Rigidity theorem for Willmore surfaces in a sphere

Abstract

Let M ² be a compact Willmore surface in the (2 + p)-dimensional unit sphere S ^{2 + p}. Denote by H and S the mean curvature and the squared length of the second fundamental form of M ², respectively. Set ρ ² = S−2H ². In this note, we proved that there exists a universal positive constant C, such that if ∥ρ ²∥₂<C, then ρ ²=0 and M ² is a totally umbilical sphere.

1 Introduction

Let M be a compact surface in the (2 + p)-dimensional unit sphere S ^{2 + p} . Choose a local orthonormal frame field {e ₁,e ₂,…,e _{2 + p} } in S ^{2 + p} such that, restricted to M, the {e ₁,e ₂} are tangent to M. The following convention of indices are used throughout.

$$1\leq i,j,k\leq 2; \qquad 3\leq \alpha,\beta,\gamma\leq 2+p. $$

Denote by H and S the mean curvature and the squared length of the second fundamental form of M, respectively. Then, we have

$$S=\sum\limits_{\alpha,i,j}(h_{ij}^{\alpha})^{2}, \quad \mathbf{ H}=\sum\limits_{\alpha}H^{\alpha} e_{\alpha}, \quad H^{\alpha}=\frac{1}{2}\sum\limits_{k} h^{\alpha}_{kk}, \quad H=|\mathbf{H}|, $$

where $h_{ij}^{\alpha }$ is the component of the second fundamental tensor of M.

Let ρ ² = S−2H ². In fact, if we set $\tilde {h}^{\alpha }_{ij}=h^{\alpha }_{ij}-\delta _{ij}H^{\alpha }$, by a direct computation, one has

$$\rho^{2}=\sum\limits_{\alpha,i,j}(\tilde{h}^{\alpha}_{ij})^{2}. $$

So, ρ ²≥0, and ρ vanishes at the umbilical points of M.

The Willmore functional is defined by

$$W(x)={\int}_{M}\rho^{2}\mathrm{d}v={\int}_{M}(S-2H^{2})\mathrm{d}v. $$

Here the integration is with respect to the area measure of M. In [3], Chen proved that this functional is invariant under conformal transformations of S ^{2 + p}.

DEFINITION

x:M→S ^{2 + p} is called a Willmore surface if it is a critical surface of the Willmore functional W(x).

It was proved by Bryant [1] and Weiner [7] that M is a Willmore surface if and only if

$$ {\Delta}^{\perp} H^{\alpha}+\sum\limits_{\beta,i,j}h^{\alpha}_{ij}h^{\beta}_{ij}H^{\beta}-2H^{2}H^{\alpha}=0, $$

(1.1)

i.e.,

$${\Delta}^{\perp} H^{\alpha}+\sum\limits_{\beta,i,j}\tilde{h}^{\alpha}_{ij}\tilde{h}^{\beta}_{ij}H^{\beta}=0, $$

where ${\Delta }^{\perp } H^{\alpha }=\sum \limits _{k}H^{\alpha }_{kk}$.

From (1.1), we know that all minimal surfaces in S ^{2 + p} are Willmore surfaces. So, the Veronese surface must be the Willmore surface. Moreover, Pinkall [4] constructed many compact non-minimal flat Willmore surfaces in S ³, and Castro and Urbano [2] constructed many compact non-minimal Willmore surfaces in S ⁴.

In [6], Li obtained the following rigidity theorem for Willmore surfaces in a unit sphere.

Theorem A.

Let M be a compact Willmore surface in S ^2+p . Then

$${\int}_{M}\rho^{2}\left( 2-\frac{2}{B}\rho^{2}\right)\mathrm{d}v\leq0, $$

where

$$B=\left\{ \begin{array}{ll} 2, &~{p}=1~,\\ \frac{4}{3}, &{p}\geq2~. \end{array} \right. $$

In particular, if

$$\rho^{2}\leq B, $$

then either ρ ² = 0 and M is totally umbilical, or ρ ² =B. In the latter case, p=2 and M is the Veronese surface or p=1 and $M=S^{1}\left (\frac {1}{\sqrt {2}}\right )\times S^{1}\left (\frac {1}{\sqrt {2}}\right )$.

Applying Theorem A and the Sobolev inequality, we proved the following result (see [9]).

Theorem B.

Let M be a compact Willmore surface in S ^2+p . There exists a positive constant $\tilde {C}(H_{0})$ , defined by

$$\tilde{C}(H_{0})=\frac{B\left( \sqrt{9+{H_{0}^{2}}}-\sqrt{1+{H_{0}^{2}}}\right)\sqrt{\pi}}{48\sqrt{3}}, $$

such that if

$$\|\rho^{2}\|_{2}<\tilde{C}(H_{0}), $$

then M is a totally umbilical surface, where H ₀ = maxx∈MH and B is defined in Theorem A.

We shall improve the constant of Theorem B and obtain the following global pinching theorem for compact Willmore surfaces in S ^{2 + p}.Main theorem. Let M be a compact Willmore surface in the unit sphere S ^{2 + p}. There exists an explicit positive constant

$$C=\frac{(\sqrt{2}-1)\sqrt{\pi}}{12\sqrt{3}}B, $$

such that if

$$\|\rho^{2}\|_{2}< C, $$

then ρ ²=0and M is a totally umbilical sphere, where B is defined in Theorem A.

Remark 1.

By a simple calculation, we know that the pinching constant in Theorem B $\tilde {C}(H_{0})\rightarrow 0$ as H→∞. But the pinching constant C in our main theorem is independent of mean curvature H. So C is superior to $\tilde {C}(H_{0})$.

2 Basic lemmas

In this section, we introduce several useful lemmas

Lemma 2.1.

Let x:M ² →S ^2+p be a surface in a unit sphere. We have the following inequality:

$$ |\nabla\rho|^{2}\leq{\sum}_{\alpha,i,j,k}(\tilde{h}_{ijk}^{\alpha})^{2}. $$

(2.1)

Proof.

We can see from $\rho ^{2}={\sum }_{\alpha ,i,j}(\tilde {h}_{ij}^{\alpha })^{2}$ and the Cauchy–Schwarz inequality that

$$ |\nabla\rho|^{2}\leq{\sum}_{\alpha,i,j,k}(\tilde{h}_{ijk}^{\alpha})^{2} $$

(2.2)

at all points where ρ ≠ 0 and hence by analyticity at all the points. □

Lemma 2.2.

[10]. Let x:M→S ^2+p be a surface. Then

$$ {\Sigma} R^{2}_{\alpha\beta12}\leq\frac{2-B}{B}\rho^{4}, $$

(2.3)

where equality holds if and only if p=1 or p≥2, ${\sum }_{\alpha }(\tilde {h}_{11}^{\alpha })^{2}={\sum }_{\alpha }(\tilde {h}_{12}^{\alpha })^{2}$ and ${\sum }_{\alpha }\tilde {h}_{11}^{\alpha }\tilde {h}_{12}^{\alpha }=0$ . HereBis defined in Theorem A.

Lemma 2.3.

Let x:M→S ^2+p be a surface.

$$ |\nabla\tilde{h}|^{2}\geq|\nabla^{\perp}\mathbf{H}|^{2}, $$

(2.4)

where $|\nabla ^{\perp }\mathbf {H}|^{2}={\sum }_{\alpha ,i}(H^{\alpha }_{i})^{2}$.

Proof.

By a simple calculation, we have

$$ |\nabla\tilde{h}|^{2}={\sum}_{\alpha,i,j,k}(\tilde{h}_{ijk}^{\alpha})^{2}={\sum}_{\alpha,i,j,k}(h^{\alpha}_{ijk})^{2}- 2|\nabla^{\perp}\mathbf{H}|^{2}. $$

(2.5)

In [6], Li proved

$$ {\sum}_{\alpha,i,j,k}(h^{\alpha}_{ijk})^{2}\geq3|\nabla^{\perp}\mathbf{H}|^{2}. $$

(2.6)

Substituting (2.6 ) into ( 2.5 ), we obtain ( 2.4). □

Lemma 2.4.

[10]. Let x:M ² →S ^2+p be a compact Willmore surface in a unit sphere. Then

$$ {\int}_{M}|\nabla^{\perp}\mathbf{H}|^{2}\mathrm{d}v\leq{\int}_{M}\rho^{2}H^{2}\mathrm{d}v. $$

(2.7)

Lemma 2.5.

Let x:M→S ^2+p be a compact surface.

$$ {\int}_{M}{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ij}H^{\alpha}_{ij}\mathrm{d}v=-{\int}_{M}|\nabla^{\perp}\mathbf{H}|^{2}\mathrm{d}v. $$

(2.8)

Proof.

By Stoke formula, we have

$$ {\int}_{M}{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ij}H^{\alpha}_{ij}\mathrm{d}v= -{\int}_{M}{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ijj}H^{\alpha}_{i}\mathrm{d}v. $$

(2.9)

$$ \begin{array}{lcl} \displaystyle{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ijj}H^{\alpha}_{i}&=& \displaystyle{\sum}_{\alpha,i,j}(h^{\alpha}_{ijj}-\delta_{ij}H^{\alpha}_{j})H^{\alpha}_{i}\\ &=&\displaystyle{\sum}_{\alpha,i,j}h^{\alpha}_{jji}H^{\alpha}_{i}-{\sum}_{\alpha,i}H^{\alpha}_{i}H^{\alpha}_{i}\\ &=&|\nabla^{\perp}\mathbf{H}|^{2}. \end{array} $$

(2.10)

We obtain (2.8 ) by putting ( 2.10 ) into ( 2.9). □

Lemma 2.6.

Let M be a compact 2-dimensional surface in S ^2+p . Then for any g∈C ¹ (M), g≥0, t>0, g satisfies

$$ \displaystyle {\int}_{M}|\nabla g|^{2}\mathrm{d}v\geq\frac{t}{A}\left( {\int}_{M}g^{4}\mathrm{d}v\right)^{\frac{1}{2}}-t^{2}{\int}_{M}g^{2}\mathrm{d}v-t{\int}_{M}\left( 1+\frac{H^{2}}{2}\right)g^{2}\mathrm{d}v, $$

(2.11)

where $A=\frac {12\sqrt {3}}{\sqrt {\pi }}$.

Proof.

From [5] and [8], we have

$$\left( {\int}_{M}g^{2}\mathrm{d}v\right)^{\frac{1}{2}}\leq A{\int}_{M}(|\nabla g|+\sqrt{1+H^{2}}g)\mathrm{d}v. $$

Replacing g by g ², we get

$$\begin{array}{@{}rcl@{}} \left( {\int}_{M}g^{4}\mathrm{d}v\right)^{\frac{1}{2}}&\leq&A{\int}_{M}(|\nabla g^{2}|+\sqrt{1+H^{2}}g^{2})\mathrm{d}v\\ &=&A{\int}_{M}(g|\nabla g|+\sqrt{1+H^{2}}g^{2})\mathrm{d}v\\ &\leq&A\left( {\kern-2.5pt}{\int}_{M}f^{2}\mathrm{d}v{\kern-2.5pt}\right)^{\frac{1}{2}}\left( {\int}_{M}|\nabla g|^{2}\mathrm{d}v{\kern-2.5pt}\right)^{\frac{1}{2}}+A{\int}_{M}\left( {\kern-2.5pt}1+\frac{H^{2}}{2}{}\right)g^{2}\mathrm{d}v\\ &\leq&At{\int}_{M}g^{2}\mathrm{d}v+\frac{A}{t}{\int}_{M}|\nabla g|^{2}\mathrm{d}v+A{\int}_{M}\left( 1+\frac{H^{2}}{2}\right)g^{2}\mathrm{d}v, \end{array} $$

where t∈R ⁺. So, we have

$${\int}_{M}|\nabla g|^{2}\mathrm{d}v\geq\frac{t}{A}\left( {\int}_{M}g^{4}\mathrm{d}v\right)^{\frac{1}{2}}-t^{2}{\int}_{M}g^{2}\mathrm{d}v-t{\int}_{M}\left( 1+\frac{H^{2}}{2}\right)g^{2}\mathrm{d}v, $$

i.e.,

$$\|\nabla g\|^{2}_{2}\geq\frac{t}{A}\|g^{2}\|_{2}-(t^{2}+t)\|g^{2}\|_{1}-\frac{t}{2}\|H^{2}g^{2}\|_{1}. $$

This proves Lemma 2.6. □

3 Proof of the main theorem

In this section, we give the proof of our main theorem. From Lemma 2.1 in [10] and (2.3), we have

$$\begin{array}{@{}rcl@{}} \frac{1}{2}{\Delta}\rho^{2}&=&{\sum}_{\alpha,i,j,k}(\tilde{h}^{\alpha}_{ijk})^{2}+ 2{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ij}H^{\alpha}_{ij} +\rho^{2}(2-\rho^{2}+2H^{2})-{\sum}_{\alpha,\beta}R^{2}_{\alpha\beta12}\\ &\geq&{\sum}_{\alpha,i,j,k}(\tilde{h}^{\alpha}_{ijk})^{2}+2{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ij}H^{\alpha}_{ij} +\rho^{2}(2-\rho^{2}+2H^{2})-\frac{2-B}{B}\rho^{4}. \end{array} $$

Integrating the above inequality and using Lemma 2.5, we get

$$ \begin{array}{lcl} 0&\geq&\displaystyle{\int}_{M}{\sum}_{\alpha,i,j,k}{\kern-2.5pt}(\tilde{h}^{\alpha}_{ijk})^{2}\mathrm{d}v{\kern-1.5pt}+{\kern-1.5pt} 2{\kern-1.5pt}{\int}_{M}{\sum}_{\alpha,i,j}\tilde{h}^{\alpha}_{ij}H^{\alpha}_{ij}\mathrm{d}v {\kern-1.5pt}+{\kern-2.5pt}{\int}_{M}\rho^{2}{\kern-1.5pt}\left[{\kern-1.5pt}2(1{\kern-1.5pt}+{\kern-1.5pt}H^{2}){\kern-1.5pt}-{\kern-1.5pt}\frac{2}{B}\rho^{2}{\kern-1.5pt}\right]{\kern-1.5pt}\mathrm{d}v\\ &=&\displaystyle{\int}_{M}|\nabla\tilde{h}|^{2}\mathrm{d}v-2{\int}_{M}|\nabla^{\perp}\mathbf{ H}|^{2}\mathrm{d}v +{\int}_{M}\rho^{2}\left[2(1+H^{2})-\frac{2}{B}\rho^{2}\right]\mathrm{d}v\\ &=&\displaystyle\eta{\int}_{M}|\nabla\tilde{h}|^{2}\mathrm{d}v+(1-\eta){\int}_{M}|\nabla\tilde{h}|^{2}\mathrm{d}v-2{\int}_{M}|\nabla^{\perp}\mathbf{ H}|^{2}\mathrm{d}v\\ &&\displaystyle+{\int}_{M}\rho^{2}\left[2(1+H^{2})-\frac{2}{B}\rho^{2}\right]\mathrm{d}v, \end{array} $$

(3.1)

where 0<η<1. From (2.1 ), ( 2.7 ) and ( 3.1), we have

$$ \begin{array}{lcl} 0&\geq&\displaystyle\eta{\int}_{M}|\nabla f_{\varepsilon}|^{2}\mathrm{d}v+(1-\eta){\int}_{M}|\nabla^{\perp}\mathbf{ H}|^{2}\mathrm{d}v-2{\int}_{M}|\nabla^{\perp}\mathbf{H}|^{2}\mathrm{d}v\\ &&\displaystyle+{\int}_{M}\rho^{2}\left[2(1+H^{2})-\frac{2}{B}\rho^{2}\right]\mathrm{d}v\\ &=&\displaystyle\eta{\int}_{M}|\nabla f_{\varepsilon}|^{2}\mathrm{d}v{\kern-1.5pt}-{\kern-1.5pt}(1{\kern-1.5pt}+{\kern-1.5pt}\eta){\kern-1.5pt}{\int}_{M}|\nabla^{\perp}\mathbf{ H}|^{2}\mathrm{d}v {\kern-1.5pt}+{\kern-2.5pt}{\int}_{M}\rho^{2}\left[{\kern-1.5pt}2(1{\kern-1.5pt}+{\kern-1.5pt}H^{2}){\kern-1.5pt}-{\kern-1.5pt}\frac{2}{B}\rho^{2}{\kern-1.5pt}\right]\mathrm{d}v\\ &\geq&\displaystyle\eta{\int}_{M}|\nabla f_{\varepsilon}|^{2}\mathrm{d}v{\kern-1.5pt}-{\kern-1.5pt}(1{\kern-1.5pt}+{\kern-1.5pt}\eta){\int}_{M}\rho^{2}H^{2}\mathrm{d}v {\kern-1.5pt}+{\kern-1.5pt}{\int}_{M}\rho^{2}\left[{\kern-1.5pt}2(1+H^{2})-\frac{2}{B}\rho^{2}{\kern-1.5pt}\right]\mathrm{d}v\\ &=&\displaystyle\eta{\int}_{M}|\nabla f_{\varepsilon}|^{2}\mathrm{d}v+(1-\eta){\int}_{M}\rho^{2}H^{2}\mathrm{d}v +2{\int}_{M}\rho^{2}\mathrm{d}v-{\int}_{M}\frac{2}{B}\rho^{4}\mathrm{d}v. \end{array} $$

(3.2)

Substituting (2.11 ) into ( 3.2), we get

$$\begin{array}{@{}rcl@{}} 0&\geq&\eta\frac{t}{A}\left( {\int}_{M}f_{\varepsilon}^{4}\right)^{\frac{1}{2}}\mathrm{d}v-\eta t^{2}{\int}_{M}f_{\varepsilon}^{2} \mathrm{d}v-\eta t{\int}_{M}\left( 1+\frac{H^{2}}{2}\right)f_{\varepsilon}^{2}\mathrm{d}v\\ &&+ (1-\eta){\int}_{M}\rho^{2}H^{2}\mathrm{d}v+2{\int}_{M}\rho^{2}\mathrm{d}v-\frac{2}{B}{\int}_{M}\rho^{4}\mathrm{d}v. \end{array} $$

As ε→0, this implies

$$\begin{array}{@{}rcl@{}} 0&\geq&\eta\frac{t}{A}\left( {\int}_{M}\rho^{4}\mathrm{d}v\right)^{\frac{1}{2}}-(\eta t^{2}+\eta t-2){\int}_{M}\rho^{2}\mathrm{d}v\\ &&+\left( 1-\eta-\frac{\eta t}{2}\right){\int}_{M}\rho^{2}H^{2}\mathrm{d}v-\frac{2}{B}{\int}_{M}\rho^{4}\mathrm{d}v. \end{array} $$

Choose $t=\displaystyle \frac {2(1-\eta )}{\eta }$, then $1-\eta -\displaystyle \frac {\eta t}{2}=0$. So we have

$$\begin{array}{@{}rcl@{}} 0&\geq&\frac{2(1-\eta)}{A}\left( {\int}_{M}\rho^{4}\mathrm{d}v\right)^{\frac{1}{2}} -\left[\frac{4(1-\eta)^{2}}{\eta}+2(1-\eta)-2\right]{\int}_{M}\rho^{2}\mathrm{d}v\\&&-\frac{2}{B}{\int}_{M}\rho^{4}\mathrm{d}v, \end{array} $$

i.e.,

$$ \begin{array}{lcl} 0&\geq&\displaystyle\frac{(1-\eta)}{A}\left( {\int}_{M}\rho^{4}\mathrm{d}v\right)^{\frac{1}{2}} -\left[\frac{2(1-\eta)^{2}}{\eta}-\eta\right]{\int}_{M}\rho^{2}\mathrm{d}v-\frac{1}{B}{\int}_{M}\rho^{4}\mathrm{d}v\\ &=&\displaystyle\frac{1-\eta}{A}\|\rho^{2}\|_{2} -\left[\frac{2(1-\eta)^{2}}{\eta}-\eta\right]\|\rho^{2}\|-\frac{1}{B}\|\rho^{2}\|_{2}^{2}\\ &=&\displaystyle\left[\frac{1-\eta}{A}-\frac{1}{B}\|\rho^{2}\|_{2}\right]\|\rho^{2}\|_{2} -\left[\frac{2(1-\eta)^{2}}{\eta}-\eta\right]\|\rho^{2}\|. \end{array} $$

(3.3)

We take $\eta =2-\sqrt {2}$. This together with (3.3) yields

$$0\geq\left\{\frac{\sqrt{2}-1}{A}-\frac{1}{B}\|\rho^{2}\|_{2}\right\}\|\rho^{2}\|_{2}, $$

which implies ∥ρ ²∥₂=0 for

$$\|\rho^{2}\|_{2}<C=\frac{(\sqrt{2}-1)B}{A}=\frac{(\sqrt{2}-1)\sqrt{\pi}}{12\sqrt{3}}B, $$

i.e., S=2H ² and M is a totally umbilical Willmore surface. This completes the proof of the main theorem.

As we all know, minimal surfaces must be Willmore surfaces, so we obtain the following corollary.

COROLLARY

Let M be a compact minimal surface in the unit sphere S ^{2 + p}. There exists a positive constant

$$C=\frac{(\sqrt{2}-1)\sqrt{\pi}}{12\sqrt{3}}B, $$

such that if

$$\|S\|_{2}< C, $$

then S = 0 and M is a totally geodesic, where B is defined in Theorem A.