Catenoid limits of singly periodic minimal surfaces with Scherk-type ends

We construct families of embedded, singly periodic minimal surfaces of any genus $g$ in the quotient with any even number $2n>2$ of almost parallel Scherk ends. A surface in such a family looks like $n$ parallel planes connected by $n-1+g$ small catenoid necks. In the limit, the family converges to an $n$-sheeted vertical plane with $n-1+g$ singular points termed nodes in the quotient. For the nodes to open up into catenoid necks, their locations must satisfy a set of balance equations whose solutions are given by the roots of Stieltjes polynomials.

The goal of this paper is to construct families of singly periodic minimal surfaces (SPMSs) of any genus in the quotient with any even number 2n > 2 of Scherk ends (asymptotic to vertical planes).Each family is parameterized by a small positive real number τ > 0. In the limit τ → 0, the Scherk ends tend to be parallel, and the surface converges to an n-sheeted vertical plane with singular points termed nodes.As τ increases, the nodes open up into catenoid necks, and the surface looks like parallel planes connected by these catenoid necks.
There are many previously known examples of such SPMSs.Scherk [Sch35] discovered the examples with genus zero and four Scherk ends, in 1835.Karcher [Kar88] generalized Scherk's surface in 1988 with any even number 2n > 2 of Scherk ends.In this paper, examples of genus 0 will be called "Karcher-Scherk saddle towers" or simply "saddle towers", and saddle towers with four Scherk ends will be called "Scherk saddle towers".Karcher also added handles between adjacent pairs of ends, producing SPMSs of genus n with 2n Scherk ends.Traizet glued Scherk saddle towers into SPMSs of genus (n 2 − 3n + 2)/2 with 2n > 2 Scherk ends because he was desingularizing simple arrangements of n > 1 vertical planes.In 2006, Martín and Ramos Batista [MRB06] replaced the ends of Costa surface by Scherk ends, thereby constructed an embedded SPMS of genus one with six Scherk ends and, for the first time, without any horizontal symmetry plane.Hauswirth, Morabito, and Rodríguez [HMR09] generalized this result in 2009, using an end-toend gluing method to replace the ends of Costa-Hoffman-Meeks surfaces by Scherk ends, thereby constructed SPMSs of higher genus with six Scherk ends.In 2010, da Silva and Ramos Batista [dSRB10] constructed a SPMS of genus two with eight Scherk ends based on Costa surface.Also, Hancco, Lobos, and Ramos Batista [HLB14] constructed SPMSs with genus 2n and 2n Scherk ends in 2014.
The examples of da Silva and Ramos Batista as well as all examples of Traizet admit catenoid limits that can be constructed using techniques in the present paper.
One motivation of this work is an ongoing project that addresses to various technical details in the gluing constructions.
Roughly speaking, given any "graph" G that embeds in the plane and minimizes the length functional, one could desingularize G × R into a SPMS by placing a saddle tower at each vertex.Previously, this was only proved for simple graphs under the assumption of an horizontal reflection plane [Tra96,Tra01].Recently, we managed to allow the graph to have parallel edges, to remove the horizontal reflection plane by Dehn twist [CT21], and to prove embeddedness by analysing the bendings of Scherk ends [Che21].
However, we still require that the vertices of G are neither "degenerate" nor "special".Here, a vertex of degree 2k is said to be degenerate (resp.special ) if k (resp.k − 1) of its adjacent edges extend in the same direction while other k (resp.k − 1) edges extend in the opposite direction.This limitation is due to the fact that a saddle tower with 2k Scherk ends can not have k − 1 ends extending in the same direction while other k − 1 ends extending in the opposite direction.Therefore, it is not possible to place a saddle tower at a degenerate or special vertex.
Nevertheless, we do know SPMSs that desingularize G × R where G is a graph with degenerate vertex.To include these in the gluing construction, we need to place catenoid limits of saddle towers, as those constructed in this paper, at degenerate vertices.From this point of view, the present paper can be seen as preparatory: The insight gained here will help us to glue saddle towers with catenoid limits of saddle towers in a future project.This paper reproduces the main result of the thesis of the third named author [Li12].Technically, the construction implemented in [Li12] was in the spirit of [Tra02b], which defines the Gauss map and the Riemann surface at the same time, and the period of the surface was assumed horizontal.Here, for the convenience of future applications, we present a construction in the spirit of [Tra08,CT21,Che21], which defines all three Weierstrass integrands by prescribing their periods, and the period of the surface is assumed vertical.In particular, we will reveal that a balance condition in [Li12] is actually a disguise of the balance of Scherk ends: The unit vectors in the directions of the ends sum up to zero.

Main result
1.1.Configuration.We consider L + 1 vertical planes, L ≥ 1, labeled by integers l ∈ [1, L+1].Up to horizontal rotations, we assume that these planes are all parallel to the xz-plane, which we identify to the complex plane C, with the x-axis (resp.z-axis) corresponding to the real (resp.imaginary) axis.We use the term "layer" for the space between two adjacent parallel planes.So there are L layers.
We want n l ≥ 1 catenoid necks on layer l, i.e. between the planes l and l + 1, 1 ≤ l ≤ L. For convenience, we adopt the convention that n l = 0 if l < 1 or l > L, and write N = n l for the total number of necks.Each neck is labeled by a pair (l, k), where 1 ≤ l ≤ L and 1 ≤ k ≤ n l .
To each neck is associated a complex number Then the positions of the necks are prescribed at ln q l,k + 2mπi, m ∈ Z. Recall that the z-axis is identified to the imaginary axis of the complex plane C, so the necks are periodic with period vector (0, 0, 2π).Note that, if we multiply q l,k 's by the same complex factor c, then the necks are all translated by ln c (mod 2πi).So we may quotient out translations by fixing q 1,1 = 1.
Moreover, each plane has two ends asymptotic to vertical planes.We label the end of plane l that expands in the −x (resp.x) direction by 0 l (resp.∞ l ).To be compatible with the language of graph theory that were used for gluing saddle towers [CT21], we use to denote the set of ends.When 0 l is used as subscript for parameter x, we write x l,0 instead of x 0 l to ease the notation; the same applies to ∞ l .
To each end is associated a real number θh , h ∈ H.They prescribe infinitesimal changes of the directions of the ends.More precisely, for small τ , we want the unit vector in the direction of the end h to have a y-component or the order τ θh +O(τ 2 ).
Remark 1. Multiplying θ by a common real constant leads to a reparameterization of the family.Adding a common real constant to θl,0 and subtracting the same constant from θl,∞ leads to horizontal rotations of the surface.
Then a configuration refers to the pair (q, θ).1.2.Force.Given a configuration (q, θ), let c l be the real numbers that solve Recall the convention n l = 0 if l < 1 or l > L, so we also adopt the convention If (2) is satisfied, the real numbers c l are determined by (1) as functions of θ.For 1 ≤ l ≤ L + 1, let ψ l be the meromorphic 1-form on the Riemann sphere Ĉ with simple poles at q l,k with residue −c l for each 1 ≤ k ≤ n l , at q l−1,k with residue c l−1 for each 1 ≤ k ≤ n l−1 , at 0 with residue θl,0 , and at ∞ with residue θl,∞ .More explicitly, We then see that Equations (1) arise from the Residue Theorem Remark 2. In the definition of configuration, we may replace θ by the parameters (c l , θl+1,0 − θl,0 ) 1≤l≤L .Then θl,0 's are defined up to an additive constant (corresponding to a rotation), θl,∞ 's are determined by (1), and (2) is automatically satisfied.To quotient out reparameterizations of the family, we may assume that c l = 1 for some 1 ≤ l ≤ L.
We define the force F l,k by (3) Or, more explicitly, l + c l ( θl+1,0 − θl,0 ).In [Li12], the force had different formula depending on the parity of l.One verifies that both are equivalent to (4).
Disregarding absolute convergence, we write this formally as Then the force is given, formally, by Recall that ln q l,k + 2mπi are the real positions of the necks.So this formal expression has an electrostatic interpretation similar to those in [Tra02b] and [Tra08].
Here, each neck interacts not only with all other necks in the same or adjacent layers, but also with background constant fields given by θ.
Remark 4 (Another electrostatic interpretation).In fact, (4)/q l,k has a similar electrostatic interpretation.But this time, the necks are seen as placed at q l,k .Each neck interacts with all other necks in the same and adjacent layers, as well as a virtual neck at 0 with "charge" c l + θl+1,0 − θl,0 .This is no surprise, as electrostatic laws are known to be preserved under conformal mappings (such as ln(z)).
Summing up all forces yields a necessary condition for the configuration to be balanced, namely Lemma 1.The Jacobian matrix ∂(Θ 1 , Θ 2 )/∂ θ has real rank 2 as long as c l ̸ = 0 for some 1 ≤ l ≤ L.
The assumption of the lemma simply says that the surface does not remain a degenerate plane to the first order.
Proof.The proposition says that the matrix has an invertible minor of size 2 × 2. Explicitly, we have This minor is invertible if and only if the ends θl,0 + θl,∞ ̸ = 0.This must be the case for at least one 1 ≤ l ≤ L because, otherwise, we have c l = 0 for all 1 ≤ l ≤ L. □ Definition 2. The configuration is rigid if the complex rank of ∂F/∂q is N − 1.
Remark 5.In fact, the complex rank of ∂F/∂q is at most N − 1.We have seen that a complex scaling of q corresponds to a translation of ln q l,k + 2mπi, m ∈ Z, which does not change the force.It then makes sense to normalize q by fixing q 1,1 = 1.
Theorem 1.Let (q, θ) be a balanced and rigid configuration such that c l ̸ = 0 for 1 ≤ l ≤ L. Then for τ > 0 sufficiently small, there exists a smooth family M τ of complete singly periodic minimal surface of genus g = N − L, period (0, 0, 2π), and 2(L + 1) Scherk ends such that, as τ → 0, • M τ converges to an (L+1)-sheeted xz-plane with singular points at ln q l,k + 2mπi, m ∈ Z. Here, the xz-plane is identified to the complex plane C, with x-axis (resp.z-axis) identified to the real (resp.imaginary) axis.• After suitable scaling and translation, each singular point opens up into a neck that converges to a catenoid.• The unit vector in the direction of each Scherk end h has the y-component Remark 6. M τ also depends smoothly on θ belonging to the local smooth manifold defined by Θ 1 = 0 and Θ 2 = 0. Up to reparameterizations of the family and horizontal rotations, we obtain families parameterized by 2L − 1 parameters.Since we have 2(L + 1) Scherk ends, this parameter count is compatible with the fact that Karcher-Scherk saddle towers with 2k ends form a family parameterized by 2k − 3 parameters.
Remark 7. If the embeddedness condition (5) is satisfied and Θ 1 = 0, the sequence θl,0 + θl,∞ is strictly monotonically decreasing, and changes sign once and only once.Then the sequence n l c l is strictly concave (i.e.n l−1 c l−1 + n l+1 c l+1 < 2n l c l for 1 ≤ l ≤ L).Hence c l , 1 ≤ l ≤ L, are strictly positive, and the condition of Lemma 1 is satisfied.
Remark 8. We could allow some c l to be negative, with the price of losing embeddedness.Even worse, with negative c l , the vertical planes in the limit will not be geometrically ordered as they are labeled.For instance, if L = 2, c 1 > 0, but c 2 < 0, then the catenoid necks, as well as the 1st and 3rd "planes", will all lie on the same side of the 2nd "plane".
Remark 9. We did not allow any c l to be 0 in Theorem 1. Otherwise, the surface might still have nodes.In that case, the claimed family might not be smooth, and the claimed genus would be incorrect.

Examples
2.1.Surfaces of genus 0. When the genus g = N − L = 0, we have n l = 1 for all 1 ≤ l ≤ L, i.e. there is only one neck on every layer.It then makes sense to drop the subscript k.For instance, the position and the force for the neck on layer l are simply denoted by q l and F l .We assume L > 1 in this part.
In this case, if Θ 1 = 0, Equation (1) can be explicitly solved by and the force can be written in the form where we changed to the parameters with the convention that Q 0 = Q L = 0. Then the forces are linear in Q and, if Θ 2 = 0, the balance condition F = 0 is uniquely solved by ( 6) Therefore, if we fix q 1 = 1, all other q l , 1 < l ≤ L are uniquely determined.
Recall from Remark 7 that, under the embeddedness condition (5), the numbers c l , 1 ≤ l ≤ L, are positive.Moreover, the summands in (6) changes sign at most once, so the sequence Q is unimodal, i.e. there exists 1 ≤ l ′ < L such that So q consists of real numbers and q l+1 /q l < 0 for all 1 ≤ l < L.
We have proved that Proposition 2. If the genus g = N −L = 0, and θ satisfies the balancing condition Θ 1 = Θ 2 = 0 as well as the embeddedness condition (5), then up to complex scalings, there exist unique values for the parameters q, depending analytically on θ, such that the configuration (q, θ) is balanced.Moreover, all such configurations are rigid.If we fix q 1 = 1, then q consist of real numbers, and we have q l > 0 (resp.< 0) if l is odd (resp.even).

Surfaces with four ends. When
Up to reparameterizations of the family, we may assume that c 1 = 1.It makes sense to drop the subscript l in other notations, and write F k for F 1,k , q k for q 1,k , and n for n 1 .The goal of this part is to prove the following classification result.
Proposition 3. Up to a complex scaling, a configuration with L = 1 and n nodes must be given by q k = exp(2πik/n), and such a configuration is rigid.
Such a configuration is an n-covering of the configuration for Scherk saddle towers.As a consequence, the arising minimal surfaces are n-coverings of Scherk saddle towers.This is compatible with the result of [MW07] that the Scherk saddle towers are the only connected SPMSs with four Scherk ends.
Proof.To find the positions q k such that (7) we use the polynomial method: Consider the polynomial Then we have For the last equation to have a polynomial solution, we must have P ′ (0) = 0. Otherwise, the left-hand side would be a polynomial of degree n − 2, but the righthand side would be a polynomial of degree n − 1.Consequently, F k = 0 if and only if zP ′′ (z) − (n − 1)P ′ (z) = 0, which, up to a complex scaling, is uniquely solved by So a balanced 4-end configuration must be given by the roots of unity We now verify that the configuration is rigid.For this purpose, we compute has real entries, has a kernel of complex dimension 1 (spanned by the all-one vector), and any of its principal submatrix is diagonally dominant.We then conclude that the matrix, as well as the Jacobian ∂F/∂q, has a complex rank n − 1.This finishes the proof of rigidity.□ Remark 10.The perturbation argument as in the proof of [Tra02b, Proposition 1] also applies here, word by word, to prove the rigidity.

2.3.
Gluing two saddle towers of different periods.We want to construct a smooth family of configurations depending on a positive real number λ such that, for small λ, the configuration looks like two columns of nodes far away from each other, one with period 2π/n 1 , and the other with period 2π/n 2 .If balanced and rigid, these configurations would give rise to minimal surfaces that look like two Scherk saddle towers with different periods that are glued along a pair of ends.The construction is in the same spirit as [Tra02b, § 2.5] and [Tra08,§ 4.3.4].
In other words, the construction only works if the configuration admits a reflection symmetry.
Remark 11.The first named author was shown a video suggesting that, when two Scherk saddle towers are glued into a minimal surface, one can slide one saddle tower with respect to the other while the surface remains minimal.The proposition above suggests that this is not possible.
In fact, the family of configurations also depends on θ belonging to the local manifold defined by Θ 1 = Θ 2 = 0 and (one equation from) (8).Up to rotations of the configuration and reparameterizations of the family of minimal surfaces, the family of configurations is parameterized, as expected, by two parameters.
Proof.Let us first study the situation at λ = 0. We compute at λ = 0 Write G l = k F l,k .Summing the above over k gives, at λ = 0, This together with Θ 1 = 0 proves (8).Now assume that (8) is satisfied.Then we have, at λ = 0 These expressions are identical to the force (7) for single layer configurations.So we know for l = 1, 2 that, at λ = 0, the configuration is balanced only if Up to complex scaling, we may fix q 1,1 = 1 so q 1,k = q 1,k .And up to reparameterization of the family (of configurations), we write q 2,1 = λ exp(iϕ).Now assume these initial values for q l,k .Then we have, at λ = 0, Seen as a power series of q 2,1 , the coefficient for q m 2,1 is It is non-zero only if m is a common multiple of n 1 and n 2 , in which case the coefficient of q m 2,1 equals n 1 n 2 .In particular, let µ = lcm(n 1 , n 2 ), then at λ = 0, Now we use the Implicit Function Theorem to find balance configurations with λ > 0. From the proof for Proposition 3, we know that ∂F l,k ∂ q l,j 2≤j,k≤n l , l = 1, 2, are invertible.Hence for λ sufficiently small, there exist unique values for ( q l,k ) l=1,2;2≤k≤n l , depending smoothly on λ, θ, and ϕ, such that (F l,k ) l=1,2;2≤k≤n l = 0.By (9), there exists a unique value for ϕ, depending smoothly on λ and θ, such that Im G 2 /λ µ = 0. Note also that Re G 2 is linear in θ.By Lemma 1, the solutions (λ, θ) to Re G 2 = 0 and Θ 1 = Θ 2 = 0 form a manifold of dimension 4 (including multiplication by common real factor on θ and rotation of the configuration).Finally, we have G 1 = 0 by the Residue Theorem, and the balance is proved.
For the rigidity of the configurations with sufficiently small λ, we need to prove that the matrix is invertible.We know that the first two blocks are invertible at λ = 0.By continuity, they remain invertible for λ sufficiently small.The last block is clearly non-zero for λ ̸ = 0 sufficiently small.□ 2.4.Surfaces with six ends of type (n,1).In this section, we investigate examples with L = 2 (hence six ends), n 1 = n, n 2 = 1.Up to a reparameterization of the family, we may assume that c 1 = 1.Up to a complex scaling, we may assume that q 2,1 = 1.We will prove that q 1,k 's are given by the roots of hypergeometric polynomials.Let us first recall their definitions.A hypergeometric function is defined by and (a) 0 = 1.The hypergeometric function w = 2 F 1 (a, b; c; z) solves the hypergeometric differential equation is a polynomial of degree n, and is referred to as a hypergeometric polynomial.
Moreover, as long as b and c are not non-positive integers, and c − b is not a non-positive integer bigger than −n, the configuration is rigid.
Proof.The force equations are where c := 1 + θ2,0 − θ1,0 .To solve F 1,k = 0 for k = 1, 2, . . ., n, we use again the polynomial method: Let P (z) = n k=1 (z − q 1,k ).Then we have So the configuration is balanced if and only if (11) For (11) to have a polynomial solution of degree n, we must have so that the leading coefficients cancel.Then (11) becomes the hypergeometric differential equation to which the only polynomial solution (up to a multiplicative constant) is given by the hypergeometric polynomial Moreover, in order for F 2,1 = 0, we must have Note that b and c are real.If b is not a non-positive integer, and c − b is not a non-positive integer bigger than −n, then all the n roots of P (z) = 2 F 1 (−n, b; c; z) are simple.Indeed, under these assumptions, we have P (0) = 1 and P (1) = (c − b) n /(c) n ̸ = 0 by the Chu-Vandermonde identity.Let z 0 be a root of P (z), then z 0 ̸ = 0, 1.In view of the hypergeometric differential equation, if z 0 is not simple, we have P (z 0 ) = P ′ (z 0 ) = 0 hence P (z) ≡ 0 by uniqueness theorem.
The rigidity means that no perturbation of q 1,k preserve the balance to the first order.To prove this fact, we use a perturbation argument similar to that in the proof of [Tra02b, Proposition 1].
Let (q 1,k (t)) 1≤k≤n be a deformation of the configuration such that q 1,k (0) = q 1,k and ( Ḟ1,k (0)) 1≤k≤n = 0, where dot denotes derivative with respect to t. Define Then we have , meaning that the coefficients from the left side are all o(t).So the coefficients of P t must satisfy Note that P t (z) is monic by definition, meaning that a n (t) ≡ 1.Since b and c are not non-positive integers, we conclude that a j (t) = o(t) for all 0 ≤ j ≤ n.The simple roots depend analytically on the coefficients, so q 1,k (t) = q 1,k + o(t).□ The simple roots of 2 F 1 (−n, b; c; z) are either real or form conjugate pairs.As a consequence, if rigid, the configurations in the proposition above will give rise to minimal surfaces with horizontal symmetry planes.Example 1.For each integer n ≥ 2, the real parameters (b, c) for which 2 F 1 (−n, b; c; z) has only real simple roots has been enumerated in [DJJ13].The results are plotted in blue in Figure 1.The embeddedness conditions (5) are θ1,0 > θ2,0 The region defined by these is plotted in red in Figure 1.Then non-integer parameters (b, c) in the intersection of red and blue regions give rise to balanced and rigid configurations with real q 1,k .
Example 2. Assume that b + c = 1 − n (hence c 2 = −2b).Then by the identity the simple roots must be symmetrically placed.That is, if z 0 is a root, so is 1/z 0 .This symmetry appears in the resulting minimal surfaces as a rotational symmetry.
If the simple roots are real, the rotation reduces to a vertical reflectional.In view of Figure 1, we obtain the following concrete examples.
• n ≥ 2 and 0 < c < 1.In this case 2 F 1 (−n, b; c; z) has n simple negative roots.See Figure 3 for an example of this type with n = 2. Figure 4  2.5.Surfaces with eight ends of type (1,n,1).Theorem 5 generalizes to the following lemma with similar proof Lemma 6.If we fix q l±1,k 's and assume that c l = 1.Then q l,k 's in a balanced configuration are given by the roots of a Stieltjes polynomial P (z) of degree n l that solves the generalized Lamé equation (a.k.a.second-order Fuchsian equation) [Mar66] where c = 1 + θl+1,0 − θl,0 , subject to conditions Moreover, the matrix (∂F l,k /∂q l,j ) 1≤j,k≤n l is nonsingular as long as b is not a nonpositive integer bigger than n l .
Indeed, a root of P (z) is simple if and only if it does not coincide with 0 or any q l±1,k .If the roots (q l,k ) of P (z) are all simple, then they solve the equations [Mar66] 1≤k̸ =j≤n l 2 q l,k − q l,j + which is exactly our balance condition; see Remark 4.Moreover, an equation system generalizing (13) has been obtained in [Hei78,§136], from which we may conclude the non-singularity of the Jacobian.In fact, there are choices of γ for which (14) has a polynomial solution of degree n l [Hei78,§135].This observation allows us to easily construct balanced and rigid configurations of type (1, n, 1): Up to reparametrizations and complex scalings, we may assume that c 2 = 1 and q 1,1 = 1.Then q 3,1 must be real, and (q 2,k ) 1≤k≤n are given by roots of a Heun polynomial.Such a configuration depends locally on four real parameters, namely q 3,1 , c 1 , c 3 and c (or b).When these are given, we have n + 1 Heun polynomials, each of which gives balanced positions of q 2,k 's.For each of the Heun polynomials P , we have θ2,0 − θ1,0 = P ′ (1) θ4,0 − θ3,0 = P ′ (q 3,1 ) P (q 3,1 ) − c 3 .
Together with the family parameter τ , the surface depends locally on five parameters, which is expected because there are eight ends.
The balance of even layers then follows from the balance of each sub-configuration.The balance of odd layers leads to θ2r,0 As expected, such a configuration depends locally on 4R real parameters, namely q 3 , and c (r) , 1 ≤ r ≤ R. We may impose symmetry by assuming that q (r) 3,1 = 1, so q 2r+1,1 = 1 for all 0 ≤ r ≤ R, and that b (r) +c (r) = 1−n (r) , so c r) , are given by the roots of 2 F 1 (−n (r) , b (r) ; c (r) ; z), for 1 ≤ l ≤ L under the condition that n 0 = n L+1 = 0.That is, the sequence (n l c l ) 1≤l≤L must be concave.For even l, the concavity implies that b (r) > −n hence c (r) < 1 for all 1 ≤ r ≤ R. We may choose, for instance, n l c l = ln(1 + l) or n l c l = (exp l − 1)/ exp(l − 1) to obtain embedded minimal surfaces.
Remark 15.We can also append a configuration of type (1, n (r) ) to the sequence of (1, n (r) , 1)-configurations to obtain a configuration of type where the q l,k , c l , θl,0 terms are defined as above.Therefore, an embedded example of any genus with any even number (> 2) of ends can be constructed.

Numerical examples.
The balance equations can be combined into one differential equation that is much easier to solve.A solution of this differential equation corresponds to a lot of balance configurations that are equivalent by permuting the locations of the nodes.
Lemma 7. Let L be a positive integer, n 1 , n 2 , . . ., n L ∈ N, and suppose {q l,k } is a configuration such that the q l,k are distinct.Let Then the configuration {q l,k } is balanced if and only if FP (z) ≡ 0.
Proof.We have seen that, Then F l,k = 0 if and only if Q l (q l,k ) = 0. Now observe that Q l (z) and Q(z) = FP (z) are polynomials with degree strictly less than Hence, Q has at least N distinct roots.Since the degree of Q is strictly less than N , we must have Q ≡ 0. □ It is relatively easy to numerically solve FP (z) ≡ 0 as long as we don't have too many levels and necks.So we use this lemma to find balanced configurations.Since all previous examples admit a horizontal reflection symmetry, we are most interested examples without this symmetry, or with no non-trivial symmetry at all. Figure 7 shows an example with L = 3, This configuration corresponds to an embedded minimal surface with eight ends and genus three in the quotient.It has no horizontal reflectional symmetry, but does have a rotational symmetry.Figure 8 shows two examples with L = 3, These configurations correspond to embedded minimal surfaces with eight ends and genus five in the quotient, with no nontrivial symmetry.
Figure 9 shows two examples with L = 3, These configurations correspond to embedded minimal surfaces with eight ends and genus eight in the quotient, with no nontrivial symmetry.

Opening nodes.
To each vertical plane is associated a punctured complex plane C × l ≃ C \ {0}, 1 ≤ l ≤ L + 1.They can be seen as Riemann spheres Ĉl ≃ C ∪ {∞} with two fixed punctures at p l,0 = 0 and p l,∞ = ∞, corresponding to the two ends.(1, 4, 3) balanced configurations with no symmetries.The circles, squares, and diamonds represent the necks at levels one, two, and three respectively.
(1, 7, 3) balanced configurations with no symmetries.The circles, squares, and diamonds represent the necks at levels one, two, and three respectively.
To each neck associated a puncture p Our initial surface at τ = 0 is the noded Riemann surface Σ 0 obtained by identifying p • l,k and p ′• l,k for 1 ≤ l ≤ L and 1 ≤ k ≤ n l .As τ increases, we open the nodes into necks as follows: Fix local coordinates w l,0 = z in the neighborhood of 0 ∈ Ĉl and w l,∞ = 1/z in the neighborhood of ∞ ∈ Ĉl .For each neck, we consider parameters (p l,k , p ′ l,k ) in the neighborhoods of (p • l,k , p ′• l,k ) and local coordinates w l,k = ln(z/p l,k ) and w ′ l,k = ln(z/p ′ l,k ) in a neighborhood of p l,k and p ′ l,k , respectively.In this paper, the branch cut of ln(z) is along the negative real axis, and we use the principal value of ln(z) with imaginary part in the interval (−π, π].
As we only open finitely many necks, we may choose δ > 0 independent of k and l such that the disks 3.2.Weierstrass data.We construct a conformal minimal immersion using the Weierstrass parameterization in the form where Φ i are meromorphic 1-forms on Σ t satisfying the conformality equation ( 15) 3.2.1.A-periods.We consider the following fixed domains in all Σ t : Let A l,k denote a small counterclockwise circle in U l,δ around p l,k ; it is then homologous in Σ t to a clockwise circle in U l+1,δ around p ′ l,k .Moreover, let A l,0 (resp.A l,∞ ) denote a small counterclockwise circle in U l,δ around 0 (resp.∞).
Recall that the vertical period vector is assumed to be (0, 0, 2π), so we need to solve the A-period problems for h ∈ H, 1 ≤ l ≤ L, and 1 ≤ k ≤ n l .Here, the orientation σ h = ±1 satisfies where the "counterclockwise rotation" ς on H is defined by ( 16) In particular, we have σ l,0 = −σ l,∞ for all 1 ≤ l ≤ L + 1.
Recall that the surface tends to an (L+1)-sheeted xz-plane in the limit τ → 0. So we define the meromorphic functions Φ 1 , Φ 2 and Φ 3 as the unique regular 1-forms on Σ t (see [Tra13,§8]) with simple poles at p h , h ∈ H, and the A-periods where Φ 2 = τ Φ 2 and, by Residue Theorem, it is necessary that for 1 ≤ l ≤ L + 1.Then the A-period problems are solved by definition.In this paper, the punctures p l,0 and p l,∞ correspond to Scherk-type ends.Hence we fix (21) α 2 h + τ 2 β 2 h ≡ 1 and γ h ≡ 0 for all h ∈ H, so that (the stereographic projection of) the Gauss map G = −(Φ 1 + iΦ 2 )/Φ 3 extends holomorphically to the punctures p h with unitary values.Then Equations ( 19) are not independent: if it is solved for 1 ≤ l ≤ L, it is automatically solved for l = L + 1.
Remark 16.The conditions (20) and ( 22) are disguises of the balance condition of Scherk ends, namely that the unit vectors in their directions should sum up to 0.
Lemma 8.For t sufficiently close to 0, the conformality condition (15) is equivalent to where A ′ l,k in (26) denotes a small counterclockwise circle in U l+1,δ around p ′ l,k (hence homologous to −A l,k ), and δ l,L = 1 if l = L and 0 otherwise.
Proof.By our choice of α h and γ h , the quadratic differential Q has at most simple poles at the 2L + 2 punctures p h , h ∈ H.The space of such quadratic differentials is of complex dimension 3(N − L) − 3 + (2L + 2) = 3N − L − 1.We will prove that is an isomorphism.We prove the claim at t = 0; then the claim follows by continuity.Consider Q in the kernel.Recall from [Tra08] that a regular quadratic differential on Σ 0 has at most double poles at the nodes p l,k and p ′ l,k .Then (24) guarantees that Q has at most simple poles at the nodes.By (25) and (26), Q may only have simple poles at p l,1 ∈ C × l , 1 ≤ l ≤ L, and p ′ L,1 ∈ C × L+1 .So, on each Riemann sphere Ĉl , Q is a quadratic differential with at most simple poles at three punctures; the other two being 0, ∞.But such a quadratic differential must be 0. □ 3.3.Using the Implicit Function Theorem.All parameters varies in a neighborhood of their central values, denoted by a superscript •.We will see that

Solving conformality problems.
Proposition 10.For τ sufficiently small and β l,k , p l,k , and p ′ l,k in a neighborhood of their central values, there exist unique values of t l,k , α l,k , and γ l,k , depending real-analytically on (τ 2 , β, p, p ′ ), such that the balance equations (17) and (19) with 1 ≤ l ≤ L and the conformality Equations (24) and (25) are solved.Moreover, at τ = 0, we have t Note that, according to this proposition, if β • l,k ̸ = 0, then t l,k > 0 for sufficiently small τ .
We then compute the partial derivatives at τ = 0 All other partial derivatives vanish.Therefore, by the Implicit Function Theorem, there exist unique values of α l,k , γ l,k (with 2 ≤ k ≤ n l ), and t l,k (with 1 ≤ k ≤ n l ) that solve the conformality Equations ( 24) and (25).Recall that α h are determined by (21).Then α l,1 and γ l,1 are uniquely determined by the linear balance equations ( 17) and (19).Moreover, Hence the total derivatives This proves the claimed partial derivatives with respect to τ 2 .□ Remark 17.We see from the computations that our local coordinates w and w ′ are chosen for convenience.Had we used other coordinates, the computations would be very difference, but ∂(α l,k − iσ l,0 γ l,k )/∂(τ 2 ) would be invariant, and ∂t l,k /∂(τ 2 ) would be rescaled to keep the conformal type of Σ t (to the first order).So the choice of local coordinates has no substantial impact on our construction.
at the central values, where F l,k is the force given by (4).Moreover, by Residue Theorem on C × l , (29) Proposition 13.Assume that the parameters t l,k , α l,k , and γ l,k are given as analytic functions of τ 2 by Proposition 10.Then F ′ l,k := τ −2 F ′ l,k extends analytically to τ = 0 with the value □ Therefore, if (q, θ) is balanced, F ′ = 0 is solved at τ = 0. Recall that we normalize the complex scaling on C × 1 by fixing p 1,1 .If (q, θ) is rigid, because Θ 2 = F l,k = 0 independent of p, the partial derivative of ( F ′ ) (l,k)̸ =(L,1) with respect to (p l,k ) (l,k)̸ =(1,1) is an isomorphism from C N −1 to C N −1 .The following proposition then follows by the Implicit Function Theorem.
Proposition 14. Assume that the parameters t l,k , α l,k , β l,k , γ l,k , p ′ l,k are given by Propositions 10, 11, and 12. Assume further that the central values q l,k = conj l p • l,k and θh = β • h form a balanced and rigid configuration (q, θ).Then for (τ, β) in a neighborhood of (0, θ) that solves (20) and (22), there exists values for p l,k , unique up to a complex scaling, depending smoothly on τ and (β h ) h∈H , such that p l,k (0, θ) = p • l,k and the conformality condition (26) is solved.3.4.Embeddedness.It remains to prove that Proposition 15.The minimal immersion given by the Weierstrass parameterization is regular and embedded.
We now prove that the immersion is an embedding, and the limit positions of the necks are as prescribed.

Figure 1 .
Figure 1.The set of b and c for which 2F1(−n, b; c; z) has only real simple roots (blue), and for which the embeddedness conditions are satisfied (red).

Figure 2 .
Figure 2. (5, 1) balanced configurations.The circles and squares represent the necks at levels one and two respectively.

Figure 4 .
Figure 4. (5, 1) balanced configurations with 0 < c < 1.The circles and squares represent the necks at levels one and two respectively.

Figure 5 .
Figure 5. (5, 1) balanced configurations with −1 < c < 0. The circles and squares represent the necks at levels one and two respectively.

Figure 7 .
Figure 7.A (1, 3, 2) balanced configuration with no horizontal reflectional symmetry.The circles, squares, and diamonds represent the necks at levels one, two, and three respectively.
Figure8.(1, 4, 3) balanced configurations with no symmetries.The circles, squares, and diamonds represent the necks at levels one, two, and three respectively.