According to Chen [1], one of the most important problems in submanifold theory is to find simple relationships between the main extrinsic invariants and the main intrinsic invariants of a submanifold. Related with famous Nash embedding theorem [2], Chen introduced a new type of Riemannian invariants, known as $\delta$-invariants [3, 4, 5]. The author's original motivation was to provide answers to a question raised by Chern concerning the existence of minimal isometric immersions into Euclidean space [6]. Therefore, Chen obtained a necessary condition for the existence of minimal isometric immersion from a given Riemannian manifold into Euclidean space and established inequalities for submanifolds in real space forms in terms of the sectional curvature, the scalar curvature and the squared mean curvature [7]. Later, he established general inequalities relating $\delta(n_1, \cdots, n_k)$ and the squared mean curvature for submanifolds in real space forms [8]. Similar inequalities also hold for Lagrangian submanifolds of complex space forms. In [9], Chen proved that, for any $\delta(n_1, \cdots, n_k)$, the equality case holds if and only if the Lagrangian submanifold is minimal. This interesting phenomenon inspired people to look for a more sharp inequality. In 2007, Oprea improved the inequality on $\delta(2)$ for Lagrangian submanifolds in complex space forms[10]. Recently, Chen and Dillen established general inequalities for Lagrangian submanifolds in complex space forms and provided some examples showing these new improved inequalities are best possible [11]. Such invariants and inequalities have many nice applications to several areas in mathematics [12].

Afterwards, many papers studied similar problems for different submanifolds in various ambient spaces, like complex space forms [13], Sasakian space forms [14], $(\kappa, \mu)$-contact space forms [15], Lorentzian manifold [16], Euclidean space [17] and locally conformal almost cosymplectic manifolds [18].

This paper is organized as follows. In Section 2, the basic elements of the theory of $\delta$-invariants are briefly presented. In Section 3, we establish general inequalities of $\delta$-invariants for submanifolds of a Riemannian manifold of quasi-constant curvature [19], which generalize a result of paper [20]. In Section 4, we obtain an inequality between the Ricci curvature and the squared mean curvature for submanifolds of the ambient space by using a algebraic lemma. Finally, in Section 5, we establish inequalities between the warping function $f$ (intrinsic structure) and the squared mean curvature (extrinsic structure) for warped product submanifolds $M_1\times_fM_2$ in a Riemannian manifold of quasi-constant curvature, as another answer of the basic problem in submanifold theory which we have mentioned in the introduction.

In [19], Chen and Yano introduced the notion of a Riemannian manifold $(N, g)$ of quasi-constant curvature as a Riemannian manifold with the curvature tensor satisfying the condition

$ \begin{aligned} \overline{R}(X, Y, Z, W)&=a[g(X, Z)g(Y, W)-g(Y, Z)g(X, W)]+b[g(X, Z)T(Y)T(W)\\ &~~~~-g(X, W)T(Y)T(Z)+g(Y, W)T(X)T(Z)-g(Y, Z)T(X)T(W)], \end{aligned} $

(2.1)

where $a, b$ are scalar functions and $T$ is a $1$-form defined by

$ g(X, P)=T(X) $

(1)

and $P$ is a unit vector field. If $b=0$, it can be easily seen that the manifold reduces to a space of constant curvature.

Decomposing the vector field $P$ on $M$ uniquely into its tangent and normal components $P^T$ and $P^{\perp}$, respectively, we have

$ P=P^T+P^{\perp}. $

(2.3)

Let $M$ be an $n$-dimensional submanifold of an $(n+p)$-dimensional Riemannian manifold of quasi-constant curvature $N^{n+p}$. The Gauss equation is given by

$ R(X, Y, Z, W)=\overline{R}(X, Y, Z, W)+g(h(X, Z), h(Y, W))-g(h(X, W), h(Y, Z)) $

(2.4)

for all X, Y, Z, W ∈ TM, where R and $\overline{R}$ are the curvature tensors of $M$ and $N^{n+p}$, respectively, and $h$ is the second fundamental form.

In $N^{n+p}$ we choose a local orthonormal frame $e_1, \cdots, e_n, e_{n+1}, \cdots, e_{n+p}, $ such that, restricting to $M^n$, $e_1, \cdots, e_n$ are tangent to $M^n$. We write $h_{ij}^{r}=g(h(e_i, e_j), e_{r})$. The mean curvature vector $\zeta$ is given by $\zeta=\sum\limits_{r=n+1}^{n+p}(\frac{1}{n}\sum\limits_{i=1}^nh_{ii}^{r})e_{r}, $ then the mean curvature $H$ is given by $H=\parallel \zeta \parallel.$

Let $K(e_i\wedge e_j), \ 1\leq i < j\leq n$, denote the sectional curvature of the plane section spanned by $e_i$ and $e_j$. Then the scalar curvature of $M^n$ is given by

$ \tau = \mathop \sum \limits_{1 \le i < j \le n} K({e_i} \wedge {e_j}). $

(2.5)

Let $L$ be an $l$-dimensional subspace of $T_xM$, $x\in M$, $l\geq2$ and $\{e_1, \cdots, e_l\}$ an orthonormal basis of $L$. We define the scalar curvature $\tau(L)$ of the $l$-plane $L$ by

$ \tau (L) = \sum\limits_{1 \le \alpha < \beta \le l} {K({e_\alpha } \wedge {e_\beta }).} $

(2.6)

For simplicity we put

$ \Psi (L) = \sum\limits_{1 \le i < j \le l} {[g{{({P^T},{e_i})}^2} + g{{({P^T},{e_j})}^2}].} $

(2.7)

For an integer $k\geq0$ we denote by $S(n, k)$ the set of $k$-tuples $(n_1, \cdots, n_k)$ of integers $\geq 2$ satisfying $n_1 < n$ and $n_1+\cdots+n_k\leq n$. We denote by $S(n)$ the set of unordered $k$-tuples with $k\geq0$ for a fixed $n$. For each $k$-tuples $(n_1, \cdots, n_k)\in S(n)$, Chen defined a Riemannian invariant $\delta(n_1, \cdots, n_k)$ as follows [8]

$ \delta(n_1, \cdots, n_k)(x)=\tau(x)-S(n_1, \cdots, n_k)(x), $

(2.8)

where $S(n_1, \cdots, n_k)(x)=\inf\{\tau(L_1)+\cdots+\tau(L_k)\}, $ and $L_1, \cdots, L_k$ run over all $k$ mutually orthogonal subspaces of $T_xM$ such that dim$L_j=n_j, \ j\in\{1, \cdots, k\}$. For each $(n_1, \cdots, n_k)\in S(n)$, we put

$ c(n_1, \cdots, n_k)=\frac{n^2\big(n+k-1-\sum\limits_{j=1}^kn_j\big)}{2\big(n+k-\sum\limits_{j=1}^kn_j\big)}, ~~ d(n_1, \cdots, n_k)=\frac{1}{2}[n(n-1)-\sum\limits_{j=1}^kn_j(n_j-1)]. $

For a differentiable function $f$ on $M$, the Laplacian $\triangle f$ of $f$ is defined by

$ \Delta f = \sum\limits_{i - 1}^n {\left[ {\left( {{\nabla _{{e_i}}}{e_i}} \right)f - {e_i}{e_i}f} \right]} . $

We shall use the following lemmas.

Lemma 2.1[7]     Let $a_1, a_2, \cdots, a_n, b$ be $(n+1)(n\geq2)$ real numbers such that

$ {(\sum\limits_{i = 1}^n {{a_i}} )^2} = (n - 1)(\sum\limits_{i = 1}^n {a_i^2} + b), $

then $2a_1a_2\geq b$, with the equality holding if and only if $a_1+a_2=a_3=\cdots=a_n$.

Lemma 2.2     Let $f(x_1, x_2, \cdots, x_n)$ be a function in $R^n$ defined by

$ f(x_1, x_2, \cdots, x_n)=x_1\sum\limits_{i=2}^nx_i. $

If $x_1+x_2+\cdots+x_n=2\lambda$, then we have $f(x_1, x_2, \cdots, x_n)\leq \lambda^2, $ with the equality holding if and only if $x_1=x_2+x_3+\cdots+x_n=\lambda.$

Proof     From $x_1+x_2+\cdots+x_n=2\lambda$, we have $\sum\limits_{i=2}^nx_i=2\lambda-x_1.$ It follows that

$ f(x_1, x_2, \cdots, x_n)=x_1(2\lambda-x_1)=-(x_1-\lambda)^2+\lambda^2, $

which represents Lemma 2.2 to prove.

Theorem 3.1     If $M^n$ $(n\geq 3)$ is a submanifold of a Riemannian manifold of quasi-constant curvature $N^{n+p}$, then we have

$ \delta(n_1, \cdots, n_k)\leq c(n_1, \cdots, n_k)H^2+d(n_1, \cdots, n_k)a+b[(n-1)\parallel P^T\parallel^2-\sum\limits_{j=1}^k\Psi(L_j)] $

(3.1)

for any $k$-tuples $(n_1, \cdots, n_k)\in S(n)$. The equality case of (3.1) holds at $x\in M^n$ if and only if there exist an orthonormal basis $\{e_1, \cdots, e_n\}$ of $T_xM$ and an orthonormal basis $\{e_{n+1}, \cdots, e_{n+p}\}$ of $T^{\perp}_xM$ such that the shape operators of $M^n$ in $N^{n+p}$ at $x$ have the following forms

$ A_{e_{n+1}}=\left( \begin{array}{ccccc} a_1& 0&\cdots&0 \\ 0 &a_2& \cdots&0 \\ \vdots&\vdots&\ddots& \vdots \\ 0&0&\cdots &a_n \\ \end{array} \right), \ \ A_{e_r}=\left( \begin{array}{ccccc} A_1^r& \cdots&0 &0 \\ \vdots& \ddots&\vdots&\vdots \\ 0&\cdots& A_k^r& 0 \\ 0&\cdots&0 &\mu_rI \\ \end{array} \right), \ \ r=n+2, \cdots, n+p, $

where $a_1, \cdots, a_n$ satisfy

$ a_1+\cdots+a_{n_1}=\cdots=a_{n_1+\cdots+n_{k-1}+1}+\cdots+a_{n_1+\cdots+n_k}=a_{n_1+\cdots+n_k+1}=\cdots=a_n $

and each $A_j^r$ is a symmetric $n_j\times n_j$ submatrix satisfying trace$(A_1^r)$ $=\cdots=$ trace$(A_k^r)=\mu_r$. $I$ is an identity matrix.

Remark 3.2     For $\delta(2)$, inequality (3.1) is due to Cihan Özgör [20, Theorem 3.1].

Proof     Let $x\in M^n$ and $\{e_1, e_2, \cdots, e_n\}$ and $\{e_{n+1}, e_{n+2}, \cdots, e_{n+p}\}$ be orthonormal basis of $T_xM^n$ and $T_x^{\perp}M^n$, respectively, such that the mean curvature vector $\zeta$ is in the direction of the normal vector to $e_{n+1}$. For convenience, we set

$ \begin{aligned} &a_i=h_{ii}^{n+1}, \ \ i=1, 2, \cdots, n, \\ &b_1=a_1, \ \ b_2=a_2+\cdots+a_{n_1}, \ \ b_3=a_{n_1+1}+\cdots+a_{n_1+n_2}, \cdots, \\ &b_{k+1}=a_{n_1+\cdots+n_{k-1}+1}+\cdots+a_{n_1+n_2+\cdots+n_{k-1}+n_k}, \\ &b_{k+2}=a_{n_1+\cdots+n_k+1}, \cdots, b_{\gamma+1}=a_n, \\ &\Delta_1=\{1, \cdots, n_1\}, \cdots, \\ &\Delta_k=\{(n_1+\cdots+n_{k-1})+1, \cdots, n_1+\cdots+n_k\}, \\ &\Delta_{k+1}=(\Delta_1\times\Delta_1)\cup\cdots\cup(\Delta_k\times\Delta_k). \end{aligned} $

Let $L_1, \cdots, L_k$ be mutually orthogonal subspaces of $T_xM$ with dim$L_j=n_j$, defined by

$ L_j={\rm Span}\{e_{n_1+\cdots+n_{j-1}+1}, \cdots, e_{n_1+\cdots+n_j}\}, \ \ j=1, \cdots, k. $

From (2.4), (2.6) and (2.7) we have

$ \tau ({L_j}) = \frac{{{n_j}({n_j} - 1)}}{2}a + b\Psi ({L_j}) + \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{{\alpha _j} < {\beta _j}} [h_{{\alpha _j}{\alpha _j}}^rh_{{\beta _j}{\beta _j}}^r - {(h_{{\alpha _j}{\beta _j}}^r)^2}], $

(3.2)

$ 2\tau=n(n-1)a+2b(n-1)\parallel P^T\parallel^2+n^2H^2-\parallel h\parallel^2. $

(3.3)

We can rewrite (3.3) as $n^2H^2=(\parallel h\parallel^2+\eta)\gamma, $ or equivalently,

$ {(\sum\limits_{i = 1}^n {h_{ii}^{n + 1}} )^2} = \gamma [\sum\limits_{i = 1}^n {{{(h_{ii}^{n + 1})}^2}} + \sum\limits_{i \ne j} {{{(h_{ij}^{n + 1})}^2}} + \sum\limits_{r = n + 2}^{n + p} {\sum\limits_{i,j = 1}^n {{{(h_{ij}^r)}^2}} } + \eta ], $

(3.4)

where

$ \begin{array}{l} \eta = 2\tau - 2c({n_1}, \cdots ,{n_k}){H^2} - n(n - 1)a - 2(n - 1)b\parallel {P^T}{\parallel ^2},\\ \gamma = n + k - \sum\limits_{j = 1}^k {{n_j}} . \end{array} $

(3.5)

From (3.4) we deduce

$ {(\sum\limits_{i = 1}^{\gamma + 1} {{b_i}} )^2} = \gamma [\eta + \sum\limits_{i = 1}^{\gamma + 1} {b_i^2} + \sum\limits_{i \ne j} {{{(h_{ij}^{n + 1})}^2}} + \sum\limits_{r = n + 2}^{n + p} {\sum\limits_{i,j = 1}^n {{{(h_{ij}^r)}^2}} } - 2\mathop \sum \limits_{{\alpha _1} < {\beta _1}} {a_{{\alpha _1}}}{a_{{\beta _1}}} - \cdots - 2\mathop \sum \limits_{{\alpha _k} < {\beta _k}} {a_{{\alpha _k}}}{a_{{\beta _k}}}], $

where $\alpha_j, \beta_j\in \Delta_j$, for all $j=1, \cdots, k$. Applying Lemma 2.1, we derive

$ \mathop \sum \limits_{j = 1}^k \mathop \sum \limits_{{\alpha _j} < {\beta _j}} {a_{{\alpha _j}}}{a_{{\beta _j}}} \ge \frac{1}{2}[\eta + \sum\limits_{i \ne j} {{{(h_{ij}^{n + 1})}^2}} + \sum\limits_{r = n + 2}^{n + p} {\sum\limits_{i,j = 1}^n {{{(h_{ij}^r)}^2}} } ], $

it follows that

$ \begin{array}{l} \mathop \sum \limits_{j = 1}^k \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{{\alpha _j} < {\beta _j}} [h_{{\alpha _j}{\alpha _j}}^rh_{{\beta _j}{\beta _j}}^r - {(h_{{\alpha _j}{\beta _j}}^r)^2}] \ge \frac{\eta }{2} + \frac{1}{2}\sum\limits_{r = n + 1}^{n + p} {\sum\limits_{(\alpha ,\beta ) \notin {\Delta _{k + 1}}} {{{(h_{\alpha \beta }^r)}^2}} } + \sum\limits_{r = n + 2}^{n + p} {\sum\limits_{{\alpha _j} \in {\Delta _j}} {{{(h_{{\alpha _j}{\alpha _j}}^r)}^2}} } \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ge \frac{\eta }{2}. \end{array} $

(3.6)

From (3.2) and (3.6) we have

$ \mathop \sum \limits_{j = 1}^k \tau {\rm{(}}{{\rm{L}}_{\rm{j}}}{\rm{)}} \ge \sum\limits_{j = 1}^k {[\frac{{{n_j}({n_j} - 1)}}{2}a + b\Psi ({L_j})] + \frac{1}{2}\eta .} $

(3.7)

Using (2.8), (3.5) and (3.7), we derive the desired inequality.

The equality case of (3.1) at a point $x\in M$ holds if and only if we have the equality in all the previous inequalities and also in the Lemma 2.1, thus, the shape operators take the desired forms.

In [21], Chen established a sharp relationship between the Ricci curvature and the squared mean curvature for any $n-$dimensional Riemannian submanifold of a real space form $R^m(c)$ of constant sectional curvature $c$ as follows

Theorem 4.1    (see [21, Theorem 4]) Let $M$ be an $n-$dimensional submanifold of a real space form $R^m(c)$. Then the following statements are true.

(1)    For each unit vector $X\in T_pM$, we have

$ \|\zeta \|^2\geq \frac{4}{n^2}[{\rm Ric}(X)-(n-1)c]. $

(4.1)

(2)    If $\zeta(p)=0$, then a unit vector $X\in T_pM$ satisfies the equality case of (4.1) if and only if $X$ belongs to the relative null space ${N}(p)$ given by

$ {N}(p)=\{X\in T_pM\mid h(X, Y)=0, \ \forall Y\in T_pM\}. $

(3)    The equality case of (4.1) holds for all unit vectors $X\in T_pM$ if and only if either $p$ is a geodesic point or $n=2$ and $p$ is an umbilical point.

Afterwards, many papers studied similar problems for different submanifolds in various ambient manifolds [22-24]. Thus, after putting an extra condition on the ambient manifold, like semi-symmetric metric connections in the case of real space forms [25] and curvature-like tensors in the case of a Riemannian manifold [26], one proves the results similar to that of Theorem 4.1.

In [20], Özgör obtained several Chen's inequalities for submanifolds of a Riemannian manifold of quasi-constant curvature. However, he didn't established an inequality between the clssical Ricci-curvature and the squared mean curvature. Under these circumstances it becomes necessary to give a theorem, which could present an inequality between the Ricci-curvature and the squared mean curvature for submanifolds in the ambient manifold.

Theorem 4.2    Let $M^n$ be an $n$-dimensional submanifold of an $(n+p)$-dimensional Riemannian manifold of quasi-constant curvature $N^{n+p}$. For each unit vector $X$ in $T_xM$ we have

$ {\rm Ric}(X)\leq (n-1)a+(n-2)bg(P^T, X)^2+b\parallel P^T\parallel^2+\frac{n^2}{4}H^2. $

(4.2)

The equality sign holds for any tangent vector $X$ in $T_xM$ if and only if either $x$ is a totally geodesic point or $n=2$ and $x$ is an umbilical point.

Remark 4.3    For $b=0$, inequality (4.2) is due to (4.1).

Remark 4.4    We should point out that our approach is different from Chen's.

Proof    Let $x\in M^n$ and $\{e_1, e_2, \cdots, e_n\}$ and $\{e_{n+1}, e_{n+2}, \cdots, e_{n+p}\}$ be orthonormal basis of $T_xM^n$ and $T_x^{\perp}M^n$, respectively, such that $X=e_1$. From the equation (2.1), (2.2), (2.3) and (2.4) it follows that

$ {R_{ijij}} = a + b\left[ {g{{({P^T},{e_i})}^2} + g{{({P^T},{e_j})}^2}} \right] + \sum\limits_{r = n + 1}^{n + p} {\left[ {h_{ii}^rh_{jj}^r - {{\left( {h_{ij}^r} \right)}^2}} \right]} . $

(4.3)

Using (4.3) one derives

$ \begin{array}{l} {\text{Ric}}(X) = \sum\limits_{i = 2}^n {{R_{1i1i}}} = (n - 1)a + (n - 1)bg{({P^T},{e_1})^2}\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; + b\mathop \sum \limits_{i = 2}^n g{({P^T},{e_i})^2} + \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{i = 2}^n [h_{11}^rh_{ii}^r - {(h_{1i}^r)^2}]\\ \;\;\;\;\;\;\;\;\;\;\; \le (n - 1)a + (n - 2)bg{({P^T},X)^2} + b\parallel {P^T}{\parallel ^2} + \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{i = 2}^n h_{11}^rh_{ii}^r. \end{array} $

(4.4)

Let us consider the quadratic forms $f_{r}:R^n\rightarrow R$, defined by

$ {f_r}(h_{11}^r,h_{22}^r, \cdots ,h_{nn}^r) = \sum\limits_{i = 2}^n {h_{11}^r} h_{ii}^r. $

We consider the problem $ \max f_r, $ subject to $\Gamma:h_{11}^{r}+h_{22}^{r}+\cdots+h_{nn}^{r}=k^r$, where $k^r$ is a real constant.

From Lemma 2.2, we see that the solution $(h_{11}^{r}, h_{22}^{r}, \cdots, h_{nn}^{r})$ of the problem in question must satisfy

$ h_{11}^r = \sum\limits_{j = 2}^n {h_{jj}^r} = \frac{{{k^r}}}{2}, $

(4.5)

which implies

$ f_{r}\leq \frac{(k^{r})^2}{4}. $

(4.6)

From (4.4) and (4.6) we have

$ {\rm Ric}(X) \leq (n-1)a+(n-2)bg(P^T, X)^2+b\parallel P^T\parallel^2+\sum\limits_{r=n+1}^{n+p}\frac{(k^{r})^2}{4}\\ \quad\quad\quad= (n-1)a+(n-2)bg(P^T, X)^2+b\parallel P^T\parallel^2+\frac{n^2}{4}H^2. $

Next, we shall study the equality case.

For each unit vector $X$ at $x$, if the equality case of inequality (4.2) holds, from (4.4), (4.5) and (4.6) we have

$ h_{1i}^{r}=0, \ \ i\neq1, \ \forall \ r, $

(4.7)

$ h_{11}^{r}+h_{22}^{r}+\cdots+h_{nn}^{r}-2h_{11}^{r}=0, \ \ \forall r. $

(4.8)

For any unit vector $X$ at $x$, if the equality case of inequality (4.2) holds, noting that $X$ is arbitrary, by computing Ric$(e_j), j=2, 3, \cdots, n$ and combining (4.7) and (4.8) we have

$ h_{ij}^{r}=0, \ \ i\neq j, \ \ \forall r;~~ h_{11}^{r}+h_{22}^{r}+\cdots+h_{nn}^{r}-2h_{ii}^{r}=0, \ \ \forall i, r. $

We can distinguish two cases:

(1) $n\neq2$, $h_{ij}^{r}=0, \ i, j=1, 2, \cdots, n, \ r=n+1, \cdots, n+p $ or

(2) $n=2$, $h_{11}^{r}=h_{22}^{r}, \ \ h_{12}^{r}=0, \ r=3, \cdots, 2+p$.

The converse is trivial.

We immediately have the following

Corollary 4.5    Let $M^n$ be an $n$-dimensional submanifold of an $(n+p)$-dimensional Riemannian manifold of quasi-constant curvature $N^{n+p}$. The equality case of inequality (4.2) holds for any tangent vector $X$ of $M^n$ if and only if either $M^n$ is a totally geodesic submanifold in $N^{n+p}$ or $n=2$ and $M^n$ is a totally umbilical submanifold.

Corollary 4.6    If $\zeta(x)=0$, then a unit vector $X\in T_xM$ satisfies the equality case of (4.2) if and only if $X$ belongs to the relative null space ${N}(x)$ given by

$ {N}(x)=\{X\in T_xM\mid h(X, Y)=0, \ \forall Y\in T_xM\}. $

Proof    Assume $\zeta(x)=0$. For each unit vector $X\in T_xM$, equality holds in (4.2) if and only if (4.5) and (4.7) hold. Then $h_{1i}^{r}=0, \ \forall i, r$, i.e., $X\in {N}(x)$.

Related with famous Nash embedding theorem[2], Chen established a general sharp inequality for wraped products in real space form [27]. Later, he studied warped products in complex hyperbolic spaces [28] and complex projective spaces [29], respectively. Afterwards, many papers studied similar prolems for different submanifolds in various ambient spaces [30-32]. In the present paper, we establish an inequality for warped product submanifolds of a Riemannian manifold of quasi-constant curvature.

The study of warped product manifolds was initiated by Bishop and O'Neill [33]. Following [33], we have

Let $(M_1, g_1)$ and $(M_2, g_2)$ be two Riemannian manifolds and $f$ a positive differentiable function on $M_1$, where dim$M_i=n_i \ (i=1, 2), \ n_1+n_2=n$. The warped product of $M_1$ and $M_2$ is the Riemannian manifold $M_1\times_f M_2=(M_1\times M_2, g), $ where $g=g_1+f^2g_2$. More explicitly, if vector fields $X$ and $Y$ tangent to $M_1\times_fM_2$ at $(x, y)$, then

$ g(X, Y)=g_1(\pi_{1\ast}X, \pi_{1\ast}Y)+f^2(x)g_2(\pi_{2\ast}X, \pi_{2\ast}Y), $

where $\pi_i(i=1, 2)$ are the canonical projections of $M_1\times_fM_2$ onto $M_1$ and $M_2$, respectively, and $\ast$ stands for derivative map.

For a warped product $M_1\times_f M_2$, we denote by $D_1$ and $D_2$ the distributions given by the vectors tangent to leaves and fibres, respectively, where $D_1$ is obtained from the tangent vectors of $M_1$ via the horizontal lift and $D_2$ by tangent vectors of $M_2$ via the vertical lift.

Let $\phi:M^n=M_1\times_f M_2\rightarrow N^{n+p}$ be an isometric immersion of a warped product $M_1\times_f M_2$ into a Riemannian manifold of quasi-constant curvature. Denote by $h$ the second fundamental form of $\phi$. Denote by tr$h_1$ and tr$h_2$ the trace of $h$ restricted to $M_1$ and $M_2$, respectively. The immersion $\phi$ is called mixed totally geodesic if $h(X, Z)=0$ for any $X$ in $D_1$ and $Z$ in $D_2$.

Since $M_1\times_f M_2$ is a warped product, we have $\nabla_XZ=\nabla_ZX=\frac{1}{f}(Xf)Z$ for any unit vector fields $X, Z$ tangent to $M_1, M_2$, respectively. It follows that

$ K(X\wedge Z)=g(\nabla_Z\nabla_XX-\nabla_X\nabla_ZX, Z)=\frac{1}{f}[(\nabla_XX)f-X^2f]. $

(5.1)

We set $\parallel P^{T}\parallel^2_{M_1}=\sum\limits_{j=1}^{n_1}g(P^{T}, e_j)^2, $ $ \parallel P^{T}\parallel^2_{M_2}=\sum\limits_{s=n_1+1}^ng(P^T, e_s)^2.$

Theorem 5.1    Let $\phi:M_1\times_f M_2 \rightarrow N^{n+p}$ be an isometric immersion of a warped product into a Riemannian manifold of quasi-constant curvature, then we have

$ \frac{\triangle f}{f}\leq \frac{n^2H^2}{4n_2}+n_1a+\frac{b}{n_2}[n_2\parallel P^{T}\parallel^2_{M_1}+n_1\parallel P^{T}\parallel^2_{M_2}], $

(5.2)

where $H^2$ is the squared mean curvature of $\phi$, and $\triangle$ is the Laplacian operator of $M_1$. The equality case of (5.2) holds if and only if $\phi$ is a mixed totally geodesic immersion with tr$h_1=$tr$h_2$.

Proof    In $N^{n+p}$ we choose a local orthonormal frame $\{e_1, \cdots, e_n, e_{n+1}, \cdots, e_{n+p}\}, $ such that $e_1, \cdots, e_{n_1}$ are tangent to $M_1$, $e_{n_1+1}, \cdots, e_{n}$ are tangent to $M_2$, $e_{n+1}$ is parallel to the mean curvature vector $\zeta$.

Using (5.1) and the definition of $\triangle f$, we get

$ \frac{{\Delta f}}{f} = \sum\limits_{j = 1}^{{n_1}} K ({e_j} \wedge {e_s}), $

(5.3)

for each $s\in \{n_1+1, \cdots, n\}$.

Using (2.1), (2.3) and (2.4) we have

$ 2\tau+\parallel h\parallel^2-n^2H^2=2b(n-1)\parallel P^T\parallel^2+(n^2-n)a. $

(5.4)

We set

$ \delta=2\tau-(n^2-n)a-2b(n-1)\parallel P^T\parallel^2-\frac{n^2}{2}H^2. $

(5.5)

Then, (5.4) can be written as

$ n^2H^2=2(\delta+\parallel h\parallel^2). $

(5.6)

If we put $a_1=h_{11}^{n+1}, a_2=\sum\limits_{i=2}^{n_1}h_{ii}^{n+1}, a_3=\sum\limits_{t={n_1}+1}^{n}h_{tt}^{n+1}$, from (5.6) we have

$ \begin{array}{l} {(\sum\limits_{i = 1}^3 {{a_i}} )^2} = 2[\delta + \sum\limits_{i = 1}^3 {a_i^2} + \sum\limits_{1 \le i \ne j \le n} {{{(h_{ij}^{n + 1})}^2}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\; + \sum\limits_{r = n + 2}^{n + p} {\sum\limits_{i,j = 1}^n {{{(h_{ij}^r)}^2}} } - \sum\limits_{2 \le j \ne k \le {n_1}} {h_{jj}^{n + 1}} h_{kk}^{n + 1} - \sum\limits_{{n_1} + 1 \le s \ne t \le n} {h_{ss}^{n + 1}} h_{tt}^{n + 1}]. \end{array} $

From Lemma 2.1 we get

$ \mathop \sum \limits_{1 \le j < k \le {n_1}} h_{jj}^{n + 1}h_{kk}^{n + 1} + \mathop \sum \limits_{{n_1} + 1 \le s < t \le n} h_{ss}^{n + 1}h_{tt}^{n + 1} \ge \frac{\delta }{2} + \mathop \sum \limits_{1 \le i < j \le n} {(h_{ij}^{n + 1})^2} + \frac{1}{2}\sum\limits_{r = n + 2}^{n + p} {\sum\limits_{i,j = 1}^n {{{(h_{ij}^r)}^2}} } , $

(5.7)

with the equality holding if and only if

$ \sum\limits_{i = 1}^{{n_1}} {h_{ii}^{n + 1}} = \sum\limits_{t = {n_1} + 1}^n {h_{tt}^{n + 1}} . $

(5.8)

From (5.3) we have

$ \begin{array}{l} \frac{{{n_2}\Delta f}}{f} = \tau - \mathop \sum \limits_{1 \le j < k \le {n_1}} K({e_j} \wedge {e_k}) - \mathop \sum \limits_{{n_1} + 1 \le s < t \le n} K({e_s} \wedge {e_t})\\ \;\;\;\;\;\;\;\;\; = \frac{{{n_1}({n_1} - 1)a}}{2} - ({n_1} - 1)b\sum\limits_{j = 1}^{{n_1}} g {({P^T},{e_j})^2} - \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{1 \le j < k \le {n_1}} [h_{jj}^rh_{kk}^r - {(h_{jk}^r)^2}]\\ \;\;\;\;\;\;\;\;\;\; - \frac{{{n_2}({n_2} - 1)a}}{2} - ({n_2} - 1)b\sum\limits_{s = {n_1} + 1}^n g {({P^T},{e_s})^2} - \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{{n_1} + 1 \le s < t \le n} [h_{ss}^rh_{tt}^r - {(h_{st}^r)^2}]\\ \;\;\;\;\;\;\;\;\; = \tau - \frac{{n(n - 1)a}}{2} + {n_1}{n_2}a - b[({n_1} - 1)\sum\limits_{j = 1}^{{n_1}} g {({P^T},{e_j})^2} + ({n_2} - 1)\sum\limits_{s = {n_1} + 1}^n g {({P^T},{e_s})^2}]\\ \;\;\;\;\;\;\;\;\; - \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{1 \le j < k \le {n_1}} [h_{jj}^rh_{kk}^r - {(h_{jk}^r)^2}] - \mathop \sum \limits_{r = n + 1}^{n + p} \mathop \sum \limits_{{n_1} + 1 \le s < t \le n} [h_{ss}^rh_{tt}^r - {(h_{st}^r)^2}]. \end{array} $

(5.9)

Combing (5.7) and (5.9) we have

$ \frac{n_2\triangle f}{f}\leq \tau-\frac{n(n-1)a}{2}+n_1n_2a-b[(n_1-1)\sum\limits_{j=1}^{n_1}g(P^T, e_j)^2+(n_2-1)\sum\limits_{s=n_1+1}^{n}g(P^T, e_s)^2]\\ \quad\quad\quad-\frac{\delta}{2}-\sum\limits_{1\leq j\leq n_1}\sum\limits_{n_1+1\leq t\leq n}(h_{jt}^{n+1})^2-\frac{1}{2}\sum\limits_{r=n+2}^{n+p}\sum\limits_{i, j=1}^n(h_{ij}^{r})^2\\ \quad\quad\quad-\sum\limits_{r=n+2}^{n+p}\sum\limits_{1\leq j < k\leq n_1}[h_{jj}^{r}h_{kk}^{r}-(h_{jk}^{r})^2]-\sum\limits_{r=n+2}^{n+p}\sum\limits_{n_1+1\leq s < t\leq n}[h_{ss}^{r}h_{tt}^{r}-(h_{st}^{r})^2].\\ \quad\quad=\tau-\frac{n(n-1)a}{2}+n_1n_2a-b[(n_1-1)\sum\limits_{j=1}^{n_1}g(P^T, e_j)^2+(n_2-1)\sum\limits_{s=n_1+1}^{n}g(P^T, e_s)^2]\\ \quad\quad\quad-\frac{\delta}{2}-\sum\limits_{r=n+1}^{n+p}\sum\limits_{1\leq j\leq n_1}\sum\limits_{n_1+1\leq t\leq n}(h_{jt}^{r})^2-\frac{1}{2}\sum\limits_{r=n+2}^{n+p}[(\sum\limits_{j=1}^{n_1}h_{jj}^{r})^2+(\sum\limits_{t=n_1+1}^nh_{tt}^{r})^2]\\ \quad\quad\leq \tau-\frac{n(n-1)a}{2}+n_1n_2a-b[(n_1-1)\sum\limits_{j=1}^{n_1}g(P^T, e_j)^2+(n_2-1)\sum\limits_{s=n_1+1}^{n}g(P^T, e_s)^2]-\frac{\delta}{2}\\ \quad\quad\quad=\frac{n^2H^2}{4}+n_1n_2a-b[(n_1-1)\sum\limits_{j=1}^{n_1}g(P^T, e_j)^2\\ \quad\quad\quad+(n_2-1)\sum\limits_{s=n_1+1}^{n}g(P^T, e_s)^2]+b(n-1)\parallel P^T\parallel^2\\ \quad\quad\quad=\frac{n^2H^2}{4}+n_1n_2a+b[n\parallel P^T \parallel^2-n_1\sum\limits_{j=1}^{n_1}g(P^{T}, e_j)^2-n_2\sum\limits_{s=n_1+1}^ng(P^T, e_s)^2], $

(5.10)

which proves inequality.

Next, we shall study the equality case.

From (5.7) and (5.10) we know that the equality case of (5.2) holds if and only if

$ h_{jt}^{r}=0, \ \ 1\leq j\leq n_1, \ \ n_1+1\leq t\leq n, \ \ n+1\leq r\leq n+p, \\ $

(5.11)

$ \sum\limits_{i=1}^{n_1}h_{ii}^{r}=\sum\limits_{t=n_1+1}^{n}h_{tt}^{r}=0, \ \ n+2\leq r\leq n+p. $

(5.12)

Obviously (5.11) is equivalent to $h(D_1, D_2)=0$, thus, the immersion $\phi$ is mixed totally geodesic. Further on, from (5.8) and (5.12), we have

$ \sum\limits_{i=1}^{n_1}h_{ii}^{r}=\sum\limits_{s=n_1+1}^{n}h_{ss}^{r}, \ \ \forall r, $

it follows that ${\text{tr}}{h_1} = {\text{tr}}{h_2}.$

Remark 5.2    If $b=0$, inequality (5.2) is due to Chen [28, Theorem 1.4].

As applications of Theorem 5.1, we have

Corollary 5.3    Under the same assumption as in Theorem 5.1, if $f$ is a harmonic function, there are no isometric minimal immersion of $M_1\times_fM_2$ into $N^{n+p}$ with $a < 0, b\leq0$.

Corollary 5.4    Under the same assumption as in Theorem 5.1, if $f$ is an eigenfunction of the Laplacian on $M_1$ with eigenvalue $\lambda>0$, there are no isometric minimal immersion of $M_1\times_fM_2$ into $N^{n+p}$ with $a < 0, b\leq0$.

Remark 5.5    In [34, Theorem 4.1], Ganchev and Mihova proved that a Riemannian manifold of quasi-constant curvature $N^{n+p}(n+p\geq 4)$ with $a < 0, b\neq 0$, can be locally $\xi$-isometric to a canal space-like hypersurface in the Minkowski space $\mathbb{R}_1^{n+p+1}$. $\xi$ is a unit vector field on $N^{n+p}$.