Lagrangian submanifolds in the homogeneous nearly Kähler ${\mathbb {S}}^3 \times {\mathbb {S}}^3$

Abstract

In this paper, we investigate Lagrangian submanifolds in the homogeneous nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. We introduce and make use of a triplet of angle functions to describe the geometry of a Lagrangian submanifold in $\mathbb {S}^3 \times \mathbb {S}^3$. We construct a new example of a flat Lagrangian torus and give a complete classification of all the Lagrangian immersions of spaces of constant sectional curvature. As a corollary of our main result, we obtain that the radius of a round Lagrangian sphere in the homogeneous nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ can only be $\frac{2}{\sqrt{3}}$ or $\frac{4}{\sqrt{3}}$.

1 Introduction

The study of Lagrangian submanifolds originates from symplectic geometry and classical mechanics. An even-dimensional manifold is called symplectic if it admits a symplectic form, which is a closed and non-degenerate two-form. A submanifold of a symplectic manifold is called Lagrangian if the symplectic form restricted to the manifold vanishes and if the dimension of the submanifold is half the dimension of the symplectic manifold. The well-known theorem of Darboux states that locally all symplectic manifolds are indistinguishable. If one considers a Lagrangian submanifold immersed in a symplectic manifold, then by the theorem of Darboux, this Lagrangian submanifold can also be locally immersed in any symplectic manifold of the same dimension. Therefore, a local classification of Lagrangian submanifolds is trivial from the symplectic point of view.

Lagrangian submanifolds can more generally be considered in almost Hermitian manifolds. Note that an almost Hermitian manifold is not necessarily symplectic. We call that a submanifold of M in an almost Hermitian manifold N Lagrangian, if the almost complex structure J interchanges the tangent and the normal spaces, and if the dimension of M is half the dimension of N. The most important class of almost Hermitian manifolds are the Kähler manifolds. Kähler manifolds admit a complex, Riemannian and symplectic structure which are all three compatible with each other. The study of Lagrangian submanifolds in Kähler manifolds is a classic topic and was initiated in the 1970’s [8]. A classification of Lagrangian submanifolds from the Riemannian point of view is far from trivial. There is no complete classification and this is too much to hope for. For this reason, it makes sense to study Lagrangian submanifolds with some additional Riemannian conditions. For instance, one can study Lagrangian submanifolds that are minimal, Hamiltonian minimal, Hamiltonian stable, or unstable (see for instance [5, 21, 22, 29, 30]) or have constant sectional curvature [15]. For a review on Riemannian geometry of Lagrangian submanifolds, we refer to [6] and the references therein.

Nearly Kähler manifolds are almost Hermitian manifolds with almost complex structure J satisfying that the covariant derivative of J is skew-symmetric. The geometry of nearly Kähler manifolds was initially studied by Gray [17, 18] in the 1970s from the point of view of weak holonomy. Nagy ([27, 28]) made further contributions to the classification of nearly Kähler manifolds using previous work in [9]. Butruille ([3, 4]) proved that the only homogeneous 6-dimensional nearly Kähler manifolds are the nearly Kähler ${\mathbb {S}}^6$, ${\mathbb {S}}^3 \times {\mathbb {S}}^3$, the complex projective space $\mathbb {CP}^3$, and the flag manifold $SU(3)/U(1)\times U(1)$. It is remarkable that Foscolo and Haskins [16] have constructed inhomogeneous nearly Kähler structures on both ${\mathbb {S}}^6$ and ${\mathbb {S}}^3 \times {\mathbb {S}}^3$ recently. In the following, when we say the nearly Kähler ${\mathbb {S}}^6$ or the nearly Kähler ${\mathbb {S}}^3 \times {\mathbb {S}}^3$, we always mean the homogeneous nearly Kähler ${\mathbb {S}}^6$ or the homogeneous nearly Kähler ${\mathbb {S}}^3 \times {\mathbb {S}}^3$. In [24], Moroianu and Semmelmann studied the infinitesimal Einstein deformations of nearly Kähler metrics. Lagrangian submanifolds of the nearly Kähler ${\mathbb {S}}^6$ are well studied by now; see, for instance, [10,11,12, 14] and [20]. We also refer to Section 18 of [6] and Chapter 19 of [7] for an overview. Moroianu and Semmelmann [25] recently gave new examples of Lagrangian immersions of round spheres and Berger spheres in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. A broader study of Lagrangian submanifolds in nearly Kähler manifolds was investigated in [32] by Schäfer and Smozcyk. It was proven in [32] that Lagrangian submanifolds in a nearly Kähler manifold behave nicely with respect to the splitting of the nearly Kähler manifold. If a nearly Kähler manifold is a product of nearly Kähler manifolds, then its Lagrangian submanifolds split into products of Lagrangian submanifolds under certain spectral assumptions on the nearly Kähler factors. Loosely speaking, this means that Lagrangian submanifolds in 6-dimensional nearly Kähler manifolds are building blocks of Lagrangian submanifolds in general nearly Kähler manifolds. This motivates the study of Lagrangian submanifolds in 6-dimensional nearly Kähler manifolds. The existence for Lagrangian submanifolds in nearly Kähler manifolds is not unobstructed. A nearly Kähler manifold is called strict nearly Kähler, if $\tilde{\nabla }_XJ\ne 0$ for all non-vanishing tangent vector field X, where $\tilde{\nabla }J$ denotes the covariant derivative of J. Schäfer and Smozcyk [32] proved that Lagrangian submanifolds in a strict nearly Kähler manifold of dimension six or a twistor space over a positive quaternionic Kähler manifold are minimal and orientable. This is different from Lagrangian submanifolds in Kähler manifolds. The reason is that there is no Darboux theorem for nearly Kähler manifolds since these manifolds are not symplectic. This is an extra reason to study these Lagrangian submanifolds from a Riemannian point of view.

In this paper, we study Lagrangian submanifolds in the nearly Kähler  $\mathbb {S}^3 \times \mathbb {S}^3$. In Sect. 2, we recall the basic properties of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ and present some properties of Lagrangian submanifolds in nearly Kähler manifolds. In Sect. 3, we show that on a Lagrangian submanifold in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ there exist a local frame and three angle functions that describe the geometry and shape of the submanifold very well. These are related to the almost product structure P introduced in [2]. We show that most of the geometry of the submanifold can be described in terms of the three angle functions. For example, the derivatives of these angle functions give information about most of the components of the second fundamental form. In Sect. 4, we present eight examples (or families of examples) of Lagrangian submanifolds in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. The examples are Lagrangian immersions of respectively round spheres, Berger spheres or a flat torus. The flat torus (see Example 4.8) is a new example. Examples 4.1–4.3 are the factors and diagonal 3-spheres which were given by Schäfer and Smozcyk in [32]. Examples 4.4–4.7 were constructed by Moroianu and Semmelmann in [25]. In Sect. 5, we classify the Lagrangian submanifolds of constant sectional curvature in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. We note that it is not possible to follow the approach introduced by Ejiri for studying Lagrangian submanifolds of constant sectional curvature in the complex space forms [15] or in the nearly Kähler 6-sphere [14]. We therefore use a new approach. The key step is Lemma 5.1 which is an identity involving the second fundament form, the almost complex structure and the almost product structure. The main result that we prove is the following:

Theorem 1.1

Let M be a Lagrangian submanifold of constant sectional curvature in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Then up to an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$, M is locally congruent to one of the following immersions:

• (1) $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,1)$, which is Example 4.1,

• (2) $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (1,u)$, which is Example 4.2,

• (3) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,u)$, which is Example 4.3,

• (4) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u\mathbf {i}u^{-1},u\mathbf {j}u^{-1})$, which is Example 4.7,

• (5) $f: {\mathbb {R}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: (u,v,w)\mapsto (p(u,w),q(u,v))$, where p and q are constant mean curvature tori in ${\mathbb {S}}^3$ given in Example 4.8.

Remark 1.2

In view of Proposition 4.4 in [25], Moroianu and Semmelmann showed that the radius of a round Lagrangian sphere in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ is necessarily of the form $\frac{k}{\sqrt{3}}$ for some integer $k\ge 2$ (note that the scaling in [25] is slightly different from ours, so the radius here has been modified to be adapted to our conventions). As a corollary of our Theorem 1.1, we obtain that the values of the integer k can only be 2 or 4.

We remark that in [35] the authors obtain the following complete classification of all the totally geodesic Lagrangian immersion in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$.

Theorem 1.3

([35]) Let M be a totally geodesic Lagrangian submanifold in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Then up to an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$, M is locally congruent to one of the following immersions:

• (1) $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,1)$, which is Example 4.1,

• (2) $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (1,u)$, which is Example 4.2,

• (3) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,u)$, which is Example 4.3,

• (4) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,u\mathbf {i})$, which is Example 4.4,

• (5) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u^{-1},u\mathbf {i}u^{-1})$, which is Example 4.5,

• (6) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u\mathbf {i}u^{-1},u^{-1})$, which is Example 4.6.

Hence, combining Theorem 1.3 with Theorem 1.1, one obtain characterizations of all the eight examples (see 4 for details of the examples) of Lagrangian submanifolds in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$.

Remark 1.4

We remark that from a very recent paper of Moruz and the second author [26], we know that the examples 1,2,3 (examples 4,5,6, respectively) are pairwise congruent to each other by applying an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Here in this paper, we mainly consider the isometries that preserve the metric of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ as well as its complex structure J and the almost product structure P. This is the reason why Theorem 1.1 and Theorem 1.3 are formulated as they are.

2 The nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ and its Lagrangian submanifolds

In this section, we recall the definition of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ from [2] and [13] and give some basic properties of Lagrangian submanifolds which will be useful for the rest of the paper.

Using the natural identification $T_{(p,q)}(\mathbb {S}^3 \times \mathbb {S}^3) \cong T_p {\mathbb {S}}^3 \oplus T_q {\mathbb {S}}^3$, we write a tangent vector at (p, q) as $Z(p,q) = \bigl ( U(p,q), V(p,q)\bigr )$ or simply $Z=(U,V)$.

The 3-sphere ${\mathbb {S}}^3$ can be regarded as the set of all the unit quaternions in ${\mathbb {H}}$. As usual we use $\mathbf {i},\,\mathbf {j},\,\mathbf {k}$ to denote the imaginary units of ${\mathbb {H}}$. Define the vector fields

$$\begin{aligned} E_1(p,q)&= (p\mathbf {i},0),&F_1(p,q)&= (0,q\mathbf {i}),\\ E_2(p,q)&= (p\mathbf {j},0),&F_2(p,q)&= (0,q\mathbf {j}),\\ E_3(p,q)&= -(p\mathbf {k},0),&F_3(p,q)&= -(0,q\mathbf {k}). \end{aligned}$$

These vector fields are mutually orthogonal with respect to the usual Euclidean product metric on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. The Lie brackets are $[E_i,E_j]=-2\varepsilon _{ijk}E_k$, $[F_i,F_j]=-2\varepsilon _{ijk}F_k$ and $[E_i,F_j]=0$, where

$$\begin{aligned} \varepsilon _{ijk}=\left\{ \begin{array}{ll} 1,&{} \quad \mathrm{if}\, \{ijk\}\ \mathrm{is\, an\, even\, permutation\, of\, \{123\}},\\ -1,&{}\quad \mathrm{if}\ \{ijk\}\ \mathrm{is\, an\, odd\, permutation\, of\, \{123\}},\\ 0,&{} \quad \mathrm{\, otherwise}. \end{array} \right. \end{aligned}$$

The almost complex structure J on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ is defined by

$$\begin{aligned} J(U,V)_{(p,q)} = \frac{1}{\sqrt{3}}\Big ( 2pq^{-1}V - U, -2qp^{-1}U + V \Big ) \end{aligned}$$

(2.1)

for $(U,V) \in T_{(p,q)}({\mathbb {S}}^3\times {\mathbb {S}}^3)$ (see [4]). Note that the definition uses the Lie group structure of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. The map

$$\begin{aligned} T_{(1,1)}\mathbb {S}^3 \times \mathbb {S}^3\rightarrow T_{(1,1)}\mathbb {S}^3 \times \mathbb {S}^3: (U,V)\mapsto \frac{1}{\sqrt{3}}\Big (2V - U, -2 V + U\Big ) \end{aligned}$$

defines an almost complex structure on the Lie algebra, the tangent space at (1, 1). By using left translations on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$, this map can be extended to an almost complex structure on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. The left translations on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ are given by left multiplications with a unit quaternion. The almost complex structure can be described as follows. The first step is to left translate a vector (U, V) at $(p,q)\in \mathbb {S}^3 \times \mathbb {S}^3$ to $(p^{-1}U, q^{-1}V)$ at the unit element (1, 1). Then this vector is mapped onto $\tfrac{1}{\sqrt{3}}\Big (2 q^{-1}V-p^{-1}U, -2p^{-1}U + q^{-1}V\Big )$ at the point (1, 1). When this vector is translated back to $T_{(p,q)}\mathbb {S}^3 \times \mathbb {S}^3$, it gives the expression (2.1).

The nearly Kähler metric on $\mathbb {S}^3 \times \mathbb {S}^3$ is the Hermitian metric associated with the usual Euclidean product metric on $\mathbb {S}^3 \times \mathbb {S}^3$:

$$\begin{aligned} g(Z,Z')&= \frac{1}{2} \Big (\langle Z, Z'\rangle + \langle JZ, JZ'\rangle \Big )\\&= \frac{4}{3} \Big (\langle U, U'\rangle + \langle V, V'\rangle \Big ) -\frac{2}{3} \Big (\langle p^{-1}U, q^{-1}V'\rangle + \langle p^{-1}U', q^{-1}V\rangle \Big ), \end{aligned}$$

where $Z=(U,V)$ and $Z'=(U',V')$. In the first line $\langle \cdot , \cdot \rangle$ stands for the usual Euclidean product metric on ${\mathbb {S}}^3\times {\mathbb {S}}^3$ and in the second line $\langle \cdot , \cdot \rangle$ stands for the usual Euclidean metric on ${\mathbb {S}}^3$. By definition the almost complex structure is compatible with the metric g. An easy calculation gives

$$\begin{aligned} g(E_i, E_j)&= 4/3 \,\delta _{ij},&g(E_i, F_j)&= -2/3 \,\delta _{ij},&g(F_i, F_j)&= 4/3 \,\delta _{ij}. \end{aligned}$$

Note that this metric differs up to a constant factor from the one introduced in [4]. Here we set everything up so that it equals the Hermitian metric associated with the usual Euclidean product metric. In [4], the factor was chosen in such a way that the standard basis $E_1,E_2,E_3,F_1,F_2,F_3$ has volume 1.

Lemma 2.1

([2]) The Levi–Civita connection $\tilde{\nabla }$ on $\mathbb {S}^3 \times \mathbb {S}^3$ with respect to the metric g is given by

$$\begin{aligned} \tilde{\nabla }_{E_i} E_j&= -\varepsilon _{ijk}E_k,&\tilde{\nabla }_{E_i} F_j&= \frac{\varepsilon _{ijk}}{3}\Big (E_k - F_k\Big ),\\ \tilde{\nabla }_{F_i} E_j&= \frac{\varepsilon _{ijk}}{3} \Big (F_k - E_k\Big ),&\tilde{\nabla }_{F_i} F_j&= -\varepsilon _{ijk}F_k. \end{aligned}$$

One easily verifies that

$$\begin{aligned} (\tilde{\nabla }_{E_i} J)E_j&= -\frac{2}{3\sqrt{3}}\varepsilon _{ijk}\Big (E_k + 2F_k\Big ),&(\tilde{\nabla }_{E_i} J)F_j&= -\frac{2}{3\sqrt{3}}\varepsilon _{ijk}\Big (E_k - F_k\Big ), \\ (\tilde{\nabla }_{F_i} J)E_j&= -\frac{2}{3\sqrt{3}}\varepsilon _{ijk}\Big (E_k - F_k\Big ),&(\tilde{\nabla }_{F_i} J)F_j&= \frac{2}{3\sqrt{3}}\varepsilon _{ijk}\Big (2E_k +F_k\Big ). \end{aligned}$$

The tensor field $G=\tilde{\nabla }J$ is skew-symmetric, i.e., $G(X,Y)+G(Y,X)=(\tilde{\nabla }_{X} J)Y+(\tilde{\nabla }_{Y} J)X=0$ for every $X,Y\in TM$, hence ($\mathbb {S}^3 \times \mathbb {S}^3$, g, J) is nearly Kähler. Moreover, G satisfies the following properties (cf. [1, 17]):

$$\begin{aligned} G(X,JY)+JG(X,Y)=0,~g(G(X,Y),Z)+g(G(X,Z),Y)=0. \end{aligned}$$

(2.2)

For unitary quaternions a, b and c, the map $F:\mathbb {S}^3 \times \mathbb {S}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3$ given by $(p,q)\mapsto (apc^{-1},bqc^{-1})$ is an isometry of $(\mathbb {S}^3 \times \mathbb {S}^3,g)$ (cf. the remark after Lemma 2.2 in [31]). Indeed, F preserves the almost complex structure J, since

$$\begin{aligned} J dF_{(p,q)}(v,w)= & {} \frac{1}{\sqrt{3}}\Big (2(apc^{-1})(cq^{-1}b^{-1})bwc^{-1} - avc^{-1},\\&\bigl . -2(bqc^{-1})(cp^{-1}a^{-1})avc^{-1}+bwc^{-1}\Big )\\= & {} dF_{(p,q)}\bigl (J(v,w)\bigr ) \end{aligned}$$

(see also [23, Proposition 3.1]) and F preserves the usual metric $\langle \cdot \,, \cdot \rangle$ as well.

Next we introduce an almost product structure on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. The (1, 1)-tensor field P is defined by

$$\begin{aligned} PZ = (pq^{-1}V, qp^{-1}U), \end{aligned}$$

(2.3)

where $Z=(U,V)$ is a tangent vector at (p, q). The definition of P also makes use of the Lie group structure of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. At the Lie algebra level the map

$$\begin{aligned} T_{(1,1)}\mathbb {S}^3 \times \mathbb {S}^3\rightarrow T_{(1,1)}\mathbb {S}^3 \times \mathbb {S}^3: (U,V)\mapsto (V,U) \end{aligned}$$

defines an almost product structure. By left translation this structure can be extended to the manifold $\mathbb {S}^3 \times \mathbb {S}^3$, similarly as was done for the almost complex structure J. We summarize the elementary properties of the almost product in the following lemma.

Lemma 2.2

([2]) The almost product structure P satisfies the following properties:

$$\begin{aligned} P^2&= \mathrm {Id}, \text { i.e., P is involutive,}\end{aligned}$$

(2.4a)

$$\begin{aligned} PJ&=-JP, \text { i.e., P and J anti-commute,} \end{aligned}$$

(2.4b)

$$\begin{aligned} g(PZ,PZ')&=g(Z,Z'), \text { i.e., P is compatible with g,} \end{aligned}$$

(2.4c)

$$\begin{aligned} g(PZ,Z')&=g(Z,PZ'), \text { i.e., P is symmetric.} \end{aligned}$$

(2.4d)

Proof

The first three equations can be verified with a direct calculation. The last equation follows from the first and third equation. $\square$

It is elementary to show that the isometries of ($\mathbb {S}^3 \times \mathbb {S}^3,~g,~J$) mentioned above also preserve the almost product structure P. Note that $PE_i = F_i$ and $PF_i=E_i$. From these equations and Lemma 2.1, it follows that

$$\begin{aligned} (\tilde{\nabla }_{E_i} P)E_j&= \frac{1}{3}\varepsilon _{ijk}\Big (E_k +2F_k\Big ),&(\tilde{\nabla }_{E_i} P)F_j = -\frac{1}{3}\varepsilon _{ijk}\Big (2E_k +F_k\Big ),\\ (\tilde{\nabla }_{E_i} P)F_j&= -\frac{1}{3}\varepsilon _{ijk}\Big (E_k +2F_k\Big ),&(\tilde{\nabla }_{F_i} P)F_j = \frac{1}{3}\varepsilon _{ijk}\Big (2E_k +F_k\Big ). \end{aligned}$$

The tensor field $\tilde{\nabla }P$ does not vanish identically, so the endomorphism P is not a product structure. However, the almost product structure P and its covariant derivative $\tilde{\nabla }P$ admit the following properties.

Lemma 2.3

([2]) For tangent vector fields X, Y on $(\mathbb {S}^3 \times \mathbb {S}^3,g,J)$ the following equations hold:

$$\begin{aligned}&PG(X,Y) + G(PX,PY) = 0, \end{aligned}$$

(2.5)

$$\begin{aligned}&(\tilde{\nabla }_X P)JY = J(\tilde{\nabla }_X P)Y, \end{aligned}$$

(2.6)

$$\begin{aligned}&G(X,PY) + PG(X,Y) = -2J(\tilde{\nabla }_X P)Y,\end{aligned}$$

(2.7)

$$\begin{aligned}&(\tilde{\nabla }_X P)PY + P(\tilde{\nabla }_X P)Y=0,\end{aligned}$$

(2.8)

$$\begin{aligned}&(\tilde{\nabla }_X P)Y + (\tilde{\nabla }_{PX} P)Y =0. \end{aligned}$$

(2.9)

The Riemannian curvature tensor $\tilde{R}$ on $(\mathbb {S}^3 \times \mathbb {S}^3,g)$ is given by (see [2])

$$\begin{aligned} \begin{aligned} \tilde{R}(U,V)W&= \frac{5}{12}\Big (g(V,W)U - g(U,W)V\Big ) \\&\quad +\frac{1}{12}\Big (g(JV,W)JU - g(JU,W)JV - 2g(JU,V)JW\Big ) \\&\quad + \frac{1}{3}\Big (g(PV,W)PU - g(PU,W)PV \\&\quad \phantom {\frac{2}{3\sqrt{3}}}\quad + g(JPV,W)JPU - g(JPU,W)JPV\Big ). \end{aligned} \end{aligned}$$

A nearly Kähler manifold is called of constant type, if there exists a nonzero constant a such that

$$\begin{aligned} g\big (G(X,Y),G(X,Y)\big ) = a\big (g(X,X) g(Y,Y) -g(X,Y)^2 -g(JX,Y)^2\big ). \end{aligned}$$

One can now show that the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ is of constant type $\frac{1}{3}$, and therefore, we have (see [2])

$$\begin{aligned} g\bigl (G(X,Y),G(Z,W)\bigr )&=\frac{1}{3} \Big (g(X,Z) g(Y,W)-g(X,W)g(Y,Z) \nonumber \\&\qquad + g(JX,Z)g(JW,Y)-g(JX,W)g(JZ,Y)\Big ), \end{aligned}$$

(2.10)

$$\begin{aligned} G\bigl (X,G(Y,Z)\bigr )&= \frac{1}{3}\Big ( g(X,Z)Y-g(X,Y)Z\nonumber \\&\qquad + g(JX,Z)JY -g(JX,Y)JZ\Big ), \end{aligned}$$

(2.11)

$$\begin{aligned} (\tilde{\nabla }G)(X,Y,Z)&= \frac{1}{3}\Big (g(X,Z) JY -g(X,Y) JZ -g(JY,Z)X\Big ). \end{aligned}$$

(2.12)

For later use, we also need the relation between the geometry of the nearly Kähler manifold $(\mathbb {S}^3 \times \mathbb {S}^3,g)$ and the product manifold $(\mathbb {S}^3 \times \mathbb {S}^3,\langle \cdot , \cdot \rangle )$, which is $\mathbb {S}^3 \times \mathbb {S}^3$ endowed with the usual Euclidean product metric. The equations in this paragraph shall be used every time we want to obtain an explicit parametrization of a submanifold in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$.

The almost product structure P can be expressed in terms of the usual product structure $QZ= Q(U,V)= (-U,V)$ and vice versa:

$$\begin{aligned} QZ&= \frac{1}{\sqrt{3}} \Big (2 PJZ - JZ\Big ), \end{aligned}$$

(2.13)

$$\begin{aligned} PZ&= \frac{1}{2}\Big (Z-\sqrt{3}QJZ\Big ). \end{aligned}$$

(2.14)

Using these equations the Euclidean product metric, $\langle \cdot , \cdot \rangle$ can be expressed in terms of g and P:

$$\begin{aligned} \langle Z, Z'\rangle = \frac{3}{8}\Big (g(Z,Z')+g(QZ,QZ')\Big ) = g(Z,Z')+ \frac{1}{2}g(Z,PZ'), \end{aligned}$$

(2.15)

and consequently

$$\begin{aligned} \langle Z, QZ'\rangle = \frac{\sqrt{3}}{2}g(Z,PJZ'). \end{aligned}$$

(2.16)

We can now show the relation between the Levi–Civita connections $\tilde{\nabla }$ of g and $\nabla ^E$ of the usual Euclidean product metric $\langle \cdot , \cdot \rangle$ on $\mathbb {S}^3 \times \mathbb {S}^3$.

Lemma 2.4

([13]) The relation between the nearly Kähler connection $\tilde{\nabla }$ and the Euclidean connection $\nabla ^E$ on $\mathbb {S}^3 \times \mathbb {S}^3$ is

$$\begin{aligned} \nabla ^E_X Y= \tilde{\nabla }_X Y +\frac{1}{2}\Big (JG(X,PY) + JG(Y,PX)\Big ). \end{aligned}$$

(2.17)

Remark 2.5

Using the above lemma and expression (2.13), one can show that $(\nabla ^E_X Q)Y = 0$ implies Eq. (2.6) and vice versa. In this sense P really is the “nearly Kähler analogue” of the Euclidean product structure Q.

In [32], Schäfer and Smoczyk gave a broader study of Lagrangian submanifolds in a nearly Kähler manifold, and they also showed that the classical result of Ejiri [14], that a Lagrangian submanifold of the nearly Kähler  ${\mathbb {S}}^6$ is always minimal and orientable, holds actually for arbitrary 6-dimensional strict nearly Kähler  manifolds (see also [19]). From now on, we will assume that M is a Lagrangian submanifold in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Hence M is 3-dimensional and the almost complex structure J maps tangent vectors to normal vectors. As for Lagrangian submanifolds of the nearly Kähler ${\mathbb {S}}^6$, the following lemma from [19] or [32] holds:

Lemma 2.6

(cf. [19, 32]) Let M be a Lagrangian submanifold of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Then M is minimal and orientable. Moreover, for X, Y tangent to M, G(X, Y) is a normal vector field on M.

If we denote the immersion by f and denote the shape operator by S, then the formulas of Gauss and Weingarten are, respectively, given by

$$\begin{aligned}&\widetilde{\nabla }_X f_*Y = f_*(\nabla _X Y) + h(X,Y), \end{aligned}$$

(2.18)

$$\begin{aligned}&\widetilde{\nabla }_X \eta = - f_*(S_\eta X) + \nabla ^\perp _X \eta , \end{aligned}$$

(2.19)

for tangent vector fields X and Y and a normal vector field $\eta$. The second fundamental form h is related to $S_\eta$ by $g(h(X,Y),\eta )=g(S_\eta X,Y)$. From (2.18) and (2.19) we find that

$$\begin{aligned}&\nabla ^\perp _X Jf_*(Y) = J f_*(\nabla _X Y) + G(f_*X,f_*Y), \end{aligned}$$

(2.20)

$$\begin{aligned}&f_*(S_{JY} X) = -J h(X,Y). \end{aligned}$$

(2.21)

The above formulas immediately imply that

$$\begin{aligned} g(h(X,Y),Jf_*Z) =g(h(X,Z),Jf_*Y), \end{aligned}$$

(2.22)

i.e., $g(h(X,Y),Jf_*Z)$ is totally symmetric. Of course as usual, whenever there is no confusion, we will drop the immersion f from the notations.

3 Lagrangian submanifolds of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$

Note that in the previous section, most of the results remain valid for Lagrangian submanifolds of arbitrary 6-dimensional strict nearly Kähler manifolds. Here, however, we will restrict ourselves to the case that the ambient space is the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. We will show how the properties of the almost product structure P, related to the product structure on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ incorporates most of the geometry of the Lagrangian submanifold. The key idea is to “decompose” the almost product structure P into a tangent part A and a normal part B.

Let M be a Lagrangian submanifold of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Since M is Lagrangian, the pullback of $T(\mathbb {S}^3 \times \mathbb {S}^3)$ to M splits into $TM\oplus JTM$. Therefore, there are two endomorphisms $A,B:TM\rightarrow TM$ such that the restriction $P|_{TM}$ of P to the submanifold M equals $A+ JB$, that is $PX = AX + JBX$ for all $X\in TM$. Note that the above formula, together with the fact that P and J anti-commute, also determines P on the normal space by $PJX = -JPX = BX- JAX$. The following lemma gives the basic properties of A and B.

Lemma 3.1

The endomorphisms A and B satisfy that (i) A and B are symmetric; (ii) A and B commute; (iii) $A^2+B^2 = \mathrm {Id}$.

Proof

The lemma follows easily from the basic properties of P and J (P is symmetric, J is compatible with g). For $X,Y\in TM$ we have $g( AX, Y)=g( PX, Y)= g( X, PY)=g( X, AY)$. Similarly one finds $g( BX, Y)=g( PJX, Y)=g( PY, JX)=g( JBY, JX)=g( BY, X)$. Since P is involutive, we also have

$$\begin{aligned} X = P^2 X = (A^2+B^2)X + J(BA-AB)X. \end{aligned}$$

Comparing the tangent and normals parts gives $A^2+B^2 = \mathrm {Id}$ and $[A,B]=0$.$\square$

As A and B are commuting symmetric operators, we know that they can be diagonalized simultaneously at a point of M. Therefore, at each point p, there is an orthonormal basis $e_1$, $e_2$, $e_3\in T_p M$ such that

$$\begin{aligned} Pe_i&= \cos 2\theta _i e_i + \sin 2\theta _i Je_i,~ i=1,2,3. \end{aligned}$$

The factor 2 in the arguments of the sines and cosines is there for convenience as it will simplify many of the expressions in the sequel.

Now we extend the orthonormal basis $e_1$, $e_2$, $e_3$ at a point p to a frame on a neighborhood of p in the Lagrangian submanifold. By Lemma 1.1-1.2 in [33], the orthonormal basis at a point can be extended to a differentiable frame $E_1$, $E_2$, $E_3$ on an open dense neighborhood where the multiplicities of the eigenvalues of A and B are constant. Taking also into account the properties of G, we know that there exists a local orthonormal frame $\{E_1,E_2,E_3\}$ on an open dense subset of M such that

$$\begin{aligned} A E_i= \cos (2 \theta _i ) E_i,~B E_i = \sin (2 \theta _i) E_i,~J G(E_i,E_j)=\tfrac{1}{\sqrt{3}} \varepsilon _{ijk} E_k. \end{aligned}$$

(3.1)

Lemma 3.2

$\theta _1 + \theta _2 + \theta _3$ is zero modulo $\pi$.

Proof

Using Eq. (2.5) and (2.4b), we get

$$\begin{aligned} PE_1 = \sqrt{3}PJG(E_2,E_3) =\sqrt{3} JG(PE_2,PE_3) \end{aligned}$$

and thus $PE_1 = \cos 2\theta _1 E_1 + \sin 2 \theta _1 JE_1$ is equal to

$$\begin{aligned} \sqrt{3}\Bigl ( \cos (2(\theta _2+\theta _3)) JG(E_2,E_3) +\sin (2(\theta _2+\theta _3)) G(E_2,E_3)\Bigr ). \end{aligned}$$

Comparing tangent and normal parts gives

$$\begin{aligned} \cos 2\theta _1&=\cos (2(\theta _2+\theta _3)), ~ \sin 2 \theta _1 = -\sin (2(\theta _2+\theta _3). \end{aligned}$$

Therefore

$$\begin{aligned} \cos (2(\theta _1+\theta _2+\theta _3))=\cos 2\theta _1 \cos (2(\theta _2+\theta _3))- \sin 2\theta _1 \sin (2(\theta _2+\theta _3))=1, \end{aligned}$$

so $\theta _1+\theta _2+\theta _3= 0 \mod \pi$.$\square$

Using the decomposition of P and the expression of the curvature tensor on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$, we can now write down the expressions for the equations of Gauss and Codazzi. We have the equation of Gauss as follows.

$$\begin{aligned} R(X,Y)Z= & {} \frac{5}{12}\Big (g(Y,Z)X - g(X,Z)Y\Big ) \nonumber \\&+ \frac{1}{3}\Big (g(AY,Z)AX - g(AX,Z)AY + g(BY,Z)BX - g(BX,Z)BY\Big )\nonumber \\&+\,S_{h(Y,Z)} X-S_{h(X,Z)} Y. \end{aligned}$$

(3.2)

Note that in view of the symmetry of the second fundamental form the above Gauss equation can be rewritten as

$$\begin{aligned} R(X,Y)Z= & {} \frac{5}{12}\Big (g(Y,Z)X - g(X,Z)Y\Big ) \nonumber \\&+\, \frac{1}{3}\Big (g(AY,Z)AX - g(AX,Z)AY + g(BY,Z)BX - g(BX,Z)BY\Big )\nonumber \\&+\, [S_{JX},S_{JY}] Z. \end{aligned}$$

(3.3)

By taking the normal part of the curvature tensor, we have that the Codazzi equation is given by

$$\begin{aligned}&(\nabla h)(X,Y,Z))-(\nabla h) (Y,X,Z)\nonumber \\&\quad =\frac{1}{3}\Big (g(AY,Z)JBX - g(AX,Z)JBY - g(BY,Z)JAX + g(BX,Z)JAY\Big ).\qquad \quad \end{aligned}$$

(3.4)

Analogously, like Lagrangian immersions of the nearly Kähler ${\mathbb {S}}^6$, we find that the Ricci equation is equivalent with the Gauss equation. Indeed from (2.20), (2.12) and the fact that G(X, Y) is a normal vector field, we get that

$$\begin{aligned} R^\perp (X,Y)JZ=J R(X,Y)Z+\tfrac{1}{3} \Big (g(X,Z)JY -g(Y,Z)JX\Big ). \end{aligned}$$

(3.5)

Therefore, by applying the Gauss equation (3.2), we recover that

$$\begin{aligned} R^\perp (X,Y)JZ= & {} \frac{1}{12}\Big (g(Y,Z)JX - g(X,Z)JY\Big ) \nonumber \\&+ \frac{1}{3}\Big (g(AY,Z)JAX - g(AX,Z)JAY \nonumber \\&+ g(BY,Z)JBX - g(BX,Z)JBY\Big )\nonumber \\&+J[S_{JX},S_{JY}] Z. \end{aligned}$$

(3.6)

Hence by taking the inner product with JW, we get the Ricci equation

$$\begin{aligned} g(R^\perp (X,Y)JZ,JW)&= g(\tilde{R}(X,Y)JZ,JW)+ g([S_{JX},S_{JY}] Z,W)\\&= g(\tilde{R}(X,Y)JZ,JW)+ g([S_{JZ},S_{JW}] X,Y). \end{aligned}$$

We now calculate the covariant derivatives of A and B.

Lemma 3.3

The covariant derivatives of the endomorphisms A and B are

$$\begin{aligned} (\nabla _X A)Y&= B S_{JX} Y - Jh(X,BY) +\frac{1}{2}\Big (JG(X,AY)-AJG(X,Y)\Big ),\\ (\nabla _X B)Y&= Jh(X,AY) -A S_{JX} Y +\frac{1}{2}\Big (JG(X,BY)-BJG(X,Y)\Big ). \end{aligned}$$

Proof

We express Eq. (2.7) in terms of A and B. By the Gauss and Weingarten formula and Lemma 2.6 we get on one hand

$$\begin{aligned} (\tilde{\nabla }_X P)Y&= \tilde{\nabla }_X AY + \tilde{\nabla }_X JBY - P\nabla _X Y - Ph(X,Y)\\&=\nabla _X AY + h(X,AY)+ J\tilde{\nabla }_X BY + G(X,BY)\\&\quad - A\nabla _X Y - JB\nabla _X Y - PJS_{JX}Y\\&= (\nabla _X A)Y + J(\nabla _X B)Y + Jh(X,BY) - BS_{JX}Y \\&\quad + h(X,AY) + JAS_{JX}Y + G(X,BY). \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \frac{1}{2}\Big (JG(X,P&Y)+JPG(X,Y)\Big ) \\&= \frac{1}{2}\Big ( JG(X,AY) +G(X,BY) - AJG(X,Y)-JBJG(X,Y)\Big ). \end{aligned}$$

Using Lemma 2.6, we can compare the tangent and normal parts in Eq. (2.7). This gives us the covariant derivatives of A and B.$\square$

It would be interesting to ask whether it is possible to prove an existence and uniqueness theorem like for submanifolds of real space forms or Lagrangian submanifolds of complex space forms. Although such a theorem would simplify some of the later proofs, it is outside the scope of the present paper.

For the Levi–Civita connection $\nabla$ on M, we introduce the functions $\omega _{ij}^k$ satisfying $\nabla _{E_i} E_j = \omega _{ij}^k E_k$ and $\omega _{ij}^k =- \omega _{ik}^j$, where we have used Einstein summation. We write $h_{ij}^k=g( h(E_i,E_j), JE_k)$. The tensor $h_{ij}^k$ is a totally symmetric tensor on the Lagrangian submanifold. The covariant derivative on the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ will be denoted by $\tilde{\nabla }$ as usual. In the following, we will use Eq. (2.7) to obtain extra information on the angles $\theta _i$ and second fundamental form $h_{ij}^k$.

Lemma 3.4

The derivatives of the angles $\theta _i$ give the components of the second fundamental form

$$\begin{aligned} E_i(\theta _j) = -h_{jj}^i \end{aligned}$$

except $h_{12}^3$. The second fundamental form and covariant derivative are related by

$$\begin{aligned} h_{ij}^k \cos (\theta _j-\theta _k) = \left( \frac{\sqrt{3}}{6}\epsilon _{ij}^k-\omega _{ij}^k \right) \sin (\theta _j-\theta _k),~\text {if}~j\ne k,\quad ~\text {where}~ \epsilon _{ij}^k:=\epsilon _{ijk}. \end{aligned}$$

Proof

We will not do all the calculations explicitly; instead, we give one calculation as an example. Choose $X=Y=E_1$ in (2.7). Then the equation $2(\tilde{\nabla }_{E_1}P)E_1 = JG(E_1,PE_1)$ gives

$$\begin{aligned} -2 \bigl ( h_{11}^1 + E_1(\theta _1)\bigr ) \sin (2 \theta _1) = 0, \\ 2 \bigl ( h_{11}^1 + E_1(\theta _1)\bigr ) \cos (2 \theta _1) = 0, \\ -2 \bigl ( h_{11}^2 \cos (\theta _1-\theta _2) + \omega _{11}^2 \sin (\theta _1-\theta _2) \bigr ) \sin (\theta _1+\theta _2) =0, \\ 2 \bigl ( h_{11}^2 \cos (\theta _1-\theta _2) + \omega _{11}^2 \sin (\theta _1-\theta _2) \bigr ) \cos (\theta _1+\theta _2) =0, \\ -2 \bigl ( h_{11}^3 \cos (\theta _1-\theta _3) + \omega _{11}^3\sin (\theta _1-\theta _3) \bigr ) \sin (\theta _1+\theta _3) = 0, \\ 2 \bigl ( h_{11}^3 \cos (\theta _1-\theta _3) + \omega _{11}^3\sin (\theta _1-\theta _3) \bigr ) \cos (\theta _1+\theta _3) = 0. \end{aligned}$$

Since the sines and cosines cannot be zero at the same time, we find that $E_1(\theta _1)=-h_{11}^1$ and two expressions relating $\omega _{ij}^k$ and $h_{ij}^k$. Doing the same calculations for $X=E_i$ and $Y=E_j$ with $i,j=1,2,3$, we get the lemma.$\square$

Note that from Lemma 3.4 we have that $E_i(\theta _j)=-h_{jj}^i$, and therefore, we also have the compatibility conditions

$$\begin{aligned} -E_k(h_{jj}^i) + E_i(h_{jj}^k)= & {} [E_k,E_i](\theta _j)\nonumber \\= & {} \sum _{\ell =1}^3 (\omega _{ki}^\ell -\omega _{ik}^\ell ) E_\ell (\theta _j)\nonumber \\= & {} \sum _{\ell =1}^3 (-\omega _{ki}^\ell +\omega _{ik}^\ell ) h_{jj}^\ell . \end{aligned}$$

(3.7)

So we have six additional independent equations. One can show, using Lemma 3.4, that the above equations are equivalent with six of the Codazzi equations. One does not obtain all the equations of Gauss and Codazzi this way, but the compatibility conditions for the $\theta _i$ are easier to calculate.

Remark 3.5

We note that from Lemma 3.4 and Lemma 3.2, we obtain that $h_{11}^i+h_{22}^i+h_{33}^i=-E_i(\theta _1+\theta _2+\theta _3)=0,~\text {for every} ~i=1,2,3$. Hence, we obtain a new proof of the fact that M is minimal (see Lemma 2.6).

Another consequence of Lemma 3.4 is

Corollary 3.6

Let M be a Lagrangian submanifold of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. If M is totally geodesic, then the angles $\theta _1$, $\theta _2$ and $\theta _3$ are constant. Conversely, if the angles are constant and $h_{12}^3=0$, then M is totally geodesic.

Remark 3.7

We must assume that $h_{12}^3 =0$ in the converse statement of Corollary 3.6. Examples 4.7–4.8 in the next section show that this assumption is necessary.

In the following, we give two other corollaries of Lemma 3.4. Lemma 3.8 gives a sufficient condition for a Lagrangian submanifold in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ to be totally geodesic. Lemma 3.9 gives a necessary condition and shows us that the converse statement of Lemma 3.8 also holds.

Lemma 3.8

If two of the angles are equal modulo $\pi$, then the Lagrangian submanifold is totally geodesic.

Proof

Without loss of generality, we may assume that $\theta _1=\theta _2 \mod \pi$. It follows from Lemma 3.4 that $h_{11}^i= h_{22}^i$ and $h_{12}^i=0$ for $i=1,2,3$. Combining the equations, together with the symmetry of h, gives $h_{11}^1= h_{22}^2=h_{11}^2=h_{22}^1=h_{12}^3=0$. Then by minimality we have $h_{33}^1=h_{33}^2=0$. The three remaining components are related by $h_{22}^3=h_{11}^3$ and $h_{33}^3=-2h_{11}^3$. The compatibility condition for $\theta _1$ with respect to $E_1$ and $E_2$ (take $k=j=1,~i=2$ in (3.7)) gives 

$$\begin{aligned} (\omega _{12}^3-\omega _{21}^3)h_{11}^3=0. \end{aligned}$$

(3.8)

Now we use the Codazzi equation (3.4) applied to $X=E_1$, $Y=E_2$, $Z=E_2$. As $\theta _1$ and $\theta _2$ are equal modulo $\pi$, the term on the right-hand side of (3.4) vanishes, and so we obtain by taking the component in the direction of $JE_2$ that

$$\begin{aligned} \Big (\frac{1}{\sqrt{3}}+\omega _{21}^3-3\omega _{12}^3\Big )h_{11}^3=0. \end{aligned}$$

(3.9)

We claim that $h_{11}^3=0$. If $h_{11}^3\ne 0$, and then from (3.8) and (3.9) we get $\omega _{21}^3=\omega _{12}^3=\frac{\sqrt{3}}{6}$. It follows from the second equation of Lemma 3.4 that $0=\Big (-\frac{\sqrt{3}}{6}-\omega _{21}^3\Big )\sin (\theta _1-\theta _3)$. Taking into account that $\omega _{21}^3=\frac{\sqrt{3}}{6}$, we obtain that $\sin (\theta _1-\theta _3)=0$, hence $\theta _1=\theta _3+a\pi$, where a is a constant integer. Then using the first equation of Lemma 3.4, we derive that $h_{33}^2=-E_3(\theta _3)=-E_3(\theta _1)=h_{11}^3$, but we have that $h_{33}^3=-2h_{11}^3$, so we get that $h_{11}^3=h_{33}^3=0$, which is a contradiction. Thus $h_{11}^3=0$ and the submanifold is totally geodesic.$\square$

Lemma 3.9

Consider a totally geodesic Lagrangian submanifold in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. After a possible permutation of the angles and the choice of the angles $2\theta _i$ at an initial point belonging to the interval $[0,2 \pi )$, we must have one of the following possibilities:

$$\begin{aligned} \begin{array}{lll} &{}(1)\ \ (2\theta _1,2\theta _2,2\theta _3) =\left( \tfrac{4\pi }{3},\tfrac{4\pi }{3},\tfrac{4\pi }{3}\right) ,\ \ \ &{}(2)\ \ (2\theta _1,2\theta _2,2\theta _3) =\left( \tfrac{2\pi }{3},\tfrac{2\pi }{3},\tfrac{2\pi }{3}\right) ,\\ &{}(3)\ \ (2\theta _1,2\theta _2,2\theta _3) = (0,0,0),\ &{}(4)\ \ (2\theta _1,2\theta _2,2\theta _3) = (0,\pi ,\pi ),\\ &{}(5)\ \ (2\theta _1,2\theta _2,2\theta _3) = \left( \tfrac{\pi }{3},\tfrac{\pi }{3},\tfrac{4\pi }{3}\right) ,\ &{}(6)\ \ (2\theta _1,2\theta _2,2\theta _3) = \left( \tfrac{2\pi }{3},\tfrac{5\pi }{3},\tfrac{5\pi }{3}\right) . \end{array} \end{aligned}$$

Proof

The Codazzi equation (3.4) gives 

$$\begin{aligned} g(AY,Z)BX - g(AX,Z)BY = g(BY,Z)AX -g(BX,Z)AY. \end{aligned}$$

Taking $X=E_i$ and $Y=Z=E_j$, this yields  $\sin \big (2(\theta _i-\theta _j)\big )=0$ for $i\ne j$. So the angles $2\theta _i$ are equal up to an integer multiple of $\pi$. Together with Lemma 3.2, we deduce that the angles need to be constant, and therefore after a choice at an initial point, one obtains the possibilities in the statement.$\square$

4 Examples of Lagrangian submanifolds in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$

In this section, we present eight examples (or families of examples) of Lagrangian submanifolds in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Example 4.8 (a flat Lagrangian torus) is a new example. Examples 4.1–4.3 are the factors and diagonal 3-spheres given by Schäfer and Smozcyk in [32]. Examples 4.4–4.7 were constructed by Moroianu and Semmelmann in [25], where they studied generalized Killing spinors on the standard sphere ${\mathbb {S}}^3$, which turn out to be related to Lagrangian embeddings in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. The first seven examples are immersions of round 3-spheres or Berger spheres. On these 3-spheres, which we regard as the set of all the unit quaternions in ${\mathbb {H}}$, we consider the left invariant tangent vector fields $X_1$, $X_2$, $X_3$ on ${\mathbb {S}}^3$ given by

$$\begin{aligned} X_1(u)=u\,\mathbf {i},~ X_2(u)=u\,\mathbf {j},~ X_3(u)=-u\,\mathbf {k}, \end{aligned}$$

(4.1)

where $u=x_1+x_2\mathbf {i}+x_3\mathbf {j}+x_4\mathbf {k}\in {\mathbb {S}}^3$ is viewed as a unit quaternion, and $\mathbf {i},\,\mathbf {j},\,\mathbf {k}$ are the imaginary units of ${\mathbb {H}}$. Obviously, $X_1,X_2,X_3$ form a basis of the tangent bundle $T{\mathbb {S}}^3$. We refer to [35] for more details of Examples 4.1–4.6.

Example 4.1

Consider the immersion: $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3:u\mapsto (u,1).$ f is a totally geodesic Lagrangian immersion, and $f({\mathbb {S}}^3)$ is isometric to a round sphere. The angles correspond to case (1) of Lemma 3.9.

Example 4.2

Consider the immersion:$f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3:u\mapsto (1,u).$ f is a totally geodesic Lagrangian immersion, and $f({\mathbb {S}}^3)$ is isometric to a round sphere. The angles correspond to case (2) of Lemma 3.9.

Example 4.3

Consider the immersion: $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3:u\mapsto (u,u).$ f is a totally geodesic Lagrangian immersion, and $f({\mathbb {S}}^3)$ is isometric to a round sphere. The angles correspond to case (3) of Lemma 3.9.

Example 4.4

Consider the immersion $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3:$ $u\mapsto (u,ub)$ with $b\in {{\mathrm{Im}}}\mathbb {H}$, $\Vert b\Vert =1$. First we note that after an isometry $(p,q)\mapsto (pa^{-1}, q a^{-1})$ and a reparametrization $u\mapsto u a$ of the 3-sphere with $a\in {{\mathrm{Im}}}\mathbb {H}$, the immersion becomes $u\mapsto (u,uaba^{-1})$. We now choose a such that $aba^{-1}=\mathbf {i}$. This is always possible, because conjugation with a unit quaternion gives a rotation of ${{\mathrm{Im}}}\mathbb {H}$ and the group of rotations acts transitively on ${{\mathrm{Im}}}\mathbb {H}$. Therefore we may always consider the immersion: $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3:u\mapsto (u,u\mathbf {i}).$ f is a totally geodesic Lagrangian immersion, $f({\mathbb {S}}^3)$ is isometric to a Berger sphere. The angles correspond to case (4) of Lemma 3.9.

Note that by changing the parametrization of ${\mathbb {S}}^3$, we can also reduce the potential immersion $f(u) = (u\mathbf {i},u)$ to the preceding example.

Example 4.5

Consider the immersion $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,u^{-1}bu)$ with $b\in {{\mathrm{Im}}}\mathbb {H}$, $\Vert b\Vert =1$. After an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ and a reparametrization of u as in the previous example, we can always consider the immersion:

$$\begin{aligned} f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3:u\mapsto (u^{-1},u\mathbf {i}u^{-1}). \end{aligned}$$

f is a totally geodesic Lagrangian immersion, $f({\mathbb {S}}^3)$ is isometric to a Berger sphere. The angles correspond to case (5) of Lemma 3.9.

Example 4.6

Consider the immersion: $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3:u\mapsto (u\mathbf {i}u^{-1},u^{-1}).$ f is a totally geodesic Lagrangian immersion, $f({\mathbb {S}}^3)$ is isometric to a Berger sphere. The angles correspond to case (6) of Lemma 3.9.

Example 4.7

Consider the immersion $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (uau^{-1},ubu^{-1})$ with unit quaternions $a,b\in {{\mathrm{Im}}}\mathbb {H}$ and $\langle a, b\rangle =0$. After an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ and a reparametrization, we can always consider the immersion

$$\begin{aligned} f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u\mathbf {i}u^{-1},u\mathbf {j}u^{-1}). \end{aligned}$$

For the tangent map, we have $df(X_1)=(0,2u\mathbf {k}u^{-1})$, $df(X_2)=(-2u\mathbf {k}u^{-1},0)$, $df(X_3)=2(-u\mathbf {j}u^{-1},u\mathbf {i}u^{-1})$. The inner products are given by $g\bigl (df(X_i),df(X_j)\bigr )=\tfrac{16}{3}\delta _{ij}$, so f is an immersion of a round sphere. We have that $Jdf(X_1)=\tfrac{2}{\sqrt{3}}\Big (2,u\mathbf {k}u^{-1}\Big )$, $Jdf(X_2)=\tfrac{2}{\sqrt{3}}\Big (u\mathbf {k}u^{-1},-2\Big )$ and $Jdf(X_3)=-\tfrac{2}{\sqrt{3}}\Big (u\mathbf {j}u^{-1},u\mathbf {i}u^{-1}\Big )$. One can now easily verify that f is a Lagrangian immersion. We also have

$$\begin{aligned}&Pdf(X_1)=(2,0)=-\tfrac{1}{2}\Big (df(X_1)-\sqrt{3}Jdf(X_1)\Big ),\\&Pdf(X_2)=(0,2)=-\tfrac{1}{2}\Big (df(X_2)+\sqrt{3}Jdf(X_2)\Big ),\\&Pdf(X_3)=-2(u\mathbf {j}u^{-1},-u\mathbf {i} u^{-1})=df(X_3). \end{aligned}$$

The angles $2\theta _i$ are thus equal to 0, $\tfrac{2\pi }{3}$ and $\tfrac{4\pi }{3}$. Therefore, by Lemma 3.9, this immersion is not totally geodesic. Since the angles are constant, $h_{12}^3$ is the only nonzero component of the second fundamental form. This example shows that we cannot omit the condition $h_{12}^3=0$ in Corollary 3.6.

Example 4.8

Consider the immersion $f: {\mathbb {R}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: (u,v,w)\mapsto (p(u,w),q(u,v))$ where p and q are constant mean curvature tori in ${\mathbb {S}}^3$ given by

$$\begin{aligned} p(u,w)&=\left( \cos \tilde{u} \cos \tilde{w},\cos \tilde{u} \sin \tilde{w}, \sin \tilde{u} \cos \tilde{w},\sin \tilde{u} \sin \tilde{w}\right) ,\\ q(u,v)&=\frac{1}{\sqrt{2}}\bigl (\cos \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) , \sin \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) \bigr . \\&\qquad \bigl . \cos \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) ,\sin \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) \bigr ), \end{aligned}$$

where in order to simplify expressions we have written $\tilde{u} = \tfrac{\sqrt{3}}{2} u$, $\tilde{v} = \tfrac{\sqrt{3}}{2} v$ and $\tilde{w} =\tfrac{\sqrt{3}}{2} w$. It follows that

$$\begin{aligned} f_u&=\left( \frac{\sqrt{3}}{2} \left( - \sin \tilde{u} \cos \tilde{w},-\sin \tilde{u} \sin \tilde{w},\cos \tilde{u} \cos \tilde{w},\cos \tilde{u} \sin \tilde{w}\right) ,\right. \\&\qquad \left. \frac{1}{2} \sqrt{\frac{3}{2}} \left( \cos \tilde{v} \left( \cos \tilde{u}-\sin \tilde{u}\right) , \sin \tilde{v} \left( \cos \tilde{u}-\sin \tilde{u}\right) , \cos \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) , \sin \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) \right) \right) ,\\ f_v&=\Biggl (\left( 0,0,0,0 \right) , \Bigr . \\&\quad \left. \frac{1}{2} \sqrt{\frac{3}{2}}\left( - \sin \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) , \cos \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) ,-\sin \tilde{v}\left( \sin \tilde{u}-\cos \tilde{u}\right) , \cos \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) \right) \right) ,\\ f_w&=\left( \frac{\sqrt{3}}{2} \left( - \cos \tilde{u} \sin \tilde{w},\cos \tilde{u} \cos \tilde{w}, -\sin \tilde{u} \sin \tilde{w},\sin \tilde{u} \cos \tilde{w}\right) ,\left( 0,0,0,0 \right) \right) , \end{aligned}$$

A straightforward computation gives that

$$\begin{aligned}&Jf_u=\left( \frac{1}{2} \left( -\sin \tilde{u} \cos \tilde{w},-\sin \tilde{u} \sin \tilde{w},\cos \tilde{u} \cos \tilde{w},\cos \tilde{u} \sin \tilde{w}\right) ,\right. \\&\qquad \left. \frac{1}{2 \sqrt{2}} \left( \cos \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) ,\sin \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) , -\cos \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) ,-\sin \tilde{v} \left( \sin \tilde{u} +\cos \tilde{u}\right) \right) \right) , \\&Jf_v=\left( \left( -\sin \tilde{u} \sin \tilde{w},\sin \tilde{u} \cos \tilde{w}, \cos \tilde{u} \sin \tilde{w},-\cos \tilde{u} \cos \tilde{w}\right) ,\right. \\&\qquad \left. \frac{1}{2 \sqrt{2}} \left( -\sin \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) , \cos \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) , \sin \tilde{v} \left( \cos \tilde{u}-\sin \tilde{u}\right) , \cos \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) \right) \right) ,\\&Jf_w=\left( \frac{1}{2} \left( \cos \tilde{u} \sin \tilde{w},-\cos \tilde{u} \cos \tilde{w},\sin \tilde{u} \sin \tilde{w},-\sin \tilde{u} \cos \tilde{w}\right) ,\right. \\&\qquad \left. \frac{1}{\sqrt{2}} \left( \sin \tilde{v} \left( \cos \tilde{u}-\sin \tilde{u}\right) ,\cos \tilde{v} \left( \sin \tilde{u}-\cos \tilde{u}\right) , \sin \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) ,-\cos \tilde{v} \left( \sin \tilde{u}+\cos \tilde{u}\right) \right) \right) . \end{aligned}$$

From this we get that f is a Lagrangian immersion and that $\{f_u,f_v,f_w\}$ is an orthonormal basis of the tangent space. Hence it is a flat Lagrangian torus. By a lengthy but straightforward computation, we also get that

$$\begin{aligned} Pf_u= f_u,~Pf_v = -\tfrac{1}{2} f_v+ \tfrac{\sqrt{3}}{2}Jf_v,~Pf_w= -\tfrac{1}{2} f_w-\tfrac{\sqrt{3}}{2} Jf_w. \end{aligned}$$

The angles $2\theta _i$ are therefore again equal to 0, $\tfrac{2\pi }{3}$ and $\tfrac{4\pi }{3}$. Therefore, by Lemma 3.9, this immersion is also not totally geodesic. Since the angles are constant, $h_{12}^3$ is again the only nonzero component of the second fundamental form. This example is another example that shows that we cannot omit the condition $h_{12}^3=0$ in Corollary 3.6.

5 Lagrangian submanifolds of constant sectional curvature

In this section, we classify all Lagrangian submanifolds of constant sectional curvature in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. We will prove that those Lagrangian submanifolds of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ are congruent to one of the examples of constant sectional curvature listed in the previous section. As a corollary, we obtain that the radius of a round Lagrangian sphere in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ can only be $\frac{2}{\sqrt{3}}$ or $\frac{4}{\sqrt{3}}$. This improves Proposition 4.4 of [25].

In order to prove the classification, the first step is to find all the components $h_{ij}^k$ of the second fundamental form. As we have already obtained the complete classification of the totally geodesic Lagrangian submanifolds in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ in [35] (see Theorem 1.3), we can assume now that the immersion is not totally geodesic. Then from Lemma 3.8 we may assume that all the angle functions are different (modulo $\pi$). Therefore, we have that there exists a local orthonormal frame $\{E_1,E_2,E_3\}$ on an open dense subset of M such that (3.1) holds.

We note that it is not possible to follow the approach introduced by Ejiri for studying Lagrangian submanifolds of constant sectional curvature in the complex space forms [15] or in the nearly Kähler 6-sphere [14]. Indeed the Gauss equations give quadratic equations for the $h_{ij}^k$ and it turns out that these are not easy to solve directly without additional information. We therefore use another approach. The next lemma gives us linear equations for the components $h_{ij}^k$. The key idea is to calculate the expression x given by

$$\begin{aligned} x= 3 \underset{{WXY}}{\mathfrak {S}} \left( (\nabla ^2 h)(W,X,Y,Z) - (\nabla ^2 h)(W,Y,X,Z)\right) , \end{aligned}$$

(5.1)

where $\mathfrak {S}$ stands for the cyclic sum, in two different ways. On the one hand, we can calculate this using the covariant derivative of the Codazzi equation (3.4), which tells us that x equals the expression (5.3). On the other hand we can rewrite x as

$$\begin{aligned} x= 3\underset{{WXY}}{\mathfrak {S}} \left( (\nabla ^2 h)(W,X,Y,Z) - (\nabla ^2 h)(X,W,Y,Z)\right) , \end{aligned}$$

(5.2)

and then by applying the Ricci identity, we obtain that this expression x vanishes.

More precisely, we have the following key lemma.

Lemma 5.1

Let M be a Lagrangian submanifold of constant sectional curvature c in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$ . Then for all tangent vector fields $W,X,Y,Z \in TM$ the expression

$$\begin{aligned} \underset{{WXY}}{\mathfrak {S}}\Bigl \{&\Bigl \{ g( JG(Y,W), AZ)+\frac{1}{2}g( JG(Y,Z), AW) -\frac{1}{2}g( JG(W,Z), AY) \Bigr . \nonumber \\&\quad +g( h(W,Z), JBY) -g( h(Y,Z), JBW) \Bigr \} JBX \nonumber \\ +&\Bigl \{ g( JG(W,Y), BZ)+\frac{1}{2}g( JG(W,Z), BY) -\frac{1}{2}g( JG(Y,Z), BW)\Bigr . \nonumber \\&\quad +g( h(W,Z), JAY) -g( h(Y,Z), JAW) \Bigr \} JAX \nonumber \\ +&g( AX, Z)\Bigl \{ JBJG(W,Y) + \frac{1}{2}G(Y,BW) - \frac{1}{2}G(W,BY) \Bigr .\nonumber \\&\qquad \qquad +\Bigl .h(W,AY) - h(Y,AW) \Bigr \} \nonumber \\ +&g( BX, Z)\Bigl \{ -JAJG(W,Y) + \frac{1}{2}G(W,AY) - \frac{1}{2}G(Y,AW) \Bigr .\nonumber \\&\qquad \qquad +\Bigl .h(W,BY) - h(Y,BW) \Bigr \}\Bigl \} \end{aligned}$$

(5.3)

is zero.

Proof

As we have mentioned before, we calculate expression (5.1) in two different ways.

First, we calculate x using the covariant derivative of the Codazzi equation (3.4), which gives us the long expression (5.3). We denote the normal part $(\tilde{R}(X,Y)Z)^\perp$ as $T_1(X,Y,Z)$. This is the right-hand side of the Codazzi equation (3.4). So we have that

$$\begin{aligned}&3((\nabla h)(X,Y,Z))-(\nabla h) (Y,X,Z))=3 T_1(X,Y,Z) \nonumber \\&\quad = g( AY, Z)JBX - g( AX, Z)JBY +g( BX, Z)JAY - g( BY, Z)JAX. \end{aligned}$$

(5.4)

Using Lemma 2.6 and (2.20), the covariant derivative $\nabla T_1$, where $\nabla$ is the covariant derivative on M, can be written as

$$\begin{aligned} 3 (\nabla T_1)(W,X,Y,Z) = T_2(W,X,Y,Z) - T_2(W,Y,X,Z), \end{aligned}$$

(5.5)

where

$$\begin{aligned} T_2(W,X,Y,Z)&=g( (\nabla _W A)Y, Z)JBX +g( (\nabla _W B)X, Z)JAY\nonumber \\&\qquad +g( AY, Z) G(W,BX) +g(BX,Z) G(W,AY)\nonumber \\&\qquad +g(AY,Z) J(\nabla _W B)X +g(BX,Z) J(\nabla _WA)Y. \end{aligned}$$

(5.6)

By Lemma 2.6 and Lemma 3.3, the tensor $T_2$ can be expressed completely in terms of A and B in the following way:

$$\begin{aligned} T_2(W,X,Y,Z)&=g(B S_{JW} Y,Z)JBX +g(h(W,BY),JZ)JBX\nonumber \\&\quad \quad -\tfrac{1}{2} g(G(W,AY),JZ) JBX +\tfrac{1}{2} g(G(W,Y), JAZ) JBX\nonumber \\&\quad \quad -g(AY,Z) h(W,AX)-g(AY,Z) J A S_{JW}X\nonumber \\&\quad \quad +\tfrac{1}{2} g(AY,Z)G(W,BX)- \tfrac{1}{2} g(AY,Z) JBJ G(W,X)\nonumber \\&\quad \quad -g(h(W,AX),JZ) JAY -g(S_{JW} X,AZ) JAY\nonumber \\&\quad \quad -\tfrac{1}{2} g(G(W,BX),JZ) JAY + \tfrac{1}{2} g(G(W,X),JBZ) JAY\nonumber \\&\quad \quad +g(BX,Z) JB S_{JW}Y +g(BX,Z) h(W,BY)\nonumber \\&\quad \quad +\tfrac{1}{2} g(BX,Z) G(W,AY) -\tfrac{1}{2} g(BX,Z) JAJ G(W,Y). \end{aligned}$$

(5.7)

Now we can compute x. From (5.1), (5.4) and (5.5), we have that

$$\begin{aligned}\begin{aligned} x&= T_2(W,X,Y,Z) + T_2(X,Y,W,Z)+ T_2(Y,W,X,Z)\\&\qquad -T_2(W,Y,X,Z)- T_2(X,W,Y,Z)- T_2(Y,X,W,Z). \end{aligned} \end{aligned}$$

Therefore, when we compute x we can omit in (5.7) all terms which are symmetric in two of the variables W, X, Y. So we get x by omitting these terms in (5.7) and by taking the cyclic sum of the difference of remainder of (5.7) with itself with two variables interchanged. Hence we can write

$$\begin{aligned}\begin{aligned} x&= T_3(W,X,Y,Z) + T_3(X,Y,W,Z)+ T_3(Y,W,X,Z)\\&\qquad -T_3(W,Y,X,Z)- T_3(X,W,Y,Z)- T_3(Y,X,W,Z), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} T_3(W,X,Y,Z)&= g(h(W,BY),JZ)JBX -\tfrac{1}{2} g(G(W,AY),JZ) JBX \\&\quad \,+\tfrac{1}{2} g(G(W,Y), JAZ) JBX-g(AY,Z) h(W,AX)\\&\quad \,+\tfrac{1}{2} g(AY,Z)G(W,BX)- \tfrac{1}{2} g(AY,Z) JBJ G(W,X)\\&\quad \,-g(h(W,AX),JZ) JAY -\tfrac{1}{2} g(G(W,BX),JZ) JAY \\&\quad \,+ \tfrac{1}{2} g(G(W,X),JBZ) JAY+g(BX,Z) h(W,BY)\\&\quad \,+\tfrac{1}{2} g(BX,Z) G(W,AY) -\tfrac{1}{2} g(BX,Z) JAJ G(W,Y). \end{aligned} \end{aligned}$$

From this we immediately get that x equals the expression (5.3).

On the other hand we can rewrite x as

$$\begin{aligned} x= 3\underset{{WXY}}{\mathfrak {S}} \left( (\nabla ^2 h)(W,X,Y,Z) - (\nabla ^2 h)(X,W,Y,Z)\right) . \end{aligned}$$

By the Ricci identity, we have that

$$\begin{aligned} x= 3 \underset{{WXY}}{\mathfrak {S}} \left( R^\perp (W,X)h(Y,Z) - h\bigl (R(W,X)Y,Z\bigr ) - h\bigl (Y,R(W,X)Z\bigr )\right) . \end{aligned}$$

Equations (2.21) and (3.5) give

$$\begin{aligned} R^\perp (W,X)&h(Y,Z) \\&= R^\perp (W,X) JS_{JY}Z \\&= JR(W,X)S_{JY}Z + \frac{1}{3}\Big (g( h(W,Y), JZ)JX - g( h(X,Y), JZ)JW\Big ). \end{aligned}$$

Since M has constant curvature c, the curvature tensor satisfies $R(X,Y)Z = c(g(Y,Z)X -g(X,Z)Y)$. An easy calculation shows that x vanishes. This completes the proof of the lemma.$\square$

We are now in a position to prove the classification result. We consider again the endomorphisms A and B that satisfy $P|_{TM}=A+JB$ and take the orthonormal basis $E_1,E_2,E_3$ such that $AE_i = \lambda _i E_i$ and $BE_i = \mu _i E_i$ for $i=1,2,3$. In the notation of the previous sections $\lambda _i= \cos 2\theta _i$ and $\mu _i=\sin 2\theta _i$. We choose $\theta _i$ such that $\theta _i\in [0,2\pi ),~i=1,2,3$. Sometimes the expressions in terms of $\lambda _i$ and $\mu _i$ are shorter, so we will not always express equations in terms of the angles $\theta _i$. Taking into account of the properties of G, we may also assume that $J G(E_1,E_2)=\frac{1}{\sqrt{3}} E_3$ by replacing $E_3$ by $-E_3$ if necessary. Thus we obtain that $J G(E_i,E_j)=\tfrac{1}{\sqrt{3}} \varepsilon _{ijk} E_k$. By taking $X=E_1$, $Y=E_2$ and $Z=W=E_3$ in formula (5.3) in Lemma 5.1, we obtain six equations, namely

$$\begin{aligned}&\bigl (\lambda _i(\lambda _j-\lambda _k)+\mu _i(\mu _j-\mu _k)\bigr ) h_{kk}^j + \bigl (\lambda _k(\lambda _i-\lambda _j)+\mu _k(\mu _i-\mu _j)\bigr ) h_{ii}^j =0, \end{aligned}$$

(5.8)

$$\begin{aligned}&\bigl (\lambda _i(\lambda _j-\lambda _k)+\mu _i(\mu _j-\mu _k)\bigr ) h_{12}^3 =0, \end{aligned}$$

(5.9)

for every positive permutation (ijk) of (123).

We now distinguish two cases: Case 1: $h_{12}^3\ne 0$ and Case 2: $h_{12}^3=0$.

Case 1: $\mathbf {h_{12}^3\ne 0}$. First we note that

$$\begin{aligned} \lambda _i(\lambda _j-\lambda _k)+\mu _i(\mu _j-\mu _k)= & {} \cos (2(\theta _i-\theta _j)) -\cos (2(\theta _i-\theta _k))\nonumber \\= & {} 2 \sin (\theta _j-\theta _k) \sin (2 \theta _i-\theta _j-\theta _k). \end{aligned}$$

(5.10)

So from Eq. (5.9), we find that $\sin (2 \theta _1 -\theta _2-\theta _3) \sin (\theta _3 -\theta _2)$, $\sin (2 \theta _2 -\theta _3-\theta _1) \sin (\theta _1 -\theta _3)$ and $\sin (2 \theta _3 -\theta _1-\theta _2) \sin (\theta _2 -\theta _1)$ have to vanish. As the immersion is not totally geodesic, from Lemma 3.8 we have that the angle functions are mutually different which in turn implies that for i different from j, we have that $\sin (\theta _j-\theta _i)$ is different from 0. Hence

$$\begin{aligned} \sin (2 \theta _1 -\theta _2-\theta _3)=\sin (2 \theta _2 -\theta _3-\theta _1)=\sin (2 \theta _3 -\theta _1-\theta _2)=0. \end{aligned}$$

So $(2 \theta _1 -\theta _2-\theta _3)$ is a multiple of $\pi$. By Lemma 3.2, $( \theta _1 +\theta _2+\theta _3)$ is also a multiple of $\pi$. Hence $\theta _1$ is a multiple of $\tfrac{\pi }{3}$. A same argument can be applied for the other angles $\theta _2$ and $\theta _3$. As the immersion is not totally geodesic, from Lemma 3.8 we have that no two angle functions are the same and therefore the angles must be different modulo $\pi$. So the only possibility for the angles are 0, $\tfrac{\pi }{3}$ and $\tfrac{2\pi }{3}$.

Since all the angles $\theta _i$ are constant, we have that all the $h_{jj}^i$ are zero except $h_{12}^3$ by using Lemma 3.4. From Lemma 3.4 it also follows that all the connection coefficients $\omega _{ij}^k$ are zero except the cases that i, j, k are pairwise different. These nonzero coefficients $\omega _{ij}^k$ can be written in terms of $h_{12}^3$ in the following way

$$\begin{aligned} \omega _{12}^3= \omega _{23}^1 = \omega _{31}^2=-\omega _{21}^3= -\omega _{32}^1 = -\omega _{13}^2= \tfrac{\sqrt{3}}{3} h_{12}^3 + \tfrac{\sqrt{3}}{6}. \end{aligned}$$

(5.11)

From the Gauss equation (3.3), it follows that the constant curvature c is related to the second fundamental form by

$$\begin{aligned} c E_2&= R(E_2,E_1)E_1\nonumber \\&=(\tfrac{5}{12}-\tfrac{1}{6}) E_2+[S_{JE_2},S_{JE_1}] E_1\nonumber \\&= \tfrac{1}{4} E_2-S_{JE_1} (h_{12}^3 E_3)\nonumber \\&=\tfrac{1}{4} E_2 -(h_{12}^3)^2 E_2. \end{aligned}$$

(5.12)

This implies that $h_{12}^3$ and therefore also $\omega _{12}^3$, $\omega _{23}^1$, $\omega _{31}^2$, $\omega _{21}^3$, $\omega _{32}^1$ and $\omega _{13}^2$are all constants. By computing the curvature from the definition, we have that

$$\begin{aligned} c E_2&=R(E_2,E_1)E_1 \nonumber \\&=\nabla _{E_2} \nabla _{E_1} E_1 - \nabla _{E_1} \nabla _{E_2} E_1 -\nabla _{[E_2,E_1]}E_1\nonumber \\&=-\omega _{21}^3 \omega _{13}^2 E_2 -(\omega _{21}^3-\omega _{12}^3) \omega _{31}^2 E_2\nonumber \\&=(\omega _{12}^3)^2 E_2\nonumber \\&=(\tfrac{\sqrt{3}}{3} h_{12}^3 + \tfrac{\sqrt{3}}{6})^2 E_2. \end{aligned}$$

(5.13)

Comparing both expressions (5.12) and (5.13), we get that $8 (h_{12}^3)^2+2h_{12}^3=1$, which implies that $h_{12}^3=\tfrac{1}{4}$ or $-\tfrac{1}{2}$. In the following, we will discuss two subcases of case 1 respectively: Case 1a: $h_{12}^3=\tfrac{1}{4}$ and Case 1b: $h_{12}^3=-\tfrac{1}{2}$.

Case 1a: $\mathbf {h_{12}^3=\tfrac{1}{4}}$. In this case, we have that $\omega _{12}^3=\omega _{23}^1=\omega _{31}^2=\frac{\sqrt{3}}{4}$ and the sectional curvature is equal to $\tfrac{3}{16}$.

In the next theorem we will prove that in this case (Case 1a) the submanifold M is locally congruent to the immersion in Example 4.7. In order to prove this, we first recall that the Berger sphere can be constructed by regarding ${\mathbb {S}}^3$ as a hypersurface of the quaternions. As before we take the frame $X_1(u)=u\mathbf {i}$, $X_2(u)=u\mathbf {j}$, $X_3(u)=-u\mathbf {k}$, which are left invariant vector fields. It follows by a straightforward calculation that

$$\begin{aligned}{}[X_1,X_2]&=-2 X_3,&[X_2,X_3]&=-2 X_1,&[X_3,X_1]&= -2 X_2. \end{aligned}$$

We now define a new metric $g_b$, depending on two constants $\tau$ and $\kappa$ on ${\mathbb {S}}^3$ by

$$\begin{aligned} g_b(X,Y)=\frac{4}{\kappa } \left( \langle X,Y\rangle +\left( \frac{4 \tau ^2}{\kappa }-1\right) \langle X,X_1\rangle \langle Y,X_1\rangle \right) . \end{aligned}$$

This implies that the vector fields $E_1=\tfrac{\kappa }{4 \tau } X_1$, $E_2 = \tfrac{\sqrt{\kappa }}{2} X_2$ and $E_3 = \tfrac{\sqrt{\kappa }}{2} X_3$ form an orthonormal basis of the tangent space with respect to $g_b$. It follows immediately from the Koszul formula that $\nabla _{E_i}E_i = 0$ and that

$$\begin{aligned} \begin{aligned}&\nabla _{E_2} E_3 = -\tau E_1, \qquad&\nabla _{E_2} E_1 = \tau E_3,\\&\nabla _{E_3} E_2 = \tau E_1, \qquad&\nabla _{E_3} E_1 = -\tau E_2,\\&\nabla _{E_1} E_2 =\Big (\tau -\frac{\kappa }{2 \tau }\Big ) E_3, \qquad&\nabla _{E_1} E_3 = \Big (-\tau +\frac{\kappa }{2 \tau }\Big )E_2. \end{aligned} \end{aligned}$$

(5.14)

Note that the following theorem of [11], which can be proved similarly to the local version of the Cartan–Ambrose–Hicks theorem (cf. the proof of Theorem 1.7.18 of [34]), then shows that a manifold admitting such vector fields is locally isometric to a Berger sphere.

Proposition 5.2

Let $M^n$ and $\tilde{M}^{n}$ be Riemannian manifolds with Levi–Civita connections $\nabla$ and $\tilde{\nabla }$. Suppose that there exists constant $c_{ij}^k$, $i,j,k \in \{1,\dots , n\}$ such that for all $p \in M$ and $\tilde{p} \in \tilde{M}$ there exist orthonormal frame fields $\{E_1,\dots , E_n\}$ around p and $\{\tilde{E}_1,\dots , \tilde{E}_n\}$ around $\tilde{p}$ such that $\nabla _{E_i} E_j = \sum _{k=1}^n c_{ij}^k E_k$ and $\nabla _{\tilde{E}_i} \tilde{E}_j = \sum _{k=1}^n c_{ij}^k \tilde{E}_k$. Then for every point $p \in M$ and $\tilde{p} \in \tilde{M}$ there exists a local isometry which maps a neighborhood of p to a neighborhood of $\tilde{p}$ and $E_i$ to $\tilde{E}_i$.

The previous proposition can of course be also applied in case that $\kappa =4 \tau ^2$. In that case we simply have a regular sphere of constant sectional curvature.

Theorem 5.3

Let M be a Lagrangian submanifold of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Assume that there exists a local orthonormal frame as in Case 1a. Then M is locally congruent to the immersion $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u\mathbf {i}u^{-1},u\mathbf {j}u^{-1})$, which is Example 4.7.

Proof

We have that $\omega _{ij}^k =\tfrac{\sqrt{3}}{4} \varepsilon _{ij}^k$ and the only non-vanishing component of the second fundamental form is $h_{12}^3= \tfrac{1}{4}$. This implies immediately that M is isometric to a space of constant sectional curvature $\tfrac{3}{16}$. Moreover, from the beginning of the discussion about Case 1, we know that the angle functions are given by $(2 \theta _1,2 \theta _2,2 \theta _3)=(0,\tfrac{2\pi }{3}, \tfrac{4\pi }{3})$. So we can find a local basis such that $\sqrt{3} JG(E_1,E_2)=E_3$ and

$$\begin{aligned} PE_1&= E_1,&PE_2&=-\tfrac{1}{2} E_2 +\tfrac{\sqrt{3}}{2} JE_2,&PE_3&= -\tfrac{1}{2} E_3 -\tfrac{\sqrt{3}}{2} JE_3. \end{aligned}$$

From this and (2.13) it follows that

$$\begin{aligned} QE_1&=-\sqrt{3} JE_1,&QE_2&= E_2,&QE_3&= -E_3. \end{aligned}$$

Applying Proposition 5.2 and comparing with (5.14) (take $\kappa =\frac{3}{4},~\tau =-\frac{\sqrt{3}}{4}$), we have that we can identify M with ${\mathbb {S}}^3$, with a proportional metric and that we may assume that

$$\begin{aligned} E_1&= -\tfrac{\sqrt{3}}{4} X_3,&E_2&= -\tfrac{\sqrt{3}}{4} X_1,&E_3&= -\tfrac{\sqrt{3}}{4} X_2. \end{aligned}$$

We write the immersion $f= (p,q)$. Let D denote the usual covariant derivative on $\mathbb H^2 = {\mathbb {R}}^8$. Assume that $df(E_i)= D_{E_i} f = (p \alpha _i, q \beta _i)$ where $\alpha _i,\beta _i$ are imaginary quaternions. In view of the above properties of Q, it immediately follows that $\beta _1= \alpha _1$, $\alpha _2 = 0$ and $\beta _3=0$.

Recall that

$$\begin{aligned} D_X Y = \nabla ^E_{X} Y -\tfrac{1}{2} \langle X,Y\rangle f -\tfrac{1}{2} \langle X,QY\rangle Qf, \end{aligned}$$

where $\nabla ^E$ denotes the usual Euclidean covariant derivative on $\mathbb {S}^3 \times \mathbb {S}^3$. Using the expression for P and the fact that $JG(E_1,E_2)=\tfrac{\sqrt{3}}{3} E_3$ we have that $\nabla ^E_{E_1}E_1=\nabla ^E_{E_2}E_2=\nabla ^E_{E_3}E_3=0$ and

$$\begin{aligned}&\nabla ^E_{E_1}E_2 =0,~\nabla ^E_{E_1} E_3=0,\\&\nabla ^E_{E_2}E_1 = - \tfrac{\sqrt{3}}{2} E_3=-\tfrac{\sqrt{3}}{2} (p \alpha _3,0),\\&\nabla ^E_{E_2} E_3=\tfrac{\sqrt{3}}{4}(E_1-QE_1)= \tfrac{\sqrt{3}}{2} (p\alpha _1,0),\\&\nabla ^E_{E_3}E_2 = -\tfrac{\sqrt{3}}{4}(E_1+QE_1)= -\tfrac{\sqrt{3}}{2} (0,q \alpha _1),\\&\nabla ^E_{E_3}E_1 = \tfrac{\sqrt{3}}{2} E_2=\tfrac{\sqrt{3}}{2} (0,q \beta _2). \end{aligned}$$

From the relation between the nearly Kähler metric and the usual Euclidean product metric (see (2.15)), we have that $E_1,E_2,E_3$ are orthogonal with respect to the induced Euclidean product metric and that their lengths are given by

$$\begin{aligned} \langle E_1,E_1\rangle =\tfrac{3}{2}, \qquad \langle E_2,E_2\rangle =\langle E_3,E_3\rangle =\tfrac{3}{4}. \end{aligned}$$

This in turn implies that $\alpha _1,\beta _2,\alpha _3$ are mutually orthogonal imaginary quaternions and

$$\begin{aligned} \vert \alpha _1 \vert ^2 =\tfrac{3}{4}, \qquad \vert \beta _2 \vert ^2= \vert \alpha _3 \vert ^2=\tfrac{3}{4}. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} D_{E_j} D_{E_i} f = (p \alpha _j \alpha _i + p E_j(\alpha _i), q \beta _j \beta _i + q E_j(\beta _i)). \end{aligned}$$

Note that for imaginary quaternions x and y, it holds that $xy=-\langle x, y\rangle +x\times y$, where $\langle \cdot ,\cdot \rangle$ is the usual Euclidean product on $\mathbb {R}^3$ and $\times$ is the vector product on $\mathbb {R}^3$. It then follows that

$$\begin{aligned} \nabla ^E_{E_j}E_i = (p (\alpha _j \times \alpha _i+E_j(\alpha _i)), q (\beta _j \times \beta _i +E_j(\beta _i)). \end{aligned}$$

Hence substituting $\alpha _2=0$, $\beta _1=\alpha _1$ and $\beta _3 =0$, it follows that

$$\begin{aligned} \beta _2 \times \alpha _1 = \tfrac{\sqrt{3}}{2} \alpha _3, \end{aligned}$$

as well as

$$\begin{aligned}&E_2(\alpha _1)=-\tfrac{\sqrt{3}}{2} \alpha _3, \qquad&E_3(\alpha _1)=\tfrac{\sqrt{3}}{2} \beta _2, \qquad&E_1(\alpha _1) = 0,\\&E_2(\beta _2)=0, \qquad&E_3(\beta _2)=-\tfrac{\sqrt{3}}{2} \alpha _1, \qquad&E_1(\beta _2) = \tfrac{\sqrt{3}}{2} \alpha _3,\\&E_2(\alpha _3)=\tfrac{\sqrt{3}}{2} \alpha _1, \qquad&E_3(\alpha _3)=0, \qquad&E_1(\alpha _3) =-\tfrac{\sqrt{3}}{2} \beta _2. \end{aligned}$$

In terms of the standard vector fields $X_1,X_2,X_3$, this gives

$$\begin{aligned}&X_1(\alpha _1)=2 \alpha _3, \qquad&X_2(\alpha _1)=- 2\beta _2, \qquad&X_3(\alpha _1) = 0,\\&X_1(\beta _2)=0, \qquad&X_2(\beta _2)=2 \alpha _1, \qquad&X_3(\beta _2) = -2 \alpha _3,\\&X_1(\alpha _3)=-2 \alpha _1, \qquad&X_2(\alpha _3)=0, \qquad&X_3(\alpha _3) = 2 \beta _2. \end{aligned}$$

We can choose a rotation (unitary quaternion h) such that

$$\begin{aligned} \beta _2(1) =\tfrac{\sqrt{3}}{2} h\mathbf {i}h^{-1},~ \alpha _3(1) = \tfrac{\sqrt{3}}{2} h\mathbf {j}h^{-1},~ \alpha _1(1) =-\tfrac{\sqrt{3}}{2} h\mathbf {k}h^{-1}, \end{aligned}$$

and we can pick the initial conditions such that $f(1)= (h\mathbf {i}h^{-1},h\mathbf {j}h^{-1})$. As the differential equations for $\alpha _i$, $\beta _i$, p and q are linear systems of differential equations with fixed initial conditions, we can apply a standard uniqueness theorem. It is therefore sufficient to give a solution which satisfies the above system with the given initial conditions.

We have that

$$\begin{aligned} \beta _2(u)&=\tfrac{\sqrt{3}}{2} hu\mathbf {i}u^{-1}h^{-1},&\alpha _3(u)&= \tfrac{\sqrt{3}}{2} hu\mathbf {j}u^{-1}h^{-1},&\alpha _1(u)&=-\tfrac{\sqrt{3}}{2} hu\mathbf {k}u^{-1}h^{-1} \end{aligned}$$

satisfy $X_1(\beta _2)=X_2(\alpha _3)=X_3(\alpha _1)=0$ and

$$\begin{aligned}&X_1(\alpha _1)=-2\tfrac{\sqrt{3}}{2} hu\mathbf {i}\mathbf {k}u^{-1}h^{-1}=2\tfrac{\sqrt{3}}{2} hu\mathbf {j}u^{-1}h^{-1}=2 \alpha _3,\\&X_2(\alpha _1)=-2\tfrac{\sqrt{3}}{2} hu\mathbf {j}\mathbf {k}u^{-1}h^{-1}=-2\tfrac{\sqrt{3}}{2} hu\mathbf {i}u^{-1}h^{-1}=-2 \beta _2,\\&X_3(\alpha _3)=2\tfrac{\sqrt{3}}{2} hu(-\mathbf {k})\mathbf {j}u^{-1}h^{-1}=2\tfrac{\sqrt{3}}{2} hu\mathbf {i}u^{-1}h^{-1}=2 \beta _2,\\&X_3(\beta _2)=2\tfrac{\sqrt{3}}{2} hu(-\mathbf {k})\mathbf {i}u^{-1}h^{-1}=-2\tfrac{\sqrt{3}}{2} hu\mathbf {j}u^{-1}h^{-1}=-2 \alpha _3,\\&X_1(\alpha _3)=2\tfrac{\sqrt{3}}{2} hu\mathbf {i}\mathbf {j}u^{-1}h^{-1}=2\tfrac{\sqrt{3}}{2} hu\mathbf {k}u^{-1}h^{-1}=-2 \alpha _1,\\&X_2(\beta _2)=2\tfrac{\sqrt{3}}{2} hu\mathbf {j}\mathbf {i}u^{-1}h^{-1}=-2\tfrac{\sqrt{3}}{2} hu\mathbf {k}u^{-1}h^{-1}=2 \alpha _1. \end{aligned}$$

Then, if we take $p=h u \mathbf {i} u^{-1} h^{-1}$ and $q = hu\mathbf {j} u^{-1} h^{-1}$, we have that

$$\begin{aligned}&D_{E_1} p= -\tfrac{\sqrt{3}}{4} D_{X_3} p=\tfrac{\sqrt{3}}{2}h u \mathbf {j} u^{-1} h^{-1} =h u \mathbf {i} u^{-1} h^{-1}(-\tfrac{\sqrt{3}}{2} hu\mathbf {k}u^{-1}h^{-1})=p \alpha _1,\\&D_{E_2} p= -\tfrac{\sqrt{3}}{4} D_{X_1} p =0 =p \alpha _2,\\&D_{E_3} p= -\tfrac{\sqrt{3}}{4} D_{X_2} p = \tfrac{\sqrt{3}}{2}h u \mathbf {k} u^{-1} h^{-1} =h u \mathbf {i} u^{-1} h^{-1}(\tfrac{\sqrt{3}}{2} hu\mathbf {j}u^{-1}h^{-1})=p \alpha _3,\\&D_{E_1} q= -\tfrac{\sqrt{3}}{4} D_{X_3} q= \tfrac{\sqrt{3}}{2}h u (-\mathbf {i}) u^{-1} h^{-1} =h u \mathbf {j} u^{-1} h^{-1}(-\tfrac{\sqrt{3}}{2} hu\mathbf {k}u^{-1}h^{-1})=q \beta _1, \\&D_{E_2} q= -\tfrac{\sqrt{3}}{4} D_{X_1} q =-\tfrac{\sqrt{3}}{2}h u \mathbf {k} u^{-1} h^{-1} =h u \mathbf {j} u^{-1} h^{-1}(\tfrac{\sqrt{3}}{2} hu\mathbf {i}u^{-1}h^{-1})=q \beta _2,\\&D_{E_3} q= -\tfrac{\sqrt{3}}{4} D_{X_2} q = 0=q \beta _3. \end{aligned}$$

After applying an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$, we complete the proof of the theorem.$\square$

Case 1b: $\mathbf {h_{12}^3=-\tfrac{1}{2}}$. In this case, all connection coefficients are zero and the submanifold M is flat. In that case we have

Theorem 5.4

Let M be a Lagrangian submanifold of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. Assume that there exists a local orthonormal frame as in Case 1b. Then M is locally congruent to the immersion $f: {\mathbb {R}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: (u,v,w)\mapsto (p(u,w),q(u,v))$, where p and q are constant mean curvature tori in ${\mathbb {S}}^3$ given in Example 4.8.

Proof

We know that all connection coefficients vanish and that the only non-vanishing component of the second fundamental form is $h_{12}^3= -\tfrac{1}{2}$. Moreover, from the beginning of the discussion about Case 1, we know that the angle functions are given by $(2 \theta _1,2 \theta _2,2 \theta _3)=(0,\tfrac{2\pi }{3}, \tfrac{4\pi }{3})$. So we can find a local basis such that $\sqrt{3} JG(E_1,E_2)=E_3$ and

$$\begin{aligned} PE_1&=E_1,&PE_2&=-\tfrac{1}{2} E_2 +\tfrac{\sqrt{3}}{2} JE_2,&PE_3&=-\tfrac{1}{2} E_3 -\tfrac{\sqrt{3}}{2} JE_3. \end{aligned}$$

From this it follows that

$$\begin{aligned} QE_1&=-\sqrt{3} JE_1,&QE_2&= E_2,&QE_3&= -E_3. \end{aligned}$$

As the connection coefficients vanish, we may identify $E_1,E_2,E_3$ with coordinate vector fields. As before we write the immersion $f= (p,q)$ and denote the coordinates by u, v, w. Therefore, we have $E_1=f_u,E_2=f_v,E_3=f_w$. It immediately follows from the above expression of Q that p does not depend on v, q does not depend on w (i.e., $p_v=q_w=0$) and that $p^{-1} p_u= q^{-1} q_u$.

Moreover, using the above expression for P and the fact that $JG(f_u,f_v)=\tfrac{\sqrt{3}}{3} f_w$, we have that $\nabla ^E_{f_u}f_u=\nabla ^E_{f_v}f_v=\nabla ^E_{f_w}f_w=0$ and

$$\begin{aligned}&\nabla ^E_{f_u}f_v=\nabla ^E_{f_v}f_u =-\tfrac{1}{2} Jf_w -\tfrac{\sqrt{3}}{4}f_w -\tfrac{1}{4} Jf_w=-\tfrac{3}{4} Jf_w -\tfrac{\sqrt{3}}{4} f_w,\\&\nabla ^E_{f_u} f_w=\nabla ^E_{f_w}f_u =-\tfrac{1}{2} Jf_v +\tfrac{\sqrt{3}}{4}f_v -\tfrac{1}{4} Jf_v=-\tfrac{3}{4} Jf_v +\tfrac{\sqrt{3}}{4} f_v,\\&\nabla ^E_{f_v}f_w =\nabla ^E_{f_w}{f_v} =-\tfrac{1}{2} Jf_u +\tfrac{1}{2} Jf_u=0. \end{aligned}$$

From the relation between the nearly Kähler metric and the usual Euclidean product metric (see (2.15)), we have that $f_u,f_v,f_w$ are also orthogonal with respect to the induced Euclidean product metric and that their lengths are given by

$$\begin{aligned} \langle f_u,f_u\rangle =\tfrac{3}{2}, \qquad \langle f_v,f_v\rangle =\langle f_w,f_w\rangle =\tfrac{3}{4}. \end{aligned}$$

Moreover, from (2.13) and (2.15), we have $\langle X,QY\rangle =-\tfrac{\sqrt{3}}{2} g(X,JPY\rangle$, hence

$$\begin{aligned} \langle f_u,Qf_v\rangle =\langle f_v,Qf_w\rangle =\langle f_u,Qf_w\rangle =0,\\ \langle f_u,Qf_u\rangle =0, \quad \langle f_v,Qf_v\rangle =\tfrac{3}{4}, \quad \langle f_w,Qf_w\rangle =-\tfrac{3}{4}. \end{aligned}$$

As $D_X Y = \nabla ^E_{X} Y -\tfrac{1}{2} \langle X,Y\rangle f -\tfrac{1}{2} \langle X,QY\rangle Qf,$ we deduce by combining the above equations that the immersion f is determined by the following system of partial differential equations:

$$\begin{aligned}&f_{uu}=-\tfrac{3}{4} f,&f_{vv}&= -\tfrac{3}{8} f - \tfrac{3}{8} Qf,&f_{ww}&= -\tfrac{3}{8} f + \tfrac{3}{8} Qf,\\&f_{vw}=0,&f_{uv}&=-\tfrac{3}{4} Jf_w -\tfrac{\sqrt{3}}{4} f_w,&f_{uw}&=-\tfrac{3}{4} Jf_v +\tfrac{\sqrt{3}}{4} f_v. \end{aligned}$$

In terms of the components p and q, this reduces to

$$\begin{aligned} p_{uu}&=-\tfrac{3}{4} p,&p_{ww}&=-\tfrac{3}{4} p,&p_{uw}&=-\tfrac{\sqrt{3}}{2} pq^{-1} q_v, \end{aligned}$$

(5.15)

$$\begin{aligned} q_{uu}&=-\tfrac{3}{4} q,&q_{vv}&=-\tfrac{3}{4} q,&q_{uv}&=\tfrac{\sqrt{3}}{2} qp^{-1} p_w. \end{aligned}$$

(5.16)

In order to simplify expressions, in the following, we will write $\tilde{u} = \tfrac{\sqrt{3}}{2} u$, $\tilde{v} = \tfrac{\sqrt{3}}{2} v$ and $\tilde{w} =\tfrac{\sqrt{3}}{2} w$.

We first look at the system of differential equations for p (see (5.15)). Solving the first two equations in (5.15), it follows that we can write

$$\begin{aligned} p=A_1 \cos (\tilde{u})\cos (\tilde{w}) +A_2 \cos (\tilde{u}) \sin (\tilde{w})+A_3 \sin (\tilde{u})\cos (\tilde{w}) +A_4 \sin (\tilde{u})\sin (\tilde{w}). \end{aligned}$$

Using the fact that $\langle p,p\rangle =1$ together with $\langle p_u,p_w\rangle =0$ (as $f_u$ and $f_w$ are mutually orthogonal with respect to the induced Euclidean product metric), it follows that by applying an isometry of SO(4) we may write $A_1=(1,0,0,0)$, $A_2=(0,1,0,0)$, $A_3=(0,0,1,0)$ and $A_4= (0,0,0,\varepsilon _1)$, where $\varepsilon _1= \pm 1$.

A similar argument is of course valid for the second map q. Also it is well known that $\mathbb {S}^3 \times \mathbb {S}^3= SU(2)\times SU(2)$ is the double cover of SO(4), so any rotation $R\in SO(4)$ can be written as $R(x) = \alpha x \beta$, where $\alpha , \beta \in {\mathbb {S}}^3$. Therefore, applying an isometry of the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$, we can write that

$$\begin{aligned}&p=\bigl (\cos (\tilde{u})\cos (\tilde{w}), \cos (\tilde{u}) \sin (\tilde{w}), \sin (\tilde{u})\cos (\tilde{w}),\varepsilon _1\sin (\tilde{u})\sin (\tilde{w})\bigr ),\\&q=\bigl (\cos (\tilde{u})\cos (\tilde{v}), \cos (\tilde{u}) \sin (\tilde{v}), \sin (\tilde{u})\cos (\tilde{v}),\varepsilon _2\sin (\tilde{u})\sin (\tilde{w})\bigr )d, \end{aligned}$$

where $\varepsilon _i=\pm 1$ and $d=(d_1,d_2,d_3,d_4)$ is a unitary quaternion. Note that taking d or $-d$ gives up to an isometry the same example. Looking now at $p^{-1}p_{uw}+\tfrac{\sqrt{3}}{2} q^{-1} q_v=0$ (see (5.15)) it immediately follows that $\varepsilon _1=\varepsilon _2=1$.

Moreover, we get from $p_{uw}=-\tfrac{\sqrt{3}}{2} pq^{-1} q_v$, $q_{uv}=\tfrac{\sqrt{3}}{2} qp^{-1} p_w$ (see (5.15) and (5.16)) that the unit quaternion d has to satisfy:

$$\begin{aligned}&d_1^2+d_2^2+d_3^2+d_4^2=1,\\&d_1 d_2 +d_3 d_4=d_1 d_4-d_2 d_3=0,\\&d_1^2+d_2^2-d_3^2-d_4^2=d_1^2-d_2^2-d_3^2+d_4^2=0,\\&-1-2 d_1 d_3+2 d_2 d_4 =1+2 d_1 d_3+2 d_2 d_4=0. \end{aligned}$$

This reduces to

$$\begin{aligned} d_3^2&= d_1^2,&d_4^2&= d_2^2,&d_1^2+d_2^2&=\tfrac{1}{2},&d_1 d_3&=-\tfrac{1}{2},&d_4 d_2&=0. \end{aligned}$$

This system has solutions $d=\Bigl (\tfrac{1}{\sqrt{2}},0,-\tfrac{1}{\sqrt{2}},0\Bigr )$ and $d=\Bigl (-\tfrac{1}{\sqrt{2}},0,\tfrac{1}{\sqrt{2}},0\Bigr )$. This completes the proof of the theorem.$\square$

Case 2: $\mathbf {h_{12}^3=0.}$ In this case, by using (5.10), we get that the general solution of (5.8) is

$$\begin{aligned} \begin{aligned} h_{ii}^j&= -2\alpha _j \sin (\theta _j-\theta _k) \sin (2 \theta _i-\theta _j-\theta _k), \\ h_{kk}^j&=\phantom {-} 2\alpha _j \sin (\theta _i-\theta _j) \sin (2 \theta _k-\theta _i-\theta _j), \end{aligned} \end{aligned}$$

(5.17)

where here and throughout the remainder of this case, (ijk) denotes a positive permutations of (123) and $\alpha _1$, $\alpha _2$ and $\alpha _3$ are some real functions. The components $h_{ii}^i$ can be calculated using the minimality of M.

As before we may assume that M is not totally geodesic. Then from Lemma 3.8 we may assume that all the angle functions are pairwise different (modulo $\pi$). Hence, $\sin {(\theta _i-\theta _j)}\ne 0,~\text {for any}~i\ne j$. We will show by contradiction that Case 2 cannot occur.

By the second equation in Lemma 3.4 and (5.17), one then can express the $\omega _{ij}^k$ in terms of the $h_{ij}^k$ (since $\sin {(\theta _j-\theta _k)}\ne 0,~\text {for any} ~j\ne k$). This gives us for all positive permutations (ijk) of (123) that

$$\begin{aligned}&\omega _{ii}^j = 2 \alpha _j \cot (\theta _j-\theta _i) \sin (\theta _k-\theta _j) \sin (2 \theta _i-\theta _j-\theta _k),\\&\omega _{kk}^j =2 \alpha _j \cot (\theta _j-\theta _k)\sin (\theta _i-\theta _j) \sin (2 \theta _k-\theta _i-\theta _j),\\&\omega _{ij}^k = -\omega _{ik}^j = \tfrac{\sqrt{3}}{6}, \end{aligned}$$

where the third equation follows from the fact that the tensor $h_{ij}^k$ is totally symmetric and the assumption of Case 2.

Using Lemma 3.4 and (5.17), the differential equations for the angles become

$$\begin{aligned} \begin{aligned} E_j(\theta _i)&= 2\alpha _j \sin (\theta _j-\theta _k) \sin (2 \theta _i-\theta _j-\theta _k),\\ E_j(\theta _k)&= -2\alpha _j \sin (\theta _i-\theta _j) \sin (2 \theta _k-\theta _i-\theta _j), \end{aligned} \end{aligned}$$

(5.18)

and

$$\begin{aligned} \begin{aligned} E_1(\theta _1)&=-\alpha _1 (\cos 2 (\theta _1-\theta _2)+\cos 2 (\theta _1-\theta _3)-2 \cos 2 (\theta _2-\theta _3)),\\ E_2(\theta _2)&=-\alpha _2 (\cos 2 (\theta _2-\theta _3)+\cos 2 (\theta _2-\theta _1)-2 \cos 2 (\theta _3-\theta _1)),\\ E_3(\theta _3)&=-\alpha _3 (\cos 2 (\theta _3-\theta _1)+\cos 2 (\theta _3-\theta _2)-2 \cos 2 (\theta _1-\theta _2)). \end{aligned} \end{aligned}$$

(5.19)

For every positive permutation (ijk) of (123), since M has constant curvature c, the curvature tensor satisfies $R(X,Y)Z = c(g(Y,Z)X -g(X,Z)Y)$, we then get that $R(E_i,E_j)E_k=0$. On the other hand, we can also calculate $R(E_i,E_j)E_k$ by using its definition (i.e., $R(E_i,E_j)E_k=\nabla _{E_i}\nabla _{E_j}E_k-\nabla _{E_j}\nabla _{E_i}E_k-\nabla _{[E_i,E_j]}E_k$) and the above expressions for $\omega _{ij}^k$. In this way, we can get the expression for $R(E_i,E_j)E_k$ involving $E_i(\alpha _k)$ and $E_j(\alpha _k)$. After straightforward calculations, we obtain a linear system of equations for $E_1(\alpha _2)$, $E_1(\alpha _3)$, $E_2(\alpha _1)$, $E_2(\alpha _3)$, $E_3(\alpha _1)$ and $E_3(\alpha _2)$.

Now we first deal with the case that $\cos (\theta _i -\theta _j) \ne 0$ and $\sin (2 \theta _i-\theta _j-\theta _k)\ne 0$ for $i,j,k ~\text {distinct}$. In this case, we can solve the linear system of equations mentioned above and get that for every positive permutation (ijk) of (123), we have

$$\begin{aligned} E_i(\alpha _j)&= -\frac{1}{6}\csc (\theta _i-\theta _j)\sin (\theta _i-\theta _k)\csc (\theta _i+\theta _j-2\theta _k)\times \\&\quad \phantom {\frac{1}{6}}\Bigl ( 6 \alpha _i \alpha _j \sin (\theta _i+\theta _j-2 \theta _k) (-7 \sin (\theta _j-\theta _k) +2 \sin (2 \theta _i -\theta _j-\theta _k) \\&\quad +\sin (2 \theta _i-3 \theta _j+\theta _k))\phantom {\frac{1}{6}} +2 \sqrt{3} \alpha _k \sin (\theta _i - 2 \theta _j + \theta _k) \Bigr ) \end{aligned}$$

and

$$\begin{aligned} E_j(\alpha _i)&= \frac{1}{6}\csc (\theta _i-\theta _j)\sin (\theta _j-\theta _k)\csc (\theta _i+\theta _j-2\theta _k)\times \\&\quad \phantom {\frac{1}{6}}\Bigl ( 6 \alpha _i \alpha _j \sin (\theta _i+\theta _j-2 \theta _k) (-7 \sin (\theta _i-\theta _k) +2 \sin (2 \theta _j -\theta _i-\theta _k) \\&\quad +\sin (2 \theta _j-3 \theta _i+\theta _k))\phantom {\frac{1}{6}} + 2 \sqrt{3} \alpha _k \sin (2 \theta _i - \theta _j - \theta _k) \Bigr ). \end{aligned}$$

Substituting these derivatives into the compatibility conditions,

$$\begin{aligned} E_i(E_j(\theta _i))-E_j(E_i(\theta _i))=(\nabla _{E_i}E_j-\nabla _{E_j}E_i)(\theta _i), \end{aligned}$$

gives the following three equations relating the functions $\alpha _1$, $\alpha _2$ and $\alpha _3$:

$$\begin{aligned} \alpha _i \bigl (\sin (2 \theta _k - \theta _i - \theta _j)\sin ( \theta _k + \theta _i - 2\theta _j)\bigr ) = 8 \sqrt{3} \alpha _j \alpha _k \sin ^3(\theta _k - \theta _j) \sin ^2(\theta _k - 2 \theta _i + \theta _j), \end{aligned}$$

(5.20)

for every positive permutation (ijk) of (123). So we have equations of the form $x_i \alpha _i = \alpha _j \alpha _k$, where

$$\begin{aligned} x_i = \frac{\sin (2 \theta _k - \theta _i - \theta _j)\sin ( \theta _k + \theta _i - 2\theta _j)}{8 \sqrt{3} \sin ^3(\theta _k - \theta _j) \sin ^2(\theta _k - 2 \theta _i + \theta _j)} \not = 0. \end{aligned}$$

As the Lagrangian submanifold M is not totally geodesic, from (5.17) we know that not all the $\alpha _i$ can vanish at the same time. Therefore, it follows from the above system of equations (5.20) that

$$\begin{aligned} \alpha _i^2 = x_j x_k= -\tfrac{\sin ^2(2 \theta _i - \theta _j - \theta _j)}{192 \sin ^3(\theta _i-\theta _j) \sin ^3(\theta _i-\theta _k) \sin (\theta _i+\theta _j-2\theta _k) \sin (\theta _i-2 \theta _j+\theta _k)}. \end{aligned}$$

Using (5.17) and the Gauss equation (3.2) to calculate the sectional curvature K of the plane spanned by $E_i$ and $E_j$, we have that $K= \tfrac{5}{12}+ \tfrac{1}{3} \cos (2(\theta _i - \theta _j))$ for all $i\ne j$. As the sectional curvature is constant, this implies that

$$\begin{aligned} \cos (2(\theta _1 - \theta _2))=\cos (2(\theta _1 - \theta _3))=\cos (2(\theta _2 - \theta _3)), \end{aligned}$$

which together with Lemma 3.2 implies that all the angles are constant, hence by Lemma 3.4 and the assumption of Case 2 that $h_{12}^3=0$, the submanifold M is totally geodesic. So we get a contradiction with the assumption that M is not totally geodesic.

Next we deal with the case that there exist some i, j, k which are distinct such that $\sin (2\theta _i-\theta _j-\theta _k) = 0$. As the sum of the angles is a multiple of $\pi$ and all the angles are determined up to a multiple of $\pi$, in this case it is sufficient to consider the case that $\theta _1= \frac{b \pi }{3}$ and $\theta _3 = a\pi -\theta _1-\theta _2=a\pi -\frac{b\pi }{3}-\theta _2$, where a and b are some constant integers and $\theta _2$ is not constant. As $\theta _2+\theta _3=a\pi -\frac{b\pi }{3}=\text {constant}$, $\theta _2$ is not constant, using (5.18), we get from

$$\begin{aligned} 0=E_1(\theta _2+\theta _3) =2\alpha _1 \left( 1-2 \cos ^2{\left( a\pi -\frac{b\pi }{3}-2 \theta _2 \right) } +(-1)^{a-b} \cos {\left( a\pi -\frac{b\pi }{3}-2\theta _2\right) } \right) \end{aligned}$$

that $\alpha _1$ has to vanish. Then from (5.17) and the Gauss equation (3.2), we obtain that the sectional curvature K of the plane spanned by $E_1$ and $E_2$ is given by

$$\begin{aligned} K=\tfrac{5}{12} + \tfrac{1}{3} \cos \left( 2 \theta _2-\frac{2b\pi }{3}\right) . \end{aligned}$$

As M has constant sectional curvature this implies that $\theta _2$, and therefore all angle functions, are constant. Hence by Lemma 3.4 and the assumption of Case 2 that $h_{12}^3=0$, the submanifold M is totally geodesic. So we get a contradiction with the assumption that M is not totally geodesic.

Finally, we deal with the case that there exist some i, j such that $\cos (\theta _i-\theta _j)=0$. As the sum of the angles is a multiple of $\pi$ and all the angles are only determined up to a multiple of $\pi$, in this case it is sufficient to consider the case that

$$\begin{aligned} \theta _2&= \theta _1 -\tfrac{b\pi }{2},&\theta _3&=a\pi -\theta _1-\theta _2 = a\pi +\tfrac{b\pi }{2}-2\theta _1, \end{aligned}$$

where a is a constant integer, b is an odd constant integer and $\theta _1$ is not constant. As $\theta _1-\theta _2=\frac{b\pi }{2}=\text {constant}$, $\theta _1$ is not constant, using (5.18)-(5.19), we get from

$$\begin{aligned} \begin{aligned}&0=E_1(\theta _1-\theta _2)=-\alpha _1\big (2\cos {(b\pi )}-3\cos {(6\theta _1)}+\cos {(b\pi -6\theta _1})\big ),\\&0=E_2(\theta _1-\theta _2)=\alpha _2\big (2\cos {(b\pi )}+\cos {(6\theta _1)}-3\cos {(b\pi -6\theta _1)}\big ),\\&0=E_3(\theta _1-\theta _2)=2\alpha _3\sin {\left( \frac{b\pi }{2}\right) }\sin {\left( \frac{3b\pi }{2}-6\theta _1\right) }, \end{aligned} \end{aligned}$$

that $\alpha _1=\alpha _2=\alpha _3=0$. Hence by (5.17), the minimality of M and the assumption of Case 2 that $h_{12}^3=0$, the submanifold M is totally geodesic. So we get a contradiction with the assumption that M is not totally geodesic.

Therefore, we have proved that Case 2 cannot occur.

Proof of Theorem 1.1:

Assume that M is a Lagrangian submanifold of constant sectional curvature in the nearly Kähler $\mathbb {S}^3 \times \mathbb {S}^3$. First, we consider the case that M is totally geodesic, then applying Theorem 1.3 obtained by Zhang–Hu–Dioos–Vrancken–Wang, we get that M is locally congruent to one of the following immersions:

• (1) $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,1)$, which is Example 4.1,

• (2) $f:{\mathbb {S}}^3 \rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (1,u)$, which is Example 4.2,

• (3) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u,u)$, which is Example 4.3.

Second, we consider the case that M is not totally geodesic. They applying our discussions for Case 1a (see Theorem 5.3), Case 1b (see Theorem 5.4) and Case 2 (we have proved that this case cannot occur), we obtain that M is locally congruent to one of the following immersions:

• (4) $f:{\mathbb {S}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: u\mapsto (u\mathbf {i}u^{-1},u\mathbf {j}u^{-1})$, which is Example 4.7,

• (5) $f: {\mathbb {R}}^3\rightarrow \mathbb {S}^3 \times \mathbb {S}^3: (u,v,w)\mapsto (p(u,w),q(u,v))$, where p and q are constant mean curvature tori in ${\mathbb {S}}^3$ given in Example 4.8.

This completes the proof of Theorem 1.1. $\square$