Totally Real Flat Minimal Surfaces in Quaternionic Projective Spaces

TL;DR

This paper classifies totally real flat minimal surfaces in quaternionic projective spaces, showing they correspond to specific surfaces in complex projective spaces, including the Clifford solution, up to symplectic congruence.

Contribution

It provides a classification of totally real flat minimal surfaces in quaternionic projective spaces, linking them to known surfaces in complex projective spaces and identifying the Clifford solution.

Findings

Classification of totally real flat minimal surfaces in $ ext{HP}^n$

Identification of surfaces as those in $ ext{CP}^n$

Inclusion of the Clifford solution

Abstract

In this paper, we study totally real minimal surfaces in the quaternionic projective space $\mathbb{H}P^n$. We prove that the linearly full totally real flat minimal surfaces of isotropy order $n$ in $\mathbb{H}P^n$ are two surfaces in $\mathbb{C}P^n$, one of which is the Clifford solution, up to symplectic congruence.

Equations

$$\mathbb{H}=\left\{a+b\text{i}+c\text{j}+d\text{k}|a,b,c,d\in \mathbb{R},{\text{i}}^2={\text{j}}^2={\text{k}}^2=\text{i}\text{j}\text{k}=-1\right\}.$$

$$\text{j}(z_1,z_2,\dots ,z_{2n+1},z_{2n+2}{)}^T=(-{\bar{z}}_2,{\bar{z}}_1,\dots ,-{\bar{z}}_{2n+2},{\bar{z}}_{2n+1}{)}^T.$$

$$\text{j}v=J_{n+1}\bar{v},$$

$$\pi :S^{4n+3}\left(\subset {\mathbb{C}}^{2n+2}\right)$$

$$q$$

$${\mathcal{V}}_q={\text{span}}_{\mathbb{R}}\left\{\text{i}q,\text{j}q,\text{k}q\right\}.$$

$${\mathcal{H}}_q={\left({\mathcal{V}}_q\right)}^{\perp }=\left\{v\in {\mathbb{C}}^{2n+2}|⟨v,q⟩=⟨v,\text{j}q⟩=0\right\},$$

$$T_q{S}^{4n+3}={\mathcal{H}}_q\oplus {\mathcal{V}}_q$$

$$\text{I}(X)={\pi }_{\ast }(\text{i}{X}^{\mathcal{H}}),\text{J}(X)={\pi }_{\ast }(\text{j}{X}^{\mathcal{H}}),\text{K}(X)={\pi }_{\ast }(\text{k}{X}^{\mathcal{H}}).$$

$${\omega }_{\text{I}}(X,Y)=g_{\mathbb{H}{P}^n}(\text{I}X,Y),{\omega }_{\text{J}}(X,Y)=g_{\mathbb{H}{P}^n}(\text{J}X,Y),{\omega }_{\text{K}}(X,Y)=g_{\mathbb{H}{P}^n}(\text{K}X,Y),$$

$$\Omega ={\omega }_{\text{I}}\wedge {\omega }_{\text{I}}+{\omega }_{\text{J}}\wedge {\omega }_{\text{J}}+{\omega }_{\text{K}}\wedge {\omega }_{\text{K}}$$

$$U\subset M$$

$$g_M={\phi }^{\ast }{g}_{\mathbb{H}{P}^n}=e^{2u}dzd\bar{z}=e^{2u}(d{x}^2+d{y}^2),$$

$$d^{\mathcal{H}}s=ds-⟨ds,s⟩s-⟨ds,\text{j}s⟩\text{j}s,$$

$${\phi }_{\ast }{e}_1$$

$${\phi }_{\ast }{e}_2=e^{-u}{\pi }_{\ast }\left[\text{i}\left(d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)-d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right)\right].$$

$$g_M(e_1,e_1)$$

$$g_M(e_2,e_2)$$

$$g_M(e_1,e_2)=e^{-2u}\text{i}\left\{⟨d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right),d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)⟩-⟨d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right),d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)⟩\right\}.$$

$$e^{2u}={\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)\right|}^2+{\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right|}^2,⟨d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right),d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)⟩=0.$$

$$\cos {\alpha }_1=\frac{{\phi }^{\ast }({\omega }_{\text{I}})}{d{\mu }_M},\cos {\alpha }_2=\frac{{\phi }^{\ast }({\omega }_{\text{J}})}{d{\mu }_M},\cos {\alpha }_3=\frac{{\phi }^{\ast }({\omega }_{\text{K}})}{d{\mu }_M},$$

$${\cos }^2\alpha =\frac{{\left({\phi }^{\ast }({\omega }_{\text{I}})\right)}^2+{\left({\phi }^{\ast }({\omega }_{\text{J}})\right)}^2+{\left({\phi }^{\ast }({\omega }_{\text{K}})\right)}^2}{{\left(d{\mu }_M\right)}^2}.$$

$$\cos {\alpha }_1$$

$$\cos {\alpha }_2$$

$$\cos {\alpha }_3$$

$${\cos }^2\alpha =\frac{{\left({\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)\right|}^2-{\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right|}^2\right)}^2+4{\left|⟨d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right),\text{j}{d}^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)⟩\right|}^2}{{\left({\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)\right|}^2+{\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right|}^2\right)}^2}.$$

$${\cos }^2\alpha =1-4\frac{{\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)\right|}^2{\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right|}^2-{\left|⟨d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right),\text{j}{d}^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)⟩\right|}^2}{{\left({\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)\right|}^2+{\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right|}^2\right)}^2}\le 1.$$

$${\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right)\right|}^2={\left|d^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)\right|}^2,⟨d^{\mathcal{H}}s\left(\frac{\partial }{\partial z}\right),\text{j}{d}^{\mathcal{H}}s\left(\frac{\partial }{\partial \bar{z}}\right)⟩=0.$$

$${\partial }_z=\frac{\partial }{\partial z},{\partial }_{\bar{z}}=\frac{\partial }{\partial \overline{z}}.$$

$${\partial }_{\bar{z}}{\partial }_{z}s-⟨{\partial }_{\bar{z}}{\partial }_{z}s,s⟩s-⟨{\partial }_{z}s,s⟩{\partial }_{\bar{z}}s-⟨{\partial }_{\bar{z}}s,s⟩{\partial }_{z}s-2{\left|⟨{\partial }_{z}s,s⟩\right|}^{2}s=0$$

$${\partial }_{\bar{z}}{A}_z=\left[A_z,A_{\bar{z}}\right]$$

Full text

Totally Real Flat Minimal Surfaces in Quaternionic Projective Spaces

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Keywords and Phrases. Minimal surfaces, Totally real, Twistor lift, Quaternionic projective spaces.  Mathematics Subject Classification (2010). 53C26,53C42.

Ling He ††† L. He

Center for Applied Mathematics, Tianjin University, Tianjin 300072, P. R. China

E-mail: heling@tju.edu.cn

and Xianchao Zhou ‡‡‡ X.C. Zhou

Department of Applied Mathematics, Zhejiang University of Technology, Hangzhou 310023, P. R. China

E-mail: zhouxianch07@zjut.edu.cn

ABSTRACT. In this paper, we study totally real minimal surfaces in the quaternionic projective space $\mathbb{H}P^{n}$. We prove that the linearly full totally real flat minimal surfaces of isotropy order $n$ in $\mathbb{H}P^{n}$ are two surfaces in $\mathbb{C}P^{n}$, one of which is the Clifford solution, up to symplectic congruence.

1 Introduction

A.Bahy-El-Dien and J.C.Wood have developed a beautiful and quite complete theory for harmonic two-spheres in the quaternionic projective space $\mathbb{H}P^{n}$(cf.[1]). All harmonic two-spheres in $\mathbb{H}P^{n}$ are generated from quaternionic Frenet pair or quaternionic mixed pair, which are direct sum of certain harmonic two-spheres in $\mathbb{C}P^{2n+1}$, by certain flag transforms called forward and backward replacements. The above theory generates a series of classification results about minimal two-spheres of constant Gauss curvature in $\mathbb{H}P^{n}$(cf.[7],[10], [11],[12]).

It is natural to consider the harmonic maps from Riemann surfaces of higher genus. One of such examples is the totally real superconformal minimal tori in $\mathbb{C}P^{2}$, H.Ma and Y.J.Ma (cf.[17]) described explicitly all these tori by Prym-theta functions. Generally, there is a family of totally real flat minimal surfaces in $\mathbb{C}P^{n}$. This family was given by K.Kenmotsu (cf.[16]) and J.Bolton and L.M.Woodward (cf.[5]) respectively. Later, G.R.Jensen and R.J.Liao discovered continuous families of noncongruent flat minimal tori in $\mathbb{C}P^{n}$ by this family of minimal surfaces(cf.[15]). We will apply this family to characterize totally real flat minimal surfaces in $\mathbb{H}P^{n}$.

It is well known that Veronese sequences are a series of harmonic two-spheres of constant Gauss curvature in $\mathbb{C}P^{n}$ (cf.[2], [4]). In [13], Y.J.He and C.P.Wang proved that the Veronese sequences in $\mathbb{R}P^{2m}$ ($2\leq 2m\leq n$) are the only totally real minimal two-spheres with constant Gauss curvature in $\mathbb{H}P^{n}$. In [21], S.Udagawa studied the classification of harmonic two-tori in $\mathbb{H}P^{2}$ and $\mathbb{H}P^{3}$.

In this paper we mainly study totally real minimal surfaces in $\mathbb{H}P^{n}$. In section 2, we define quaternionic Kähler angle with respect to an isometric immersion from Riemann surface to $\mathbb{H}P^{n}$, which gives a measure of the failure of the immersion to be a totally complex map or a totally real map. In section 3, we study totally real minimal surfaces in $\mathbb{H}P^{n}$ by the method of twistor lift, and get some properties with respect to harmonic sequence (see Theorem 3.4, Proposition 3.5 and Proposition 3.6). In section 4, we give a classification theorem about the linearly full totally real flat minimal surfaces of isotropy order $n$ in $\mathbb{H}P^{n}$ (see Theorem 4.2).

2 Surfaces in quaternionic projective spaces

For any $N=1,2,\ldots,$ let $\left\langle,\right\rangle$ denote the standard Hermitian inner product on $\mathbb{C}^{N}$ defined by $\left\langle z,w\right\rangle=z_{1}\overline{w}_{1}+\ldots+z_{N}\overline{w}_{N}$, where $z=(z_{1},\ldots,z_{N})^{T},w=(w_{1},\ldots,w_{N})^{T}\in{\mathbb{C}}^{N}$ and $\bar{}$ denotes complex conjugation. Let $\mathbb{H}$ denote the division ring of quaternions, i.e.

[TABLE]

Since $a+b\texttt{i}+c\texttt{j}+d\texttt{k}=(a+b\texttt{i})+(c+d\texttt{i})\texttt{j}$, then it gives an identification of ${\mathbb{C}}^{2}$ with $\mathbb{H}$. Let $n\in\left\{1,2,\cdots\right\}$, we have a corresponding identification of ${\mathbb{C}}^{2n+2}$ with ${\mathbb{H}}^{n+1}$. The inner product on ${\mathbb{H}}^{n+1}$ is given by $g_{{\mathbb{H}}^{n+1}}=\text{Re}\left\langle,\right\rangle$. For any $z_{1}+z_{2}\texttt{j}\in\mathbb{H}$, the left multiplication by j is given by $\texttt{j}(z_{1}+z_{2}\texttt{j})=-\bar{z}_{2}+\bar{z}_{1}\texttt{j}$. Then j induces a conjugate linear map from ${\mathbb{C}}^{2n+2}$ to ${\mathbb{C}}^{2n+2}$, also denoted by j, i.e.

[TABLE]

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

[TABLE]

where J_{n+1}=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+1}. By the above, we immediately have the following lemma (cf. [10]).

Lemma 2.1

*The operator j has the following properties:

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

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

(iii) $\partial_{z}\circ{\texttt{j}}=\texttt{j}\circ{\partial}_{\bar{z}}$,  ${\partial}_{\bar{z}}\circ\texttt{j}=\texttt{j}\circ\partial_{z}$;

(iv) $\texttt{j}(\lambda v)=\overline{\lambda}\texttt{j}v$ for any $\lambda\in{\mathbb{C}}$, $v\in{\mathbb{C}}^{2n+2}$.*

The quaternionic projective space $\mathbb{H}P^{n}$ is the set of all one-dimensional quaternionic subspaces of ${\mathbb{H}}^{n+1}$. Analogous to the Fubini-Study metric on $\mathbb{C}P^{n}$, $\mathbb{H}P^{n}$ also carries a natural Riemannian metric denoted by $g_{\mathbb{H}P^{n}}$. Now we give the canonical quaternionic Kähler structure compatible with $g_{\mathbb{H}P^{n}}$, locally denoted by $\left\{\texttt{I},\texttt{J},\texttt{K}\right\}$. Let

[TABLE]

be the Riemannian submersion with fiber $S^{3}$, i.e., for $q=(z_{1},z_{2},\cdots,z_{2n+1},z_{2n+2})$, $\pi(q)=[z_{1}+z_{2}\texttt{j},\cdots,z_{2n+1}+z_{2n+2}\texttt{j}]$, where $|q|^{2}=\sum_{\alpha=1}^{2n+2}|z_{\alpha}|^{2}=1$. For any $q\in S^{4n+3}$, the tangent space of fiber is given by

[TABLE]

We define the horizontal space $\mathcal{H}_{q}$ by

[TABLE]

which is isomorphic to $\mathbb{H}^{n}$. Then

[TABLE]

and the tangent map $\pi_{*}:\mathcal{H}_{q}\rightarrow T_{\pi(q)}\mathbb{H}P^{n}$ is an isometric. For any $X\in T_{\pi(q)}\mathbb{H}P^{n}$, denote the horizontal lift of $X$ by $X^{\mathcal{H}}=\left(\pi_{*}\right)^{-1}(X)$,

[TABLE]

Let $\left\{\omega_{\texttt{I}},\omega_{\texttt{J}},\omega_{\texttt{K}}\right\}$ be the two-forms related to the quaternionic K$\ddot{\text{a}}$hler structure $\left\{\texttt{I},\texttt{J},\texttt{K}\right\}$, defined as follows

[TABLE]

where $X,Y\in T_{\pi(q)}\mathbb{H}P^{n}$. We know every two-form of $\left\{\omega_{\texttt{I}},\omega_{\texttt{J}},\omega_{\texttt{K}}\right\}$ is defined locally, but

[TABLE]

is a globally defined, non-degenerate four-form on $\mathbb{H}P^{n}$(cf.[18],[3], [14]). It is usually called the fundamental four-form of $\mathbb{H}P^{n}$.

Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be an isometric immersion from Riemann surface $M$ to $\mathbb{H}P^{n}$. Then on some open set $U$ of $M$, $\varphi$ has a natural local lift $s:U\rightarrow S^{4n+3}$. The explicit description is that the following diagram commutes:

[TABLE]

That is, if $\varphi=[z_{1}+z_{2}\texttt{j},\cdots,z_{2n+1}+z_{2n+2}\texttt{j}]^{T}$ with $\sum_{\alpha=1}^{2n+2}|z_{\alpha}|^{2}=1$, then

$s=(z_{1},z_{2},\cdots,z_{2n+1},z_{2n+2})^{T}$. Obviously it satisfies $\varphi=\pi\circ s$.

Let $z=x+\texttt{i}y$ be the local coordinate of $M$ such that the metric induced by $\varphi$ is given by

[TABLE]

where $u$ is a real function. Set $\theta_{1}=e^{u}dx,~{}\theta_{2}=e^{u}dy$, $e_{1}=e^{-u}\frac{\partial}{\partial x}$, $e_{2}=e^{-u}\frac{\partial}{\partial y}$. Then $\left\{e_{1},e_{2}\right\}$ is a standard orthogonal basis of $TM$, and $\left\{\theta_{1},\theta_{2}\right\}$ is its dual basis. For given isometric immersion $\varphi$ and the corresponding lift $s$, we define a new differential operator $d^{\mathcal{H}}$ called horizontal differential operator by

[TABLE]

where $d$ is the usual differential operator. The lift $s$ is called a horizontal lift if $s$ satisfies $ds=d^{\mathcal{H}}s$, that is, $\left\langle ds,s\right\rangle=\left\langle ds,\texttt{j}s\right\rangle=0$. In the following, all the tangent maps such as $ds,~{}\pi_{*}$ are extended by complex linearity. A straightforward calculation shows

[TABLE]

Similarly, we have

[TABLE]

Since $\varphi$ and $\pi_{*}\mid_{\mathcal{H}_{q}}$ are isometric, then we have by (2.6)

[TABLE]

Similarly, we have by (2.6) and (2.7)

[TABLE]

and

[TABLE]

From $g_{M}(e_{1},e_{1})=g_{M}(e_{2},e_{2})=1$ and $g_{M}(e_{1},e_{2})=0$, we get by (2.8)-(2.10)

[TABLE]

In term of the three two-forms $\left\{\omega_{\texttt{I}},\omega_{\texttt{J}},\omega_{\texttt{K}}\right\}$, we can define three angles ${\alpha_{1}},{\alpha_{2}},{\alpha_{3}}$ as follows,

[TABLE]

where $d\mu_{M}=\theta_{1}\wedge\theta_{2}$ is the volume form of the metric $g_{M}$. Here ${\alpha_{1}},{\alpha_{2}},{\alpha_{3}}$ are locally defined. But since $\Omega=\omega_{\texttt{I}}\wedge\omega_{\texttt{I}}+\omega_{\texttt{J}}\wedge\omega_{\texttt{J}}+\omega_{\texttt{K}}\wedge\omega_{\texttt{K}}$ is a globally defined, non-degenerate four-form on $\mathbb{H}P^{n}$, then we can define a global function ${\alpha}:M\rightarrow[0,\pi]$ by

[TABLE]

We call $\alpha$ the quaternionic Kähler angle with respect to the isometric immersion $\varphi$. Obviously $\cos^{2}{\alpha}=\cos^{2}{\alpha_{1}}+\cos^{2}{\alpha_{2}}+\cos^{2}{\alpha_{3}}$. In the following we compute the explicit expression of $\cos^{2}{\alpha}$. A straightforward calculation shows

[TABLE]

[TABLE]

and

[TABLE]

Then we get by (2.14)-(2.16)

[TABLE]

From (2.17), we find

[TABLE]

We know there are two kinds of typical immersed surfaces in $\mathbb{H}P^{n}$: totally complex and totally real(cf. [8][9][19]). Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be an isometric immersion of a Riemann surface $M$ into $\mathbb{H}P^{n}$. Then quaternionic Kähler angle gives a measure of the failure of $\varphi$ to be a totally complex map or a totally real map. Indeed $\varphi$ is totally complex if and only if $\alpha(p)=0$ for all $p\in M$, while $\varphi$ is totally real if and only if $\alpha(p)=\frac{\pi}{2}$ for all $p\in M$. From the above discussions, we get the following conclusion, which firstly appeared in ([13], Lemma 1).

Lemma 2.2

Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be a totally real isometric immersion. Let $s$ be a local lift of $\varphi$ satisfying $|s|=1$. Then

[TABLE]

3 Totally real minimal surfaces

We suppose that $(M,~{}ds^{2}_{M})$ is a simply connected domain in the complex plane $\mathbb{C}$ with local coordinate $\{z\}$. Denote

[TABLE]

The following lemmas will be used in this section.

Lemma 3.1

A map $[s]:M\rightarrow\mathbb{C}P^{2n+1}$ is an isometric minimal immersion satisfying $|s|^{2}=1$ if and only if

[TABLE]

holds.

Proof: Let $s$ be a column vector valued in $\mathbb{C}P^{2n+1}$ satisfying $|s|^{2}=1$. Set $f=ss^{*}$, where $*$ denotes conjugate transpose. We consider $\mathbb{C}P^{2n+1}$ as a totally geodesic submanifold of $U(2n+2)$ by Cartan imbedding $\tau(f)=2f-I\in U(2n+2)$ for all $f\in\mathbb{C}P^{2n+1}$. Here the bi-invariant metric on $U(n+1)$ is given by $ds_{U(n+1)}^{2}=\frac{1}{8}\text{tr}\omega\omega^{*}$, where $\omega=g^{-1}dg$ is the Maurer-Cartan form of $U(n+1)$. The metric on $\mathbb{C}P^{2n+1}$ induced by $\tau$ is the Fubini-Study metric of constant holomorphic sectional curvature $4$.

Since $\tau$ is totally geodesic, $[s]$ is an isometric minimal immersion if and only if $\tau\circ[s]$ is an isometric minimal immersion, if and only if the following equation

[TABLE]

holds (cf.[20]), where $A_{z}=(2f-I)\partial_{z}f$, $A_{\bar{z}}=(2f-I)\partial_{\bar{z}}f$, and $\left[,\right]$ is Lie bracket of $\text{gl}(2n+2,\mathbb{C})$.

Let $s^{\perp}(\partial_{z}s)=\partial_{z}s-\left\langle\partial_{z}s,~{}s\right\rangle s$, $s^{\perp}(\partial_{\bar{z}}s)=\partial_{\bar{z}}s-\left\langle\partial_{\bar{z}}s,~{}s\right\rangle s$. Then,

[TABLE]

Substituting (3.3) into (3.2), we obtain

[TABLE]

which is equivalent to

[TABLE]

Thus (3.1) follows from (LABEL:lem3-01-6).   $\Box$

Lemma 3.2

A map $[s]:M\rightarrow\mathbb{C}P^{2n+1}$ is a totally real isometric immersion satisfying $|s|^{2}=1$ if and only if

[TABLE]

holds.

Proof: The Kähler angle of $[s]$ is a function $\theta:M\rightarrow[0,\pi]$ given in terms of the complex coordinate $z$ on $M$ by (cf. [4]) $\tan\frac{\theta}{2}=\frac{\left|s^{\perp}(\partial_{\bar{z}}s)\right|}{\left|s^{\perp}(\partial_{z}s)\right|}.$ Since $[s]$ is totally real, then $\theta=\frac{\pi}{2}$, which implies $\left|s^{\perp}(\partial_{\bar{z}}s)\right|^{2}=\left|s^{\perp}(\partial_{z}s)\right|^{2}.$ It is equivalent to (3.6) by $|s|^{2}=1$.   $\Box$

Similar to Lemma 3.1, we have the following conclusion, which firstly appeared in ([13], Proposition 1).

Lemma 3.3

*Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be an isometric immersion. Let $s$ be a local lift of $\varphi$ satisfying $|s|=1$, then $\varphi$ is minimal if and only if *

[TABLE]

holds.

Proof: Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be an isometric immersion. Let $s$ be a column vector valued in $\mathbb{C}^{2n+2}$ satisfying $|s|=1$, which is a natural local lift of $\varphi$. Set $\varphi=ss^{*}+(\texttt{j}s)(\texttt{j}s)^{*}$. We consider $\mathbb{H}P^{n}$ as a totally geodesic submanifold of $U(2n+2)$ by Cartan imbedding $\tau(\varphi)=2\varphi-I\in U(2n+2)$ for all $\varphi\in\mathbb{H}P^{n}$. Then $\varphi$ is minimal if and only if $\tau\circ\varphi$ is minimal. It follows that (3.2) holds, here

[TABLE]

and

[TABLE]

Thus (3.7) follows from (3.2).   $\Box$

The complex projective space $\mathbb{C}P^{2n+1}$ is the twistor space of $\mathbb{H}P^{n}$. The twistor map $t:\mathbb{C}P^{2n+1}\rightarrow\mathbb{H}P^{n}$, is given by $t\left([z_{1},z_{2},\cdots,z_{2n+1},z_{2n+2}]\right)=[z_{1}+z_{2}\texttt{j},\cdots,z_{2n+1}+z_{2n+2}\texttt{j}].$ Then $t$ is a Riemann submersion and the horizontal distribution is given by $\sum_{i=1}^{n+1}\left(z_{2i-1}dz_{2i}-z_{2i}dz_{2i-1}\right)=0.$

A part of the following result appeared in [13]. For completeness we will give the whole proof.

Theorem 3.4

Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real isometric minimal immersion, then there exists a totally real isometric horizontal minimal lift $[s]:M\rightarrow\mathbb{C}P^{2n+1}$.

Proof: Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real isometric minimal immersion. Let $[s]:U\subset M\rightarrow\mathbb{C}P^{2n+1}$ be a natural local lift of $\varphi$ satisfying $|s|=1$. Here $s$ is a column vector valued in $\mathbb{C}^{2n+2}$. Set

[TABLE]

then $D$ is a $2\times 2$-matrix. Then we have

[TABLE]

where $\left(d^{\mathcal{H}}s({\partial}_{z}),\texttt{j}d^{\mathcal{H}}s({\partial}_{\bar{z}})\right)^{*}\left(s,\texttt{j}s\right)=\left(d^{\mathcal{H}}s({\partial}_{\bar{z}}),\texttt{j}d^{\mathcal{H}}s({\partial}_{z})\right)^{*}\left(s,\texttt{j}s\right)=0$.

From (3.9) and the identity $\partial_{z}\partial_{\bar{z}}=\partial_{\bar{z}}\partial_{z}$, we get

[TABLE]

Since $\varphi$ is totally real, then by Lemma 2.18 we have

[TABLE]

Let $\tilde{s}$ be another local lift of $\varphi$ satisfying $|\tilde{s}|=1$, then

[TABLE]

where $T:U\rightarrow SU(2)$ is to be determined such that $\tilde{s}$ is horizontal, i.e. $\left(\tilde{s},\texttt{j}\tilde{s}\right)^{*}d\left(\tilde{s},\texttt{j}\tilde{s}\right)=0$. Such $T$ is a solution of the linear PDE

[TABLE]

The integrable condition of (3.12) is just (3.10), so it has a unique solution locally on $M$ for any given initial value. Let $T$ be a solution of (3.12) with the initial value $T(0)\in SU(2)$. From (3.12) we have $d(T^{*}T)=0$ and $d|T|=0$, so $T\in SU(2)$. Hence there exists a local horizontal lift of $\varphi$, also denoted by $s$.

Let $s$ and $\tilde{s}$ be two horizontal lifts of $\varphi$ on $U$, then by (3.8) we have $D=0$. We define $T$ by (3.11). It follows from (3.12) that $T:U\rightarrow SU(2)$ is a constant map. Since $M$ is simply connected, we can extend $s$ to a global horizontal lift of $\varphi$ on $M$.

Since $\varphi$ is minimal , then from Lemma 3.3 we know (3.7) holds. Since $s$ is horizontal, then we have

[TABLE]

Substituting (3.13) into (3.7) and using (2.18), we have

[TABLE]

From (3.14) we find (3.6) holds, then $[s]$ is totally real by Lemma 3.2. From (3.13) and (3.14) we find (3.1) holds, then $s$ is minimal by Lemma 3.1. Since $\varphi$ is isometric, then

[TABLE]

where in the last equation we use the fact that $s$ is horizontal. It verifies that $s$ is isometric. So we get our conclusions.   $\Box$

Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real isometric harmonic map. By [1] and [6], we know that $\varphi$ belongs to the following harmonic sequence in $G(2,2n+2)$,

[TABLE]

where $\underline{\varphi}_{-k}={\texttt{j}}\underline{\varphi}_{k}$ are $2$-dimensional harmonic subbundle of the trivial bundle $M\times\mathbb{C}^{2n+2}$. Here $\mathbb{H}P^{n}$ is seen as the totally geodesic submanifold of $G(2,2n+2)$.

Proposition 3.5

Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real isometric harmonic map generating the harmonic sequence (3.16). If its totally real horizontal minimal lift $[s]$ generates the following harmonic sequence in $\mathbb{C}P^{2n+1}$,

[TABLE]

where $\underline{f}_{-k},\underline{f}_{k}$ are line bundles, then for $k=0,1,\cdots,m,\cdots$,

[TABLE]

and

[TABLE]

Proof: It follows that

[TABLE]

Since $\left\langle\partial_{z}s,~{}\texttt{j}s\right\rangle=\left\langle\partial_{\bar{z}}s,~{}\texttt{j}s\right\rangle=\left\langle\partial_{z}s,~{}\texttt{j}\partial_{\bar{z}}s\right\rangle=0$ and $f_{1}=\partial_{z}{s},~{}f_{-1}=\partial_{\bar{z}}{s},$ then we have

[TABLE]

which implies

[TABLE]

By differentiating with respect to $z$ in $\left\langle f_{1},~{}\texttt{j}f_{0}\right\rangle=0$, $\left\langle f_{1},~{}\texttt{j}f_{-1}\right\rangle=0$, $\left\langle\texttt{j}f_{-1},~{}f_{0}\right\rangle=0$ and $\left\langle\texttt{j}f_{-1},~{}f_{1}\right\rangle=0$ respectively, using (3.20) we obtain

[TABLE]

Then differentiating with respect to $\bar{z}$ in $\left\langle f_{2},~{}\texttt{j}f_{-1}\right\rangle=0$ we have

[TABLE]

by (3.20) and (3.22). It follows from (3.22) and (3.23) that

[TABLE]

In the following, we prove (3.18) and (3.19) by induction on $k$. When $k=1$ and $k=2$ the conclusions hold by (3.20) -(3.24). Suppose the conclusion is true for $k-1$. Consider the case of $k$. By induction hypotheses we have

[TABLE]

and

[TABLE]

By differentiating with respect to $z$ in $\left\langle f_{k-1},~{}\texttt{j}f_{-(k-2)}\right\rangle=0$, $\left\langle f_{k-1},~{}\texttt{j}f_{-(k-1)}\right\rangle=0$, $\left\langle\texttt{j}f_{-(k-1)},~{}f_{k-2}\right\rangle=0$ and $\left\langle\texttt{j}f_{-(k-1)},~{}f_{k-1}\right\rangle=0$ respectively, using (LABEL:prop3-eq+4) we obtain

[TABLE]

Then differentiating with respect to $\bar{z}$ in $\left\langle f_{k},~{}\texttt{j}f_{-(k-1)}\right\rangle=0$ we have

[TABLE]

by (LABEL:prop3-eq+4) and (3.27). So the case of $k$ holds, which implies (3.18) holds.

It follows from the properties of harmonic sequence (3.16) that

[TABLE]

Then (3.18) verifies (3.19). So we finish our proof.   $\Box$

Proposition 3.6

Let $\varphi:M\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real isometric harmonic map of isotropy order $n$. Then its totally real horizontal minimal lift $[s]:M\rightarrow\mathbb{C}P^{2n+1}$ has isotropy order $n$ or $2n+1$.

Proof: Since the isotropy order of $\varphi$ is $n$, then the harmonic sequence (3.16) is cyclic, i.e., $\underline{\varphi}_{p}\perp\underline{\varphi}_{q}~{}(0\leq p<q\leq n),~{}\underline{\varphi}_{n+1}=\underline{\varphi}_{0}.$ From Proposition 3.5, $\underline{\varphi}_{k}=\underline{f}_{k}\oplus\texttt{j}\underline{f}_{-k}~{}(0\leq k\leq n+1).$ If $\left\langle f_{n+1},~{}f_{0}\right\rangle\neq 0$, then the isotropy order of $\underline{f}_{0}=[s]$ is $n$. If $\left\langle f_{n+1},~{}f_{0}\right\rangle=0$, then $\underline{f}_{n+1}=\texttt{j}\underline{f}_{0}$ and $\texttt{j}\underline{f}_{-(n+1)}=\underline{f}_{0}$ by $\underline{\varphi}_{n+1}=\underline{\varphi}_{0}$. It follows that the isotropy order of $\underline{f}_{0}=[s]$ is $2n+1$.   $\Box$

4 Totally real flat minimal surfaces

The following result appears in [15], see also [16] and [5].

Theorem 4.1

If $f:\mathbb{C}\rightarrow\mathbb{C}P^{n}$ is a linearly full totally real harmonic map with induced metric $f^{*}ds^{2}=2dzd\overline{z}$. Then up to a unitary equivalence, $f=\left[V_{0}^{(n)}\right]$, where

[TABLE]

and $a_{k}=e^{\texttt{i}\theta_{k}},~{}\xi^{k}=\sqrt{r_{k}}~{}(r_{k}>0)$ for $k=0,1,\cdots,n$, satisfying $0=\theta_{0}<\theta_{1}<\cdots<\theta_{n}<2\pi,~{}r_{0}+r_{1}+\cdots+r_{n}=1$.

In the following we will study linearly full totally real flat minimal surfaces in $\mathbb{H}P^{n}$. Let $\varphi:\mathbb{C}\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real flat minimal surface, from Theorem 3.4, we know there exists a totally real flat minimal surface $[s]:\mathbb{C}\rightarrow\mathbb{C}P^{2n+1}$ as the horizontal lift of $\varphi$. Here $[s]$ may not be linearly full in $\mathbb{C}P^{2n+1}$, then it lies in $\mathbb{C}P^{m}\subset\mathbb{C}P^{2n+1}$ for $n\leq m\leq 2n+1$, where $m\geq n$ by $\varphi$ being linearly full in $\mathbb{H}P^{n}$. Using Theorem 4.1, we know $s$ is given by (4.1), up to $U(2n+2)$. But the isometry group of $\mathbb{H}P^{n}$ is $Sp(n+1)=\left\{U\in U(2n+2),~{}UJ_{n+1}U^{T}=J_{n+1}\right\}$, a subgroup of $U(2n+2)$, then in order to give the explicit expression of $\varphi$, we need to characterize $s$, up to $Sp(n+1)$.

Theorem 4.2

Let $\varphi:\mathbb{C}\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real flat minimal surface of isotropy order $n$ with the induced metric $\varphi^{*}ds^{2}=2dzd\overline{z}$, then up to a symplectic isometry of $\mathbb{H}P^{n}$, $\varphi$ lies in $\mathbb{C}P^{n}~{}(\subset\mathbb{H}P^{n})$ given by

[TABLE]

where $a_{k}=e^{\texttt{i}\frac{2k\pi}{n+1}}~{}(k=0,1,\cdots,n)$ (the Clifford solution in $\mathbb{C}P^{n}$), or $a_{k}=e^{\texttt{i}\frac{k\pi}{n+1}}~{}(k=0,1,\cdots,n)$.

Proof: Let $\varphi:\mathbb{C}\rightarrow\mathbb{H}P^{n}$ be a linearly full totally real flat minimal surface of isotropy order $n$ with the induced metric $\varphi^{*}ds^{2}_{\mathbb{H}P^{n}}=2dzd\overline{z}$. From Theorem 3.4 and Proposition 3.6, there exists a totally real horizontal minimal lift $\underline{f}_{0}=[s]:\mathbb{C}\rightarrow\mathbb{C}P^{2n+1}$ of isotropy order $n$ or $2n+1$ with the induced metric $f_{0}^{*}ds^{2}_{\mathbb{C}P^{2n+1}}=2dzd\overline{z}$, such that

[TABLE]

Then by Theorem 4.1, there exists a unitary matrix $U\in U(2n+2)$ such that $s=U{V}_{0}^{(m)}~{}(n\leq m\leq 2n+1)$ satisfying $\left\langle\partial^{k}_{z}s,~{}s\right\rangle=\left\langle\partial^{k}_{z}s,~{}\texttt{j}s\right\rangle=\left\langle\partial^{k}_{\bar{z}}s,~{}\texttt{j}s\right\rangle=0~{}(k=1,\cdots,n)$. Here, $V_{0}^{(m)}$ is a column vector of order $2n+2$ as follows,

[TABLE]

Since $\texttt{j}s=J_{n+1}\bar{s}$, setting $W=U^{T}J_{n+1}U=(w_{ij})$, we know $W$ is an anti-symmetric unitary matrix of order $2n+2$. Then for any $k=1,\cdots,n$,

[TABLE]

For a given $W$, define a set

[TABLE]

The following can be easily checked,

(i) $\forall~{}A\in Sp(n+1),~{}U\in G_{W}$, we have that $AU\in G_{W}$;

(ii) $\forall~{}U,V\in G_{W},~{}\exists~{}A=UV^{*}\in Sp(n+1)~{}s.t.~{}U=AV$.

In the following we discuss $W$ in two cases of the isotropy order of $\underline{f}_{0}$ being $n$ or $2n+1$.

If the isotropy order of $\underline{f}_{0}$ is $n$, then by Proposition 3.6,

[TABLE]

where $\lambda$ is a unit complex and $a(\neq 0),b$ are complexes satisfying $|a|^{2}+|b|^{2}=1$. It follows from (4.5) that $f_{n+1}=\lambda f_{-(n+1)}$, which implies by $f_{n+1}=\partial_{z}^{n+1}s$ and $f_{-(n+1)}=\partial_{\bar{z}}^{n+1}s$,

[TABLE]

Since $a_{0}=1$ and $a_{i}\neq a_{j}~{}(i\neq j)$, then from (4.6), $a_{j}^{2n+2}=1~{}(j=0,1,\cdots,m)$.

In this case, if $n\leq m<2n+1$ or $m=2n+1$($n$ is odd), then we claim that $W$ belongs to the following type

[TABLE]

In fact if $\mathbf{\forall~{}i,j=0,1,\cdots,m,~{}a_{i}+a_{j}\neq 0}$, then we have $w_{ij}=0$ by (4.4), which implies that $W$ is just the type of (4.7). If $\mathbf{\forall~{}i,~{}\exists~{}j~{}s.t.~{}a_{i}+a_{j}=0}$, then $m$ must be odd. Set $m=2p+1$. Then, for any $i=0,1,\cdots,p$, without of generality, we may assume that $a_{i}+a_{p+1+i}=0$. Applying it in (4.4), we obtain that for any $i=0,1,\cdots,m$, $w_{ij}=0~{}(j\neq p+1+i~{}(\text{mod}~{}m+1))$, and $\sum_{i=0}^{p}w_{i,p+1+i}\xi^{i}\xi^{p+1+i}a_{i}^{2\tau+1}=\sum_{i=0}^{p}w_{i,p+1+i}\xi^{i}\xi^{p+1+i}\bar{a}_{i}^{2\tau+1}=0~{}(\tau=0,1,\cdots,q)$, where $2q+1$ is the maximum odd number not greater than $n$. It follows that

[TABLE]

Since $a_{j}^{2n+2}=1$, then $\bar{a}_{j}=a_{j}^{2n+1}$. It means that the coefficient matrix of the above equation is a Vandermonde matrix of $(2q+2)\times(p+1)$, whose rank is $p+1$. Here $2q+2\geq p+1$ by $n\leq m<2n+1$ or $m=2n+1$($n$ is odd). So, we have $w_{i,p+1+i}=0~{}(i=0,1,\cdots,p)$. Thus $W$ is also just the type of (4.7). If $\mathbf{\exists~{}i~{}s.t.\forall~{}j,~{}a_{i}+a_{j}\neq 0~{}and~{}\exists~{}k,l~{}s.t.~{}a_{k}+a_{l}=0}$, by the similar discussion as the above, we get $W$ is also just the type of (4.7).

If $n<m\leq 2n+1$ then $W$ in (4.7) is a degenerate matrix, hence there doesn’t exist this case. Thus $m=n$ and

[TABLE]

From (4.8), the corresponding matrix $U\in G_{W}$ can be expressed as follow:

[TABLE]

Then we get the horizontal lift of $\varphi$

[TABLE]

which implies

[TABLE]

So, the image of $\varphi$ lies in $\mathbb{C}P^{n}$. Since the isotropy order of $\varphi$ is $n$, then it follows from ([15], Proposition 4.1) that $\varphi$ is just the Clifford solution of (4.2) in $\mathbb{C}P^{n}$, up to $U(n+1)$ ($\subset Sp(n+1)$).

Now we discuss the case of $m=2n+1$($n$ is even). Noticing $2q+2<n+1$ in this case, we can’t follow the above arguments. Since $a_{j}^{2n+2}=1$ and $a_{j}=e^{\texttt{i}\theta_{j}}$ with $0=\theta_{0}<\theta_{1}<\cdots<\theta_{2n+1}<2\pi$, then $a_{j}=e^{\texttt{i}\frac{j\pi}{n+1}}~{}(j=0,1,\cdots,2n+1)$, which implies $a_{\tau}+a_{n+1+\tau}=0~{}(\tau=0,1,\cdots,n)$. Using $f_{n+1}=af_{0}+b\texttt{j}f_{0}$ again, we obtain $(a_{i}^{n+1}-a)e^{a_{i}z-\bar{a}_{i}\bar{z}}\xi^{i}=b\sum_{j=0}^{2n+1}\bar{w}_{ij}e^{\bar{a}_{j}\bar{z}-a_{j}z}\xi^{j}~{}(i=0,1,\cdots,2n+1)$, which implies that for any $i=0,1,\cdots,2n+1$, $w_{ij}=0~{}(j\neq n+1+i~{}(\text{mod}~{}2n+2))$, and

[TABLE]

Set $n=2p$, from (4.10) we get $a=\frac{r_{0}-r_{n+1}}{r_{0}+r_{n+1}}=(-1)^{\tau}\frac{r_{\tau}-r_{n+1+\tau}}{r_{\tau}+r_{n+1+\tau}}~{}(\tau=1,\cdots,n)$, which implies $\frac{r_{n+1}}{r_{0}}=\frac{r_{n+1+2\zeta}}{r_{2\zeta}}=\frac{r_{2\zeta-1}}{r_{n+2\zeta}}~{}(\zeta=1,\cdots,p)$, where $r_{0}\neq r_{n+1}$ by $a\neq 0$. Substituting it into (4.10) and the first equation of (4.4) respectively, a straightforward calculation shows ${w}_{\tau,n+1+\tau}=(-1)^{\tau}w_{0,n+1}~{}(\tau=1,\cdots,n)$ and $r_{0}+\sum_{\tau=1}^{n}\left(a_{2\tau}^{k}r_{2\tau}\right)=0~{}(k=1,\cdots,n),$ which implies by the fact of $\sum_{\tau=1}^{n}a_{2\tau}^{k}=-1$,

[TABLE]

Since the determinant of coefficient matrix of the above equation is $a_{2}a_{4}\cdots a_{2n}\prod_{1\leq j<i\leq n}(a_{2i}-a_{2j})\neq 0$, we have $r_{2\tau}=r_{0}$, which implies $r_{2\tau+1}=r_{1}~{}(\tau=1,\cdots,n)$. It follows that

[TABLE]

where $|w_{0,n+1}|^{2}=1$. By the similar calculation we can get the horizontal lift of $\varphi$

[TABLE]

which implies by $a_{\tau}+a_{n+1+\tau}=0,~{}\xi^{2\tau}=\xi^{0},~{}\xi^{2\tau+1}=\xi^{1}~{}(\tau=0,1,\cdots,n)$,

[TABLE]

After removing the projective factor $(\xi^{0}-w_{0,n+1}\xi^{1}\texttt{j})$, $\varphi$ is the surface (4.2) with $a_{k}=e^{\texttt{i}\frac{k\pi}{n+1}}$ and $n$ being even, up to $Sp(n+1)$.

If the isotropy order of $\underline{f}_{0}$ is $2n+1$, then $\underline{f}_{0}$ is the Clifford solution (up to congruence) in $\mathbb{C}P^{2n+1}$ by ([15], Proposition 4.1), that is, $s=U{V}_{0}^{(2n+1)}$ with

[TABLE]

where $a_{k}=e^{\texttt{i}\frac{k\pi}{n+1}},~{}\xi^{k}=\sqrt{\frac{1}{2n+2}}~{}(k=0,1,\cdots,2n+1)$. Since $\underline{f}_{n+1}=\texttt{j}\underline{f}_{0}$, then

[TABLE]

By the similar computation we can get the horizontal lift of $\varphi$

[TABLE]

which implies by $a_{k}=e^{\texttt{i}\frac{k\pi}{n+1}}$ and $\xi^{k}=\sqrt{\frac{1}{2n+2}}$,

[TABLE]

After removing the projective factors $\xi^{k}$ and $(1+\texttt{j})$ or $(1-\texttt{j})$, $\varphi$ is the surface (4.2) with $a_{k}=e^{\texttt{i}\frac{k\pi}{n+1}}$, up to $Sp(n+1)$. In summary, we finish our proofs.   $\Box$

Remark 4.3

Two surfaces of isotropy order $n$ in $\mathbb{H}P^{n}$ given by Theorem 4.2 have different horizontal lifts. For the surface (4.2) with $a_{k}=e^{\texttt{i}\frac{2k\pi}{n+1}}$, the horizontal lift (4.9) is not linearly full in $\mathbb{C}P^{2n+1}$, whose isotropy order is $n$. However for the surface (4.2) with $a_{k}=e^{\texttt{i}\frac{k\pi}{n+1}}$, the horizontal lifts (4.11)($n$ is even) and (4.12) are both linearly full in $\mathbb{C}P^{2n+1}$, whose isotropy order is $n$ and $2n+1$ respectively.

Acknowledgments  The authors would like to appreciate the referee for carefully reading the manuscript and pointing out a mistake in the proof of Proposition 3.5. The authors were supported by NSF in China (No. 11501548, 11501505).