Erratum to: Extrinsic diameter of immersed flat tori in $S^3$

Abstract

In the paper referred to in the title, we proved that any immersed flat tori in $S^3$ whose mean curvature does not change sign has extrinsic diameter $\pi$. Although the main result there is correct, there is a gap of the proof of this fact. The purpose here is to correct the previous paper’s argument and clarify the statement.

1 Introduction

Let $S^3$ be the unit sphere in $\mathbf{R}^4=\mathbf{C}^2$. The Clifford torus in $S^3$ given by

$$\begin{aligned} M_\theta :=\biggl \{(z,w)\in C^2\,;\, |z|^2=\cos ^2 \theta ,\quad |w|^2=\sin ^2 \theta \biggr \}, \quad 0 < \theta < \pi /2, \end{aligned}$$

is a flat embedded torus of constant mean curvature. We denote by $\iota _\theta :M_\theta \rightarrow S^3$ the canonical inclusion, and by $ds^2_\theta$ is the induced metric on $M_\theta$. In the previous paper [1, Theorem 0.2], we stated the following assertion:

Theorem

Let $f :(M_\theta ,ds^2_\theta ) \rightarrow S^3$ be an isometric immersion whose mean curvature function does not change sign on $M_\theta$, then $f$ is congruent to $\iota _\theta$.

Unfortunately, there was a gap in the proof of Proposition 4.7, which was a key assertion to prove the theorem. In fact, in [1, Claim 5 (Page 133)], we stated that the subset $\mathcal{L }$ is closed in the subarc $C|_{[A,B]}$ of $C$. However, this is not true if $s_\infty$ does not coincide with $s_{X_\infty }^{}$. In fact, $s=s_\infty$ might not be the first point where $\varphi _s(S'_2)$ meets the union of the arcs $[AP)\cup [BP)$, where $[AP)$ and $[BP)$ are subarcs of $\gamma _1$ defined in [1, Page 132]. For example, if we consider the case as in Fig. 1 , the set $\mathcal{L }$ is not closed, since the shell is tangent to $(PB]$ before attaching to $[AP)$. The purpose of this paper is to correct the proof of Proposition 4.7.

Fig. 1

A typical case where the previous argument fails

2 A key proposition

The following assertion will play a crucial role in the new proof of Proposition 4.7. (The definition of positive shell and negative shell is given in [1, Page 118].)

Proposition 1.1

Let $\sigma :[0,a]\rightarrow S^2$ be a positive shell (respectively negative shell) whose geodesic curvature $\kappa (t)\, (0\le t\le a)$ satisfies $\kappa (t)>\kappa _0$ (respectively $\kappa (t)<-\kappa _0$), where $\kappa _0$ is a positive constant. Let $\Gamma$ be the circle of geodesic curvature $\kappa _0 \,(respectively -\kappa _0)$ which is tangent to $\sigma (t)$ at $t=b\, (0\le b \le a)$. Then $\sigma ([0,a])\setminus \{\sigma (b)\}$ lies in the interior $\Delta _{\Gamma }$ of the circle $\Gamma$ (the definition of $\Delta _{\Gamma }$ is given in [1, Page 118]).

To prove this, we show the following lemma, which is a special case of the proposition.

Lemma 1.2

Let $\sigma :[0,a]\rightarrow S^2$ be a positive shell whose geodesic curvature $\kappa (t)\, (0\le t\le a)$ satisfies $\kappa (t)>\kappa _0$, where $\kappa _0$ is a positive constant. Let $\Gamma$ be the circle of geodesic curvature $\kappa _0$ which is tangent to $\sigma (t)$ at $t=0$ or $t=a$. Then $\sigma ((0,a))$ lies in the interior $\Delta _{\Gamma }$ of the circle $\Gamma$.

Proof

We consider the case $t=0$. (The case $t=a$ is proved similarly.) Suppose that the image $\sigma ((0,a))$ does not lie in $\Delta _{\Gamma }$. The condition $\kappa (t)>\kappa _0$ yields that $\sigma (t)$ lies in $\Delta _{\Gamma }$ for sufficiently small $t>0$. Since $\sigma$ is a positive shell, $\sigma (a-t)$ also lies in $\Delta _{\Gamma }$ for sufficiently small $t>0$. Thus, there exist two points $s_1,s_2\in (0,a)$ satisfying the following three properties (see Fig. 2, left):

• $0<s_1\le s_2<a$,

• $\sigma (s_1)$ and $\sigma (s_2)$ lie on $\Gamma$,

• $\sigma ((0,s_1))$ and $\sigma ((s_2,a))$ are contained in the domain $\Delta _{\Gamma }$.

Applying [1, Lemma 4.5] to the arc $\sigma ([0,s_1])$, we can conclude that $\sigma (s_1)$ lies in the past part of $\Gamma$ with respect to the node $P:=\sigma (0)=\sigma (a)$. Since $\sigma$ is a simple closed arc, $\sigma (s_2)$ must also lie in the past part of $\Gamma$ with respect to $P$. On the other hand, applying [1, Lemma 4.5] to another arc $\sigma ([s_2, a])$, we can conclude that $\sigma (s_2)$ must lie in the future part of $\Gamma$ with respect to $P$, which is a contradiction. $\square$

Fig. 2

Three impossible arrangements of $\sigma$

We now prove Proposition 1.1 as follows: Reversing the orientation of the curve $\sigma$ if necessary, we may assume that $\sigma$ is a positive shell. Since Lemma 1.2 proves the case of $b=0$ or $b=a$, we may assume that $0< b < a$. Suppose also that the image $\sigma ([0,a])\setminus \{\sigma (b)\}$ does not lie in $\Delta _{\Gamma }$. The center and right of Fig.  2 are typical such cases. Suppose that both $\sigma ([0,b))$ and $\sigma ((b,a])$ do not lie in $\Delta _\Gamma$. Then, there exist two points $s_1,s_2\in [0,a]$ satisfying the following three properties:

• $0\le s_1<b<s_2\le a$,

• $\sigma (s_1)$ and $\sigma (s_2)$ lie on $\Gamma$,

• $\sigma ((s_1,b))$ and $\sigma ((b,s_2))$ are contained in the domain $\Delta _{\Gamma }$.

We can make a contradiction applying [1, Lemma 4.5] for two arcs $\sigma ([s_1,b])$ and $\sigma ([b,s_2])$, respectively, like as in the proof of Lemma 1.2. Thus, either $\sigma ([0,b))$ or $\sigma ((b,a])$ lies in $\Delta _\Gamma$. In particular, $P$ lies in $\Delta _\Gamma$. We now consider the case that $\sigma ([0,b))$ lies in $\Delta _\Gamma$. (The case $\sigma ((b,a])\subset \Delta _\Gamma$ also makes a contradiction using the same argument.) Then, there exists a point $s\in (b,a)$ such that $\sigma (s)$ lies on $\Gamma$ and $\sigma ((b,s))$ lies in $\Delta _\Gamma$ (see Fig. 3, left). Let $\Gamma '$ be the circle of geodesic curvature $\kappa _0$ which is tangent to $\sigma (t)$ at $t=0$. $\Gamma '$ intersects $\Gamma$ with two points. Let $Q$ be one of two such intersection points, which lies in the past part of $\Gamma$ with respect to $\sigma (b)$. By Lemma 1.2, $\sigma ((0,a))$ lies in $\Delta _{\Gamma '}$ (cf. Fig. 3, right). In particular, $\sigma (b)$ and $\sigma (s)$ lie on $\Gamma \cap \Delta _{\Gamma '}$. Then, we can conclude that $\sigma (b)$ and $\sigma (s)$ lie in the future part of $\Gamma$ with respect to $Q$. On the other hand, applying [1, Lemma 4.5] to the arc $\sigma ([b,s])$, we can conclude that $\sigma (s)$ lies in the past part of $\Gamma$ with respect to $\sigma (b)$. Hence $\sigma (s)$ must lie in the subarc $\alpha$ of $\Gamma$ bounded by $Q$ and $\sigma (b)$ (see Fig. 3, right). However, this implies that $\sigma ((b,s))$ must meet the arc $\sigma ([0,b])$, which contradicts the fact that $\sigma$ is a simple closed arc. This proves the proposition.

Fig. 3

The case $\sigma ((0,b))\subset \Delta _\Gamma$

3 A modification of the definition of $r$-diamonds

In [1, Definition 4.8], we gave a definition of $r$-diamonds. However, we need to modify it for the new proof of Proposition 4.7. More precisely, the condition (v) of $r$-diamonds should be replaced by the following condition:

• (v$^{\prime }$) Let $C$ be the circle of radius $r$ centered at $O$, which is tangent to $\gamma _1$ at $A$ and $B$. We denote by $[AP]$ (respectively $[BP]$) the subarc of $\gamma _1$ bounded by $A$ (respectively $B$) and $P$. We also use another notion $[AP)$ (respectively $[BP)$) which contains the end point $A$ (respectively $B$) but does not contain the other end point $P$. Fix $X\in [AP)$ and $Y\in [BP)$. Then there exists a circle $C_X$ (respectively $C_Y$) of radius $\delta _\mu$ which is tangent to $[AP)$ at $X$ (respectively $[BP)$ at $Y$) and $(XP]$ (respectively $(YP]$) lies in $\Delta _{C_X}$ (respectively $\Delta _{C_Y}$), where $\Delta _{C_X}$ and $\Delta _{C_Y}$ are the interior domains of the circles $C_X$ and $C_Y$, respectively.

From now on, we replace condition (v) with this new modification for the definition of $r$-diamonds. (All the other conditions (i)–(iv) remain as before. The previous condition (v) is the special case of the new condition (${v}^{\prime }$) by putting $X=A$ and $Y=B$.) Thus, our $r$-diamonds satisfy the old definition of $r$-diamonds in [1], but the converse is not true. By this change, two points $A,B$ can move not only on $S_1$ but also whole on $\gamma _1$ (see [1, (4.9)] for the definition of $S_1$).

In Claim 1-3 in [1, Section 4], we proved that there is a $\delta _\mu ^*$-diamond $\diamondsuit _{OAPB}$. This proof can apply for our new definition of $r$-diamonds by a suitable modification as follows: Since the curvature radius of each point of the spherical curve $\gamma _1$ is less than $\delta _\mu$ (cf. [1, (4.7)]), the condition (${v}^{\prime }$) holds for the $4$-gon $OAPB$ as in [1, Fig. 17] for sufficiently small $r>0$. By replacing $S_1$ to $\gamma _1$, the proofs of Claim 1-2 are valid even if $A,B\not \in S_1$. We now pay attention to the sequence of $r_n$-diamonds $\{\diamondsuit _{O_n A_n P B_n}\}_{n=1,2,\ldots }$ in the proof of Claim 3. For the sake of simplicity, we set $A:=A_\infty$ and $B:=B_\infty$. From now on, we show that $[A P]$ and $[B P]$ are both simple arcs: In fact, there exist $c_1,c_2\in S^1:=\varvec{R}/\varvec{Z}$ such that $\gamma _1(c_1)=P$ and $\gamma _1(c_2)=A$. Since $\diamondsuit _{O_n A_n P B_n}$ is a diamond, $[A_n P]$ is a simple arc. So if we suppose that $[A P]$ has a self-intersection, then there exists $m\in (c_1,c_2)$ such that $A=\gamma _1(m)$. Since all of the crossings on $\gamma _1$ are transversal, either $\gamma _1(m-\varepsilon )$ or $\gamma _1(m+\varepsilon )$ does not lie in the interior domain $\Delta _{C_{A_n}}$ for a sufficiently small $\varepsilon >0$ and for sufficiently large $n$. However, it contradicts the condition ($\mathrm{v}'$) of the diamond $\diamondsuit _{O_n A_n P B_n}$, since $\gamma _1(m)$ lies in $[A_n P)$ for sufficiently large $n$. Similarly, $[B P]$ is also a simple arc.

We now suppose that the limit 4-gon $O_\infty A P B$ is not a $\rho$-diamond ($\rho :=\delta ^*_\mu$).

Then, by the condition ($\mathrm{v}'$), there exists $X\in [A P)$ or $Y\in [B P)$ such that either $(XP]$ meets $C_{X}$ or $(YP]$ meets $C_{Y}$. Without loss of generality, we may consider the case $(XP]$ meets $C_{X}$. If $X\ne A$, then $X\in (A_nP]$ holds for a sufficiently large $n$. Then, the property (${v}^{\prime }$) of the $r_n$-diamond $\diamondsuit _{OA_nPB_n}$ yields that $C_X$ does not meet $(XP]$, a contradiction.

So we may assume that $X=A$. Now it is sufficient to make a contradiction under the assumption that $(A P]$ meets $C_{A}$. Since $(A_n P]$ lies in the domain $\Delta _{C_{A_n}}$, by taking limit $n\rightarrow \infty$, there exists a point $Q$ on $(AP]$ such that $Q$ lies on the circle $C_A$. Without loss of generality, we may assume that $Q$ is the first such point, namely, $(AQ)$ does not meet $C_A$. (see [1, Fig. 20, left and right]). We give an orientation of $C_A$ so that it has positive geodesic curvature. Since $C_A$ is of radius $\delta _\mu$, the geodesic curvature of $\gamma _1$ is greater than that of $C_A$. Then [1, Lemma 4.5] implies that $Q$ lies in the future part of $C_A$ with respect to $A$.

We first consider the case $Q=P$. Consider the subarc of $C_{A}$ defined by

$$\begin{aligned} \mathfrak{a }:=C_{A} \cap \overline{\Delta _{C_{B}}}. \end{aligned}$$

Then $\mathfrak{a }$ lies in the past part of $C_A$ with respect to $A$. Since $Q$ lies in the subarc $\mathfrak{a }$, it contradicts the fact that $Q$ lies in the future part of $C_A$.

We next consider the case $Q\ne P$. We denote by $C_{A}^+$ (respectively $C_{A}^-$) the future part (respectively the past part) of $C_{A}$ with respect to $A$. Since $Q\in C_{A}^+$, we can consider the subarc $\mathfrak{b }$ of $C_{A}^+$ bounded by $Q$ and $A$. Since $\mathfrak{a }\subset C_{A}^-$, the two circular arcs $\mathfrak{a }$ and $\mathfrak{b }$ are disjoint. Since the geodesic curvature of $\gamma _1(t)$ at $Q$ is greater than $k_0$, the subarc $[QP]$ of $\gamma _1$ lies on the domain $D$ bounded by $\mathfrak{b }$ and $[AQ)$ (see [1, Fig. 20], right). Since $P$ lies in $\overline{D}$ and $[BP)$ never meets $[AP)$, there is a point $R$ on $[BP)\cap \mathfrak{b }$. On the other hand, since $[BP) \subset \overline{\Delta _{C_{B}}}$, we have that

$$\begin{aligned} R\in \mathfrak{a }, \end{aligned}$$

which contradicts that $\mathfrak{a }$ and $\mathfrak{b }$ are disjoint.

Remark 2.1

In [1, Figure 9], we showed the two possibilities of the behavior of the curve $\gamma _1$. However, the case as in Fig.  4 was missing in [1, Figure 9] as the third possibility. Fortunately, this is just the case of $\Theta _C$ as in [1, Figure 11], the statement and the proof of [1, Corollary 3.11] themselves do not need any corrections.

Fig. 4

A figure which should added in [1, Figure 9]

4 The proof of Proposition 4.7

For the correction of the proof of Proposition 4.7, we also need the following renewal of the arguments given in [1, Page 131 Line 15–Page 133 Line 21]: We set $\rho :=\delta _\mu ^*$. Let $\diamond _{OAPB}$ be the $\rho$-diamond as in Section 2, and $C$ the circle of radius $\rho$ centered at $O$. By [1, Corollary 3.7], there exists a negative shell $S_2$ on $\gamma _2$. We place $S_2$ so that it lies inside of the circle C of radius $\rho$, and the node of $S_2$ is tangent to the arc $[BP)$ at $B$ (see Fig. 5, left). By Proposition 1.1, $S_2$ actually lies inside of the circle $C$. We slide the shell $S_2$ along the arc $[BP)$. We denote by $S_2(Y)$ this sliding shell whose node lies at $Y$ for each $Y\in [BP)$. When $Y=B$, $S_2(Y)$ coincides with $S_2$.

Fig. 5

The sliding operation of $S_2$

Consider the baroon-like closed domain $\bar{\Omega }$ bounded by three arcs $[AP],\, [BP]$ and $AB$, where $AB$ is the major arc on $C$ bounded by $A,B$. By Proposition 1.1, the shell $S_2$ lies in $\bar{\Omega }$ (cf. Fig. 5, left). Let $Z$ be the first point at which $S_2(Y)\setminus \{Y\}$ meets the boundary of $\bar{\Omega }$. If $Z$ lies on the circle $C$, then whole of $S_2(Y)\setminus \{Z\}$ lies in $\Delta _C$ by Proposition 1.1, which contradicts the fact that $Z \ne Y$ and $Y\in [BP)$. So $Z$ lies in $[AP)$ or $[BP)$. If $Z$ lies in $[AP)$, then $(Y,Z)$ gives an admissible bi-tangent (cf. [1, Definition 2.1.] and Fig. 5, right), which proves Proposition 4.7. So it is sufficient to make a contradiction if $Z$ lies in $[PB)$.

We first consider the case $Z \in (YP]$. By Proposition 1.1, there exists a circle $\Gamma$ of radius $\rho$ which is tangent to $[BP)$ at $Y$ such that $S_2(Y)\setminus \{Y\}$ lies in $\Delta _{\Gamma }$. In particular, $Z\in \Delta _{\Gamma }$. Since $(YP]$ lies in the interior $\Delta _{C_Y}$ of the circle $C_Y$ by the condition (${v}^{\prime }$) of diamonds, we can conclude that $\Delta _{C_Y}\cap \Delta _{\Gamma }$ is an empty set. Thus $Z\not \in \Delta _{C_Y}$. However, this contradicts the fact that $(YP]$ lies in $\Delta _{C_Y}$.

We next consider the other case $Z \in [BY)$. By Proposition 1.1, there exists a circle $\Gamma '$ of radius $\rho$ which is tangent to $[BP)$ at $Z$ such that $S_2(Y)\setminus \{Z\}$ lies in the domain $\Delta _{\Gamma '}$. In particular, $Y\in \Delta _{\Gamma '}$. Since $(ZP]$ is contained in $\Delta _{C_Z}$ by the condition (${v}^{\prime }$) of diamonds, $\Delta _{C_Z}\cap \Delta _{\Gamma '}$ is an empty set. Thus $Y\not \in \Delta _{C_Z}$, which contradicts the fact that $(ZP]$ lies in $\Delta _{C_Z}$.

All of the other arguments in [1] are correct without any modifications.