Classification of Four-Dimensional CR Submanifolds of the Homogenous Nearly Kähler ${\mathbb{S}}^{\mathbf{3}}\mathbf{\times }{\mathbb{S}}^{\mathbf{3}}$ Which Almost Complex Distribution Is Almost Product Orthogonal on Itself

Abstract

The product manifold ${\mathbb{S}}^3\times {\mathbb{S}}^3$, which belongs to the homogenous six-dimensional nearly Kähler manifolds, admits two structures, the almost complex structure J and the almost product structure P. The investigation of embeddings of different classes of CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ was started some time ago by investigating three-dimensional CR submanifolds. It resulted that the almost product structure P is very important for the study of CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, since submanifolds characterized by different actions of the almost product structure on base vector fields often appear as a result of the study of some specific types of CR submanifolds. Therefore, the investigation of four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ is initiated in this article. The main result is the classification of four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, whose almost complex distribution ${\mathcal{D}}_1$ is almost product orthogonal on itself. First, it was proved that such submanifolds have a non-integrable almost complex distribution, and then it was proved that these submanifolds are locally product manifolds of curves and three-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ of the same type, and they were therefore constructed in this way.

1. Introduction

The almost complex structure J of a differentiable manifold is an endomorphism of its tangent bundle for which $J^2=-Id$. A manifold endowed with a Riemannian metric for which the almost complex structure J is an isometry is called an almost Hermitian manifold. In terms of the condition satisfied by a tensor field $\tilde{\nabla }J$, there are several different types of almost Hermitian manifolds. A nearly Kähler manifold is an almost Hermitian manifold which entails that $\left({\tilde{\nabla }}_{X}J\right)X=0$, for any vector field X on that manifold. Jean-Baptiste Butruille in [1] has shown that the product manifold ${\mathbb{S}}^3\times {\mathbb{S}}^3$ is one of the four six-dimensional homogeneous strict nearly Kähler manifolds, besides the sphere ${\mathbb{S}}^6$, the complex projective space ${\mathbb{C}P}^3$ and the flag manifold ${\mathbb{F}}^3$. The action of the almost complex structure J on a tangent space of the submanifold M can be different. If a $TM={\mathcal{D}}_1\oplus {\mathcal{D}}_1^{\perp },$ where ${\mathcal{D}}_1$ is an almost complex distribution $(J{\mathcal{D}}_1={\mathcal{D}}_1)$ and ${\mathcal{D}}_1^{\perp }$ is a totally real distribution $(J{\mathcal{D}}_1^{\perp }\subseteq T{M}^{\perp }),$ such a submanifold is called a CR submanifold. The research of the different types of submanifolds of the nearly Kähler ${\mathbb{S}}^3\times {\mathbb{S}}^3$ has recently begun. The theory of differentiable manifolds can be found in [2,3,4], while more results on nearly Kähler manifolds, as well as on ${\mathbb{S}}^3\times {\mathbb{S}}^3$ itself and its submanifolds can be found in [5,6,7,8,9,10,11,12].

Along a proper four-dimensional CR submanifold M of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, the tangent bundle of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ can be split by three two-dimensional distributions in the following way ${\mathcal{D}}_1\oplus {\mathcal{D}}_2\oplus {\mathcal{D}}_3$, where ${\mathcal{D}}_1$ is an almost complex distribution, ${\mathcal{D}}_2={\mathcal{D}}_1^{\perp }$ is a totally real distribution of the submanifold M and the distribution ${\mathcal{D}}_3=J{\mathcal{D}}_2$ corresponds to the normal bundle. In the investigation of CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, an almost product structure P plays a very important role and the first obtained classifications refer to the behavior of a tangent bundle under the action of the almost product structure P. These classifications are important for all further research, since studies of some other types of CR submanifolds lead to them. For example, there is a class of four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with null distribution which belongs to the submanifolds obtained in this article. Some examples of the submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ that are classified with respect to the action of the almost product structure P on their tangent bundle or with respect to the action of the almost product structure P on particular vector fields can be found in [13,14,15]. This article is a continuation of previous research, but it is much more comprehensive than the previous one. The action of the almost product structure on the tangent bundle has already been studied by other mathematicians, but this is the first paper that investigates the action on four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$. Moreover, the resulting class is one of the most extensive classes among the studied different types of submanifolds on six-dimensional nearly Kähler manifolds, which are a very popular research topic recently. Besides an action of the almost product structure on the base vector fields, the integrability of the almost complex distribution ${\mathcal{D}}_1$ plays an important role in the classification of the different types of CR submanifolds.

The main topic of this article is the study of four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, whose almost complex distribution ${\mathcal{D}}_1$ is almost product orthogonal on itself. In the case of three-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$, an almost complex distribution can be integrable or non-integrable, but it will turn out that for our investigation an already obtained class of three-dimensional CR submanifolds with non-integrable almost complex distribution is important. In this article, we obtain that an arbitrary four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, whose almost complex distribution is almost product orthogonal on itself is foliated by a corresponding three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ of the same type.

In order to obtain a classification of the four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$, we will first analyze the integrability of the almost complex distribution ${\mathcal{D}}_1$ and in the article we will prove that there is no four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ that has an integrable almost complex distribution ${\mathcal{D}}_1$ such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$. Furthermore, we will study submanifold M, whose almost complex distribution ${\mathcal{D}}_1$ contains an arbitrary vector field $E_1$ such that $P{E}_1\in T{M}^{\perp }$. The main result of this article is the classification theorem that proves that a four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$ holds, belongs to previously obtained submanifolds.

The integrability of an almost complex distribution of a CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ can be related to its behavior under the action of an almost product structure P on it. Whether M is a three-dimensional or four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, the condition $PTM=TM$ implies that the almost complex distribution is integrable, which means that the CR submanifold has a foliation corresponding to an almost complex surface of the nearly Kähler ${\mathbb{S}}^3\times {\mathbb{S}}^3$, which is almost product invariant. These surfaces are investigated in [13]. On the other hand, the condition $P{\mathcal{D}}_1\perp ({\mathcal{D}}_1\oplus {\mathcal{D}}_1^{\perp }\oplus J{\mathcal{D}}_1^{\perp })$ in the case of three-dimensional CR submanifolds, as well as the condition $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$ in the case of four-dimensional CR submanifolds, imply that the almost complex distribution ${\mathcal{D}}_1$ is non-integrable. The conclusion is that these extreme positions of the almost complex distribution ${\mathcal{D}}_1$ and the tangent bundle of the submanifold M under the action of the almost complex structure P can be directly related to its integrability. In contrast to the almost complex distribution, the totally real distribution is never integrable in the case of four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, which has the consequence that four-dimensional CR submanifold is not a product manifold of an almost complex and totally real surface of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, but in most cases is a product of curve and Lagrangian or three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$. The product of a manifolds, as well as various types of submanifold embedded into a manifold, have been very actively researched in recent years and are becoming increasingly important. More about CR embedded submanifolds of CR manifolds could be found in [16]. We have found that the four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$ is a locally product manifold of a curve, which is an integral curve of the vector field belonging to the totally real distribution, and the three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, with a non-integrable almost complex distribution ${\mathcal{D}}_1$ which is P-orthogonal on itself, but this four-dimensional CR submanifold is not even a double-twisted product manifold, so in this article we are working with a very broad new class of CR submanifolds which leads us to a number of still unexplored four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$.

2. Preliminaries

A unit sphere ${\mathbb{S}}^3$ in the space ${\mathbb{R}}^4$ can be identified with the space of unit quaternions $\mathbb{H}$, which is spanned by $1,i,j$ and k. Recall that the vector fields $X_1=pi$, $X_2=pj$ and $X_3=-pk$ form an orthonormal moving frame on ${\mathbb{S}}^3$ and some arbitrary vector tangent at point p on a sphere ${\mathbb{S}}^3$ can be represented as $p·\alpha \in T_p{\mathbb{S}}^3$, where $\alpha \in Im\mathbb{H}$. Therefore, using the isomorphism of the spaces $T_{(p,q)}({\mathbb{S}}^3\times {\mathbb{S}}^3)\cong T_p{\mathbb{S}}^3\oplus T_q{\mathbb{S}}^3$, an arbitrary tangent vector at a point $(p,q)\in {\mathbb{S}}^3\times {\mathbb{S}}^3$ is represented by $Z(p,q)=(p\alpha ,q\beta )$, where $\alpha$ and $\beta$ are imaginary quaternions. Let us also recall a multiplication of quaternions given in the following way: $ij=k=-ji$, $jk=i=-kj$, $ki=j=-ik$, and $i^2=j^2=k^2=-1$. Moreover, we have that $\alpha ·\beta =-⟨\alpha ,\beta ⟩+\alpha \times \beta$, where $\alpha ,\beta \in Im\mathbb{H}$.

It is shown that ${\mathbb{S}}^3\times {\mathbb{S}}^3$ is six-dimensional homogeneous strict nearly Kähler manifold. An almost complex structure and other relations that mentioned further are obtained in [17] and in [18]. The almost complex structure J on ${\mathbb{S}}^3\times {\mathbb{S}}^3$ is defined as

$$J\left(Z(p,q)\right)={\displaystyle \frac{1}{\sqrt{3}}}(p(2\beta -\alpha ),q(-2\alpha +\beta )),$$

(1)

where $Z(p,q)=(p\alpha ,q\beta )\in T_{(p,q)}{\mathbb{S}}^3\times {\mathbb{S}}^3$. J is an isometry with regard to the Hermitian metric g and for vector fields $Z,Z^{\prime }\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$, the Hermitian metric g in terms of the Euclidean product metric of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ induced from ${\mathbb{R}}^4\times {\mathbb{R}}^4$ is given by

$$g(Z,Z^{\prime })={\displaystyle \frac{1}{2}}(⟨Z,Z^{\prime }⟩+⟨JZ,J{Z}^{\prime }⟩).$$

(2)

For these tensor fields, the manifold $({\mathbb{S}}^3\times {\mathbb{S}}^3,g,J)$ becomes a nearly Kähler manifold. Let us denote by $G=\tilde{\nabla }J$ a new tensor field, where $\tilde{\nabla }$ is the Levi-Civita connection of the metric g. Recall that the tensor field G defined in this way further satisfies

$$\begin{array}{ccc}G(X,Y)+G(Y,X)=0,\hfill & & \end{array}$$

(3)

$$\begin{array}{ccc}& & g\left(G\right(X,Y),Z)+g\left(G\right(X,Z),Y)=0,\hfill \end{array}$$

(4)

$$\begin{array}{ccc}& & \left(\tilde{\nabla }G\right)(X,Y,Z)={\displaystyle \frac{1}{3}}(g(X,Z)JY-g(X,Y)JZ-g(JY,Z)X),\hfill \end{array}$$

(5)

$$\begin{array}{ccc}& & g(G(X,Y),G(Z,W))={\displaystyle \frac{1}{3}}(g(X,Z)g(Y,W)-g(X,W)g(Y,Z)\hfill \\ & & +g(JX,Z)g(JW,Y)-g(JX,W)g(JZ,Y)),\hfill \end{array}$$

(6)

for all $X,Y,Z,W\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$. For the vectors $X(p,q)=(p{\gamma }_1,q{\delta }_1)$, $Y(p,q)=(p{\gamma }_2,q{\delta }_2)\in T_{(p,q)}{\mathbb{S}}^3\times {\mathbb{S}}^3$, we have that

$$\begin{array}{ccc}\hfill G\left(X\right(p,q),Y(p,q\left)\right)& =& {\displaystyle \frac{2}{3\sqrt{3}}}(p({\delta }_1\times {\gamma }_2+{\gamma }_1\times {\delta }_2+{\gamma }_1\times {\gamma }_2-2{\delta }_1\times {\delta }_2),\hfill \\ & & q(-{\gamma }_1\times {\delta }_2-{\delta }_1\times {\gamma }_2+2{\gamma }_1\times {\gamma }_2-{\delta }_1\times {\delta }_2)).\hfill \end{array}$$

(7)

Besides an almost complex structure, an almost product structure P is also defined on ${\mathbb{S}}^3\times {\mathbb{S}}^3$, too. The difference is that an almost product structure is not directly related to the definition of a nearly Kähler manifold. The almost product structure P is a tensor field of type $(1,1)$ such that $P^2=Id$. Here, on ${\mathbb{S}}^3\times {\mathbb{S}}^3$, an almost product structure P is defined in a natural way, by

$$P\left(Z\right(p,q\left)\right)=(p\beta ,q\alpha ),$$

(8)

where $Z(p,q)=(p\alpha ,q\beta )\in T_{(p,q)}{\mathbb{S}}^3\times {\mathbb{S}}^3$. It holds that $PJ=-JP$, $g(PZ,P{Z}^{\prime })=g(Z,Z^{\prime })$, $g(PZ,Z^{\prime })=g(Z,P{Z}^{\prime })$, for arbitrary vector fields $Z,Z^{\prime }\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$. We can conclude that P is isometric with regard to g. It follows that the Euclidean product metric of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ induced from ${\mathbb{R}}^4\times {\mathbb{R}}^4$ is given in terms of the Hermitian metric in the following way

$$⟨Z,Z^{\prime }⟩=g(Z,Z^{\prime })+{\displaystyle \frac{1}{2}}g(Z,P{Z}^{\prime }),$$

(9)

where $Z,Z^{\prime }\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$. The tensor fields J, P, and G also satisfy the following relations:

$$\begin{array}{ccc}& & G(X,PY)+PG(X,Y)+2J\left(\left({\tilde{\nabla }}_{X}P\right)Y\right)=0,\hfill \end{array}$$

(10)

for a vector fields $X,Y\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$. Besides the almost product structure, there is an usual product structure. Note that the usual product structure $Q\left(Z\right(p,q\left)\right)=(-p\alpha ,q\beta )$, where $Z(p,q)=(p\alpha ,q\beta )\in T_{(p,q)}{\mathbb{S}}^3\times {\mathbb{S}}^3$, can be expressed in terms of the almost product structure P by $QZ={\displaystyle \frac{1}{\sqrt{3}}}(2PJZ-JZ)$. Using this relation, for a tangent vector $Z(p,q)=(p\alpha ,q\beta )\in T_{(p,q)}{\mathbb{S}}^3\times {\mathbb{S}}^3$, we can obtain its projections onto the tangent spaces of both spheres, $T_p{\mathbb{S}}^3$ and $T_q{\mathbb{S}}^3$, in the following way

$$(p\alpha ,0)={\displaystyle \frac{1}{2}}(Z(p,q)-Q\left(Z(p,q)\right)),(0,q\beta )={\displaystyle \frac{1}{2}}(Z(p,q)+Q\left(Z(p,q)\right)).$$

(11)

Using these structures, a curvature tensor of the metric g of the homogenous nearly Kähler ${\mathbb{S}}^3\times {\mathbb{S}}^3$ is expressed as

$$\begin{array}{ccc}\hfill & & R(X,Y)Z={\displaystyle \frac{5}{12}}(g(Y,Z)X-g(X,Z)Y)+{\displaystyle \frac{1}{12}}(g(JY,Z)JX-g(JX,Z)JY-2g(JX,Y)JZ)\hfill \\ & & +{\displaystyle \frac{1}{3}}(g(PY,Z)PX-g(PX,Z)PY+g(JPY,Z)JPX-g(JPX,Z)JPY)\hfill \end{array}$$

and the following equation holds

$$R(X,Y)Z-({\tilde{\nabla }}_X{\tilde{\nabla }}_{Y}Z-{\tilde{\nabla }}_Y{\tilde{\nabla }}_{X}Z-{\tilde{\nabla }}_{[X,Y]}Z)=0,$$

(12)

for $\forall X,Y,Z\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$.

Let M be an n-dimensional submanifold of the m-dimensional manifold $\tilde{M}$, whose metric is induced by the Riemannian metric on $\tilde{M}$. Let ∇ and $\tilde{\nabla }$ be the corresponding Levi-Civita connections on M and $\tilde{M}$, respectively. The tangent bundle on $\tilde{M}$ can be decomposed by $T\tilde{M}=TM\oplus T{M}^{\perp }$. The bundles $TM$ and $T{M}^{\perp }$ are tangent and normal bundles on M. Then for any $X,Y\in TM$ and $\xi \in T{M}^{\perp }$, it holds that

$$\begin{array}{ccc}& & {\tilde{\nabla }}_{X}Y={\nabla }_{X}Y+h(X,Y),\left(Gaussformula\right)\hfill \\ & & {\tilde{\nabla }}_X\xi =-A_{\xi }X+{{\nabla }^{\perp }}_X\xi ,\left(Weingartenformula\right)\hfill \end{array}$$

where h is the second fundamental form and $A_{\xi }$ is shape operator and they are related in the following way: $g(h(X,Y),\xi )=g(A_{\xi }X,Y),$ while ${\nabla }^{\perp }$ is the normal connection on M.

In [19], it was shown that the relation between the Euclidean connection ${\nabla }^E$ of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ and the Levi-Civita connection $\tilde{\nabla }$ of the nearly Kähler ${\mathbb{S}}^3\times {\mathbb{S}}^3$ is given by

$${\nabla }_X^{E}Y={\tilde{\nabla }}_{X}Y+{\displaystyle \frac{1}{2}}(JG(X,PY)+JG(Y,PX)),$$

(13)

for $\forall X,Y\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$. For the Euclidean connection D of ${\mathbb{R}}^4\times {\mathbb{R}}^4$, we have that $D_{E_i}{E}_j={\nabla }_{E_i}^E{E}_j+\overline{h}(E_i,E_j),$ where $E_i,E_j\in T{\mathbb{S}}^3\times {\mathbb{S}}^3$, where the second fundamental form is given by $\overline{h}(E_i,E_j)=(-⟨{\alpha }_i,{\alpha }_j⟩p,-⟨{\beta }_i,{\beta }_j⟩q)$ and the induced connection of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ in ${\mathbb{R}}^4\times {\mathbb{R}}^4$ is

$$\begin{array}{ccc}& & {\nabla }_{E_i}^E{E}_j=(p({\alpha }_i\times {\alpha }_j+E_i\left({\alpha }_j\right)),q({\beta }_i\times {\beta }_j+E_i\left({\beta }_j\right))).\hfill \end{array}$$

(14)

2.1. Preliminaries on Product Manifolds and Isometries

Now suppose that ${\mathcal{D}}^{\prime }$ is an n-distribution on an $(n+p)$-dimensional differentiable manifold $\tilde{M}$ and ${\mathcal{D}}^{\prime \prime }$ is a p-distribution complementary to ${\mathcal{D}}^{\prime },$ i.e., $T\tilde{M}={\mathcal{D}}^{\prime }\oplus {\mathcal{D}}^{\prime \prime }.$ If $Q^{\prime }$ and $Q^{\prime \prime }$ are the projection morphisms of $T\tilde{M}$ onto ${\mathcal{D}}^{\prime }$ and ${\mathcal{D}}^{\prime \prime },$ respectively, then the tensor field $F^{\prime }$ of type $(1,1)$ defined by $F^{\prime }=Q^{\prime }-Q^{\prime \prime }$ is an almost product structure on $\tilde{M}$, that is, $F^{\prime }$ satisfies $F^{\prime 2}=I$ and the manifold $(\tilde{M},{\mathcal{D}}^{\prime },{\mathcal{D}}^{\prime \prime })$ is an almost product manifold. The following theorem is proved in [20].

Theorem 1.

Let $(\tilde{M},{\mathcal{D}}^{\prime },{\mathcal{D}}^{″})$ be an almost product manifold. If both ${\mathcal{D}}^{\prime }$ and ${\mathcal{D}}^{\prime \prime }$ are integrable, then every point $x^{*}\in \tilde{M}$ has a neighborhood ${\nu }^{*}={\nu }^{\prime }\times {\nu }^{″},$ where ${\nu }^{\prime }$ and ${\nu }^{\prime \prime }$ are open submanifolds of leaves of ${\mathcal{D}}^{\prime }$ and ${\mathcal{D}}^{\prime \prime }$ through $x^{*}.$

By using the previous theorem, it is proven that four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$ holds is a locally product manifold of a curve and a three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, with a non-integrable almost complex distribution ${\mathcal{D}}_1$ which is P-orthogonal on itself.

Moreover, the following lemma, see [21], shows that there is an isometry mapping between two manifolds that have the same Levi-Civita connection.

Lemma 1.

Let $M^n$ and ${\tilde{M}}^n$ be Riemannian manifolds with Levi-Civita connections ∇ and $\tilde{\nabla }$. Suppose that there exist $c_{ij}^k$, $i,j,k\in \{1,\dots ,n\}$ such that for all $p\in M$ and $\tilde{p}\in \tilde{M}$ there exist orthonormal frame fields $\left\{E_i\right\}$ around p and $\left\{{\tilde{E}}_i\right\}$ around $\tilde{p}$, such that ${\nabla }_{E_i}{E}_j={\sum }_k{c}_{ij}^k{E}_k$ and ${\tilde{\nabla }}_{{\tilde{E}}_i}{\tilde{E}}_j={\sum }_k{c}_{ij}^k{\tilde{E}}_k$ for all $i,j$. Then, for every $p\in M$ and $\tilde{p}\in \tilde{M}$, there exists a local isometry f which maps a neighborhood of p onto a neighborhood of $\tilde{p}$ and $E_i$ on ${\tilde{E}}_i$.

Let us recall here from [18] the notion of a Berger sphere $({\mathbb{S}}^3,g_1)$, where the metric $g_1$ has the expression $g_1(X,Y)={\displaystyle \frac{4}{\kappa }}(⟨X,Y⟩+({\displaystyle \frac{4{\tau }^2}{\kappa }}-1)⟨X,X_1⟩⟨Y,X_1⟩)$ and $\tau$, $\kappa$ are constants. Then, the vector fields ${\overline{E}}_1={\displaystyle \frac{\kappa }{4\tau }}{X}_1,$ ${\overline{E}}_2={\displaystyle \frac{\sqrt{\kappa }}{2}}{X}_2,$ ${\overline{E}}_3={\displaystyle \frac{\sqrt{\kappa }}{2}}{X}_3$ are orthonormal with respect to the metric $g_1$. The Levi-Civita connection $\overline{\nabla }$ of the metric $g_1$ is then given by

$$\begin{array}{ccc}\hfill & & {\overline{\nabla }}_{{\overline{E}}_i}{\overline{E}}_i=0,fori\in \{1,2,3\},\hfill \\ & & {\overline{\nabla }}_{{\overline{E}}_1}{\overline{E}}_2=(\tau -{\displaystyle \frac{\kappa }{2\tau }}){\overline{E}}_3,{\overline{\nabla }}_{{\overline{E}}_1}{\overline{E}}_3=-(\tau -{\displaystyle \frac{\kappa }{2\tau }}){\overline{E}}_2,\hfill \\ & & {\overline{\nabla }}_{{\overline{E}}_2}{\overline{E}}_3=-{\overline{\nabla }}_{{\overline{E}}_3}{\overline{E}}_2=-\tau {\overline{E}}_1,\hfill \\ & & {\overline{\nabla }}_{{\overline{E}}_3}{\overline{E}}_1=-\tau {\overline{E}}_2,{\overline{\nabla }}_{{\overline{E}}_2}{\overline{E}}_1=\tau {\overline{E}}_3.\hfill \end{array}$$

(15)

2.2. Preliminaries on Isometries of Homogenous Nearly Kähler ${\mathbb{S}}^3\times {\mathbb{S}}^3$

Isometries of the homogenous nearly Kähler product manifold ${\mathbb{S}}^3\times {\mathbb{S}}^3$ are the maps in which points $(p,q)\in {\mathbb{S}}^3\times {\mathbb{S}}^3$ are mapped in the following way ${\mathcal{F}}_{a,b,c}(p,q)=(ap\overline{c},bq\overline{c})$, ${\mathcal{F}}_1(p,q)=(q,p)$, ${\mathcal{F}}_2(p,q)=(\overline{p},q\overline{p})$, where the curves $a,b,c$ are unit quaternions. The maps ${\mathcal{F}}_{a,b,c}$ preserve the almost product structure P, but the group generated by the isometries ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$ need not preserve the P. In [22], we obtained relations among different almost product structures with respect to these isometries. They proved that on ${\mathbb{S}}^3\times {\mathbb{S}}^3$, there are three different almost product structures related in the following way: $P_1=P$, $P_2=-{\displaystyle \frac{1}{2}}P-{\displaystyle \frac{\sqrt{3}}{2}}JP$, $P_3=-{\displaystyle \frac{1}{2}}P+{\displaystyle \frac{\sqrt{3}}{2}}JP$. The behavior of these almost product structures under isometries ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$ is given by

$$\begin{array}{ccc}\hfill & & \mathrm{d}{\mathcal{F}}_1\circ P_1=P_1\circ \mathrm{d}{\mathcal{F}}_1,\mathrm{d}{\mathcal{F}}_2\circ P_1=P_3\circ \mathrm{d}{\mathcal{F}}_2,\hfill \\ & & \mathrm{d}{\mathcal{F}}_1\circ P_2=P_3\circ \mathrm{d}{\mathcal{F}}_1,\mathrm{d}{\mathcal{F}}_2\circ P_2=P_2\circ \mathrm{d}{\mathcal{F}}_2,\hfill \\ & & \mathrm{d}{\mathcal{F}}_1\circ P_3=P_2\circ \mathrm{d}{\mathcal{F}}_1,\mathrm{d}{\mathcal{F}}_2\circ P_3=P_1\circ \mathrm{d}{\mathcal{F}}_2.\hfill \end{array}$$

(16)

An arbitrary tangent vector $X(p,q)=(p\alpha ,q\beta )$ at a point $(p,q)\in {\mathbb{S}}^3\times {\mathbb{S}}^3$ under the differential of isometries ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$ maps in the following way

$$\begin{array}{ccc}\hfill & & \mathrm{d}{\mathcal{F}}_1\left(X(p,q)\right)=(q\beta ,p\alpha ),\mathrm{d}{\mathcal{F}}_2\left(X(p,q)\right)=(\overline{p}\left(p(-\alpha )\overline{p}\right),q\overline{p}\left(p(\beta -\alpha )\overline{p}\right)).\hfill \end{array}$$

(17)

The differential $\mathrm{d}{\mathcal{F}}_{a,b,c}$ commutes with J, since the differentials $\mathrm{d}{\mathcal{F}}_1$ and $\mathrm{d}{\mathcal{F}}_2$ anticommutes with J, implying that J-invariant distributions are still J-invariant under the action of these differentials.

3. Four-Dimensional CR Submanifolds of the Homogenous Nearly Kähler ${\mathbb{S}}^{\mathbf{3}}\mathbf{\times }{\mathbb{S}}^{\mathbf{3}}$

First, we will construct a local moving frame suitable for computation. We can take unit vector fields $E_1$ and $E_2=J{E}_1$ that span ${\mathcal{D}}_1$ and mutually orthogonal unit vector fields $E_3$ and $E_4$ that span the distribution ${\mathcal{D}}_2$. Then, the distribution ${\mathcal{D}}_3=J{\mathcal{D}}_2$ is spanned by the vector fields $E_5=J{E}_3$ and $E_6=J{E}_4$. In a straightforward manner, we obtain the following lemmas. By using the fact that G vanishes on any two-dimensional almost complex distribution and by using the relations (3), (4), and (6), we obtain the following expression of the tensor field G.

Lemma 2.

The tensor field G can be expressed in the frame $E_1,\dots E_6$ in the following way: $G(E_i,E_i)=0,$$i\in \overline{1,6}$ and

$$\begin{array}{cccc}\hfill & G(E_1,E_2)=0,\hfill & \hfill & G(E_1,E_3)={\displaystyle \frac{cost}{\sqrt{3}}}{E}_4+{\displaystyle \frac{sint}{\sqrt{3}}}{E}_6,\hfill \\ \hfill & G(E_1,E_4)=-{\displaystyle \frac{cost}{\sqrt{3}}}{E}_3-{\displaystyle \frac{sint}{\sqrt{3}}}{E}_5,\hfill & \hfill & G(E_1,E_5)={\displaystyle \frac{sint}{\sqrt{3}}}{E}_4-{\displaystyle \frac{cost}{\sqrt{3}}}{E}_6,\hfill \\ \hfill & G(E_1,E_6)=-{\displaystyle \frac{sint}{\sqrt{3}}}{E}_3+{\displaystyle \frac{cost}{\sqrt{3}}}{E}_5,\hfill & \hfill & G(E_2,E_3)={\displaystyle \frac{sint}{\sqrt{3}}}{E}_4-{\displaystyle \frac{cost}{\sqrt{3}}}{E}_6,\hfill \\ \hfill & G(E_2,E_4)=-{\displaystyle \frac{sint}{\sqrt{3}}}{E}_3+{\displaystyle \frac{cost}{\sqrt{3}}}{E}_5,\hfill & \hfill & G(E_2,E_5)=-{\displaystyle \frac{cost}{\sqrt{3}}}{E}_4-{\displaystyle \frac{sint}{\sqrt{3}}}{E}_6,\hfill \\ \hfill & G(E_2,E_6)={\displaystyle \frac{cost}{\sqrt{3}}}{E}_3+{\displaystyle \frac{sint}{\sqrt{3}}}{E}_5,\hfill & \hfill & G(E_3,E_4)={\displaystyle \frac{cost}{\sqrt{3}}}{E}_1+{\displaystyle \frac{sint}{\sqrt{3}}}{E}_2,\hfill \\ \hfill & G(E_3,E_5)=0,\hfill & \hfill & G(E_3,E_6)={\displaystyle \frac{sint}{\sqrt{3}}}{E}_1-{\displaystyle \frac{cost}{\sqrt{3}}}{E}_2,\hfill \\ \hfill & G(E_4,E_5)=-{\displaystyle \frac{sint}{\sqrt{3}}}{E}_1+{\displaystyle \frac{cost}{\sqrt{3}}}{E}_2,\hfill & \hfill & G(E_4,E_6)=0,\hfill \\ \hfill & G(E_5,E_6)=-{\displaystyle \frac{cost}{\sqrt{3}}}{E}_1-{\displaystyle \frac{sint}{\sqrt{3}}}{E}_2,\hfill \end{array}$$

(18)

where $t\in [0,2\pi )$ is an angle function.

Now, let us denote by ${\Gamma }_{ij}^k=g({\tilde{\nabla }}_{E_i}{E}_j,E_k)$, for $1\le i,j,k\le 4$, $h_{ij}^k=g({\tilde{\nabla }}_{E_i}{E}_j,E_{k+4})$, for $1\le i,j\le 4$, $1\le k\le 2$ and $b_{ij}^k=g({\tilde{\nabla }}_{E_i}{E}_{j+4},E_{k+4})$, for $1\le i\le 4$, $1\le j,k\le 2$. By using the equations ${\tilde{\nabla }}_X\left(JY\right)=G(X,Y)+J\left({\tilde{\nabla }}_{X}Y\right)$ and ${\tilde{\nabla }}_{X}G(Y,Z)=\left(\tilde{\nabla }G\right)(X,Y,Z)+G({\tilde{\nabla }}_{X}Y,Z)+G(Y,{\tilde{\nabla }}_{X}Z)$, where $\tilde{\nabla }G$ is expressed by (5), it straightforwardly follows that these coefficients satisfy the following relations given by the following two lemmas.

Lemma 3.

The coefficients ${\Gamma }_{ij}^k$, $h_{ij}^k$, $b_{ij}^k$ satisfy

$$\begin{array}{ccc}\hfill {\Gamma }_{12}^3& =& -h_{11}^1,{\Gamma }_{12}^4=-h_{11}^2,{\Gamma }_{11}^3=h_{12}^1,{\Gamma }_{11}^4=h_{12}^2,{\Gamma }_{22}^3=-h_{12}^1,\hfill \\ \hfill {\Gamma }_{22}^4& =& -h_{12}^2,{\Gamma }_{21}^3=h_{22}^1,{\Gamma }_{21}^4=h_{22}^2,{\Gamma }_{32}^3=-h_{13}^1,\hfill \\ \hfill {\Gamma }_{32}^4& =& -h_{13}^2-{\displaystyle \frac{cost}{\sqrt{3}}},{\Gamma }_{31}^4=h_{23}^2+{\displaystyle \frac{sint}{\sqrt{3}}},{\Gamma }_{31}^3=h_{23}^1,{\Gamma }_{42}^4=-h_{14}^2,\hfill \\ \hfill {\Gamma }_{41}^4& =& h_{24}^2,{\Gamma }_{42}^3=-h_{14}^1+{\displaystyle \frac{cost}{\sqrt{3}}},h_{34}^1=h_{33}^2,{\Gamma }_{41}^3=h_{24}^1-{\displaystyle \frac{sint}{\sqrt{3}}},\hfill \\ \hfill h_{44}^1& =& h_{34}^2,h_{14}^1=h_{13}^2-{\displaystyle \frac{cost}{\sqrt{3}}},h_{24}^1=h_{23}^2-{\displaystyle \frac{sint}{\sqrt{3}}},b_{31}^2={\Gamma }_{33}^4,\hfill \\ \hfill b_{41}^2& =& {\Gamma }_{43}^4,b_{21}^2={\Gamma }_{23}^4-{\displaystyle \frac{cost}{\sqrt{3}}},b_{11}^2={\Gamma }_{13}^4+{\displaystyle \frac{sint}{\sqrt{3}}}.\hfill \end{array}$$

(19)

Lemma 4.

It holds that

$$\begin{array}{ccc}\hfill E_1\left(t\right)& =& -h_{13}^1-h_{14}^2-{\Gamma }_{11}^2,E_2\left(t\right)=-h_{23}^1-h_{24}^2-{\Gamma }_{21}^2,\hfill \\ \hfill E_3\left(t\right)& =& -h_{33}^1-h_{34}^2-{\Gamma }_{31}^2,E_4\left(t\right)=-h_{33}^2-h_{44}^2-{\Gamma }_{41}^2.\hfill \end{array}$$

We can express an action of the almost product structure P in the following way.

Lemma 5.

On an open dense subset of the four-dimensional CR submanifold M of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, we can choose the orthonormal frame such that the almost product structure P in the frame $E_1,\dots ,E_6$ is given by

$$\begin{array}{ccc}\hfill P{E}_1& =& cos\theta cosm{E}_1+cos\theta sinm{E}_2+a_{1}sin\theta E_3+a_{2}sin\theta E_4+a_{3}sin\theta E_5\hfill \\ & & +a_{4}sin\theta E_6,\hfill \\ \hfill P{E}_2& =& cos\theta sinm{E}_1-cos\theta cosm{E}_2+a_{3}sin\theta E_3+a_{4}sin\theta E_4-a_{1}sin\theta E_5\hfill \\ & & -a_{2}sin\theta E_6,\hfill \\ \hfill P{E}_3& =& a_{1}sin\theta E_1+a_{3}sin\theta E_2+((a_2^2-a_4^2)cos\left(2t\right)+(-a_1^2+a_3^2)cos\theta cosm\hfill \\ & & -2a_1{a}_{3}cos\theta sinm+2a_2{a}_{4}sin\left(2t\right))E_3-((cos\theta cosm+cos\left(2t\right))·\hfill \\ & & (a_1{a}_2-a_3{a}_4)+(a_2{a}_3+a_1{a}_4)(cos\theta sinm+sin\left(2t\right)))E_4-(2a_2{a}_4·\hfill \\ & & cos\left(2t\right)+2a_1{a}_{3}cos\theta cosm-(a_1^2-a_3^2)cos\theta sinm-(a_2^2-a_4^2)·\hfill \\ & & sin\left(2t\right))E_5+((a_1{a}_2-a_3{a}_4)(cos\theta sinm-sin\left(2t\right))+(a_2{a}_3+a_1{a}_4)·\hfill \\ & & (cos\left(2t\right)-cos\theta cosm))E_6,\hfill \\ \hfill P{E}_4& =& a_{2}sin\theta E_1+a_{4}sin\theta E_2-((a_1{a}_2-a_3{a}_4)(cos\theta cosm+cos\left(2t\right))\hfill \\ & & +(a_2{a}_3+a_1{a}_4)(cos\theta sinm+sin\left(2t\right)))E_3+((a_1^2-a_3^2)cos\left(2t\right)-(a_2^2\hfill \\ & & -a_4^2)cos\theta cosm-2a_2{a}_{4}cos\theta sinm+2a_1{a}_{3}sin\left(2t\right))E_4+\left(\right(a_1{a}_2\hfill \\ & & -a_3{a}_4)(cos\theta sinm-sin\left(2t\right))+(a_2{a}_3+a_1{a}_4)(cos\left(2t\right)-cos\theta ·\hfill \\ & & cosm\left)\right)E_5+(-2a_1{a}_{3}cos\left(2t\right)-2a_2{a}_{4}cos\theta cosm+(a_2^2-a_4^2)cos\theta ·\hfill \\ & & sinm+(a_1^2-a_3^2)sin\left(2t\right))E_6,\hfill \\ \hfill P{E}_5& =& a_{3}sin\theta E_1-a_{1}sin\theta E_2+(-2a_2{a}_{4}cos\left(2t\right)-2a_1{a}_{3}cos\theta cosm+(a_1^2\hfill \\ & & -a_3^2)cos\theta sinm+(a_2^2-a_4^2)sin\left(2t\right))E_3+\left((a_1{a}_2-a_3{a}_4)\right(cos\theta sinm\hfill \\ & & -sin\left(2t\right))+(a_2{a}_3+a_1{a}_4)(cos\left(2t\right)-cos\theta cosm))E_4-((a_2^2-a_4^2)·\hfill \\ & & cos\left(2t\right)-(a_1^2-a_3^2)cos\theta cosm-2a_1{a}_{3}cos\theta sinm+2a_2{a}_{4}sin\left(2t\right))E_5\hfill \\ & & +((a_1{a}_2-a_3{a}_4)(cos\left(2t\right)+cos\theta cosm)+(a_2{a}_3+a_1{a}_4)(cos\theta sinm\hfill \\ & & +sin\left(2t\right)\left)\right)E_6,\hfill \\ \hfill P{E}_6& =& a_{4}sin\theta E_1-a_{2}sin\theta E_2+((a_1{a}_2-a_3{a}_4)(cos\theta sinm-sin\left(2t\right))-(cos\theta ·\hfill \\ & & cosm-cos\left(2t\right))(a_2{a}_3+a_1{a}_4))E_3+(-2a_1{a}_{3}cos\left(2t\right)-2a_2{a}_{4}cos\theta ·\hfill \\ & & cosm+(a_2^2-a_4^2)cos\theta sinm+(a_1^2-a_3^2)sin\left(2t\right))E_4+((a_1{a}_2-a_3{a}_4)·\hfill \\ & & (cos\left(2t\right)+cos\theta cosm)+(a_2{a}_3+a_1{a}_4)(cos\theta sinm+sin\left(2t\right)))E_5\hfill \\ & & +((-a_1^2+a_3^2)cos\left(2t\right)+(a_2^2-a_4^2)cos\theta cosm+2a_2{a}_{4}cos\theta sinm\hfill \\ & & -2a_1{a}_{3}sin\left(2t\right))E_6,\hfill \end{array}$$

for some differentiable functions $\theta ,m,a_1,a_2,a_3$ and $a_4$ such that $a_1^2+a_2^2+a_3^2+a_4^2=1$ and $\theta \in [0,{\displaystyle \frac{\pi }{2}}]$, $m\in [0,2\pi )$.

Proof. 

For an arbitrary vector field $E_1$ there is an unit vector field $F_1$ orthogonal to ${\mathcal{D}}_1$ such that the representation of $P{E}_1$ is

$$P{E}_1=cos\theta cosm{E}_1+cos\theta sinm{E}_2+sin\theta F_1.$$

Then, for $F_2=J{F}_1$ we have that

$$P{E}_2=cos\theta sinm{E}_1-cos\theta cosm{E}_2-sin\theta F_2$$

and also

$$\begin{array}{ccc}& & P{F}_1=sin\theta E_1-cos\theta cosm{F}_1+cos\theta sinm{F}_2,\hfill \\ & & P{F}_2=-sin\theta E_2+cos\theta sinm{F}_1+cos\theta cosm{F}_2.\hfill \end{array}$$

(20)

Further on, in the case when $sin\theta \ne 0$, we denote by $F_3={\displaystyle \frac{\sqrt{3}}{sin\theta }}G(E_1,P{E}_1)=\sqrt{3}G(E_1,F_1)$. Straightforward computations show that $F_3$ and $F_4=J{F}_3$ are unit vector fields, orthogonal on $E_1$, $E_2$, $F_1$, $F_2$, such that

$$P{F}_3=F_3,P{F}_4=-F_4.$$

(21)

If $sin\theta =0$ then $E_1,E_2$ are eigenvector fields for P, then in the distribution ${\mathcal{D}}_1^{\perp }$ invariant for P and J, we can choose $F_1$, $F_2=J{F}_1$, $F_3$, $F_4=J{F}_3$ such that (20) and (21) hold.

For a fixed vector field $F_1$, there exist differentiable functions $a_1$, $a_2$, $a_3$, $a_4$ such that

$$F_1=a_1{E}_3+a_2{E}_4+a_3{E}_5+a_4{E}_6$$

and we can write $a_1^2+a_2^2+a_3^2+a_4^2=1$. By a straightforward computation, we obtain

$$\begin{array}{ccc}\hfill F_2& =& -a_3{E}_3-a_4{E}_4+a_1{E}_5+a_2{E}_6,\hfill \\ \hfill F_3& =& \sqrt{3}G(E_1,F_1)=(-a_{2}cost-a_{4}sint)E_3+(a_{1}cost+a_{3}sint)E_4\hfill \\ & & +(-a_{2}sint+a_{4}cost)E_5+(a_{1}sint-a_{3}cost)E_6,\hfill \\ \hfill F_4& =& J{F}_3=(a_{2}sint-a_{4}cost)E_3+(-a_{1}sint+a_{3}cost)E_4\hfill \\ & & +(-a_{2}cost-a_{4}sint)E_5+(a_{1}cost+a_{3}sint)E_6.\hfill \end{array}$$

Now, the expressions for the tensor P in the frame $E_1$, $E_2$, $F_1$, $F_2$, $F_3$, $F_4$ are easily transformed into the expressions for the frame $E_1,\dots ,E_6$. □

By applying a rotation in the almost complex distribution ${\mathcal{D}}_1$ for an angle function $n\in [0,2\pi )$, if we take for new base vector fields $\widetilde{E_1}=cosn{E}_1+sinn{E}_2$, $\widetilde{E_2}=-sinn{E}_1+cosn{E}_2$, this rotation has no effect on the remaining base vector fields, including the vector field $E_6=J{E}_4$ and the new coefficient $h_{13}^2$ becomes

$$\begin{array}{ccc}& & \widetilde{h_{13}^2}=g({\tilde{\nabla }}_{\widetilde{E_1}}{E}_3,E_6)=cosn{h}_{13}^2+sinn{h}_{23}^2\hfill \end{array}$$

(22)

implying that the coefficient $h_{13}^2$ vanishes for an appropriate angle n. The almost product structure P can also be expressed by using the following lemma.

Lemma 6.

On an open dense subset of the four-dimensional CR submanifold M of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, we can choose the orthonormal frame for an almost complex distribution ${\mathcal{D}}_1$ and a totally real distribution ${\mathcal{D}}_2$ such that the almost product structure P in the frame $E_1,\dots ,E_6$ is given by

$$\begin{array}{ccc}\hfill P{E}_1& =& cos\theta E_1+a_{1}sinmsin\theta E_3+a_{2}sinmsin\theta E_4+cosmsin\theta E_5,\hfill \\ \hfill P{E}_2& =& -cos\theta E_2+cosmsin\theta E_3-a_{1}sinmsin\theta E_5-a_{2}sinmsin\theta E_6,\hfill \\ \hfill P{E}_3& =& a_{1}sinmsin\theta E_1+cosmsin\theta E_2+((a_2^{2}cos\left(2t\right)-a_1^{2}cos\theta ){sin}^{2}m+\hfill \\ & & {cos}^{2}mcos\theta )E_3-(a_{1}sinm(cos\theta +cos\left(2t\right))+cosmsin\left(2t\right))a_2·\hfill \\ & & sinm{E}_4+sinm(-2a_{1}cos\theta cosm+a_2^{2}sinmsin\left(2t\right))E_5\hfill \\ & & -a_{2}sinm(cosm(cos\theta -cos\left(2t\right))+a_{1}sinmsin\left(2t\right))E_6,\hfill \\ \hfill P{E}_4& =& a_{2}sinmsin\theta E_1-a_{2}sinm(a_{1}sinm(cos\left(2t\right)+cos\theta )+sin\left(2t\right)·\hfill \\ & & cosm)E_3+(a_{1}sin\left(2m\right)sin\left(2t\right)-{cos}^{2}mcos\left(2t\right)+{sin}^{2}m(a_1^{2}cos\left(2t\right)\hfill \\ & & -a_2^{2}cos\theta \left)\right)E_4-sinm(cosm(cos\theta -cos\left(2t\right))+a_{1}sinmsin\left(2t\right))·\hfill \\ & & a_2{E}_5-2(a_{1}costsinm+cosmsint)(cosmcost-a_{1}sinmsint)E_6,\hfill \\ \hfill P{E}_5& =& cosmsin\theta E_1-a_{1}sinmsin\theta E_2+sinm(a_2^{2}sinmsin\left(2t\right)-2cosm·\hfill \\ & & a_{1}cos\theta )E_3-a_{2}sinm(a_{1}sinmsin\left(2t\right)+cosm(cos\theta -cos\left(2t\right)))E_4\hfill \\ & & -({cos}^{2}mcos\theta +(a_2^{2}cos\left(2t\right)-a_1^{2}cos\theta ){sin}^{2}m)E_5\hfill \\ & & +a_{2}sinm(a_{1}sinm(cos\left(2t\right)+cos\theta )+cosmsin\left(2t\right))E_6,\hfill \\ \hfill P{E}_6& =& -a_{2}sinmsin\theta E_2-sinm(a_{1}sinmsin\left(2t\right)+cosm(cos\theta -cos\left(2t\right)))·\hfill \\ & & a_2{E}_3+(-a_{1}cos\left(2t\right)sin\left(2m\right)-{cos}^{2}msin\left(2t\right)+a_1^{2}sin\left(2t\right){sin}^{2}m)E_4\hfill \\ & & +a_{2}sinm(a_{1}sinm(cos\left(2t\right)+cos\theta )+cosmsin\left(2t\right))E_5+({cos}^{2}m·\hfill \\ & & cos\left(2t\right)-a_{1}sin\left(2m\right)sin\left(2t\right)+(a_2^{2}cos\theta -a_1^{2}cos\left(2t\right)){sin}^{2}m)E_6,\hfill \end{array}$$

for some angle functions $\theta ,m,a_1,a_2$ such that $a_1^2+a_2^2=1$ and $\theta ,m\in [0,{\displaystyle \frac{\pi }{2}}]$.

Proof. 

The function $X_{(p,q)}\mapsto g(P{X}_{(p,q)},X_{(p,q)})$ attains the maximum for a some unit vector $X_{(p,q)}\in {\mathcal{D}}_1(p,q)$ at every point $(p,q)$ of the submanifold. Since we have the freedom to rotate vector fields belonging to the distribution ${\mathcal{D}}_1$, we can assume that this vector field is $E_1$. Then the differentiable function $f\left(n\right)=g(P(cosn{E}_1+sinn{E}_2),cosn{E}_1+sinn{E}_2)$ attains the maximum for $n=0$. Moreover, the equality $f^{\prime }\left(0\right)=0$ reduces to $2g(P{E}_1,E_2)=0$. We also have that $g(P{E}_2,E_2)=-g(P{E}_1,E_1)$. Therefore, we denote by $cos\theta =g(P{E}_1,E_1)$ for $\theta \in [0,{\displaystyle \frac{\pi }{2}}]$.

Let us first assume that $sin\theta \ne 0$. Then, there exists a unit vector field $F_1$ orthogonal to ${\mathcal{D}}_1$ such that

$$P{E}_1=cos\theta E_1+sin\theta F_1.$$

Then, for $F_2=J{F}_1$ it holds that

$$P{E}_2=-cos\theta E_2-sin\theta F_2$$

and furthermore

$$P{F}_1=sin\theta E_1-cos\theta F_1,P{F}_2=-sin\theta E_2+cos\theta F_2.$$

(23)

Further on, we denote by $F_3={\displaystyle \frac{\sqrt{3}}{sin\theta }}G(E_1,P{E}_1)=\sqrt{3}G(E_1,F_1)$. Straightforward computation shows that $F_3$ and $F_4=J{F}_3$ are unit vector fields, orthogonal on $E_1$, $E_2$, $F_1$, $F_2$, such that

$$P{F}_3=F_3,P{F}_4=-F_4.$$

(24)

If $sin\theta =0$, then $E_1$, $E_2$ are eigenvector fields for P, so in the distribution ${\mathcal{D}}_1^{\perp }$ invariant for P and J we can choose $F_1$, $F_2=J{F}_1$, $F_3$, $F_4=J{F}_3$ such that (23) and (24) hold.

For the fixed vector field $F_1$, there exists the vector field $E_3$ such that a projection of $F_1$ onto the distribution ${\mathcal{D}}_3$ has the direction of the vector field $E_5$. If we denote by $cosm=g(F_1,E_5)$, for $m\in [0,{\displaystyle \frac{\pi }{2}}]$, then there exist differentiable functions $a_1,a_2$ such that $a_1^2+a_2^2=1$ and we can write

$$F_1=a_{1}sinm{E}_3+a_{2}sinm{E}_4+cosm{E}_5.$$

By a straightforward computation we obtain

$$\begin{array}{ccc}\hfill F_2& =& -cosm{E}_3+a_{1}sinm{E}_5+a_{2}sinm{E}_6,\hfill \\ \hfill F_3& =& \sqrt{3}G(E_1,F_1)=-a_{2}sinmcost{E}_3+(cosmsint+a_{1}sinmcost)E_4\hfill \\ & & -a_{2}sinmsint{E}_5+(-cosmcost+a_{1}sinmsint)E_6,\hfill \\ \hfill F_4& =& J{F}_3=a_{2}sinmsint{E}_3+(cosmcost-a_{1}sinmsint)E_4-a_{2}sinm·\hfill \\ & & cost{E}_5+(cosmsint+a_{1}sinmcost)E_6.\hfill \end{array}$$

The expression for the tensor field P in the frame $E_1,E_2,F_1,F_2,F_3,F_4$ can now be easily transformed into the expression in the frame $E_1,\dots ,E_6$. □

We have that J-invariant distributions ${\mathcal{D}}_1$, $Span\{E_3,E_5\}$ and $Span\{E_4,E_6\}$ are still J-invariant under the action of differentials of isometries ${\mathcal{F}}_{a,b,c}$, ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$. Note that a four-dimensional CR submanifold under the these isometries maps to another four-dimensional CR submanifold, as the almost complex distribution ${\mathcal{D}}_1$ and the totally real distribution ${\mathcal{D}}_2$ map to an almost complex distribution and totally real distribution on ${\mathcal{F}}_i\left(M\right)$, because there is no four-dimensional almost complex submanifold of a six-dimensional compact strict NK manifold, which was proved in [23]. In the following lemma, we will use the relations (16) to obtain the relations amongst the coefficients of the almost product structure P expressed by Lemma 6, in the case when an almost complex distribution ${\mathcal{D}}_1$ contains a vector field $E_1$ such that $P{E}_1\in {\mathcal{D}}_3$, under the action of the isometries ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$. The almost product structure P expressed by Lemma 6, for $\theta ={\displaystyle \frac{\pi }{2}}$, $m=0$ is determined by the variable t, which determines the tensor field G.

Lemma 7.

Let M be a four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ such that an almost complex distribution ${\mathcal{D}}_1$ has a vector field $E_1$ such that $P{E}_1\in {\mathcal{D}}_3$ and the base vector field $E_5$ is chosen in a way that $E_5=P{E}_1$. If we denote by $\widehat{t}\in [0,2\pi )$ and $\tilde{t}\in [0,2\pi )$, respectively, variables of the tensor field G under the isometries ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$, we find that:

$$\begin{array}{ccc}& & \widehat{t}=2\pi -t,fort\in (0,2\pi ),\hfill \\ & & \widehat{t}=0,fort=0\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \tilde{t}={\displaystyle \frac{2\pi }{3}}-t,fort\in [0,{\displaystyle \frac{2\pi }{3}}],\hfill \\ & & \tilde{t}={\displaystyle \frac{8\pi }{3}}-t,fort\in ({\displaystyle \frac{2\pi }{3}},2\pi ).\hfill \end{array}$$

Proof. 

The isometry ${\mathcal{F}}_1$. The vector field $E_1\in {\mathcal{D}}_1$ of the corresponding frame on the manifold M is chosen in a way that $P{E}_1\in {\mathcal{D}}_3$ and the base vector field $E_5$ is such that $E_5=P{E}_1$. For the rest of the base vector fields, it holds that $E_2=J{E}_1$, $E_3=-J{E}_5$ and the vector field $E_4$ is orthogonal on the vector field $E_3$ in a totally real distribution ${\mathcal{D}}_2$ and $E_6=J{E}_4$. Recall from (16) that $P\circ \mathrm{d}{\mathcal{F}}_1=\mathrm{d}{\mathcal{F}}_1\circ P$ and for the new base vector fields ${\widehat{E}}_1=\mathrm{d}{\mathcal{F}}_1\left(E_1\right)$ and ${\widehat{E}}_5=\mathrm{d}{\mathcal{F}}_1\left(E_5\right)$ on ${\mathbb{S}}^3\times {\mathbb{S}}^3$ along ${\mathcal{F}}_1\left(M\right)$ chosen in this way, we have that

$$P{\widehat{E}}_1=P\circ \mathrm{d}{\mathcal{F}}_1\left(E_1\right)=\mathrm{d}{\mathcal{F}}_1\left(P\left(E_1\right)\right)=\mathrm{d}{\mathcal{F}}_1\left(E_5\right)={\widehat{E}}_5.$$

By using the relation $\mathrm{d}{\mathcal{F}}_1\circ J=-J\circ \mathrm{d}{\mathcal{F}}_1$, we obtain ${\widehat{E}}_1=\mathrm{d}{\mathcal{F}}_1\left(E_1\right)$, ${\widehat{E}}_2=J{\widehat{E}}_1=-\mathrm{d}{\mathcal{F}}_1\left(E_2\right)$, ${\widehat{E}}_3=-\mathrm{d}{\mathcal{F}}_1\left(E_3\right)$, ${\widehat{E}}_4=\mathrm{d}{\mathcal{F}}_1\left(E_4\right)$, ${\widehat{E}}_5=J{\widehat{E}}_3=\mathrm{d}{\mathcal{F}}_1\left(E_5\right)$, ${\widehat{E}}_6=J{\widehat{E}}_4=-\mathrm{d}{\mathcal{F}}_1\left(E_6\right)$ for new base vector fields. From the expression of the tensor field G given by (18), it holds that $G({\widehat{E}}_1,{\widehat{E}}_3)={\displaystyle \frac{cos\widehat{t}}{\sqrt{3}}}{\widehat{E}}_4+{\displaystyle \frac{sin\widehat{t}}{\sqrt{3}}}{\widehat{E}}_6$. The relation $\mathrm{d}{\mathcal{F}}_1\left(G(X,Y)\right)=-G(\mathrm{d}{\mathcal{F}}_1\left(X\right),\mathrm{d}{\mathcal{F}}_1\left(Y\right)),$ for the vector fields $X=E_1$ and $Y=E_3$ implies that:

$$\begin{array}{ccc}\hfill G({\widehat{E}}_1,{\widehat{E}}_3)& =& G(d{\mathcal{F}}_1\left(E_1\right),-\mathrm{d}{\mathcal{F}}_1\left(E_3\right))=\mathrm{d}{\mathcal{F}}_1\left(G(E_1,E_3)\right)\hfill \\ & =& {\displaystyle \frac{cost}{\sqrt{3}}}\mathrm{d}{\mathcal{F}}_1\left(E_4\right)+{\displaystyle \frac{sint}{\sqrt{3}}}\mathrm{d}{\mathcal{F}}_1\left(E_6\right)={\displaystyle \frac{cost}{\sqrt{3}}}{\widehat{E}}_4-{\displaystyle \frac{sint}{\sqrt{3}}}{\widehat{E}}_6.\hfill \end{array}$$

The above expressions imply that $cos\widehat{t}=cost$ and $sin\widehat{t}=-sint$, implying further that $\widehat{t}=2\pi -t$, for $t\ne 0$ and $\widehat{t}=0$, for $t=0$. The same relations between the variables $\widehat{t}$ and t hold for other combinations of the vector fields X and Y, too.

The isometry ${\mathcal{F}}_2$. The vector field ${\tilde{E}}_1$$\in {\tilde{\mathcal{D}}}_1=d{\mathcal{F}}_2\left({\mathcal{D}}_1\right)$ of the corresponding frame is chosen in a way that $P{\tilde{E}}_1$ belongs to the distribution ${\tilde{\mathcal{D}}}_3=d{\mathcal{F}}_2\left({\mathcal{D}}_3\right)$. Then, for an angle function $\beta \in [0,2\pi )$, we have that ${\tilde{E}}_1=cos\beta \mathrm{d}{\mathcal{F}}_2\left(E_1\right)+sin\beta \mathrm{d}{\mathcal{F}}_2\left(E_2\right)$. Recall again from (16) that $P\circ \mathrm{d}{\mathcal{F}}_2=\mathrm{d}{\mathcal{F}}_2\circ (-{\displaystyle \frac{1}{2}}P+{\displaystyle \frac{\sqrt{3}}{2}}JP)$ and straightforwardly it follows that

$$\begin{array}{ccc}\hfill P{\tilde{E}}_1& =& P(cos\beta \mathrm{d}{\mathcal{F}}_2\left(E_1\right)+sin\beta \mathrm{d}{\mathcal{F}}_2\left(E_2\right))\hfill \\ & =& sin({\displaystyle \frac{4\pi }{3}}+\beta )\mathrm{d}{\mathcal{F}}_2\left(E_3\right)+cos({\displaystyle \frac{4\pi }{3}}+\beta )\mathrm{d}{\mathcal{F}}_2\left(E_5\right).\hfill \end{array}$$

In this case, the vector field $P{\tilde{E}}_1$ belongs to the distribution ${\tilde{\mathcal{D}}}_3$ for $\beta ={\displaystyle \frac{2\pi }{3}}$, so we can write

$${\tilde{E}}_1=-{\displaystyle \frac{1}{2}}\mathrm{d}{\mathcal{F}}_2\left(E_1\right)+{\displaystyle \frac{\sqrt{3}}{2}}\mathrm{d}{\mathcal{F}}_2\left(E_2\right)$$

and as $\mathrm{d}{\mathcal{F}}_2$ and J anticomutes, straightforward we have ${\tilde{E}}_2={\displaystyle \frac{\sqrt{3}}{2}}\mathrm{d}{\mathcal{F}}_2\left(E_1\right)+{\displaystyle \frac{1}{2}}\mathrm{d}{\mathcal{F}}_2\left(E_2\right)$, ${\tilde{E}}_3=-\mathrm{d}{\mathcal{F}}_2\left(E_3\right)$, ${\tilde{E}}_4=\mathrm{d}{\mathcal{F}}_2\left(E_4\right)$, ${\tilde{E}}_5=\mathrm{d}{\mathcal{F}}_2\left(E_5\right)$, ${\tilde{E}}_6=-\mathrm{d}{\mathcal{F}}_2\left(E_6\right)$. From the expression of the tensor field G given by (18) it holds that $G({\tilde{E}}_1,{\tilde{E}}_3)={\displaystyle \frac{cos\tilde{t}}{\sqrt{3}}}{\tilde{E}}_4+{\displaystyle \frac{sin\tilde{t}}{\sqrt{3}}}{\tilde{E}}_6$. Using the relation $\mathrm{d}{\mathcal{F}}_2\left(G(X,Y)\right)=-G(\mathrm{d}{\mathcal{F}}_2\left(X\right),\mathrm{d}{\mathcal{F}}_2\left(Y\right)),$ evaluated for the vector fields $(X,Y)=(E_1,E_3)$, we have that:

$$\begin{array}{ccc}\hfill G({\tilde{E}}_1,{\tilde{E}}_3)& =& G(-{\displaystyle \frac{1}{2}}\mathrm{d}{\mathcal{F}}_2\left(E_1\right)+{\displaystyle \frac{\sqrt{3}}{2}}\mathrm{d}{\mathcal{F}}_2\left(E_2\right),-\mathrm{d}{\mathcal{F}}_2\left(E_3\right))\hfill \\ & =& \mathrm{d}{\mathcal{F}}_2(-{\displaystyle \frac{1}{2}}G(E_1,E_3)+{\displaystyle \frac{\sqrt{3}}{2}}G(E_2,E_3))\hfill \\ & =& {\displaystyle \frac{1}{\sqrt{3}}}cos({\displaystyle \frac{2\pi }{3}}-t){\tilde{E}}_4+{\displaystyle \frac{1}{\sqrt{3}}}sin({\displaystyle \frac{2\pi }{3}}-t){\tilde{E}}_6.\hfill \end{array}$$

The above relation implies $cos\tilde{t}=cos({\displaystyle \frac{2\pi }{3}}-t)$ and $sin\tilde{t}=sin({\displaystyle \frac{2\pi }{3}}-t)$, implying further that $\tilde{t}={\displaystyle \frac{2\pi }{3}}-t$, for $t\in [0,{\displaystyle \frac{2\pi }{3}}]$ and $\tilde{t}={\displaystyle \frac{8\pi }{3}}-t$, for $t\in ({\displaystyle \frac{2\pi }{3}},2\pi )$. The same relations between the variables $\widehat{t}$ and t hold for other combinations of the vector fields X and Y as well. □

4. Existence of Four-Dimensional CR Submanifolds of ${\mathbb{S}}^{\mathbf{3}}\mathbf{\times }{\mathbb{S}}^{\mathbf{3}}$ with an Integrable Almost Complex Distribution ${\mathcal{D}}_{\mathbf{1}}$ Such That $\mathit{P}{\mathcal{D}}_{\mathbf{1}}\mathbf{\perp }{\mathcal{D}}_{\mathbf{1}}$

A lot of results about integrability conditions for almost complex and totally real distributions of CR submanifolds were found in [24], including the following theorem.

Theorem 2.

Let M be a CR submanifold of a nearly Kähler manifold $\tilde{M}$. Then, the almost complex distribution ${\mathcal{D}}_1$ is integrable if and only if the following conditions are satisfied $h(X,JY)=h(JX,Y)$ and $[J,J](X,Y)\in {\mathcal{D}}_1$, for any X, Y$\in {\mathcal{D}}_1$.

Now, we will prove a theorem that enables us to exclude a large subclass of four-dimensional CR submanifolds with an almost complex distribution that is P orthogonal to itself, that are submanifolds with an integrable almost complex distribution.

Theorem 3.

There is no four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ that has an integrable almost complex distribution ${\mathcal{D}}_1$ such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$.

Proof. 

For a CR submanifold that has an integrable almost complex distribution, we have from Theorem 2 that $h(E_1,J{E}_2)=h(J{E}_1,E_2)$, from which we obtain the following relations $h_{22}^1=-h_{11}^1$ and $h_{22}^2=-h_{11}^2$. In the proof, there is the use of an almost product structure P expressed by Lemma 5 for an angle $\theta ={\displaystyle \frac{\pi }{2}}$. Further, evaluating by Equation (12), we can express the derivatives of some Levi-Civita connection coefficients. If we evaluate this equation for the vector fields $(X,Y,Z)=(E_1,E_2,E_1)$, in the direction of the vector fields $E_2$, $E_3$, $E_4$, $E_5$, $E_6$, we can express $E_1\left({\Gamma }_{21}^2\right)$, $E_2\left(h_{12}^1\right)$, $E_2\left(h_{12}^2\right)$, $E_1\left(h_{12}^1\right)$, $E_1\left(h_{12}^2\right)$. From the same equation for the vector fields $(X,Y,Z)=(E_1,E_2,E_3)$, in the direction of the vector fields $E_4$, $E_5$, $E_6$, the derivatives $E_1\left({\Gamma }_{23}^4\right)$, $E_1\left(h_{23}^1\right)$, $E_1\left(h_{23}^2\right)$ are obtained. Evaluating by the equation for the vector fields $(X,Y,Z)=(E_1,E_2,E_4)$, along the vector field $E_6$, we obtain $E_1\left(h_{24}^2\right)$. Further, from the evaluation for the vector fields $(X,Y,Z)=(E_1,E_3,E_1)$, in the direction of the vector fields $E_2$, $E_3$, $E_4$, $E_5$, $E_6$, we can express $E_1\left({\Gamma }_{31}^2\right)$, $E_2\left(h_{13}^1\right)$, $E_2\left(h_{13}^2\right)$, $E_1\left(h_{13}^1\right)$, $E_1\left(h_{13}^2\right)$ and for the vector fields $(X,Y,Z)=(E_1,E_3,E_3)$ in the direction of the vector fields $E_4$, $E_5$, $E_6$, we can obtain $E_1\left({\Gamma }_{33}^4\right)$, $E_1\left(h_{33}^1\right)$, $E_1\left(h_{33}^2\right)$. The evaluation for the vector fields $(X,Y,Z)=(E_1,E_3,E_4)$, along the vector field $E_6$ implies that we can express $E_1\left(h_{34}^2\right)$. From the evaluation for the vector fields $(X,Y,Z)=(E_2,E_3,E_1)$, in the direction of the vector fields $E_2$, $E_3$, $E_4$, we can express $E_2\left({\Gamma }_{31}^2\right)$, $E_2\left(h_{23}^1\right)$, $E_2\left(h_{23}^2\right)$ and from the evaluation for the vector fields $(X,Y,Z)=(E_2,E_3,E_3)$ in the direction of the vector fields $E_4$, $E_5$, $E_6$, we obtain $E_2\left({\Gamma }_{33}^4\right)$, $E_2\left(h_{33}^1\right)$, $E_2\left(h_{33}^2\right)$. Further on, from the same equation evaluated for the vector fields $(X,Y,Z)=(E_2,E_3,E_4)$, along the vector field $E_6$, we obtain $E_2\left(h_{34}^2\right)$. From the evaluation for the vector fields $(X,Y,Z)=(E_1,E_4,E_1)$, in the direction of the vector fields $E_2$, $E_3$, $E_4$, $E_5$, $E_6$, the derivatives $E_1\left({\Gamma }_{41}^2\right)$, $E_3\left(h_{12}^2\right)$, $E_2\left(h_{14}^2\right)$, $E_3\left(h_{11}^2\right)$, $E_1\left(h_{14}^2\right)$ could also be expressed, for the vector fields $(X,Y,Z)=(E_1,E_4,E_3)$, in the direction of the vector fields $E_4$, $E_5$, $E_6$, the derivatives $E_1\left({\Gamma }_{43}^4\right)$, $E_3\left(h_{13}^2\right)$, $E_3\left(h_{14}^2\right)$ and from the evaluation for the vector fields $(X,Y,Z)=(E_1,E_4,E_4)$, in the direction of the vector field $E_6$, we can express $E_1\left(h_{44}^2\right)$. From the same equation, which is evaluated for the vector fields $(X,Y,Z)=(E_2,E_4,E_1)$, along the vector fields $E_2$, $E_4$, the derivatives $E_2\left({\Gamma }_{41}^2\right)$, $E_2\left(h_{24}^2\right)$ can be obtained. Furthermore, for the vector fields $(X,Y,Z)=(E_2,E_4,E_3)$, along the vector fields $E_4$, $E_5$, $E_6$, the derivatives $E_2\left({\Gamma }_{43}^4\right)$, $E_3\left(h_{23}^2\right)$, $E_3\left(h_{24}^2\right)$ and for the vector fields $(X,Y,Z)=(E_2,E_4,E_4)$, along the vector field $E_6$, we can express $E_2\left(h_{44}^2\right)$. From the evaluation for the vector fields $(X,Y,Z)=(E_3,E_4,E_1)$, along the vector field $E_2$, we can express $E_3\left({\Gamma }_{41}^2\right)$ and from the evaluation for the vector fields $(X,Y,Z)=(E_3,E_4,E_3)$, along the vector fields $E_4$, $E_5$, $E_6$, we obtain $E_3\left({\Gamma }_{43}^4\right)$, $E_3\left(h_{33}^2\right)$, $E_3\left(h_{34}^2\right)$, and finally, from this equation, which is evaluated for the vector fields $(X,Y,Z)=(E_3,E_4,E_4)$, along the vector field $E_6$, we can express $E_3\left(h_{44}^2\right)$. Further, from Equation (12) evaluated for vector fields $(X,Y,Z)=(E_2,E_4,E_1)$, in the direction of vector field $E_3$, we obtain the equation

$$\begin{array}{ccc}\hfill & & 6(h_{11}^2{h}_{12}^1-h_{11}^1{h}_{12}^2+h_{13}^2{h}_{23}^1-h_{13}^1{h}_{23}^2+h_{14}^2{h}_{23}^2-h_{13}^2{h}_{24}^2)+2(a_2{a}_3\hfill \\ & & -a_1{a}_4)-\sqrt{3}(3h_{23}^1+h_{24}^2)cost+\sqrt{3}(3h_{13}^1+h_{14}^2)sint=0.\hfill \end{array}$$

(25)

From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_2,E_1),(E_1,E_1),(E_3,E_1),(E_3,E_1),(E_4,E_1),(E_4,E_1)\}$, along vector fields $E_1$, $E_1$, $E_1$, $E_2$, $E_1$, $E_2$, the equations

$$\begin{array}{ccc}& & h_{12}^1{a}_1+h_{12}^2{a}_2+h_{11}^1{a}_3+h_{11}^2{a}_4=0,\hfill \end{array}$$

(26)

$$\begin{array}{ccc}& & h_{11}^1{a}_1+h_{11}^2{a}_2-h_{12}^1{a}_3-h_{12}^2{a}_4=0,\hfill \end{array}$$

(27)

$$\begin{array}{ccc}& & -4a_1{h}_{13}^1-4a_2{h}_{13}^2+4a_3{h}_{23}^1+4a_4{h}_{23}^2-{\displaystyle \frac{2}{\sqrt{3}}}{a}_{2}cost+{\displaystyle \frac{2}{\sqrt{3}}}{a}_{4}sint=0,\hfill \end{array}$$

(28)

$$\begin{array}{ccc}& & 6a_1{h}_{23}^1+6a_2{h}_{23}^2+6a_3{h}_{13}^1+6a_4{h}_{13}^2+\sqrt{3}{a}_{4}cost+\sqrt{3}{a}_{2}sint=0,\hfill \end{array}$$

(29)

$$\begin{array}{ccc}& & -4a_1{h}_{13}^2-4a_2{h}_{14}^2+4a_3{h}_{23}^2+4a_4{h}_{24}^2+2\sqrt{3}{a}_{1}cost-2\sqrt{3}{a}_{3}sint=0,\hfill \end{array}$$

(30)

$$\begin{array}{ccc}& & -2a_3{h}_{13}^2-2a_4{h}_{14}^2-2a_1{h}_{23}^2-2a_2{h}_{24}^2+\sqrt{3}{a}_{3}cost+\sqrt{3}{a}_{1}sint=0\hfill \end{array}$$

(31)

are directly obtained and solution of the system of Equations (28) and (29), in the case when $a_1^2+a_3^2\ne 0$, and solution of the system of Equations (30) and (31), in the case when $a_2^2+a_4^2\ne 0$, is

$$\begin{array}{ccc}& & h_{13}^1=-{\displaystyle \frac{(6h_{13}^2+\sqrt{3}cost)(a_1{a}_2+a_3{a}_4)+(a_2{a}_3-a_1{a}_4)(6h_{23}^2+\sqrt{3}sint)}{6(a_1^2+a_3^2)}},\hfill \\ & & h_{23}^1={\displaystyle \frac{(6h_{13}^2+\sqrt{3}cost)(a_2{a}_3-a_1{a}_4)-(a_1{a}_2+a_3{a}_4)(6h_{23}^2+\sqrt{3}sint)}{6(a_1^2+a_3^2)}},\hfill \\ & & h_{14}^2={\displaystyle \frac{(-2h_{13}^2+\sqrt{3}cost)(a_1{a}_2+a_3{a}_4)+(a_2{a}_3-a_1{a}_4)(2h_{23}^2-\sqrt{3}sint)}{2(a_2^2+a_4^2)}},\hfill \\ & & h_{24}^2={\displaystyle \frac{(-2h_{13}^2+\sqrt{3}cost)(a_2{a}_3-a_1{a}_4)-(a_1{a}_2+a_3{a}_4)(2h_{23}^2-\sqrt{3}sint)}{2(a_2^2+a_4^2)}}.\hfill \end{array}$$

(32)

Note that cases $a_1^2+a_3^2=0$ and $a_2^2+a_4^2=0$, from a system of Equations (26) and (27), imply that ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$ and ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2=0$, respectively, which will be covered by the following three side cases: ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2=0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$; ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2\ne 0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$; ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2=0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2\ne 0$.

Let us look at the main case now, when ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2\ne 0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2\ne 0$. If we suppose that $a_2{a}_3-a_1{a}_4=0$, applying rotation in an almost complex distribution ${\mathcal{D}}_1$, we can suppose that $a_4=0$, implying further that $a_3=0$, which after a few steps implies a contradiction. Let us now assume that $a_2{a}_3-a_1{a}_4\ne 0$. We still have freedom as regards rotation in an almost complex distribution ${\mathcal{D}}_1$ and the expression (22) implies that we can use it to vanish coefficient $h_{13}^2$. For the expressions given by (32), Equation (25) implies the following new equation

$$\begin{array}{ccc}\hfill & & 12(h_{11}^2{h}_{12}^1-h_{11}^1{h}_{12}^2)(a_1^2+a_3^2)(a_2^2+a_4^2)-(a_2{a}_3-a_1{a}_4)·\hfill \\ & & (3-12{\left(h_{23}^2\right)}^2-4(a_1^2+a_3^2)(a_2^2+a_4^2))+4\sqrt{3}{h}_{23}^2·\hfill \\ & & (2cost(a_1{a}_2+a_3{a}_4)-(a_2{a}_3-a_1{a}_4)sint)=0.\hfill \end{array}$$

(33)

As we already suppose that ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2\ne 0$, by solving the system of Equations (26) and (27) by variables $a_1$, $a_3$ we have that

$$\begin{array}{ccc}& & a_1=-{\displaystyle \frac{(h_{11}^1{h}_{11}^2+h_{12}^1{h}_{12}^2)a_2+(h_{11}^2{h}_{12}^1-h_{11}^1{h}_{12}^2)a_4}{{\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2}},\hfill \\ & & a_3=-{\displaystyle \frac{(h_{11}^1{h}_{12}^2-h_{11}^2{h}_{12}^1)a_2+(h_{11}^1{h}_{11}^2+h_{12}^1{h}_{12}^2)a_4}{{\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2}}.\hfill \end{array}$$

(34)

Also, for ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2\ne 0$, a solution of the system of Equations (26) and (27) by variables $a_2$, $a_4$ is

$$\begin{array}{ccc}& & a_2=-{\displaystyle \frac{(h_{11}^1{h}_{11}^2+h_{12}^1{h}_{12}^2)a_1+(h_{11}^1{h}_{12}^2-h_{11}^2{h}_{12}^1)a_3}{{\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2}},\hfill \\ & & a_4=-{\displaystyle \frac{(h_{11}^2{h}_{12}^1-h_{11}^1{h}_{12}^2)a_1+(h_{11}^1{h}_{11}^2+h_{12}^1{h}_{12}^2)a_3}{{\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2}}.\hfill \end{array}$$

(35)

If we start from the expression $a_2{a}_3-a_1{a}_4$, by using the expressions (34) and (35), respectively, these relations further imply that

$$\begin{array}{ccc}\hfill (h_{11}^2{h}_{12}^1-h_{11}^1{h}_{12}^2)(a_2^2+a_4^2)& =& (a_2{a}_3-a_1{a}_4)({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2),\hfill \\ \hfill (h_{11}^2{h}_{12}^1-h_{11}^1{h}_{12}^2)(a_1^2+a_3^2)& =& (a_2{a}_3-a_1{a}_4)({\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2)\hfill \end{array}$$

and if we use these two relations in Equation (33), we respectively obtain two new equations

$$\begin{array}{ccc}& & 8\sqrt{3}cost{h}_{23}^2(a_1{a}_2+a_3{a}_4)+(a_2{a}_3-a_1{a}_4)(-3+12{\left(h_{23}^2\right)}^2\hfill \\ & & -4\sqrt{3}sint{h}_{23}^2+4(a_1^2+a_3^2)(3{\left(h_{11}^1\right)}^2+3{\left(h_{12}^1\right)}^2+a_2^2+a_4^2))=0,\hfill \end{array}$$

(36)

$$\begin{array}{ccc}& & 8\sqrt{3}cost{h}_{23}^2(a_1{a}_2+a_3{a}_4)+(a_2{a}_3-a_1{a}_4)(-3+12{\left(h_{23}^2\right)}^2\hfill \\ & & -4\sqrt{3}sint{h}_{23}^2+4(a_2^2+a_4^2)(3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+a_1^2+a_3^2))=0,\hfill \end{array}$$

(37)

that have the same variables $a_1^2+a_3^2$ and $a_2^2+a_4^2$ connected with the relation $a_1^2+a_3^2+a_2^2+a_4^2=1$. If we solve the quadratic Equation (36) by variable $a_2^2+a_4^2$, we obtain the following two solutions

$$\begin{array}{ccc}& & a_2^2+a_4^2={\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2+{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}}),\hfill \\ & & a_2^2+a_4^2={\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2-{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}}),\hfill \end{array}$$

where $de{t}_3=(a_2{a}_3-a_1{a}_4)(8\sqrt{3}cost{h}_{23}^2(a_1{a}_2+a_3{a}_4)+(a_2{a}_3-a_1{a}_4)({(3{\left(h_{11}^1\right)}^2+3{\left(h_{12}^1\right)}^2+1)}^2-3$  $+12{\left(h_{23}^2\right)}^2-4\sqrt{3}sint{h}_{23}^2))$. Further on, solutions of the quadratic Equation (37) by a variable $a_2^2+a_4^2$ are

$$\begin{array}{ccc}& & a_2^2+a_4^2={\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}}),\hfill \\ & & a_2^2+a_4^2={\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2-{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}}),\hfill \end{array}$$

where $de{t}_4=(a_2{a}_3-a_1{a}_4)(8\sqrt{3}cost{h}_{23}^2(a_1{a}_2+a_3{a}_4)+(a_2{a}_3-a_1{a}_4)({(3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+1)}^2-3+12{\left(h_{23}^2\right)}^2-$$4\sqrt{3}sint{h}_{23}^2))$. If $a_2{a}_3-a_1{a}_4>0$, then ${\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2+{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}})>{\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2-{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}})$ and ${\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}})>{\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2-{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}})$. Also, if $a_2{a}_3-a_1{a}_4<0$, then ${\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2+{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}})<{\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2-{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}})$ and ${\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}})<{\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2-{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}})$. These solutions are actually the solutions of equivalent Equations (36) and (37) obtained from the same Equation (33) and as we cannot lose solutions, all four solutions are a solution of (33) and they should match. An appropriate solution must be the same, so from the equations ${\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2+{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}})={\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}})$ and ${\displaystyle \frac{1}{2}}(1-3{\left(h_{11}^1\right)}^2-3{\left(h_{12}^1\right)}^2-{\displaystyle \frac{\sqrt{de{t}_3}}{a_2{a}_3-a_1{a}_4}})={\displaystyle \frac{1}{2}}(1+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2-{\displaystyle \frac{\sqrt{de{t}_4}}{a_2{a}_3-a_1{a}_4}})$, we obtain the following two equations, ${\displaystyle \frac{1}{a_2{a}_3-a_1{a}_4}}\sqrt{de{t}_3}=3{\left(h_{11}^1\right)}^2+3{\left(h_{12}^1\right)}^2+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+{\displaystyle \frac{1}{a_2{a}_3-a_1{a}_4}}\sqrt{de{t}_4}$ and $3{\left(h_{11}^1\right)}^2+3{\left(h_{12}^1\right)}^2+3{\left(h_{11}^2\right)}^2+3{\left(h_{12}^2\right)}^2+{\displaystyle \frac{1}{a_2{a}_3-a_1{a}_4}}\sqrt{de{t}_3}={\displaystyle \frac{1}{a_2{a}_3-a_1{a}_4}}\sqrt{de{t}_4}$, that imply ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2+{\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$, which is a contradiction with respect to the assumption that ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2\ne 0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2\ne 0$.

Case ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2=0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$. We still have freedom as regards rotation in an almost complex distribution ${\mathcal{D}}_1$ and the expression (22) implies that we can use it to vanish coefficient $h_{13}^2$. By evaluating Equation (12) for the vector fields $(X,Y,Z)=(E_1,E_3,E_1)$, along vector fields $E_4$, $E_6$, we obtain the equations

$$\begin{array}{ccc}& & -(h_{13}^1+h_{14}^2)cost+(h_{23}^1+h_{24}^2)sint=0,\hfill \end{array}$$

(38)

$$\begin{array}{ccc}& & 3({\Gamma }_{13}^4(h_{13}^1-h_{14}^2)+h_{23}^2(2h_{14}^2-{\Gamma }_{11}^2))+\sqrt{3}sint(h_{13}^1+2h_{14}^2)\hfill \\ & & +a_2{a}_3-a_1{a}_4=0.\hfill \end{array}$$

(39)

Also, by evaluating Equation (12) for the vector fields $(X,Y,Z)=(E_1,E_4,E_1)$ in the direction of vector field $E_5$ and by evaluating the same equation for the vector fields $(X,Y,Z)=(E_2,E_4,E_1)$, in the direction of $E_3$, we obtain the following equations

$$\begin{array}{ccc}& & 3({\Gamma }_{13}^4(h_{13}^1-h_{14}^2)+h_{23}^2(2h_{13}^1-{\Gamma }_{11}^2))-\sqrt{3}sint(5h_{13}^1+2h_{14}^2)\hfill \\ & & -2\sqrt{3}cost{h}_{24}^2-a_2{a}_3+a_1{a}_4=0,\hfill \end{array}$$

(40)

$$\begin{array}{ccc}& & 6h_{23}^2(h_{13}^1-h_{14}^2)+\sqrt{3}(3h_{23}^1+h_{24}^2)cost-\sqrt{3}(3h_{13}^1+h_{14}^2)sint\hfill \\ & & -2a_2{a}_3+2a_1{a}_4=0,\hfill \end{array}$$

(41)

which further imply that

$$\begin{array}{ccc}& & 0=\left(39\right)-\left(40\right)=2(-3h_{23}^2(h_{13}^1-h_{14}^2)+\sqrt{3}(3h_{13}^1+2h_{14}^2)sint\hfill \\ & & +\sqrt{3}cost{h}_{24}^2+a_2{a}_3-a_1{a}_4)\hfill \end{array}$$

(42)

and

$$\begin{array}{ccc}& & 0=\left(41\right)+\left(42\right)=3\sqrt{3}((h_{23}^1+h_{24}^2)cost+(h_{13}^1+h_{14}^2)sint).\hfill \end{array}$$

(43)

Finally, from the homogenous system of Equations (38) and (43), we obtain that sums $h_{13}^1+h_{14}^2$ and $h_{23}^1+h_{24}^2$ vanish, implying that $h_{14}^2=-h_{13}^1$ and $h_{24}^2=-h_{23}^1$, and for these values Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_2,E_4)$, in the direction of vector field $E_6$, implies a contradiction $a_1^2+a_2^2+a_3^2+a_4^2=0$.

Case ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2\ne 0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$. We still have freedom as regards rotation in an almost complex distribution ${\mathcal{D}}_1$ and the expression (22) implies that we can use it to vanish coefficient $h_{13}^2$. From the homogenous system of Equations (26) and (27), which the determinant $-{\left(h_{11}^1\right)}^2-{\left(h_{12}^1\right)}^2$ does not vanish, we have that $a_1=0$ and $a_3=0$. For this values, the system of Equations (28) and (29) and the system of Equations (30) and (31), with non vanishing determinants, have the solutions $h_{23}^2=-{\displaystyle \frac{sint}{2\sqrt{3}}}$, $cost=0$ and $h_{14}^2=0$, $h_{24}^2=0$. Now, we will split the proof into a two cases.

Case $t={\displaystyle \frac{\pi }{2}}$. By evaluating Equation (12) for the vector fields $(X,Y,Z)=(E_1,E_2,E_6)$, in the direction of vector fields $E_4$, $E_5$, we obtain directly that ${\Gamma }_{13}^4=-{\displaystyle \frac{\sqrt{3}}{2}}$ and ${\Gamma }_{11}^2=-h_{13}^1$. By evaluating Equation (12) for the vector fields $(X,Y,Z)=(E_1,E_2,E_6)$, in the direction of vector fields $E_1$, $E_2$, $E_3$, we obtain the following relations ${\Gamma }_{23}^4{h}_{11}^1=0$, ${\Gamma }_{23}^4{h}_{12}^1=0$ and $6{\Gamma }_{23}^4{h}_{13}^1+\sqrt{3}({\Gamma }_{21}^2+h_{23}^1)=0$. As the coefficients $h_{11}^1$ and $h_{12}^1$ cannot vanish at the same time, we obtain that ${\Gamma }_{23}^4=0$, which further implies ${\Gamma }_{21}^2=-h_{23}^1$. From the Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_3,E_4)$, in the direction of vector fields $E_5$, $E_2$, $E_1$, we obtain the following relation ${\Gamma }_{31}^2=-h_{33}^1-h_{34}^2$ and the system of equations

$$\begin{array}{ccc}& & {\Gamma }_{33}^4{h}_{11}^1+{\displaystyle \frac{1}{\sqrt{3}}}{h}_{13}^1+h_{12}^1{h}_{33}^2=0,\hfill \\ & & {\Gamma }_{33}^4{h}_{12}^1+{\displaystyle \frac{1}{\sqrt{3}}}{h}_{23}^1-h_{11}^1{h}_{33}^2=0.\hfill \end{array}$$

(44)

Further, from Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_4,E_6)$, in the direction of vector fields $E_1$, $E_2$, we obtain the following system of equations

$$\begin{array}{ccc}& & {\Gamma }_{43}^4{h}_{11}^1+h_{12}^1{h}_{34}^2=0,\hfill \\ & & 3{\Gamma }_{43}^4{h}_{12}^1-3h_{11}^1{h}_{34}^2-2=0.\hfill \end{array}$$

(45)

An unique solutions of systems (44) and (45) are ${\Gamma }_{33}^4={\displaystyle \frac{-h_{11}^1{h}_{13}^1-h_{12}^1{h}_{23}^1}{\sqrt{3}({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}$, $h_{33}^2={\displaystyle \frac{-h_{12}^1{h}_{13}^1+h_{11}^1{h}_{23}^1}{\sqrt{3}({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}$ and ${\Gamma }_{43}^4={\displaystyle \frac{2h_{12}^1}{3({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}$, $h_{34}^2=-{\displaystyle \frac{2h_{11}^1}{3({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}$. By using these expressions in the equation obtained from Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_3,E_4,E_1)$, in the direction of vector field $E_6$, we have that ${\Gamma }_{41}^2={\displaystyle \frac{h_{12}^1{h}_{13}^1-h_{11}^1{h}_{23}^1}{\sqrt{3}({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}-h_{44}^2$. From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_1),(E_2,E_1),(E_3,E_1),(E_4,E_1)\}$ in the direction of vector fields $E_4$, $E_6$ for all of the chosen combinations of $(X,Y)$, we obtain the following derivatives $E_1\left(a_4\right)=h_{13}^1{a}_2$, $E_1\left(a_2\right)=-h_{13}^1{a}_4$; $E_2\left(a_4\right)=h_{23}^1{a}_2$, $E_2\left(a_2\right)=-h_{23}^1{a}_4$; $E_3\left(a_4\right)=h_{33}^1{a}_2$, $E_3\left(a_2\right)=-h_{33}^1{a}_4$; $E_4\left(a_4\right)={\displaystyle \frac{a_2(h_{11}^1{h}_{23}^1-h_{12}^1{h}_{13}^1)}{\sqrt{3}({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}$, $E_4\left(a_2\right)={\displaystyle \frac{a_4(h_{12}^1{h}_{13}^1-h_{11}^1{h}_{23}^1)}{\sqrt{3}({\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2)}}$ and from Equation (10) evaluated for $(X,Y)=(E_1,E_1)$ in the direction of vector fields $E_3$, $E_5$, we obtain the following system

$$\begin{array}{ccc}& & 2(a_2^2-a_4^2)h_{11}^1+4h_{12}^1{a}_2{a}_4-{\displaystyle \frac{2}{\sqrt{3}}}{a}_4=0,\hfill \\ & & 2(a_2^2-a_4^2)h_{12}^1-4h_{11}^1{a}_2{a}_4+{\displaystyle \frac{2}{\sqrt{3}}}{a}_2=0,\hfill \end{array}$$

(46)

of which a unique solution is $h_{11}^1=-{\displaystyle \frac{1}{\sqrt{3}}}{a}_4(a_4^2-3a_2^2)$, $h_{12}^1=-{\displaystyle \frac{1}{\sqrt{3}}}{a}_2(a_2^2-3a_4^2)$. Further, by evaluating Equation (12) for the vector fields $(X,Y,Z)=(E_2,E_4,E_3)$, in the direction of vector fields $E_6$, $E_1$, we obtain that $h_{23}^1{a}_2(a_2^2-3a_4^2)=0$ and $h_{44}^2{a}_2(a_2^2-3a_4^2)=0$. With respect to these relations, the following cases will be examined: the case of $a_2=0$, the case $a_2\ne 0$ and $a_2^2-3a_4^2=0$ and the case when $a_2\ne 0$, $a_2^2-3a_4^2\ne 0$ and $h_{23}^1=0$, $h_{44}^2=0$.

Case $a_2=0$. In this case, it holds that $a_4=\pm 1$ and from the system of equations given by (46), it follows that $h_{12}^1=0$ and $h_{11}^1=-{\displaystyle \frac{1}{\sqrt{3}{a}_4}}$. From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_1),(E_2,E_1),(E_3,E_1)\}$, in the direction of vector field $E_6$, we directly obtain that $h_{13}^1=0$, $h_{23}^1=0$ and $h_{33}^1=0$. If we evaluate Equation (12) for $(X,Y,Z)=(E_1,E_3,E_1)$, in the direction of vector field $E_3$, we obtain a contradiction $0={\displaystyle \frac{1}{3}}(-7+{\displaystyle \frac{1}{a_4^2}})=-2$.

Case $a_2^2-3a_4^2=0$ and $a_2\ne 0$. By using the relation $a_2^2+a_4^2=1$, we obtain that $a_2=\pm {\displaystyle \frac{\sqrt{3}}{2}}$ and $a_4=\pm {\displaystyle \frac{1}{2}}$ and for any combination of $a_2$ and $a_4$ by evaluating the Equation (10) for the vector fields $(X,Y)\in \{(E_1,E_4),(E_2,E_4),(E_3,E_4)\}$, in the direction of vector field $E_1$, we obtain that $h_{13}^1=0$, $h_{23}^1=0$ and $h_{33}^1=0$ and from Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_3,E_1)$, in the direction of vector field $E_3$, we obtain a contradiction $-2=0$.

Case $h_{23}^1=0$, $h_{44}^2=0$ and $a_2^2-3a_4^2\ne 0$, $a_2\ne 0$. By evaluating Equation (12) for the vector fields $(X,Y,Z)=(E_2,E_4,E_4)$, in the direction of $E_6$, we obtain a contradiction with respect to a condition $a_2(a_2^2-3a_4^2)\ne 0$.

Case $t={\displaystyle \frac{3\pi }{2}}$. In this case, a contradiction is obtained by using the same procedure as in the previous case.

Case ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2=0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2\ne 0$ is solved in a similar way as the previous case when ${\left(h_{11}^1\right)}^2+{\left(h_{12}^1\right)}^2\ne 0$ and ${\left(h_{11}^2\right)}^2+{\left(h_{12}^2\right)}^2=0$.

The conclusion is that there are no four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with an integrable almost complex distribution ${\mathcal{D}}_1$ for which it holds that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$. □

We can conclude that the expressions $h_{11}^1+h_{22}^1$ and $h_{11}^2+h_{22}^2$ cannot vanish at the same time.

5. Some Special Types of Four-Dimensional CR Submanifolds with Respect to $\mathit{P}{\mathit{E}}_{\mathbf{1}}$

In order to classify four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$, we should first look at some special cases with respect to the position of $P{E}_1$, for some arbitrary vector field $E_1\in {\mathcal{D}}_1$.

Theorem 4.

Let M be a four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ such that an almost complex distribution ${\mathcal{D}}_1$ has an arbitrary vector field $E_1$ such that $P{E}_1\in {\mathcal{D}}_3$ holds; then, M is a locally product manifold of a curve γ, which is an integral curve of the vector field $E_3$ belonging totally real distribution, such that $J{E}_3=P{E}_1$ and the three-dimensional CR submanifold $M^3$ of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, with a non-integrable almost complex distribution ${\mathcal{D}}_1$, which is P-orthogonal on itself. The three-dimensional submanifold $M^3$ and the curve γ are respectively locally congruent to one of the following three immersions $f_i$ and ${\gamma }_i$, $i\in \overline{1,3}$:

$$\begin{array}{ccc}\hfill f_1& =& (u,u{\displaystyle \frac{\sqrt{3}+i}{2}}),\hfill \\ \hfill {\gamma }_1& =& \left(a\left(x\right)\right({\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1+i)+{\displaystyle \frac{1}{4}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))·\hfill \\ & & ((\sqrt{3}-1)j+(\sqrt{3}+1)k))\overline{h}\left(x\right),b\left(x\right)({\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1+i)\hfill \\ & & +{\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(-j+k)\left)\overline{h}\left(x\right)\right);\hfill \end{array}$$

(47)

$$\begin{array}{ccc}\hfill f_2& =& (u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{u},\overline{u}),\hfill \\ \hfill {\gamma }_2& =& \left(b\left(x\right)\right({\displaystyle \frac{1}{4}}(3+sin\left(\sqrt{2}x\right))+{\displaystyle \frac{\sqrt{3}}{4}}(1-sin\left(\sqrt{2}x\right))i-{\displaystyle \frac{1}{4}}cos\left(\sqrt{2}x\right)j\hfill \\ & & -{\displaystyle \frac{\sqrt{3}}{4}}cos\left(\sqrt{2}x\right)k)\overline{a}\left(x\right),h\left(x\right)({\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1-i)\hfill \\ & & +{\displaystyle \frac{1}{4}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))((1-\sqrt{3})j-(\sqrt{3}+1)k)\left)\overline{a}\left(x\right)\right);\hfill \end{array}$$

(48)

$$\begin{array}{ccc}\hfill f_3& =& ({\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u},u{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}),\hfill \\ \hfill {\gamma }_3& =& \left(h\left(x\right)\right({\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1-i)+{\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))·\hfill \\ & & (j-k))\overline{b}\left(x\right),a\left(x\right)({\displaystyle \frac{1}{4}}(3+sin\left(\sqrt{2}x\right))-{\displaystyle \frac{\sqrt{3}}{4}}(1-sin\left(\sqrt{2}x\right))i\hfill \\ & & +{\displaystyle \frac{1}{4}}cos\left(\sqrt{2}x\right)j+{\displaystyle \frac{\sqrt{3}}{4}}cos\left(\sqrt{2}x\right)k\left)\overline{b}\left(x\right)\right);\hfill \end{array}$$

(49)

where $u\in {\mathbb{S}}^3$, ${\mathbb{S}}^3$ is Berger sphere with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$ and $a\left(x\right)$, $b\left(x\right)$ and $h\left(x\right)$ are unit quaternion curves.

Proof. 

For a vector field $E_5$ that belongs to the normal distribution ${\mathcal{D}}_3$, the vector field $P{E}_1$ could be chosen due to the condition of $P{E}_1\in {\mathcal{D}}_3$. Other relations between vector fields that form an orthonormal moving frame are the same. In the proof, there is the use of an appropriate almost product structure P expressed by Lemma 6 for $\theta ={\displaystyle \frac{\pi }{2}}$, $m=0$. For these values, an almost product structure P is expressed as $P{E}_1=E_5$, $P{E}_2=E_3$, $P{E}_3=E_2$, $P{E}_4=-cos\left(2t\right)E_4-sin\left(2t\right)E_6$, $P{E}_5=E_1$ and $P{E}_6=-sin\left(2t\right)E_4+cos\left(2t\right)E_6$. From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_1),(E_3,E_1),(E_4,E_1)\}$, along vector fields $E_1$, $E_2$, $E_5$ for all of the mentioned combinations of $(X,Y)$, we obtain that $h_{12}^1=0$, $h_{11}^1=0$, ${\Gamma }_{11}^2=-h_{13}^1$; $h_{23}^1=0$, $h_{13}^1=0$, ${\Gamma }_{31}^2=-h_{33}^1$; $h_{23}^2={\displaystyle \frac{\sqrt{3}}{2}}sint$, $h_{13}^2={\displaystyle \frac{\sqrt{3}}{2}}cost$, ${\Gamma }_{41}^2=-h_{33}^2$ and evaluating by the same equation for the vector fields $(X,Y)=(E_2,E_1)$, along vector fields $E_1$ and $E_5$, we obtain that $h_{22}^1=0$, ${\Gamma }_{21}^2=0$. By evaluating Equation (10) for the vector fields $(X,Y)\in \{(E_3,E_1),(E_4,E_1),(E_1,E_1)\}$, in the direction of vector fields $E_4,E_6$ for all of the previously mentioned combinations, we obtain the following relations ${\Gamma }_{33}^4={\displaystyle \frac{2}{\sqrt{3}}}cos\left(3t\right)$, $h_{33}^2={\displaystyle \frac{2}{\sqrt{3}}}sin\left(3t\right)$; ${\Gamma }_{43}^4=h_{14}^{2}cos\left(2t\right)-h_{24}^{2}sin\left(2t\right)$, $h_{34}^2=h_{14}^{2}sin\left(2t\right)+h_{24}^{2}cos\left(2t\right)$ and a system

$$\begin{array}{ccc}& & 2h_{11}^{2}cos\left(2t\right)-2h_{12}^{2}sin\left(2t\right)-{\displaystyle \frac{1}{\sqrt{3}}}sint-2{\Gamma }_{13}^4=0,\hfill \end{array}$$

$$\begin{array}{ccc}& & 2h_{12}^{2}cos\left(2t\right)+2h_{11}^{2}sin\left(2t\right)-{\displaystyle \frac{2}{\sqrt{3}}}cost=0,\hfill \end{array}$$

with an unique solution $h_{11}^2={\Gamma }_{13}^{4}cos\left(2t\right)+{\displaystyle \frac{1}{4\sqrt{3}}}(sint+3sin\left(3t\right))$ and $h_{12}^2=-{\displaystyle \frac{cost}{6}}(\sqrt{3}-3\sqrt{3}cos\left(2t\right)+12{\Gamma }_{13}^{4}sint)$. Finally, by evaluating Equation (10) for the vector fields $(X,Y)=(E_2,E_1)$, in the direction of vector fields $E_4,E_6$, we obtain the equations

$$\begin{array}{ccc}& & 2({\Gamma }_{23}^4+h_{22}^{2}sin\left(2t\right))+{\Gamma }_{13}^{4}sin\left(4t\right)-{\displaystyle \frac{3cos\left(5t\right)+cos\left(3t\right)+8cost}{4\sqrt{3}}}=0,\hfill \end{array}$$

(50)

$$\begin{array}{ccc}& & {\Gamma }_{13}^4(cos\left(4t\right)-1)+2h_{22}^{2}cos\left(2t\right)+{\displaystyle \frac{3sin\left(5t\right)+sin\left(3t\right)-10sint}{4\sqrt{3}}}=0.\hfill \end{array}$$

(51)

Case $cos\left(2t\right)\ne 0$. From Equation (51) we can express $h_{22}^2=-{\displaystyle \frac{sec\left(2t\right)}{24}}(12{\Gamma }_{13}^4(cos\left(4t\right)-1)+\sqrt{3}(3sin\left(5t\right)+sin\left(3t\right)-10sint))$ and from Equation (50) we have that ${\Gamma }_{23}^4={\displaystyle \frac{\sqrt{3}}{2}}cos\left(3t\right)sec\left(2t\right)-{\Gamma }_{13}^{4}tan\left(2t\right)$. From Equation (12), derivatives of some Levi-Civita connection coefficients could be expressed. Evaluating by Equation (12) for the vector fields $(X,Y,Z)\in \{(E_1,E_2,E_4),(E_1,E_3,E_1),(E_1,E_3,E_1),(E_1,E_3,E_4),(E_2,E_3,E_1),(E_2,E_3,E_4),$$(E_1,E_4,E_1),(E_1,E_4,E_1),(E_1,E_4,E_4),(E_2,E_4,E_1),(E_2,E_4,E_4),(E_3,E_4,E_1),(E_3,E_4,E_4)\}$, in the direction of vector fields $E_6$, $E_2$, $E_6$, $E_6$, $E_2$, $E_6$, $E_4$, $E_6$, $E_6$, $E_4$, $E_6$, $E_2$, $E_6$, the following derivatives $E_1\left(h_{24}^2\right)$, $E_1\left(h_{33}^1\right)$, $E_3\left({\Gamma }_{13}^4\right)$, $E_3\left(h_{14}^2\right)$, $E_2\left(h_{33}^1\right)$, $E_3\left(h_{24}^2\right)$, $E_2\left(h_{14}^2\right)$, $E_1\left(h_{14}^2\right)$, $E_1\left(h_{44}^2\right)$, $E_2\left(h_{24}^2\right)$, $E_2\left(h_{44}^2\right)$, $E_4\left(h_{33}^1\right)$, $E_3\left(h_{44}^2\right)$ could be expressed. Further, from Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_2,E_3,E_2)$, in the direction of vector fields $E_4$ and $E_6$, we obtain that the following equations hold up to a nonzero factor $sec\left(2t\right)$

$$\begin{array}{ccc}& & (h_{33}^1-2h_{24}^{2}cos\left(2t\right))(-6{\Gamma }_{13}^4+\sqrt{3}sint)+2h_{14}^{2}cost(-5\sqrt{3}\hfill \\ & & +9\sqrt{3}cos\left(2t\right)+12{\Gamma }_{13}^{4}sint)=0,\hfill \end{array}$$

(52)

$$\begin{array}{ccc}& & cost(\sqrt{3}(7h_{24}^2(1+cos\left(4t\right))+h_{33}^1-3cos\left(2t\right)(2h_{24}^2+h_{33}^1))+8\sqrt{3}·\hfill \\ & & cost{sin}^{3}t{h}_{14}^2+12{\Gamma }_{13}^{4}sint(h_{33}^1-2h_{24}^{2}cos\left(2t\right)-2h_{14}^{2}sin\left(2t\right)))=0,\hfill \end{array}$$

(53)

which further imply that

$$\begin{array}{ccc}\hfill 0& =& 2sintcost\left(52\right)+\left(53\right)\hfill \\ \hfill & =& 2\sqrt{3}cost(1-2cos\left(2t\right))(h_{33}^1-4h_{24}^{2}cos\left(2t\right)-4h_{14}^{2}sin\left(2t\right))\hfill \end{array}$$

(54)

and with respect to this equation the proof will be split into a few cases.

If we suppose that $1-2cos\left(2t\right)=0$, we have that $t\in \{{\displaystyle \frac{\pi }{6}},{\displaystyle \frac{5\pi }{6}},{\displaystyle \frac{7\pi }{6}},{\displaystyle \frac{11\pi }{6}}\}$.

Case $t={\displaystyle \frac{\pi }{6}}$. From Equation (12) evaluated for the vector fields $(X,Y,Z)\in$$\{(E_1,E_4,E_1),(E_2,E_4,E_1)\}$, along vector field $E_3$, directly is obtained $h_{14}^2=0$, $h_{24}^2=0$ and from Equation (10) evaluated for the vector fields $(X,Y)=(E_4,E_4)$, in the direction of $E_4$, we have that $h_{44}^2=0$. By evaluating again Equation (12) for $(X,Y,Z)=(E_1,E_4,E_3)$, along vector field $E_6$, we obtain ${\Gamma }_{13}^4={\displaystyle \frac{1}{4\sqrt{3}}}$ and further all equations are satisfied. The Levi-Civita connection is

$$\begin{array}{ccc}& & {\tilde{\nabla }}_{E_1}{E}_1={\displaystyle \frac{1}{\sqrt{3}}}{E}_6,{\tilde{\nabla }}_{E_1}{E}_2=-{\displaystyle \frac{1}{\sqrt{3}}}{E}_4,{\tilde{\nabla }}_{E_1}{E}_3={\displaystyle \frac{1}{4\sqrt{3}}}{E}_4+{\displaystyle \frac{3}{4}}{E}_6,\hfill \\ & & {\tilde{\nabla }}_{E_1}{E}_4={\displaystyle \frac{1}{\sqrt{3}}}{E}_2-{\displaystyle \frac{1}{4\sqrt{3}}}{E}_3+{\displaystyle \frac{1}{4}}{E}_5,{\tilde{\nabla }}_{E_2}{E}_1={\displaystyle \frac{1}{\sqrt{3}}}{E}_4,{\tilde{\nabla }}_{E_2}{E}_2={\displaystyle \frac{1}{\sqrt{3}}}{E}_6,\hfill \\ & & {\tilde{\nabla }}_{E_2}{E}_3=-{\displaystyle \frac{1}{4}}{E}_4+{\displaystyle \frac{\sqrt{3}}{4}}{E}_6,{\tilde{\nabla }}_{E_2}{E}_4=-{\displaystyle \frac{1}{\sqrt{3}}}{E}_1+{\displaystyle \frac{1}{4}}{E}_3+{\displaystyle \frac{1}{4\sqrt{3}}}{E}_5,\hfill \\ \hfill & & {\tilde{\nabla }}_{E_3}{E}_1=-h_{33}^1{E}_2+{\displaystyle \frac{5}{4\sqrt{3}}}{E}_4+{\displaystyle \frac{3}{4}}{E}_6,{\tilde{\nabla }}_{E_3}{E}_2=h_{33}^1{E}_1-{\displaystyle \frac{5}{4}}{E}_4+{\displaystyle \frac{\sqrt{3}}{4}}{E}_6,\hfill \\ & & {\tilde{\nabla }}_{E_3}{E}_3=h_{33}^1{E}_5+{\displaystyle \frac{2}{\sqrt{3}}}{E}_6,{\tilde{\nabla }}_{E_3}{E}_4=-{\displaystyle \frac{5}{4\sqrt{3}}}{E}_1+{\displaystyle \frac{5}{4}}{E}_2+{\displaystyle \frac{2}{\sqrt{3}}}{E}_5,\hfill \\ & & {\tilde{\nabla }}_{E_4}{E}_1=-{\displaystyle \frac{2}{\sqrt{3}}}{E}_2-{\displaystyle \frac{1}{4\sqrt{3}}}{E}_3+{\displaystyle \frac{1}{4}}{E}_5,{\tilde{\nabla }}_{E_4}{E}_2={\displaystyle \frac{2}{\sqrt{3}}}{E}_1+{\displaystyle \frac{1}{4}}{E}_3+{\displaystyle \frac{1}{4\sqrt{3}}}{E}_5,\hfill \\ & & {\tilde{\nabla }}_{E_4}{E}_3={\displaystyle \frac{1}{4\sqrt{3}}}{E}_1-{\displaystyle \frac{1}{4}}{E}_2+{\displaystyle \frac{2}{\sqrt{3}}}{E}_5,{\tilde{\nabla }}_{E_4}{E}_4=0.\hfill \end{array}$$

(55)

In this case, the almost product structure is expressed as $P{E}_4=-{\displaystyle \frac{1}{2}}{E}_4-{\displaystyle \frac{\sqrt{3}}{2}}{E}_6$ and $P{E}_6=-{\displaystyle \frac{\sqrt{3}}{2}}{E}_4+{\displaystyle \frac{1}{2}}{E}_6$. If we look at the chosen orthonormal frame, we see that the vector fields $E_1,E_2,E_4$ span a tangent bundle on a three-dimensional submanifold $M^3$ of an obtained four-dimensional CR submanifold, on which the Levi-Civita connection induced from the Levi-Civita connection of four-dimensional CR submanifold matches connection of a Berger sphere given by (15) with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$, and by Lemma 1, these two three-dimensional manifolds are locally isometric. The J-relation between vector fields that span a tangent bundle on a three-dimensional submanifold $M^3$ implies that this is actually a three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with the same almost complex distribution ${\mathcal{D}}_1$, which is non-integrable and it is P-orthogonal on itself. The investigated four-dimensional CR submanifold is by Theorem 1 a locally product manifold of the above mentioned three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ and an integral curve of vector field $E_3$. From the Levi-Civita connection, it directly follows that this three-dimensional submanifold $M^3$ is not a totally umbilical foliation of the investigated four-dimensional CR submanifold and furthermore it is implied that this four-dimensional CR submanifold is not a double-twisted product manifold or any of its subtypes.

Cases $t\in \{{\displaystyle \frac{5\pi }{6}},{\displaystyle \frac{7\pi }{6}},{\displaystyle \frac{11\pi }{6}}\}$ are solved as a previous case.

Case $t={\displaystyle \frac{5\pi }{6}}$. We have that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$ and ${\Gamma }_{13}^4={\displaystyle \frac{1}{4\sqrt{3}}}$.

Case $t={\displaystyle \frac{7\pi }{6}}$. In this case, we have that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$ and ${\Gamma }_{13}^4=-{\displaystyle \frac{1}{4\sqrt{3}}}$.

Case $t={\displaystyle \frac{11\pi }{6}}$. In this case, we have that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$ and ${\Gamma }_{13}^4=-{\displaystyle \frac{1}{4\sqrt{3}}}$.

Now, we will treat a next case when $cost=0$, etc. $t\in \{{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$. Case $t={\displaystyle \frac{\pi }{2}}$. From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_4),(E_2,E_4),(E_4,E_4)\}$, in the direction of vector field $E_4$, we have that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$. For these values, Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_4,E_2)$, in the direction of vector field $E_6$, implies that ${\Gamma }_{13}^4={\displaystyle \frac{1}{2\sqrt{3}}}$.

Case $t={\displaystyle \frac{3\pi }{2}}$. We have that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$ and ${\Gamma }_{13}^4=-{\displaystyle \frac{1}{2\sqrt{3}}}$.

In a same way as in the case $t={\displaystyle \frac{\pi }{6}}$, in all these cases, we obtain that the investigated four-dimensional CR submanifold is by Theorem 1 a locally product manifold of three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, for which it holds that an almost complex distribution is non-integrable and it is P-orthogonal on itself, which is further locally isometric to Berger sphere with $\kappa =2$ and $\tau =\pm {\displaystyle \frac{1}{\sqrt{3}}}$ and an integral curve of vector field $E_3$.

If we suppose that $1-2cos\left(2t\right)\ne 0$ and $cost\ne 0$, from Equation (54) we straightforwardly have that $h_{33}^1=4h_{24}^{2}cos\left(2t\right)+4h_{14}^{2}sin\left(2t\right)$ and from Equation (12) evaluated for the vector fields $(X,Y,Z)\in \{(E_1,E_3,E_1),(E_2,E_3,E_1)\}$, along vector field $E_2$, we obtain the following equations

$$\begin{array}{ccc}& & 4{\Gamma }_{13}^4{h}_{44}^2-(-{\displaystyle \frac{5}{6}}+4{\left(h_{14}^2\right)}^2+4{\left(h_{24}^2\right)}^2)cos\left(2t\right)-{\displaystyle \frac{13}{6}}cos\left(4t\right)\hfill \\ & & +2\sqrt{3}{h}_{44}^{2}sint-6\sqrt{3}{\Gamma }_{13}^{4}sin\left(3t\right)=0,\hfill \end{array}$$

(56)

$$\begin{array}{ccc}& & {\displaystyle \frac{1}{6}}((5-24{\left(h_{14}^2\right)}^2-24{\left(h_{24}^2\right)}^2)cos\left(2t\right)+67cos\left(4t\right)+4(10+6{\Gamma }_{13}^4{h}_{44}^2\hfill \\ & & +3\sqrt{3}{h}_{44}^{2}sint-9\sqrt{3}{\Gamma }_{13}^{4}sin\left(3t\right)\left)\right)tan\left(2t\right)=0,\hfill \end{array}$$

(57)

which, for $sin\left(2t\right)\ne 0$, imply the following equation $0=\left(56\right)-cot\left(2t\right)\left(57\right)=-{\displaystyle \frac{20}{3}}(1+2cos\left(4t\right))$, which further implies that $t\in \{{\displaystyle \frac{\pi }{6}},{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}},{\displaystyle \frac{5\pi }{6}},{\displaystyle \frac{7\pi }{6}},{\displaystyle \frac{4\pi }{3}},{\displaystyle \frac{5\pi }{3}},{\displaystyle \frac{11\pi }{6}}\}$, but a condition $1-2cos\left(2t\right)\ne 0$ further implies that $t\in \{{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}},{\displaystyle \frac{4\pi }{3}},{\displaystyle \frac{5\pi }{3}}\}$. In all these cases, from Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_4),(E_2,E_4),(E_4,E_4)\}$, in the direction of vector field $E_4$, it is directly obtained that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$ and for these values, Equation (12) evaluated for vector fields $(X,Y,Z)=(E_1,E_3,E_1)$, along vector field $E_2$, implies a contradiction ${\displaystyle \frac{2}{3}}=0$.

Now, we will treat a special case. Let us suppose that $sin\left(2t\right)=0$ when $sint=0$, etc. $t\in \{0,\pi \}$, as the cases $t\in \{{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$ are treated already. Case $t=0$. From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_4),(E_2,E_4),(E_4,E_4)\}$, in the direction of vector field $E_4$, it is directly obtained that $h_{14}^2=0$, $h_{24}^2=0$, $h_{44}^2=0$ and further, Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_4,E_1)$, along the vector field $E_4$, implies a contradiction ${\displaystyle \frac{2}{3}}=0$. In case $t=\pi$, a contradiction is obtained in the same way.

Case $cos\left(2t\right)=0$. In this case, we have the following subcases $t\in \{{\displaystyle \frac{\pi }{4}},{\displaystyle \frac{3\pi }{4}},{\displaystyle \frac{5\pi }{4}},{\displaystyle \frac{7\pi }{4}}\}$. Let us assume that $t={\displaystyle \frac{\pi }{4}}$. By evaluating Equation (10) for the vector fields $(X,Y)=(E_2,E_3)$, in the direction of vector fields $E_4$ and $E_6$, we obtain that ${\Gamma }_{23}^4={\displaystyle \frac{1}{2\sqrt{6}}}-h_{22}^2$, ${\Gamma }_{13}^4=-{\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{3}{2}}}$ and by evaluating one more time the same equation for the vector fields $(X,Y)\in \{(E_1,E_4),(E_2,E_4),(E_4,E_4)\}$, in the direction of vector field $E_6$, we have that $h_{14}^2=0$, $h_{24}^2=0$ and $h_{44}^2=0$. For these values, from Equation (12) evaluated for the vector fields $(X,Y,Z)=(E_1,E_3,E_1)$, along vector fields $E_6$ and $E_2$, we obtain $h_{33}^1=0$ and ${\displaystyle \frac{4}{3}}+E_1\left(h_{33}^1\right)=0$, which implies a contradiction. For other values of t, for which it holds $cos\left(2t\right)=0$, a contradiction is obtained in a same way.

Construction. It has been proven that a four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with an almost complex distribution ${\mathcal{D}}_1$, which has an arbitrary vector field $E_1$ such that $P{E}_1\in {\mathcal{D}}_3$ holds, is a locally product manifold of an integral curve of a totally real vector field $E_3$, such that $J{E}_3=P{E}_1$ and the three-dimensional submanifold $M^3$, of which the tangent bundle is spanned by vector fields $E_1,E_2,E_4$ and which is actually a three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ of the same type. This three-dimensional submanifold $M^3$ is locally isometric to Berger sphere with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$ or $\tau =-{\displaystyle \frac{1}{\sqrt{3}}}$. An almost product structure P is expressed as $P{E}_1=E_5$, $P{E}_2=E_3$, $P{E}_3=E_2$, $P{E}_4=-cos\left(2t\right)E_4-sin\left(2t\right)E_6$, $P{E}_5=E_1$ and $P{E}_6=-sin\left(2t\right)E_4+cos\left(2t\right)E_6$, for $t\in \{{\displaystyle \frac{\pi }{6}},{\displaystyle \frac{5\pi }{6}},{\displaystyle \frac{7\pi }{6}},{\displaystyle \frac{11\pi }{6}},{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$. If we express the base vector fields by using quaternions $E_i=(p{\alpha }_i,q{\beta }_i)$, for $i\in \{1,\dots 6\}$ and (1) and (8), the J and P relations between $E_3$, $E_4$ and the rest of the base vector fields imply that

$$\begin{array}{ccc}& & E_1={\displaystyle \frac{1}{\sqrt{3}}}(p({\beta }_3-2{\alpha }_3),q(2{\beta }_3-{\alpha }_3)),E_2=(p{\beta }_3,q{\alpha }_3),E_3=(p{\alpha }_3,q{\beta }_3),\hfill \\ & & E_4=(p{\alpha }_4,q{\beta }_4),E_5={\displaystyle \frac{1}{\sqrt{3}}}(p(2{\beta }_3-{\alpha }_3),q({\beta }_3-2{\alpha }_3)),\hfill \\ & & E_6={\displaystyle \frac{1}{\sqrt{3}}}(p(2{\beta }_4-{\alpha }_4),q({\beta }_4-2{\alpha }_4)).\hfill \end{array}$$

(58)

Further, by using the expressions (11) and (9), it is directly obtained that for the quaternions ${\alpha }_i$ and ${\beta }_i$, where ${\beta }_2={\alpha }_3$ and ${\beta }_3={\alpha }_2$, it holds that $<{\alpha }_1,{\alpha }_1>=<{\alpha }_2,{\alpha }_2>=<{\alpha }_3,{\alpha }_3>={\displaystyle \frac{1}{2}}$, $<{\alpha }_1,{\alpha }_2>=<{\alpha }_1,{\alpha }_4>=<{\alpha }_2,{\alpha }_4>=0$ and $<{\alpha }_1,{\alpha }_3>$$=-{\displaystyle \frac{\sqrt{3}}{4}}$, $<{\alpha }_2,{\alpha }_3>={\displaystyle \frac{1}{4}}$, $<{\alpha }_4,{\alpha }_3>=0$, $<{\alpha }_4,{\alpha }_4>={\displaystyle \frac{1}{4}}(1-2cos\left(2t\right))$ and at the same time we have that $<{\beta }_1,{\beta }_1>=<{\beta }_2,{\beta }_2>=<{\beta }_3,{\beta }_3>={\displaystyle \frac{1}{2}}$, $<{\beta }_1,{\beta }_2>=<{\beta }_1,{\beta }_4>=<{\beta }_2,{\beta }_4>=0$ and $<{\beta }_1,{\beta }_3>={\displaystyle \frac{\sqrt{3}}{4}}$, $<{\beta }_2,{\beta }_3>={\displaystyle \frac{1}{4}}$, $<{\beta }_4,{\beta }_3>=0$, $<{\beta }_4,{\beta }_4>={\displaystyle \frac{1}{4}}(2-cos\left(2t\right)-\sqrt{3}sin\left(2t\right))$. Now, one of these submanifolds will be constructed and the rest of them will be obtained by using isometries.

• Case $t={\displaystyle \frac{3\pi }{2}}$. The almost product structure P is expressed as $P{E}_4=E_4$ and $P{E}_6=-E_6$. From the Levi-Civita connection, we can conclude that vector fields $E_1,E_2,E_4$ span a tangent bundle on a three-dimensional submanifold $M^3$ of an investigated four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ and the J-relation between them implies that this is actually a three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with the same almost complex distribution ${\mathcal{D}}_1$. From the relation $P{E}_4=E_4$, it follows that ${\beta }_4={\alpha }_4$. From the expression of the tensor field G given by (18) we have that $G(E_2,E_3)=-{\displaystyle \frac{1}{\sqrt{3}}}{E}_4$, which further along with the relation (7) evaluated for the vector fields $(X,Y)=(E_2,E_3)$, imply that ${\alpha }_4=2{\alpha }_3\times {\beta }_3$. The quaternions ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_4$ are mutually orthogonal and we can use a rotation of $Im\mathbb{H}$, expressed by an unit quaternion $h=h_0+h_{1}i+h_{2}j+h_{3}k$, that imaginary quaternions j, $-k$, i obtain directions of ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_4$. The lengths of these quaternions imply that ${\alpha }_1={\displaystyle \frac{1}{\sqrt{2}}}hj\overline{h}$, ${\alpha }_2=-{\displaystyle \frac{1}{\sqrt{2}}}hk\overline{h}$ and ${\alpha }_4={\displaystyle \frac{\sqrt{3}}{2}}hi\overline{h}$, because of $<{\alpha }_4,{\alpha }_4>={\displaystyle \frac{3}{4}}$ and using the relations given by (58) we have that ${\beta }_3={\alpha }_2=-{\displaystyle \frac{1}{\sqrt{2}}}hk\overline{h}$, ${\beta }_2={\alpha }_3=-{\displaystyle \frac{1}{2\sqrt{2}}}h(k+\sqrt{3}j)\overline{h}$, ${\beta }_1={\displaystyle \frac{1}{2\sqrt{2}}}h(-\sqrt{3}k+j)\overline{h}$ and ${\beta }_4={\alpha }_4={\displaystyle \frac{\sqrt{3}}{2}}hi\overline{h}$. From the obtained Levi-Civita connection given by (55), by using the relation (13) and (14), the following derivatives are obtained

  $$\begin{array}{ccc}& & E_1\left({\alpha }_1\right)=0,E_1\left({\alpha }_2\right)=0,E_1\left({\alpha }_4\right)=0,\hfill \\ & & E_2\left({\alpha }_1\right)=0,E_2\left({\alpha }_2\right)=0,E_2\left({\alpha }_4\right)=0,\hfill \\ & & E_3\left({\alpha }_1\right)=-h_{33}^1{\beta }_3-2\sqrt{3}{\alpha }_3\times {\beta }_3,\hfill \\ & & E_3\left({\alpha }_2\right)={\displaystyle \frac{1}{\sqrt{3}}}{h}_{33}^1({\beta }_3-2{\alpha }_3)-2{\alpha }_3\times {\beta }_3,E_3\left({\alpha }_4\right)=3({\beta }_3-{\alpha }_3),\hfill \\ & & E_4\left({\alpha }_1\right)=0,E_4\left({\alpha }_2\right)=0,E_4\left({\alpha }_4\right)=0.\hfill \end{array}$$

  (59)
  For new expressions of quaternions ${\alpha }_1$, ${\alpha }_2$, and ${\alpha }_4$ from the derivatives (59), we obtain the following new relations:

  $$\begin{array}{ccc}& & E_3\left({\alpha }_1\right)=\sqrt{2}Im\left(E_3\left(h\right)j\overline{h}\right)={\displaystyle \frac{1}{\sqrt{2}}}h(-{\displaystyle \frac{3}{\sqrt{2}}}i+h_{33}^{1}k)\overline{h},\hfill \\ & & E_3\left({\alpha }_2\right)=-\sqrt{2}Im\left(E_3\left(h\right)k\overline{h}\right)={\displaystyle \frac{1}{\sqrt{2}}}h(-{\displaystyle \frac{\sqrt{3}}{\sqrt{2}}}i+h_{33}^{1}j)\overline{h},\hfill \\ & & E_3\left({\alpha }_4\right)=\sqrt{3}Im\left(E_3\left(h\right)i\overline{h}\right)={\displaystyle \frac{3}{2\sqrt{2}}}h(\sqrt{3}j-k)\overline{h},\hfill \end{array}$$

  which imply that

  $$E_3\left(h\right)=h({\displaystyle \frac{1}{2}}{h}_{33}^{1}i+{\displaystyle \frac{\sqrt{3}}{2\sqrt{2}}}j+{\displaystyle \frac{3}{2\sqrt{2}}}k)$$

  and for $i\in \{1,2,4\}$, we have that

  $$\begin{array}{ccc}& & E_i\left({\alpha }_1\right)=\sqrt{2}Im\left(E_i\left(h\right)j\overline{h}\right)=0,\hfill \\ & & E_i\left({\alpha }_2\right)=-\sqrt{2}Im\left(E_i\left(h\right)k\overline{h}\right)=0,\hfill \\ & & E_i\left({\alpha }_4\right)=\sqrt{3}Im\left(E_i\left(h\right)i\overline{h}\right)=0.\hfill \end{array}$$

  (60)
  From the system (60), it is directly obtained that $E_i\left(h_j\right)=0$, for $i\in \{1,2,4\}$ and $j\in \{0,1,2,3\}$, implying that $E_1\left(h\right)=0$, $E_2\left(h\right)=0$ and $E_4\left(h\right)=0$.
  The vector fields $E_1,E_2,E_4$ span a tangent bundle on a three-dimensional CR submanifold $M^3$ of ${\mathbb{S}}^3\times {\mathbb{S}}^3$. Moreover, note that for vector fields chosen as $({\displaystyle \frac{\sqrt{3}}{2}}{X}_1,{\displaystyle \frac{1}{\sqrt{2}}}{X}_2,{\displaystyle \frac{1}{\sqrt{2}}}{X}_3)$ and $(E_4,E_1,E_2)$, it holds that the Levi-Civita connection of $M^3$ induced from the Levi-Civita connection of the four-dimensional CR submanifold M matches the connection of a Berger sphere given by (15) with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$ and by Lemma 1, these two three-dimensional manifolds are locally isometric. An idea is to find the isometry $\mathcal{F}:u\in {\mathbb{S}}^3\mapsto \tilde{p}\in M^3$, such that its differential maps vector $X_i\left(u\right)$, $i\in \{1,2,3\}$ to an appropriate vector from a triple, collinear with $E_j\left(\tilde{p}\right)$, $j\in \{1,2,4\}$. On $M^3$, we will use the following notation, $\widetilde{E_1}=E_1$, $\widetilde{E_2}=E_2$, $\widetilde{E_3}=E_4$ and $\widetilde{E_1}=(p{\tilde{\alpha }}_1,q{\tilde{\beta }}_1)$, $\widetilde{E_2}=(p{\tilde{\alpha }}_2,q{\tilde{\beta }}_2)$ and $\widetilde{E_3}=(p{\tilde{\alpha }}_3,q{\tilde{\beta }}_3)$, at the point $(p,q)\in M^3$ and we have that $\widetilde{E_i}\left({\tilde{\alpha }}_j\right)=0$ and $\widetilde{E_i}\left({\tilde{\beta }}_j\right)=0$, for $i\in \{1,2,3\}$ and $j\in \{1,2,3\}$. We can use the following expressions, ${\tilde{\alpha }}_1={\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}j\overline{\tilde{h}}$, ${\tilde{\alpha }}_2=-{\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}k\overline{\tilde{h}}$ and ${\tilde{\alpha }}_3={\displaystyle \frac{\sqrt{3}}{2}}\tilde{h}i\overline{\tilde{h}}$, ${\tilde{\beta }}_1={\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}(-\sqrt{3}k+j)\overline{\tilde{h}}$, ${\tilde{\beta }}_2=-{\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}(k+\sqrt{3}j)\overline{\tilde{h}}$ and ${\tilde{\beta }}_3={\displaystyle \frac{\sqrt{3}}{2}}\tilde{h}i\overline{\tilde{h}}$, where $\tilde{h}$ is a constant unit quaternion.
  The preceding relations imply that at the point $\tilde{p}=(p,q)\in M^3\subset {\mathbb{S}}^3\times {\mathbb{S}}^3$, it holds that $X_1\left(p\right)={\displaystyle \frac{2}{\sqrt{3}}}\widetilde{E_3}(p,0)$, $X_2\left(p\right)=\sqrt{2}\widetilde{E_1}(p,0)$, $X_3\left(p\right)=\sqrt{2}\widetilde{E_2}(p,0)$ and $X_1\left(q\right)={\displaystyle \frac{2}{\sqrt{3}}}\widetilde{E_3}(0,q)$, $X_2\left(q\right)=\sqrt{2}\widetilde{E_1}(0,q)$, $X_3\left(q\right)=\sqrt{2}\widetilde{E_2}(0,q)$. At an arbitrary point $(p,q)\in {\mathbb{S}}^3\times {\mathbb{S}}^3$, it holds that

  $$\begin{array}{ccc}& & X_1\left(p\right)={\displaystyle \frac{2}{\sqrt{3}}}\widetilde{E_3}(p,0)={\displaystyle \frac{2}{\sqrt{3}}}p{\tilde{\alpha }}_3=p\tilde{h}i\overline{\tilde{h}}.\hfill \end{array}$$
  For the point $u\in {\mathbb{S}}^3$, if we suppose that $p=\tilde{h}u\overline{\tilde{h}}$, we have that

  $$\begin{array}{ccc}& & X_1\left(p\right)=\tilde{h}{X}_1\left(u\right)\overline{\tilde{h}}=\tilde{h}ui\overline{\tilde{h}}=\tilde{h}u\overline{\tilde{h}}\tilde{h}i\overline{\tilde{h}}=p\tilde{h}i\overline{\tilde{h}}.\hfill \end{array}$$
  In the same way, if we suppose that $p=\tilde{h}u\overline{\tilde{h}}$, we see that everything agrees.

  $$\begin{array}{ccc}& & X_2\left(p\right)=\sqrt{2}\widetilde{E_1}(p,0)=\sqrt{2}p{\tilde{\alpha }}_1=p\tilde{h}j\overline{\tilde{h}},\hfill \\ & & X_2\left(p\right)=\tilde{h}{X}_2\left(u\right)\overline{\tilde{h}}=\tilde{h}uj\overline{\tilde{h}}=\tilde{h}u\overline{\tilde{h}}\tilde{h}j\overline{\tilde{h}}=p\tilde{h}j\overline{\tilde{h}}\hfill \end{array}$$

  and

  $$\begin{array}{ccc}& & X_3\left(p\right)=\sqrt{2}\widetilde{E_2}(p,0)=\sqrt{2}p{\tilde{\alpha }}_2=-p\tilde{h}k\overline{\tilde{h}},\hfill \\ & & X_3\left(p\right)=\tilde{h}{X}_3\left(u\right)\overline{\tilde{h}}=-\tilde{h}uk\overline{\tilde{h}}=-\tilde{h}u\overline{\tilde{h}}\tilde{h}k\overline{\tilde{h}}=-p\tilde{h}k\overline{\tilde{h}}.\hfill \end{array}$$
  In the same way, it holds that

  $$\begin{array}{ccc}& & X_1\left(q\right)={\displaystyle \frac{2}{\sqrt{3}}}\widetilde{E_3}(0,q)={\displaystyle \frac{2}{\sqrt{3}}}q{\tilde{\beta }}_3=q\tilde{h}i\overline{\tilde{h}},\hfill \\ & & X_2\left(q\right)=\sqrt{2}\widetilde{E_1}(0,q)=\sqrt{2}q{\tilde{\beta }}_1=q\tilde{h}{\displaystyle \frac{j-\sqrt{3}k}{2}}\overline{\tilde{h}},\hfill \\ & & X_3\left(q\right)=\sqrt{2}\widetilde{E_2}(0,q)=\sqrt{2}q{\tilde{\beta }}_2=-q\tilde{h}{\displaystyle \frac{\sqrt{3}j+k}{2}}\overline{\tilde{h}}\hfill \end{array}$$

  and a straightforward computation shows us that $q=\tilde{h}u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{\tilde{h}}$ satisfies previous relations and the immersion is given by ${\tilde{f}}_1=(\tilde{h}u\overline{\tilde{h}},\tilde{h}u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{\tilde{h}})$. The three-dimensional manifold $M^3$, which forms a product manifold, is locally congruent to the immersion $f_1=(u,u{\displaystyle \frac{\sqrt{3}+i}{2}})$, where $u\in {\mathbb{S}}^3$, ${\mathbb{S}}^3$ is the Berger sphere with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$.
  We still should obtain an immersion of curve $\gamma$, which is an integral curve of the vector field $E_3(p,q)=(p{\alpha }_3,q{\beta }_3)$.
  Let us suppose that the curve $({\gamma }_p\left(x\right),{\gamma }_q\left(x\right))$ is an integral curve of the vector field $X(p,q)=(-p{\displaystyle \frac{\sqrt{3}j+k}{2\sqrt{2}}},-q{\displaystyle \frac{1}{\sqrt{2}}}k)$, determined with imaginary quaternions $\tilde{\alpha }=-{\displaystyle \frac{\sqrt{3}j+k}{2\sqrt{2}}}$ and $\tilde{\beta }=-{\displaystyle \frac{1}{\sqrt{2}}}k$. For the curve ${\gamma }_p\left(x\right)=A_0\left(x\right)+A_1\left(x\right)i+A_2\left(x\right)j+A_3\left(x\right)k$, we have that the corresponding coordinate vector field along the curve is ${\gamma }_p^{\prime }\left(x\right)=A_0^{\prime }\left(x\right)+A_1^{\prime }\left(x\right)i+A_2^{\prime }\left(x\right)j+A_3^{\prime }\left(x\right)k$. At the same time, the previous relation between vector fields implies that ${\gamma }_p^{\prime }\left(x\right)=-{\gamma }_p\left(x\right){\displaystyle \frac{\sqrt{3}j+k}{2\sqrt{2}}}$, that further implies a system of following differential equations

  $$\begin{array}{ccc}& & A_0^{\prime }\left(x\right)={\displaystyle \frac{1}{2\sqrt{2}}}(A_3\left(x\right)+\sqrt{3}{A}_2\left(x\right)),A_1^{\prime }\left(x\right)={\displaystyle \frac{-1}{2\sqrt{2}}}(A_2\left(x\right)-\sqrt{3}{A}_3\left(x\right)),\hfill \\ & & A_2^{\prime }\left(x\right)={\displaystyle \frac{1}{2\sqrt{2}}}(A_1\left(x\right)-\sqrt{3}{A}_0\left(x\right)),A_3^{\prime }\left(x\right)={\displaystyle \frac{-1}{2\sqrt{2}}}(A_0\left(x\right)+\sqrt{3}{A}_1\left(x\right)),\hfill \end{array}$$

  (61)

  which further imply that $A_i^{″}\left(x\right)=-{\displaystyle \frac{1}{2}}{A}_i\left(x\right)$, for $i\in \{0,1,2,3\}$ and the general solution is

  $$\begin{array}{ccc}& & A_i\left(x\right)=A_i^{1}cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+A_i^{2}sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right),fori\in \{0,1,2,3\}.\hfill \end{array}$$
  For a previous solution, Equations (61) imply that $A_2^1=-{\displaystyle \frac{1}{2}}{A}_1^2+{\displaystyle \frac{\sqrt{3}}{2}}{A}_0^2$, $A_2^2={\displaystyle \frac{1}{2}}{A}_1^1-{\displaystyle \frac{\sqrt{3}}{2}}{A}_0^1$ and $A_3^1={\displaystyle \frac{1}{2}}{A}_0^2+{\displaystyle \frac{\sqrt{3}}{2}}{A}_1^2$, $A_3^2=-{\displaystyle \frac{1}{2}}{A}_0^1-{\displaystyle \frac{\sqrt{3}}{2}}{A}_1^1$. For the initial condition ${\gamma }_p\left(0\right)={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}i+{\displaystyle \frac{\sqrt{3}-1}{4}}j+{\displaystyle \frac{\sqrt{3}+1}{4}}k$, the curve ${\gamma }_p\left(x\right)$ is unique and given by

  $$\begin{array}{ccc}\hfill {\gamma }_p\left(x\right)& =& {\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1+i)\hfill \\ & & +{\displaystyle \frac{1}{4}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))((\sqrt{3}-1)j+(\sqrt{3}+1)k).\hfill \end{array}$$
  In the same way, for the curve ${\gamma }_q\left(x\right)=B_0\left(x\right)+B_1\left(x\right)i+B_2\left(x\right)j+B_3\left(x\right)k$, we have that the corresponding coordinate vector field along the curve is ${\gamma }_q^{\prime }\left(x\right)=B_0^{\prime }\left(x\right)+B_1^{\prime }\left(x\right)i+B_2^{\prime }\left(x\right)j+B_3^{\prime }\left(x\right)k$. At the same time, the previous relation between vector fields implies that ${\gamma }_q^{\prime }\left(x\right)=-{\gamma }_q\left(x\right){\displaystyle \frac{1}{\sqrt{2}}}k$, which further implies a system of the following differential equations:

  $$\begin{array}{ccc}& & B_0^{\prime }\left(x\right)={\displaystyle \frac{1}{\sqrt{2}}}{B}_3\left(x\right),B_1^{\prime }\left(x\right)=-{\displaystyle \frac{1}{\sqrt{2}}}{B}_2\left(x\right),\hfill \\ & & B_2^{\prime }\left(x\right)={\displaystyle \frac{1}{\sqrt{2}}}{B}_1\left(x\right),B_3^{\prime }\left(x\right)=-{\displaystyle \frac{1}{\sqrt{2}}}{B}_0\left(x\right),\hfill \end{array}$$

  (62)

  which further imply that $B_i^{″}\left(x\right)=-{\displaystyle \frac{1}{2}}{B}_i\left(x\right)$, for $i\in \{0,1,2,3\}$, and the general solution is

  $$\begin{array}{ccc}& & B_i\left(x\right)=B_i^{1}cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+B_i^{2}sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right),fori\in \{0,1,2,3\}.\hfill \end{array}$$
  For a previous solution, Equations (62) imply that $B_2^1=-B_1^2$, $B_2^2=B_1^1$, and $B_3^1=B_0^2$, $B_3^2=-B_0^1$. For the initial condition ${\gamma }_q\left(0\right)={\displaystyle \frac{1}{2}}(1+i-j+k)$, the curve ${\gamma }_q\left(x\right)$ is unique and given by

  $$\begin{array}{ccc}& & {\gamma }_q\left(x\right)={\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1+i)-{\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(j-k).\hfill \end{array}$$
  As any two regular curves on ${\mathbb{S}}^3\times {\mathbb{S}}^3$ are locally isometric, the curve ${\gamma }_1$, which is an integral curve of the vector field $E_3$, and an integral curve of the vector field X are locally isometric. The isometry ${\mathcal{F}}_{a,b,c}={\mathcal{F}}_{a,b,h}$, where $a\left(x\right)$ and $b\left(x\right)$ are unit quaternion curves, maps the curve $({\gamma }_p,{\gamma }_q)$ to the curve ${\gamma }_1=(a{\gamma }_p\overline{h},b{\gamma }_q\overline{h})$. An integral curve of the vector field $E_3$ is locally congruent to the immersion

  $$\begin{array}{ccc}\hfill {\gamma }_1\left(x\right)& =& \left(a\left(x\right)\right({\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1+i)+{\displaystyle \frac{1}{4}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-\hfill \\ & & sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))((\sqrt{3}-1)j+(\sqrt{3}+1)k))\overline{h}\left(x\right),b\left(x\right)\left({\displaystyle \frac{1}{2}}\right(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)+\hfill \\ & & sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(1+i)+{\displaystyle \frac{1}{2}}(cos\left({\displaystyle \frac{1}{\sqrt{2}}}x\right)-sin\left({\displaystyle \frac{1}{\sqrt{2}}}x\right))(-j+k)\left)\overline{h}\left(x\right)\right),\hfill \end{array}$$

  where $a^{\prime }\left(x\right)=a\left(x\right){\gamma }_p\left(x\right)\overline{h}\left(x\right)h^{\prime }\left(x\right)\overline{{\gamma }_p}\left(x\right)$ and $b^{\prime }\left(x\right)=b\left(x\right){\gamma }_q\left(x\right)\overline{h}\left(x\right)h^{\prime }\left(x\right)\overline{{\gamma }_q}\left(x\right)$.
  The remaining cases could be obtained from the previous one. If we apply a different combination of the isometries ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$, we map the obtained four-dimensional CR submanifold M to another four-dimensional CR submanifold, for which it also holds that $P{E}_1=E_5$.
• Case $t={\displaystyle \frac{5\pi }{6}}$. In this case, the almost product structure is expressed as $P{E}_4=-{\displaystyle \frac{1}{2}}{E}_4+{\displaystyle \frac{\sqrt{3}}{2}}{E}_6$ and $P{E}_6={\displaystyle \frac{\sqrt{3}}{2}}{E}_4+{\displaystyle \frac{1}{2}}{E}_6$. From the Levi-Civita connection, we can conclude that vector fields $E_1,E_2,E_4$ span a tangent bundle on a three-dimensional submanifold $M^3$ of an investigated four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ and the J-relation between them implies that this is actually a three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with the same almost complex distribution ${\mathcal{D}}_1$. From the expression of the almost product structure P in the base vector fields, it follows that ${\alpha }_4=0$. From the expression of the tensor field G given by (18), we have that $G(E_2,E_3)={\displaystyle \frac{1}{2\sqrt{3}}}{E}_4+{\displaystyle \frac{1}{2}}{E}_6$, which, along with the relation (7) evaluated for the vector fields $(X,Y)=(E_2,E_3)$, implies that ${\beta }_4=\sqrt{3}{\beta }_1\times {\beta }_2$. From the obtained Levi-Civita connection, by using the relations (13) and (14), the following derivatives are obtained

  $$\begin{array}{ccc}& & E_1\left({\beta }_1\right)=0,E_1\left({\beta }_2\right)=-{\displaystyle \frac{2}{\sqrt{3}}}{\beta }_4,E_1\left({\beta }_3\right)=-{\displaystyle \frac{1}{\sqrt{3}}}{\beta }_4,E_1\left({\beta }_4\right)=\sqrt{3}{\beta }_2,\hfill \\ & & E_2\left({\beta }_1\right)={\displaystyle \frac{2}{\sqrt{3}}}{\beta }_4,E_2\left({\beta }_2\right)=0,E_2\left({\beta }_3\right)={\beta }_4,E_2\left({\beta }_4\right)=-\sqrt{3}{\beta }_1,\hfill \\ & & E_3\left({\beta }_1\right)=-h_{33}^1{\beta }_2+{\displaystyle \frac{1}{\sqrt{3}}}{\beta }_4,E_3\left({\beta }_2\right)=h_{33}^1{\beta }_1+{\beta }_4,\hfill \\ & & E_3\left({\beta }_3\right)={\displaystyle \frac{1}{2}}{h}_{33}^1{\beta }_1-{\displaystyle \frac{\sqrt{3}}{2}}{h}_{33}^1{\beta }_2+{\beta }_4,E_3\left({\beta }_4\right)=-{\displaystyle \frac{\sqrt{3}}{2}}{\beta }_1-{\displaystyle \frac{3}{2}}{\beta }_2,\hfill \\ & & E_4\left({\beta }_1\right)=-\sqrt{3}{\beta }_2,E_4\left({\beta }_2\right)=\sqrt{3}{\beta }_1,E_4\left({\beta }_3\right)={\displaystyle \frac{\sqrt{3}}{2}}{\beta }_1-{\displaystyle \frac{3}{2}}{\beta }_2,E_4\left({\beta }_4\right)=0.\hfill \end{array}$$

  (63)
  If we apply an isometry ${\mathcal{F}}_1\circ {\mathcal{F}}_2$ to the submanifold M corresponding to the case $t={\displaystyle \frac{3\pi }{2}}$, by Lemma 7, the obtained four-dimensional CR submanifold corresponds to the case $t={\displaystyle \frac{5\pi }{6}}$. Moreover, note that for vector fields chosen as $({\displaystyle \frac{\sqrt{3}}{2}}{X}_1,{\displaystyle \frac{1}{\sqrt{2}}}{X}_2,{\displaystyle \frac{1}{\sqrt{2}}}{X}_3)$, and $(E_4,E_1,E_2)$, it holds that the Levi-Civita connection of $M^3$, induced from the Levi-Civita connection of the four-dimensional CR submanifold M, matches the connection of a Berger sphere given by (15) with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$, and that these two three-dimensional manifolds are locally isometric according to Lemma 1. On $M^3$, we will use the following notation $\widetilde{E_1}=E_1$, $\widetilde{E_2}=E_2$, $\widetilde{E_3}=E_4$ and $\widetilde{E_1}=(p{\tilde{\alpha }}_1,q{\tilde{\beta }}_1)$, $\widetilde{E_2}=(p{\tilde{\alpha }}_2,q{\tilde{\beta }}_2)$ and $\widetilde{E_3}=(p{\tilde{\alpha }}_3,q{\tilde{\beta }}_3)$ at an arbitrary point $(p,q)\in M^3$. Note that the derivatives $\widetilde{E_i}\left({\tilde{\beta }}_j\right)$, where $i,j\in \{1,2,3\}$, on $M^3$ can also be obtained and we have that they satisfy the same relations that hold for the corresponding derivatives $E_i\left({\beta }_j\right)$, where $i,j\in \{1,2,4\}$, on M given by (63). The isometry ${\mathcal{F}}_1\circ {\mathcal{F}}_2$ maps the immersion ${\tilde{f}}_1$ to the immersion ${\tilde{f}}_2=(\tilde{h}u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{u}\overline{\tilde{h}},\tilde{h}\overline{u}\overline{\tilde{h}})$ and by using the relations (17) its differential maps the corresponding vector fields to the vector fields

  $$\begin{array}{ccc}& & \widetilde{E_1}=(\left(\tilde{h}u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{u}\overline{\tilde{h}}\right)(-{\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(\sqrt{3}k+j)\overline{u}\overline{\tilde{h}}),\left(\tilde{h}\overline{u}\overline{\tilde{h}}\right)(-{\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uj\overline{u}\overline{\tilde{h}})),\hfill \\ & & \widetilde{E_2}=(\left(\tilde{h}u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{u}\overline{\tilde{h}}\right)\left({\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(-\sqrt{3}j+k)\overline{u}\overline{\tilde{h}}\right),\left(\tilde{h}\overline{u}\overline{\tilde{h}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uk\overline{u}\overline{\tilde{h}}\right)),\hfill \\ & & \widetilde{E_3}=(0,\left(\tilde{h}\overline{u}\overline{\tilde{h}}\right)(-{\displaystyle \frac{\sqrt{3}}{2}}\tilde{h}ui\overline{u}\overline{\tilde{h}})).\hfill \end{array}$$
  For the immersion ${\tilde{f}}_2$, the relations between the vector fields $({\displaystyle \frac{\sqrt{3}}{2}}{X}_1,{\displaystyle \frac{1}{\sqrt{2}}}{X}_2,{\displaystyle \frac{1}{\sqrt{2}}}{X}_3)$ and $(\widetilde{E_3},\widetilde{E_1},\widetilde{E_2})$ are satisfied. Note also that by using these relations between the vector fields, the derivatives $\widetilde{E_i}\left({\tilde{\beta }}_j\right)$, where $i,j\in \{1,2,3\}$, are satisfied for ${\tilde{\beta }}_1=-{\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uj\overline{u}\overline{\tilde{h}}$, ${\tilde{\beta }}_2={\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uk\overline{u}\overline{\tilde{h}}$ and ${\tilde{\beta }}_3=-{\displaystyle \frac{\sqrt{3}}{2}}\tilde{h}ui\overline{u}\overline{\tilde{h}}$ on $M^3$. In the same way, we can prove that the derivatives of quaternions ${\tilde{\alpha }}_1=-{\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(\sqrt{3}k+j)\overline{u}\overline{\tilde{h}}$, ${\tilde{\alpha }}_2={\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(-\sqrt{3}j+k)\overline{u}\overline{\tilde{h}}$ and ${\tilde{\alpha }}_3=0$ on $M^3$ are satisfied. We obtain that the three-dimensional submanifold $M^3$, which forms a product manifold, is locally congruent to the immersion $f_2=(u{\displaystyle \frac{\sqrt{3}+i}{2}}\overline{u},\overline{u})$, where $u\in {\mathbb{S}}^3$, ${\mathbb{S}}^3$ is the Berger sphere with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$. In the same way, the curve $\gamma$, the integral curve of the vector field $E_3$, is locally isometric to the curve ${\gamma }_2={\mathcal{F}}_1\left({\mathcal{F}}_2\left({\gamma }_1\right)\right)$.
• Case $t={\displaystyle \frac{\pi }{6}}$. In this case, the almost product structure is expressed as $P{E}_4=-{\displaystyle \frac{1}{2}}{E}_4-{\displaystyle \frac{\sqrt{3}}{2}}{E}_6$ and $P{E}_6=-{\displaystyle \frac{\sqrt{3}}{2}}{E}_4+{\displaystyle \frac{1}{2}}{E}_6$. From the Levi-Civita connection, we can conclude that vector fields $E_1,E_2,E_4$ span a tangent bundle on a three-dimensional submanifold $M^3$ of an investigated four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ and the J-relation between them implies that this is actually a three-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ with the same almost complex distribution ${\mathcal{D}}_1$. From the expression of the almost product structure P in the base vector fields, it follows that ${\beta }_4=0$. From the expression of the tensor field G given by (18), we have that $G(E_2,E_3)={\displaystyle \frac{1}{2\sqrt{3}}}{E}_4-{\displaystyle \frac{1}{2}}{E}_6$, which further along with the relation (7) evaluated for the vector fields $(X,Y)=(E_2,E_3)$, implies that ${\alpha }_4=\sqrt{3}{\alpha }_1\times {\alpha }_2$. From the obtained Levi-Civita connection, by using the relation (13) and (14), the following derivatives are obtained

  $$\begin{array}{ccc}& & E_1\left({\alpha }_1\right)=0,E_1\left({\alpha }_2\right)=-{\displaystyle \frac{2}{\sqrt{3}}}{\alpha }_4,E_1\left({\alpha }_3\right)=-{\displaystyle \frac{1}{\sqrt{3}}}{\alpha }_4,E_1\left({\alpha }_4\right)=\sqrt{3}{\alpha }_2,\hfill \\ & & E_2\left({\alpha }_1\right)={\displaystyle \frac{2}{\sqrt{3}}}{\alpha }_4,E_2\left({\alpha }_2\right)=0,E_2\left({\alpha }_3\right)=-{\alpha }_4,E_2\left({\alpha }_4\right)=-\sqrt{3}{\alpha }_1,\hfill \\ & & E_3\left({\alpha }_1\right)=-h_{33}^1{\alpha }_2+{\displaystyle \frac{1}{\sqrt{3}}}{\alpha }_4,E_3\left({\alpha }_2\right)=h_{33}^1{\alpha }_1-{\alpha }_4,\hfill \\ & & E_3\left({\alpha }_3\right)={\displaystyle \frac{1}{2}}{h}_{33}^1{\alpha }_1+{\displaystyle \frac{\sqrt{3}}{2}}{h}_{33}^1{\alpha }_2-{\alpha }_4,E_3\left({\alpha }_4\right)=-{\displaystyle \frac{\sqrt{3}}{2}}{\alpha }_1+{\displaystyle \frac{3}{2}}{\alpha }_2,\hfill \\ & & E_4\left({\alpha }_1\right)=-\sqrt{3}{\alpha }_2,E_4\left({\alpha }_2\right)=\sqrt{3}{\alpha }_1,E_4\left({\alpha }_3\right)={\displaystyle \frac{\sqrt{3}}{2}}{\alpha }_1+{\displaystyle \frac{3}{2}}{\alpha }_2,E_4\left({\alpha }_4\right)=0.\hfill \end{array}$$

  (64)
  If we apply an isometry ${\mathcal{F}}_2\circ {\mathcal{F}}_1$ to the submanifold M corresponding to the case $t={\displaystyle \frac{3\pi }{2}}$, by Lemma 7, the obtained four-dimensional CR submanifold corresponds to the case $t={\displaystyle \frac{\pi }{6}}$. Moreover, note that for vector fields chosen as $({\displaystyle \frac{\sqrt{3}}{2}}{X}_1,{\displaystyle \frac{1}{\sqrt{2}}}{X}_2,{\displaystyle \frac{1}{\sqrt{2}}}{X}_3)$ and $(E_4,E_1,E_2)$, it holds that the Levi-Civita connection of $M^3$, induced from the Levi-Civita connection of the four-dimensional CR submanifold M, matches the connection of a Berger sphere given by (15) with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$, and that these two three-dimensional manifolds are locally isometric according to Lemma 1. On $M^3$, we will use the following notation $\widetilde{E_1}=E_1$, $\widetilde{E_2}=E_2$, $\widetilde{E_3}=E_4$ and $\widetilde{E_1}=(p{\tilde{\alpha }}_1,q{\tilde{\beta }}_1)$, $\widetilde{E_2}=(p{\tilde{\alpha }}_2,q{\tilde{\beta }}_2)$ and $\widetilde{E_3}=(p{\tilde{\alpha }}_3,q{\tilde{\beta }}_3)$ at an arbitrary point $(p,q)\in M^3$. Note that the derivatives $\widetilde{E_i}\left({\tilde{\alpha }}_j\right)$, where $i,j\in \{1,2,3\}$, on $M^3$ can also be obtained and we have that they satisfy the same relations that hold for the corresponding derivatives $E_i\left({\alpha }_j\right)$, where $i,j\in \{1,2,4\}$, on M given by (64). The isometry ${\mathcal{F}}_2\circ {\mathcal{F}}_1$ maps the immersion ${\tilde{f}}_1$ to the immersion ${\tilde{f}}_3=(\tilde{h}{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}},\tilde{h}u{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}})$ and by using the relations (17), its differential maps the corresponding vector fields to the vector fields

  $$\begin{array}{ccc}& & \widetilde{E_1}=(\left(\tilde{h}{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}}\right)(-{\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uj\overline{u}\overline{\tilde{h}}),\left(\tilde{h}u{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}}\right)\left({\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(\sqrt{3}k-j)\overline{u}\overline{\tilde{h}}\right)),\hfill \\ & & \widetilde{E_2}=(\left(\tilde{h}{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}}\right)\left({\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uk\overline{u}\overline{\tilde{h}}\right),\left(\tilde{h}u{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}}\right)\left({\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(\sqrt{3}j+k)\overline{u}\overline{\tilde{h}}\right)),\hfill \\ & & \widetilde{E_3}=(\left(\tilde{h}{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u}\overline{\tilde{h}}\right)(-{\displaystyle \frac{\sqrt{3}}{2}}\tilde{h}ui\overline{u}\overline{\tilde{h}}),0).\hfill \end{array}$$
  For the immersion ${\tilde{f}}_3$, the relations between the vector fields $({\displaystyle \frac{\sqrt{3}}{2}}{X}_1,{\displaystyle \frac{1}{\sqrt{2}}}{X}_2,{\displaystyle \frac{1}{\sqrt{2}}}{X}_3)$ and $(\widetilde{E_3},\widetilde{E_1},\widetilde{E_2})$ are satisfied. Note also that by using these relations between the vector fields, the derivatives $\widetilde{E_i}\left({\tilde{\alpha }}_j\right)$, where $i,j\in \{1,2,3\}$, are satisfied for ${\tilde{\alpha }}_1=-{\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uj\overline{u}\overline{\tilde{h}}$, ${\tilde{\alpha }}_2={\displaystyle \frac{1}{\sqrt{2}}}\tilde{h}uk\overline{u}\overline{\tilde{h}}$ and ${\tilde{\alpha }}_3=-{\displaystyle \frac{\sqrt{3}}{2}}\tilde{h}ui\overline{u}\overline{\tilde{h}}$ on $M^3$. In the same way, we can prove that the derivatives of quaternions ${\tilde{\beta }}_1={\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(\sqrt{3}k-j)\overline{u}\overline{\tilde{h}}$, ${\tilde{\beta }}_2={\displaystyle \frac{1}{2\sqrt{2}}}\tilde{h}u(\sqrt{3}j+k)\overline{u}\overline{\tilde{h}}$ and ${\tilde{\beta }}_3=0$ on $M^3$ are satisfied. We obtain that the three-dimensional submanifold $M^3$, which forms a product manifold, is locally congruent to the immersion $f_3=({\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u},u{\displaystyle \frac{\sqrt{3}-i}{2}}\overline{u})$, where $u\in {\mathbb{S}}^3$, ${\mathbb{S}}^3$ is the Berger sphere with $\kappa =2$ and $\tau ={\displaystyle \frac{1}{\sqrt{3}}}$. In the same way, the curve $\gamma$, the integral curve of the vector field $E_3$, is locally isometric to the curve ${\gamma }_3={\mathcal{F}}_2\left({\mathcal{F}}_1\left({\gamma }_1\right)\right)$.
• Cases $t={\displaystyle \frac{\pi }{2}}$, $t={\displaystyle \frac{7\pi }{6}}$ and $t={\displaystyle \frac{11\pi }{6}}$. Note that if the vector fields $-E_4$ and $-E_6$ are chosen instead of the vector fields $E_4$ and $E_6$, the relations given by (18) are valid for the new variable $t^{\prime }$, which is related to the variable t in the following way $t^{\prime }=t+\pi$, for $t\in [0,\pi )$ and $t^{\prime }=t-\pi$, for $t\in [\pi ,2\pi )$, which means that the cases $t={\displaystyle \frac{\pi }{2}}$ and $t={\displaystyle \frac{3\pi }{2}}$, $t={\displaystyle \frac{\pi }{6}}$ and $t={\displaystyle \frac{7\pi }{6}}$, and the cases $t={\displaystyle \frac{5\pi }{6}}$ and $t={\displaystyle \frac{11\pi }{6}}$ can be connected with each other. If the vector fields $-E_4$ and $-E_6$ are selected for the base vector fields in the cases $t={\displaystyle \frac{\pi }{2}}$, $t={\displaystyle \frac{7\pi }{6}}$ and $t={\displaystyle \frac{11\pi }{6}}$, the result is that the Levi-Civita connections and the expressions of the almost product structure P and the tensor field G in the new base vector fields correspond to the Levi-Civita connections and the expressions of the almost product structure P and the tensor field G in the old base vector fields in the cases $t={\displaystyle \frac{3\pi }{2}}$, $t={\displaystyle \frac{\pi }{6}}$ and $t={\displaystyle \frac{5\pi }{6}}$. The three-dimensional submanifold $M^3$ and $\gamma$, the integral curve of the vector field $E_3$, which form a four-dimensional product manifold M in the cases $t={\displaystyle \frac{\pi }{2}}$ and $t={\displaystyle \frac{3\pi }{2}}$ are respectively locally congruent to immersions $f_1$ and ${\gamma }_1$, given by (47), in the cases $t={\displaystyle \frac{5\pi }{6}}$ and $t={\displaystyle \frac{11\pi }{6}}$, which are respectively locally congruent to immersions $f_2$ and ${\gamma }_2$, given by (48) and in the cases $t={\displaystyle \frac{\pi }{6}}$ and $t={\displaystyle \frac{7\pi }{6}}$, which are respectively locally congruent to immersions $f_3$ and ${\gamma }_3$, given by (49).

This finalizes our proof. □

Lemma 8.

Let M be a four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, such that an almost complex distribution ${\mathcal{D}}_1$ has an arbitrary vector field $E_1$ such that $P{E}_1\in {\mathcal{D}}_2$ holds, then M matches one of the submanifolds obtained by Theorem 4.

Proof. 

Under the assumption that $P{E}_1\in {\mathcal{D}}_2$, the J-relation between the distributions ${\mathcal{D}}_2$ and ${\mathcal{D}}_3$ implies that $P{E}_2\in {\mathcal{D}}_3$, implying that a condition of Theorem 4 is satisfied and it holds. □

6. The Four-Dimensional CR Submanifolds of ${\mathbb{S}}^{\mathbf{3}}\mathbf{\times }{\mathbb{S}}^{\mathbf{3}}$ Such That $\mathit{P}{\mathcal{D}}_{\mathbf{1}}\mathbf{\perp }{\mathcal{D}}_{\mathbf{1}}$

Now, we will use previous theorems to prove that four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$, all belong to the same type. In order to solve the problem, first we will inspect one more special type of four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$.

Lemma 9.

If an almost complex distribution ${\mathcal{D}}_1$ contains an arbitrary vector field $E_1$ such that $P{E}_1=cosmY+sinmJY$, for an angle function $m\in [0,2\pi )$ and for some arbitrary vector field Y belonging to the totally real distribution ${\mathcal{D}}_2$, then an almost complex distribution ${\mathcal{D}}_1$ contains the vector field $X\in {\mathcal{D}}_1$ such that $PX=JY$ and this submanifold matches one of a submanifolds obtained by Theorem 4.

Proof. 

If we choose for a base vector field $E_3\in {\mathcal{D}}_2$ the above mentioned vector field Y, we have that $P{E}_1=cosm{E}_3+sinm{E}_5$, where $m\in [0,2\pi )$ and for $E_2=J{E}_1$, it holds that $P{E}_2=sinm{E}_3-cosm{E}_5$. For an arbitrary vector field $X=cosn{E}_1+sinn{E}_2$, where $n\in [0,2\pi )$, that belongs to an almost complex distribution ${\mathcal{D}}_1$, $PX=cos(m-n)E_3+sin(m-n)E_5$ is obtained and an angle function n could be chosen in a such way that it holds $PX=E_5$. □

Further, we will use a previous lemma to obtain a full classification of this type of submanifold.

Theorem 5.

Let M be a four-dimensional CR submanifold of ${\mathbb{S}}^3\times {\mathbb{S}}^3$ such that $P{\mathcal{D}}_1\perp {\mathcal{D}}_1$ holds, then M belongs to a submanifolds obtained by Theorem 4. M is a locally product manifold of a curve γ and a three-dimensional CR submanifold $M^3$ of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, with a non-integrable almost complex distribution ${\mathcal{D}}_1$ which is P-orthogonal on itself. The three-dimensional submanifold $M^3$ and the curve γ are locally congruent to one of the following three immersions $f_i$ and ${\gamma }_i$, $i\in \overline{1,3}$, given by (47)–(49).

Proof. 

In the proof, the almost product structure P could be expressed in the frame $E_1,\dots ,E_6$ by using Lemma 6, for $\theta ={\displaystyle \frac{\pi }{2}}$. From Equation (10) evaluated for the vector fields $(X,Y)\in \{(E_1,E_1),(E_2,E_1)\}$, along vector fields $E_1$, $E_2$ for all of the mentioned combinations of $(X,Y)$, it follows that

$$\begin{array}{ccc}& & h_{12}^{1}cosm-(h_{11}^1{a}_1+h_{11}^2{a}_2)sinm=0,\hfill \end{array}$$

(65)

$$\begin{array}{ccc}& & h_{11}^{1}cosm+(h_{12}^1{a}_1+h_{12}^2{a}_2)sinm=0,\hfill \end{array}$$

(66)

$$\begin{array}{ccc}& & h_{22}^{1}cosm-(h_{12}^1{a}_1+h_{12}^2{a}_2)sinm=0,\hfill \end{array}$$

(67)

$$\begin{array}{ccc}& & h_{12}^{1}cosm+(h_{22}^1{a}_1+h_{22}^2{a}_2)sinm=0.\hfill \end{array}$$

(68)

The cases $P{E}_1\in {\mathcal{D}}_2$ and $P{E}_1\in {\mathcal{D}}_3$ are examined in Lemma 8 and Theorem 4. Let us now examine the case when $sinm\ne 0$ and $cosm\ne 0$. Equation (66) + (67) = $(h_{11}^1+h_{22}^1)cosm=0$, implies $h_{22}^1=-h_{11}^1$. Further, according to Theorem 3, an almost complex distribution is non-integrable, implying that $h_{11}^2+h_{22}^2\ne 0$. Then, from Equation (68) − (65) = $(h_{11}^2+h_{22}^2)a_{2}sinm=0$, it follows that $a_2=0$, implying that $a_1=\pm 1$. From the expression $P{E}_1=a_{1}sinm{E}_3+cosm{E}_5$, by using Lemma 9, we can conclude that this submanifold belongs to submanifolds given by Theorem 4 and their construction is given in the previous section. □

7. Discussion

In this paper, a very large class of four-dimensional CR submanifolds is obtained. Four-dimensional CR submanifolds are classified for which the almost complex distribution is orthogonal on itself under the action of the almost product structure P. The classification is complete and without any additional restrictions, which makes it one of the largest classifications of submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$. This classification is characterized by the non-integrability of the almost complex distribution. A natural next step would be to study in more detail the conditions for the integrability of the almost complex distribution of four-dimensional CR submanifolds of ${\mathbb{S}}^3\times {\mathbb{S}}^3$, as well as the integrability of the almost product structure P on ${\mathbb{S}}^3\times {\mathbb{S}}^3$. It would also be very interesting to find a physical model on which this theory of different actions of the almost product structure P on the tangent bundle could be realized.