Rigidity theorems of Lagrangian submanifolds in the homogeneous nearly K\"ahler $\mathbb{S}^6(1)$

TL;DR

This paper establishes rigidity theorems for Lagrangian submanifolds in the nearly Kähler 6-sphere, deriving integral inequalities and characterizing cases of equality as totally geodesic spheres or Berger spheres.

Contribution

It introduces a Simons' type integral inequality for compact Lagrangian submanifolds in the nearly Kähler 6-sphere and characterizes the equality cases.

Findings

Derived a Simons' type integral inequality involving the second fundamental form.

Characterized the equality case as either totally geodesic or Berger spheres.

Provided rigidity results for Lagrangian submanifolds in the nearly Kähler 6-sphere.

Abstract

In this paper, we study Lagrangian submanifolds of the homogeneous nearly K\"ahler $6$-dimensional unit sphere $\mathbb{S}^6(1)$. As the main result, we derive a Simons' type integral inequality in terms of the second fundamental form for compact Lagrangian submanifolds of $\mathbb{S}^6(1)$. Moreover, we show that the equality sign occurs if and only if the Lagrangian submanifold is either the totally geodesic $\mathbb{S}^3(1)$ or the Dillen-Verstraelen-Vrancken's Berger sphere $S^3$ discribed in J Math Soc Japan, 42: 565-584, 1990.

Equations

$$\int_{M^{3}}\|h\|^{2}\big{(}\|h\|^{2}-\tfrac{5}{4}-\tfrac{3}{2}\Theta^{2}\big{)}dM\geq 0,$$

$$g((\nabla h)(v,v,v),Jv)=0.$$

$$J_{x}U=x\times U,$$

$$G(X,Y)=({\bar{\nabla }}_{X}J)Y,$$

$$G(X,Y)+G(Y,X)=0,$$

$$G(X,JY)+JG(X,Y)=0,$$

$$g(G(X,Y),Z)+g(G(X,Z),Y)=0,$$

$$({\bar{\nabla }}_{X}G)(Y,Z)=g(Y,JZ)X+g(X,Z)JY-g(X,Y)JZ,$$

$$g(G(X,Y),G(Z,W))=$$

$$+g(JX,Z)g(Y,JW)-g(JX,W)g(Y,JZ),$$

$${\bar{\nabla }}_{X}Y={\nabla }_{X}Y+h(X,Y),{\bar{\nabla }}_X\xi =-A_{\xi }X+{\nabla }_X^{\perp }\xi ,$$

$$g(h(X,Y),\xi )=g(A_{\xi }X,Y).$$

$${\nabla }_X^{\perp }JY=G(X,Y)+J{\nabla }_{X}Y,A_{JX}Y=-Jh(X,Y).$$

$$g((\nabla h)(W,X,Z),JY)-g((\nabla h)(W,X,Y),JZ)=g(h(W,X),G(Y,Z)),$$

$$1\le i,j,k,l,\cdots \le 3\mathrm{and}{i}^{\ast }=3+i\mathrm{for}i=1,2,3.$$

$$\{\begin{array}{rl}& dx=\sum _i{\omega }_i{e}_i,\\ & d{e}_i=\sum _j{\omega }_{ij}{e}_j+\sum _{j,k}{h}_{ij}^{k^{\ast }}{\omega }_j{e}_{k^{\ast }}-{\omega }_{i}x,{\omega }_{ij}+{\omega }_{ji}=0,\\ & d{e}_{i^{\ast }}=-\sum _{j,k}{h}_{jk}^{i^{\ast }}{\omega }_j{e}_k+\sum _j{\omega }_{i^{\ast }{j}^{\ast }}{e}_{j^{\ast }},{\omega }_{i^{\ast }{j}^{\ast }}+{\omega }_{i^{\ast }{j}^{\ast }}=0,\end{array}$$

$$\{\begin{array}{rl}& d{\omega }_i=\sum _j{\omega }_{ij}\wedge {\omega }_j,\\ & d{\omega }_{ij}-\sum _k{\omega }_{ik}\wedge {\omega }_{kj}:=-\frac{1}{2}\sum _{k,l}{R}_{ijkl}{\omega }_k\wedge {\omega }_l,\\ & d{\omega }_{i^{\ast }{j}^{\ast }}-\sum _k{\omega }_{i^{\ast }{k}^{\ast }}\wedge {\omega }_{k^{\ast }{j}^{\ast }}:=-\frac{1}{2}\sum _{k,l}{R}_{i^{\ast }{j}^{\ast }kl}{\omega }_k\wedge {\omega }_l,\\ & \sum _l{h}_{ij,l}^{k^{\ast }}{\omega }_l:=d{h}_{ij}^{k^{\ast }}+\sum _l{h}_{il}^{k^{\ast }}{\omega }_{lj}+\sum _l{h}_{lj}^{k^{\ast }}{\omega }_{li}+\sum _l{h}_{ij}^{l^{\ast }}{\omega }_{l^{\ast }{k}^{\ast }},\end{array}$$

$$R_{ijkl}={\delta }_{ik}{\delta }_{jl}-{\delta }_{il}{\delta }_{jk}+\sum _p(h_{ik}^{p^{\ast }}{h}_{jl}^{p^{\ast }}-h_{il}^{p^{\ast }}{h}_{jk}^{p^{\ast }}),$$

$$h_{ij,l}^{k^{\ast }}=h_{il,j}^{k^{\ast }},$$

$$R_{i^{\ast }{j}^{\ast }kl}=\sum _p(h_{ik}^{p^{\ast }}{h}_{jl}^{p^{\ast }}-h_{il}^{p^{\ast }}{h}_{jk}^{p^{\ast }}).$$

$$R_{ij}=3{\delta }_{ij}-\sum _{k,p}{h}_{ik}^{p^{\ast }}{h}_{kj}^{p^{\ast }},\tau =6-\Vert h{\Vert }^2,$$

$$h_{ij,kl}^{p^{\ast }}-h_{ij,lk}^{p^{\ast }}=\sum _m{h}_{mi}^{p^{\ast }}{R}_{mjkl}+\sum _m{h}_{mj}^{p^{\ast }}{R}_{mikl}+\sum _m{h}_{ij}^{m^{\ast }}{R}_{m^{\ast }{p}^{\ast }kl},$$

$$\sum _l{h}_{ij,kl}^{p^{\ast }}{\omega }_l:=d{h}_{ij,k}^{p^{\ast }}+\sum _l{h}_{lj,k}^{p^{\ast }}{\omega }_{li}+\sum _l{h}_{il,k}^{p^{\ast }}{\omega }_{lj}+\sum _l{h}_{ij,l}^{p^{\ast }}{\omega }_{lk}+\sum _l{h}_{ij,k}^{l^{\ast }}{\omega }_{l^{\ast }{p}^{\ast }}.$$

$$\Psi :S^3\to {\mathbb{S}}^6(1):(y_1,y_2,y_3,y_4)\mapsto (x_1,x_2,x_3,x_4,x_5,x_6,x_7),$$

$$\{\begin{array}{rl}& x_1=\frac{1}{9}(5y_1^2+5y_2^2-5y_3^2-5y_4^2+4y_1),x_2=-\frac{2}{3}{y}_2,\\ & x_3=\frac{2\sqrt{5}}{9}(y_1^2+y_2^2-y_3^2-y_4^2-y_1),x_4=\frac{\sqrt{3}}{9\sqrt{2}}(-10y_1{y}_3-2y_3-10y_2{y}_4),\\ & x_5=\frac{\sqrt{3}\sqrt{5}}{9\sqrt{2}}(2y_1{y}_4-2y_4-2y_2{y}_3),x_6=\frac{\sqrt{3}\sqrt{5}}{9\sqrt{2}}(2y_1{y}_3-2y_3+2y_2{y}_4),\\ & x_7=\frac{\sqrt{3}}{9\sqrt{2}}(10y_1{y}_4+2y_4-10y_2{y}_3).\end{array}$$

$$\{\begin{array}{rl}& X_1(y_1,y_2,y_3,y_4)=(y_2,-y_1,y_4,-y_3),\\ & X_2(y_1,y_2,y_3,y_4)=(y_3,-y_4,-y_1,y_2),\\ & X_3(y_1,y_2,y_3,y_4)=(y_4,y_3,-y_2,-y_1).\end{array}$$

$$E_1=\frac{3}{2}{X}_1,E_2=\frac{\sqrt{3}}{2\sqrt{2}}{X}_2,E_3=-\frac{\sqrt{3}}{2\sqrt{2}}{X}_3$$

$$⟨R(X,Y)W,Z⟩=$$

$$+\frac{20}{16}(⟨X^{\perp },Z^{\perp }⟩⟨Y^{\perp },W^{\perp }⟩-⟨X^{\perp },V^{\perp }⟩⟨Y^{\perp },Z^{\perp }⟩),$$

$$\{\begin{array}{rl}& h(E_1,E_1)=\frac{\sqrt{5}}{2}J{E}_1,h(E_1,E_2)=-\frac{\sqrt{5}}{4}J{E}_2,h(E_1,E_3)=-\frac{\sqrt{5}}{4}J{E}_3,\\ & h(E_2,E_2)=-\frac{\sqrt{5}}{4}J{E}_1,h(E_3,E_3)=-\frac{\sqrt{5}}{4}J{E}_1,h(E_2,E_3)=0.\end{array}$$

$$\frac{1}{2}\Delta \Vert h{\Vert }^2=\sum _{i,j,k}(h_{ij,k}^{l^{\ast }}{)}^2+3\Vert h{\Vert }^2-\sum _{i,j}N(H_i{H}_j-H_j{H}_i)-\sum _{i,j}{S}_{ij}^2.$$

Full text

2010 Mathematics Subject Classification. Primary 53D12; Secondary 53C24, 53C42.

This project was supported by NSF of China, Grant Number 11771404.

1. Introduction

It is well-known that the $6$-dimensional unit sphere $\mathbb{S}^{6}(1)$ with the standard metric $g$ of constant sectional curvature $1$ admits a canonical nearly Kähler structure $J$, which can be constructed by using the Cayley number system. A $3$-dimensional Riemannian submanifold $M^{3}$ of $\mathbb{S}^{6}(1)$ is called Lagrangian if $J(TM^{3})=T^{\perp}M^{3}$, where $TM^{3}$ and $T^{\perp}M^{3}$ denote, respectively, the tangent and normal bundle of $M^{3}$ in $\mathbb{S}^{6}(1)$. Butruille [2] proved that the only Riemannian homogeneous $6$-dimensional nearly Kähler manifolds are $\mathbb{S}^{6}$, $\mathbb{S}^{3}\times\mathbb{S}^{3}$, $\mathbb{C}P^{3}$ and $SU(3)/U(1)\times U(1)$. However, Foscolo and Haskins [11] have proved the existence of at least one exotic (cohomogeneity one) nearly Kähler structure on both $\mathbb{S}^{6}$ and $\mathbb{S}^{3}\times\mathbb{S}^{3}$. In this paper, we consider $\mathbb{S}^{6}(1)$ restricted to its canonical homogeneous nearly Kähler structure.

For compact Lagrangian submanifolds of the nearly Kähler $\mathbb{S}^{6}(1)$, the rigidity phenomena with respect to the sectional curvature $K$, the Ricci curvature $Ric$ and the scalar curvature $\tau$ have been previously studied in [1, 5, 6, 7, 12, 15].

Regarding the pinching theorems for the sectional curvature, we have

Theorem 1.1** ([5, 6]).**

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ whose sectional curvatures $K$ satisfy $K>\tfrac{1}{16}$. Then $M^{3}$ is totally geodesic, and thus $K=1$.

Theorem 1.2** ([6, 7]).**

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ whose sectional curvatures $K$ satisfy $\tfrac{1}{16}\leq K<\tfrac{21}{16}$, then $M^{3}$ is totally geodesic with $K=1$, or $M^{3}$ has constant sectional curvature $\tfrac{1}{16}$.

Notice that Lagrangian submanifolds of $\mathbb{S}^{6}(1)$ with constant sectional curvature were classified by Ejiri [10]. Each such submanifold is either totally geodesic or congruent to an equivariant immersion of $\mathbb{S}^{3}(1/16)$ in $\mathbb{S}^{6}(1)$ (the immersion can be realized by using harmonic polynomials of degree $6$ and an explicitly expression is given in [6, 7]). Also there exists another equivariant immersion of ${\rm SU}(2)$, equipped with a suitable left invariant metric, see [7, 17], of which at every point all sectional curvatures satisfy $\tfrac{21}{16}>K\geq\tfrac{1}{16}$. Therefore the above-mentioned theorems are the best possible pinching results for the sectional curvatures.

Regarding the pinching theorems for the Ricci curvature, we have

Theorem 1.3** ([15]).**

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ and assume that all Ricci curvatures $Ric$ satisfy ${\rm Ric}(v)>\tfrac{53}{64}$. Then $M^{3}$ is totally geodesic, and thus ${\rm Ric}=2$ on $M^{3}$.

An improved version of Theorem 1.3 was obtained by Antić-Djorić-Vrancken [1]:

Theorem 1.4** ([1]).**

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ and assume that all Ricci curvatures $Ric$ satisfy ${\rm Ric}(v)\geq\tfrac{3}{4}$. Then $M^{3}$ is totally geodesic.

Since a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ must be minimal ([10]), the squared length $\|h\|^{2}$ of the second fundamental form $h$ and the scalar curvature $\tau$ is related by, the Gauss equation, $\tau=6-\|h\|^{2}$. In [3], the authors classified the Lagrangian submanifolds of $\mathbb{S}^{6}(1)$ with constant scalar curvature that realize the Chen’s inequality. As far as the pinching theorem for the scalar curvature are concerned, we have

Theorem 1.5** ([12]).**

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Assume that $\|h\|^{2}<\tfrac{5}{2}$, then $M^{3}$ is totally geodesic.

Remark 1.1*.*

Although the result of Theorem 1.5 is not optimal, it is still significant. In fact, it stands for a very interesting improvement of the following results: If $M^{3}$ is a compact minimal submanifold of the round sphere $\mathbb{S}^{6}(1)$, then A. M. Li and J. M. Li [14] proved the result if $\|h\|^{2}\leq 2$, while J. Simons [20] and Chern-do Carmo-Kobayashi [4] earlier achieved the same result provided $\|h\|^{2}\leq 9/5$.

On the other hand, we noticed that next to the totally geodesic Lagrangian immersion $\mathbb{S}^{3}(1)\hookrightarrow\mathbb{S}^{6}(1)$ for which we have $\|h\|^{2}=0$, the isometric Lagrangian immersion $\mathbb{S}^{3}(1/16)\hookrightarrow\mathbb{S}^{6}(1)$ has the property that $\|h\|^{2}=45/8$. Thus, Theorem 1.5 and Theorems 1.1 and 1.2 motivate us to consider the following problem:

Problem. Try to characterize the compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ whose second fundamental form $h$ has an optimal value of length next to that of the totally geodesic one.

In this paper, we have solved the above problem. More specifically, for compact Lagrangian submanifolds of $\mathbb{S}^{6}(1)$, we will derive an optimal Simons’ type integral inequality in terms of the second fundamental form. Our main result is the following

Main Theorem. Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Then it holds the Simons’ type integral inequality

[TABLE]

where $\Theta(p)=\max_{u\in U_{p}M^{3}}g(h(u,u),Ju)$ for $p\in M^{3}$.

Moreover, the equality sign in (1.1) holds if and only if $M^{3}$ is either the totally geodesic $\mathbb{S}^{3}(1)$ with $\|h\|^{2}\equiv 0$, or the Dillen-Verstraelen-Vrancken’s Berger sphere $\Psi(S^{3})$ defined by (3.1) which satisfies $\|h\|^{2}=\tfrac{5}{4}+\tfrac{3}{2}\Theta^{2}$ with $\|h\|^{2}\equiv\tfrac{25}{8}$ and $\Theta\equiv\tfrac{\sqrt{5}}{2}$.

As direct consequence of the Main Theorem, we have

Corollary 1.1**.**

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. If $\|h\|^{2}\leq\tfrac{5}{4}+\tfrac{3}{2}\Theta^{2}$, then either $\|h\|^{2}\equiv 0$ and $M^{3}$ is totally geodesic, or $\|h\|^{2}=\tfrac{5}{4}+\tfrac{3}{2}\Theta^{2}$ with $\|h\|^{2}\equiv{25}/8$ and $\Theta\equiv\sqrt{5}/2$ and $M^{3}$ is the Dillen-Verstraelen-Vrancken’s Berger sphere $\Psi(S^{3})$ that is defined by (3.1).

Remark 1.2*.*

Generalizing the observation that a parallel Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ is totally geodesic in [9], it was shown in [21] that, in any $6$-dimensional strict nearly Kähler manifold, Lagrangian submanifolds with parallel second fundamental form are always totally geodesic. On the other hand, M. Djorić and L. Vrancken [9] considered Lagrangian submanifolds of the nearly Kähler $\mathbb{S}^{6}(1)$ which satisfy the following condition, namely for any tangent vector $v$ it holds

[TABLE]

Lagrangian submanifolds satisfying the above condition were called $J$-parallel. It is worth pointing out that if the equality sign of (1.1) holds then $M^{3}$ is $J$-parallel, and that the $J$-parallel Lagrangian submanifolds of $\mathbb{S}^{6}(1)$ have been classified in [9]. In this respect, see also [13] for a complete classification of the $J$-parallel Lagrangian submanifolds of the homogeneous nearly Kähler manifold $\mathbb{S}^{3}\times\mathbb{S}^{3}$.

2. The nearly Kähler $\mathbb{S}^{6}(1)$ and its Lagrangian submanifolds

In this section, we review some aspects of the nearly Kähler manifold $\mathbb{S}^{6}(1)$ and its Lagrangian submanifolds. More details can be found in [19] and [7, 9].

By considering $\mathbb{R}^{7}$ as the imaginary Cayley numbers, the Cayley multiplication induces a vector product on $\mathbb{R}^{7}$. On $S^{6}:=\mathbb{S}^{6}(1)$ with the standard metric $g$ we now define a $(1,1)$-tensor field $J$ by

[TABLE]

for $x\in S^{6}$ and $U\in T_{x}S^{6}$. It is well defined (i.e., $J_{x}U\in T_{x}S^{6}$) and determines an almost complex structure on $\mathbb{S}^{6}(1)$. Furthermore, let $G$ be the $(2,1)$-tensor field on $S^{6}$ defined by

[TABLE]

where $\bar{\nabla}$ is the Levi-Civita connection on $\mathbb{S}^{6}(1)$. Then we have (cf. [7, 10]):

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

where $X,Y,Z,W$ are vector fields on $S^{6}$. Here, (2.2) and (2.6) imply that $(S^{6},g,J)$ is a strict nearly Kähler manifold.

Let $x:M^{3}\rightarrow\mathbb{S}^{6}(1)$ be a Lagrangian isometric immersion. We denote the Levi-Civita connection of $M^{3}$ by $\nabla$ and the normal connection in the normal bundle $T^{\perp}M^{3}$ (defined by the orthogonal projection of $\bar{\nabla}$ on $T^{\perp}M^{3}$) by $\nabla^{\perp}$. The shape operator $A_{\xi}$ in the direction of a normal vector field $\xi$ on $M^{3}$ and $T^{\perp}M^{3}$-valued second fundamental form $h$ are defined by the following Gauss-Weingarten formulas

[TABLE]

where $X,Y$ are tangent vector fields of $M^{3}$, and $h$ is related to $A_{\xi}$ by

[TABLE]

From (2.1) and (2.7) we compute that

[TABLE]

After having the results for the nearly Kähler $\mathbb{S}^{6}(1)$, the following two lemmas have been proved for all $6$-dimensional strict nearly Kähler manifold.

Lemma 2.1** ([10, 18]).**

Let $M^{3}$ be a Lagrangian submanifold of a $6$-dimensional strict nearly Kähler manifold. Then

1. (1)

$M^{3}$* is orientable and minimal,* 2. (2)

$M^{3}$* has volume form $\omega(X,Y,Z)=g(G(X,Y),JZ)$,* 3. (3)

If $X,Y$ are tangent vector fields of $M^{3}$, then $G(X,Y)$ is a normal vector field.

Lemma 2.2** ([21]).**

Let $M^{3}$ be a Lagrangian submanifold of a $6$-dimensional strict nearly Kähler manifold. Then we have

[TABLE]

for any tangent vector fields $X,Y,Z,W$ on $M^{3}$.

Let $x:M^{3}\to\mathbb{S}^{6}(1)\hookrightarrow\mathbb{R}^{7}$ be a Lagrangian submanifold of $\mathbb{S}^{6}(1)$. From now on, we agree on the following index ranges:

[TABLE]

We choose $\{e_{1},e_{2},e_{3},e_{1^{*}},e_{2^{*}},e_{3^{*}}\}$ to be a local orthonormal frame field of the tangent bundle $TS^{6}$ such that $e_{i}$ lies in $TM^{3}$ and $e_{i^{*}}=Je_{i}$ lies in $T^{\perp}M^{3}$. Let $\{\omega_{1},\omega_{2},\omega_{3},\omega_{1^{*}},\omega_{2^{*}},\omega_{3^{*}}\}$ be the associated dual frame field so that restricted to $M^{3}$ it holds that $\omega_{1^{*}}=\omega_{2^{*}}=\omega_{3^{*}}=0$. With respect to $\{e_{1},e_{2},e_{3},e_{1^{*}},e_{2^{*}},e_{3^{*}}\}$, let $\omega_{ij}$ and $\omega_{i^{*}j^{*}}$ denote the connection $1$-forms of $TM^{3}$ and $T^{\perp}M^{3}$, respectively. Then the structure equations of $x:M^{3}\to\mathbb{S}^{6}(1)$ are:

[TABLE]

where $h_{ij}^{k^{*}}=h_{ji}^{k^{*}}=h_{ik}^{j^{*}}$ for any $i,j,k$, and $h=\sum_{i,j,k}h_{ij}^{k^{*}}\omega_{i}\omega_{j}e_{k^{*}}$. Taking exterior differentiation of (2.10) we get

[TABLE]

where $R_{ijkl},\ R_{i^{*}j^{*}kl}$ and $h^{k^{*}}_{ij,l}$ are components of the curvature tensor of the tangent bundle, the normal bundle and the first covariant derivative of the second fundamental form of $M^{3}$, and they satisfy the Gauss-Codazzi-Ricci equations:

[TABLE]

[TABLE]

[TABLE]

From (2.12), the Ricci curvature $R_{ij}$ and the scalar curvature $\tau$ of $M^{3}$ satisfy

[TABLE]

where $\|h\|^{2}=\sum_{i,j,k}(h^{k^{*}}_{ij})^{2}$ is the squared length of the second fundamental form.

Exterior differentiation of the last equation of (2.11) we get the Ricci identity

[TABLE]

where, $h^{p^{*}}_{ij,kl}$ is the components of the second covariant derivative of $h$:

[TABLE]

3. Dillen-Verstraelen-Vrancken’s Berger sphere in $\mathbb{S}^{6}(1)$

Consider the unit sphere $S^{3}:=\{(y_{1},y_{2},y_{3},y_{4})\in\mathbb{R}^{4}\,|\,y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}=1\}$ in $\mathbb{R}^{4}$. There are many Lagrangian immersions from the topological three-sphere $S^{3}$ into the nearly Kähler unit $6$-sphere that have nice properties. Indeed, besides that of constant sectional curvature appeared in Theorem 1.2, immersions of Berger $3$-spheres are also introduced and geometrically characterized in [7] and [3] (see also [16]). For our purpose, we particularly mention that, in [7] (cf. also [9] and [16]), Dillen, Verstraelen and Vrancken constructed an embedding from the topological three-sphere into the nearly Kähler unit $6$-sphere, defined by

[TABLE]

where

[TABLE]

To make calculation of the mapping $\Psi:S^{3}\to\mathbb{S}^{6}(1)$, let $X_{1},\ X_{2},\ X_{3}$ be the vector fields on $S^{3}$, defined by

[TABLE]

Then $X_{1},X_{2}$ and $X_{3}$ form a basis of tangent vector fields to $S^{3}$, and it holds that $[X_{1},X_{2}]=2X_{3},\ [X_{2},X_{3}]=2X_{1}$ and $[X_{3},X_{1}]=2X_{2}$.

We define a Berger metric $\langle\cdot,\cdot\rangle$ on $S^{3}$ such that $X_{1},X_{2}$ and $X_{3}$ are orthogonal and such that $\langle X_{1},X_{1}\rangle=4/9$ and $\langle X_{2},X_{2}\rangle=\langle X_{3},X_{3}\rangle=8/3$. Then

[TABLE]

form an orthonormal frame field on $(S^{3},\langle\cdot,\cdot\rangle)$. Moreover, direct calculations give the following results.

Lemma 3.1** ([7]).**

The curvature tensor of the Berger sphere $(S^{3},\langle\cdot,\cdot\rangle)$ has the following expression

[TABLE]

where $V^{\perp}$ denotes the orthogonal complement of a vector $V$ with respect to $E_{1}$. Moreover, $(S^{3},\langle\cdot,\cdot\rangle)$ has constant scalar curvature $23/16$.

Lemma 3.2** ([7, 9]).**

The above mapping $\Psi:S^{3}\to\mathbb{S}^{6}(1)$ is an isometric Lagrangian embedding from $(S^{3},\langle\cdot,\cdot\rangle)$ into $\mathbb{S}^{6}(1)$. Moreover, with respect to the globally defined orthonormal tangent vector fields $\{E_{1},E_{2},E_{3}\}$, it holds that $G(E_{2},E_{3})=JE_{1}$, and the second fundamental form $h$ of $\Psi:S^{3}\to\mathbb{S}^{6}(1)$ takes the following form

[TABLE]

Remark 3.1*.*

1. (1)

Let $\sigma$ be any plane in the tangent space of $S^{3}$. Then we have an orthonormal basis $\{X,Y\}$ of $\sigma$ such that $X=\cos\theta E_{2}+\sin\theta E_{3}$ and $Y=\sin\varphi E_{1}-\cos\varphi\sin\theta E_{2}+\cos\varphi\cos\theta E_{3}$, where $\theta,\varphi\in\mathbb{R}$. Thus the sectional curvature of the plane $\sigma$ is given by $R(X,Y,Y,X)=K(\sigma)=1/16+20/16\cos^{2}\varphi$. It follows that $1/16\leqq K(\sigma)\leqq 21/16$, where $1/16$ is attained for every plane which contains $E_{1}$, and where $21/16$ is attained only for the plane spanned by $E_{2}$ and $E_{3}$. 2. (2)

Lemma 3.2 implies that the second fundamental form of the Lagrangian embedding $\Psi:S^{3}\to\mathbb{S}^{6}(1)$ has constant squared norm. Indeed, it holds that $\|h\|^{2}(p)=\tfrac{25}{8}$, $\Theta(p)=\max_{u\in U_{p}S^{3}}\langle h(u,u),Ju\rangle=\tfrac{\sqrt{5}}{2}$ for any $p\in S^{3}$. 3. (3)

Due to Lemmas 3.1 and 3.2, we will call the embedding $\Psi:S^{3}\to\mathbb{S}^{6}(1)$ defined by (3.1) as the Dillen-Verstraelen-Vrancken’s Berger sphere.

4. Lemmas and Proof of the Main Theorem

First, thanks to that Lagrangian submanifolds of the nearly Kähler $\mathbb{S}^{6}(1)$ are minimal, and applying for the Gauss-Codazzi-Ricci equations (2.12)–(2.14) and the Ricci identity (2.16), we have the following well known result.

Lemma 4.1** ([4, 14]).**

Let $M^{3}$ be a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Then, in terms the notations in section 2 and put $H_{i}=(h_{jk}^{i^{*}})$, we have the following formula for the Laplacian of $\|h\|^{2}$:

[TABLE]

Here, $S_{ij}={\rm trace}(H_{i}H_{j})$ and $N(A)={\rm trace}(AA^{t})=\sum_{i,j}(a_{ij})^{2}$ for $A=(a_{ij})$.

Next, to calculate the invariant $\sum_{i,j}N(H_{i}H_{j}-H_{j}H_{i})+\sum_{i,j}S_{ij}^{2}$, we will choose a canonical orthonormal bases following the standard way of N. Ejiri [10].

Let $M^{3}$ be a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Let $UM^{3}$ be the unit tangent bundle over $M^{3}$ such that $U_{q}M^{3}=\{u\in T_{q}M^{3}\mid g(u,u)=1\}$ for any $q\in M^{3}$. We define a function $f_{q}$ on $U_{q}M^{3}$ by $f_{q}(u)=g(h(u,u),Ju)$. Since $U_{q}M^{3}$ is compact, there is an element $e_{1}\in U_{q}M^{3}$ such that $f_{q}(e_{1})=\max_{u\in U_{q}M^{3}}f_{q}(u)$. Actually, we have the following lemma.

Lemma 4.2** ([1, 9]).**

Let $M^{3}$ be a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Then, for all $q\in M^{3}$, there exists an orthonormal basis $\{e_{1},e_{2},e_{3}\}$ of $T_{q}M^{3}$ such that

[TABLE]

where

[TABLE]

Lemma 4.3**.**

If (4.2) holds, then by notations of Lemma 4.1 we have

[TABLE]

[TABLE]

Proof.

If (4.2) holds, then we can write $H_{k}=(h^{k^{*}}_{ij})$ in more explicit form:

[TABLE]

[TABLE]

[TABLE]

From (4.6)–(4.8), we have the following computations

[TABLE]

[TABLE]

[TABLE]

It follows that

[TABLE]

[TABLE]

[TABLE]

Next, by direct calculation of $S_{ij}=\sum_{k,l}h^{i^{*}}_{kl}h^{j^{*}}_{kl}$, we get

[TABLE]

From the above computations we immediately verify (4.4) and (4.5). ∎

Next, for a Lagrangian submanifold $M^{3}$ of the nearly Kähler $\mathbb{S}^{6}(1)$, we introduce a $T^{\perp}M^{3}$-valued tensor $\mathbb{T}:TM^{3}\times TM^{3}\times TM^{3}\to T^{\perp}M^{3}$ by

[TABLE]

where F(X,Y,Z)=\tfrac{1}{4}\big{[}G(X,A_{JZ}Y)+G(Y,A_{JX}Z)+G(Z,A_{JY}X)\big{]}.

The tensor $\mathbb{T}$ has important properties that we state as the following lemmas.

Lemma 4.4**.**

Let $M^{3}$ be a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Then we have

[TABLE]

Proof.

Let $\{e_{1},e_{2},e_{3}\}$ be a local orthonormal basis of the tangent bundle of $M^{3}$ as assumed in section 2. From (2.9), we have $A_{Je_{i}}e_{j}=-Jh(e_{i},e_{j})=\sum_{k}h_{ij}^{k^{*}}e_{k}$. It follows that

[TABLE]

Then, by using the minimality of $M^{3}$ and (2.6), which gives that

[TABLE]

we can easily verify by direct calculations that

[TABLE]

Next, by definition $(\nabla h)(e_{k},e_{i},e_{j})=\sum_{l}h_{ij,k}^{l^{*}}Je_{l}$, applying Lemma 2.2 and (2.13) we get

[TABLE]

Using (2.4), (2.13) and (4.19), we have the following calculation:

[TABLE]

Now, by using (2.3) and (2.4), we have

[TABLE]

Combining (4.21) and (4.17), then inserting the results into (4.20), we get

[TABLE]

From (4.18), (4.22) and the fact

[TABLE]

we finally verify the assertion (4.16). ∎

Lemma 4.5**.**

A Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$ satisfies $\mathbb{T}=0$ if and only if it is $J$-parallel, namely (1.2) holds.

Proof.

By using (2.2)–(2.4) and (2.9), we get the calculations:

[TABLE]

It follows that

[TABLE]

Therefore, $\mathbb{T}=0$ if and only if

[TABLE]

This is equivalent to that the submanifold is $J$-parallel (cf. (24) of [9]). ∎

Lemma 4.5 allows us to apply for Theorem A and Theorem 1 of [9] so that we can obtain the following

Lemma 4.6**.**

Let $M^{3}$ be a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. If $M^{3}$ satisfies $\mathbb{T}=0$, then, for each point $q\in M^{3}$, there exists an orthonormal basis $\{e_{1},e_{2},e_{3}\}$ of $T_{q}M^{3}$ such that either

1. (a)

$h(e_{1},e_{1})=h(e_{2},e_{2})=h(e_{3},e_{3})=h(e_{1},e_{2})=h(e_{1},e_{3})=h(e_{2},e_{3})=0$, i.e., $M^{3}$ is totally geodesic with $K=1$; or 2. (b)

$h(e_{1},e_{1})=\tfrac{\sqrt{5}}{2}Je_{1}$,  $h(e_{1},e_{2})=-\tfrac{\sqrt{5}}{4}Je_{2}$,  $h(e_{1},e_{3})=-\tfrac{\sqrt{5}}{4}Je_{3}$,

$h(e_{2},e_{2})=-\tfrac{\sqrt{5}}{4}Je_{1}+\tfrac{\sqrt{10}}{4}Je_{2}$,  $h(e_{3},e_{3})=-\tfrac{\sqrt{5}}{4}Je_{1}-\tfrac{\sqrt{10}}{4}Je_{2}$,

$h(e_{2},e_{3})=-\tfrac{\sqrt{10}}{4}Je_{3}$.

Moreover, $M^{3}$ has constant sectional curvature $\tfrac{1}{16}$; or 3. (c)

$h(e_{1},e_{1})=\tfrac{\sqrt{5}}{2}Je_{1}$,  $h(e_{1},e_{2})=-\tfrac{\sqrt{5}}{4}Je_{2}$, $h(e_{1},e_{3})=-\tfrac{\sqrt{5}}{4}Je_{3}$,

$h(e_{2},e_{2})=-\tfrac{\sqrt{5}}{4}Je_{1}$,  $h(e_{3},e_{3})=-\tfrac{\sqrt{5}}{4}Je_{1}$,  $h(e_{2},e_{3})=0$.

Moreover, $M^{3}$ is locally congruent to Dillen-Verstraelen-Vrancken’s Berger sphere $\Psi(S^{3})$, defined by (3.1).

The Completion of Main Theorem’s Proof.

Let $M^{3}$ be a compact Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Now, we apply for Lemma 4.2 and make calculation at an arbitrary fixed point $q\in M^{3}$ with the orthonormal basis $\{e_{1},e_{2},e_{3}\}$ of $T_{q}M^{3}$. Then, from Lemmas 4.1, 4.3 and 4.4, we have

[TABLE]

Noticing that $\lambda_{1}+\lambda_{2}=\max_{u\in UM^{3}}g(h(u,u),Ju)=\Theta$, from Lemma 4.2, (4.24) and the arbitrariness of $q\in M^{3}$, by applying for the divergence theorem, we get

[TABLE]

The equality sign in (4.25) holds if and only if $\mathbb{T}=0$ and that, either $M^{3}$ is totally geodesic, or $\mu_{1}=\mu_{2}=0$ and $\lambda_{1}=\lambda_{2}\not=0$ on $M^{3}$. In the latter case, according to Lemma 4.6, $M^{3}$ is locally congruent to the Dillen-Verstraelen-Vrancken’s Berger sphere $\Psi(S^{3})$, defined by (3.1). It follows from Lemma 3.2 that $\|h\|^{2}\equiv{25}/8$ and $\Theta\equiv\sqrt{5}/2$. This shows that $\|h\|^{2}=\tfrac{5}{4}+\tfrac{3}{2}\Theta^{2}$.∎

Finally, in conclusion we state the following locally rigidity theorem which is of independent meaning.

Theorem 4.1**.**

Let $M^{3}$ be a Lagrangian submanifold of the nearly Kähler $\mathbb{S}^{6}(1)$. Then it holds that

[TABLE]

Moreover, (4.26) holds identically on $M^{3}$ if and only if one of the following three cases occurs:

1. (a)

$M^{3}$* is totally geodesic $(K=1\ {\rm and}\ h=0)$, or* 2. (b)

$M^{3}$* has constant sectional curvature $\tfrac{1}{16}$ and $\|h\|^{2}=\tfrac{45}{8}$, or* 3. (c)

$M^{3}$* is locally congruent to an open part of the Dillen-Verstraelen-Vrancken’s Berger sphere $\Psi(S^{3})$ defined by (3.1) with $\|h\|^{2}=\tfrac{25}{8}$.*

Proof.

This is a direct consequence of Theorem A of [9], Lemmas 4.4 and 4.5. ∎