Constant Mean Curvature Trinoids with one Irregular End
CCONSTANT MEAN CURVATURE TRINOIDS WITH ONEIRREGULAR END
MARTIN KILIAN, EDUARDO MOTA, AND NICHOLAS SCHMITT
Abstract.
We construct a new five parameter family of constant mean curvature tri-noids with two asymptotically Delaunay ends and one irregular end.
Introduction
The generalised Weierstrass representation [4] constructs conformal constant mean cur-vature (CMC) immersions in Euclidean 3-space from a holomorphic 1-form ξ on a Riemannsurface 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 casethat Σ is the thrice-punctured Riemann sphere and ξ has three regular singular points atthe punctures, the resulting three-parameter family of CMC trinoids have asymptoticallyDelaunay ends [12]. In this paper we extend this result by replacing one of the regular sin-gularities with an irregular singularity of rank 1. The corresponding second-order scalarODE is the confluent Heun equation (CHE) with five free parameters. The monodromycan be computed by an asymptotic formula for the connection matrix between solutions atthe two regular singular points given by Sch¨afke-Schmidt [10]. In this way, we construct anew five-parameter family of CMC trinoids with two Delaunay ends and one irregular end.1. The Generalised Weierstrass Representation and the MonodromyProblem
Let us briefly recall the generalized Weierstrass representation [4] to set the notationand 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 → sl ( C ). The potential ξ has a simple polein its upper right entry in the loop parameter λ at λ = 0, and has no other poles in theunit λ disk. Moreover, the upper-right entry of ξ is non-zero on Σ. Let Φ be a solution of(1) dΦ = Φ ξ . (ii) Let Φ = F B be the pointwise Iwasawa factorization on the universal cover (cid:101)
Σ.(iii) Then f = F (cid:48) F − is an associated family of conformal CMC immersions (cid:101) Σ → su ∼ = R . The prime denotes differentiation with respect to θ , where λ = e iθ .The gauge action is defined by ξ.g := g − ξg + g − d g . If dΦ = Φ ξ , then Φ g solvesdΨ = Ψ( ξ.g ). If g is a map on Σ with values in a positive loop group of SL C , then ξ.g is again a potential. Mathematics Subject Classification.
Primary 53A10.
Key words and phrases.
Constant mean curvature surfaces, irregular singularities, period problem. a r X i v : . [ m a t h . DG ] M a r M. KILIAN, E. MOTA, AND N. SCHMITT
Now let Σ = C \ { z , z , z ∞ } be the thrice-punctured Riemann sphere with punctures z = 0 , z = 1 and z ∞ = ∞ . Let ∆ denote the group of deck transformations of theuniversal cover ˜Σ. Let ξ be a holomorphic potential on Σ. Then ξ ( τ ( z ) , λ ) = ξ ( z, λ ) forall τ ∈ ∆. Let Φ be a solution of dΦ = Φ ξ . Note that in general Φ it is only defined on˜Σ. We define the monodromy M τ with respect to τ by(2) M τ = Φ( τ ( z ) , λ ) Φ( z, λ ) − . The period problem cannot be solved simultaneously for the whole associated family, sowe contend ourselves with solving it for the member of the associated family λ = 1. For i ∈ { , , ∞} let γ i be a loop around the puncture z i and M i the monodromy of Φ alongthis loop. The period problem consists of the following three conditions [12]: For all i ∈ { , , ∞} (3) M i takes values in the unitary loop group ,M i | λ =1 = ± Id ,∂ λ M i | λ =1 = 0 . The Associated Second Order
ODEConsider a holomorphic potential(4) η = (cid:18) S ( z, λ ) R ( z, λ ) T ( z, λ ) − S ( z, λ ) (cid:19) dz on Σ = C \ { z , z , z ∞ } . Set(5) g = (cid:18) e − f ( z,λ ) e f ( z,λ ) (cid:19) , where f ( z, λ ) = (cid:90) zz ∗ S ( w, λ ) dw. Then g is a positive loop and gauging the holomorphic potential η by g yields(6) η := η .g = (cid:18) e f R ( z, λ ) e − f T ( z, λ ) 0 (cid:19) dz. Hence we can assume without loss of generality that our potential is off-diagonal. Let usthen consider a potential, denoted η , of the form(7) η = (cid:18) ν ( z, λ ) ρ ( z, λ ) 0 (cid:19) dz. For the purpose of our work it will be convenient to work with the associated scalar secondorder ODE corresponding to the 2 × Lemma . Solutions of dΦ = Φ η are of the form (8) (cid:18) y (cid:48) /ν y y (cid:48) /ν y (cid:19) where y and y are a fundamental system of the scalar ODE(9) y (cid:48)(cid:48) − ν (cid:48) ν y (cid:48) − ρ ν y = 0 . MC TRINOIDS WITH ONE IRREGULAR END 3 Two Regular Singular Points
Our goal is to construct CMC trinoids for which two ends are regular and one end isirregular. We will assume that at the two ends z and z , the potential is a holomorphicperturbation of a Delaunay potential, and hence regular singular there. In addition, ourpotentials will have a singularity of rank 1 at z ∞ , making it an irregular end. Thesechoices will determine the form of the associated scalar ODE (9).Note that in this and the following sections, we omit the dependence on the spectralparameter λ . Let ϑ , ϑ ∈ C \ Z be parameters, and a and b functions on Σ such that a is holomorphic at z and z , and b is allowed to have simple poles at z and z . Define ξ := (cid:18) Q ( z ) 0 (cid:19) dz,Q ( z ) := ϑ ( ϑ − z − z ) + ϑ ( ϑ − z − z ) + b ( z ) . The points z and z are regular singular points of ξ . These double poles can be gaugedto simple poles by(10) g := (cid:18) G (cid:19) , where G ( z ) = ϑ z − z + ϑ z − z + a ( z ) , to obtain(11) η := ξ. (cid:0) g − (cid:1) = A dzz − z + A dzz − z + B dz where B is holomorphic at z and z and(12) A k := H k ∆ k H − k , ∆ k := diag( − ϑ k , ϑ k ) , H k := (cid:18) ∗ (cid:19) , k ∈ { , } . It is only left to see that a general potential like in (4), the one we started with, isequivalent to a potential of the form in (11) by the gauge(13) g = (cid:18) S / S − / (cid:19) . The z A P lemma. Suppose dΦ = Φ η for which η = A dzz − z k + O ( z ) dz has a simplepole at z = z k and Delaunay residue A . A standard result in the theory of regularsingularities states that under certain conditions on the eigenvalues of A , there exists asolution of the form Φ = z A P = exp ( A log z ) P , where P extends holomorphically to z = z k (see [12, Lemma 14]). Theorem 3.1 summarizes these ideas for our context. Lemma . For k ∈ { , } , the ODE dΦ = Φ ξ has solutions (14) Φ k ( z ) = exp (cid:0) ∆ k log (cid:0) ( − k ( z − z k ) (cid:1)(cid:1) P k ( z ) g ( z ) , at z k , where P k ( z ) is holomorphic at z = z k and P k ( z k ) = H − k . Proof.
The potential η has a simple pole at z k with residue A k . Since ϑ k / ∈ Z \ { } ,by the theory of regular singular points, there exists a solution to the ODE d Ψ = Ψ η ofthe form Ψ k ( z ) = exp ( A k log ( z − z k )) Q k ( z ) , M. KILIAN, E. MOTA, AND N. SCHMITT where Q k ( z ) is holomorphic at z = z k and Q k ( z k ) = . Since ξ = η.g , thenˆΦ k ( z ) = H − k exp ( A k log ( z − z k )) Q k ( z ) g ( z ) , is a solution to the ODE dΦ = Φ ξ . Since A k H k = H k ∆ k , thenˆΦ k ( z ) = exp (∆ k log ( z − z k )) H − k Q k ( z ) g ( z ) . The theorem follows with P k := H − k Q k , Φ := ˆΦ and Φ := diag (cid:0) ( − − ϑ , ( − ϑ (cid:1) ˆΦ . (cid:3) The Confluent Heun Equation
We have illustrated in sections 2 and 3 why we want to prescribe two regular singularitiesand one irregular singularity in our potential and, consequently, in the scalar ODE (9)associated to the initial value problem (1). Let us do this now explicitly.The simplest second order ODE with two regular singularities and one irregular sin-gularity is the confluent Heun equation (CHE), which in its non-symmetrical canonicalform is written as(15) y (cid:48)(cid:48) + (cid:18) p + γz + δz − (cid:19) y (cid:48) + 4 pαz − σz ( z − y = 0where the parameters p, γ, δ, α, σ ∈ C . This equation arises as a result of the confluenceof two regular singular points in the Heun equation [1], and counts the points z = 0 and z = 1 as regular singularities and z ∞ = ∞ as an irregular singular point of rank 1. Forthe purpose of this work we are going to consider the CHE written as in [10], that is y (cid:48)(cid:48) + (cid:18) a + 1 − µ z + 1 − µ z − (cid:19) y (cid:48) + (cid:18) z [ a (2 − µ − µ ) − ( r + r )] + ( µ µ − a (1 − µ ) − ( µ + µ ) + 2 r + 1) z ( z − (cid:19) y = 0 , (16)The parameters µ , µ , a, r , r are again complex, but we will need to restrict our choiceslater on.Consider again the off-diagonal potential in (7) with functions ν := e f R and ρ := e − f T , where f = (cid:82) S ( w ) dw , obtained in (6). Plugging ν and ρ into equation (9), it iseasy to find expressions for the functions R , S and T . In particular, one way to prescribethe CHE (16) in the potential is with the following relations: R ( z ) := ( z − µ − z µ − ,S ( z ) := − a, (17) T ( z ) := − z − µ ( z − − µ [ z ( a (2 − µ − µ ) − ( r + r ))+ 12 ( µ µ − a (1 − µ ) − ( µ + µ ) + 2 r + 1)] . Using these functions in the potential (6) and doing the gauge considered in section 2 weend up with an off-diagonal potential which has associated (16) as scalar ODE . Also,
MC TRINOIDS WITH ONE IRREGULAR END 5 the correspondence with the parameters and functions used in the gauge in section 3 isas follows: set z = 0 and z = 1 and for k ∈ { , } let ϑ k := 1 − µ k ,a ( z ) := a, (18) b ( z ) := r z + r z − a . Sch¨afke and Schmidt give in [10] an asymptotic formula for the connection coefficientsbetween a set of two solutions of equation (16) around z = 0 and another set of twosolutions of equation (16) around z = 1, in terms of their series expansion coefficients.From the two equations appearing in Proposition 2.14 in [10], we only need the first onein order to write down our matrix relationship C = Φ Φ − , as in what follows we willonly consider the unique solution at z = 0. These connection coefficients can be writtenin a matrix form allowing us to define the connection matrix between two solutions Φ and Φ . Let(19) y ( z ) = Γ(1 − µ ) ∞ (cid:88) k =0 c k z k be the series expansion of the unique solution to equation (16) which is holomorphic at z = 0 and satisfies y (0) = 1. Theorem . [10] The connection matrix C := Φ Φ − is (20) C = (cid:18) Γ( µ )Γ(1 − µ )
00 1 (cid:19) (cid:18) q ( µ , − µ ) q ( µ , µ ) q ( − µ , − µ ) q ( − µ , µ ) (cid:19) (cid:18) Γ( µ )Γ(1 − µ ) (cid:19) , where the asymptotic formula by which q can be calculated explicitly is (21) q ( µ , µ ) = Γ(1 − µ )Γ(1 − µ ) lim k →∞ Γ( k + 1)Γ( k − µ ) c k . Proof.
The proof amounts to converting the notation of Sch¨afke-Schmidt to our no-tation. Define(22) ˜Φ k = G − D − k Φ k g − , D k := diag (cid:0) Γ( µ k ) , ( − k Γ(1 − µ k ) (cid:1) , k ∈ { , } . By [10, Proposition 2.14, Theorem 2.15], the connection matrix ˜ C = ˜Φ ˜Φ − ∈ GL C is(23) ˜ C = Γ( µ )Γ(1 − µ ) (cid:18) ˜ q ( µ , − µ ) − ˜ q ( µ , µ )˜ q ( − µ , − µ ) − ˜ q ( − µ , µ ) (cid:19) , where(24) ˜ q ( µ , µ ) = lim k →∞ Γ( k + 1)Γ( k − µ ) c k . The theorem follows by the relations between our notation and that of [10]:(25) q ( µ , µ ) = Γ(1 − µ )Γ(1 − µ )˜ q ( µ , µ ) and C = D ˜ CD − . (cid:3) M. KILIAN, E. MOTA, AND N. SCHMITT
By standard methods (see [1, Part B, 2.2]), the sequence { c k } of coefficients defined byequation (19) satisfy the 3-term recurrence(26) U ( k ) c k +1 = V ( k ) c k + W ( k ) c k − , c − = 0 , c = 1where U ( k ) := (1 + k )(1 + k − µ ) ,V ( k ) := k ( k + 1 − a − µ − µ ) + 12 ( µ − µ −
1) + a ( µ −
1) + r , (27) W ( k ) := a (2 k − µ − µ ) − ( r + r ) . The recurrence in (26) and its polynomials in (27) will be used to obtain unitarisabilityin section 5. 5.
Unitarisability of the Monodromy
The proof of the next proposition is deferred to the appendix.
Proposition . Let M , M ∈ SL ( C ) \ {± } be irreducible and individually uni-tarisable. Let ϕ, ϕ (cid:48) ∈ C P and ψ, ψ (cid:48) ∈ C P be the respective eigenlines of M and M .Then M and M are simultaneously unitarisable if and only if the cross-ratio (28) [ ϕ, ψ, ϕ (cid:48) , ψ (cid:48) ] ∈ R − . In our context, the two matrices to be unitarised are of the form ∆ and C ∆ C − ,where ∆ and ∆ are diagonal. The unitarisability criterion in Theorem 5.1 for this casecan be expressed as follows. Proposition . Consider two matrices M := ∆ and M := C ∆ C − , where ∆ , ∆ ∈ SL ( C ) \ {± } are diagonal matrices and (29) C =: (cid:18) a bc d (cid:19) ∈ SL ( C ) . (i) M and M are irreducible if and only if a, b, c and d are non-zero.(ii) If M and M are irreducible and individually unitarisable, then M and M aresimultaneously unitarisable if and only if the ratio bcad ∈ R − . Proof.
For (i), since M is diagonal and not ± , then M and M are reducible if andonly if M is upper or lower triangular. Writing ∆ = diag( β, β − ), compute(30) M = C ∆ C − = (cid:18) adβ − bcβ − − ab ( β − β − ) cd ( β − β − ) adβ − − bcβ (cid:19) . Then M is upper or lower triangular if and only if at least one of a, b, c or d vanishes.To prove (ii), since M is diagonal, its eigenlines in C P are ϕ = 0 and ϕ = ∞ . Since M C = C ∆ , the eigenlines of M in C are the columns of C , so its eigenlines in C P are ψ = a/c and ψ = b/d . Then(31) [ ϕ , ψ , ϕ , ψ ] = ψ ψ = bcad . By Theorem 5.1, M and M are simultaneously unitarisable if and only if this cross ratiois in R − . (cid:3) MC TRINOIDS WITH ONE IRREGULAR END 7
We will apply the above criterion to the monodromy of an ODE as follows. Considera potential ξ with singular points z and z . Let τ and τ be the deck transformationscorresponding to closed paths around z and z respectively. Let Φ and Φ be solutionsto the ODE dΦ = Φ ξ chosen so that their respective monodromies∆ := Φ ( τ ( z ))Φ − and ∆ := Φ ( τ ( z ))Φ − are diagonal. Let C = Φ Φ − be the connection matrix between these two solutions withentries as in (29). The monodromies of Φ at z and z are respectively M = ∆ and M = Φ ( τ ( z ))Φ − = C Φ ( τ ( z ))Φ − C − = C ∆ C − . Suppose M and M are irreducible and individually unitarisable. By Theorem 5.2, M and M are simultaneously unitarisable if and only if bcad ∈ R − . For the remainder of thepaper we choose all parameters in the CHE to be real. Also, we assume a >
0. Is easy tocheck that the coefficients U ( k ), V ( k ) and W ( k ) of the CHE recurrence (26) are positivefor all sufficient large k . Under this assumption, the signs of the entries of the connectionmatrix C in (20) can be computed as follows. Proposition . Suppose µ j < for j ∈ { , } and that there exists k ∈ N such that U ( k ) > , V ( k ) > and W ( k ) > for all k ≥ k . If c k − > and c k > , then q definedin equation (21) satisfies q ≥ . Similarly, if c k − < and c k < , then q ≤ . Proof.
Since µ j < j ∈ { , } and the function Γ( x ) > x >
0, we havethat Γ(1 − µ j ) > j ∈ { , } , Γ( k + 1) > k − µ ) > k ≥ k .By hypothesis U ( k ), V ( k ) and W ( k ) are positive for all k ≥ k . Hence, in the case of c k − > c k >
0, the third term c k +1 must be positive as well. Then, by induction, { c k } ∞ k = k − are all positive coefficients. This implies that q ≥ c k − < c k < { c k } ∞ k = k − are negative and therefore q ≤ (cid:3) The criterion for unitarisability in Theorem 5.2, the asymptotic formula for the con-nection matrix in Theorem 4.1, and the recurrence relation for the CHE solution in equa-tion (26) yield the following sufficient condition for the unitarisability of the monodromy.Write the parameters for the CHE as a 5-tuple χ := ( µ , µ , r , r , a ) ∈ R . Define thefinite integer(32) m ( χ ) := min k ∈ N { U ( k, χ ) > , V ( k, χ ) > W ( k, χ ) > k ≥ k } , and the sets S + := { χ ∈ R | there exists (cid:96) ≥ m ( χ ) such that c (cid:96) − > c (cid:96) > } , (33) S − := { χ ∈ R | there exists (cid:96) ≥ m ( χ ) such that c (cid:96) − < c (cid:96) < } . Proposition . If each of the four 5-tuples ( ± µ , ± µ , r , r , a ) ∈ R lies in S + ∪ S − ,then the monodromy is unitarisable if and only if an odd number of these tuples lie in S + . Proof.
By Theorem 5.3, if χ ∈ S + then q ( χ ) ≥
0, and if χ ∈ S − then q ( χ ) ≤
0. Anodd number of the tuples lie in S + if and only if(34) q ( µ , µ ) q ( − µ , − µ ) q ( µ , − µ ) q ( − µ , µ ) ∈ R − , that is, by the remarks after Theorem 5.2, if and only if the monodromy is unitarisable. (cid:3) M. KILIAN, E. MOTA, AND N. SCHMITT Construction of New Trinoids
The trinoid potential.
To construct trinoids we choose the potential(35) ξ T = (cid:18) λ − λQ T (cid:19) dz, where(36) Q T := t (cid:18) − w z − w z − + ˆ r z + ˆ r z − p (cid:19) , t := − λ − ( λ − , and w , w , ˆ r , ˆ r , p ∈ R are free parameters. The parameters w and w will be theasymptotic end weights of the Delaunay ends at 0 and 1. The parameters ˆ r and ˆ r affectthe weight of the irregular end, and p the shape of the trinoid. Note that, defining λ = e iθ with θ ∈ [0 , π ], the value of t = − λ − ( λ − = sin θ ∈ [0 , λ / , λ − / ), the gauged potential(37) ξ T . (Λ − ) = (cid:18) Q T (cid:19) dz has the form of the CHE potential defined in sections 3 and 4, where the coefficients( µ , µ , r , r , a ) in the CHE equation are related to the parameters ( w , w , ˆ r , ˆ r , p ) inthe trinoid potential ξ T by(38) µ k = √ − w k t, r k = ˆ r k t, a = p t, k ∈ { , } . The monodromy of the trinoid potential is unitarisable along S if and only if that of thegauged potential ξ T . (Λ − ) is.6.2. Construction of trinoids.
Theorem . Let ξ T be a trinoid potential with unitarisable monodromy on S minusa finite set. Let Φ be a solution of dΦ = Φ ξ T . Then there exists a positive dressing h suchthat the CMC immersion induced by h Φ via the generalized Weierstrass representation onthe universal cover descends to the three-punctured sphere. The ends at z = 0 and z = 1 are asymptotic to Delaunay surfaces. Proof.
By [12], there exists a positive loop h : D → GL C such that the monodromyof h Φ is unitary. The local unitary monodromies M and M satisfy the closing conditions M k (0) = ± and M (cid:48) k (0) = 0, for k ∈ { , } . Hence h Φ induces an immersion of the three-punctured sphere via the GL C version of the generalised Weierstrass representation. Theends at z = 0 and z = 1 are asymptotic to Delaunay cylinders with respective weights w and w by [7]. (cid:3) Remark . Due to the structure of ξ T , any trinoid T constructed from ξ T in factlies in a one-parameter family of trinoids T κ with monotonically varying Delaunay endweights. If ( µ , µ , r , r , a ) are the parameters for T , then the parameters for the familyof trinoids T κ are ( κw , κw , κ ˆ r , κ ˆ r , √ κp ) with κ ranging over the interval (0 , MC TRINOIDS WITH ONE IRREGULAR END 9
Unitarisability of the monodromy.
It remains to find values of the 5 parametersin ξ T so that the monodromy is unitarisable. An algorithm to test the hypotheses ofTheorem 5.4 is as follows. For a 5-tuple χ = ( µ , µ , r , r , a ), consider the ( k + 1)-coefficient of the recurrence in (26), which is given by(39) c k +1 ( χ ) = V ( k, χ ) c k ( χ ) + W ( k, χ ) c k − ( χ ) U ( k, χ ) . The radicals appearing in c (cid:96) +1 ( χ ) can be eliminated, reducing the problem to showingthat a polynomial is positive in an interval. Let Θ = ( w , w , ˆ r , ˆ r , p ) ∈ R be a choice ofparameters for ξ T . Note that c (cid:96) +1 ( y , y , ˆ r x , ˆ r x , px ) defines a rational function P (cid:96) ( y , y , ˆ r x , ˆ r x , px ) Q (cid:96) ( y , y , ˆ r x , ˆ r x , px )for some polynomials P (cid:96) , Q (cid:96) ∈ R [ x, y , y ] depending on (cid:96) and Θ. Define the polynomialfunctions F k , G k ∈ R [ x, y , y ] F k ( x, y , y ) := P k ( y , y , ˆ r x , ˆ r x , px ) , (40) G k ( x, y , y ) := F k ( x, y , y ) F k ( x, y , − y ) F k ( x, − y , y ) F k ( x, − y , − y ) . Since G k is even in y and in y , then the function f k depending on k and Θ(41) f k ( x ) := G k (cid:16) x, (cid:112) − w x , (cid:112) − w x (cid:17) is in R [ x ]. Proposition . Let
Θ := ( w , w , ˆ r , ˆ r , p ) ∈ R be a choice of parameters for thetrinoid potential ξ T and let (42) χ ±± := ( ± µ , ± µ , r , r , a ) = (cid:16) ±√ − w t, ±√ − w t, ˆ r t, ˆ r t, p √ t (cid:17) . Let k ∈ N be such that for each of the four choices of signs, U ( k, χ ±± ) > , V ( k, χ ±± ) > and W ( k, χ ±± ) > for all k ≥ k . Suppose(i) f k − ( x ) (cid:54) = 0 and f k ( x ) (cid:54) = 0 along x ∈ (0 , ,(ii) for each of the four choices χ ±± , and some t ∈ (0 , , (43) sign c k − ( χ ±± ( t )) = sign c k ( χ ±± ( t )) , (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 (cid:96) +1 ( x ) has a zero along x ∈ (0 ,
1) if and only if at least oneof the four functions c (cid:96) +1 ( χ ±± ( t )) has a zero along t ∈ (0 , c k − ( χ ±± ( t )) and c k ( χ ±± ( t )) has a zero along t ∈ (0 , χ ±± ∈ S + ∪ S − , where S + and S − are the sets defined in (33). The monodromy isunitarisable by Theorem 6.3(iii) and Theorem 5.4. (cid:3) The examples in figure 1 were computed by Theorem 6.1 and the unitarisability criterionin Theorem 6.3.
Proposition . The conditions of Theorem 6.3 are satisfied for the choice of param-eters (44) Θ = (cid:18) , , − , , (cid:19) . Figure 1.
Trinoids with one irregular end and two Delaunay ends. Graphics wereproduced with CMCLab [11]. The regular end weights are either or − while theirregular end weights vary. The parameters ( w , w , ˆ r , ˆ r , p ) used to construct each ofthem are (cid:0) , , − , , (cid:1) , (cid:0) , − , − , , (cid:1) , and (cid:0) − , − , − , , (cid:1) . Proof.
For this choice of Θ, we compute f ( x ) = x (cid:0) − x + 243712 x − x +26368 x − x + 1008 x − x + 27 x (cid:1) × (cid:0) − x − x − x +1267968 x − x + 189328 x − x + 6859 x (cid:1) f ( x ) = x (cid:0) − x + 6455033856 x − x +762642432 x − x + 41717760 x − (45) 5871616 x + 1031424 x − x + 12240 x − x + 81 x (cid:1) × (cid:0) − x − x +358612992 x + 24553881600 x − x +7316135936 x − x + 570771456 x − x + 16092368 x − x + 130321 x (cid:1) Sturm’s theorem applied to f and f shows that these two polynomials have no zeroon the interval (0 , i ). The conditions ( ii ) and ( iii ) areverified by computing at t = , for k ∈ { , } , yielding c k ( χ ++ ( t )) < c k ( χ + − ( t )) > c k ( χ − + ( t )) > c k ( χ −− ( t )) > (cid:3) Theorem . There exists a five parameter family of
CMC trinoids with two Delaunayends and one irregular end.
Proof.
The coefficients U ( k ), V ( k ) and W ( k ) of the recurrence in (26) depend holo-morphically on the parameters of the choice Θ. Each term of the sequence c k +1 ( χ ) isa 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 holomorphicallyon the parameters. Note that under the assumptions made for the CHE parameters, inparticular µ <
1, the polynomial U ( k ) is never zero. Therefore, the function f k definedin (41) also depends holomorphically on the parameters. Let Θ be as in Theorem 6.4, for MC TRINOIDS WITH ONE IRREGULAR END 11 which we have checked that the conditions of Theorem 6.3 are satisfied. The polynomials f and f have a zero of order 4 at x = 0. A calculation shows that this order is preservedunder a small perturbation of Θ ∈ R . For i ∈ { , } , let us denote f := f i . We havethat f ( x ) = x g ( x ), where g has no zeros on [0 , ε := inf x ∈ [0 , | g ( x ) | >
0. Wecan make a small perturbation of g by choosing ˜ g such that | g − ˜ g | [0 , < ε . If we consider˜ f ( x ) := x ˜ g ( x ), then we obtain that | ˜ f ( x ) | = x | ˜ g ( x ) | = x | g ( x ) − ( g ( x ) − ˜ g ( x )) |≥ x || g ( x ) | − | g ( x ) − ˜ g ( x ) || (46) > x | ε − ε | = 0 . Hence, ˜ f has no zeros on (0 , x = 0 is not a zero of theperturbation ˜ g :(47) | ˜ g (0) | = | g (0) − ( g (0) − ˜ g (0)) | ≥ || g (0) | − | g (0) − ˜ g (0) || > | ε − ε | = 0 . It follows that the condition in Theorem 6.3( i ) is preserved under small perturbations ofΘ. Also conditions ( ii ) and ( iii ) are trivially preserved under such perturbations. Henceby Theorem 6.3, the monodromy of ξ T is unitarisable in a small neighborhood of Θ ∈ R .Theorem 6.1 constructs a five parameter family of trinoids with one irregular end. (cid:3) End weights.
We conclude by computing the weight at the irregular end of a trinoid,in a similar fashion as in [8]. Let(48) ξ = (cid:18) tQ (cid:19) dz be a potential where Q is holomorphic on the circle | z | = 2 with Laurent series Q = (cid:80) ∞ k = −∞ a k z k . The first few terms of the series of the monodromy M of the solution ofdΦ = Φ ξ , Φ(2) = along the circle | z | = 2 is as follows. Proposition . Let M be the monodromy with respect to the curve γ ( s ) = 2 e is , s ∈ [0 , π ] of Φ for the trinoid potential ξ T with Φ(2) = . Define λ = e iθ and let (cid:80) ∞ k =0 M k θ k be the series expansion of M along | z | = 2 . Then M = , M = 0 and (49) M = 2 πi (cid:18) a − − a − a − − a − − a − a − a − − a − (cid:19) . Proof.
By the gauge Λ := diag( λ / , λ − / ) we obtain(50) ξ = ξ T . (Λ − ) = ξ + tξ , ξ = (cid:18) α (cid:19) , ξ = β (cid:18) (cid:19) , where(51) α = dz, β = Q dz and t = − λ − ( λ − = sin θ . Let Ψ = Ψ + Ψ t + O ( t ) be the solution to the initial value problem(52) d Ψ = Ψ ξ, Ψ(2) = , and let P = P + P t + O ( t ) be the monodromy of Ψ. To compute P and P , equatethe coefficients as powers of t in(53) d Ψ + d Ψ t + O ( t ) = (cid:0) Ψ + Ψ t + O ( t ) (cid:1) ( ξ + ξ t ) to obtain the two equations(54) d Ψ = Ψ ξ , Ψ (2) = , (55) d (cid:0) Ψ Ψ − (cid:1) = Ψ ξ Ψ − , Ψ (2) = 0 . The solution to (54) is(56) Ψ = (cid:18) (cid:82) α (cid:19) , where the path integral is along a path based at 2. Since (cid:82) α = z −
2, then (cid:82) γ α = 0 for γ ( s ) = 2 e is , s ∈ [0 , π ]. Hence P = . Solve (55) by computing(57) Ψ Ψ − = (cid:90) Ψ ξ Ψ − = (cid:90) β (cid:18)(cid:82) α − (cid:0)(cid:82) α (cid:1) − (cid:82) α (cid:19) = (cid:90) β (cid:18) z − − ( z − − z (cid:19) . We are integrating along a path enclosing the two singular points so, by the residuetheorem, we obtain(58) P = (cid:18)(cid:90) γ (cid:0) Ψ ξ Ψ − (cid:1)(cid:19) P = 2 πi (cid:18) a − − a − a − − a − − a − a − a − − a − (cid:19) . The series for the monodromy M of Φ now follows from M = Λ − P Λ and expressing theseries in terms of θ as defined above. (cid:3) The force associated to an element in the fundamental group [9, 3, 5] is the matrix A ∈ su in the series expansion of the monodromy(59) M = + Aθ + O ( θ )where λ = e iθ . The force is a homomorphism from the fundamental group to su ∼ = R .Its length | A | = √ det A is the weight of the end. Proposition . The weights of a trinoid constructed from ξ T with parameters ( w , w , ˆ r , ˆ r , p ) at z = 0 , , ∞ are respectively w , w , w ∞ where (60) w ∞ = π (cid:112) ( w + w ) + 8ˆ r ( w − r ) − r w . Proof.
By Theorem 6.6, the weight of the irregular end of a trinoid constructed usingits potential (35) is given by(61) π (cid:113) a − − a − a − where a k are the Laurent coefficients of Q as before. The result follows by a computationof these coefficients. (cid:3) Remark . The three weights (lengths of the weight vectors) determine the threeweight vectors. It is unknown to the authors if a notion of end axis can be establishedfor an irregular end. In the case of all three ends being regular, a necessary conditionfor the unitarisability of the monodromy comes from the balancing formula [5, 12]: Ifthe monodromy is unitary, the weight vectors W , W , W ∞ ∈ R satisfy W + W + W ∞ = 0. The weights are w k = ±| W k | . It follows that | w i | ≤ | w j | + | w k | for allpermutations of (0 , , ∞ ). A counterexample can be found in the presence of one irrgularend: For the parameters ( w , w , ˆ r , ˆ r , p ) = (cid:0) , , , , (cid:1) the resulting monodromy isunitarisable, but the balancing formula does not hold for all permutations of (0 , , ∞ ). MC TRINOIDS WITH ONE IRREGULAR END 13
This counterexample questions the suitability of equation (60) as the way of measuringthe 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 Theorem 5.1, which give a criterion for the simultaneous unitaris-ability of two matrices in SL ( C ) in terms of their eigenlines.A.1. Hyperbolic -space. Hyperbolic 3-space H can be identified with the quotientSL ( C ) / SU . For X ∈ SL ( C ), let [ X ] ∈ H denote the left coset(A-1) [[ X ]] := { XU | U ∈ SU } .M ∈ SL ( C ) acts isometrically on the hyperbolic 3-space H by(A-2) [[ X ]] (cid:55)→ [[ M X ]] . The fixed point set of this action is(A-3) fix( M ) = { [[ X ]] ∈ H | X − M X ∈ SU } . The fixed point set fix( M ), if non-empty, is called the axis of M . Hyperbolic 3-space canbe extended to include the sphere at infinity as follows. Let(A-4) GU := { X ∈ M × C | XX ∗ = x , x (cid:54) = 0 } be the group of unitary similitudes. Let N := M × C ∗ and Ξ := N / GU . Then Ξ = A (cid:116) B where(A-5) A := { [[ X ]] ∈ Ξ | det X (cid:54) = 0 } , B := { [[ X ]] ∈ Ξ | det X = 0 } . Then A = GL C / GU = SL ( C ) / SU = H . To show B = C P , note that any X ∈ Nwith det X = 0 can be written in the form(A-6) X = (cid:18) xy (cid:19) (cid:0) a b (cid:1) with ( x, y ) t , ( a, b ) ∈ C \ { } . The first factor ( x, y ) t is unique up to multiplication by anelement of C ∗ , so the map φ : B → C P given by(A-7) (cid:20)(cid:20)(cid:18) xy (cid:19) (cid:0) a b (cid:1)(cid:21)(cid:21) (cid:55)→ [ x, y ]is well defined, and a bijection, since GU acts transitively on C \ { } .A.2. Unitarisability.
Definition
A.1 . For the sake of clarification and to fix notation, we say that • M ∈ SL ( C ) is unitarisable if there exists X ∈ SL ( C ) such that X − M X ∈ SU . • M , . . . , M n ∈ SL ( C ) are simultaneously unitarisable if there exists X ∈ SL ( C )such that X − M k X ∈ SU for k ∈ { , . . . , n } . • M ∈ SL ( C ) is irreducible if it is not similar via a permutation to a block uppertriangular matrix.The next proposition follows immediately from (A-3). Proposition
A.2 . M ∈ SL ( C ) is unitarisable if and only if it has an axis. Proposition
A.3 . Individually unitarisable matrices M , . . . , M n ∈ SL ( C ) are simul-taneously unitarisable if and only if their axes intersect in a common point. Proof. M , . . . , M n ∈ SL ( C ) are simultaneously unitarisable if and only if there exists X ∈ SL ( C ) such that X − M k X ∈ SU for all k ∈ { , . . . , n } . This is equivalent to X ∈ fix( M k ) for all k ∈ { , . . . , n } . (cid:3) A.3.
Eigenlines.
The fixed point set fix( M ) of M ∈ SL ( C ) \ {± } is a disjoint unionof a (possibly empty) component in H and a component on the sphere at infinity:(A-8) fix( M ) = (fix( M ) ∩ A ) (cid:116) (fix( M ) ∩ B ) . The part in H , if non-empty, is the axis of M . The part on the sphere at infinity is theset of eigenlines of M , as the following proposition shows. Note that this part consists ofexactly one or two points, since M has one or two eigenlines. Proposition
A.4 . For M ∈ SL ( C ) \ {± } , the set φ (fix( M ) ∩ B ) ⊂ C P is the setof eigenlines of M . Proof.
Note that fix( M ) ∩ B is the set of elements X ∈ M × C with det X = 0 suchthat [[ M X ]] = [[ X ]]. Since det X = 0, X is of the form (A-6) with ( x, y ) t , ( a, b ) ∈ C \{ } .Since φ is a bijection, [[ M X ]] = [[ X ]] if and only if φ ([[ M X ]]) = φ ([[ X ]]). That is, if andonly if in C P (A-9) (cid:20)(cid:20) M (cid:18) xy (cid:19)(cid:21)(cid:21) = (cid:20)(cid:20)(cid:18) xy (cid:19)(cid:21)(cid:21) . That is, if and only if ( x, y ) t is an eigenline of M . (cid:3) A.4.
The Klein model of H . The Klein model of H is the unit ball B = { x ∈ R |(cid:107) x (cid:107) ≤ } . The sphere at infinity is its boundary ∂ B = { x ∈ R | (cid:107) x (cid:107) = 1 } . The map K : H → B is defined as the map [[ X ]] (cid:55)→ XX ∗ followed by the map(A-10) (cid:18) a + b c + idc − id a − b (cid:19) = 1 a ( b, c, d ) . We need some well-known results about this model. For details, see [6, Sections II.5, VIII]and [2, Sections A.4, A.5]. Non-trivial isometries in H are identified with elements ofSL ( C ) \ {± } . In particular, we have the following Proposition
A.5 . M ∈ SL ( C ) \ {± } is unitarisable if and only if M is elliptic asan isometry of H . Proposition
A.6 . If M ∈ SL ( C ) \ {± } is unitarisable, then the axis of M in theKlein model is a Euclidean straight line segment in B with two distinct endpoints on ∂ B . Proof.
Since M ∈ SL ( C ) is unitarisable, by Theorem A.2, f ix ( M ) (cid:54) = ∅ . Following[6, Section VIII.11], M has two fixed endpoints on ∂ B and the axis through these is ageodesic. Since geodesics are straight line segments, also the axis of M is a straight linesegment. (cid:3) MC TRINOIDS WITH ONE IRREGULAR END 15
A.5.
The Cross Ratio.
The cross ratio of four distinct points in C P = C ∪ {∞} isdenoted by(A-11) [ a, b, c, d ] := ( b − c )( d − a )( b − a )( d − c ) . The cross ratio is chosen so that [0 , , ∞ , x ] = x . The cross ratio is invariant under M¨obiustransformations. The cross ratio [ a, b, c, d ] of four distinct points is real if and only if thepoints lie on a circle. Proposition
A.7 . Let a , b , c and d be distinct points in C P , and suppose [ a, b, c, d ] ∈ R \ { } , so a , b , c , d lie on a circle C .(1) [ a, b, c, d ] ∈ R + if and only if b and d lie in the same connected component of C \ { a, c } .(2) [ a, b, c, d ] ∈ R − if and only if b and d lie on different connected components of C \ { a, c } . Proof.
There exists a unique M¨obius transformation taking a , b and c to 0, 1 and ∞ respectively, taking circles to circles, and preserving or reversing the order of a , b , c and d on the circle. Thus we may assume a = 0, b = 1, c = ∞ . Thus [ a, b, c, d ] = [0 , , ∞ , d ] = d ∈ R \ { , } . Then a , b , c and d lie on the circle C = R ∪ {∞} , and the two connectedcomponents of C \ { a, c } are R − and R + . The theorem follows by an examination of thetwo cases d ∈ R + \ { } and d ∈ R − . (cid:3) A.6.
Unitarisability of two matrices.
We conclude by proving Theorem 5.1, whichgives a characterization for the simultaneous unitarisability of two matrices in SL ( C ). Proof of Theorem 5.1 . Since M and M are individually unitarisable then, by Theo-rem A.2, fix( M ) (cid:54) = ∅ and fix( M ) (cid:54) = ∅ . By Theorem A.3, M and M are simultaneuoslyunitarisable if and only if fix( M ) ∩ fix( M ) (cid:54) = ∅ .First suppose that fix( M ) ∩ fix( M ) (cid:54) = ∅ . By Theorem A.6, fix( M ) and fix( M ) arestraight line segments. And since they intersect, they lie in a unique Euclidean plane P ⊂ R . Then C := P ∩ ∂ B is a circle. By Theorem A.4, the endpoints of fix( M ) andfix( M ) are ϕ, ϕ (cid:48) and ψ, ψ (cid:48) respectively. Since fix( M ) and fix( M ) intersect at a pointinside the disk P ∩ B bounded by C , it follows that ψ and ψ (cid:48) lie in different connectedcomponents of C \ { ϕ, ϕ (cid:48) } . By Theorem A.7, [ ϕ, ψ, ϕ (cid:48) , ψ (cid:48) ] ∈ R − .Conversely, suppose [ ϕ, ψ, ϕ (cid:48) , ψ (cid:48) ] ∈ R − . By Theorem A.7 ϕ, ψ, ϕ (cid:48) and ψ (cid:48) lie on somecircle C ⊂ ∂ B , and ψ and ψ (cid:48) lie on different connected components of C \ { ϕ, ϕ (cid:48) } . Let P bethe unique Euclidean plane containing C . By Theorem A.4 and Theorem A.6, fix( M ) isthe straight line segment with endpoints ϕ and ϕ (cid:48) , and fix( M ) is the straight line segmentwith endpoints ψ and ψ (cid:48) . Hence fix( M ) and fix( M ) lie on P and intersect in the disk P ∩ B bounded by C . Thus fix( M ) ∩ fix( M ) (cid:54) = ∅ . (cid:50) References
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