Simons-Type Equation for $f$-Minimal Hypersurfaces and Applications

Abstract

We derive the Simons-type equation for $f$-minimal hypersurfaces in weighted Riemannian manifolds and apply it to obtain a pinching theorem for closed $f$-minimal hypersurfaces immersed in the product manifold $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$ with $f=\frac{t^2}{4}$. Also, we classify closed $f$-minimal hypersurfaces with $L_f$-index one immersed in $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$ with the same $f$ as above.

1 Introduction

In [18], Simons proved an identity, called Simons’s equation, for the Laplacian of $|A|^2$, the square of the norm of the second fundamental form of minimal hypersurfaces in Riemannian manifolds. Simons’s equation plays an important role in the study of minimal hypersurfaces. For self-shrinkers for the mean curvature flow in the Euclidean space $\mathbb {R}^{n+1}$, the Simons-type equation also holds ([7, 10] Sect. 10.2). By applying it, Huisken [10] proved that an embedded closed self-shrinker of nonnegative mean curvature must be a sphere of radius $\sqrt{2n}$, and recently Colding and Minicozzi [7] classified complete embedded self-shrinkers of nonnegative mean curvature with polynomial volume growth. Later, Le and Sesum [12] and Cao and Li [2] used it to obtain gap theorems for self-shrinkers.

Both minimal hypersurfaces and self-shrinkers are special cases of $f$-minimal hypersurfaces in the weighted Riemannian manifolds. See the definition of $f$-minimal hypersurfaces in Sect. 2 and more examples in [4]. An $f$-minimal hypersurface $\varSigma$ is not only a critical point of the weighted volume functional $\int _\varSigma e^{-f}d\sigma$ of $\varSigma$, where $d\sigma$ denotes the volume element of $(\varSigma , g)$, but also a minimal hypersurface in $(M, \tilde{g})$, where the new metric $\tilde{g}=e^{-\frac{2}{n}f}\overline{g}$ of $M$ is conformal to $\overline{g}$. Recently, Lott [15] and Magni et al. [16] showed that $f$-minimal hypersurfaces arise in the study of the mean curvature flow of a hypersurface in an ambient manifold evolving by Ricci flow. Especially the mean curvature soliton (for the mean curvature flow of a hypersurface in a gradient Ricci soliton solution) introduced by Lott [15] is just an $f$-minimal hypersurface, where $f$ is the potential function of the ambient gradient Ricci soliton.

Recently, Liu [14] studied stable $f$-minimal hypersurfaces in manifolds with nonnegative Bakry–Émery Ricci curvature and gave a partial classification of the ambient space when the dimension is $3$ and $f$ is bounded. The present authors [4] studied the stability condition and compactness of $f$-minimal surfaces. Li and Wei [13] gave eigenvalue estimates for closed $f$-minimal hypersurfaces in a compact manifold with positive $m$-Bakry–Émery curvature. We [3] also obtained similar eigenvalue estimates for some closed $f$-minimal hypersurfaces in a complete manifold with positive Bakry–Émery curvature. These estimates have been used to prove compactness theorems for closed $f$-minimal surfaces.

In this paper, we will prove a Simons-type equation for $f$-minimal hypersurfaces in a smooth metric measure space $(M, \overline{g}, e^{-f}d\mu )$, that is, an identity for the weighted Laplacian $\Delta _f$ of $|A|^2$ of $f$-minimal hypersurfaces, involving the Bakry–Émery Ricci curvature $\overline{\mathrm{Ric}}_f$ (see Theorem 3). Also, we derive the equations for the weighted Laplacian $\Delta _f$ of some other geometric quantities on $f$-minimal hypersurfaces, like the mean curvature $H$, etc.; see, for instance, Propositions 1 and 2. Since these equations involve $\overline{\mathrm{Ric}}_f$, we naturally would like to consider the cases in which the ambient manifolds are gradient Ricci solitons, that is, $M$ satisfies $\overline{\mathrm{Ric}}_f=C\overline{g}$; see Corollaries 2, 3 and 4. Further, we apply the equations mentioned above to the special case of $f$-minimal hypersurfaces in the cylinder shrinking soliton $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$ with $f=\frac{t^2}{4}$, where $t$ is the coordinate of the second factor $\mathbb {R}$. Namely, we obtain the following pinching theorem.

Theorem 1

Let $\varSigma ^{n}$ be a closed immersed $f$-minimal hypersurface in the product manifold $\mathbb {S}^{n}\bigr (\sqrt{2(n-1)}\bigr )\times \mathbb {R}$, $n\ge 3$, with $f=\frac{t^2}{4}$, $t\in \mathbb {R}$. If the square of the norm of the second fundamental form of $\varSigma$ satisfies

$$\begin{aligned} \frac{1}{4}\biggr (1-\sqrt{1-\frac{8}{n-1}\alpha ^2(1-\alpha ^2)}\biggr )\le |A|^2\le \frac{1}{4}\biggr (1+\sqrt{1-\frac{8}{n-1}\alpha ^2(1-\alpha ^2)}\biggr ), \end{aligned}$$

then $\varSigma$ is $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$, where $\alpha =\langle \nu ,\frac{\partial }{\partial t}\rangle$, $\nu$ is the outward unit normal to $\varSigma$ and $t$ denotes the coordinate of the factor $\mathbb {R}$ of $\mathbb {S}^{n}\bigr (\sqrt{2(n-1)}\bigr )\times \mathbb {R}$.

Observe that $n\ge 3$ implies that $\frac{8}{n-1}\alpha ^2(1-\alpha ^2)\le 1$ and hence the inequalities in Theorem 1 make sense. Theorem 1 implies that

Corollary 1

There is no closed immersed $f$-minimal hypersurface in the product manifold $\mathbb {S}^{n}\bigr (\sqrt{2(n-1)}\bigr )\times \mathbb {R}$, $n\ge 3$, with $f=\frac{t^2}{4}$, $t\in \mathbb {R}$ so that the square of its norm of the second fundamental form satisfies

$$\begin{aligned} \frac{1}{4}\biggr (1-\sqrt{1-\frac{2}{n-1}}\biggr )\le |A|^2\le \frac{1}{4}\biggr (1+\sqrt{1-\frac{2}{n-1}}\biggr ). \end{aligned}$$

Next, we discuss, as another application, the eigenvalues and the index of the operator $L_f$ on $f$-minimal hypersurfaces. The eigenvalues of the $L$-operator for self-shrinkers were discussed in [7] and recently, Hussey [11] studied the index of the $L$-operator for self-shrinkers in $\mathbb {R}^{n+1}$. Observe that the $L$-operator is just the $L_f$ operator for self-shrinkers (see Example 1). In this paper, we classify closed $f$-minimal hypersurfaces in the cylinder shrinking soliton $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$ whose $L_f$ operators have index one and prove that

Theorem 2

Let $\varSigma ^{n}$ be a closed immersed $f$-minimal hypersurface in the product manifold $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$ with $f=\frac{t^2}{4}$. Then $L_f$-$\hbox {ind}(\varSigma )\ge 1$. Moreover, the equality holds if and only if $\varSigma =\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$.

For complete noncompact $f$-minimal hypersurfaces in the cylinder shrinking soliton $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$ with $f=\frac{t^2}{4}$, the first and third authors [6] proved the results corresponding to Theorems 1 and 2. In this case complete noncompact $f$-minimal hypersurfaces are assumed to have polynomial volume growth, which is equivalent to properness of immersion, or finiteness of weighted volume (see [5] and [4]).

The rest of this paper is organized as follows: In Sect. 2 some definitions and notation are given. In Sect. 3 we prove the Simons-type equation and the equations for $\Delta _f$ of other geometric quantities. In Sect. 4, we calculate the index of the $L_f$ operator on closed $f$-minimal hypersurfaces in the cylinder shrinking soliton $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$. In Sect. 5, we prove Theorem 1 for closed $f$-minimal hypersurfaces in the cylinder shrinking soliton $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$.

2 Definitions and Notation

Let $(M^{n+1}, \overline{g}, e^{-f}d\mu )$ be a smooth metric measure space, which is an $(n+1)$-dimensional Riemannian manifold $(M^{n+1}, \overline{g})$ together with a weighted volume form $e^{-f}d\mu$ on $M$, where $f$ is a smooth function on $M$ and $d\mu$ the volume element induced by the metric $\overline{g}$. In this paper, unless otherwise specified, we denote by a bar all quantities on $(M, \overline{g})$, for instance by $\overline{\nabla }$ and $\overline{\mathrm{Ric}}$, the Levi-Civita connection and the Ricci curvature tensor of $(M, \overline{g})$ respectively. For $(M, \overline{g}, e^{-f}d\mu )$, the $\infty$-Bakry–Émery Ricci curvature tensor $\overline{\mathrm{Ric}}_{f}$ (for simplicity, Bakry–Émery Ricci curvature), which is defined by

$$\begin{aligned} \overline{\mathrm{Ric}}_{f }:=\overline{\mathrm{Ric}}+\overline{\nabla }^{2}f, \end{aligned}$$

where $\overline{\nabla }^{2}f$ is the Hessian of $f$ on $M$. If $f$ is constant, $\overline{\mathrm{Ric}}_{f}$ is the Ricci curvature $\overline{\mathrm{Ric}}$.

Now let $i: \varSigma ^{n}\rightarrow M^{n+1}$ be an $n$-dimensional smooth immersion. Then $i$ induces a metric $g=i^*\overline{g}$ on $\varSigma$ so that $i: (\varSigma ^{n}, g) \rightarrow (M, \overline{g})$ is an isometric immersion. We will denote for instance by ${\nabla }$, $\mathrm{Ric}$, $\Delta$ and $d\sigma$, the Levi-Civita connection, the Ricci curvature tensor, the Laplacian, and the volume element of $(\varSigma , g)$ respectively.

The restriction of $f$ on $\varSigma$, still denoted by $f$, yields a weighted measure $e^{-f}d\sigma$ on $\varSigma$ and hence an induced smooth metric measure space $(\varSigma ^{n}, g, e^{-f}d\sigma )$. The associated weighted Laplacian ${\Delta }_{f}$ on $\varSigma$ is defined by

$$\begin{aligned} {\Delta }_{f}u:={\Delta } u-\langle {\nabla } f,{\nabla } u\rangle . \end{aligned}$$

The second-order operator ${\Delta }_f$ is a self-adjoint operator on $L^2(e^{-f}d\sigma )$, the space of square integrable functions on $\varSigma$ with respect to the measure $e^{-f}d\sigma$.

We define the second fundamental form $A$ of $\varSigma$ by

$$\begin{aligned} A: T_p\varSigma \times T_p\varSigma \rightarrow \mathbb {R}, \quad A(X,Y)=-\langle \overline{\nabla }_XY, \nu \rangle , \end{aligned}$$

where $p\in \varSigma , X, Y\in T_p \varSigma$, $\nu$ is a unit normal vector at $p$.

In a local orthonormal system $\{e_i\}, i=1,\ldots , n$ of $\varSigma$, the components of $A$ are denoted by $a_{ij}=A(e_i,e_j)=\langle \overline{\nabla }_{e_i}\nu , e_j\rangle$. The shape operator $A$ and the mean curvature $H$ of $\varSigma$ are defined by

$$\begin{aligned} A: T_p\varSigma \rightarrow T_p\varSigma , \quad AX=\overline{\nabla }_X\nu , X\in T_p\varSigma ; H=\hbox {tr}A=\displaystyle \sum _{i=1}^na_{ii}. \end{aligned}$$

With the above notation, we have the following

Definition 1

The weighted mean curvature $H_f$ of the hypersurface $\varSigma$ is defined by

$$\begin{aligned} H_f=H-\langle \overline{\nabla } f,\nu \rangle . \end{aligned}$$

$\varSigma$ is called an $f$-minimal hypersurface if it satisfies

$$\begin{aligned} H=\langle \overline{\nabla } f,\nu \rangle . \end{aligned}$$

(1)

Definition 2

The weighted volume of $\varSigma$ is defined by

$$\begin{aligned} V_f(\varSigma ):=\int _\varSigma e^{-f}d\sigma . \end{aligned}$$

(2)

It is known that $\varSigma$ is $f$-minimal if and only if it is a critical point of the weighted volume functional. On the other hand, we can view it in another manner: $\varSigma$ being $f$-minimal in $(M,\overline{g})$ is equivalent to $(\varSigma , i^*\tilde{g})$ being minimal in $(M,\tilde{g})$, where the conformal metric $\tilde{g}=e^{-\frac{2f}{n}}\overline{g}$ (cf. [4]).

Now we assume that $\varSigma$ is a two-sided hypersurface, that is, there is a globally defined unit normal $\nu$ on $\varSigma$.

Definition 3

For a two-sided hypersurface $\varSigma$, the $L_f$ operator on $\varSigma$ is given by

$$\begin{aligned} L_f:=\Delta _f+|A|^2+\overline{\mathrm{Ric}}_f(\nu ,\nu ), \end{aligned}$$

where $|A|^2$ denotes the square of the norm of the second fundamental form $A$ of $\varSigma$.

The operator $L_{f}=\Delta _{f}+|A|^2+\overline{\mathrm{Ric}}_{f}(\nu ,\nu )$ is called the $L_{f}$-stability operator of $\varSigma$.

Example 1

For self-shrinkers in $\mathbb {R}^{n+1}$, the operator $L_f$, where $f=\frac{|x|^2}{4}$, is just the $L$ operator in [7]:

$$\begin{aligned} L=\Delta -\frac{1}{2}\langle x,\nabla \cdot \rangle +|A|^{2}+\frac{1}{2}. \end{aligned}$$

Definition 4

A two-sided $f$-minimal hypersurface $\varSigma$ is said to be $L_{f}$-stable if for any compactly supported smooth function $\varphi \in C_o^{\infty }(\varSigma )$, it holds that

$$\begin{aligned} -\int _{\varSigma }\varphi L_f\varphi e^{-f}d\sigma \ge 0, \end{aligned}$$

(3)

or equivalently,

$$\begin{aligned} \int _{\varSigma }\big [|\nabla \varphi |^2-(|A|^{2}+\overline{\mathrm{Ric}}_ {f}(\nu ,\nu ))\varphi ^2\big ]e^{-f}d\sigma \ge 0. \end{aligned}$$

(4)

$L_f$-stability means that the second variation of the weighted volume of $f$-minimal hypersurface $\varSigma$ is nonnegative. Further, one has the definition of the $L_f$-index of $f$-minimal hypersurfaces. Since $\Delta _f$ is self-adjoint in the weighted space $L^2(e^{-f}d\sigma )$, we may define a symmetric bilinear form on the space $C_o^{\infty }(\varSigma )$ of compactly supported smooth functions on $\varSigma$ by, for $\varphi , \psi \in C_o^{\infty }(\varSigma )$,

$$\begin{aligned} B_f(\varphi , \psi ){:}&= -\int _\varSigma \varphi L_f\psi e^{-f}d\sigma \nonumber \\&= \int _{\varSigma }[\langle \nabla \varphi ,\nabla \psi \rangle -(|A|^{2}+\overline{\mathrm{Ric}}_{f}(\nu ,\nu ))\varphi \psi ]e^{-f}d\sigma . \end{aligned}$$

(5)

Definition 5

The $L_{f}$-index of $\varSigma$, denoted by $L_{f}$-$\hbox {ind}(\varSigma )$, is defined to be the maximum of the dimensions of negative definite subspaces for $B_f$.

In particular, $\varSigma$ is $L_f$-stable if and only if $L_{f}-\hbox {ind}(\varSigma )=0$. The $L_f$-index of $\varSigma$ has the following equivalent definition: Consider the Dirichlet eigenvalue problems of $L_f$ on a compact domain $\Omega \subset \varSigma$:

$$\begin{aligned} L_fu=-\lambda u, u\in \Omega ; \quad u|_{\partial \Omega }=0. \end{aligned}$$

$L_{f}$-$\hbox {ind}(\varSigma )$ is defined to be the supremum over compact domains of $\varSigma$ of the number of negative (Dirichlet) eigenvalues of $L_f$ (cf. [9]).

It is known that an $f$-minimal hypersurface $(\varSigma ,{g})$ is $L_f$-stable if and only if $(\varSigma , i^*\tilde{g})$ is stable as a minimal surface in $(M,\tilde{g})$. Further, the Morse index of $L_f$ on $(\varSigma ,g)$ is equal to the Morse index of the Jacobi operator on minimal hypersurface $(\varSigma , i^*\tilde{g})$ (see [4]).

We will take the following convention for tensors. For instance, under a local frame field on $M$, suppose that $T=(T_{j_1,\ldots , j_r})$ is an $(r,0)$-tensor on $M$. The components of the covariant derivative $\overline{\nabla }T$ are denoted by $T_{j_1,\ldots , j_r;i}$, that is,

$$\begin{aligned} T_{j_1,\ldots , j_r;i} =(\overline{\nabla }_{e_i}T)(e_{j_1},\ldots ,e_{j_r})=(\overline{\nabla }T)(e_i, e_{j_1},\ldots ,e_{j_r}). \end{aligned}$$

Meanwhile, under a local frame field on $\varSigma$, suppose that $S=(S_{k_1,\ldots , k_s})$ is an $(s,0)$-tensor on $\varSigma$. The components of the covariant derivative ${\nabla }S$ are denoted by $S_{k_1,\ldots , k_s,l}$, that is,

$$\begin{aligned} S_{k_1\ldots k_s,l} =({\nabla }_{e_l}S)(e_{k_1},\ldots ,e_{k_s})=({\nabla }S)(e_l, e_{k_1},\ldots ,e_{k_s}). \end{aligned}$$

Throughout this paper, we assume that the $f$-minimal hypersurfaces are orientable and without boundary. For a closed hypersurface, we choose $\nu$ to be the outer unit normal. Finally, we refer the interested reader to [1, 3, 4, 8] and the references therein for more details about $f$-minimal hypersurfaces.

3 Simons-Type Equation for $f$-Minimal Hypersurfaces

First, we calculate the weighted Laplacian $\Delta _f$ for mean curvature $H$ of $f$-minimal hypersurfaces.

Proposition 1

Let $(\varSigma ^{n}, g)$ be an $f$-minimal hypersurface isometrically immersed in a smooth metric measure space $(M,\overline{g},e^{-f}d\mu )$. Then the mean curvature $H$ of $\varSigma$ satisfies

$$\begin{aligned} \Delta _fH&=2\hbox {tr}_g(\overline{\nabla }^3f(\cdot ,\nu , \cdot )|_{\varSigma })-\hbox {tr}_g(\overline{\nabla }^3f(\nu , \cdot , \cdot )|_{\varSigma })\nonumber \\&\qquad {}+2\langle A, \overline{\nabla }^2f|_{\varSigma }\rangle _g -\overline{\mathrm{Ric}}_{f}(\nu ,\nu )H-|A|^{2}H, \end{aligned}$$

(6)

or equivalently,

$$\begin{aligned} \Delta _fH&=2\sum _{i=1}^{n}(\overline{\nabla }^3f)_{i\nu i}-\sum _{i=1}^n(\overline{\nabla }^3f)_{\nu i i}+2\sum _{i,j=1}^{n}a_{ij}(\overline{\nabla }^2f)_{ij}\nonumber \\&\qquad {}-\overline{\mathrm{Ric}}_{f}(\nu ,\nu )H-|A|^{2}H, \end{aligned}$$

(7)

where $\{e_1,\ldots , e_n\}$ is a local orthonormal frame field on $\varSigma$, $\nu$ denotes the unit normal to $\varSigma$, and $|_{\varSigma }$ denotes the restriction to $\varSigma$.

Proof

We choose a local orthonormal frame field $\{e_i\}_{i=1}^{n+1}$ for $M$ so that, restricted to $\varSigma$, $\{e_i\}_{i=1}^{n}$ are tangent to $\varSigma$, and $e_{n+1}=\nu$ is the unit normal to $\varSigma$. Throughout this paper, for simplicity of notation, we substitute $\nu$ for the subscript $n+1$ in the components of the tensors on $M$, for instance, $\overline{R}_{\nu ikj}=\overline{Rm}(\nu , e_i, e_k, e_j)$, $(\overline{\nabla }^2f)_{\nu i}=(\overline{\nabla }^2f)(\nu , e_i)$. Differentiating the mean curvature $H=\langle \overline{\nabla }f,\nu \rangle$, we have, for $1\le i\le n$,

$$\begin{aligned} e_iH&=e_i\langle \overline{\nabla }f,\nu \rangle \nonumber \\&=\langle \overline{\nabla }_{e_i}(\overline{\nabla }f), \nu \rangle +\langle \overline{\nabla }f,\overline{\nabla }_{e_i}\nu \rangle \nonumber \\&=\overline{\nabla }^2f(\nu , e_i)+\sum _{k=1}^na_{ik}\langle \overline{\nabla }f,e_k\rangle . \end{aligned}$$

(8)

Then for $1\le i, j\le n$,

$$\begin{aligned} e_je_i(H)=e_j(\overline{\nabla }^2f(\nu , e_i))+\sum _{k=1}^ne_j(a_{ik})f_k+\sum _{k=1}^na _{ik}(e_j\langle \overline{\nabla }f,e_k\rangle ). \end{aligned}$$

(9)

For a fixed point $p\in \varSigma$, we may further choose the local orthonormal frame $\{e_1,\ldots ,e_n\}$ so that $\nabla _{e_i}e_j( p)=(\overline{\nabla }_{e_i}e_j)^{\top }(p)=0$, $1\le i,j\le n$. Then at $p$, for $1\le i, j\le n$,

$$\begin{aligned} e_j[\overline{\nabla }^2f(\nu , e_i)]&=\overline{\nabla }_{e_j}[\overline{\nabla }^2f(\nu , e_i)]\nonumber \\&=\overline{\nabla }_{e_j}(\overline{\nabla }^2f)(\nu , e_i)+\overline{\nabla }^2f(\overline{\nabla }_{e_j}\nu ,e_i)+ \overline{\nabla }^2f(\nu ,\overline{\nabla }_{e_j}e_i)\nonumber \\&=\overline{\nabla }^3f(e_j,\nu ,e_i)+\sum _{k=1}^na_{jk}\overline{\nabla }^2f(e_k,e_i)+ \overline{\nabla }^2f(\nu , \langle \overline{\nabla }_{e_j}e_i,\nu )\rangle \nu )\nonumber \\&=(\overline{\nabla }^3f)_{j\nu i}+\sum _{k=1}^na_{jk}(\overline{\nabla }^2f)_{ki}-a_{ji} (\overline{\nabla }^2f)_{\nu \nu }. \end{aligned}$$

(10)

In the third equality of (10), we used the assumption: $\nabla _{e_j}e_i( p)=0$, $1\le i,j\le n$. Also by this assumption and the Codazzi equation, we have at $p$, for $1\le i,j\le n,$

$$\begin{aligned} e_{j}(a_{ik})&=a_{ik,j}=a_{ij,k}+\overline{R}_{\nu ikj}\\ e_j\langle \overline{\nabla }f,e_k\rangle&=\langle \overline{\nabla } _{e_j}(\overline{\nabla }f),e_k\rangle +\langle \overline{\nabla }f, \overline{\nabla } _{e_j}e_k\rangle =(\overline{\nabla }^2f)_{jk}-a_{jk}{f}_\nu \\ (\nabla ^2 H)({e_j}, {e_i})&=e_je_iH. \end{aligned}$$

Substituting these equalities and (10) into (9), we have at $p$, for $1\le i,j\le n$,

$$\begin{aligned} (\nabla ^2H)(e_j,e_i)&=(\overline{\nabla }^3f)_{j\nu i}+\sum _{k=1}^na_{jk}(\overline{\nabla }^2f)_{ki}-a_{ji} (\overline{\nabla }^2f)_{\nu \nu }\nonumber \\&\qquad {}+\sum _{k=1}^{n}a_{ij,k}f_k+\sum _{k=1}^{n}\overline{R}_{\nu ikj}f_k\nonumber \\&\qquad {}+\sum _{k=1}^{n}a_{ik}(\overline{\nabla }^2f)_{jk} -\sum _{k=1}^{n}a_{ik}a_{jk}{f}_\nu . \end{aligned}$$

(11)

On the other hand, it holds that on $\varSigma$,

$$\begin{aligned} (\overline{\nabla }^3f)_{i\nu j}&=(\overline{\nabla }^2f)_{\nu j;i}=(\overline{\nabla }^2f)_{j\nu ;i}\\&=(\overline{\nabla }^2f)_{ji;\nu }+\sum _{k=1}^{n+1}{f}_{k}\overline{R}_{k j\nu i}\\&=(\overline{\nabla }^3f)_{\nu ji}+{f}_{\nu }\overline{R}_{\nu i \nu j}+\sum _{k=1}^nf_k\overline{R}_{\nu ikj}. \end{aligned}$$

So

$$\begin{aligned} \sum _{k=1}^nf_k\overline{R}_{\nu ikj}=(\overline{\nabla }^3f)_{i\nu j}-(\overline{\nabla }^3f)_{\nu ji}-{f}_{\nu }\overline{R}_{\nu i \nu j}. \end{aligned}$$

(12)

Substituting (12) into (11) and noting that $f_{\nu }=H$, we have at $p$, for $1\le i,j\le n$,

$$\begin{aligned} (\nabla ^2H)(e_j,e_i)&=(\overline{\nabla }^3f)_{j\nu i}+(\overline{\nabla }^3f)_{i\nu j}-(\overline{\nabla }^3f)_{\nu ji}\nonumber \\&\quad {}+\sum _{k=1}^na_{jk}(\overline{\nabla }^2f)_{ki} +\sum _{k=1}^na_{ik}(\overline{\nabla }^2f)_{jk} +\sum _{k=1}^{n}a_{ij,k}f_k\nonumber \\&\quad {}-H\overline{R}_{i\nu j\nu }-a_{ji}(\overline{\nabla }^2f)_{\nu \nu }-\sum _{k=1}^{n}a_{ik}a_{kj}H. \end{aligned}$$

(13)

Taking the trace, we have that at $p$,

$$\begin{aligned} \Delta H&=2\sum _{i=1}^{n}(\overline{\nabla }^3f)_{i\nu i}-\sum _{i=1}^{n}(\overline{\nabla }^3f)_{\nu ii}+2\sum _{i,k=1}^na_{ik}(\overline{\nabla }^2f)_{ki}\nonumber \\&\qquad {}+\langle \nabla f,\nabla H\rangle -\overline{\mathrm{Ric}}_{f}(\nu ,\nu )H-|A|^{2}H. \end{aligned}$$

(14)

Since $p\in \varSigma$ is arbitrary and (14) is independent of the choice of frame, (14) holds on $\varSigma$. By (14) and $\Delta _f=\Delta -\langle \nabla f,\nabla H\rangle$, we obtain (7) and also the equivalent identity (6).$\square$

Proposition 1 yields the following

Corollary 2

With the same assumption and notation as in Proposition 1,

$$\begin{aligned} L_fH =2{\hbox {tr}}_g(\overline{\nabla }^3f(\cdot ,\nu , \cdot )|_{\varSigma })-{\hbox {tr}}_g(\overline{\nabla }^3f(\nu , \cdot , \cdot )|_{\varSigma }) +2\langle A, \overline{\nabla }^2f|_{\varSigma }\rangle _g, \end{aligned}$$

(15)

or equivalently,

$$\begin{aligned} L_fH=2\sum _{i=1}^{n}(\overline{\nabla }^3f)_{i\nu i}-\sum _{i=1}^n(\overline{\nabla }^3f)_{\nu i i}+2\sum _{i,j=1}^{n}a_{ij}(\overline{\nabla }^2f)_{ij}. \end{aligned}$$

(16)

Next we will derive the Simons-type equation for $f$-minimal hypersurfaces.

Theorem 3

Let $(\varSigma ^{n},g)$ be an $f$-minimal hypersurface isometrically immersed in $(M,\overline{g},e^{-f}d\mu )$. Then the square of the norm of the second fundamental form of $\varSigma$ satisfies

$$\begin{aligned} \frac{1}{2}\Delta _{f}|A|^{2}&=|\nabla A|^{2}+2\sum _{i,j,k=1}^{n} a_{ij}a_{ik}(\overline{\mathrm{Ric}}_f)_{jk}-(\overline{\mathrm{Ric}}_f)_{ \nu \nu }|A|^{2}-|A|^{4}\nonumber \\&\quad {}+2\sum _{i,j=1}^na_{ij}(\overline{\mathrm{Ric}}_f)_{i\nu ;j} -\sum _{i,j=1}^{n}a_{ij}(\overline{\mathrm{Ric}}_f)_{ij;\nu } +\sum _{i,j=1}^{n}a_{ij}\overline{R}_{i\nu j\nu ;\nu }\nonumber \\&\quad -2\sum _{i,j,k=1}^{n}a_{ij}a_{ik}\overline{R}_{j\nu k\nu }-2\sum _{i,j,k,l=1}^{n}a_{ij}a_{lk}\overline{R}_{iljk}, \end{aligned}$$

(17)

where the notation is the same as in Proposition 1.

Proof

Simons [18] proved the following identity (see, e.g., [17] (1.20)) under a local orthonormal frame $e_1,\ldots , e_n$ of $\varSigma$:

$$\begin{aligned} \Delta a_{ij}&=\nabla ^2H(e_j,e_i)+\sum _{k=1}^n\overline{R}_{\nu kik;j}+\sum _{k=1}^n\overline{R}_{\nu ijk;k}\nonumber \\&\quad {}-H\overline{R}_{\nu ij\nu }-\overline{\mathrm{Ric}}_{\nu \nu }a_{ij} +H\sum _{k=1}^na_{ik}a_{kj}-|A|^{2}a_{ij}\nonumber \\&\quad {}+\sum _{k,l=1}^n(a_{ik}\overline{R}_{kljl}+a_{jk} \overline{R}_{klil}+2a_{kl}\overline{R}_{lijk}). \end{aligned}$$

(18)

Observe that

$$\begin{aligned} \sum _{k=1}^{n}\overline{R}_{\nu kik;j}&= (\overline{\mathrm{Ric}})_{\nu i;j}=(\overline{\mathrm{Ric}})_{i\nu ;j},\end{aligned}$$

(19)

$$\begin{aligned} \sum _{k=1}^{n}\overline{R}_{\nu ijk;k}&= \sum _{k=1}^{n}\overline{R}_{jk\nu i;k}=-\sum _{k=1}^{n}\overline{R}_{jkik;\nu }-\sum _{k=1}^{n}\overline{R}_{jkk\nu ;i}\nonumber \\&= -(\overline{\mathrm{Ric}})_{ij;\nu }+\overline{R}_{i\nu j\nu ;\nu }+(\overline{\mathrm{Ric}})_{j\nu ;i},\end{aligned}$$

(20)

$$\begin{aligned} \sum _{k,l=1}^{n}a_{ik}\overline{R}_{kljl}&= \sum _{k=1} ^{n}a_{ik}(\overline{\mathrm{Ric}})_{kj}-\sum _{k=1}^{n}a_{ik}\overline{R}_{k\nu j\nu },\end{aligned}$$

(21)

$$\begin{aligned} \sum _{k,l=1}^{n}a_{jk}\overline{R}_{klil}&= \sum _{ k=1}^{n}a_{jk}(\overline{\mathrm{Ric}})_{ki}-\sum _{k=1}^{n}a_{jk}\overline{R}_{k\nu i\nu }. \end{aligned}$$

(22)

Using the same local frame as in the proof of Proposition 1 and substituting (13), (19), (20), (21) and (22) into (18), we have that at $p$, for $1\le i,j\le n$,

$$\begin{aligned} \Delta a_{ij}&=(\overline{\mathrm{Ric}}_{f})_{i\nu ;j}+ (\overline{\mathrm{Ric}}_{f})_{j\nu ;i}- (\overline{\mathrm{Ric}}_{f})_{ij;\nu }+\overline{R}_{i\nu j\nu ;\nu }\nonumber \\&\quad {}+\sum _{k=1}^na_{ik}(\overline{\mathrm{Ric}}_{f})_{kj}+ \sum _{k=1}^na_{jk}(\overline{\mathrm{Ric}}_{f})_{ki}\nonumber \\&\quad {}-\sum _{k=1}^na_{ik}\overline{R}_{k\nu j\nu }-\sum _{k=1} ^na_{jk}\overline{R}_{k\nu i\nu }-2\sum _{l,k=1}^na_{lk}\overline{R}_{likj}\nonumber \\&\quad {}+\sum _{k=1}^na_{ij,k}f_{k}-(\overline{\mathrm{Ric}} _{f})_{\nu \nu }a_{ij}-|A|^{2}a_{ij}. \end{aligned}$$

(23)

Multiply (23) by $a_{ij}$ and take the trace. Then it holds that at $p$,

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&=\frac{1}{2}\Delta |A|^2-\frac{1}{2}\langle \nabla f,\nabla |A|^2\rangle \nonumber \\&= \sum _{i,j=1}^{n}a_{ij}\Delta a_{ij}+\sum _{i,j,k=1}^{n}a_{ij,k}^2-\sum _{i,j,k=1}^na_{ij}f_ka_{ij,k}\nonumber \\&=|\nabla A|^{2}+2\sum _{i,j,k=1}^{n}a_{ij}a_{ik}(\overline{\mathrm{Ric}}_{f}) _{kj}-(\overline{\mathrm{Ric}}_f)_{\nu \nu }|A|^{2}-|A|^{4}\nonumber \\&\quad {}+2\sum _{i,j=1}^{n}a_{ij}(\overline{\mathrm{Ric}}_{f})_{i\nu ;j} -\sum _{i,j=1}^{n}a_{ij}(\overline{\mathrm{Ric}}_f)_{ij;\nu }\nonumber \\&\quad {}+\sum _{i,j=1}^{n}a_{ij}\overline{R}_{i\nu j\nu ;\nu }\nonumber \\&\quad {}-2\sum _{i,j,k=1}^{n}a_{ij}a_{ik}\overline{R}_ {k\nu j\nu }-2\sum _{i,j,k,l=1}^{n}a_{ij}a_{lk}\overline{R}_{iljk}. \end{aligned}$$

(24)

Thus

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&=|\nabla A|^{2}+2\langle A^2, \overline{\mathrm{Ric}}_{f}|_{\varSigma }\rangle -(\overline{\mathrm{Ric}}_f)_{\nu \nu }|A|^{2}-|A|^{4}\nonumber \\&\quad {}+2\langle A, \overline{\nabla } (\overline{\mathrm{Ric}}_ {f})(\cdot ,\nu , \cdot )|_{\varSigma }\rangle -\langle A, \overline{\nabla }(\overline{\mathrm{Ric}}_{f})(\nu , \cdot ,\cdot )|_{\varSigma }\rangle \nonumber \\&\quad {}+\langle A, \overline{\nabla } (\overline{Rm}) (\nu , \cdot ,\nu ,\cdot ,\nu )|_{\varSigma }\rangle \nonumber \\&\quad {}-2\langle A^2, \overline{Rm}(\cdot ,\nu , \cdot ,\nu )|_{\varSigma }\rangle -2\sum _{i,j,k,l=1}^{n}a_{ij}a_{lk}\overline{R}_{iljk}. \end{aligned}$$

(25)

Since (25) is independent of the choice of the coordinates, (17) holds on $\varSigma$.$\square$

When the ambient space $M$ has the property $\overline{\mathrm{Ric}}_f=C\overline{g}$, i.e., $M$ is a gradient Ricci soliton, the Simons-type equation for $f$-minimal hypersurfaces is the following (26).

Corollary 3

Let $(M^{n+1},\overline{g}, e^{-f}d\mu )$ be a smooth metric measure space satisfying $\overline{\mathrm{Ric}}_f=C\overline{g}$, where $C$ is a constant. If $(\varSigma ,g)$ is an $f$-minimal hypersurface isometrically immersed in $M$, then it holds that on $\varSigma$

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&=|\nabla A|^{2}+C|A|^{2}-|A|^{4} +\sum _{i,j=1}^{n}a_{ij}\overline{R}_{i\nu j\nu ;\nu }\nonumber \\&\qquad {}-2\sum _{i,j,k=1}^{n}a_{ij}a_{ik}\overline{R}_{j\nu k\nu }-2\sum _{i,j,k,l=1}^{n}a_{ij}a_{lk}\overline{R}_{iljk}, \end{aligned}$$

(26)

where the notation is the same as in Theorem 3

Finally, in this section, we prove the following identity for $L_f$ operator, which is useful for the study of the eigenvalues of $L_f$ and $L_f$-index of $f$-minimal hypersurfaces.

Proposition 2

Let $(M^{n+1},\overline{g},e^{-f}d\mu )$ be a smooth metric measure space and $X$ a parallel vector field on $M$. If $(\varSigma ^{n},g)$ is an $f$-minimal hypersurface isometrically immersed in $M$, then the function $\alpha :\varSigma \rightarrow \mathbb {R}$ defined by $\alpha =\langle X,\nu \rangle$ satisfies

$$\begin{aligned} \Delta _f\alpha&=\overline{\mathrm{Ric}}_{f}(X,\nu )-|A|^{2 }\alpha -(\overline{\mathrm{Ric}}_{f})_{\nu \nu }\alpha ,\end{aligned}$$

(27)

$$\begin{aligned} L_{f}\alpha&=\overline{\mathrm{Ric}}_{f}(X,\nu ), \end{aligned}$$

(28)

where the notation is the same as in Theorem 3.

Proof

Choose a local field of orthonormal frame $\{e_{1},\ldots ,e_{n},e_{n+1}\}$ on $M$ as in the proof of Proposition 1. Then for $1\le i\le n$,

$$\begin{aligned} e_i\alpha&=\langle X,\overline{\nabla }_{e_{i}}\nu \rangle \nonumber \\&=\sum _{j=1}^na_{ij}\langle X,e_{j}\rangle . \end{aligned}$$

(29)

Note that $\nabla _{e_i}e_j(p)=0$, $1\le i,j\le n$. The Hessian of $\alpha$ at $p$ is given by

$$\begin{aligned} (\nabla ^2\alpha )(e_k,e_i)=e_ke_i\alpha&= \sum _{j=1}^na_{ij,k}\langle X,e_{j}\rangle +\sum _{j=1}^na_{ij}\langle X,\overline{\nabla }_{e_{k}}e_{j}\rangle \nonumber \\&= \sum _{j=1}^na_{ij,k}\langle X,e_{j}\rangle -\sum _{j=1}^na_{ij}a_{kj}\langle X,\nu \rangle \nonumber \\&= \sum _{j=1}^na_{ik,j}\langle X,e_{j}\rangle +\sum _{j=1}^n\overline{R}_{\nu ijk}\langle X,e_{j}\rangle -\sum _{j=1}^na_{ij}a_{jk}\alpha .\qquad \end{aligned}$$

(30)

Take the trace in (30). Then

$$\begin{aligned} \Delta \alpha =\langle \nabla H,X\rangle +\sum _{j=1}^n(\overline{\mathrm{Ric}})_{\nu j}\langle X,e_{j}\rangle -|A|^{2}\alpha . \end{aligned}$$

(31)

Also, from (8) and (30) we have at $p$,

$$\begin{aligned} \langle \nabla H,X\rangle&=\sum _{j=1}^n(\overline{\nabla }^2{f})_{\nu j}\langle X,e_{j}\rangle +\sum _{j,k=1}^na_{jk}f_{k}\langle X,e_{j}\rangle \\&=\sum _{j=1}^n(\overline{\nabla }^2{f})_{\nu j}\langle X,e_{j}\rangle +\sum _{k=1}^nf_{k}\alpha _{k}\\&=\sum _{j=1}^n(\overline{\nabla }^2{f})_{\nu j}\langle X,e_{j}\rangle +\langle \nabla f,\nabla \alpha \rangle . \end{aligned}$$

Substituting the above identity into (31), we have

$$\begin{aligned} \Delta \alpha&=\sum _{j=1}^n(\overline{\mathrm{Ric}}_{f})_{\nu j}\langle X,e_{j}\rangle +\langle \nabla f,\nabla \alpha \rangle -|A|^{2}\alpha \nonumber \\&=\overline{\mathrm{Ric}}_{f}(X,\nu )+\langle \nabla f,\nabla \alpha \rangle -|A|^{2}\alpha -(\overline{\mathrm{Ric}}_{f})_{\nu \nu }\alpha , \end{aligned}$$

(32)

that is,

$$\begin{aligned} \Delta _f\alpha =\overline{\mathrm{Ric}}_{f}(X,\nu )-|A|^{2} \alpha -(\overline{\mathrm{Ric}}_{f})_{\nu \nu }\alpha . \end{aligned}$$

(33)

Since $p$ is arbitrary and (33) is independent of the frame, we have proved (27) and then (28).$\square$

When the ambient manifold is a gradient Ricci soliton, we obtain

Corollary 4

Let $(M^{n+1},\overline{g}, e^{-f}d\mu )$ be a smooth metric measure space satisfying $\overline{\mathrm{Ric}}_f=C\overline{g}$, where $C$ is a constant. Suppose that $X$ is a parallel vector field on $\varSigma$. If $(\varSigma ,g)$ is an $f$-minimal hypersurface isometrically immersed in $M$, then $\alpha =\langle X,\nu \rangle$ satisfies that on $\varSigma$,

$$\begin{aligned} L_{f}\alpha =C\alpha . \end{aligned}$$

(34)

Example 2

[7] Let $M=\mathbb {R}^{n+1}$ and $f=\frac{|x|^{2}}{4}$. The $f$-minimal hypersurfaces are self-shrinkers. Suppose that $\varSigma$ is a self-shrinker. Then

$$\begin{aligned} \frac{1}{2}\Delta _fH^{2}&=|\nabla H|^{2}+(\frac{1}{2}-|A|^{2})H^{2},\\ \frac{1}{2}\Delta _f|A|^{2}&=|\nabla A|^{2}+(\frac{1}{2}-|A|^{2})|A|^{2},\\ L_{f}H&=H. \end{aligned}$$

If $V$ is a constant vector in $\mathbb {R}^{n+1}$ and $\nu$ is the unit normal to $\varSigma$, then

$$\begin{aligned} L_f\langle V,\nu \rangle =\frac{1}{2}\langle V,\nu \rangle . \end{aligned}$$

4 $L_f$-Index of $f$-Minimal Hypersurfaces

In this section, we study the $L_{f}$-index of closed $f$-minimal hypersurfaces immersed in the product manifold $\mathbb {S}^{n}(a)\times \mathbb {R}$, $n\ge 2$, with $f(x,t)=\frac{(n-1)t^2}{4a^2}$, where $(x,t)\in \mathbb {S}^{n}(a)\times \mathbb {R}$ and $\mathbb {S}^{n}(a)$ denotes the round sphere in $\mathbb {R}^{n+1}$ of radius $a$. For simplicity of notation, we only consider $a=\sqrt{2(n-1)}$ and hence $f=\frac{t^2}{4}$. The cases of other $a$ are analogous. $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$ has the metric

$$\begin{aligned} \overline{g}=g_{\mathbb {S}^{n}(\sqrt{2(n-1)})}+dt^{2}, \end{aligned}$$

where $g_{\mathbb {S}^{n}(\sqrt{2(n-1)})}$ denotes the canonical metric of $\mathbb {S}^{n}(\sqrt{2(n-1)})$. Let $\{\overline{e}_{1},\ldots ,\overline{e}_{n+1}\}$ be a local orthonormal frame on $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$. By a straightforward computation, one has the components of the curvature tensor and Ricci curvature tensor of $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$ given by, for $1\le i,j,k,l\le n+1$,

$$\begin{aligned} \overline{R}_{ijkl}&=\frac{1}{2(n-1)}\biggr (\delta _{ik} \delta _{jl}-\delta _{il}\delta _{jk}-\langle \overline{e}_{j},\frac{\partial }{\partial t}\rangle \langle \overline{e}_{l},\frac{\partial }{\partial t}\rangle \delta _{ik} \nonumber \\&\qquad {}-\langle \overline{e}_{i},\frac{\partial }{\partial t}\rangle \langle \overline{e}_{k},\frac{\partial }{\partial t}\rangle \delta _{jl}+\langle \overline{e}_{j},\frac{\partial }{\partial t}\rangle \langle \overline{e}_{k},\frac{\partial }{\partial t}\rangle \delta _{il}+\langle \overline{e}_{i},\frac{\partial }{\partial t}\rangle \langle \overline{e}_{l}, \frac{\partial }{\partial t}\rangle \delta _{jk})\biggr ), \end{aligned}$$

(35)

and

$$\begin{aligned} (\overline{\mathrm{Ric}})_{ik}=\frac{1}{2}\bigg (\delta _{ik} -\langle \overline{e}_{i},\frac{\partial }{\partial t}\rangle \langle \overline{e}_{k},\frac{\partial }{\partial t}\rangle \bigg ). \end{aligned}$$

(36)

On the other hand,

$$\begin{aligned} \overline{\nabla }f=\frac{t}{2}\frac{\partial }{\partial t}, \end{aligned}$$

$$\begin{aligned} (\overline{\nabla }^2f)_{ik}=\frac{1}{2}\langle \overline{e}_{i},\frac{\partial }{\partial t}\rangle \langle \overline{e}_{k},\frac{\partial }{\partial t}\rangle . \end{aligned}$$

(37)

By (36) and (37), we have

$$\begin{aligned} (\overline{\mathrm{Ric}})_{ik}+(\overline{\nabla }^2f) _{ik}=\frac{1}{2}\overline{g}_{ik}. \end{aligned}$$

(38)

Hence $(\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R},\overline{g},e^{-f}d\mu )$ is a smooth metric measure space with $\overline{\mathrm{Ric}}_{f}=\frac{1}{2}\overline{ g}$. In addition, in the theory of Ricci flow, $(\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R},\overline{g},f)$ is a shrinking gradient soliton.

For an $f$-minimal hypersurface $\varSigma$ immersed in $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$,

$$\begin{aligned} 0=H_f=H-\frac{t}{2}\langle \frac{\partial }{\partial t},\nu \rangle =H-\frac{t}{2}\alpha , \end{aligned}$$

where $\alpha =\langle \frac{\partial }{\partial t},\nu \rangle$. So $\varSigma$ satisfies

$$\begin{aligned} H=\frac{t\alpha }{2}. \end{aligned}$$

The operator $L_f$ on $\varSigma$ is

$$\begin{aligned} L_f=\Delta -\frac{t}{2}\langle \Big (\frac{\partial }{\partial t}\Big )^{T},\nabla \cdot \rangle +|A|^2+\frac{1}{2}. \end{aligned}$$

(39)

Lemma 1

The slice $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$ is an $f$-minimal hypersurface in $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$. Moreover, a complete $f$-minimal hypersurface $\varSigma$ is immersed in a horizontal slice $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{t\}$, where $t\in \mathbb {R}$ is fixed, if and only if $\varSigma$ is $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$.

Proof

The unit normal $\nu$ to $\varSigma$ satisfies $\nu =\frac{\partial }{\partial t}$ and hence $AX=\overline{\nabla }_X\nu =0, X\in T\varSigma$. Thus $\varSigma$ is totally geodesic. Meanwhile,

$$\begin{aligned} \langle \overline{\nabla }f,\nu \rangle =\frac{t}{2}, \end{aligned}$$

$$\begin{aligned} H_{f}=H-\langle \overline{\nabla }f,\nu \rangle =-\frac{t}{2}. \end{aligned}$$

(40)

It follows that $\varSigma$ is $f$-minimal if and only if $t=0$. Further, by Gauss’s equation we know $\varSigma$ has constant positive section curvature and hence is closed. Since the closed $\varSigma$ has dimension $n$ and $\mathbb {S}^n$ is simply connected, $\varSigma$ must be $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$.$\square$

We will prove that

Lemma 2

$L_f$-$\hbox {ind}(\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\})=1$.

Proof

On $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$, we have $\nabla f=(\overline{\nabla }f)^{\top }=0$, $|A|^{2}=0$. Hence,

$$\begin{aligned} L_{f}=\Delta _{\mathbb {S}^{n}(\sqrt{2(n-1)})}+\frac{1}{2}. \end{aligned}$$

(41)

Thus the eigenvalues of $L_{f}$ are

$$\begin{aligned} \mu _{k}=\lambda _{k}-\frac{1}{2}, \end{aligned}$$

where $\lambda _{k}=\frac{k(k+n-1)}{2(n-1)}, k=0,1,\ldots$, are the eigenvalues of the Laplacian $\Delta _{\mathbb {S}^{n}(\sqrt{2(n-1)})}$. Observe that

$$\begin{aligned} \mu _{0}&= -\frac{1}{2},\\ \mu _{k}&> 0,\qquad \mathrm{for\,all} \quad k\ge 1, \end{aligned}$$

that is, $L_{f}$ has a unique negative eigenvalue with multiplicity one. Therefore, the $L_f$-index of $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$ is $1$.$\square$

We will prove Theorem 2, which says that $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$ is the unique closed $f$-minimal hypersurface of $L_f$-index one.

.

Proof of Theorem 2

On $\varSigma$,

$$\begin{aligned} \nabla t=(\overline{\nabla }t)^{\top }=\frac{\partial }{\partial t}-\langle \frac{\partial }{\partial t},\nu \rangle \nu . \end{aligned}$$

So

$$\begin{aligned} |\nabla t|^{2}=1-\langle \frac{\partial }{\partial t},\nu \rangle ^{2}=1-\alpha ^2. \end{aligned}$$

(42)

Since $\varSigma$ is closed, there is a point $p\in \varSigma$ such that $t(p)=\max _{\varSigma } t$ and $|\nabla t|(p)=0$. By equation (42), we have

$$\begin{aligned} 0=|\nabla t|^{2}(p)=1-\alpha ^{2}(p). \end{aligned}$$

(43)

Hence $\alpha (p)=\pm 1$ and so $\alpha \not \equiv 0$. Since $\frac{\partial }{\partial t}$ is a parallel vector field on $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$ and $\overline{\mathrm{Ric}}_{f}=\frac{1}{2}\overline{g}$, Proposition 2 implies that

$$\begin{aligned} L_{f}\alpha =\overline{\mathrm{Ric}}_{f}(\frac{\partial }{\partial t},\nu )=\frac{1}{2}\alpha . \end{aligned}$$

Thus $\alpha$ is an eigenfunction of $L_f$ with eigenvalue $-\frac{1}{2}$ and this implies that $L_f$-$\mathrm ind (\varSigma )\ge 1$.

Now we consider the equality case. Lemma 2 says that $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$ has $L_f$-index one. Conversely, if $L_f$-$\mathrm ind (\varSigma )=1$, then $-\frac{1}{2}$ is the first eigenvalue. Then the corresponding eigenfunction $\alpha$ cannot change sign. We may assume that $\alpha >0$. On the other hand, $L_{f}\alpha =\frac{1}{2}\alpha$ and $\overline{\mathrm{Ric}}_f=\frac{1}{2}\overline{g}$ imply

$$\begin{aligned} \Delta _{f}\alpha +|A|^{2}\alpha =0. \end{aligned}$$

Hence

$$\begin{aligned} \Delta _{f}\alpha \le 0. \end{aligned}$$

(44)

By the maximum principle, $\alpha$ is constant on $\varSigma$. On the other hand, by  (43), there is a point $p\in \varSigma$ such that $\alpha (p)=\pm 1$. Since $\alpha$ is positive, $\alpha \equiv 1$. Hence $\nabla t=0$ and thus $\varSigma$ is in a horizontal slice $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{t\}$. By Lemma 1, $\varSigma$ must be $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$.$\square$

5 Pinching Theorem

First, we will derive various identities, including a Simons-type equation (see (47)) for $f$-minimal hypersurfaces immersed in $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$. Next, we apply them to obtain a pinching result for $f$-minimal hypersurfaces. We use the same notation as in Sect. 4.

Proposition 3

Let $\varSigma$ be an $f$-minimal hypersurface immersed in the product manifold $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$ with $f=\frac{t^{2}}{4}$. Then

$$\begin{aligned} \frac{1}{2}\Delta _f\alpha ^{2}&=|\nabla \alpha |^{2}-|A|^{2}\alpha ^{2},\end{aligned}$$

(45)

$$\begin{aligned} \frac{1}{2}\Delta _fH^{2}&=|\nabla H|^{2}-(|A|^{2}+\frac{1}{2})H^{2}+\frac{1}{2}\langle \nabla \alpha ^{2},\nabla f\rangle ,\end{aligned}$$

(46)

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&= |\nabla A|^{2}+|A|^{2}(\frac{1}{2}-|A|^{2})-\frac{1}{n-1}(|\nabla \alpha |^{2}-\alpha ^{2}|A|^{2})\nonumber \\&\qquad {}-\frac{1}{n-1}(\alpha ^{2}f-\langle \nabla \alpha ^{2},\nabla f\rangle ). \end{aligned}$$

(47)

Proof

Choose a local orthonormal frame field $\{e_i\}_{i=1}^{n+1}$ for $M$ so that, restricted to $\varSigma$, $\{e_i\}_{i=1}^{n}$ are tangent to $\varSigma$, and $e_{n+1}=\nu$ is the unit normal to $\varSigma$. Recall that Proposition 2 states that

$$\begin{aligned} \Delta _f\alpha =\overline{\mathrm{Ric}}_f(\nu ,\frac{\partial }{\partial t})-|A|^{2}\alpha -\overline{\mathrm{Ric}}_f(\nu ,\nu )\alpha . \end{aligned}$$

(48)

Substituting (38) in (48), we have

$$\begin{aligned} \Delta _f\alpha =-|A|^{2}\alpha . \end{aligned}$$

(49)

(49) implies (45):

$$\begin{aligned} \frac{1}{2}\Delta _f\alpha ^{2}&= |\nabla \alpha |^{2}+\alpha \Delta _f\alpha \\&= |\nabla \alpha |^{2}-|A|^{2}\alpha ^{2}. \end{aligned}$$

Now we prove (46). Note $f=\frac{1}{4}t^2$. $\overline{\nabla }^3 f=0$. This and Proposition 1 yield

$$\begin{aligned} \Delta _fH=2\sum _{i,j=1}^{n}a_{ij}(\overline{\nabla }^2 f)_{ij}-\overline{\mathrm{Ric}}_{f}(\nu ,\nu )H-|A|^{2}H. \end{aligned}$$

(50)

Substituting (37) and (38) into (50), we have

$$\begin{aligned} \Delta _fH&=\sum _{i,j=1}^{n}a_{ij}\langle e_i,\frac{\partial }{\partial t}\rangle \langle e_j,\frac{\partial }{\partial t}\rangle -\frac{1}{2} H-|A|^{2}H\\&=\langle \nabla \alpha ,\frac{\partial }{\partial t}\rangle -\frac{1}{2} H-|A|^{2}H. \end{aligned}$$

Then,

$$\begin{aligned} \frac{1}{2}\Delta _fH^{2}&=|\nabla H|^{2}+H\Delta _fH\\&=|\nabla H|^{2}+H\langle \nabla \alpha ,\frac{\partial }{\partial t}\rangle -(|A|^{2}+\frac{1}{2})H^{2}\\&=|\nabla H|^{2}+\frac{ t\alpha }{2}\langle \nabla \alpha ,\frac{\partial }{\partial t}\rangle -(|A|^{2}+\frac{1}{2})H^{2}\\&=|\nabla H|^{2}+\frac{1}{2}\langle \nabla \alpha ^{2},\nabla f\rangle -(|A|^{2}+\frac{1}{2})H^{2}. \end{aligned}$$

In the above, we used $H=\frac{t\alpha }{2}$ and $\overline{\nabla }f=\frac{t}{2}\frac{\partial }{\partial t}$. Thus (46) holds. Finally we prove (47). Since $\mathbb {S}^n(\sqrt{2(n-1)})\times \mathbb {R}$ is a symmetric space, $\overline{\nabla }R=0$. By the Simons-type equation (Corollary 3), it holds that

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&= |\nabla A|^{2}+|A|^{2}(\frac{1}{2}-|A|^{2})\nonumber \\&\,-\,2\sum _{i,j,k=1}^{n}a_{ij}a_{ik}\overline{R}_{k\nu j\nu }-2\sum _{i,j,k,l=1}^{n}a_{ij}a_{lk}\overline{R}_{iljk}. \end{aligned}$$

(51)

Substituting the curvature tensors (35) into (51) and computing directly, we obtain

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&=|\nabla A|^{2}+|A|^{2}(\frac{1}{2}-|A|^{2}) -\frac{1}{n-1}\biggr (H^{2}-\alpha ^{2}|A|^{2}\\&\qquad {}-2H\sum _{i,j=1}^{n}a_{ij}\langle e_i,\frac{\partial }{\partial t}\rangle \langle e_j,\frac{\partial }{\partial t}\rangle +\sum _{i,j,k=1}^{n}a_{ij}a_{ik}\langle e_j,\frac{\partial }{\partial t}\rangle \langle e_k,\frac{\partial }{\partial t}\rangle \biggr ). \end{aligned}$$

Note that the function $\alpha$ satisfies $\alpha _i=\displaystyle \sum _{j=1}^{n}a_{ij}\langle e_j,\frac{\partial }{\partial t}\rangle$. Hence,

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&=|\nabla A|^{2}+|A|^{2}(\frac{1}{2}-|A|^{2})\\&\qquad {}-\frac{1}{n-1}\bigr (H^{2}-\alpha ^{2}|A|^ {2}-2H\langle \nabla \alpha ,\frac{\partial }{\partial t}\rangle +|\nabla \alpha |^{2}\bigr )\\&=|\nabla A|^{2}+|A|^{2}(\frac{1}{2}-|A|^{2})-\frac{1}{n-1}(|\nabla \alpha |^{2}-\alpha ^{2}|A|^{2})\\&\qquad {}-\frac{1}{n-1}(H^{2}-2H\langle \nabla \alpha ,\frac{\partial }{\partial t}\rangle ). \end{aligned}$$

Using $\overline{\nabla }f=\frac{t}{2}\frac{\partial }{\partial t}$ and $H=\frac{ t\alpha }{2}$, we obtain (47):

$$\begin{aligned} \frac{1}{2}\Delta _f|A|^{2}&=|\nabla A|^{2}+|A|^{2}(\frac{1}{2}-|A|^{2})-\frac{1}{n-1} (|\nabla \alpha |^{2}-\alpha ^{2}|A|^{2})\\&\qquad {}-\frac{1}{n-1}(\alpha ^{2}f-\langle \nabla \alpha ^{2},\nabla f\rangle ). \end{aligned}$$

$\square$

Proposition 3 implies the following equations:

Lemma 3

If $\varSigma$ is a closed orientable $f$-minimal hypersurface immersed in $M=\mathbb {S}^{n}(\sqrt{2(n-1)})\times \mathbb {R}$, then

$$\begin{aligned} \int _{\varSigma }|\nabla \alpha |^2e^{-f}&-\int _{\varSigma } \alpha ^2|A|^2e^{-f}=0,\end{aligned}$$

(52)

$$\begin{aligned} -\int _{\varSigma }|\nabla H|^{2}e^{-f}+\int _{\varSigma } H^{2}|A|^{2}e^{-f}&+\frac{1}{4}\int _{\varSigma }\alpha ^{2} (1-\alpha ^{2})e^{-f}=0,\end{aligned}$$

(53)

$$\begin{aligned} \int _{\varSigma }|\nabla A|^{2}e^{-f}+\int _{\varSigma }|A|^{2}(\frac{1}{2}- |A|^{2})e^{-f}&-\frac{1}{2(n-1)}\int _{\varSigma } \alpha ^{2}(1-\alpha ^{2})e^{-f}=0. \end{aligned}$$

(54)

Proof

(52) can be obtained by integrating (45) directly. Now we prove (53). Since

$$\begin{aligned} \Delta f&=\hbox {tr}\nabla ^2f=\sum _{i=1}^n[(\overline{\nabla }^2f)_{ii}-a_{ii}f_{\nu }]\\&=\frac{1}{2}\sum _{i=1}^{n}\langle e_{i},\frac{\partial }{\partial t}\rangle ^{2}-H{f}_\nu \\&=\frac{1}{2}(1-\alpha ^{2})-\langle \overline{\nabla }f,\nu \rangle ^2, \end{aligned}$$

$$\begin{aligned} \Delta _ff&=\frac{1}{2}(1-\alpha ^{2})-\langle \nabla f,\nabla f\rangle -\langle \overline{\nabla }f,\nu \rangle ^2\nonumber \\&=\frac{1}{2}(1-\alpha ^{2})-|\overline{\nabla }f|^{2}\nonumber \\&=\frac{1}{2}(1-\alpha ^{2})-f. \end{aligned}$$

(55)

Integrating (46) and using (55), we obtain

$$\begin{aligned}&-\int _{\varSigma }|\nabla H|^{2}e^{-f}+\int _{\varSigma }(|A|^{2}+\frac{1}{2})H^{2}e^{-f}\\&\quad =\frac{1}{2}\int _{\varSigma }\langle \nabla \alpha ^{2},\nabla f\rangle e^{-f}\\&\quad =-\frac{1}{2}\int _{\varSigma }\alpha ^{2}(\Delta _{f}f)e^{-f}\\&\quad =-\frac{1}{4}\int _{\varSigma }\alpha ^{2}(1-\alpha ^{2})e^ {-f}+\frac{1}{2}\int _{\varSigma }H^{2}e^{-f} \end{aligned}$$

In the above we have used $\int _\varSigma \Delta _fH^2e^{-f}=0$ and $H^2=\alpha ^2f$. Thus, we get (53). Finally we prove (54). Integrating (47) and using (52) and (55), we have

$$\begin{aligned}&\int _{\varSigma }|\nabla A|^{2}e^{-f}+\int _{\varSigma }|A|^{2}(\frac{1}{2}-|A| ^{2})e^{-f}\nonumber \\&\quad =\frac{1}{n-1}\int _{\varSigma }(\alpha ^{2}f-\langle \nabla \alpha ^{2},\nabla f\rangle )e^{-f}\\&\quad =\frac{1}{n-1}\int _{\varSigma }(\alpha ^{2}f+\alpha ^{2}\Delta _ff)e^{-f}\\&\quad =\frac{1}{2(n-1)}\int _{\varSigma }\alpha ^{2}(1-\alpha ^{2})e^{-f}. \end{aligned}$$

$\square$

Using Lemma 3, we may prove Theorem 1.

.

Proof of Theorem 1

Observe that, for $n\ge 3$,

$$\begin{aligned}&|A|^{2}(\frac{1}{2}-|A|^{2})-\frac{1}{2(n-1)}\alpha ^ {2}(1-\alpha ^{2})\\&\quad =-(|A|^2-\frac{1}{4})^2+(\frac{1}{4}) ^2[1-\frac{8}{n-1}\alpha ^2(1-\alpha ^2)]\ge 0 \end{aligned}$$

if and only if

$$\begin{aligned} \frac{1}{4}\biggr (1-\sqrt{1-\frac{8}{n-1}\alpha ^2(1-\alpha ^2)}\biggr )\le |A|^2\le \frac{1}{4}\biggr (1+\sqrt{1-\frac{8}{n-1}\alpha ^2(1-\alpha ^2)}\biggr ). \end{aligned}$$

So (54) implies that on $\varSigma$, for $n\ge 3$,

$$\begin{aligned} \nabla A\equiv 0, \end{aligned}$$

and

$$\begin{aligned} |A|^{2}\left( \frac{1}{2}-|A|^{2}\right) -\frac{1}{2(n-1)}\alpha ^{2}(1-\alpha ^{2})=0. \end{aligned}$$

Hence, $|A|^{2}$ and $H$ are constants. Substituting in (53), we obtain

$$\begin{aligned} \int _{\varSigma }|A|^{2}H^{2}e^{-f}+\frac{1}{4} \int _{\varSigma }\alpha ^{2}(1-\alpha ^{2})e^{-f}=0. \end{aligned}$$

So

$$\begin{aligned} \alpha ^{2}(1-\alpha ^{2})=0. \end{aligned}$$

This implies that on $\varSigma$,

$$\begin{aligned} \alpha \equiv 0\qquad \mathrm{or}\qquad \alpha ^2\equiv 1. \end{aligned}$$

Since $\varSigma$ is closed, $\alpha ^2\equiv 1$. Without loss of generality, we choose $\alpha \equiv 1$. So $\varSigma$ is in a horizontal slice $\mathbb {S}^n(\sqrt{2(n-1)})\times \{t\}$. By Lemma 1, we conclude that $\varSigma$ is $\mathbb {S}^{n}(\sqrt{2(n-1)})\times \{0\}$.$\square$