Energy Estimates for the Tracefree Curvature of Willmore Surfaces and Applications

Abstract

We prove an $\varepsilon$-regularity result for the tracefree curvature of a Willmore surface with bounded second fundamental form. For such a surface, we obtain a pointwise control of the tracefree second fundamental form from a small control of its $L^2$-norm. Several applications are investigated. Notably, we derive a gap statement for surfaces of the aforementioned type. We further apply our results to deduce regularity results for conformal minimal spacelike immersions into the de Sitter space ${\mathbb {S}}^{4,1}$.

Appendix

1.1 Variational Bubbles are Conformally Minimal

In Theorem H of [26], the authors obtain as a byproduct that any branched sphere that appears as a bubble must be conformally minimal. The Proof of Theorem H in [26] is quite involved due to the general assumptions used by the authors. For the sake of completeness of the present paper, we give an elementary argument to obtain the same result.

Lemma 3.1

Let ${\Phi }_k :{\mathbb {D}}\rightarrow {\mathbb {R}}^3$ be a sequence of conformal Willmore immersions, $a_k\in {\mathbb {D}}$ a sequence of points converging to some $a_\infty$, $\mu _k$ converging to 0, and $T_k$ a sequence of conformal transformations of ${\mathbb {R}}^3$. Suppose that

$$\begin{aligned} \tilde{{\Phi }}_k:=T_k\circ {\Phi }_k(a_k+\mu _k \, \cdot ) \rightarrow \omega \hbox { on } {\mathbb {R}}^2\setminus {S}, \end{aligned}$$

where S is the finite set and $\omega$ a branched Willmore sphere. Then $\omega$ is conformally minimal.

Proof

Following Bryant’s work, in order to prove that a sphere is conformally minimal, it suffices to prove that its associated quartic form $Q_\omega$ vanishes (see Theorem E in [7]). Let $Q_{\tilde{{\Phi }}_k}$ be the quartic form associated with ${\Phi }_k$. Per Theorem B in [7], $Q_{\tilde{{\Phi }}_k}$ is holomorphic. Let $p\in R^2$ be a branch point. Our strong convergence hypothesis away from branch points guarantees that the quartic form $Q_\omega$ is holomorphic around p, because $Q_{\tilde{{\Phi }}_k}$ is bounded.  owing to the maximum principle. Hence $Q_\omega$ may have at most one pole at infinity. Letting $\tilde{Q}= Q_{\omega }\left( \frac{1}{z} \right)$, by Theorem 3.1 of [16], the order of the pole of $\tilde{Q}$ at 0 is at most 2. One easily checks that $\tilde{Q}(z)=O\left( \frac{1}{\vert z\vert ^8}\right)$, so that $Q_\omega \equiv 0$ by Liouville’s Theorem, thereby concluding the proof. $\square$

1.2 Convenient Reformulation of the Gauss–Codazzi Equation

Recall that $\nabla \vec {n}= -e^{-2\lambda } A \nabla \Phi$, that is,

$$\begin{aligned} -\vec {n}_x= & {} \left( H + \left( \frac{l-n}{2} \right) e^{-2\lambda } \right) \Phi _x + m e^{-2\lambda } \Phi _y \\{} & {} \text {and}\\ -\vec {n}_y= & {} m e^{-2\lambda } \Phi _x + \left( H - \left( \frac{l-n}{2} \right) e^{-2\lambda } \right) \Phi _y. \end{aligned}$$

Differentiating yields

$$\begin{aligned} - \vec {n}_{xy}= & {} \left( H_y+ \left( \frac{l-n}{2} \right) _y e^{-2\lambda } + \lambda _y H - \lambda _y \left( \frac{l-n}{2} \right) e^{-2\lambda } - \lambda _x m e^{-2\lambda } \right) \Phi _x \\{} & {} + \left( m_y e^{-2\lambda } - \lambda _y m e^{-2 \lambda } + \lambda _x \left( H + \left( \frac{l-n}{2} \right) \right) \right) \Phi _y + \left( \dots \right) \vec {n}\\{} & {} \text {and}\\ -\vec {n}_{yx}= & {} \left( m_x e^{-2\lambda } - \lambda _x m e^{-2\lambda } + \lambda _y \left( H - \left( \frac{l-n}{2} \right) e^{-2\lambda } \right) \right) \Phi _x \\{} & {} + \left( H_x - \left( \frac{l-n}{2} \right) _x e^{-2\lambda } + \lambda _x H + \lambda _x \left( \frac{l-n}{2} \right) - \lambda _y m e^{-2\lambda } \right) + \left( \dots \right) \vec {n}. \end{aligned}$$

Identifying $\vec {n}_{xy}$ and $\vec {n}_{yx}$, one finds that

$$\begin{aligned} e^{2\lambda } H_x= & {} \left( \frac{l-n}{2} \right) _x + m_y \nonumber \\{} & {} \text {and}\nonumber \\ e^{2\lambda } H_y= & {} - \left( \frac{l-n}{2} \right) _y + m_x. \end{aligned}$$

(54)

1.3 Brief Proof of Theorem 1.6

In order to contrast 1.5 from 1.6, we sketch a Proof of Theorem 1.6.

Proof

A weakly harmonic application u satisfies

$$\begin{aligned} \Delta u + \langle \nabla u , \nabla u \rangle _{4,1} u =0. \end{aligned}$$

Equivalently, the latter may be recast in the conservative form

$$\begin{aligned} \textrm{div}( \nabla u u^T - u \nabla u^T ) = 0. \end{aligned}$$

Accordingly, on ${\mathbb {D}}$, there exists a matrix B such that $\nabla ^\perp B = \nabla u u^T - u \nabla u^T$. As a consequence, B satisfies $\nabla B = u \nabla ^\perp u^T - \nabla ^\perp u u^T$ which implies

$$\begin{aligned} \Delta B = 2 \nabla u \nabla ^\perp u^T. \end{aligned}$$

(55)

On the other hand, if we denote by $\varepsilon$ the signature matrix of ${\mathbb {R}}^{4,1}$, then for any two vectors a, b, we have $\langle a, b \rangle _{4,1}= a^T \varepsilon b$. Then

$$\begin{aligned} \nabla ^\perp B \varepsilon \nabla u= & {} \nabla u u^T \varepsilon \nabla u -u \nabla u^T \varepsilon \nabla u = \langle u, \nabla u \rangle _{4,1} \nabla u - \langle \nabla u , \nabla u \rangle _{4,1} u \\= & {} - \langle \nabla u , \nabla u \rangle _{4,1} u = \Delta u, \end{aligned}$$

since, given that $u\in {\mathbb {S}}^{4,1}$, we have $\langle u, \nabla u \rangle _{4,1}=0$. Combining this to (55) yields the system:

$$\begin{aligned} \Delta u= & {} \nabla ^\perp B \nabla (\varepsilon u ) \\ \Delta B= & {} 2 \nabla u \nabla ^\perp u^T. \end{aligned}$$

Provided that $\left\| \nabla u \right\| _{L^2 } \le \varepsilon _0$, the statement (14) now follows from classical integration by compensation techniques, such as those presented in [12]. $\square$