Complete Lagrangian self-shrinkers in ${\mathbf {R}}^4$

Abstract

The purpose of this paper is to study complete self-shrinkers of mean curvature flow in Euclidean spaces. In the paper, we give a complete classification for 2-dimensional complete Lagrangian self-shrinkers in Euclidean space ${\mathbb {R}}^4$ with constant squared norm of the second fundamental form.

1 Introduction

Let $X: M\rightarrow {\mathbb {R}}^{n+p}$ be an n-dimensional submanifold in the ($n+p$)-dimensional Euclidean space ${\mathbb {R}}^{n+p}$. A family of n-dimensional submanifolds $X(\cdot , t):M\rightarrow {\mathbb {R}}^{n+p}$ is called a mean curvature flow if they satisfy $X(\cdot , 0)=X(\cdot )$ and

$$\begin{aligned} \dfrac{\partial X(p,t)}{\partial t}=\vec H(p,t), \quad (p,t)\in M\times [0,T), \end{aligned}$$

(1.1)

where $\vec H(p,t)$ denotes the mean curvature vector of submanifold $M_t=X(M,t)$ at point X(p, t). The mean curvature flow has been used to model various things in material sciences and physics such as cell, bubble growth and so on. The study of the mean curvature flow from the perspective of partial differential equations commenced with Huisken’s paper [16] on the flow of convex hypersurfaces. One of the most important problems in the mean curvature flow is to understand the possible singularities that the flow goes through. A key starting point for singularity analysis is Huisken’s monotonicity formula, the monotonicity implies that the flow is asymptotically self-similar near a given type I singularity. Thus, it is modeled by self-shrinking solutions of the flow.

An n-dimensional submanifold $X: M\rightarrow {\mathbb {R}}^{n+p}$ in the $(n+p)$-dimensional Euclidean space ${\mathbb {R}}^{n+p}$ is called a self-shrinker if it satisfies

$$\begin{aligned} \vec H+ X^{\perp }=0, \end{aligned}$$

where $X^{\perp }$ denotes the normal part of the position vector X. It is known that self-shrinkers play an important role in the study of the mean curvature flow because they describe all possible blow-ups at a given singularity of the mean curvature flow.

For complete self-shrinkers with co-dimension 1, Abresch and Langer [1] classified closed self-shrinker curves in ${\mathbb {R}}^2$ and showed that the round circle is the only embedded self-shrinker. Huisken [15, 17], Colding and Minicozzi [11] have proved that if $X: M\rightarrow {\mathbb {R}}^{n+1}$ is an n-dimensional complete embedded self-shrinker in ${\mathbb {R}}^{n+1}$ with mean curvature $H\ge 0$ and with polynomial volume growth, then $X: M\rightarrow {\mathbb {R}}^{n+1}$ is isometric to ${\mathbb {R}}^{n}$, or the round sphere $S^{n}(\sqrt{n})$, or a cylinder $S^m (\sqrt{m})\times {\mathbb {R}}^{n-m}$, $1\le m\le n-1$. Halldorsson in [14] proved that there exist complete self-shrinking curves $\Gamma$ in ${\mathbb {R}}^2$, which is contained in an annulus around the origin and whose image is dense in the annulus. Furthermore, Ding and Xin [12], Cheng and Zhou [10] proved that a complete self-shrinker has polynomial volume growth if and only if it is proper. Thus, the condition on polynomial volume growth in [15] and [11] is essential since these complete self-shrinking curves $\Gamma$ of Halldorsson [14] are not proper and for any integer $n>0$, $\Gamma \times {\mathbb {R}}^{n-1}$ is a complete self-shrinker without polynomial volume growth in ${\mathbb {R}}^{n+1}$.

As for the study on the rigidity of complete self-shrinkers, many important works have been done (cf. [4, 7,8,9, 12, 13, 22] and so on). In particular, Cheng and Peng in [8] proved that for an n-dimensional complete self-shrinker $X:M^n\rightarrow {\mathbb {R}}^{n+1}$ with $\inf H^2>0$, if the squared norm S of the second fundamental form is constant, then $M^n$ is isometric to one of the following:

• (1) $S^n(\sqrt{n})$,

• (2) $S^m(\sqrt{m})\times {\mathbb {R}}^{n-m}\subset {\mathbb {R}}^{n+1}$.

Furthermore, Ding and Xin [13] studied 2-dimensional complete self-shrinkers with polynomial volume growth and with constant squared norm S of the second fundamental form. They have proved that a 2-dimensional complete self-shrinker $X: M\rightarrow {\mathbb {R}}^{3}$ with polynomial volume growth is isometric to one of the following:

• (1) ${\mathbb {R}}^{2}$,

• (2) $S^1 (1)\times {\mathbb {R}}$

• (3) $S^{2}(\sqrt{2})$,

if S is constant. Recently, Cheng and Ogata [7] have removed both the assumption on polynomial volume growth in the above theorem of Ding and Xin [13] and the assumption $\inf H^2>0$ in the theorem of Cheng and Peng [8] for $n=2$.

It is natural to ask the following problems:

Problem 1. To classify 2-dimensional complete self-shrinkers in ${\mathbb {R}}^4$ if the squared norm S of the second fundamental form is constant.

It is well-known that the unit sphere $S^2(1)$, the Clifford torus $S^1(1)\times S^1(1)$ , the Euclidean plane ${\mathbb {R}}^2$ and the cylinder $S^1(1)\times {\mathbb {R}}^{1}$ are the canonical self-shrinkers in ${\mathbb {R}}^4$. Besides the standard examples, there are many examples of complete self-shrinkers in ${\mathbb {R}}^4$. For examples, compact minimal surfaces in the sphere $S^3(2)$ are compact self-shrinkers in ${\mathbb {R}}^4$. Further, Anciaux [2], Lee and Wang [21], Castro and Lerma [5] constructed many compact self-shrinkers in ${\mathbb {R}}^4$ (cf. Sect. 3). Except the canonical self-shrinkers in ${\mathbb {R}}^4$, the known examples of complete self-shrinkers in ${\mathbb {R}}^4$ do not have the constant squared norm S of the second fundamental form.

Since the above problem is very difficult, one may consider the special case of complete Lagrangian self-shrinkers in ${\mathbb {R}}^4$ first. Here we have identified ${\mathbb {R}}^{2n}$ with ${\mathbb {C}}^n$ and let us recall the definition of Lagrangian submanifolds. A submanifold $X: M\rightarrow {\mathbb {R}}^{2n}$ is called a Lagrangian submanifold if $J(T_pM)=T^{\perp }_pM$, for any $p\in M$, where J is the complex structure of ${\mathbb {R}}^{2n}$, $T_pM$ and $T_p^{\perp }M$ denote the tangent space and the normal space at p.

It is known that the mean curvature flow preserves the Lagrangian property, which means that, if the initial submanifold $X: M\rightarrow {\mathbb {R}}^{2n}$ is Lagrangian, then the mean curvature flow $X(\cdot , t):M\rightarrow {\mathbb {R}}^{2n}$ is also Lagrangian. Lagrangian submanifolds are a class of important submanifolds in geometry of submanifolds and they also have many applications in many other fields of differential geometry. For instance, the existence of special Lagrangian submanifolds in Calabi-Yau manifolds attracts a lot of attention since it plays a critical role in the T-duality formulation of Mirror symmetry of Strominger-Yau-Zaslow [28]. In particular, recently, the study on complete Lagrangian self-shrinkers of mean curvature flow has attracted much attention. Many important examples of compact Lagrangian self-shrinkers are constructed (see Sect. 3 and cf. [2, 5, 21]). It was proved by Smoczyk [26] that there are no Lagrangian self-shrinkers, which are topological spheres, in ${\mathbb {R}}^{2n}$. In [6], Castro and Lerma gave a classification of Hamiltonian stationary Lagrangian self-shrinkers in ${\mathbb {R}}^{4}$ and in [5], they proved that Clifford torus $S^1(1)\times S^1(1)$ is the only compact Lagrangian self-shrinker with $S\le 2$ in ${\mathbb {R}}^{4}$ if the Gaussian curvature does not change sign. Here, it is noticeable that compactness is important since the Gauss–Bonnet theorem is the key in their proof. In fact, Since $X: M^2\rightarrow {\mathbb {R}}^{4}$ is compact, according to the Gauss–Bonnet theorem, we have

$$\begin{aligned} 8\pi (1-g)=2\int _MKdA=\int _M(H^2-S)dA=\int _M(2-S)dA. \end{aligned}$$

Hence, $X: M^2\rightarrow {\mathbb {R}}^{4}$ is a torus and $K\equiv 0$, $S\equiv 2$. Recently, Li and Wang [22] have removed the condition on Gaussian curvature. They proved that Clifford torus $S^1(1)\times S^1(1)$ is the only compact Lagrangian self-shrinker with $S\le 2$ in ${\mathbb {R}}^{4}$. Furthermore, they proved that Clifford torus $S^1(1)\times S^1(1)$ is the only compact Lagrangian self-shrinker with constant squared norm S of the second fundamental form in ${\mathbb {R}}^{4}$. The Gauss–Bonnet theorem is still the key in their proof. Since the Euclidean plane ${\mathbb {R}}^2$ and the cylinder $S^1(1)\times {\mathbb {R}}^{1}$ are complete and non-compact Lagrangian self-shrinkers with $S=$ constant in ${\mathbb {R}}^{4}$, we may ask the following problem:

Problem 2. Let $X: M^2\rightarrow {\mathbb {R}}^{4}$ be a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$. If the squared norm S of the second fundamental form is constant, is $X: M^2\rightarrow {\mathbb {R}}^{4}$ isometric to one of the following

• (1) ${\mathbb {R}}^2$,

• (2) $S^1(1)\times {\mathbb {R}}^{1}$,

• (3) $S^1(1)\times S^1(1)$?

It is our motivation to solve the above problem. In fact, we prove the following:

Theorem 1.1

Let $X: M^2\rightarrow {\mathbb {R}}^{4}$ be a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$. If the squared norm S of the second fundamental form is constant, then $X: M^2\rightarrow {\mathbb {R}}^{4}$ is isometric to one of

• (1) ${\mathbb {R}}^2$,

• (2) $S^1(1)\times {\mathbb {R}}^{1}$,

• (3) $S^1(1)\times S^1(1)$.

Remark 1.1

We should remark the condition that S is constant is essential. In fact, from examples of Lee-Wang in Sect. 3, we know

$$\begin{aligned} \dfrac{3m^2+n^2}{n(m+n)}\le S \le \dfrac{m^2+3n^2}{m(n+m)}, \end{aligned}$$

for $m\le n$. By taking $n=m+1$ and letting $m\rightarrow \infty$, we have

$$\begin{aligned} \dfrac{3m^2+n^2}{n(m+n)}<2 , \ \ \dfrac{m^2+3n^2}{m(n+m)}>2 \end{aligned}$$

and

$$\begin{aligned} \lim _{m\rightarrow \infty }\dfrac{3m^2+n^2}{n(m+n)}=\lim _{m\rightarrow \infty }\dfrac{m^2+3n^2}{m(n+m)}=2. \end{aligned}$$

Since we do not assume that Lagrangian self-shrinkers are compact, we can not use Gauss–Bonnet theorem. Hence, in this paper, in place of the powerful Gauss–Bonnet theorem, we use the generalized maximum principle and moving frame methods.

In order to prove our theorem, we need to compute the supremum and infimum of mean curvature about 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$. Thus, a very precise computation is needed. Therefore, we must give a precise estimate of the squared norm of the second covariant derivative of the second fundamental form.

This paper is organized as follows.

In Sect. 2, in order to get a precise estimate of the squared norm of the second covariant derivative of the second fundamental form of 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$, we need to compute ${\mathcal {L}}\sum _{i, j,k,p}(h_{ijk}^{p^{*}})^{2}$ in two ways, which is a long computation.

In Sect. 3, we give several examples of 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$, which show that the condition of $S=$ constant is indispensable.

In Sect. 4, we prove our theorem. In order to do it, we make use of the generalized maximum principle. We choose a special frame fields at points, which we consider. We need to prove $h_{12}^*=\lambda =0$. This assertion is the key in our proof. Thus, a precise and detailed computation is needed.

2 Preliminaries

Let $X: M\rightarrow {\mathbb {R}}^{2n}$ be an n-dimensional connected submanifold of the 2n-dimensional Euclidean space ${\mathbb {R}}^{2n}$. We choose a local orthonormal frame field $\{e_A\}_{A=1}^{2n}$ in ${\mathbb {R}}^{2n}$ with dual coframe field $\{\omega _A\}_{A=1}^{2n}$, such that, restricted to M, $e_1,\ldots , e_n$ are tangent to $M^n$. Here we have identified ${\mathbb {R}}^{2n}$ with ${\mathbb {C}}^n$.

For a Lagrangian submanifold $X: M\rightarrow {\mathbb {R}}^{2n}$, we choose an adapted Lagrangian frame field

$$\begin{aligned} e_1, e_2,\ldots , e_n \ \ \text { and } \ e_{1^*}=Je_1, e_{2^*}=Je_2,\ldots , e_{n^*}=Je_n. \end{aligned}$$

From now on, we use the following conventions on the ranges of indices:

$$\begin{aligned} 1\le i,j,k,l\le n, \quad 1\le \alpha ,\beta ,\gamma \le n \end{aligned}$$

and $\sum _{i}$ means taking summation from 1 to n for i. Then we have

$$\begin{aligned} dX= & {} \sum _i \omega _i e_i, \\ de_i= & {} \sum _j \omega _{ij}e_j+\sum _{\alpha } \omega _{i\alpha ^{*}}e_{\alpha ^{*}}, \\ de_{\alpha ^{*}}= & {} \sum _i\omega _{\alpha ^{*} i}e_i+\sum _\beta \omega _{\alpha ^{*}\beta ^{*}}e_{\beta ^{*}} , \end{aligned}$$

where $\omega _{ij}$ is the Levi–Civita connection of M, $\omega _{\alpha ^{*}\beta ^{*}}$ is the normal connection of $T^{\perp }M$.

By restricting these forms to M, we have

$$\begin{aligned} \omega _{\alpha ^{*}}=0 \quad \text {for}\quad 1\le \alpha \le n \end{aligned}$$

(2.1)

and the induced Riemannian metric of M is written as $ds^2_M=\sum _i\omega ^2_i$. Taking exterior derivatives of (2.1), we have

$$\begin{aligned} 0=d\omega _{\alpha ^{*}}=\sum _i \omega _{\alpha ^{*} i}\wedge \omega _i. \end{aligned}$$

By Cartan’s lemma, we have

$$\begin{aligned} \omega _{i\alpha ^{*}}=\sum _j h^{\alpha ^{*}}_{ij}\omega _j,\quad h^{\alpha ^{*}}_{ij}=h^{\alpha ^{*}}_{ji}. \end{aligned}$$

Since $X: M\rightarrow {\mathbb {R}}^{2n}$ is a Lagrangian submanifold, we have

$$\begin{aligned}&h^{p^{*}}_{ij}=h^{p^{*}}_{ji}=h^{i^{*}}_{pj}, \ \ \text { for any } \ i, j, p.\nonumber \\&h=\sum _{i,j,p}h^{p^{*}}_{ij}\omega _i\otimes \omega _j\otimes e_{p^{*}} \end{aligned}$$

(2.2)

and

$$\begin{aligned} \vec {H}=\sum _{p} H^{p^{*}} e_{p^{*}}=\sum _{p} \sum _i h^{p^{*}}_{ii}e_{p^{*}} \end{aligned}$$

are called the second fundamental form and the mean curvature vector field of $X: M\rightarrow {\mathbb {R}}^{2n}$, respectively. Let $S=\sum _{i,j,p} (h^{p^{*}}_{ij})^2$ be the squared norm of the second fundamental form and $H=|\vec {H}|$ denote the mean curvature of $X: M\rightarrow {\mathbb {R}}^{2n}$. The induced structure equations of M are given by

$$\begin{aligned} d\omega _{i}= & {} \sum _j \omega _{ij}\wedge \omega _j, \quad \omega _{ij}=-\omega _{ji},\\ d\omega _{ij}= & {} \sum _k \omega _{ik}\wedge \omega _{kj}-\frac{1}{2}\sum _{k,l} R_{ijkl} \omega _{k}\wedge \omega _{l}, \end{aligned}$$

where $R_{ijkl}$ denotes components of the curvature tensor of M. Hence, the Gauss equations are given by

$$\begin{aligned} R_{ijkl}=\sum _{p}\left( h^{p^{*}}_{ik}h^{p^{*}}_{jl}-h^{p^{*}}_{il}h^{p^{*}}_{jk}\right) ,\ \ R_{ik}=\sum _{p} H^{p^{*}} h^{p^{*}}_{ik}-\sum _{p,j}h^{p^{*}}_{ij}h^{p^{*}}_{jk}. \end{aligned}$$

(2.3)

Letting $R_{p^{*}q^{*} ij}$ denote the curvature tensor of the normal connection $\omega _{p^{*}q^{*}}$ in the normal bundle of $X:M\rightarrow {\mathbb {R}}^{2n}$, then Ricci equations are given by

$$\begin{aligned} R_{p^{*}q^{*}kl}=\sum _i\left( h^{p^{*}}_{ik}h^{q^{*}}_{il}-h^{p^{*}}_{il}h^{q^{*}}_{ik}\right) . \end{aligned}$$

(2.4)

Defining the covariant derivative of $h^{p^{*}}_{ij}$ by

$$\begin{aligned} \sum _{k}h^{p^{*}}_{ijk}\omega _k=dh^{p^{*}}_{ij}+\sum _kh^{p^{*}}_{ik}\omega _{kj} +\sum _k h^{p^{*}}_{kj}\omega _{ki}+\sum _{q} h^{q^{*}}_{ij}\omega _{p^{*}q^{*}}, \end{aligned}$$

(2.5)

we obtain the Codazzi equations

$$\begin{aligned} h_{ijk}^{p^{*}}=h_{ikj}^{p^{*}}=h^{i^{*}}_{pjk}. \end{aligned}$$

(2.6)

By taking exterior differentiation of (2.5), and defining

$$\begin{aligned} \sum _lh^{p^{*}}_{ijkl}\omega _l=dh^{p^{*}}_{ijk}+\sum _lh^{p^{*}}_{ljk}\omega _{li} +\sum _lh^{p^{*}}_{ilk}\omega _{lj}+\sum _l h^{p^{*}}_{ijl}\omega _{lk} +\sum _{q} h^{q^{*}}_{ijk}\omega _{{q^{*}p^{*}}}, \end{aligned}$$

(2.7)

we have the following Ricci identities:

$$\begin{aligned} h^{p^{*}}_{ijkl}-h^{p^{*}}_{ijlk}=\sum _m h^{p^{*}}_{mj}R_{mikl}+\sum _m h^{p^{*}}_{im}R_{mjkl}+\sum _{q} h^{q^{*}}_{ij}R_{{q^{*}p^{*}}kl}. \end{aligned}$$

(2.8)

Defining

$$\begin{aligned} \begin{aligned} \sum _mh_{ijklm}^{p^{*}}\omega _m&=dh_{ijkl}^{p^{*}}+\sum _mh_{mjkl}^{p^{*}}\omega _{mi} +\sum _mh_{imkl}^{p^{*}}\omega _{mj}+\sum _mh_{ijml}^{p^{*}}\omega _{mk}\\&\quad +\sum _mh_{ijkm}^{p^{*}}\omega _{ml} +\sum _{m}h_{ijkl}^{m^{*}}\omega _{m^{*}p^{*}} \end{aligned} \end{aligned}$$

(2.9)

and taking exterior differentiation of (2.7), we get

$$\begin{aligned} \begin{aligned} h_{ijkln}^{p^{*}}-h_{ijknl}^{p^{*}}&=\sum _{m} h_{mjk}^{p^{*}}R_{miln} + \sum _{m}h_{imk}^{p^{*}}R_{mjln}+ \sum _{m}h_{ijm}^{p^{*}}R_{mkln}\\&\quad +\sum _{m}h_{ijk}^{m^{*}}R_{m^{*}p^{*}ln}. \end{aligned} \end{aligned}$$

(2.10)

For the mean curvature vector field $\vec {H}=\sum _{p} H^{p^{*}} e_{p^{*}}$, we define

$$\begin{aligned}&\sum _i H^{p^{*}}_{,i}\omega _i=dH^{p^{*}}+\sum _{q} H^{q^{*}}\omega _{q^{*}p^{*}}, \end{aligned}$$

(2.11)

$$\begin{aligned}&\sum _j H^{p^{*}}_{,ij}\omega _j=dH^{p^{*}}_{,i}+\sum _j H^{p^{*}}_{,j}\omega _{ji}+\sum _{q} H^{q^{*}}_{,i}\omega _{q^{*}p^{*}}, \end{aligned}$$

(2.12)

$$\begin{aligned}&|\nabla ^{\perp }\vec H|^2=\sum _{i,p }(H^{p^{*}}_{,i})^2,\ \ \ \ \Delta ^\perp H^{p^{*}}=\sum _i H^{p^{*}}_{,ii}. \end{aligned}$$

(2.13)

For a smooth function f, the ${\mathcal {L}}$-operator is defined by

$$\begin{aligned} {\mathcal {L}}f=\Delta f-\langle X,\nabla f\rangle , \end{aligned}$$

(2.14)

where $\Delta$ and $\nabla$ denote the Laplacian and the gradient operator, respectively.

Formulas in the following Lemma 2.1 may be found in several papers, for examples, [4, 8, 22, 23]. Since many calculations in their proof are used in this paper, we also provide the proofs for reader’s convenience.

If $X:M^2\rightarrow {\mathbb {R}}^4$ is a self-shrinker, then we have

$$\begin{aligned} H^{p^{*}}=-\langle X,e_{p^{*}}\rangle , \ \ p=1, 2. \end{aligned}$$

(2.15)

From (2.15), we can get

$$\begin{aligned} H^{p^*}_{,i}=\nabla _{i}H^{p^{*}}=-\nabla _{i}\langle X,e_{p^{*}}\rangle =\sum _{j}h_{ij}^{p^{*}}\langle X,e_{j}\rangle . \end{aligned}$$

(2.16)

Since

$$\begin{aligned} \nabla _{i}|X|^{2}=2\langle X,e_{i}\rangle , \end{aligned}$$

(2.17)

we have the following equations from (2.15)

$$\begin{aligned}&\begin{aligned} \nabla _{j}\nabla _{i}|X|^{2}&=2\langle e_{i},e_{j}\rangle +2\langle X,X_{ij}\rangle \\&=2\delta _{ij}+2\langle X,\sum _{p}h_{ij}^{p^{*}}e_{p^{*}}\rangle \\&=2\delta _{ij}-2 \sum _{p}h_{ij}^{p^{*}}H^{p^{*}}, \end{aligned} \end{aligned}$$

(2.18)

$$\begin{aligned}&\begin{aligned} \nabla _{j}\nabla _{i}H^{p^{*}}&=\nabla _{j}(\sum _{k}h_{ik}^{p^{*}}\langle X,e_{k}\rangle )\\&=\sum _{k}h_{ikj}^{p^{*}}\langle X,e_{k}\rangle +h_{ij}^{p^{*}} +\sum _{k}h_{ik}^{p^{*}}\sum _{q}h_{jk}^{q^{*}}\langle X,e_{q^{*}}\rangle \\&=\sum _{k}h_{ikj}^{p^{*}}\langle X,e_{k}\rangle +h_{ij}^{p^{*}}-\sum _{k,q}h_{ik}^{p^{*}}h_{jk}^{q^{*}}H^{q^{*}}. \end{aligned} \end{aligned}$$

(2.19)

By a direct calculation, from (2.15) and (2.19), we have

$$\begin{aligned} {\mathcal {L}}H^{p^{*}}=\sum _{k}H_{,kk}^{p^{*}}-\langle X,\sum _{k}H_{,k}^{p^{*}}e_{k}\rangle =H^{p^{*}} -\sum _{i,j,q}h_{ij}^{p^{*}}h_{ij}^{q^{*}}H^{q^{*}}. \end{aligned}$$

(2.20)

From the definition of the self-shrinker, we get

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}} |X|^{2}=2-H^{2}-\langle X,\sum _{i}\langle X,e_{i}\rangle e_{i}\rangle =2-H^{2}-|X^{\top }|^{2}=2-|X|^{2}. \end{aligned} \end{aligned}$$

(2.21)

Since $X:M^2\rightarrow {\mathbb {R}}^4$ is a 2-dimensional Lagrangian self-shrinker, we know

$$\begin{aligned} R_{ijkl}=K(\delta _{ik}\delta _{jl}-\delta _{il}\delta _{jk})=R_{i^{*}j^{*}kl}, \end{aligned}$$

(2.22)

where $K=\dfrac{1}{2}(H^2-S)$ is the Gaussian curvature of $X:M^2\rightarrow {\mathbb {R}}^4$.

According to (2.3), (2.6), (2.8), (2.22), we have

$$\begin{aligned} \begin{aligned} {\mathcal {L}} h_{ij}^{p^{*}}&=\sum _{k}h_{ijkk}^{p^{*}}-\langle X,\sum _{k}\nabla _{k}h_{ij}^{p^{*}}e_{k}\rangle \\&=\sum _{k}h_{ikjk}^{p^{*}}-\sum _{k}h_{ijk}^{p^{*}}\langle X,e_{k}\rangle \\&=\sum _{m,k}h_{mk}^{p^{*}}R_{mijk}+ \sum _{m,k}h_{im}^{p^{*}}R_{mkjk}\\&\quad + \sum _{q,k}h_{ik}^{q^{*}}R_{q^{*}p^{*}jk} +\sum _{k}h_{ikkj}^{p^{*}}-\sum _{k}h_{ijk}^{p^{*}}\langle X,e_{k}\rangle \\&=K \sum _{m,k}h_{mk}^{p^{*}}(\delta _{mj}\delta _{ik}-\delta _{mk}\delta _{ij}) + K \sum _{m,k}h_{im}^{p^{*}}(\delta _{mj}\delta _{kk}-\delta _{mk}\delta _{kj})\\&\quad +K\sum _{q,k} h_{ik}^{q^{*}}(\delta _{qj}\delta _{pk}-\delta _{qk}\delta _{pj}) +H_{,ij}^{p^{*}}-\sum _{k}h_{ijk}^{p^{*}}\langle X,e_{k}\rangle \\&=K(h_{ij}^{p^{*}}-H^{p^{*}}\delta _{ij})+K(2h_{ij}^{p^{*}}-h_{ij}^{p^{*}}) +K(h_{ij}^{p^{*}}-\sum _{k}h_{kk}^{i^{*}}\delta _{pj})\\&\quad +\sum _{k}h_{ijk}^{p^{*}}\langle X,e_{k}\rangle +h_{ij}^{p^{*}}-\sum _{q,k}h_{ik}^{p^{*}}h_{jk}^{q^{*}}H^{q^{*}}-\sum _{k}h_{ijk}^{p^{*}}\langle X,e_{k}\rangle \\&=(3K+1)h_{ij}^{p^{*}}-K(H^{p^{*}}\delta _{ij}+H^{i^{*}}\delta _{pj})-\sum _{q,k}h_{ik}^{p^{*}}h_{jk}^{q^{*}}H^{q^{*}}. \end{aligned} \end{aligned}$$

(2.23)

Hence, we get

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}}S&=\frac{1}{2}\sum _{k}\nabla _{k}\nabla _{k}\sum _{i,j,p}(h_{ij}^{p^{*}})^{2} -\frac{1}{2}\langle X,\sum _{k}\nabla _{k}S e_{k}\rangle \\&=\sum _{k}\nabla _{k}(\sum _{i,j,p} h_{ijk}^{p^{*}}h_{ij}^{p^{*}}) -\langle X,\sum _{i,j,k,p}h_{ijk}^{p^{*}}h_{ij}^{p^{*}}e_{k}\rangle \\&=\sum _{i,j,p} h_{ij}^{p^{*}} {\mathcal {L}} h_{ij}^{p^{*}}+\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2} \\&=\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}\\&\quad +\sum _{i,j,p} h_{ij}^{p^{*}}\Big [(3K+1)h_{ij}^{p^{*}} -K(H^{p^{*}}\delta _{ij}+H^{i^{*}}\delta _{pj})-\sum _{k,q} h_{ik}^{p^{*}}h_{jk}^{q^{*}} H^{q^{*}}\Big ]\\&=\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+(3K+1)S-K(H^{2}+H^{2}) -\sum _{i,j,k,p,q} h_{ik}^{p^{*}}h_{ij}^{p^{*}}h_{jk}^{q^{*}} H^{q^{*}} \\&=\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+S(1-\frac{3}{2}S)+\frac{5}{2}H^{2}S-H^{4} -\sum _{i,j,k,p,q}h_{ik}^{p^{*}}h_{ij}^{p^{*}}h_{jk}^{q^{*}} H^{q^{*}}. \end{aligned} \end{aligned}$$

(2.24)

From (2.6) and (2.22), we get

$$\begin{aligned} \sum _{p,i} h_{ik}^{p^{*}}h_{ji}^{p^{*}}-\sum _{p} H^{p^{*}}h_{jk}^{p^{*}} =K(\delta _{kj}-2\delta _{jk}) \end{aligned}$$

and

$$\begin{aligned} \sum _{p,i} h_{ik}^{p^{*}}h_{ji}^{p^{*}}=-K\delta _{jk}+\sum _{p}H^{p^{*}}h_{jk}^{p^{*}}. \end{aligned}$$

Since

$$\begin{aligned} \sum _{j,k}(K\delta _{jk}-\sum _{p}H^{p^{*}}h_{jk}^{p^{*}})\sum _{q}h_{jk}^{q^{*}} H^{q^{*}}= KH^{2}-\sum _{j,k}\sum _{p}(H^{p^{*}}h_{jk}^{p^{*}})\sum _{q}(H^{q^{*}}h_{jk}^{q^{*}}), \end{aligned}$$

we obtain from (2.24)

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}}S =&\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+S(1-\frac{3}{2}S)+\frac{5}{2}H^{2}S-H^{4}\\&+\sum _ {j,k}(K\delta _{jk}-\sum _{p}H^{p^{*}}h_{jk}^{p^{*}})\sum _{q}h_{jk}^{q^{*}} H^{q^{*}}\\ =&\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+S(1-\frac{3}{2}S)+2H^{2}S-\frac{1}{2}H^{4} -\sum _{j,k,p,q} H^{p^{*}}h_{jk}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}. \end{aligned} \end{aligned}$$

From (2.20), we have

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}}H^{2}&= \frac{1}{2}{\mathcal {L}}\sum _{p}(H^{p^{*}})^{2} =\sum _{i,p}(\nabla _{i}H^{p^{*}})^{2}+\sum _{p}H^{p^{*}}{\mathcal {L}}H^{p^{*}}\\&=|\nabla ^{\perp }{\vec H}|^{2}+H^{2}-\sum _{i,j,p,q}H^{p^{*}}h_{ij}^{p^{*}}H^{q^{*}}h_{ij}^{q^{*}}. \end{aligned} \end{aligned}$$

Thus, we conclude the following lemma

Lemma 2.1

Let $X:M^2\rightarrow {\mathbb {R}}^4$ is a 2-dimensional Lagrangian self-shrinker in ${\mathbb {R}}^4$. We have

$$\begin{aligned} \frac{1}{2}{\mathcal {L}}S= & {} \sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+S(1-\frac{3}{2}S)+2H^{2}S-\frac{1}{2}H^{4} -\sum _{j,k,p,q}H^{p^{*}}h_{jk}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}.\nonumber \\ \end{aligned}$$

(2.25)

$$\begin{aligned} \frac{1}{2}{\mathcal {L}}H^{2}= & {} |\nabla ^{\perp }{\vec H}|^{2} +H^{2}-\sum _{i,j,p,q}H^{p^{*}}h_{ij}^{p^{*}}\cdot H^{q^{*}}h_{ij}^{q^{*}}. \end{aligned}$$

(2.26)

Next, we will prove the following lemma, by making use of a long calculation:

Lemma 2.2

Let $X:M^2\rightarrow {\mathbb {R}}^4$ is a 2-dimensional Lagrangian self-shrinker in ${\mathbb {R}}^4$. Then

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i, j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}+(10K+2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} -5K|\nabla ^{\perp } {\vec H}|^{2}+3\langle \nabla K,\nabla S\rangle \\&\qquad -\frac{3K}{4}\langle \nabla S,\nabla |X|^{2}\rangle -\langle \nabla K,\nabla H^{2}\rangle -3\sum _{j,l,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}}\\&\qquad -2\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{kl}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}}\\&-\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}} \end{aligned} \end{aligned}$$

(2.27)

holds.

Proof

We have the following equation from the Ricci identities (2.10).

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}h_{ijk}^{p^{*}}=\sum _{l}h_{ijkll}^{p^{*}}-\langle X,\sum _{l}\nabla _{l}h_{ijk}^{p^{*}}e_{l}\rangle \\&\quad =\sum _{l}h_{ijlkl}^{p^{*}}+\sum _{l,m}(h_{mj}^{p^{*}}R_{mikl} +h_{im}^{p^{*}}R_{mjkl}+h_{ij}^{m^{*}}R_{m^{*}p^{*}kl})_{,l}-\langle X,\sum _{l}\nabla _{l}h_{ijk}^{p^{*}}e_{l}\rangle \\&\quad =\sum _{l}h_{ijllk}^{p^{*}}+\sum _{l,m}h_{mjl}^{p^{*}}R_{mikl} +\sum _{l,m}h_{iml}^{p^{*}}R_{mjkl}+\sum _{l,m}h_{ijm}^{p^{*}}R_{mlkl}\\&\qquad +\sum _{l,m}h_{ijl}^{m^{*}}R_{m^{*}p^{*}kl} +\sum _{l,m}h_{mjl}^{p^{*}}R_{mikl}\\&\qquad +\sum _{l,m}h_{iml}^{p^{*}}R_{mjkl}+\sum _{l,m}h_{ijl}^{m^{*}}R_{m^{*}p^{*}kl} +\sum _{l,m}h_{mj}^{p^{*}}R_{mikl,l}\\&\qquad +\sum _{l,m}h_{im}^{p^{*}}R_{mjkl,l} +\sum _{l,m}h_{ij}^{m^{*}}R_{m^{*}p^{*}kl,l} -\langle X,\sum _{l}\nabla _{l}h_{ijk}^{p^{*}}e_{l}\rangle . \end{aligned} \end{aligned}$$

(2.28)

From (2.23), we have

$$\begin{aligned} \begin{aligned} \sum _{l}h_{ijllk}^{p^{*}}&=\Big [(3K+1)h_{ij}^{p^{*}}-K(H^{p^{*}}\delta _{ij}+H^{i^{*}}\delta _{pj})\\&\quad -\sum _{l,q}h_{il}^{p^{*}}h_{jl}^{q^{*}}H^{q^{*}} +\langle X,\sum _{l}h_{ijl}^{p^{*}}e_{l}\rangle \Big ]_{,k} \\&=3K_{,k}h_{ij}^{p^{*}}+(3K+1)h_{ijk}^{p^{*}}-K_{,k}(H^{p^{*}}\delta _{ij}+H^{i^{*}}\delta _{pj}) \\&\quad -K(H_{,k}^{p^{*}}\delta _{ij}+H_{,k}^{i^{*}}\delta _{pj})\\&\quad -\sum _{l,q}h_{ilk}^{p^{*}}h_{jl}^{q^{*}}H^{q^{*}} -\sum _{l,q}h_{il}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}}-\sum _{l,q}h_{il}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}\\&\quad +h_{ijk}^{p^{*}}+\langle X,\sum _{l,q}h_{ijl}^{p^{*}}h_{lk}^{q^{*}}e_{q^{*}}\rangle +\langle X,\sum _{l}h_{ijlk}^{p^{*}}e_{l}\rangle . \end{aligned} \end{aligned}$$

(2.29)

From (2.22), we obtain

$$\begin{aligned}&\begin{aligned}&(a)^{p^{*}}_{ijk}:=\sum _{l,m}h_{mjl}^{p^{*}}R_{mikl}+\sum _{l,m}h_{iml}^{p^{*}}R_{mjkl} +\sum _{l,m}h_{ijm}^{p^{*}}R_{mlkl}+\sum _{l,m}h_{ijl}^{m^{*}}R_{m^{*}p^{*}kl}\\&\qquad +\sum _{l,m}h_{mjl}^{p^{*}}R_{mikl}+\sum _{l,m}h_{iml}^{p^{*}}R_{mjkl}+\sum _{l,m}h_{ijl}^{m^{*}}R_{m^{*}p^{*}kl}\\&\quad =2K\sum _{l,m}\Big [h_{mjl}^{p^{*}}(\delta _{mk}\delta _{il}-\delta _{ml}\delta _{ik})+h_{iml}^{p^{*}}(\delta _{mk}\delta _{jl}-\delta _{ml}\delta _{jk})\\&\qquad +h_{ijl}^{m^{*}}(\delta _{mk}\delta _{pl}-\delta _{ml}\delta _{pk})\Big ]+\sum _{l,m}Kh_{ijm}^{p^{*}}(\delta _{mk}\delta _{ll} -\delta _{ml}\delta _{lk})\\&\quad =2K\Big [h_{ijk}^{p^{*}}-H_{,j}^{p^{*}}\delta _{ik} +h_{ijk}^{p^{*}}-H_{,i}^{p^{*}}\delta _{jk}+h_{ijk}^{p^{*}}-H_{,j}^{i^{*}}\delta _{pk} +h_{ijk}^{p^{*}}-\frac{1}{2}h_{ijk}^{p^{*}}\Big ]\\&\quad =2K\Big [\frac{7}{2}h_{ijk}^{p^{*}}-H_{,j}^{p^{*}}\delta _{ik}-H_{,i}^{p^{*}}\delta _{jk}-H_{,j}^{i^{*}}\delta _{pk}\Big ],\\ \end{aligned}\\&\begin{aligned}&(b)^{p^{*}}_{ijk}:=\sum _{l,m}h_{mj}^{p^{*}}R_{mikl,l}+\sum _{l,m}h_{im}^{p^{*}}R_{mjkl,l}+\sum _{l,m}h_{ij}^{m^{*}}R_{m^{*}p^{*}kl,l}\\&\quad =\sum _{l,m}K_{,l}\Big [h_{mj}^{p^{*}}(\delta _{mk}\delta _{il}-\delta _{ml}\delta _{ik}) +h_{im}^{p^{*}}(\delta _{mk}\delta _{jl}-\delta _{ml}\delta _{jk})\\&\qquad + h_{ij}^{m^{*}}(\delta _{mk}\delta _{pl}-\delta _{ml}\delta _{pk})\Big ]\\&=K_{,i}h_{jk}^{p^{*}}-\sum _{l}K_{,l}h_{lj}^{p^{*}}\delta _{ik}+K_{,j}h_{ik}^{p^{*}}-\sum _{l}K_{,l}h_{il}^{p^{*}}\delta _{jk}+ K_{,p}h_{ij}^{k^{*}}-\sum _{l}K_{,l}h_{ij}^{l^{*}}\delta _{pk} \end{aligned} \end{aligned}$$

and

$$\begin{aligned}&(c)^{p^{*}}_{ijk}:=\sum _{l}\langle X,e_{l}\rangle h_{ijlk}^{p^{*}}-\sum _{l}\langle X,e_{l}\rangle h_{ijkl}^{p^{*}}\\&\quad =\sum _{l,m}\langle X,e_{l}\rangle \Big [h_{mj}^{p^{*}}R_{milk}+h_{im}^{p^{*}}R_{mjlk} +h_{ij}^{m^{*}}R_{m^{*}p^{*}lk}\Big ] \\&\quad =K\sum _{l,m}\langle X,e_{l}\rangle \Big [h_{mj}^{p^{*}}(\delta _{ml}\delta _{ik} -\delta _{mk}\delta _{il})+h_{im}^{p^{*}}(\delta _{ml}\delta _{jk}-\delta _{mk}\delta _{jl})\\&\qquad +h_{ij}^{m^{*}}(\delta _{ml}\delta _{pk}-\delta _{mk}\delta _{pl})\Big ]\\&\quad =K\sum _{l}\langle X,e_{l}\rangle \Big [h_{lj}^{p^{*}}\delta _{ik}-h_{kj}^{p^{*}}\delta _{il} +h_{il}^{p^{*}}\delta _{jk}-h_{ik}^{p^{*}}\delta _{jl} +h_{ij}^{l^{*}}\delta _{pk}-h_{ij}^{k^{*}}\delta _{pl}\Big ] \\&\quad =K\Big [\sum _{l}\langle X,e_{l}\rangle h_{lj}^{p^{*}}\delta _{ik}-\langle X,e_{i}\rangle h_{kj}^{p^{*}}+\sum _{l}\langle X,e_{l}\rangle h_{il}^{p^{*}}\delta _{jk}-\langle X,e_{j}\rangle h_{ik}^{p^{*}}\\&\qquad +\sum _{l}\langle X,e_{l}\rangle h_{ij}^{l^{*}}\delta _{pk}-\langle X,e_{p}\rangle h_{ij}^{k^{*}}\Big ]. \end{aligned}$$

We conclude

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,p}h_{ijk}^{p^{*}}\cdot ((a)^{p^{*}}_{ijk}+(b)^{p^{*}}_{ijk})\\&\quad =2K\Big [\frac{7}{2}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}-\sum _{j,p}H_{,j}^{p^{*}}H_{,j}^{p^{*}} -\sum _{i,p}H_{,i}^{p^{*}}H_{,i}^{p^{*}}-\sum _{j,i}H_{,j}^{i^{*}}H_{,j}^{i^{*}}\Big ]\\&\qquad +\sum _{i,j,k,p}K_{,i}h_{jk}^{p^{*}}h_{ijk}^{p^{*}}-\sum _{l,j,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}} +\sum _{i,j,k,p}K_{,j}h_{ik}^{p^{*}}h_{ijk}^{p^{*}}\\&\qquad -\sum _{l,i,p}K_{,l}h_{il}^{p^{*}}H_{,i}^{p^{*}} +\sum _{i,j,k,p}K_{,p}h_{ij}^{k^{*}}h_{ijk}^{p^{*}}-\sum _{l,i,j}K_{,l}h_{ij}^{l^{*}}H_{,j}^{i^{*}}\\&\quad =2K\Big [\frac{7}{2}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}-3|\nabla ^{\perp } {\vec H}|^{2}] + \frac{3}{2}\langle \nabla K,\nabla S\rangle -3\sum _{l,j,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}} \end{aligned} \end{aligned}$$

(2.30)

and

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,p}h_{ijk}^{p^{*}}\cdot (c)^{p^{*}}_{ijk}\\&\quad =K\Big [\sum _{l,j,p}\langle X,e_{l}\rangle h_{lj}^{p^{*}}H_{,j}^{p^{*}} +\sum _{l,i,p}\langle X,e_{l}\rangle h_{li}^{p^{*}}H_{,i}^{p^{*}} +\sum _{l,i,j}\langle X,e_{l}\rangle h_{ij}^{l^{*}}H_{,j}^{i^{*}}\\&\qquad -\frac{1}{2}\sum _{i}\langle X,e_{i}\rangle \nabla _{i}S -\frac{1}{2}\sum _{i}\langle X,e_{i}\rangle \nabla _{i}S -\frac{1}{2}\sum _{i}\langle X,e_{i}\rangle \nabla _{i}S\Big ]\\&\quad =K\Big (3\sum _{l,j,p}\langle X,e_{l}\rangle h_{lj}^{p^{*}}H_{,j}^{p^{*}} -\frac{3}{4}\langle \nabla S,\nabla \langle X,X\rangle \rangle \Big )\\&\quad =3K(|\nabla ^{\perp } {\vec H}|^{2}-\frac{1}{4}\langle \nabla S,\nabla |X|^{2}\rangle ). \end{aligned} \end{aligned}$$

(2.31)

From (2.29), we have

$$\begin{aligned}&\sum _{i,j,k,l,p}h_{ijk}^{p^{*}}(h_{ijll,k}^{p^{*}}-\langle X,h_{ijlk}^{p^{*}}e_{l}\rangle )\nonumber \\&\quad =3\sum _{i,j,k,p}K_{,k}h_{ij}^{p^{*}}h_{ijk}^{p^{*}} +(3K+1)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}-\sum _{k,p}K_{,k}(H^{p^{*}}H_{,k}^{p^{*}}+H^{p^{*}}H_{,k}^{p^{*}})\nonumber \\&\qquad -\sum _{k,p}K(H_{,k}^{p^{*}}H_{,k}^{p^{*}}+H_{,k}^{p^{*}}H_{,k}^{p^{*}}) -\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ilk}^{p^{*}}h_{jl}^{q^{*}}H^{q^{*}}\nonumber \\&\qquad -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}\nonumber \\&\qquad +\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} -\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{lk}^{q^{*}}H^{q^{*}}\nonumber \\&\quad =\frac{3}{2}\langle \nabla K,\nabla S\rangle +(3K+2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} -\langle \nabla K,\nabla H^{2}\rangle - 2K|\nabla ^{\perp } {\vec H}|^{2}\nonumber \\&\qquad -2\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{kl}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}.\nonumber \\ \end{aligned}$$

(2.32)

From the above equations, we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} =\sum _{i,j,k,p}h_{ijk}^{p^{*}}{\mathcal {L}}h_{ijk}^{p^{*}}+\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}+2K[\frac{7}{2}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} -3|\nabla ^{\perp } {\vec H}|^{2}]+\frac{3}{2}\langle \nabla K,\nabla S\rangle \\&\qquad -3\sum _{l,j,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}} +3K\Big [|\nabla ^{\perp } {\vec H}|^{2}-\frac{1}{4}\langle \nabla S,\nabla |X|^{2}\rangle \Big ]\\&\qquad +\frac{3}{2}\langle \nabla K,\nabla S\rangle +(3K+2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}-\langle \nabla K,\nabla H^{2}\rangle -2K|\nabla ^{\perp } {\vec H}|^{2}\\&\qquad -2\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{kl}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}} \\&\qquad -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}+(10K+2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} -5K|\nabla ^{\perp } {\vec H}|^{2}\\&\qquad +3\langle \nabla K,\nabla S\rangle -\frac{3K}{4}\langle \nabla S,\nabla |X|^{2}\rangle -\langle \nabla K,\nabla H^{2}\rangle \\&\qquad -3\sum _{l,j,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}} -2\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{kl}^{q^{*}}H^{q^{*}}\\&\qquad -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}. \end{aligned} \end{aligned}$$

(2.33)

It completes the proof of the lemma. $\square$

Lemma 2.3

Let $X:M^2\rightarrow {\mathbb {R}}^4$ be a 2-dimensional Lagrangian self-shrinker in ${\mathbb {R}}^4$. If S is constant, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}+(10K+2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}-5K|\nabla ^{\perp } {\vec H}|^{2} \\&\qquad -\langle \nabla K,\nabla H^{2}\rangle -3\sum _{l,j,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}} -2\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{kl}^{q^{*}}H^{q^{*}}\\&\qquad -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}} \end{aligned} \end{aligned}$$

(2.34)

and

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =(H^{2}-2S)(|\nabla ^{\perp } {\vec H}|^{2}+H^{2})+\dfrac{1}{2}|\nabla H^2|^{2}\\&\qquad +(3K+2-H^{2}+2S)\sum _{i,j,p,q}H^{p^{*}}h_{ij}^{p^{*}} H^{q^{*}}h_{ij}^{q^{*}}\\&\qquad -K(H^{4}+\sum _{j,k,p}H^{k^{*}}H^{j^{*}}H^{p^{*}}h_{jk}^{p^{*}}) -\sum _{i,j,k,l,p,q,r}H^{r^{*}}H^{q^{*}}h_{jk}^{r^{*}}h_{jk}^{p^{*}}h_{il}^{p^{*}}h_{il}^{q^{*}}\\&\qquad +2\sum _{i,j,k,p,q}H_{,i}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}h_{ijk}^{p^{*}} -\sum _{i,j,k,p,q,r}H^{p^{*}}H^{q^{*}}H^{r^{*}}h_{ik}^{p^{*}}h_{ji}^{q^{*}}h_{jk}^{r^{*}}\\&\qquad +\sum _{i,j,k}\Big (\sum _{q}(H_{,i}^{q^{*}}h_{jk}^{q^{*}}+ H^{q^{*}}h_{ijk}^{q^{*}})\Big )\cdot \Big (\sum _{p}(H_{,i}^{p^{*}}h_{jk}^{p^{*}}+H^{p^{*}}h_{ijk}^{p^{*}})\Big ). \end{aligned}\nonumber \\ \end{aligned}$$

(2.35)

Proof

Since S is constant, we have the following equation from (2.33)

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}+(10K+2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} -5K|\nabla ^{\perp } {\vec H}|^{2}-\langle \nabla K,\nabla H^{2}\rangle \\&\qquad -3\sum _{j,l,p}K_{,l}h_{lj}^{p^{*}}H_{,j}^{p^{*}}-2\sum _{i,j,k,l,p,q}h_{ijk}^{p^{*}}h_{ijl}^{p^{*}}h_{kl}^{q^{*}}H^{q^{*}}\\&\qquad -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jlk}^{q^{*}}H^{q^{*}} -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}. \end{aligned} \end{aligned}$$

Now, we prove the formula (2.35). From (2.25) in Lemma 2.1, we obtain

$$\begin{aligned} 0=\frac{1}{2}{\mathcal {L}}S =\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+S(1-\frac{3}{2}S)+2H^{2}S-\frac{1}{2}H^{4} -\sum _{j,k,p,q} H^{p^{*}}h_{jk}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}.\nonumber \\ \end{aligned}$$

(2.36)

Hence, we have

$$\begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} =-S{\mathcal {L}}H^{2}+\frac{1}{4}{\mathcal {L}}H^{4} +\frac{1}{2}{\mathcal {L}}\sum _{j,k,p,q}H^{p^{*}}h_{jk}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}\nonumber \\&\quad =-2S\Bigg (|\nabla ^{\perp } {\vec H}|^{2}+H^{2}-\sum _{i,j,p,q}H^{p^{*}}h_{ij}^{p^{*}} H^{q^{*}}h_{ij}^{q^{*}}\Bigg ) +\frac{1}{2}H^{2}{\mathcal {L}}H^{2}\nonumber \\&\qquad +\frac{1}{2}|\nabla H^{2}|^{2}+\sum _{j,k}(\sum _{q} H^{q^{*}}h_{jk}^{q^{*}}){\mathcal {L}}(\sum _{p} H^{p^{*}}h_{jk}^{p^{*}})\nonumber \\&\qquad +\sum _{i,j,k}\nabla _{i}(\sum _{q} H^{q^{*}}h_{jk}^{q^{*}})\cdot \nabla _{i}(\sum _{p} H^{p^{*}}h_{jk}^{p^{*}})\nonumber \\&\quad =(H^{2}-2S)\Bigg (|\nabla ^{\perp } {\vec H}|^{2}+H^{2}-\sum _{i,j,p,q} H^{p^{*}}h_{ij}^{p^{*}} H^{q^{*}}h_{ij}^{q^{*}}\Bigg )\nonumber \\&\qquad +\dfrac{1}{2}|\nabla H^2|^{2}+\sum _{j,k}(\sum _{q} H^{q^{*}}h_{jk}^{q^{*}}){\mathcal {L}}(\sum _{p} H^{p^{*}}h_{jk}^{p^{*}})\nonumber \\&\qquad +\sum _{i,j,k,p,q}(H_{,i}^{q^{*}}h_{jk}^{q^{*}} + H^{q^{*}}h_{ijk}^{q^{*}})\cdot (H_{,i}^{p^{*}}h_{jk}^{p^{*}}+H^{p^{*}}h_{ijk}^{p^{*}}). \end{aligned}$$

(2.37)

Since

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}\sum _{p}(H^{p^{*}}h_{jk}^{p^{*}})\\&\quad =\sum _{i,p}\nabla _{i}\nabla _{i}(H^{p^{*}}h_{jk}^{p^{*}}) -\langle X,\sum _{i,p}\nabla _{i}(H^{p^{*}}h_{jk}^{p^{*}})e_{i}\rangle \\&\quad =\sum _{i,p}\nabla _{i}(H_{,i}^{p^{*}}h_{jk}^{p^{*}} +H^{p^{*}}h_{ijk}^{p^{*}}) -\langle X,\sum _{i,p}(\nabla _{i}H^{p^{*}}h_{jk}^{p^{*}}+H^{p^{*}}\nabla _{i}h_{jk}^{p^{*}})e_{i}\rangle \\&\quad =\sum _{p}h_{jk}^{p^{*}}{\mathcal {L}}H^{p^{*}}+\sum _{p}H^{p^{*}}{\mathcal {L}}h_{jk}^{p^{*}} +2\sum _{i,p}H_{,i}^{p^{*}}h_{ijk}^{p^{*}} \end{aligned} \end{aligned}$$

(2.38)

and from (2.20) and (2.23),

$$\begin{aligned}&\sum _{p}h_{jk}^{p^{*}}{\mathcal {L}}H^{p^{*}}=\sum _{p}h_{jk}^{p^{*}}H^{p^{*}}-\sum _{i,l,p,q}h_{jk}^{p^{*}}h_{il}^{p^{*}}h_{il}^{q^{*}}H^{q^{*}}, \end{aligned}$$

(2.39)

$$\begin{aligned}&\sum _{p}H^{p^{*}}{\mathcal {L}} h_{jk}^{p^{*}}\nonumber \\&\quad =(3K+1)\sum _{p}H^{p^{*}}h_{jk}^{p^{*}}-K\sum _{p}((H^{p^{*}})^{2}\delta _{kj}+H^{p^{*}}H^{k^{*}}\delta _{pj})\nonumber \\&\qquad -\sum _{i,p,q}h_{ik}^{p^{*}}H^{p^{*}}h_{ji}^{q^{*}}H^{q^{*}}\nonumber \\&\quad =(3K+1)\sum _{p}H^{p^{*}}h_{jk}^{p^{*}}-K(H^{2}\delta _{kj}+H^{k^{*}}H^{j^{*}}) -\sum _{i,p,q}h_{ik}^{p^{*}}H^{p^{*}}h_{ji}^{q^{*}}H^{q^{*}}, \end{aligned}$$

(2.40)

we get

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}\sum _{p}(H^{p^{*}}h_{jk}^{p^{*}})\\&\quad =\sum _{p}h_{jk}^{p^{*}}H^{p^{*}}-\sum _{i,l,p,q}h_{jk}^{p^{*}}h_{il}^{p^{*}}h_{il}^{q^{*}}H^{q^{*}} +2\sum _{i,p}H_{,i}^{p^{*}}h_{ijk}^{p^{*}}\\&\qquad +(3K+1)\sum _{p}H^{p^{*}}h_{jk}^{p^{*}}-K(H^{2}\delta _{kj}+H^{k^{*}}H^{j^{*}}) -\sum _{i,p,q}h_{ik}^{p^{*}}H^{p^{*}}h_{ji}^{q^{*}}H^{q^{*}} \end{aligned} \end{aligned}$$

(2.41)

and

$$\begin{aligned}&\sum _{j,k}(\sum _{q} H^{q^{*}}h_{jk}^{q^{*}}){\mathcal {L}}(\sum _{p} H^{p^{*}}h_{jk}^{p^{*}}) \nonumber \\&\quad =-\sum _{i,j,k,l,p,q,r}H^{r^{*}}h_{jk}^{r^{*}}h_{jk}^{p^{*}}h_{il}^{p^{*}}h_{il}^{q^{*}}H^{q^{*}} +2\sum _{i,j,k,p,q}H_{,i}^{p^{*}}h_{ijk}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}\nonumber \\&\qquad +(3K+2)\sum _{j,k}\Big (\sum _q H^{q^{*}}h_{jk}^{q^{*}}\Big )\Big (\sum _p H^{p^{*}}h_{jk}^{p^{*}}\Big )\nonumber \\&\qquad -K(H^{4}+\sum _{j,k,p}H^{k^{*}}H^{j^{*}}H^{p^{*}}h_{jk}^{p^{*}}) -\sum _{i,j,k,p,q,r}H^{p^{*}}H^{q^{*}}H^{r^{*}}h_{ik}^{p^{*}}h_{ji}^{q^{*}}h_{jk}^{r^{*}}.\nonumber \\ \end{aligned}$$

(2.42)

From the above equations, we conclude

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =(H^{2}-2S)(|\nabla ^{\perp } {\vec H}|^{2}+H^{2})+\dfrac{1}{2}|\nabla H^2|^{2}\\&\qquad +(3K+2-H^{2}+2S)\sum _{i,j,p,q}H^{p^{*}}h_{ij}^{p^{*}} H^{q^{*}}h_{ij}^{q^{*}}\\&\qquad -K(H^{4}+\sum _{j,k,p}H^{k^{*}}H^{j^{*}}H^{p^{*}}h_{jk}^{p^{*}}) -\sum _{i,j,k,l,p,q,r}H^{r^{*}}H^{q^{*}}h_{jk}^{r^{*}}h_{jk}^{p^{*}}h_{il}^{p^{*}}h_{il}^{q^{*}}\\&\qquad +2\sum _{i,j,k,p,q}H_{,i}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}}h_{ijk}^{p^{*}} -\sum _{i,j,k,p,q,r}H^{p^{*}}H^{q^{*}}H^{r^{*}}h_{ik}^{p^{*}}h_{ji}^{q^{*}}h_{jk}^{r^{*}}\\&\qquad +\sum _{i,j,k}\Big (\sum _{q}(H_{,i}^{q^{*}}h_{jk}^{q^{*}}+ H^{q^{*}}h_{ijk}^{q^{*}})\Big )\cdot \Big (\sum _{p}(H_{,i}^{p^{*}}h_{jk}^{p^{*}}+H^{p^{*}}h_{ijk}^{p^{*}})\Big ). \end{aligned} \end{aligned}$$

$\square$

Lemma 2.4

Let $X:M^2\rightarrow {\mathbb {R}}^4$ be a 2-dimensional Lagrangian self-shrinker in ${\mathbb {R}}^4$. Then we have

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}\\&\quad =4(h_{1122}^{1^{*}})^{2}+6(h_{2211}^{1^{*}})^{2}+6(h_{1122}^{2^{*}})^{2}+4(h_{2211}^{2^{*}})^{2} +4(h_{2222}^{1^{*}})^{2}+4(h_{1111}^{2^{*}})^{2}\\&\qquad +(h_{1111}^{1^{*}})^{2}+(h_{2222}^{2^{*}})^{2}+(h_{1112}^{1^{*}})^{2}+(h_{2221}^{2^{*}})^{2}\\&\quad \ge 2(h_{1122}^{1^{*}}-h_{2211}^{1^{*}})^{2}+2(h_{1122}^{2^{*}}-h_{2211}^{2^{*}})^{2} +\frac{1}{2}(h_{2222}^{1^{*}}-h_{2221}^{2^{*}})^{2}. \\ \end{aligned} \end{aligned}$$

(2.43)

Proof

Since

$$\begin{aligned}&\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}=\sum _{i,j,k,l}(h_{ijkl}^{1^{*}})^{2}+\sum _{i,j,k,l}(h_{ijkl}^{2^{*}})^{2}, \end{aligned}$$

(2.44)

$$\begin{aligned} \sum _{i,j,k,l}(h_{ijkl}^{1^{*}})^{2}= & {} (h_{1111}^{1^{*}})^{2}+(h_{1112}^{1^{*}})^{2}+(h_{2221}^{1^{*}})^{2}+(h_{2222}^{1^{*}})^{2}\nonumber \\&+3(h_{1121}^{1^{*}})^{2}+3(h_{1122}^{1^{*}})^{2} +3(h_{2211}^{1^{*}})^{2}+3(h_{2212}^{1^{*}})^{2} \nonumber \\= & {} (h_{1111}^{1^{*}})^{2}+3\Big [(h_{1122}^{1^{*}})^{2}+(h_{2211}^{1^{*}})^{2}\Big ]+3(h_{1122}^{2^{*}})^{2}\nonumber \\&+(h_{1112}^{1^{*}})^{2}+(h_{2211}^{2^{*}})^{2}+3(h_{1111}^{2^{*}})^{2}+(h_{2222}^{1^{*}})^{2}, \end{aligned}$$

(2.45)

$$\begin{aligned}&\sum _{i,j,k,l}(h_{ijkl}^{2^{*}})^{2} =(h_{1111}^{2^{*}})^{2}+(h_{1112}^{2^{*}})^{2}+(h_{2222}^{2^{*}})^{2}+(h_{2221}^{2^{*}})^{2}\nonumber \\&\qquad +3(h_{1121}^{2^{*}})^{2}+3(h_{1122}^{2^{*}})^{2} +3(h_{2211}^{2^{*}})^{2}+3(h_{2212}^{2^{*}})^{2} \nonumber \\&\quad =(h_{1111}^{2^{*}})^{2}+(h_{2222}^{2^{*}})^{2}+(h_{2221}^{2^{*}})^{2}+(h_{1122}^{1^{*}})^{2}\nonumber \\&\qquad +3(h_{2211}^{1^{*}})^{2}+3(h_{1122}^{2^{*}})^{2} +3(h_{2211}^{2^{*}})^{2}+3(h_{2222}^{1^{*}})^{2}, \end{aligned}$$

(2.46)

we get

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}\\&\quad =4(h_{1122}^{1^{*}})^{2}+6(h_{2211}^{1^{*}})^{2}+6(h_{1122}^{2^{*}})^{2}+(h_{1111}^{1^{*}})^{2} +(h_{1112}^{1^{*}})^{2}+3(h_{1111}^{2^{*}})^{2}\\&\qquad +(h_{2222}^{1^{*}})^{2}+(h_{1111}^{2^{*}})^{2}+(h_{2222}^{2^{*}})^{2}+(h_{2221}^{2^{*}})^{2}+ 3(h_{2222}^{1^{*}})^{2}+4(h_{2211}^{2^{*}})^{2}\\&\quad =4(h_{1122}^{1^{*}})^{2}+6(h_{2211}^{1^{*}})^{2}+6(h_{1122}^{2^{*}})^{2}+4(h_{2211}^{2^{*}})^{2} +4(h_{2222}^{1^{*}})^{2}+4(h_{1111}^{2^{*}})^{2}\\&\qquad +(h_{1111}^{1^{*}})^{2}+(h_{2222}^{2^{*}})^{2}+(h_{1112}^{1^{*}})^{2}+(h_{2221}^{2^{*}})^{2}\\&\quad \ge 2(h_{1122}^{1^{*}}-h_{2211}^{1^{*}})^{2}+2(h_{1122}^{2^{*}}-h_{2211}^{2^{*}})^{2} +\frac{1}{2}(h_{2222}^{1^{*}}-h_{2221}^{2^{*}})^{2} \\&\qquad +2(h_{1122}^{1^{*}}+h_{2211}^{1^{*}})^{2}+2(h_{1122}^{2^{*}}+h_{2211}^{2^{*}})^{2} +\frac{1}{2}(h_{2222}^{1^{*}}+h_{2221}^{2^{*}})^{2} \\&\quad \ge 2(h_{1122}^{1^{*}}-h_{2211}^{1^{*}})^{2}+2(h_{1122}^{2^{*}}-h_{2211}^{2^{*}})^{2} +\frac{1}{2}(h_{2222}^{1^{*}}-h_{2221}^{2^{*}})^{2}. \\ \end{aligned} \end{aligned}$$

(2.47)

$\square$

If ${\vec H}\ne 0$ at p, we can choose a local orthogonal frame $\{e_{1}, e_{2} \}$ such that

$$\begin{aligned} e_{1^{*}}=\frac{{\vec H}}{|{\vec H}|}, \ \ H^{1^{*}}=H=|{\vec H}|,\ \ H^{2^{*}}=h_{11}^{2^{*}}+h_{22}^{2^{*}}=0. \end{aligned}$$

(2.48)

Defining $\lambda =h_{12}^{1^{*}}$, $\lambda _1=h_{11}^{1^{*}}$ and $\lambda _2=h_{22}^{1^{*}}$, we have $h_{22}^{2^{*}}=-\lambda$.

Lemma 2.5

Let $X:M^2\rightarrow {\mathbb {R}}^4$ be a 2-dimensional Lagrangian self-shrinker in ${\mathbb {R}}^4$. If S is constant, ${\vec H}(p)\ne 0$ and $\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}(p)=0$, then we have, at p,

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}=\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2} \end{aligned} \end{aligned}$$

(2.49)

and

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =H^{2}\Big [H^{2}-2S+\frac{1}{2}H^{4}-H^2\lambda ^2-KH^{2}-K^{2}\Big ]\\&\qquad +H^{2}(S+2-\frac{3}{2}H^{2}-\lambda _{1}^{2} -\lambda _{2}^{2}-2\lambda ^2)(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2). \end{aligned} \end{aligned}$$

(2.50)

Proof

Since ${\vec H}\ne 0$ at p, we can choose a local orthogonal frame $\{e_{1}, e_{2} \}$ such that

$$\begin{aligned} e_{1^{*}}=\frac{{\vec H}}{|{\vec H}|}, \ \ H^{1^{*}}=H=|{\vec H}|,\ \ H^{2^{*}}=h_{11}^{2^{*}}+h_{22}^{2^{*}}=0. \end{aligned}$$

(2.51)

By the definition of $\lambda =h_{12}^{1^{*}}$, $\lambda _1=h_{11}^{1^{*}}$ and $\lambda _2=h_{22}^{1^{*}}$, we have $h_{22}^{2^{*}}=-\lambda$. Since S is constant and $h_{ijk}^{p^{*}}=0$ at p, we obtain from (2.34) of Lemma 2.3,

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}. \end{aligned} \end{aligned}$$

(2.52)

Furthermore, by making use of

$$\begin{aligned} S=\lambda _{1}^{2}+3\lambda _{2}^{2}+4\lambda ^2, \ \ H\lambda _{1}=K+\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2, \end{aligned}$$

(2.53)

from (2.35) in Lemma 2.3, we have the following equations, at p,

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =(H^{2}-2S)H^{2}+(\frac{1}{2}H^{2}+\frac{1}{2}S+2)H^{2}(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2)\\&\qquad -K(H^{4}+H^{3}\lambda _{1})-H^{2}\Big \{(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2)^{2}+\lambda ^2H^2\Big \}\\&\qquad -H^{3}(\lambda _{1}^{3}+\lambda _{2}^{3}+3H\lambda ^2)\\&\quad =H^{2}\Bigg [H^{2}-2S+(\frac{1}{2}H^{2} +\frac{1}{2}S+2-\lambda _{1}^{2}-\lambda _{2}^{2}-2\lambda ^2)(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2)\\&\qquad -K(H^{2}+K+\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2)-H^{2}(\lambda _{1}^{2}+\lambda _{2}^{2}-\lambda _{1}\lambda _{2}+4\lambda ^2)\Bigg ]\\&\quad =H^{2}\Bigg [H^{2}-2S+(S+2-\lambda _{1}^{2}-\lambda _{2}^{2}-2\lambda ^2)(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2) -K(H^{2}+K)\\&\qquad -H^{2}(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2) +\frac{1}{2}H^{4}-H^2\lambda ^2-\frac{1}{2}H^{2}(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2)\Bigg ]\\&\quad =H^{2}\Big [H^{2}-2S+\frac{1}{2}H^{4}-H^2\lambda ^2-KH^{2}-K^{2}\Big ]\\&\qquad +H^{2}(S+2-\frac{3}{2}H^{2}-\lambda _{1}^{2} -\lambda _{2}^{2}-2\lambda ^2)(\lambda _{1}^{2}+\lambda _{2}^{2}+2\lambda ^2). \end{aligned} \end{aligned}$$

(2.54)

This finishes the proof. $\square$

In order to prove our results, we need the following important generalized maximum principle for ${\mathcal {L}}$-operator on self-shrinkers which was proved by Cheng and Peng in [8]:

Lemma 2.6

(Generalized maximum principle for ${\mathcal {L}}$-operator ) Let $X: M^n\rightarrow {\mathbb {R}}^{n+p}$ ($p\ge 1$) be a complete self-shrinker with Ricci curvature bounded from below. Let f be any $C^2$-function bounded from above on this self-shrinker. Then, there exists a sequence of points $\{p_m\}\subset M^n$, such that

$$\begin{aligned} \lim _{m\rightarrow \infty } f(X(p_m))=\sup f,\quad \lim _{m\rightarrow \infty } |\nabla f|(X(p_m))=0,\quad \limsup _{m\rightarrow \infty }{\mathcal {L}} f(X(p_m))\le 0. \end{aligned}$$

3 Examples of Lagrangian self-shrinkers in ${\mathbb {R}}^4$

It is known that the Euclidean plane ${\mathbb {R}}^2$, the cylinder $S^1(1)\times {\mathbb {R}}^{1}$ and the Clifford torus $S^1(1)\times S^1(1)$ are the canonical Lagrangian self-shrinkers in ${\mathbb {R}}^4$. Apart from the standard examples, there are many other examples of complete Lagrangian self-shrinkers in ${\mathbb {R}}^4$.

Example 3.1

Let $\Gamma _1(s)=(x_1(s), y_1(s))^T$, $0\le s<L_1$ and $\Gamma _2(t)=(x_2(t), y_2(t))^T$, $0\le t<L_2$ be two self-shrinker curves in ${\mathbb {R}}^2$ with arc length as parameters, respectively. We consider Riemannian product $\Gamma _1(s)\times \Gamma _2(t)$ of $\Gamma _1(s)$ and $\Gamma _2(t)$ defined by

$$\begin{aligned} X(s,t)=\begin{pmatrix}x_1(s)\\ x_2(t)\\ y_1(s)\\ y_2(t) \end{pmatrix}. \end{aligned}$$

We can prove $\Gamma _1(s)\times \Gamma _2(t)$ is a Lagrangian self-shrinker in ${\mathbb {R}}^4$ and the Gaussian curvature K of $\Gamma _1(s)\times \Gamma _2(t)$ satisfies $K\equiv 0$.

In [1], Abresch and Langer classfied closed self-shrinking curves. For two closed self-shrinking curves $\Gamma _1(s)$ and $\Gamma _2(t)$ of Abresch and Langer in ${\mathbb {R}}^2$, $\Gamma _1(s)\times \Gamma _2(t)$ is a compact Lagrangian self-shrinker in ${\mathbb {R}}^4$, which is called Abresch-Langer torus. It is known that complete and non-compact self-shrinking curves exist in ${\mathbb {R}}^2$ (see [14]). Consequently, there are many complete and non-compact Lagrangian self-shrinkers with zero Gaussian curvature in ${\mathbb {R}}^4$.

Example 3.2

For a closed curve $\gamma (t)=(x_1(t), x_2(t))^T$, $t\in I$, such that its curvature $\kappa _{\gamma }$ satisfy

$$\begin{aligned} \kappa _{\gamma }=E\dfrac{e^{\frac{|\gamma |^2}{2}}}{|\gamma |^2}(|\gamma |^2-1), \ \ E^2= |\gamma |^4(1-(\dfrac{d|\gamma |}{dt})^2)e^{|\gamma |^2}, \end{aligned}$$

where E is a positive constant. In [2], Anciaux proved that

$$\begin{aligned} \begin{aligned}&X(s,t)=\begin{pmatrix}x_1(t)\cos s\\ x_1(t)\sin s\\ x_2(t)\cos s\\ x_2(t)\sin s\end{pmatrix} \end{aligned} \end{aligned}$$

defines a compact Lagrangian self-shrinker in ${\mathbb {R}}^4$, which is called Anciaux torus, and the squared norm S of the second fundamental form satisfies

$$\begin{aligned} S=E^2\dfrac{e^{\frac{|\gamma |^6}{2}}}{|\gamma |^2}(|\gamma |^4-2|\gamma |^2+4). \end{aligned}$$

Example 3.3

For positive integers m, n with $(m,n)=1$, define $X^{m,n}: {\mathbb {R}}^2\rightarrow {\mathbb {R}}^4$ by

$$\begin{aligned} \begin{aligned} X^{m,n}(s,t)=\sqrt{m+n}\begin{pmatrix}\dfrac{\cos s}{\sqrt{n}}\cos \sqrt{\frac{n}{m}}t\\ \dfrac{\sin s}{\sqrt{m}}\cos \sqrt{\frac{m}{n}}t\\ \dfrac{\cos s}{\sqrt{n}}\sin \sqrt{\frac{n}{m}}t\\ \dfrac{\sin s}{\sqrt{m}}\sin \sqrt{\frac{m}{n}}\end{pmatrix}. \end{aligned} \end{aligned}$$

Lee and Wang [21] proved $X^{m,n}: {\mathbb {R}}^2\rightarrow {\mathbb {R}}^4$ is a Lagrangian self-shrinker in ${\mathbb {R}}^4$. It is not difficult to prove that the squared norm S and the Gauss curvature K of $X^{m,n}: {\mathbb {R}}^2\rightarrow {\mathbb {R}}^4$, for $m\le n$, satisfy

$$\begin{aligned}&\dfrac{3m^2+n^2}{n(m+n)}\le S \le \dfrac{m^2+3n^2}{m(n+m)},\\&- \dfrac{n(n-m)}{m(m+n)}\le K \le \dfrac{m(n-m)}{n(n+m)}. \end{aligned}$$

4 Proofs of the main results

First of all, we prove the following:

Theorem 4.1

Let $X: M^2\rightarrow {\mathbb {R}}^{4}$ be a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$. If the squared norm S of the second fundamental form is constant, then $S\le 2$.

Proof

Since S is constant, from the Gauss equations, we know that the Ricci curvature of $X: M^2\rightarrow {\mathbb {R}}^{4}$ is bounded from below. We can apply the generalized maximum principle for ${\mathcal {L}}$-operator to the function $-|X|^2$. Thus, there exists a sequence $\{p_m\}$ in $M^2$ such that

$$\begin{aligned} \lim _{m\rightarrow \infty } |X|^2(p_m)=\inf |X|^2,\quad \lim _{m\rightarrow \infty } |\nabla |X|^2(p_m)|=0,\quad \liminf _{m\rightarrow \infty }{\mathcal {L}} |X|^2(p_m)\ge 0. \end{aligned}$$

Since $|\nabla |X|^2|^2=4\sum _{i=1}^2\langle X, e_i\rangle ^2$ and

$$\begin{aligned} \dfrac{1}{2}{\mathcal {L}}|X|^2=2-|X|^2, \end{aligned}$$

we have

$$\begin{aligned} \lim _{m\rightarrow \infty }\sum _j\langle X, e_j\rangle ^2(p_m)=0 \ \ \text { and} \ \ 2-\inf |X|^2\ge 0. \end{aligned}$$

(4.1)

Since $X: M^2\rightarrow {\mathbb {R}}^{4}$ is a self-shrinker, we know

$$\begin{aligned} H^{p^{*}}_{,i}=\sum _kh^{p^{*}}_{ik} \langle X,e_k \rangle , \quad i, p=1, 2. \end{aligned}$$

(4.2)

From the definition of the self-shrinker, (4.1) and (4.2), we get

$$\begin{aligned} \inf |X|^2=\lim _{m\rightarrow \infty }H^2(p_m)\le 2, \quad \lim _{m\rightarrow \infty }|\nabla ^{\perp } \vec H|^2(p_m)=0. \end{aligned}$$

(4.3)

Since $S=\sum _{i, j, p^{*}}(h^{p^{*}}_{ij})^2$ is constant, from (2.25) in Lemma 2.1, we know $\{h^{p^{*}}_{ij}(p_m)\}$ and $\{h^{p^{*}}_{ijl}(p_m)\}$ are bounded sequences for any i, j, l, p. Thus, we can assume

$$\begin{aligned} \lim _{m\rightarrow \infty }h^{p^{*}}_{ijl}(p_m)={\bar{h}}^{p^{*}}_{ijl}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ij}(p_m)={\bar{h}}^{p^{*}}_{ij}, \end{aligned}$$

for $i, j, l, p=1, 2$.

Therefore, we have

$$\begin{aligned} {\bar{h}}^{p^{*}}_{11j}+{\bar{h}}^{p^{*}}_{22j}=0, \ \ \text { for} \ j, p=1, 2 \end{aligned}$$

(4.4)

and

$$\begin{aligned} \sum _{i,j,p}h^{p^{*}}_{ij}h^{p^{*}}_{ijk}=0, \ \ \text {for } \ k=1, 2, \end{aligned}$$

because of S constant. Since $X: M^2\rightarrow {\mathbb {R}}^{4}$ is a Lagrangian self-shrinker,

$$\begin{aligned} h^{1^{*}}_{11}h^{1^{*}}_{11k}+3h^{1^{*}}_{12}h^{1^{*}}_{12k} +3h^{2^{*}}_{12}h^{2^{*}}_{12k}+h^{2^{*}}_{22}h^{2^{*}}_{22k}=0, \ \ \text {for } \ k=1, 2 \end{aligned}$$

holds. Thus, we conclude

$$\begin{aligned} {\bar{h}}^{1^{*}}_{11}{\bar{h}}^{1^{*}}_{11k}+3{\bar{h}}^{1^{*}}_{12}{\bar{h}}^{1^{*}}_{12k} +3{\bar{h}}^{2^{*}}_{12}{\bar{h}}^{2^{*}}_{12k}+{\bar{h}}^{2^{*}}_{22}{\bar{h}}^{2^{*}}_{22k}=0, \ \ \text {for } \ k=1, 2. \end{aligned}$$

(4.5)

If $\lim _{m\rightarrow \infty }H^2(p_m)=0$, we get

$$\begin{aligned} {\bar{h}}^{1^{*}}_{11}+{\bar{h}}^{1^{*}}_{22}=0, \ \ \ {\bar{h}}^{2^{*}}_{11}+{\bar{h}}^{2^{*}}_{22}=0. \end{aligned}$$

Consequently, from (4.4) and (4.5), we have the following equations for $k=1, 2$,

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0,\\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}4{\bar{h}}^{1^{*}}_{11}{\bar{h}}^{1^*}_{11k}+4{\bar{h}}^{2^{*}}_{11}{\bar{h}}^{2^*}_{11k}=0. \end{array}\right. } \end{aligned}$$

Hence, we obtain $S=0$ or ${\bar{h}}^{p^{*}}_{ijk}=0$ for any i, j, k and p. According to (2.25) in Lemma 2.1, we have $S=0$ or $S=\dfrac{2}{3}$.

If $\lim _{m\rightarrow \infty }H^2(p_m)={\bar{H}}^2\ne 0$, without loss of the generality, at each point $p_m$, we choose $e_1$, $e_2$ such that

$$\begin{aligned} \vec H=H^{1^{*}}e_{1^{*}}. \end{aligned}$$

Then we have

$$\begin{aligned} {\bar{h}}^{1^*}_{11}+ {\bar{h}}^{1^*}_{22}={\bar{H}}, \ \ {\bar{h}}^{2^{*}}_{11}+{\bar{h}}^{2^{*}}_{22}=0 \end{aligned}$$

and

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0,\\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}({\bar{h}}^{1^*}_{11}-3{\bar{h}}^{1^*}_{22}){\bar{h}}^{1^*}_{11k}+4{\bar{h}}^{2^{*}}_{11}{\bar{h}}^{2^*}_{11k}=0. \end{array}\right. } \end{aligned}$$

(4.6)

If ${\bar{h}}^{1^{*}}_{11}=3{\bar{h}}^{1^{*}}_{22}$ and ${\bar{h}}^{2^{*}}_{11}=0$, we know

$$\begin{aligned} \lim _{m\rightarrow \infty }H^2(p_m)=({\bar{\lambda }}_1+{\bar{\lambda }}_2)^2=16{\bar{\lambda }}_2^2\le 2 \ \text { and } \ S=12{\bar{\lambda }}_2^2\le \dfrac{3}{2}. \end{aligned}$$

If ${\bar{h}}^{1^{*}}_{11}\ne 3{\bar{h}}^{1^{*}}_{22}$ or ${\bar{h}}^{2^{*}}_{11}\ne 0$, we have ${\bar{h}}^{p^{*}}_{ijk}=0$ for any i, j, k, p from (4.6). Thus, from (2.25) in Lemma 2.1, we get

$$\begin{aligned} \begin{aligned}&0=S(1-\dfrac{3}{2}S)+2{\bar{H}}^2S-\dfrac{1}{2}{\bar{H}}^4-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2+2{\bar{\lambda }}^2 )\\&=S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2-\dfrac{1}{2}{\bar{H}}^2({\bar{\lambda }}_1-{\bar{\lambda }}_2)^2-2{\bar{H}}^2{\bar{\lambda }}^2. \end{aligned} \end{aligned}$$

Then we conclude

$$\begin{aligned} S\le 2. \end{aligned}$$

This completes the proof of Theorem 4.1. $\square$

Since S is constant, from the result of Cheng and Peng in [8], we know that $S=0$ or $S=1$ if $S\le 1$ . Thus, we only need to prove the following

Theorem 4.2

There are no 2-dimensional complete Lagrangian self-shrinkers $X: M^2\rightarrow {\mathbb {R}}^{4}$ with constant squared norm S of the second fundamental form and $1<S< 2$.

The following lemma is key in this paper.

Lemma 4.1

If $X: M^2\rightarrow {\mathbb {R}}^{4}$ is a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$ with $S=$constant and $1\le S\le 2$, there exists a sequence $\{p_m\}$ in M such that

$$\begin{aligned} \lim _{m\rightarrow \infty } H^2(p_m)=\sup H^2, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ijl}(p_m)={\bar{h}}^{p^{*}}_{ijl}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ij}(p_m)={\bar{h}}^{p^{*}}_{ij}, \end{aligned}$$

for $i, j, l,p=1, 2$, and one can choose an orthonormal frame $e_1, e_2$ at $p_m$ such that ${\bar{\lambda }}={\bar{h}}^{1^{*}}_{12}=0$.

Proof

From (2.25) and (2.26) in Lemma 2.1, we have

$$\begin{aligned} \begin{aligned} \dfrac{1}{2}{\mathcal {L}}H^2&=|\nabla ^{\perp } \vec H|^2+H^2-\sum _{i,j,k,p}(h^{p^{*}}_{ijk})^2-S(1-\dfrac{3}{2}S) -2H^2S+\dfrac{1}{2}H^4\\&=|\nabla ^{\perp } \vec H|^2-\sum _{i,j,k,p}(h^{p^{*}}_{ijk})^2+\dfrac{1}{2}(H^2-S)(H^2-3S+2). \end{aligned} \end{aligned}$$

If, at $p\in M$, $H=0$, we have $H^2<S$. If $H\ne 0$ at $p\in M$, we choose $e_1$, $e_2$ such that

$$\begin{aligned} \vec H=H^{1^{*}}e_{1^{*}}. \end{aligned}$$

From $2ab\le \epsilon a^2+\dfrac{1}{\epsilon }b^2$, we obtain

$$\begin{aligned} S=\lambda _1^2 +3\lambda _2^2+4\lambda ^2, \ \ H^2=(\lambda _1+\lambda _2)^2\le \dfrac{4}{3}(\lambda _1^2 +3\lambda _2^2)\le \dfrac{4}{3}S, \end{aligned}$$

where we denote $\lambda _1= h^{1^{*}}_{11}$, $\lambda _2= h^{1^{*}}_{22}$ and $\lambda =h^{1^{*}}_{12}$. Hence, we have on M

$$\begin{aligned} H^2\le \dfrac{4}{3}S \end{aligned}$$

and the equality holds if and only if $\lambda _1=3\lambda _2$ and $\lambda =0$. Thus, by applying the generalized maximum principle of Cheng and Peng [8] to $H^2$, there exists a sequence $\{p_m\}$ in $M^2$ such that

$$\begin{aligned} \lim _{m\rightarrow \infty } H^2(p_m)=\sup H^2,\quad \lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0,\quad \limsup _{m\rightarrow \infty }{\mathcal {L}}H^2(p_m)\le 0. \end{aligned}$$

Since $X: M^2\rightarrow {\mathbb {R}}^{4}$ is a self-shrinker, we have

$$\begin{aligned} H^{p^{*}}_{,i}=\sum _k h^{p^{*}}_{ik} \langle X,e_k \rangle , \quad i, p=1, 2. \end{aligned}$$

(4.7)

According to $1\le S\le 2$, we know $\sup H^2>0$. Hence, without loss of the generality, at each point $p_m$, we can assume $H(p_m)\ne 0$ and choose $e_1$, $e_2$ such that

$$\begin{aligned} \vec H=H^{1^{*}}e_{1^{*}}. \end{aligned}$$

From (2.25) in Lemma 2.1, Lemma 2.3 and the definition of S, we know that $\{h^{p^{*}}_{ij}(p_m)\}$, $\{h^{p^{*}}_{ijl}(p_m)\}$ and $\{h^{p^{*}}_{ijkl}(p_m)\}$, for any i, j, k, l, p, are bounded sequences. We can assume

$$\begin{aligned} \lim _{m\rightarrow \infty }h^{p^{*}}_{ijl}(p_m)={\bar{h}}^{p^{*}}_{ijl}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ij}(p_m)={\bar{h}}^{p^{*}}_{ij}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ijkl}(p_m)={\bar{h}}^{p^{*}}_{ijkl}, \end{aligned}$$

for $i, j, k, l,p=1, 2$

and get

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}\lim _{m\rightarrow \infty } H^2(p_m)=\sup H^2={\bar{H}}^2,\quad \lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0,\\ &{}0\ge \lim _{m\rightarrow \infty } |\nabla ^{\perp } \vec H|^2(p_m)-\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2+\dfrac{1}{2}({\bar{H}}^2-S)({\bar{H}}^2-3S+2). \end{aligned} \end{array}\right. } \end{aligned}$$

(4.8)

From $\lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0$ and $|\nabla H^2|^2=4\sum _i(\sum _{p^*}H^{p^{*}}H^{p^{*}}_{,i})^2$, we have

$$\begin{aligned} {\bar{H}}^{1^*}_{,k}=0. \end{aligned}$$

(4.9)

From (4.7), we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{\lambda }}_1\lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m)+{\bar{\lambda }} \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0,\\ &{}{\bar{\lambda }}\lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m) +{\bar{\lambda }}_2 \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(4.10)

We will then prove ${\bar{\lambda }}=0$.

Let us assume ${\bar{\lambda }} \ne 0$, we will get a contradiction. The proof is divided into three cases.

Case 1: ${\bar{\lambda }}_2=0$.

Since ${\bar{H}}^2\ne 0$, we have ${\bar{\lambda }}_1\ne 0$. From (4.10), we get

$$\begin{aligned} \lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m)= \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0. \end{aligned}$$

Thus, for $k=1, 2$, from (4.5) and (4.7),

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0, \\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}{\bar{\lambda }}_1{\bar{h}}^{1^{*}}_{11k}+4{\bar{\lambda }} {\bar{h}}^{2^{*}}_{11k}=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(4.11)

We can draw a conclusion, for any i, j, k, p,

$$\begin{aligned} {\bar{h}}^{p^*}_{ijk}=0. \end{aligned}$$

From (4.8), we know $S\le {\bar{H}}^2$, which is in contradiction to $S={\bar{H}}^2+4{\bar{\lambda }}^2>{\bar{H}}^2$ .

Case 2: ${\bar{\lambda }}_1=0$.

In this case, we have

$$\begin{aligned} {\bar{\lambda }}_2\ne 0, \ \ {\bar{H}}^2={\bar{\lambda }}_2^2, \ \ S=3{\bar{\lambda }}_2^2+4{\bar{\lambda }}^2=3{\bar{H}}^2+4{\bar{\lambda }}^2. \end{aligned}$$

From (4.10), we obtain

$$\begin{aligned} \lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m)= \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0. \end{aligned}$$

Therefore, we infer

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0, \\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}3{\bar{\lambda }}_2{\bar{h}}^{1^{*}}_{22k}+4{\bar{\lambda }} {\bar{h}}^{2^{*}}_{11k}=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(4.12)

By solving the above system of equations, we have for any i, j, k, p,

$$\begin{aligned} {\bar{h}}^{p^*}_{ijk}=0. \end{aligned}$$

From (4.8), we know

$$\begin{aligned} ({\bar{H}}^2-S)({\bar{H}}^2-3S+2)=(2{\bar{H}}^2+ 4{\bar{\lambda }}^2)(2S-2+2{\bar{H}}^2+ 4{\bar{\lambda }}^2)\le 0, \end{aligned}$$

it is impossible since $S\ge 1$.

Case 3: ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$.

From (4.10), we have

$$\begin{aligned} ({\bar{\lambda }}_1{\bar{\lambda }}_2 -{\bar{\lambda }}^2)\lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0. \end{aligned}$$

(4.13)

If ${\bar{\lambda }}_1{\bar{\lambda }}_2 = {\bar{\lambda }}^2$, we get, for $k=1, 2$, in view of (4.5) and (4.9),

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0, \\ &{}({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}^{1^{*}}_{11k} +3{\bar{\lambda }} {\bar{h}}^{2^{*}}_{11k}-{\bar{\lambda }} {\bar{h}}^{2^{*}}_{22k}=0. \end{aligned} \end{array}\right. } \end{aligned}$$

By solving the above system of equations, we have

$$\begin{aligned} ({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^2{\bar{h}}^{1^*}_{111}=-4{\bar{\lambda }}^2{\bar{h}}^{2^*}_{222}. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} \begin{aligned}&{\bar{h}}^{1^*}_{111}=\dfrac{-4{\bar{\lambda }}^2}{({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^2}{\bar{h}}^{2^*}_{222},\\&{\bar{h}}^{1^*}_{221}=-{\bar{h}}^{1^*}_{111},\\&{\bar{h}}^{1^*}_{222}=-{\bar{h}}^{2^*}_{111}={-} \dfrac{{\bar{\lambda }}({\bar{\lambda }}_1-3{\bar{\lambda }}_2)}{({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^2}{\bar{h}}^{2^*}_{222}. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} \lim _{m\rightarrow \infty } |\nabla ^{\perp } \vec H|^2(p_m)=({\bar{h}}^{2^*}_{112}+{\bar{h}}^{2^*}_{222})^2= \dfrac{(10 {\bar{\lambda }}^2+{\bar{\lambda }}_1^2+9{\bar{\lambda }}_2^2)^2}{({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^4}({\bar{h}}^{2^*}_{222})^2 \end{aligned}$$

and

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=7({\bar{h}}^{1^*}_{111})^2+8({\bar{h}}^{2^*}_{111})^2+({\bar{h}}^{2^*}_{222})^2= \dfrac{(10 {\bar{\lambda }}^2+{\bar{\lambda }}_1^2+9{\bar{\lambda }}_2^2)^2}{({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^4}({\bar{h}}^{2^*}_{222})^2, \end{aligned}$$

we get the following inequality from (4.8)

$$\begin{aligned} ({\bar{H}}^2-S)({\bar{H}}^2-3 S+2)\le 0, \end{aligned}$$

that is,

$$\begin{aligned} S\le {\bar{H}}^2\le 3S-2. \end{aligned}$$

It is impossible because of $S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2+4{\bar{\lambda }}^2>{\bar{\lambda }}_1^2+ {\bar{\lambda }}_2^2+2{\bar{\lambda }}^2={\bar{H}}^2$. Hence, we obtain ${\bar{\lambda }}_1{\bar{\lambda }}_2 \ne {\bar{\lambda }}^2$.

From (4.10) and (4.13), we have

$$\begin{aligned} \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=\lim _{m\rightarrow \infty } \langle X,e_1 \rangle (p_m)=0. \end{aligned}$$

Thus, we know from (4.7)

$$\begin{aligned} {\bar{H}}^{p^*}_{,k}=0, \end{aligned}$$

for any $k, p=1, 2$. Hence we infer

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{p^*}_{11k}+{\bar{h}}^{p^*}_{22k}=0,\\ &{}({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}^{1^*}_{11k}+4{\bar{\lambda }}{\bar{h}}^{2^*}_{11k}=0. \end{array}\right. } \end{aligned}$$

Through the above system, we have

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0. \end{aligned}$$

From (4.8) and (2.25) in Lemma 2.1, we get

$$\begin{aligned} \begin{aligned}&S\le {\bar{H}}^2\le 3S-2,\\&S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}{\bar{H}}^4-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2+2{\bar{\lambda }}^2)=0. \end{aligned} \end{aligned}$$

(4.14)

From Lemma 2.5 and taking limit,

$$\begin{aligned} \begin{aligned} 0&\le \sum _{i,j,k,l,p}({\bar{h}}_{ijkl}^{p^{*}})^{2} =\frac{1}{2}\lim _{m\rightarrow \infty } {\mathcal {L}}\sum _{i,j,k,p}( h_{ijk}^{p^{*}})^{2}(p_m)\\&={\bar{H}}^{2}\Big [{\bar{H}}^{2}-2S+\frac{1}{2}{\bar{H}}^{4}-{\bar{K}}{\bar{H}}^{2}-{\bar{K}}^{2}\Big ]-{\bar{\lambda }}^2{\bar{H}}^{4}\\&\quad +{\bar{H}}^{2}(S+2-\frac{3}{2}{\bar{H}}^{2}-{\bar{\lambda }}_{1}^{2} -{\bar{\lambda }}_{2}^{2}-2{\bar{\lambda }}^2)({\bar{\lambda }}_{1}^{2}+{\bar{\lambda }}_{2}^{2}+2{\bar{\lambda }}^2)\\&<{\bar{H}}^{2}\Big [{\bar{H}}^{2}-2S+\frac{1}{2}{\bar{H}}^{4}-{\bar{K}}{\bar{H}}^{2}-{\bar{K}}^{2}\Big ]\\&\quad +{\bar{H}}^{2}(S+2-\frac{3}{2}{\bar{H}}^{2}-{\bar{\lambda }}_{1}^{2} -{\bar{\lambda }}_{2}^{2}-2{\bar{\lambda }}^2)({\bar{\lambda }}_{1}^{2}+{\bar{\lambda }}_{2}^{2}+2{\bar{\lambda }}^2). \end{aligned} \end{aligned}$$

According to (4.14), we have

$$\begin{aligned} \begin{aligned}&0<{\bar{H}}^{2}\Big [{\bar{H}}^{2}-2S+\frac{1}{2}{\bar{H}}^{4}-{\bar{K}}{\bar{H}}^{2}-{\bar{K}}^{2}\Big ]\\&\qquad + \Bigg (S+2-\frac{3}{2}{\bar{H}}^{2}-\dfrac{1}{{\bar{H}}^2}\Big (S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}{\bar{H}}^4\Big )\Bigg )\\&\qquad \times \Bigg (S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}{\bar{H}}^4\Bigg )\\&\quad =\dfrac{1}{4{\bar{H}}^2}\Bigg ({\bar{H}}^8-2S{\bar{H}}^6-6S(S-1){\bar{H}}^4+2S(2-3S)^2{\bar{H}}^2-(2-3S)^2S^2\Bigg )\le 0. \end{aligned} \end{aligned}$$

This is a contradiction. In fact, we consider a function f(t) defined by

$$\begin{aligned} \begin{aligned}&f(t)=t^4-2St^3-6S(S-1)t^2+2S(2-3S)^2t-(2-3S)^2S^2, \end{aligned} \end{aligned}$$

(4.15)

for $S\le t\le 3S-2$. Thus, we have

$$\begin{aligned} f^{'}(t)=4t^3-6St^2-12S(S-1)t+2S(2-3S)^2,\ \ f^{''}(t)=12(t^2-St-S(S-1)),\nonumber \\ \end{aligned}$$

(4.16)

$f^{''}(t)<0$ for $t\in (S,\frac{S+\sqrt{S^2+4S(S-1)}}{2})$, $f^{''}(t)>0$ for $t\in (\frac{S+\sqrt{S^2+4S(S-1)}}{2}, 3S-2)$. Hence, $f^{'}(t)$ is a decreasing function for $t\in (S,\frac{S+\sqrt{S^2+4S(S-1)}}{2})$ and $f^{'}(t)$ is an increasing function for $t\in (\frac{S+\sqrt{S^2+4S(S-1)}}{2}, 3S-2)$. According to

$$\begin{aligned} f(S)=2(S-1)(S-2)S^2\le 0,\ \ f(3S-2)=2(3S-2)^2(S-1)(S-2)\le 0, \end{aligned}$$

(4.17)

we conclude $f(t)\le 0$ for $t\in (S, 3S-2)$ because of $f^{'}(S)=4S(S-1)(S-2)\le 0$. Hence, we obtain ${\bar{\lambda }}=0$ and

$$\begin{aligned} {\bar{h}}^{1^{*}}_{ij}={\bar{\lambda }}_{i}\delta _{ij}, \ \ {\bar{H}}={\bar{\lambda }}_1+{\bar{\lambda }}_2, \ \ S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2. \end{aligned}$$

$\square$

Since the proof of Theorem 4.2 is very long, we will divide the proof into three steps.

In the first step, we prove the following:

Proposition 4.1

If $X: M^2\rightarrow {\mathbb {R}}^{4}$ is a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$ with $S=$constant and $1<S<2$, there exists a sequence $\{p_m\}$ in M and at $p_m$, we can choose an orthonormal $e_1, e_2$ such that

$$\begin{aligned} \lim _{m\rightarrow \infty }h^{p^{*}}_{ijl}(p_m)={\bar{h}}^{p^{*}}_{ijl}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ij}(p_m)={\bar{h}}^{p^{*}}_{ij}, \end{aligned}$$

for $i, j, l,p=1, 2$, ${\bar{\lambda }} =0$ and the following holds, either

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0, \ \ \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \\ &{} S<\sup H^2={\bar{H}}^2 \le 3S-2 \ \ \text { and } \ \ S<\sup H^2<\dfrac{4}{3}S, \end{aligned} \end{array}\right. } \end{aligned}$$

(4.18)

or

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{\lambda }}_1=3{\bar{\lambda }}_2, \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \ \ {\bar{h}}^{p^{*}}_{11k}+{\bar{h}}^{p^{*}}_{22k}=0\\ &{}\ \sup H^2=\dfrac{4}{3}S, \ \ S{\ge }\dfrac{6}{5}, \end{aligned} \end{array}\right. } \end{aligned}$$

(4.19)

for $k, p=1, 2$, where we denote ${\bar{\lambda }}_1={\bar{h}}^{1^{*}}_{11}$, ${\bar{\lambda }}_2={\bar{h}}^{1^{*}}_{22}$ and ${\bar{\lambda }}={\bar{h}}^{1^{*}}_{12}$.

Proof

By making use of the same assertion as in the proof of Lemma 4.1, there exists a sequence $\{p_m\}$ in $M^2$ such that

$$\begin{aligned} \lim _{m\rightarrow \infty }h^{p^{*}}_{ijl}(p_m)={\bar{h}}^{p^{*}}_{ijl}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ij}(p_m)={\bar{h}}^{p^{*}}_{ij}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ijkl}(p_m)={\bar{h}}^{p^{*}}_{ijkl}, \end{aligned}$$

for $i, j, k, l,p=1, 2$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}\lim _{m\rightarrow \infty } H^2(p_m)=\sup H^2={\bar{H}}^2,\quad \lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0,\\ &{}0\ge \lim _{m\rightarrow \infty } |\nabla ^{\perp } \vec H|^2(p_m)-\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2+\dfrac{1}{2}({\bar{H}}^2-S)({\bar{H}}^2-3S+2), \end{aligned} \end{array}\right. } \end{aligned}$$

(4.20)

with ${\bar{\lambda }}=0$. From $\lim _{k\rightarrow \infty } |\nabla H^2(p_m)|=0$ and $|\nabla H^2|^2=4\sum _i(\sum _{p^*}H^{p^{*}}H^{p^{*}}_{,i})^2$, we have

$$\begin{aligned} {\bar{H}}^{1^*}_{,k}=0. \end{aligned}$$

(4.21)

From (4.7) and ${\bar{\lambda }}=0$, we have

$$\begin{aligned} \begin{aligned}&{\bar{\lambda }}_1\lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m)=0, \ \ {\bar{\lambda }}_2 \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0, \end{aligned} \end{aligned}$$

it means that,

$$\begin{aligned} {\bar{\lambda }}_i\lim _{m\rightarrow \infty } \langle X,e_i \rangle (p_m)=0. \end{aligned}$$

According to $S={\bar{\lambda }}_1^2 +3{\bar{\lambda }}_2^2>1$ and $\sup H^2=({\bar{\lambda }}_1+{\bar{\lambda }}_2)^2$, if ${\bar{\lambda }}_2=0$, we have

$$\begin{aligned} {\bar{\lambda }}_1\ne 0, \ \ S=\sup H^2, \ \ {\bar{H}}^{2^*}_{,k}=0 \end{aligned}$$

because of $H^{p^{*}}_{,i}=\sum _k h^{p^{*}}_{ik} \langle X,e_k \rangle$. Hence, by using the same calculations as in (4.6), we have

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0,\\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}{\bar{\lambda }}_1{\bar{h}}^{1^*}_{11k}=0. \end{array}\right. } \end{aligned}$$

(4.22)

Then we obtain

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0. \end{aligned}$$

From $S=\sup H^2$ and (2.25), we get $S=1$ or $S=0$. It is impossible. If ${\bar{\lambda }}_1=0$, we have

$$\begin{aligned} {\bar{\lambda }}_2\ne 0, \ \ S=3\sup H^2, \ \ {\bar{H}}^{2*}_{,1}=0. \end{aligned}$$

In this way, by using the same calculations as in (4.6), we get

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0,\\ &{}{\bar{h}}^{2^*}_{111}+{\bar{h}}^{2^*}_{221}=0,\\ &{}3{\bar{\lambda }}_2{\bar{h}}^{1^*}_{22k}=0. \end{array}\right. } \end{aligned}$$

So, we know

$$\begin{aligned} {\bar{h}}^{p^{*}}_{ijk}=0, \ \ \text { except} \ \ i=j=k=p^{*}=2 \end{aligned}$$

and

$$\begin{aligned} |\nabla ^{\perp } \vec H|^2=\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2 \ \ \text {and}\ \ \dfrac{1}{2}(H^2-S)(H^2-3S+2)\le 0. \end{aligned}$$

Hence $S\le \dfrac{3}{4}$. This is also impossible.

We get ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$.

Because of

$$\begin{aligned} H^{1^{*}}_{,i}=\sum _k h^{1^{*}}_{ik} \langle X,e_k \rangle , \ \ H^{2^{*}}_{,i}=\sum _k h^{2^{*}}_{ik} \langle X,e_k \rangle , \end{aligned}$$

for $i=1, 2$, we obtain $\lim _{m\rightarrow \infty } \langle X,e_i \rangle (p_m)=0$ from ${\bar{H}}^{1^*}_{,i}=0$ and ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$. Thus, we have ${\bar{H}}^{2^*}_{,i}=0$, then we get from (4.5), for $i=1, 2$,

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11i}+{\bar{h}}^{1^*}_{22i}=0,\\ &{}{\bar{h}}^{2^*}_{11i}+{\bar{h}}^{2^*}_{22i}=0,\\ &{}({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}^{1^*}_{11i}=0. \end{array}\right. } \end{aligned}$$

(4.23)

If ${\bar{\lambda }}_1 \ne 3{\bar{\lambda }}_2$, we have

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0. \end{aligned}$$

Therefore, from (4.8) and (2.25), we get

$$\begin{aligned} S < \sup H^2 \le 3S-2. \end{aligned}$$

If ${\bar{\lambda }}_1 = 3{\bar{\lambda }}_2$, we have $\sup H^2=\dfrac{4}{3}S$ and $\lim _{m\rightarrow \infty } |\nabla ^{\perp } \vec H|^2(p_m)=0$. From (2.25), we know

$$\begin{aligned} \begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2&=-S(1-\dfrac{1}{2}S)+(S-\sup H^2)^2 -\dfrac{1}{2}(\sup H^2)^2+{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)\\&=(\frac{5}{6}S-1)S. \end{aligned} \end{aligned}$$

(4.24)

Hence, we get $S\ge \dfrac{6}{5}$. When $S= \dfrac{6}{5}$, from (4.24), we have $\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0$. It finishes the proof of Proposition 4.1. $\square$

In the step 2, we prove the following:

Proposition 4.2

Under the assumptions of Proposition 4.1, the formula (4.19) in Proposition 4.1 does not occur.

Proof

If the formula (4.19) holds, we have

$$\begin{aligned} {\bar{\lambda }}=0, \ \ {\bar{\lambda }}_1 = 3{\bar{\lambda }}_2, \ \ S{\ge }\dfrac{6}{5}, \ \ \sup H^2={\bar{H}}^2=\dfrac{4}{3}S \end{aligned}$$

and

$$\begin{aligned} {\bar{H}}_{,k}^{p^{*}}={\bar{h}}^{p^{*}}_{11k}+{\bar{h}}^{p^{*}}_{22k}=0, \ \text { for}\ k, p=1, 2. \end{aligned}$$

(4.25)

From (2.6) and (4.25), we have

$$\begin{aligned} \begin{aligned}&\sum _{i,j}({\bar{h}}^{1^{*}}_{ij1})^2=({\bar{h}}^{1^{*}}_{111})^2+({\bar{h}}^{1^{*}}_{221})^2+2({\bar{h}}^{1^{*}}_{121})^2 =2({\bar{h}}^{1^{*}}_{111})^2+2({\bar{h}}^{1^{*}}_{112})^2\\&\sum _{i,j}({\bar{h}}^{2^{*}}_{ij1})^2=({\bar{h}}^{2^{*}}_{111})^2+({\bar{h}}^{2^{*}}_{221})^2+2({\bar{h}}^{2^{*}}_{121})^2 =2({\bar{h}}^{1^{*}}_{111})^2+2({\bar{h}}^{1^{*}}_{112})^2\\&\sum _{i,j,p}({\bar{h}}^{p^{*}}_{ij1})^2 =\sum _{p}\Big [({\bar{h}}^{p^{*}}_{111})^2+({\bar{h}}^{p^{*}}_{221})^2+2({\bar{h}}^{p^{*}}_{121})^2\Big ] =\sum _{p}\Big [2({\bar{h}}^{p^{*}}_{111})^2+2({\bar{h}}^{p^{*}}_{112})^2\Big ]\\&\sum _{i,j,p}({\bar{h}}^{p^{*}}_{ij2})^2 =\sum _{p}\Big [({\bar{h}}^{p^{*}}_{112})^2+({\bar{h}}^{p^{*}}_{222})^2+2({\bar{h}}^{p^{*}}_{122})^2\Big ] =\sum _{p}\Big [2({\bar{h}}^{p^{*}}_{111})^2+2({\bar{h}}^{p^{*}}_{112})^2\Big ]\\&\sum _{i,j,p}({\bar{h}}^{p^{*}}_{ijk})^2 =\sum _{i,j,p}({\bar{h}}^{p^{*}}_{ij1})^2+\sum _{i,j,p}({\bar{h}}^{p^{*}}_{ij2})^2 =8({\bar{h}}^{1*}_{111})^2+8({\bar{h}}^{1*}_{112})^2. \end{aligned} \end{aligned}$$

From (2.25) in Lemma 2.1 and ${\bar{H}}^2=\dfrac{4}{3}S$, we get

$$\begin{aligned} \sum _{i,j,p^{*}}({\bar{h}}^{p^{*}}_{ijk})^2=(\frac{5}{6}S-1)S. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \sum _{i,j}({\bar{h}}^{1^{*}}_{ij1})^2=\sum _{i,j}({\bar{h}}^{2^{*}}_{ij1})^2=\dfrac{1}{4}(\frac{5}{6}S-1)S \end{aligned}$$

and

$$\begin{aligned} \sum _{i,j,p^{*}}({\bar{h}}^{p^{*}}_{ij1})^2=\sum _{i,j,p^{*}}({\bar{h}}^{p^{*}}_{ij2})^2=\dfrac{1}{2}(\frac{5}{6}S-1)S. \end{aligned}$$

Since

$$\begin{aligned}&{\bar{H}}{\bar{\lambda }}_1=\frac{{\bar{H}}^2-S}{2}+{\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2=\dfrac{{\bar{H}}^2}{2}+\dfrac{S}{3}=S,\\&{\bar{H}}({\bar{\lambda }}_1^3+ {\bar{\lambda }}_2^3)={\bar{H}}^2({\bar{\lambda }}_1^2 -{\bar{\lambda }}_1{\bar{\lambda }}_2+{\bar{\lambda }}_2^2)=\dfrac{7S^2}{9}, \end{aligned}$$

according to (2.35) in Lemma 2.3, we get

$$\begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p^{*}}(h^{p^{*}}_{ijk})^2(p_m)\nonumber \\&\quad =({\bar{H}}^2-2S){\bar{H}}^2+(\dfrac{1}{2}{\bar{H}}^2+\dfrac{S}{2}+2){\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)\nonumber \\&\qquad -\dfrac{{\bar{H}}^2-S}{2}({\bar{H}}^4+{\bar{H}}^3{\bar{\lambda }}_1)-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)^2 -{\bar{H}}^3({\bar{\lambda }}_1^3+{\bar{\lambda }}_2^3)+{\bar{H}}^2\sum _{i,j,k}({\bar{h}}^{1^{*}}_{ijk})^2\nonumber \\&\quad =({\bar{H}}^2-2S){\bar{H}}^2+(\dfrac{1}{2}{\bar{H}}^2+\dfrac{S}{2}+2){\bar{H}}^2\dfrac{5S}{6}\nonumber \\&\qquad -\dfrac{{\bar{H}}^2-S}{2}\Big ({\bar{H}}^4+{\bar{H}}^2S\Big ) -{\bar{H}}^2(\dfrac{5S}{6})^2-{\bar{H}}^2\dfrac{7S^2}{9}+{\bar{H}}^2\sum _{i,j,k}({\bar{h}}^{1^{*}}_{ijk})^2\nonumber \\&\quad =S^2(\dfrac{2}{3}-\dfrac{17}{27}S)<0. \end{aligned}$$

(4.26)

On the other hand, from (2.34) in Lemma 2.3, we have

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p^{*}} (h^{p^{*}}_{ijk})^2(p_m)\\&\quad =\sum _{i,j,k,l,p^{*}}({\bar{h}}^{p^{*}}_{ijkl})^2+(5{\bar{H}}^2-5S+2)\sum _{i,j,k,p^{*}}({\bar{h}}^{p^{*}}_{ijk})^2\\&\qquad -2{\bar{H}}\sum _{i,j,k,p^{*}}({\bar{h}}^{p^{*}}_{ijk})^2{\bar{h}}_{kk}^{1^{*}} -{\bar{H}}\sum _{i,j,k}({\bar{h}}^{1*}_{ijk})^2{\bar{h}}_{kk}^{1^{*}}\\&\quad =\sum _{i,j,k,l,p^{*}}({\bar{h}}^{p^{*}}_{ijkl})^2+2\sum _{i,j,k,p^{*}}({\bar{h}}^{p^{*}}_{ijk})^2{\ge } 0. \end{aligned} \end{aligned}$$

(4.27)

Hence, we conclude that (4.26) is in contradiction to (4.27). It completes the proof of Proposition 4.2. $\square$

In the step 3, we prove the following:

Proposition 4.3

Under the assumptions of Proposition 4.1, the formula (4.18) in Proposition 4.1 does not occur either.

In this case, we have $\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0$, ${\bar{\lambda }}=0$ and ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$.

Since ${\bar{H}}={\bar{\lambda }}_1+{\bar{\lambda }}_2$ and S=${\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2$, we get

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}\pm \sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}\mp \sqrt{4S-3{\bar{H}}^2}}{4}. \end{aligned}$$

Lemma 4.2

Under the assumptions of Proposition 4.1, if

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0, \ \ {\bar{\lambda }}=0, \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \\ &{} S<\sup H^2={\bar{H}}^2 \le 3S-2 \ \ \text { and } \ \ S<\sup H^2<\dfrac{4}{3}S \end{aligned} \end{array}\right. } \end{aligned}$$

(4.28)

is satisfied, then

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}+ \sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}- \sqrt{4S-3{\bar{H}}^2}}{4}. \end{aligned}$$

do not occur.

Proof

If

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}+ \sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}-\sqrt{4S-3{\bar{H}}^2}}{4} \end{aligned}$$

hold, we have

$$\begin{aligned} {\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2=\dfrac{{\bar{H}}^2+2S+ \sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}. \end{aligned}$$

(4.29)

Due to ${\bar{\lambda }}_1\ne 3{\bar{\lambda }}_2$, we know ${\bar{H}}^2<\dfrac{4}{3}S$ and $\dfrac{4S}{3} \le 3S-2$ if and only if $S\ge \dfrac{6}{5}$. According to (2.25) and $\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0$, we have

$$\begin{aligned} \begin{aligned} S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}({\bar{H}}^2)^2-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)=0. \end{aligned} \end{aligned}$$

(4.30)

We get from (4.29)

$$\begin{aligned} \begin{aligned} S(1-\dfrac{3}{2}S)+\dfrac{3}{2}S{\bar{H}}^2-\dfrac{3}{4}{\bar{H}}^4-\dfrac{{\bar{H}}^2\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}=0. \end{aligned} \end{aligned}$$

(4.31)

We consider function

$$\begin{aligned} f(x)=S(1-\dfrac{3}{2}S)+\dfrac{3}{2}Sx-\dfrac{3}{4}x^2-\dfrac{x\sqrt{(4S-3x)x}}{4} \end{aligned}$$

for $S<x< \dfrac{4}{3}S$. We know that

$$\begin{aligned} f(S)=S(1-S)<0 \end{aligned}$$

since $1<S<2$,

$$\begin{aligned} \begin{aligned} f^{\prime }(x)=\dfrac{df(x)}{dx}&=\dfrac{3}{2}S-\dfrac{3}{2}x-\dfrac{\sqrt{(4S-3x)x}}{4}-\dfrac{x(2S-3x)}{4\sqrt{(4S-3x)x}}\\&=\dfrac{3}{2}(S-x)-\dfrac{3x(S-x)}{2\sqrt{(4S-3x)x}}\\&=\dfrac{3}{2}(S-x)\Big (1-\dfrac{x}{\sqrt{(4S-3x)x}}\Big )>0 \end{aligned} \end{aligned}$$

(4.32)

since $S<x$ and $x>\sqrt{(4S-3x)x}$. Thus, f(x) is an increasing function of x.

If $S\ge \dfrac{6}{5}$, then $\dfrac{4}{3}S\le 3S-2$. Hence, we have $S<{\bar{H}}^2< \dfrac{4}{3}S$.

Since

$$\begin{aligned} f(\dfrac{4}{3}S)=S(1-\dfrac{3}{2}S)+\dfrac{3}{2}S\dfrac{4}{3}S-\dfrac{3}{4}(\dfrac{4}{3}S)^2=S(1-\dfrac{5}{6}S) \le 0, \end{aligned}$$

we conclude $f(x)<0$ for any $x \in (S, \dfrac{4}{3} S)$, which is in contradiction to (4.31). Thus, we must have $S<\dfrac{6}{5}$. In this case, $\dfrac{4S}{3} >3S-2$ and

$$\begin{aligned} \begin{aligned} f(3S-2)&=S(1-\dfrac{3}{2}S)+\dfrac{3}{2}S(3S-2)-\dfrac{3}{4}(3S-2)^2\\&\quad -\dfrac{(3S-2)\sqrt{(4S-3(3S-2))(3S-2)}}{4}\\&=(3S-2)\Bigg (\dfrac{3}{2}-\dfrac{5}{4}S-\dfrac{\sqrt{(6-5S)(3S-2)}}{4}\Bigg )\\&=(3S-2)\dfrac{\sqrt{6-5S}}{4}\Big (\sqrt{6-5S}-\sqrt{3S-2}\Big )<0. \end{aligned} \end{aligned}$$

Therefore, it is also impossible. It finishes the proof of Lemma 4.2.

$\square$

Lemma 4.3

Under the assumptions of Proposition 4.1, if

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0, \ \ {\bar{\lambda }}=0, \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \\ &{} S<\sup H^2={\bar{H}}^2 \le 3S-2 \ \ \text { and } \ \ S<\sup H^2<\dfrac{4}{3}S, \end{aligned} \end{array}\right. } \end{aligned}$$

(4.33)

is satisfied, then we have

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}- \sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}+\sqrt{4S-3{\bar{H}}^2}}{4}. \end{aligned}$$

and $S\ge \dfrac{6}{5}$.

Proof

According to Lemma 4.2, we must have

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}-\sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}+\sqrt{4S-3{\bar{H}}^2}}{4}. \end{aligned}$$

Thus,

$$\begin{aligned} {\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2=\dfrac{{\bar{H}}^2+2S- \sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}. \end{aligned}$$

(4.34)

If $S< \dfrac{6}{5}$ holds, then we get ${\bar{\lambda }}_1\ne 3{\bar{\lambda }}_2$ and $\dfrac{4S}{3} > 3S-2$. According to (2.25), we have

$$\begin{aligned} \begin{aligned} S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}({\bar{H}}^2)^2-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)=0. \end{aligned} \end{aligned}$$

we obtain from (4.34)

$$\begin{aligned} \begin{aligned} S(1-\dfrac{3}{2}S)+\dfrac{3}{2}S{\bar{H}}^2-\dfrac{3}{4}{\bar{H}}^4+\dfrac{{\bar{H}}^2\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}=0. \end{aligned} \end{aligned}$$

(4.35)

Now we consider function

$$\begin{aligned} f_1(x)=S(1-\dfrac{3}{2}S)+\dfrac{3}{2}Sx-\dfrac{3}{4}x^2+\dfrac{x\sqrt{(4S-3x)x}}{4} \end{aligned}$$

for $S<x\le 3S-2$. Since

$$\begin{aligned} \begin{aligned} f_1^{\prime }(x)=\dfrac{df_1(x)}{dx}&=\dfrac{3}{2}S-\dfrac{3}{2}x+\dfrac{\sqrt{(4S-3x)x}}{4}+\dfrac{x(2S-3x)}{4\sqrt{(4S-3x)x}}\\&=\dfrac{3}{2}(S-x)+\dfrac{3x(S-x)}{2\sqrt{(4S-3x)x}}\\&=\dfrac{3}{2}(S-x)\Big (1+\dfrac{x}{\sqrt{(4S-3x)x}}\Big )<0 \end{aligned} \end{aligned}$$

for $S<x$, $f_1(x)$ is a decreasing function of x on $(S,3S-2)$.

$$\begin{aligned} \begin{aligned} f_1(3S-2)&=S(1-\dfrac{3}{2}S)+\dfrac{3}{2}S(3S-2)-\dfrac{3}{4}(3S-2)^2\\&\quad +\dfrac{(3S-2)\sqrt{(4S-3(3S-2))(3S-2)}}{4}\\&=(3S-2)\Bigg (\dfrac{3}{2}-\dfrac{5}{4}S+\dfrac{\sqrt{(6-5S)(3S-2)}}{4}\Bigg )\\&=(3S-2)\dfrac{\sqrt{6-5S}}{4}\Big (\sqrt{6-5S}+\sqrt{3S-2}\Big )>0 \end{aligned} \end{aligned}$$

since $S< \dfrac{6}{5}$. Thus $f_1(x)>0$ for any $x \in (S, 3S-2]$, which is in contradiction to (4.35). $\square$

Lemma 4.4

Under the assumptions of Proposition 4.1, if

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0, \ \ {\bar{\lambda }}=0, \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \\ &{} S<\sup H^2={\bar{H}}^2 \le 3S-2 \ \ \text { and } \ \ S<\sup H^2<\dfrac{4}{3}S, \end{aligned} \end{array}\right. } \end{aligned}$$

(4.36)

are satisfied, then we have $1.89\le S<2$.

Proof

According to Lemmas 4.2 and 4.3, we know $2>S\ge \dfrac{6}{5}$ and

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}- \sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}+\sqrt{4S-3{\bar{H}}^2}}{4}. \end{aligned}$$

In this case, $\dfrac{4S}{3}\le 3S-2$. Hence, we have

$$\begin{aligned} \dfrac{6}{5} \le S<2, \ \ S<{\bar{H}}^2<\dfrac{4}{3}S, \ \ \sum _{i,j,k,p^{*}}({\bar{h}}^{p^{*}}_{ijk})^2=0 \end{aligned}$$

and

$$\begin{aligned} {\bar{\lambda }}_1=\dfrac{3{\bar{H}}-\sqrt{4S-3{\bar{H}}^2}}{4}, \ \quad {\bar{\lambda }}_2=\dfrac{{\bar{H}}+\sqrt{4S-3{\bar{H}}^2}}{4}. \end{aligned}$$

From (2.35) of Lemma 2.3 in Sect. 2, we get

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p}(h^{p^{*}}_{ijk})^2(p_m)\\&\quad =({\bar{H}}^2-2S){\bar{H}}^2+(\dfrac{1}{2}{\bar{H}}^2+\dfrac{S}{2}+2){\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)\\&\qquad -\dfrac{{\bar{H}}^2-S}{2}({\bar{H}}^4+H^3{\bar{\lambda }}_1)-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)^2 -{\bar{H}}^3({\bar{\lambda }}_1^3+{\bar{\lambda }}_2^3). \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} H\lambda _1=\frac{H^2-S}{2}+\lambda _1^2+\lambda _2^2, \ \ \lambda _1^3 + \lambda _2^3=H(\lambda _1^2-\lambda _1\lambda _2+\lambda _2^2), \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p} (h^{p^{*}}_{ijk})^2(p_m)\\&\quad ={\bar{H}}^2\Big [{\bar{H}}^2-2S-\dfrac{({\bar{H}}^2-S)^2}{4}-\dfrac{{\bar{H}}^2({\bar{H}}^2-S)}{2}+\dfrac{{\bar{H}}^4}{2} \Big ]\\&\qquad +{\bar{H}}^2\Big (S-\dfrac{3{\bar{H}}^2}{2} +2-{\bar{\lambda }}_1^2-{\bar{\lambda }}_2^2\Big )({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2). \end{aligned} \end{aligned}$$

By making use of

$$\begin{aligned} {\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2=\dfrac{{\bar{H}}^2+2S- \sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}, \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p} ( h^{p^{*}}_{ijk})^2(p_m)\\&\quad ={\bar{H}}^2\Bigg [-\dfrac{{\bar{H}}^4}{2}+\dfrac{3{\bar{H}}^2}{2} -S +\dfrac{({\bar{H}}^2-1)\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{2} \Bigg ].\\ \end{aligned} \end{aligned}$$

(4.37)

On the other hand, from (2.34) and (2.35)

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p} (h^{p^{*}}_{ijk})^2(p_m) =\lim _{m\rightarrow \infty }\sum _{i,j,k,l,p}({\bar{h}}^{p^{*}}_{ijkl})^2(p_m)\\&\quad \ge 2({\bar{h}}_{1122}^{1^{*}}-{\bar{h}}_{2211}^{1^{*}})^{2}+2({\bar{h}}_{1122}^{2^{*}}-{\bar{h}}_{2211}^{2^{*}})^{2} +\frac{1}{2}({\bar{h}}_{2222}^{1^{*}}-{\bar{h}}_{2221}^{2^{*}})^{2}. \\ \end{aligned} \end{aligned}$$

From Gauss equation and Ricci identities, we have

$$\begin{aligned} h_{2222}^{1^{*}}-h_{2221}^{2^{*}}= & {} h_{2212}^{2^{*}}-h_{2221}^{2^{*}}\\= & {} \sum _mh_{m2}^{2^{*}}R_{m212}+\sum _mh_{2m}^{2^{*}}R_{m212}+\sum _mh_{22}^{m^{*}}R_{m212}\\= & {} (h_{12}^{2^{*}}+h_{21}^{2^{*}}+h_{22}^{1^{*}})R_{1212}\\= & {} 3\lambda _{2}K(\delta _{11}\delta _{22}-\delta _{12}\delta _{21})\\= & {} 3\lambda _{2}K,\\ h_{1122}^{1^{*}}-h_{2211}^{1^{*}}= & {} h_{1212}^{1^{*}}-h_{1221}^{1^{*}}\\= & {} \sum _mh_{m2}^{1^{*}}R_{m112}+\sum _mh_{1m}^{1^{*}}R_{m212}+\sum _mh_{12}^{m^{*}}R_{m112}\\= & {} \lambda _{2}R_{2112}+\lambda _{1}R_{1212}+\lambda _{2}R_{2112}\\= & {} (\lambda _{1}-2\lambda _{2})K. \end{aligned}$$

From the above equations, we obtain

$$\begin{aligned} \begin{aligned}&2(h_{1122}^{1^{*}}-h_{2211}^{1^{*}})^{2}+2(h_{1122}^{2^{*}}-h_{2211}^{2^{*}})^{2}+\frac{1}{2}(h_{2222}^{1^{*}}-h_{2221}^{2^{*}})^{2} \\&\quad =2(\lambda _{1}-2\lambda _{2})^{2}K^{2}+\frac{1}{2}(3\lambda _{2}K)^{2}\\&\quad =K^{2}[2\lambda _{1}^{2}-8\lambda _{1}\lambda _{2}+\frac{25}{2}\lambda _{2}^{2}]\\&\quad =K^{2}[2S-4(H^{2}-\lambda _{1}^{2}-\lambda _{2}^{2})+\frac{13}{2}\lambda _{2}^{2}]\\&\quad =K^{2}(6S-4H^{2}-\frac{3}{2}\lambda _{2}^{2}). \end{aligned} \end{aligned}$$

Thus, we have

$$\begin{aligned} \begin{aligned}&\dfrac{1}{2}\lim _{m\rightarrow \infty }{\mathcal {L}}\sum _{i,j,k,p^{*}} (h^{p^{*}}_{ijk})^2(p_m) =\lim _{m\rightarrow \infty }\sum _{i,j,k,l,p^{*}}({\bar{h}}^{p^{*}}_{ijkl})^2(p_m)\\&\quad \ge 2({\bar{h}}_{1122}^{1^{*}}-{\bar{h}}_{2211}^{1^{*}})^{2}+2({\bar{h}}_{1122}^{2^{*}}-{\bar{h}}_{2211}^{2^{*}})^{2} +\frac{1}{2}({\bar{h}}_{2222}^{1^{*}}-{\bar{h}}_{2221}^{2^{*}})^{2} \\&\quad \ge \dfrac{({\bar{H}}^2-S)^2}{4}\Big (6S-4{\bar{H}}^2-\dfrac{3{\bar{\lambda }}_2^2}{2} \Big )\\&\quad =\dfrac{({\bar{H}}^2-S)^2}{4}\Big [(6-\dfrac{3}{8})S-(4-\dfrac{3}{16}){\bar{H}}^2-\dfrac{3\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{16} \Big ] \end{aligned} \end{aligned}$$

Hence, we obtain, in view of (4.37) and (4.38)

$$\begin{aligned} \begin{aligned}&{\bar{H}}^2\Big [-\dfrac{{\bar{H}}^4}{2}+\dfrac{3{\bar{H}}^2}{2} -S +\dfrac{({\bar{H}}^2-1)\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{2} \Big ]\\&\quad \ge \dfrac{({\bar{H}}^2-S)^2}{4}\Big [(6-\dfrac{3}{8})S-(4-\dfrac{3}{16}){\bar{H}}^2 -\dfrac{3\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{16} \Big ]. \end{aligned} \end{aligned}$$

From (2.25) and $\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0$, we know

$$\begin{aligned} \dfrac{{\bar{H}}^2\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}=-S(1-\dfrac{3}{2}S)-\dfrac{3}{2}S{\bar{H}}^2+\dfrac{3}{4}{\bar{H}}^4. \end{aligned}$$

Therefore, we conclude

$$\begin{aligned} \begin{aligned}&{\bar{H}}^6-3{\bar{H}}^4S +3{\bar{H}}^2S^2-3S^2+2S\\&\quad \ge \dfrac{({\bar{H}}^2-S)^2}{4}\Big [(6+\dfrac{3}{4})S-(4+\dfrac{3}{8}){\bar{H}}^2 -\dfrac{3S(3S-2)}{8{\bar{H}}^2} \Big ]. \end{aligned} \end{aligned}$$

(4.38)

Since $-\dfrac{3}{4}x-\dfrac{3S(3S-2)}{8x}$ is a decreasing function of x, for $S<x<\dfrac{4S}{3}$, we have

$$\begin{aligned} -\dfrac{3}{4}{\bar{H}}^2-\dfrac{3S(3S-2)}{8{\bar{H}}^2} > -S-\dfrac{9(3S-2)}{32}. \end{aligned}$$

Hence, we get

$$\begin{aligned} \begin{aligned}&{\bar{H}}^6-3{\bar{H}}^4S +3{\bar{H}}^2S^2-3S^2+2S\\&\quad \ge \dfrac{({\bar{H}}^2-S)^2}{4}\Big [(5-\dfrac{3}{32})S-(4-\dfrac{3}{8}){\bar{H}}^2+\dfrac{9}{16}\Big ]. \end{aligned} \end{aligned}$$

(4.39)

We consider a function $g=g(x)$ of x defined by

$$\begin{aligned} g(x)= & {} x^3-3x^2S +3xS^2-3S^2+2S- \dfrac{(x-S)^2}{4}\Big [(5-\dfrac{3}{32})S-(4-\dfrac{3}{8})x+\dfrac{9}{16} \Big ].\\ g^{\prime }(x)= & {} 3x^2-6xS +3S^2+\dfrac{(x-S)^2}{4}(4-\dfrac{3}{8})\\&- \dfrac{(x-S)}{2}\Big [(5-\dfrac{3}{32})S-(4-\dfrac{3}{8})x+\dfrac{9}{16}\Big ]\\= & {} (x-S)\Big [(6-\dfrac{9}{32})x-(6+\dfrac{23}{64})S-\dfrac{9}{32}\Big ]. \end{aligned}$$

Hence, g(x) attains its minimum at $(6-\dfrac{9}{32})x-(6+\dfrac{23}{64})S-\dfrac{9}{32}=0$.

$$\begin{aligned} g(S)=S(S-1)(S-2)<0, \ \ g(\dfrac{4S}{3})= (1+\dfrac{121}{36\cdot 96})S^3-\dfrac{193}{64}S^2+2S<0 \end{aligned}$$

if $\dfrac{6}{5}\le S<1.89$. We have $g(x)<0$ for $S<x<\dfrac{4S}{3}$, which is in contradiction to (4.39). Hence, S satisfies

$$\begin{aligned} 1.89\le S<2. \end{aligned}$$

$\square$

Lemma 4.5

Under the assumptions of Proposition 4.1, if

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0, \ \ {\bar{\lambda }}=0, \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \\ &{} S<\sup H^2={\bar{H}}^2 \le 3S-2 \ \ \text { and } \ \ S<\sup H^2<\dfrac{4}{3}S, \end{aligned} \end{array}\right. } \end{aligned}$$

is satisfied, then we have

$$\begin{aligned} S<{\bar{H}}^2\le S+\dfrac{1}{5}S. \end{aligned}$$

Proof

Since ${\bar{H}}^2=\sup H^2$, if $S+\dfrac{S}{5}<{\bar{H}}^2<S+\dfrac{S}{3}$, we consider a function $f_2=f_2(x)$ of x defined by

$$\begin{aligned} f_2(x)=S(1-\dfrac{3}{2}S)+\dfrac{3}{2}Sx-\dfrac{3}{4}x^2+\dfrac{x\sqrt{(4S-3x)x}}{4} \end{aligned}$$

for $\dfrac{6}{5}S<x\le \dfrac{4}{3}S$. We know that

$$\begin{aligned} f_2(\dfrac{6S}{5})=S(1- \dfrac{39-6\sqrt{3}}{50}S)<0 \end{aligned}$$

since $1.89<S<2$,

$$\begin{aligned} \begin{aligned} f_2^{\prime }(x)=\dfrac{df(x)}{dx}&=\dfrac{3}{2}S-\dfrac{3}{2}x+\dfrac{\sqrt{(4S-3x)x}}{4}+\dfrac{x(2S-3x)}{4\sqrt{(4S-3x)x}}\\&=\dfrac{3}{2}(S-x)+\dfrac{3x(S-x)}{2\sqrt{(4S-3x)x}}\\&=\dfrac{3}{2}(S-x)\Big (1+\dfrac{x}{\sqrt{(4S-3x)x}}\Big )<0. \end{aligned} \end{aligned}$$

$f_2(x)$ is a decreasing function of x and we can not have

$$\begin{aligned} \begin{aligned} S(1-\dfrac{3}{2}S)+\dfrac{3}{2}S{\bar{H}}^2-\dfrac{3}{4}{\bar{H}}^4+\dfrac{{\bar{H}}^2\sqrt{(4S-3{\bar{H}}^2){\bar{H}}^2}}{4}=0. \end{aligned} \end{aligned}$$

Hence, we must have

$$\begin{aligned} S<{\bar{H}}^2<\dfrac{6}{5}S. \end{aligned}$$

$\square$

Proof of Proposition 4.3

According to Lemma 4.4 and Lemma 4.5, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=0, \ \ {\bar{\lambda }}=0, \ \ {\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0, \\ &{} S<\sup H^2<\dfrac{6}{5}S, \ \ 1.89<S<2. \end{aligned} \end{array}\right. } \end{aligned}$$

We obtain from (4.38)

$$\begin{aligned} \begin{aligned}&{\bar{H}}^6-3{\bar{H}}^4S +3{\bar{H}}^2S^2-3S^2+2S\\&\quad \ge \dfrac{({\bar{H}}^2-S)^2}{4}\Big [(6+\dfrac{3}{4})S-(4+\dfrac{3}{8}){\bar{H}}^2 -\dfrac{3S(3S-2)}{8{\bar{H}}^2} \Big ]. \end{aligned} \end{aligned}$$

Since $-\dfrac{3}{4}{\bar{H}}^2-\dfrac{3S(3S-2)}{8{\bar{H}}^2}$ is a decreasing function of ${\bar{H}}^2$, for $S<{\bar{H}}^2<\dfrac{6S}{5}$, we have

$$\begin{aligned} -\dfrac{3}{4}{\bar{H}}^2-\dfrac{3S(3S-2)}{8{\bar{H}}^2} > -\dfrac{9}{10}S-\dfrac{5(3S-2)}{16}. \end{aligned}$$

$$\begin{aligned} \begin{aligned}&{\bar{H}}^6-3{\bar{H}}^4S +3{\bar{H}}^2S^2-3S^2+2S\\&\quad \ge \dfrac{({\bar{H}}^2-S)^2}{4}\Big [(5-\dfrac{7}{80})S-(4-\dfrac{3}{8}){\bar{H}}^2+\dfrac{5}{8} \Big ]. \end{aligned} \end{aligned}$$

(4.40)

We consider function

$$\begin{aligned} f_3(x)= & {} x^3-3x^2S +3xS^2-3S^2+2S- \dfrac{(x-S)^2}{4}\Big [(5-\dfrac{7}{80})S-(4-\dfrac{3}{8})x+\dfrac{5}{8} \Big ].\\ f_3^{\prime }(x)= & {} 3x^2-6xS +3S^2+\dfrac{(x-S)^2}{4}(4-\dfrac{3}{8})\\&- \dfrac{(x-S)}{2}\Big [(5-\dfrac{7}{80})S-(4-\dfrac{3}{8})x+\dfrac{5}{8}\Big ]\\= & {} (x-S)\Big [(6-\dfrac{9}{32})x-(6+\dfrac{29}{80})S-\dfrac{5}{16}\Big ]. \end{aligned}$$

Hence, $f_3(x)$ attains its minimum at $(6-\dfrac{9}{32})x-(6+\dfrac{29}{80})S-\dfrac{5}{16}=0$.

$$\begin{aligned} f_3(S)=S(S-1)(S-2)<0, \ \ f_3(\dfrac{6S}{5})= (1+\dfrac{1}{125}-\dfrac{9}{1600})S^3-(3\!+\dfrac{1}{160})S^2\!+\!2S\!<0 \end{aligned}$$

if $\dfrac{6}{5}\le S<2$. This is in contradiction to (4.40). Therefore, we conclude that the formula (4.18) in Proposition 4.1 does not occur either. $\square$

Proof of Theorem 4.2

According to Propositions 4.1, 4.2 and 4.3, we know that there are no 2-dimensional complete Lagrangian self-shrinkers $X: M^2\rightarrow {\mathbb {R}}^{4}$ with constant squared norm S of the second fundamental form and $1<S<2$. $\square$

Theorem 4.3

Let $X: M^2\rightarrow {\mathbb {R}}^{4}$ be a 2-dimensional Lagrangian self-shrinker in ${\mathbb {R}}^4$. If $S\equiv 2$ or $S\equiv 1$, then the mean curvature H satisfies $H\ne 0$ on $M^2$.

Proof

If there exists a point $p \in M^2$ such that $H=0$ at p, then we know $H^{1^*}=H^{2^*}=0$. Thus, at p, we have

$$\begin{aligned} H=0, \ \ H^{1^*}=\lambda _1+\lambda _2=0, \ \ \lambda =h_{12}^{1^*}=-h_{22}^{2^*}. \end{aligned}$$

From

$$\begin{aligned} H^{p^*}_{,i}=\sum _{k}h_{ik}^{p^*}\langle X, e_k\rangle , \ \ \text { for} \ \ i, p=1, 2, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&h_{111}^{1^*}+h_{221}^{1^*}=H^{1^*}_{,1}=\lambda _1\langle X, e_1\rangle +\lambda \langle X, e_2\rangle ,\\&h_{112}^{1^*}+h_{222}^{1^*}=H^{1^*}_{,2}=\lambda \langle X, e_1\rangle -\lambda _1\langle X, e_2\rangle ,\\&h_{112}^{2^*}+h_{222}^{2^*}=H^{2^*}_{,2}=-\lambda _1\langle X, e_1\rangle -\lambda \langle X, e_2\rangle \end{aligned} \end{aligned}$$

and

$$\begin{aligned} 0=\dfrac{1}{2}\nabla _i S=\lambda _1(h_{11i}^{1^*}-3h_{22i}^{1^*})+\lambda (3h_{11i}^{2^*}-h^{2^{*}}_{22i}), \ \ \text {for } \ i=1, 2, \end{aligned}$$

it means that,

$$\begin{aligned} \lambda _1(h_{11i}^{1^*}-3h_{22i}^{1^*})+\lambda (3h_{11i}^{2^*}-h^{2^{*}}_{22i})=0, \ \ \text {for } \ i=1, 2 \end{aligned}$$

since S is constant. Thus, we get a system of linear equations

$$\begin{aligned} \begin{pmatrix} &{}- \lambda &{} -3 \lambda _1 &{}3 \lambda &{} \lambda _1 &{}0\\ &{}0&{}-\lambda &{} -3{\bar{\lambda }}_1 &{}3 \lambda &{} \lambda _1 \\ &{}0&{}0&{}1&{}0&{}1\\ &{}0&{}1&{}0&{}1&{}0\\ &{}1&{}0&{}1&{}0&{}0 \end{pmatrix} \begin{pmatrix} h_{222}^{2^*}\\ h_{222}^{1^*}\\ h_{122}^{1^*}\\ h_{112}^{1^*}\\ h_{111}^{1^*} \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ \lambda _1\langle X, e_1\rangle +\lambda \langle X, e_2\rangle \\ \lambda \langle X, e_1\rangle -\lambda _1\langle X, e_2\rangle \\ -\lambda _1\langle X, e_1\rangle -\lambda \langle X, e_2\rangle \end{pmatrix}. \end{aligned}$$

(4.41)

From $S=4\lambda ^2+4\lambda _1^2$, we know $\lambda _1=-\lambda _2$ and $2\lambda _1^2+2\lambda ^2=\dfrac{1}{2}S\ne 0$. Hence, by solving the above system, we get

$$\begin{aligned} \begin{aligned} h_{222}^{2^*}&=\frac{-5\lambda _1}{4}\langle X, e_1\rangle -\frac{3\lambda }{4}\langle X, e_2\rangle , \\ h_{222}^{1^*}&=\frac{3\lambda }{4}\langle X, e_1\rangle -\frac{\lambda _1}{4}\langle X, e_2\rangle , \\ h_{221}^{1^*}&=\frac{\lambda _1}{4}\langle X, e_1\rangle -\frac{\lambda }{4}\langle X, e_2\rangle , \\ h_{211}^{1^*}&=\frac{\lambda }{4}\langle X, e_1\rangle -\frac{3\lambda _1}{4}\langle X, e_2\rangle , \\ h_{111}^{1^*}&=\frac{3\lambda _1}{4}\langle X, e_1\rangle +\frac{5\lambda }{4}\langle X, e_2\rangle . \\ \end{aligned} \end{aligned}$$

With a direct calculation, we obtain

$$\begin{aligned} \begin{aligned} |\nabla ^{\perp } \vec H|^2&=\sum _{i,p}(H^{p^*}_{,i})^2\\&=\sum _i(h_{11i}^{1^*}+h_{22i}^{1^*})^2+\sum _i(h_{11i}^{2^*}+h_{22i}^{2^*})^2\\&=2(\lambda ^2+\lambda _1^2)\langle X, e_1\rangle ^2+2(\lambda ^2+\lambda _1^2)\langle X, e_2\rangle ^2 =2(\lambda ^2+\lambda _1^2) |X|^2\\ \end{aligned} \end{aligned}$$

and

$$\begin{aligned}&\sum _{i,j,k,p}(h^{p^*}_{ijk})^2\nonumber \\&\quad =\sum _p(h_{111}^{p^*})^2+\sum _p(h_{222}^{p^*})^2+3\sum _p(h_{221}^{p^*})^2 +3\sum _p(h_{112}^{p^*})^2\nonumber \\&\quad =\sum _{p}(h_{11p}^{1^*})^2+(h_{222}^{1^*})^2+(h_{222}^{2^*})^2 +3\sum _p(h_{22p}^{1^*})^2+3(h_{112}^{1^*})^2+3(h_{112}^{2^*})^2\nonumber \\&\quad =\dfrac{5}{2}\Big \{(\lambda _1^2+\lambda ^2)\langle X, e_1\rangle ^2+(\lambda _1^2+\lambda ^2)\langle X, e_2\rangle ^2\Big \}\nonumber \\&\quad =\dfrac{5}{2}(\lambda _1^2+\lambda ^2)|X|^2=\dfrac{5}{4}|\nabla ^{\perp } \vec H|^2. \end{aligned}$$

(4.42)

From the Ricci identity (2.8), we have

$$\begin{aligned}&h^{2^*}_{1122}-h^{2^*}_{2211}=3\lambda K,\\&h^{1^*}_{1112}-h^{1^*}_{1121}=-3\lambda K,\\&h^{2^*}_{1112}-h^{2^*}_{1121}=3\lambda _{1}K,\\&h^{2^*}_{2212}-h^{2^*}_{2221}=-3\lambda _{1}K. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}\\&\quad =4(h_{1122}^{1^{*}})^{2}+6(h_{2211}^{1^{*}})^{2}+6(h_{1122}^{2^{*}})^{2}+4(h_{2211}^{2^{*}})^{2} +4(h_{2222}^{1^{*}})^{2}+4(h_{1111}^{2^{*}})^{2}\\&\qquad +(h_{1111}^{1^{*}})^{2}+(h_{2222}^{2^{*}})^{2}+(h_{1112}^{1^{*}})^{2}+(h_{2221}^{2^{*}})^{2}\\&\quad =2(h_{1122}^{1^{*}}-h_{2211}^{1^{*}})^{2}+2(h_{1122}^{1^{*}}+h_{2211}^{1^{*}})^{2}\\&\qquad +2(h_{1122}^{2^{*}}-h_{2211}^{2^{*}})^{2}+2(h_{1122}^{2^{*}}+h_{2211}^{2^{*}})^{2}\\&\qquad +\frac{1}{2}(h_{2222}^{1^{*}}-h_{2221}^{2^{*}})^{2}+\frac{1}{2}(h_{2222}^{1^{*}}+h_{2221}^{2^{*}})^{2}\\&\qquad +\dfrac{1}{2}(h_{1121}^{1^{*}}+h_{1112}^{1^{*}})^{2}+\dfrac{1}{2}(h_{1121}^{1^{*}}-h_{1112}^{1^{*}})^{2}\\&\qquad +\dfrac{1}{2}(h_{1111}^{1^{*}}+h_{2211}^{1^{*}})^{2}+\dfrac{1}{2}(h_{1111}^{1^{*}}-h_{2211}^{1^{*}})^{2}\\&\qquad +\frac{1}{2}(h_{2222}^{2^{*}}-h_{1122}^{2^{*}})^{2}+\frac{1}{2}(h_{2222}^{2^{*}}+h_{1122}^{2^{*}})^{2}\\&\qquad +\dfrac{1}{2}(h_{2211}^{1^{*}}-h_{2222}^{1^{*}})^{2}+\dfrac{1}{2}(h_{2211}^{1^{*}}+h_{2222}^{1^{*}})^{2}\\&\qquad +(h_{1122}^{2^{*}})^2+3(h_{1111}^{2^{*}})^2+2(h_{2222}^{1^{*}})^{2} \\&\quad \ge 18\lambda _1^2K^2+\dfrac{9}{2}\lambda _1^2K^2+18\lambda ^2K^2+\dfrac{9}{2}\lambda ^2K^2\\&\quad = \dfrac{45}{2}(\lambda _1^2+\lambda ^2)K^2\\&\quad =\dfrac{45}{32}S^3 \end{aligned} \end{aligned}$$

(4.43)

because of $S=4(\lambda _1^2+\lambda ^2)$ and $K=-\dfrac{1}{2}S$.

On the other hand, since S is constant and $H=0$ at p, from (2.35) in Lemma 2.3 and (4.42), we obtain, at p,

$$\begin{aligned} \begin{aligned} \frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} =-\dfrac{3S}{2}|\nabla ^{\perp } {\vec H}|^{2}=-\dfrac{3}{5}S^2(3S-2). \end{aligned} \end{aligned}$$

(4.44)

According to $h_{11k}^{p^*}+h_{22k}^{p^*}=H^{p^*}_{,k}$ and $H^{p^*}=0$ for $p, k=1, 2$, by a direct calculation, we have

$$\begin{aligned} \sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}=\dfrac{S}{4}|\nabla ^{\perp } {\vec H}|^{2}. \end{aligned}$$

From (2.34) in Lemma 2.3, we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}-(5S-2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2}+\dfrac{5S}{2}|\nabla ^{\perp } {\vec H}|^{2}\\&\qquad -\sum _{i,j,k,l,p,q}h_{il}^{p^{*}}h_{ijk}^{p^{*}}h_{jl}^{q^{*}}H_{,k}^{q^{*}}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}-(5S-2)\sum _{i,j,k,p}(h_{ijk}^{p^{*}})^{2} +\dfrac{9S}{4}|\nabla ^{\perp } {\vec H}|^{2}\\&\quad =\sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}-(5S-2)\dfrac{S(3S-2)}{2}+\dfrac{9}{10}S^2(3S-2).\\ \end{aligned} \end{aligned}$$

Thus, we have from (4.44)

$$\begin{aligned} \begin{aligned} \sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}-(5S-2)\dfrac{S(3S-2)}{2}+\dfrac{9}{10}S^2(3S-2)=-\dfrac{3}{5}S^2(3S-2), \end{aligned} \end{aligned}$$

namely,

$$\begin{aligned} \begin{aligned} \sum _{i,j,k,l,p}(h_{ijkl}^{p^{*}})^{2}=S(S-1)(3S-2), \end{aligned} \end{aligned}$$

which is in contradiction to (4.43) for $S\equiv 2$ or $S\equiv 1$. Hence, we conclude that $H\ne 0$ on $M^2$. $\square$

Proposition 4.4

Let $X: M^2\rightarrow {\mathbb {R}}^{4}$ be a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$. If the squared norm S of the second fundamental form satisfies $S\equiv 1$ or $S\equiv 2$, then $\sup H^2=S$.

Proof

In terms of Lemma 4.1, there exists a sequence $\{p_m\}$ in $M^2$ such that

$$\begin{aligned} \lim _{m\rightarrow \infty } H^2(p_m)=\sup H^2, \ \ \lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0, \ \ \limsup _{m\rightarrow \infty }{\mathcal {L}} H^2(p_m)\le 0 \end{aligned}$$

and

$$\begin{aligned} {\bar{\lambda }}=0, \ \ {\bar{h}}^{1^{*}}_{ij}={\bar{\lambda }}_i\delta _{ij}. \end{aligned}$$

(1) Case for $S\equiv 2$. By making use of the same assertion as in the proof of Proposition 4.1, we have for $k=1, 2$,

$$\begin{aligned} {\left\{ \begin{array}{ll} {\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0,\\ {\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ ({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}^{1^*}_{11k}=0 \end{array}\right. } \end{aligned}$$

(4.45)

with ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$.

If ${\bar{\lambda }}_1=3{\bar{\lambda }}_2$, we get

$$\begin{aligned} \lim _{m\rightarrow \infty }H^2(p_m)={\bar{H}}^2=({\bar{\lambda }}_1+{\bar{\lambda }}_2)^2=16{\bar{\lambda }}_2^2=\dfrac{4S}{3}. \end{aligned}$$

By making use of the same assertion as in the proof of Proposition 4.2, we can know that this is impossible.

Thus, we get ${\bar{\lambda }}_1\ne 3{\bar{\lambda }}_2$. In this case, we obtain ${\bar{h}}^{p^{*}}_{ijk}=0$ for any i, j, k, p from (4.45). Hence, we have from (2.25) in Lemma 2.1,

$$\begin{aligned} \begin{aligned}&0=S(1-\dfrac{3}{2}S)+2{\bar{H}}^2S-\dfrac{1}{2}{\bar{H}}^4-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)\\&=-(S-{\bar{H}}^2)^2-\dfrac{1}{2}{\bar{H}}^2({\bar{\lambda }}_1-{\bar{\lambda }}_2)^2. \end{aligned} \end{aligned}$$

(4.46)

We conclude

$$\begin{aligned} \sup H^2={\bar{H}}^2=S=2. \end{aligned}$$

(2) Case for $S\equiv 1$. Since $S=1$, we have $\sup H^2>0$. From $\lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0$ and $|\nabla H^2|^2=4\sum _i(\sum _{p^*}H^{p^{*}}H^{p^{*}}_{,i})^2$, we get

$$\begin{aligned} {\bar{H}}^{1^*}_{,i}=0. \end{aligned}$$

From (4.7) and (4.10), we have

$$\begin{aligned} {\bar{\lambda }}_i\lim _{m\rightarrow \infty } \langle X,e_i \rangle (p_m)=0. \end{aligned}$$

Next, we take the following three cases into consideration.

(a) If ${\bar{\lambda }}_1=0$, in this case, ${\bar{\lambda }}_2\ne 0$, $3{\bar{H}}^2=S=1$. Since ${\bar{H}}^{1^*}_{,i}=0$ and $S=1$, we get

$$\begin{aligned} {\bar{h}}_{11k}^{1^{*}}+{\bar{h}}_{22k}^{1^{*}}=0,\ \ {\bar{h}}_{22k}^{1^{*}}=0,\ \ k=1,2. \end{aligned}$$

Therefore,

$$\begin{aligned} {\bar{h}}_{111}^{1^{*}}={\bar{h}}_{112}^{1^{*}}={\bar{h}}_{122}^{1^{*}}={\bar{h}}_{222}^{1^{*}}=0 \end{aligned}$$

and

$$\begin{aligned} |\nabla ^{\perp } \overrightarrow{{\bar{H}}}|^2=\sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^2. \end{aligned}$$

From $\limsup _{m\rightarrow \infty }{\mathcal {L}} |H|^2(p_m)\le 0$ and (4.8), we obtain

$$\begin{aligned} \frac{1}{2}({\bar{H}}^2-1)^2\le 0, \end{aligned}$$

it means that, ${\bar{H}}^2=1$. It is a contradiction.

(b) If ${\bar{\lambda }}_2=0$, in this case, ${\bar{\lambda }}_1\ne 0$, $\sup H^2={\bar{H}}^2=S=1$.

(c) If ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$, in this case, for $k=1, 2$,

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0,\\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}^{1^*}_{11k}=0. \end{array}\right. } \end{aligned}$$

If ${\bar{\lambda }}_1\ne 3{\bar{\lambda }}_2$, from the above equations, we know

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^2=0,\ \ i, j, k, p=1, 2. \end{aligned}$$

From (4.8), we get

$$\begin{aligned} 0\ge \frac{1}{2}({\bar{H}}^2-1)^2. \end{aligned}$$

Hence, we have

$$\begin{aligned} \sup {\bar{H}}^2=1=S. \end{aligned}$$

If ${\bar{\lambda }}_1=3{\bar{\lambda }}_2$, we have ${\bar{H}}^2=\frac{4}{3}S=\frac{4}{3}$ and $1=S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2=12{\bar{\lambda }}_2^2$. From (2.25), we get

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^2=-\frac{1}{6}<0. \end{aligned}$$

It is impossible. From the above arguments, we conclude that, for $S=2$ or $S=1$,

$$\begin{aligned} \sup H^2=S. \end{aligned}$$

$\square$

Theorem 4.4

Let $X: M^2\rightarrow {\mathbb {R}}^{4}$ be a 2-dimensional complete Lagrangian self-shrinker in ${\mathbb {R}}^4$. If the squared norm S of the second fundamental form satisfies $S\equiv 1$ or $S\equiv 2$, then $H^2=S$ is constant.

Proof

We can apply the generalized maximum principle for ${\mathcal {L}}$-operator to the function $-H^2$. Thus, there exists a sequence $\{p_m\}$ in $M^2$ such that

$$\begin{aligned} \lim _{m\rightarrow \infty } H^2(p_m)=\inf H^2, \ \ \lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0, \ \ \liminf _{m\rightarrow \infty }{\mathcal {L}} H^2(p_m)\ge 0. \end{aligned}$$

By making use of the similar assertion as in the proof of Lemma 4.1, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}\lim _{m\rightarrow \infty } H^2(p_m)=\inf H^2={\bar{H}}^2,\quad \lim _{m\rightarrow \infty } |\nabla H^2(p_m)|=0,\\ &{} \lim _{m\rightarrow \infty } |\nabla ^{\perp }\vec H|^2(p_m)-\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2+\dfrac{1}{2}({\bar{H}}^2-S)({\bar{H}}^2-3S+2)\ge 0. \end{aligned} \end{array}\right. } \end{aligned}$$

(4.47)

By taking the limit and making use of the same assertion as in Theorem 4.3, we can prove $\inf H^2\ne 0$. Hence, without loss of the generality, at each point $p_m$, we choose $e_1$, $e_2$ such that

$$\begin{aligned} \vec H=H^{1^{*}}e_{1^{*}} \end{aligned}$$

and we can assume

$$\begin{aligned} \lim _{m\rightarrow \infty }h^{p^{*}}_{ijl}(p_m)={\bar{h}}^{p^{*}}_{ijl}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ij}(p_m)={\bar{h}}^{p^{*}}_{ij}, \quad \lim _{m\rightarrow \infty }h^{p^{*}}_{ijkl}(p_m)={\bar{h}}^{p^{*}}_{ijkl}, \end{aligned}$$

for $i, j, k, l,p\!=\!1, 2$. From $\lim _{k\rightarrow \infty } |\nabla H^2(p_m)|\!=\!0$ and $|\nabla H^2|^2\!=\!4\sum _i(\sum _{p^*}H^{p^{*}}H^{p^{*}}_{,i})^2$, we have

$$\begin{aligned} {\bar{H}}^{1^*}_{,k}=0. \end{aligned}$$

(4.48)

From (4.7) and (4.48), we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{\lambda }}_1\lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m)+{\bar{\lambda }} \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0,\\ &{}{\bar{\lambda }}\lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m) +{\bar{\lambda }}_2 \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(4.49)

If ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne {\bar{\lambda }}^2$ and ${\bar{\lambda }}\ne 0$, we get

$$\begin{aligned} \lim _{m\rightarrow \infty } \langle X,e_1\rangle (p_m)= \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)=0. \end{aligned}$$

Thus, we know, for $k=1, 2$,

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0, \\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0,\\ &{}{\bar{\lambda }}_1{\bar{h}}^{1^{*}}_{11k}+3{\bar{\lambda }} {\bar{h}}^{1^{*}}_{12k} +3{\bar{\lambda }}_2 {\bar{h}}^{2^{*}}_{12k}-{\bar{\lambda }}{\bar{h}}^{2^{*}}_{22k}=0. \end{aligned} \end{array}\right. } \end{aligned}$$

(4.50)

We conclude, for any i, j, k, p,

$$\begin{aligned} {\bar{h}}^{p^*}_{ijk}=0. \end{aligned}$$

From (4.47) and (2.25) in Lemma 2.1, we have

$$\begin{aligned} \begin{aligned}&S\ge \inf H^2= {\bar{H}}^2,\\&S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}{\bar{H}}^4-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2+2{\bar{\lambda }}^2)=0. \end{aligned} \end{aligned}$$

(4.51)

From Lemma 2.5 and taking limit,

$$\begin{aligned} \begin{aligned} 0&\le \sum _{i,j,k,l,p}({\bar{h}}_{ijkl}^{p^{*}})^{2} =\frac{1}{2}\lim _{m\rightarrow \infty } {\mathcal {L}}\sum _{i,j,k,p}( h_{ijk}^{p^{*}})^{2}(p_m)\\&={\bar{H}}^{2}\Big [{\bar{H}}^{2}-2S+\frac{1}{2}{\bar{H}}^{4}-{\bar{K}}{\bar{H}}^{2}-{\bar{K}}^{2}\Big ]-{\bar{\lambda }}^2{\bar{H}}^{4}\\&\quad +{\bar{H}}^{2}(S+2-\frac{3}{2}{\bar{H}}^{2}-{\bar{\lambda }}_{1}^{2} -{\bar{\lambda }}_{2}^{2}-2{\bar{\lambda }}^2)({\bar{\lambda }}_{1}^{2}+{\bar{\lambda }}_{2}^{2}+2{\bar{\lambda }}^2)\\&<{\bar{H}}^{2}\Big [{\bar{H}}^{2}-2S+\frac{1}{2}{\bar{H}}^{4}-{\bar{K}}{\bar{H}}^{2}-{\bar{K}}^{2}\Big ]\\&\quad +{\bar{H}}^{2}(S+2-\frac{3}{2}{\bar{H}}^{2}-{\bar{\lambda }}_{1}^{2} -{\bar{\lambda }}_{2}^{2}-2{\bar{\lambda }}^2)({\bar{\lambda }}_{1}^{2}+{\bar{\lambda }}_{2}^{2}+2{\bar{\lambda }}^2). \end{aligned} \end{aligned}$$

(4.52)

According to (4.51), we have

$$\begin{aligned} \begin{aligned}&{\bar{H}}^{2}\Big [{\bar{H}}^{2}-2S+\frac{1}{2}{\bar{H}}^{4}-{\bar{K}}{\bar{H}}^{2}-{\bar{K}}^{2}\Big ]\\&\qquad + \Bigg (S+2-\frac{3}{2}{\bar{H}}^{2}-\dfrac{1}{{\bar{H}}^2}\Big (S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}{\bar{H}}^4\Big )\Bigg )\\&\qquad \times \Bigg (S(1-\dfrac{1}{2}S)-(S-{\bar{H}}^2)^2+\dfrac{1}{2}{\bar{H}}^4\Bigg )\\&\quad =\dfrac{1}{4{\bar{H}}^2}\Bigg ({\bar{H}}^8-2S{\bar{H}}^6-6S(S-1){\bar{H}}^4+2S(2-3S)^2{\bar{H}}^2-(2-3S)^2S^2\Bigg ). \end{aligned} \end{aligned}$$

We consider a function f(t) defined by

$$\begin{aligned} \begin{aligned}&f(t)=t^4-2St^3-6S(S-1)t^2+2S(2-3S)^2t-(2-3S)^2S^2, \end{aligned} \end{aligned}$$

(4.53)

for $0<t\le S$. Thus, we get

$$\begin{aligned} f^{'}(t)=4t^3-6St^2-12S(S-1)t+2S(2-3S)^2,\ \ f^{''}(t)=12(t^2-St-S(S-1)),\nonumber \\ \end{aligned}$$

(4.54)

$f^{''}(t)<0$ for $t\in (0, S)$. Hence, $f^{'}(t)$ is a decreasing function for $t\in (0, S)$. Since $f^{'}(S)=4S(S-1)(S-2)=0$, f(t) is a increasing function for $t\in (0, S)$. According to

$$\begin{aligned} f(S)=2(S-1)(S-2)S^2=0, \end{aligned}$$

(4.55)

we conclude $f(t)<0$ for $t\in (0, S)$. This is a contradiction.

Hence, we have ${\bar{\lambda }}_1{\bar{\lambda }}_2\ne 0$ and ${\bar{\lambda }}=0$. In this case, we get for $k=1, 2$,

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0\\ &{}{\bar{h}}^{2^*}_{11k}+{\bar{h}}^{2^*}_{22k}=0\\ &{}({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}^{1^*}_{11k}=0. \end{array}\right. } \end{aligned}$$

(4.56)

If ${\bar{\lambda }}_1=3{\bar{\lambda }}_2$, we obtain

$$\begin{aligned} \inf H^2=\lim _{m\rightarrow \infty }H^2(p_m)=({\bar{\lambda }}_1+{\bar{\lambda }}_2)^2=16{\bar{\lambda }}_2^2=\dfrac{4S}{3}, \end{aligned}$$

which is impossible from Proposition 4.4. Thus, we get ${\bar{\lambda }}_1\ne 3{\bar{\lambda }}_2$. In this case, we have

$$\begin{aligned} {\bar{h}}^{p^{*}}_{ijk}=0 \end{aligned}$$

for any i, j, k, p from (4.56).

From (2.19), we know

$$\begin{aligned} {\bar{H}}_{,ij}^{p^{*}}={\bar{h}}_{ij}^{p^{*}}-\sum _k{\bar{h}}_{ik}^{p^{*}}{\bar{h}}_{jk}^{1^{*}}{\bar{H}} \end{aligned}$$

because of $H^{1^*}=H$ and $H^{2^*}=0$. Thus, we get

$$\begin{aligned} \begin{aligned}&{\bar{h}}_{1111}^{1^{*}}+{\bar{h}}_{2211}^{1^{*}}={\bar{\lambda }}_1-{\bar{\lambda }}_1^2{\bar{H}},\\&{\bar{h}}_{1122}^{1^{*}}+{\bar{h}}_{2222}^{1^{*}}={\bar{\lambda }}_2-{\bar{\lambda }}_2^2{\bar{H}},\\&{\bar{h}}_{1112}^{1^{*}}+{\bar{h}}_{2212}^{1^{*}}={\bar{h}}_{1121}^{1^{*}}+{\bar{h}}_{2221}^{1^{*}}=0,\\&{\bar{h}}_{1122}^{2^{*}}+{\bar{h}}_{2222}^{2^{*}}=0,\\&{\bar{h}}_{1121}^{2^{*}}+{\bar{h}}_{2221}^{2^{*}}={\bar{\lambda }}_2-{\bar{\lambda }}_1{\bar{\lambda }}_2{\bar{H}}. \end{aligned} \end{aligned}$$

(4.57)

From Ricci identities (2.8), we obtain

$$\begin{aligned} \begin{aligned}&{\bar{h}}_{1122}^{2^{*}}={\bar{h}}_{2211}^{2^{*}}, \ {\bar{h}}_{1112}^{1^{*}}={\bar{h}}_{1121}^{1^{*}},\\&{\bar{h}}_{1112}^{2^{*}}-{\bar{h}}_{1121}^{2^{*}}=({\bar{\lambda }}_1-2{\bar{\lambda }}_2){\bar{K}}, \\&{\bar{h}}_{2212}^{2^{*}}-{\bar{h}}_{2221}^{2^{*}}=3{\bar{\lambda }}_2{\bar{K}}. \end{aligned} \end{aligned}$$

(4.58)

On the other hand, since S is constant, we know, for $k,l=1,2$,

$$\begin{aligned} 0=-\sum _{i,j,p}{\bar{h}}_{ijl}^{p^{*}}{\bar{h}}_{ijk}^{p^{*}}=\sum _{i,j,p}{\bar{h}}_{ij}^{p^{*}}{\bar{h}}_{ijkl}^{p^{*}} ={\bar{h}}_{11}^{1^{*}}{\bar{h}}_{11kl}^{1^{*}}+3{\bar{h}}_{12}^{2^{*}}{\bar{h}}_{12kl}^{2^{*}}={\bar{\lambda }}_1{\bar{h}}_{11kl}^{1^{*}}+3{\bar{\lambda }}_2{\bar{h}}_{22kl}^{1^{*}}.\nonumber \\ \end{aligned}$$

(4.59)

From (4.57) and (4.59), we have

$$\begin{aligned} \begin{aligned}&({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}_{2211}^{1^{*}}={\bar{\lambda }}_1^2-{\bar{\lambda }}_1^3{\bar{H}},\\&({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}_{2222}^{1^{*}}={\bar{\lambda }}_1{\bar{\lambda }}_2-{\bar{\lambda }}_1{\bar{\lambda }}_2^2{\bar{H}},\\&({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}_{1122}^{1^{*}}=-3{\bar{\lambda }}_2^2+3{\bar{\lambda }}_2^3{\bar{H}}\\&({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}_{2212}^{1^{*}}=0. \end{aligned} \end{aligned}$$

(4.60)

Hence, we conclude from (4.57) and (4.60)

$$\begin{aligned} {\bar{h}}_{2212}^{1^{*}}={\bar{h}}_{1112}^{1^{*}}=0 \end{aligned}$$

because of ${\bar{\lambda }}_1\ne 3{\bar{\lambda }}_2$. According to (4.60), we obtain

$$\begin{aligned} ({\bar{\lambda }}_1-3{\bar{\lambda }}_2)({\bar{h}}_{1112}^{2^{*}}-{\bar{h}}_{1121}^{2^{*}}) =-S+(3{\bar{\lambda }}_2^3+{\bar{\lambda }}_1^3){\bar{H}} \end{aligned}$$

(4.61)

because of $S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2$.

For the case $S\equiv 1$, from (4.58), we know

$$\begin{aligned} \begin{aligned}&({\bar{\lambda }}_1-3{\bar{\lambda }}_2)({\bar{h}}_{1112}^{2^{*}}-{\bar{h}}_{1121}^{2^{*}})\\&\quad =({\bar{\lambda }}_1-3{\bar{\lambda }}_2)({\bar{\lambda }}_1-2{\bar{\lambda }}_2){\bar{K}}\\&\quad =(1+3{\bar{\lambda }}_2^2-5{\bar{\lambda }}_1{\bar{\lambda }}_2)({\bar{\lambda }}_1{\bar{\lambda }}_2-{\bar{\lambda }}_2^2)\\&\quad ={\bar{\lambda }}_1{\bar{\lambda }}_2-{\bar{\lambda }}_2^2+8{\bar{\lambda }}_1{\bar{\lambda }}_2^3 -3{\bar{\lambda }}_2^4-5{\bar{\lambda }}_1^2{\bar{\lambda }}_2^2\\&\quad ={\bar{\lambda }}_1{\bar{\lambda }}_2-2{\bar{\lambda }}_2^2+8{\bar{\lambda }}_1{\bar{\lambda }}_2^3 -4{\bar{\lambda }}_1^2{\bar{\lambda }}_2^2.\\ \end{aligned} \end{aligned}$$

(4.62)

By a direct calculation and by using $S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2=1$, we get

$$\begin{aligned} -1+(3{\bar{\lambda }}_2^3+{\bar{\lambda }}_1^3){\bar{H}}-\Big \{{\bar{\lambda }}_1{\bar{\lambda }}_2-{\bar{\lambda }}_2^2 +8{\bar{\lambda }}_1{\bar{\lambda }}_2^3 -3{\bar{\lambda }}_2^4-5{\bar{\lambda }}_1^2{\bar{\lambda }}_2^2\Big \} =-8{\bar{\lambda }}_1{\bar{\lambda }}_2^3\ne 0. \end{aligned}$$

From (4.61) and (4.62), it is impossible.

For the case $S\equiv 2$, we have from (2.25) in Lemma 2.1,

$$\begin{aligned} \begin{aligned}&0=S(1-\dfrac{3}{2}S)+2{\bar{H}}^2S-\dfrac{1}{2}{\bar{H}}^4-{\bar{H}}^2({\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2)\\&=-(S-{\bar{H}}^2)^2-\dfrac{1}{2}{\bar{H}}^2({\bar{\lambda }}_1-{\bar{\lambda }}_2)^2. \end{aligned} \end{aligned}$$

(4.63)

We conclude from Proposition 4.4

$$\begin{aligned} \inf H^2={\bar{H}}^2=S=\sup H^2. \end{aligned}$$

Thus, we know that $H^2=S$ is constant.

From now on, we consider the case ${\bar{\lambda }}_1{\bar{\lambda }}_2={\bar{\lambda }}^2$. In this case, we have

$$\begin{aligned} S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2+4{\bar{\lambda }}^2 =({\bar{\lambda }}_1+{\bar{\lambda }}_2)({\bar{\lambda }}_1+3{\bar{\lambda }}_2)={\bar{H}}({\bar{\lambda }}_1+3{\bar{\lambda }}_2). \end{aligned}$$

If $S\equiv 1$, from (2.25), we get

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=\dfrac{1}{2}({\bar{H}}^2-1)(3{\bar{H}}^2-1)\ge 0. \end{aligned}$$

Hence, either ${\bar{H}}^2\ge 1$, or ${\bar{H}}^2\le \dfrac{1}{3}$. If ${\bar{H}}^2\ge 1$, then we have $H^2\equiv 1=S$ since $\inf H^2={\bar{H}}^2\le \sup H^2=1$ in view of Proposition 4.4. According to $S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2+4{\bar{\lambda }}^2$ and $H^2={\bar{\lambda }}_1^2+{\bar{\lambda }}_2^2+2{\bar{\lambda }}^2$, we know ${\bar{\lambda }}=0$ and ${\bar{\lambda }}_2=0$.

If ${\bar{H}}^2\le \dfrac{1}{3}$, from $S={\bar{H}}({\bar{\lambda }}_1+3{\bar{\lambda }}_2)=1$, we obtain $({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^2\ge 3$, which implies ${\bar{\lambda }}_1={\bar{\lambda }}=0$ because of $({\bar{\lambda }}_1+3{\bar{\lambda }}_2)^2= {\bar{\lambda }}_1^2+9{\bar{\lambda }}_2^2+6{\bar{\lambda }}^2\le 3{\bar{\lambda }}_1^2+9{\bar{\lambda }}_2^2+12{\bar{\lambda }}^2=3S=3$. Hence, we have $\inf H^2={\bar{\lambda }}_2^2=\dfrac{S}{3}\ne 0$,

$$\begin{aligned} {\bar{H}}^{1^*}_{,k}=0, \ \ {\bar{H}}^{2^*}_{,1}=0 \end{aligned}$$

because of $H^{p^{*}}_{,i}=\sum _k h^{p^{*}}_{ik} \langle X,e_k \rangle$. Hence, we have, by using the same calculations as in (4.6),

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0\\ &{}{\bar{h}}^{2^*}_{111}+{\bar{h}}^{2^*}_{221}=0\\ &{}3{\bar{\lambda }}_2{\bar{h}}^{1^*}_{22k}=0. \end{array}\right. } \end{aligned}$$

(4.64)

Hence, we get

$$\begin{aligned} {\bar{h}}^{p^{*}}_{ijk}=0, \ \ \text { except} \ \ i=j=k=p^{*}=2. \end{aligned}$$

If ${\bar{h}}^{2^{*}}_{222}\ne 0$, since ${\bar{\lambda }}_2\ne 0$, $3{\bar{H}}^2=3{\bar{\lambda }}_2^2=S=1$, we have

$$\begin{aligned} 0=\sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^2+S(1-\frac{3}{2}S)+2{\bar{H}}^{2}S-\frac{1}{2}{\bar{H}}^{4} -{\bar{H}}^{2}{\bar{\lambda }}_2^2=\sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^2>0. \end{aligned}$$

It is impossible. Hence, we know

$$\begin{aligned} {\bar{h}}^{p^{*}}_{ijk}=0, \end{aligned}$$

for any i, j, k, p. From (2.19), we get

$$\begin{aligned} {\bar{H}}_{,ij}^{p^{*}}={\bar{h}}_{ij}^{p^{*}}-\sum _k{\bar{h}}_{ik}^{p^{*}}{\bar{h}}_{jk}^{1^{*}}{\bar{H}}. \end{aligned}$$

We obtain

$$\begin{aligned} {\bar{h}}_{1121}^{2^{*}}+{\bar{h}}_{2221}^{2^{*}}={\bar{\lambda }}_2\ne 0. \end{aligned}$$

From (2.34) of Lemma 2.3, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^{2}=\sum _{i,j,k,l,p}({\bar{h}}_{ijkl}^{p^{*}})^{2}>0.\\ \end{aligned} \end{aligned}$$

(4.65)

From (2.35) of Lemma 2.3, we get

$$\begin{aligned} \begin{aligned}&\frac{1}{2}{\mathcal {L}}\sum _{i,j,k,p}({\bar{h}}_{ijk}^{p^{*}})^{2}\\&\quad =({\bar{H}}^{2}-2S){\bar{H}}^{2}+(3{\bar{K}}+2-{\bar{H}}^{2}+2S)\sum _{i,j}{\bar{H}}^2{\bar{h}}_{ij}^{1^{*}} {\bar{h}}_{ij}^{1^{*}}\\&\qquad -{\bar{K}}({\bar{H}}^{4}+{\bar{H}}^3{\bar{h}}_{11}^{1^{*}}) -\sum _{i,j,k,l,p}{\bar{H}}^2{\bar{h}}_{jk}^{1^{*}}{\bar{h}}_{jk}^{p^{*}}{\bar{h}}_{il}^{p^{*}}{\bar{h}}_{il}^{1^{*}} -\sum _{i,j,k}{\bar{H}}^3 {\bar{h}}_{ik}^{1^{*}}{\bar{h}}_{ji}^{1^{*}}{\bar{h}}_{jk}^{1^{*}}\\&\quad =({\bar{H}}^{2}-2S){\bar{H}}^{2}+(3{\bar{K}}+2-{\bar{H}}^{2}+2S)\sum _{i,j}{\bar{H}}^2{\bar{\lambda }}_2^2 -{\bar{K}}{\bar{H}}^{4}-{\bar{H}}^2{\bar{\lambda }}_2^4-{\bar{H}}^3 {\bar{\lambda }}_2^3 \\&\quad ={\bar{H}}^6-3{\bar{H}}^4=-\frac{8}{27}, \end{aligned} \end{aligned}$$

which is in contradiction to (4.65). Hence, we get $\inf H^2=S$, that is, $H^2=S$ is constant from Proposition 4.4.

For the case $S\equiv 2$, first of all, we will prove ${\bar{\lambda }}=0$. If not, we have $S=2$ and ${\bar{\lambda }}_1{\bar{\lambda }}_2={\bar{\lambda }}^2\ne 0$. By making use of the same assertion as in the proof of Theorem 4.3, we have

$$\begin{aligned} \begin{pmatrix} &{}- {\bar{\lambda }} &{} 3 {\bar{\lambda }}_2 &{}3{\bar{\lambda }}&{} {\bar{\lambda }}_1 &{}0\\ &{}0&{}-{\bar{\lambda }} &{} 3{\bar{\lambda }}_2 &{}3 {\bar{\lambda }}&{} {\bar{\lambda }}_1 \\ &{}0&{}0&{}1&{}0&{}1\\ &{}0&{}1&{}0&{}1&{}0\\ &{}1&{}0&{}1&{}0&{}0 \end{pmatrix} \begin{pmatrix} {\bar{h}}_{222}^{2^*}\\ {\bar{h}}_{222}^{1^*}\\ {\bar{h}}_{122}^{1^*}\\ {\bar{h}}_{112}^{1^*}\\ {\bar{h}}_{111}^{1^*} \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0\\ A \end{pmatrix}, \end{aligned}$$

(4.66)

where

$$\begin{aligned} \begin{aligned} A&={\bar{\lambda }}_2 \lim _{m\rightarrow \infty } \langle X,e_1 \rangle (p_m) -{\bar{\lambda }} \lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m)\\&={\bar{H}}\lim _{m\rightarrow \infty } \langle X,e_1 \rangle (p_m)\\&=-\dfrac{H{\bar{\lambda }}}{{\bar{\lambda }}_1}\lim _{m\rightarrow \infty } \langle X,e_2 \rangle (p_m). \end{aligned} \end{aligned}$$

(4.67)

Solving this system of linear equations, we have

$$\begin{aligned} \begin{aligned} \mu {\bar{h}}_{222}^{2^*}&=\Big \{12{\bar{\lambda }}^2+({\bar{\lambda }}_1-3{\bar{\lambda }}_2)^2\Big \}A , \\ {\bar{h}}_{222}^{1^*}&=-{\bar{h}}_{211}^{1^*},\\ {\bar{h}}_{221}^{1^*}&=-{\bar{h}}_{111}^{1^*}, \\ \mu {\bar{h}}_{211}^{1^*}&={\bar{\lambda }}({\bar{\lambda }}_1-3{\bar{\lambda }}_2)A, \\ \mu {\bar{h}}_{111}^{1^*}&=-4{\bar{\lambda }}^2A \\ \end{aligned} \end{aligned}$$

(4.68)

with $\mu =16{\bar{\lambda }}^2+({\bar{\lambda }}_1-3{\bar{\lambda }}_2)^2$.

$$\begin{aligned} \lim _{m\rightarrow \infty }|\nabla ^{\perp } \vec H|^2= & {} \sum _{i,p}({\bar{H}}^{p^*}_{,i})^2=({\bar{h}}_{112}^{2^*}+{\bar{h}}_{222}^{2^*})^2=A^2 \end{aligned}$$

(4.69)

$$\begin{aligned}&\sum _{i,j,k,p}({\bar{h}}^{p^*}_{ijk})^2=\sum _p({\bar{h}}_{111}^{p^*})^2+\sum _p({\bar{h}}_{222}^{p^*})^2+3\sum _p({\bar{h}}_{221}^{p^*})^2 +3\sum _p({\bar{h}}_{112}^{p^*})^2\nonumber \\&\quad =\sum _{p}({\bar{h}}_{11p}^{1^*})^2+({\bar{h}}_{222}^{1^*})^2+({\bar{h}}_{222}^{2^*})^2 +3\sum _p({\bar{h}}_{22p}^{1^*})^2+3({\bar{h}}_{112}^{1^*})^2+3({\bar{h}}_{112}^{2^*})^2\nonumber \\&\quad = \lim _{m\rightarrow \infty }|\nabla ^{\perp } \vec H|^2. \end{aligned}$$

(4.70)

Since $S= 2$ and ${\bar{\lambda }}_1{\bar{\lambda }}_2={\bar{\lambda }}^2\ne 0$, we obtain

$$\begin{aligned} S={\bar{\lambda }}_1^2+3{\bar{\lambda }}_2^2+4{\bar{\lambda }}_3^2={\bar{H}}({\bar{\lambda }}_1+3{\bar{\lambda }}_2)=2, \end{aligned}$$

From (2.25), we have

$$\begin{aligned} \sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=4-4{\bar{H}}^2+\frac{3}{2}{\bar{H}}^4=(2-H^2)^2+\dfrac{1}{2}{\bar{H}}^4>0. \end{aligned}$$

(4.71)

Since $S=2$ is constant, we get, for $k,l=1,2$,

$$\begin{aligned} -\sum _{i,j,p}{\bar{h}}_{ijl}^{p^{*}}{\bar{h}}_{ijk}^{p^{*}}=\sum _{i,j,p}{\bar{h}}_{ij}^{p^{*}}{\bar{h}}_{ijkl}^{p^{*}} ={\bar{\lambda }}_1{\bar{h}}_{11kl}^{1^{*}} +3{\bar{\lambda }}{\bar{h}}_{12kl}^{1^{*}}+3{\bar{\lambda }}_2{\bar{h}}_{22kl}^{1^{*}}-{\bar{\lambda }}{\bar{h}}_{22kl}^{2^{*}}, \end{aligned}$$

namely,

$$\begin{aligned} \begin{aligned}&{\bar{\lambda }}_1 {\bar{h}}_{1122}^{1^{*}}+3{\bar{\lambda }} {\bar{h}}_{1222}^{1^{*}}+3{\bar{\lambda }}_2 {\bar{h}}_{2222}^{1^{*}} -{\bar{\lambda }} {\bar{h}}_{2222}^{2^{*}}\\&\quad =-({\bar{h}}_{112}^{1^{*}})^2-3({\bar{h}}_{112}^{2^{*}})^2-3({\bar{h}}_{212}^{2^{*}})^2-({\bar{h}}_{222}^{2^{*}})^2\\&\quad =-\dfrac{12{\bar{\lambda }}^2+({\bar{\lambda }}_1-3{\bar{\lambda }}_2)^2}{\mu }A^2,\\&{\bar{\lambda }}_1{\bar{h}}_{1112}^{1^{*}}+3{\bar{\lambda }}{\bar{h}}_{1212}^{1^{*}} +3{\bar{\lambda }}_2 {\bar{h}}_{2212}^{1^{*}}-{\bar{\lambda }} {\bar{h}}_{2212}^{2^{*}}\\&\quad =-{\bar{h}}_{111}^{1^{*}}{\bar{h}}_{112}^{1^{*}}-3{\bar{h}}_{111}^{2^{*}}{\bar{h}}_{112}^{2^{*}} -3{\bar{h}}_{121}^{2^{*}}{\bar{h}}_{122}^{2^{*}}-{\bar{h}}_{221}^{2^{*}}{\bar{h}}_{222}^{2^{*}}\\&\quad =\dfrac{{\bar{\lambda }}({\bar{\lambda }}_1-3{\bar{\lambda }}_2)}{\mu }A^2.\\ \end{aligned} \end{aligned}$$

(4.72)

From (2.19) and taking limit, we know, for $\ i, j, p=1, 2$

$$\begin{aligned} {\bar{h}}_{11ij}^{p^{*}}+{\bar{h}}_{22ij}^{p^{*}}=\sum _k{\bar{h}}_{ikj}^{p^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m) +{\bar{h}}_{ij}^{p^{*}}-\sum _k{\bar{h}}_{ik}^{p^{*}}{\bar{h}}_{jk}^{1^{*}}{\bar{H}}, \end{aligned}$$

it means that,

$$\begin{aligned} \begin{aligned}&{\bar{h}}_{1122}^{1^{*}}+{\bar{h}}_{2222}^{1^{*}}= \sum _k{\bar{h}}_{2k2}^{1^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m) + {\bar{\lambda }}_2 -({\bar{\lambda }}_2^2+{\bar{\lambda }}^2) {\bar{H}}, \\&{\bar{h}}_{1112}^{1^{*}}+{\bar{h}}_{2212}^{1^{*}}= \sum _k{\bar{h}}_{1k2}^{1^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m)+{\bar{\lambda }} -{\bar{\lambda }}{\bar{H}}^2, \\&{\bar{h}}_{1122}^{2^{*}}+{\bar{h}}_{2222}^{2^{*}}=\sum _k{\bar{h}}_{2k2}^{2^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m)-{\bar{\lambda }}, \\ \end{aligned} \end{aligned}$$

Since, from (4.49) and (4.67) and $S={\bar{H}}({\bar{\lambda }}_1+3{\bar{\lambda }}_2)=2$,

$$\begin{aligned}&\sum _k{\bar{h}}_{2k2}^{1^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m)=\dfrac{{\bar{\lambda }}_1A^2}{\mu },\\&\sum _k{\bar{h}}_{2k2}^{2^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m)= \dfrac{3{\bar{\lambda }}}{\mu }A^2- \dfrac{{\bar{\lambda }}_1^2+3{\bar{\lambda }}^2}{2 {\bar{\lambda }}}A^2,\\&\sum _k{\bar{h}}_{1k2}^{1^{*}}\lim _{m\rightarrow \infty } \langle X,e_k \rangle (p_m) =-\dfrac{3{\bar{\lambda }}}{\mu }A^2, \end{aligned}$$

we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} &{}{\bar{\lambda }}_1 {\bar{h}}_{1122}^{1^{*}}+3{\bar{\lambda }} {\bar{h}}_{1222}^{1^{*}}+3{\bar{\lambda }}_2 {\bar{h}}_{2222}^{1^{*}} -{\bar{\lambda }} {\bar{h}}_{2222}^{2^{*}}=-\dfrac{12{\bar{\lambda }}^2+({\bar{\lambda }}_1-3{\bar{\lambda }}_2)^2}{\mu }A^2,\\ &{}{\bar{\lambda }}_1{\bar{h}}_{1112}^{1^{*}}+3{\bar{\lambda }}{\bar{h}}_{1212}^{1^{*}} +3{\bar{\lambda }}_2 {\bar{h}}_{2212}^{1^{*}}-{\bar{\lambda }} {\bar{h}}_{2212}^{2^{*}} =\dfrac{{\bar{\lambda }}({\bar{\lambda }}_1-3{\bar{\lambda }}_2)}{\mu }A^2.\\ &{} {\bar{h}}_{1122}^{1^{*}}+{\bar{h}}_{2222}^{1^{*}}= \dfrac{{\bar{\lambda }}_1A^2}{\mu }+ {\bar{\lambda }}_2 -({\bar{\lambda }}_2^2+{\bar{\lambda }}^2) {\bar{H}}, \\ &{}{\bar{h}}_{1112}^{1^{*}}+{\bar{h}}_{2212}^{1^{*}}=- \dfrac{3{\bar{\lambda }}}{\mu }A^2+{\bar{\lambda }} -{\bar{\lambda }}{\bar{H}}^2, \\ &{} {\bar{h}}_{1122}^{2^{*}}+{\bar{h}}_{2222}^{2^{*}}=\dfrac{3{\bar{\lambda }}}{\mu }A^2- \dfrac{{\bar{\lambda }}_1^2+3{\bar{\lambda }}^2}{2 {\bar{\lambda }}}A^2-{\bar{\lambda }}. \\ \end{aligned} \end{array}\right. } \end{aligned}$$

(4.73)

Taking covariant differentiation of (2.25) and using (4.47) and (4.48), we obtain

$$\begin{aligned} \begin{aligned} 0&={\bar{h}}_{111}^{1^{*}} {\bar{h}}_{1112}^{1^{*}}+4{\bar{h}}_{111}^{2^{*}} {\bar{h}}_{1122}^{1^{*}}+6{\bar{h}}_{122}^{1^{*}}{\bar{h}}_{1222}^{1^{*}} +4{\bar{h}}_{222}^{1^{*}}{\bar{h}}_{2222}^{1^{*}}+{\bar{h}}_{222}^{2^{*}}{\bar{h}}_{2222}^{2^{*}}\\&\quad -{\bar{H}}^2 \Big \{{\bar{\lambda }} ({\bar{h}}_{112}^{2^{*}}+{\bar{h}}_{222}^{2^{*}}) +{\bar{\lambda }}_1 {\bar{h}}_{111}^{2^{*}}+2{\bar{\lambda }}{\bar{h}}_{122}^{1^{*}}+{\bar{\lambda }}_2{\bar{h}}_{222}^{1^{*}}\Big \}\\&\end{aligned} \end{aligned}$$

Since

$$\begin{aligned}&\begin{aligned}&{\bar{H}}^2 \Big \{{\bar{\lambda }} ({\bar{h}}_{112}^{2^{*}}+{\bar{h}}_{222}^{2^{*}}) +{\bar{\lambda }}_1 {\bar{h}}_{111}^{2^{*}}+2{\bar{\lambda }}{\bar{h}}_{122}^{1^{*}}+{\bar{\lambda }}_2{\bar{h}}_{222}^{1^{*}}\Big \}={\bar{H}}^2 \dfrac{{\bar{\lambda }} A}{\mu }(2+\mu ), \end{aligned} \\&\begin{aligned}&{\bar{h}}_{111}^{1^{*}} {\bar{h}}_{1112}^{1^{*}}+4{\bar{h}}_{111}^{2^{*}} {\bar{h}}_{1122}^{1^{*}}+6{\bar{h}}_{122}^{1^{*}}{\bar{h}}_{1222}^{1^{*}} +4{\bar{h}}_{222}^{1^{*}}{\bar{h}}_{2222}^{1^{*}}+{\bar{h}}_{222}^{2^{*}}{\bar{h}}_{2222}^{2^{*}}\\&\quad =\dfrac{A}{\mu }\Bigg (\Big \{ -16{\bar{\lambda }}^2+({\bar{\lambda }}_1-3{\bar{\lambda }}_2)^2\Big \}{\bar{h}}_{2222}^{2^{*}}-8{\bar{\lambda }}({\bar{\lambda }}_1-3{\bar{\lambda }}_2){\bar{h}}_{2222}^{1^{*}}\Bigg )\\&\qquad +\dfrac{{\bar{\lambda }} A^3}{\mu }\Big \{\dfrac{84{\bar{\lambda }}^2+4\lambda _1^2}{\mu }-14({\bar{\lambda }}_1^2+3{\bar{\lambda }}^2)\Big \} +\dfrac{4{\bar{\lambda }} A}{\mu }(-7{\bar{\lambda }}^2-3\lambda _2^2+3{\bar{\lambda }}_2^2{\bar{H}}^2),\\&\end{aligned} \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&{\bar{H}}^2 \dfrac{{\bar{\lambda }} A}{\mu }(2+\mu )\\&\quad =\dfrac{A}{\mu }\Bigg (\Big \{-4{\bar{\lambda }}\dfrac{21{\bar{\lambda }}^2+{\bar{\lambda }}_1^2}{\mu } +\dfrac{({\bar{\lambda }}_1^2+3{\bar{\lambda }}^2)(18{\bar{\lambda }}^2-{\bar{\lambda }}_1^2-9{\bar{\lambda }}_2^2)}{2{\bar{\lambda }}}\Big \}A^2\\&\qquad +10{\bar{\lambda }}^3-6{\bar{\lambda }}{\bar{\lambda }}_2^2+4{\bar{\lambda }}^3{\bar{H}}^2\Bigg )\\&\qquad +\dfrac{{\bar{\lambda }} A^3}{\mu }\Big \{\dfrac{84{\bar{\lambda }}^2+4\lambda _1^2}{\mu }-14({\bar{\lambda }}_1^2+3{\bar{\lambda }}^2)\Big \} +\dfrac{4{\bar{\lambda }} A}{\mu }(-7{\bar{\lambda }}^2-3\lambda _2^2+3{\bar{\lambda }}_2^2{\bar{H}}^2)\\&\quad =\dfrac{ A^3}{\mu }\dfrac{({\bar{\lambda }}_1^2+3{\bar{\lambda }}^2)(-10{\bar{\lambda }}^2-{\bar{\lambda }}_1^2-9{\bar{\lambda }}_2^2)}{2{\bar{\lambda }}} +\dfrac{{\bar{\lambda }} A}{\mu }(-18{\bar{\lambda }}^2-18\lambda _2^2+4({\bar{\lambda }}^2+3{\bar{\lambda }}_2^2){\bar{H}}^2), \end{aligned} \end{aligned}$$

which is impossible because of $2+\mu =2{\bar{\lambda }}_1^2+12{\bar{\lambda }}_2^2+14{\bar{\lambda }}^2$. Hence, we have ${\bar{\lambda }}=0$, that is, ${\bar{\lambda }}_1 {\bar{\lambda }}_2=0$.

If ${\bar{\lambda }}_2=0$, we get $\inf H^2=S=\sup H^2$ from Proposition 4.4. Namely, $H^2=S$ is constant.

If ${\bar{\lambda }}_1=0$, we have

$$\begin{aligned} {\bar{\lambda }}_2\ne 0, \ \ S=3\inf H^2, \ \ {\bar{H}}^{1*}_{,k}=0, \ k=1,2. \end{aligned}$$

Hence, we have, by using the same calculations as in (4.6),

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\bar{h}}^{1^*}_{11k}+{\bar{h}}^{1^*}_{22k}=0\\ &{}{\bar{h}}^{2^*}_{111}+{\bar{h}}^{2^*}_{221}=0\\ &{}3{\bar{\lambda }}_2{\bar{h}}^{1^*}_{22k}=0. \end{array}\right. } \end{aligned}$$

(4.74)

Hence, we have

$$\begin{aligned} {\bar{h}}^{p^{*}}_{ijk}=0, \ \ \text { except} \ \ i=j=k=p^{*}=2. \end{aligned}$$

If ${\bar{h}}^{2^{*}}_{222}=0$, we get

$$\begin{aligned} {\bar{h}}^{p^{*}}_{ijk}=0, \end{aligned}$$

for any i, j, k, p. According to Lemma 2.1, we have

$$\begin{aligned} 0=\sum _{i,j,k,p} (h_{ijk}^{p^{*}})^{2}+S(1-\frac{3}{2}S)+2H^{2}S-\frac{1}{2}H^{4} -\sum _{j,k,p,q}H^{p^{*}}h_{jk}^{p^{*}}H^{q^{*}}h_{jk}^{q^{*}} =-2. \end{aligned}$$

This is impossible.

If ${\bar{h}}^{2^{*}}_{222}\ne 0$, from Lemma 2.1, we obtain

$$\begin{aligned} |\nabla ^{\perp } \vec H|^2=\sum _{i,j,k,p}({\bar{h}}^{p^{*}}_{ijk})^2=({\bar{h}}^{2^{*}}_{222})^2=2. \end{aligned}$$

Since $S=\sum _{i,j,p}(h_{ij}^{p^{*}})^2$ is constant, we have

$$\begin{aligned} \sum _{i,j,p}h_{ij}^{p^{*}}h_{ijk}^{p^{*}}=0, \ k=1, 2 \end{aligned}$$

and

$$\begin{aligned} \sum _{i,j,p}h_{ijl}^{p^{*}}h_{ijk}^{p^{*}}+\sum _{i,j,p}h_{ij}^{p^{*}}h_{ijkl}^{p^{*}}=0, \ k, l=1, 2. \end{aligned}$$

Then, for $k, l=1, 2$, we get

$$\begin{aligned} \sum _{i,j,p}{\bar{h}}_{ijl}^{p^{*}}{\bar{h}}_{ijk}^{p^{*}}=-\sum _{i,j,p}{\bar{h}}_{ij}^{p^{*}}{\bar{h}}_{ijkl}^{p^{*}} =-{\bar{h}}_{22}^{1^{*}}{\bar{h}}_{22kl}^{1^{*}}-2{\bar{h}}_{12}^{2^{*}}{\bar{h}}_{12kl}^{2^{*}}=-3{\bar{\lambda }}_2{\bar{h}}_{22kl}^{1^{*}}. \end{aligned}$$

If $k=l=1$, we have

$$\begin{aligned} {\bar{h}}_{2211}^{1^{*}}=0. \end{aligned}$$

(4.75)

From (2.19), we know

$$\begin{aligned} {\bar{H}}_{,ij}^{p^{*}}={\bar{h}}_{ij}^{p^{*}}-\sum _k{\bar{h}}_{ik}^{p^{*}}{\bar{h}}_{jk}^{1^{*}}{\bar{H}}. \end{aligned}$$

Let $p=i=2, j=1$, we get

$$\begin{aligned} {\bar{h}}_{1121}^{2^{*}}+{\bar{h}}_{2221}^{2^{*}}={\bar{\lambda }}_2. \end{aligned}$$

From (4.75), we obtain

$$\begin{aligned} {\bar{h}}_{2221}^{2^{*}}={\bar{\lambda }}_2\ne 0. \end{aligned}$$

On the other hand, from Lemma 2.1, we have

$$\begin{aligned} 2\sum _{i,j,k,p}{\bar{h}}_{ijk}^{p^{*}}{\bar{h}}_{ijk1}^{p^{*}}=(\sum _{j,k,p,q}{\bar{H}}^{p^{*}}{\bar{h}}_{jk}^{p^{*}}{\bar{H}}^{q^{*}}{\bar{h}}_{jk}^{q^{*}})_{,1}=0 \end{aligned}$$

because ${\bar{H}}_{,i}^{1^{*}}=0, h_{ij1}^{q^{*}}=0$. Since

$$\begin{aligned} 2\sum _{i,j,k,p}{\bar{h}}_{ijk}^{p^{*}}{\bar{h}}_{ijk1}^{p^{*}}=2{\bar{h}}_{222}^{2^{*}}{\bar{h}}_{2221}^{2^{*}}=2{\bar{\lambda }}_2{\bar{h}}_{222}^{2^{*}} \ne 0, \end{aligned}$$

it is a contradiction. Thus, we know that $H^2=S$ is constant. $\square$

Proof of Theorem 1.1

From Theorem 4.1 and Theorem 4.2, we know $S=0$, $S=1$ or $S=2$. According to the result of Cheng and Peng [8], we only consider the case $S\equiv 2$ and $S\equiv 1$. Therefore, the mean curvature $H^2=S$ is constant from Theorem 4.4.

If $H^2=S=2$, from (2.25) in Lemma 2.1, we have

$$\begin{aligned} \sum _{i,j,k,p}(h^{p^{*}}_{ijk})^2\equiv 0, \ \ \lambda _1=\lambda _2\ne 0. \end{aligned}$$

According to

$$\begin{aligned} H^{p^*}_{,i}=\sum _{k}h_{ik}^{p^*}\langle X, e_k\rangle , \ \ \text { for} \ \ i, p=1, 2, \end{aligned}$$

we know, at any point,

$$\begin{aligned} 0=h_{11i}^{1^*}+h_{22i}^{1^*}=H^{1^*}_{,i}=\lambda _i\langle X, e_i\rangle . \end{aligned}$$

Hence, we get $\langle X, e_i\rangle =0$ for $i=1, 2$ at any point. Thus, $|X|^2$ is constant. According to

$$\begin{aligned} \dfrac{1}{2}{\mathcal {L}} |X|^2=2-|X|^2 \end{aligned}$$

we obtain

$$\begin{aligned} |X|^2\equiv 2, \end{aligned}$$

it means that, $X: M^2\rightarrow {\mathbb {R}}^{4}$ becomes a complete surface in the sphere $S^3(\sqrt{2})$. Because $S=2$, it is easy to prove that $X: M^2\rightarrow S^3(\sqrt{2})$ is minimal and its Gaussian curvature is zero. Thus, we conclude that $X: M^2\rightarrow {\mathbb {R}}^{4}$ is the Clifford torus $S^1(1)\times S^1(1)$.

If $H^2=S=1$, from (2.25) in Lemma 2.1, we have

$$\begin{aligned} \sum _{i,j,k,p}(h^{p^{*}}_{ijk})^2\equiv 0, \ \ \lambda =0, \ \ \lambda _2= 0. \end{aligned}$$

From the results of Yau in [29], we know that $X:M^2\rightarrow {\mathbb {R}}^4$ is $S^1(1)\times {\mathbb {R}}^1$. It completes the proof of Theorem 1.1. $\square$