A Free Boundary Minimal Surface via a 6-Sweepout

Abstract

We prove that the Almgren–Pitts 6-width of the unit 3-ball is less than $2\pi$. We also prove that there exists a free boundary minimal surface in the unit 3-ball that has genus at most 1, index at most 5, area less than $2\pi$, and is not the equatorial disk or the critical catenoid.

Appendices

Appendix A: Proof of Proposition 3.5

The following proof is due to the MathOverflow user fedja [16]. Let M denote the saddle in ${\mathbb {R}}^3$ given by $x^2-y^2+z=0$. Then to prove Proposition 3.5, it suffices, by rescaling, to show that for any ball B with center $(x_0,y_0,z_0)$ and radius $R>0$, the area of $M\cap B$ is less than $2\pi R^2$.

Recall that M is foliated by straight lines: It can be parametrized by $\textbf{x}(s,t)=(s+t,s-t,4st)$. Then the Jacobian of $\textbf{x}$ is $2\sqrt{1+8s^2+8t^2}$. Thus we have

$$\begin{aligned} \textrm{area}(M\cap B)<\iint _{\{(s,t):\textbf{x}(s,t)\in B\}}(2\sqrt{1+8s^2}+2\sqrt{1+8t^2})dsdt. \end{aligned}$$

(20)

Now, for each fixed s, let $L_s$ be the corresponding coordinate line segment in $M\cap B$. Letting d be the distance between $L_s$ and the center of B, we have

$$\begin{aligned} d^2\ge \min _{t\in {\mathbb {R}}}[(s+t-x_0)^2+(s-t-y_0)^2]= 2\left( s-\frac{x_0+y_0}{2}\right) ^2. \end{aligned}$$

(21)

Note that $L_s$ is parameterized by a time interval of length

$$\begin{aligned} \frac{\textrm{length}{(L_s)}}{\left\Vert {\partial _t\textbf{x}}\right\Vert }=\frac{2\sqrt{(R^2-d^2)^+}}{\sqrt{2(1+8s^2)}}, \end{aligned}$$

(22)

where $^+$ denotes the positive part. It follows that, using (21) and (22),

$$\begin{aligned}{} & {} \iint _{\{(s,t):\textbf{x}(s,t)\in B\}}2\\{} & {} \quad \sqrt{1+8s^2}dtds\le \int _{s\in {\mathbb {R}}}2\sqrt{2}\sqrt{\left[ R^2-2\left( s-\frac{x_0+y_0}{2}\right) ^2\right] ^+} ds=\pi R^2. \end{aligned}$$

The second integral in (20) can be similarly bounded, by integrating with respect to s first. So $\textrm{area}(M\cap B)<2\pi R^2$, finishing the proof of Proposition 3.5.

Appendix B: First Variation Formula

Lemma B.1

Let $\Omega$ be a compact $(n+1)$-dimensional region with smooth boundary in ${\mathbb {R}}^{n+1}$, $\{\Sigma _s\}$ a 1-parameter family of hypersurfaces without boundary in ${\mathbb {R}}^3$, and V a deformation vector field of $\{\Sigma _s\}$. Then

$$\begin{aligned} \frac{d}{ds}\textrm{area}(\Sigma _s\cap \Omega )=-\int _{\Sigma _s\cap \Omega }{} \textbf{H}\cdot V-\int _{\partial (\Sigma _s\cap \Omega )}\frac{\textbf{n}\cdot w}{{\nu }\cdot w}\;V\cdot \textbf{n}, \end{aligned}$$

where $\textbf{n}$ is a chosen unit normal vector field of $\Sigma _s$, w the outward unit normal of $\partial \Omega$, and $\nu$ the outward unit conormal of $\Sigma _s$ on $\partial \Omega$.

Proof

We first smoothly extend w to a unit vector field on a neighborhood of $\partial \Omega$ in $\Omega$, and $\nu$ to a unit tangent vector field on a neighborhood of $\partial \Sigma _s$ in $\Sigma _s$. Let $\epsilon >0$, and $\Omega _\epsilon \subset \Omega$ be where the distance from $\partial \Omega$ is at least $\epsilon$. Then by using the function $\textrm{dist}(\cdot ,\partial \Omega )$ on $\Omega$, with suitable smoothening, we can approximate the indicator function $\chi _\Omega$ by a smooth function $\chi ^\epsilon _\Omega$ that is 0 outside $\Omega$ and 1 on $\Omega _\epsilon$, with $\nabla \chi ^\epsilon _\Omega =-|\nabla \chi ^\epsilon _\Omega |w$ in between.

Now, using the first variation formula (4.2) in [15, p.49],

$$\begin{aligned} \frac{d}{ds}\int _{\Sigma _s}\chi ^\epsilon _\Omega =\int _{\Sigma _s}-\chi ^\epsilon _\Omega \textbf{H}\cdot V+\nabla \chi ^\epsilon _\Omega \cdot \textbf{n}\; V\cdot \textbf{n}. \end{aligned}$$

Note that

$$\begin{aligned} \nabla \chi _\Omega ^\epsilon \cdot \textbf{n}\; V\cdot \textbf{n}=-|\nabla \chi _\Omega ^\epsilon |\;w\cdot \textbf{n}\; V\cdot \textbf{n}=\nabla \chi _\Omega ^\epsilon \cdot \nu \;\frac{w\cdot \textbf{n}}{w\cdot \nu }\; V\cdot \textbf{n}. \end{aligned}$$

Denoting $g:=\frac{w\cdot \textbf{n}}{w\cdot \nu }\; V\cdot \textbf{n}$, we then have

$$\begin{aligned} \int _{\Sigma _s}\nabla \chi _\Omega ^\epsilon \cdot \textbf{n}\; V\cdot \textbf{n}= & {} \int _{\Sigma _s}\nabla \chi _\Omega ^\epsilon \cdot g\nu \\= & {} -\int _{\Sigma _s}\chi _\Omega ^\epsilon \textrm{div}(g\nu )\xrightarrow {\epsilon \rightarrow 0}-\int _{\Sigma _s\cap \Omega }\textrm{div}(g\nu )\\= & {} -\int _{\partial (\Sigma _s\cap \Omega )}g, \end{aligned}$$

in which the second and the third equality are due to divergence theorem. Hence, $\frac{d}{ds}\textrm{area}(\Sigma _s\cap \Omega )$ is equal to

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\frac{d}{ds}\int _{\Sigma _s}\chi ^\epsilon _\Omega= & {} \lim _{\epsilon \rightarrow 0}\int _{\Sigma _s}-\chi ^\epsilon _\Omega \textbf{H}\cdot V+\nabla \chi ^\epsilon _\Omega \cdot \textbf{n}\; V\cdot \textbf{n}\\= & {} -\int _{\Sigma _s\cap \Omega }{} \textbf{H}\cdot V-\int _{\partial (\Sigma _s\cap \Omega )}g. \end{aligned}$$

$\square$

Appendix C: A Lemma About Cubic Polynomials

Lemma C.1

There exists $h>0$ such that the following is true. For any a, b, c such that $a^2+b^2+c^2=1$, define $f:[-\frac{1}{2}, \frac{1}{2}]\rightarrow {\mathbb {R}}$ by $f(x)=ax^3+bx+c$. Then there exists some interval of length $\frac{1}{8}$ in $[-\frac{1}{2},\frac{1}{2}]$ on which $|f|>h$.

Proof

Assume, by contradiction, that for each positive integer n there exists a cubic function $f_n(x)=a_nx^3+b_nx+c_n$, with $a^2_n+b^2_n+c^2_n=1$ such that there is no interval of length $\frac{1}{8}$ in $[-\frac{1}{2},\frac{1}{2}]$ on which $|f|>\frac{1}{n}$. For each n, let $x_i$, for i runs from 1 to at most 3, be the roots of $f_n(x)=0$, and $I_i\subset [-\frac{1}{2},\frac{1}{2}]$ be the maximal interval containing $x_i$ on which $|f_n|<\frac{1}{n}$. Then $[-\frac{1}{2},\frac{1}{2}]\backslash (I_1\cup I_2\cup I_3)$ is a union of at most 4 intervals, each of which has length at most $\frac{1}{8}$. Thus, $I_1\cup I_2\cup I_3$ has length at least $1-4\cdot \frac{1}{8}=\frac{1}{2}$, and on it $|f_n|<\frac{1}{n}$. Then it follows easily that $\sup _{x\in [-\frac{1}{2},\frac{1}{2}]}|f'_n(x)|\rightarrow 0$ as $n\rightarrow \infty$. Since $f'_n(x)=3a_nx^2+b_n$, we must have $a_n\rightarrow 0$ and $b_n\rightarrow 0$ too, which forces $c_n\rightarrow 1$ since $a^2_n+b^2_n+c^2_n=1$. But then $f_n$ is very close to 1 on $[-\frac{1}{2},\frac{1}{2}]$, contradicting that $|f_n|<\frac{1}{n}$ on a set of length at least $\frac{1}{2}$. $\square$