1. Introduction
In the study of submanifolds in an ambient Riemannian manifold, the classification of hypersurfaces in terms of the Ricci tensor is an interesting topic. Recall that the Ricci operator
$\textrm {Ric}$
of a Riemannian manifold
$M$
is said to be recurrent if there exists a
$1$
-form
$\omega$
on
$M$
such that, for any vector fields
$X,Y$
of
$M$
, it holds
\begin{equation} (\nabla _X\textrm {Ric})Y=\omega (X)\textrm {Ric}(Y), \end{equation}
where
$\nabla$
is the Levi-Civita connection on
$M$
. Obviously, (1.1) is weaker than that
$M$
has parallel Ricci tensor as the latter is equivalent to
$\omega$
vanishing identically.
Studies under condition (1.1) have had many results. Amongst them, we know from [Reference Cui and Hu3, Reference Loo14, Reference Niebergall and Ryan17] that in the
$n$
-dimensional nonflat complex space forms
$\bar {M}^{n}(c)$
of constant holomorphic sectional curvature
$c$
, there do not exist real hypersurfaces with recurrent Ricci tensor for
$n\geq 2$
; and also, it was shown in [Reference Hu, Yao and Zhang9, Reference Zhang22] that there do not exist Hopf hypersurfaces with recurrent Ricci tensor in both the
$n$
-dimensional complex quadric
$Q^n$
and the
$n$
-dimensional complex hyperbolic quadric
$Q^{n\ast }$
for
$n\geq 3$
. Here, a connected orientable real hypersurface
$M$
of an almost Hermitian manifold
$(\bar {M}, J)$
with unit normal vector field
$N$
is called a Hopf hypersurface if the structure vector field
$\xi$
defined by
$\xi \,:\!=\,-JN$
is a principal vector field of
$M$
.
Notice that, as Einstein manifolds, the complex quadric
$Q^2$
with Einstein constant
$1$
and the complex hyperbolic quadric
$Q^{2\ast }$
with Einstein constant
$-1$
are holomorphically isometric to the Kähler surfaces
$\mathbb{S}^2\times \mathbb{S}^2$
and
$\mathbb{H}^2\times \mathbb{H}^2$
, respectively. Thus, instead of
$Q^{2}$
and
$Q^{2\ast }$
, in this paper, we come to consider equivalently both
$\mathbb{S}^2\times \mathbb{S}^2$
and
$\mathbb{H}^2\times \mathbb{H}^2$
.
The Riemannian product
$\mathbb{S}^2\times \mathbb{S}^2$
of two unit
$2$
-spheres and the Riemannian product
$\mathbb{H}^2\times \mathbb{H}^2$
of two hyperbolic planes of curvature
$-1$
are important Hermitian symmetric 4-manifolds. In recent years, there have appeared many interesting studies on submanifolds of
$\mathbb{S}^2\times \mathbb{S}^2$
and
$\mathbb{H}^2\times \mathbb{H}^2$
, see, for example, the references [Reference Castro and Urbano1, Reference Gao, Van der Veken, Wijffels and Xu6, Reference Li, Ma, Van der Veken, Vrancken and Wang11, Reference Torralbo and Urbano19, Reference Torralbo and Urbano20] on surfaces, and [Reference Gao, Hu, Ma and Yao4, Reference Gao, Ma and Yao5, Reference Hu and Lu7, Reference Hu, Lu, Yao and Zhang8, Reference Hu and Zhang10, Reference Li, Vrancken, Wang and Yao12, Reference Lu, Wang and Wang16, Reference Urbano21, Reference Zhang, Gao, Hu and Yao23] on hypersurfaces. In particular, the homogeneous and isoparametric hypersurfaces of
$\mathbb{S}^2\times \mathbb{S}^2$
and
$\mathbb{H}^2\times \mathbb{H}^2$
were classified by Urbano [Reference Urbano21] and Gao, Ma and Yao [Reference Gao, Ma and Yao5], respectively.
For convenience, we denote by
$M^2(\epsilon )$
,
$\epsilon =\pm 1$
, the unit 2-sphere
$\mathbb{S}^2$
endows with the canonical metric of constant curvature 1 when
$\epsilon =1$
, or the hyperbolic plane
$\mathbb{H}^2$
endows with the canonical metric of constant curvature
$-1$
when
$\epsilon =-1$
.
Notice that, giving
$a, b \in \mathbb{S}^2$
, Urbano [Reference Urbano21] defined a hypersurface of
$\mathbb{S}^2\times \mathbb{S}^2$
with constant sectional curvature as follows:
\begin{align*} \hat {M}_{a,b}=\{(p,q)\in \mathbb{S}^2\times \mathbb{S}^2| \left \langle p,a\right \rangle ^2+\left \langle q,b\right \rangle ^2=1 \}. \end{align*}
As the important advancement in the research of hypersurfaces of
$\mathbb{S}^2\times \mathbb{S}^2$
and
$\mathbb{H}^2\times \mathbb{H}^2$
, Li, Vrancken, Wang, and Yao [Reference Li, Vrancken, Wang and Yao12, Reference Li, Vrancken, Wang and Yao13] classified hypersurfaces of both
$\mathbb{S}^2\times \mathbb{S}^2$
and
$\mathbb{H}^2\times \mathbb{H}^2$
with constant sectional curvature. Moreover, Zhang, Gao, Hu, and Yao [Reference Zhang, Gao, Hu and Yao23] established the following theorem:
Theorem 1.1 ([Reference Zhang, Gao, Hu and Yao23]). Let
$M$
be a Hopf hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor, where
$\epsilon =\pm 1$
. Then, up to isometries of
$M^2(\epsilon )\times M^2(\epsilon )$
, there exists a curve
$\Gamma$
in
$M^2(\epsilon )$
such that
$M$
is an open part of
$\Gamma \times M^2(\epsilon )$
.
In this paper, by removing the Hopf condition in Theorem1.1, we extend the result and obtain a full classification of the hypersurfaces of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor. It turns out that all new examples are hypersurfaces of constant sectional curvature and thus have parallel Ricci tensor. Our main result can be stated as follows:
Theorem 1.2.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor, where
$\epsilon =\pm 1$
. Then the Ricci tensor of
$M$
is parallel and up to isometries of
$M^2(\epsilon )\times M^2(\epsilon )$
, one of the following two cases occurs:
-
(1)
$M$
is an open part of
$\Gamma \times M^2(\epsilon )$
, where
$\Gamma$
is a curve of
$M^2(\epsilon )$
; or
-
(2)
$M$
is a hypersurface with constant sectional curvature
$\tfrac {\epsilon }{2}$
.
Remark 1.1.
We would mention that hypersurfaces of
$M^2(\epsilon )\times M^2(\epsilon )$
with constant sectional curvature have been classified in [Reference Li, Vrancken, Wang and Yao12] for
$\epsilon =1$
and [Reference Li, Vrancken, Wang and Yao13] for
$\epsilon =-1$
, respectively.
Remark 1.2.
We know from [Reference Hu, Yao and Zhang9, Reference Zhang22] that there do not exist Hopf hypersurfaces with recurrent Ricci tensor in both the complex quadric
$Q^n$
and the complex hyperbolic quadric
$Q^{n\ast }$
for
$n\geq 3$
. Thus, combining with Theorem
1.2
, an interesting question can be raised as follows: “Do there exist non-Hopf hypersurfaces in
$Q^n$
or
$Q^{n\ast }$
with recurrent Ricci tensor for
$n\geq 3$
?”
The rest of this paper is organized as follows. In Section 2, we review the basic materials about
$M^2(\epsilon )\times M^2(\epsilon )$
and its hypersurfaces. In Section 3, we give several lemmas on Riemannian manifold with recurrent Ricci tensor. In Section 4, we prove that hypersurfaces in
$M^2(\epsilon )\times M^2(\epsilon )$
with
$0\lt |C|\lt 1$
and recurrent Ricci tensor are of constant sectional curvature. In Section 5, we complete the proof of Theorem1.2.
2. Preliminaries
2.1. Geometry of the Kähler surface
$M^2(\epsilon )\times M^2(\epsilon )$
In this subsection, we recall the necessary materials of
$M^2(\epsilon )\times M^2(\epsilon )$
from [Reference Gao, Ma and Yao5, Reference Urbano21, Reference Zhang, Gao, Hu and Yao23]. We denote by
$M^2(\epsilon )$
, the
$2$
-dimensional unit sphere
$\mathbb{S}^2=\{x\in \mathbb{R}^3|x_1^2+x_2^2+x_3^2=1\}$
with standard metric
$\langle \cdot, \cdot \rangle$
of constant sectional curvature
$1$
when
$\epsilon =1$
, and the hyperbolic plane
$\mathbb{H}^2=\{x\in \mathbb{R}^3_1|-x_1^2+x_2^2+x_3^2=-1,{}x_1\gt 0\}$
is of constant sectional curvature
$-1$
when
$\epsilon =-1$
, where
$\mathbb{R}^3_1$
is the
$3$
-dimensional Minkowski space with the Lorentzian metric
$\langle \cdot, \cdot \rangle _1$
.
The complex structure
$J$
on
$M^2(1)$
is defined by
\begin{align*} J_pv=p\wedge v, \end{align*}
for any
$p\in M^2(1)$
and
$v\in T_p(M^2(1))$
, where
$\wedge$
stands for the cross product in
$\mathbb{R}^3$
. The complex structure
$J$
on
$M^2(-1)$
is defined by
\begin{align*} J_xu=x\boxtimes u, \end{align*}
for any
$x\in M^2(-1)$
and
$u\in T_x(M^2(-1))$
, where
$\boxtimes$
denotes the Lorentzian cross product defined by
\begin{align*} (a_1,a_2,a_3)\boxtimes (b_1,b_2,b_3)=(a_3b_2-a_2b_3,a_3b_1-a_1b_3,a_1b_2-a_2b_1). \end{align*}
Considering
$M^2(\epsilon )\times M^2(\epsilon )$
as naturally embedded in
$\mathbb{R}^3\times \mathbb{R}^3\cong \mathbb{R}^6$
when
$\epsilon =1$
and in
$\mathbb{R}_1^3\times \mathbb{R}_1^3\cong \mathbb{R}_2^6$
when
$\epsilon =-1$
, with the induced product metric, still denoted by
$\langle \cdot, \cdot \rangle$
, we have two complex structures
\begin{align*} J_{1}=(J, J), \quad J_{2}=(J,-J), \end{align*}
which define two Kähler structures on
$M^2(\epsilon )\times M^2(\epsilon )$
.
The product structure
$P$
on
$M^2(\epsilon )\times M^2(\epsilon )$
is defined by
$P\,:\, T(M^2(\epsilon )\times M^2(\epsilon ))\rightarrow T(M^2(\epsilon )\times M^2(\epsilon ))$
such that
\begin{align*} P(X_1,X_2)=(X_1,-X_2), \quad \forall \, X_1, X_2\in T(M^2(\epsilon )). \end{align*}
Obviously, we have
$P=-J_1J_2=-J_2J_1$
,
$P^2=\mathrm{Id}$
and
\begin{align*} \langle PX, Y\rangle =\langle X,PY\rangle, \quad \forall \,X,Y\in T(M^2(\epsilon )\times M^2(\epsilon )). \end{align*}
Moreover, it holds that
$\textrm {tr}\ P=0$
,
$\bar {\nabla } P=0$
, where
$\bar {\nabla }$
is the Levi-Civita connection on
$M^2(\epsilon )\times M^2(\epsilon )$
.
The Riemannian curvature tensor
$\bar {R}$
of
$M^2(\epsilon )\times M^2(\epsilon )$
with the standard Riemannian product metric
$\langle \cdot, \cdot \rangle$
is given by
\begin{equation} \bar {R}(X,Y) Z=\tfrac {\epsilon }{2}\big \{\langle Y,Z\rangle X- \langle X,Z\rangle Y+\langle PY,Z\rangle PX-\langle PX,Z\rangle PY\big \}, \end{equation}
where
$X, Y, Z \in T(M^2(\epsilon )\times M^2(\epsilon ))$
. Thus,
$M^2(\epsilon )\times M^2(\epsilon )$
is an Einstein manifold with scalar curvature
$4\epsilon$
and its sectional curvature is non-negative when
$\epsilon =1$
, and non-positive when
$\epsilon =-1$
.
2.2. Hypersurfaces of
$M^2(\epsilon )\times M^2(\epsilon )$
Let
$M$
be an orientable hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with
$N$
a unit normal vector field and
$g$
the induced metric on
$M$
. Then, with respect to the product structure
$P$
, we define by
\begin{align*} C\,:\!=\,\left \langle PN,N\right \rangle =\left \langle J_1 N,J_2 N\right \rangle, \quad V\,:\!=\,PN-CN, \end{align*}
the product angle function
$C\,:\,M\rightarrow \mathbb{R}$
and a vector field
$V$
tangent to
$M$
. It is clear that
$-1\leq C\leq 1$
and
$|V|^2=1-C^2$
.
For any tangential vector field
$X$
of
$M$
, acting by the complex structure
$J_1$
and the product structure
$P$
, we have the following decompositions:
\begin{align*} J_1X=\phi X+\eta (X)N, \quad PX=T X+\mu (X)N, \end{align*}
where
$\phi X$
and
$\eta (X)N$
(resp.
$T X$
and
$\mu (X)N$
) are the tangential and normal part of
$J_1 X$
(resp.
$PX$
), respectively. Thus
$\phi$
and
$T$
are tensorial fields of type
$(1,1)$
,
$\eta$
and
$\mu$
are
$1$
-forms over
$M$
. Moreover,
$\eta (X)=g(X,\xi )$
,
$\mu (X)=\langle PX, N\rangle =g(X,V)$
. Here,
$\xi =-J_1 N$
is called the structure vector field of
$M$
. It is also easy to see that
\begin{align*} g(TX, Y)=g(X, TY). \end{align*}
From the fact
$J_1^2X=-X$
, by taking the tangential and normal parts, we obtain
\begin{equation} \begin{aligned} &\eta (X)=g(X,\xi ),\ \eta (\phi X)=0, \ \phi ^2 X=-X+\eta (X) \xi, \\ &g(\phi X,Y)=-g(X,\phi Y),\ g(\phi X,\phi Y)=g(X,Y)-\eta (X)\eta (Y). \end{aligned} \end{equation}
It follows that
$(\phi, \xi, \eta, g)$
defines an almost contact metric structure on
$M$
.
Let
$\nabla$
be the Levi-Civita connection compatible with the induced metric
$g$
on
$M$
. Then, the Gauss and Weingarten formulae are given by
\begin{align*} \bar {\nabla }_X Y=\nabla _X Y+g(AX,Y)N \quad \textrm {and} \quad \bar {\nabla }_XN=-AX, \end{align*}
where
$A$
is the shape operator of
$M$
.
By
$\bar {\nabla }J_1=0$
, we have
\begin{equation} \nabla _X \xi =\phi AX, \quad \text{for any } X\in TM. \end{equation}
By (2.1), the Gauss and Codazzi equations of
$M$
are given by
\begin{equation} \begin{aligned} R(X, Y)Z=\ &\tfrac {\epsilon }{2}\big [g(Y,Z)X-g(X,Z)Y +g(TY,Z)TX-g(TX,Z)TY\big ]\\ &+g(AY,Z)AX-g(AX,Z)AY, \end{aligned} \end{equation}
and
\begin{equation} (\nabla _XA)Y-(\nabla _Y A)X=\tfrac {\epsilon }{2}\big [g(X,V)TY-g(Y, V)TX\big ], \end{equation}
where
$X, Y, Z\in TM$
, and
$R$
denotes the Riemannian curvature tensor of
$M$
.
Let
$H$
be the trace of
$A$
, from (2.4), the Ricci operator is given by
\begin{equation} \textrm {Ric}(X)=\tfrac \epsilon 2\big [X-CTX+g(X,V)V\big ]+HAX-A^2X. \end{equation}
According to Lemma2.1 of [Reference Gao, Ma and Yao5] and Lemma 1 of [Reference Urbano21], we have the following lemma.
Lemma 2.1 ([Reference Gao, Ma and Yao5, Reference Urbano21]). Let
$\Phi \,:\,M\rightarrow M^2(\epsilon )\times M^2(\epsilon )$
be an orientable hypersurface and
$A$
the shape operator associated to the unit normal field
$N$
. Then the gradient of
$C$
and the covariant derivative of
$V$
are given by
\begin{align*} \nabla C=-2AV, \quad \nabla _{X}V=CAX-TA X. \end{align*}
2.3. On hypersurfaces of
$M^2(\epsilon )\times M^2(\epsilon )$
with
$|C|\lt 1$
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with
$|C|\lt 1$
. According to [Reference Gao, Hu, Ma and Yao4] and [Reference Zhang, Gao, Hu and Yao23], we can choose the local orthonormal frame fields
$\{E_1, E_2, E_3\}$
as follows:
\begin{equation} E_1=J_1 N, \ \ E_2=\tfrac {J_2N-CJ_1 N}{\sqrt {1-C^2}},\ \ E_3=\tfrac {V}{\sqrt {1-C^2}}, \end{equation}
and then the frame has the following properties:
\begin{align} J_1 E_1&=-N,\quad J_1 E_2=-E_3, J_1 E_3=E_2,\nonumber\\ \phi E_1&=0,\quad \phi E_2=-E_3, \phi E_3=E_2,\nonumber\\ J_2 E_1&=-\sqrt {1-C^{2}} E_3-C N, J_2 E_2=C E_3-\sqrt {1-C^{2}} N, \nonumber\\ J_2 E_3&=\sqrt {1-C^{2}} E_1-C E_2, T E_1=C E_1+\sqrt {1-C^{2}} E_2, \nonumber\\ T E_2&=\sqrt {1-C^{2}} E_1-C E_2, T E_3=-C E_3. \end{align}
Now, we assume
\begin{equation} \begin{split} &A E_1=\alpha E_1+\rho E_2+\theta E_3,\\ &A E_2=\rho E_1+\lambda E_2+\beta E_3, \\ &A E_3=\theta E_1+\beta E_2+\gamma E_3, \end{split} \end{equation}
where
$\alpha$
,
$\rho$
,
$\theta$
,
$\lambda$
,
$\beta$
,
$\gamma$
are smooth functions on
$M$
. Then we have the following lemma.
Lemma 2.2.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with
$|C|\lt 1$
. Under the frame
$\{E_{1},E_{2},E_{3}\}$
defined by
(2.7)
and relations
(2.9)
, it holds
\begin{align*} \begin{aligned} &E_1(C)=-2\theta \sqrt {1-C^2},\ E_2( C)=-2\beta \sqrt {1-C^2},\ E_3(C)=-2\gamma \sqrt {1-C^2},\\ &\nabla _{E_{1}}E_{1}=-\theta E_{2}+\rho E_{3},\ \nabla _{E_{2}}E_{1}=-\beta E_{2}+\lambda E_{3},\ \nabla _{E_{3}}E_{1}=-\gamma E_{2}+\beta E_3,\\ &\nabla _{E_{1}}E_{2}=\theta E_{1}+\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^{2}}}\right)\!E_{3},\ \nabla _{E_{2}}E_{2}=\beta E_{1}+\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^{2}}}\right)\!E_{3},\\ &\nabla _{E_{3}}E_{2}=\gamma E_{1}+\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^{2}}}\right)\!E_{3},\ \nabla _{E_{1}}E_{3}=-\rho E_{1}-\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^{2}}}\right)\!E_{2},\\ &\nabla _{E_{2}}E_{3}=-\lambda E_{1}-\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^{2}}}\right)\!E_{2},\ \nabla _{E_{3}} E_{3}=-\beta E_{1}-\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^{2}}}\right)E_{2}. \end{aligned} \end{align*}
Proof. By using (2.9) and Lemma2.1, we have
\begin{align*} E_1(C)=g(\nabla C, E_1)=-2\sqrt {1-C^2}g(AE_3, E_1)=-2\theta \sqrt {1-C^2}. \end{align*}
By a similar way, we can obtain
$E_2(C)$
and
$E_3(C)$
.
By using (2.3) and (2.8), we have
\begin{align*} \nabla _{E_{1}}E_{1}=-\phi AE_1=-\theta E_2+\rho E_3. \end{align*}
By a similar way, we can obtain
$\nabla _{E_{2}}E_{1}$
and
$\nabla _{E_{3}}E_{1}$
.
By using Lemma2.1, we have
\begin{align*} \begin{split} \nabla _{E_{1}}E_{3}&=E_1\left(\tfrac {1}{\sqrt {1-C^2}}\right)V+\tfrac {1}{\sqrt {1-C^2}} \nabla _{E_1}V\\ &=-\tfrac {2C\theta }{1-C^2}V+\tfrac {1}{\sqrt {1-C^2}} (CAE_1-TAE_1)\\ &=-\rho E_1-\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)\!E_2. \end{split} \end{align*}
$\nabla _{E_{2}}E_{3}$
and
$\nabla _{E_{3}}E_{3}$
can be calculated similarly.
\begin{align*} \begin{split} \nabla _{E_{1}}E_{2}&=g(\nabla _{E_{1}}E_{2},E_1)E_1+g(\nabla _{E_{1}}E_{2},E_3)E_3\\ &=-g(E_{2},\nabla _{E_{1}}E_1)E_1-g(E_{2},\nabla _{E_{1}}E_3)E_3\\ &=\theta E_{1}+\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^{2}}}\right)\!E_{3}. \end{split} \end{align*}
Similarly, we can verify the expressions of
$\nabla _{E_{2}}E_{2}$
and
$\nabla _{E_{3}}E_{2}$
.
This completes the proof of Lemma2.2.
By using Lemma2.2 and Codazzi equation, we have
Lemma 2.3. Under the assumptions of Lemma 2.2 , we have
\begin{equation} E_1(\rho )-E_2(\alpha )=\ \theta \left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right) -3(\lambda \theta -\rho \beta )+\alpha \theta, \end{equation}
\begin{equation} E_1(\lambda )-E_2(\rho )= 2\beta \left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right) -\theta \left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)+2\theta \rho -\alpha \beta, \end{equation}
\begin{equation} E_1(\beta )-E_2(\theta )=\ \theta ^{2}+\beta ^2 +\gamma \left(\alpha -\lambda -\tfrac {2\rho C}{\sqrt {1-C^2}}\right), \end{equation}
\begin{equation} E_{1}(\theta )-E_3(\alpha )=\ 3(\rho \gamma -\theta \beta ) -\rho \left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)-\alpha \rho -\tfrac 12\epsilon C\sqrt {1-C^2}, \end{equation}
\begin{equation} \begin{split} E_3(\rho )-E_1(\beta )=&-\tfrac {2\beta \theta C}{\sqrt {1-C^2}} +\tfrac {2\rho \gamma C}{\sqrt {1-C^2}}+\lambda \left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)\\ &+\rho ^2-\lambda \gamma +\beta ^2+\tfrac {1}{2}\epsilon (1-C^2), \end{split} \end{equation}
\begin{equation} E_3(\theta )-E_1(\gamma )= \beta \left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right) +\theta \rho, \end{equation}
\begin{equation} \begin{split} E_3(\lambda )-E_2(\beta )=&\ 2\beta \left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right) +(\lambda -\gamma )(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}})\\ &+\rho (\lambda +\gamma )-\tfrac {1}{2}\epsilon C\sqrt {1-C^2} +\gamma \rho -\theta \beta, \end{split} \end{equation}
\begin{equation} E_2(\gamma )-E_3(\beta )= -\beta \left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right) -\theta (\lambda +\gamma )-\gamma \left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right). \end{equation}
Proof. On the one hand, by (2.5), we have
\begin{align*} (\nabla _{E_1}A)E_2-(\nabla _{E_2}A)E_1=0. \end{align*}
On the other hand, by using Lemma2.2, we have
\begin{align*} \begin{split} (\nabla _{E_1}A)E_2-(\nabla _{E_2}A)E_1=\ &\nabla _{E_1}(AE_2) -A\nabla _{E_1}E_2-\nabla _{E_2}(AE_1)+A\nabla _{E_2}E_1\\ =\ &[E_1(\rho )-E_2(\alpha )-\ \theta (\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}) +3(\lambda \theta -\rho \beta )\\ &-\alpha \theta ]E_1+[E_1(\lambda )-E_2(\rho )-2\beta (\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}})\\ &+\theta (\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}})-2\theta \rho +\alpha \beta ]E_2+[E_1(\beta )\\ &-E_2(\theta )-\theta ^{2}-\beta ^2 -\gamma (\alpha -\lambda -\tfrac {2\rho C}{\sqrt {1-C^2}})]E_3. \end{split} \end{align*}
Comparing the above two equations, we obtain (2.10)–(2.12). Similarly, substituting
$(X,Y)=(E_1,E_3)$
,
$(E_2,E_3)$
into (2.5), respectively, we can obtain (2.13)–(2.17).
Let
$K_{ij}$
be the sectional curvature of the plane spanned by
$\{E_i,E_j\}, 1\leq i,j\leq 3$
, by using (2.4) and (2.8) we obtain that
\begin{equation} \begin{split} K_{12}&=\alpha \lambda -\rho ^2,\\ K_{13}&=\tfrac 12\epsilon (1-C^2)+\alpha \gamma -\theta ^2,\\ K_{23}&=\tfrac 12\epsilon (1+C^2)+\lambda \gamma -\beta ^2. \end{split} \end{equation}
Substituting
$X$
into (2.6) by
$E_i$
,
$i=1,2,3$
, and letting
\begin{equation} \begin{split} &f_{1}\,:\!=\,\alpha \beta -\rho \theta, \ f_{2}\,:\!=\,-\tfrac 12\epsilon C\sqrt {1-C^2} +\rho \gamma -\theta \beta, \\ &f_{3}\,:\!=\,\theta \lambda -\rho \beta, \ f_{4}\,:\!=\,K_{12},\ f_{5}\,:\!=\,K_{13},\ f_{6}\,:\!=\,K_{23}, \end{split} \end{equation}
we have
\begin{equation} \begin{split} &\textrm {Ric}\,(E_1)=(f_{4}+f_{5}) E_1+f_{2} E_2+f_{3} E_3,\\ &\textrm {Ric}\,(E_2)=f_{2} E_1+(f_{4}+f_{6}) E_2+f_{1} E_3, \\ &\textrm {Ric}\,(E_3)=f_{3} E_1+f_{1} E_2+(f_{5}+f_{6}) E_3. \end{split} \end{equation}
Then the scalar curvature of
$M$
is
$r= 2(f_{4}+f_{5}+f_{6})$
.
3. On Riemannian manifolds with recurrent Ricci tensor
Let
$M$
be a Riemannian manifold with Riemannian curvature tensor
$R$
. Recall that a
$(1, s)$
-tensor field
$F$
on
$M$
is said to be semi-parallel (cf. [Reference Cho, Hamada and Inoguchi2]) if
$R(X,Y)\cdot F=0$
, where
$\cdot$
means that
$R(X,Y)$
acts as a derivation on
$F$
, or equivalently,
\begin{align*} R(X,Y)F(X_1,\dots, X_s) -\sum _{i=1}^{s}F(X_1,\dots, R(X,Y)X_i,\dots, X_s)=0, \end{align*}
for any vector fields
$X,Y,X_1,\dots, X_s$
of
$M$
.
For a symmetric
$(1,1)$
-tensor field
$F$
, from Theorem 20 of [Reference Loo15] and its proof, we have the following lemma.
Lemma 3.1 ([Reference Loo15]). Let
$M$
be a connected Riemannian manifold and let
$F$
be a non-vanish symmetric endomorphism on
$TM$
. If
$F$
is recurrent, that is, there exists a
$1$
-form
$\omega$
on
$M$
such that
$(\nabla _XF)Y=\omega (X)F(Y)$
for any vector fields
$X,Y$
of
$M$
. Then
$\omega$
is closed (i.e.,
$d\omega =0$
) and
$F$
is semi-parallel.
Recall that a Riemannian manifold
$M$
is called Ricci-pseudosymmetric if there is a smooth function
$\kappa$
on
$M$
such that
\begin{equation} (R(X,Y)\cdot \textrm {Ric})Z= \kappa ((X\wedge Y)\cdot \textrm {Ric})Z,\ \textrm {for any } X,Y,Z \in TM, \end{equation}
where
$((X\wedge Y)\cdot \textrm {Ric})Z=(X\wedge Y)\textrm {Ric}(Z)-\textrm {Ric}((X\wedge Y)Z)$
and the endomorphism field
$X\wedge Y$
is defined by
$ (X\wedge Y)Z=g(Y,Z)X-g(X,Z)Y$
. Obviously, when
$\kappa =0$
, Ricci-pseudosymmetric reduces to Ricci-semiparallel.
Now, following the similar idea of Cho, Hamada, and Inoguchi as in their proof of Corollary 2.2 in [Reference Cho, Hamada and Inoguchi2], we give the following lemma.
Lemma 3.2.
Let
$M$
be a Riemannian manifold of dimension
$n$
. If the Ricci operator Ric of
$M$
is pseudosymmetric such that
(3.1)
holds, and Ric admits one eigenvalue
$\sigma$
of multiplicity one, then
$\sigma =(n-1)\kappa$
.
Proof.
For any point
$x\in M$
, taking a local frame field
$\{e_1,e_2,\dots, e_n\}$
defined on a neighborhood of
$x$
such that
$\textrm {Ric}\,(e_i)=b_ie_i$
. Substituting
$(X,Y,Z)=(e_i,e_j,e_j)$
into (3.1) for
$i\neq j$
, we get
\begin{equation} b_jR(e_i,e_j)e_j-\textrm {Ric}\,(R(e_i,e_j)e_j)=\kappa (b_j-b_i)e_i,\ i\neq j. \end{equation}
Taking the inner product of (3.2) with
$e_i$
, we get
\begin{equation} (b_j-b_i)(g(R(e_i,e_j)e_j,e_i)-\kappa )=0,\ i\neq j. \end{equation}
Supposing
$e_n$
is the eigenvector of the Ricci operator with eigenvalue
$\sigma$
, that is,
$\textrm {Ric}\,(e_n)=\sigma e_n$
, and substituting
$e_i=e_n$
into (3.3), we obtain
\begin{align*} (b_j-\sigma )(g(R(e_n,e_j)e_j,e_n)-\kappa )=0,\ j\not =n. \end{align*}
If
$\sigma$
is of multiplicity one, then
$g(R(e_n,e_j)e_j,e_n)=\kappa$
,
$j\neq n$
. It follows that
\begin{align*} \sigma =g(\textrm {Ric}\,(e_n),e_n) =\sum _{j=1}^{n-1} g(R(e_n,e_j)e_j,e_n)=(n-1)\kappa . \end{align*}
This completes the proof of Lemma3.2.
The following lemma can be regarded as a corollary of Lemma3.2.
Lemma 3.3.
Let
$M$
be a
$3$
-dimensional Riemannian manifold with recurrent Ricci tensor and non-constant sectional curvature. Then up to renumbering the eigenvalues
$\{\sigma _1,\sigma _2,\sigma _3\}$
of the Ricci operator satisfy
\begin{align*} \sigma _1=\sigma _2\neq 0,\ \sigma _3=0. \end{align*}
Proof.
By Lemma3.1, the Ricci operator Ric is semi-parallel. Let
$\{\sigma _1,\sigma _2,\sigma _3\}$
denote the three eigenvalues of the Ricci operator of
$M$
. As
$M$
does not have constant sectional curvature, at least two of
$\{\sigma _1,\sigma _2,\sigma _3\}$
are distinct. Without loss of generality, we assume that
$\sigma _3$
is of multiplicity one. Then, by Lemma3.2 and the fact that Ric is semi-parallel, we get
$\sigma _3=0$
. If Ric has three distinct eigenvalues, by (3.3), we get that
$M$
has constant sectional curvature, a contradiction to the assumption. Thus Ric has exactly two distinct eigenvalues and
$\sigma _1=\sigma _2\neq \sigma _3=0$
.
Next, for any
$3$
-dimensional Riemannian manifold with recurrent Ricci tensor and non-constant sectional curvature, based on Lemma3.3, we prove the following lemma which is crucial to the proof of Theorem1.2.
Lemma 3.4.
Let
$(M,g)$
be a
$3$
-dimensional Riemannian manifold with recurrent Ricci tensor and non-constant sectional curvature. Then the eigenvector
$X_3$
corresponding to the Ricci eigenvalue
$0$
satisfies
\begin{align*} \omega (X_3)=0. \end{align*}
Proof.
By Lemma3.1, the Ricci operator Ric of
$M$
is semi-parallel. Denote by
$r$
the scalar curvature of
$M$
, and
$\{X_1,X_2,X_3\}$
the unit orthogonal eigenvectors of Ric. By using Lemma3.3, without loss of generality, we can assume that
\begin{equation} \textrm {Ric}\,(X_1)=\tfrac {r}{2}X_1,\ \textrm {Ric}\,(X_2)=\tfrac {r}{2}X_2, \ \textrm {Ric}\,(X_3)=0. \end{equation}
Now we calculate
$X_3(r)$
in two different ways.
On the one hand, by the definition of
$r$
, we have
\begin{equation} \begin{split} X_3(r)&=X_3(\textrm {tr}(\textrm {Ric}))=\textrm {tr}(\nabla _{X_3}\textrm {Ric})\\ &=\sum _{i=1}^{3} g((\nabla _{X_3}\textrm {Ric})X_i,X_i)\\ &=\sum _{i=1}^{3} g(\omega (X_3)\textrm {Ric}(X_i),X_i)\\ &=\omega (X_3)r. \end{split} \end{equation}
On the other hand, by using the contracted Bianchi second identity (cf. Proposition 3.1.5 of [Reference Petersen18]), we obtain that
\begin{equation} \begin{split} X_3(r)&=dr(X_3)=2\sum _{i=1}^{3} g((\nabla _{X_i}\textrm {Ric})X_3,X_i)\\ &= 2\sum _{i=1}^{3} g(\omega (X_i)\textrm {Ric}(X_3),X_i)\\ &= 0, \end{split} \end{equation}
where we have used the third term of (3.4).
Comparing (3.5) and (3.6), we obtain
$\omega (X_3)r=0$
. Since
$M$
is not of constant sectional curvature, from (3.4), we get
$r\neq 0$
. Thus
$\omega (X_3)=0$
holds on
$M$
.
4. Hypersurfaces of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and satisfying
$0\lt |C|\lt 1$
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor. Since the function
$C$
satisfies
$0\leq |C|\leq 1$
, we can put
$M=M_0\cup M_{1}\cup M_{c}$
, where:
\begin{align*} \begin{split} &M_0=\{p\in M\,:\,C(p)=0\},\\ &M_{1}=\{p\in M\,:\,C^2(p)=1\},\\ &M_{c}=\{p\in M\,:\,0\lt |C(p)|\lt 1\}, \end{split} \end{align*}
and
$M_{c}$
is an open subset of
$M$
.
In this section, assuming that
$M_{c}$
is non-empty, we will prove that the three Ricci eigenvalues of
$M_{c}$
must be equal, or in other words,
$M_{c}$
is Einstein. Firstly, we have the following lemma:
Lemma 4.1.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and non-empty
$M_0\cup M_c$
. Then, with respect to the frame
$\{E_{1},E_{2},E_{3}\}$
as defined by
(2.7)
on
$M_0\cup M_c$
with assuming
(2.9)
and
(2.19)
, the following equations hold:
\begin{equation} E_1(f_{3})=\rho (f_{6}-f_{4})-\theta f_{1} -\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)f_{2}+\omega (E_1)f_3, \end{equation}
\begin{equation} E_2(f_{3})=\lambda (f_{6}-f_{4})-\beta f_{1} -\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)f_{2}+\omega (E_2)f_3, \end{equation}
\begin{equation} E_3(f_{3})=\beta (f_{6}-f_{4})-\gamma f_{1} -\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)f_{2}+\omega (E_3)f_3, \end{equation}
\begin{equation} E_1(f_{1})=-\rho f_{2}+\theta f_{3}+\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right) (f_{5}-f_{4})+\omega (E_1)f_1, \end{equation}
\begin{equation} E_2(f_{1})=-\lambda f_{2}+\beta f_{3} +\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)(f_{5}-f_{4})+\omega (E_2)f_1, \end{equation}
\begin{equation} E_3(f_{1})=-\beta f_{2}+\gamma f_{3} +\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)(f_{5}-f_{4})+\omega (E_3)f_1, \end{equation}
\begin{equation} E_1(f_{2})=\theta (f_{5}-f_{6})+\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)f_{3} +\rho f_{1}+\omega (E_1)f_2, \end{equation}
\begin{equation} E_2(f_{2})= \beta (f_{5}-f_{6}) +\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)f_{3} +\lambda f_{1}+\omega (E_2)f_2, \end{equation}
\begin{equation} E_3(f_{2}) =\gamma (f_{5}-f_{6})+\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)f_{3} +\beta f_{1}+\omega (E_3)f_2, \end{equation}
\begin{equation} E_1(f_{4})=2\rho f_{3}+2\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_1)f_4, \end{equation}
\begin{equation} E_2(f_{4})=2\lambda f_{3} +2\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_2)f_4, \end{equation}
\begin{equation} E_3(f_{4})=2\beta f_{3} +2\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_3)f_4, \end{equation}
\begin{equation} E_1(f_{5}) =-2\theta f_{2} -2\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_1)f_5, \end{equation}
\begin{equation} E_2(f_{5}) =-2\beta f_{2}-2\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_2)f_5, \end{equation}
\begin{equation} E_3(f_{5})=-2\gamma f_{2} -2\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_3)f_5, \end{equation}
\begin{equation} E_1(f_{6}) =2\theta f_{2}-2\rho f_{3}+\omega (E_1)f_6, \end{equation}
\begin{equation} E_2(f_{6})=2\beta f_{2}-2\lambda f_{3}+\omega (E_2)f_6, \end{equation}
\begin{equation} E_3(f_{6})=2\gamma f_{2}-2\beta f_{3}+\omega (E_3)f_6. \end{equation}
Proof. On the one hand, by using (2.20) and Lemma2.2, we have
\begin{align*} \begin{split} (\nabla _{E_1}\textrm {Ric})E_1=\, &\nabla _{E_1}(\textrm {Ric}E_1)-\textrm {Ric}(\nabla _{E_1}E_1)\\ =\, &[E_1(f_4+f_5)+2\theta f_2-2\rho f_3]E_1\\ &+[E_1(f_2)+\theta (f_6-f_5)-\rho f_1-\left(\alpha -\tfrac {2C\rho }{\sqrt {1-C^2}}\right)f_3]E_2\\ &+[E_1(f_3)+\rho (f_4-f_6)+\theta f_1+\left(\alpha -\tfrac {2C\rho }{\sqrt {1-C^2}}\right)f_2]E_3. \end{split} \end{align*}
On the other hand, by (1.1), we have
\begin{align*} (\nabla _{E_1}\textrm {Ric})E_1 =\omega (E_1)[(f_{4}+f_{5}) E_1+f_{2} E_2+f_{3} E_3]. \end{align*}
From the above two equations, we obtain (4.1), (4.7) and
\begin{equation} E_1(f_{4}+f_{5}) =-2\theta f_{2}+2\rho f_{3}+\omega (E_1)(f_4+f_5). \end{equation}
Similarly, taking
$(X,Y)=(E_i,E_j)$
in
$(\nabla _{X}\textrm {Ric})Y=\omega (X)\textrm {Ric}(Y)$
for
$i,j\in \{1,2,3\}$
and
$(i,j)\ne (1,1)$
, then combining with (2.20) and Lemma2.2, we can derive (4.2)–(4.6), (4.8), (4.9) and the following:
\begin{equation} \begin{split} &E_2(f_{4}+f_{5}) =-2\beta f_{2}+2\lambda f_{3}+\omega (E_2)(f_4+f_5),\\ &E_3(f_{4}+f_{5}) =-2\gamma f_{2}+2\beta f_{3}+\omega (E_3)(f_4+f_5), \end{split} \end{equation}
\begin{equation} E_1(f_{5}+f_{6})=-2\rho f_{3} -2\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_1 )(f_5+f_6), \end{equation}
\begin{equation} E_2(f_{5}+f_{6})=-2\lambda f_{3} -2\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_2)(f_5+f_6), \end{equation}
\begin{equation} E_3(f_{5}+f_{6})=-2\beta f_{3} -2\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_3)(f_5+f_6), \end{equation}
\begin{equation} \begin{split} &E_1(f_{4}+f_{6}) =2\theta f_{2}+2\left(\alpha -\tfrac {2\rho C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_1)(f_4+f_6),\\ &E_2(f_{4}+f_{6}) =2\beta f_{2}+2\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_2)(f_4+f_6),\\ &E_3(f_{4}+f_{6}) =2\gamma f_{2}+2\left(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}}\right)f_{1}+\omega (E_3)(f_4+f_6). \end{split} \end{equation}
From the equations (4.19) up to (4.24), we immediately obtain (4.10)–(4.18).
Let
$\Omega \subseteq M_c$
be an open subset with two distinct Ricci eigenvalues. By using Lemma3.1, we obtain the following lemma:
Lemma 4.2.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and non-empty
$\Omega$
. Then, with respect to the frame
$\{E_{1},E_{2},E_{3}\}$
as defined by
(2.7)
on
$M_0\cup M_c$
with assuming
(2.9)
and
(2.19)
, it holds the following equations on
$\Omega$
:
\begin{equation} f_{1}f_{2}+f_{3}f_{5}=0, \end{equation}
\begin{equation} f_{1}f_{6}+f_{2}f_{3}=0, \end{equation}
\begin{equation} f_{1}f_{3}+f_{2}f_{4}=0, \end{equation}
\begin{equation} f_{1}^2=f_{4}f_{5}, \end{equation}
\begin{equation} f_{3}^2=f_{4}f_{6}, \end{equation}
\begin{equation} f_{2}^2=f_{5}f_{6}. \end{equation}
Proof.
First of all, by Lemma3.1, we know that the Ricci tensor of
$M$
is semi-parallel, that is,
$R(X,Y)\cdot \textrm {Ric}=0$
. Focusing on
$\Omega$
, taking
$(X,Y,Z,W)=(E_i,E_j,E_k,E_l)$
in
\begin{align*} g((R(X,Y)\cdot \textrm {Ric}) Z,W)=0 \end{align*}
for
$i,j,k, l\in \{1,2,3\}$
, and combining with (2.20) and Lemma2.2, we can obtain the equations (4.25)–(4.27) and
\begin{equation} f_{3}^{2}-f_{1}^{2}+f_{4}(f_{5}-f_{6})=0, \end{equation}
\begin{equation} f_{2}^{2}-f_{1}^{2}+f_{5}(f_{4}-f_{6})=0, \end{equation}
\begin{equation} f_{2}^{2}-f_{3}^{2}+f_{6}(f_{4}-f_{5})=0. \end{equation}
Multiplying both sides of (4.25) by
$f_{1}$
and using (4.27), we get
\begin{equation} f_{2}(f_{1}^2-f_{4}f_{5})=0. \end{equation}
Multiplying both sides of (4.27) by
$f_1$
and using (4.25), we get
\begin{equation} f_{3}(f_{1}^2-f_{4}f_{5})=0. \end{equation}
Multiplying both sides of (4.25) by
$f_2$
and using (4.26), we get
\begin{equation} f_{1}(f_{2}^2-f_{5}f_{6})=0. \end{equation}
To show that (4.28) holds on
$\Omega$
, we suppose on the contrary that there exists a point
$p_0\in \Omega$
such that
$f_{1}^2-f_{4}f_{5}\neq 0$
at
$p_0$
. Then there is an open subset
$\tilde {\Omega }_0$
containing
$p_0$
such that
$f_{1}^2-f_{4}f_{5}\neq 0$
on
$\tilde {\Omega }_0$
. From (4.34) and (4.35), we have
$f_{2}=f_{3}=0$
on
$\tilde {\Omega }_0$
. These further imply from (4.31) and (4.32) that
\begin{equation} f_{5}f_{6}\neq 0,\ f_{4}f_{6}\neq 0\ \text{on}\ \tilde {\Omega }_0. \end{equation}
It follows that
$f_{4}\neq 0$
,
$f_{5}\neq 0$
, and
$f_{6}\neq 0$
, and (4.36) gives
$f_{1}=0$
on
$\tilde {\Omega }_0$
. Moreover, from the fact
$f_{1}=f_{2}=f_{3}=0$
on
$\tilde {\Omega }_0$
, (4.31), (4.32), and (4.33), we obtain
$f_{4}=f_{5}=f_{6}$
on
$\tilde {\Omega }_0$
. It follows from (2.20) that all Ricci eigenvalues of
$\tilde {\Omega }_0$
equal to
$2f_4$
, a contradiction to the assumption. Hence, as claimed, (4.28) holds on
$\Omega$
.
Finally, substituting (4.28) into (4.31) and (4.32), we obtain that (4.29) and (4.30) hold on
$\Omega$
. This completes the proof of Lemma4.2.
Now, we put
$\Omega =\Omega _1\cup \Omega _2$
, where
\begin{align*} \Omega _1=\{p\in \Omega \,:\,\beta (p)\ne 0\}, \ \ \Omega _2=\{p\in \Omega \,:\,\beta (p)=0\}. \end{align*}
Then we have the following lemma.
Lemma 4.3.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and non-empty
$\Omega$
. Then, with respect to the frame
$\{E_{1},E_{2},E_{3}\}$
as defined by
(2.7)
on
$M_0\cup M_c$
with assuming
(2.9)
and
(2.19)
, it holds that
$\Omega _1$
is non-empty and
$\Omega _{2}$
does not contain any non-empty open subset of
$\Omega$
.
Proof.
To prove the first assertion, we suppose on the contrary that
$\Omega =\Omega _2$
, that is,
$\beta =0$
on
$\Omega$
. From (2.17), (4.3), (4.8), (4.17) and (4.26), the following equations hold on
$\Omega$
:
\begin{equation} E_2(\gamma )=-\theta (2\gamma +\lambda ), \end{equation}
\begin{equation} E_3(\theta \lambda ) =\tfrac 12\epsilon C\sqrt {1-C^2}\theta +\omega (E_3)\theta \lambda, \end{equation}
\begin{equation} E_2(-\tfrac 12\epsilon C\sqrt {1-C^2}+\rho \gamma ) =-\tfrac {2C\theta \lambda ^2}{\sqrt {1-C^2}} +\omega (E_2)(-\tfrac 12\epsilon C\sqrt {1-C^2}+\rho \gamma ), \end{equation}
\begin{equation} E_2(\tfrac 12\epsilon (1+C^2)+\lambda \gamma ) =-2\theta \lambda ^2+\omega (E_2)(\tfrac 12\epsilon (1+C^2)+\lambda \gamma ), \end{equation}
\begin{equation} \theta \big [\rho (1+C^2)+\lambda C\sqrt {1-C^2}\big ]=0. \end{equation}
Claim 1.
$\theta =0$
on
$\Omega$
.
In fact, if there is a point
$p_1\in \Omega$
such that
$\theta (p_1)\neq 0$
, then there is a neighborhood
$\tilde {\Omega }\subseteq \Omega$
containing
$p_1$
such that
$\theta \neq 0$
holds on
$\tilde {\Omega }$
. Now, we claim that there exists at least one point
$p_2\in \tilde {\Omega }$
such that
$\gamma (p_2)\neq 0$
. In fact, if
$\gamma \equiv 0$
on
$\tilde {\Omega }$
, by (4.38), we have
$\lambda =0$
on
$\tilde {\Omega }$
. From (4.39), we have
$\epsilon C\sqrt {1-C^2}\theta =0$
on
$\tilde {\Omega }$
, a contradiction to
$\theta \ne 0$
and
$0\lt |C|\lt 1$
. Thus, there exists at least a point
$p_2\in \tilde {\Omega }$
such that
$\gamma \neq 0$
. It follows that there is a neighborhood
$\tilde {\Omega }_1\subseteq \tilde {\Omega }$
containing
$p_2$
such that
$\gamma \neq 0$
holds on
$\tilde {\Omega }_1$
.
Now, focusing on
$\tilde {\Omega }_1$
, from (4.38) and (4.40), we have
\begin{equation} E_2(\rho )=-\frac {2\lambda ^2\theta C}{\gamma \sqrt {1-C^2}} +\frac {\rho \theta (2\gamma +\lambda )}{\gamma } +\frac {\omega (E_2)\left(-\frac{\epsilon C\sqrt {1-C^2}}{2}+\rho \gamma \right)}{\gamma }. \end{equation}
From (4.38) and (4.41), we have
\begin{equation} E_2(\lambda )=\frac{\theta \lambda (2\gamma -\lambda )}{\gamma } +\frac{\omega (E_2)\left(\frac{\epsilon (1+C^2)}{2}+\gamma \lambda \right)}{\gamma }. \end{equation}
From (4.42) and
$\theta \ne 0$
, we have
\begin{align*} \rho =-\tfrac {C\sqrt {1-C^2}}{1+C^2}\lambda \ \ \text{on}\ \tilde {\Omega }_1. \end{align*}
Taking the derivative of the above equation along
$E_2$
and combining with (4.43) and (4.44), we have
\begin{align*} 0=E_2(\rho )+\tfrac {C\sqrt {1-C^2}}{1+C^2}E_2(\lambda ) =\tfrac {-4C\lambda ^{2}\theta }{\gamma (\sqrt {1-C^2})(1+C^2)}\ \text{on}\ \tilde {\Omega }_1. \end{align*}
It follows from
$\theta \ne 0$
that
$\lambda =0$
on
$\tilde {\Omega }_1$
. From (4.39), we have
$\epsilon \tfrac { C\sqrt {1-C^2}}{2}\theta =0$
on
$\tilde {\Omega }_1$
, a contradiction. Hence
$\theta =0$
on
$\Omega$
and Claim 1 are verified.
Claim 2.
$\rho \neq 0$
on
$\Omega$
.
Indeed, if there is a point
$p_3\in \Omega$
such that
$\rho (p_3)=0$
, by (2.19) and
$\beta (p_3)=\theta (p_3)=0$
, we obtain at
$p_3$
\begin{align*} \begin{split} &f_1=f_3=0,\ \ f_2=-\tfrac 12\epsilon C\sqrt {1-C^2},\ \ f_4=\alpha \lambda, \\ &f_5=\tfrac {\epsilon }{2}(1-C^2)+\alpha \gamma, \ f_6=\tfrac {\epsilon }{2}(1+C^2)+\lambda \gamma . \end{split} \end{align*}
From (2.12), we have
\begin{equation} \gamma (\alpha -\lambda )=0 \ \text{at}\ p_3. \end{equation}
If
$\gamma (p_3)=0$
, from (4.30), we obtain
\begin{align*} C^2(1-C^2)=1-C^4 \ \text{at}\ p_3, \end{align*}
a contradiction to
$0\lt |C(p_3)|\lt 1$
. Thus
$\gamma (p_3)\ne 0$
. It follows from (4.45) that
$\alpha =\lambda$
at
$p_3$
. Substituting
$f_1=f_3=0$
into (4.27), we have
\begin{align*} 0=f_2f_4=-\tfrac 12\epsilon C\sqrt {1-C^2}\alpha ^2 \ \text{at}\ p_3. \end{align*}
It follows that
$\alpha (p_3)=0$
. By (4.30), we have
$C^2(1-C^2)=1-C^4$
, a contradiction to
$0\lt |C(p_3)|\lt 1$
. Hence
$\rho \neq 0$
on
$\Omega$
and Claim 2 is verified.
Since
$\beta =\theta =0$
on
$\Omega$
, we have
$f_1=f_3=0$
on
$\Omega$
. By (2.12), we have
\begin{equation} \gamma \left(\alpha -\lambda -\tfrac {2C\rho }{\sqrt {1-C^2}}\right)=0 \ \text{on}\ \Omega . \end{equation}
If there is a point
$p_4\in \Omega$
such that
$\gamma (p_4)=0$
, from (4.30), we obtain
$1-C^2=0$
at
$p_4$
, a contradiction. Thus
$\gamma \neq 0$
on
$\Omega$
. From (4.46), we have
\begin{equation} \alpha -\tfrac {2 C\rho }{\sqrt {1-C^2}}=\lambda \ \text{on}\ \Omega . \end{equation}
Substituting
$f_1=f_3=0$
into (4.27), we have
\begin{align*} f_{2}f_{4}=0\ \text{on}\ \Omega . \end{align*}
Now, we claim that
$f_{4}=0$
on
$\Omega$
. Indeed, if there is a point
$p_5\in \Omega$
such that
$f_{4}(p_5)\ne 0$
, then
$f_{2}(p_5)=0$
. Since
$f_1=0$
holds on
$\Omega$
, from (4.1) and Claim 2, we have
$f_6(p_5)=f_4(p_5)\ne 0$
. By
$f_3=0$
and (4.29), we get
$f_4^2(p_5)=0$
, a contradiction. Thus
$f_{4}=0$
holds on
$\Omega$
.
As
$f_{1}=f_{3}=0$
on
$\Omega$
, from (4.1) and (4.47), we have
\begin{equation} 0=\rho f_6-\lambda f_2=\rho (1+C^2)+\lambda C\sqrt {1-C^2} \ \text{on}\ \Omega . \end{equation}
It follows that
$\lambda =-\tfrac {(1+C^2)}{C\sqrt {1-C^2}}\rho$
on
$\Omega$
.
We now claim that
$\alpha \neq 0$
on
$\Omega$
. In fact, if there is a point
$p_6\in \Omega$
such that
$\alpha (p_6)=0$
, from (4.47), we have
$\lambda =-\tfrac {2 C\rho }{\sqrt {1-C^2}}$
at
$p_6$
. This, together with (4.48), implies
$\rho (p_6)=0$
, which is a contradiction to Claim 2. Thus
$\alpha \neq 0$
on
$\Omega$
.
Multiplying both sides of (4.48) by
$\alpha$
and combining with
$\alpha \lambda =\rho ^{2}$
,
$\rho \neq 0$
on
$\Omega$
, we have
\begin{align*} \alpha (1+C^2)+\rho C\sqrt {1-C^2}=0\ \text{on}\ \Omega . \end{align*}
It follows that
$\alpha =-\tfrac { C\sqrt {1-C^2}}{1+C^2}\rho$
on
$\Omega$
. Substituting
$\lambda =-\tfrac {(1+C^2)}{C\sqrt {1-C^2}}\rho$
and
$\alpha =-\tfrac {C\sqrt {1-C^2}}{1+C^2}\rho$
into (4.47), we can obtain
\begin{align*} \tfrac {1-C^2}{C(1+C^2)}\rho =0\ \text{on}\ \Omega . \end{align*}
Thus
$\rho =0$
on
$\Omega$
, which is a contradiction to Claim 2.
In conclusion, we have proved the first assertion of Lemma4.3 that
$\Omega _1$
is non-empty.
Next, if
$\Omega _{2}$
contains a non-empty open subset
$\hat {\Omega }$
of
$\Omega$
, replacing
$\Omega$
by
$\hat {\Omega }$
in the preceding proof for the first assertion, we can verify the second assertion of Lemma4.3.
This completes the proof of Lemma4.3.
Now, we put
$\Omega _1=\Omega _{11}\cup \Omega _{12}$
, where
\begin{align*} \Omega _{11}=\{p\in \Omega _1\,:\,f_4(p)\ne 0\},\ \ \Omega _{12}=\{p\in \Omega _1\,:\,f_4(p)=0\}. \end{align*}
Then we have the following lemma:
Lemma 4.4.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and non-empty
$\Omega$
. Then, with respect to the frame
$\{E_{1},E_{2},E_{3}\}$
as defined by
(2.7)
on
$M_0\cup M_c$
with assuming
(2.9)
and
(2.19)
, it holds that
$\Omega _{11}$
is non-empty and
$\Omega _{12}$
does not contain any non-empty open subset of
$\Omega _1$
.
Proof. From (2.19) and (4.27), we have
\begin{equation} \begin{split} 0&=-(\alpha \beta -\rho \theta )(\theta \lambda -\beta \rho )- \left(-\tfrac {\epsilon C\sqrt {1-C^2}}{2}+\rho \gamma -\theta \beta \right)(\alpha \lambda -\rho ^2)\\ &=\alpha \beta ^2\rho +\rho \lambda \theta ^2-2\beta \theta \rho ^2 -\rho \gamma f_4+\tfrac 12\epsilon C\sqrt {1-C^2}f_4\\ &=\rho (\beta f_1+\theta f_3-\gamma f_4)+\tfrac 12\epsilon C\sqrt {1-C^2}f_4 \ \text{on}\ \Omega _1. \end{split} \end{equation}
Similarly, from (4.28), (4.29), and (4.30), we can derive that on
$\Omega _1$
:
\begin{equation} \alpha (\beta f_1+\theta f_3-\gamma f_4)-\tfrac {\epsilon (1-C^2)}{2}f_4=0, \end{equation}
\begin{equation} \lambda (\beta f_1+\theta f_3-\gamma f_4)-\tfrac {\epsilon (1+C^2)}{2}f_4=0, \end{equation}
\begin{equation} \gamma (\beta f_1+\theta f_3-\gamma f_4)-\tfrac {\epsilon (1-C^2)}{2}f_6 -\tfrac {\epsilon (1+C^2)}{2}f_5-\epsilon C\sqrt {1-C^2}f_2+\tfrac {1-C^2}{4}=0. \end{equation}
From (4.49), (4.50), and (4.51), we can obtain that
\begin{equation} \begin{split} &\left(\tfrac {1+C^2}{2}\alpha -\tfrac {1-C^2}{2}\lambda \right)f_4=0,\\ &\left(\tfrac {C\sqrt {1-C^2}}{2}\alpha +\tfrac {1-C^2}{2}\rho \right)f_4=0, \end{split} \end{equation}
hold on
$\Omega _1$
.
Now, suppose on the contrary that
$\Omega _{11}$
is empty, that is,
$f_4\equiv 0$
on
$\Omega _{1}$
. From (4.28) and (4.29), we have
$f_1=f_3=0$
on
$\Omega _1$
. Then, (4.52) reduces to
\begin{equation} \epsilon \left(\tfrac {1-C^2}{2}f_6+\tfrac {1+C^2}{2}f_5+C\sqrt {1-C^2}f_2\right) -\tfrac {1-C^2}{4}=0\ \text{on}\ \Omega _1. \end{equation}
Taking the derivative of (4.54) with respect to
$E_i,\ i=1,2$
, and combining with (4.7), (4.8), and (4.13)–(4.18), we obtain
\begin{equation} \begin{split} 0=&\ \epsilon \left(\tfrac{1-C^2}{2}f_6+\tfrac{1+C^2}{2}f_5+C\sqrt{1-C^2}f_2\right)\omega (E_i)\\ &+\tfrac{E_i(C)}{2}\left[\epsilon C\big(f_5-f_6\big)+2\epsilon \sqrt{1-C^2}f_2+C\right] \ \text{on}\ \Omega _1. \end{split} \end{equation}
We claim that
$E_1(C)\neq 0$
on
$\Omega _1$
. In fact, if there is a point
$p_7\in \Omega _1$
such that
$E_1(C)=0$
at
$p_7$
, by
$E_1(C)=-2\theta \sqrt {1-C^2}$
, we have
$\theta (p_7)=0$
. Since
$f_1=f_3=0$
on
$\Omega _1$
, we obtain
$\alpha (p_7)=\rho (p_7)=0$
, where we have used
$\beta \ne 0$
on
$\Omega _1$
. When (4.6) takes the value at
$p_7$
, with the use of
$\alpha (p_7)=\rho (p_7)=\theta (p_7)=0$
and
$\beta (p_7)\ne 0$
, we obtain
\begin{align*} 0=f_2+\tfrac {2C}{\sqrt {1-C^2}}f_5=\tfrac 12\epsilon C\sqrt {1-C^2}\ \text{at}\ p_7, \end{align*}
a contradiction to
$0\lt |C(p_7)|\lt 1$
. Thus
$E_1(C)\neq 0$
and
$\theta \ne 0$
on
$\Omega _1$
.
From Lemma4.3 and
$E_2(C)=-2\beta \sqrt {1-C^2}$
, we have
$E_2(C)\neq 0$
on
$\Omega _1$
. Then, by (4.54) and (4.55), we obtain that
\begin{equation} \epsilon (f_5+f_6)-\tfrac {C(1-C^2)}{2}\tfrac {\omega (E_i)}{E_i(C)} -\tfrac {1+C^2}{2}=0, \ i=1,2 \ \text{on}\ \Omega _1. \end{equation}
It follows that
\begin{equation} \omega (E_1)=\tfrac {E_1(C)}{E_2(C)}\omega (E_2)=\tfrac {\theta }{\beta }\omega (E_2) \ \text{on}\ \Omega _1. \end{equation}
Finally, we put
$\Omega _1=\tilde {\Omega }_{11}\cup \tilde {\Omega }_{12}$
, where
\begin{align*} \tilde {\Omega }_{11}=\{p\in \Omega _1\,:\,\omega (E_1)(p)\ne 0\},\ \ \tilde {\Omega }_{12}=\{p\in \Omega _1\,:\,\omega (E_1)(p)=0\}. \end{align*}
We claim that
$\tilde {\Omega }_{11}$
is non-empty.
In fact, if
$\tilde {\Omega }_{11}$
is empty, by (4.57) and
$E_1(C)\ne 0$
on
$\Omega _1$
, we have
$\omega (E_1)=\omega (E_2)=0$
on
$\Omega _1$
. By (4.56), we get
\begin{equation} \epsilon (f_5+f_6)-\tfrac {1+C^2}{2}=0 \ \text{on}\ \Omega _1. \end{equation}
Taking the derivative of (4.58) along
$E_2$
and combining with (4.22), we have
\begin{align*} 2 C\sqrt {1-C^2}\beta =0\ \ \text{on}\ \ \Omega _1, \end{align*}
a contradiction to Lemma4.3. Thus
$\tilde {\Omega }_{11}$
is non-empty.
By (2.20) and Lemma4.2, it can be checked that
$\textrm {Ric}(-f_2E_1+f_5E_2)=0$
on
$\tilde {\Omega }_{11}$
. By using Lemma3.4, we have
\begin{align*} 0=-f_2\omega (E_1)+f_5\omega (E_2)\ \ \text{on}\ \ \tilde {\Omega }_{11}. \end{align*}
Substituting (4.57) into the above equation and using
$\beta \neq 0$
,
$\omega (E_2)\neq 0$
,
$f_1=0$
on
$\tilde {\Omega }_{11}$
, we have
\begin{equation} 0=-\theta f_2+\beta f_5 =\varepsilon \tfrac {\sqrt {1-C^2}}{2}(C\theta +\sqrt {1-C^2}\beta ) \ \text{on}\ \tilde {\Omega }_{11}. \end{equation}
We claim that
$f_2\ne 0$
on
$\tilde {\Omega }_{11}$
. Indeed, if there is a point
$p_8\in \tilde {\Omega }_{11}$
such that
$f_2(p_8)=0$
, by (4.59) and
$\beta \ne 0$
on
$\tilde {\Omega }_{11}$
, we obtain
$f_5(p_8)=0$
. Since
$f_1=f_3=f_4=0$
on
$\Omega _1$
, by (4.3), we have
\begin{align*} 0=\beta f_6-(\theta -\tfrac {2\beta C}{\sqrt {1-C^2}})f_2 \ \text{on}\ \Omega _{1}. \end{align*}
From the above equation, we get
$f_6(p_8)=0$
. It follows from (2.20) that the Ricci eigenvalues are all zero at
$p_8$
, a contradiction to the assumption. Thus
$f_2\ne 0$
on
$\tilde {\Omega }_{11}$
. It follows that
$f_5\ne 0$
and
$f_6\ne 0$
on
$\tilde {\Omega }_{11}$
.
Multiplying both sides of
$\theta f_2=\beta f_5$
by
$f_2$
and combining with
$f_2^2=f_5f_6$
and
$f_5\ne 0$
on
$\tilde {\Omega }_{11}$
, we obtain
\begin{equation} 0=\theta f_6-\beta f_2=\tfrac {\varepsilon }{2}((1+C^2)\theta +C\sqrt {1-C^2}\beta )\ \text{on}\ \tilde {\Omega }_{11}. \end{equation}
From (4.59) and (4.60), we have
$\theta =0$
on
$\tilde {\Omega }_{11}$
, a contradiction to
$E_1(C)\ne 0$
on
$\Omega _1$
. Hence
$\Omega _{11}$
is non-empty and the first assertion of Lemma4.4 is proved.
Moreover, if
$\Omega _{12}$
contains a non-empty open subset
$\hat {\Omega }_1$
of
$\Omega _1$
. Replacing
$\Omega _1$
by
$\hat {\Omega }_1$
in the preceding proof, we can also get a contradiction. Hence
$\hat {\Omega }_1$
does not exist.
This completes the proof of Lemma4.4.
In addition, we can solve
$\omega$
on
$\Omega _{11}$
as follows.
Lemma 4.5.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and non-empty
$\Omega$
. Then, with respect to the frame
$\{E_{1},E_{2},E_{3}\}$
as defined by
(2.7)
on
$M_0\cup M_c$
with assuming
(2.9)
and
(2.19)
,
$\omega$
satisfies the following relations on
$\Omega _{11}$
:
\begin{equation} \omega (E_1)=4\left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right), \end{equation}
\begin{equation} \omega (E_2)=4\theta, \end{equation}
\begin{equation} \omega (E_3)=\frac {4(1-C^2)}{\alpha }\left[\left(\frac{1+C^2}{1-C^2}\theta +\frac{C\beta }{\sqrt {1-C^2}}\right)(\beta +\frac{2C\theta}{\sqrt {1-C^2}}) +\theta (\beta +\frac{C\theta }{\sqrt {1-C^2}})\right]. \end{equation}
Proof.
Focusing on
$\Omega _{11}$
, from (4.53), we obtain
\begin{equation} \lambda =\frac {1+C^2}{1-C^2}\alpha, \ \ \rho =-\frac {C\alpha }{\sqrt {1-C^2}}. \end{equation}
It follows that
$\alpha -\tfrac {2 C\rho }{\sqrt {1-C^2}}=\lambda$
and the following hold on
$\Omega _{11}$
:
\begin{equation} f_1=\alpha \left(\beta +\frac{C\theta}{\sqrt {1-C^2}}\right), \end{equation}
\begin{equation} f_3=\alpha \left(\frac{1+C^2}{1-C^2}\theta +\frac{C\beta}{\sqrt{1-C^2}}\right), \end{equation}
\begin{equation} f_4=\frac{\alpha ^2}{1-C^2}. \end{equation}
By using Lemma4.2 and (4.65), (4.66), (4.67), we obtain
\begin{equation} f_2=-(1-C^2)\left(\beta +\frac{C\theta}{\sqrt{1-C^2}}\right)\left(\frac{1+C^2}{1-C^2}\theta +\frac{C\beta}{\sqrt{1-C^2}}\right), \end{equation}
\begin{equation} f_5=(1-C^2)\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right)^2. \end{equation}
Taking the derivative of (4.67) with respect to
$E_1$
and using
$E_1(C)=-2\theta \sqrt {1-C^2}$
, we obtain
\begin{equation} E_1(f_4)=\tfrac {2(1-C^2)\alpha E_1(\alpha )-4C\alpha ^2\theta \sqrt {1-C^2}}{(1-C^2)^2}. \end{equation}
Substituting (4.64) into (4.10), by direct calculations, we obtain
\begin{equation} E_1(f_4)=\tfrac {2\alpha ^2\beta }{1-C^2}+\omega (E_1)f_4. \end{equation}
Comparing (4.70) and (4.71), we obtain
\begin{equation} E_1(\alpha )=\alpha \beta +\tfrac {2C\alpha \theta }{\sqrt {1-C^2}}+\tfrac 12w(E_1)\alpha . \end{equation}
Taking the derivative of
$ \rho =-\tfrac {C}{\sqrt {1-C^2}}\alpha$
with respect to
$E_1$
and using (4.72), we obtain
\begin{equation} E_1(\rho )=2\alpha \theta -\tfrac {C}{\sqrt {1-C^2}}\left(\alpha \beta +\frac{1}{2}w(E_1)\alpha\right). \end{equation}
Taking the derivative of (4.67) with respect to
$E_2$
and using
$E_2 (C)=-2\beta \sqrt {1-C^2}$
, we obtain
\begin{equation} E_2(f_4)=\tfrac {2(1-C^2)\alpha E_2(\alpha )-4C\alpha ^2\beta \sqrt {1-C^2}}{(1-C^2)^2}. \end{equation}
Substituting (4.64) into (4.11), we obtain
\begin{equation} E_2(f_4)=\tfrac {2\alpha ^2}{1-C^2}\left(\theta -\frac{2\beta C}{\sqrt {1-C^2}} +\frac{1}{2}\omega (E_2)\right). \end{equation}
Comparing (4.74) and (4.75), we obtain
\begin{equation} E_2(\alpha )=\alpha (\theta +\tfrac 12\omega (E_2)). \end{equation}
From (4.73) and (4.76), we have
\begin{equation} E_1(\rho )-E_2(\alpha )=\alpha \left(\theta -\frac{C\beta}{\sqrt {1-C^2}} -\frac{C}{2\sqrt{1-C^2}}\omega (E_1)-\frac{1}{2}\omega (E_2)\right). \end{equation}
Substituting (4.64) into (2.10), we obtain
\begin{equation} E_1(\rho )-E_2(\alpha )=-\tfrac {1+3C^2}{\sqrt {1-C^2}}\alpha \theta -\tfrac {3C}{\sqrt {1-C^2}}\alpha \beta . \end{equation}
Comparing (4.77) and (4.78), we obtain
\begin{equation} \alpha \left[\theta (1+C^2)+\beta C\sqrt {1-C^2}-\tfrac {C\sqrt {1-C^2}}{4}\omega (E_1) -\tfrac {1-C^2}{4}\omega (E_2)\right]=0. \end{equation}
Taking the derivative of
$\lambda =\tfrac {1+C^2}{1-C^2}\alpha$
with respect to
$E_1$
, with the use of (4.72) and
$E_1(C)=-2\theta \sqrt {1-C^2}$
, we obtain
\begin{equation} E_1(\lambda )=-8\frac{C\sqrt {1-C^2}}{(1-C^2)^2}\alpha \theta +\frac{1+C^2}{1-C^2}\left(\alpha \beta +\frac{2C}{\sqrt {1-C^2}}\alpha \theta +\frac{1}{2}\omega (E_1)\alpha\right). \end{equation}
Taking the derivative of
$\rho =-\tfrac {C}{\sqrt {1-C^2}}\alpha$
with respect to
$E_2$
, with the use of (4.76) and
$E_2(C)=-2\beta \sqrt {1-C^2}$
, we obtain
\begin{equation} E_2(\rho )=\tfrac {2}{1-C^2}\alpha \beta -\tfrac {C}{\sqrt {1-C^2}}(\alpha \theta +\tfrac 12\omega (E_2)\alpha ). \end{equation}
From (4.80) and (4.81), we have
\begin{equation} \begin{split} E_1(\lambda )-E_2(\rho )=\ &\alpha \left[-\frac {5C\theta }{(1-C^2)\sqrt {1-C^2}} +\frac {C^3 \theta }{(1-C^2)\sqrt {1-C^2}}\right.\\ &\left.-\beta +\frac {1+C^2}{2(1-C^2)}\omega (E_1) +\frac {C}{2\sqrt {1-C^2}}\omega (E_2)\right]. \end{split} \end{equation}
By (2.11), we have
\begin{equation} E_1(\lambda )-E_2(\rho ) =\frac {1+3C^2}{1-C^2}\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right)\alpha . \end{equation}
From (4.82) and (4.83), we obtain
\begin{equation} \alpha \left[(1+C^2)\beta +\frac { C(3+C^2)\theta }{\sqrt {1-C^2}} -\frac {1+C^2}{4}\omega (E_1)-\frac {C\sqrt {1-C^2}}{4}\omega (E_2)\right]=0. \end{equation}
From (4.79) and (4.84), we have
\begin{equation} \alpha (\theta -\frac 14\omega (E_2))=0. \end{equation}
We claim that
$\alpha \not =0$
on
$\Omega _{11}$
. In fact, if there is a point
$p_9\in \Omega _{11}$
such that
$\alpha (p_9)=0$
, by (4.64), we have
$\rho (p_9)=\lambda (p_9)=0$
and hence
$f_4(p_9)=0$
. This contradicts to the assumption. Hence
$\alpha \not =0$
on
$\Omega _{11}$
.
From (4.84) and (4.85), we obtain (4.61) and (4.62).
From (4.72) and (4.76), it holds on
$\Omega _{11}$
that
\begin{equation} E_1(\alpha )=3\alpha \left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right) =\frac {3}{4}\alpha \omega (E_1), \end{equation}
\begin{equation} E_2(\alpha )=3\alpha \theta =\frac 34\alpha \omega (E_2). \end{equation}
Now, applying (2.20) and Lemma4.2, we can easily show that
\begin{align*} \textrm {Ric}\,(f_3E_1+f_1E_2-f_4E_3)=0\ \ \text{on}\ \Omega _{11}. \end{align*}
By Lemma3.4, we get
\begin{equation} \omega (f_3E_1+f_1E_2-f_4E_3)=0\ \ \text{on}\ \Omega _{11}. \end{equation}
Substituting (4.61) and (4.62) into (4.88), with the help of (4.65), (4.66) and (4.67), we obtain
\begin{align*} \omega (E_3)=\frac {4(1-C^2)}{\alpha }\left[\left(\frac {1+C^2}{1-C^2}\theta +\frac {C\beta }{\sqrt {1-C^2}}\right)\left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right) +\theta \left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right)\right]\ \ \text{on}\ \Omega _{11}. \end{align*}
We have completed the proof of Lemma4.5.
Applying Lemma4.5, we can prove the following proposition.
Proposition 4.1.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor and non-empty
$M_c$
. Then
$M_c$
is Einstein.
Proof.
First of all, we prove that
$\Omega$
is empty. Suppose on the contrary that
$\Omega$
is non-empty. By Lemma4.4, we see that
$\Omega _{11}$
is non-empty. Then, focused on
$\Omega _{11}$
, we have the following discussions:
Taking the derivative of (4.67) with respect to
$E_3$
and using
$E_3(C)=-2\gamma \sqrt {1-C^2}$
, we obtain
\begin{equation} E_3(f_4)=\frac{2(1-C^2)\alpha E_3(\alpha ) -4C\alpha ^2\gamma \sqrt {1-C^2}}{(1-C^2)^2}. \end{equation}
Substituting (4.63) into (4.12), with the help of (4.65), (4.66), and (4.67), we obtain
\begin{equation} E_3(f_4)=\alpha \left(\tfrac{1+C^2}{1-C^2}\theta +\tfrac{C\beta}{\sqrt {1-C^2}}\right)\left(6\beta +\tfrac{8C\theta}{\sqrt{1-C^2}}\right) +\alpha \left(\beta +\tfrac{C\theta}{\sqrt{1-C^2}}\right)\left(6\theta -\tfrac{4C\beta}{\sqrt{1-C^2}}\right). \end{equation}
It follows from (4.89) and (4.90) that
\begin{align*} \begin{split} E_3(\alpha )=\ &(1-C^2)\left [\left(\frac {1+C^2}{1-C^2}\theta +\frac {C\beta }{\sqrt {1-C^2}}\right)\left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right)\right. \\ &\left.+\frac {(5-C^2)\theta }{1-C^2}\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right) +\frac {2C\alpha \gamma }{(1-C^2)\sqrt {1-C^2}}\right]. \end{split} \end{align*}
By (4.64), we obtain
\begin{equation} \begin{split} E_3(\alpha )+E_3(\lambda )=\ &2\left[\left(\frac {1+C^2}{1-C^2}\theta +\frac {C\beta }{\sqrt {1-C^2}}\right)\left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right)\right.\\ &\left.+\frac {(5-C^2)\theta }{1-C^2}\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right) +\frac {2C\alpha \gamma }{(1-C^2)\sqrt {1-C^2}}\right] -\frac {8C\alpha \gamma }{(1-C^2)\sqrt {1-C^2}}. \end{split} \end{equation}
Since
$M$
has recurrent Ricci tensor, by Lemma3.1, we have
$d\omega =0$
. Substituting (4.61) and (4.62) into the following equation:
\begin{align*} 0=d\omega (E_1,E_2)=E_1(\omega (E_2))-E_2(\omega (E_1)) -\omega (\nabla _{E_1}E_2-\nabla _{E_2}E_1), \end{align*}
and combining with Lemma2.2, we obtain
\begin{equation} 0=E_1(\theta )-E_2\left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right) -\theta \left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right)-\theta \beta . \end{equation}
By using (4.65), we obtain
\begin{equation} \begin{split} E_2\left(\beta +\frac{2C\theta}{\sqrt {1-C^2}}\right) &=2E_2\left(\beta +\frac{C\theta }{\sqrt{1-C^2}}\right)-E_2(\beta )\\ &=\frac{2}{\alpha}\left[E_2(f_1)-\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right)E_2(\alpha )\right]-E_2(\beta ). \end{split} \end{equation}
Substituting (4.93) into (4.92), we have
\begin{equation} \begin{split} 0=\ &E_1(\theta )+E_2(\beta )-\frac {2}{\alpha }\left[E_2(f_1) -\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right)E_2(\alpha )\right]\\ &-\theta \left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right)-\theta \beta . \end{split} \end{equation}
Subtracting (2.16) from (2.13), and using (4.91), we obtain
\begin{equation} \begin{split} E_1(\theta )+E_2(\beta ) =\ &2\left [\left(\frac {1+C^2}{1-C^2}\theta +\frac {C\beta }{\sqrt {1-C^2}}\right) \left(\beta +\frac {2C\theta }{\sqrt {1-C^2}}\right)\right.\\ &\left.+\frac {(5-C^2)\theta }{1-C^2}\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right) +\frac {2C\alpha \gamma }{(1-C^2)\sqrt {1-C^2}}\right ]-\frac {8C\alpha \gamma }{(1-C^2)\sqrt {1-C^2}}\\ &+3(\rho \gamma -\theta \beta )-\rho \left(\alpha -\frac {2\rho C}{\sqrt {1-C^2}}\right)-\alpha \rho \\ &-\frac {\epsilon C\sqrt {1-C^2}}{2}-\left[2\beta \left(\theta -\frac {2\beta C}{\sqrt {1-C^2}}\right) +(\lambda -\gamma )\left(\rho -\frac {2\lambda C}{\sqrt {1-C^2}}\right)\right.\\ &\left.+\rho (\lambda +\gamma )-\frac {\epsilon C\sqrt {1-C^2}}{2}+\gamma \rho -\theta \beta \right]. \end{split} \end{equation}
Substituting (4.67), (4.68), and (4.69) into (4.5), we obtain
\begin{equation} \begin{split} E_2(f_1)=&\ (1+C^2)\alpha \left(\beta +\tfrac {C\theta }{\sqrt {1-C^2}}\right)\left(\tfrac {1+C^2}{1-C^2}\theta +\tfrac {C\beta }{\sqrt {1-C^2}}\right) +\beta \alpha \left(\tfrac {1+C^2}{1-C^2}\theta +\tfrac {C\beta }{\sqrt {1-C^2}}\right)\\ &+\left(\rho -\tfrac {2\lambda C}{\sqrt {1-C^2}}\right)\left((1-C^2)\left(\beta +\tfrac {C\theta }{\sqrt {1-C^2}}\right)^2 -\tfrac {\alpha ^2}{1-C^2}\right)\\ &+4\theta \alpha \left(\beta +\tfrac {C\theta }{\sqrt {1-C^2}}\right). \end{split} \end{equation}
Now, substituting (4.87), (4.95), and (4.96) into (4.94), we obtain
\begin{align*} 0=\frac {8C}{(1-C^2)\sqrt {1-C^2}} \left [\left(C\theta +\sqrt {1-C^2}\beta \right)^2+\theta ^2-\alpha \gamma \right ]. \end{align*}
As
$C\neq 0$
on
$\Omega _{11}$
, we obtain that
\begin{equation} 0=\left(C\theta +\sqrt {1-C^2}\beta \right)^2+\theta ^2-\alpha \gamma . \end{equation}
By substituting
$f_5=\tfrac {\epsilon (1-C^2)}{2}+\alpha \gamma -\theta ^2$
, (4.65) and (4.67) into
$f_1^2=f_4f_5$
, we obtain
\begin{align*} 0=\alpha ^2\left(\beta +\frac {C\theta }{\sqrt {1-C^2}}\right)^2 -\frac {\alpha ^2}{1-C^2}\left(\frac{1}{2}\epsilon (1-C^2)+\alpha \gamma -\theta ^2\right). \end{align*}
Since
$\alpha \neq 0$
on
$\Omega _{11}$
, we obtain
\begin{equation} 0=\left(C\theta +\sqrt {1-C^2}\beta \right)^2-\left(\frac 12\epsilon (1-C^2)+\alpha \gamma -\theta ^2\right). \end{equation}
From (4.97) and (4.98), we obtain
$\epsilon (1-C^2)=0$
on
$\Omega _{11}$
, which contradicts to the assumptions
$0\lt |C|\lt 1$
and
$\epsilon =\pm 1$
. Thus
$\Omega$
is empty.
Thus, at any point of
$M_c$
, all the Ricci eigenvalues are equal, that is,
$M_c$
is Einstein. This completes the proof of Proposition4.1.
5. Proof of Theorem 1.2
We are now ready to complete the proof of Theorem1.2.
Let
$M$
be a hypersurface of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor. Write
$M=M_0\cup M_{1}\cup M_{c}$
, where
$M_0$
,
$M_{1}$
, and
$M_{c}$
are defined as in section 4. Denote by int
$(M_1)$
and int
$(M_0)$
the interior of
$M_1$
and
$M_0$
, respectively. We can split the following discussions into several cases:
Case I.
$M_c$
is empty.
By continuity of the function
$C$
, Case I can be split into two subcases as below:
Case I-(i).
$C^2\equiv 1$
on
$M$
.
In this case, without loss of generality, we suppose that
$C=1$
. We know from [Reference Gao, Ma and Yao5, Reference Urbano21] that
$M$
is locally congruent to
$\Gamma \times M^2(\epsilon )$
, where
$\Gamma$
is a curve in
$M^2(\epsilon )$
. Let
$TM=\{J_1N\}\oplus \{J_1N\}^\bot$
and let
$X_1=(X_{11},X_{12})$
,
$X_2=(X_{21},X_{22})\in \{J_1N\}^\bot$
be the orthonormal vector fields. As
$C=1$
, we have
$N=(N_1,0)$
,
$X_1=(0,X_{12})$
and
$X_2=(0,X_{22})$
. Hence
$T|_{\{J_1N\}^\bot }=-Id$
,
$A|_{\{J_1N\}^\bot }=0$
and
$M$
is a Hopf hypersurface. By (2.6), we get
\begin{equation} \textrm {Ric}\,(J_1N)=0,\ \textrm {Ric}\,(X_1)=\epsilon X_1,\ \textrm {Ric}\,(X_2)=\epsilon X_2. \end{equation}
By (2.3) and
$A|_{\{J_1N\}^\bot }=0$
, for any
$X,Y\in TM$
, we have
\begin{align*} \nabla _XJ_1N=-\phi AX=0,\ \nabla _{X}Y\in \{J_1N\}^\bot . \end{align*}
By (5.1) and the above equation, for any
$X,Y\in \{J_1N\}^\bot$
, we have
\begin{align*} \begin{split} &(\nabla _X\textrm {Ric})Y=\nabla _X\textrm {Ric}(Y)-\textrm {Ric}(\nabla _XY)=0,\\ &(\nabla _X\textrm {Ric})J_1N=0, \ (\nabla _{J_1N}\textrm {Ric})J_1N=0, \ (\nabla _{J_1N}\textrm {Ric})X=0. \end{split} \end{align*}
It follows that
$M$
has parallel Ricci tensor.
Case I-(ii).
$C\equiv 0$
on
$M$
.
By (2.7) and Lemma2.2, we have
$E_1=J_1N$
,
$E_2=J_2N$
,
$E_3=V$
, and
$\theta =\beta =\gamma =0$
. From (2.19) and (2.20), we have
\begin{equation} \begin{split} &\textrm {Ric}\,(E_1)=\left(\frac {\epsilon }{2}+\alpha \lambda -\rho ^{2}\right)E_1,\\ &\textrm {Ric}\,(E_2)=\left(\frac {\epsilon }{2}+ \alpha \lambda -\rho ^{2}\right)E_2,\\ &\textrm {Ric}\,(E_3)=\epsilon E_3. \end{split} \end{equation}
If
$M$
has two distinct Ricci eigenvalues, by Lemma3.3 and (5.2), we have
$\epsilon =0$
, a contradiction. Thus
$M$
is an Einstein manifold with Einstein constant
$\epsilon$
. It follows that
$M$
has parallel Ricci tensor.
Case II.
$M_c$
is non-empty.
In this case, we first prove that
$M_c\cup M_0$
is Einstein. In fact, by Proposition4.1, we see that
$M_c$
is Einstein. If
$M_0$
is empty, then
$M_c\cup M_0=M_c$
is Einstein. If int
$(M_0)$
is non-empty, in each component of int
$(M_0)$
, by Case I-(ii), we know that at any point of int
$(M_0)$
all the Ricci eigenvalues are equal. By continuity of the Ricci eigenvalues, we get that
$M_c\cup M_0$
has exactly one Ricci eigenvalue. If
$M_0$
is non-empty and without interior, by continuity of the Ricci eigenvalues, we also get that
$M_c\cup M_0$
has exactly one Ricci eigenvalue. Thus
$M_c\cup M_0$
is Einstein.
Now, Case II can be split into two subcases as below:
Case II-(i).
$M_1$
is empty.
In this case, we have
$M=M_c\cup M_0$
. From the above discussions, we obtain that
$M$
is Einstein.
Case II-(ii).
$M_1$
is non-empty.
If int
$(M_1)$
is non-empty, in each component of int
$(M_1)$
, by (5.1), we get that int
$(M_1)$
has two distinct constant Ricci eigenvalues. But
$M_c\cup M_0$
is Einstein. This contradicts to the continuity of the Ricci eigenvalues. Hence this case does not occur.
If int
$(M_1)$
is empty, by the fact that
$M_c\cup M_0$
has only one Ricci eigenvalue and the continuity of Ricci eigenvalues, we obtain that
$M_1$
has one Ricci eigenvalue. It follows that
$M$
is Einstein.
To sum up, we obtain that a hypersurface
$M$
of
$M^2(\epsilon )\times M^2(\epsilon )$
with recurrent Ricci tensor has parallel Ricci tensor. Moreover,
$M$
is locally congruent to an open subset of
$\Gamma \times M^2(\epsilon )$
, where
$\Gamma$
is a curve in
$M^2(\epsilon )$
or an Einstein hypersurface. The latter case is equivalent to say that
$M$
has constant sectional curvature. The complicated class of constant sectional curvature hypersurfaces in
$M^2(\epsilon )\times M^2(\epsilon )$
has been classified in [Reference Li, Vrancken, Wang and Yao12] for
$\epsilon =1$
, and in [Reference Li, Vrancken, Wang and Yao13] for
$\epsilon =-1$
, respectively.
We have completed the proof of Theorem1.2.
Acknowledgements
The authors would like to express their thanks to the referee and editor for helpful comments and suggestions to improve the presentation of the paper.
This work was supported by NSF of China (Grant Nos. 12171437 and 12401061). Z. Yao was also supported by Young Teachers Research Foundation of South China Normal University (Grant No. 23KJ23).
Conflict of interests
The authors declare no competing interests.
