A symmetry condition for genus zero free boundary minimal surfaces attaining the first eigenvalue of one

Assume $D_{1}\subset D_{2}$. Take a coordinate plane $\Pi$ that meets $\partial\Sigma$ transversely and meets $(\partial\Sigma)_{1}$ and $(\partial\Sigma)_{2}$. Let $\{(\partial\Sigma)_{1},\dots(\partial\Sigma)_{m}\}$ be the boundary components of $\Sigma$ that has an intersection with $\Pi$, and we denote $(\partial\Sigma)_{i}\cap\Pi$ by $v_{i}^{1}$ and $v_{i}^{2}$ for all $i=1,2,\dots,m$.

We consider a graph whose vertex set is

and the edges are the arcs in $\Pi\cap\Sigma$ whose endpoints are contained in $\partial\Sigma$. By the transversality property (Lemma 2.3 (1)), every vertex is contained in exactly one edge. Furthermore, we give a direction on the edge set by following the procedure.

1. (1)

   Take the edge containing $v_{1}^{2}$ and give a direction from $v_{1}^{2}$. Let $v$ be the other vertex in the edge.

2. (2)

   Take the vertex $w$ which is contained in the same boundary component with $v$ but not identical to $v$.

3. (3)

   Repeat the first and second steps until the vertex $v_{1}^{1}$ comes out.

By the two-piece property (Theorem 2.1) with the fact that $\Sigma$ has genus zero, every edge has a direction.

We claim that the directed graph from $v_{1}^{2}$ escapes the convex cone over $D_{2}\cap\Pi$ by an edge. Note the fact that a small part of an arc of $\Pi\cap\Sigma$ from each of vertices $v_{i}^{1}$ and $v_{i}^{2}$ is contained outside of the convex cone over $D_{i}\cap\Pi$ by Lemma 2.4 (2). Since we can give directions on every edge, the directed graph from $v_{1}^{2}$ escapes the convex cone over $D_{2}\cap\Pi$. On the other hand, the embeddedness of $\Sigma$ implies that for every $i$, we have either $v_{i}^{1},v_{i}^{2}\in\text{int}(D_{2})$ or $v_{i}^{1},v_{i}^{2}\notin\text{int}(D_{2})$. Therefore, the directed graph escapes by an edge that meets one of the line segments $ov_{2}^{1}$ and $ov_{2}^{2}$ transversely by Lemma 2.3 (1), which completes that proof.

By the previous claim and the fact in the proof, there is a ray from $o$ that meets $\Sigma$ at least twice. It contracts the radial graphicality of $\Sigma$ (Lemma 2.3 (2)).

∎

Suppose $f_{i}=f_{j}$ for $i\neq j$. If $D_{i}\cap D_{j}=\emptyset$, we can find a coordinate plane that separates $D_{i}$ and $D_{j}$. Then, $f_{i}\neq f_{j}$, a contradiction. On the other hand, we consider $D_{i}\cap D_{j}\neq\emptyset$. Since $\Sigma$ is embedded, we have either $D_{i}\subsetneq D_{j}$ or $D_{j}\subsetneq D_{i}$, which contradicts Theorem 3.1. ∎

It is clear that $R_{\Pi}((\partial\Sigma)_{i})=(\partial\Sigma)_{j}$ implies $R_{\Pi}(f_{i})=f_{j}$. Now suppose $R_{\Pi}(f_{i})=f_{j}$. We note that $R_{\Pi}((\partial\Sigma)_{i})$ is a boundary component of $\Sigma$, say $(\partial\Sigma)_{k}$. Then, $f_{k}=R_{\Pi}(f_{i})$, and by the assumption, $f_{k}=f_{j}$. By Corollary 3.2, we have $k=j$ and $R_{\Pi}((\partial\Sigma)_{i})=(\partial\Sigma)_{j}$. ∎

Suppose the first statement is not true. We consider a graph with vertices $\{f_{1},\dots,f_{n}\}$ and edges $(f_{i}f_{j}),1\leq i,j\leq n$ if there is a plane $\Pi\in\{\Pi_{1},\dots,\Pi_{n}\}$ such that $R_{\Pi}(f_{i})=f_{j}$. By the assumption, this graph has at least two components. Take one of the smallest components with $k$ vertices. Then, $k\leq n-k$.

We show that $k\geq 2$. Suppose $k=1$ and let $f_{1}$ be such vertex. Then, every reflection, $R_{\Pi_{1}},\dots,R_{\Pi_{n}}$, maps $f_{1}$ into $f_{1}$. This $n$ reflections map $f_{2}$ into $\{f_{1},\dots,f_{n}\}\setminus\{f_{1}\}$. If $f_{2}\neq-f_{1}$, by Corollary 3.2, only one coordinate plane passes through $f_{1}$ and $f_{2}$. It implies that there is at most one reflection in $\{R_{\Pi_{1}},\dots,R_{\Pi_{n}}\}$ that map $f_{2}$ into $f_{2}$. In addition, using Lemma 2.2 (1), there are at most $n-2$ reflections that map $f_{2}$ into $\{f_{1},\dots,f_{n}\}\setminus\{f_{1},f_{2}\}$. Thus, there are at most $n-1$ distinct reflections in $\{R_{\Pi_{1}},\dots,R_{\Pi_{n}}\}$, a contradiction. Thus, $f_{2}=-f_{1}$. In a similar argument, we can show that $f_{2}=f_{3}=\cdots=f_{n}$, which is $-f_{1}$. It contradicts Corollary 3.2. Therefore, $k\geq 2$.

Let $\{f_{1},\dots,f_{k}\}$ be the vertex set of the smallest component. By Lemma 2.2 (1), there are at most $k-1$ reflection planes in $\{\Pi_{1},\dots,\Pi_{n}\}$ that do not pass through $f_{1}$. Therefore, there must be at least $n-k+1$ reflection planes in $\{\Pi_{1},\dots,\Pi_{n}\}$ that pass through $f_{1}$. Suppose $k=2$ and $f_{2}=-f_{1}$. In this case, there is exactly one reflection plane in $\{\Pi_{1},\dots,\Pi_{n}\}$ that does not pass through $f_{1}$. Because every reflection plane passing through $f_{1}$ maps $f_{3}$ into $\{f_{3},\dots,f_{n}\}$, Lemma 2.2 further implies that there can be at most $n-2$ distinct reflection planes passing through $f_{1}$ in $\{\Pi_{1},\dots,\Pi_{n}\}$. Then, there are at most $n-1$ reflections in $\{\Pi_{1},\dots,\Pi_{n}\}$, contrary to Corollary 3.2. Therefore we can take $f_{2}\neq-f_{1}$, and we can assume $f_{2}$ is connected to $f_{1}$ by an edge. By Lemma 2.2 (1), the reflection planes in $\{\Pi_{1},\dots,\Pi_{n}\}$ that pass through $f_{1}$ maps $f_{2}$ one-to-one into $\{f_{2},\dots,f_{k}\}$. Thus, there are at most $k-1$ such reflections. However, $n-k+1>k-1$ leads to a contradiction.

The second statement follows from the first statement with an argument in the proof of Corollary 3.3. ∎

The case for $n=1$ is trivial by the classical work by Nitsche [18]. For $n=2$, our theorem is known in [19, Theorem 1.2]. Thus, we only need to consider $n>2$. We prove this theorem by proof by contradiction. We assume $\sigma_{1}(\Sigma)<1$. We will obtain a contradiction by following the steps.

Step 1. Let $\Pi_{1},\dots,\Pi_{n}$ be the given reflection planes of $\Sigma$. We show that for each $1\leq i,j\leq n$, there is $\Pi\in\{\Pi_{1},\dots,\Pi_{n}\}$ such that $R_{\Pi}((\partial\Sigma)_{i})=(\partial\Sigma)_{j}$. By Corollary 3.3 and Lemma 2.2 (1), it is sufficient to prove that there is exactly one reflection plane in $\{\Pi_{1},\dots,\Pi_{n}\}$ that passes through each $f_{i}$. Suppose there are no reflection planes in $\{\Pi_{1},\dots,\Pi_{n}\}$ that pass through $f_{i}$. Then, by Lemma 2.2 (1), there are at most $n-1$ distinct reflections in $\{R_{\Pi_{1}},\dots,R_{\Pi_{n}}\}$, a contradiction. If there are two reflections in $\{R_{\Pi_{1}},\dots,R_{\Pi_{n}}\}$ that pass through $f_{i}$, by Lemma 2.7, a first eigenfunction $u$ has a constant sign on $(\partial\Sigma)_{i}$. By Lemma 3.4 with Corollary 3.3 and Lemma 2.6, $u$ has a constant sign on $\partial\Sigma$. It contradicts Lemma 2.5 (2).

By Step 1, for each $1\leq i,j\leq n$, we can define $\Pi_{ij}\in\{\Pi_{1},\dots,\Pi_{n}\}$ and $R_{ij}:=R_{\Pi_{ij}}$ such that

In addition, we show that each of the boundary components of $\Sigma$ has exactly two nodal points of $u$. By Lemma 2.6 and Lemma 2.5 (2), $u$ changes its sign on each of the boundary components. In addition, every $(\partial\Sigma)_{i}$ has the same number of nodal points of $u$, and the number is at least two. By Lemma 2.4 (4), the number should be two.

Let $p_{i}^{1},p_{i}^{2}$ be the nodal point of $u$ in $(\partial\Sigma)_{i}$. Then, by Lemma 2.6, $R_{ii}(p_{i}^{1})=p_{i}^{2}$. Otherwise, $R_{ii}(p_{i}^{1})=p_{i}^{1}$ and $R_{ii}(p_{i}^{2})=p_{i}^{2}$. Then, by Lemma 2.6 again, $u$ cannot change its sign on $(\partial\Sigma)_{i}$, a contradiction. These nodal points divide each of the boundary components into two arcs.

Step 2. We show that the two arcs of $(\partial\Sigma)_{i}\setminus\{p_{i}^{1},p_{i}^{2}\}$ are not congruent. Suppose not. Then, either a nontrivial rotation or a reflection that maps one component of $(\partial\Sigma)_{i}\setminus\{p_{i}^{1},p_{i}^{2}\}$ into the other and vice versa. If there is such a reflection, it cannot be $R_{ii}$. By Lemma 2.7, $u$ has a constant sign on $(\partial\Sigma)_{i}$, a contradiction. If there is such a rotation $A$, $A(f_{i})=f_{i}$ and $A(\{p_{i}^{1},p_{i}^{2}\})=\{p_{i}^{1},p_{i}^{2}\}$. For the case of $A(p_{i}^{1})=p_{i}^{1}$ and $A(p_{i}^{2})=p_{i}^{2}$, the rotation axis for $A$ passes through $f_{i},p_{i}^{1},$ and $p_{i}^{2}$. Since the axis meets $\mathbb{S}^{2}$ at exactly two points, $f_{i}$ should be either $p_{i}^{1}$ or $p_{i}^{2}$. However, $f_{i}$ lies in the interior of $D_{i}$ [19, Proposition 3.5], a contradiction. For the case of $A(p_{i}^{1})=p_{i}^{2}$ and $A(p_{i}^{2})=p_{i}^{1}$, we note that $A$ is the 180°rotation through the axis passing through the origin and $f_{i}$. Together with $R_{ii}$, $(\partial\Sigma)_{i}$ has a reflection symmetry distinct from $R_{ii}$. Lemma 2.7 again gives a contradiction.

We prove the following step using an argument on arcs.

Step 3. We show that $\Pi_{1},\dots,\Pi_{n}$ share a common axis. By the previous step, let $\widehat{p_{1}^{1}p_{1}^{2}}$ and $\widehat{p_{1}^{1}p_{1}^{2}}^{\prime}$ be the components of $(\partial\Sigma)_{1}\setminus\{p_{1}^{1},p_{i}^{2}\}$. Furthermore, we define an arc $\widehat{p_{i}^{1}p_{i}^{2}}$ of $(\partial\Sigma)_{i}$ that is congruent to $\widehat{p_{1}^{1}p_{1}^{2}}$. Clearly,

Then, we have

For simplicity, let $A:=(R_{2i}\circ R_{12})^{-1}\circ(R_{1i}\circ R_{11})$ and let $q_{1}:=\widehat{p_{1}^{1}p_{1}^{2}}\cap\Pi_{11}$. Then, $A(q_{1})=q_{1}$ and $A(p_{1}^{1})$ is either $p_{1}^{1}$ or $p_{1}^{2}$. Note that $R_{2i}\circ R_{12}$, $R_{1i}\circ R_{11}$, and $A$ are all rotations along axes passing through the origin. Then, $A$ is a rotation along the axis passing through the origin and $q_{1}$. Suppose $A$ is not the identity map. Since the axis and $\widehat{p_{1}^{1}p_{1}^{2}}$ are invariant under $A$ and $R_{11}$, each of components of $\widehat{p_{1}^{1}p_{1}^{2}}\setminus\{q_{1}\}$ lies in a plane that passes through the axis. Similarly, $\widehat{p_{1}^{1}p_{1}^{2}}^{\prime}$ lies in two planes passing through the axis. Since $(\partial\Sigma)_{1}$ is real analytic (see [13, Section 6]), this boundary component lies in a plane, so it is a circle. By [19, Remark 6.2], $\Sigma$ is congruent to one of an equatorial disk and a critical catenoid. It contradicts $n>2$. Thus $A$ is the identity map. Then, the rotation axes of $R_{2i}\circ R_{12}$ and $R_{1i}\circ R_{11}$ are identical. Since each of these axes is the intersection of the two reflection planes, $\Pi_{11},\Pi_{12},$ and $\Pi_{1i}$ contain a common axis. Since $i$ is arbitrary and

we obtained our desired conclusion.

Step 4. Let $x_{3}$-axis be the common axis of $\Pi_{1},\dots,\Pi_{n}$ as in the previous step. By Step 1 and Lemma 2.6, every nodal point of $u$ in $\partial\Sigma$ has the same $x_{3}$-coordinates, say $c$. Moreover, using (2.2), we have

By Lemma 2.4 (3), $u$ does not change its sign on each of $\partial\Sigma\cap\{x_{3}>c\}$ and $\partial\Sigma\cap\{x_{3}<c\}$. Thus, we have

which contradicts Lemma 2.5 (2).

Throughout the steps, we obtained a contradiction by assuming $\sigma_{1}(\Sigma)<1$. By Lemma 2.5 (1), we have $\sigma_{1}(\Sigma)=1$.

∎