2.1. Sasakian structure $(\varphi,\,\xi,\,\eta,\,g)$ of the unit sphere $\mathbb {S}^{2n+1}$

As a real hypersurface of the complex Euclidean space $\mathbb {C}^{n+1}$ with canonical complex structure $J$, the $(2n+1)$-dimensional unit sphere $\mathbb {S}^{2n+1}$ naturally admits a Sasakian structure $(\varphi,\,\xi,\,\eta,\,g)$: $\xi =J{\bar N}$ is the structure vector field with the unit normal vector field ${\bar N}$ of the inclusion $\mathbb {S}^{2n+1} \hookrightarrow \mathbb {C}^{n+1}$; $g$ is the induced metric on $\mathbb {S}^{2n+1}$; $\eta (X) =g(X,\,\xi )$ and $\varphi X=JX-\langle JX,\,{\bar N}\rangle {\bar N}$ for any tangent vector field $X$ on $\mathbb {S}^{2n+1}$, where $\langle \cdot,\,\cdot \rangle$ denotes the standard Hermitian metric on $\mathbb {C}^{n+1}$. In particular, for any tangent vector fields $X,\,Y$ on $\mathbb {S}^{2n+1}$, the Sasakian structure $(\varphi,\,\xi,\,\eta,\,g)$ of $\mathbb {S}^{2n+1}$ satisfies the following properties:

(2.1)\begin{equation} \left\{ \begin{aligned} & g(\varphi X,\varphi Y)=g(X,Y)-\eta(X)\eta(Y),\\ & \varphi\xi=0,\quad \eta(\varphi X)=0,\quad {\rm rank}\,(\varphi)=2n,\\ & \varphi^2X={-}X+\eta(X)\xi,\quad {\rm d}\eta(X,Y)=g(X,\varphi Y),\\ & \bar{\nabla}_{X}\xi={-}\varphi X,\quad (\bar{\nabla}_{X}\varphi)Y =g(X,Y)\xi-\eta(Y)X, \end{aligned}\right. \end{equation}

where $\bar {\nabla }$ is the Levi-Civita connection with respect to the induced metric $g$ on $\mathbb {S}^{2n+1}$.

2.2. Local theory of Legendrian submanifolds in the unit sphere $\mathbb {S}^{2n+1}$

Let $M^n$ be a Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$, i.e. the contact form $\eta$ restricted to $M^n$ vanishes. Consequently, $\xi$ is a normal vector field over $M^n$. Denote by $N$ a unit normal vector field along $M^n$, and by $U,\,X,\,Y,\,Z$ the tangent vector fields on $M^n$ in the subsequent paragraphs. Then, we have the Gauss and Weingarten formulas:

(2.2)\begin{equation} \bar\nabla_XY=\nabla_XY+h(X,Y),\quad \bar\nabla_XN={-}A_NX+\nabla_X^\perp N, \end{equation}

where $\nabla$ is the Levi-Civita connection of the induced metric on $M^n$, still denoted by $g$, $h$ (resp. $A_N$) is the second fundamental form (resp. the shape operator with respect to $N$) of $M^n\to \mathbb {S}^{2n+1}$, and $\nabla ^\bot$ is the normal connection in the normal bundle $T^\perp M^n$. Then, by means of (2.2) it can be verified that

(2.3)\begin{equation} g(h(X,Y),N)=g(A_NX,Y). \end{equation}

Note from the facts $\eta (X)=0$ and ${\rm d}\eta (X,\,Y)=g(X,\,\varphi Y)$ that $\varphi$ maps the tangent vector fields of $M^n$ to the normal vector fields in $T^\perp M^n$. Applying (2.2), we further have

(2.4)\begin{equation} \nabla^{\bot}_{X}\varphi Y=\varphi\nabla_XY+g(X,Y)\xi,\quad A_{\varphi X}Y={-}\varphi h(X,Y)=A_{\varphi Y}X, \end{equation}

and thus $g(h(X,\,Y),\,\varphi Z)$ is totally symmetric in $X$, $Y$, and $Z$:

(2.5)\begin{equation} g(h(X,Y),\varphi Z)=g(h(X,Z),\varphi Y)=g(h(Y,Z),\varphi X). \end{equation}

It follows from (2.1), (2.3), and the Weingarten formula that

(2.6)\begin{equation} g(h(X,Y),\xi)=g(A_\xi X,Y)=0. \end{equation}

Moreover, the equations of Gauss, Ricci, and Codazzi are respectively given by

(2.7)\begin{align} & R(X,Y)Z=g(Y,Z)X-g(X,Z)Y+[A_{\varphi X},A_{\varphi Y}]Z, \end{align}

(2.8)\begin{align} & R^\perp(X,Y)\varphi Z=\varphi[A_{\varphi X},A_{\varphi Y}]Z, \end{align}

(2.9)\begin{align} & (\bar\nabla h)(X,Y,Z)=(\bar\nabla h)(Y,X,Z), \end{align}

where, by definitions:

\begin{align*} & [A_{\varphi X},A_{\varphi Y}]=A_{\varphi X}A_{\varphi Y}-A_{\varphi Y}A_{\varphi X},\\ & R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z,\\ & R^\perp(X,Y)\varphi Z=\nabla^\perp_X\nabla^\perp_Y\varphi Z -\nabla^\perp_Y\nabla^\perp_X\varphi Z-\nabla^\perp_{[X,Y]}\varphi Z,\\ & (\bar\nabla h)(X,Y,Z)=\nabla^\perp_Xh(Y,Z)-h(\nabla_XY,Z)-h(Y,\nabla_XZ). \end{align*}

From now on, we assume that $M^n$ is a minimal Legendrian submanifold in the sphere $\mathbb {S}^{2n+1}$, unless otherwise stated. Contracting the Gauss equation (2.7) twice, we have

(2.10)\begin{equation} n(n-1)\chi=n(n-1)-S,\quad S=\|h\|^2, \end{equation}

where $\chi$ is the normalized scalar curvature and $\|\cdot \|^2$ denotes the squared norm relative to the metric $g$. Furthermore, the Ricci identity reads:

(2.11)\begin{equation} \begin{aligned} & (\bar\nabla^2 h)(U,X,Y,Z)-(\bar\nabla^2 h)(X,U,Y,Z) =(\bar{R}\cdot h)(U,X,Y,Z)\\ & =R^\perp(U,X)h(Y,Z)-h(R(U,X)Y,Z)-h(Y,R(U,X)Z), \end{aligned} \end{equation}

where $\bar {R}$ is the curvature tensor of the Van der Waerden–Bortolotti connection and

(2.12)\begin{equation} \begin{aligned} (\bar\nabla^2 h)(U,X,Y,Z)= & \nabla_U^{\bot}((\bar\nabla h)(X,Y,Z))-(\bar\nabla h)(\nabla_UX,Y,Z)\\ & -(\bar\nabla h)(X,\nabla_UY,Z)-(\bar\nabla h)(X,Y,\nabla_UZ). \end{aligned} \end{equation}

As usual, the so-called local Legendre frame $\{e_1,\,\ldots,\,e_n,\,e_{1^*},\,\ldots,\,e_{n^*},\, e_{2n+1}\}$ along $M^n$ can be chosen such that, restricted to $M^n$, the vector fields $e_1,\, e_2,\,\ldots,\,e_n$ are orthonormal and tangent to $M^n$, whereas $\{e_{1^*}=\varphi e_1,\,\ldots,\, e_{n^*}=\varphi e_n,\,e_{2n+1}=\xi \}$ are the orthonormal normal vector fields of $M^n$ in $\mathbb {S}^{2n+1}$. In the sequel, we shall make the following convention on range of indices:

\begin{align*} i,j,k,\ell,m,p& =1,\ldots,n;\quad \alpha,\beta=1,\ldots,n+1,\\ i^*,j^*,k^*,\ell^*,m^*,p^*& =n+1,\ldots,2n;\quad \alpha^*=\alpha+n, \beta^*=\beta+n. \end{align*}

Set $h^{k^*}_{ij}=g(h(e_i,\,e_j),\,\varphi e_k)$ and $h^{2n+1}_{ij}=g(h(e_i,\,e_j),\,e_{2n+1})$. Denote by $h^{\alpha ^*}_{ij,\ell }$ and $h^{\alpha ^*}_{ij,\ell m}$ the first and the second covariant derivatives of $h^{\alpha ^*}_{ij}$ with respect to $\bar \nabla$, respectively. Let $R_{ijk\ell }=g(R(e_i,\,e_j)e_\ell,\,e_k)$ and $R_{ij\alpha ^*\beta ^*}=g(R^\perp (e_i,\, e_j)e_{\beta ^*},\,e_{\alpha ^*})$ be the components of the curvature tensors of $\nabla$ and $\nabla ^\perp$. Denote by $R_{ij}=\sum _kg(R(e_i,\,e_k)e_k,\,e_j)$ the components of the Ricci tensor of $g$. As $M^n$ is minimal in $\mathbb {S}^{2n+1}$, it is known from (2.5)–(2.9) that

(2.13)\begin{align} & R_{ijk\ell}=\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk} +\sum_m(h_{ik}^{m^*}h_{j\ell}^{m^*}-h_{i\ell}^{m^*}h_{jk}^{m^*}),\quad R_{ijk^*(2n+1)}=0, \end{align}

(2.14)\begin{align} & R_{ij}=(n-1)\delta_{ij}-\sum_{k,\ell}h^{\ell^*}_{ik}h^{\ell^*}_{jk},\quad R_{ijk^*\ell^*}=\sum_m(h_{ik}^{m^*}h_{j\ell}^{m^*} -h_{i\ell}^{m^*}h_{jk}^{m^*}), \end{align}

(2.15)\begin{align} & h^{k^*}_{ij}=h^{j^*}_{ik}=h^{i^*}_{kj},\ \ h^{2n+1}_{ij}=0,\quad h^{\alpha^*}_{ij,k}=h^{\alpha^*}_{ik,j}, \sum_ih^{\alpha^*}_{ii}=0. \end{align}

In this situation, the Ricci identity (2.11) can be rewritten as follows:

(2.16)\begin{equation} h^{\alpha^*}_{ij,\ell p}-h^{\alpha^*}_{ij,p\ell} =\sum_mh^{\alpha^*}_{mj}R_{mi\ell p}+\sum_mh^{\alpha^*}_{im}R_{mj\ell p} +\sum_\beta h^{\beta^*}_{ij}R_{\ell p\beta^*\alpha^*}. \end{equation}

Finally, the following uniqueness theorem for the Legendrian submanifolds in the unit sphere $\mathbb {S}^{2n+1}$ is also needed.

Theorem 2.1 (cf. [Reference Hu, Li and Xing12])

Let $f$ and $\bar {f}:M^n\to \mathbb {S}^{2n+1}$ be two Legendrian immersions of a connected manifold $M^n$ into the unit sphere $\mathbb {S}^{2n+1}$ with the second fundamental forms $h$ and $\bar {h}$, respectively. Assume that

(2.17)\begin{equation} g(f_*X,f_*Y)=g(\bar{f}_*X,\bar{f}_*Y),\quad g(h(X,Y),\varphi f_*Z)=g(\bar{h}(X,Y),\varphi\bar{f}_*Z),\end{equation}

where $X,\,Y,\,Z$ are any tangent vector fields on $M^n$. Then there exists an isometry $\tau$ of $\mathbb {S}^{2n+1}$ such that $f=\tau \circ \bar {f}$.

2.3. Equivalent properties of parallel or semi-parallel tensor $K$

Recall that, for the Legendrian submanifold $M^n$ in the unit sphere $\mathbb {S}^{2n+1}$, its second fundamental form $h$ is said to be parallel if $\bar \nabla h=0$ on $M^n$, and as a direct generalization, $h$ is said to be semi-parallel if $\bar {R}\cdot {h} =0$ on $M^n$, where $\bar {R}$ denotes the curvature tensor corresponding to the Van der Waerden–Bortolotti connection. For the latter one, it is easy to see from the Ricci identity (2.11) that, for tangent vector fields $U,\,X,\,Y,\,Z$ on $M^n$,

(2.18)\begin{equation} (\bar\nabla^2 h)(U,X,Y,Z)=(\bar\nabla^2 h)(X,U,Y,Z),\end{equation}

which, under the Legendre frame $\{e_1,\,\ldots,\,e_n,\,e_{1^*},\,\ldots,\,e_{n^*},\,e_{2n+1}\}$ on $M^n$, is equivalent to

(2.19)\begin{equation} h^{\alpha^*}_{ij,\ell p}=h^{\alpha^*}_{ij,p\ell},\end{equation}

where $1\leq i,\,j,\,\ell,\,p\leq n$ and $\alpha ^*=\alpha +n$ for $1\leq \alpha \leq n+1$.

In addition, associated with $\bar \nabla$ and $\xi$, a covariant differentiation $\bar \nabla ^{\xi }$ can be defined such that it acts on $h$ as

(2.20)\begin{equation} (\bar\nabla^\xi h)(X,Y,Z)=(\bar\nabla h)(X,Y,Z)-g(h(Y,Z),\varphi X)\xi. \end{equation}

Under the Legendre frame on $M^n$, setting $(\bar \nabla ^{\xi }h)(e_k,\,e_i,\,e_j) =\sum _\ell {\tilde {h}}^{\ell ^*}_{ij,k}e_{\ell ^*}$ and

(2.21)\begin{equation} (\bar\nabla^\xi_{e_p}(\bar\nabla^\xi h))(e_\ell,e_i,e_j) =((\bar\nabla^\xi)^2h)(e_p,e_\ell,e_i,e_j) =\sum_k\tilde{h}^{k^*}_{ij,\ell p}e_{k^*}, \end{equation}

we obtain the following relations (cf. [Reference Hu and Yin17]):

(2.22)\begin{equation} \tilde{h}^{\ell^*}_{ij,k}=h^{\ell^*}_{ij,k},\quad h^{k^*}_{ij,\ell p}=\tilde{h}^{k^*}_{ij,\ell p}-h^{\ell^*}_{ij}\delta_{kp},\ \ h^{2n+1}_{ij,\ell p}=2h^{\ell^*}_{ij,p}.\end{equation}

In particular, $h$ is called $C$-parallel if it satisfies $\bar \nabla ^\xi h=0$ on $M^n$.

For our purposes, we introduce the $(1,\,2)$-tensor $K:TM^n\times TM^n\to TM^n$ defined by $K:=-\varphi h$ satisfying $h(X,\,Y)=\varphi K(X,\,Y)$. A straightforward calculation shows that

Furthermore, it can be checked by using (2.20) and (2.21) that

(2.23)\begin{equation} \begin{aligned} & (\bar\nabla^\xi h)(X,Y,Z)=\varphi(\nabla K)(X,Y,Z),\\ & ((\bar\nabla^\xi)^2 h)(U,X,Y,Z)-((\bar\nabla^\xi)^2 h)(X,U,Y,Z)=\varphi(R\cdot K)(U,X,Y,Z). \end{aligned} \end{equation}

Consequently, $\nabla K=0$ if and only if $\bar \nabla ^\xi h=0$, and $R\cdot K=0$ if and only if

(2.24)\begin{equation} ((\bar\nabla^\xi)^2 h)(U,X,Y,Z)=((\bar\nabla^\xi)^2 h)(X,U,Y,Z), \end{equation}

which, under the Legendre frame $\{e_1,\,\ldots,\,e_n,\,e_{1^*},\,\ldots,\,e_{n^*},\, e_{2n+1}\}$ on $M^n$, is equivalent to

(2.25)\begin{equation} \tilde{h}^{k^*}_{ij,\ell p}=\tilde{h}^{k^*}_{ij,p\ell}.\end{equation}

Here, we shall call the tensor $K$ parallel (resp. semi-parallel ) if $\nabla K=0$ (resp. $R\cdot K=0$) holds on $M^n$.

Assume that the tensor $K$ of Legendrian submanifold $M^n$ does not vanish at some point $x\in M^n$. We shall consider $U_{x}M^n=\{v\in T_xM^n\,|\,g(v,\,v)=1\}$ and then define a function $F$ on $U_xM^n$ by $F(u):=g(K(u,\,u),\,u)=g(A_{\varphi u}u,\,u)$ for $u\in U_xM^n$. Since $U_xM^n$ is compact, there exists a unit vector $e_1\in U_xM^n$ at which the function $F(u)$ attains an absolute maximum, denoted by $\lambda _1$ and $\lambda _1>0$. As a result, it holds that:

(2.26)\begin{equation} g(K_{e_1}e_1,u)=0,\quad g(K_{e_1}e_1,e_1)\geq2g(K_{e_1}u,u),\ u\perp e_1,\ u\in U_{x}M^n. \end{equation}

2.4. Conformally flat Riemannian manifolds and product manifolds

Let $(M^n,\,g)$ be an $n$-dimensional connected Riemannian manifold with the normalized scalar curvature $\chi$ and the Levi-Civita connection $\nabla$ of the metric $g$. The Schouten tensor $P$ of $(1,\,1)$-type defined by

(2.27)\begin{equation} P=\frac{Q}{n-2}-\frac{n\chi}{2(n-2)}{\rm id} \end{equation}

is a self-adjoint operator with respect to the metric $g$, where $Q$ and id denote the Ricci operator and the identity transformation, respectively. By definition there holds:

(2.28)\begin{equation} g(X,QY)={\rm Ric}(X,Y)=g(QX,Y),\end{equation}

where ${\rm Ric}$ denotes the Ricci tensor of $M^n$ and $X,\,Y$ are vector fields tangent to $M^n$.

Recall that $(M^n,\,g)$ is said to be conformally flat if around each point of $M^n$ there exists a neighbourhood which can be conformally immersed into the Euclidean space $\mathbb {R}^n$. When $n\ge 4$, it is known that $(M^n,\,g)$ is conformally flat if and only if its Weyl curvature tensor vanishes. In this situation, the curvature tensor $R$ of $g$ can be rewritten as below:

(2.29)\begin{equation} R(X,Y)Z=g(Y,Z)PX-g(X,Z)PY+g(PY,Z)X-g(PX,Z)Y, \end{equation}

where $X,\,Y,\,Z$ are tangent vector fields of $M^n$, and the Schouten tensor $P$ is Codazzi, i.e.:

(2.30)\begin{equation} (\nabla_XP)Y=(\nabla_YP)X. \end{equation}

When $n=3$, we should remark that the Weyl curvature tensor vanishes automatically, and $(M^3,\,g)$ is conformally flat if and only if $P$ is a Codazzi tensor as above.

Moreover, the Ricci tensor ${\rm Ric}$ of $M^n$ is called parallel or semi-parallel if it satisfies $\nabla {\rm Ric}=0$ or $R\cdot {\rm Ric}=0$. In the latter case, by definition one has:

(2.31)\begin{equation} (R\cdot{\rm Ric})(X,Y)=R(X,Y){\rm Ric}=\nabla_X\nabla_Y{\rm Ric} -\nabla_Y\nabla_X{\rm Ric}-\nabla_{[X,Y]}{\rm Ric}.\end{equation}

In addition, we call the Riemannian manifold $(M^n,\,g)$ quasi-Einstein if its Ricci operator $Q$ admits exactly two distinct eigenvalues at each point, one of which is simple, and the traceless Ricci tensor $\tilde {\rm Ric}$ of $M^n$ is defined by

(2.32)\begin{equation} \tilde{\rm Ric}(X,Y)={\rm Ric}(X,Y)-(n-1)\chi g(X,Y). \end{equation}

Finally, we recall the following theorem for later use.

4.1. Minimal Legendrian submanifolds

For our purposes, we first calculate the Laplacian of $S$ to derive the following:

Lemma 4.1 Let $M^n$ $(n\ge 2)$ be a minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$. Then, it holds the identity:

(4.1)\begin{equation} \frac{1}{2}\Delta S =\|\bar\nabla^\xi h\|^2-\|{\rm Rie}\|^2-\|{\rm Ric}\|^2+n(n^2-1)\chi, \end{equation}

where $\|{\rm Rie}\|^2$ denotes the squared norm of the Riemannian curvature tensor with respect to the metric $g$ on $M^n$. Moreover, if it is of semi-parallel tensor $K$, we further have:

(4.2)\begin{align} & n(n-1)\Delta\chi+2\|\bar\nabla^\xi h\|^2=0, \end{align}

(4.3)\begin{align} & n(n^2-1)\chi=\|{\rm Rie}\|^2+\|{\rm Ric}\|^2. \end{align}

Proof. Choose the local Legendre frame $\{e_1,\,\ldots,\,e_n,\,e_{1^*},\,\ldots,\,e_{n^*},\,e_{2n+1}\}$ along $M^n$ as in § 2. By definition, we have

(4.4)\begin{equation} \frac{1}{2}\Delta S =\frac{1}{2}\Delta\bigg(\sum_{i,j,\alpha}(h^{\alpha^*}_{ij})^2\bigg) =\sum_{i,j,\ell,\alpha}(h^{\alpha^*}_{ij,\ell})^2 +\sum_{i,j,\alpha}h^{\alpha^*}_{ij}\Delta h^{\alpha^*}_{ij}. \end{equation}

Applying the Ricci identity (2.16), we deduce from (2.15) that

(4.5)\begin{equation} \begin{aligned} \Delta h^{\alpha^*}_{ij} & =\sum_\ell h^{\alpha^*}_{ij,\ell\ell}=\sum_\ell h^{\alpha^*}_{i\ell,j\ell}\\ & =\sum_\ell h^{\alpha^*}_{i\ell,\ell j} +\sum_{\ell,m}h^{\alpha^*}_{m\ell}R_{mij\ell} +\sum_{\ell,m}h^{\alpha^*}_{im}R_{m\ell j\ell} +\sum_{\ell,\beta}h^{\beta^*}_{i\ell}R_{j\ell\beta^*\alpha^*}\\ & =\sum_\ell h^{\alpha^*}_{\ell\ell,ij} +\sum_{\ell,m}h^{\alpha^*}_{m\ell}R_{mij\ell} +\sum_mh^{\alpha^*}_{im}R_{mj} +\sum_{\ell,m}h^{m^*}_{i\ell}R_{j\ell m^*\alpha^*}. \end{aligned} \end{equation}

This together with (2.13)–(2.15) yields that

(4.6)\begin{equation} \begin{aligned} \sum_{i,j,\alpha}h^{\alpha^*}_{ij}\Delta h^{\alpha^*}_{ij} & =\sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{k^*}_{m\ell}R_{mij\ell} +\sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{m^*}_{i\ell}R_{j\ell mk}\\ & \quad+\sum_{i,j,k,m}h^{k^*}_{ij}h^{k^*}_{im}R_{mj}-S. \end{aligned} \end{equation}

To go on from (4.6), we make use of the relation

(4.7)\begin{equation} \sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{m^*}_{i\ell}R_{j\ell mk} ={-}\sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{m^*}_{i\ell}R_{kmj\ell} ={-}\sum_{i,j,k,\ell,m}h^{k^*}_{mj}h^{k^*}_{i\ell}R_{mij\ell}, \end{equation}

and (2.13) to calculate that:

(4.8)\begin{equation} \sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{k^*}_{m\ell}R_{mij\ell} +\sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{m^*}_{i\ell}R_{j\ell mk} ={-}\sum_{i,j,\ell,m}(R_{imj\ell})^2+2n(n-1)\chi. \end{equation}

On the contrary, with the help of (2.14), it is easily seen that:

(4.9)\begin{equation} \sum_{i,j,k,m}h^{k^*}_{ij}h^{k^*}_{im}R_{mj} ={-}\sum_{i,j}(R_{ij})^2+n(n-1)^2\chi. \end{equation}

Substituting (4.8) and (4.9) into (4.6), we then conclude that:

(4.10)\begin{equation} \sum_{i,j,\alpha}h^{\alpha^*}_{ij}\Delta h^{\alpha^*}_{ij} ={-}\sum_{i,j,k,\ell}(R_{ijk\ell})^2-\sum_{i,j}(R_{ij})^2+n(n^2-1)\chi-S. \end{equation}

As $h_{ij,\ell }^{2n+1}=h_{ij}^{\ell ^*}$ and $h^{k^*}_{ij,\ell }=\tilde {h}^{k^*}_{ij,\ell }$ for any $i,\,j,\,k,\,\ell$, it follows that

(4.11)\begin{equation} \sum_{i,j,\ell,\alpha}(h^{\alpha^*}_{ij,\ell})^2 =\sum_{i,j,k,\ell}(h^{k^*}_{ij,\ell})^2+\sum_{i,j,\ell}(h^{\ell^*}_{ij})^2 =\|\bar\nabla^\xi h\|^2+S, \end{equation}

and thus we obtain (4.1) by substituting (4.10) and (4.11) into (4.4) immediately.

Now, if assuming that $M^n$ has semi-parallel tensor $K$, then

(4.12)\begin{equation} 0=\tilde{h}^{k^*}_{ij,\ell p}-\tilde{h}^{k^*}_{ij,p\ell} =\sum_mh^{k^*}_{mj}R_{mi\ell p} +\sum_mh^{k^*}_{im}R_{mj\ell p} +\sum_mh^{m^*}_{ij}R_{mk\ell p},\end{equation}

where we used $h^{k^*}_{ij,\ell p}=\tilde {h}^{k^*}_{ij,\ell p}-h^{\ell ^*}_{ij}\delta _{kp}$ and $h^{2n+1}_{ij,\ell p}=2h^{\ell ^*}_{ij,p}$. Similarly, we obtain that

(4.13)\begin{equation} \sum_{i,j,\alpha}h^{\alpha^*}_{ij}\Delta h^{\alpha^*}_{ij} ={-}\sum_{i,j,k,\ell,m}h^{k^*}_{ij}h^{m^*}_{i\ell}(\delta_{jm}\delta_{\ell k} -\delta_{jk}\delta_{\ell m})={-}S, \end{equation}

which combining with (4.11) shows that

(4.14)\begin{equation} \frac{1}{2}\Delta S=\|\bar\nabla^\xi h\|^2. \end{equation}

This gives (4.2) by (2.10), and substituting (4.14) into (4.1) finally yields (4.3).

Furthermore, recalling the components of the Weyl curvature tensor $W$ of $M^n$ satisfy

(4.15)\begin{equation} W_{ijk\ell}=R_{ijk\ell}-\frac{1}{n-2}(\delta_{ik}R_{j\ell} +\delta_{j\ell}R_{ik}-\delta_{i\ell}R_{jk}-\delta_{jk}R_{i\ell}) +\frac{n\chi}{n-2}(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}),\end{equation}

we have the following expression for $n\ge 3$ (cf. [Reference Hu and Xing15, Reference Xing and Yin32]):

(4.16)\begin{equation} \|{\rm Rie}\|^2=\|W\|^2+\frac{4}{n-2}\|{\rm Ric}\|^2-\frac{2n^2(n-1)}{n-2}\chi^2, \end{equation}

where $\|W\|^2=\sum _{i,j,k,\ell }(W_{ijk\ell })^2$. From $\tilde {R}_{ij}=R_{ij} -(n-1)\chi \delta _{ij}$, it is easy to see that

(4.17)\begin{equation} \|{\rm Ric}\|^2=\|\tilde{\rm Ric}\|^2+n(n-1)^2\chi^2. \end{equation}

Here, $\tilde {\rm Ric}$ is the traceless part of ${\rm Ric}$ and $\|\tilde {\rm Ric}\|^2 =\sum _{i,j}(\tilde {R}_{ij})^2$. From the combination of (2.10), (4.16), and (4.17), we then derive by using (4.1) the following result:

Proposition 4.3 Let $M^n$ $(n\ge 3)$ be a minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$. Then, it holds the identity:

(4.18)\begin{equation} \frac{1}{2}\Delta S=\|\bar\nabla^\xi h\|^2-\|W\|^2 -\frac{n+2}{n-2}\|\tilde{\rm Ric}\|^2+(n+1)S\chi. \end{equation}

4.2. Conformally flat Legendrian submanifolds

Assume that $M^n$ $(n\ge 3)$ is a conformally flat Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$. Then, we shall prove that

Lemma 4.4 Let $M^n$ $(n\ge 3)$ be a conformally flat Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$. Then, for the tensor $K$ and Schouten tensor $P$ of $M^n$, it holds that

(4.19)\begin{equation} \begin{aligned} & \mathop{\mathfrak{S}}_{U,X,Y}(g(U,Z)K(PX,Y)-g(X,Z)K(PU,Y)\\ & \quad+g(PX,K(Y,Z))U-g(PU,K(Y,Z))X)=0, \end{aligned} \end{equation}

where $U,\,X,\,Y,\,Z$ are vector fields tangent to $M^n$ and $\mathop {\mathfrak {S}}$ denotes the cyclic summation.

Proof. We begin with taking the covariant derivative of the Codazzi equation for $K$ along a vector field $U$ tangent to $M^n$:

(4.20)\begin{equation} (\nabla^2K)(U,X,Y,Z)-(\nabla^2K)(U,Y,X,Z)=(\nabla G)(U,X,Y,Z)=0, \end{equation}

where, according to lemma 2.2 (3), $G(X,\,Y,\,Z):=(\nabla K)(X,\,Y,\,Z)-(\nabla K) (Y, X,\,Z)=0$ for tangent vector fields $X,\,Y,\,Z$ on $M^n$. It is obvious from (4.20) that

(4.21)\begin{equation} \mathop{\mathfrak{S}}_{U,X,Y}((\nabla^2K)(U,X,Y,Z)-(\nabla^2K)(U,Y,X,Z))=0.\end{equation}

Furthermore, direct calculations by using the Ricci identity show that

(4.22)\begin{equation} \begin{aligned} 0 & =\mathop{\mathfrak{S}}_{U,X,Y}((\nabla^2K)(U,X,Y,Z) -(\nabla^2K)(U,Y,X,Z))\\ & =\mathop{\mathfrak{S}}_{U,X,Y}((\nabla^2K)(U,X,Y,Z) -(\nabla^2K)(X,U,Y,Z))\\ & =\mathop{\mathfrak{S}}_{U,X,Y}(R(U,X)K(Y,Z) -K(R(U,X)Y,Z)-K(Y,R(U,X)Z)). \end{aligned} \end{equation}

Finally, the assertion immediately follows by substituting (2.29) into (4.22).

Choosing an orthonormal frame $\{e_i\}^n_{i=1}$ over $M^n$ such that $e_i$ is the eigenvector field of the Ricci operator $Q$ with $\mu _i$ the corresponding eigenvalue, by (2.27) we easily see that $Pe_i=\nu _ie_i$ and $\nu _i={\mu _i}/{(n-2)} -{n\chi }/{(2(n-2))}$. Without loss of generality, we shall suppose that $Q$ has $t$ distinct eigenvalues $\mu _1,\,\dots,\,\mu _t$ with multiplicities $n_1,\,\dots,\,n_t$, respectively. Let $\mathfrak {D}(\mu _s)$ (resp. $\mathfrak {D}(\nu _s)$) denote the distribution such that $\mathfrak {D}(\mu _s)(x)$ (resp. $\mathfrak {D}(\nu _s)(x)$) is the eigenspace of $\mu _s(x)$ (resp. $\nu _s(x)$) at an arbitrary point $x\in M^n$ for $1\leq s\leq t$, and $n_1+\cdots +n_t=n$. For simplicity of notations, we also make the convention that, for $i\le t$ and $j\ge t+1$, if $\mu _j=\mu _i$ we shall write $n_j=n_i$, $\mathfrak {D}(\mu _j)=\mathfrak {D}(\mu _i)$ and $\mathfrak {D}(\nu _j) =\mathfrak {D}(\nu _i)$.

Proof. We begin with taking the product of equation (4.19) with vector field $V$ and setting $U=e_k$, $X=e_i$, $Y=e_j$, $Z=e_\ell$, and $V=e_m$ to derive the relation:

(4.23)\begin{equation} \begin{aligned} 0 & =(\nu_i-\nu_j)(K^\ell_{ij}\delta_{km}+K^m_{ij}\delta_{k\ell})\\ & \quad+(\nu_j-\nu_k)(K^\ell_{jk}\delta_{im}+K^m_{jk}\delta_{i\ell})\\ & \quad+(\nu_k-\nu_i)(K^\ell_{ik}\delta_{jm}+K^m_{ik}\delta_{j\ell}), \end{aligned} \end{equation}

where $1\leq i,\,j,\,k,\,\ell,\,m\leq n$, and $K^m_{ij}:=g(K(e_i,\,e_j),\,e_m)$.

First of all, we assume that $\nu _i\neq \nu _j=\nu _k$ for distinct $i,\,j,\,k$, and then $e_i \in \mathfrak {D}(\nu _i)$ and $e_j,\,e_k\in \mathfrak {D}(\nu _j)$ for $n_j\geq 2$. Taking $m\neq j,\,k$, we therefore obtain from (4.23) that

(4.24)\begin{equation} K^m_{ij}\delta_{k\ell}-K^m_{ik}\delta_{j\ell}=0. \end{equation}

Taking $\ell =k$ yields that $K^m_{ij}=0$, by which we see that $K(e_i,\,e_j)\in \mathfrak {D} (\nu _j)$ for $n_j\geq 2$. Similarly, $K(e_i,\,e_j)\in \mathfrak {D}(\nu _i)$ for $n_i\geq 2$. Combining with the assumption $\nu _i\neq \nu _j$, we can conclude that $K(e_i,\,e_j)=0$ provided that $n_i,\,n_j\geq 2$. Hence, we get assertion (1).

Next, if $n_i=1$ and $n_j\geq 2$, some $e_k\in \mathfrak {D}(\nu _j)$ different from $e_j$ can be chosen to satisfy $\nu _i\neq \nu _j=\nu _k$ for distinct $i,\,j,\,k$. In this situation, we take $m=k$ in (4.23) to deduce that

(4.25)\begin{equation} K^\ell_{ij}+K^k_{ij}\delta_{k\ell}-K^k_{ik}\delta_{j\ell}=0. \end{equation}

It follows that $K^i_{jj}=K^i_{kk}$ for $\ell =j$, $K^k_{ij}=0$ for $\ell =k$, and moreover $K^\ell _{ij}=0$ for $\ell \neq j,\,k$. Consequently, assertion (2) follows by putting $\lambda ^i_j=K^i_{jj}$, where $\lambda ^i_j$ does not depend on the choice of $e_k\in \mathfrak {D}(\nu _j)$.

Finally, for $n_i=1$ and $\nu _j\neq \nu _k$ with $n_j,\,n_k\geq 2$, by taking $\ell =j$ in (4.23) and applying assertion (1), we easily get:

(4.26)\begin{equation} (\nu_i-\nu_j)K^j_{ij}\delta_{km} +(\nu_j-\nu_k)K^j_{jk}\delta_{im} +(\nu_k-\nu_i)(K^j_{ik}\delta_{jm}+K^m_{ik})=0. \end{equation}

Thus, taking $m=k$, and noting $\lambda ^i_j=K^i_{jj}=K_{ij}^j$ and $\lambda ^i_k=K^i_{kk} =K_{ik}^k$, we further have

(4.27)\begin{equation} (\nu_i-\nu_j)\lambda^i_j=(\nu_i-\nu_k)\lambda^i_k=:\bar\lambda_i.\end{equation}

This verifies assertion (3). Hence, lemma 4.6 has been proved.

4.3. Conformally flat minimal Legendrian submanifolds with $R\cdot K=0$

In this subsection, we shall consider that $M^n$ $(n\ge 3)$ is a conformally flat minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$ with semi-parallel tensor $K$. Therefore, we obtain from lemma 2.2 (4) that

(4.28)\begin{equation} \begin{aligned} 0 & =(R\cdot K)(X,Y,Z,U)\\ & =R(X,Y)K(Z,U)-K(R(X,Y)Z,U)-K(Z,R(X,Y)U), \end{aligned} \end{equation}

where $X,\,Y,\,Z,\,U$ are vector fields tangent to $M^n$. According to lemma 2.3, we then present the following lemma involving the number $r$ of distinct eigenvalues of the Schouten tensor $P$ of $M^n$.

Proof. First of all, by taking $X=e_i$, $Y=e_j$, $Z=e_k$, and $U=e_\ell$ in (4.28), we can apply (2.29) to calculate the relation:

(4.29)\begin{equation} \begin{aligned} 0 & =(\nu_i+\nu_j)[g(K(e_k,e_\ell),e_j)e_i-g(K(e_k,e_\ell),e_i)e_j]\\ & \quad+(\nu_i+\nu_j)[\delta_{ik}K(e_j,e_\ell)-\delta_{jk}K(e_i,e_\ell)]\\ & \quad+(\nu_i+\nu_j)[\delta_{i\ell}K(e_j,e_k)-\delta_{j\ell}K(e_i,e_k)]. \end{aligned} \end{equation}

For $k=\ell =i\neq j$ in (4.29), it is easy to see that:

(4.30)\begin{equation} (\nu_i+\nu_j)[g(K(e_i,e_i),e_j)e_i-g(K(e_i,e_i),e_i)e_j+2K(e_i,e_j)]=0. \end{equation}

Taking the inner product of (4.30) with $e_i$, we then deduce from lemma 2.2 that:

(4.31)\begin{equation} (\nu_i+\nu_j)g(K(e_i,e_i),e_j)=0,\quad \forall\, i\neq j. \end{equation}

Similarly, interchanging the roles of $e_i$ and $e_j$ in (4.31) gives

(4.32)\begin{equation} (\nu_j+\nu_i)g(K(e_j,e_j),e_i)=0,\quad \forall\,i\neq j. \end{equation}

Furthermore, by taking the inner product of (4.30) with $e_j$, we obtain that

(4.33)\begin{equation} (\nu_i+\nu_j)g(K(e_i,e_i),e_i)=0,\quad \forall\,i\neq j, \end{equation}

which together with (4.30) and (4.31) implies that

(4.34)\begin{equation} (\nu_i+\nu_j)K(e_i,e_j)=0,\quad \forall\,i\neq j. \end{equation}

On the contrary, for $k=i\neq j=\ell$ in (4.29), we easily get:

(4.35)\begin{equation} \begin{aligned} 0 & =(\nu_i+\nu_j)[g(K(e_i,e_j),e_j)e_i-g(K(e_i,e_j),e_i)e_j]\\ & \quad+(\nu_i+\nu_j)[K(e_j,e_j)-K(e_i,e_i)]. \end{aligned} \end{equation}

This combining with (4.31) and (4.32) yields that

(4.36)\begin{equation} (\nu_i+\nu_j)[K(e_i,e_i)-K(e_j,e_j)]=0,\quad \forall\,i\neq j. \end{equation}

Next, we shall continue with the proof here by proving the following three claims.

Claim 1. If $\nu _i\neq 0$ and $n_i\geq 2$, then $g(K(u,\,v),\,\omega )=0$ for any $u,\,v,\,\omega \in \mathfrak {D}(\nu _i)$.

If $\nu _i\neq 0$ and $n_i\geq 2$, taking $\nu _i=\nu _j$ and $e_i=u$ in (4.33), then we easily conclude that $g(K(u,\,u),\,u)=0$ for any unit vector field $u\in \mathfrak {D} (\nu _i)$. Consequently, the assertion follows from the symmetry given in lemma 2.2.

Claim 2. If $\nu ^2_i\neq \nu ^2_j$, then $K(u,\,v)=0$ for any $u,\,v\in \mathfrak {D} (\nu _i)\oplus \mathfrak {D}(\nu _j)$.

If $\nu ^2_i\neq \nu ^2_j$, it is obvious that $(\nu _i+\nu _j)(\nu _i-\nu _j)\neq 0$ and we then obtain from (4.33) that $g(K(u,\,u),\,u)=0$ for any unit vector field $u\in \mathfrak {D}(\nu _i)$. From the result of the symmetry in lemma 2.2, we have $K(u,\,u)\notin \mathfrak {D}(\nu _i)$. Similarly, $g(K(v,\,v),\,v)=0$ for any unit vector field $v\in \mathfrak {D}(\nu _j)$ and thus $K(v,\,v)\notin \mathfrak {D}(\nu _j)$. Moreover, with the help of (4.34) and (4.36), it can be checked that $K(u,\,v)=0$ and $K(u,\,u)=K(v,\,v)\notin \mathfrak {D}(\nu _i)\oplus \mathfrak {D}(\nu _j)$. When $r=2$, claim 2 holds immediately. When $r\geq 3$, for an arbitrary eigenvalue $\nu _k$ different from $\nu _i$ and $\nu _j$, it is known from $\nu ^2_i\neq \nu ^2_j$ that either $\nu _i+\nu _k\neq 0$ or $\nu _j+\nu _k\neq 0$. In either case, we can apply (4.31) to obtain $K(u,\,u)=K(v,\,v)=0$. Hence, claim 2 has been proved.

Claim 3. If $r\geq 2$, there exist two distinct eigenvalues $\nu _i$ and $\nu _j$ such that ${\nu _i+\nu _j=0}$.

If $r\geq 2$, we suppose on the contrary that $\nu _i+\nu _j\neq 0$ holds for any $\nu _i \neq \nu _j$, namely $\nu ^2_i\neq \nu ^2_j$. It then follows from claim 2 that $K=0$ on $M^n$ and hence $h=0$ by definition. This together with the Gauss equation shows that $M^n$ has constant sectional curvature, which is a contradiction to $r\geq 2$, and thus we have verified claim 3.

Now, according to claim 3, we denote by $\nu _1$ and $\nu _2$ the two distinct eigenvalues of $P$ such that $\nu _1+\nu _2=0$. In this situation, we further claim that $r\leq 2$. Otherwise, if $r\geq 3$, then $(\nu _i+\nu _1)(\nu _i+\nu _2)\neq 0$ for an arbitrary eigenvalue $\nu _i$ different from $\nu _1$ and $\nu _2$. Therefore, it satisfies that $\nu ^2_i\neq \nu ^2_1$ and $\nu ^2_i\neq \nu ^2_2$, and so we obtain from claim 2 that:

(4.37)\begin{equation} \begin{aligned} & K(u,u)=K(v,v)=K(\omega,\omega)=0,\quad \omega\in\mathfrak{D}(\nu_i),\\ & K(u,\omega)=K(v,\omega)=0,\ \ u\in\mathfrak{D}(\nu_1),\ \ v\in\mathfrak{D}(\nu_2). \end{aligned} \end{equation}

Furthermore, we see that $K(u,\,v)=0$ in terms of the arbitrariness of $\nu _i$, meaning that $h=0$ identically. This contradiction implies that the number $r\le 2$.

Assume that $r=1$. It is obvious that $M^n$ has constant sectional curvature for $n\ge 3$.

Assume that $r=2$. We will denote these two distinct eigenvalues of $P$ by $\nu _1$ and $\nu _2$, whose multiplicities are $n_1$ and $n_2$, respectively. Together with claim 3, it follows that $\nu _2=-\nu _1\neq 0$. In what follows, we shall argue by contradiction and suppose that $n_1\geq 2$ and $n_2\geq 2$. Let $X^i_1,\,\ldots,\,X^i_{n_i}$ be the orthonormal eigenvector fields of $P$ that span the distribution $\mathfrak {D}(\nu _i)$, $i=1,\,2$. Thus, we see from claim 1 and (4.36) that

(4.38)\begin{equation} g(K(X^i_p,X^i_q),X^i_s)=0,\ \ K(X^i_p,X^i_p)=K(X^i_q,X^i_q),\ \ \forall\, p,q,s,\ \ i=1,2. \end{equation}

Since $M^n$ is minimal in $\mathbb {S}^{2n+1}$, by means of (4.38) we have

(4.39)\begin{equation} \begin{aligned} 0 & =\sum^{n_1}_{p=1}g(K(X^1_p,X^1_p),v)+\sum^{n_2}_{q=1}g(K(X^2_q,X^2_q),v)\\ & =\sum^{n_1}_{p=1}g(K(X^1_p,X^1_p),v)=n_1g(K(X^1_1,X^1_1),v),\ \ v\in\mathfrak{D}(\nu_2), \end{aligned} \end{equation}

which implies that $g(K(u,\,u),\,v)=0$ for any $u\in \mathfrak {D}(\nu _1)$ and $v\in \mathfrak {D} (\nu _2)$. Similarly, there holds that $g(K(v,\,v),\,u)=0$ and therefore $K(u,\,v)=0$. According to this and (4.38), we get $h=0$ identically, a contradiction to $r=2$. Hence, lemma 4.7 has been proved.

4.4. Conformally flat minimal Legendrian submanifolds with $R\cdot Q=0$

In this subsection, assuming that $M^n$ $(n\geq 3)$ is a conformally flat minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$ such that the Ricci tensor Ric is semi-parallel, by definition we have

(4.40)\begin{equation} \begin{aligned} 0 & =(R\cdot{\rm Ric})(X,Y,U,Z)=(R(X,Y){\rm Ric})(U,Z)\\ & ={-}{\rm Ric}(R(X,Y)U,Z)-{\rm Ric}(U,R(X,Y)Z)\\ & =g(R(X,Y)QZ,U)-g(QR(X,Y)Z,U)\\ & =g((R(X,Y)Q)Z,U)=g((R\cdot Q)(X,Y,Z),U), \end{aligned} \end{equation}

where $U,\,X,\,Y,\,Z$ are tangent vector fields on $M^n$. This implies that Ric is semi-parallel if and only if the Ricci operator $Q$ is semi-parallel, i.e. $R\cdot Q=0$.

Proof. Under the orthonormal frame $\{e_i\}^n_{i=1}$ on $M^n$ as in § 4.2, direct calculations by using (2.29) yield that

(4.41)\begin{equation} \begin{aligned} 0 & =g((R(e_i,e_j)Q)e_j,e_i)\\ & =g(R(e_i,e_j)Qe_j,e_i)-g(QR(e_i,e_j)e_j,e_i)\\ & =(n-2)(\nu_j+\nu_i)(\nu_j-\nu_i),\quad 1\leq i\neq j\leq n, \end{aligned} \end{equation}

where we used the relation $\mu _j-\mu _i=(n-2)(\nu _j-\nu _i)$. Consequently, the number $r$ of distinct eigenvalues of the Schouten tensor $P$ is at most $2$, and if $r=1$, then we conclude that $M^n$ has constant sectional curvature.

If $r=2$, then we obtain from (4.41) that $\nu _1+\nu _2=0$ by expressing the two distinct eigenvalues of $P$ by $\nu _1$ and $\nu _2$, respectively. Now, it suffices to prove that either of $\nu _1$ and $\nu _2$ must be simple. For this purpose, we shall suppose on the contrary that $n_i\geq 2$, where $n_i$ is the multiplicity of $\nu _i$ for $i=1,\,2$. Let $\{X_i\}^{n_1}_{i=1}$ (resp. $\{Y_j\}^{n_2}_{j=1}$) be the orthonormal frame of $\mathfrak {D}(\nu _1)$ (resp. $\mathfrak {D}(\nu _2)$). It then follows from lemma 4.6 (1) that

(4.42)\begin{equation} K(X_i,X_i)\in\mathfrak{D}(\nu_1),\quad K(Y_j,Y_j)\in\mathfrak{D}(\nu_2),\quad K(X_i,Y_j)=0,\ \forall\,i,j. \end{equation}

Applying lemma 2.2 (1), we deduce from the Gauss equation (2.7) that

(4.43)\begin{equation} g(R(X_i,Y_j)Y_j,X_i)=1. \end{equation}

Since $M^n$ is conformally flat, it is easy to see from (2.29) that

(4.44)\begin{equation} g(R(X_i,Y_j)Y_j,X_i)=\nu_1+\nu_2,\end{equation}

which together with (4.43) yields that

(4.45)\begin{equation} \nu_1+\nu_2=1. \end{equation}

This contradicts with the fact $\nu _1+\nu _2=0$, and thus lemma 4.9 has been proved.

4.5. Conformally flat Legendrian submanifolds with quasi-Einstein metric

In the following, according to lemmas 4.7 and 4.9, we shall deal with the case when $M^n$ $(n\ge 3)$ is a conformally flat Legendrian submanifold in $\mathbb {S}^{2n+1}$, such that it is quasi-Einstein with $\nu _1$ and $\nu _2$ the distinct eigenvalues of its Schouten tensor, and $\mathfrak {D}(\nu _1)$ and $\mathfrak {D}(\nu _2)$ the corresponding distributions of eigenspace, where $\nu _1+\nu _2=0$ and $\nu _1$ is simple. Then, for a unit vector field $E_1$ of $\mathfrak {D}(\nu _1)$ and an orthonormal frame $\{E_i\}^n_{i=2}$ of $\mathfrak {D}(\nu _2)$, we see from (2.27) and (2.29) that

(4.46)\begin{equation} \begin{aligned} & R(E_1,E_i)E_1=R(E_1,E_i)E_j=R(E_i,E_j)E_1=0,\\ & R(E_i,E_j)E_k=2\nu_2(\delta_{jk}E_i-\delta_{ik}E_j),\quad 2\leq i,j,k\leq n,\\ & QE_1=0,\ QE_i=n\chi E_{i},\quad \nu_2={-}\nu_1=\frac{n}{2(n-2)}\chi\neq0,\\ \end{aligned} \end{equation}

where we used the facts tr $Q=n(n-1)\chi$ and $\nu _i={\mu _i}/{(n-2)}-{n\chi }/{(2(n-2))}$ for $i=1,\,2$.

Lemma 4.10 Let $M^n$ $(n\geq 3)$ be a conformally flat Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$. Assume that $M^n$ is quasi-Einstein with $\nu _1$ and $\nu _2$ the distinct eigenvalues of its Schouten operator such that $\nu _1+\nu _2=0$, where $\nu _1$ is simple. Then, it holds that

(4.47)\begin{equation} \begin{aligned} & \nabla_{E_1}E_1=0,\quad\nabla_{E_i}E_1={-}\alpha E_i,\ \ \alpha=\frac{E_1(\nu_2)}{\nu_2-\nu_1},\\ & E_1(\alpha)=\alpha^2,\ \ E_i(\alpha)=E_i(\nu_1)=E_i(\nu_2)=0,\ \ 2\leq i\leq n. \end{aligned} \end{equation}

Proof. For any tangent vector $V$ on $M^n$, we denote by $V^\ell$ the projection of $V$ onto $\mathfrak {D}(\nu _\ell )$ for $\ell =1,\,2$, respectively. A straightforward computation shows that

(4.48)\begin{equation} \begin{aligned} & (\nabla_{E_1}P)E_i=E_1(\nu_2)E_i+(\nu_2-\nu_1)(\nabla_{E_1}E_i)^1,\\ & (\nabla_{E_i}P)E_1=E_i(\nu_1)E_1+(\nu_1-\nu_2)(\nabla_{E_i}E_1)^2, \end{aligned} \end{equation}

where $2\leq i\leq n$. As $M^n$ is conformally flat, we have $(\nabla _{E_1}P)E_i =(\nabla _{E_i}P)E_1$ and then it follows that:

(4.49)\begin{equation} E_1(\nu_2)E_i-E_i(\nu_1)E_1-(\nu_1-\nu_2) [(\nabla_{E_1}E_i)^1+(\nabla_{E_i}E_1)^2]=0. \end{equation}

Multiplying the above equation with $E_1$ and further with $E_k$ for $k\geq 2$, we easily get:

(4.50)\begin{align} & E_i(\nu_1)=(\nu_1-\nu_2)g(\nabla_{E_1}E_1,E_i), \end{align}

(4.51)\begin{align} & E_1(\nu_2)g(E_i,E_k)=(\nu_1-\nu_2)g(\nabla_{E_i}E_1,E_k), \end{align}

where $2\leq i,\,k\leq n$. Similarly, we deduce from $(\nabla _{E_i}P)E_j=(\nabla _{E_j} P)E_i$ that

(4.52)\begin{equation} E_i(\nu_2)E_j-E_j(\nu_2)E_i+(\nu_1-\nu_2) [(\nabla_{E_j}E_i)^1-(\nabla_{E_i}E_j)^1]=0, \end{equation}

where $2\leq i\neq j\leq n$, and thus there holds

(4.53)\begin{equation} E_i(\nu_2)E_j-E_j(\nu_2)E_i+(\nu_1-\nu_2) [(\nabla_{E_j}E_i)^1-(\nabla_{E_i}E_j)^1]=0, \end{equation}

which together with the fact $\nu _1\neq \nu _2$ immediately yields that

(4.54)\begin{equation} g(\nabla_{E_i}E_j-\nabla_{E_j}E_i,E_1)=0,\quad E_i(\nu_2)=0,\quad 2\leq i\neq j\leq n. \end{equation}

Next, with the help of $\nu _1\neq \nu _2$, it is easy to see from (4.50) and (4.54) that

(4.55)\begin{equation} \nabla_{E_1}E_1=0,\ \ (\nabla_{E_i}E_j-\nabla_{E_j}E_i)\in\mathfrak{D}(\nu_2),\ \ E_i(\nu_1)=E_i(\nu_2)=0,\ \ 2\leq i\leq n. \end{equation}

On the contrary, by applying (4.51) we further have

(4.56)\begin{equation} \nabla_{E_i}E_1={-}\alpha E_i,\ \quad \alpha=\frac{E_1(\nu_2)}{\nu_2-\nu_1}. \end{equation}

Therefore, it can be checked from (4.46), (4.55) and (4.56) that

(4.57)\begin{equation} \begin{aligned} & 0=R(E_i,E_1)E_1=E_1(\alpha)E_i-\alpha^2E_i,\\ & 0=R(E_i,E_j)E_1=E_j(\alpha)E_i-E_i(\alpha)E_j, \end{aligned} \end{equation}

where we made use of the definition of the curvature tensor $R$ for $2\leq i\neq j \leq n$. Hence, $E_1(\alpha )=\alpha ^2$ and $E_i(\alpha )=0$ for $2\le i\le n$. This completes the proof of lemma 4.10.

5.1. Completion of the proof of theorem 1.3

$M^n$ $(n\ge 3)$ be a conformally flat minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$ with semi-parallel tensor $K$. By means of lemma 4.7, we see that either $M^n$ is of constant sectional curvature, or it is quasi-Einstein with $\nu _1$ and $\nu _2$ the distinct eigenvalues of its Schouten operator such that $\nu _1+\nu _2=0$, where $\nu _1$ is simple. In the former case, theorem 1.1 states that $M^n$ is the totally geodesic sphere or the flat Clifford torus. In the latter case, according to proposition 5.1 and theorem 5.2, we conclude that $M^n$ is locally congruent to the Calabi torus. Conversely, these calculations in § 3 guarantee that examples (a)–(c) are all conformally flat minimal Legendrian submanifolds in $\mathbb {S}^{2n+1}$ with semi-parallel tensor $K$.

5.2. Completion of the proof of corollary 1.5

Let $M^n$ $(n\ge 3)$ be a minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$ with semi-parallel tensor $K$. By means of lemma 4.1 and proposition 4.3, we calculate that

(5.35)\begin{equation} \|\bar\nabla^\xi h\|^2=\frac{1}{2}\Delta S=\|\bar\nabla^\xi h\|^2-\|W\|^2 -\frac{n+2}{n-2}\|\tilde{\rm Ric}\|^2+(n+1)S\chi.\end{equation}

Consequently, we have

(5.36)\begin{equation} (n+1)S\chi =\frac{n+2}{n-2}\|\tilde{\rm Ric}\|^2+\|W\|^2\ge\frac{n+2}{n-2}\|\tilde{\rm Ric}\|^2, \end{equation}

by which we obtain (1.3) and find that the equality holds on $M^n$ if and only if the Weyl curvature tensor of $M^n$ vanishes identically. When $n\ge 4$, the assertion immediately follows from theorem 1.3 and proposition 3.1. When $n=3$, according to lemma 4.7 and remark 4.8, either $M^3$ has constant sectional curvature, or $M^3$ is quasi-Einstein with $\nu _1$ and $\nu _2$ the distinct eigenvalues of its Schouten operator so that $\nu _1+\nu _2=0$, where $\nu _1$ is simple. In the former case, we obtain from theorem 1.1 that $M^3$ is the totally geodesic sphere or the flat Clifford torus. In the latter case, by remark 4.8 and claim 1, we deduce from (4.34) and (4.36) that

(5.37)\begin{equation} \begin{aligned} & g(K(E_i,E_j),E_k)=0,\ \ 2\leq i,j,k\leq 3,\\ & K(E_i,E_i)=K(E_j,E_j),\ \ K(E_i,E_j)=0,\ \ 2\leq i\neq j\leq 3, \end{aligned} \end{equation}

where $E_1$ is a unit vector field of $\mathfrak {D}(\nu _1)$ and $\{E_2,\,E_3\}$ is an arbitrary orthonormal frame of $\mathfrak {D}(\nu _2)$. As $M^3$ is minimal in $\mathbb {S}^7$, we further have

(5.38)\begin{equation} \begin{aligned} & g(K(E_1,E_1),E_1)={-}2g(K(E_i,E_i),E_1),\ \ 2\le i\le 3,\\ & g(K(E_1,E_1),E_i)={-}\sum^3_{j=2}g(K(E_j,E_j),E_i)=0,\ \ 2\le i\le 3. \end{aligned} \end{equation}

Therefore, similar calculations as in the proof of proposition 5.1 show that

(5.39)\begin{equation} \begin{aligned} & h(E_1,E_1)=\frac{2}{\sqrt{3}}\varphi E_1,\ \ h(E_1,E_i)={-}\frac{1}{\sqrt{3}}\varphi E_i,\\ & h(E_i,E_j)={-}\frac{1}{\sqrt{3}}\delta_{ij}\varphi E_1,\ \ 2\le i,j\le 3. \end{aligned} \end{equation}

Finally, combining with (3.4) and (5.39), we derive from theorem 2.1 that $M^n$ is locally congruent to the Calabi torus. Conversely, by § 3 these examples (a)–(c) for $n=3$ are all minimal Legendrian submanifolds in $\mathbb {S}^7$ with semi-parallel tensor $K$.

5.3. Completion of the proof of corollary 1.6

Let $M^n$ $(n\ge 3)$ be a closed minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$ with vanishing Weyl curvature tensor. By calculation we apply (4.18) to obtain that

(5.40)\begin{equation} \frac{1}{2}\Delta S=\|\bar\nabla^\xi h\|^2-\frac{n+2}{n-2}\|\tilde{\rm Ric}\|^2 +(n+1)S\chi\geq{-}\frac{n+2}{n-2}\|\tilde{\rm Ric}\|^2+(n+1)S\chi. \end{equation}

Now, by using the compactness of $M^n$, we can integrate inequality (5.40) to obtain the integral inequality in (1.4), according to the divergence theorem, where the equality holds on $M^n$ if and only if $M^n$ is of $C$-parallel second fundamental form, i.e. $\bar \nabla ^\xi h=0$, which implies that $R\cdot K=0$. When $n\ge 4$, the assertion follows from theorem 1.3 and proposition 3.1 immediately. When $n=3$, Theorem 1.3 and (1.5) of Xing–Zhai [Reference Xing and Zhai33] state that $M^3$ is locally congruent to one of the examples (a)–(c) for $n=3$. Conversely, by § 3 the examples (a)–(c) for $n=3$ are closed minimal Legendrian submanifolds in $\mathbb {S}^7$ with $C$-parallel second fundamental form.

5.4. Completion of the proof of theorem 1.7

Let $M^n$ $(n\ge 3)$ be a conformally flat minimal Legendrian submanifold in the unit sphere $\mathbb {S}^{2n+1}$ with semi-parallel Ricci tensor. Together with (4.40), it then follows from lemma 4.9 that either $M^n$ has constant sectional curvature, or $M^n$ is quasi-Einstein with $\nu _1$ and $\nu _2$ the distinct eigenvalues of its Schouten operator such that $\nu _1+\nu _2=0$, where $\nu _1$ is simple. In the former case, either $M^n$ is the totally geodesic sphere, or $M^n$ is the flat Clifford torus, in terms of theorem 1.1. In the latter case, applying proposition 5.1 and theorem 5.2, we derive that $M^n$ is locally congruent to the Calabi torus. Conversely, it is easy to see from proposition 3.1 that examples (a)–(c) are all conformally flat minimal Legendrian submanifolds in $\mathbb {S}^{2n+1}$ with semi-parallel Ricci tensor.