On conformal minimal immersions of $S^2$ in $HP^n$ with parallel second fundamental form

Abstract

In this paper, we determine all conformal minimal immersions of 2-spheres in quaternionic projective spaces $HP^n$ with parallel second fundamental form.

1 Introduction

In 1976, Nakagawa and Takagi studied some properties about Kähler imbeddings of compact Hermitian symmetric spaces in the complex projective space $CP^n$ and gave a classification of Kähler submanifolds in $CP^n$ with parallel second fundamental form (cf. [8]). In 1984, Ros decided the compact Einstein Kähler submanifold in $CP^n$ with parallel second fundamental form (cf. [9]). In 1985, Tsukada classified $2n$-dimensional totally complex submanifolds in $HP^n$ with parallel second fundamental form into eight types (cf. [10, 11]). Recently, we studied conformal minimal immersions of 2-spheres in $CP^n$ and $G(k,N)$ with parallel second fundamental form, and obtained some geometric properties of them (cf. [6, 7]).

In this paper, our interest is to study classification of conformal minimal immersions from $S^2$ to the quaternionic projective space $HP^n$ with parallel second fundamental form by the theory of harmonic maps. Here, we regard $HP^n$ as a totally geodesic totally real submanifolds in complex Grassmann manifolds $G(2,2n+2)$ and obtain the following classification theorem:

Theorem 1.1

Let $\varphi :S^2\rightarrow HP^n$ be a linearly full conformal minimal immersion, and let $K$ and $B$ be its Gauss curvature and second fundamental form respectively. If $B$ is parallel, then, up to a symplectic isometry of $HP^n$, it belongs to one of the following minimal immersions.

• (1) $\varphi =\left[ (\sqrt{3}\overline{z}+\sqrt{3}\overline{z}^2j,~-1+ 2z\overline{z}-2\overline{z}j-z\overline{z}^2j)^T\right] : S^2\rightarrow HP^1$, with $K=\frac{2}{3},\Vert B\Vert ^2=\frac{8}{3}$;

• (2) $\varphi =\left[ (1,~z)^T\right] :S^2\rightarrow CP^1\subset HP^1$, with $K=2, ~B=0$;

• (3) $\varphi =\left[ (1,~\sqrt{2}z,z^2)^T\right] :S^2\rightarrow CP^2\subset HP^2$, with $K=1, ~\Vert B\Vert ^2=2$;

• (4) $\varphi =\left[ (1-\frac{1}{2}\overline{z}^3j, ~\sqrt{3}z+ \frac{\sqrt{3}}{2}\overline{z}^2j, ~\frac{3}{2}z^2,~\frac{\sqrt{3}}{2}z^3)^T\right] : S^2\rightarrow HP^3$, with $K=\frac{2}{3}, ~\Vert B\Vert ^2=\frac{1}{3}$;

• (5) $\varphi =\left[ (-2\overline{z},~\sqrt{2}-\sqrt{2}z\overline{z},~2z)^T\right] :S^2\rightarrow CP^2\subset HP^2$, with $K=\frac{1}{2}, ~B=0$;

• (6) $\varphi =\left[ (6\overline{z}^2,~-6\overline{z}+6z\overline{z}^2, ~\sqrt{6}-4\sqrt{6}z\overline{z}+ \sqrt{6}z^2\overline{z}^2,~6z-6z^2\overline{z}, ~6z^2)^T\right] :S^2\rightarrow CP^4\subset HP^4$, with $K=\frac{1}{6}, ~\Vert B\Vert ^2 = \frac{2}{3}$.

Further, no two of the above six cases are congruent, i.e., there is no symplectic isometry which transforms one case into another.

2 Preliminaries

(A) For any $N=1,2,\ldots ,$ let $\left\langle ,\right\rangle$ denote the standard Hermitian inner product on $\mathbb {C}^{N}$ defined by $\left\langle z, w\right\rangle = z_1\overline{w}_1 +\cdots + z_N\overline{w}_N$, where $z=(z_1, \ldots , z_N)^T, w=(w_1, \ldots , w_N)^T\in {\mathbb {C}}^N$ and $\bar{\,}$ denote complex conjugation. Let $\mathbb {H}$ denote the division ring of quaternions. Let $j$ be a unit quaternion with $j^2=-1$. Then, we have an identification of ${\mathbb {C}}^2$ with $\mathbb {H}$ given by making $(a,b)\in {\mathbb {C}}^2$ correspond to $a+bj\in \mathbb {H}$; let $n \in \left\{ 1,2,\ldots \right\}$, and we have a corresponding identification of ${\mathbb {C}}^{2n}$ with ${\mathbb {H}}^n$. For any $a+bj\in \mathbb {H}$, the left multiplication by $j$ is given by $j(a+bj)=-\overline{b}+\overline{a}j$; the conjugation is given by $\overline{a+bj}=\overline{a}-bj$; the positive-definite inner product is given by $\left\langle x,y\right\rangle _{\mathbb {H}}=Re(x\overline{y})$ for any $x,y\in \mathbb {H}$.

Let $\mathbf {J}:{\mathbb {C}}^{2n} \rightarrow {\mathbb {C}}^{2n}$ be the conjugate linear map given by left multiplication by $j$, i.e.,

$$\begin{aligned} {\mathbf {J}}(z_1, z_2, \ldots , z_{2n-1}, z_{2n})^T= (-\bar{z}_2, \bar{z}_1, \ldots , -\bar{z}_{2n}, \bar{z}_{2n-1})^T. \end{aligned}$$

Then, ${\mathbf {J}}^2=-id$ where $id$ denotes the identity map on ${\mathbb {C}}^{2n}$. In fact, for any $v\in {\mathbb {C}}^{2n}$,

$$\begin{aligned} {\mathbf {J}}v=J_n \bar{v}, \end{aligned}$$

where $J_n=diag \underbrace{\left\{ \left( \begin{array} {lr} 0 &{} -1 \\ 1 &{} 0 \\ \end{array}\right) , \ldots , \left( \begin{array} {lr} 0 &{} -1 \\ 1 &{} 0 \\ \end{array}\right) \right\} }_{n}$.

By the above, we immediately have the following lemma (cf. [1]).

Lemma 2.1

The operator $\mathbf {J}$ has the following properties:

• (i) $\left\langle {\mathbf {J}}v, {\mathbf {J}}w\right\rangle =\left\langle w, v\right\rangle$ for all $v,w\in {\mathbb {C}}^{2n}$;

• (ii) $\left\langle {\mathbf {J}}v, v\right\rangle =0$ for all $v\in {\mathbb {C}}^{2n}$;

• (iii) $\partial \circ {\mathbf {J}}=\mathbf {J} \circ \overline{\partial }$, $\overline{\partial } \circ \mathbf {J}=\mathbf {J} \circ \partial$;

• (iv) $\mathbf {J}(\lambda v)=\overline{\lambda } \mathbf {J} v$ for any $\lambda \in {\mathbb {C}}$, $v\in {\mathbb {C}}^{2n}$.

Let $G(2,2n+2)$ denote the Grassmann manifold of all complex $2$-dimensional subspaces of ${\mathbb {C}}^{2n+2}$ with its standard Kähler structure. The quaternionic projective space $HP^{n}$ is the set of all one-dimensional quaternionic subspaces of ${\mathbb {H}}^{n+1}$. Throughout the above, we shall regard $HP^{n}$ as the totally geodesic submanifold of $G(2,2n+2)$ given by

$$\begin{aligned} HP^{n}=\left\{ V\in {G(2,2n+2)}: {\mathbf {J}}V=V\right\} . \end{aligned}$$

Let $Sp(n+1)=\left\{ g\in GL(n+1;\mathbb {H}), \ g^*g=I_{n+1}\right\}$ be the symplectic isometry group of $HP^{n}$, here $I_{n+1}$ is the identity matrix. The explicit description is that the following diagram commutes:

where $i_1,i_2$ are inclusions and $\pi _1,\pi _2$ are projections, and $i_1(g)=U$, for $1\le a,b\le n+1$

$$\begin{aligned} {\left\{ \begin{array}{ll} U_{2b-1}^{2a-1}=A_{b}^{a},\quad U_{2b}^{2a-1}=-\overline{D}_{b}^{a},\\ U_{2b-1}^{2a}=D_{b}^{a},\quad U_{2b}^{2a}=\overline{A}_{b}^{a}, \end{array}\right. } \end{aligned}$$

where $A=(A_{b}^{a}),~D=(D_{b}^{a})\in M_{n+1}(\mathbb {C}),~g=A+\overline{D}j\in Sp(n+1);$

Here, we take the standard metric on $G(2,2n+2)$ as described in section 2 of [7]; then, the metric induced by $i_2$ is twice as much as the standard metric on $HP^{n}$.

Thus, a harmonic map from $S^2$ to $HP^n$ is precisely a harmonic map from $S^2$ to $G(2,2n+2)$ which has image in $i_2(HP^n)$.

Then, for any $g\in Sp(n+1)$, the action of $g$ on $HP^{n}$ induces an action of $i_1(g)$ on $CP^{2n+1}$, where $i_1(g)\in U(2n+2)$ commutes with ${\mathbf {J}}$. We shall retain $g$ to also denote $i_1(g)$. Then

$$\begin{aligned} Sp(n+1)=\left\{ g\in U(2n+2), \ g\circ {\mathbf {J}}={\mathbf {J}} \circ g\right\} =\left\{ g\in U(2n+2), \ g J_{n+1} g^{T}=J_{n+1}\right\} . \end{aligned}$$

In the following, we deal with the symplectic isometry of $HP^{n}$ through the corresponding symplectic isometry of $CP^{2n+1}$.

(B) In this section, we give general expression of some geometric quantities about conformal minimal immersions from $S^2$ to $HP^n$ (cf. [7]).

Let $M$ be a simply connected domain in the unit sphere $S^2$ and let $(z,\overline{z})$ be complex coordinates on $M$. We take the metric $ds^2_M=dzd\overline{z}$ on $M$. Denote

$$\begin{aligned} \partial =\frac{\partial }{\partial z}, \quad \overline{\partial }=\frac{\partial }{\partial \overline{z}}. \end{aligned}$$

We consider the complex Grassmann manifold $G(2,N)$ as the set of Hermitian orthogonal projections from ${\mathbb {C}}^N$ onto a $2$-dimensional subspace in ${\mathbb {C}}^N$. Then, a map $\psi :S^2\rightarrow G(2,N)$ is a Hermitian orthogonal projection onto a $2$-dimensional subbundle $\underline{\psi }$ of the trivial bundle ${\underline{\mathbb {C}}}^N=M\times {\mathbb {C}}^N$ given by setting the fibre of $\underline{\psi }$ at $x$, ${\underline{\psi }}_x$, equal to $\psi (x)$ for all $x\in M$. We say that $\underline{\psi }$ is a harmonic subbundle if $\psi$ is harmonic (cf. [3]).

Let $\varphi : S^2 \rightarrow HP^{n}$ be a conformal minimal immersion. The map $i_2\circ \varphi :S^2 \rightarrow G(2,2n+2)$ may be represented via the local sections of the subbundle $\underline{Im}(i_2\circ \varphi )$ by the projection map (cf. [7], (2.10)):

$$\begin{aligned} i_2\circ \varphi =XX^*+({\mathbf {J}}X)({\mathbf {J}}X)^*, \end{aligned}$$

where $X\in \underline{Im}(i_2\circ \varphi )$ is a unit column vector in $\mathbb {C}^{2n+2}$, and $X$ and ${\mathbf {J}}X$ are naturally orthogonal.

Denote $i_2\circ \varphi$ by $\varphi _0$ (in the following, we will use this notation all the time). Suppose that the metric induced by $\varphi _0$ is $ds^2=\lambda ^2dzd\overline{z}$. Let $K$ and $B$ be its Gauss curvature and second fundamental form, respectively. From section 2 and 3 of [7], we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda ^2=tr\partial {\varphi _0}\overline{\partial }{\varphi _0},\\ K=-\frac{2}{\lambda ^2} \partial \overline{\partial }\log \lambda ^2,\\ \Vert B \Vert ^2=4trPP^*. \end{array}\right. } \end{aligned}$$

(2.1)

where $P=\partial \left( {A_z}/{\lambda ^2}\right)$ with $A_z=\left( 2\varphi _0-I\right) \partial {\varphi _0}$, and $I$ is the identity matrix, then $P^*=\overline{\partial }\left( {A_z^*}/{\lambda ^2}\right) ,~A_z^*=-A_{\overline{z}}$.

3 The proof of main theorem

We recall that an immersion of $S^2$ in $HP^n$ is conformal and minimal if and only if it is harmonic (cf. [4], Sec 10.6). Thus, we shall consider the immersive harmonic maps from $S^2$ to $HP^n$ with parallel second fundamental form for the reducible and irreducible cases to give the proof of Theorem 1.1 in Sect. 1. At first, we state a conclusion about parallel minimal immersions of 2-spheres in $G(k,N)$ as follows:

Lemma 3.1

([7]) Let $\varphi :S^2 \rightarrow G(k,N)$ be a conformal minimal immersion with the second fundamental form $B$. Then $B$ is parallel if and only if the equation

$$\begin{aligned} \frac{\lambda ^2}{16}\Vert B \Vert ^2\left( 8K+\Vert B \Vert ^2\right) +2tr[A_z, P][A_z^*, P^*]-5tr[A_z, A_z^*][P, P^*]=0 \end{aligned}$$

(3.1)

holds.

(I) Let $\varphi :S^2\rightarrow HP^n$ be a linearly full reducible harmonic map, then by ([1], Proposition 3.7) we know that $\varphi$ is a quaternionic mixed pair or a quaternionic Frenet pair. In the following, we discuss the two cases of $\varphi$ with parallel second fundamental form, respectively.

(Ia) If $\varphi$ is a linearly full quaternionic Frenet pair, then

$$\begin{aligned} \underline{\varphi }_0=\underline{f}_n^{(2n+1)} \oplus \underline{f}_{n+1}^{(2n+1)}, \end{aligned}$$

(3.2)

where $\underline{f}_0^{(2n+1)},~\underline{f}_1^{(2n+1)}, \ldots , ~\underline{f}_{2n+1}^{(2n+1)}:S^2 \rightarrow CP^{2n+1}$ is the harmonic sequence generated by a linearly full totally $\mathbf {J}$-isotropic map $\underline{f}_0^{(2n+1)}$.

Firstly, we recall ([1], §3) that a full holomorphic map $\underline{f}_0^{(2n+1)} : S^2 \rightarrow CP^{2n+1}$ in the following harmonic sequence satisfying ${\underline{f}}_{2n+1}^{(2n+1)}={\mathbf {J}}\underline{f}_0^{(2n+1)}$ is said to be totally $\mathbf {J}$-isotropic,

$$\begin{aligned} 0\mathop {\longleftarrow }\limits ^{A_0''} {\underline{f}} _0^{(2n+1)} \mathop {\longrightarrow }\limits ^{A_0'}\cdots \mathop {\longrightarrow }\limits ^{A_{n-1}'} {\underline{f}}_{n}^{(2n+1)} \mathop {\longrightarrow }\limits ^{A_{n}'} {\underline{f}}_{n+1}^{(2n+1)} \mathop {\longrightarrow }\limits ^{A_{n+1}'} \cdots \mathop {\longrightarrow }\limits ^{A_{2n}'} {\underline{f}}_{2n+1}^{(2n+1)} \mathop {\longrightarrow }\limits ^{A_{2n+1}'} 0, \end{aligned}$$

where $A'_{j}(v)=\pi _{{f_{j}^{(2n+1)}}^{\perp }}(\partial v), \quad A''_{j}(v)=\pi _{{f_{j}^{(2n+1)}}^{\perp }}(\overline{\partial } v)$ for $v\in C^{\infty }({\underline{f}}_{j}^{(2n+1)})$, here $\pi _{{f_{j}^{(2n+1)}}^{\perp }}$ denotes orthogonal projection onto bundle ${{\underline{f}}_{j}^{(2n+1)}}^{\perp }$ and $C^{\infty }({\underline{f}}_{j}^{(2n+1)})$ denotes the vector space of smooth sections of bundle ${\underline{f}}_{j}^{(2n+1)}$, $j=0,\ldots ,2n+1$.

Let $f_0^{(2n+1)}$ be a holomorphic section of ${\underline{f}} _0^{(2n+1)}$, i.e., $\overline{\partial } f_0 ^{(2n+1)} = 0$, and let $f_j^{(2n+1)}$ be a local section of ${\underline{f}}_{j}^{(2n+1)}$ such that

$$\begin{aligned} f_j^{(2n+1)} = \pi _{{f_{j-1}^{(2n+1)}}^{\perp }}\left( \partial f_{j-1} ^{(2n+1)} \right) \end{aligned}$$

for $j=1, \ldots ,2n+1$. Then, we have some formulas as follows (cf. [2]):

$$\begin{aligned} \partial f_j ^{(2n+1)}&= f_{j+1}^{(2n+1)} + \partial \log |f_j ^{(2n+1)}|^2 f_j ^{(2n+1)}, \ j= 0,\ldots , 2n,\\ \overline{\partial } f_j^{(2n+1)}&= - l_{j-1} ^{(2n+1)} f_{j-1} ^{(2n+1)}, \ j = 1, \ldots , 2n+1,\\ \partial \overline{\partial } \log |f_j ^{(2n+1)}|^2&= l_j^{(2n+1)}-l_{j-1}^{(2n+1)},\\ \partial \overline{\partial } \log l_j^{(2n+1)}&= l_{j+1} ^{(2n+1)}- 2 l_j^{(2n+1)} + l_{j-1}^{(2n+1)}, \ j =0, \ldots , 2n, \end{aligned}$$

where $l_j ^{(2n+1)}= |f_{j+1}^{(2n+1)}|^2/ |f_j^{(2n+1)}|^2$ for $j =0, \ldots , 2n+1$, and $l_{-1} ^{(2n+1)}= l_{2n+1}^{(2n+1)}=0$.

Since ${\underline{f}} _0^{(2n+1)}$ is totally $\mathbf {J}$-isotropic, in a similar fashion to ([2], Lemma 7.1) we obtain

$$\begin{aligned} l_j^{(2n+1)}=l_{2n-j}^{(2n+1)}. \end{aligned}$$

(3.3)

And set $\mathbf {J} f_0^{(2n+1)}=\tau _0 f_{2n+1}^{(2n+1)}$ , then

$$\begin{aligned} |\tau _0|^2=\frac{|f_0^{(2n+1)}|^2}{|f_{2n+1}^{(2n+1)}|^2},\quad \mathbf {J} f_j^{(2n+1)}=(-1)^j \tau _0 \frac{|f_{2n+1}^{(2n+1)}|^2}{|f_{2n+1-j}^{(2n+1)}|^2} f_{2n+1-j}^{(2n+1)}, \end{aligned}$$

where $j=0, \ldots , n$.

Obviously, $\varphi _0$ belongs to the following harmonic sequence (cf. [3])

$$\begin{aligned} 0 \mathop {\longleftarrow }\limits ^{A_{0}''}\underline{f}_0^{(2n+1)} \mathop {\longleftarrow }\limits ^{A_{1}''}\cdots \mathop {\longleftarrow }\limits ^{A_{n-1}''}\underline{f}_{n-1}^{(2n+1)} \mathop {\longleftarrow }\limits ^{A_{\varphi _0}''}\underline{\varphi }_0 \mathop {\longrightarrow }\limits ^{A_{\varphi _0}'}\underline{f}_{n+2}^{(2n+1)} \mathop {\longrightarrow }\limits ^{A_{n+2}'}\cdots \mathop {\longrightarrow }\limits ^{A_{2n}'}\underline{f}_{2n+1}^{(2n+1)} \mathop {\longrightarrow }\limits ^{A_{2n+1}'}0, \end{aligned}$$

(3.4)

where $A'_{\varphi _0}(v)=\pi _{\varphi _0^{\perp }}(\partial v), \quad A''_{\varphi _0}(v)=\pi _{\varphi _0^{\perp }}(\overline{\partial } v)$ for $v\in C^{\infty }(\underline{\varphi }_0)$, here $\pi _{\varphi _0^{\perp }}$ denotes orthogonal projection onto bundle $\underline{\varphi }_0^{\perp }$ and $C^{\infty }(\underline{\varphi }_0)$ denotes the vector space of smooth sections of bundle $\underline{\varphi }_0$.

From (3.2), we have $\varphi _0=\frac{{f}_n^{(2n+1)}({f}_n^{(2n+1)})^*}{|{f}_n^{(2n+1)}|^2}+\frac{{f}_{n+1}^{(2n+1)}({f}_{n+1}^{(2n+1)})^*}{|{f}_{n+1}^{(2n+1)}|^2}$. Then by (2.1), (3.3) and a series of calculations, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda ^2=2l_{n-1}^{(2n+1)},\\ K=2-\frac{l_{n}^{(2n+1)}+l_{n-2}^{(2n+1)}}{l_{n-1}^{(2n+1)}},\\ \Vert B \Vert ^2=2\frac{l_{n}^{(2n+1)}+l_{n-2}^{(2n+1)}}{l_{n-1}^{(2n+1)}},\\ tr[A_z, P][A_z^*, P^*]=-l_{n}^{(2n+1)},\\ tr[A_z, A_z^*][P, P^*]=\frac{1}{2}l_{n-2}^{(2n+1)}. \end{array}\right. } \end{aligned}$$

(3.5)

Now, we prove that if $\varphi :S^2 \rightarrow HP^n$ is a linearly full quaternionic Frenet pair with parallel second fundamental form, then, up to $Sp(n+1)$, it belongs to the following case: (1) $\varphi =\left[ (\sqrt{3}\overline{z}+\sqrt{3}\overline{z}^2j,-1+2z\overline{z}-2\overline{z}j-z\overline{z}^2j)^T\right] :S^2\rightarrow HP^1,$ with $K=\frac{2}{3},||B||^2=\frac{8}{3}$.

If $\varphi$ is a linearly full quaternionic Frenet pair with parallel second fundamental form, then applying Lemma 3.1 and substituting (3.5) into (3.1), we get

$$\begin{aligned} 3l_{n-2}^{(2n+1)}l_{n-1}^{(2n+1)}+4l_{n-1}^{(2n+1)}l_{n}^{(2n+1)}-3 \left( l_{n}^{(2n+1)}+l_{n-2}^{(2n+1)}\right) ^2=0. \end{aligned}$$

(3.6)

Since the second fundamental form of the map $\varphi$ is parallel, its Gauss curvature is a constant (cf. [7], Theorem 4.5). We know up to $U(2n+2)$, $\underline{f}_0^{(2n+1)}$ is a Veronese surface by ([5], Lemma 4.1). Then from [2], we have $\underline{f}_0^{(2n+1)}, \underline{f}_1^{(2n+1)}, \ldots , \underline{f}_{2n+1}^{(2n+1)}$ is the Veronese sequence in $CP^{2n+1}$, up to $U(2n+2)$. So, from ([2], Section 5), we get

$$\begin{aligned} |f_i^{(2n+1)}|^2=\frac{(2n+1)!i!}{(2n+1-i)!}(1+z\overline{z})^{2n+1-2i}, ~l_j ^{(2n+1)}=\frac{(j+1)(2n+1-j)}{(1+z\overline{z})^2}, \end{aligned}$$

(3.7)

where $i=0,\ldots ,2n+1,~j=0,\ldots ,2n$.

Substituting (3.7) into (3.6), we get

$$\begin{aligned} (n-1)(n+3)(5n^2+10n-4)=0, \end{aligned}$$

which implies $n=1$, since $n$ is a positive integer. Hence,

$$\begin{aligned} \underline{\varphi }_0=U\underline{V}_1^{(3)} \oplus {\mathbf {J}} U\underline{V}_1^{(3)}, \end{aligned}$$

(3.8)

where $\underline{V}_1^{(3)}$ is a Veronese surface in $CP^3\subset CP^{2n+1}$ with the standard expression given in ([2], §5), and $U\in U(2n+2)$ satisfies $J_{n+1} \overline{U} \overline{V}_0^{(3)}=\lambda U V_3^{(3)}$ ($\lambda$ is a parameter).

Set $U^{T}J_{n+1}U=\overline{W}$, then we immediately get

$$\begin{aligned} \overline{W}V_0^{(3)}=\overline{\lambda }\overline{V}_3^{(3)},~W^{T}=-W,~W^*W=I, \end{aligned}$$

(3.9)

where $I$ is the identity matrix.

Define a set

$$\begin{aligned} G_W\triangleq \left\{ U\in U(2n+2), \ UWU^{T}=J_{n+1}\right\} . \end{aligned}$$

For a given $W$, the following can be easily checked

• (i) $\forall \ g\in Sp(n+1), \ U\in G_W$, we have that $gU\in G_W$;

• (ii) $\forall \ U, V \in G_W, \ \exists \ g=UV^* \in Sp(n+1) \ s.t. \ U=gV$.

Then, we discuss the type of $W$ to get the type of the corresponding $U$. From ([2], section 5), we get

$$\begin{aligned} V^{(3)}_0&= (1, \ \sqrt{3} z,\ \sqrt{3} z^2,\ z^3,\ 0,\ldots ,\ 0)^T,\nonumber \\ V^{(3)}_3&= \frac{6}{(1+z\overline{z})^3}(-\overline{z}^3, \ \sqrt{3}\overline{z}^2, \ -\sqrt{3}\overline{z},\ 1,\ 0,\ldots ,\ 0)^T. \end{aligned}$$

(3.10)

Then, by (3.9) and (3.10) we get the type of $\overline{W}$ as follows:

(3.11)

From $\overline{W}\overline{U}^T = U^T J_{n+1}$, the corresponding $U=\left[ e_1,e_2,\ldots ,e_{2n+1},e_{2n+2}\right] ^T$ satisfy

$$\begin{aligned} e_{2p}=\overline{W}\overline{e}_{2p-1},\quad p=1,\ldots ,n+1, \end{aligned}$$

(3.12)

where $e_i$ are unit column vectors in $\mathbb {C}^{2n+2}$.

Without loss of generality, in this case we choose

$$\begin{aligned} {\left\{ \begin{array}{ll} e_1=(1,0,0,0,\ldots ,0)^T,\\ e_3=(0,1,0,0,\ldots ,0)^T. \end{array}\right. } \end{aligned}$$

(3.13)

By(3.11)–(3.13), we get $e_2=\overline{W}\overline{e}_1=(0,0,0,-1,\ldots ,0)^T$ and $e_4=\overline{W}\overline{e}_3=(0,0,1,0,\ldots ,0)^T$, obviously $\{e_1,e_2,e_3,e_4\}$ are mutually orthogonal. Next, we choose a unit column vector $e_5=(0,0,0,0,*)^T\in \mathbb {C}^{2n+2}$, which satisfies $\{e_1,e_2,e_3,e_4,e_5\}$ are mutually orthogonal. Set $e_6=\overline{W}\overline{e}_5$, then $\{e_1,e_2,e_3,e_4,e_6\}$ are mutually orthogonal. Since $\left\langle e_6,~e_5\right\rangle =e_5^TW^Te_5=-tr(e_5e_5^TW)=0$, then $\{e_1,e_2,e_3,e_4,e_5,e_6\}$ are mutually orthogonal.

Generally, suppose $\{e_1,e_2,\ldots ,e_{2p-3},e_{2p-2}=\overline{W}\overline{e}_{2p-3}\}~(p\ge 3)$ are mutually orthogonal, we choose a unit column vector $e_{2p-1}=(0,0,0,0,*)^T\in \mathbb {C}^{2n+2}$ such that $\{e_1,e_2,\ldots ,e_{2p-3},e_{2p-2},e_{2p-1}\}$ are mutually orthogonal. Set $e_{2p}=\overline{W}\overline{e}_{2p-1}$, then

$$\begin{aligned} \left\langle e_{2p},~e_{2p-1}\right\rangle =e_{2p-1}^TW^Te_{2p-1}=-tr(e_{2p-1}e_{2p-1}^TW)=0, \end{aligned}$$

and for any $2\le q\le p$,

$$\begin{aligned}&\left\langle e_{2p},~e_{2q-3}\right\rangle ={e}_{2p-1}^TW^Te_{2q-3} =-{e}_{2p-1}^TWe_{2q-3} =-{e}_{2p-1}^T\overline{e}_{2q-2}\\&\quad =-\left\langle e_{2p-1},~e_{2q-2}\right\rangle =0,\left\langle e_{2p},~e_{2q-2}\right\rangle \\&\quad ={e}_{2p-1}^TW^T\overline{W}\overline{e}_{2q-3} ={e}_{2p-1}^T\overline{e}_{2q-3} =\left\langle e_{2p-1},~e_{2q-3}\right\rangle =0. \end{aligned}$$

Thus $\{e_1,e_2,\ldots ,e_{2p-3},e_{2p-2},e_{2p-1},e_{2p}\}$ are mutually orthogonal.

So, we can choose $n-1$ proper unit column vectors $e_{2p+1}=(0,0,0,0,*)^T\in \mathbb {C}^{2n+2}~(2\le p\le n)$ such that $\{e_1,e_2,\ldots ,e_{2n+1},e_{2n+2}=\overline{W}\overline{e}_{2n+1}\}$ are mutually orthogonal, and the type of the corresponding $U$ is as follows:

(3.14)

Thus, we have

$$\begin{aligned}&UV_1^{(3)}=\frac{-\sqrt{3}}{1+z\overline{z}}(\sqrt{3}\overline{z}, \ \sqrt{3}z^2, \ -1+2z\overline{z},\ -2z+z^2\overline{z},\ 0,\ldots ,\ 0)^T,\\&{\mathbf {J}}UV_1^{(3)}=\frac{-\sqrt{3}}{1+z\overline{z}}(-\sqrt{3}\overline{z}^2, \ \sqrt{3}z,\ 2\overline{z}-z\overline{z}^2, \ -1+2z\overline{z},\ 0,\ldots ,\ 0)^T. \end{aligned}$$

Obviously, in this case $\varphi$ is congruent to the case (1) with $K=\frac{2}{3},~\Vert B \Vert ^2=\frac{8}{3}$.

(Ib) If $\varphi$ is a linearly full quaternionic mixed pair, then

$$\begin{aligned} \underline{\varphi }_0=\underline{f}_0^{(m)} \oplus {\mathbf {J}} \underline{f}_0^{(m)}, \end{aligned}$$

(3.15)

where $\underline{f}_0^{(m)} :S^2 \rightarrow CP^{m} \subseteq CP^{2n+1}$ ($n\le m\le 2n+1$) is holomorphic and ${\underline{f}}_1^{(m)} \perp {\mathbf {J}}{\underline{f}_0^{(m)}}$.

Obviously, $\varphi _0$ belongs to the following harmonic sequence

$$\begin{aligned} 0\mathop {\longleftarrow }\limits ^{A_{m}''} \mathbf {J}{{\underline{f}}_m^{(m)}} \mathop {\longleftarrow }\limits ^{A_{m-1}''} \ldots \mathop {\longleftarrow }\limits ^{A_{1}''} \mathbf {J}{{\underline{f}}_1^{(m)}} \mathop {\longleftarrow }\limits ^{A_{\varphi _0}''} \underline{\varphi }_0 \mathop {\longrightarrow }\limits ^{A_{\varphi _0}'} {\underline{f}}_1^{(m)} \mathop {\longrightarrow }\limits ^{A_{1}'} \ldots \mathop {\longrightarrow }\limits ^{A_{m-1}'} {\underline{f}}_m^{(m)} \mathop {\longrightarrow }\limits ^{A_{m}'} 0. \end{aligned}$$

(3.16)

As in the case (Ia), let $f_0^{(m)}$ be a holomorphic section of ${\underline{f}} _0^{(m)}$, i.e., $\overline{\partial } f_0 ^{(m)} = 0$, and $f_j^{(m)}$ $~(j=1,\ldots ,m)$ satisfy the corresponding formulas. From (3.15), we have $\varphi _0=\frac{{f}_0^{(m)}({f}_0^{(m)})^*}{|{f}_0^{(m)}|^2}+\frac{({\mathbf {J}} {f}_0^{(m)})({\mathbf {J}} {f}_0^{(m)})^*}{|{f}_{0}^{(m)}|^2}$. Then by (2.1) and a series of calculations, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda ^2=2l_{0}^{(m)},\\ K=2-\frac{l_{1}^{(m)}}{l_{0}^{(m)}},\\ \Vert B \Vert ^2=2\frac{l_{1}^{(m)}}{l_{0}^{(m)}},\\ tr[A_z, P][A_z^*, P^*]=-\frac{1}{4}\frac{\big |\left\langle f_2^{(m)},~\mathbf {J}{f_1^{(m)}}\right\rangle \big |^2}{|f_1^{(m)}|^4},\\ tr[A_z, A_z^*][P, P^*]=\frac{1}{2}\left( l_{1}^{(m)}-\frac{\big |\left\langle f_2^{(m)},~\mathbf {J}{f_1^{(m)}}\right\rangle \big |^2}{|f_1^{(m)}|^4}\right) . \end{array}\right. } \end{aligned}$$

(3.17)

Now, we prove that if $\varphi :S^2 \rightarrow HP^n$ is a linearly full quaternionic mixed pair with parallel second fundamental form, then, up to $Sp(n+1),$ it belongs to one of the following three cases: (2) $\varphi =\left[ (1,~z)^T\right] :S^2\rightarrow CP^1\subset HP^1$, with $K=2, ~B=0;$

(3) $\varphi =\left[ (1,~\sqrt{2}z,z^2)^T\right] :S^2\rightarrow CP^2\subset HP^2$, with $K=1, ~||B||^2=2$;

(4) $\varphi =\left[ \!(1-\frac{1}{2}\overline{z}^3j,~\sqrt{3}z+\frac{\sqrt{3}}{2}\overline{z}^2j,\frac{3}{2}z^2,\frac{\sqrt{3}}{2}z^3)^T\!\right] :S^2\rightarrow \textit{HP}^3,$ with $K=\frac{2}{3}, ~||B||^2=\frac{1}{3}$.

If $\varphi$ is a linearly full quaternionic mixed pair with parallel second fundamental form, then applying Lemma 3.1 and substituting (3.17) into (3.1), we get

$$\begin{aligned} \frac{\Big |\left\langle f_2^{(m)}, ~\mathbf {J}{f_1^{(m)}}\right\rangle \Big |^2}{|f_1^{(m)}|^4}=\frac{3}{4}l_1 ^{(m)}\left( \frac{l_1 ^{(m)}}{l_0^{(m)}}-1\right) . \end{aligned}$$

(3.18)

Since the metric $ds^2=2l_{0}^{(m)}dzd\overline{z}$ induced by $\varphi$ is of constant curvature, and the metric induced by $\underline{f}_0^{(m)}$ is $ds^2=l_{0}^{(m)}dzd\overline{z}$, then it follows from ([2], Theorem 5.4) that $\underline{f}_0^{(m)}, \underline{f}_1^{(m)}, \ldots , \underline{f}_{m}^{(m)}$ is the Veronese sequence in $CP^{m}\subset CP^{2n+1}$, up to $U(2n+2)$. Then from (3.7) and (3.18) we get

$$\begin{aligned} \Big |\left\langle f_2^{(m)}, ~\mathbf {J}{f_1^{(m)}}\right\rangle \Big |^2=\frac{3m(m-1)(m-2)}{2}(1+z\overline{z})^{2m-6}. \end{aligned}$$

(3.19)

We denote by $r$ the isotropy order of $\varphi$ (cf. [3], §3A). If $r$ is finite, then $r=2s~(1\le s\le n+1)$ by ([1], Proposition 3.2). Otherwise, $r = \infty$, in which case $\varphi$ is called strongly isotropic (cf. [1], section 2C).

If $m=1$, observing (3.17), we find $K=2, ~B=0$. It belongs to the case of totally geodesic. If $m=2$, since $r\ge 2$, which implies ${\underline{f}}_2^{(2)} \perp {\mathbf {J}}{\underline{f}_0^{(2)}}$ by (3.16), then we have $\left\langle f_2^{(2)}, ~\mathbf {J}{f_1^{(2)}}\right\rangle =\partial \left\langle f_2^{(2)}, ~\mathbf {J}{f_0^{(2)}}\right\rangle =0$, which implies (3.19) holds. Hence, its second fundamental form is parallel. In fact, the above two cases are both strongly isotropic.

If $m\ge 3$, from (3.19) we find $\left\langle f_3^{(m)}, ~\mathbf {J}{f_0^{(m)}}\right\rangle =-\left\langle f_2^{(m)}, ~\mathbf {J}{f_1^{(m)}}\right\rangle \ne 0$, which implies in this case $r=2$. In the following, we discuss the above three cases, respectively.

Case Ib1, $m=1$.

In this case, we have

$$\begin{aligned} \underline{\varphi }_0=U\underline{V}_0^{(1)} \oplus {\mathbf {J}} U\underline{V}_0^{(1)}, \end{aligned}$$

(3.20)

where $\underline{V}_0^{(1)}$ is a Veronese surface in $CP^1\subset CP^{2n+1}$ with the standard expression given in ([2], section 5), and $U\in U(2n+2)$ satisfies $tr\Big (V_1^{(1)} {V_0^{(1)}}^{T} U^{T}J_{n+1}U\Big )=0$, as this expresses the orthogonality of $Jf_0^{(1)}$ and $f_1^{(1)}$.

Similarly, we get the type of $\overline{W}=U^{T}J_{n+1}U\in U(2n+2)$ as follows:

(3.21)

As in case (1a), by (3.12),(3.13), and (3.21), we get

$$\begin{aligned} {\left\{ \begin{array}{ll} e_2=\overline{W}\overline{e}_1=\left( 0,0,-a_{13},-a_{14},\ldots ,-a_{1,2n+2}\right) ^T,\\ e_4=\overline{W}\overline{e}_3=\left( 0,0,-a_{23},-a_{24},\ldots ,-a_{2,2n+2}\right) ^T. \end{array}\right. } \end{aligned}$$

Since $\overline{W}$ in (3.21) is a unitary matrix, $\{e_1,e_2,e_3,e_4\}$ are mutually orthogonal. Similarly, we can choose $n-1$ proper unit column vectors $e_{2p+1}=(0,0,*)^T\in \mathbb {C}^{2n+2}~(2\le p\le n)$ such that $\{e_1,e_2,\ldots ,e_{2n+1},e_{2n+2}=\overline{W}\overline{e}_{2n+1}\}$ are mutually orthogonal, and the type of the corresponding $U$ is as follows:

(3.22)

Thus, we have

$$\begin{aligned}&UV_0^{(1)}=(1,~0,~z,~0,~0,\ldots ,0)^T,\\&{\mathbf {J}}UV_0^{(1)}=(0,1,~0,~\overline{z},~0,\ldots ,0)^T. \end{aligned}$$

Obviously, in this case $\varphi$ is congruent to the case (2) with $K=2,~B=0$.

Case Ib2, $m=2$.

In this case, we have

$$\begin{aligned} \underline{\varphi }_0=U\underline{V}_0^{(2)} \oplus {\mathbf {J}} U\underline{V}_0^{(2)}, \end{aligned}$$

(3.23)

where $\underline{V}_0^{(2)}$ is a Veronese surface in $CP^2\subset CP^{2n+1}$ with the standard expression given in ([2], §5), and $U\in U(2n+2)$ satisfies $tr\Big ( V_1^{(2)} {V_0^{(2)}}^{T} U^{T}J_{n+1}U\Big )=0$, as this expresses the orthogonality of $Jf_0^{(2)}$ and $f_1^{(2)}$.

Similarly, we get the type of $\overline{W}=U^{T}J_{n+1}U\in U(2n+2)$ as follows:

(3.24)

And the type of the corresponding $U$ is as follows:

(3.25)

Thus, we have

$$\begin{aligned} UV_0^{(2)}=(1,~0,~\sqrt{2}z,~0,~z^2,~0,~0,\ldots ,0)^T,\\ {\mathbf {J}}UV_0^{(2)}=(0,~1,~0,~\sqrt{2}\overline{z},~0,~\overline{z}^2,~0,\ldots ,0)^T. \end{aligned}$$

Obviously, in this case $\varphi$ is congruent to the case (3) with $K=1,~\Vert B \Vert ^2=2$.

Case Ib3, $m\ge 3$.

In this case, the trivial bundle $S^2 \times \mathbb {C}^{2n+2}$ over $S^2$ has a corresponding decomposition $S^2 \times \mathbb {C}^{2n+2} = S^2 \times \mathbb {C}^{m+1} \oplus S^2 \times \mathbb {C}^{2n-m+1}$. From (3.16) we set $\mathbf {J}{f_0^{(m)}}=x_3f_3^{(m)}+x_4f_4^{(m)}+\cdots +x_mf_m^{(m)}+V$, where $x_i~(i=3,\ldots ,m)$ are complex coefficients and bundle $\underline{V}\subset S^2 \times \mathbb {C}^{2n-m+1}$. Then, it follows from $\partial {\mathbf {J}{f_0^{(m)}}}=0$ that

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial x_3+x_3\partial \log |f_3^{(m)}|^2=0,\\ \partial x_i+x_{i-1}+x_i\partial \log |f_i^{(m)}|^2=0,~(i=4,\ldots ,m),\\ \partial V=0. \end{array}\right. } \end{aligned}$$

(3.26)

And we have $\left\langle f_3^{(m)}, ~\mathbf {J}{f_0^{(m)}}\right\rangle =\overline{x}_3 |f_3^{(m)}|^2$. Then, from ([2], §5) and (3.19) we get

$$\begin{aligned} |x_3|^2|f_3^{(m)}|^4=\frac{3m(m-1)(m-2)}{2}(1+z\overline{z})^{2m-6}. \end{aligned}$$

(3.27)

By (3.26) and (3.27), we find $\partial \overline{\partial }\log \left( |x_3|^2|f_3^{(m)}|^4\right) =\frac{2m-6}{(1+z\overline{z})^2}=0$, which implies $m=3$, i.e.,

$$\begin{aligned} \underline{\varphi }_0=U\underline{V}_0^{(3)} \oplus {\mathbf {J}} U\underline{V}_0^{(3)}, \end{aligned}$$

(3.28)

where $\underline{V}_0^{(3)}$ is a Veronese surface in $CP^3\subset CP^{2n+1}$ with the standard expression given in ([2], §5), and $U\in U(2n+2)$ satisfies $tr\Big ( V_1^{(3)} {V_0^{(3)}}^{T} U^{T}J_{n+1}U\Big )=0$ and $tr\Big ( V_3^{(3)} {V_0^{(3)}}^{T} U^{T}J_{n+1}U\Big )\ne 0$, as these express $r=2$.

Similarly, we get the type of $\overline{W}=U^{T}J_{n+1}U\in U(2n+2)$ as follows:

(3.29)

where $a_{14}\ne 0$.

In this case, if $|a_{14}|^2=1$, as in the case (Ia) we choose (3.13), then choose $n-1$ proper unit column vectors $e_{2p+1}=(0,0,0,0,*)^T\in \mathbb {C}^{2n+2}~(2\le p\le n)$ such that the type of the corresponding $U\in U(2n+2)$ is as follows:

(3.30)

Thus, we have

$$\begin{aligned}&UV_0^{(3)}=(1,~-a_{14}z^3,~\sqrt{3}z,~\sqrt{3}a_{14}z^2,~0,~0,\ldots ,0)^T,\\&{\mathbf {J}}UV_0^{(3)}=(\overline{a}_{14}\overline{z}^3,~1,~-\sqrt{3}\overline{a}_{14}\overline{z}^2,~\sqrt{3}\overline{z},~0,~0,\ldots ,0)^T. \end{aligned}$$

If $|a_{14}|^2\ne 1$, then we choose

$$\begin{aligned} {\left\{ \begin{array}{ll} e_1=(1,0,0,0,0,\ldots ,0)^T,\\ e_3=(0,1,0,0,0,\ldots ,0)^T,\\ e_5=\frac{1}{\sqrt{1-|a_{14}|^2}}(0,0,1-|a_{14}|^2,0,\overline{a}_{14}a_{25},\ldots ,\overline{a}_{14}a_{2,2n+2})^T,\\ e_7=\frac{1}{\sqrt{1-|a_{14}|^2}}(0,0,0,1-|a_{14}|^2,-\overline{a}_{14}a_{15},\ldots ,-\overline{a}_{14}a_{1,2n+2})^T. \end{array}\right. } \end{aligned}$$

(3.31)

And we choose $n-3$ proper unit column vectors $e_{2p+1}=(0,0,0,0,*)^T\in \mathbb {C}^{2n+2}~(4\le p\le n)$ such that the type of the corresponding $U\in U(2n+2)$ is as follows:

(3.32)

Thus, we have

$$\begin{aligned} UV_0^{(3)}{=}(1,~-a_{14}z^3,~\sqrt{3}z,~\sqrt{3}a_{14}z^2,\sqrt{3-3|a_{14}|^2}z^2,~0,\sqrt{1-|a_{14}|^2}z^3,~0,~0,\ldots ,0)^T,\\ {\mathbf {J}}UV_0^{(3)}{=}(\overline{a}_{14}\overline{z}^3,1,~-\sqrt{3}\overline{a}_{14}\overline{z}^2,~\sqrt{3}\overline{z},~0,\sqrt{3-3|a_{14}|^2}\overline{z}^2,~0,~\sqrt{1-|a_{14}|^2}\overline{z}^3,~0,\ldots ,0)^T. \end{aligned}$$

From (3.30) and (3.32), we have $\left\langle f_3^{(3)}, ~\mathbf {J}{f_0^{(3)}}\right\rangle =\left\langle UV_3^{(3)}, ~\mathbf {J}{U V_0^{(3)}}\right\rangle =6a_{14}$. On the other hand, from (3.27) we get $\left| \left\langle f_3^{(3)}, ~\mathbf {J}{f_0^{(3)}}\right\rangle \right| ^2=9$. So $a_{14}=\frac{1}{2}e^{\sqrt{-1}\theta }~(0\le \theta \le 2\pi )$. Hence, in this case $\varphi$ is congruent to the case (4) with $K=\frac{2}{3},~\Vert B \Vert ^2=\frac{1}{3}$.

(II) Let $\varphi :S^2 \rightarrow HP^n$ be an irreducible linearly full harmonic map. At first, we state a conclusion about parallel minimal immersions of 2-spheres in $G(k,N)$ as follows:

Lemma 3.2

([7]) Let $\varphi :S^2 \rightarrow G(k,N)$ be a conformal minimal immersion with the second fundamental form $B$. Suppose that $B$ is parallel, then the following equations

$$\begin{aligned} \left\{ \begin{array}{l} \lambda ^2(2K+||B||^2)A_{z}^*+4\left[ [A_z,A_z^*],A_{z}^*\right] =0,\\ \lambda ^2(\frac{||B||^2}{4}-K)P+\left[ [A_z,A_z^*],P\right] =0 \end{array}\right. \end{aligned}$$

(3.33)

hold.

By [1], we know $\varphi _0$ belongs to the following harmonic sequence

$$\begin{aligned} 0{\longleftarrow } \cdots \mathop {\longleftarrow }\limits ^{A''_{\varphi _2}} \underline{\varphi }_{-2} \mathop {\longleftarrow }\limits ^{A''_{\varphi _1}} \underline{\varphi }_{-1} \mathop {\longleftarrow }\limits ^{A''_{\varphi _0}} \underline{\varphi }_0 \mathop {\longrightarrow }\limits ^{A'_{\varphi _0}} \underline{\varphi }_{1} \mathop {\longrightarrow }\limits ^{A'_{\varphi _1}} \underline{\varphi }_{2} \mathop {\longrightarrow }\limits ^{A'_{\varphi _2}} \cdots {\longrightarrow } 0, \end{aligned}$$

(3.34)

where $\underline{\varphi }_0=\underline{i_2\circ \varphi },~\underline{\varphi }_{-1}={\mathbf {J}}\underline{\varphi }_{1},~\underline{\varphi }_{-2}={\mathbf {J}}\underline{\varphi }_{2}$.

We choose a unit column vector $X\in \underline{\varphi }_0$, then we have

$$\begin{aligned} \underline{\varphi }_0=\underline{X} \oplus \mathbf {J} \underline{X},\quad \underline{\varphi }_1=span\Big \{\underline{X}_1, \underline{Y}_1\Big \}, \end{aligned}$$

(3.35)

where $X_1=\partial {X}-\left\langle \partial {X},X\right\rangle X-\left\langle \partial {X},\mathbf {J}X\right\rangle \mathbf {J}X$ and $Y_1=\partial {\mathbf {J}X}-\left\langle \partial {\mathbf {J}X},X\right\rangle X-\left\langle \partial {\mathbf {J}X},\mathbf {J}X\right\rangle \mathbf {J}X$. Here, $X_1$ and $Y_1$ are not orthogonal in general.

Let

$$\begin{aligned} E=[X,\mathbf {J}X]^*\partial [X,\mathbf {J}X]. \end{aligned}$$

(3.36)

Then, from (3.35), we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial [X,\mathbf {J}X]=[X,\mathbf {J}X]E+[X_1,Y_1],\\ \overline{\partial }[X,\mathbf {J}X]=-[X,\mathbf {J}X]E^*+[-\mathbf {J}Y_1,\mathbf {J}X_1], \end{array}\right. } \end{aligned}$$

(3.37)

where $[X_1,Y_1]^*[X,\mathbf {J}X]=[-\mathbf {J}Y_1,\mathbf {J}X_1]^*[X,\mathbf {J}X]=0$.

From (3.37) and the identity $\partial \overline{\partial }=\overline{\partial }\partial$, we get

(3.38)

From (3.35), we have $\varphi _0=XX^*+({\mathbf {J}} X)({\mathbf {J}} X)^*$. Then by (2.1), a straightforward calculation shows

$$\begin{aligned} {\left\{ \begin{array}{ll} \lambda ^2=2(|X_1|^2+|Y_1|^2),\\ A_{z}=(\mathbf {J}X)(\mathbf {J}X_1)^*-X(\mathbf {J}Y_1)^*-X_1X^*-Y_1(\mathbf {J}X)^*. \end{array}\right. } \end{aligned}$$

(3.39)

Since $\varphi _0$ is harmonic, the corresponding equivalent condition $\overline{\partial }A_z+A_zA_z^*-A_z^*A_z=0$ (cf. [12]) implies

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\left\langle \overline{\partial }X_1,X_1\right\rangle }{|X_1|^2}=-\frac{\left\langle \overline{\partial }Y_1,Y_1\right\rangle }{|Y_1|^2}=\left\langle \overline{\partial }X,X\right\rangle ,\\ \frac{\left\langle \overline{\partial }X_1,Y_1\right\rangle }{|Y_1|^2}=\left\langle \overline{\partial }X,\mathbf {J}X\right\rangle ,\quad \frac{\left\langle \overline{\partial }Y_1,X_1\right\rangle }{|X_1|^2}=\left\langle \overline{\partial }\mathbf {J}X,X\right\rangle . \end{array}\right. } \end{aligned}$$

(3.40)

Now, we prove that if $\varphi :S^2 \rightarrow HP^n$ is an irreducible linearly full harmonic map with parallel second fundamental form, then, up to $Sp(n+1),$ it belongs to one of the following two cases: (5) $\varphi =\left[ (-2\overline{z},~\sqrt{2}-\sqrt{2}z\overline{z},~2z)^T\right] :S^2\rightarrow CP^2\subset HP^2$, with $K=\frac{1}{2}, ~B=0;$ (6) $\varphi =\left[ (6\overline{z}^2,~-6\overline{z}+6z\overline{z}^2,~\sqrt{6}-4\sqrt{6}z\overline{z}+\sqrt{6}z^2\overline{z}^2,~6z-6z^2\overline{z}, ~6z^2)^T\right] :S^2\rightarrow CP^4\subset HP^4,$ with $K=\frac{1}{6}, ~\Vert B\Vert ^2=\frac{2}{3}$.

If $\varphi :S^2 \rightarrow HP^n$ is an irreducible linearly full harmonic map with parallel second fundamental form, then applying Lemma 3.2 and substituting (3.39) into the first equation of (3.33), we get

$$\begin{aligned} \lambda ^2=4|X_1|^2,\quad 2K+||B||^2=1,\quad \left\langle X_1,Y_1\right\rangle =0,\quad |X_1|^2=|Y_1|^2. \end{aligned}$$

(3.41)

From (3.38) and (3.41) we have

$$\begin{aligned} \overline{\partial }E+\partial E^*+[E,E^*]=0. \end{aligned}$$

(3.42)

Let $\widetilde{X}\in \underline{\varphi }_0$ be another unit column vector such that $\underline{\varphi }_0=\underline{\widetilde{X}} \oplus \mathbf {J} \underline{\widetilde{X}}$, then

$$\begin{aligned}{}[\widetilde{X},\mathbf {J}\widetilde{X}]=[X,\mathbf {J}X]T, \end{aligned}$$

(3.43)

where $T:S^2\rightarrow SU(2)$ is to be determined such that $\widetilde{X}$ satisfies $[\widetilde{X},\mathbf {J}\widetilde{X}]^*d[\widetilde{X},\mathbf {J}\widetilde{X}]=0$. Such $T$ is a solution of the linear PDE

$$\begin{aligned} dT+(Edz-E^*d\overline{z})T=0. \end{aligned}$$

(3.44)

The integrability condition of (3.44) is just (3.42), so it has a unique solution locally on $S^2$ for any given initial value. Let $T$ be a solution of (3.44) with the initial value $T(0)\in SU(2)$. From (3.44) we have $d(T^*T)=0$ and $d|T|=0$, so $T\in SU(2)$.

Now, we choose a unit column vector $X\in \underline{\varphi }_0$ such that $\underline{\varphi }_0=\underline{X} \oplus \mathbf {J} \underline{X}$ and

$$\begin{aligned}{}[X,\mathbf {J}X]^*d[X,\mathbf {J}X]=0. \end{aligned}$$

(3.45)

It follows from (3.40) and (3.45) that

$$\begin{aligned} \partial \overline{\partial }X=-|X_1|^2X. \end{aligned}$$

(3.46)

Let $\underline{f}=[X]:S^2\rightarrow CP^{2n+1}$ be a smooth immersion. Similarly, by calculating the equivalent condition of harmonic, we find $\underline{f}$ is harmonic by (3.45) and (3.46). Of course, $\mathbf {J}\underline{f}=[\mathbf {J}X]:S^2\rightarrow CP^{2n+1}$ is also harmonic. So $\underline{\varphi }_0=\underline{f}\oplus \mathbf {J}\underline{f}$, where $\underline{f}$ belongs to the following harmonic sequence

$$\begin{aligned} 0{\longrightarrow } \cdots \mathop {\longrightarrow }\limits ^{A_{p-1}''} \underline{f}_{p-1} \mathop {\longrightarrow }\limits ^{A_{p}''} \underline{f}_p=\underline{f} \mathop {\longrightarrow }\limits ^{A_{p}'} \underline{f}_{p+1} \mathop {\longrightarrow }\limits ^{A_{p+1}'} \cdots {\longrightarrow } 0. \end{aligned}$$

(3.47)

As in the case (Ia), let $f_0$ be a holomorphic section of ${\underline{f}} _0$, i.e. $\overline{\partial } f_0 = 0$, and $f_p$ satisfy the corresponding formulas. From (3.41), we know $l_{p-1}=l_p$, which implies that $\underline{f}_p$ is totally real by ([2], Theorem 7.3), i.e. $\underline{f}_p=\underline{f}_m^{(2m)}:S^2\rightarrow RP^{2m}\subset CP^{2m}\subset CP^{2n+1}$, where $2\le 2m\le 2n+1$. Let $f_p={f}_m^{(2m)}$ satisfy the corresponding formulas, then in harmonic sequence (3.34) by (3.40) and (3.45) we have

$$\begin{aligned} \underline{\varphi }_0=\underline{f}_m^{(2m)} \oplus {\mathbf {J}} \underline{f}_m^{(2m)},\quad \underline{\varphi }_1=\underline{f}_{m+1}^{(2m)} \oplus {\mathbf {J}} \underline{f}_{m-1}^{(2m)},\quad \underline{\varphi }_2=\underline{f}_{m+2}^{(2m)}\oplus {\mathbf {J}} \underline{f}_{m-2}^{(2m)}, \end{aligned}$$

(3.48)

where $l_i^{(2m)}=l_{2m-1-i}^{(2m)}\ (i=0,\ldots ,m-1)$ and $\underline{\varphi }_0,\ \underline{\varphi }_1,\ \underline{\varphi }_2$ are mutually orthogonal.

At this time, from (3.48), we have $\varphi _0=\frac{{f}_m^{(2m)}({f}_m^{(2m)})^*}{|{f}_m^{(2m)}|^2}+\frac{({\mathbf {J}} {f}_m^{(2m)})({\mathbf {J}} {f}_m^{(2m)})^*}{|{f}_{m}^{(2m)}|^2}$. Then by (2.1) and a series of calculations, we obtain

$$\begin{aligned} \left\{ \begin{array}{l} \lambda ^2=4l_m^{(2m)},\\ K=\frac{1}{2}-\frac{l_{m+1}^{(2m)}}{2l_{m}^{(2m)}},\\ A_{z}=\frac{(\mathbf {J}f_{m}^{(2m)})(\mathbf {J}f_{m+1}^{(2m)})^*}{|f_{m}^{(2m)}|^2}+\frac{(\mathbf {J}f_{m-1}^{(2m)})(\mathbf {J}f_{m}^{(2m)})^*}{|f_{m-1}^{(2m)}|^2} -\frac{f_{m+1}^{(2m)}{f_{m}^{(2m)}}^*}{|f_{m}^{(2m)}|^2}-\frac{f_{m}^{(2m)}{f_{m-1}^{(2m)}}^*}{|f_{m-1}^{(2m)}|^2},\\ P=\frac{1}{4}\left[ \frac{l_{m-2}^{(2m)}}{|f_{m}^{(2m)}|^2}f_{m}^{(2m)}{f_{m-2}^{(2m)}}^*+\frac{(\mathbf {J}f_{m}^{(2m)})(\mathbf {J}f_{m+2}^{(2m)})^*}{|f_{m+1}^{(2m)}|^2} -\frac{f_{m+2}^{(2m)}{f_{m}^{(2m)}}^*}{|f_{m+1}^{(2m)}|^2}-\frac{l_{m-2}^{(2m)}}{|f_{m}^{(2m)}|^2}(\mathbf {J}f_{m-2}^{(2m)})(\mathbf {J}f_{m}^{(2m)})^*\right] ,\\ ||B||^2=\frac{l_{m+1}^{(2m)}}{l_{m}^{(2m)}}. \end{array}\right. \quad \end{aligned}$$

(3.49)

Then applying Lemma 3.2 and substituting (3.49) into the second equation of (3.33), we get $m=1$ or

$$\begin{aligned} \left\langle \mathbf {J}f_{m-2}^{(2m)},f_{m-1}^{(2m)}\right\rangle =\left\langle f_{m+2}^{(2m)},\mathbf {J}f_{m+1}^{(2m)}\right\rangle =0,\ 3l_{m+1}^{(2m)}=2l_{m}^{(2m)}. \end{aligned}$$

(3.50)

In the latter case, since $l_{m+1}^{(2m)}=\frac{(m+2)(m-1)}{(1+z\overline{z})^2}$ and $l_{m}^{(2m)}=\frac{(m+1)m}{(1+z\overline{z})^2}$ by ([2], §5), we have $m=2$ by (3.50). Hence, in the following, we discuss the above two cases of $m=1$ and $m=2$ respectively.

Case II1, $m=1$.

In this case, by (3.50) we have

$$\begin{aligned} \underline{\varphi }_0=U\underline{V}_1^{(2)} \oplus {\mathbf {J}} U\underline{V}_1^{(2)}, \end{aligned}$$

(3.51)

where $\underline{V}_1^{(2)}$ is a Veronese surface in $CP^2\subset CP^{2n+1}$ with the standard expression given in ([2], Section 5) and $U\in U(2n+2)$ satisfies $tr\Big ( V_2^{(2)} {V_0^{(2)}}^{T} U^{T}J_{n+1}U\Big )=0$, as this expresses the orthogonality of $Jf_0^{(2)}$ and $f_2^{(2)}$.

By calculating, we find in this case $\overline{W}=U^{T}J_{n+1}U\in U(2n+2)$ is the same type as (3.24). Then, the type of the corresponding $U\in U(2n+2)$ is the same as (3.25). Thus, we have

$$\begin{aligned}&UV_1^{(2)}=\frac{1}{1+z\overline{z}}(-2\overline{z},~0,~\sqrt{2}-\sqrt{2}z\overline{z},~0,~2z,~0,~0,\ldots ,0)^T,\\&{\mathbf {J}}UV_1^{(2)}=\frac{1}{1+z\overline{z}}(0,~-2{z},~0,~\sqrt{2}-\sqrt{2}z\overline{z},~0,~2\overline{z},~0,\ldots ,0)^T. \end{aligned}$$

In this case, it is easy to check that the corresponding map $\varphi$ is totally geodesic. Obviously, it is congruent to the case (5) with $K=\frac{1}{2},~B=0$.

Case II2, $m=2$.

In this case, by (3.50) we have

$$\begin{aligned} \underline{\varphi }_0=U\underline{V}_2^{(4)} \oplus {\mathbf {J}} U\underline{V}_2^{(4)}, \end{aligned}$$

(3.52)

where $\underline{V}_2^{(4)}$ is a Veronese surface in $CP^4\subset CP^{2n+1}$ with the standard expression given in ([2], Section 5) and $U\in U(2n+2)$ satisfies $tr\Big ( V_4^{(4)} {V_0^{(4)}}^{T} U^{T}J_{n+1}U\Big )=0$, as this expresses the orthogonality of $Jf_0^{(4)}$ and $f_4^{(4)}$.

Similarly, we get the type of $\overline{W}=U^{T}J_{n+1}U\in U(2n+2)$ as follows:

And the type of the corresponding $U\in U(2n+2)$ is as follows:

(3.53)

Thus, we have

$$\begin{aligned} UV_2^{(4)}&= \frac{2}{(1+z\overline{z})^2}(6\overline{z}^2,0,-6\overline{z}+6z\overline{z}^2,0,\sqrt{6}-4\sqrt{6}z\overline{z} \\&+\,\,\sqrt{6}z^2\overline{z}^2,0,6z-6z^2\overline{z},0,6z^2,0,\ldots ,0)^T,\\ {\mathbf {J}}UV_2^{(4)}&= \frac{2}{(1+z\overline{z})^2}(0,~6{z}^2,0,-6z+6z^2\overline{z},0,\sqrt{6}-4\sqrt{6}z\overline{z} \\&+ \,\,\sqrt{6}z^2\overline{z}^2,0,6\overline{z}-6z\overline{z}^2,0,6\overline{z}^2,\ldots ,0)^T. \end{aligned}$$

In this case, it is easy to check that the Eq. (3.1) holds, which shows the second fundamental form of the corresponding map $\varphi$ is parallel. Obviously, it is congruent to the case (6) with $K=\frac{1}{6},~||B||^2=\frac{2}{3}$.

It is easy to check that no two of the above six cases are congruent, i.e., we cannot transform any one into any other by left multiplication by $Sp(n+1)$. To sum up, we get Theorem 1.1.