2}{(1-u_3)^2}\right).\notag
\end{eqnarray}

On the other hand, assume that $u=\mathcal S(f)$ where $f(z)=f_1(z)+if_2(z)$ is a complex (irreducible) function, $g(z)=g_1(z)+ig_2(z)$. Then the tangent map of $\mathcal S$ is given by
\begin{eqnarray}\label{e:ds-holomorphic}
D\mathcal S_f(g)=\left(\begin{array}{c}
                2\dfrac{-f_1^{2}+f_2^{2}+1}{\left( f_1^{2}+f_2^{2}+1\right) ^{2}}g_1-\dfrac{4f_1f_2}{\left( f_1^{2}+f_2^{2}+1\right) ^{2}}g_2 \\
                 2\dfrac{f_1^{2}-f_2^{2}+1}{\left( f_1^{2}+f_2^{2}+1\right) ^{2}}g_2-\dfrac{4f_1f_2}{\left( f_1^{2}+f_2^{2}+1\right) ^{2}}g_1 \\
                \dfrac{4f_1}{\left( f_1^{2}+f_2^{2}+1\right) ^{2}}g_1+\dfrac{4f_2}{\left( f_1^{2}+f_2^{2}+1\right) ^{2}}g_2
              \end{array}\right).
\end{eqnarray}
We should point out that if $g$ is holomorphic with large orders of singularities, then $D\mathcal S_f(g)$ becomes unbounded. Whereas, if the order of $g$ is small but still larger than $f$, $D\mathcal S_f(g)$ may be bounded. Let us consider the following examples.

\begin{example}
1. Let $u(\cdot,a,b)=\mathcal S(\frac{1}{z-(a+ib)})$. Then
\begin{eqnarray}
u(z,a,b)=\left(
         \begin{array}{c}
           \dfrac{2(x-a)}{a^{2}-2ax+b^{2}-2by+x^{2}+y^{2}+1} \\
           -\dfrac{2(y-b)}{a^{2}-2ax+b^{2}-2by+x^{2}+y^{2}+1} \\
           -\dfrac{a^{2}-2ax+b^{2}-2by+x^{2}+y^{2}-1}{
a^{2}-2ax+b^{2}-2by+x^{2}+y^{2}+1}\\
         \end{array}
       \right).\notag
\end{eqnarray}
Letting $a=b=0$, we find that
\begin{eqnarray}
v(z)=\left.\frac{d}{da}u(z;a,b)\right|_{a=0,b=0}=\left(
                     \begin{array}{c}
                       -2\dfrac{-x^{2}+y^{2}+1}{\left(
x^{2}+y^{2}+1\right) ^{2}} \\
                       \dfrac{-4xy}{\left(
x^{2}+y^{2}+1\right) ^{2}} \\
                       \dfrac{4x}{\left(x^{2}+y^{2}+1\right)
^{2}} \\
                     \end{array}
                   \right),\notag
\end{eqnarray}
The generate function of $\frac{d}{da}u(z;0,0)$ is
\begin{eqnarray}
D\mathcal S^{-1}_u(v(z))=\frac{x^{2}-y^{2}}{\left(
x^{2}+y^{2}\right) ^{2}}+i( \dfrac{-2x y}{\left(x^{2}+y^{2}\right) ^{2}})=\dfrac{1}{z^2}.\notag
\end{eqnarray}

2. Let $u(\cdot,a,b)=\mathcal S(\frac{1}{(z-(a+ib))^2})$. Direct calculations yield that
\begin{eqnarray}
v(z)=\left.\frac{d}{da}u(z,a)\right|_{a=0,b=0}=\left(
                            \begin{array}{c}
                              \dfrac{-4x(-x^{4}+2x^{2}y^{2}+3y^{4}+1)}{(x^{4}+2x^{2}y^{2}+y^{4}+1)^2} \\
                              \dfrac{-4y(3x^{4}+2x^{2}y^{2}-y^{4}-1)}{(x^{4}+2x^{2}y^{2}+y^{4}+1)^2} \\
                              \dfrac{8x(x^{2}+y^{2})}{(x^{4}+2x^{2}y^{2}+y^{4}+1)^2}\\
                            \end{array}
                          \right).\notag
\end{eqnarray}
Hence
\begin{eqnarray}
D\mathcal S^{-1}_u(v(z))=\dfrac{2x^{3}-6xy^{2}}{\left( x^{2}+y^{2}\right) ^{3}}+i\dfrac{2y^{3}-6x^{2}y}{\left( x^{2}+y^{2}\right) ^{3}}=\dfrac{2}{z^3}.\notag
\end{eqnarray}

3. Let $u=\mathcal S(f)$ where $f$ is a complex rational function. Compute
\begin{eqnarray}
D\mathcal S^{-1}_u(v)&=&\frac{u_1u_2+i(1-u_3-u_1^2)}{(1-u_3)^2}\notag\\
&=&f_1f_2+\frac{i}{2}(f_2^2-f_1^2+1)=-\frac{i}{2}f^2+\frac{i}{2}.\notag
\end{eqnarray}

\end{example}

We turn to compute the kernel of the linearized operators $L_{1,u}$.
\begin{prop}\label{p:cr}
Assume that $u=\mathcal S(f)$ is a harmonic map with degree $m\ge 0$, where a irreducible rational function is given by
$f(z)=\frac{q(z)}{p(z)}.$
Here $p,\,q$ are two polynomials with order $l,\,n$, respectively.
Then all bounded kernel maps of ${\rm ker}L_{1,u}$ are generated by changing of coefficients of $p$ and $q$.
In particular,
the real dimension of ${\rm ker}L_{1,u}$ is
$4m+2$.
\end{prop}
\begin{proof}
1. Assume ${\rm deg}(u)=m$. Then $m=l$ or $m=n$. We first consider $l=m$. Assume that
\begin{eqnarray}
p(z)=a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m,\quad q(z)=b_0z^n+b_1z^{n-1}+\cdots+b_{n-1}z+b_n,\notag
\end{eqnarray}
where $a_0,\cdots,a_m$ and $b_0,\cdots,b_n$ are complex numbers with $a_0\ne 0$ and $b_0\ne 0$. Let
\begin{eqnarray}
a_j=a_j^1+ia_j^2,\quad j=0,\cdots,m,\notag
\end{eqnarray}
\begin{eqnarray}
b_k=b_k^1+ib_k^2,\quad k=0,\cdots,n,\notag
\end{eqnarray}
where $a_j^1, a_j^2$ ($j=0,\cdots,m$) and $b_k^1,b_k^2$ ($k=0,\cdots,n$) are real numbers.
Set
\begin{eqnarray}
&&F(a_0,\cdots,a_m;b_0,\cdots,b_n;c_1,\cdots,c_{m-n})\notag\\
&=&\frac{c_1z^m+\cdots+c_{m-n}z^{n+1}+b_0z^n+b_1z^{n-1}+\cdots+b_{n-1}z+b_n}{a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m},\notag
\end{eqnarray}
where $c_1=c_1^1+ic_1^2,\cdots, c_{m-n}=c_{m-n}^1+ic_{m-n}^2$ are complex numbers.
Hence
\begin{eqnarray}\label{e:harmonic-family}
w(a_0,\cdots,a_m;b_0,\cdots,b_n;c_1,\cdots,c_{m-n}):=\mathcal S(F(a_0,\cdots,a_m;b_0,\cdots,b_n;c_1,\cdots,c_{m-n}))
\end{eqnarray}
is a family of harmonic maps with
$$\mathcal S(F(a_0,\cdots,a_m;b_0,\cdots,b_n;0,\cdots,0))=u.$$
Taking derivative on $w$ with respect to the real and imaginary part of the complex coefficients, from (\ref{e:ds-holomorphic}) we fine the following bounded kernels
\begin{eqnarray}
v^{1,c_{\nu}}(z)=D\mathcal S_u\left[\frac{z^{m-\nu+1}}{a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m}\right], \notag
\end{eqnarray}
\begin{eqnarray}
v^{2,c_{\nu}}(z)=D\mathcal S_u\left[i\frac{z^{m-\nu+1}}{a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m}\right],\notag
\end{eqnarray}
with $\nu=1,\cdots, m-n$, and
\begin{eqnarray}
v^{1,b_{k}}(z)=D\mathcal S_u\left[\frac{z^{n-k}}{a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m}\right],\notag
\end{eqnarray}
\begin{eqnarray}
v^{2,b_{k}}(z)=D\mathcal S_u\left[i\frac{z^{n-k}}{a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m}\right],\notag
\end{eqnarray}
with $k=0,\cdots,n$, and
\begin{eqnarray}
v^{1,a_{j}}(z)=D\mathcal S_u\left[-\frac{z^{m-j}}{(a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m)^2}\right],\notag
\end{eqnarray}
\begin{eqnarray}
v^{2,a_{j}}(z)=D\mathcal S_u\left[-i\frac{z^{m-j}}{(a_0z^m+a_1z^{m-1}+\cdots+a_{m-1}z+a_m)^2}\right],\notag
\end{eqnarray}
with $j=0,\cdots,m$. Note that $v^{1,b_{k}}$, $v^{2,b_{k}}$ and can be linearly represented by $v^{1,a_{0}},\cdots,v^{1,a_{m}},v^{2,a_{0}},\cdots,v^{2,a_{m}}$. Therefore, we obtain that
\begin{eqnarray}
{\rm span}\left\{\begin{array}{c}
v^{1,c_{1}},\cdots,v^{1,c_{m-n}};
v^{2,c_{1}},\cdots,v^{2,c_{m-n}} \\
v^{1,b_{0}},\cdots, v^{1,b_{n-1}};
v^{2,b_{0}},\cdots, v^{2,b_{n-1}} \\
v^{1,a_{0}},\cdots, v^{1,a_{m}};
v^{2,a_{0}},\cdots, v^{2,a_{m}}
\end{array}
 \right\}\subset{\rm ker}L_{1,u}.\notag
\end{eqnarray}

Similarly, we can compute the bounded kernels generated by the harmonic maps near $u$ of form (\ref{e:harmonic-family}) for $n=m$.

2. Next, we prove that for all $v\in {\rm ker}L_{1,u}$, $D\mathcal S^{-1}_u(v)$ satisfies Cauchy-Riemann equation in $\mathbb C\setminus\{z_1,\cdots,z_K\}$ where $z_1,\cdots,z_K$ are poles of $f$. That is, we want to verify that
\begin{eqnarray}
\left\{\begin{array}{c}
         \partial_x\left(\dfrac{u_1v_3-u_3v_1+v_1}{(1-u_3)^2}\right)=\partial_y\left(\dfrac{u_2v_3-u_3v_2+v_2}{(1-u_3)^2}\right) \\
         \partial_y\left(\dfrac{u_1v_3-u_3v_1+v_1}{(1-u_3)^2}\right)=-\partial_x\left(\dfrac{u_2v_3-u_3v_2+v_2}{(1-u_3)^2}\right)
       \end{array}
\right..\notag
\end{eqnarray}
That is,
\begin{eqnarray}
\left\{\begin{array}{l}
         \partial_x\left(\dfrac{(v\times u)_2+v_1}{(1-u_3)^2}\right)=\partial_y\left(\dfrac{(u\times v)_1+v_2}{(1-u_3)^2}\right) \\
         \partial_y\left(\dfrac{(v\times u)_2+v_1}{(1-u_3)^2}\right)=-\partial_x\left(\dfrac{(u\times v)_1+v_2}{(1-u_3)^2}\right)
       \end{array}
\right..\notag
\end{eqnarray}
So it is sufficient to check that
\begin{eqnarray}\label{e:cr2}
\left\{\begin{array}{l}
         \partial_x[(v\times u)_2+v_1](1-u_3)+2[(v\times u)_2+v_1]\partial_x u_3\\
         \quad\quad=\partial_y[(u\times v)_1+v_2](1-u_3)+2[(u\times v)_1+v_2]\partial_y u_3\\
        \partial_y[(v\times u)_2+v_1](1-u_3)+2[(v\times u)_2+v_1]\partial_y u_3\\
        \quad\quad=-\partial_x[(u\times v)_1+v_2](1-u_3)-2[(u\times v)_1+v_2]\partial_x u_3\\
       \end{array}
\right.
\end{eqnarray}

In fact, note that
\begin{eqnarray}\label{e:uxuy}
u_x=u\times u_y\quad \mbox{ and }\quad u_y=-u\times u_x.
\end{eqnarray}
Moreover, since $v\in {\rm ker}L_{1,u}$, we have that
\begin{eqnarray}
v_x-v\times u_y-u\times v_y=0,\quad
v_y+v\times u_x+u\times v_x=0.\notag
\end{eqnarray}
Using Lagrange identity and Jacobi identity for cross product, we obtain that
\begin{eqnarray}
\partial_x[(v\times u)_2+v_1]&=&(v_x\times u)_2+(v\times u_x)_2+(v_1)_x\notag\\
&=&2(v\times u_x)_2+(v_2)_y+(v_1)_x,\notag
\end{eqnarray}
\begin{eqnarray}
\partial_y[(u\times v)_1+v_2]&=&(u_y\times v)_1+(u\times v_y)_1+(v_2)_y\notag\\
&=&2(u_y\times v)_1+(v_1)_x+(v_2)_y.\notag
\end{eqnarray}
Hence
\begin{eqnarray}
&&\partial_x[(v\times u)_2+v_1](1-u_3)+2[(v\times u)_2+v_1]\partial_x u_3\notag\\
        && \quad\quad-\partial_y[(u\times v)_1+v_2](1-u_3)-2[(u\times v)_1+v_2]\partial_y u_3\notag\\
        &=&2[(v\times u_x)_2-(u_y\times v)_1](1-u_3)+2[(v\times u)_2+v_1]\partial_x u_3-2[(u\times v)_1+v_2]\partial_y u_3\notag.
\end{eqnarray}
Further,
\begin{eqnarray}
&&[(v\times u_x)_2-(u_y\times v)_1](1-u_3)\notag\\
&=&[(1-u_3)(-(u_3)_x)]v_1+[(1-u_3)(-(u_3)_y)]v_2+[((u_1)_x-(u_2)_y)(1-u_3)]v_3,\notag
\end{eqnarray}
and
\begin{eqnarray}
&&[(v\times u)_2+v_1]\partial_x u_3-[(u\times v)_1+v_2]\partial_y u_3\notag\\
&=&(1-u_3)(u_3)_x v_1+(1-u_3)(u_3)_y v_2 +(u_1 (u_3)_x-u_2 (u_3)_y)v_3.\notag
\end{eqnarray}
By (\ref{e:uxuy}), we find that
\begin{eqnarray}
&&((u_1)_x-(u_2)_y)(1-u_3)+u_1 (u_3)_x-u_2 (u_3)_y\notag\\
&=&(u_1)_x-(u_2)_y-(u_1)_xu_3+(u_2)_yu_3+u_1 (u_3)_x-u_2 (u_3)_y\notag\\
&=&(u_1)_x-(u_2)_y-(u\times u_y)_1-(u\times u_x)_2=0.\notag
\end{eqnarray}
Similar computations yield the second equation of (\ref{e:cr2}).

3. By using Equation (\ref{e:ds-holomorphic}) and Step 1 and 2, we have
all bounded kernels must be generated by harmonic maps near $u$.
This completes the proof.
\end{proof}

\begin{remark}
For the harmonic maps with negative degree, replacing holomorphic maps by anti-holomorphic ones, we can show the nondegeneracy similarly.
\end{remark}

\begin{proof}[Proof of Theorem \ref{t:general}]
Combining Proposition \ref{l:Lu-L1u} and Proposition \ref{p:cr}, we obtain the result in Theorem \ref{t:general}.
\end{proof}

As a consequence of Theorem \ref{t:general}, we can give an explicit formula for the bounded kernel of linearized operator.
\begin{proof}[Proof of Corollary \ref{t:main}]
Let $g_{k}(z)=z^{m-k}$, $k=0,1,\cdots,m$.
From (\ref{e:ds-holomorphic}),
\begin{eqnarray}
D\mathcal S_{U_m}(ig_k)=-2E_{k1},\quad D\mathcal S_{U_m}(g_k)=-2E_{k2}.\notag
\end{eqnarray}
They are generated by the families of form $u_t(z)=z^m+tz^{m-k}$, $t\in \mathbb C$.
Let $\tilde g_{\nu}(z)=z^{m+\nu}$ ($\nu=1,\cdots, m$). Then
\begin{eqnarray}
D\mathcal S_{U_m}(i\tilde g_\nu)=-2\tilde E_{\nu1},\quad D\mathcal S_{U_m}(\tilde g_\nu)=-2\tilde E_{\nu2}.\notag
\end{eqnarray}
These kernels are generated by $u_t(z)=\frac{z^m}{tz^{\nu}+1}$, $t\in \mathbb C$.
\end{proof}

\appendix

\section{}
In this appendix, we give some details of computations and an alternative proof of Corollary \ref{t:main} by ODE method.
\begin{lemma}\label{l:first-cr}
Map $u$ satisfies (\ref{e:JH}) if and only if
\begin{eqnarray}\label{e:cr}
\mathcal S^{-1}(u)=\dfrac{u_1+iu_2}{1-u_3}
\end{eqnarray}
satisfies Cauchy-Riemann equations in $\mathbb C\setminus\mathcal S^{-1}(N)$.
\end{lemma}
\begin{proof}
1. The equation (\ref{e:JH}) is equivalent to
\begin{eqnarray}
\left\{\begin{array}{c}
         \partial_{x}u_1 = u_2\partial_{y}u_3-u_3\partial_{y}u_2\\
         \partial_{x}u_2 = u_3\partial_{y}u_1-u_1\partial_{y}u_3 \\
         \partial_{x}u_3 = u_1\partial_{y}u_2-u_2\partial_{y}u_2
       \end{array}
\right.\quad \mbox{and}\quad u_{y}=-u\times u_{x}.\notag
\end{eqnarray}
Hence
\begin{eqnarray}
\left\{\begin{array}{c}
         \partial_{y}u_1 = u_3\partial_{x}u_2-u_2\partial_{x}u_3\\
         \partial_{y}u_2 = u_1\partial_{x}u_3-u_3\partial_{x}u_1 \\
         \partial_{y}u_3 = u_2\partial_{x}u_2-u_1\partial_{x}u_2
       \end{array}
\right..\notag
\end{eqnarray}
Therefore we find that
\begin{eqnarray}\label{e:Cauchy-Riemann}
\left\{\begin{array}{c}
                \dfrac{\partial_x u_1 + u_1\partial_x u_3-u_3\partial_x u_1}{(1-u_3)^2}=\dfrac{\partial_y u_2 + u_2\partial_y u_3-u_3\partial_{ x_2} u_2}{(1-u_3)^2} \\
                \dfrac{\partial_y u_1 + u_1\partial_y u_3-u_3\partial_y u_1}{(1-u_3)^2}=-\dfrac{\partial_x u_2 + u_2\partial_x u_3-u_3\partial_x u_2}{(1-u_3)^2}
              \end{array}.\right.
\end{eqnarray}
This yields that (\ref{e:cr}) satisfies Cauchy-Riemann equations in $\mathbb C\setminus\mathcal S^{-1}(N)$.

2. Conversely, Cauchy-Riemann equation yields (\ref{e:Cauchy-Riemann}).
Then after a stereographic projection,
\begin{eqnarray}
  D\mathcal S^{-1}_{u}(u_{x})= \left(
                                   \begin{array}{c}
                                     \dfrac{u_1\partial_{x}u_3-u_3\partial_{x}u_1+\partial_{x}u_1}{(1-u_3)^2} \\
                                     \dfrac{u_2\partial_{x}u_3-u_3\partial_{x}u_2+\partial_{x}u_2}{(1-u_3)^2} \\
                                   \end{array}
                                 \right),\notag
\end{eqnarray}
and
\begin{eqnarray}
&&D\mathcal S^{-1}_{u}(u\times u_{y})\notag\\
&=&\left(
\begin{array}{c}
\dfrac{u_1(u_1\partial_{y}u_2-u_2\partial_yu_1)-u_3(u_2\partial_yu_3-u_3\partial_yu_2)+u_2\partial_yu_3 - u_3\partial_yu_2}{(1-u_3)^2}\\
\dfrac{u_2(u_1\partial_yu_2-u_2\partial_yu_1)-u_3(u_3\partial_yu_1-u_1\partial_yu_3) +u_3\partial_yu_1-u_1\partial_yu_3}{(1-u_3)^2}  \\
                                   \end{array}
                                 \right) \notag\\
 &=& \left(
 \begin{array}{c}
 \dfrac{u_2\partial_yu_3-u_3\partial_yu_2+\partial_yu_2}{(1-u_3)^2} \\
 -\dfrac{u_1\partial_yu_3-u_3\partial_yu_1+\partial_yu_1}{(1-u_3)^2} \\
 \end{array}
 \right) = D\mathcal S^{-1}_{u}(u_{x}).\notag
\end{eqnarray}
Hence $u_x=u\times u_y$.
This completes the proof.
\end{proof}

\begin{remark}\label{r:anti-cr}
A similar computation yields that $u$ satisfies (\ref{e:anti-JH}) if and only if
\begin{eqnarray}\label{e:cr}
\mathcal S^{-1}(u)=\dfrac{u_1+iu_2}{1-u_3}
\end{eqnarray}
satisfies the anti-version of Cauchy-Riemann equations in $\mathbb C\setminus\mathcal S^{-1}(N)$.
\end{remark}

\begin{lemma}\label{l:equivalence}
$u$ satisfies (\ref{e:harmonic-map}) if and only if $u$ satisfies (\ref{e:JH}) or (\ref{e:anti-JH}).
\end{lemma}
\begin{proof}
1. ``$\Longrightarrow$'': Since $u_x=J^uu_y$, we find that
\begin{eqnarray}
u_{xx}=u_x\times u_y+u\times u_{xy}.\notag
\end{eqnarray}
Further note that $u_y=-J^u u_x$, then
\begin{eqnarray}
u_{yy}=-u_y\times u_x-u\times u_{xy}.\notag
\end{eqnarray}
It follows that
\begin{eqnarray}
\Delta u=2u_x\times u_y.\notag
\end{eqnarray}
Recall that it is the standard H-bubble equation.
On the other hand, from (\ref{e:JH}) we have that
\begin{eqnarray}
2(u_x\times u_y)\cdot u=-|\nabla u|^2.\notag
\end{eqnarray}
Therefore, (\ref{e:harmonic-map}) holds.

2. ``$\Longleftarrow$'': Compute
\begin{eqnarray}
\partial_{xx}u\cdot \partial_x u=\dfrac{1}{2}\partial_x |\partial_x u|^2,\notag
\end{eqnarray}
\begin{eqnarray}
\partial_{yy}u\cdot \partial_x u&=&\partial_x (\partial_{yy}u\cdot u)-(\partial_x\partial_{yy}u)\cdot u\notag\\
&=&\partial_x(\partial_y(\partial_yu\cdot u)-\partial_y u\cdot \partial_y u)-(\partial_y(\partial_{xy}u\cdot u)-\partial_{xy}u\cdot \partial_y u)\notag\\
&=&-\partial_x|\partial_y u|^2-\partial_y(\partial_{xy}u\cdot u)+\partial_{xy}u\cdot \partial_y u\notag\\
&=&-\partial_x|\partial_y u|^2+\partial_y(\partial_{x}u\cdot \partial_y u)+\partial_y(\partial_{x}u\cdot \partial_y u)-\partial_x u\cdot \partial_{yy}u\notag
\end{eqnarray}
It follows that
\begin{eqnarray}
\partial_{yy}u\cdot \partial_x u=-\dfrac{1}{2}\partial_x|\partial_y u|^2+\partial_y(\partial_{x}u\cdot \partial_y u).\notag
\end{eqnarray}
By (\ref{e:harmonic-map}), we have that
\begin{eqnarray}
0=\partial_{xx}u\cdot \partial_x u+\partial_{yy}u\cdot \partial_x u=\dfrac{1}{2}\partial_x( |\partial_x u|^2-|\partial_y u|^2)+\partial_y(\partial_{x}u\cdot \partial_y u).\notag
\end{eqnarray}
Similarly,
\begin{eqnarray}
0=\partial_{xx}u\cdot \partial_y u+\partial_{yy}u\cdot \partial_y u=\dfrac{1}{2}\partial_y( |\partial_y u|^2-|\partial_x u|^2)+\partial_x(\partial_{x}u\cdot \partial_y u).\notag
\end{eqnarray}
It follows that
\begin{eqnarray}
\Delta (\partial_{x}u\cdot \partial_y u)=0.\notag
\end{eqnarray}
From $|\partial_{x}u\cdot \partial_y u|\le \dfrac{1}{2}|\nabla u|^2$, mean value property and $\mathcal E(u)<\infty$, it holds that
\begin{eqnarray}
\partial_{x}u\cdot \partial_y u=0.\notag
\end{eqnarray}
Then recalling that $u_x\cdot u=0$ and $u_y\cdot u=0$, we may assume that
\begin{eqnarray}
ku_x=u\times u_y.\notag
\end{eqnarray}
That means
\begin{eqnarray}
ku_y=-u\times u_x.\notag
\end{eqnarray}
Note that
\begin{eqnarray}
ku_x\cdot u_x=(u\times u_y)\cdot u_x\quad \mbox{ and }\quad ku_y\cdot u_y=-(u\times u_x)\cdot u_y.\notag
\end{eqnarray}
Then we find that
\begin{eqnarray}
|u_x|=|u_y|.\notag
\end{eqnarray}
Therefore, $|k|=1$. This completes the proof.
\end{proof}

We now give a proof of Corollary \ref{t:main} by ODE.

Let
\[
u=\left(
\begin{array}
[c]{c}
u_{1}\\
u_{2}\\
u_{3}
\end{array}
\right)=\left(
\begin{array}
[c]{c}
\cos\phi\sin Q\\
\sin\phi\sin Q\\
\cos Q
\end{array}
\right).
\]
where $\phi$ and $Q$ are two real functions on $\mathbb R^2$. In polar coordinates,
the energy functional becomes
\begin{eqnarray}\label{e:energy}
\mathcal E(u)=\mathcal E\left(\phi,Q\right)  =\int_{\mathbb R^2}\left(  \sin^{2}Q\left\vert \nabla\phi\right\vert
^{2}+\left\vert \nabla Q\right\vert ^{2}\right)  .
\end{eqnarray}
Hence if $\left(\phi,Q\right)$ is a critical point, then it satisfies the following system:
\begin{equation}\label{e:phi-q}
\left\{\begin{array}{l}
         \operatorname{div}\left( \sin^{2}Q\nabla\phi\right) =0, \\
         -2\Delta Q+\sin\left(  2Q\right)  \left\vert \nabla\phi\right\vert ^{2}=0.
       \end{array}\right.
\end{equation}
Choose $\phi=\phi_m=m\theta$, $Q=Q_m=\pi-2\arctan r^m$.
The linearized system of (\ref{e:phi-q}) at $(\phi_m,Q_m)$ is
\begin{equation}\label{e:phi-q2}
\left\{
\begin{array}
{l}
\left(\dfrac{1}{2}\tan Q_m\right)\Delta\xi+(Q_m)_{r}\xi_{r}+\dfrac{m}{r^{2}}\eta_{\theta}=0,\\
-\Delta\eta+\dfrac{m^{2}}{r^{2}}\cos\left(  2Q_m\right)  \eta+\sin\left(
2Q_m\right)  \dfrac{m}{r^{2}}\xi_{\theta}=0.
\end{array}
\right.
\end{equation}
By some direct computations, (\ref{e:phi-q2}) becomes
\begin{equation}\label{e:ODE}
\left\{
\begin{array}
{l}
- \xi_{rr}-\dfrac{1}{r}\xi_{r}-\dfrac{1}{r^{2}
}\partial_{\theta}^{2}\xi -\dfrac{2m(1-r^{2m})}{r(1+r^{2m})}\xi_{r}+\dfrac{m(1-r^{2m})}{r^{m+2}
}\eta_{\theta}=0,\\
-\eta_{rr}-\dfrac{\eta_{r}}{r}-\dfrac{1}{r^{2}}\partial_{\theta}^{2}\eta
+\dfrac{r^{4m}-6r^{2m}+1}{(1+r^{2m})^2}\dfrac{m^2}{r^{2}} \eta-\dfrac{4r^m(1-r^{2m})}{(1+r^{2m})^2}\dfrac{m}{r^{2}}\xi_{\theta}=0.
\end{array}
\right.
\end{equation}
In $\mathbb R^3$, the kernel maps are given by
\begin{eqnarray}\label{e:e-general-form}
E&=&\dfrac{d}{dt}\left.\left(
              \begin{array}{c}
                \cos (\phi_m+t\xi)\sin (Q_m+t\eta) \\
                \sin(\phi_m+t\xi) \sin (Q_m+t\eta)\\
                \cos (Q_m+t\eta) \\
              \end{array}
            \right)\right|_{t=0}\notag\\
            &=&\left(
                                   \begin{array}{c}
                                     -\xi\sin\phi_m\sin Q_m+\eta\cos \phi_m \cos Q_m \\
                                     \xi\cos\phi_m\sin Q_m+\eta\sin \phi_m \cos Q_m \\
                                     -\eta\sin Q_m \\
                                   \end{array}
                                 \right).
\end{eqnarray}

Assume that $\xi(r,\theta)=\xi_1(r)\cos(k\theta)+\xi_2(r)\sin(k\theta)$ and $\eta(r,\theta)=\eta_1(r)\cos(k\theta)+\eta_2(r)\sin(k\theta)$ with $k\in \mathbb N$. Then we have
\begin{equation}\label{e:general-m}
\left\{
\begin{array}
{l}
- \xi_{1rr}-\dfrac{1}{r}\xi_{1r}+\dfrac{k^2}{r^{2}}\xi_{1}
-\dfrac{2m\left(  1-r^{2m}\right)  }{r\left(  1+r^{2m}\right)  }\xi_{1r}
+\dfrac{mk\left(  1-r^{2m}\right)  }{r^{m+2}}\eta_{2}=0,\\
-\xi_{2rr}-\dfrac{1}{r}\xi_{2r}+\dfrac{k^2}{r^{2}}\xi_{2}
-\dfrac{2m\left(  1-r^{2m}\right)  }{r\left(  1+r^{2m}\right)  }\xi_{2r}
-\dfrac{mk\left(  1-r^{2m}\right)  }{r^{m+2}}\eta_{1}=0,\\
-\eta_{1rr}-\dfrac{\eta_{1r}}{r}+\dfrac{k^2}{r^{2}}\eta_{1}+\dfrac{r^{4m}-6r^{2m}+1}{(1+r^{2m})^2}\dfrac{m^2}{r^{2}}\eta
_{1}+\dfrac{-4mk r^{m-2}\left(  1-r^{2m}\right)  }{\left(  1+r^{2m}\right)  ^{2}}\xi
_{2}=0,\\
-\eta_{2rr}-\dfrac{\eta_{2r}}{r}+\dfrac{k^2}{r^{2}}\eta_{2}+\dfrac{r^{4m}-6r^{2m}+1}{(1+r^{2m})^2}\dfrac{m^2}{r^{2}}\eta
_{2}-\dfrac{-4mk r^{m-2}\left(  1-r^{2m}\right)  }{\left(  1+r^{2m}\right)  ^{2}}\xi
_{1}=0.
\end{array}
\right.
\end{equation}
Therefore, the solutions of these systems are of form
\begin{equation*}
\left(\begin{array}{c}
        \xi \\
        \eta
      \end{array}
\right)=C_1\left(
          \begin{array}{c}
            \xi_1 \cos k\theta \\
            \eta_2 \sin k\theta \\
          \end{array}
        \right)+C_2\left(
          \begin{array}{c}
            \xi_2 \sin k\theta \\
            \eta_1 \cos k\theta \\
          \end{array}
        \right),
\end{equation*}
where $C_1,\,C_2$ are two arbitrary constants. By (\ref{e:e-general-form}), we have that
\begin{eqnarray}\label{e:E-general}
E&=&C_1\left[\xi_1\cos k\theta \sin Q_m E_1+\eta_2\sin k\theta E_2
\right]
+C_2\left[\xi_2\sin k\theta \sin Q_m E_1+\eta_1\cos k\theta E_2
\right]\\
&=& C_1\left[\xi_1\dfrac{2r^m}{1+r^{2m}}\cos k\theta  E_1+\eta_2\sin k\theta E_2
\right]
+C_2\left[\xi_2\dfrac{2r^m}{1+r^{2m}}\sin k\theta E_1+\eta_1\cos k\theta E_2
\right].\notag
\end{eqnarray}

System (\ref{e:general-m}) can be solved by different cases of $k$.
If $k=0$, then
\begin{equation*}
\xi_1(r)=C_1+C_2\dfrac{r^{4m}+4m\,r^{2m}\ln r-1}{2m\, r^{2m}},
\end{equation*}
\begin{equation*}
\eta_1(r)=C_3\dfrac{r^m}{1+r^{2m}}+C_4\dfrac{r^{4m}+4m\,r^{2m}\ln r-1}{r^m(1+r^{3m})}.
\end{equation*}
If $k\ne m$, then the solutions of (\ref{e:general-m}) can be given by
\begin{multline*}
\xi_1(r)=C_1\left(-\dfrac{1}{2r^k}\right)+C_2\dfrac{r^{k}}{2}\\
-C_3\dfrac{r^{k+4m}+\dfrac{4k+4m}{k}r^{k+2m}+\dfrac{k+m}{k-m}r^{k}}{2 r^{2m}}+C_4\dfrac{-r^{4m}+\dfrac{4 k-4m}{k}r^{2m}+\dfrac{k-m}{k+m}}{r^{k+2m}},
\end{multline*}
\begin{equation*}
\eta_2(r)=C_1\dfrac{1}{(1+r^{2m})r^{k-m}}+C_2\dfrac{r^{k+m}}{1+r^{2m}}
+C_3\dfrac{r^{k+4m}+\dfrac{k+m}{k-m}r^{k}}{(1+r^{2m})r^{m}}+C_4\dfrac{2r^{4m}+\dfrac{2k-2m}{k+m}}{(1+r^{2m})r^{k+m}},
\end{equation*}
and
\begin{multline*}
\xi_2(r)=C_5\left(-\dfrac{1}{2r^k}\right)+C_6\dfrac{r^{k}}{2}\\
-C_7\dfrac{r^{k+4m}+\dfrac{4k+4m}{k}r^{k+2m}+\dfrac{k+m}{k-m}r^{k}}{2 r^{2m}}+C_8\dfrac{-r^{4m}+\dfrac{4 k-4m}{k}r^{2m}+\dfrac{k-m}{k+m}}{r^{k+2m}},
\end{multline*}
\begin{equation*}
\eta_1(r)=-C_5\dfrac{1}{(1+r^{2m})r^{k-m}}-C_6\dfrac{r^{k+m}}{1+r^{2m}}
-C_7\dfrac{r^{k+4m}+\dfrac{k+m}{k-m}r^{k}}{(1+r^{2m})r^{m}}-C_8\dfrac{2r^{4m}+\dfrac{2k-2m}{k+m}}{(1+r^{2m})r^{k+m}}.
\end{equation*}
If $k=m$, the solutions are
\begin{eqnarray}
\xi_1(r)&=&C_1\dfrac{r^{2m}-1}{2r^{m}}-C_2\dfrac{1}{4m\, r^m}\notag\\
&&+C_3\dfrac{-r^{6m}+4m\, r^{4m}\ln r-7r^{4m}-4m\,r^{2m}\ln r-13r^{2m}-1}{4m\, r^{3m}}\notag\\
&&+C_4\dfrac{4m\,r^{4m}\ln r-7r^{2m}-1}{4m r^{3m}},\notag
\end{eqnarray}
\begin{multline*}
\eta_2(r)=C_1+C_2\dfrac{1}{2m(1+r^{2m})}+C_3\dfrac{r^{6m}+4m\,r^{4m}\ln r+r^{4m}+4m\,r^{2m}\ln r+5r^{2m}-1}{2m r^{2m}(1+r^{2m})}\\
+C_4\dfrac{4m \,r^{4m}\ln r-r^{2m}-1}{2m r^{2m}(1+r^{2m})},
\end{multline*}
and
\begin{eqnarray}
\xi_2(r)&=&C_5\dfrac{r^{2m}-1}{2r^{m}}-C_6\dfrac{1}{4m\, r^m}\notag\\
&&+C_7\dfrac{-r^{6m}+4m\, r^{4m}\ln r-7r^{4m}-4m\,r^{2m}\ln r-13r^{2m}-1}{4m\, r^{3m}}\notag\\
&&+C_8\dfrac{4m\,r^{4m}\ln r-7r^{2m}-1}{4m r^{3m}},\notag
\end{eqnarray}
\begin{multline*}
\eta_1(r)=-C_5-C_6\dfrac{1}{2m(1+r^{2m})}-C_7\dfrac{r^{6m}+4m\,r^{4m}\ln r+r^{4m}+4m\,r^{2m}\ln r+5r^{2m}-1}{2m r^{2m}(1+r^{2m})}\\
-C_8\dfrac{4m \,r^{4m}\ln r-r^{2m}-1}{2m r^{2m}(1+r^{2m})}.
\end{multline*}
Investigating these explicit solutions and using (\ref{e:E-general}),
we find that the bounded kernel maps are linearly combinations (\ref{e:ek1}) (\ref{e:ek3}) and (\ref{e:ek5}) .


TITLE: As $n$ goes to infinity, can she expect her average measurement to converge to the actual distance $d$?
QUESTION [0 upvotes]: Alice is an astronomer performing many measurements which she averages to estimate the true distance $d$ of a star. Assume the measurements are independent, each with mean $d$. Alice is slowly getting more and more tired. Hence her $n$th measurement has standard deviation $n^{1/3}$. Let $n$ be the number of times Alice repeats measuring this distance. As $n$ goes to infinity, can she expect her average measurement to converge to the actual distance $d$?
I'm pretty stuck for this question, because it seems like I would want to invoke the Strong Law of Large Numbers, but the conditions do not hold since each measurement is not identically distributed. Do I have to manipulate these variables somehow to be able to use the SLLN, or am I missing something else entirely. Another idea I had, was that maybe it was possible that I would only need to use some form of the Weak Law of Large Numbers, which would be more easy to use, but I'm not sure if the WLLN would actually be a strong enough conclusion anyway for this quesiton (is this right?). Any help would be massively appreciated

REPLY [0 votes]: Let the $k$th measurement be $X_k$. To see that this converges to $d$, we can see that
$$
\text{Var}\left(\frac1n\sum_{k=1}^nX_k\right)\to0
$$
as $n\to\infty$.
We have
$$
\begin{align*}
\text{Var}\left(\frac1n\sum_{k=1}^nX_k\right)&=\frac1{n^2}\text{Var}\left(\sum_{k=1}^nX_k\right)\\
&=\frac1{n^2}\sum_{k=1}^n\text{Var}(X_k)\tag{by independence}\\
&=\frac1{n^2}\sum_{k=1}^nk^{\frac23}\\
&\leq\frac1{n^2}\sum_{k=1}^nn^{\frac23}\\
&=\frac1{n^2}n\cdot n^{\frac23}\\
&=n^{-\frac13}\to0.
\end{align*}
$$


TITLE: In applications where heating is not the required from of energy, current is to be kept low as per Joules law. What energy is considered useful?
QUESTION [1 upvotes]: As per this paragraph from Wikipedia
"Joule heating is referred to as ohmic heating or resistive heating because of its relationship to Ohm's Law. It forms the basis for the large number of practical applications involving electric heating.
However, in applications where heating is an unwanted by-product of current use the diversion of energy is often referred to as resistive loss."
My question is, "If not in the form of heat, then how else is the use of energy justified? How else is the energy being used?"

REPLY [0 votes]: Wikipedia is referring to any use of electric power other than resistive heating. I suppose I could list a few:

Electric motors
Computers/circuitry
LED/fluorescent lighting
Radio/communication
Lasers
Radar/sonar
Refrigerators/AC
Microwave ovens

...must I go on?


TITLE: Show that $f_n(x) = \frac{\sin{(nx+3)}}{\sqrt{n+1}}$ converges pointwise
QUESTION [0 upvotes]: I need to show that this sequence function converges pointwise with the help of ($\delta-\epsilon$) definition.
I tried to first split this sequence by sandwich theorem then I am stuck at applying pointwise convergence definition that how should I choose $\epsilon$ and $N$.

REPLY [0 votes]: $|f_n(x)| \le \frac{1}{\sqrt{n+1}}$ for all $x$ and all $n$. Hence $(f_n)$ converges uniformly (and hence pointwise) on $ \mathbb  R$ to $0$.
Can you proceed ?


text:
\section{Application: Distributed Model Predictive Control}
\label{sec:d-mpc}

\begin{figure*}[h!]
\centering
\subfloat[The sum of the inputs of all rooms, $\sum^{M}_{i = 1} u_{i}(t)$.]{
  \includegraphics[width=8cm]{figures/inputs_5sys_96steps.pdf}
  \label{fig:close loop inputs tracking using DMPC}
}\qquad
\subfloat[The temperature of room $1$.]{
  \includegraphics[width=8cm]{figures/temp_5sys_96steps.pdf}
  \label{fig:close loop temperature of room 1 using DMPC}
}
\caption{Simulation results}
\end{figure*}

In this section, we demonstrate the integration of customized local differential privacy in distributed model predictive control. We solve a problem that considers regulating the temperature in a set of rooms in a building, while making sure the cumulative control signal tracks a reference value set by a grid operator in a demand response scheme.
In this example, we consider a linearized discrete-time dynamical model for each sub-system (room):

\begin{equation}
x_{i}(t+1) = A_{i}x_{i}(t)+B_{i}u_{i}(t)\enspace .
\end{equation}

Each sub-system has two states $x_i(t)\in\mathbb{R}^{2}$ and one input $u_i(t)\in\mathbb{R}^{1}$. The first state represents the temperature of room $i$ and the second state is a virtual state that is the result of linearization around an intended operating point.

The input $u_i$ represents the power input of the heating equipment in room $i$. The control goal of this example is keep of the temperature of each room at a desired level of $23^\circ$C and let the power consumption of all rooms to track a signal given by a higher level planner, denoted by $r(t)$. The dynamical matrices $A_i$ and $B_i$ are assumed to be controllable. We formulate the following problem:

\begin{problem}[Distributed Model Predictive Control]
\label{pr: MPC}
\begin{align*}
\min_{x,u}  & \quad \sum^{N}_{t=0} \sum^{M}_{i=1}  x_{i}(t)^{T} Q_i x_i(t) + u_{i}(t)^{T}  R_i  u_i(t) \\
&\quad\quad+ \alpha_{tr} (\sum^{M}_{i=1} u_i(t) - r(t))^{2} \\
s.t. \quad &x_{i}(t+1) = A_{i}x_{i}(t)+B_{i}u_{i}(t)\\
& u_i(t) \in \mathbb{U}_i, \; x_i(0) = \bar{x}_i, \quad i=1,2,\cdots ,M\enspace.
\end{align*}
\end{problem}
\noindent Here, the weight matrices are set as $Q_i = \mbox{diag}[1 0]\succeq 0$ and $R_i = \mbox{diag}[1 1]\succ 0$, and the tracking parameter $\alpha_{tr} = 10$.
The horizon for the MPC problem is set to $N = 13$.
The state and input sequences along the horizon of agent $i$ are denoted by $x_{i}=[x^{T}_{i}(0),x^{T}_{i}(1),\cdots,x^{T}_{i}(N)]^{T}$ and $u_{i}=[u^{T}_{i}(0),u^{T}_{i}(1),\cdots,u^{T}_{i}(N)]^{T}$.
The input constraint $\mathbb{U}_i$ for sub-system $i$ is set to $\mathbb{U}_i = \{u_i| -0.2047 \leq u_i(t)\leq 0.7409\}$. This constraint guarantees that the linearized model is a reasonable approximation of the original non-linear model.
The linearization allows us to formulate a distributed QP of the form in Problem~\ref{pr:dist_opt} with local variables $z_0 =  [u^{T}_{1},u^{T}_{2},\cdots,u^{T}_{M}]^{T}$, and $z_i = u_i$ for $i = 1, \ldots, M$. The global variable $v= [u^{T}_{1},u^{T}_{2},\cdots,u^{T}_{M}]^{T}$. The $i$-th component of $v$ is equal to $[v]_i = [u_{i}]$.
Matrix $H_i$ is dense and positive definite, since the dynamical matrices $A_i$ and $B_i$ are controllable and the matrix $R_i$ is chosen to be positive definite.
The local input constraints are $\mathbb{C}_i = \mathbb{U}_i$ for $i = 1, \ldots, M$, and $\mathbb{C}_0 = \mathbb{R}^M$.

We consider the tracking signal $r(t)$ to be a confidential signal designed by the utility operator, containing sensitive market information.
Therefore, in the resulted distributed QP, Problem~\ref{pr:dist_quad_opt}, the vector $h_i$ in the first objective is the private parameter to protect.
By using the sampling rule of Lemma~\ref{le:sampling_rule} and the analytical upper bound (discussed in supplementary material), we can compute an upper-bound of the sensitivity of the distributed QP problem as $\Theta_i \leq 0.9756$.
We test Algorithm~\ref{al: Distributed optimization with errors} for solving this distributed QP.
At each time-step $t$, we utilize a warm-starting strategy to solve the MPC problems by taking the solution of the previous step to be the initial solution for the current step, and we set the number of iterations and the noise variance to be $K = 10$ and $\sigma = 0.1$.
We compare the performance of the closed-loop system of the sub-optimal solutions given by Algorithm~\ref{al: Distributed optimization with errors} to the optimal solutions, over a period of $24$ hours ($96$ steps, $15$ mins per step). Fig.~\ref{fig:close loop inputs tracking using DMPC} shows the tracking performance of the sum of the inputs $u_i(t)$ of all rooms. Fig.~\ref{fig:close loop temperature of room 1 using DMPC} illustrates how the temperature of room $1$ varies.

config: arxiv


TITLE: Is it true that $s_{n+1} - s_n \to 0$ in a convergent sequence?
QUESTION [2 upvotes]: Is it correct to say that if $s_n$ is a convergent sequence, then $s_{n+1} - s_n \to 0$. It feels right but I don't know how to test it rigorously. I am asking this because I am using this to prove a few questions.
Let $t_n$ be a sequence with:
$$t_1 = 1$$
$$t_n = \sqrt{t_{n-1} +1}$$
To find if the sequence converges to something, I used $t_{n+1} - t_n \to 0$, or $t_n \to t_{n+1}$.
Thus, setting $t_n = \sqrt{t_{n-1} +1} = t_{n-1}$, (again, not sure how to justify this rigorously), but this gives a quadratic on squaring both sides, which gives $\frac{1+\sqrt{5}}{2}$ as the possible value of $t_{n-1}$ where this equality happens, and thus I infer that that is the limiting value. But this seems a really bad proof. Any advice for improvement?

REPLY [2 votes]: For your second question, you asked if
$$t_n=\sqrt{t_{n-1}+1}$$
implies $t_n\to \frac{1+\sqrt{5}}{2}=\phi$? It does not, but you are on the right track. The result you are using implicitly here is the following: if

$x_n=f(x_{n-1},x_{n-2},...,x_{n-m})$
$\lim_{n\to\infty}x_n=L$
$f(y_1,y_2,...,y_m)$ is continuous at $f(L,L,...,L)$

then
$$L=f(L,L,...,L)$$
In your case

$f(x)=\sqrt{x+1}$
$t_n=f(t_{n-1})=\sqrt{t_{n-1}+1}$

We have to show that $\lim_{n\to\infty}t_n=L> -1$ since $\sqrt{x+1}$ is continuous for all $x>-1$. To show that the limit exists and is positive (clearly greater than $-1$) we will show two things: it is increasing and it is bounded. This is enough to show that it converges to a positive number. Specifically, we will show $t_{n+1}-t_{n}>0$ and it is bounded above by $\phi$ and below by $1$. We will do this by two induction proofs:
Base case 1: For $n=1$
$$t_1\geq 1$$
Induction 1: Assume $t_{n-1}>1$. Then
$$t_n=\sqrt{t_{n-1}+1}>\sqrt{1+1}=\sqrt{2}>1$$
Base case 2: For $n=1$ we have
$$t_2-t_1=\sqrt{2}-1>0$$
$$t_2=\sqrt{2}<\phi$$
Induction 2: Assume the proposition holds for some $n-1$. Then
$$t_{n}=\sqrt{t_{n-1}+1}<\sqrt{2+1}=\sqrt{3}<2$$
$$t_{n+1}-t_{n}=\sqrt{t_n+1}-t_n$$
Now, by our inductive assumption and our previous induction proof we know $1\leq t_n<\phi$. What can we say about the function $\sqrt{x+1}-x$ in the interval $[1,\phi)$? Well
$$\frac{d}{dx}\left[\sqrt{x+1}-x\right]=\sqrt{x+2}-\sqrt{x+1}-1$$
For $1\leq x\leq 2$ this is
$$\sqrt{x+2}-\sqrt{x+1}-1<\sqrt{2+2}-\sqrt{1+1}-1=1-\sqrt{2}<0$$
We conclude on the interval $[1,\phi)$ that $\sqrt{x+1}-x$ is decreasing. Since
$$\sqrt{\phi+1}-\phi=0$$
we may conclude
$$\sqrt{t_n+1}-t_n>0$$
and we are done.
Having shown that $t_n\to L$ for some $L>0$, we may now finish the proof using the theorem stated above. We have
$$L=\sqrt{L+1}\Rightarrow L=\phi$$
We conclude
$$\lim_{n\to\infty}t_n=\frac{1+\sqrt{5}}{2}$$


TITLE: One third of Lyapunov exponents are zero? What does it mean?
QUESTION [3 upvotes]: This may be quite a straightforward question, but I have a dynamical system with a high dimensional phase-space. I calculated the Lyapunov spectrum for it and saw that one third of my Lyapunov exponents are approximately zero (which is a lot and was quite unexpected).
What can I conclude from this? These do not signify stable manifolds in the phase-space as you would need negative Lypaunov exponents for this I guess. Does this signify like a giant oscillatory motion in my phase-space? What can I deduce from this abundance of zero modes in the spectrum?
Sidenote: This is a Hamiltonian system so that all Lyapunov exponents sum to zero

REPLY [3 votes]: What first comes to my mind is that you're probably sampling invariant quasiperiodic tori, which are typically neutrally stable.
The associated motion is regular, but not periodic, in that the phase space trajectory comes arbitrarily close to previous states, but never exactly repeats itself (hence quasiperiodic).
Especially if the system's nonintegrable term is relatively weak, its phase space is bound to be full of these so-called KAM surfaces.


TITLE: Determine if the following subset is a subspace of the indicated vector space
QUESTION [1 upvotes]: $W=\{\text {A|A in} M_{22}, \text {A is not invertible}\}$ $\in M_{22}$
Since 2 non-invertible matrix $\in M_{22}$ can be added up to an invertible matrix, it is not closed under addition, therefore it is not a subspace of $M_{22}$.

REPLY [1 votes]: You have the right idea.  Here is an example that you may use to be more explicit in your statement that $W$ is not closed under addition:
$$ \begin{bmatrix}
0 & 0 \\
0 & 1 \\
\end{bmatrix} + \begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix} = I_2 \notin W, $$
but each of the matrices in the sum is an element of $W$. You may be so inclined to check, however, that $W$ is closed under scalar multiplication.


TITLE: In mechanical systems, it is easier to understand concept of natural frequency, could somebody explain natural frequency in electronic elements?
QUESTION [0 upvotes]: In mechanical systems, natural frequency simply means the frequency at which the body will oscillate when it is disturbed(assuming the body will suffer zero resistance in motion) but wihle studying electronics it has become harder to visualise this thing.

REPLY [0 votes]: The same holds for an electronic system. The current keeps oscillating back and forth after the system has been disturbed. This answer contains a nice visualization of what happens in an LRC circuit.


TITLE: What do I call a 2 dimensional surface bound by edges that are line segments or circular arcs?
QUESTION [3 upvotes]: What do I call a 2d surface similar to a polygon, but more general in the sense that edges can be circular arcs as well?
The generalized name should allow for:

Arbitrary number of countable edges
Edge lengths may vary
Arcs and Lines can be used in any combination
Surface may be concave

REPLY [6 votes]: Circular-arc polygons. E.g.,

Wang, Z. J., Lin, X., Fang, M. E., Yao, B., Guan, H., & Guo, M. (2012). RE2L: An Efficient Output-sensitive Algorithm for Computing Boolean Operation on Circular-arc Polygons. arXiv:1211.0729.

Same definition used in other papers, e.g.,

Howell, Louis H. "Numerical conformal mapping of circular arc polygons." Journal of computational and applied mathematics 46.1-2 (1993): 7-28.


TITLE: On the space $L^0$ and $\lim_{p \to 0} \|f\|_p$
QUESTION [5 upvotes]: For $0 < p < \infty$, the definitions of the spaces $L^p$ are very natural. Then, we of course want $L^\infty$ and $L^0$ to be some kind of limits of $L^p$ spaces.
What does the parameter $p$ tell about a function $f \in L^p$? If $f \in L^p$ for very large $p$, it means that $f$ behaves very well near its singularities. If $f \in L^p$ for very small $p$, it means that $f$ decreases fast at infinity.
The limiting case $p \to \infty$ then of course means that $f \in L^\infty$ if $f$ has no singularities in some sense (essentially bounded). Then we define $\|f\|_\infty = \operatorname{ess sup}_{x\in X} |f(x)|$, which is a norm in $L^\infty$. In some cases it even holds that $\|f\|_\infty = \lim_{p \to \infty} \|f\|_p$.
How about $p \to 0$? I've never really seen anything said about this space, so I guess it's not very interesting. But I still started to think about it. Now, $f \in L^0$ should mean that $f$ is extremely bounded at infinity, meaning that $f$ is compactly supported. Of course because we are talking about measure and integration theory, the notion of compact support has to be modified accordingly. I guess we want to define that $f \in L^0$ if the support of $f$ is compact modulo sets of measure $0$. Is this how we should think about $L^0$? If it is, then we probably want to define $\|f\|_0 = \mu(\operatorname{spt} f)$, or something like that.
So, what is the number $\lim_{p \to 0} \|f\|_p$? For measurable compactly supported bounded functions it is easy to see that
\begin{align}
\lim_{p \to 0} \int_X |f|^p = \mu(\operatorname{spt} f)\,.
\end{align}
But if we want to take the limit of $\|f\|_p$, the integral should be raised to the power $1/p$, which tends to infinity, which complicates things.
So my questions are: what should the space $L^0$ be, what should be its "norm" $\|f\|_0$ and when is $\|f\|_0$ the limit of $\|f\|_p$?
Also, why does the wikipedia page on $L^p$ spaces refer to $L^0$ as the space of measurable functions? (In the case $\mu(X) < \infty$ it would make sense since $L^0$ restricts the behavior of its elements at infinity, and in the $\mu(X)< \infty$ case there is no "behavior at infinity".)

REPLY [2 votes]: If $\mu(X)=1$, there exists $p>0$ with $f\in L^p(\mu)$, and if we agree that $\log(0)=-\infty$ and $\exp(-\infty)=0$ then $$\lim_{p\to0}||f||_p=\exp\left(\int\log|f|\,d\mu\right), $$the geometric mean of $|f|$. Hence some people sometimes write $||f||_0$ for the geometric mean.
Assume $f\ge0$. First, Jensen's inequality with $\phi(t)=e^{pt}$ shows that $$||f||_p\ge||f||_0\quad(p>0).$$
Note that $$\log(t)\le c_pt^p\quad(t\ge0).$$So if $f\in L^p$ then $\int\log^+f<\infty$, and hence $\int\log f$ exists (as an element of $[-\infty,\infty)$). Fix $q>0$ such that $$\int f^q\log^2(f)<\infty.$$
Now $$e^t\le 1+t+\frac{t^2}{2}e^t\quad(t\ge0)$$and $$e^t\le 1+t+\frac{t^2}2\quad(t\le 0).$$
Suppose $0<p<q$. If $f=0$ on a set of positive measure then Holder shows that $||f||_p\to0$. Assume that $f>0$ almost everywhere. Let $f=e^g$. Let $E$ be the set where $g\ge0$ and $F$ the set where $g<0$. Now
$$\int_Ef^p=\int_Ee^{pg}\le\int_E\left(1+pg+\frac{p^2g^2}{2}f^p\right)\le\int_E\left(1+pg+\frac{p^2g^2}{2}f^q\right),$$while$$\int_F f^p\le\int_F\left(1+pg+\frac{p^2g^2}{2}\right).$$
Hence $$\int f^p\le 1+p\int\log f+cp^2,$$where $c=\frac12\int(1+f^q)\log^2f$.
Hence $\limsup_{p\to0}||f||_p\le||f||_0$.


TITLE: Characterization of the family of simple groups PSL(2,q) by tensor multiplicity
QUESTION [9 upvotes]: Let $G$ be a finite group and $(\chi_i)$ its irreducible characters. Then $\forall i,j,k, \exists!n_{i,j}^k \in \mathbb{N}_{\ge 0}$ such that     $$\chi_i \chi_j = \sum_k n_{i,j}^k \chi_k.$$
Let the tensor multiplicity of $G$ be $m(G):= \max_{i,j,k} (n_{i,j}^k)$; for ex. $m(A_5) = 2$ and $m(A_6) = 3$.
Theorem: Let $q > 5$ be a prime-power, then $$m(\mathrm{PSL}(2,q)) = \left\{
    \begin{array}{ll}
        2 & \text{ if } q \text{ even,} \\
        3 & \text{ if } q \text{ odd.}
    \end{array}
\right.$$
proof: see (for example) the generic computation of $(n_{i,j}^k)$ for $\mathrm{PSL}(2,q)$ in my talk at 30:00. $\square$
This post is about the converse of above theorem.
Let $G$ be a non-abelian finite simple group with $m(G) \le 3$.
Question: Is it true that $G \simeq \mathrm{PSL}(2,q)$ for some prime-power $q$?
It was checked by GAP for $|G|<10^7$ (see Appendix).
Remark: The classification below suggests that the family $(\mathrm{PSL}(2,q))$ is the only one, among the usual infinite families of (non-abelian) finite simple groups, where the tensor multiplicity is bounded above. This would provide an other interesting characterization of this family.

Appendix
We improved our way to check as suggested by Mikko Korhonen in comment. Let $G$ be a finite group, $d_+(G)$ be the maximum among the degrees of complex irreducible characters of $G$, and $\Sigma(G)$ their sum. They are exactly $27$ non-abelian finite simple groups $|G|<10^7$ not isomorphic to some $\mathrm{PSL(2,q)}$. The classification below lists them ordered according to $d_+^2(G)/\Sigma(G)$, which is less than or equal to $m(G)$.
Computation
Each element of the list reads as $d_+^2(G)/\Sigma(G)$, its numerical approximation, $G$ and $|G|$.
gap> classification(10000000);
[ [ 256/63, 4.06349, PSU(3,3), 6048 ],
[ 1125/208, 5.40865, PSU(3,4), 62400 ],
[ 13/2, 6.5, PSL(3,3), 5616 ],
[ 18432/2107, 8.74798, PSU(3,7), 5663616 ],
[ 175/18, 9.72222, A7, 2520 ],
[ 2048/203, 10.0887, PSL(3,4), 20160 ],
[ 175/16, 10.9375, A8, 20160 ],
[ 279/25, 11.16, PSL(3,5), 372000 ],
[ 729/64, 11.3906, PSp(4,3), 25920 ],
[ 3025/218, 13.8761, M11, 7920 ],
[ 8281/484, 17.1095, Sz(8), 29120 ],
[ 10368/553, 18.7486, PSU(3,5), 126000 ],
[ 7225/271, 26.6605, PSp(4,4), 979200 ],
[ 321489/11210, 28.6788, PSU(3,8), 5515776 ],
[ 43681/1464, 29.8367, J_1, 175560 ],
[ 704/21, 33.5238, M12, 95040 ],
[ 23328/683, 34.1552, A9, 181440 ],
[ 5472/155, 35.3032, PSL(3,7), 1876896 ],
[ 28224/709, 39.8082, J_2, 604800 ],
[ 16384/319, 51.3605, PSp(6,2), 1451520 ],
[ 321489/5356, 60.0241, A10, 1814400 ],
[ 152100/2519, 60.3811, PSp(4,5), 4680000 ],
[ 21175/258, 82.0736, M22, 443520 ],
[ 173056/1875, 92.2965, G(2, 3), 4245696 ],
[ 20800/207, 100.483, PSL(4,3), 6065280 ],
[ 775/7, 110.714, PSL(5,2), 9999360 ],
[ 28672/227, 126.308, PSU(4,3), 3265920 ] ]

Below is the alternative classification suggested by Goeff Robinson in comment, considering $d_+(G)/c(G)$, with $c(G)$ the class number of $G$:
gap> classification2(10000000);
[ [ 16/7, 2.28571, PSU(3,3), 6048 ],
[ 13/4, 3.25, PSL(3,3), 5616 ],
[ 75/22, 3.40909, PSU(3,4), 62400 ],
[ 35/9, 3.88889, A7, 2520 ],
[ 81/20, 4.05, PSp(4,3), 25920 ],
[ 5, 5., A8, 20160 ],
[ 11/2, 5.5, M11, 7920 ],
[ 31/5, 6.2, PSL(3,5), 372000 ],
[ 32/5, 6.4, PSL(3,4), 20160 ],
[ 192/29, 6.62069, PSU(3,7), 5663616 ],
[ 91/11, 8.27273, Sz(8), 29120 ],
[ 72/7, 10.2857, PSU(3,5), 126000 ],
[ 176/15, 11.7333, M12, 95040 ],
[ 12, 12., A9, 181440 ],
[ 340/27, 12.5926, PSp(4,4), 979200 ],
[ 209/15, 13.9333, J_1, 175560 ],
[ 16, 16., J_2, 604800 ],
[ 256/15, 17.0667, PSp(6,2), 1451520 ],
[ 81/4, 20.25, PSU(3,8), 5515776 ],
[ 228/11, 20.7273, PSL(3,7), 1876896 ],
[ 390/17, 22.9412, PSp(4,5), 4680000 ],
[ 189/8, 23.625, A10, 1814400 ],
[ 385/12, 32.0833, M22, 443520 ],
[ 1040/29, 35.8621, PSL(4,3), 6065280 ],
[ 832/23, 36.1739, G(2, 3), 4245696 ],
[ 224/5, 44.8, PSU(4,3), 3265920 ],
[ 1240/27, 45.9259, PSL(5,2), 9999360 ] ]

Code
classification:=function(n)  #for Geoff way: classification2:=function(n)
    local it,LL,g,A,L,l,dmax,S,c,cc;
    it:=SimpleGroupsIterator(2520,n);; #2520=|A_7|
    LL:=[];;
    for g in it do
        A:=StructureDescription(g);;
        if Length(A)<6 or List([1..6],i->A[i])<>"PSL(2," then #PSL(2,q) excluded
            L:=CharacterDegrees(g);;
            l:=Length(L);;
            dmax:=L[l][1];;
            S:=Sum(L,i->i[1]*i[2]); #Goeff way: Sum(L,i->i[2]);
                    cc:=dmax^2/S;;
            Add(LL,[cc,Float(cc),g,Order(g)]);;
        fi;
    od;
    Sort(LL);
    return LL;
end;;

Recall that $A_7$ is the smallest non-abelian finite simple group not isomorphic to some $\mathrm{PSL}(2,q)$.

REPLY [5 votes]: The check of $\textrm{max}\{\textrm{deg}(\chi)\mid \chi \in \textrm{Irr}(G)\} > 3   k$ could be extended to more finite simple groups (including all sporadics) by using the character tables which are available in GAP:
simpnames:= AllCharacterTableNames( IsSimple, true, IsAbelian, false);
LL := [];;
for nam in simpnames do
  Print(nam," ");
  t := CharacterTable(nam);
  d := Maximum(List(Irr(t),Degree));
  k := NrConjugacyClasses(t);
  Print([d,k,Float(d/k)],"\n");
  Add(LL, [d/k,Float(d/k),nam, Size(t),d,k]);
od;
Sort(LL, {a,b} -> a[4] <= b[4]);
Filtered(LL, a-> a[1] <= 3);

From the result I would guess the following statement:
G is non-abelian simple with max. character degree $\leq 3k$ iff  G = PSU(3,3) or PSL(2,q).
Sketch of proof:

G sporadic: checked above (GAP, resp. ATLAS)

G alternating: (note that $A_5\cong PSL(2,4)$ and $A_6 \cong PSL(2,9)$ are among the exceptions) Symmetric group $S_n$ has representation for partition
(2,2,2, ...., 2,2 or 1)
and the hook formula and Stirling approximation of $n!$ yield that
its degree grows with $n$ like $e^n$, while the number of conjugacy classes
of $S_n$ (= number of partitions of $n$) grows like $e^{const \sqrt{n}}$.
(The numbers for the alternating group differ at most by a factor $2$).

G of Lie type of rank l:

they always have the Steinberg representation of degree $q^{N}$ where $N$ is the number of postive roots of $G$,  which is roughly $N \sim l^2$
their number of conjugacy classes is bounded by $< c q^l$ for some constant c

(so only rank $l=1$, $PSL_2(q)$, or very small $l$ and $q$ will give exceptions)

So, asymptotically the statement looks ok, one needs to be a bit more precise to get the list of small exceptions.


TITLE: Interval for unique solution for first order I.V.P.
QUESTION [1 upvotes]: Find largest interval in which the I.V.P $$\frac{dy}{dx}=y^{2}+\cos^{2}x,x>0,y(0)=0.$$ has unique solution.
My attempt is as follows
Firstly I apply Picard Theorem for the rectangle $\{(x,y):|x|\leq a,|y|\leq b\}$
$M=b^{2}+1, h=\min\{\frac{b}{b^{2}+1},a\}$,  I found $h=1/2$ and hence our required interval is $[0,1/2]$
But my problem is that the function $f(x,y)=y^{2}+\cos^{2}x$ is globally Lipschiz on the region of the type $[0,\infty)\times [-k,k]$ so can i say that the given I.V.P. has unique solution on $[0,\infty)?$
Please suggest where I am wrong? Am I using the wrong result in last one? What is the way to find largest interval of uniqueness of solution on the above I.V.P.? Thank you.

REPLY [2 votes]: The function $f(x,y)=y^{2}+\cos^{2}x$ is of class $C^{1}$ so it is locally
Lipschitz in a neighborhood of every point $(x_{0},y_{0})$. Hence, as long as
the solution exists, the solution is unique. Since $f$ is continuous, the
solution exists at least for a short time. Let $[0,T)$ be the maximal time of
existence on the right (the case $x<0$ is similar). We want to prove that
$T<\infty$. Assume by contradiction that $y(x)$ exists for all times. Since
$f\geq0$, we have that $y(x)$ is increasing. Actually, it is strictly
increasing, since $y^{\prime}(0)=1>0$ and so for $x>0$ sufficiently small, say
$0<x\leq a$ (for some small $a>0$), we have that $y(x)>y(0)=0$. But then
$y^{\prime}(x)\geq y^{2}(x)\geq y^{2}(a)>0$ for all $x\geq a$ and so $y(x)$ is
actually strictly increasing for all $x>0$. Then for $x>0$, $y^{\prime}(x)\geq
y^{2}(x)>0$ and so
$$
\frac{y^{\prime}}{y^{2}}\geq1
$$
for all $x>0$. Integrating both sides over the interval $[1,x]$ we get
$$
\frac{1}{y(1)}\geq-\frac{1}{y(x)}+\frac{1}{y(1)}\geq x-1.
$$
Letting $x\rightarrow\infty$ we get a contradiction because the left-hand side
is bounded and the right-hand side goes to infinity. This means that
$T<\infty$. Thus the solution exists in $[0,T)$ and since it is strictly
increasing
$$
\lim_{x\rightarrow T^{-}}y(x)=\infty.
$$
You can do the same for $x<0$.


TITLE: Is there a general notion of definability/expressibility?
QUESTION [0 upvotes]: If we want to talk about uncountable sets, then we can distinguish between objects in the set that we can actually "specify" and those that we can't. An example of objects that can be specified are $5$, $\sup_xf(x)$, and Chaitin's constant. I can't give any examples of objects that can't be specified, since I would have specified them in doing so.
This notion is formalized in some specific contexts: E.g. a definable real number, or a definable set (of natural numbers). Based on a few quick google searches I could only find these two examples. I'd just like to get a pointer to some literature that I can read about it.
Is there a general notion of "definability" of arbitrary mathematical objects?

In particular, in classical mathematics. In constructive mathematics the notion of definability collapses to computability afaik.

I'm particularly interested if it's in type-theoretic foundations.

I can imagine there are some difficulties related to logical paradoxes when we try to define a notion of "definability" in full generality?

REPLY [2 votes]: For any structure $\mathfrak{M}$ there is a general notion of (first-order) definable subset/function/relation/element of $\mathfrak{M}$, provided by the (first-order) formulas in the language of $\mathfrak{M}$. We can also talk about definability with parameters, and so on. See here for a quick summary of this.
Now note the "first-order" bit above. This refers to the specific choice of first-order logic as our framework for making and evaluating definitions. There are other logics out there - second-order logic, infinitary logic(s), etc. - and each logic gives rise to its own definability notion. If you're interested, I strongly recommend the (legally freely available online!) wonderful collection Model-theoretic logics.
As to meta-issues, for any "reasonable" logic $\mathcal{L}$ there is a general obstacle (usually phrased for first-order logic, but much more generally applicable than that) to talking about the $\mathcal{L}$-definable phenomena in a structure $\mathfrak{M}$ within that structure itself. The existence of pointwise-definable models of set theory provides a particularly powerful example of how strong this limitation truly is! However, this doesn't prevent us from performing this analysis from outside $\mathcal{M}$. So we just have to be careful what statement is being made where.


TITLE: Better estimate on binomial coefficient $\binom{n}{k}$ when $k=(1-o(1))n$?
QUESTION [3 upvotes]: For binomial coefficient $\binom{n}{k}$, I know one inequality saying that
$$\binom{n}{k}\le (en/k)^k.$$
While for $k=(1-\epsilon)n$ if $\epsilon$ is an $o(1)$-term, where we assume $n$ tends to infinity, the above inequality gives an exponential bound. I am wondering is there a better estimate on $\binom{n}{k}$ in this case?

REPLY [3 votes]: Note that $\binom{a}{b}=\binom{a}{a-b}$ as long as $b<a$. This means
$$\binom{n}{(1-\epsilon)n}=\binom{n}{\epsilon n}$$
So we may as well call $\epsilon n\to\epsilon$. (We are only diddling around with orders of magnitude of $\epsilon$.)
Remember
$$\binom{n}{k}=\frac{n!}{(n-k)!k!}$$
So using the Gamma function,
$$\binom{n}{x}=\frac{\Gamma(1+n)}{\Gamma(1+n-x)\Gamma(1+x)}$$
We can already make some progress using the recursive properties of Gamma and Euler's reflection formula. Note
$$\Gamma(1+n-x)=\Gamma(1-x)\prod_{k=0}^{n-1}(n-k-x)$$
Combining this with Euler's reflection formula, we get
$$\binom{n}{x}=n!\operatorname{sinc}(\pi x)\prod_{k=0}^{n-1}\frac{1}{(n-k-x)}$$
Writing the factorial as a product,
$$\binom{n}{x}=\operatorname{sinc}(\pi x)\prod_{k=0}^{n-1}\frac{(n-k)}{(n-k-x)} \\ =\operatorname{sinc}(\pi x)\prod_{k=0}^{n-1}\frac{1}{1-\frac{x}{n-k}} \\ =\operatorname{sinc}(\pi x)\left(\prod_{k=0}^{n-1}\left(1-\frac{x}{n-k}\right)\right)^{-1}$$
So, we need to figure out how to estimate
$$P_n(x)=\prod_{k=0}^{n-1}\left(1-\frac{x}{n-k}\right)$$

Lemma: Let $$P_n(x)=\prod_{k=0}^{n-1}\left(1-\frac{x}{n-k}\right)$$ be a degree $n$ polynomial with coefficients $a_0,a_1,\dots ,a_n$. Then clearly $a_1=-H_n$, where $H_n$ is the $n$th harmonic number.

Proof.
We proceed with induction. Clearly the $n=1$ case is true. Now:
$$P_{n+1}(x)=\prod_{k=0}^{n}\left(1-\frac{x}{n+1-k}\right)  \\ =\left(1-\frac{x}{n+1}\right)\prod_{k=1}^{n}\left(1-\frac{x}{n-(k-1)}\right)\\  =\left(1-\frac{x}{n+1}\right)P_n(x) \\ =\left(1-\frac{x}{n+1}\right)\big(1-H_n x+a_2x^2+\dots\big) \\ =1-H_n x-\frac{x}{n+1}+\text{higher order terms} \\ =1-H_{n+1}x+\text{higher order terms}$$
$\blacksquare$
Finally, taking $H_n\asymp \gamma+\log(n)$ where $\gamma$ is the Euler-Mascheroni constant, we have
$$\boxed{\binom{n}{x}\approx \operatorname{sinc}(\pi x)\frac{1}{1-(\gamma+\log n)x}} \\ \text{as}~n\to\infty~\text{and}~x\to 0$$
You can of course take $\operatorname{sinc}(z)\approx 1$ for small $z$ if you'd like.
This approximation is a bit crude, for sure. If you want more accuracy, unfortunately you will have to crunch more coefficients of $P_n(x)$. This is by no means easy. But I have a sneaking suspicion it has to do with polygamma functions.

EDIT 1: On the coefficients of $P_n(x)$.
We have the exact expression
$$\binom{n}{x}=\operatorname{sinc}(\pi x)\frac{1}{P_n(x)}$$
Where $P_n(x)$ are polynomials given by
$$P_n(x)=\prod_{k=0}^{n-1}\left(1-\frac{x}{n-k}\right)$$
They can be alternatively defined by the recurrence
$$P_{n}(x)=\left(1-\frac{x}{n}\right)P_{n-1}(x) \\ P_1(x)=1-x$$
We let $0\leq k\leq n$ and consider the $k$th coefficient of $P_n(x)$. Call it $a_{n,k}$. We have
$$P_n(x)=\left(1-\frac{x}{n}\right)\left(\sum_{k=0}^{n-1} a_{n-1,k}x^k\right) \\ =\sum_{k=0}^{n-1}a_{n-1,k}x^k-\sum_{k=0}^{n-1}\frac{1}{n}a_{n-1,k}x^{k+1}$$
Playing around with the summations,
$$P_n(x)=\sum_{k=0}^{n-1}a_{n-1,k}x^k-\sum_{k=1}^{n}\frac{1}{n}a_{n-1,k-1}x^{k} \\ =a_{n-1,0}-\frac{1}{n}a_{n-1,n-1}x^n+\sum_{k=1}^{n-1}\left(a_{n-1,k}-\frac{1}{n}a_{n-1,k-1}\right)x^k \\ =a_{n,0}+a_{n,n}x^n+\sum_{k=1}^{n-1}a_{n,k}x^k$$
From which we get the equations
$$a_{n,0}=1 \\ a_{n,n}=(-1)^n n! \\ \boxed{a_{n,k}=a_{n-1,k}-\frac{1}{n}a_{n-1,k-1}}$$
Good  luck!

EDIT 2: An exact expression for the coefficients of $P_n(x)$.
Remember that
$$P_n(x)=\prod_{k=0}^{n-1}\left(1-\frac{x}{n-k}\right)$$
Which can be reshuffled a little bit:
$$P_n(x)=\frac{1}{n!}\prod_{k=0}^{n-1}(n-k-x) \\ =\frac{(-1)^n}{n!}\prod_{k=0}^{n-1}(x-n+k) \\ =\frac{(-1)^n}{n!}\prod_{k=1}^{n}(x-k) \\=\frac{(-1)^n}{n!}\frac{x-n}{x}\prod_{k=0}^{n-1}(x-k)$$
But $\prod_{k=0}^{n-1}(x-k)$ is a falling factorial, and can be expressed using signed Stirling numbers of the first kind:
$$\prod_{k=0}^{n-1}(x-k)=\sum_{k=0}^ns(n,k)x^k$$
So
$$P_n(x)=\frac{(-1)^n}{n!}\left(1-\frac{n}{x}\right)\sum_{k=0}^{n}s(n,k)x^k$$
However, it is known that $s(n,0)=0~\forall n\in\Bbb N$, so we can rewrite this as
$$P_n(x)=\frac{(-1)^n}{n!}\left(\sum_{k=1}^n s(n,k)x^k-\frac{n}{x}\sum_{k=0}^{n-1}s(n,k+1)x^{k+1}\right) \\ =\frac{(-1)^n}{n!}\left(\sum_{k=1}^n s(n,k)x^k-\sum_{k=0}^{n-1}n~s(n,k+1)x^k\right) \\ =\frac{(-1)^n}{n!}\left(-n~s(n,1)+n~s(n,n+1)x^n+\sum_{k=1}^n s(n,k)x^k-\sum_{k=1}^{n}n~s(n,k+1)x^k\right)$$
Combining the sums and using $s(n,n+1)=0$ and $s(n,1)=(-1)^{n+1}(n-1)!$ we get
$$P_n(x)=\frac{(-1)^n}{n!}\left(-n(-1)^{n+1}(n-1)!+\sum_{k=1}^n \left(s(n,k)-n~s(n,k+1)\right)x^k\right)$$
So
$$P_n(x)=1+\sum_{k=1}^n \frac{(-1)^n}{n!}\big(s(n,k)-n~s(n,k+1)\big)x^k$$
But - this is amazing! The signed Stirling numbers of the first kind obey the recurrence
$$s(n+1,k)=-n~s(n,k)+s(n,k-1)$$
Which can be rearranged to give
$$s(n+1,k+1)=s(n,k)-n~s(n,k+1)$$
Which can be inserted into our expression for $P_n(x)$ to give
$$P_n(x)=1+\sum_{k=1}^n\frac{(-1)^n}{n!}s(n+1,k+1) x^k$$
Or alternatively we can introduce the unsigned Stirling numbers of the first kind,
$$\begin{bmatrix}n\\k\end{bmatrix}=(-1)^{n-k}s(n,k)$$
Verified for e.g $n=7$:

Hence
$$\boxed{P_n(x)=1+\sum_{k=1}^n \frac{1}{n!}\begin{bmatrix}n+1\\k+1\end{bmatrix}(-1)^k x^k}$$

EDIT 3
Computation of Stirling numbers is generally speaking very hard, but there are a few known formulas,
$$\begin{bmatrix}n+1\\2\end{bmatrix}=n!H_n \\ \begin{bmatrix}n+1\\3\end{bmatrix}=\frac{n!}{2}\left((H_n)^2-H^{(2)}_n\right) \\ \begin{bmatrix}n+1\\4\end{bmatrix}=\frac{n!}{3!}\left((H_n)^3-3H_nH^{(2)}_n+2H^{(3)}_n\right) \\ \text{and} \\ \begin{bmatrix}n\\n-1\end{bmatrix}=\binom{n}{2} \\ \begin{bmatrix}n\\n-2\end{bmatrix}=\frac{3n-1}{4}\binom{n}{3} \\ \begin{bmatrix}n\\n-3\end{bmatrix}=\binom{n}{2}\binom{n}{4}$$
Where $H^{(m)}_n=\sum_{k=1}^n\frac{1}{k^m}$ is a generalized harmonic number.
But things get very complicated very fast.


TITLE: Pointwise convergence at every rational but not everywhere
QUESTION [3 upvotes]: I don't think such a sequence exists.
I have a proof (given in class) that if the sequence is equicontinuous and converges point-wise (non-uniformly) for every rational on $[a,b]$ then it converges uniformly on all of $[a,b]$.
I also have a proof (written by me) that if a sequence of continuous functions defined on $[a,b]$ converges point-wise for every rational on $[a,b]$, then it  converges.

EDIT: "uniformly" change to "pointwise"

on all of $[a,b]$.
But I was told that equicontinuity is essential. I'm having hard time seeing why. Can someone give me a counterexample to demonstrate the need for equicontinuity?
My understanding is at the level of Rudin's principles [...] chapters 1-6 and most of chapter 7.

REPLY [0 votes]: Define the sequence of functions $(f_n)_n$ as follows: for any $n\geq 1$,

$f_n(x) = 0$ for $x\in[0,\frac{1}{\sqrt{2}}-\frac{1}{10n}]\cup [\frac{1}{\sqrt{2}}+\frac{1}{10n},1]$;
$f_n(\frac{1}{\sqrt{2}}) = n$;
$f_n$ is piecewise affine and continuous;

so that $f_n$ is a continuous function looking like a very narrow triangle centered at $\frac{1}{\sqrt{2}}$.
Then $f_n(x) \xrightarrow[n\to\infty]{}0$ for any $x\in[0,1]\setminus\{\frac{1}{\sqrt{2}}\}$, and a fortiori at all rationals. But there is not pointwise convergence on $[0,1]$, since $f_n(\frac{1}{\sqrt{2}}) \xrightarrow[n\to\infty]{}\infty$.


TITLE: Integral domain ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$
QUESTION [3 upvotes]: Exercise: Show that the smallest subdomain of complex numbers containing the element $\alpha=\frac{\sqrt{5} - 1}{2}$ is ${\mathbf Z}[\alpha] = \{a + b\alpha\ |\ a,\ b \in {\mathbf Z}\}$.
I thought I could first show that $\mathbf{Z}[\alpha]$ is indeed an integral domain by verifying all axioms (that's  easy) and then I would show that if there's a smaller domain $N$, $x \in \mathbf{Z}[\alpha]$ exists such that $x \notin N$. This $x = a + b\alpha$ for some $a,\ b \in \mathbf{Z}$ (as it is in $\mathbf{Z}[\alpha]$). But from the axioms it follows that for integral domain $N$, we have $\mathbf{Z} \in N$ and $\alpha \in N$, so $a + b\alpha \in N$, a contradiction.
It's easy to show that ${\mathbf Z}[\alpha]$ is a commutative ring, but I have trouble showing that $a,\ b \in {\mathbf Z}[\alpha],\ a,b \neq 0 \implies ab \neq 0$.
Have I chosen a sensible way or is there a better approach?
Thanks.

REPLY [1 votes]: A subring of a field is automatically an integral domain: zero divisors in it would be zero divisors in the field.
The set $\{a+b\alpha:a,b\in\mathbb{Z}\}$ is surely contained in any subring of $\mathbb{C}$ containing $\alpha$. So you just have to prove it's a subring.
Closure under addition and opposite is clear; moreover $0=0+0\alpha$ and $1=1+0\alpha$. For closure under multiplication:
$$
(a+b\alpha)(c+d\alpha)=ac+(bc+ad)\alpha+bd\alpha^2
$$
But
$$
\alpha^2=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}=1-\frac{\sqrt{5}-1}{2}
=1-\alpha
$$
so you can conclude.


TITLE: The number of perfect squares which can occur in an arithmetic progression of length n
QUESTION [4 upvotes]: This is a similar question to https://math.stackexchange.com/questions/2023399/the-maximum-number-of-perfect-squares-that-can-be-in-an-arithmetic-progression/3693487#3693487
Let f(n) be the maximum number of squares in an AP (arithmetic progression) of length n. For example, $f(3)=3$, as 1, 25, 49 is a 3-term arithmetic progression with three squares, and $f(4)=3$, as there are no 4 term arithmetic progressions of squares. Also, $f(5)=4$, with the AP 49, 169, 289, 409, 529 as a small example.
Trivially, f is monotone increasing, as adding terms onto an existing AP cannot reduce the number of squares. Also, $f(a+b) \leq f(a)+f(b)$, by concatenation of sequences. It seems to me that the easiest way to find upper bounds on f is to constrain configurations of squares. Let $(0, a, b, c)$ (with $0<a<b<c$ denote a configuration of squares of the form: $M, M+aK, M+bk, M+ck$, where $k>0$. The configuration $(0, 1, 2, 3)$ is a four term arithmetic progression, which we already know is ruled out. Using elliptic curves you can show that $(0, 1, 3, 4)$ and $(0, 1, 4, 5)$ are also impossible (and it looks like many more are impossible as well. On the positive end, there are solutions for any configuration $(0, a, b, c)$ when $c \neq a+b$ (I'm working on a parametric solution).
By eliminating those configurations, I have found that $f(6)=4$, $4 \leq f(7) \leq 5$, and $f(8)=5$ with $1, 25, 49, 73, 97, 121, 145, 169$ as an example.
Up to what $n$ is $f(n)$ known? Specifically, are f(9) and f(10) known?

REPLY [3 votes]: It's tabulated out to $f(52)=12$ at the Online Encyclopedia of Integer Sequences.
A reference is given to Enrique González-Jiménez and Xavier Xarles, On a conjecture of Rudin on squares in Arithmetic Progressions.
A conjecture is presented which gives a very simple form for $f(n)$ that works for all $n\le52$ except $n=5$.


TITLE: Solving theoretical least squares for simple linear regression
QUESTION [0 upvotes]: Let $X$ and $Y$ be random variables. On solving the theoretical least squares for simple linear regression case, I came across the following step (while differentiating the expected mean square error with respect to the first parameter):
$$\frac{\partial}{\partial{a}} \mathbb{E}(Y-aX-b)^2 = \mathbb{E}\big[\frac{\partial}{\partial{a}}(Y-aX-b)^2\big]$$
I have the following doubts on this step:

Is this always true? Can we interchange derivative and expectation
operator like this? Or do we need to check any condition before doing so?
Since we are differentiating the term $\mathbb{E}(Y-aX-b)^2$ with respect to $a$ (keeping $b$ constant), the term $\mathbb{E}(Y-aX-b)^2$ is now a function of two random variables $X, Y$ or is it a function of $a, b$ ?

REPLY [0 votes]: Is this always true? Can we interchange derivative and expectation
operator like this? Or do we need to check any condition before doing so?

Yes.  Expectation is a linear operation over the random variables, and here $a$ is a free variable.

Since we are differentiating the term $\mathbb{E}(Y-aX-b)^2$ with respect to $a$ (keeping $b$ constant), the term $\mathbb{E}(Y-aX-b)^2$ is now a function of two random variables $X, Y$ or is it a function of $a, b$ ?

$\mathbb E((Y-aX-b)^2)$ is a function of free variables $a$ and $b$, for exactly the reason $\Bbb E(X)$ is a constant rather than a function of random variable $X$.

In general, when the domain is not a function of $a$, we have for summation over a discrete domain:
$$\dfrac{\partial ~~}{\partial a}\sum_{x\in\mathcal D} g(a,x)=\sum_{x\in\mathcal D} \dfrac{\partial g(a,x)}{\partial a}$$
And analogously for integrals over a continuous domain:
$$\dfrac{\partial ~~}{\partial a}\int_{\mathcal D} g(a,x)\,\mathrm d x=\int_{\mathcal D} \dfrac{\partial g(a,x)}{\partial a}\,\mathrm d x$$
.
Remark: The partial differentiation in the RHS expressions do also treat $x$ as an independent variable.