Hypersurfaces of the homogeneous nearly K\"ahler $\mathbb{S}^6$ and $\mathbb{S}^3\times\mathbb{S}^3$ with anticommutative structure tensors

TL;DR

This paper characterizes hypersurfaces in homogeneous nearly Kähler spheres, showing that certain anticommutative tensor conditions imply the hypersurface is totally geodesic or does not exist in specific cases.

Contribution

It provides a complete characterization of hypersurfaces satisfying an anticommutative tensor condition in homogeneous nearly Kähler spheres, revealing their geometric properties.

Findings

Hypersurfaces in $ abla$-Kähler $ ext{S}^6$ satisfying $A+ A=0$ are totally geodesic.

No hypersurfaces in $ abla$-Kähler $ ext{S}^3 imes ext{S}^3$ satisfy $A+ A=0$.

The condition $A+ A=0$ characterizes special geometric structures in these manifolds.

Abstract

Each hypersurface of a nearly K\"ahler manifold is naturally equipped with two tensor fields of $(1,1)$-type, namely the shape operator $A$ and the induced almost contact structure $\phi$. In this paper, we show that, in the homogeneous NK $\mathbb{S}^6$ a hypersurface satisfies the condition $A\phi+\phi A=0$ if and only if it is totally geodesic; moreover, similar as for the non-flat complex space forms, the homogeneous nearly K\"ahler manifold $\mathbb{S}^3\times\mathbb{S}^3$ does not admit a hypersurface that satisfies the condition $A\phi+\phi A=0$.

Equations

$$JZ(p,q)=\frac{1}{\sqrt{3}}(2p{q}^{-1}V-U,-2q{p}^{-1}U+V).$$

$$\begin{array}{rl}g(Z,Z^{\prime })& =\frac{1}{2}(⟨Z,Z^{\prime }⟩+⟨JZ,J{Z}^{\prime }⟩)\\ & =\frac{4}{3}(⟨U,U^{\prime }⟩+⟨V,V^{\prime }⟩)-\frac{2}{3}(⟨p^{-1}U,q^{-1}{V}^{\prime }⟩+⟨p^{-1}{U}^{\prime },q^{-1}V⟩),\end{array}$$

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

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

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

$$\displaystyle\begin{aligned} g(G(X,Y),G(Z,W))=&\tfrac{1}{3}\big{[}g(X,Z)g(Y,W)-g(X,W)g(Y,Z)\\ &\quad+g(JX,Z)g(JW,Y)-g(JX,W)g(JZ,Y)\big{]}.\end{aligned}$$

$$PZ=(p{q}^{-1}V,q{p}^{-1}U),\forall Z=(U,V)\in T_{(p,q)}({\mathbb{S}}^3\times {\mathbb{S}}^3).$$

$$2({\tilde{\nabla }}_{X}P)Y=JG(X,PY)+JPG(X,Y).$$

$$\begin{split}\tilde{R}(X,Y)Z=&\tfrac{5}{12}\big{[}g(Y,Z)X-g(X,Z)Y\big{]}\\ &+\tfrac{1}{12}\big{[}g(JY,Z)JX-g(JX,Z)JY-2g(JX,Y)JZ\big{]}\\ &+\tfrac{1}{3}\Big{[}g(PY,Z)PX-g(PX,Z)PY\\ &\qquad+g(JPY,Z)JPX-g(JPX,Z)JPY\Big{]}.\end{split}$$

$$JX=\varphi X+f(X)\xi ,$$

$$\{\begin{array}{rl}& f(X)=g(X,U),f(\varphi X)=0,{\varphi }^{2}X=-X+f(X)U,\\ & g(\varphi X,Y)=-g(X,\varphi Y),g(\varphi X,\varphi Y)=g(X,Y)-f(X)f(Y),\end{array}$$

$$\begin{array}{r}{\tilde{\nabla }}_{X}Y={\nabla }_{X}Y+h(X,Y),{\tilde{\nabla }}_X\xi =-AX,\forall X,Y\in TM,\end{array}$$

$${\nabla }_{X}U=\varphi AX-G(X,\xi ).$$

$$\begin{split}R(X,Y)Z=&\tfrac{5}{12}\big{[}g(Y,Z)X-g(X,Z)Y\big{]}\\ &+\tfrac{1}{12}\big{[}g(\phi Y,Z)\phi X-g(\phi X,Z)\phi Y-2g(\phi X,Y)\phi Z\big{]}\\ &+\tfrac{1}{3}\Big{[}g(PY,Z)(PX)^{\top}-g(PX,Z)(PY)^{\top}\\ &\qquad\,\ +g(JPY,Z)(JPX)^{\top}-g(JPX,Z)(JPY)^{\top}\Big{]}\\ &+g(AZ,Y)AX-g(AZ,X)AY,\end{split}$$

$$\begin{split}(\nabla_{X}A)Y-(\nabla_{Y}A)X=&\tfrac{1}{12}\big{[}g(X,U)\phi Y-g(Y,U)\phi X-2g(\phi X,Y)U\big{]}\\ &+\tfrac{1}{3}\Big{[}g(PX,\xi)(PY)^{\top}-g(PY,\xi)(PX)^{\top}\\ &\qquad+g(PX,U)(JPY)^{\top}-g(PY,U)(JPX)^{\top}\Big{]},\end{split}$$

$$\frac{1}{6}$$

$$=g((\mu I-A)G(X,\xi ),Y)+g(G((\mu I-A)X,\xi ),Y)$$

$$-\mu g((A\varphi +\varphi A)X,Y)+2g(A\varphi AX,Y),X,Y\in \{U{\}}^{\perp },$$

$$({\nabla }_{X}A)U=X(\mu )U+(\mu I-A)(-G(X,\xi )+\varphi AX).$$

$$g(({\nabla }_{X}A)Y,U)=g(({\nabla }_{X}A)U,Y)=g((\mu I-A)(-G(X,\xi )+\varphi AX),Y).$$

$$g(({\nabla }_{X}A)Y-({\nabla }_{Y}A)X,U)=$$

$$-g(G((\mu I-A)X,\xi ),Y)+\mu g((A\varphi +\varphi A)X,Y).$$

$$g(({\nabla }_{X}A)Y-({\nabla }_{Y}A)X,U)$$

$$=-\frac{1}{6}g(\varphi X,Y)+\frac{2}{3}(g(PX,\xi )g(PY,U)-g(PX,U)g(PY,\xi )).$$

$$\mathfrak{D}(p):=\mathrm{Span}\{\xi (p),U(p),P\xi (p),PU(p)\}.$$

$$P\xi =a\xi +bU+c{e}_1,a^2+b^2+c^2=1.$$

$$\{\begin{array}{rl}& P\xi =a\xi +c{e}_1+b{e}_5,P{e}_1=c\xi -a{e}_1-b{e}_2,\\ & P{e}_2=c{e}_5-b{e}_1+a{e}_2,P{e}_3=e_3,\\ & P{e}_4=-e_4,P{e}_5=b\xi +c{e}_2-a{e}_5.\end{array}$$

$$P\xi =a\xi +bU,a^2+b^2=1.$$

$$({\nabla }_{X}A)Y=({\nabla }_{Y}A)X,\forall X,Y\in TM.$$

$$g((\mu I-A)$$

$$-\mu g((A\varphi +\varphi A)X,Y)+2g(A\varphi AX,Y)=0,X,Y\in TM.$$

$$\{\begin{array}{rl}& A{X}_1=\tan (\theta +r)X_1,A{X}_2=\tan (\theta -r)X_2,A{X}_3=-\cot r{X}_3,\\ & A{X}_4=-\cot r{X}_4,A{X}_5=-\cot r{X}_5,X_5=U.\end{array}$$

$$A\varphi X_1=-\tan (\theta +r)\varphi X_1,A\varphi X_2=-\tan (\theta -r)\varphi X_2,A\varphi X_3=\cot r\varphi X_3.$$

$$\{\begin{array}{rl}& A{X}_1=\tan (\theta -r)X_1,A{X}_2=-\cot r{X}_2,A{X}_3=-\cot r{X}_3,\\ & A{X}_4=-\cot r{X}_4,A{X}_5=-\cot r{X}_5,X_5=U.\end{array}$$

Full text

2010 Mathematics Subject Classification.  53B35, 53C30, 53C42.

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

1. Introduction

The nearly Kähler (abbrev. NK) manifold $\mathbb{S}^{3}\times\mathbb{S}^{3}$ is one of the only four homogeneous $6$-dimensional nearly Kähler spaces (with the remaining three the NK $6$-sphere $\mathbb{S}^{6}$, the complex projective space $\mathbb{C}P^{3}$ and the flag manifold $SU(3)/U(1)\times U(1)$, cf. [5, 6]). Ever since the groundbreaking research of Bolton-Dillen-Dioos-Vrancken [4], people become increasingly interested in the study of submanifolds of this homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$, and many beautiful results have been established. For details of the study, besides [4], we would refer the readers to [8, 12] on almost complex surfaces, to [1, 2, 9, 13, 18] on Lagrangian submanifolds, and to [11] on hypersurfaces. It is worth mentioning that Foscolo and Haskins [10] have recently constructed cohomogeneity one NK structure on both $\mathbb{S}^{6}$ and $\mathbb{S}^{3}\times\mathbb{S}^{3}$. Thus, in order to avoid confusion, from now on in this paper, when we say NK $\mathbb{S}^{6}$ and NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$, we mean always $\mathbb{S}^{6}$ and $\mathbb{S}^{3}\times\mathbb{S}^{3}$ equipped with the homogeneous NK structures that were elaborate described in [7] (cf. references therein) and [4], respectively.

In the present paper, continuing with our research starting from [11], we will focus mainly on hypersurfaces of the NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$. Recall that given a hypersurface $M$ of an almost Hermitian manifold with almost complex structure $J$, it appears on $M$ two naturally defined tensor fields of $(1,1)$-type: a submanifold structure represented by the shape operator $A$, and an almost contact structure $\phi$ induced from $J$. Then, it is an interesting problem to study hypersurfaces with special relations between $A$ and $\phi$. The first problem one might study is that the shape operator $A$ and the induced almost contact structure $\phi$ satisfy the commutativity condition $A\phi=\phi A$. Indeed, Okumura [17] and Montiel-Romero [16] considered real hypersurfaces of the non-flat complex space forms, and they obtained the classification of such real hypersurfaces satisfying $A\phi=\phi A$ for complex projective space [17] and complex hyperbolic space [16], respectively. Moreover, it was shown that hypersurfaces of the homogeneous NK $\mathbb{S}^{6}$ satisfy $A\phi=\phi A$ if and only if they are geodesic hyperspheres (cf. Theorem 2 of [15] and Remark 2.1 of [11]). Then following this approach, we have considered a similar situation for the NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ [11], our result is the following classification theorem.

Theorem 1.1** (cf. [11]).**

Let $M$ be a hypersurface of the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ that satisfies the condition $A\phi=\phi A$. Then $M$ is locally given by one of the following immersions $f_{1}$, $f_{2}$ and $f_{3}$:

1. (1)

$f_{1}:\ \ \mathbb{S}^{3}\times\mathbb{S}^{2}\rightarrow\mathbb{S}^{3}\times\mathbb{S}^{3}\ \ {\rm defined\ by}\ \ (x,y)\mapsto(x,y)$; 2. (2)

$f_{2}:\ \ \mathbb{S}^{3}\times\mathbb{S}^{2}\rightarrow\mathbb{S}^{3}\times\mathbb{S}^{3}\ \ {\rm defined\ by}\ \ (x,y)\mapsto(y,x)$; 3. (3)

$f_{3}:\ \ \mathbb{S}^{3}\times\mathbb{S}^{2}\rightarrow\mathbb{S}^{3}\times\mathbb{S}^{3}\ \ {\rm defined\ by}\ \ (x,y)\mapsto(\bar{x},y\bar{x})$,

here, $x\in\mathbb{S}^{3}$, $y\in\mathbb{S}^{2}$, and as usual $\mathbb{S}^{3}$ (resp. $\mathbb{S}^{2}$) is regarded as the set of the unit (resp. imaginary) quaternions in the quaternion space $\mathbb{H}$.

One might realize that the next simplest relation between the shape operator $A$ and the induced almost contact structure $\phi$ is the anti-commutativity condition $A\phi+\phi A=0$. In this respect, to our knowledge only Ki-Suh have shown that (cf. Lemma 2.1 and Proposition 2.2 of [14]), by denoting $\bar{M}^{n}(c)$ the $n$-dimensional complex space form of constant holomorphic sectional curvature $c$, if there exists a real hypersurface $M$ of $\bar{M}^{n}(c)$ that satisfies the condition $A\phi+\phi A=0$, then $c=0$ and $M$ is cylindrical. To see how about other ambient spaces, in this paper, we consider the question for two important $6$-dimensional homogeneous NK manifolds, namely that the homogeneous NK $\mathbb{S}^{6}$ and the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$. Our first result is the following

Theorem 1.2**.**

The totally geodesic hypersurfaces of the homogeneous NK $\mathbb{S}^{6}$ are the only hypersurfaces of $\mathbb{S}^{6}$ satisfying the condition $A\phi+\phi A=0$.

For the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$, however, in Theorem 1.1 of [11], we have shown that it admits neither totally umbilical hypersurfaces nor hypersurfaces having parallel second fundamental form. Now, as the second result of this paper, a further nonexistence theorem can be proved that is stated as below.

Theorem 1.3**.**

The homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ does not admit a hypersurface that satisfies the condition $A\phi+\phi A=0$.

2. Preliminaries

2.1. The homogeneous NK structure on $\mathbb{S}^{3}\times\mathbb{S}^{3}$

In this subsection, we review some elementary notions and results from [4].

By the natural identification $T_{(p,q)}(\mathbb{S}^{3}\times\mathbb{S}^{3})\cong T_{p}\mathbb{S}^{3}\oplus T_{q}\mathbb{S}^{3}$, we can write a tangent vector at $(p,q)\in\mathbb{S}^{3}\times\mathbb{S}^{3}$ as $Z(p,q)=(U_{(p,q)},V_{(p,q)})$ or simply $Z=(U,V)$. Then the well-known almost complex structure $J$ on $\mathbb{S}^{3}\times\mathbb{S}^{3}$ is given by

[TABLE]

Define the Hermitian metric $g$ on $\mathbb{S}^{3}\times\mathbb{S}^{3}$ by

[TABLE]

where $Z=(U,V),\ Z^{\prime}=(U^{\prime},V^{\prime})$ are tangent vectors, and $\langle\cdot,\cdot\rangle$ is the standard product metric on $\mathbb{S}^{3}\times\mathbb{S}^{3}$. Then $\{g,J\}$ give the homogeneous NK structure on $\mathbb{S}^{3}\times\mathbb{S}^{3}$.

As usual let $G$ be the (1,2)-tensor field defined by $G(X,Y):=(\tilde{\nabla}_{X}J)Y$, where $\tilde{\nabla}$ is Levi-Civita connection of $g$. Then, the following further formulas hold:

[TABLE]

An almost product structure $P$ on $\mathbb{S}^{3}\times\mathbb{S}^{3}$ is introduced by:

[TABLE]

Then we have the following formula for $\tilde{\nabla}P$:

[TABLE]

The curvature tensor $\tilde{R}$ of the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ is given by:

[TABLE]

2.2. Hypersurfaces of the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$

Let $M$ be a hypersurface of the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ with $\xi$ its unit normal vector field. For any vector field $X$ tangent to $M$, we have the decomposition

[TABLE]

where $\phi X$ and $f(X)\xi$ are, respectively, the tangent and normal parts of $JX$. Then $\phi$ is a tensor field of type (1,1), and $f$ is a $1$-form on $M$. By definition, $\phi$ and $f$ satisfy the following relations:

[TABLE]

where $U:=-J\xi$, which is called the structure vector field of $M$. The equations (2.11) show that $(\phi,U,f)$ determines an almost contact structure over $M$.

Let $\nabla$ be the induced connection on $M$ with $R$ its Riemannian curvature tensor. The formulas of Gauss and Weingarten state that

[TABLE]

where $h$ is the second fundamental form, and it is related to the shape operator $A$ by $h(X,Y)=g(AX,Y)\xi$. Here, using the formulas of Gauss and Weingarten, we have

[TABLE]

The Gauss and Codazzi equations of $M$ are given by

[TABLE]

[TABLE]

where $\cdot^{\top}$ denotes the tangential part.

Following the usual terminology, we call a hypersurface $M$ of the NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ the Hopf hypersurface if the integral curves of the structure vector field $U$ are geodesics of $M$, that is $\nabla_{U}U=0$. It is also equivalent that the structure vector field $U$ is a principal direction, with principal curvature function denoted by $\mu$. A basic lemma for Hopf hypersurfaces of the NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ is stated as follows:

Lemma 2.1**.**

Let $M$ be a Hopf hypersurface in the homogeneous NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$. Then we have

[TABLE]

where $\{U\}^{\bot}$ denotes a distribution of $TM$ that is orthogonal to $U$, and $I$ denotes the identity transformation.

Proof.

A direct calculation of $(\nabla_{X}A)U$, with using $AU=\mu U$, (2.13), we have

[TABLE]

It follows that, for $\forall\,X,Y\in\{U\}^{\bot}$,

[TABLE]

Thus, we have

[TABLE]

On the other hand, by using the Codazzi equation (2.15), we get

[TABLE]

From (2.19) and (2.20), we immediately get (2.16). ∎

Before concluding this section, following that in [11] we introduce the distribution $\mathfrak{D}$. When we study hypersurfaces of the NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$, the consideration of $\mathfrak{D}$ is very helpful for the choice of a canonical frame. Precisely, for each point $p\in M$, we define

[TABLE]

Since $P$ is anti-commutative with $J$, it is clear that $\mathfrak{D}$ defines a distribution on $M$ with dimension $2$ or $4$, and that it is invariant under the action of both $J$ and $P$. Along $M$, let $\mathfrak{D}^{\bot}$ denote the distribution in $T(\mathbb{S}^{3}\times\mathbb{S}^{3})$ that is orthogonal to $\mathfrak{D}$ at each $p\in M$.

If $\dim\mathfrak{D}=4$ holds in an open set, then there exists a unit tangent vector field $e_{1}\in\mathfrak{D}$ and functions $a,b,c$ with $c>0$ such that

[TABLE]

Put $e_{2}=Je_{1}$. From the fact $\dim\,\mathfrak{D}^{\bot}=2$ and that $\mathfrak{D}^{\bot}$ is invariant under the action of both $J$ and $P$, we can choose a local unit vector field $e_{3}\in\mathfrak{D}^{\bot}$ such that $Pe_{3}=e_{3}$. Put $e_{4}=Je_{3}$ and $e_{5}=U$. Then $\{e_{i}\}_{i=1}^{5}$ is a well-defined orthonormal basis of $TM$ and, acting by $P$, it has the following properties:

[TABLE]

If $\dim\mathfrak{D}=2$ holds in an open set, then we can write

[TABLE]

Now, $\mathfrak{D}^{\bot}$ is a $4$-dimensional distribution that is invariant under the action of both $J$ and $P$. Hence, we can choose unit vector fields $e_{1},\,e_{3}\in\mathfrak{D}^{\bot}$ such that $Pe_{1}=e_{1},\,Pe_{3}=e_{3}$. Put $e_{2}=Je_{1},\,e_{4}=Je_{3}$ and $e_{5}=U$. In this way, we obtain an orthonormal basis $\{e_{i}\}_{i=1}^{5}$ of $TM$. However, we would remark that such choice of $\{e_{1},e_{3}\}$ (resp. $\{e_{2},e_{4}\}$) is unique up to an orthogonal transformation.

3. Proof of Theorem 1.2

For basic results of the well-known NK $\mathbb{S}^{6}$, i.e., the six-dimensional unit sphere $\mathbb{S}^{6}$ equipped with a homogeneous NK structure $(J,g)$, of which $J$ is the almost complex structure defined by using the vector cross product of purely imaginary Cayley numbers $\mathbb{R}^{7}$ and $g$ is the metric induced from the Euclidean space $\mathbb{R}^{7}$, we refer to [7] and the references therein.

Let $M$ be an orientable hypersurface of the NK $\mathbb{S}^{6}$ with $\xi$ its unit normal vector field. Then, the equations from (2.10) up to (2.13) in subsection 2.2 also hold, so that $M$ admits an almost contact metric structure $(\phi,U,f,g)$ induced from the NK structure of $\mathbb{S}^{6}$, whereas the Codazzi equation becomes

[TABLE]

For the NK $\mathbb{S}^{6}$, totally geodesic hypersurfaces do exist and they trivially satisfy the relation $A\phi+\phi A=0$.

Now, we assume that $M$ is an orientable hypersurface of the NK $\mathbb{S}^{6}$ that satisfies the condition $A\phi+\phi A=0$. Then, by definition $\phi U=0$, we have $AU=\mu U$, i.e., $M$ is a Hopf hypersurface and, $\mu$ is the principal curvature function corresponding to the structure vector field $U$. Moreover, if $X\in\{U\}^{\bot}$ is a principal vector field with principal curvature function $\lambda$, then $A\phi X=-\phi AX=-\lambda\phi X$ implies that $\phi X$ is also a principal vector field with principal curvature function $-\lambda$.

Recall that Berndt-Bolton-Woodward (Theorem 2 of [3]) proved that a connected Hopf hypersurface of the NK $\mathbb{S}^{6}$ is an open part of either a geodesic hypersphere of $\mathbb{S}^{6}$ or a tube around an almost complex curve in the NK $\mathbb{S}^{6}$, and the principal curvature function $\mu$ is constant (Lemma 2 of [3]).

Similar to the proof of Lemma 2.1, for Hopf hypersurfaces of the NK $\mathbb{S}^{6}$, we can easily show that, by using (2.13), the following basic equation holds:

[TABLE]

If $M$ is a geodesic hypersphere, then $M$ is totally umbilical and we have a function $\lambda$ on $M$ such that $AX=\lambda X,\forall\,X\in TM$. This together with $A\phi+\phi A=0$ implies that $\lambda=0$. Hence, $M$ is a totally geodesic hypersurface.

If $M$ is a tube around an almost complex curve $\Gamma$ with radius $r$ in $\mathbb{S}^{6}$, then, according to the proof of Proposition 2 and subsequent Remark in [3], we have $AU=-\cot r\,U$, and the remaining principal curvatures on the distribution $\{U\}^{\bot}$ are $\tan(\theta+r),\,\tan(\theta-r)$ and $-\cot r$ for $\theta\in[0,\frac{\pi}{2})$ which is a function on $M$. Moreover, as [3] has pointed out, the hypersurface $M$ has exactly two or three distinct principal curvatures at each point. We denote by $\nu,\ 2\leq\nu\leq 3,$ the maximum number of distinct principal curvatures on $M$, then the set M_{\nu}=\{x\in M|$$M has exactly $\nu$ distinct principal curvatures at $x$} is a non-empty open subset of $M$. By the continuity of the principal curvature function, each connected component of $M_{\nu}$ is an open subset, and the multiplicities of distinct principal curvatures remain unchanged on each connected component of $M_{\nu}$, so we can find a local smooth frame field with respect to the principal curvatures. The following discussion will be divided into two cases, depending on the value of $\nu$.

Case I. $\nu=3$.

In this case, on each connected component of $M_{3}$, the multiplicities of the distinct principal curvatures, namely $\tan(\theta+r)$, $\tan(\theta-r)$ and $-\cot r$, should be $1,1$ and $3$, respectively. Then we have an orthonormal frame field $\{X_{i}\}_{i=1}^{5}$ such that

[TABLE]

Applying the condition $A\phi+\phi A=0$, we have

[TABLE]

Taking $X=X_{1}$ and $Y=\phi X_{1}$ in (3.2), and using $A\phi+\phi A=0$, we get $\tan(\theta+r)=0$. Analogously, taking $X=X_{2}$ and $Y=\phi X_{2}$ in (3.2), we get $\tan(\theta-r)=0$, which is a contradiction with $\tan(\theta+r)\neq\tan(\theta-r)$. Thus, Case I does not occur.

Case II. $\nu=2$.

In this case, $M$ has exactly two distinct principal curvatures, that is, two of the three principal curvatures $\tan(\theta+r)$, $\tan(\theta-r)$ and $-\cot r$ are equal. Without loss of generality, we assume that $\tan(\theta+r)=-\cot r$, so that the multiplicities of the distinct principal curvatures, namely $\tan(\theta-r)$ and $-\cot r$, are $1$ and $4$, respectively. Then, we have an orthonormal frame field $\{X_{i}\}_{i=1}^{5}$ such that

[TABLE]

Applying $A\phi+\phi A=0$, we get $A\phi X_{1}=-\tan(\theta-r)\phi X_{1}$ and $A\phi X_{2}=\cot r\phi X_{2}$. Then taking in (3.2) that $(X,Y)=(X_{1},\phi X_{1})$ and $(X,Y)=(X_{2},\phi X_{2})$, respectively, we immediately get $\tan(\theta-r)=-\cot r=0$. This is again a contradiction.

This completes the proof of Theorem 1.2.∎

4. Proof of Theorem 1.3

To give the proof, we assume that $M$ is a hypersurface of the NK $\mathbb{S}^{3}\times\mathbb{S}^{3}$ which satisfies the condition $A\phi+\phi A=0$. Then, by the fact $\phi U=0$, we see that $M$ is a Hopf hypersurface with $AU=\mu U$. Moreover, if $X\in\{U\}^{\bot}$ is a principal vector field with principal curvature function $\lambda$, i.e., $AX=\lambda X$, then $A\phi X=-\phi AX=-\lambda\phi X$ implies that $\phi X$ is also a principal vector field with principal curvature function $-\lambda$. We denote $\lambda,\,-\lambda,\,\beta,\,-\beta$ with $\lambda\geq 0$ and $\beta\geq 0$ the four principal curvatures on distribution $\{U\}^{\bot}$. Since the only possible dimension of $\mathfrak{D}$ is $2$ or $4$, we will divide the proof of Theorem 1.3 into the proofs of two Lemmas. First, we have the following Lemma.

Lemma 4.1**.**

The case $\dim\mathfrak{D}=4$ does not occur.

Proof.

Suppose that $\dim\mathfrak{D}=4$ does occur on some point of $M$. We denote by $\Omega=\{x\in M|$ the dimension of $\mathfrak{D}$ is 4 at $x$}, then $\Omega$ is an open set of $M$. Since $A\phi+\phi A=0$, we can write (2.16) on $\Omega$ as

[TABLE]

We denote by $\nu$ $(\nu\leq 5)$ the maximum number on $\Omega$ of distinct principal curvatures, then the set $\Omega_{\nu}:=\{x\in\Omega\,|\,M$ has exactly $\nu$ distinct principal curvatures at $x$} is a non-empty open subset of $M$. By the continuity of the principal curvature function, each connected component of $\Omega_{\nu}$ is an open subset, the multiplicities of distinct principal curvatures remain unchanged on each connected component of $\Omega_{\nu}$, so we can find a local smooth frame field with respect to the principal curvatures. From Theorem 1.1 of [11], we know that $M$ can not be totally umbilical, even locally. So the following discussion will be divided into four cases, depending on the value of $\nu,\ 2\leq\nu\leq 5$.

Case I. $\nu=5$.

In this case, on each connected component of $\Omega_{5}$, we can have an orthonormal frame field $\{X_{i}\}_{i=1}^{5}$ such that

[TABLE]

where $X_{3}=JX_{1},\,X_{4}=JX_{2},\,X_{5}=U$. As $\nu=5$, we have $\lambda>0,\,\beta>0,\,\lambda\neq\beta$ and $\mu\not\in\{\lambda,-\lambda,\,\beta,-\beta\}$. Let $\{e_{i}\}_{i=1}^{5}$ be the frame field as described in (2.22). Then, by assuming that $X_{i}=\sum_{j=1}^{4}a_{ij}e_{j}$ for $1\leq i\leq 4$, we have $(a_{ij})\in SO(4)$, and by the choice of $\{e_{i}\}_{i=1}^{5}$ it holds that

[TABLE]

First, taking $X=X_{i}$ and $Y=X_{j}$ in (4.1) for $1\leq i<j\leq 4$, using (2.3)–(2.5) and (2.22), we can derive the following equations:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The equations (4.6) and (4.7), (4.8) and (4.9) imply that

[TABLE]

From (2.3), (2.4) and (2.5) we see that, for $1\leq i\leq 4$, it holds $g(G(X_{1},X_{2}),X_{i})=0$. Thus, $G(X_{1},X_{2})\in{\rm Span}\,\{\xi,U\}$. On the other hand, from (2.6), we have

[TABLE]

It follows from (4.10) that $\mu=0$.

Second, from the fact $AU=0$, we have

[TABLE]

On the other hand, applying (2.22) to the Codazzi equation (2.15), we can get

[TABLE]

[TABLE]

Then, from (4.12) and (4.13), calculating the $U$-component of both the right hand sides, we can get $ac=0$. Analogously, from (4.12) and (4.14), we can get $bc=0$. Therefore, according to (2.21), we have $a=b=0$ and $c=1$.

Third, in order to apply the Codazzi equations, we need to calculate the connections $\{\nabla_{X_{i}}X_{j}\}$. Put $\nabla_{X_{i}}X_{j}=\sum\Gamma_{ij}^{k}X_{k}$ with $\Gamma_{ij}^{k}=-\Gamma_{ik}^{j}$, $1\leq i,j,k\leq 5$. Assume that

[TABLE]

Then (4.11) and the fact $G(X_{1},X_{2})\in{\rm Span}\,\{\xi,U\}$ show that $k^{2}+l^{2}=\frac{1}{3}$.

By definition and the Gauss and Weingarten formulas, we have the calculation

[TABLE]

However, according to (4.15), we also have $G(X_{1},\xi)=-kX_{2}+lX_{4}$. Hence, we obtain

[TABLE]

Similarly, taking $(X,Y)=(X_{i},\xi)$ in $G(X,Y)=(\tilde{\nabla}_{X}J)Y$ for $2\leq i\leq 4$, and by use of (4.15), we further obtain

[TABLE]

Moreover, by using (4.15) and the Gauss and Weingarten formulas, we get

[TABLE]

It follows that

[TABLE]

Finally, we will calculate the expressions $(\nabla_{U}A){e_{i}}-(\nabla_{e_{i}}A)U$ for $1\leq i\leq 4$.

On one hand, for each $1\leq i\leq 4$, we directly calculate $(\nabla_{U}A)e_{i}-(\nabla_{e_{i}}A)U$, with the use of $e_{i}=\sum_{j=1}^{4}{a_{ji}}X_{j}$ and the preceding results (4.16) and (4.17). Then we get an expression for $(\nabla_{U}A){e_{i}}-(\nabla_{e_{i}}A)U$ in terms of the frame field $\{X_{i}\}_{i=1}^{4}$.

On the other hand, for each $1\leq i\leq 4$, we calculate $(\nabla_{U}A)e_{i}-(\nabla_{e_{i}}A)U$ by the Codazzi equation (2.15). Then, by using (2.22) and $e_{i}=\sum_{j=1}^{4}{a_{ji}}X_{j}$, we get another expression of $(\nabla_{U}A){e_{i}}-(\nabla_{e_{i}}A)U$ in terms of the frame field $\{X_{i}\}_{i=1}^{4}$.

In this way, comparing both calculations of $(\nabla_{U}A)e_{i}-(\nabla_{e_{i}}A)U$ for each $1\leq i\leq 4$, we get a matrices equation $C=(a_{ij})^{T}B$, where

[TABLE]

[TABLE]

Thus, $B=(a_{ij})C:=(B_{ij})$. Using (4.3), it is straightforward to verify that $B=(a_{ij})C$ is skew-symmetric. From the facts $B_{12}+B_{21}=0$ and $\lambda\neq\beta$, we have $\Gamma_{51}^{2}=\frac{k}{2}$. Moreover, from the facts $B_{34}+B_{43}=0$ and $\lambda\neq\beta$, we have $\Gamma_{53}^{4}=-\frac{k}{2}$. Combining these with (4.19) we get $k=0$. Analogously, from the facts $B_{23}+B_{32}=0$, $B_{14}+B_{41}=0$, $\lambda+\beta\neq 0$ and (4.19), we can further get $l=0$. Thus, we get a contradiction to $k^{2}+l^{2}=\tfrac{1}{3}$. This implies that Case I does not occur.

Case II. $\nu=4$.

In this case, on a connected component of $\Omega_{4}$, without loss of generality, we are sufficient to consider the following two subcases:

II-(i): $\lambda\neq\beta,\ \lambda>0,\ \beta>0$ and $\mu\in\{\lambda,\beta,-\lambda,-\beta\}$.

II-(ii): $\lambda=0,\ \beta>0$ and $\mu\not\in\{0,\beta,-\beta\}$.

For both of the above two subcases, following similar arguments as the discussion of Case I from (4.2) up to (4.11), we can also get $\mu=0$. This is a contradiction, showing that Case II does not occur.

Case III. $\nu=3$.

In this case, on a connected component of $\Omega_{3}$, without loss of generality, we are sufficient to consider the following three subcases:

III-(i): $\lambda=0,\ \beta>0$ and $\mu\in\{\beta,-\beta\}$.

III-(ii): $\lambda=\mu=0$ and $\beta>0$.

III-(iii): $\lambda=\beta>0$ and $\mu\not\in\{\lambda,-\lambda\}$.

In case III-(i), similar arguments as the discussion of Case I from (4.2) up to (4.11), we can get $\mu=0$. Thus, we get a contradiction.

In case III-(ii), taking an orthonormal frame field $\{X_{i}\}_{i=1}^{5}$ satisfying (4.2), we still have the equations from (4.4) up to (4.14). Then we can get $c=1$. By calculating (4.4)+(4.5) and that $(a_{ij})\in SO(4)$, we further have the conclusion

[TABLE]

By $\lambda=0$, we have $\beta=\frac{\sqrt{6}}{6}$. Then (4.4) and (4.5) give that

[TABLE]

On the other hand, making the summation $\eqref{eq:4.6}^{2}+\eqref{eq:4.8}^{2}$, we easily see that

[TABLE]

which is a contradiction to (4.21).

In case III-(iii), taking an orthonormal frame field $\{X_{i}\}_{i=1}^{5}$ satisfying (4.2), we can also derive the equations from (4.4) up to (4.11), thus we have $\mu=0$. Then, similarly, we have the equations from (4.12) up to (4.14), so we get in (2.22) that $a=b=0$ and $c=1$, and by calculating (4.4)+(4.5), we get $\lambda=\beta=\tfrac{\sqrt{3}}{6}$. It follows from (4.4), (4.5) and (4.6) that

[TABLE]

Let us put $a_{11}=\tfrac{1}{\sqrt{2}}\cos\theta_{1}$, $a_{12}=\tfrac{1}{\sqrt{2}}\sin\theta_{1}$, $a_{21}=\tfrac{1}{\sqrt{2}}\cos\theta_{2}$ and $a_{22}=\tfrac{1}{\sqrt{2}}\sin\theta_{2}$. Then $0=a_{11}a_{21}+a_{12}a_{22}=\tfrac{1}{2}\cos(\theta_{1}-\theta_{2})$ implies that $\theta_{1}-\theta_{2}=\tfrac{\pi}{2}(2k+1)$, $k\in\mathbb{Z}$. Therefore, we have either $(a_{21},a_{22})=(a_{12},-a_{11})$ or $(a_{21},a_{22})=(-a_{12},a_{11})$. If necessary by taking $-X_{2}$ instead of $X_{2}$, we are sufficient to consider the case that $a_{21}=a_{12}$ and $a_{22}=-a_{11}$.

From (4.22) and that $(a_{ij})\in SO(4)$, we further have

[TABLE]

This implies that, similar to the preceding paragraph, $(a_{23},a_{24})=(a_{14},-a_{13})$ or $(a_{23},a_{24})$ $=(-a_{14},a_{13})$. If $a_{23}=a_{14}$ and $a_{24}=-a_{13}$, then $X_{2}=-X_{3}$, which is impossible. Thus, $a_{23}=-a_{14}$ and $a_{24}=a_{13}$ hold.

For simplicity, we put $m=-\tfrac{2\sqrt{6}}{3}a_{13}a_{14}$ and $n=\tfrac{\sqrt{6}}{3}(a_{14}^{2}-a_{13}^{2})$. Then $m^{2}+n^{2}=\tfrac{1}{6}$.

Now, from (2.22) we can express $\{PX_{i}\}_{i=1}^{4}$ as follows:

[TABLE]

Then, applying the Codazzi equation (2.15), we get

[TABLE]

[TABLE]

Let $\nabla_{X_{i}}X_{j}=\sum\Gamma_{ij}^{k}X_{k}$ with $\Gamma_{ij}^{k}=-\Gamma_{ik}^{j}$, $1\leq i,j,k\leq 5$. Then, from (4.24) and (4.25), after calculating the left hand sides of (4.24) and (4.25), we get

[TABLE]

Next, (4.8) gives that $g(G(X_{1},X_{2}),\xi)=\tfrac{\sqrt{3}}{3}$, and so that $g(G(X_{1},X_{2}),U)=0$ from (4.11). Then by the relations (2.3)–(2.5) we can easily solve $G(X_{1},\xi)=-\tfrac{\sqrt{3}}{3}X_{2}$. Thus, by the Gauss and Weingarten formulas, a direct calculation gives that

[TABLE]

Hence, we have

[TABLE]

By (4.26) and (4.28), we obtain

[TABLE]

Now, using that $G(X_{1},X_{2})=\tfrac{\sqrt{3}}{3}\xi$ and $G(X_{1},\xi)=-\tfrac{\sqrt{3}}{3}X_{2}$, $a_{11}^{2}+a_{12}^{2}=\tfrac{1}{2}$ and $m^{2}+n^{2}=\frac{1}{6}$, (4.23) and (4.29), by direct calculations of both sides of

[TABLE]

we obtain the following equations:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Then, carrying the computations $\eqref{eq:4.30}\times a_{12}-\eqref{eq:4.31}\times a_{11}$ and $\eqref{eq:4.32}\times m+\eqref{eq:4.33}\times n$, respectively, we get

[TABLE]

If $a_{11}=0$, we get $a_{12}^{2}=\tfrac{1}{2}$, $m=0$ and $n^{2}=\tfrac{1}{6}$. Inserting these into (4.32), we obtain $\Gamma_{12}^{1}=0$. Then by (4.31), we have $n=0$. This yields a contradiction.

If $a_{11}\neq 0$, it holds that $a_{11}^{2}=\tfrac{1}{2}$, $a_{12}=0$, $m^{2}=\tfrac{1}{6}$ and $n=0$. Then by (4.30) and (4.33), we have $\tfrac{\sqrt{2}m}{a_{11}}=\Gamma_{12}^{1}=-\tfrac{\sqrt{2}a_{11}}{3m}$. This contradicts to the facts $a_{11}^{2}=\tfrac{1}{2}$ and $m^{2}=\tfrac{1}{6}$.

Thus, Case III does not occur.

Case IV. $\nu=2$.

In this case, we restrict the discussion on a connected component of $\Omega_{2}$. It is easily seen that we are sufficient, without loss of generality, to consider the following two subcases:

IV-(i): $\lambda=\beta>0,\ \mu\in\{\lambda,-\lambda\}$.

IV-(ii): $\lambda=\beta=0,\ \mu\neq 0$.

Actually, for both of the above two subcases, following similar arguments as in the discussion of Case I from (4.2) up to (4.11), we can also get $\mu=0$. This is a contradiction, showing that Case IV does not occur.

We have completed the proof of Lemma 4.1. ∎

Next, we have the following Lemma.

Lemma 4.2**.**

The case $\dim\mathfrak{D}=2$ does not occur either.

Proof.

In this case, we denote still by $\nu,\ \nu\leq 5$, the maximum number of distinct principal curvatures of $M$. Then the set $M_{\nu}=\{x\in M\,|\,M$ has exactly $\nu$ distinct principal curvatures at $x\}$ is a non-empty open subset of $M$. By the continuity of the principal curvature function, each connected component of $M_{\nu}$ is an open subset, the multiplicities of distinct principal curvatures remain unchanged on each connected component of $M_{\nu}$. So we can choose a local smooth frame field with respect to the principal curvatures.

Now, by assumption $A\phi+\phi A=0$ and Lemma 2.1, we can write (2.16) as:

[TABLE]

In a connected component of $M_{\nu}$, we take a local orthonormal frame field $\{X_{i}\}_{i=1}^{5}$ of $M$ such that

[TABLE]

where $X_{3}=JX_{1},\,X_{4}=JX_{2},\,X_{5}=U$. Then, taking $(X,Y)=(X_{1},\phi X_{1})$ in (4.34), with using $AX_{1}=\lambda X_{1}$ and $A\phi X_{1}=-\lambda\phi X_{1}$, we get $-\tfrac{1}{6}=2\lambda^{2}$, this is impossible and hence, we have proved Lemma 4.2. ∎

Finally, from Lemmas 4.1, 4.2 and the fact that $\dim\,\mathfrak{D}$ can only be $2$ or $4$ at each point of $M$, we get immediately the assertion of Theorem 1.3. ∎