FIRST EIGENVALUE CHARACTERISATION OF CLIFFORD HYPERSURFACES AND VERONESE SURFACES

Abstract

We give a sharp estimate for the first eigenvalue of the Schrödinger operator $L:=-\Delta -\sigma $ which is defined on the closed minimal submanifold $M^{n}$ in the unit sphere $\mathbb {S}^{n+m}$, where $\sigma $ is the square norm of the second fundamental form.

1 Introduction

The study of rigidity theorems plays an important role in the theory of minimal submanifolds. There has been extensive research on rigidity theorems for minimal submanifolds in spheres since the pioneering results obtained by Simons [9], Lawson [3] and Chern et al. [2]. Let $\sigma $ denote the square norm of the second fundamental form and let $M^n$ be a compact minimal submanifold in a unit sphere $\mathbb {S}^{n+m}$ . From this work, if $0\leq \sigma \leq {n}/{(2-{1}/{m})}$ , then either $\sigma =0$ or $\sigma ={n}/{(2-{1}/{m})}$ , and M is the Clifford hypersurface or the Veronese surface in $\mathbb {S}^4$ . Later, Li [4] and Chen and Xu [1] improved the pinching number ${n}/{(2-{1}/{m})}$ to ${2n}/{3}$ . They showed that if $0\leq \sigma \leq {2n}/{3}$ , then either $\sigma =0$ or $\sigma ={2n}/{3}$ , and M is the Veronese surface in $\mathbb {S}^4$ . Recently, Lu generalised this result and proved the following rigidity theorem. Here, $\lambda _2$ denotes the second largest eigenvalue of the fundamental matrix (see Definition 2.5).

Theorem 1.1 (Lu [5]).

Let $ 0\leq \sigma +\lambda _2\leq n. $ Then either M is totally geodesic or is one of the Clifford hypersurfaces $M_{r,n-r}$ ( $1\leq r\leq n$ ) in $\mathbb {S}^{n+m}, m\geq 1$ , or a Veronese surface in $\mathbb {S}^{2+m}, m \geq 2 $ .

Remark 1.2. Lu suggests that the quantity $\sigma +\lambda _2$ might be the right object for studying pinching theorems.

Using Lu’s inequality [5, Lemma 2] (see Lemma 2.4), we investigate the first eigenvalue of the Schrödinger operator $L:=-\Delta +q$ , where $\,q $ is a continuous function on M. If there is a nonzero $f\in C^{\infty }(M)$ satisfying $Lf=\mu f$ , we call $\mu $ an eigenvalue of L. Since $\Delta $ is elliptic, so is L and the set of eigenvalues can be written as

$$ \begin{align*} \operatorname{\mathrm{Spec}}(L)=\{\mu_{i} : \mu_{1}<\mu_{2}\leq\mu_{3}\leq\cdots\}. \end{align*} $$

We call $\mu _{1}$ the first eigenvalue of L.

The pinching theorems cited above give a characterisation of Clifford hypersurfaces and Veronese surfaces. The proofs make use of Simons’ identity. Similar arguments lead to estimates of the first eigenvalue of the Schrödinger operator, which gives another way of characterising Clifford hypersurfaces and Veronese surfaces. Simons [9] studied the Schrödinger operator $L_{I}:= -\Delta -\sigma $ of minimal hypersurfaces ${M^{n} \to \mathbb {S}^{n+1}} $ and proved that its first eigenvalue $\mu _{1}^{I}\leq -n$ if M is not totally geodesic. Later, Wu [10] and Perdomo [7] independently proved that if $\mu _{1}^{I}\geq -n$ , then M is either totally geodesic or a Clifford hypersurface. Define $L_{II}:= -\Delta -(2-{1}/{m})\sigma $ on the minimal submanifold $M^{n} \to \mathbb {S}^{n+m} $ and $L_{III}:= -\Delta -\tfrac 32\sigma $ on the minimal submanifold $M^{n} \to \mathbb {S}^{n+m}, m\geq 2 $ , and denote by $\mu _{1}^{II}$ and $\mu _{1}^{III}$ their respective first eigenvalues. For $L_{II}$ , Wu [10] proved that $\mu _{1}^{II}\leq -n$ if M is not totally geodesic, and if $\mu _{1}^{II}\geq -n$ , then M is either totally geodesic, or $\mu _{1}^{II}=-n$ and M is either a Clifford hypersurface or a Veronese surface. Also, for $L_{III}$ , $\mu _{1}^{III}\leq -n$ if M is not totally geodesic, and if $\mu _{1}^{III}\geq -n$ , then M is either totally geodesic, or $\mu _{1}^{III}=-n$ and M is a Veronese surface. Similar results hold in the Legendrian case. Using a pinching rigidity result in [6], Yin and Qi [11] gave a sharp estimate for the first eigenvalue of the Schrödinger operator defined on a minimal Legendrian submanifold $M^{3} \to \mathbb {S}^{7} $ .

Based on the correspondence between pinching theorems and estimates of the first eigenvalue of certain Schrödinger operators, one expects to find the same phenomenon for Lu’s rigidity Theorem 1.1. That observation leads to our main theorem. Define the Schrödinger operator $L:=-\Delta -\sigma $ and denote the first eigenvalue of L by $\mu _{1}$ .

Theorem 1.3 (Main Theorem).

Let $M^n$ be a closed minimal submanifold in $\mathbb {S}^{n+m}(1)$ . If M is not totally geodesic, then

$$ \begin{align*} \mu_{1} \leq -n + \max_{p\in M}\lambda_{2}. \end{align*} $$

Moreover, if $\mu _{1} \geq -n+\max _{p\in M}\lambda _{2}$ , then either $\mu _{1}=0$ and M is totally geodesic, or $\mu _{1}=-n+\max _{p\in M}\lambda _{2}$ and M is the Clifford hypersurface in $\mathbb {S}^{n+m}(1)$ or the Veronese surface in $\mathbb {S}^{2+m}(1)$ .

2 Preliminaries and Lu’s inequality

Let $M^n$ be a compact minimal submanifold in a unit sphere $\mathbb {S}^{n+m}$ . We shall make use of the following convention on the range of indices:

$$ \begin{align*} 1\leq A,B,C,\ldots\leq n+m;\quad 1\leq i,j,k,\ldots\leq n;\quad n+1\leq \alpha,\beta,\gamma,\ldots\leq n+m. \end{align*} $$

We choose a local field of orthonormal frames $\{e_{1},e_{2},\dots ,e_{n+m}\}$ in $\mathbb {S}^{n+m}$ such that when restricted to M, $\{e_{1},e_{2},\dots ,e_{n}\}$ are tangent to M and $\{e_{n+1},e_{n+2},\dots ,e_{n+m}\}$ are normal to M. Also, $\{\omega _{1},\dots ,\omega _{n+m}\}$ is the corresponding dual frame. It is well known that

(2.1) $$ \begin{align} \omega_{\alpha i} =\sum_{j} h^{\alpha}_{ij}\omega_{j}, \quad h^{\alpha}_{ij} & =h^{\alpha}_{ji}, \quad h =\sum_{\alpha,i,j}h^{\alpha}_{ij}\omega_{i}\otimes\omega_{j}\otimes e_{\alpha},\quad H=\frac{1}{n}\sum_{\alpha,i}h^{\alpha}_{ii}e_{\alpha},\nonumber \\ R_{ijkl} & =\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}+ \sum_{\alpha}(h^{\alpha}_{ik}h^{\alpha}_{jl}-h^{\alpha}_{il}h^{\alpha}_{jk}), \end{align} $$

(2.2) $$ \begin{align} \ \ \kern-0.1ptR_{\alpha\beta kl} & =\sum_{i}(h^{\alpha}_{ik}h^{\beta}_{il}-h^{\alpha}_{il}h^{\beta}_{ik}),\qquad\qquad\qquad \end{align} $$

(2.3) $$ \begin{align} \kern-1.7pt\!\!\!h^{\alpha}_{ijk} & =h^{\alpha}_{ikj}, \qquad\qquad\qquad\qquad\qquad\end{align} $$

where $h, H, R_{ijkl}, R_{\alpha \beta kl},$ are respectively the second fundamental form, the mean curvature vector, the curvature tensor and the normal curvature tensor of M. We define

$$ \begin{align*} \sigma=|h|^{2}, \quad A_{\alpha}=(h^{\alpha}_{ij})_{n\times n}. \end{align*} $$

Denote by $h_{ijk}^\alpha $ the component of the covariant derivative of $h_{ij}^\alpha $ , defined by

(2.4) $$ \begin{align} h_{ijk}^\alpha\omega_k=dh_{ij}^\alpha- \sum_{l}h_{il}^\alpha\omega_{lj}-\sum_{l}h_{lj}^\alpha\omega_{li}+ \sum_{\beta}h_{ij}^\beta\omega_{\alpha\beta}. \end{align} $$

From the Gauss–Codazzi–Ricci equations (2.1)–(2.3), the well-known Simons identity follows:

(2.5) $$ \begin{align} \sum_{i,j}h_{ij}^{\alpha}\Delta h_{ij}^{\alpha}=n\,|A_{\alpha}|^{2} +\sum_{\beta}\mathrm{ Tr}(A_{\alpha}A_{\beta}-A_{\beta}A_{\alpha})^2-\sum_{\beta}(\mathrm{Tr}\,A_{\alpha}A_{\beta})^2. \end{align} $$

Now, we introduce Lu’s inequality [5, Lemma 2] (see Lemma 2.4), which is the main tool in the proof of Theorems 1.1 and 1.3. The proof of Lu’s inequality relies on an algebraic inequality [5, Lemma 1]. We use the Lagrange multiplier method to give another proof and find that there are more cases when the equality holds. Consequently, we restate Lu’s lemma [5, Lemma 1] as the following lemma.

Lemma 2.1. Suppose $\eta _1,\ldots ,\eta _n$ are real numbers, $\eta _1+\cdots +\eta _n=0$ and $\eta _1^2+\cdots +\eta _n^2=1$ . Let $r_{ij}\geq 0$ be nonnegative numbers for $i<j$ . Then

(2.6) $$ \begin{align} \sum_{i<j}(\eta_i-\eta_j)^2 r_{ij}\leq\sum_{i<j}r_{ij}+\max(r_{ij}). \end{align} $$

If $\eta _1\geq \cdots \geq \eta _n$ and $r_{ij}$ are not simultaneously zero, then equality holds in (2.6) only in one of the following cases. Fix an integer k with $k \in \{1,\dots ,n-1\}$ .

1. (1) $r_{ij}=0$ if $2\leq i<j$ , $r_{12}=\dots =r_{1k}=0, r_{1\,k+1}=\dots =r_{1n}>0$ ,

   $$ \begin{align*}\eta_{1}=\frac{\sqrt{n-k}}{\sqrt{n-k+1}},\quad \eta_{2}=\dots=\eta_{k}=0,\quad \eta_{k+1}=\dots=\eta_{n}=\frac{-1}{\sqrt{(n-k+1)(n-k)}}.\end{align*} $$

2. (2) $r_{ij}=0$ if $i<j<n$ , $r_{n-1\,n}=\dots =r_{n-k+1\,n}=0, r_{n-k\,n}=\dots =r_{1n}>0$

   $$ \begin{align*} \eta_{n}=\frac{-\sqrt{n-k}}{\sqrt{n-k+1}},\quad & \eta_{n-1}=\eta_{n-2}\dots=\eta_{n-k+1}=0, \\ & \eta_{n-k}=\dots=\eta_{1}=\frac{1}{\sqrt{(n-k+1)(n-k)}}. \end{align*} $$

Remark 2.2. We prove the lemma in two steps. The first step is the same as Lu’s original proof of [5, Lemma 1] which reduces the problem to proving the inequality

$$ \begin{align*} \sum_{1<j}(\eta_1-\eta_j)^2 r_{1j}\leq\sum_{1<j} r_{1j}+\max_{1<j} (r_{1j}). \end{align*} $$

Then, we apply the Lagrange multiplier method to prove this inequality.

Proof. First step. Assume $ \eta _1\geq \cdots \geq \eta _n. $ If $\eta _1-\eta _n\leq 1$ or $n=2$ , then (2.6) is trivial. So assume $n>2$ and $ \eta _1-\eta _n>1. $ Observe that $\eta _i-\eta _j< 1$ for $2\leq i<j\leq n-1$ . Otherwise,

$$ \begin{align*} 1\geq\eta_1^2+\eta_n^2+\eta_i^2+\eta_j^2\geq\tfrac 12 ((\eta_1-\eta_n)^2+(\eta_i-\eta_j)^2)>1, \end{align*} $$

which is a contradiction.

Using the same reasoning, if $\eta _1-\eta _{n-1}>1$ , then $\eta _2-\eta _{n}\leq 1$ ; and if $\eta _2-\eta _{n}>1$ , then $\eta _1-\eta _{n-1}\leq 1$ . Replacing $\eta _1,\ldots ,\eta _n$ by $-\eta _n,\ldots ,-\eta _1$ if necessary, we can always assume that $\eta _2-\eta _{n}\leq 1$ . Thus, $\eta _i-\eta _j\leq 1$ if $2\leq i<j$ , and (2.6) is implied by the inequality

(2.7) $$ \begin{align} \sum_{1<j}(\eta_1-\eta_j)^2 r_{1j}\leq\sum_{1<j} r_{1j}+\max_{1<j} (r_{1j}). \end{align} $$

Before proving (2.7), we observe that if equality holds in (2.6), we must have ${\eta _1-\eta _n>1}$ . Otherwise,

$$ \begin{align*} \sum_{i<j}(\eta_i-\eta_j)^2 r_{ij}\leq\sum_{i<j}r_{ij}<\sum_{i<j}r_{ij}+\max (r_{ij}), \end{align*} $$

which is a contradiction.

Notice that when $\eta _1-\eta _n>1$ , by the discussion above, $\eta _i-\eta _j<1$ for $2\leq i<j<n$ . So, $r_{ij}=0$ for $2\leq i<j<n$ . Otherwise, by (2.7), equality cannot hold in (2.6). Thus, when discussing equality in (2.6), we only need to analyse the inequality (2.7).

Second step. Let $s_j=r_{1j}$ , where $j=2,\ldots ,n$ . We write (2.7) as

$$ \begin{align*} \sum_{1<j}(\eta_1-\eta_j)^2 s_{j}\leq\sum_{1<j} s_{j}+\max_{1<j} (s_{j}). \end{align*} $$

Write

$$ \begin{align*}f(\eta_1,\eta_2,\dots,\eta_{n-1},\eta_{n})= \sum_{1<j}(\eta_1-\eta_j)^2 s_{j}.\end{align*} $$

We apply the Lagrange multiplier method to f with constraints

$$ \begin{align*} \eta_1+\cdots+\eta_n=0,\quad \eta_1^2+\cdots+\eta_n^2-1=0. \end{align*} $$

Consider the function

$$ \begin{align*}\Phi(\eta_1,\eta_2,\dots,\eta_{n-1},\eta_{n})= \sum_{1<j}(\eta_1-\eta_j)^2 s_{j}+\lambda\,(\eta_1+\cdots+\eta_n)+ \mu\,(\eta_1^2+\cdots+\eta_n^2-1), \end{align*} $$

where $\lambda $ and $\mu $ are the Lagrange multipliers. Setting the partial derivatives with respect to each variable to zero gives the equations

(2.8) $$ \begin{align} \frac{\partial \Phi}{\partial \eta_1}&= \sum_{1<j}2(\eta_1-\eta_j) s_{j}+\lambda+2\mu\,\eta_1=0,\nonumber\\ \frac{\partial \Phi}{\partial \eta_j}&= -2(\eta_1-\eta_j) s_{j}+\lambda+2\mu\,\eta_{j}=0\quad \text{for } j=2,\dots,n-1,n. \end{align} $$

Now

$$ \begin{align*} \sum_{i=1}^n \frac{\partial \Phi}{\partial \eta_i}=n\lambda=0, \quad \sum_{i=1}^n \eta_{i}\frac{\partial \Phi}{\partial \eta_i}= 2\sum_{1<j}(\eta_1-\eta_j)^2 s_{j}+2\mu=0,\end{align*} $$

and so

$$ \begin{align*}\lambda=0, \quad \sum_{1<j}(\eta_1-\eta_j)^2 s_{j}=-\mu.\end{align*} $$

Hence, the critical values of f are given by $-\mu $ .

Assume $-\mu \neq 0$ . We can also assume that $\mu + \max _{1<j} (s_{j})< 0$ . Otherwise,

$$ \begin{align*} -\mu=\sum_{1<j}(\eta_1-\eta_j)^2 s_{j}\leq \max_{1<j} (s_{j})<\sum_{1<j} s_{j}+ \max_{1<j} (s_{j}).\end{align*} $$

Then by (2.8),

(2.9) $$ \begin{align} \eta_{j}=\frac{\eta_{1}\,s_{j}}{\mu+s_j},\quad j=2,\dots,n-1,n. \end{align} $$

Substituting (2.9) into $\eta _1+\cdots +\eta _n=0$ gives

$$ \begin{align*}1+\sum_{1<j}\frac{\,s_{j}}{\mu+s_j}=0.\end{align*} $$

Hence,

(2.10) $$ \begin{align} 0=1+\sum_{1<j}\frac{\,s_{j}}{\mu+s_j}\geq1+\sum_{1<j}\frac{\,s_{j}}{\mu+ \max_{1<i} (s_{i})}. \end{align} $$

Multiplying both sides of (2.10) by $\mu + \max _{1<i} (s_{i})$ gives

(2.11) $$ \begin{align} -\mu=\sum_{1<j}(\eta_1-\eta_j)^2 s_{j}\leq\sum_{1<j} s_{j}+ \max_{1<i} (s_{i}). \end{align} $$

Notice that if $s_{j}>0$ for any j, then

$$ \begin{align*}0>\frac{\,s_{j}}{\mu+s_j}\geq\frac{\,s_{j}}{\mu+ \max_{1<i} (s_{i})}\end{align*} $$

and that equality in (2.11) is equivalent to equality in (2.10). If equality holds in (2.10), then for each $j>1$ ,

$$ \begin{align*}\frac{\,s_{j}}{\mu+s_j}=\frac{\,s_{j}}{\mu+ \max_{1<i} (s_{i})},\end{align*} $$

which means that either $s_{j}=0$ or the nonzero $s_{j}= \max _{1<i} (s_{i})$ and so all nonzero $s_{j}$ are equal.

Thus, from (2.9) and the assumption above, there are $n-1$ cases, one for each k with $k \in \{1,\dots ,n-1\}$ , namely,

$$ \begin{align*}\eta_{1}=\frac{\sqrt{n-k}}{\sqrt{n-k+1}},\quad\eta_{2}=\dots=\eta_{k}=0,\quad \eta_{k+1}=\dots=\eta_{n}=-\frac{1}{\sqrt{(n-k+1)(n-k)}}.\end{align*} $$

Case (2) in the statement of the lemma is just a permutation of Case (1) under a different assumption at the beginning. This completes the proof.

Remark 2.3. The values $k=n-1$ and $k=1$ in Lemma 2.1(1) correspond to [5, Cases (1) and (2) in Lemma 1], respectively.

The new version of [5, Lemma 1] changes [5, Lemma 2], but Lu’s rigidity theorem still holds, as we discuss later.

Define the inner product of two $n\times n$ matrices $A,B$ by $\langle A,B\rangle =\mathrm {Tr}\, AB^{\top }$ and let $||A||^2=\langle A,A\rangle =\sum _{i,j}a_{ij}^{2}$ , where $(a_{ij})$ are the entries of A. The next lemma gives the revised version of Lu’s inequality [5, Lemma 2].

Lemma 2.4. Let $A_{1}$ be an $n\times n$ diagonal matrix of norm $1$ . Let $A_2,\ldots , A_m$ be symmetric matrices such that:

1. (1) $\langle A_\alpha ,A_\beta \rangle =0$ if $\alpha \neq \beta $ ;

2. (2) $||A_2||\geq \cdots \geq ||A_m||$ .

Then,

(2.12) $$ \begin{align} \sum_{\alpha=2}^m||[A_1,A_\alpha]||^2\leq \sum_{\alpha=2}^m ||A_\alpha||^2+||A_2||^2. \end{align} $$

Equality holds in (2.12) if and only if, after an orthonormal base change and up to a sign, and for each integer k with $k \in \{1,\dots ,n-1\}$ , $A_1$ is the diagonal matrix

(2.13) $$ \begin{align} A_1=\mathrm{diag}\bigg( \frac{\sqrt{k}}{\sqrt{k+1}}, -\frac{1}{\sqrt{k(k+1)}}, -\frac{1}{\sqrt{k(k+1)}}, \ldots, -\frac{1}{\sqrt{k(k+1)}}, 0, \ldots, 0\bigg), \end{align} $$

with $k$ entries $-1/\sqrt {k(k+1)}$ and $n-k-1$ entries $0$ , and $A_{i}$ is $\mu $ times the matrix whose only nonzero entries are $1$ at the $(1,i)$ and $(i,1)$ places, where $i=2,\ldots ,k+1$ and $A_{k+2}=\cdots =A_m=0$ .

Next, we briefly review the proof of Lu’s rigidity theorem to set up the notation and state some formulae for later use.

Definition 2.5. The fundamental matrix S of M is an $m\times m$ matrix-valued function defined by $S=(a_{{\alpha }{\beta }})$ , where

$$ \begin{align*} a_{{\alpha}{\beta}}=\langle A_{\alpha}, A_{\beta}\rangle. \end{align*} $$

We denote the eigenvalues of the fundamental matrix S by $\lambda _1\geq \cdots \geq \lambda _m$ . In particular, $\lambda _1$ is the largest eigenvalue and $\lambda _2$ is the second largest eigenvalue of the matrix S, and r is defined by

$$ \begin{align*} \lambda_1=\cdots=\lambda_r>\lambda_{r+1}\geq\cdots\geq\lambda_m. \end{align*} $$

Using this notation, the trace of the fundamental matrix is $\sigma =\lambda _1+\cdots +\lambda _m$ . For a positive integer $p\geq 2$ , we define

$$ \begin{align*} f_p:=\mathrm{Tr}\,(S^p)=\sum_{{\alpha_1},\dots,{\alpha_p}} a_{{\alpha_1 }{\alpha_2 }} \, a_{{\alpha_2 }{\alpha_3 }}\cdots \, a_{{\alpha_p }{\alpha_1 }} \end{align*} $$

and $g_p:=(f_p)^{{1}/{p}}$ . Using the Simons identity (2.5) and Lemma 2.4, Lu derived the following inequalities.

Proposition 2.6 (Lu [5]).

With the notation as above,

(2.14) $$ \begin{align} \!|\nabla f_p|^2&\leq p^2 f_p \sum_{k,\alpha}\lambda_\alpha^{p-2}(\nabla_{{\partial}/{\partial x_k}} a_{\alpha\alpha})^2,\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{align} $$

(2.15) $$ \begin{align} \Delta g_p&=\frac 1p f_p^{1/p -1}\Delta f_p+\frac 1p \bigg(\frac 1p-1\bigg) f_p^{1/p -2}|\nabla f_p|^2 \notag\\ &\geq 2f_p^{1/p-1}\sum_\alpha\bigg(\lambda_\alpha^{p-1}\sum_{i,j,k}(h_{ijk}^\alpha)^2\bigg) \notag\\& \quad+2f_p^{1/p-1}\bigg( r||A_{1}||^{2p}\bigg(n-||A_{1}||^2-\sum_{\alpha=2}^m||A_{\alpha}||^2-\lambda_2\bigg)-3mn\lambda_{r+1}^{p}\bigg). \end{align} $$

By integrating both sides of (2.15) and letting $p \to \infty $ , since $\lambda _{r+1}^p/f_p\to 0$ as p tends to $\infty $ , Lu derived

(2.16) $$ \begin{align} \int_M\sum_{i,j,k}\sum_{\alpha\leq r}(h^\alpha_{ijk})^2+||A_1||^2\bigg(n-||A_1||^2-\sum_{\alpha=2}^m||A_\alpha||^2-\lambda_2\bigg) \leq 0. \end{align} $$

If equality holds in (2.16), then equality holds in (2.12), so $A_{\alpha }$ takes the form in Lemma 2.4. Using the structure equation case by case, Lu proved Theorem 1.1.

Remark 2.7. Although we have found more cases when equality holds in (2.12), we can rule out the new cases using similar arguments to those in the original proof. To be precise, if $n>k+1, j\geq k+2$ , then from $0=d h_{1j}^{n+1}=h_{11}^{n+1}\omega _{1j}$ , we conclude $\omega _{1j}=0$ . Similarly, by computing $d h_{ij}^{n+1}$ for $i = 2,\dots, k+1$ , we also have $\omega _{2j}=\dots =\omega _{k+1\,j}=0$ for $j\geq k+2$ . By the structure equations, $ 0=d\omega _{1j}=\omega _1\wedge \omega _j, $ which is a contradiction if $n>k+1$ . Thus, Theorem 1.1 is still correct.

3 Proof of the main theorem

Let $g_{\epsilon }=(g_{p}+\epsilon )^{{1}/{2}}$ , where $\epsilon>0$ is a constant. We first prove the inequality in the main theorem.

Proposition 3.1. If $M^n$ is a closed nontotally geodesic minimal submanifold in $\mathbb {S}^{n+m}(1)$ , then

$$ \begin{align*} \mu_{1} \leq -n + \max_{p\in M}\lambda_{2} -\frac{2}{n+2}\frac{\int_{M} [\frac{1}{r}\,\sum_{i,j,k}\sum_{{\alpha}\leq n+r}(h^{\alpha}_{ijk})^2]}{\int_{M}\lambda_{1}}. \end{align*} $$

Proof. By direct computation, using (2.15),

$$ \begin{align*} \begin{aligned} \Delta g_{\epsilon}&= \frac{1}{2}\,(g_{p}+\epsilon)^{-{1}/{2}}\,\Delta g_{p}-\frac{1}{4}\,|\nabla g_{p}|^{2}\,(g_{p}+\epsilon)^{-{3}/{2}}\\ &=\frac{1}{2}\,(g_{p}+\epsilon)^{-{1}/{2}}\bigg(\frac 1p f_p^{1/p -1}\Delta f_p+\frac 1p \bigg(\frac 1p-1\bigg) f_p^{1/p -2}|\nabla f_p|^2\bigg)-\frac{1}{4}\,|\nabla g_{p}|^{2}\,(g_{p}+\epsilon)^{-{3}/{2}}\\ &\geq\frac{1}{2}\,(g_{p}+\epsilon)^{-{1}/{2}} \bigg( 2f_p^{1/p-1}\sum_{\alpha}\bigg(\lambda_{\alpha}^{p-1}\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg) \\ &\quad +2f_p^{1/p-1}\bigg( r||A_{1}||^{2p}\bigg(n+1-||A_{1}||^2-\sum_{{\alpha}=2}^m||A_{\alpha}||^2-\lambda_2\bigg)\bigg) -6nm \,f_p^{1/p-1}\lambda_{r+1}^{p}\bigg) \\ &\quad -\frac{1}{4}\,|\nabla g_{p}|^{2}\,(g_{p}+\epsilon)^{-{3}/{2}}\\ &\geq\underbrace{(g_{p}+\epsilon)^{-{3}/{2}}\bigg[(g_{p}+\epsilon)\,f_p^{1/p-1}\sum_{\alpha}\bigg(\lambda_{\alpha}^{p-1}\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg)-\frac{1}{4}\,|\nabla g_{p}|^{2} \bigg]}_{I}\\ &\quad +\underbrace{(g_{p}+\epsilon)^{-{1}/{2}}\bigg[f_p^{1/p-1}\bigg( r||A_{1}||^{2p}\bigg(n-||A_{1}||^2-\sum_{{\alpha}=2}^m||A_{\alpha}||^2-\lambda_2\bigg)\bigg) -3nm\, f_p^{1/p-1}\lambda_{r+1}^{p}\bigg]}_{II}. \end{aligned} \end{align*} $$

To deal with I, we use the next lemma which follows from [8, (1.9) and (1.11) in Proposition 1].

Lemma 3.2 (Shen [8]).

If $M^n$ is a closed minimal submanifold in $\mathbb {S}^{n+m}(1)$ , then

$$ \begin{align*} |\nabla(|A_{\alpha}|^{2})|^{2} \leq \frac{4n}{n+2}|A_{\alpha}|^{2}\, \bigg[\sum_{i,j,k}(h_{ijk}^{\alpha})^{2}\bigg]. \end{align*} $$

Applying Lemma 3.2 to (2.14) yields

$$ \begin{align*} |\nabla g_{p}|^{2}=\frac{1}{p^{2}}f_p^{{2}/{p}-2}|\nabla f_{p}|^{2}\leq f_p^{{2}/{p}-1}\sum_{\alpha}\lambda_{\alpha}^{p-2}|\nabla \lambda_{\alpha}|^{2} \leq f_p^{{2}/{p}-1}\sum_{\alpha}\,\frac{4n}{n+2}\bigg(\lambda_{\alpha}^{p-1}\,\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg). \end{align*} $$

Thus,

$$ \begin{align*} I & \geq (g_{p}+\epsilon)^{-{3}/{2}}\bigg[(g_{p}+\epsilon)\,f_p^{1/p-1}\sum_{\alpha}\bigg(\lambda_{\alpha}^{p-1}\,\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg)-\frac{1}{4}\frac{1}{p^{2}} f_p^{{2}/{p}-2}|\nabla f_{p}|^{2} \bigg]\\ & \geq(g_{p}+\epsilon)^{-{3}/{2}}\bigg[(g_{p}+\epsilon)f_p^{1/p-1}\sum_{\alpha}\bigg(\lambda_{\alpha}^{p-1}\,\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg) \\ &\quad -\frac{1}{4}f_p^{{2}/{p}-1} \frac{4n}{n+2}\sum_{\alpha}\bigg(\lambda_{\alpha}^{p-1}\,\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg)\bigg]\\ & \geq\frac{2}{n+2}(g_{p}+\epsilon)^{-{1}/{2}}f_p^{{1}/{p}-1} \bigg[\sum_{\alpha}\bigg(\lambda_{\alpha}^{p-1}\,\sum_{i,j,k}(h_{ijk}^{\alpha})^2\bigg)\bigg]\\ & \geq 0. \end{align*} $$

Inserting the definition of $g_{\epsilon }$ into $\mu _{1}=\inf _{f\in C^{\infty }(M)}{\int _{M}L(f)f}/{\int _{M}f^{2}}$ , yields

$$ \begin{align*} \mu_{1} {\int_{M}g_{\epsilon}^{2}} & \leq {\int_{M}L(g_{\epsilon})g_{\epsilon}}={\int_{M}-g_{\epsilon}\Delta \,g_{\epsilon} -\sigma\,g_{\epsilon}^{2}} = {\int_{M}-g_{\epsilon}(I+II) -\sigma\,g_{\epsilon}^{2}} \\ & \leq {\int_{M}-\frac{2}{n+2}f_p^{{1}/{p}-1} \bigg[\sum\limits_{\alpha} \bigg(\lambda_{\alpha}^{p-1}\,\sum\limits_{i,j,k}(h_{ijk}^{\alpha})^2\bigg)\bigg]} \\ &\quad +{\int_{M}-\bigg[f_p^{1/p-1}\bigg( r||A_{1}||^{2p}\bigg(n-||A_{1}||^2-\sum\limits_{{\alpha}=2}^{m}||A_{\alpha}||^2-\lambda_2\bigg)\bigg) -3nm f_p^{1/p-1}\lambda_{r+1}^{p}\bigg]} \\ &\quad -{\int_{M}\sigma\,g_{\epsilon}^{2}}. \end{align*} $$

Then, letting $p \to \infty $ and $\epsilon \to 0$ , and using the fact that $\lambda _{r+1}^p/f_p\to 0$ almost everywhere when $p\to \infty $ completes the proof.

Proof of Theorem 1.3.

From the proof of Proposition 3.1, if

$$ \begin{align*} \mu_{1}\geq-n+ \max_{p\in M}\lambda_{2}, \end{align*} $$

then either M is totally geodesic so $\mu _{1}=0$ or $ \mu _{1}=-n+ \max _{p\in M}\lambda _{2}$ and

$$ \begin{align*} \frac{1}{r}\,\sum\limits_{i,j,k}\sum_{{\alpha}\leq n+r}(h^{\alpha}_{ijk})^2=0. \end{align*} $$

We claim that $\sigma $ is a constant. By Lemma 2.4, there are two cases.

Case 1. $A_{1} \neq 0$ and $A_{2}=A_{3}=\dots =A_{m}=0$ . By Lemma 3.2, $\sigma =||A_{1}||^{2}=\lambda _{1}$ is a constant.

Case 2. There is a positive integer k with $1\leq k\leq n-1$ such that $A_1$ is $\lambda $ times the diagonal matrix in (2.13), $A_i$ is $\mu /\sqrt {k(k+1)}$ times the matrix whose only nonzero entries are $1$ at the $(1,i)$ and $(i,1)$ places for $2\le i\le k+1$ , and $A_{k+2}=\dots =A_{m}=0$ . Since $\sum _{i,j,k}(h^{n+1}_{ijk})^2=0$ , by (2.4), it follows that $\lambda $ is constant. Also, $\mu $ is constant since $\lambda _{2}=\max _{p\in M}\lambda _{2}$ . Thus, $\sigma $ is constant.

Since $\sigma $ is constant when $\mu _{1}=-n+ \max _{p\in M}\lambda _{2}$ and the first eigenvalue of L is $-\sigma $ , it follows that $\sigma +\lambda _{2}=n$ . Then, by Theorem 1.1, M is either one of the Clifford hypersurfaces or the Veronese surface.