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\maketitle

\noindent\textbf{Abstract}: In this paper we completely classify the homogeneous two-spheres, especially, the minimal homogeneous ones
in the quaternionic projective space $\textbf{HP}^n$. According to our classification, more minimal constant curved two-spheres in $\textbf{HP}^n$ are obtained than Ohnita conjectured in \cite{[Oh]}.

\vspace{0.2cm}
\noindent\textbf{Keywords}: minimal two-sphere; Gauss curvature; quaternionic projective space.

\vspace{0.2cm}
\noindent\textbf{Mathematics Subject Classification(2010):} 53C42; 53C55

\tableofcontents

\section{Introduction}
The study of minimal constant curved two-spheres in symmetric spaces has a long history. E.Calabi \cite{[Ca]} and do Carmo and Wallach \cite{[DW]} proved that
any isometric minimal immersion from $S^2_K$ into $S^n(1)$ is
congruent to a linearly full one in $S^{2m}(1)$ with
$K=2/m(m\!+\!1)$ for some positive integer $m$. S. Bando, Y.
Ohnita \cite{[BO]} and J. Bolton, G.R.
Jensen, M. Rigoli, L. M. Woodward \cite{[BJRW]} proved that the linearly full minimal $S^2_K$ in
$\textbf{CP}^n$ must be one of the Veronese surfaces with
$K=4/(n\!+\!2j(n\!-\!j))$ for some integer $0\leq\! j\!\leq n$. These perfect works lead to the study of minimal two-spheres with constant curvature in symmetric spaces. For example,
Zh.Q.Li and Zh.H.Yu \cite{[LY]} classified the minimal constant curved two-spheres in the complex Grassmann manifold $G(2,4)$, X.X.Jiao and C.K.Peng \cite{[JP]} classified the holomorphic ones in $G(2,5)$. L.Delisle, V.Hussin and W.J.Zakrzewski proved \cite{[DHZ]} \cite{[DHZ2]} these two results again from the viewpoint of the Grassmannian sigma-model, and they also proposed conjectures w.r.t. the holomorphic constant curved two-spheres in complex Grassmannians. Recently, C.K.Peng and X.W.Xu \cite{[PX]} completely classified the homogenous minimal two-spheres in the complex Grassmann manifold $G(2,n)$, and C.K.Peng, J.Wang and X.W.Xu \cite{[PJX]} completely classified the homogenous minimal ones in the complex hyperquadric $Q_n$.

Naturally, one can consider the corresponding problem when the target space is quaternionic projective space $\textbf{HP}^n$. We call a minimal immersion $f$ from $S^2$ into $\textbf{HP}^n$ is \emph{proper} means that $f(S^2)$ is not contained in any totally geodesic submanifold of $\textbf{HP}^n$. Without loss of generality, we only consider the proper ones in this paper. Base on the works of homogeneous harmonic maps into projective space, Y.Ohnita \cite{[Oh]} proposed the following conjecture:
\begin{conj}
Let $f$ be a proper minimal constant curved immersion from $S^2$ into $\textbf{\emph{HP}}^n$, then $f$ is congruent to one of $\{f_\lambda\;|\;\lambda=1,3,\ldots,2n+1\}$, where $f_\lambda$ is defined in the remark of Proposition $4.1$.
\end{conj}

By using the theory of harmonic sequences, L.He and X.X.Jiao \cite{[HJ]} proved that this conjecture is true for $n=2$ under the totally unramified condition. So far, there is no breakthrough towards this conjecture for general $n>2$. Notice that the homogeneous submanifold is a class of important submanifolds. So, the classification of homogenous two-spheres in quaternionic projective space are helpful in understanding this conjecture. In this paper we completely classify the homogeneous two-spheres in $\textbf{HP}^n$. In particular, we obtain all minimal homogeneous two-spheres in $\textbf{HP}^n$. Namely, we have
\begin{theo}
Let $f:S^2\longrightarrow\textbf{\emph{HP}}^n$ be a proper homogeneous minimal immersion. Then, in terms of homogeneous coordinates, $f$ is congruent to one of the following\emph{:}

\emph{(1)} $f_\lambda:[a,b]\mapsto [\phi_{\lambda,n+1}],$ $\lambda\in\{3,5,\ldots,2n+1\}$, and the Gauss curvature $K=8/[(2n+2)^2-(\lambda^2+1)];$

\emph{(2)} $f_1:[a,b]\mapsto[\phi_{1,n+1}]$ and the Gauss curvature $K=4/[n(n+2)];$

\emph{(3)} $f_{\lambda,m,t}:[a,b]\mapsto[\cos t\phi_{\lambda,m},\texttt{\emph{i}}\sin t\phi_{\lambda,m}]$ for some positive weight $\lambda\in\{3,5,\ldots,n\}$, $t\in(0,\pi/2)$, $2m=n+1$, and the Gauss curvature $K=8/[(n+1)^2-(\lambda^2+1)];$

\emph{(4)} $f_{m_1,m_2}:[a,b]\mapsto[\sqrt{m_1/(m_1+m_2)}\;\phi_{1,m_1},
\texttt{\emph{i}}\sqrt{m_2/(m_1+m_2)}\;\phi_{1,m_2}]$ for some positive $m_1\leq m_2$ so that $m_1+m_2=n+1$ is an even, and the Gauss curvature $K=4/(m_1^2+m_2^2-1);$

\emph{(5)} $f'_{m_1,m_2}:[a,b]\mapsto[\sqrt{m_1/(m_1+m_2)}\;\phi_{1,m_1},
\sqrt{m_2/(m_1+m_2)}\;\phi_{1,m_2}]$ for some positive $m_1<m_2$ so that $m_1+m_2=n+1$ is an odd, and the Gauss curvature $K=4/(m_1^2+m_2^2-1)$,
where $\phi_{\lambda,m}$ is defined in the end of Section 2.1.
\end{theo}
\noindent All these minimal two-spheres obtained are not congruent to each other, up to a rigidity of $\textbf{HP}^n$. So, by comparing Conjecture 1.1 with Theorem 1.2, we obtain three families of minimal two-spheres more than Y.Ohnita conjectured. Particularly, there is a family of minimal two-spheres in $\textbf{HP}^n$ which depend on a parameter $t\in(0,\pi/2)$. This phenomenon appeared in Theorem 1.2 is quite different from the case of complex projective space (see \cite{[BO]}, \cite{[BJRW]}), due to the non-commutativity of quaternion.

In this paper we first show that there exists a quaternionic representation $\rho$ of $SU(2)$ associated to each homogeneous immersion $f$ from $S^2$ into $\textbf{HP}^n$ so that $f(S^2)$ is a $\rho(SU(2))$-orbit in $\textbf{HP}^n$. Then, in order to characterize the geometry of such orbit, we find the "best" base point in it. That is, we prove that such an orbit must contain a point $P_0$
spanned by vectors belonging to a weight space of the quaternionic representation $\rho$. To study the minimal orbits, we obtain a minimality criterion by using the method of moving frame which is inspired from S.S.Chern and J.G.Wolfson's work \cite{[CW2]}.
By using this criterion, up to a rigidity of $\textbf{HP}^n$, we show that the representation
$\rho$ is a direct sum of two different irreducible representations at the most. Thus, we completely classify the homogenous and minimal homogeneous two-spheres in $\textbf{HP}^n$, see Theorem 3.3, Theorem 4.4 respectively.

\section{Preliminaries}
\subsection{The quaternionic representation of $SU(2)$}\label{sec2}
For completeness, we review some facts on the unitary and quaternionic representations of the special unitary group $SU(2)$.

The special unitary group $SU(2)$ is defined by
\begin{equation*}
SU(2)=\left\{g=\left(
          \begin{array}{cc}
            a & b \\
            -\bar{b} & \bar{a} \\
          \end{array}
        \right)
 : |a|^2+|b|^2=1,\;\; a,b\in \textbf{C} \right\}.
\end{equation*}
Its Lie algebra $\mathfrak{su}(2)$ is real spanned by
\begin{equation*}
    \varepsilon_1=\left(
                     \begin{array}{cc}
                        \texttt{i}&0\\
                         0&-\texttt{i}\\
                     \end{array}
                  \right),\;
    \varepsilon_2=\left(
                     \begin{array}{cc}
                         0&1\\
                         -1&0\\
                     \end{array}
                  \right),\;
    \varepsilon_3=\left(
                     \begin{array}{cc}
                         0&\texttt{i}\\
                         \texttt{i}&0\\
                     \end{array}
                  \right),
\end{equation*}
where $\texttt{i}^2=-1$.
The Maurer-Cartan forms of $SU(2)$ is determined by
\begin{equation}\label{2.1}
\Theta:=dgg^{-1}=\left(
                         \begin{array}{cc}
                           \texttt{i}\omega & \varphi\\
                           -\bar{\varphi}  &-\texttt{i}\omega \\
                         \end{array}
                       \right),
\end{equation}
where $\omega$, $\varphi$ are real and complex one-forms
respectively. The Maurer-Cartan equation is given by
\begin{equation*}
d\Theta=\Theta\wedge\Theta,
\end{equation*}
which implies
\begin{equation}\label{2.2}
d(\texttt{i}\omega)=-\varphi\wedge\bar{\varphi},\;\;\;d\varphi=(2\texttt{i}\omega)\wedge\varphi.
\end{equation}
Let
$T=\big\{\mbox{diag}\{e^{\texttt{i}t},e^{\texttt{-i}t}\}\;|\;t\in\mathbb{R}\big\}$
be a maximal torus subgroup of $SU(2)$. It is known that the set
$SU(2)/T=\{[g]=Tg\;|\;g\in SU(2)\}$ is the complex projective space
$\textbf{CP}^1$, which is diffeomorphic to $S^2$. The canonical
metric $\varphi\bar\varphi$ on $\textbf{CP}^1$ has constant curvature $4$ by
(\ref{2.2}).

Let $V_n$ be the $(n\!+\!1)$-dimensional complex vector space of all
complex homogeneous polynomials of degree $n$ w.r.t. the two complex
variables $z_0$ and $z_1$. We define a Hermitian inner product
$(\;,\;)$ on $V_n$ by
\begin{equation*}
( f,g):=\sum\limits_{k=0}^n\;k!(n-k)!\;a_k\bar{b}_k,
\end{equation*}
for $f=\sum\limits_{k=0}^na_kz_0^kz_1^{n-k}$, $g=\sum\limits_{k=0}^nb_kz_0^kz_1^{n-k}\in V_n$. So, we know
$\big\{\textbf{v}_{k,n}=z_0^kz_1^{n-k}/\sqrt{k!(n-k)!}\;\big|\;0\leq k\leq
n\big\}$
is a unitary basis for $V_n$. A unitary representation $\rho_n$ of
$SU(2)$ on $V_n$ is defined by
\begin{equation*}
\rho_n(g)f(z_0,z_1):=f((z_0,z_1)g^{-1})=
f(\bar{a}z_0+\bar bz_1,-bz_0+az_1)
\end{equation*}
for $g\in SU(2)$ and $ f\in V_n$. Under the basis $\{\textbf{v}_{k,n}:0\leq k\leq n\}$, we get a matrix representation $\rho_n:SU(2)\longrightarrow U(n+1)$, $g\mapsto \rho_n(g)$, and $\rho_n(g)$ is described by
$\textbf{v}_{k,n}\rho_n(g):=\sum\limits_{k=0}^n\Lambda_{l\;k}(g)\textbf{v}_{l,n}$,
where
$$\Lambda_{l\;k}(g)=\sqrt{\frac{l!(n-l)!}{k!(n-k)!}}\sum\limits_{p+q=l}
\left(
  \begin{array}{c}
    k\\
    p \\
  \end{array}
\right)\left(
         \begin{array}{c}
           n-k \\
           q \\
         \end{array}
       \right)a^{n-k-q}(-b)^{q}\bar{a}^p\bar{b}^{k-p}
$$
which satisfies $\Lambda_{l\;k}(g)=(-1)^{k+l}\Lambda_{n-l\;n-k}(g)$.
The action of
$\mathfrak{su}(2)$ on $V_n$ is as follows:
\begin{equation}\label{2.3}
\textbf{v}_{k,n}d\rho_{n}(\varepsilon):=\frac{d}{dt}
\big(\rho_n(\exp{t\varepsilon})(\textbf{v}_{k,n})\big)\big|_{t=0},
\end{equation}
for $0\leq k\leq n$ and any element
$\varepsilon\in\mathfrak{su}(2)$. In particular, when $\varepsilon$ takes
$\varepsilon_1$, $\varepsilon_2$, $\varepsilon_3$ respectively, we
have
\begin{equation}\label{2.4}
\textbf{v}_{k,n}d\rho_{n}(\varepsilon_1)=(n-2k)\texttt{i}\textbf{v}_{k,n},
\end{equation}
\begin{equation}\label{2.5}
\textbf{v}_{k,n}d\rho_{n}(\varepsilon_2)=-a_{k-1,n}\textbf{v}_{k-1,n}
+a_{k,n}\textbf{v}_{k+1,n},
\end{equation}
\begin{equation}\label{2.6}
\textbf{v}_{k,n}d\rho_{n}(\varepsilon_3)=
-a_{k-1,n}\texttt{i}\textbf{v}_{k-1,n}-a_{k,n}\texttt{i}
\textbf{v}_{k+1,n},
\end{equation}
for $0\leq k\leq n$ and $a_{k,n}=\sqrt{(k+1)(n-k)}$. Set
$$\lambda_k:=n-2k,\;\;\;\;\;\textbf{v}_{\lambda_k,n}:=\textbf{v}_{k,n},\;\;\;\;
\;a_{\lambda_k,n}=a_{k,n}=\sqrt{[(n+1)^2-(\lambda_k-1)^2]}\big/2.$$
An element in $\Delta_n=\{\lambda_0,\ldots,\lambda_n\}$ is called a weight of $\rho_n$, and $\lambda_0$ is called the highest weight. The representation space $V_n$ has an orthonormal decomposition w.r.t. the Hermitian inner product $(\;,\;)$, i.e., $V_n=V_{\lambda_0}\oplus\cdots\oplus V_{\lambda_n}$, where $V_{\lambda_k}=\mbox{Span}_{\textbf{C}}\{\textbf{v}_{\lambda_k,n}\}$ is called the weight space w.r.t. the weight $\lambda_k$.

Suppose $n=2m-1\in\textbf{N}^+$ is an odd number. Define $\textbf{u}_{\lambda_k,n}$, $\textbf{u}_{-\lambda_k,n}$ as follows
\begin{equation}\label{2.7}
\textbf{u}_{\lambda_k,n}:=\textbf{v}_{\lambda_k,n},\;\;\;
\textbf{u}_{-\lambda_k,n}:=(-1)^k\texttt{i}\textbf{v}_{-\lambda_k,n},
\end{equation}
where $0\leq k\leq m-1$. Under the basis $\{\textbf{u}_{\lambda_0,n},\ldots,\textbf{u}_{\lambda_{m-1},n},
\textbf{u}_{-\lambda_0,n},\ldots,
\textbf{u}_{-\lambda_{m-1},n}\}$, for every $g\in SU(2)$, one can check that $\rho_n(g)$ satisfies
\begin{equation}\label{2.8}
J\;\rho_n(g)=\overline{\rho_n(g)}\;J,\;\;\;\mbox{for}\;\;\;J=\left(
                                               \begin{array}{cc}
                                                 0 & -I_m \\
                                                 I_m & 0 \\
                                               \end{array}
                                             \right).
\end{equation}
We identify $V_{2m-1}$ with $\textbf{H}^m$ by
\begin{equation}\label{2.9}
\textbf{v}=\sum\limits_{k=0}^{m-1}(a_k\textbf{u}_{\lambda_k,n}+b_k
\textbf{u}_{-\lambda_k,n})\mapsto\big(a_0+b_0\texttt{j},\ldots,a_{m-1}+
b_{m-1}\texttt{j}\big),
\end{equation}
where $\texttt{j}\in\textbf{H}$ and $\texttt{j}^2=-1$. From this identification, it is convenient to define
\begin{equation}\label{2.10}
{\texttt{j}\textbf{u}_{\lambda_k,n}:=\textbf{u}_{-\lambda_k,n}}.
\end{equation}
 Thus, an element in $V_{2m-1}$ can be written as $\textbf{v}=\sum\limits_{k=0}^{{m-1}}(a_k+b_k{
 \texttt{j}}){\textbf{u}_{\lambda_k,n}}$. When $V_{2m-1}$ is viewed as a quaternionic linear space, the property (\ref{2.8}) ensures us obtain a quaternionic representation of $SU(2)$ denoted by $\rho_m$. In terms of matrix, we have
\begin{equation}\label{2.11}
\rho_m:SU(2)\longrightarrow Sp(m),\;\;\;g\mapsto \Big(\Xi_{k\;l}(g):=\Lambda_{k\;l}(g)+(-1)^{k+1}\texttt{i}\Lambda_{k\;2m-1-l}(g)\texttt{j}\Big),
\end{equation}
where $0\leq k,l\leq m-1$ and $Sp(m)=\big\{A\in M(m,\textbf{H}):AA^*=I_m\big\}$ is the unitary symplectic group.

It's well known that $\big\{(V_n,\rho_n):n=0,1,2,\cdots\big\}$ are all inequivalent unitary representations of $SU(2)$. By the Theorem 6.3 in \cite{[BD]}, we know that $\big\{(V_{2m-1},\rho_m):m=1,2,\ldots\big\}$ are all inequivalent proper unitary quaternionic representations of $SU(2)$.
Since any quaternionic representation $(V,\rho)$ is completely reducible, for a proper one, we can write
\begin{equation}\label{2.12}
V=\bigoplus\limits_{\alpha=1}^s\;V_{2m_\alpha-1},\;\;\;\rho=\bigoplus
\limits_{\alpha=1}^s\;\rho_{m_\alpha}.
\end{equation}
In terms of weights, under the natural inner product, we also have the orthogonal decomposition
\begin{equation}\label{2.13}
V=\bigoplus\limits_{\lambda>0}\;U_\lambda=\bigoplus\limits_{\lambda>0}
\Big(\bigoplus\limits_\alpha\;U_{\lambda,\alpha}\Big),\;\;\;U_{\lambda,\alpha}
=V_{\lambda,2m_\alpha-1}\oplus V_{-\lambda,2m_\alpha-1}.
\end{equation}
Here, the "orthogonal decomposition" is related to the inner product $(\;,\;)$ defined by
\begin{equation*}
(\textbf{z},\textbf{w}):=\sum\limits_{k=0}^n\;p_k\bar q_k,
\end{equation*}
for $\textbf{z}=\sum\limits_{k=0}^np_k\textbf{u}_{\lambda_k,n}$,
$\textbf{w}=\sum\limits_{k=0}^nq_k\textbf{u}_{\lambda_k,n}$.
Notice that $d\rho$ is a real representation of $\mathfrak{su}(2)$, we extend naturally $d\rho$ to be a complex representation of $\mathfrak{sl}(2;\textbf{C})$, which is still denoted by $d\rho$. Set
$$\sigma_1=\left(
             \begin{array}{cc}
               1 & 0 \\
               0 & -1 \\
             \end{array}
           \right),\;\;\;\sigma_2=\left(
                                    \begin{array}{cc}
                                      0 & 1 \\
                                      0 & 0 \\
                                    \end{array}
                                  \right),\;\;\;
                                  \sigma_3=\left(
                      \begin{array}{cc}
                        0 & 0 \\
                        1 & 0 \\
                      \end{array}
                    \right),$$
which forms a basis of $\mathfrak{sl}(2;\textbf{C})$. In the sequel, we denote $d\rho(\sigma_1),d\rho(\sigma_2),d\rho(\sigma_3)\in End(V)$ by $H,A,B$ respectively. Alternatively, they are determined by
\begin{equation}\label{2.14}
H=-\texttt{i}d\rho(\varepsilon_1),\;\;\;
A=\frac{d\rho(\varepsilon_2)-\texttt{i}d\rho(\varepsilon_3)}{2},\;\;\;
B=-\frac{d\rho(\varepsilon_2)+\texttt{i}d\rho(\varepsilon_3)}{2},
\end{equation}
which satisfy
\begin{equation}\label{2.15}
[H,A]=2A,\;\;\;[H,B]=-2B,\;\;\;[A,B]=H.
\end{equation}
It follows from (\ref{2.4})-(\ref{2.7}), (\ref{2.10}) and (\ref{2.14}), we have
\begin{eqnarray}\label{2.16}
\textbf{u}_{\lambda_k,n_\alpha}H=\lambda_k\textbf{u}_{\lambda_k,n_\alpha},\;\;\;
(\texttt{j}\textbf{u}_{\lambda_k,n_\alpha})H=-\lambda_k(\texttt{j}
\textbf{u}_{\lambda_k,n_\alpha}),\;\;\mbox{for}\;\;0\leq k\leq m_\alpha-1,\nonumber\\
\textbf{u}_{\lambda_k,n_\alpha}A=-a_{\lambda_k,n_\alpha}\textbf{u}_{\lambda_k-2,
n_\alpha},\;\;\mbox{for}\;\;0\leq k\leq m_\alpha-2,\nonumber\\
{\textbf{u}_{\lambda_{m_\alpha-1},n_\alpha}A=-(-1)^{m_\alpha}\texttt{i}
a_{\lambda_{m_\alpha-1},n_\alpha}(\texttt{j}\textbf{u}_{\lambda_{m_\alpha-1},n_\alpha})},
\hspace{3.3cm}\nonumber\\
(\texttt{j}\textbf{u}_{\lambda_k,n_\alpha})A=a_{-\lambda_k,n_\alpha}(\texttt{j}
\textbf{u}_{\lambda_k,n_\alpha}),\;\;\mbox{for}\;\;0\leq k\leq m_\alpha-1,\\
\textbf{u}_{\lambda_k,n_\alpha}B=-a_{\lambda_k+2,n_\alpha}\textbf{u}_{\lambda_k+2,
n_\alpha},\;\;\mbox{for}\;\;0\leq k\leq m_\alpha-1,\nonumber\\
(\texttt{j}\textbf{u}_{\lambda_k,n_\alpha})B=a_{-\lambda_k+2,n_\alpha}
(\texttt{j}\textbf{u}_{\lambda_k-2,n_\alpha}),\;\;\mbox{for}\;\;0\leq k \leq m_\alpha-2,\nonumber\\
{(\texttt{j}\textbf{u}_{\lambda_{m_\alpha-1},n_\alpha})B=(-1)^{m_\alpha}
\texttt{i}a_{\lambda_{m_\alpha-1},n_\alpha} \textbf{u}_{\lambda_{m_\alpha-1},n_\alpha}}\nonumber.\hspace{3.3cm}
\end{eqnarray}
The pull-back of Maurer-Cartan forms of $Aut(V)$ is
\begin{equation}\label{2.17}
d\rho\rho^{-1}=(\texttt{i}\omega)H+\varphi A-\bar{\varphi}B.
\end{equation}
It follows that
\begin{equation}\label{2.-17}
\textbf{z}d\rho=\big((\texttt{i}\omega)(\textbf{z}H)+\varphi(\textbf{z} A)-\bar{\varphi}(\textbf{z}B)\big)\rho.
\end{equation}

Throughout this paper we will agree on the following conventions:
\begin{itemize}
\item $n_\alpha=2m_\alpha-1$, for $m_\alpha\in\textbf{N}^+$;

    \item
    $\Delta_\alpha:=\{n_\alpha,n_\alpha-2,\cdots,-(n_\alpha\!-\!2),-n_\alpha\}$;

    \item$a_{\lambda,\alpha}:=a_{\lambda,n_\alpha}$, and
$a_{\lambda,\alpha}=0$ if $\lambda\notin\Delta_\alpha$;

\item $\textbf{u}_{\lambda,\alpha}:=\textbf{u}_{\lambda,n_\alpha}$,
$V_{\lambda,\alpha}=\mbox{Span}_{\mathbb{C}}\{\textbf{u}_{\lambda,\alpha}\}$,
and $V_{\lambda,\alpha}=\{0\}$ if $\lambda\notin\Delta_\alpha$;

\item {$\phi_{\lambda,m}:=\textbf{u}_{\lambda,n}\rho_m(g)$, $g\in SU(2)$, for $\lambda>0$}.

  \end{itemize}

\subsection{Geometry of surfaces in quaternionic projective space $\textbf{HP}^n$}
Throughout this subsection we agree on the following ranges of indices:
\begin{equation*}
0\leq A,B,\ldots\leq n,\;\; 1\leq \alpha,\beta,\ldots\leq n.
\end{equation*}
The standard inner product $(\;,\;)$ of $\textbf{H}^{n+1}$ is defined by
\begin{equation*}
(\textbf{z},\textbf{w})=\sum\limits_{A}z_A\bar w_A=\sum\limits_{A}
z_Aw_{\bar A},
\end{equation*}
for $\textbf{z}=(z_0,z_1,\ldots,z_n)$, $\textbf{w}=(w_0,w_1,\ldots,w_n)\in\textbf{H}^{n+1}$. A unitary frame of $\textbf{H}^{n+1}$ is an ordered set of $n+1$ linearly independent vectors $Z_0,Z_1,\ldots,Z_n$ satisfying
\begin{equation*}
(Z_A,Z_B)=\delta_{AB}.
\end{equation*}
Taking the exterior derivative of $Z_A$, we have
\begin{equation}\label{2.18}
dZ_A=\sum\limits_{B}\omega_{A\bar B}\,Z_B,
\end{equation}
where $\omega_{A\bar B}$ are $\textbf{H}$-valued one forms. We identify the space of unitary frames with the unitary symplectic group $Sp(n+1)$. Then, $\omega_{A\bar B}$ are the Maurer-Cartan forms of $Sp(n+1)$, which satisfy the Maurer-Cartan equations:
\begin{equation}\label{2.19}
d\omega_{A\bar B}=\sum\limits_{C}\omega_{A\bar C}\wedge\omega_{C\bar B}.
\end{equation}

The quaterionic projective space $\textbf{HP}^n$ is the set of all one-dimensional subspaces in $\textbf{H}^{n+1}$. An element of $\textbf{HP}^n$ can be defined by the unitary vector $Z_0$,  up to a factor of absolute 1. Its orthogonal vectors $Z_\alpha$ are defined up to a transformation of $Sp(n)$. So, $\textbf{HP}^{n}$ has a $Sp(1)\times Sp(n)$-structure. Therefore, the form
\begin{equation}\label{2.20}
ds^2=\sum\limits_{\alpha}\omega_{0\bar\alpha}\,\omega_{\bar 0\alpha}
\end{equation}
defines a Riemannian metric on $\textbf{HP}^{n}$.

Let $M$ be an oriented Riemnnian surface and let $f:M\longrightarrow
\textbf{HP}^{n}$ be a smooth immersion. The induced metric on $M$ by
$ds^2_M=\varphi\bar\varphi$, where $\varphi$ is complex-valued one-form. $\varphi$ is defined up to a complex factor of absolute value 1.
The first structure equation of $ds^2_M$ is given by
\begin{equation}\label{2.21}
d\varphi=\eta\wedge\varphi,
\end{equation}
where the one-form $\eta$ is purely imaginary. The Gauss curvature $K$ is described by
\begin{equation}\label{2.22}
d\eta=-\frac{K}{2}\varphi\wedge\bar{\varphi}.
\end{equation}
Locally, we have
\begin{equation}\label{2.23}
f^*\omega_{0\bar{\alpha}}=\varphi\,a_{\alpha}+\bar\varphi\,
b_{\alpha},
\end{equation}
where $a_{\alpha}$, $b_{\alpha}$ are local $\textbf{H}$-valued functions. Taking the exterior derivative on both sides of (\ref{2.23}) and using (\ref{2.19}), (\ref{2.21}) and (\ref{2.22}), we obtain
\begin{equation}\label{2.24}
\varphi\wedge Da_{\alpha}+\bar{\varphi}\wedge Db_{\alpha}=0,
\end{equation}
where
\begin{equation}\label{2.25}
Da_{\alpha}=da_{\alpha}+\eta a_{\alpha}-\theta_{0\bar 0}a_{\alpha}-\theta'_{0\bar 0}b_{\alpha}+a_{\beta}\omega_{\beta\bar\alpha},
\end{equation}
\begin{equation}\label{2.26}
Db_{\alpha}=db_{\alpha}-\eta b_{\alpha}-\theta'_{0\bar 0}a_{\alpha}-\theta_{0\bar 0}b_{\alpha}+b_{\beta}\omega_{\beta\bar\alpha},
\end{equation}
and
\begin{equation}\label{2.27}
\theta_{0\bar 0}=\frac{\omega_{0\bar 0}-\texttt{i}\omega_{0\bar 0}\texttt{i}}{2},\;\;\;\theta'_{0\bar 0}=\frac{\omega_{0\bar 0}+\texttt{i}\omega_{0\bar 0}\texttt{i}}{2}.
\end{equation}
Because $\textbf{HP}^{n}$ is non-Kahlerian, the definition
of $Da_{\alpha}$ (resp. $Db_{\alpha}$) involves $b_{\alpha}$ (resp. $a_{\alpha}$). By (\ref{2.24}), we can set
\begin{equation}\label{2.28}
Da_{\alpha}=\varphi\,p_{\alpha}+\bar{\varphi}\,q_{\alpha},
\;\;\;\;\;
Db_{\alpha}=\varphi\,q_{\alpha}+\bar{\varphi}\,r_{\alpha}.
\end{equation}
Similar to \cite{[CW2]}, $f$ is minimal if and only if
$q_{\alpha}=0$, i.e.,
\begin{equation}\label{2.29}
Da_{\alpha}\equiv0,\mod\varphi\;\;\;\;\text{or}\;\;\;\;
Db_{\alpha}\equiv0,\mod\bar\varphi.
\end{equation}
Set the local $\textbf{H}^{n+1}$-valued functions $X,Y$ as
\begin{equation}\label{2.30}
X=\sum\limits_{\alpha}a_{\alpha}Z_\alpha,\;\;\;\;
Y=\sum\limits_{\alpha}b_{\alpha}Z_\alpha,
\end{equation}
and define the covariant derivative of $X$ and $Y$ by
\begin{equation}\label{2.31}
DX=dX+\eta X-\theta_{0\bar 0}X-\theta'_{0\bar 0}Y,
\end{equation}
\begin{equation}\label{2.32}
DY=dY-\eta Y-\theta_{0\bar 0}X-\theta'_{0\bar 0}Y.
\end{equation}
We give a criterion to measure the minimality of $f$ in terms of $DX$ or $DY$, that is
\begin{prop}\label{proposition2.1}
The smooth immersion $f$ is minimal if and only if one of the following holds:

\emph{(a)} $DX\equiv0,\;\;\;\mod Z_0,\;\varphi;$

\emph{(b)} $DY\equiv0,\;\;\;\mod Z_0,\;\bar\varphi.$
\end{prop}

\emph{Proof}.
It follows from the facts
\begin{equation*}
DX=\sum\limits_{\alpha}Da_{\alpha}\;Z_\alpha+\sum\limits_{\beta}
a_{\beta}\omega_{\beta\bar 0}\;Z_0,\;\;\;
DY=\sum\limits_{\alpha}Db_{\alpha}\;Z_\alpha+\sum\limits_{\beta}
b_{\beta}\omega_{\beta\bar 0}\;Z_0.
\end{equation*}
This completes the proof. \hfill$\Box$

\section{Homogeneous two-spheres in $\textbf{HP}^n$}\label{sec3}
In this section we first determine the relations between the homogeneous two-spheres and the two-dimensional $\rho(SU(2))$-orbits in $\textbf{HP}^n$, where $\rho$ is a quaternionic representation of $SU(2)$. Then, all the two-dimensional
$\rho(SU(2))$-orbits in $\textbf{HP}^n$ are determined.

\begin{theo}\label{theorem 3.1}
Let $f:S^2\longrightarrow \textbf{\emph{HP}}^n$ be a homogeneous immersion, then
there exists a quaternionic representation $\rho$ of $SU(2)$ such that $f(S^2)$ is a
two-dimensional $\rho(SU(2))$-orbit in $\textbf{\emph{HP}}^n$.
\end{theo}

\emph{Proof}. It is similar to the proof of Theorem 3.2 in \cite{[FJXX]} or Theorem 3.3 in \cite{[PJX]}, we
omit the details.\hfill$\Box$

\vspace{0.3cm}

In general, a $\rho(SU(2))$-orbit in $\textbf{HP}^n$ is a principle orbit,
i.e., it is a three-dimensional orbit. For the two-dimensional
$\rho(SU(2))$-orbits, we have
\begin{lem}\label{lemma3.2}
Let $M$ be a $\rho(SU(2))$-orbit in $ \textbf{\emph{HP}}^n$, then $M$ is
two-dimensional if and only if there exists a point $P_0\in M$ s.t.
$P_0$ is invariant under the $\rho(T)$-action.
\end{lem}

\emph{Proof}. Let $P$ be a point in $M$. Since $dim M=2$, the isotropy group
$G$ at $P$ is a 1-dimensional subgroup of $SU(2)$. Therefore, there exists a nonzero
element $\varepsilon\in\mathfrak{su}(2)$ s.t.
$T_{I_2}G=\mbox{Span}_{\mathbb{R}}\{\varepsilon\}$. Notice that
$\varepsilon\in\mathfrak{su}(2)$, there exist a fixed $g\in
SU(2)$ s.t. $g^{-1}\varepsilon g= c \varepsilon_1$ for some
nonzero real number $c$. It follows that the isotropy group at
$P_0=P\rho(g)\in M$ contains the maximal torus subgroup $T$,
i.e., $P_0$ is invariant under $\rho(T)$-action. The sufficiency
follows from the fact that there is no two-dimensional subgroup in $SU(2)$.
\hfill$\Box$

\vspace{0.3cm}

\noindent\textbf{\emph{Remark}}. By taking derivative, the condition "$P_0$ is invariant under $\rho(T)$" is equivalent to
\begin{equation}\label{3.1}
H(P_0)\subseteq P_0,
\end{equation}
where $P_0$ is viewed as a linear subspace and $H$ is defined in (\ref{2.14}).

\vspace{0.3cm}

Let $\rho$ be a quaternonic representation of $SU(2)$. Then $\rho(SU(2))$ acts on $\textbf{HP}^n$ by
\begin{equation}\label{3.2}
SU(2)\times \textbf{HP}^n\longrightarrow \textbf{HP}^n,\;(g,P)\mapsto P\rho(g).
\end{equation}
Therefore, for a two-dimensional $\rho(SU(2))$-orbit
$M$ in $\textbf{HP}^n$, we can obtain an immersion $f$ from $S^2$ into
$\textbf{HP}^n$ as follows
\begin{equation}\label{3.3}
f:S^2\simeq SU(2)/T\longrightarrow \textbf{HP}^n,\; [g]\mapsto P_0\rho(g),
\end{equation}
where $P_0$ is a fixed point in $M$. According to Theorem \ref{theorem
3.1}, we only need to study the two-dimensional $\rho(SU(2))$-orbits
in $\textbf{HP}^n$ if we study the homogeneous immersions from
$S^2$ into $\textbf{HP}^n$.

\vspace{0.3cm}

Let $\rho=\rho_{m_1}\oplus\cdots\oplus\rho_{m_s}$ be a quaternionic
representation of $SU(2)$. Let $M$ be a linearly full two-dimensional $\rho(SU(2))$-orbit in $\textbf{HP}^n$. Suppose $P_0=[\textbf{z}]$ is a point obtained in
Lemma \ref{lemma3.2}. According to the decomposition
(\ref{2.13}), we can write
$$\textbf{z}=\sum\limits_{\lambda>0}\sum\limits_{\alpha}(c_{\lambda,\alpha}
+d_{\lambda,\alpha}\texttt{j})\textbf{u}_{\lambda,\alpha},\;\;\mbox{for}\;\;
c_{\lambda,\alpha},d_{\lambda,\alpha}\in\textbf{C}.$$
Without loss of generality, we further assume $d_{\lambda_0,\alpha_0}=0$ for some $\lambda_0$ and $\alpha_0$. If not, we use $(\bar{c}_{\lambda_0,\alpha_0}-d_{\lambda_0,\alpha_0}\texttt{j})\textbf{z}$ instead of $\textbf{z}$. The condition (\ref{3.1}) reduces to
\begin{equation}\label{3.4}
\textbf{z}H=p\textbf{z},
\end{equation}
for some fixed $p=p_1+p_2\texttt{j}\in \textbf{H}$. In terms of coefficients, (\ref{3.4}) is equivalent to
\begin{equation}\label{3.5}
\lambda c_{\lambda,\alpha}=p_1c_{\lambda,\alpha}-p_2\bar{d}_{\lambda,\alpha},\;\;\;
-\lambda d_{\lambda,\alpha}=p_1d_{\lambda,\alpha}+p_2\bar{c}_{\lambda,\alpha},
\end{equation}
for all $\lambda,\alpha$. Notice that $d_{\lambda_0,\alpha_0}=0$, it follows from (\ref{3.5}) we have $p_1=\lambda_0$, $p_2=0$. This implies
\begin{equation}\label{3.6}
\lambda=p_1=\lambda_0,\;\;
d_{\lambda,\alpha}=0, \;\;\mbox{for all}\;\; \lambda,\alpha,
\end{equation}
by the linearly full assumption. Thus, we have
\begin{equation}\label{3.7}
\textbf{z}=\sum\limits_{\alpha}c_\alpha\textbf{u}_{\lambda,\alpha},
\;\;\;c_\alpha\in\textbf{C},
\end{equation}
for a positive weight $\lambda$.

In summery, we completely classify all homogeneous two-spheres in $\textbf{HP}^{n}$. That is
\begin{theo}\label{theorem3.3}
Let $\rho=\rho_{m_1}\oplus\cdots\oplus\rho_{m_s}$, $m_1+\cdots+m_s=n+1$, be a quaternionic representation of $SU(2)$. If $M$ is a two-dimensional $\rho(SU(2))$-orbit in $\textbf{\emph{HP}}^{n}$, then there exists a point $[\textbf{\emph{z}}]\in M$ and a positive weight $\lambda$ s.t.
$$\textbf{\emph{z}}=\sum\limits_{\alpha}c_\alpha
\textbf{\emph{u}}_{\lambda,\alpha},\;\;\;c_\alpha\in\textbf{\emph{C}}.$$

\end{theo}

\vspace{0.3cm}

\section{Minimal homogeneous two-spheres in $\textbf{HP}^n$}\label{sec4}

Let $\rho=\rho_{m_1}\oplus\cdots\oplus\rho_{m_s}$, $m_1+\cdots+m_s=n+1$ be a quaternionic representation of $SU(2)$, and let $M$ be a two-dimensional $\rho(SU(2))$-orbit in $\textbf{HP}^n$. Suppose $[\textbf{z}]\in M$ is the point we get in Theorem \ref{theorem3.3}, i.e., $\textbf{z}=\sum\limits_{\alpha}\;c_\alpha\textbf{u}_{\lambda,\alpha}$, $c_\alpha\in\textbf{C}$, for some positive weight $\lambda$. We further assume $|\textbf{z}|^2=1$. Set
\begin{equation}\label{4.1}
Z_0=\textbf{z}\rho.
\end{equation}
Taking the exterior derivative of (\ref{4.1}) and using (\ref{2.16}), (\ref{2.17}), we have
\begin{eqnarray}\label{4.2}
dZ_0=\big(\lambda(\texttt{i}\omega)\textbf{z}+\varphi\textbf{z}A-\bar{\varphi}\textbf{z}B\big)\rho.
\end{eqnarray}
We define
\begin{equation}\label{4.3}
dZ_0\equiv \varphi X+\bar{\varphi} Y,\;\;\; \mod Z_0.
\end{equation}
Notice that $\textbf{z}\in U_{\lambda}$, $\textbf{z}A\in U_{\lambda-2}$ for $\lambda>1$, $\textbf{z}A\in U_{\lambda}$ for $\lambda=1$,  $\textbf{z}B\in U_{\lambda+2}$, we have
\begin{equation}\label{4.4}
X=\textbf{z}A\rho-\ell\textbf{z}\rho,\;\;\;
Y=-\textbf{z}B\rho,
\end{equation}
where $\ell=(\textbf{z}A,\textbf{z})$. It's clear that $\ell=0$ when $\lambda>1$, and $\ell$ has the possibility to be nonzero when $\lambda=1$.
The fact $(X,Y)=0$ means that the orbit $M$ is conformal in $\textbf{HP}^n$, and the induced metric is
\begin{equation}\label{4.5}
(|X|^2+|Y|^2)\;\varphi\bar\varphi.
\end{equation}
Notice that $|X|^2+|Y|^2$ is a constant and the metric $\varphi\bar{\varphi}$ has the curvature 4, we know that the induced metric (\ref{4.5}) has constant curvature
\begin{equation}\label{4.6}
K=\frac{4}{|X|^2+|Y|^2}.
\end{equation}

Now we give a simple expression of the criteria (a) in Proposition \ref{proposition2.1} for later use. Notice that
$\eta=2\texttt{i}\omega$ by (\ref{2.2}), and substituting (\ref{4.2}) into (\ref{2.27}), we have $\theta_{0\bar 0}=\texttt{i}\omega$, $\theta'_{0\bar 0}=\varphi\ell$. From the definition (\ref{2.31}), we get
\begin{equation}\label{4.7}
DX=\varphi\Big(\textbf{z}A^2-(\ell\textbf{z})A-(\ell\textbf{z})B)\Big)\rho
-\bar\varphi\Big(\textbf{z}AB-(\ell\textbf{z})B\Big)\rho.
\end{equation}
Thus, according to Proposition \ref{proposition2.1}, $M$ is minimal in
$\textbf{HP}^{n}$ if and only if
\begin{equation}\label{4.8}
\textbf{z}AB-(\ell\textbf{z})B=p\textbf{z}
\end{equation}
for fixed $p=p_1+p_2\texttt{j}\in\textbf{H}$. We will study the geometry of  minimal orbits in following two cases: $\lambda>1$ and $\lambda=1$.

\vspace{0.3cm}

\noindent\textbf{Case I.} \hspace{0.3cm} Suppose that $\lambda>1$, and hence $\ell=0$. From (\ref{2.16}), by comparing the coefficients, (\ref{4.7}) is equivalent to
\begin{equation}\label{4.9}
a_{\lambda,\alpha}^2=p,\;\;\;\mbox{for all}\;\; \alpha.
\end{equation}
This implies that $n_1=\cdots=n_s=:n_0=2m_0-1$. So, the associated quaternionic representation $\rho$ takes the form $\rho=\rho_{m_0}\oplus\cdots\oplus\rho_{m_0}$, and the corresponding minimal immersion is given by
\begin{equation}\label{4.10}
[a,b]\mapsto[c_1\phi_{\lambda,m_0},\ldots,c_s\phi_{\lambda,m_0}],
\end{equation}
where $c_\alpha\in\textbf{C}$ are nonzero.
\begin{prop}\label{proposition4.1}
Up to an isometry of $\textbf{\emph{HP}}^{n}$, the immersion \emph{(\ref{4.10})} is congruent to
\begin{equation}\label{4.11}
f_{\lambda,m_0,\;t}:[a,b]\mapsto [\cos t\;\phi_{\lambda,m_0},\texttt{\emph{i}}\sin t\;\phi_{\lambda,m_0},0,\ldots,0],
\end{equation}
where $t\in [0,\pi/2).$
\end{prop}

\emph{\textbf{Remark}}. If the parameter $t=0$ and $f$ is proper, we obtain the immersion
\begin{equation*}
f_{\lambda}:S^2\longrightarrow \textbf{HP}^n,\hspace{0.5cm} [a,b]\mapsto[\phi_{\lambda,n+1}],
\end{equation*}
which firstly appeared in \cite{[Oh]}.

\vspace{0.3cm}

\emph{Proof}. Set $\textbf{c}:=(c_1,\ldots,c_s)$. Notice that the choice of $\textbf{c}$ is unique up to a factor (multiplied on the left) of absolute 1, without loss of generality, we can find a matrix $\textbf{T}\in SO(s)$ s.t. $\textbf{c}\;\textbf{T}=(\cos t,\texttt{i}\sin t,0,\ldots,0)$. Then the proposition holds by choosing the isometry $\textbf{T}\otimes I_{m_0}$ of $\textbf{HP}^{n}$. \hfill$\Box$

\vspace{0.3cm}
Next, we calculate the Gauss curvature of the immersion (\ref{4.11}). By (\ref{2.16}) and (\ref{4.4}), we obtain
$$|X|^2+|Y|^2=|\textbf{z}A|^2+|\textbf{z}B|^2=a^2_{\lambda,n_0}+a^2_{\lambda+2,n_0}.$$
So by (\ref{4.6}), the Gauss curvature is
\begin{equation}\label{4.12}
K=\frac{4}{a^2_{\lambda,n_0}+a^2_{\lambda+2,n_0}}=\frac{8}{(n_0+1)^2-(\lambda^2+1)}.
\end{equation}
\vspace{0.3cm}

\noindent\textbf{Case II.} \hspace{0.3cm}Suppose that $\lambda=1$. Since $\textbf{z}=\sum\limits_{\alpha=1}^sc_\alpha
\textbf{u}_{1,\alpha}$, $c_\alpha\in\textbf{C}$, we have $$\textbf{z}A=\sum\limits_{\alpha=1}^s\big((-1)^{m_\alpha+1}\texttt{i}c_{\alpha}a_{1,\alpha}
\texttt{j}\big)\textbf{u}_{1,\alpha}$$ by (\ref{2.16}), and then,
\begin{equation}\label{4.13}
(\textbf{z}A,\textbf{z})=\Big(\sum\limits_{\alpha=1}^s(-1)^{m_\alpha+1}
a_{1,\alpha}c_\alpha^2\texttt{i}\Big)\texttt{j}.
\end{equation}
From (\ref{4.13}), we find $\big((e^{\texttt{i}\tau}\textbf{z})A,e^{\texttt{i}\tau}\textbf{z}\big)=
e^{2\texttt{i}\tau}(\textbf{z}A,\textbf{z})$. So, we can further assume
$\ell=(\textbf{z}A,\textbf{z})=(\ell'\texttt{i})\texttt{j}$, where $\ell'\in\textbf{R}$. In terms of components, (\ref{4.8}) is equivalent to
\begin{equation}\label{4.14}
a_{1,\alpha}^2c_\alpha+(-1)^{m_\alpha}\ell'a_{1,\alpha}\bar c_\alpha
=p_1c_\alpha,\;\;\;\mbox{for all}\;\;\alpha.
\end{equation}
Splitting $c_\alpha$, $p_1$ into real and imaginary parts, i.e., $c_\alpha=
c'_\alpha+c''_{\alpha}\texttt{i}$, $p_1=p'_1+p''_1\texttt{i}$. Then, (\ref{4.14}) are equivalent to
\begin{equation}\label{4.15}
(a_{1,\alpha}^2+(-1)^{m_\alpha}\ell'a_{1,\alpha}-p_1')c_\alpha'+p_1''c_\alpha''=0,
\end{equation}
\begin{equation}\label{4.16}
-p_1''c_\alpha'+(a_{1,\alpha}^2-(-1)^{m_\alpha}\ell'a_{1,\alpha}-p_1')c_\alpha''=0,
\end{equation}
for all $\alpha$. Notice that $c_\alpha\neq 0$, we know that the determinant of the coefficient matrix of the system (\ref{4.15}) and
(\ref{4.16}) is equal to zero, which follows that $a_{1,\alpha}^2$ satisfy the real quadratic equation
\begin{equation*}
x^2-\big((\ell')^2+2p_1'\big)x+|p_1|^2=0.
\end{equation*}
This means that there are two different $m_\alpha$ at most. So, the associated representation $\rho$ can be written as
\begin{equation}\label{4.17}
\rho=\rho_{m_1}\oplus\cdots\oplus\rho_{m_1},
\end{equation}
or
\begin{equation}\label{4.18}
\rho=\rho_{m_1}\oplus\cdots\oplus\rho_{m_1}\oplus\rho_{m_2}
\oplus\cdots\oplus\rho_{m_2},\;\;\;m_1<m_2.
\end{equation}

\textbf{(II.I)} \hspace{0.3cm}If the associate representation $\rho$ takes form of (\ref{4.17}), in terms of homogeneous coordinates, the corresponding immersion is given by
\begin{equation}\label{4.19}
[a,b]\mapsto[c_1\phi_{1,m_1},\ldots,c_s\phi_{1,m_1}],
\end{equation}
where $c_\alpha\in\textbf{C}$.

\begin{prop}\label{proposition4.2}
Up to an isometry of $\textbf{\emph{HP}}^{n}$, the minimal immersion \emph{(\ref{4.19})}
is congruent to
\begin{equation}\label{4.20}
f_1:[a,b]\mapsto[\phi_{1,m_1},0,\ldots,0]
\end{equation}
or
\begin{equation}\label{4.21}
f_{m_1,\;m_1}:[a,b]\mapsto\frac{\sqrt{2}}{2}[\phi_{1,m_1},\texttt{\emph{i}}\;\phi_{1,m_1},
0,\ldots,0].
\end{equation}
\end{prop}

\emph{Proof}. For this subcase, we have $\ell'=(-1)^{m_1+1}a_{1,n_1}\sum\limits_{\alpha}c_{\alpha}^2$. Then the minimality condition (\ref{4.14}) can be written as
\begin{equation}\label{4.22}
(a_{1,n_1}^2-p_1)c_\alpha=a_{1,n_1}^2(\sum\limits_{\alpha}c_{\alpha}^2)\bar{c}_\alpha
\end{equation}
for all $\alpha$, where $\sum\limits_{\alpha}c_{\alpha}\bar{c}_\alpha=1$. Multiplying by $c_{\alpha}$ on both sides of (\ref{4.22}) and summating for all $\alpha$, we obtain $$(a_{1,n_1}^2-p_1)\sum\limits_{\alpha}c_{\alpha}^2=a_{1,n_1}^2(\sum\limits_{\alpha}c_{\alpha}^2),$$
which implies $p_1=0$ or $\sum\limits_{\alpha}c_{\alpha}^2=0$.

If $p_1=0$, then from (\ref{4.22}), we get $c_\alpha=(\sum\limits_{\alpha}c_{\alpha}^2)\bar{c}_\alpha$. Since we assume $\ell'$ is real, then $\sum\limits_{\alpha}c_{\alpha}^2$ is also real. Therefore, we must have $\sum\limits_{\alpha}c_{\alpha}^2=1$ and $c_\alpha$ are real for all $\alpha$. Now we can find a matrix $\textbf{T}\in SO(s)$ s.t. $(c_1,\cdots,c_s)\;\textbf{T}=(1,0,\ldots,0)$. Then the immersion (\ref{4.19})
is congruent to (\ref{4.20}) by choosing the isometry $\textbf{T}\otimes I_{m_1}$ of $\textbf{HP}^{n}$.

If $\sum\limits_{\alpha}c_{\alpha}^2=0$, i.e. $\ell'=0$, then the choosing of $\textbf{z}$ is still allowed a $U(1)$-transformation. Similar to the proof of Proposition \ref{proposition4.1}, the immersion (\ref{4.19}) is congruent to
\begin{equation*}
[a,b]\mapsto[\cos t\;\phi_{1,m_1},\texttt{i}\sin t\;\phi_{1,m_1},0,\ldots,0],
\end{equation*}
where $t\in[0,\pi)$. It follows from $\sum\limits_{\alpha}c_{\alpha}^2=0$ that $t=\pi/4$. Thus we complete the proof. \hfill$\Box$

\vspace{0.3cm}
Next, we calculate the Gauss curvatures of the immersions (\ref{4.20}) and (\ref{4.21}). For the immersion (\ref{4.20}), by (\ref{2.16}), (\ref{4.4}) and straightforward computation, we obtain
$|X|^2+|Y|^2=|\textbf{z}B|^2=a^2_{3,n_1}.$
So, by (\ref{4.6}), the Gauss curvature is
\begin{equation}\label{4.23}
K=\frac{4}{a^2_{3,n_1}}=\frac{16}{(n_1+1)^2-4}.
\end{equation}

For the immersion (\ref{4.21}), by $\ell'=0$, we have $|X|^2+|Y|^2=|\textbf{z}A|^2+|\textbf{z}B|^2=a^2_{1,n_1}+a^2_{3,n_1}.$
Then the Gauss curvature is
\begin{equation}\label{4.24}
K=\frac{4}{a^2_{1,n_1}+a^2_{3,n_1}}=\frac{8}{(n_1+1)^2-2}.
\end{equation}

\vspace{0.3cm}

\textbf{(II.II)}\hspace{0.3cm} If the associate representation $\rho$ takes form of (\ref{4.18}), we split $$\textbf{z}=\sum\limits_{\alpha=1}^{s_1}c_{\alpha}\textbf{u}_{1,n_1}+\sum\limits_{\beta=s_1+1}^sc_{\beta}\textbf{u}_{1,n_2},$$
where $c_\alpha,c_\beta\in\textbf{C}$ and $\sum\limits_{\alpha}c_{\alpha}\bar{c}_\alpha+\sum\limits_{\beta}c_{\beta}\bar{c}_\beta=1$. Then the corresponding immersion is given by
\begin{equation}\label{4.25}
[a,b]\mapsto[c_1\phi_{1,m_1},\ldots,c_{s_1}\phi_{1,m_1},
c_{s_1+1}\phi_{1,m_2},\ldots,c_{s}\phi_{1,m_2}].
\end{equation}

\begin{prop}\label{proposition4.3}
Up to an isometry of $\textbf{\emph{HP}}^{n}$, the minimal immersion \emph{(\ref{4.25})}
is congruent to
\begin{equation}\label{4.26}
f_{m_1,m_2}:[a,b]\mapsto\bigg[\sqrt{\frac{m_1}{m_1+m_2}}\;\phi_{1,m_1},\texttt{\emph{i}}
\sqrt{\frac{m_2}{m_1+m_2}}\;\phi_{1,m_2},
0,\ldots,0\bigg]
\end{equation}
when $m_1+m_2$ is an even,
or
\begin{equation}\label{4.27}
f'_{m_1,m_2}:[a,b]\mapsto\bigg[\sqrt{\frac{m_1}{m_1+m_2}}\;\phi_{1,m_1},
\sqrt{\frac{m_2}{m_1+m_2}}\phi_{1,m_2},0,\ldots,0\bigg]
\end{equation}
when $m_1+m_2$ is an odd, respectively, where $m_1<m_2$.
\end{prop}

\emph{Proof}. For this subcase, the minimality condition (\ref{4.14}) can be written as
\begin{eqnarray}
(a_{1,n_1}^2-p_1)c_\alpha=(-1)^{m_1+1}\ell'a_{1,n_1}\bar{c}_\alpha, \label{4.28} \\
(a_{1,n_2}^2-p_1)c_\beta=(-1)^{m_2+1}\ell'a_{1,n_2}\bar{c}_\beta, \label{4.29}
\end{eqnarray}
for all $\alpha,\ \beta$, where
\begin{equation}\label{4.30}
\ell'=(-1)^{m_1+1}a_{1,n_1}\sum\limits_{\alpha}c_{\alpha}^2+(-1)^{m_2+1}a_{1,n_2}\sum\limits_{\beta}c_{\beta}^2.
\end{equation}
We first claim that $\ell'\neq 0$. Otherwise, if $\ell'= 0$, then from (\ref{4.28}) and (\ref{4.29}), we get $p_1=a_{1,n_1}^2=a_{1,n_2}^2$, i.e. $m_1=m_2$, which leads to a contradiction.
Multiplying by $\bar{c}_{\alpha}$ and $\bar{c}_{\beta}$ on both sides of (\ref{4.28}) and (\ref{4.29}), respectively, and summating for all $\alpha$ and $\beta$, we obtain
\begin{eqnarray}
(a_{1,n_1}^2-p_1)\sum_{\alpha}c_\alpha\bar{c}_{\alpha}=(-1)^{m_1+1}\ell'a_{1,n_1}\sum_{\alpha}\bar{c}^2_\alpha, \label{4.31} \\
(a_{1,n_2}^2-p_1)\sum_{\beta}c_\beta\bar{c}_{\beta}=(-1)^{m_2+1}\ell'a_{1,n_2}\sum_{\beta}\bar{c}^2_\beta, \label{4.32}
\end{eqnarray}
By using the assumption $\ell'$ is real and adding (\ref{4.31}) and (\ref{4.32}), we have
$$p_1=a_{1,n_1}^2\sum_{\alpha}c_\alpha\bar{c}_{\alpha}+a_{1,n_2}^2\sum_{\beta}c_\beta\bar{c}_{\beta}-(\ell')^2,$$
so $p_1$ is also real. Therefore, it follows from (\ref{4.28}) and (\ref{4.29}) that $c_\alpha=\pm\bar{c}_\alpha$ for all $\alpha$ and $c_\beta=\pm\bar{c}_\beta$ for all $\beta$. Moreover, we have
$$a_{1,n_1}^2-p_1=\pm(-1)^{m_1+1}\ell'a_{1,n_1},\ \ a_{1,n_2}^2-p_1=\pm(-1)^{m_2+1}\ell'a_{1,n_2}.$$
Without lose of generality, we first consider $c_\alpha=\bar{c}_\alpha$ for all $\alpha$ and $c_\beta=\bar{c}_\beta$ for all $\beta$, i.e. $c_\alpha,c_\beta$ are all real. Then $\sum\limits_{\alpha}c_{\alpha}^2+\sum\limits_{\beta}c_{\beta}^2=1$ and
\begin{eqnarray*}
a_{1,n_1}^2-p_1=(-1)^{m_1+1}\ell'a_{1,n_1}, \ \ a_{1,n_2}^2-p_1=(-1)^{m_2+1}\ell'a_{1,n_2},
\end{eqnarray*}
which gives
$$a_{1,n_1}^2- a_{1,n_2}^2=\ell'\big((-1)^{m_1+1}a_{1,n_1}-(-1)^{m_2+1}a_{1,n_2})\big),$$
so
\begin{eqnarray}\label{4.33}
\ell'=(-1)^{m_1+1}a_{1,n_1}+(-1)^{m_2+1}a_{1,n_2}.
\end{eqnarray}
Combining (\ref{4.30}) and (\ref{4.33}), we obtain
$$\sum_{\beta}c_\beta^2=(-1)^{m_1+m_2+1}\frac{a_{1,n_2}}{a_{1,n_1}}\sum_{\alpha}c_\alpha^2>0,$$
which follows that $m_1+m_2$ is an odd number and
\begin{eqnarray}
\sum_{\alpha}c_\alpha^2=\frac{a_{1,n_1}}{a_{1,n_1}+a_{1,n_2}}=\frac{m_1}{m_1+m_2},\ \ \sum_{\beta}c_\beta^2=\frac{a_{1,n_2}}{a_{1,n_1}+a_{1,n_2}}=\frac{m_2}{m_1+m_2}.\label{4.34}
\end{eqnarray}
Therefore, by similar argument as the proof of Proposition \ref{proposition4.1}, the immersion (\ref{4.25}) is congruent to (\ref{4.27}).

Next, we only need to consider $c_\alpha=\bar{c}_\alpha$ for all $\alpha$ and $c_\beta=-\bar{c}_\beta$ for all $\beta$, i.e. $c_\alpha$ are all real and $c_\beta$ are all purely imaginary. Then $\sum\limits_{\alpha}c_{\alpha}^2-\sum\limits_{\beta}c_{\beta}^2=1$ and
\begin{eqnarray*}
a_{1,n_1}^2-p_1=(-1)^{m_1+1}\ell'a_{1,n_1}, \ \ a_{1,n_2}^2-p_1=-(-1)^{m_2+1}\ell'a_{1,n_2},
\end{eqnarray*}
which gives
$$a_{1,n_1}^2- a_{1,n_2}^2=\ell'\big((-1)^{m_1+1}a_{1,n_1}+(-1)^{m_2+1}a_{1,n_2})\big),$$
so
\begin{eqnarray}\label{4.35}
\ell'=(-1)^{m_1+1}a_{1,n_1}-(-1)^{m_2+1}a_{1,n_2}.
\end{eqnarray}
Combining (\ref{4.30}) and (\ref{4.35}), we obtain
$$\sum_{\beta}c_\beta^2=(-1)^{m_1+m_2+1}\frac{a_{1,n_2}}{a_{1,n_1}}\sum_{\alpha}c_\alpha^2<0,$$
which follows that $m_1+m_2$ is an even number and
\begin{eqnarray}\label{4.36}
\sum_{\alpha}c_\alpha^2=\frac{a_{1,n_1}}{a_{1,n_1}+a_{1,n_2}}=\frac{m_1}{m_1+m_2},\ \ \sum_{\beta}c_\beta^2=-\frac{a_{1,n_2}}{a_{1,n_1}+a_{1,n_2}}=-\frac{m_2}{m_1+m_2}.
\end{eqnarray}
Therefore, by similar argument as the proof of Proposition \ref{proposition4.1}, the immersion (\ref{4.25}) is congruent to (\ref{4.26}). Thus we complete the proof. \hfill$\Box$

\vspace{0.2cm}

Now, we calculate the Gauss curvatures of the immersions (\ref{4.26}) and (\ref{4.27}). For the immersion (\ref{4.27}), by (\ref{2.16}), (\ref{4.4}), (\ref{4.34}) and straightforward computation, we obtain
\begin{eqnarray*}
|X|^2=|\textbf{z}A-\ell\textbf{z}|^2&=&a^2_{1,n_2}\sum_{\alpha}c_{\alpha}^2+a^2_{1,n_1}\sum_{\beta}c_{\beta}^2\\
                                    &=&m_2^2\frac{m_1}{m_1+m_2}+m_1^2\frac{m_2}{m_1+m_2}=m_1m_2,
\end{eqnarray*}
and
\begin{eqnarray*}
|Y|^2=|\textbf{z}B|^2&=&a^2_{3,n_1}\sum_{\alpha}c_{\alpha}^2+a^2_{3,n_2}\sum_{\beta}c_{\beta}^2\\
                                    &=&(m_1-1)^2\frac{m_1}{m_1+m_2}+(m_2-1)^2\frac{m_2}{m_1+m_2}=m_1^2-m_1m_2+m_2^2-1.
\end{eqnarray*}
So by (\ref{4.6}), the Gauss curvature is
\begin{equation}\label{4.37}
K=\frac{4}{m_1^2+m_2^2-1}.
\end{equation}
For the immersion (\ref{4.26}), we can obtain the expression of the Gauss curvature by similar calculation, which is also given by (\ref{4.37}).

\vspace{0.3cm}

In summary, we completely classify the minimal homogeneous 2-spheres in quaternionic projective space $\textbf{HP}^n$. That is

\begin{theo}\label{theorem4.4}
Let $f:S^2\longrightarrow\textbf{\emph{HP}}^n$ be a linearly full homogeneous minimal immersion. Then, in terms of homogeneous coordinates, $f$ is congruent to one of the following\emph{:}

\emph{(1)} $f_\lambda:[a,b]\mapsto [\phi_{\lambda,n+1}],$ $\lambda\in\{3,5,\ldots,2n+1\}$, and the Gauss curvature $K=8/[(2n+2)^2-(\lambda^2+1)];$

\emph{(2)} $f_1:[a,b]\mapsto[\phi_{1,n+1}]$ and the Gauss curvature $K=4/[n(n+2)];$

\emph{(3)} $f_{\lambda,m,t}:[a,b]\mapsto[\cos t\phi_{\lambda,m},\texttt{\emph{i}}\sin t\phi_{\lambda,m}]$ for some positive weight $\lambda\in\{3,5,\ldots,n\}$, $t\in(0,\pi/2)$, $2m=n+1$, and the Gauss curvature $K=8/[(n+1)^2-(\lambda^2+1)];$

\emph{(4)} $f_{m_1,m_2}:[a,b]\mapsto[\sqrt{m_1/(m_1+m_2)}\;\phi_{1,m_1},
\texttt{\emph{i}}\sqrt{m_2/(m_1+m_2)}\;\phi_{1,m_2}]$ for some positive $m_1\leq m_2$ so that $m_1+m_2=n+1$ is an even, and the Gauss curvature $K=4/(m_1^2+m_2^2-1);$

\emph{(5)} $f'_{m_1,m_2}:[a,b]\mapsto[\sqrt{m_1/(m_1+m_2)}\;\phi_{1,m_1},
\sqrt{m_2/(m_1+m_2)}\;\phi_{1,m_2}]$ for some positive $m_1<m_2$ so that $m_1+m_2=n+1$ is an odd, and the Gauss curvature $K=4/(m_1^2+m_2^2-1).$

\end{theo}
\emph{Proof}. From the Proposition \ref{proposition4.1}, \ref{proposition4.2} and \ref{proposition4.3}, we know that $f$ is congruent to $f_\lambda$, or $f_1$, or $f_{\lambda,m,t}$, or $f_{m_1,m_2}$, or $f'_{m_1,m_2}$. The expressions for the Gauss curvature $K$ follows from (\ref{4.12}), (\ref{4.23}), (\ref{4.24}) and (\ref{4.37}), respectively. \hfill$\Box$

\vspace{0.3cm}

We should point out the congruence of $f_{\lambda,m,t}$ w.r.t. the parameter $t$, that is
\begin{prop}\label{proposition4.5}
$f_{\lambda,m,t_1}$ is not congruent to $f_{\lambda,m,t_2}$ for any $t_1,t_2\in(0,\pi/2)$, $t_1<t_2$.
\end{prop}
\emph{Proof}. Suppose that $f_{\lambda,m,t_1}$ is congruent to $f_{\lambda,m,t_2}$ for $t_1,t_2\in(0,\pi/2)$, then there exists
fixed $p\in Sp(1)$, $P\in Sp(2m)$ such that
\begin{equation}\label{4.38}
p(\cos t_1,\texttt{i}\sin t_1)\rho_m\oplus\rho_m(g)=(\cos t_2,\texttt{i}\sin t_2)\rho_m\oplus\rho_m(g)P
\end{equation}
for all $g\in SU(2)$. Set $p=p'+p''\texttt{j}$ and $P=(p'_{AB})+(p''_{AB})\texttt{j}$. Since $\{\Lambda_{AB}\;|\;0\leq A,B\leq 2m-1\}$ are linearly independent in $L^2(SU(2))$ by Peter-Weyl's theorem,
we obtain
\begin{equation}\label{4.39}
p''=p''_{AB}=p'_{s\;l}=p'_{s\;m+l}=p'_{m+s\;l}=p'_{m+s\;m+l}=0,\;\;s\neq l,
\end{equation}
and
\begin{equation}\label{4.40}
(p'+\bar p')\cos t_1=2\cos t_2\; p'_{l\;l},\;\;\;
(p'-\bar p')\cos t_1=2\texttt{i}\sin t_2\; p'_{m+l\;l},
\end{equation}
\begin{equation}\label{4.41}
(p'-\bar p')\sin t_1=-2\texttt{i}\cos t_2\; p'_{l\;m+l},\;\;\;
(p'+\bar p')\sin t_1=2\sin t_2\; p'_{m+l\;m+l},
\end{equation}
by comparing the coefficients of $\Lambda_{AB}$. The identities (\ref{4.40}) and (\ref{4.41}) imply that $t_1=t_2$. \hfill$\Box$

\vspace{0.3cm}

Base on the study of homogeneous harmonic maps into projective space, Ohnita \cite{[Oh]} conjectured that all proper minimal constant curved minimal two-spheres in $\textbf{HP}^n$ are $\{f_\lambda:\lambda=1,3,\ldots,2n+1\}$, up to a rigidity of $\textbf{HP}^n$.
Notice that $\textbf{HP}^n$ is a rank one compact symmetric space,
together with Theorem \ref{theorem4.4},
we propose the following conjecture, still called Ohnita's conjecture:
\begin{conj}\label{conjecture4.5}
Let $f$ be a proper minimal constant curved immersion from $S^2$ into $\textbf{\emph{HP}}^n$, then $f$ is congruent to one of $f_\lambda$, $f_{\lambda,m,t}$, $f_{m_1,m_2}$, $f'_{m_1,m_2}$, where $\lambda\in\{1,3,\ldots,2n+1\}$.
\end{conj}

\vspace{0.3cm}
\noindent\textbf{Acknowledgments}. This work was supported by NSFC (Nos. 11471299, 11401481, 11331002).


TITLE: Operator norm induced by Frobenius norm
QUESTION [5 upvotes]: Suppose $T \colon \mathbb{R}^n \to \mathbb{R}^n$ is a linear operator. In following, we will consider $T$ as a matrix. The operator norm induced by $2$-norm on $\mathbb R^n$ is given by $\|T\|_{op,2} = \max_{\|x\|_2 = 1} \|Tx\|_2$. Let $T$ act by matrix multiplication on the vector space $\mathcal{M}(n \times m, \mathbb{R})$, i.e., the space of $n \times m$ matrices. What is operator norm $\|T\|_{op,F}$ induced by Frobenius norm $\| \cdot \|_F$ on $\mathcal {M}(n \times m)$?
Here is my thought: For a given matrix $A \in \mathcal{M}(n \times m, \mathbb{R})$, let $A = (a_1, \dots, a_m)$ where $a_j \in \mathbb R^n$
\begin{align*}
\| TA\|_F^2 &= \|(Ta_1, \dots, Ta_m)\|_F^2 \\
&= \| Ta_1\|_2^2 + \dots \|Ta_m\|_2^2 \le \|T\|_{op,2}^2 \|a_1\|_2^2 + \dots + \|T\|_{op,2}^2 \|a_m\|_2^2 \\&= \|T\|_{op,2}^2 \|A\|_F^2.
\end{align*}
It seems like the operator norm $\|T\|_{op,F}$ should be upper bounded by $\|T\|_{op,2}$. Are they indeed equal? Thanks.

EDIT: @erfink points to this link. Indeed, I am asking a quite different question from this. I am not asking the Frobenius norm of $T$ but the operator norm induced by Frobenius norm of $\mathcal{M}(n \times m)$ when $T$ acts on this space by matrix multiplication. $T$ itself even is not an element of space but an elemnt of $\mathcal{M} (n \times n)$.

REPLY [1 votes]: Yes, $\max_{\|A\|_F=1}\|TA\|_F=\|T\|_2$, because
$$
\|TA\|_F^2
=\sum_j \|T\mathbf a_j\|_2^2
\le\sum_j \|T\|_2^2\|\mathbf a_j\|_2^2
=\|T\|_2^2\|A\|_F^2
$$
and equality occurs when $A=[\mathbf x|0|\cdots|0]$, where $\mathbf x$ is a unit singular vector corresponding to the largest singular value of $T$.
Alternatively, if you vectorise $A$ and $TA$, the mapping $A\mapsto TA$ is equivalent to $\operatorname{vec}(A)\mapsto(I\otimes T)\operatorname{vec}(A)$. Hence
$$
\max_{\|A\|_F=1}\|TA\|_F = \max_{\|\operatorname{vec}(A)\,\|_2=1}\|(I\otimes T)\operatorname{vec}(A)\|_2=\|I\otimes T\|_2=\|T\|_2.
$$


TITLE: Rotate corner of triangle around opposite edge
QUESTION [1 upvotes]: I have two triangles in $R^3$:

$p_1$, $p_2$, $p_3$
$p_1$, $p_2$, $p_4$

The triangles share points $p_1$ and $p_2$ and thus edge $p_2 - p_1$.
I would like to rotate $p_4$ such that it will be diametrically with respect to $p_3$, i.e. the angle between $p_3$ and $p_4$ should be $180$ degrees or $\pi$.
I can derive the current angle between $p_3$ and $p_4$:
$d_1 = (p_3 - p_1) \times (p_2 - p_1)$
$d_2 = (p_4 - p_1) \times (p_2 - p_1)$
$rad = \arccos(d_1 / |d_1| \cdot d_2 / |d_2|)$
The next step would be to rotate $p_4$ around edge ($p_2 - p_1$) by $\pi - rad$. However I do not know how to rotate a corner of a triangle around the opposite edge. Therefore I was wondering if anybody would know how to accomplish this.

REPLY [1 votes]: You have the angle $\phi$ and the axis of rotation $\hat{n} = \frac{p_2-p_1}{|p_2-p_1|}$. You only need Rodrigues' rotation formula:
$$p_{\text{rot}} = p_4 \cos\phi + (\hat{n}\times p_4) \sin\phi + \hat{n}(\hat{n}\cdot p_4)(1-\cos\theta).$$


TITLE: Decide whether a graph is planar or not
QUESTION [0 upvotes]: A simple connected graph has 20 vertices and 22 edges.Decide whether given graph is planar or not(if answer is yes prove it otherwise give counter example)

REPLY [0 votes]: Hint By Kuratowski-Wagner it suffices to check, whether the graph contains a subdivision of $K_5$ or $K_{3,3}$. Hence you may as well reduce to consider such subdivisions. Can you construct a subdivision of $K_5$ or $K_{3,3}$ on $n$ vertices by using at most $n+2$ edges?


TITLE: Prove G has at most 2 commponents
QUESTION [0 upvotes]: Let G be a graph of order n.  Prove that if $deg(u)+deg(v)\geq n-2$ for every pair $u,v$ of nonadjacent vertices of G, then G has at most two components.
MY ATTEMP:
Assume to the contrary that G has 3 components, thus since all the components are disconnected, we can say that $deg(v_i)\leq\frac{n}{3}-1$ for some natural number $i$.  Therefore $deg(v_i)+deg(v_j)\leq \frac{2n}{3}-2$.  From here I am stuck I am unsure where to go from here to get a contadiction.

REPLY [1 votes]: Sure, contradiction works.
Suppose that there are at least three components, pick vertices $u$ and $v$ inside two different connected components, suppose they have $a$ and $b$ vertices each, clearly $a+b<n$.
We also have $d(u)\leq a-1$ and $d(v)\leq b-1$, it follows that $d(u)+d(v)\leq a+b-2<n-2$. Contradiction.


TITLE: Probability theory (heads and tails)
QUESTION [1 upvotes]: A coin was tossed $N$ times. The number of heads is 10% more than the number of tails. At what $N$ can we say that the coin is "dishonest" (heads and tails fall out with different probability)?

REPLY [1 votes]: Sample size for a dichotomous variable with $p$ proportion of success:
$$N = p(1-p)\bigg({Z \over E}\bigg)^2$$
where,
$Z$ is the $z$-value corresponding to your confidence interval (eg. $z$-value $1.96$ for $95\%$ confidence interval),
$E$ is the margin of error (eg: $0.05$ for $5\%$ MoE)
For the given example, let $p$ be the proportion of tails.
$$\therefore 1 - p = 1.1 \times p$$
$$\implies 1 = 2.1p \implies p = {10 \over 21}$$
Substituting, we get
$$N = \bigg({10 \over 21}\bigg)\bigg(1 - {10 \over 21}\bigg)\bigg({1.96 \over 0.05}\bigg)^2 = 383.28 \approx 384$$
So, 384 trials will tell you with 5% margin of error in 95% Confidence Interval.
This will change if you change your CI and margin of error parameters.


TITLE: Historically first axiomatization of the inner product
QUESTION [3 upvotes]: When was the dot/scalar/inner (however you want to call it) product historically first abstractly introduced as a mapping that had to fulfill certain axioms ?

REPLY [23 votes]: According to http://www-history.mcs.st-andrews.ac.uk/history/HistTopics/Abstract_linear_spaces.html, the scalar product was introduced by Grassmann in 1844, in a work that was rejected by other mathematicians for many years. Peano was apparently the first to give a clear  and precise account of Grassmann's ideas in his 1888 book Geometrical Calculus (see http://www-history.mcs.st-andrews.ac.uk/Biographies/Peano.html)


TITLE: Descending chain condition and an identity
QUESTION [4 upvotes]: Let $R$ be a ring with descending chain condition on right ideals. Suppose $l(R)=0$ (the left annihilator of $R$) and $\exists c\in R$ with $r(c)=0$ (the right annihilator of $c$).
Show that $R$ has identity.
I think I can prove that there is a left identity, although I am not confident about it. I cannot seem to find a way to construct an identity, so if someone could point me down a path, that would be great!
Cheers

REPLY [3 votes]: That's a nice question. Thanks for sharing it.
For $a \in R,\,$ $aR := \{ar\mid r \in R\}$ is a right idal (note: it is not necessarily the ideal generated by $a$). Hence
$$cR \supseteq c^2R\supseteq \cdots$$
is a decending chain of right ideals and by decending chain condition the chain stabilizes, i.e. $c^nR = c^{n+1}R$ for some $n > 0$. In particular, there is $e \in R$ s.t. $$c^n \cdot c = c^{n+1}\cdot e.$$
So $c^n(c-ce)=0$ and since the right annulator of $c$ is zero, we obtain $c-ce=0.$ Multiplying from the right yields for all $r \in R:\,\, 0=cr-cer=c(r-er)$ i.e. $r-er$ is a right annulator of $c$. Hence $r=er$, and  $e$  a left identity.
Fix $r \in R$. Then for all $s \in R$:
$$(r-re)s=rs - r(es)=rs - rs =0$$
Thus $r-re$ annulates $R$ from the left and since the left annulator of $R$ is zero we find $r=re$ and $e$ is also a right identity.


TITLE: How can the universe by infinite when it has known start and finite time since?
QUESTION [2 upvotes]: Just watched the 'How Big is the Universe?' TV program and it indicated that the universe is believed to be infinite in size. They used an approach involving measuring the internal angles of triangles. Or something like that anyway.
But hang on, the Universe has a known starting time and size followed by a finite time expanding at a finite speed. So surely, by definition, it cannot be infinite? I can accept it is honking big but by definition it cannot be infinite. Are they misleading me?

REPLY [0 votes]: There is difference between what there is and what we observe. We do observe finite amount of matter and energy exactly because it all started somewhere and moves away, while speed of light is pretty constant. In that, most useful, sense universe is finite. We can speculate about size of what "there is", but it still will be baseless theories.


TITLE: If $a, b, c$ are integers with $a^2 + b^2 = c^2$, then $a$ and $b$ cannot both be odd
QUESTION [1 upvotes]: If $a, b, c$ are integers with $a^2 + b^2 = c^2$, it's true that $a$ and $b$ cannot both be odd.
But how can we prove it

REPLY [1 votes]: We can prove this by contradiction.
$\text{Let }\;\; a = 2m+1, b = 2n + 1 $
So, the LHS is:
$$
\begin{align}
a^2 + b^2 &= 4(m^2+n^2) + 4(m+n) + 2 \\\\
          &= 2\left( 2(m^2+n^2) + 2(m+n)+1 \right) \\\\
          &= 2\times\text{an odd number}
\end{align}
$$
This can not be a perfect square, as it doesn't have an even power of $2$.


TITLE: Understanding the physical meaning of interatomic potential and the meaning of its sign
QUESTION [1 upvotes]: I have been learning about lennard jones potential as a model of interatomic bonds:
$$V(r) = 4ε\left[\left(\frac{σ}{r}\right)^q - \left(\frac{σ}{r}\right)^p\right]$$
But, I've been struggling to understand the physical significance of $V(r)$. I understand that it is a form of energy stored in the bond. What I don't understand is what is meant by its sign: what does it mean to have negative energy? I have also heard $V(r)$ described as the energy required to fuse the atoms together, and as the energy required to bring them an infinite distance apart, but I don't see how these definitions are reconcilable.

REPLY [1 votes]: By convention we set the energy when the atoms are very far away as zero. As we have to supply energy to separate the atoms of the molecule, the equilibrium must correspond to a negative value of energy.


TITLE: Prove that $G(B\cap C)=G(B)\cap G(C)$.
QUESTION [0 upvotes]: Let G be a graph and let B be a subclass of the domain of G. We use the symbol G(B) to designate
the class $$G(B)=\{y\mid \exists x\in B\ \text{such that} \ (x,y)\in G\} $$.
The one way containment is pretty straightforward.
But for the coming one, we take $ y \in G (B) \cap G (C) $, then $ y \in G (B) $ and $ y \in G (C) $. Then, there exists $ x \in B $ such that $ (x, y) \in G $, and there exists $ z \in C $, such that $ (z, y) \in G $, to get that $ y \in G (B \cap C) $ would have to guarantee that $ z = x $, but how do I do that?

REPLY [1 votes]: This identity doesn't hold. Define $G = \{(0, 0), (1, 1), (2, 1)\}$. Then for $A = \{0, 1\}$ and $B = \{0, 2\}$, we see that $G(A) = G(B) = \{0, 1\}$ but $G(A \cap B) = G(\{0\}) = \{0\}$.


TITLE: How to prove the maximum possible number of elements of $S$ is $48$?
QUESTION [4 upvotes]: Let set $S\subseteq \{1,2,3,\cdots,100\}$,for any two different $a,b\in S$,there exist postive integer $k$ and $c,d\in S(c<d)$,($c,d$ can equal to $a$ or $b$),such $$a+b=c^k\cdot d$$
  show that
  $$\max|S|=48$$

I have found this example is such it,and it is not easy to prove when $$S=\{1,2,3,4,5,7,9,11,13,15,17,19,21,23,\cdots79,81,83,85,87,89,91\}\Longrightarrow |S|=48$$
Because $a,b$ are odd(even),then
$$4\le a+b\le 89+91$$
Let $c=2,d=\dfrac{a+b}{2^k}$
if $a$ odd,$b$ is even,then
$$3\le a+b\le 95$$
take $c=1,k=1,d=a+b$
hold.But How to prove why the maximum the number is $48$?

REPLY [1 votes]: As noted in @YuvalFilmus' comment, we have an upper bound of $50$ because we need to avoid a sum of $101$, so of all $50$ distinct pairs in range that sum to this value, we can have at most one of that pair.
However having observed that restriction, it seems that we can in fact select $50$ numbers as members of $S$, giving $\max|S| = 50$ in contradiction to the claim. All the odd numbers except $99$, and $2$.
$$\{1,2,3,5,7,9,11,...,95,97\}$$
Now for $a,b$ both odd, we take $c=2$, and $k$ as the multiplicity of $2$ as a factor of $a+b$, to leave an odd number in range as $d$. We know that $d$ is part of $S$ because $\frac{a+b}{2} < 97 = \max S$.
For $\{a,b\} = \{2,97\} $ we can take (for example) $c=9, k=1, d=11$
For any other sum involving $2$, we can take $c=a+b, k=1, d=1$.


TITLE: Find the diameter of the new sphere assuming that the volume of a sphere is proportional to the cube of its diameter
QUESTION [0 upvotes]: Find the diameter of the new sphere assuming that the volume of  a sphere is proportional to the cube of its diameter. I know that diameter is equal to the twice of radius. How can you possibly solve this if the radius in the formula of the sphere is radius cube.

REPLY [0 votes]: What is the "new" sphere? Anyway, if $D$ is the diameter and $R$ the radius and if you have $V=\alpha R^3$
You will have $V=\alpha (\dfrac{D}{2})^3$...


TITLE: A game of square root modulo n
QUESTION [2 upvotes]: recently there was a CTF competition with a challenge which basically goes like this:
Choose two primes $p, q > 2$. The opponent then selects a random number $r \  \epsilon \  [0, n]$ with $n = pq$ and calculates $s = r^2\ mod\ n$.
Now, given $s$ we have to tell a number $\hat{r}$ such that $s = \hat{r}^2 \ mod \ n$
The opponent now calculates $\hat{p} = GCD (n, (r - \hat{r})\ mod\ n)$ and $\hat{q} = n / \hat{p}$ (GCD being the greatest common divisor).
You lose if $\hat{p} > 1$ and $\hat{q} > 1$ (i.e. your opponent is able to factorize n). Otherwise you win.
The tasks were to a) always win and b) always lose.
Task a) was not that hard: Simply by choosing $ p = q $ the possible $\hat{r}$ shrink down to 2 which always allow your opponent to factorize n.
Calculating $\hat{r}$ boils down to using the tonelli shanks algorithm alongside the chinese remainder theorem.
However I have no idea for task b). Can someone point me in a direction?
Also note that the original opponent uses a miller rabin primality test to check if you are cheating. I was wondering if one could use this to cheat and select non-primes which pass the test? Though this did not seem feasible as the primes had to be very large.
Here is the link to the original challenge

REPLY [1 votes]: 1. General Strategy for Task b.
If $p,q$ are two distinct odd, large primes, then with very high probability you will have $\gcd(r,n)=1$ so you may as well assume that.
In such cases, there will be 4 distinct solutions in $[1,n]$ to
$$
x^2\equiv s\pmod n
$$
Two of the solutions will factor $n$ and there is no way to know which are the two without knowing $r$. (If $r\equiv u\pmod p,r\equiv v\pmod q$, then $x=r$ and $x\equiv -u\pmod p,x\equiv -v\pmod q$ are the two that fails to factor. $(u,-v)$ and $(-u,v)$ works.) So your chances to lose is only $50\%$. You need to lose $0.9$ of the $42$ runs, so the odds are very low.
This means that to solve task (b) you must cheat the primality test.
If you are able to use $n$ such that it comprises of $k$ different distinct odd primes, then there are $2^k$ possible solutions. Once again we can show that a factorization fails for only $2$ of those solutions, so the chances to lose increases to $(2^k-2)/2^{k}$.
If you can find many $p,q$ each a composite of 2 primes that passes the primality test, then $k=4$ and the probability is $87.5\%$, already almost $90\%$. You'll probably pass the challenge if given a few tries.

2. Passing the Primality Test
Looking at the code, the Miller-Rabin test uses base in the range $[1,314]$. An integer that passes the MR test with base $a$ is known as a Strong Pseudoprime to base $a$, so you are looking for integers that are Strong Pseudoprime to bases $1$ to $314$.
One example that you can find online that is Strong Pseudoprime to bases up to $306$ is
$$
\begin{align*}
n =  \;&28871482380507712126714295971303939919776094592797227009265160241974323037991527\\
& 33116328983144639225941977803110929349655578418949441740933805615113979999421542\\
& 41693397290542371100275104208013496673175515285922696291677532547504444585610194\\
& 94042000399044321167766199496295392504526987193290703735640322737012784538991261\\
& 20309244841494728976885406024976768122077071687938121709811322297802059565867
\end{align*}
$$
If you try the primality test in the python code on this number, you'll see that it passes frequently, failing only for $a=307,311$ and $313$.

3. Generating Strong Pseudoprimes to Several Bases
So the problem boils down to constructing the Strong Pseudoprimes. One such algorithm can be found in Arnault's paper: "Rabin-Miller Primality Test: Composite Numbers Which Pass It" and it seems to be available publicly. However I'm not sure if it's efficient enough to solve the current problem without trying it. You may need a better algorithm if it finds solutions too slowly.
The idea of the algorithm is to find primes $p_1,p_2$ such that

(Arnault Strong Pseudoprime) Let $p_1,p_2$ be odd primes satisfying
  $$p_1=2q+1,\;\; p_2=4q+1$$
  for some positive integer $q$. For a given integer $b>1$ (the base), if $b$ is a square but not 4th power modulo $p_2$ and $b$ is not a square modulo $p_1$, then $n=p_1p_2$ is a Strong Pseudoprime to base $b$.

Hence we can search for $p=p_1p_2$ being a Strong Pseudoprime to all bases $b<314$, after which it can be used for cheating.
We can brute force and try all $q$'s, using Jacobi Symbols to test for squares, but that would be inefficient. The paper gives steps to finding them faster using some Number Theory.


TITLE: What is the Probability of a Very Large Digit Constant Number to be Exactly Divided by 19?
QUESTION [0 upvotes]: Let's just put the 40s and 8s side by side randomly to create a 4002 digit number. We have a chance to do this once. Is it possible to talk about the probability of dividing the created constant by 19 exactly? A number is either a multiple of 19 or it isn't. So I thought the answer to this question was 1/2. But such a large number probably won't be divisible by 19. Would it be more accurate to say that the probability of such a situation cannot be defined?

REPLY [1 votes]: You can define a sample space of all the $4002$ digit numbers that can be made by putting substrings of $40$ and $8$ together.  It is a finite collection.  You can then compute how many of them are divisible by $19$.  If you sample your space randomly the chance that the number you get is divisible by $19$ is just the second divided by the first.  I would guess it is not far from $1/19$


TITLE: Is the rising factorial function a convex function?
QUESTION [4 upvotes]: Let $(x)_p=x(x+1)\dots(x+p-1)$ be the rising factorial function.
My question is: Is $(x)_p$ a convex function or not? And how to proof?
And what is about the falling factorial function $(x)^p=x(x-1)\dots(x-p+1)$

REPLY [5 votes]: Well, not on the whole real line. $x\mapsto (x)_p$ is a polynomial of degree $p$ with the roots $0,-1,\dots,-p+1$ and has positive and negative values in between this roots. Hence, it is not convex there.
However, as already stated by Robert Israel, it is convex for $x\geq 0$. To see this you could also argue that it is the product of convex and non-decreasing functions.

REPLY [4 votes]: $(x)_p$ is convex for (at least) $x > 0$, and $(x)^p$ is convex for (at least) $x > p-1$.  To prove it, just take the second derivative using the product rule, and note that all terms are positive.


TITLE: Giving Formula For Recursion given a1 and a2
QUESTION [0 upvotes]: Give a closed formula in terms of $n$ for: $a_n = 2a_{n-1}+3a_{n-2}, a_1 = 1, a_2 = 1$.
In class, we learned how to give a closed formula given $a_0$ and $a_1$, but not for $a_1$ and $a_2$. Please show me the right way to tackle this problem.

REPLY [0 votes]: Observe that for any $n \in \Bbb N$ $$\begin{pmatrix} a_{2n} \\ a_{2n-1} \end{pmatrix} = {\begin{pmatrix} 7 & 6 \\ 2 & 3 \end{pmatrix}}^{n-1} \begin{pmatrix} a_2 \\ a_1 \end{pmatrix}$$ and that the matrix $$\begin{pmatrix} 7 & 6 \\ 2 & 3 \end{pmatrix}$$ is diagonalizable.
So $${\begin{pmatrix} 7 & 6 \\ 2 & 3 \end{pmatrix}}^{n-1} = P {\begin{pmatrix} 9 & 0 \\ 0 & 1 \end{pmatrix}}^{n-1} P^{-1}$$ where $$P = \begin{pmatrix} 3 & 1 \\ 1 & -1 \end{pmatrix}$$ is a matrix consisting of column vectors which are respectively the eigenvectors of $$\begin{pmatrix} 7 & 6 \\ 2 & 3 \end{pmatrix}$$ corresponding to the eigenvalues $9$ and $1.$ Can you finish it from here?


TITLE: How to test given group is special?
QUESTION [3 upvotes]: A group $G$ is said to be special if it can be decomposed into direct product of two groups $A$ and $B$ Where $B$ is a abelian group which is a direct product of $Z_2$'s ( cyclic group of order 2 ) and group $A$ is non abelian and it contains an element of order $x$. One more thing known about $A$ is it has $y$ number of conjugacy classes.
$$G = A \times B$$
Now given a group $G$ by table representation i want to find whether given group is special or not
Question : What could be the brute force way to test if group is special or not?
I am thinking like first finding the sylow subgroups and then from those try to to do something.
Any high level idea or technique will be appreciated.

REPLY [0 votes]: Say we have the group $G$ and we pick an element $c$ of order 2. It is easy to check if $<c>$ is a normal group.
. An element in $G$ and not in $<c>$ can be written as $a=p_1p_2...p_{i-1}cp_i...p_n$. But because $<c>$ is normal  we have $a=p_1p_2...p_{i-1}p_ip_i^{-1}cp_i...p_n=p_1p_2...p_{i-1}p_icp_{i+1}...p_n$ and thereofe every element can be written $a=p_1p_2,....,p_nc$ where $p_i$ can't contain a product of $c$ (unless it is $c\cdot c^{-1}=1$).
If we denote as $N$ all the elements that cannot be written as
 $\text{something}\cdot c$ then because of the previous paragraph $G=N<c>$ (that is the set with all the possible combination of elements)and because $N$ is normal and $N\cap <c>=1$ we have that $G=N\times <c>$.
We now do the following procedure to $N$ until we have exausted all the $\mathbb{Z_2}$.


TITLE: Confused about length contraction and aberration
QUESTION [1 upvotes]: Einstein states that a moving object's shape changes due to the length contraction. But people had done such an experiment: Taking photos to a moving object, but they didn't find Lorentz contraction, instead, the object on picture just takes a rotation, which is the aberration. I'm confused about the two effect. Is there self-contradiction？

REPLY [2 votes]: The Lorentz transformation gives the relationship between the time and space coordinates of two inertial reference frames in relative motion.
As such, it is coordinate length contraction and coordinate time dilation that results.
Thus, in SR, observing (or measuring) length contraction does not mean to photograph length contraction.  Rather, it means recording events:   "the front of the rod was at this location at the same (coordinate) time as the rear of the rod was at this other location; the rod's (coordinate) length is thus..."
From the article "Photography in Special Relativity":

Observation vs Photography
In Special Relativity there is a very specific meaning to the word
  observer.  An observer is one of an infinite collection, in space and
  time, of robots whose sole task in life is to record the time and
  location of each detected event. When we say that an observation was
  made we simply mean that the space and time coordinates of an event
  have been recorded.
A photograph is a record of all of the photons that were received at
  the focal point of the camera at the time the photograph was taken.
These are fundamentally different processes and they give us very
  different information about the universe.


TITLE: Pontrjagin numbers and exotic spheres
QUESTION [10 upvotes]: Hi everyone, im reading Milnor's article "On manifolds homeomorphic to the 7-sphere", in which he constructs the first example of an exotic structure, id like to know if there's a particular reason why Pontrjagin classes become so relevant in defining the invariant that "detects" exotic smoothness. More generally, is there a way to interpret the Pontrjagin classes that connects them with the way the smooth structure of the manifold behaves? Or are they just happy coincidences and the most "geometric" meaning you can get from them is the obstruction of finding n-k-2 linearly independent sections on the tangent bundle?
References are much appreciated!
Thanx in advance

REPLY [12 votes]: The Pontryagin classes of the tangent bundle are not easy to interpret (witness the Novikov conjecture), but one geometric datum that can be extracted from them is the rational cobordism class of a manifold. According to Thom, the Pontryagin numbers $p_1^{a_1}\cdots p_k^{a_k}[X]$ of a closed, oriented, smooth $4n$-manifold $X$ vanish iff there's a compact, oriented $(4n+1)$-dimensional manifold bounding a disjoint union of copies of $X$. From Thom's theorem follows Hirzebruch's, expressing the (cobordism-invariant) signature $\sigma(X)$ as a certain Pontryagin number $L(p_1,\dots,p_n)[X]$. When $n=2$,
$$\sigma(X)=(-p_1^2+7p_2)[X]/45.$$
One striking thing about this formula is that $\sigma(X)$ is an integer, while $(-p_1^2+7p_2)[X]/45$ is a priori only a rational number. Milnor's observation is that an integrality theorem for a characteristic class of closed $4n$-manifolds gives rise to an invariant of those $(4n-1)$-manifolds which bound $4n$-manifolds. This is a useful principle, a variant of which also underlies Chern-Simons theory.
We can define an invariant for homotopy 7-spheres $S$ by taking $\kappa(S)=45\sigma(Y)+p_1^2[Y,\partial Y] \mod 7$; if $Y'$ is another such bounding manifold, the difference between their invariants will be $45 \sigma(X)+p_1^2(X)$ for the closed manifold $X= (-Y') \cup_S Y$, hence a multiple of 7 by the signature theorem. (Milnor prefers the invariant $\lambda=2\kappa \mod 7$.)
In his later work with Kervaire ("Groups of homotopy spheres I"), Milnor identifies two different reasons why a homotopy-sphere may not be (h-cobordant to) a standard sphere: (i) it may not bound a parallelizable manifold; or (ii) it may bound a parallelizable manifold, but not one which is also contractible. A homotopy 7-sphere which bounds a parallelizable 8-manifold is in fact standard, but this is not true of homotopy 8-spheres. The invariant $\lambda$ of homotopy 7-spheres is an obstruction to bounding a parallelizable 8-manifold; not a complete invariant, since Kervaire-Milnor show that there are exactly 28 h-cobordism classes.


TITLE: Uniform convergence of difference quotient function implies differentiability in $\mathbb{R}^n$
QUESTION [2 upvotes]: This arises in the context of the proof of the Rademacher theorem.
Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is a Lipschitz function. It therefore has a distributional gradient $L$ and suppose that $x$ is a point in the Lebesgue set of $L$.
We can define $f_r(y) = \frac{f(x+ry)-f(x)}{r}$ and the claim is that if $f_r(y) \to \langle L(x), y\rangle$ uniformly as $r \to 0$ then $f$ is differentiable at $x$. Why is this true and why is uniform convergence necessary rather than just point wise convergence?

REPLY [2 votes]: In order for the existence of all directional derivatives to imply full differentiability, you need

the directional derivative $Df|_x(y) = \lim_{r \to 0}f_r(y)$ to be linear in $y$, and
the directional difference quotients to converge uniformly in $y$.

You can find some discussion of this on this Wikipedia page, including an example of what can go wrong when the convergence is not uniform in $y$: a multivariable function with vanishing directional derivatives at a point that is not differentiable there.
When both conditions are satisfied, the proof is fairly straightforward. We know the directional derivative $Df|_x (y) = \langle L, y\rangle$ is linear, so it is a suitable candidate to be the derivative. The uniform convergence of the quotients can be written $|f_r(y) - \langle L,y \rangle| \le\omega(r)$ for some function satisfying $\lim_{r \to 0} \omega(r) = 0$.
Thus we calculate $$f(x+y)-f(x)-\langle L,y\rangle=|y|f_{|y|}(\frac{y}{|y|})-\langle L,y\rangle;$$ so we have $$\frac{|f(x+y)-f(x)-\langle L,y\rangle|}{|y|}=\left|f_{|y|}(\frac{y}{|y|}) -
 \langle L, \frac{y}{|y|}\rangle\right| \le\omega(|y|)$$which converges to zero as $y \to 0$ as desired.