An optimal pinching theorem of minimal Legendrian submanifolds in the unit sphere

Abstract

In this paper, we study the rigidity theorem of closed minimally immersed Legendrian submanifolds in the unit sphere. Utilizing the maximum principle, we obtain a new characterization of the Calabi torus in the unit sphere which is the minimal Calabi product Legendrian immersion of a point and the totally geodesic Legendrian sphere. We also establish an optimal Simons’ type integral inequality in terms of the second fundamental form of three-dimensional closed minimal Legendrian submanifolds in the unit sphere. Our optimal rigidity results for minimal Legendrian submanifolds in the unit sphere are new and also can be applied to minimal Lagrangian submanifolds in the complex projective space.

Appendix A. An application to lagrangian submanifolds in the nearly Kähler \(\mathbb {S}^6\)

Here we give a slight improvement of the main theorem in [15] as follows, by similar arguments used in the proof of Theorem 3.1.

Theorem Appendix A.1

Let M be a closed Lagrangian submanifold in the homogeneous nearly Kähler $\mathbb {S}^6$. Then we have

$$\begin{aligned} \int _{M}\left|\mathbf {B}\right|^2\left( \left|\mathbf {B}\right|^2-\frac{75}{56}-\frac{10}{7}\Theta ^2\right) \ge 0. \end{aligned}$$

(4.5)

Moreover, the equality in (4.5) holds if and only if M is either the totally geodesic sphere, or the Dillen-Verstraelen-Vrancken’s Berger sphere (see [9, Theorem 5.1]) which satisfies $\left|\mathbf {B}\right|^2=\frac{75}{56}+\frac{10}{7}\Theta ^2$ with $\left|\mathbf{B}\right|^2\equiv \frac{25}{8}$ and $\Theta \equiv \frac{\sqrt{5}}{2}$.

Proof

Here we only give a brief sketch. For more details please see [15]. We identify $\mathbb {R}^7$ as the imaginary Cayley numbers. The Cayley multiplication induces a cross product $``\times "$ on $\mathbb {R}^7$. The almost complex structure J on $\mathbb {S}^6\subset \mathbb {R}^7$ is then given by

$$\begin{aligned} JX{:}{=}x\times X,\quad \forall X\in T_x\mathbb {S}^6. \end{aligned}$$

Let $\bar{\nabla }$ be the Levi-Civita connection on $\mathbb {S}^6$, then $\left( \bar{\nabla }_XJ\right) X=0$ for all $X\in T\mathbb {S}^6$. Then $\omega _{ijk}=\left\langle \left( \bar{\nabla }_{e_i}J\right) e_j,Je_k\right\rangle$ is the volume form of M. Since M is Lagrangian, i.e., $JTM\subset T^{\bot }M$, we have ([26, Lemma 3.2])

$$\begin{aligned} \mathbf {B}\left( e_i,\left( \bar{\nabla }_{e_j}J\right) e_k\right) =J\left( \bar{\nabla }_{\mathbf {B}\left( e_i,e_j\right) }J\right) e_k+J\left( \bar{\nabla }_{e_j}J\right) \mathbf {B}\left( e_i,e_k\right) , \end{aligned}$$

which implies that M is minimal (cf. [12, Theorem 1]). We have the following Simons’ identity (cf. [7])

$$\begin{aligned} \dfrac{1}{2}\Delta \left|\mathbf {B}\right|^2=\left|\nabla ^{\bot }\mathbf {B}\right|^2+3\left|\mathbf {B}\right|^2-\sum _{\alpha ,\beta =1}^3\left\langle \mathbf {A}^{\nu _{\alpha }},\mathbf {A}^{\nu _{\beta }}\right\rangle ^2-\sum _{\alpha ,\beta =1}^3\left|\left[ \mathbf {A}^{\nu _{\alpha }},\mathbf {A}^{\nu _{\beta }}\right] \right|^2. \end{aligned}$$

Here $\left\{ \nu _{\alpha }\right\}$ is a local orthonormal frame of $T^{\bot }M$. Set

$$\begin{aligned} \sigma _{ijk}=\left\langle \mathbf {B}\left( e_i,e_j\right) ,Je_k\right\rangle , \end{aligned}$$

then $\sigma$ is a tri-linear symmetric tensor. One can check that

$$\begin{aligned} \sigma _{ijk,l}=\left\langle \left( \nabla _{e_i}^{\bot }\mathbf {B}\right) \left( e_j,e_k\right) ,Je_l\right\rangle . \end{aligned}$$

Introduce

$$\begin{aligned} u_{ijkl}{:}{=}&\dfrac{1}{4}\left( \sigma _{ijk,l}+\sigma _{jkl,i}+\sigma _{kli,j}+\sigma _{lij,k}\right) \\ =&\sigma _{ijk,l}-\dfrac{1}{4}\left( \sigma _{jkm}\omega _{lim}+\sigma _{ikm}\omega _{ljm}+\sigma _{ijm}\omega _{lkm}\right) . \end{aligned}$$

One can check that u is a four-linear symmetric tensor and $\sum _{i}u_{iij,k}=0$. By using the fact $\sigma _{ijk,l}=\sigma _{ijl,k}+\sigma _{ijm}\omega _{lkm}$, a direct calculation yields (cf. [15, emma 4.4])

$$\begin{aligned} \left|\nabla ^{\bot }\mathbf {B}\right|^2=\left|u\right|^2+\dfrac{3}{4}\left|\mathbf {B}\right|^2. \end{aligned}$$

We therefore obtain

$$\begin{aligned} \dfrac{1}{2}\Delta \left|\sigma \right|^2=&\left|u\right|^2+\dfrac{15}{4}\left|\sigma \right|^2-\sum _{i,j=1}^3\left\langle \sigma _i,\sigma _j\right\rangle ^2-\sum _{i,j=1}^3\left|[\sigma _i,\sigma _j]\right|^2, \end{aligned}$$

where $\sigma _i=\left( \sigma _{ijk}\right) _{1\le j,k\le n}$. Then, similarly as in the proof of Theorem 3.1, we obtain

$$\begin{aligned} \frac{1}{2}\Delta \left|\mathbf {B}\right|^2\ge \frac{14}{5}\left( \frac{75}{56}+\dfrac{10}{7}\Theta ^2-\left|\mathbf {B}\right|^2\right) \left|\mathbf {B}\right|^2. \end{aligned}$$

The rest of the proof follows from that in [15]. $\square$