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Constant Mean Curvature Trinoids with one Irregular End

We construct a new five parameter family of constant mean curvature trinoids with two asymptotically Delaunay ends and one irregular end.


Introduction
The generalised Weierstrass representation [4] constructs conformal CMC immersions in Euclidean 3-spaces from a holomorphic 1-form ξ on a Riemann surface Σ. The associated period problem involves showing that the monodromy group of ξ is pointwise unitarisable along the unit circle of the spectral parameter λ. For the case that Σ is the thrice-punctured Riemann sphere and ξ has three regular singular points at the punctures, the resulting three-parameter family of CMC trinoids have asymptotically Delaunay ends [12]. In this paper we extend this result by replacing one of the regular singularities with an irregular singularity of rank 1. The corresponding second-order scalar ODE is the confluent Heun equation (CHE) with five free parameters. The monodromy can be computed by an asymptotic formula for the connection matrix between solutions at the two regular singular points given by Schäfke-Schmidt [10]. In this way, we construct a new five-parameter family of trinoids with two Delaunay ends and one irregular end.

The Generalised Weierstrass Representation and the Monodromy Problem
Let us briefly recall the generalized Weierstrass representation [4] to set the notation and conventions adapted from [12]. It consists of the following three steps: (i) On a connected Riemann surface Σ, let ξ be a holomorphic 1-form, called a potential, with values in the loop algebra of maps S 1 → sl 2 C. The potential ξ has a simple pole in its upper right entry in the loop parameter λ at λ = 0, and has no other poles in the unit λ disk. Moreover, the upper-right entry of ξ is non-zero on Σ. Let Φ be a solution of (1.1) dΦ = Φ ξ .
(ii) Let Φ = F B be the pointwise Iwasawa factorization on the universal cover Σ.
(iii) Then f = F F −1 is an associated family of conformal CMC immersions Σ → su 2 ∼ = R 3 . The prime denotes differentiation with respect to θ, where λ = e iθ .
The period problem cannot be solved simultaneously for the whole associated family, so we contend ourselves with solving it for the member of the associated family λ = 1. For i ∈ {0, 1, ∞} let γ i be a 2. The Associated Second Order ODE Consider a holomorphic potential Then g is a positive loop and gauging the holomorphic potential η 0 by g yields Hence we can assume without loss of generality that our potential is off-diagonal. Let us then consider a potential, denoted η, of the form For the purpose of our work it will be convenient to work with the associated scalar second order ODE corresponding to the 2 × 2 system (1.1), given by the following straightforward where y 1 and y 2 are a fundamental system of the scalar ODE (2.6) y − ν ν y − ρ ν y = 0.

Two Regular Singular Points
Our goal is to construct CMC trinoids for which two ends are regular and one end is irregular. We will assume that at the two ends z 0 and z 1 , the potential is a holomorphic perturbation of a Delaunay potential, and hence regular singular there. In addition, our potentials will have a singularity of rank 1 at z ∞ , making it an irregular end. These choices will determine the form of the associated scalar ODE (2.6).
Note that in this and the following sections, we omit the dependence on the spectral parameter λ. Let ϑ 0 , ϑ 1 ∈ C \ 1 2 Z be parameters, and a and b functions on Σ such that a is holomorphic at z 0 and z 1 , and b is allowed to have simple poles at z 0 and z 1 . Define The points z 0 and z 1 are regular singular points of ξ. These double poles can be gauged to simple poles by where B is holomorphic at z 0 and z 1 and It is only left to see that a general potential like in (2.1), the one we started with, is equivalent to a potential of the form in (3.2) by the gauge 3.1. The z A P lemma. Suppose dΦ = Φη for which η = A dz z−z k +O(z 0 ) dz has a simple pole at z = z k and Delaunay residue A. A standard result in the theory of regular singularities states that under certain conditions on the eigenvalues of A, there exists a solution of the form Φ = z A P = exp (A log z) P , where P extends holomorphically to z = z k (see [12,Lemma 14]). Lemma 3.1 summarizes these ideas for our context.
at z k , where P k (z) is holomorphic at z = z k and P k (z k ) = H −1 k .
Proof. The potential η has a simple pole at z k with residue A k . Since ϑ k / ∈ 1 2 Z \ {0}, by the theory of regular singular points, there exists a solution to the ODE dΨ = Ψη of the form where Q k (z) is holomorphic at z = z k and Q k (z k ) = 1. Since ξ = η.g 0 , then

The Confluent Heun Equation
We have illustrated in sections 2 and 3 why we want to prescribe two regular singularities and one irregular singularity in our potential and, consequently, in the scalar ODE (2.6) associated to the initial value problem (1.1). Let us do this now explicitly.
The simplest second order ODE with two regular singularities and one irregular singularity is the confluent Heun equation (CHE), which in its non-symmetrical canonical form is wirtten as where the parameters p, γ, δ, α, σ ∈ C. This equation arises as a result of the confluence of two regular singular points in the Heun equation [1], and counts the points z 0 = 0 and z 1 = 1 as regular singularities and z ∞ = ∞ as an irregular singular point of rank 1. For the porpouse of this work we are going to consider the CHE written as in [10], that is The parameters µ 0 , µ 1 , a, r 0 , r 1 are again complex, but we will need to restrict our choices later on.
Consider again the off-diagonal potential in (2.4) with functions ν := e 2f R and ρ := e −2f T , where f = S(w) dw, obtained in (2.3). Plugging ν and ρ into equation (2.6), it is easy to find expressions for the functions R, S and T . In particular, one way to prescribe the CHE (4.2) in the potential is with the following relations: Using these functions in the potential (2.3) and doing the gauge considered in section 2 we end up with an off-diagonal potential which has associated (4.2) as scalar ODE . Also, the correspondence with the parameters and functions used in the gauge in section 3 is as follows: set z 0 = 0 and z 1 = 1 and for k ∈ {0, 1} let Schäfke and Schmidt [10] give an asymptotic formula for the connection matrix between two sets of solutions of (4.2) in terms of their series expansion coefficients. Let be the series expansion of the unique solution to (4.2) which is holomorphic at z = 0 and satisfies y 0 (0) = 1.
where the asymptotic formula by which q can be calculated explicitly is Proof. The proof amounts to converting the notation of Schäfke-Schmidt to our notation. Define By [10, Proposition 2.14, Theorem 2.15], the connection matrixC =Φ 0Φ −1 The theorem follows by the relations between our notation and that of [10]: By standard methods (see [1, Part B, 2.2]), the sequence {c k } of coefficients defined by (4.5) satisfy the 3-term recurrence The recurrence in (4.12) and its polynomials (4.13) will be used to obtain unitarisability in section 5.

Unitarisability of the Monodromy
The proof of the next proposition is deferred to the appendix.
In our context, the two matrices to be unitarised are of the form ∆ 0 and C∆ 1 C −1 , where ∆ 0 and ∆ 1 are diagonal. The unitarisability criterion in Proposition 5.1 for this case can be expressed as follows.
Proposition 5.2. Consider two matrices M 0 := ∆ 0 and M 1 : are diagonal matrices and Then M 1 is upper or lower triangular if and only if at least one of a, b, c or d vanishes.
To prove (ii), since M 0 is diagonal, its eigenlines in CP 1 are ϕ 1 = 0 and ϕ 2 = ∞. Since M 1 C = C∆ 1 , the eigenlines of M 1 in C 2 are the columns of C, so its eigenlines in CP 1 are ψ 1 = a/c and ψ 2 = b/d. Then We will apply the above criterion to the monodromy of an ODE as follows. Consider a potential ξ with singular points z 0 and z 1 . Let τ 0 and τ 1 be the deck transformations corresponding to closed paths around z 0 and z 1 respectively. Let Φ 0 and Φ 1 be solutions to the ODE dΦ = Φξ chosen so that their respective monodromies be the connection matrix between these two solutions with entries as in 5.2. The monodromies of Φ 0 at z 0 and z 1 are respectively ad ∈ R − . For the remainder of the paper we choose all parameters in the CHE to be real. Also, we assume a > 0. Is easy to check that the coefficients U (k), V (k) and W (k) of the CHE recurrence (4.12) are positive for all sufficient large k. Under this assumption, the signs of the entries of the connection matrix C in (4.6) can be computed as follows.
The criterion for unitarisability in Proposition 5.2, the asymptotic formula for the connection matrix in Theorem 4.1, and the recurrence relation for the CHE solution in (4.12) yield the following sufficient condition for the unitarisability of the monodromy. Write the parameters for the CHE as a 5-tuple χ := (µ 0 , µ 1 , r 0 , r 1 , a) ∈ R 5 . Define the finite integer that is, by the remarks after Proposition 5.2, if and only if the monodromy is unitarisable.
6. Construction of New Trinoids 6.1. The trinoid potential. To construct trinoids we choose the potential and w 0 , w 1 ,r 0 ,r 1 , p ∈ R are free parameters. The parameters w 0 and w 1 will be the asymptotic end weights of the Delaunay ends at 0 and 1. The parametersr 0 andr 1 affect the weight of the irregular end, and p the shape of the trinoid. Note that, defining λ = e iθ with θ ∈ [0, 2π], the value of has the form of the CHE potential defined in 3 and 4, where the coefficients (µ 0 , µ 1 , r 0 , r 1 , a) in the CHE equation are related to the parameters (w 0 , w 1 ,r 0 ,r 1 , p) in the trinoid potential ξ T by The monodromy of the trinoid potential is unitarisable along S 1 if and only if that of the gauged potential ξ T .(Λ −1 ) is. Remark 6.2. Due to the structure of ξ T , any trinoid T constructed from ξ T in fact lies in a oneparameter family of trinoids T κ with monotonically varying Delaunay end weights. If (µ 0 , µ 1 , r 0 , r 1 , a) are the parameters for T , then the parameters for the family of trinoids T κ are (κw 0 , κw 1 , κr 0 , κr 1 , √ κp) with κ ranging over the interval (0, 1]. 6.3. Unitarisability of the monodromy. It remains to find values of the 5 parameters in ξ T so that the monodromy is unitarisable. An algorithm to test the hypotheses of Proposition 5.4 is as follows. For a 5-tuple χ = (µ 0 , µ 1 , r 0 , r 1 , a), consider the (k + 1)-coefficient of the recurrence (4.12), which is given by The radicals appearing in c +1 (χ) can be eliminated, reducing the problem to showing that a polynomial is positive in an interval. Let Θ = (w 0 , w 1 ,r 0 ,r 1 , p) ∈ R 5 be a choice of parameters for ξ T . Note Trinoids with one irregular end and two Delaunay ends. Graphics were produced with CMCLab [11].
The regular end weights are either 1 2 or − 1 2 while the irregular end weights vary.
Since G k is even in y 0 and in y 1 , then the function f k depending on k and Θ Proposition 6.3. Let Θ := (w 0 , w 1 ,r 0 ,r 1 , p) ∈ R 5 be a choice of parameters for the trinoid potential ξ T and let (6.8) Let k 0 ∈ N be such that for each of the four choices of signs, U (k, for each of the four choices χ ±± , and some t 0 ∈ (0, 1), (6.9) sign c k 0 −1 (χ ±± (t 0 )) = sign c k 0 (χ ±± (t 0 )), (iii) of the four signs in (ii) and odd number are + and an odd number are −.
Then, the monodromy with parameters Θ is unitarisable.
Proof. By its definition, f +1 (x) has a zero along x ∈ (0, 1) if and only if at least one of the four functions c +1 (χ ±± (t)) has a zero along t ∈ (0, 1). Thus by (i), none of the eight functions c k 0 −1 (χ ±± (t)) and c k 0 (χ ±± (t)) has a zero along t ∈ (0, 1). By (ii) and continuity, all χ ±± ∈ S + ∪ S − , where S + and S − are the sets defined in (5.6). The monodromy is unitarisable by Proposition 6.3(iii) and The examples in figure 1 were computed by Proposition 6.1 and the unitarisability criterion in Proposition 6.3.
Proof. The coefficients U (k), V (k) and W (k) of the recurrence (4.12) depend holomorphically on the parameters of the choice Θ. Each term of the sequence c k+1 (χ) is a rational function P k (χ) Q k (χ) , where P k and Q k are polynomials in W (0), . . . , W (k) and in V (0), . . . , V (k) and U (0), . . . , U (k) respectively. Thus they also depend holomorphically on the parameters. Note that under the assumptions made for the CHE parameters, in particular µ 0 < 1, the polynomial U (k) is never zero. Therefore, the function f k defined in (6.7) also depends holomorphically on the parameters.
Let Θ be as in 6.4, for which we have checked that the conditions of 6.3 are satisfied. The polynomials f 1 and f 2 have a zero of order 4 at x = 0. A calculation shows that this order is preserved under a small perturbation of Θ ∈ R 5 . For i ∈ {1, 2}, let us denote f := f i . We have that f (x) = x 4 g(x), where g has no zeros on [0, 1], that is, ε := inf x∈[0,1] |g(x)| > 0. We can make a small perturbation of g by choosingg such that |g −g| [0,1] < ε. If we considerf (x) := x 4g (x), then we obtain that Hence,f has no zeros on (0, 1]. It is also easy to check that x = 0 is not a zero of the perturbationg: It follows that the condition in Proposition 6.3(i) is preserved under small perturbations of Θ. Also conditions (ii) and (iii) are trivially preserved under such perturbations. Hence by Proposition 6.3, the monodromy of ξ T is unitarisable in a small neighborhood of Θ ∈ R 5 . Theorem 6.1 constructs a five parameter family of trinoids with one irregular end.
6.4. End weights. We conclude by computing the weight at the irregular end of a trinoid, in a similar fashion as in [8]. Let (6.14) ξ = 0 1 tQ 0 dz be a potential where Q is holomorphic on the circle |z| = 2 with Laurent series Q = ∞ k=−∞ a k z k .
The first few terms of the series of the monodromy M of the solution of dΦ = Φξ, Φ(2) = 1 along the circle |z| = 2 is as follows.
Proof. By the gauge Λ := diag(λ 1/2 , λ −1/2 ) we obtain Let Ψ = Ψ 0 + Ψ 1 t + O(t 2 ) be the solution to the initial value problem (6.18) dΨ = Ψξ, Ψ(2) = 1, and let P = P 0 + P 1 t + O(t 2 ) be the monodromy of Ψ. To compute P 0 and P 1 , equate the coefficients os powers of t in to obtain the two equations (6.20) where the path integral is along a path based at 2. Since α = z − 2, then γ α = 0 for γ(s) = 2e is , s ∈ [0, 2π]. Hence P 0 = 1. Solve (6.21) by computing (6.23) We are integrating along a path enclosing the two singular points so, by the residue theorem, we obtain (6.24) The series for the monodromy M of Φ now follows from M = Λ −1 P Λ and expressing the series in terms of θ as defined above.
The force associated to an element in the fundamental group [9,3,5] is the matrix A ∈ su 2 in the series expansion of the monodromy (6.25) where λ = e iθ . The force is a homomorphism from the fundamental group to su 2 ∼ = R 3 . Its length |A| = √ det A is the weight of the end.
Proposition 6.7. The weights of a trinoid constructed from ξ T with parameters (w 0 , w 1 ,r 0 ,r 1 , p) at z = 0, 1, ∞ are respectively w 0 , w 1 , w ∞ where Proof. By Proposition 6.6, the weight of the irregular end of a trinoid constructed using its potential (6.1) is given by where a k are the Laurent coefficients of Q as before. The result follows by a computation of these coefficients.
Remark 6.8. The three weights (lenghts of the weight vectors) determine the three weight vectors. It is unknown to the authors if a notion of end axis can be established for an irregular end. In the case of all three ends being regular, a necessary condition for the unitarisability of the monodromy comes from the balancing formula [5,12]: If the monodromy is unitary, the weight vectors W 0 , W 1 , W ∞ ∈ R 3 satisfy W 0 + W 1 + W ∞ = 0. The weights are w k = ±|W k |. It follows that |w i | ≤ |w j | + |w k | for all permutations of (0, 1, ∞). A counterexample can be found in the presence of one irrgular end: For the parameters (w 0 , w 1 ,r 0 ,r 1 , p) = 1 2 , 1 4 , 1 4 , 17 128 , 1 8 the resulting monodromy is unitarisable, but the balancing formula does not hold for all permutations of (0, 1, ∞). This counterexample questions the suitability of (6.26) as the way of measuring the end weight at ∞. Since with this definition the notion of balancing is not preserved, it might be interesting if this could be modified so that balancing still holds.
Appendix A. Appendix: Geometry of Unitarisability It remains to prove Proposition 5.1, which give a criterion for the simultaneous unitarisability of two matrices in SL 2 C in terms of their eigenlines.
Then A = GL 2 C/GU 2 = SL 2 C/SU 2 = H 3 . To show B = CP 1 , note that any X ∈ N with det X = 0 can be written in the form with (x, y) t , (a, b) ∈ C 2 \ {0}. The first factor (x, y) t is unique up to multiplication by an element of C * , so the map φ : B → CP 1 given by is well defined, and a bijection, since GU 2 acts transitively on C 2 \ {0}.
Definition A.1. For the sake of clarification and to fix notation, we say that Proof. M 1 , . . . , M n ∈ SL 2 C are simultaneously unitarisable if and only if there exists X ∈ SL 2 C such that X −1 M k X ∈ SU 2 for all k ∈ {1, . . . , n}. This is equivalent to X ∈ fix(M k ) for all k ∈ {1, . . . , n}.
We Proof. Since M ∈ SL 2 C is unitarisable, by Proposition A.