---
abstract: 'The classical H surfaces of H. A. Schwarz form a 1-parameter family of triply periodic minimal surfaces (TPMS) that are usually described as close relatives to his more famous P surface.  However, a crucial distinction between these surfaces is that the P surface belongs to a 5-dimensional smooth family of embedded TPMS of genus three discovered by W. Meeks, while the H surfaces are among the few known examples outside this family.  We construct a 2-parameter family of embedded TPMS of genus three that contains the H family and meets the Meeks family.  In particular, we prove that H surfaces can be deformed continuously within the space of TPMS of genus three into Meeks surfaces.' 
address:
- 'Georg-August-Universität Göttingen, Institut für Numerische und Angewandte Mathematik'
- 'Indiana University, Department of Mathematics'
author:
- Hao Chen
- Matthias Weber
bibliography:
- 'References.bib'
title: |
    An orthorhombic deformation family\
    of Schwarz’ H surfaces
---

[^1]

Introduction
============

This is the second of two papers dealing with new 2-dimensional families of embedded triply periodic minimal surfaces (TPMS) of genus three whose 1-dimensional “intersection” with the well-known Meeks family exhibits singularities in the moduli space of TPMS. 

Among the many TPMS discovered by H. A. Schwarz [@schwarz1890] is a 1-parameter family $\H$ that can be constructed by extending the Plateau solution for the boundaries of the two triangular faces of a triangular prism.  Such a Plateau solution does not exist for all heights of the prism.  For small heights, there are two distinct solutions.  One of them limits in the most symmetric singly periodic Scherk surfaces with 6 annular ends.  The other degenerate to a foliation of ${\mathbb{R}}^3$ by horizontal parallel planes that are joined by catenoidal nodes, placed in a hexagonal lattice. 

This family is remarkable because it does not belong to the 5-dimensional Meeks family ${\mathcal{M}}$ of TPMS of genus 3 [@meeks1990].  Members of that family have the eight branched values of the Gauss map forming four antipodal pairs, while for an $\H$ surface, they are located at the north and south poles of the 2-sphere and the six vertices of a triangular prism.  The only other known TPMS of genus 3 outside ${\mathcal{M}}$ are the Gyroid-Lidinoid family [@schoen1970; @lidin1990; @fogden1993; @fogden1999; @weyhaupt2006; @weyhaupt2008], and the recently discovered $\tDD$ family [@chenweber1]. 

We will exhibit a 2-parameter family $\oH$ of embedded TPMS of genus $3$ that can be understood as an orthorhombic deformation family of Schwarz’ $\H$ surfaces.  The closure of this family has a 1-dimensional intersection with the $\oP$ family, a classical 2-parameter orthorhombic deformation family of Schwarz $\P$ surface.  However, surfaces in $\oH$ are not in ${\mathcal{M}}$. This is, after the $\oDD$ family in [@chenweber1], another 2-parameter non-Meeks family of TPMS of genus 3. 

![ Fundamental Piece and Translational Fundamental Piece[]{data-label="fig:fundpieces"}](figures/fundpiece-1.pdf "fig:"){width="48.00000%"} ![ Fundamental Piece and Translational Fundamental Piece[]{data-label="fig:fundpieces"}](figures/transpiece.png "fig:"){width="48.00000%"}

Consider an embedded minimal surfaces $S$ inside an axes parallel box $[-A,A]\times[-B,B]\times[-1,1]$ such that

-   $S$ satisfies free boundary condition on the vertical planes $x=\pm A$ and $y=\pm B$, and fixed (Plateau) boundary condition on the horizontal segments $\{(x,0,\pm 1) \mid -A\le x \le A\}$.

-   $S$ intersects the edges of the box in eight vertices, but disjoint from the vertical lines with $(x,y)=\pm(+A,+B)$. Hence, apart from the four ends of the fixed boundaries, $S$ intersects the vertical lines $(x,y)=\pm(+A,-B)$ in two vertices each. 

-   $S$ is symmetric under the inversion in the origin. 

Therefore $S$ is a right-angled minimal octagon, with its inversion center at the origin.  See Figure \[fig:fundpieces\] (left) for an example.  If the vertices are labeled as in this figure, then the fixed boundaries are the segments $V_8V_1$ and $V_4V_5$.

Because the two horizontal segments are in the middle of the top and bottom faces of the box, rotations about them and reflections in the lateral faces of the box extend $S$ to a TPMS $\tilde \Sigma$. More specifically, $\tilde
\Sigma$ is invariant under the lattice $\Lambda$ spanned by $(A,0,1)$, $(-A,0,1)$ and $(0,2B,0)$. In the 3-torus ${\mathbb{R}}^3 / \Lambda$, $\tilde \Sigma /
\Lambda$ is a compact surface of genus $3$. In Figure \[fig:fundpieces\] (right) we show part of $\tilde \Sigma$ consisting of eight copies of $S$.

Such a surface $\tilde \Sigma$ must be embedded: As the genus is 3, the degree of the Gauss map is 2, the only points with vertical normal in the octagon are the vertices $V_2$, $V_3$, $V_6$ and $V_7$. Thus the octagon boundary is graphical over an embedded curve in the rectangle $[-A,A]\times [-B,B]$, and so is the unique Plateau solution for that boundary. 

For crystallographers, the orthorhombic lattice spanned by $(2A,0,0)$, $(0,2B,0)$ and $(0,0,2)$ is more convenient.  This is responsible for the letter “o” in our naming.  The part shown in Figure \[fig:fundpieces\] is actually a translational fundamental domain of the orthorhombic lattice. 

![ Surfaces in $\oP$ and $\H$ near catenoidal limits[]{data-label="fig:nodes"}](figures/oPnodes.png "fig:"){width="48.00000%"} ![ Surfaces in $\oP$ and $\H$ near catenoidal limits[]{data-label="fig:nodes"}](figures/Hnodes.png "fig:"){width="48.00000%"}

We use $\cO$ to denote the set of all TPMS obtained in this way.  Two classical families of surfaces in $\cO$ were already known to Schwarz [@schwarz1890]. 

Surfaces in the first family have an additional reflectional symmetry in the plane $z=0$. Then, because of the inversional symmetry in the origin, these surfaces must also contain the $z$-axis, which serves as the axis of an order-2 rotational symmetry.  This $2$-parameter family contains Schwarz’ $P$ surface and belongs to the Meeks’ family ${\mathcal{M}}$ [@meeks1990].  It is known as $\oP
b$ in the literature to distinguish from another orthorhombic deformation family $\oP a$; see [@fischer1989; @fogden1992].  In this paper, we simply write $\oP$ in place of $\oP b$. An example of $\oP$ with small $B$ is shown in Figure \[fig:nodes\] (left). 

Surfaces in the second family has an additional order-3 rotational symmetry about a line in the $y$-direction.  The rotational axis necessarily passes through an end of fixed boundary.  This $1$-parameter family is Schwarz’ $\H$ family.  An example, again with small $B$, is shown in Figure \[fig:nodes\] (right). 

The main purpose of this paper is to establish the existence of a new 2-parameter family described in the following theorem, and study its properties. 

\[thm:main\] There exists a 2-parameter family $\oH$ in $\cO$ that contains Schwarz’ $\H$ surfaces as a subfamily.  Surfaces in $\oH$ do not belong to the Meeks family.  That is, the branched values of the Gauss map do not form four antipodal pairs.  In fact, the only Meeks surfaces in $\cO$ are the $\oP$ surfaces.  However, the closure of $\oH$ intersects $\oP$ in a $1$-parameter family of TPMS. 

The intersection $\overline\oH \cap \oP$ will be explicitly described in terms of elliptic integrals. 

According to [@traizet2015], the $\oH$ and $\oP$ surfaces correspond to solutions to the hollow vortex problem in fluid mechanics. 

We now provide the motivation that leads to the discovery of $\oH$.

By opening nodes among 2-tori, Traizet [@traizet2008] constructed TPMS that looks like horizontal planes connected by catenoidal necks.  In the degenerate limit, the catenoidal necks become nodes whose positions have to satisfy a balancing equation, formulated in terms of elliptic functions, and a non-degeneracy condition. 

For surfaces of genus 3, one needs to open two nodes between two tori.  In the limit, it degenerates to a two-sheeted torus with two singular points.  Let $(T_1, T_2) \in {\mathbb{C}}^2$ be vectors that span the limit torus, and write $T_3 =
-T_1-T_2$. Assume that limit positions of the two nodes are $p_1$ and $p_2$, respectively.  Up to a translation, we may assume that $p_1 = 0$. Write $p_2 =
x T_1 + y T_2$ with $(x,y) \in [0,1]^2$. Then $p_1, p_2$ form a balance configuration if (see [@traizet2008 §4.3.3]) $$\label{eq:balanced}
    \zeta(p_2) = x \eta_1 + y \eta_2 \ ,$$ where $\zeta$ is Weierstrass Zeta function, which is quasi-periodic in the sense that $\zeta(z+T_i)-\zeta(z) = \eta_i = 2 \zeta(T_i/2)$, $i=1,2,3$. Traizet proved that, if such a balance configuration is non-degenerate, then there exists a family of triply periodic minimal surfaces that limits in this configuration. 

Because $\eta_1+\eta_2+\eta_3=0$, one such balanced configuration is given by $x=y=1/2$ for any $T_1, T_2$. In other words, if $p_2$ is a 2-division point, the balance equation is automatically solved for any torus.  In particular, when $|T_1|=|T_2|$, this simple configuration is the Traizet limit of the $\oP$ family. 

Another special case is the rhombic $60^\circ$ torus.  The solution set $(x,y)$ for the real and imaginary parts of are show in Figure \[fig:sol60\].  The intersection at $x=y=1/2$ regenerates to a Traizet limit of $\oP$ surfaces (in Meeks).  The hexagonal symmetry implies other intersections at $x=y=1/3$ and $x=y=2/3$, which is the Traizet limit of the $\H$ family (non-Meeks).  See [@chen2018] for an analytic proof of the balancedness at the non-trivial intersections. 

![ Solutions to the real and imaginary parts of the balance equation on the rhombic $60^\circ$ torus.[]{data-label="fig:sol60"}](figures/sol60.pdf){width="40.00000%"}

More generally, we consider on rhombic tori spanned by $T_{1,2}=\exp(\pm i
\theta/2)$ the balance configurations with $x=y$. The symmetry suggests that these are Traizet limit of $\cO$ surfaces with $B \to 0$. To see this, just rotate the configuration to place $T_3$ in the $z$-direction and open nodes in the $x$-direction.  The balance equation for such a configuration is given by $$\label{eq:traizet1}
    x \zeta(T_3/2) = \zeta(x T_3)/2$$ Our choice of conjugate vectors $T_1, T_2$ guarantees real values on both sides.  The solution set to this balanced equation is shown in Figure \[fig:diagsols\].  The vertical line $x=1/2$ is the trivial locus, giving the Traizet limit of the $\oP$ surfaces. 

But we also see a second, non-trivial locus, which is the motivation of the current project.  This “exotic” locus has been noticed independently by both authors, and probably by many other people in the minimal surface community.  We will see that balanced configurations on this locus are all non-degenerate, so they are indeed Traizet limits.  In fact, they are the Traizet limits of the $\oH$ family.  Our discovery of $\oH$ is actually the result of an attempt to push TPMS away from these Traizet limits. 

The two loci of Traizet limits intersect at $x=1/2$ and $\theta=\theta^*\approx
1.23409 \approx 70.7083^\circ$. The balance configuration at the intersection is degenerate. 

The torus ${\mathbb{T}}^*$ at the intersection is of particular significance.  It is the only rhombic torus on which there exists a meromorphic 1-form with double order pole at $0$ and double order $0$ at $T_3$ and only real periods.  This was exploited for the construction of translation invariant helicoids with handles [@hoffman1999; @weber2009].  This meromorphic 1-form can be constructed geometrically as follows (see Figure \[fig:slit70\]): Take the complex plane, and slit it along the interval $[-1,1]$ on the real axis.  Then identify the top (resp.  bottom) edge of $[-1,0]$ with the bottom (resp.  top) edge of $[0,1]$. The result is a torus carrying a cone metric with two cone points, of cone angle $6\pi$ at the point identified with $\{-1,0,1\}$, and of cone angle $-2\pi$ at $\infty$. The corresponding 1-form has thus a double order pole at $0$, and a doubly order zero at $\infty$. Its periods are obviously real, and the symmetry of the slit ensures that the torus is rhombic.  The same torus with flat metric is nothing but ${\mathbb{T}}^*$.

We will revisit the Traizet limit in the framework of our parametrization of $\oH$ in section \[sec:traizet\].  We first prove that the non-trivial locus is non-degenerate, and unique in the sense that for every $0<\theta<\theta^*$, has a unique solution $0 < x < 1/2$. Then will be reformulated in terms of elliptic integrals, leading to an explicit formula for the non-trivial locus.  We will also recover $\theta^*$, not only as the end point of the Traizet limit of $\oH$, but also as the Traizet limit of the intersection $\overline\oH \cap \oP$.

![ Solution set $(x,\theta)$ to the balance equation  on the diagonal of rhombic tori.[]{data-label="fig:diagsols"}](figures/diagsols.pdf){width="40.00000%"}

![ Model for the torus ${\mathbb{T}}^*$ []{data-label="fig:slit70"}](figures/slit70.pdf){width="40.00000%"}

Our paper is organized as follows:

In Section \[sec:period\], we describe the Weierstrass data for surfaces in $\cO$ and formulate the period problem, depending on three real positive parameters $a, b$ and $t$. The case $a=b$ corresponds to the $\oP$ surfaces, where the period problem is automatically solved.  Away from $a=b$, the period problem becomes 1-dimensional. 

In Section \[sec:gauss\] we show that, if $a \ne b$, the branched values of the Gauss map can *not* be antipodal.  This proves that $\cO \cap {\mathcal{M}}=
\oP$, and that any solution with $a \ne b$ (namely $\oH$) lies outside ${\mathcal{M}}$.

Section \[sec:exist\] is dedicated to the existence proof of $\oH$. We show that for any choice of $a \ne b$, there is a value of $t$ that solves the period problem.  This is accomplished through a careful asymptotic analysis of the period integrals.  We also conjecture the uniqueness of $t$ based on numerical experiments. 

To prove that $\oP$ and the closure of $\oH$ have a non-empty intersection, we consider in Section \[sec:elliptic\] a modified period problem that eliminates the trivial solutions coming from $\oP$. It turns out that this period problem can be solved explicitly in terms of elliptic integrals. 

In section \[sec:traizet\] we study the Traizet limit of $\oH$. In particular, the loci of  will receive another explicit description in terms of elliptic integrals, and the intersection of the loci will be recovered in two different ways.  We also locate the Traizet limit of $\H$ family on the locus.  It is then possible to find a continuous deformation path within the space of TPMS of genus three, starting from an $\H$ surface and ending with an $\oP$ surface, that passes through a sufficiently small neighborhood of the Traizet limit. 

Despite different appearances, motivations and focus points, our parametrization of $\oH$, as well as many computations, share similarities with our previous work on $\oDD$ [@chenweber1].  So we will, whenever appropriate, refer the readers to [@chenweber1] for details.  We also omit technical details in Sections \[sec:elliptic\] and \[sec:traizet\], where integral tables in [@byrd1971] are used for the computations involving elliptic integrals. 

Weierstrass Data and the Period Problem {#sec:period}
=======================================

We parameterise a surface in $\cO$ using a Weierstrass representation defined on the upper half plane such that the real axis is mapped to the boundary of the octagon $S$. Let the vertices of $S$ be labeled by $V_1, V_2, \cdots, V_8$ as in Figure \[fig:fundpieces\] (left).  Denote the preimage of $V_k$ by $v_k
\in {\mathbb{R}}$, and assume that $v_1<v_2<\ldots<v_8$.

![ Images of a fundamental piece under $\Phi_1$ and $\Phi_2$.[]{data-label="fig:phidomains"}](figures/phidomains.pdf){width="80.00000%"}

Given an $\cO$ surface, denote by $dh$ its height differential and by $G$ its Gauss map.  Let $\phi_1:=dh \cdot G$ and $\phi_2:=dh / G$. The assumed boundary symmetries of the surface imply that $\Phi_j: z \mapsto \int^z \phi_j$ ($j=1$ or $2$) map the upper half plane to “right angled” Euclidean octagons.  The interior angle is $270^\circ$ at $\Phi_1(v_2)$, $\Phi_1(v_6)$ and $\Phi_2(v_3)$, $\Phi_2(v_7)$. Indeed, the Gauss map is vertical at $V_2$, $V_3$, $V_6$ and $V_7$, hence these vertices are the poles and the zeros of $G$. Interior angles at all other vertices are $90^\circ$; see Figure \[fig:phidomains\]. 

We may assume that the inversion is represented by the transform $z \mapsto
-1/z$, hence the inversion center of the minimal octagon at the origin is represented by $i$ in the upper half plane.  Then we assume the eight points $v_i$ to be $-t < -a < -1/b < -1/t < 1/t < 1/a < b < t$ for $t>1$.

Such maps are given by Schwarz-Christoffel maps.  More specifically, we have $$\begin{aligned}
    \phi_1 :=  - \rho \, & \frac{ (z+a)^{+1/2} (z+1/b)^{-1/2}  (z-1/a)^{+1/2}(z-b)^{-1/2}}
    {(z+t)^{1/2} (z+1/t)^{1/2}(z-1/t)^{1/2}(z-t)^{1/2}}\, dz,\\
    \phi_2 := \frac{1}{\rho} & \frac{ (z+a)^{-1/2} (z+1/b)^{+1/2} (z-1/a)^{-1/2}(z-b)^{+1/2}}
    {(z+t)^{1/2} (z+1/t)^{1/2}(z-1/t)^{1/2}(z-t)^{1/2}}\, dz,\\
    dh := & \frac{i} {(z+t)^{1/2} (z+1/t)^{1/2}(z-1/t)^{1/2}(z-t)^{1/2}}\, dz \ .\end{aligned}$$ Here, the real positive Lop[é]{}z-Ros factor $\rho$ determines the scaling of the image domains.  The Gauss map is given by $$G(z)=\rho i (z-1/a)^{+1/2}(z+a)^{+1/2} (z+1/b)^{-1/2}(z-b)^{-1/2}\ .$$

Up to congruence and dilation, the image of the upper half plane under the map $$\label{eq:weierstrass}
        z \mapsto {\operatorname{Re}}\int^z (\omega_1, \omega_2, \omega_3) = {\operatorname{Re}}\int^z \left(
        \frac{1}{2}(\phi_2-\phi_1), \frac{i}{2}(\phi_2+\phi_1), dh \right)$$ is *almost* the fundamental octagon of an $\cO$ surface in the following sense:

-   The intervals $v_8v_1$ and $v_4v_5$ are mapped to straight segments in the $x$-direction, but not necessarily in the $y=0$ plane. 

-   The other intervals are mapped to planar symmetry curves in vertical planes.  More specifically:

    -   the interval $v_1v_2$ (resp.  $v_5v_6$) is mapped into the plane $x=+A$ (resp.  $-A$);

    -   the interval $v_2v_3$ (resp.  $v_6v_7$) is mapped into the plane $y=-B$ (resp.  $+B$);

    -   the interval $v_3v_4$ (resp.  $v_7v_8$) is mapped into the plane $x=+A'$ (resp.  $-A'$). 

-   The image is symmetric under the inversion in the image of $i$.

The proof is a straightforward modification of the proof in [@chenweber1]. 

For such a minimal octagon to lie in $\cO$, we must have have $A=A'$ so that the curves $V_1V_2$ and $V_3V_4$ are coplanar, hence the image of  is contained in an axis parallel box centered at the origin.  Moreover, $V_8V_1$ and $V_4V_5$ must lie in the middle of, respectively, the top and bottom faces of the box.  We now express these conditions in terms of the periods of $\phi_1$ and $\phi_2$. To this end, we introduce notations for the edge lengths of the Euclidean octagons $$I_k := \left| \int_{v_k}^{v_{k+1}} \phi_1 \right|, \qquad
    J_k := \left| \int_{v_k}^{v_{k+1}} \phi_2 \right|$$ for $1 \le k \le 7$. These are positive real numbers that depend analytically on the parameters $a, b, t$ and $\rho$. Note that by the inversional symmetry, we have $$\label{eq:symmetry}
    I_k=I_{k+4} \quad\text{and}\quad J_k=J_{k+4}$$ for $1 \le k \le 3$.

The image of the upper half plane under the Weierstrass representation  is the fundamental octagon of a surface in $\cO$ if and only if the following period conditions are satisfied: $$\begin{aligned} \label{eq:periodcond}
            I_1+I_3 &= J_1+J_3 \ , \\
            I_2&= J_2 \ . 
        \end{aligned}$$

The curves $V_1V_2$ and $V_3V_4$ are coplanar if and only if $${\operatorname{Re}}\int_{v_2}^{v_3} \omega_1 = 0 \ .$$ This is equivalent to $${\operatorname{Re}}\int_{v_2}^{v_3} (\phi_1 - \phi_2) = 0.$$ Observe that on $v_2v_3$, the integrands of $\phi_1$ and $\phi_2$ are both negative real.  So the equation above can be written as $I_2 = J_2$, which is the second period condition. 

The top segment $V_8V_1$ lies in the middle of the top face if and only if $${\operatorname{Re}}\int_{v_1}^{v_2} \omega_2 = {\operatorname{Re}}\int_{v_7}^{v_8} \omega_2 \ .$$ This is equivalent to $${\operatorname{Im}}\int_{v_1}^{v_2} (\phi_1+\phi_2) = {\operatorname{Im}}\int_{v_7}^{v_8} (\phi_1+\phi_2) \ .$$ Observe on $v_1v_2$ that the integrand in $\phi_1$ (resp.  $\phi_2$) is negative (resp.  positive) imaginary, and on $v_7v_8$ that the integrand in $\phi_1$ (resp.  $\phi_2$) is positive (resp.  negative) imaginary.  So the equation above can be written as $$I_1 - J_1 = J_7 -I_7 = J_3 - I_3,$$ where the second equation follows from the symmetry .  This proves the first period condition. 

If the period conditions are satisfied, then by the inversional symmetry, the curves $V_5V_6$ and $V_7V_8$ must be coplanar, and the segment $V_4V_5$ must lie in the middle of the bottom box. 

We can eliminate $\rho$ by taking the quotient of the two equations, therefore:

\[cor:periodcondition\] If $$Q_I := \frac{I_1+I_3}{I_2} = \frac{J_1+J_3}{J_2} =: Q_J$$ or, equivalently, if $$\label{eq:quotient}
        Q := Q_I - Q_J = \frac{I_1+I_3}{I_2} - \frac{J_1+J_3}{J_2} = 0$$ for some choice of $a,b,t$, then $\rho \in {\mathbb{R}}_{>0}$ can be uniquely adjusted so that the period conditions  are satisfied. 

Thus we have expressed the period condition as a single equation $Q=0$, where $Q$ depends on three parameters $a,b,t$.

We note that when $a=b$, the period conditions  are satisfied automatically with $\rho=1$. In this case, the involution $z \mapsto
-\overline{z}$ induces an order-2 rotation of the surface in a vertical axis.  This can be seen by noting that $\omega_1$ and $\omega_2$ change sign but $\omega_3$ keeps sign under the involution.  Indeed, on the imaginary axis which is fixed by the involution, $\phi_1$ and $\phi_2$ are conjugate and $dh$ is real.  Hence the positive imaginary axis is mapped by the Weierstrass representation  to the vertical straight segment between the middle points of $V_4V_5$ and of $V_8V_1$, which serves as the axis of the order-2 rotation.  This shows that the surface is in $\oP$.

To simplify our computations in the following sections, we employ the substitution $\zeta=z-1/z$, which is monotone on the positive real axis.  We also replace $a-1/a$ by $\alpha$, $b-1/b$ by $\beta$, and $t-1/t$ by $\tau$ so that $-\tau < -\alpha < \beta < \tau$. Then the 1-forms $\phi_1$ and $\phi_2$ become $$\begin{aligned}
    \varphi_1 = {}& -\rho (\zeta + \alpha)^{+1/2} (\zeta - \beta)^{-1/2} (\zeta^2 - \tau^2)^{-1/2} (\zeta^2 + 4)^{-1/2}\, d\zeta \ ,\\
    \varphi_2 = {}& \frac{1}{\rho} (\zeta +\alpha)^{-1/2} (\zeta -\beta)^{+1/2} (\zeta^2 - \tau^2)^{-1/2} (\zeta^2 + 4)^{-1/2}\, d\zeta \ ,\end{aligned}$$ and the Gauss map is simplified to $$\label{eq:altGauss}
    G(\zeta)=\rho i (\zeta + \alpha)^{+1/2} (\zeta-\beta)^{-1/2}.$$

In the rest of the paper, the original parametrization is understood whenever Latin letters $a,b,t,z$ are used, and the simplified parametrization is understood whenever Greek letters $\alpha,\beta,\tau,\zeta$ are used.  This should not cause any confusion. 

Branched Values of the Gauss Map {#sec:gauss}
================================

In this section, we will show that the branched values of the Gauss map are never antipodal with $a \ne b$. As a consequence, the only surfaces in $\cO$ that belong to the Meeks family ${\mathcal{M}}$ are the surfaces in $\oP$. The arguments don’t require the period condition to be satisfied and are purely algebraic. 

The branched values of the Gauss map of a surface in $\cO$ are antipodal if and only if $a=b$.

We begin by locating the branched points of the Gauss map in the fundamental octagon.  By a result of Meeks [@meeks1990], the branched points of a TPMS of genus 3 are precisely the inversion centers of the surface.  They are situated, in the fundamental octagon, at the center of the octagon and at the end points of the fixed boundary segments. 

The octagon center corresponds to $i$ in the upper half plane, so that $G(i)$ is a branched value.  Three more branched points and values are obtained after extending the octagon by reflections.  We then have four branched values, namely $\pm G(i)$ and $\pm \overline{G(i)}$. Their stereographic images on the 2-sphere lie at the vertices of a horizontal rectangle, symmetric in the planes $x=0$ and $y=0$.

The end points of the fixed boundary segments correspond to $\pm t$ and $\pm
    1/t$ in the parameter domain.  Because of the inversional symmetry, they provide only two branched values $G(t)$ and $G(-t)$. These both lie on the positive imaginary axis, and their stereographic images on the 2-sphere lie on the upper half-circle with $y>0$ and $x=0$. Extending the octagon by reflections gives two more branched values at $-G(t)$ and $-G(-t)$, whose stereographic images lie on the lower half-circle.  The stereographic images of these four branched values then form a quadrilateral in the plane $x=0$ symmetric to the plane $y=0$.

We show an example for the location of the eight branched values in Figure \[fig:gaussbranched\]. 

![ Branched values of the Gauss map[]{data-label="fig:gaussbranched"}](figures/gaussbranched.pdf){width="40.00000%"}

In order that these eight branched values occur in antipodal pairs, the first quadrilateral must lie in the plane $z=0$, while the second rectangle must be a rectangle.  These conditions mean, in terms of the Gauss map, that $|G(i)|=1$ and $G(t)G(-t)=-1$. We then compute from  that $$\rho^2 \sqrt\frac{\alpha^2+4}{\beta^2+4} = 1\quad\text{and}\quad
        \rho^2 \sqrt\frac{\tau^2-\alpha^2}{\tau^2-\beta^2} = 1,$$ which forces $\alpha = \pm\beta$, hence $a=b$ under the constraint $1/t < 1/a
    < b < t$.

The reader might be curious about the parameter values for the Schwarz $\H$ surfaces within this representation.  These are difficult to determine explicitly.  But we know that, among the eight branched values of the Gauss map of an $\H$ surface, there is one and only one antipodal pair.  This implies either $G(t)=G(-1/t)=i$ or $G(-t)=G(1/t)=i$. We then obtain from  a necessary condition of the parameters for $\H$, namely $$\rho^2 \frac{\tau + \alpha}{\tau - \beta} = 1 \quad\text{or}\quad \rho^2 \frac{\tau - \alpha}{\tau + \beta} = 1.$$ In view of Conjecture \[conj:unique\] in the next section, we believe that this condition is also sufficient. 

We see from  that, for any reals $\alpha,\beta,\rho$ with $-\alpha < \beta$ and $\rho > 0$, there is a unique $\zeta^* \in
(\infty, -\alpha) \cup (\beta, \infty) \cup \infty$ such that $G(\zeta^*) = i$. In other words, there must be a point on the boundary of the octagon, namely the image of $\zeta^*$ under , where the normal vector points in the $y$ direction.  For the $\oP$ family, $\rho = 1$ and $\alpha =
\beta$, hence $\zeta^* = \infty$. For the $\H$ family, our calculation above shows that $\zeta^* = \pm \tau$. So $\cO$ is divided in two parts, depending on the image of $\zeta^*$ being on the fixed boundary (as $\oP$) or on the free boundary of the octagon; Schwarz $\H$ family lies on the interface. 

Using the order-$3$ rotational symmetry of the $\H$ surfaces, a computer algebra system gives the explicit expressions $$\begin{aligned}
        a &= \frac{t^3-15 t+15 t^{-1} - t^{-3} +\left(t+t^{-1}\right) \sqrt{t^4-60 t^2+134 -60 t^{-2}+t^{-4}}}{2 \left(7 t^2-10 - t^{-2}\right)} \\
        b &= \frac{t^3-15 t+15 t^{-1} - t^{-3} +\left(t+t^{-1}\right) \sqrt{t^4-60 t^2+134 -60 t^{-2}+t^{-4}}}{2 \left(t^2+10-7t^{-2}\right)} 
    \end{aligned}$$ for the parameters $a$ and $b$ in terms of $t$. Then the period problem seems automatically solved, at least numerically. 

Existence of Non-Trivial Solutions {#sec:exist}
==================================

Recall that $0 < 1/t < 1/a < b < t$, and the periodic condition  as we copy below $$Q(a,b;t) = \frac{I_1+I_3}{I_2} - \frac{J_1+J_3}{J_2} = 0.$$ The quantity $Q$ is our focus in the remaining sections of this paper.  From now on, we will ignore the Lopéz-Ros factor $\rho$ in our calculations, since $Q$ is independent of this factor. 

We now prove the main theorem of this paper. 

\[thm:exist\] If $a = b$, the period condition  is solved for any choice of $t$.

If $a < b$, then there exists a value of $t$ that solves the period condition . 

The case $a = b$ has been discussed in Section \[sec:period\].  The case $a <
b$, as well as the existence of $\oH$, follows from the continuity of $Q$ in $t$, and the following proposition. 

\[prop:lim1\] If $1/t<1/a<b<t$ and $a<b$ then $$\begin{aligned}
        \lim_{t \to b+} Q(a,b;t) &< 0 \ ,\label{eq:ttob}\\
        \lim_{t \to +\infty} Q(a,b;t) &= +\infty \ .\label{eq:ttoi}
    \end{aligned}$$

The remainder of this section is devoted to the proof of this proposition. 

The argument in [@chenweber1] applies with slight modification.  As $t
    \to b+$, all periods have finite positive limits, with the exceptions $\lim_{t \to b+} J_3=0$ and $\lim_{t \to b+} I_2$ diverges to $+\infty$. Thus $$\lim_{t\to b+} \frac{I_1+I_3}{I_2} = 0 \quad\text{and}\quad \lim_{t\to b+} \frac{J_1+J_3}{J_2} >0 \ ,$$ and follows. 

The proof is similar to the argument in [@chenweber1].  But it is much easier to work with the simplified 1-forms $\varphi_1$ and $\varphi_2$. Recall that the substitution $\zeta=z-1/z$ is monotone increasing for $z>0$, and write $\alpha=a-1/a$, $\beta=b-1/b$, $\tau=t-1/t$ as before. 

For the periods in the denominators, we first note that $$\begin{aligned}
        \lim_{\tau \to \infty} \tau \cdot I_2(\alpha,\beta;\tau) &= \int_{-\alpha}^{\beta} \frac{1}{\sqrt{\zeta^2+4}} \sqrt\frac{\alpha+\zeta}{\beta-\zeta}\, d\zeta,\\
        \lim_{\tau \to \infty} \tau \cdot J_2(\alpha,\beta;\tau) &= \int_{-\alpha}^{\beta} \frac{1}{\sqrt{\zeta^2+4}} \sqrt\frac{\beta-\zeta}{\alpha+\zeta}\, d\zeta
    \end{aligned}$$ are all bounded.  Their difference $$\begin{aligned}
        \lim_{\tau \to \infty} \tau \cdot (I_2-J_2) &= \int_{-\alpha}^{\beta} \frac{2\zeta+\alpha-\beta}{\sqrt{(\zeta^2+4)(\alpha+\zeta)(\beta-\zeta)}}\, d\zeta\\
        &= \int_{-\gamma}^{\gamma} \frac{2\xi\, d\xi}{\sqrt{((\xi-\alpha/2+\beta/2)^2+4)(\gamma^2-\xi^2)}}\\
        &= \int_{0}^{\gamma} \frac{2\xi\, d\xi}{\sqrt{\gamma^2-\xi^2}}\Big(\frac{1}{\sqrt{(\xi-\alpha/2+\beta/2)^2+4}}-\frac{1}{\sqrt{(\xi+\alpha/2-\beta/2)^2+4}}\Big)\ ,
    \end{aligned}$$ where $\gamma = (\alpha+\beta)/2$ and $\xi = \zeta - (\beta - \alpha)/2$, is negative when $\alpha < \beta$. Hence we have $$\label{eq:denominator}
        \lim_{\tau \to \infty} \tau I_2 < \lim_{\tau \to \infty} \tau J_2$$ for all $0<\alpha<\beta$.

The periods in the numerators have logarithmic asymptotics.  For instance, as $\tau \to \infty$, $$\begin{aligned}
        \tau \cdot J_3(\alpha,\beta;\tau)
        &= \int_\beta^\tau \frac{\tau}{\sqrt{\zeta^2+4}\sqrt{\tau^2-\zeta^2}} \sqrt\frac{\zeta-\beta}{\zeta+\alpha}\, dz \\
        &\sim \int_\beta^\tau \frac{\tau}{\zeta\sqrt{\tau^2-\zeta^2}} \, d\zeta\\
        &\sim \log t,
    \end{aligned}$$ hence $\tau \cdot I_3(\alpha,\beta;\tau)$ diverges to $+\infty$ as $\tau \to
    \infty$. Fortunately, the integrals $I_1$ and $J_1$ (and $I_3$ and $J_3$) have the same logarithmic singularities.  By the dominant convergence theorem, we obtain the following limits: $$\label{eq:numerator}
        \begin{aligned} 
            \lim_{\tau\to\infty} \tau \cdot (I_1-J_1)
            &= -\lim_{\tau\to\infty} \int_{-\tau}^{-\alpha} \frac{\tau(\alpha+\beta)}{\sqrt{(\tau^2-\zeta^2)(\zeta^2+4)(\beta-\zeta)(-\alpha-\zeta)}}\, d\zeta\\
            &= -\int_{-\infty}^{-\alpha} \frac{\alpha+\beta}{\sqrt{\zeta^2+4}\sqrt{\beta-\zeta}\sqrt{-\alpha-\zeta}}\, d\zeta,\\
            \lim_{\tau\to\infty} \tau \cdot (I_3-J_3)
            &= \lim_{\tau\to\infty} \int_{\beta}^{\tau} \frac{\tau(\alpha+\beta)}{\sqrt{\tau^2-\zeta^2}\sqrt{\zeta^2+4}\sqrt{\zeta-\beta}\sqrt{\zeta+\alpha}}\, d\zeta\\
            &= \int_{\beta}^\infty \frac{\alpha+\beta}{\sqrt{\zeta^2+4}\sqrt{\zeta-\beta}\sqrt{\zeta+\alpha}}\, d\zeta. 
        \end{aligned}$$ Note that both integrals are finite and non-zero. 

Finally, we write $$Q(\alpha,\beta;\tau) = \frac{\tau (I_1 - J_1) + \tau (I_3 - J_3)}{\tau I_2 } + \tau (J_1 + J_3) \Big[\frac{1}{\tau I_2 } - \frac{1}{\tau J_2} \Big].$$ The part in the square bracket is positive by .  As $\tau\to\infty$, the first fraction is bounded by , and $\tau (J_1+J_3) \to +\infty$. This concludes the proof of the proposition. 

Before ending this section, we propose the following uniqueness conjecture based on numeric experiments:

\[conj:unique\] If $a < b$, then there exists a unique $t$ that solves the period condition . 

Intersection with the Meeks-Locus {#sec:elliptic}
=================================

We show in this section that $\overline\oH$ intersects $\oP$ in a 1-parameter family.  To make this precise, we use on $\cO$ the topology induced by the space of possible Weierstrass data, which are determined by the four real parameters $a, b, t$ and $\rho$. Clearly, the convergence of Weierstrass data implies the locally uniform convergence of the minimal surfaces. 

The goal is to explicitly determine the intersection of the Meeks locus $$\oP = \{(a,b,t): Q(a,b;t) = 0, a=b, 0<1/t<1/a<b<t\}$$ with the closure of the non-Meeks locus $$\oH = \{(a,b,t): Q(a,b;t) = 0, a \ne b, 0<1/t<1/a<b<t\}.$$ Without loss of generality, we will focus on the case $a<b$ hence $\alpha <
\beta$. The idea is to divide the function $Q(\alpha,\beta;\tau)$ by $\beta-\alpha$ and take the limit for $\alpha \to \beta-$ to eliminate solutions in the Meeks locus.  We claim:

\[thm:elliptic\] The intersection $\overline\oH \cap \oP$ is described by the equation $$\label{eq:intersection}
        {\bar{K}}(m_1) E(m_2) + m_2 {\bar{E}}(m_1) K(m_2) = {\bar{K}}(m_1) K(m_2),$$ where $$\begin{aligned}
    K(m) &= \int_0^{\pi/2} \frac{1}{\sqrt{1-m \sin^2(\theta)}}\, d\theta,  \\
    E(m) &= \int_0^{\pi/2} {\sqrt{1-m \sin^2(\theta)}}\, d\theta
    \end{aligned}$$ are complete elliptic integrals of the first and the second kind, ${\bar{K}}(m)=K(1-m)$ and ${\bar{E}}(m)=E(1-m)$ are the associated elliptic integrals, using the moduli $$m_1 =\frac{\alpha^2+4}{\tau^2+4}, \qquad
    m_2 =\frac{\alpha^2}{\tau^2}\frac{\tau^2+4}{\alpha^2+4}.$$ Note that $0<m_1,m_2<1$.

![  Solution set $(\alpha,\tau)$ to the period condition describing the intersection $\overline\oH \cap
        \oP$. []{data-label="fig:plotintersect"}](figures/PlotMeeks.pdf){width="40.00000%"}

The theorem follows from the following proposition:

\[prop:tildeQ\] The function $$\tilde Q(\alpha,\beta;\tau) = \frac{Q(\alpha,\beta;\tau)}{\beta-\alpha}$$ extends analytically to $\alpha=\beta$ by $$\tilde Q(\alpha,\alpha;\tau) = \frac{\tau^2}{\alpha^2}\frac{\tau}{\tau^2-\alpha^2}\sqrt{\frac{\alpha^2+4}{\tau^2+4}}  \frac{{\bar{K}}(m_1) E(m_2) + m_2 {\bar{E}}(m_1) K(m_2) - {\bar{K}}(m_1) K(m_2)}{K(m_2)^2}.$$

With the help of the integral tables in [@byrd1971], we obtain the following explicit evaluation of the periods.  $$\begin{aligned}
        (I_1+I_3)(\alpha,\alpha;\tau) &= (J_1+J_3)(\alpha,\alpha;\tau) = \frac{2}{\sqrt{\tau^2+4}}{\bar{K}}(m_1) \ ,\\
        I_2(\alpha,\alpha;\tau) &= J_2(\alpha,\alpha;\tau) = \frac{\alpha}{\tau}\frac{2}{\sqrt{\alpha^2+4}} K(m_2) \ . 
    \end{aligned}$$

Then we evaluate the derivatives $$I'_k(\alpha,\alpha;\tau) = \frac{\partial}{\partial \beta} \Bigr|_{\beta=\alpha} I_k(\alpha,\beta;\tau),\qquad
        J'_k(\alpha,\alpha;\tau) = \frac{\partial}{\partial \beta} \Bigr|_{\beta=\alpha} J_k(\alpha,\beta;\tau),$$ and obtain $$\begin{aligned}
        (I'_1 + I'_3)(\alpha,\alpha;\tau) &= 0 \ ,\\
        I'_2(\alpha,\alpha;\tau) &= \frac{1}{\tau\sqrt{\alpha^2+4}} K(m_2) \ ,\\
        (J'_1 + J'_3)(\alpha,\alpha;\tau) &= \frac{\alpha}{\tau^2-\alpha^2}\frac{2}{\sqrt{\tau^2+4}}\Big( {\bar{K}}(m_1) - \frac{\tau^2+4}{\alpha^2+4} {\bar{E}}(m_1) \Big) \ ,\\
        J'_2(\alpha,\alpha;\tau) &= \frac{2}{\tau \sqrt{\alpha^2+4}} \Big( \frac{\tau^2}{\tau^2-\alpha^2} E(m_2) - \frac{1}{2} K(m_2) \Big) \ . 
    \end{aligned}$$

Finally, by l’Hôpital,

$$\lim_{a \to b} \tilde Q(\alpha,\beta;\tau)
        = \frac{\partial Q}{\partial \beta} \Bigr|_{\alpha=\beta}
        = \frac{\tau^2}{\alpha^2}\frac{\tau}{\tau^2-\alpha^2}\sqrt{\frac{\alpha^2+4}{\tau^2+4}}  \frac{{\bar{K}}(m_1) E(m_2) + m_2 {\bar{E}}(m_1) K(m_2) - {\bar{K}}(m_1) K(m_2)}{K(m_2)^2}.$$

Now note that the function $\tilde Q$ can be extended to a holomorphic function of complex arguments $\alpha, \beta, \tau$. The computation above shows that it remains bounded for $\alpha=\beta$, and hence extends holomorphically to $\alpha=\beta$. In particular, the extension for real arguments is real analytic. 

![ Two surfaces in the intersection of $\overline\oH \cap \oP$[]{data-label="fig:intersect"}](figures/intersect-1.png "fig:"){width="48.00000%"} ![ Two surfaces in the intersection of $\overline\oH \cap \oP$[]{data-label="fig:intersect"}](figures/intersect-2.png "fig:"){width="48.00000%"}

The solution set to  is shown in Figure \[fig:plotintersect\].  In Figure \[fig:intersect\], we show two surfaces in the intersection with extreme values of $\alpha$. In the next section, we will analyze the Traizet limit on the right (with small $\alpha$).  The left image strongly suggest that, in the limit of large $\alpha$, the family tends to a doubly periodic Karcher-Meeks-Rosenberg surface of genus 1 [@karcher1988; @karcher1989; @meeks1989]. 

Revisiting the Traizet limit {#sec:traizet}
============================

While we did not manage to prove uniqueness Conjecture \[conj:unique\], we can however prove the uniqueness at the Traizet limit. 

\[thm:1\] For $0 < \theta < \theta^* \approx 1.23409$, there is a unique solution $0 <
    x < 1/2$ that solves the balance equation .  This solution is non-degenerate in Traizet’s sense, hence is the Traizet limit for a family of TPMS. 

We consider the function $$f(x,\theta) = x \eta_3(\theta) - \zeta(x T_3(\theta);\theta).$$ The uniqueness follows from the following behaviors of $f$ for any fixed $0<\theta<\theta^*$:

1.   $f \to +\infty$ as $x\to 0$. To see this, note that the lattice is spanned by conjugate vectors $T_1, T_2$, so $\zeta(z)$ is real for real $z$. By definition, it has residue $+1$ at $0$. The claim follows. 

2.   $\partial_x f > 0$ for $x = 1/2$. This can be seen by noting that $$\frac{\partial f}{\partial x}\Bigr|_{x=1/2} = T_3\wp(T_3/2) + \eta_3$$ is positive if $\theta < \theta^* \approx 1.23409$. By now we already proved the existence of non-trivial solution. 

3.   $\partial_x^2 f>0$ for $0<x<1/2$. This can be seen by noting that $$\frac{\partial^2 f}{\partial x^2} = T_3^2 \wp'(x T_3)$$ is clearly non-zero.  So $f$ must be convex in $x$. This proves the uniqueness of the non-trivial solution. 

For the non-degeneracy, we consider the function $$F(x,y;\theta) = (x+y) \eta_1(\theta) + (x-y) \eta_2(\theta) -
        \zeta((x+y)T_1(\theta) + (x-y)T_2(\theta);\theta) \ .$$ Note that $F(x,0;\theta) = -f(x,\theta)$. So the convexity of $f$ in $x$ implies that $\partial_x F(x,0;\theta) \ne 0$ for $0<x<1/2$. It remains to prove that $$\frac{\partial F}{\partial y}\Bigr|_{y=0} = \eta_1 - \eta_2 + (T_1-T_2) \wp
        (x T_3) \ne 0$$ at a solution of .  This can be seen by noting that $$\frac{\partial}{\partial x} \frac{\partial F}{\partial y}\Bigr|_{y=0} =
        (T_2^2-T_1^2) \wp' (x T_3)$$ is negative purely imaginary for $0<x<1/2$, and $\partial_y F$ is positive purely imaginary at $(x,y)=(1/2,0)$.

Note that, as $\theta$ approaches $\theta^*$, the non-trivial solution tends to $1/2$, and $T_3 \wp(T_3/2)+\eta_3$ tends to $0$. So the balance configuration at $(x,\theta)=(1/2,\theta^*)$ is degenerate. 

With our parametrization of $\oH$, the Traizet limit, with infinitesimally small catenoid nodes, corresponds to the limit $ab\to 1$ or $\alpha+\beta\to
0$. In this limit, the Gauss map $G(z)=i$ except at two singular points $z=1/a=b$ and $z=-a=-1/b$. So the octagon degenerates into the plane $y=0$ as expected.  The angle of the limit rhombic torus can be computed as $$\label{eq:rhombicangle}
    \tan\frac{\theta}{2} = \frac{|\int_{-1/t}^{1/t}dh|}{|\int_{1/t}^{t}dh|} =
    \frac{K'(m)}{K(m)},$$ where $m=\tau^2/(\tau^2+4)$.

From , we can already recover the intersection of the two loci of .  First note that $m_2 \to 0$ when $\alpha =
\beta \to 0$. Divide both sides of by $m_2$ to eliminate the trivial but meaningless solution at $\alpha=\beta=0$. Recall that [@byrd1971] $(K(m)-E(m))/m \to \pi/4$ and $K(m) \to \pi/2$ as $m \to
0$. Hence we obtain for the intersection $$2E(m) = K(m),$$ where $m = \tau^2/(\tau^2+4)$. This is solved with $\tau \approx 4.35932$ or $t \approx 4.57777$. By putting these parameters into , we recover $\theta^* \approx 1.23409 \approx 70.7083^\circ$.

Alternatively, we may recover the loci of  in terms of our parametrization. 

The Traizet limit of $\oH$ is described by the equation $$\label{eq:traizet2}
        (2\beta^2-\tau^2+4)K(m) = 2(\beta^2+4)\Pi(n,m),$$ where $\Pi(n,m)$ is the complete elliptic integral of the third kind, with the characteristic $n = \tau^2/(\tau^2-\beta^2)>0$, and the modulus $m=\tau^2/(\tau^2+4)$.

![  Solution set $(\beta,\tau)$ to the period condition describing the Traizet limit of $\oH$, together with the trivial locus $\beta
        = 0$ describing the Traizet limit of $\oP$. To compare with Figure \[fig:diagsols\]. []{ data-label="fig:plottraizet2"}](figures/PlotTraizet2.pdf){width="40.00000%"}

Again, with the help of the integral tables in [@byrd1971], we obtain $I_2(-\beta,\beta;\tau)=J_2(-\beta,\beta;\tau)=0$ and $$(I_1+I_3)(-\beta,\beta;\tau) =
        (J_1+J_3)(-\beta,\beta;\tau)
        = \sqrt{1-m} K(m) \ ,$$ and the derivatives up to order $2$ with respect to $\alpha$ at $\alpha = -\beta$ $$\begin{aligned}
        (I'_1 + I'_3)(\alpha,\beta;\tau) = (J'_1+J'_3)(\alpha,\beta;\tau) &= \frac{\beta}{\tau^2-\beta^2} \frac{1}{\sqrt{\tau^2+4}} \Pi(n,m) \ ,\\
        I'_2(\alpha,\beta;\tau) = J'_2(\alpha,\beta;\tau) &= \frac{\pi}{2}\frac{1}{\sqrt{(\tau^2-\beta^2)(\beta^2+4)}} \ ,\\
        I''_2(\alpha,\beta;\tau) = J''_2(\alpha,\beta;\tau)
        &= \frac{\pi}{2}\frac{\beta}{\sqrt{(\tau^2-\beta^2)(\beta^2+4)}}
        \Big(\frac{1}{\tau^2-\beta^2}-\frac{1}{\beta^2+4}\Big). 
    \end{aligned}$$

We look at a modified period condition, namely $$\hat Q=\frac{1/Q_I-1/Q_J}{(\alpha+\beta)^2} = 0 \ .$$ The evaluations above suffice to compute, by l’Hôpital, that $$\begin{aligned}
        &\lim_{\alpha \to -\beta+} \hat Q(\alpha,\beta;\tau)
        = \frac{\partial^2}{\partial \alpha^2} \Bigr|_{\alpha = -\beta} (1/Q_I-1/Q_J)\\
        &= \frac{\pi\beta\sqrt{\tau^2+4}}{4(\beta^2+4)^{3/2}(\tau^2-\beta^2)^{3/2}}\frac{(2\beta^2-\tau^2+4)K(m) - 2(\beta^2+4)\Pi(n,m)}{K(m)^2}. 
    \end{aligned}$$

Hence $\hat Q$ extends analytically to $\alpha + \beta = 0$. Under the constraint $\tau > \beta$, we notice indeed two loci: $\beta=0$ for the Traizet limit of $\oP$, and the Traizet limit of $\oH$ must be described by . 

And must be describing the unique nontrivial locus of .  Its solution set is plotted in Figure \[fig:plottraizet2\].  Alternatively, can also be written in the forms $$(\tau^2+4)K(m) = 2(\beta^2+4)\Pi(n',m)$$ where $n' = (\tau^2-\beta^2)/(\tau^2+4)$, or $$\Big(8\frac{\tau^2}{n''}+(\tau^2-4)(\beta^2+4)\Big)K(m) = 8\Big(\frac{\tau^2}{n''}-(\beta^2+4)\Big)\Pi(n'',m)$$ where $n'' = \beta^2/(\beta^2+4)$.

To find the intersection with the trivial locus, let $\beta \to 0+$. For the three forms of , we recall, respectively, that $$\begin{aligned}
    \lim_{n\to 1+} \Pi(n,m) &= K(m) - \frac{E(m)}{1-m} & \text{\cite[(19.6,6)]{DLMF};}\\
    \lim_{n'\to m} \Pi(n',m) &= \frac{E(m)}{1-m} & \text{\cite[(19.6.1)]{DLMF};}\\
    \lim_{n''\to 0+} \frac{\Pi(n'',m)-K(m)}{n''} &= \frac{K(m)-E(m)}{m} & \text{cf.\ \cite[(733.00)]{byrd1971}.}\end{aligned}$$ Any one of these leads to, again, $$2E(m) = K(m).$$

The magic equation $2E(m) = K(m)$ also appeared in [@chenweber1] for locating the bifurcation point in the $\tD$ family. 

Let $k_r$ denote the $r$-th *elliptic integral singular values*, i.e.$K'(k_r^2)/K(k_r^2)=\sqrt{r}$. A table of $k_r$ can be found in [@bowman1961 p. 95] and [@borwein1987 (4.6.10)]. 

It was calculated by Legendre (see [@whittaker1962 §22.81]) that $k_3 =
(\sqrt{6}-\sqrt{2})/4$, hence $K'(m)/K(m) = 1/\sqrt{3}$ when $m = 1-k_3^2 =
(2+\sqrt{3})/4$. We then see from  that the rhombic torus with $\theta = 60^\circ$ occurs when $\tau = 2(2+\sqrt{3})$. Then  is solved, very conveniently, with $\beta = 2$. One then verifies that the singular point at $\beta$ is mapped to one third of the height of the box.  These are then explicit parameters for the Traizet limit of Schwarz’ $\H$ family. 

We are now ready to prove:

Schwarz $\H$ surfaces can be deformed within the set of TPMS of genus three into Meeks surfaces. 

Within a sufficiently small neighborhood of a Traizet limit, Traizet’s construction actually implies a homeomorphism between the space of TPMS of genus three and the space of $3$-tori.  This was not explicitly stated in [@traizet2008], but follows from his design of the Weierstrass data and the uniqueness in the implicit function theorem, as argued in [@traizet2002].  Let $U$ be such a neighborbood of the Traizet limit of $H$. In particular, $U \cap \oH$ is connected. 

Now fix $\epsilon > 0$. We consider the $\oH$ surfaces with $\alpha + \beta
    = \epsilon$. The period condition for such surfaces is $\tilde
    Q_\epsilon(\beta,\tau) = \tilde Q(\epsilon-\beta, \beta; \tau) = 0$, defined on the region $\{(\beta, \tau) \in {\mathbb{R}}_+^2 \colon \tau > \beta \ge
    \epsilon/2\}$. We have shown that $\tilde Q_\epsilon(\beta; \tau)$ is negative as $\tau$ approaches $\beta$, and positive as $\tau$ tends to infinity.  This holds, in particular, also for $\alpha = \beta = \epsilon/2$. Hence in the real analytic solution set of $\tilde Q_\epsilon = 0$, there must be a real analytic curve $\gamma$ that separates the line $\tau = \beta$ from $\tau = \infty$. If $\epsilon$ is sufficiently small, the curve $\gamma$ passes through $U$.

So we deform an $\H$ surface first along the $\H$ family into $U$, then within $U \cap \oH$ onto the curve $\gamma$, finally along $\gamma$ until an $\oP$ surface.  The latter belongs to Meeks, which is connected.  Note that this deformation path is within $\oH$ until hitting $\oP$.

It is easy to find $k_1=1/\sqrt{2}$, hence $K'(m)/K(m)=1$ when $m = 1-k_1^2 =
    1/2$. We then see from  that the rhombic torus becomes square when $\tau = 2$. In this case, has no solution with $\beta < \tau$. So the only balance configuration is with $\beta = 0$. This is the Traizet limit of the tetragonal deformation family $\tP$ of Schwarz’ $\P$ surface. 

[^1]: H. Chen is supported by Individual Research Grant from Deutsche Forschungsgemeinschaft within the project “Defects in Triply Periodic Minimal Surfaces”, Projektnummer 398759432.