A rigidity theorem of $\xi$-submanifolds in ${\mathbb {C}}^{2}$

Abstract

In this paper, we first introduce the concept of $\xi$-submanifold which is a natural generalization of self-shrinkers for the mean curvature flow and also an extension of $\lambda$-hypersurfaces to the higher codimension. Then, as the main result, we prove a rigidity theorem for Lagrangian $\xi$-submanifold in the complex 2-plane ${\mathbb C}^2$.

1 Introduction

Let $x: M^n\rightarrow {\mathbb {R}}^{n+p}$ be an n-dimensional submanifold in the $(n+p)$-dimensional Euclidean space ${\mathbb {R}}^{n+p}$. Then x is called a self-shrinker (to the mean curvature flow) in ${\mathbb {R}}^{n+p}$ if its mean curvature vector field H satisfies

$$\begin{aligned} H+x^{\bot }=0, \end{aligned}$$

(1.1)

where $x^\bot$ is the orthogonal projection of the position vector x to the normal space $T^\bot M^n$ of x.

It is well known that the self-shrinker plays an important role in the study of the mean curvature flow. Not only self-shrinkers correspond to self-shrinking solutions to the mean curvature flow, but also they describe all possible Type I blow ups at a given singularity of the flow. Up to now, there have been a plenty of research papers on self-shrinkers among which are many that provide various results of classification or rigidity theorems. In particular, there are also interesting results about the Lagrangian self-shrinkers in the complex Euclidean n-space ${\mathbb C}^n$. For example, in [1], Anciaux gives new examples of self-shrinking and self-expanding Lagrangian solutions to the mean curvature flow. In [3], the authors classify all Hamiltonian stationary Lagrangian surfaces in the complex plane ${\mathbb C}^2$, which are self-similar solutions of the mean curvature flow and, in [4], several rigidity results for Lagrangian mean curvature flow are obtained. As we know, a canonical example of the compact Lagrangian self-shrinker in ${\mathbb {C}}^2$ is the Clifford torus ${\mathbb {S}}^{1}(1)\times {\mathbb {S}}^{1}(1)$.

Recently in [13], Li and Wang prove a rigidity theorem which improves a previous theorem by Castro and Lerma [4].

Theorem 1.1

(cf. [4, 13]). Let $x:M^2\rightarrow {\mathbb C}^2$ be a compact oriented Lagrangian self-shrinker with h its second fundamental form. If $|h|^{2}\le 2$, then $|h|^{2}=2$ and $x(M^{2})$ is the Clifford torus ${\mathbb {S}}^{1}(1)\times {\mathbb {S}}^{1}(1)$, up to a holomorphic isometry on ${\mathbb C}^2$.

Remark 1.1

Castro and Lerma also proved Theorem 1.1 in [4] under the additional condition that the Gauss curvature K of $M^{2}$ is either non-negative or non-positive.

To make an extension of hypersurface self-shrinkers, Cheng and Wei recently introduce in [7] the definition of $\lambda$-hypersurface of weighted volume-preserving mean curvature flow in Euclidean space, and classify complete $\lambda$-hypersurfaces with polynomial area growth and $H-\lambda \ge 0$, which are generalizations of the results due to Huisken [12] and Colding-Minicozzi [9]. According to [7], a hypersurface $x: M^{n}\rightarrow {\mathbb {R}}^{n+1}$ is called a $\lambda$-hypersurface if its mean curvature $H_0$ satisfies

$$\begin{aligned} H_{0}+\langle x, N\rangle =\lambda \end{aligned}$$

(1.2)

for some constant $\lambda$, where N is the unit normal vector of x. Some rigidity or classification results for $\lambda$-hypersurfaces are obtained, for example, in [6, 8, 11]; for the rigidity theorems for space-like $\lambda$-hypersurfaces see [15].

As a natural generalization of both self-shrinkers and $\lambda$-hypersurfaces, we introduce the concept of $\xi$-submanifolds. Precisely, an immersed submanifold $x: M^{n}\rightarrow {\mathbb {R}}^{n+p}$ is called a $\xi$-submanifold if there is a parallel normal vector field $\xi$ such that the mean curvature vector field H satisfies

$$\begin{aligned} H+x^{\bot }=\xi . \end{aligned}$$

(1.3)

Obviously, the Clifford tori ${\mathbb {S}}^{1}(a)\times {\mathbb {S}}^{1}(b)$ with positive numbers a and b are examples of Lagrangian $\xi$-submanifold in ${\mathbb C}^2$. Similar examples in higher dimensions can be listed as those in [5] for self-shrinkers. In this paper, we focus on the rigidity of compact Lagrangian $\xi$-submanifolds in ${\mathbb {C}}^2$, and our main theorem is as follows:

Theorem 1.2

Let $x: M^{2}\rightarrow {\mathbb {C}}^{2}$ be a compact oriented Lagrangian $\xi$-submanifold with the second fundamental form h and mean curvature vector H. Assume that

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}\le |\xi |^{2}+4. \end{aligned}$$

Then $|h|^{2}+|H-\xi |^{2}\equiv |\xi |^{2}+4$ and $x(M^2)=T^2$ is a topological torus.

Furthermore, if $\langle H, \xi \rangle$ is constant and one of the following four conditions holds:

$$\begin{aligned} (1)\ |h|^{2}\ge 2,\quad (2)\ |H|^{2}\ge 2,\quad (3)\ |h|^{2}\ge \langle H, H-\xi \rangle ,\quad (4)\ \langle H, \xi \rangle \ge 0, \end{aligned}$$

(1.4)

then, up to a holomorphic isometry on ${\mathbb {C}}^{2}$, $x(M^{2})={\mathbb {S}}^{1}(a)\times {\mathbb {S}}^{1}(b)$ is a standard torus, where a and b are positive numbers satisfying $a^{2}+b^{2}\ge 2a^{2}b^{2}$.

Corollary 1.3

Let $x: M^{2}\rightarrow {\mathbb {C}}^{2}$ be a compact oriented Lagrangian self-shrinker. If

$$\begin{aligned} |h|^{2}+|H|^{2}\le 4, \end{aligned}$$

then $|h|^{2}+|H|^{2}\equiv 4$ and $x(M^{2})={\mathbb {S}}^{1}(1)\times {\mathbb {S}}^{1}(1)$ up to a holomorphic isometry on ${\mathbb {C}}^{2}$.

Clearly, Corollary 1.3 can be viewed as a different new version of Theorem 1.1.

Remark 1.2

We believe that the last condition (1.4) in Theorem 1.2 can be removed. On the other hand, the condition that $\langle H,\xi \rangle$ is constant may also be removed. In fact, as suggested by the referee, we can use (3.11) and the compactness of $M^2$ to show that $|x|^2$ is constant when either $\langle H, H-\xi \rangle \le 2$ or $\langle H, H-\xi \rangle \ge 2$. Then by the argument at the end of the paper, we can simplify Theorem 1.2 as follows:

Theorem 1.4

Let $x: M^{2}\rightarrow {\mathbb {C}}^{2}$ be a compact oriented Lagrangian $\xi$-submanifold with the second fundamental form h and mean curvature vector H. Assume that

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}\le |\xi |^{2}+4. \end{aligned}$$

Then $|h|^{2}+|H-\xi |^{2}\equiv |\xi |^{2}+4$ and $x(M^2)=T^2$ is a topological torus.

Furthermore, if either $\langle H, H-\xi \rangle \le 2$ or $\langle H, H-\xi \rangle \ge 2$, then, up to a holomorphic isometry on ${\mathbb {C}}^{2}$, $x(M^{2})={\mathbb {S}}^{1}(a)\times {\mathbb {S}}^{1}(b)$ is a standard torus for some $a,b>0$.

Remark 1.3

Cheng and Wei have introduced in [7] a weighted area functional ${\mathcal {A}}$ and derived a related variation formula. Besides the relation between $\lambda$-hypersurfaces and the weighted volume preserving mean curvature flow, they also prove that $\lambda$-hypersurfaces are the critical points of the weighted area functional. Based on this, we believe that similar conclusions will be valid for the $\xi$-submanifolds defined above. Furthermore, We reasonably believe that, if self-shrinkers and $\lambda$-hypersurfaces take the places of minimal submanifolds and constant mean curvature hypersurfaces, respectively, then $\xi$-submanifolds must take the place of submanifolds of parallel mean curvature vector.

2 Lagrangian submanifolds in ${\mathbb C}^n$ and their Maslov class

Let ${\mathbb {C}}^{n}$ be the complex Euclidean n-space with the canonical complex structure J. Through out this paper, $x:M^{n}\rightarrow {\mathbb C}^n$ always denotes an n-dimensional Lagrangian submanifold, and $\nabla$, D, $\nabla ^{\bot }$ denote, respectively, the Levi-Civita connections on $M^{n}$, ${\mathbb {C}}^{n}$, and the normal connection on the normal boundle $T^\bot M^n$. The formulas of Gauss and Weingarten are given by

$$\begin{aligned} D_{X}Y=\nabla _{X}Y+h(X, Y),\quad D_{X}\eta =-A_{\eta }X+\nabla ^{\bot }_{X}\eta , \end{aligned}$$

where X, Y are tangent vector fields on $M^n$ and $\eta$ is a normal vector field of x. The Lagrangian condition implies that

$$\begin{aligned} \nabla ^{\bot }_{X}JY=J\nabla _{X}Y, \quad A_{JX}Y=-Jh(X,Y)=A_{JY}X, \end{aligned}$$

where h and A are the second fundamental form and the shape operator of x, respectively. In particular, $\langle h(X, Y), JZ \rangle$ is totally symmetric as a 3-form, namely

$$\begin{aligned} \langle h(X, Y), JZ\rangle = \langle h(X, Z), JY\rangle = \langle h(Y, Z), JX\rangle . \end{aligned}$$

(2.1)

From now on, we agree with the following convention on the ranges of indices:

$$\begin{aligned} 1\le i,j,\cdots \le n,\quad n+1\le \alpha ,\beta ,\cdots \le 2n,\quad 1\le A,B,\cdots \le 2n,\quad i^*=n+i. \end{aligned}$$

For a Lagrangian submanifold $x:M^{n}\rightarrow {\mathbb {C}}^{n}$, there are orthonormal frame fields of the form $\{e_i,e_{i^*}\}$ for ${\mathbb C}^n$ along x, where $e_i\in TM^n$ and $e_{i^*}=Je_i$. Such a frame is called an adapted Lagrangian frame field in the literature. The dual frame field is always denoted by $\{\theta _i,\theta _{i^*}\}$, where $\theta _{i^*}=-J\theta _i$. Write

$$\begin{aligned} h=\sum h^{k^*}_{ij}\theta _i\theta _je_{k^*},\quad \text{ where } h^{k^*}_{ij}=\langle h(e_i,e_j),e_{k^*}\rangle , \end{aligned}$$

or equivalently,

$$\begin{aligned} h(e_{i}, e_{j})=\sum _{k}h_{ij}^{k^{*}}e_{k^{*}},\quad \text{ for } \text{ all } e_i,e_j. \end{aligned}$$

Then (2.1) is equivalent to

$$\begin{aligned} h_{ij}^{k^{*}}=h_{kj}^{i^{*}}=h_{ik}^{j^{*}}, \quad 1\le i, j, k\le n. \end{aligned}$$

(2.2)

If $\theta _{ij}$ and $\theta _{i^*j^*}$ denote the connection forms of $\nabla$ and $\nabla ^\bot$, respectively, then the components $h_{ij,l}^{k^{*}}$, $h_{ij,lp}^{k^{*}}$ of the covariant derivatives of h are given respectively by

$$\begin{aligned} \sum _{l}h_{ij,l}^{k^{*}}\theta _{l}= & {} dh_{ij}^{k^{*}}+\sum _{l}h_{lj}^{k^{*}}\theta _{li} +\sum _{l}h_{il}^{k^{*}}\theta _{lj}+\sum _{m}h_{ij}^{m^{*}}\theta _{m^{*}k^{*}}; \end{aligned}$$

(2.3)

$$\begin{aligned} \sum _{p}h_{ij,lp}^{k^{*}}\theta _{p}= & {} dh_{ij,l}^{k^{*}}+\sum _{p}h_{pj,l}^{k^{*}}\theta _{pi} +\sum _{p}h_{ip,l}^{k^{*}}\theta _{pj}+\sum _{p}h_{ij,p}^{k^{*}}\theta _{pl} +\sum _{p}h_{ij,l}^{p^{*}}\theta _{p^{*}k^{*}}.\quad \qquad \end{aligned}$$

(2.4)

Moreover, the equations of motion are as follows:

$$\begin{aligned} dx= & {} \sum _{i}\theta _{i}e_{i},\quad de_{i}=\sum _{j}\theta _{ij}e_{j}+\sum _{k,j}h_{ij}^{k^{*}}\theta _{j}e_{k^{*}},\end{aligned}$$

(2.5)

$$\begin{aligned} de_{k^{*}}= & {} -\sum _{i,j}h_{ij}^{k^{*}}\theta _{j}e_{i}+\sum _{l}\theta _{k^{*}l^{*}}e_{l^{*}}. \end{aligned}$$

(2.6)

Let $R_{ijkl}$ and $R_{i^{*}j^{*}kl}$ denote the components of curvature operators of $\nabla$ and $\nabla ^{\bot }$, respectively. Then the equations of Gauss, Codazzi and Ricci are as follows:

$$\begin{aligned} R_{mijk}= & {} \sum _{l}(h_{mk}^{l^{*}}h_{ij}^{l^{*}}-h_{mj}^{l^{*}}h_{ik}^{l^{*}}), \quad 1\le m,i, j, k\le n, \end{aligned}$$

(2.7)

$$\begin{aligned} h_{ij,l}^{k^{*}}= & {} h_{il,j}^{k^{*}}, \quad 1\le i, j, k, l\le n, \end{aligned}$$

(2.8)

$$\begin{aligned} R_{i^{*}j^{*}kl}= & {} \sum _{m}(h_{ml}^{i^{*}}h_{mk}^{j^{*}}-h_{mk}^{i^{*}}h_{ml}^{j^{*}}), \quad 1\le i, j, k, l\le n. \end{aligned}$$

(2.9)

The scalar curvature of $\nabla$ is

$$\begin{aligned} R=|H|^{2}-|h|^{2}\quad \text{ with } |H|^{2}=\sum _{k}\left( \sum _{i}h^{k^*}_{ii}\right) ^2,\quad |h|^{2}=\sum _{i,j,k}(h_{ij}^{k^{*}})^{2}, \end{aligned}$$

(2.10)

where the mean curvature vector field H is defined by

$$\begin{aligned} H=\sum _{k}H^{k^{*}}e_{k^{*}}=\sum _{i,k} h_{ii}^{k^{*}}e_{k^{*}}. \end{aligned}$$

Combining (2.2) and (2.8), we know that $h_{ij,l}^{k^{*}}$ is totally symmetric, namely

$$\begin{aligned} h_{ij,l}^{k^{*}}=h_{jl,k}^{i^{*}}=h_{lk,i}^{j^{*}}=h_{ki,j}^{l^{*}},\quad 1\le i,j,k,l\le n, \end{aligned}$$

(2.11)

and the Ricci identities are as follows:

$$\begin{aligned} h_{ij,lp}^{k^{*}}-h_{ij,pl}^{k^{*}}=\sum _{m}h_{mj}^{k^{*}}R_{imlp} +\sum _{m}h_{im}^{k^{*}}R_{jmlp}+\sum _{m}h_{ij}^{m^{*}}R_{k^{*}m^{*}lp}. \end{aligned}$$

(2.12)

Note that, with respect to the adapted Lagrangian frame $\{e_i,e_{i^*}\}$, the connection forms $\theta _{i^*j^*}=\theta _{ij}$. It follows that

$$\begin{aligned} R_{m^{*}i^{*}jk}=R_{mijk},\quad \forall m,i,j,k. \end{aligned}$$

(2.13)

Furthermore, the first and second derivatives $H^{k^*}_{,i}$, $H^{k^*}_{,ij}$ of the mean curvature vector field H are given as

$$\begin{aligned} H^{k^*}_{,i}=\sum _jh^{k^*}_{jj,i},\quad H^{k^*}_{,ij}=\sum _lh^{k^*}_{ll,ij}. \end{aligned}$$

(2.14)

For any smooth function f on $M^{n}$, the covariant derivatives $f_{,i}$, $f_{,ij}$ of f, the Laplacian of f are respectively defined as follows:

$$\begin{aligned} df=\sum _{i}f_{,i}\theta _{i}, \quad \sum _{j}f_{,ij}\theta _{j}=df_{,i}-\sum _{j}f_{,j}\theta _{ij},\quad \triangle f=\sum _{i}f_{,ii}. \end{aligned}$$

(2.15)

Finally, we also need to introduce the Lagrangian angles, Maslov form and Maslov class of a Lagrangian submanifold in ${\mathbb C}^n$ which we shall make use of later.

Let $(z^1,\ldots ,z^n)$ be the standard complex coordinates on ${\mathbb C}^n$. Then $\Omega =dz^1\wedge \cdots \wedge dz^n$ is a globally defined holomorphic volume form which is clearly parallel. For a Lagrangian submanifold $x:M^n\rightarrow {\mathbb C}^n$, the Lagrangian angle of x is by definition a multi-valued function $\beta :M^n\rightarrow {\mathbb {R}}/2\pi {\mathbb {Z}}$ given by

$$\begin{aligned} \Omega _M:=x^*\Omega =e^{\sqrt{-1}\beta }dV_M. \end{aligned}$$

As one knows, although the Lagrangian angle $\beta$ can not be determined globally in general, its gradient $\nabla \beta$ is clearly a well-defined vector field on $M^n$, or the same, $\alpha :=d\beta$ is a globally defined 1-form which is called the Maslov form of x. Clearly, $\alpha$ is closed and thus represents a cohomology class $[\alpha ]\in H^1(M^n)$ called the Maslov class.

In [16], the author proved an important formula by which the mean curvature and the Lagrangian angle of a Lagrangian submanifold are linked to each other; A. Arsie has extended this result in [2] to Lagrangian submanifolds in a general Calabi-Yau manifold.

Theorem 2.1

([16]) Let $x:M^n\rightarrow {\mathbb C}^n$ be a Lagrangian submanifold and J be the canonical complex structure of ${\mathbb C}^n$. Then the mean curvature vector H and the Lagrangian angle $\beta$ meet the following formula:

$$\begin{aligned} x_*(\nabla \beta )=-JH. \end{aligned}$$

(2.16)

Corollary 2.2

([4, 17]) Let $x:M^n\rightarrow {\mathbb C}^n$ be a compact and oriented Lagrangian self-shrinkers. Then the Maslov class $[\alpha ]$ can not be trivial. In particular, there does not exist any Lagrangian self-shrinker in ${\mathbb C}^n$ with the topology of a sphere.

Remark 2.1

For our use in this paper, it is necessary to show that Corollary 2.2 is still true if we replace the self-shrinker by a $\xi$-submanifold. Precisely, we need

Proposition 2.3

Let $x: M^{n}\rightarrow {\mathbb {C}}^{n}$ be a Lagrangian $\xi$-submanifold. If M is compact and orientable, then $[\alpha ]\ne 0$; Consequently, there does not exist any Lagrangian $\xi$-submanifold in ${\mathbb C}^n$ with the topology of a sphere.

Proof

By the definition of a $\xi$-submanifold, we have $x=x^{\top }+\xi -H$. By Gauss and Weingarten formulas it follows that, for any $v\in TM^n$,

$$\begin{aligned} A_{H}v= & {} -D_{v}H+\nabla ^{\bot }_{v}H =-D_{v}(\xi -x^{\bot })+\nabla ^{\bot }_{v}H\\= & {} D_{v}x^{\bot }-D_{v}\xi +\nabla ^{\bot }_{v}H =D_{v}x-D_{v}x^{\top }-D_{v}\xi +\nabla ^{\bot }_{v}H\\= & {} v-\nabla _{v}x^{\top }+A_\xi (v)-h(v,x^{\top })+\nabla ^{\bot }_{v}H, \end{aligned}$$

where $A_{H}$ and $A_\xi$ are Weingarten transformations with respect to H and $\xi$, respectively. Thus

$$\begin{aligned} A_{H}v=v-\nabla _{v}x^{\top }+A_\xi (v),\quad \nabla _{v}^{\bot }H=h(v,x^{\top }). \end{aligned}$$

So that

$$\begin{aligned} \mathrm{div\,}JH= & {} \sum _{i}\langle \nabla _{e_{i}}JH,e_{i}\rangle =\sum _{i}\langle J\nabla _{e_{i}}JH, Je_{i} \rangle =\sum _{i}\langle -\nabla _{e_{i}}^{\bot }H, Je_{i}\rangle \nonumber \\= & {} \sum _{i}\langle -h(e_{i}, x^{\top }), Je_{i}\rangle =\sum _{i}-\langle h(e_{i},e_{i}), Jx^{\top }\rangle \nonumber \\= & {} \sum _{i}\langle Jh(e_{i},e_{i}), x^{\top }\rangle =\langle JH, x^{\top }\rangle , \end{aligned}$$

(2.17)

where $\mathrm{div\,}$ is the divergence operator. By (2.16) and (2.17) we obtain

$$\begin{aligned} \Delta \beta =\langle \nabla \beta ,x^\top \rangle =\frac{1}{2}\langle \nabla \beta , \nabla |x|^{2}\rangle . \end{aligned}$$

(2.18)

If $[\alpha ]= 0$, then there exists a globally defined Lagrangian angle $\beta$ such that $\alpha =-d\beta$, implying (2.18) holds globally on $M^n$. Then the compactness assumption and the maximum principle for a second linear elliptic partial equation (see [10], for example) assure that $\beta$ must be constant. Hence $H=x_*(J\nabla \beta )\equiv 0$, contradicting to the fact that there are no compact minimal submanifolds in Euclidean space. This contradiction proves that $[\alpha ]\ne 0$.

Since the first homology of a sphere $S^n$ vanishes for $n>1$, there can not be any Lagrangian $\xi$-submanifolds with the topology of a sphere. $\square$

3 Proof of the main theorem

Let $x: M^n\rightarrow {\mathbb {C}}^n$ be a Lagrangian $\xi$-submanifold without boundary. Then, with respect to an orthonormal frame field $\{e_i\}$, the defining equation (1.3) is equivalent to

$$\begin{aligned} H^{k^{*}}=-\langle x, e_{k^{*}}\rangle +\xi ^{k^{*}}, \quad 1 \le k \le n. \end{aligned}$$

(3.1)

where $\xi =\sum \xi ^{k^*}e_{k^*}$ is a given parallel normal vector field. From now on, we always assume that $n=2$ if no other specification is given.

We start with a well-known operator ${\mathcal {L}}$ acting on smooth functions defined by

$$\begin{aligned} {\mathcal {L}}=\triangle -\langle x, \nabla \cdot \rangle =e^{\frac{|x|^{2}}{2}}\mathrm{div\,}(e^{-\frac{|x|^{2}}{2}}\nabla \cdot ), \end{aligned}$$

(3.2)

which was first introduced by Colding and Minicozzi [9] to the study of self-shrinkers. Since then, the operator ${\mathcal {L}}$ has been one of the most effect tools adapted by many authors. In particular, the following is a fundamental lemma related to ${\mathcal {L}}$:

Lemma 3.1

([14]) Let $x: M^n\rightarrow {\mathbb {R}}^{n+p}$ be a complete immersed submanifold. If u and v are $C^2$-smooth functions with

$$\begin{aligned} \int _M(|u\nabla v|+|\nabla u||\nabla v|+|u{\mathcal {L}} v|)e^{-\frac{|x|^2}{2}}dV_M< \infty , \end{aligned}$$

then it holds that

$$\begin{aligned} \int _M u{\mathcal {L}}v e^{-\frac{|x|^2}{2}}dV_M=-\int _M\langle \nabla u, \nabla v\rangle e^{-\frac{|x|^2}{2}}dV_M. \end{aligned}$$

Now, to make the whole argument more readable, we divide our proof into the following lemmas and propositions:

Lemma 3.2

(cf. [13]) Let $x: M^{2}\rightarrow {\mathbb {C}}^{2}$ be a Lagrangian $\xi$-submanifold. Then

$$\begin{aligned} H_{,i}^{k^{*}}= & {} \sum _{j}h_{ij}^{k^{*}}\langle x, e_{j}\rangle , \quad 1\le i,k\le 2, \end{aligned}$$

(3.3)

$$\begin{aligned} H_{,ij}^{k^{*}}= & {} \sum _{m}h_{im,j}^{k^{*}}\langle x, e_{m}\rangle +h_{ij}^{k^{*}}-\sum _{m,p}(H-\xi )^{p^{*}}h_{im}^{k^{*}}h_{mj}^{p^{*}},\quad 1\le i,j,k\le 2. \end{aligned}$$

(3.4)

Lemma 3.3

It holds that

$$\begin{aligned} \frac{1}{2}{\mathcal {L}}(|h|^{2}+|H-\xi |^{2})= & {} |\nabla h|^{2}+|\nabla ^{\bot }H|^{2}+|h|^{2}\nonumber \\&-\,\frac{1}{2}(|h|^{2}-|H|^{2})(3|h|^{2}-2|H|^{2}+\langle H, H-\xi \rangle )\nonumber \\&+\,\langle H,H-\xi \rangle -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{k^{*}}(H-\xi )^{l^{*}}\nonumber \\&-\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}H^{k^{*}}(H-\xi )^{l^{*}}. \end{aligned}$$

(3.5)

Proof

By a direct computation using Lemma 3.2 we find (cf. [13])

$$\begin{aligned} \frac{1}{2}{\mathcal {L}}|h|^{2}= & {} |\nabla h|^{2}+|h|^{2}-\frac{3}{2}|h|^{4} +\frac{5}{2}|H|^{2}|h|^{2}-|H|^{4}\nonumber \\&+\frac{1}{2}\langle H, H-\xi \rangle (|H|^{2}-|h|^{2})- \sum _{i,j,k,l}H^{k^{*}}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{l^{*}}; \end{aligned}$$

(3.6)

$$\begin{aligned} \frac{1}{2}{\mathcal {L}}(|H-\xi |^{2})= & {} \frac{1}{2}\triangle (|H-\xi |^{2}) -\frac{1}{2}\langle x, \nabla |H-\xi |^{2}\rangle \nonumber \\= & {} \sum _{i,k}(H-\xi )^{k^{*}}H_{,ii}^{k^{*}}+|\nabla ^{\bot }H|^{2} -\sum _{i.k}(H-\xi )^{k^{*}}H_{,i}^{k^{*}}\langle x, e_{i}\rangle \nonumber \\= & {} \langle H- \xi , H\rangle +|\nabla ^{\bot }H|^{2} -\sum _{i,j,k,l}(H-\xi )^{k^{*}}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{l^{*}}. \end{aligned}$$

(3.7)

By taking the sum we obtain (3.5). $\square$

Lemma 3.4

It holds that

$$\begin{aligned} \frac{1}{2}\triangle (|x^{\top }|^{2})= & {} \sum _{i,j,k}h_{ij}^{k^{*}}\langle x,e_{i}\rangle \langle x, e_{j}\rangle (\xi -H)^{k^{*}}-\sum _{i,j,k,l}h_{il}^{k^{*}}h_{lj}^{k^{*}} \langle x, e_{i}\rangle \langle x, e_{j}\rangle \nonumber \\&+\,2-2\langle H,H-\xi \rangle +\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{k^{*}}(H-\xi )^{l^{*}}. \end{aligned}$$

(3.8)

Proof

We find

$$\begin{aligned} \frac{1}{2}\triangle (|x^{\top }|^{2})= & {} \frac{1}{2}\sum _{i,j}\langle x, e_{j}\rangle ^{2}_{,ii}=\sum _{i,j}(\langle x, e_{j}\rangle \langle x, e_{j}\rangle _{i})_{,i}\\= & {} \sum _{i,j}\left( \langle x, e_{j}\rangle \langle x_{i}, e_{j}\rangle \right) _{,i}+\sum _{i,j,k}\left( \langle x, e_{j}\rangle \langle x, h_{ji}^{k^{*}}e_{k^{*}}\rangle \right) _{,i}\\= & {} 2+2\sum _{i,k}h_{ii}^{k^{*}}\langle x, e_{k^{*}}\rangle +\sum _{j,k}H_{,j}^{k^{*}}\langle x, e_{j}\rangle \langle x, e_{k^{*}}\rangle \\&+\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\langle x, e_{l^{*}}\rangle \langle x, e_{k^{*}}\rangle -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{il}^{k^{*}}\langle x, e_{j}\rangle \langle x, e_{l}\rangle \\= & {} 2-2\langle H,H-\xi \rangle +\sum _{i,j,k}h_{ij}^{k^{*}}\langle x, e_{i}\rangle \langle x, e_{j}\rangle (\xi -H)^{k^{*}}\\&+\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{l^{*}}(H-\xi )^{k^{*}} -\sum _{i,j,k,l}h_{il}^{k^{*}}h_{lj}^{k^{*}}\langle x, e_{i}\rangle \langle x, e_{j}\rangle , \end{aligned}$$

and the lemma is proved. $\square$

Lemma 3.5

It holds that

$$\begin{aligned} \triangle (\langle H, \xi \rangle )= & {} \sum _{i,j,k}h_{ij}^{k^{*}}\langle x, e_{i}\rangle \langle x, e_{j}\rangle \xi ^{k^{*}}+\langle H, \xi \rangle -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\xi ^{k^{*}}(H-\xi )^{l^{*}},\end{aligned}$$

(3.9)

$$\begin{aligned} {\mathcal {L}}(\langle H, \xi \rangle )= & {} \langle H, \xi \rangle -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\xi ^{k^{*}}(H-\xi )^{l^{*}}. \end{aligned}$$

(3.10)

Proof

By (3.3) and (3.4),

$$\begin{aligned} \triangle (\langle H, \xi \rangle )= & {} \sum _{i,k}(H^{k^{*}}\xi ^{k^{*}})_{,ii}=\sum _{i,k}H^{k^{*}}_{,ii}\xi ^{k^{*}}\\= & {} \sum _{i,k,l,m}(h_{im,i}^{k^{*}}\langle x, e_{m}\rangle +h_{ii}^{k^{*}}-(H-\xi )^{^{l^{*}}}h_{im}^{k^{*}}h_{mi}^{l^{*}})\xi ^{k^{*}}\\= & {} \sum _{i,k}H_{,i}^{k^{*}}\langle x, e_{i}\rangle \xi ^{k^{*}}+\langle H, \xi \rangle - \sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\xi ^{k^{*}}(H-\xi )^{l^{*}}\\= & {} \sum _{i,j,k}h_{ij}^{k^{*}}\langle x, e_{i}\rangle \langle x, e_{j}\rangle \xi ^{k^{*}}+\langle H, \xi \rangle -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\xi ^{k^{*}}(H-\xi )^{l^{*}};\\ \langle x, \nabla \langle H, \xi \rangle \rangle= & {} \sum _{i}\langle H, \xi \rangle _{,i}\langle x, e_{i}\rangle =\sum _{i,j,k}h_{ij}^{k^{*}}\langle x, e_{i}\rangle \langle x, e_{j}\rangle \xi ^{k^{*}}. \end{aligned}$$

Thus, by adding them up, we get (3.10). $\square$

Lemma 3.6

(cf. [5, 9]; also [13]) It holds that

$$\begin{aligned} \frac{1}{2}\triangle (|x|^{2})= & {} 2-\langle H, H-\xi \rangle , \end{aligned}$$

(3.11)

$$\begin{aligned} \frac{1}{2}{\mathcal {L}}(|x|^{2})= & {} |\xi |^{2}+2-(|x|^{2}+\langle H, \xi \rangle ). \end{aligned}$$

(3.12)

Proof

From (3.1), we find

$$\begin{aligned} \frac{1}{2}\triangle (|x|^{2})= & {} 2+\langle x, \triangle x\rangle =2+\sum _{k}H^{k^{*}}\langle x, e_{k^{*}}\rangle =2-\langle H,H-\xi \rangle ,\\ \frac{1}{2}{\mathcal {L}}(|x|^{2})= & {} \frac{1}{2}\triangle (|x|^{2})-\frac{1}{2}\langle x, \nabla |x|^{2}\rangle =2-|H|^{2}+\langle H,\xi \rangle -|x^{\top }|^{2}\\= & {} 2+|\xi |^{2}-(|x|^{2}+\langle H, \xi \rangle ). \end{aligned}$$

$\square$

Proposition 3.7

Let $M^2$ be oriented and compact. If

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}\le |\xi |^{2}+4, \end{aligned}$$

then

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}\equiv |\xi |^{2}+4 \end{aligned}$$

(3.13)

and $x(M^{2})$ is a topological torus.

Proof

By Lemma 3.6,

$$\begin{aligned} \int _{M}|H-\xi |^{2}dV_{M}= & {} \int _{M}(|\xi |^{2}+2(|H|^{2}-\langle H, \xi \rangle )-|H|^{2})dV_{M}\nonumber \\= & {} \int _{M}(|\xi |^{2}+4-|H|^{2})dV_{M}. \end{aligned}$$

(3.14)

Let K be the Gauss curvature of $M^2$. Then the Gauss equation gives that

$$\begin{aligned} 2K= |H|^{2}-|h|^{2}. \end{aligned}$$

Denote by $\mathrm{gen}(M^2)$ the genus of $M^2$. Then from the Gauss-Bonnet theorem and (3.14) it follows that

$$\begin{aligned} 8\pi (1-\mathrm{gen}(M^2))= & {} 2\int _{M}KdV_{M}=\int _{M}(|H|^{2}-|h|^{2})dV_{M} \nonumber \\= & {} \int _{M}(|\xi |^{2}+4-(|h|^{2}+|H-\xi |^{2}))dV_{M}\ge 0, \end{aligned}$$

(3.15)

implying that ${\mathrm{gen}}(M^2)\le 1$. So $M^2$ is topologically either a 2-sphere or a torus. But Proposition 2.3 excludes the first possibility. So ${\mathrm{gen}}(M^2)=1$ and (3.13) is proved. $\square$

Lemma 3.8

Let $p_{0}\in M^2$ be a point where $|x|^{2}$ attains its minimum on $M^2$. If $M^2$ is orientable, compact and

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}=\mathrm{const}, \end{aligned}$$

then

$$\begin{aligned} \nabla ^{\bot }H(p_{0})=0, \quad (\nabla h)(p_{0})=0. \end{aligned}$$

(3.16)

Proof

Since $(|x|^{2})_{,j}=0,1\le j\le 2$ at $p_{0}$, it holds that $\langle x, e_{j}\rangle (p_{0})=0$, $1\le j\le 2$. So by (3.3) we have

$$\begin{aligned} H^{k^{*}}_{,i}=0, \ 1\le i,k \le 2,\quad |H-\xi |^{2}_{,i}=2\sum _{k}(H-\xi )^{k^{*}}H_{,i}^{k^{*}} =0, \quad 1\le i\le 2 \quad \text{ at } \ p_{0} \nonumber \\ \end{aligned}$$

(3.17)

where the first set of equalities are exactly $\nabla ^\bot H(p_0)=0$, which give

$$\begin{aligned} h_{11,1}^{1^{*}}+h_{22,1}^{1^{*}}=0,\quad h_{11,2}^{1^{*}}+h_{22,2}^{1^{*}}=0,\quad h_{11,1}^{2^{*}}+h_{22,1}^{2^{*}}=0,\quad h_{11,2}^{2^{*}}+h_{22,2}^{2^{*}}=0.\nonumber \\ \end{aligned}$$

(3.18)

On the other hand, from

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}=\mathrm{const}, \end{aligned}$$

(3.19)

we obtain

$$\begin{aligned} |h|^{2}_{,k}+|H-\xi |^{2}_{,k}\equiv 0,\quad 1\le k\le 2, \end{aligned}$$

(3.20)

which with (3.17) implies that

$$\begin{aligned} (|h|^{2})_{,k}=0,\quad 1\le k\le 2\quad \text {at } p_{0}. \end{aligned}$$

Since

$$\begin{aligned} |h|^{2}=(h_{11}^{1^{*}})^{2}+2(h_{12}^{1^{*}})^{2}+(h_{22}^{1^{*}})^{2} +(h_{11}^{2^{*}})^{2}+2(h_{12}^{2^{*}})^{2}+(h_{22}^{2^{*}})^{2}, \end{aligned}$$

we find that

$$\begin{aligned}&h_{11}^{1^{*}}h_{11,1}^{1^{*}}+2h_{12}^{1^{*}}h_{12,1}^{1^{*}}+h_{22}^{1^{*}}h_{22,1}^{1^{*}} +h_{11}^{2^{*}}h_{11,1}^{2^{*}}+2h_{12}^{2^{*}}h_{12,1}^{2^{*}} +h_{22}^{2^{*}}h_{22,1}^{2^{*}}=0, \end{aligned}$$

(3.21)

$$\begin{aligned}&h_{11}^{1^{*}}h_{11,2}^{1^{*}}+2h_{12}^{1^{*}}h_{12,2}^{1^{*}}+h_{22}^{1^{*}}h_{22,2}^{1^{*}} +h_{11}^{2^{*}}h_{11,2}^{2^{*}}+2h_{12}^{2^{*}}h_{12,2}^{2^{*}} +h_{22}^{2^{*}}h_{22,2}^{2^{*}}=0 \end{aligned}$$

(3.22)

hold at $p_{0}$. From (2.11) and (3.18) we get

$$\begin{aligned} h_{22,1}^{1^{*}}=-h_{11,1}^{1^{*}},\quad h_{22,2}^{1^{*}}=-h_{11,2}^{1^{*}},\quad h_{22,2}^{2^{*}}=h_{11,1}^{1^{*}}\quad \text {at }p_{0}. \end{aligned}$$

(3.23)

Since, by (2.2) and (2.11), both $h_{ij}^{k^{*}}$ and $h_{ij,l}^{k^{*}}$ are totally symmetric, we obtain by (3.23), (3.21) and (3.22) that

$$\begin{aligned}&(h_{11}^{1^{*}}-3h_{22}^{1^{*}})h_{11,1}^{1^{*}}- (h_{22}^{2^{*}}-3h_{11}^{2^{*}})h_{11,2}^{1^{*}}=0, \end{aligned}$$

(3.24)

$$\begin{aligned}&(h_{22}^{2^{*}}-3h_{11}^{2^{*}})h_{11,1}^{1^{*}}+ (h_{11}^{1^{*}}-3h_{22}^{1^{*}})h_{11,2}^{1^{*}}=0\quad \text {at }p_{0}. \end{aligned}$$

(3.25)

We claim that

$$\begin{aligned} (\nabla h)(p_{0})=0. \end{aligned}$$

(3.26)

Otherwise, we should have $(h_{11,1}^{1^{*}})^{2}+(h_{11,2}^{1^{*}})^{2}\ne 0$ at $p_0$. Then from (3.24) and (3.25) it follows that

$$\begin{aligned} (h_{11}^{1^{*}}-3h_{22}^{1^{*}})^{2}+(h_{22}^{2^{*}}-3h_{11}^{2^{*}})^{2}=0\quad \text {at }p_{0}. \end{aligned}$$

Thus

$$\begin{aligned} |h|^{2}(p_{0})=\frac{4}{3}((h_{11}^{1^{*}})^{2}+(h_{22}^{2^{*}})^{2}),\quad |H|^{2}(p_{0})=\frac{16}{9}((h_{11}^{1^{*}})^{2}+(h_{22}^{2^{*}})^{2}). \end{aligned}$$

(3.27)

Now by the definition of $p_0$ and Lemma 3.6,

$$\begin{aligned} 0\le \frac{1}{2}\triangle {|x|^{2}}(p_{0}) =2-\langle H, H-\xi \rangle (p_{0}). \end{aligned}$$

It follows that

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}= & {} (|h|^{2}+|H-\xi |^{2})(p_{0})\nonumber \\= & {} \frac{3}{4}|H|^{2}(p_{0})+2\langle H, H-\xi \rangle (p_{0}) -|H|^{2}(p_{0})+|\xi |^{2}\nonumber \\= & {} |\xi |^{2}+2\langle H, H-\xi \rangle (p_{0})-\frac{1}{4}|H|^{2}(p_{0}) \end{aligned}$$

(3.28)

$$\begin{aligned}\le & {} |\xi |^{2}+4. \end{aligned}$$

(3.29)

Therefore, by Proposition 3.7, $|h|^{2}+|H-\xi |^{2}=|\xi |^{2}+4$. But it is easy to see that the equality in (3.29) holds if and only if $|H|^{2}(p_{0})=0$ and $\langle H, H-\xi \rangle (p_{0})=2$, which is of course not possible! This contradiction proves the above claim and completes the proof of Lemma 3.8. $\square$

Remark 3.1

Our main observation here is that, if $p_{0}\in M^{2}$ is a minimum point of $|x|^{2}$ then

$$\begin{aligned} x^{\top }(p_{0})=\sum _{i}\langle x, e_{i}\rangle e_{i}(p_{0})=0, \end{aligned}$$

implying

$$\begin{aligned} \nabla ^{\bot }H(p_{0})=\nabla ^{\bot }(H-\xi )(p_{0})=0. \end{aligned}$$

In particular, $p_{0}$ is also a minimum point of $|x^{\top }|^{2}$.

Proposition 3.9

Let $x: M^{2}\rightarrow {\mathbb {C}}^{2}$ be a compact and oriented Lagrangian $\xi$-submanifold. Suppose that

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}=|\xi |^{2}+4 \end{aligned}$$

and $\langle H, \xi \rangle$ is constant. If one of the followings holds,

$$\begin{aligned} (1)\ |h|^{2}\ge 2,\quad (2)\ |H|^{2}\ge 2,\quad (3)\ |h|^{2}\ge \langle H, H-\xi \rangle ,\quad (4)\ \langle H, \xi \rangle \ge 0, \end{aligned}$$

(3.30)

then $|x|^{2}$ is a constant.

Proof

As above, let $p_0$ be a minimum point of $|x|^2$. Then, by Lemma 3.3 and Lemma 3.8, it holds at $p_{0}$ that

$$\begin{aligned} 0= & {} \frac{1}{2}{\mathcal {L}}(|h|^{2}+|H-\xi |^{2})\nonumber \\= & {} |h|^{2}-\frac{1}{2}(|h|^{2}-|H|^{2})(3|h|^{2}-2|H|^{2} +\langle H,H-\xi \rangle )+\langle H,H-\xi \rangle \nonumber \\&-\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{k^{*}}(H-\xi )^{l^{*}} -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}H^{k^{*}}(H-\xi )^{l^{*}}. \end{aligned}$$

(3.31)

Furthermore, form Lemma 3.4 and Lemma 3.5 it follows that, at $p_{0}$

$$\begin{aligned} 0\le & {} \frac{1}{2}\triangle (|x^{\top }|^{2})=2-2\langle H,H-\xi \rangle + \sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{k^{*}}(H-\xi )^{l^{*}}, \end{aligned}$$

(3.32)

$$\begin{aligned} 0= & {} \frac{1}{2}{\mathcal {L}}(\langle H,\xi \rangle ) =\frac{1}{2}(\langle H,\xi \rangle - \sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\xi ^{k^{*}}(H-\xi )^{l^{*}}), \end{aligned}$$

(3.33)

implying

$$\begin{aligned} -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{k^{*}}(H-\xi )^{l^{*}} \le 2-2\langle H,H-\xi \rangle , \end{aligned}$$

(3.34)

and

$$\begin{aligned}&-\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{l^{*}}H^{k^{*}}\\&\quad =-\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}(H-\xi )^{k^{*}}(H-\xi )^{l^{*}} -\sum _{i,j,k,l}h_{ij}^{k^{*}}h_{ij}^{l^{*}}\xi ^{k^{*}}(H-\xi )^{l^{*}} \\&\quad \le 2-2\langle H,H-\xi \rangle -\langle H,\xi \rangle =2-\langle H,H-\xi \rangle -|H|^{2}. \end{aligned}$$

Consequently, we have at $p_{0}$

$$\begin{aligned} 0= & {} \frac{1}{2}{\mathcal {L}}(|h|^{2}+|H-\xi |^{2})\\\le & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(3|h|^{2}-2|H|^{2} +\langle H,H-\xi \rangle )\\&+\,|h|^{2}-|H|^{2}+2(2-\langle H,H-\xi \rangle ). \end{aligned}$$

On the other hand, from

$$\begin{aligned} |h|^{2}+|H-\xi |^{2}=|\xi |^{2}+4, \end{aligned}$$

we know that

$$\begin{aligned} |h|^{2}-|H|^{2}=2(2-\langle H,H-\xi \rangle )\ge 0\quad \text {at } p_{0}. \end{aligned}$$

(3.35)

Thus, if one of (3.30) holds, then at $p_0$

$$\begin{aligned} 0= & {} \frac{1}{2}{\mathcal {L}}(|h|^{2}+|H-\xi |^{2}) \\\le & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(2|h|^{2}-|H|^{2}+4 -\langle H,H-\xi \rangle )+2(|h|^{2}-|H|^{2}) \\= & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(2|h|^{2} -|H|^{2}-\langle H,H-\xi \rangle ) \\= & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(|h|^{2}-|H|^{2}+|h|^{2} -\langle H,H-\xi \rangle ) \\= & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(|h|^{2}-|H|^{2}+|h|^{2} -2+2-\langle H,H-\xi \rangle ) \\= & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(2(|h|^{2}-|H|^{2}) +\langle H,\xi \rangle ) \\= & {} -\frac{1}{2}(|h|^{2}-|H|^{2})(2(|h|^{2}-|H|^{2})+|H|^{2} -2+2-\langle H,H-\xi \rangle ) \le 0. \end{aligned}$$

Consequently

$$\begin{aligned} |h|^{2}-|H|^{2}=2-\langle H,H-\xi \rangle =0\quad \text {at } p_{0}. \end{aligned}$$

(3.36)

It follows that

$$\begin{aligned} |x|^{2}+\langle H,\xi \rangle\ge & {} |x|^{2}(p_{0})+\langle H,\xi \rangle (p_{0})\\= & {} |H-\xi |^{2}(p_{0})+\langle H,\xi \rangle (p_{0})\\= & {} \langle H,H-\xi \rangle (p_{0})+|\xi |^{2}\\= & {} |\xi |^{2}+2. \end{aligned}$$

This together with Lemma 3.1 (for $u=1$, $v=|x|^2$) and 3.6 gives that

$$\begin{aligned} 0=\int _{M}\frac{1}{2}{\mathcal {L}}(|x|^{2})e^{-\frac{|x|^{2}}{2}}dV_{M} =\int _{M}(|\xi |^{2}+2-(|x|^{2}+\langle H,\xi \rangle ) )e^{-\frac{|x|^{2}}{2}}dV_{M}\le 0 \end{aligned}$$

implying that $|x|^{2}+\langle H,\xi \rangle =|\xi |^{2}+2$. In particular, $|x|^{2}=\mathrm{const}$. $\square$

Proposition 3.10

Let $x: M^{n}\rightarrow N^{n}$ be a Lagrangian submanifold in a $K\ddot{a}hler$ manifold $N^{n}$. If both $M^{n}$ and $N^{n}$ are flat, then around each point $p\in M^{n}$, there exists some orthonormal frame field $\{e_{i},e_{i^*}\}$ with $e_{i^{*}}=Je_{i}$ ($1\le i\le n$), such that

$$\begin{aligned} h_{ij}^{k^{*}}:=\langle h(e_{i},e_{j}),e_{k^{*}}\rangle =\lambda _{i}^{k^{*}}\delta _{ij}, \quad 1\le i,j,k\le n. \end{aligned}$$

Proof

For $p\in M^{n}$, we pick an orthonormal tangent frame $\{\bar{e}_{i}\}$ and an orthonormal normal frame $\{\bar{e}_{\alpha }\}_{n+1\le \alpha \le 2n}$. Define

$$\begin{aligned} \bar{h}_{ij}^{\alpha }=\langle h(\bar{e}_{i},\bar{e}_{j}),\bar{e}_{\alpha }\rangle . \end{aligned}$$

Since $M^n$ is flat, x is Lagrangian and $N^n$ is Kähler, $T^{\bot }M^{n}$ is also flat with respect to the normal connection. By the Ricci equation and the flatness of $N^n$,

$$\begin{aligned} 0=\langle R^{^{\bot }}(e_{i},e_{j})e_{\alpha },e_{\beta }\rangle =\sum _k\left( h_{ik}^{\beta }h_{jk}^{\alpha }-h_{ik}^{\alpha }h_{jk}^{\beta }\right) . \end{aligned}$$

Hence we can choose another orthonormal tangent frame $\{e_{i}\}$ such that

$$\begin{aligned} h_{ij}^{\alpha }:=\langle h(e_{i},e_{j}),e_{\alpha }\rangle =\mu _{i}^{\alpha }\delta _{ij}. \end{aligned}$$

Write $e_{k^{*}}=\sum _{\alpha }a_{k^{*}}^{\alpha }e_{\alpha }$. Then

$$\begin{aligned} h_{ij}^{k^{*}}= & {} \langle h(e_{i},e_{j}),e_{k^{*}}\rangle =\big \langle h(e_{i},e_{j}),\sum _{\alpha }a_{k^{*}}^{\alpha }e_{\alpha }\big \rangle =\sum _{\alpha }a_{k^{*}}^{\alpha }\langle h(e_{i},e_{j}),e_{\alpha }\rangle \nonumber \\= & {} \sum _{\alpha }a_{k^{*}}^{\alpha }h_{ij}^{\alpha } =\sum _{\alpha }a_{k^{*}}^{\alpha }\mu _{i}^{\alpha }\delta _{ij} =\lambda _{i}^{k^{*}}\delta _{ij}\ \text{ with } \lambda _{i}^{k^{*}}:=\sum _{\alpha }a_{k^{*}}^{\alpha }\mu _{i}^{\alpha }. \end{aligned}$$

(3.37)

Thus Proposition 3.10 is proved. $\square$

Proof of Theorem

Since $|x|^2=\mathrm{const}$, $|H|^{2}-|h|^{2}\le 0$ on $M^2$ by (3.35). Then it follows from (3.15) that $|H|^{2}-|h|^{2}\equiv 0$ which with the Gauss equation shows that $M^{2}$ is flat. Therefore, due to Proposition 3.10, we can choose $\{e_{1},e_{2}\}$ such that

$$\begin{aligned} h_{12}^{1^{*}}=h_{12}^{2^{*}}=0. \end{aligned}$$

(3.38)

It follows that

$$\begin{aligned} h_{22}^{1^{*}}=h_{11}^{2^{*}}=0. \end{aligned}$$

(3.39)

On the other hand, since $\nabla h\equiv 0$, we have

$$\begin{aligned} 0=\sum h _{ijl}^{k^{*}}\theta _{l}=dh_{ij}^{k^{*}}-\sum h_{lj}^{k^{*}}\theta _{il} -\sum h_{il}^{k^{*}}\theta _{jl}+\sum h_{ij}^{p^{*}}\theta _{p^{*}k^{*}}. \end{aligned}$$

It follows that

• (1) $i=j=1$, $k=2$, we get $h_{11}^{1^{*}}\theta _{1^{*}2^{*}}=0$,

• (2) $i=j=2$, $k=1$, we get $h_{22}^{2^{*}}\theta _{2^{*}1^{*}}=0$,

• (3) $i=j=k=1$, $0=dh_{11}^{1^{*}}+h_{11}^{1^{*}}\theta _{1^{*}1^{*}}=dh_{11}^{1^{*}}$, we get $h_{11}^{1^{*}}=\mathrm{const}$,

• (4) $i=j=k=2$, $0=dh_{22}^{2^{*}}+h_{22}^{2^{*}}\theta _{2^{*}2^{*}}=dh_{22}^{2^{*}}$, we get $h_{22}^{2^{*}}=\mathrm{const}$.

Since x can not be totally geodesic, $(h_{11}^{1^{*}})^{2}+(h_{22}^{2^{*}})^{2}=|h|^2\ne 0$ by (3.38) and (3.39). Without loss of generality we can assume that $h_{11}^{1^{*}}\ne 0$. Thus, by (1), we have $\theta _{1^{*}2^{*}}=0$ which with $\nabla J=0$ shows that $\theta _{12}=0$. Now let

$$\begin{aligned} \tilde{e}_{1}=e_{1}\cos \theta -e_{2}\sin \theta , \quad \tilde{e}_{2}=e_{1}\sin \theta +e_{2}\cos \theta \end{aligned}$$

be another frame field such that

$$\begin{aligned} \tilde{h}_{ij}^{k^{*}}:=\langle h(\tilde{e}_{i},\tilde{e}_{j}),\tilde{e}_{k^{*}}\rangle =\tilde{\lambda }_{i}^{k^{*}}\delta _{ij}. \end{aligned}$$

Then a direct computation shows that

$$\begin{aligned} \sin \theta \cos \theta (h_{11}^{1^{*}}\cos \theta +h_{22}^{2^{*}}\sin \theta ) =\sin \theta \cos \theta (h_{11}^{1^{*}}\sin \theta -h_{22}^{2^{*}}\cos \theta )=0. \end{aligned}$$

Since $(h_{11}^{1^{*}})^{2}+(h_{22}^{2^{*}})^{2}\ne 0$, we have $\sin 2\theta =0$, that is $\theta =0,$ or $\frac{\pi }{2}$ or $\pi$. Clearly, by choosing $\theta =\frac{\pi }{2}$, we can change the sign of $h_{11}^{1^{*}}h_{22}^{2^{*}}$; while by choosing $\theta =\pi$, we can change the sign of both $h_{11}^{1^{*}}$ and $h_{22}^{2^{*}}$. Thus we can always assume that $h_{11}^{1^{*}}>0$ and $h_{22}^{2^{*}}\ge 0$. It then follows that $\{e_{1},e_{2}\}$ can be uniquely determined and, in particular, is globally defined.

Now we claim that $h_{22}^{2^{*}}>0$. In fact, if $h_{22}^{2^{*}}=0$, then $\theta _{22^{*}}=0$. This with $\theta _{12}=\theta _{21^{*}}=0$ shows that $e_2$ is constant in ${\mathbb C}^2$ along M which means that M contains a family of parallel straight lines, contradicting the assumption that M is compact.

Define

$$\begin{aligned} V_{1}=\mathrm{Span\,}_{\mathbb R}\{e_{1},e_{1^*}\}=\mathrm{Span\,}_{{\mathbb {C}}}\{e_{1}\}, \quad V_{2}=\mathrm{Span\,}_{\mathbb R}\{e_2,e_{2^*}\}=\mathrm{Span\,}_{{\mathbb {C}}}\{e_{2}\}. \end{aligned}$$

(3.40)

Since

$$\begin{aligned} de_{1}= & {} \nabla e_{1}+\sum h_{1j}^{k^{*}}\theta _{j}e_{k^{*}}=h_{11}^{1^{*}}\theta _{1}e_{1^{*}}\in V_{1},\\ de_{1^{*}}= & {} Jde_{1}=h_{11}^{1^{*}}\theta _{1}Je_{1^{*}}=-h_{11}^{1^{*}}\theta _{1}e_{1}\in V_{1}, \end{aligned}$$

we know that $V_{1}$ is a 1-dimensional constant complex subspace of ${\mathbb {C}}^{2}$.

Similarly, $V_{2}$ is also a 1-dimensional constant complex subspace of ${\mathbb {C}}^{2}$. Furthermore, $V_1$ and $V_2$ are clearly orthogonal. So, up to a holomorphic isometry on ${\mathbb {C}}^{2}$, we can assume that $V_{1}={\mathbb {C}}^{1}$, $V_{2}={\mathbb {C}}^{1}$ so that ${\mathbb {C}}^{2}=V_{1}\times V_{2}$. Write

$$\begin{aligned} x=(x^1,x^2)\in V_1\times V_2\equiv {\mathbb C}^2. \end{aligned}$$

Then

$$\begin{aligned} 0=e_{i}(|x|^{2})=e_{i}(|x^1|^{2})+e_{i}(|x^2|^{2}),\quad i=1,2, \end{aligned}$$

which with the definitions (3.40) of $V_1$ and $V_2$ shows that

$$\begin{aligned} e_{1}(|x^{1}|^{2})=e_{2}(|x^{1}|^{2})=0,\quad e_{1}(|x^{2}|^{2})=e_{2}(|x^{2}|^{2})=0, \end{aligned}$$

that is,

$$\begin{aligned} |x^{1}|^{2}=\mathrm{const},\quad |x^{2}|^{2}=const. \end{aligned}$$

It is easily seen that both $|x^{1}|^{2}$ and $|x^{2}|^{2}$ are positive since x is non-degenerate. Thus we can write $|x^{1}|^{2}=a^{2}>0$, $|x^{2}|^{2}=b^{2}>0$. It then follows that $M^{2}={\mathbb {S}}^{1}(a)\times {\mathbb {S}}^{1}(b)$.

Finally, by the assumption (1.4), $|h|^{2}\ge 2$, it should holds that $a^{2}+b^{2}\ge 2a^{2}b^{2}$. $\square$