Interpolation by maximal surfaces and minimal surfaces
aa r X i v : . [ m a t h . DG ] F e b INTERPOLATION BY MAXIMAL SURFACES AND MINIMALSURFACES
RUKMINI DEY AND RAHUL KUMAR SINGH
Abstract.
In this article, we interpolate a given real analytic spacelike curve a in Lorentz-Minkowski space L to another real analytic spacelike curve c , which is“close” enough to a in a certain sense, by a maximal surface using inverse functiontheorem for Banach spaces. Using the same method we also interpolate a givenreal analytic curve a in Euclidean space E to another real analytic curve c , whichis “close” enough to a in a certain sense, by a minimal surface. The Björlingproblem and Schwartz’s solution to it play an important role. Introduction
The classical problem of Plateau asks the following question: “Given a Jordancurve γ in E , does there exist a minimal surface having its boundary as γ ”. In a solution was obtained by J. Douglas ([6]) and simultaneously by T. Radó ([20]).In fact, in [7], Douglas also solved the problem of Plateau for the case of two closedcurves (non-intersecting) in E n .In ([22]), I.N. Vekua uses implicit function theorem for Banach spaces in order tosolve Plateau’s problem for all curves which lies sufficiently close to a given planecurve in E .In ([9]), in example (5.4.13) Hamilton solves the Plateau’s problem for minimalsurfaces when the surface is a graph using Nash-Moser inverse function theorem.A version of the interpolation for minimal surfaces using inverse function theoremfor Banach spaces was studied in an article jointly by the authors and P. Kumar,currently in the arxiv ([4]). In the present article we address the problem somewhatdifferently from ([4]) and include the case of the maximal surfaces in L , but someof the preliminary steps are similar to ([4]).In ([3]), we had solved the problem of interpolating a given space-like closed curveand a point by a maximal surface in L , such that this point is a singularity. Subject classification: 53A35, 53B30Keywords: Minimal surfaces, maximal surfaces, Björling problem.
The problem of “interpolating” curves by maximal surfaces for specific exampleshave been considered for instance by Fujimori, Rossman, Umehara, Yang and Ya-mada, ([8]).The existence of spacelike constant mean curvature surfaces spanning two circularcontours has been considered by López in ([15]).The Plateau’s problem for maximal surfaces has been considered for instance inpseudo-hyperbolic spaces by Labourie, Toulisse and Wolf, ([13]) where the boundarydata is at infinity.In this article, using the inverse function theorem for Banach spaces we interpolatea given real analytic spacelike curve a in L to another real analytic spacelike curve c ,which is “close” enough to a in a certain sense, by a maximal surface X = Re(a + id) where a, d : Ω ⊂ C → C are analytic maps such that < ( a + id ) ′ , ( a + id ) ′ > L = 0 and X is a spacelike immersion, see Theorem 2.4. Here < · , · > L stands for theLorentzian inner product. The Björling problem plays an important role.Using the same method we also interpolate a given real analytic curve a in E toanother real analytic curve c , which is “close” enough to a in a certain sesne, by aminimal surface X = Re(a − id) where a, d : Ω ⊂ C → C are analytic maps suchthat < ( a − id ) ′ , ( a − id ) ′ > E = 0 and X is an immersion, see Theorem 3.2. Notethat here < · , · > E stands for the Euclidean inner product. The Björling problemand Schwartz’s solution to it play an important role.2. Interpolation by Maximal Surfaces
We fix some notations which we will be using throughout this article.
Notations 2.1. (1)
A map a : I → R will always denote a real analytic curvewhich can be extended analytically on a disc Ω containing I , an open orclosed interval in R and this extension is differentiable on Ω . (2) For a bounded function f : Ω → C n , where f = ( f j ) ; j = 1 , , .., n , we defineits norm as k f k C n := max j { sup z ∈ Ω | f j ( z ) |} . Our starting point is the Björling problem for maximal surfaces. Mathematicianshave considered the Björling’s problem for a maximal surface in Lorentz-Minkowski -space, L , for instance see ([1]). Here L is ( R , < · , · > L ) and < v, w > L = v w + v w − v w for v = ( v , v , v ) ∈ R and w = ( w , w , w ) ∈ R . Note that × L is our notation for the cross product in L .Let a : I → L , ( I an open interval) be a real analytic spacelike curve with a ′ ( t ) = 0 , t ∈ I . Let a be spacelike curve in L and n : I → L be a real analyticmap such that h n ( t ) , a ′ ( t ) i L = 0 , t ∈ I and n ( t ) + n ( t ) − n ( t ) = − , t ∈ I , where n = ( n , n , n ) . Then n is called the unit normal vector field along a . Note that itis timelike. a ′ ( t ) can be zero only at isolated points t ∈ I . NTERPOLATION 3
The Björling problem is to find a maximal surface X : Ω → L with I ⊂ Ω ⊂ C , Ω simply connected, such that the following conditions are satisfied: X ( u,
0) = a ( u ) , N ( u,
0) = n ( u ) , ∀ u ∈ I , where N : Ω → L is a normal to the surface X . Thesolution to the Björling’s problem is given by [1],(2.1) X ( u, v ) = Re ( a(w) + i Z wu n( ˜w) × L a ′ ( ˜w)d ˜w ) ; w = u + iv , u ∈ I , where n ( w ) and a ( w ) are the complex analytic extensions of n ( u ) and a ( u ) over Ω .Observe that the map, w a ( w )+ i R wu n ( ˜ w ) × L a ′ ( ˜ w ) d ˜ w , from Ω to C is an isotropiccurve , i.e. if we write d = R wu n ( ˜ w ) × L a ′ ( ˜ w ) d ˜ w , then < ( a + id ) ′ , ( a + id ) ′ > L = 0 .This can be shown by an easy calculation. Note that here we are using × L for theextended Lorentzian cross product and < · , · > L for the extended Lorentzian innerproduct also. Some Defintions and Notations:
Let Ω ⊂ C be a disc which contains an interval I ⊂ R as in the Notations 2.1. For n ≥ , let C ω (Ω , C n ) denote the space of all complex analytic maps on Ω which can becontinuously extended to the boundary of Ω , i.e., till Ω . This forms a Banach spaceunder the norm k f k C n := max i { sup z ∈ Ω | f i ( z ) |} , where f ( z ) = ( f j ( z )) , j = 1 , , .., n. Let C ω (Ω , Ω) denote the space of Ω -valued complex analytic functions definedon Ω which are automorphisms of Ω , i.e. they are invertible. Since invertibility isan open condition, this is an open subset of C ω (Ω , C ) and is a Banach manifold.Let γ ( z ) = z . Then we take ˜ C ⊂ C ω (Ω , Ω) such that its tangent space at γ is C ⊂ C ω (Ω , C ) where C consists of all V ∈ C ω (Ω , C ) such that V ( u ) = 0 , where u ∈ I . Since C is a Banach space, ˜ C has a Banach manifold structure at γ .Let W ⊂ C ω (Ω , C ) consist of all those s ∈ C ω (Ω , C ) such that s ( u ) is real for u ∈ I and s ( u ) = 0 , for u ∈ I . This is a closed subspace of C ω (Ω , C ) and henceis a Banach space and let ˜ W denotes the set of those s ∈ C ω (Ω , C ) such that s ( u ) is real on u ∈ I . This also forms a Banach space.Let L denotes the space of l : I → L such that it has a differentiable extension l : Ω → C and is analytic on Ω .Let S denote the space of all space-like vectors in L s.t v = ( v , v , v ) ∈ S iff v + v > v . Let S p ⊂ L be the set of all c = ( c , c , c ) such that c ′ + c ′ − c ′ > on I , where ′ denotes the derivative. Proposition 2.1. S p is an open subset of L in subspace sup-norm topology.Proof. Let c ∈ S p . We will show that there exists an open neighbourhood U of c such that U ⊂ S p .Let us first take the inclusion map f : I × S p → I × L given by ( u, c ) ( u, c ) .Next we take the map g : I ×L → R given by g ( u, ˜ c ) = c ′ ( u )+ c ′ ( u ) − c ′ ( u ) which isan evaluation map and hence continuous. Therefore, the map φ = g ◦ f : I × S p → R is continuous. Next φ − ( R + ) = V ⊂ I × S p . V is an open set in I × S p since φ is RUKMINI DEY AND RAHUL KUMAR SINGH continuous and R + is an open set of R . Now since ( c ′ + c ′ − c ′ )( u ) > for all u ∈ I , I × c ⊂ V . Then there exists a set U ⊂ π ( V ) , containing c such that U isopen in π ( V ) . Since the latter is open in S p , U is open in S p . It is easy to checkthat φ ( I × U ) ⊂ R + . (cid:3) Let < · , · > E and < · , · > L be Euclidean and Lorentzian inner products respec-tively. Let ⊥ E denote perpendicular w.r.t. < · , · > E and ⊥ L denote perpendicularw.r.t. < · , · > L . Let × E denote cross-product in Euclidean sense and × L denote thesame in Lorentzian sense.Let a ∈ S p , i.e. it is as in Notation 2.1 and is spacelike on I .We let n be such that n = n ( u ) is real and real analytic on I and n l + n − n = − and < n, a ′ > L = 0 .Let d n ∈ W be such that d n ′ = n × L a ′ . One can show that d n ′ is spacelike on I , ( [16]). Moreover, < d n ′ , d n ′ > L = < a ′ , a ′ > L , < d n ′ , a ′ > L = 0 . In other words, ( a + id n ) is L-isotropic in the sense that < ( a + id n ) ′ , ( a + id n ) ′ > L = 0 .Let us denote by N the timelike normal vector field on I , n = ( n , n , n ) suchthat n + n − n = − and n ⊥ L a ′ = 0 . This is closed in sup-norm topology.Let J = ∪ n ∈N J n where J n = { c ∈ C ω (Ω , C ) | Re ( c ′ ( u )) ⊥ E ∆ n ( u ) ∀ u ∈ I } , i.e.Re ( c ′ )( u ) lie in the plane of a ′ ( u ) and d n ′ ( u ) for all u ∈ I (whenever d n ′ ( u ) is notidentically zero). J is a Banach manifold with the topology that V is open in J iff V n = V ∩ J n is open in J n in sup-norm topology. Let us call this n -topology.Let p = a + id be such that p ∈ ∩ J n for all n . This is ensured by the fact that d ( u ) is real for u ∈ I . We will give a proof of the following proposition later inProposition 2.8. Proposition 2.2.
Let V be an open set containing p ∈ J in the n -topology of J .There is an open ball B ǫ ( p ) in sup-norm topology in C ω (Ω , C ) such that B ǫ ( p ) ⊂ V . Note:
We use the notion of smooth maps and derivations of maps between Banachspaces in the same sense as in ([12]). For more details see ([4]).
Proposition 2.3.
Given a ∈ S p , the map H : C ω (Ω , Ω) × C ω (Ω , C ) → C ω (Ω , C ) defined by H ( γ, d ) = (( a + id ) ◦ γ ) is smooth. Let γ ∈ C ω (Ω , Ω) , γ ( z ) = z and d ∈ C ω (Ω , C ) . Then the derivative of H at the point ( γ , d ) , DH ( γ ,d ) : C ω (Ω , C ) × C ω (Ω , C ) → C ω (Ω , C ) is given by DH ( γ ,d ) ( V, d ) = (( a ′ + id ′ ) V + id ) .Proof. As in ([4]) one can show that the map H ( γ, d ) = ( a + id ) ◦ γ is smooth. Let P = ( γ , d ) and h = ( V, d ) , where V ∈ C ω (Ω , C ) and d ∈ W . Then DH P ( h ) = lim t → t { H ( P + th ) − H ( P ) } NTERPOLATION 5
A straightforward calculation similar to ([4]) shows that DH ( γ ,d ) ( V, d ) = ( a ′ + id ′ ) V + id. (cid:3) Notation:
Recall ˜ C ⊂ C ω (Ω , Ω) such that its tangent space at γ = id is C ⊂ C ω (Ω , C ) consisting of all V such that V ( u ) = 0 .Let H n be H with the domain restricted to ˜ C × W and range restricted to J n . Inother words, H n : H − ( J n ) → J n defined by H n ( γ, d ) = ( a + id ) ◦ γ . H − ( J n ) isnonempty. This is because for all u ∈ I , Re [(( a + id n ) ◦ γ )( u )] ′ = a ′ ( u ) is ⊥ E to ∆ n ( u ) since ∆ n ( u ) = d n ′ ( u ) × E a ′ ( u ) . Thus ( γ , d l ) ∈ H − ( J n ) for all n .Let T = { ( V, ˜ d ) ∈ C × W such that on I , Re (( a ′ ( u ) + id n ′ ( u )) V ( u ) + i ˜ d ( u )) ′ ⊥ E ∆ n ( u ) } . (We shall prove that this is the tangent space to H − ( J n ) at ( γ , d n ) ).Let H n : T → J n be a map defined by H n ( V, ˜ d ) = ( a ′ + id n ′ ) V + i ˜ d .We show that H n is invertible. (As a consequence of this, it will follow that DH ( γ ,d n ) is invertible). Proposition 2.4.
Fix γ ∈ C ω (Ω , Ω) to be γ ( z ) = z and d n ∈ W as before. Let p = ( γ , d n ) . Then the map H n : T ⊂ C × W → J n is invertible.Proof. We know that H n is given by the map ( V, ˜ d ) → ( a ′ + id n ′ ) V + i ˜ d and there isan additional constraint that Re (( a ′ + id n ′ ) V + i ˜ d )) ′ ⊥ E ∆ n on I since ( V, ˜ d ) ∈ T .We want to show that H n is invertible, i.e., given a fixed s = ( s , s , s ) ∈ J n there exists a unique ˜ d ∈ W and V such that H n ( V, ˜ d )( z ) = s ( z ) , i.e. s ( z ) = ( a ′ ( z ) + id n ′ ( z )) V ( z ) + i ˜ d ( z ) . Since s ∈ J n , Re s ′ ⊥ E ∆ n on I . Let τ ( z ) = s ′ ( z ) and thus Re τ ( u ) ⊥ E ∆ n on I .As a first step, we want to solve for ˜ V and ˜˜ d such that τ ( z ) = s ′ ( z ) = ( a ′ ( z ) + id n ′ ( z )) ˜ V ( z ) + i ˜˜ d ( z )) such that ˜ V ∈ C ω (Ω , C ) and ˜˜ d ∈ ˜ W .This gives us a set of three equations ( A ( z ) ˜ V ( z ) + X ( z )) = τ ( z ) , ( B ( z ) ˜ V ( z ) + Y ( z )) = τ ( z ) , ( C ( z ) ˜ V ( z ) + Z ( z )) = τ ( z ) , (2.2)where ( A ( z ) , B ( z ) , C ( z )) = ( a ′ + d n ′ , a ′ + id n ′ , a ′ + id n ′ )( z ) , ( X ( z ) , Y ( z ) , Z ( z )) =( i ˜˜ d , i ˜˜ d , i ˜˜ d )( z ) and ˜ V = ˜ V + i ˜ V , ˜ V , ˜ V the real and imaginary parts. In otherwords, RUKMINI DEY AND RAHUL KUMAR SINGH X ( z ) = τ ( z ) − A ( z ) ˜ V ( z ) ,Y ( z ) = τ ( z ) − B ( z ) ˜ V ( z ) ,Z ( z ) = τ ( z ) − C ( z ) ˜ V ( z ) (2.3)We first solve for ˜ V and ˜ V . We have an additional constraint that ˜˜ d = ˜˜ d ( u, isreal on I .We show that the constraint is ensured by the fact that Re τ ( u ) ⊥ E ∆ n ( u ) on I where ∆ n = d n ′ ( u ) × E a ′ ( u ) is a normal to the plane of a ′ ( u ) and d n ′ . This can beshowed as below.For u ∈ I we need to solve the following equation: − Im ˜˜ d ( u ) = Re X ( u ) = Re ( τ ( u ) − A ( u ) ˜ V ( u )) , − Im ˜˜ d ( u ) = Re Y ( u ) = Re ( τ ( u ) − B ( u ) ˜ V ( u )) , − Im ˜˜ d ( u ) = Re Z ( u ) = Re ( τ ( u ) − C ( u ) ˜ V ( u )) (2.4) ∆ n = d n ′ × E a ′ = (∆ n , ∆ n , ∆ n ) is not zero on I . At a u ∈ I , suppose w.l.g., itsthird cooordinate is non-zero, i.e. ∆ n ( u ) = ( a ′ d n ′ − a ′ d n ′ )( u ) = 0 at some u ∈ I At that u ∈ I , writing ˜ V ( u ) = ˜ V ( u ) + i ˜ V ( u ) with V and V real, we have fromthe first two equations: ˜ V ( u ) = Re τ d n ′ − Re τ d n ′ ∆ n ( u ) , ˜ V ( u ) = Re τ a ′ − Re τ a ′ − ∆ n ( u ) . (2.5)Substituting in the third equation, we getRe ( τ ( u )) = Re ( C ( u ) ˜ V ( u )) = ( a ′ ˜ V − d n ′ ˜ V )( u ) Re ( τ ( u )) = a ′ Re τ d n ′ − Re τ d n ′ ∆ n ( u ) + d n ′ Re τ a ′ − Re τ a ′ ∆ n ( u ) . (2.6)In other words, ∆ n ( u ) Re ( τ ( u )) = ( − ∆ n Re τ − ∆ n Re τ )( u ) (2.7)where on I , ∆ n = a ′ d n ′ − a ′ d n ′ , ∆ n = a ′ d n ′ − a ′ d n ′ and ∆ n = a ′ d n ′ − a ′ d n ′ are thecomponents of ∆ n = d n ′ × E a ′ .In other words, (∆ n Re ( τ ) + ∆ n Re ( τ ) + ∆ n Re ( τ ))( u ) = 0 for this u ∈ I .This equation is valid for all u ∈ I since τ satisfies the constraint. It is symmetricin all coordinates. NTERPOLATION 7
Thus ˜˜ d ( u, is real on I . It is unique as is clear from the equations.Thus we have(2.8) τ ( z ) = s ′ ( z ) = ( a ′ ( z ) + id n ′ ( z )) ˜ V ( z ) + i ˜˜ d ( z ) with ˜ V ∈ C ω (Ω , C ) and ˜˜ d ∈ ˜ W We have s ′ = ( a ′ + id n ′ ) ˜ V + i ˜˜ d We need to solve for ( V, ˜ d ) ∈ T ⊂ C × W such that s = ( a ′ + id n ′ ) V + i ˜ d .We need to solve for ( V, ˜ d ) such that s ′ = ( a ′ + id n ′ ) ˜ V + i ˜˜ d = (( a ′ + id n ′ ) V + i ˜ d ) ′ .We claim that there is a unique solution ( V, ˜ d ) ∈ T to the above equation.We need s ′ = ( a ′′ + id n ′′ ) V + ( a ′ + id n ′ ) V ′ + i ˜ d ′ .Re s ′ ( u ) = ( a ′′ ( u ) V ( u ) − d n ′′ ( u ) V ( u ))+( a ′ ( u ) V ′ ( u ) − d n ′ ( u ) V ′ ( u )) ⊥ E ∆ n ( u ) wherewe write V = V + iV , V , V are the real and imaginary parts of V .Recall that on I , Re s ′ ( u ) ⊥ E ∆ n ( u ) .Since ( a ′ ( u ) V ′ ( u ) − d n ′ ( u ) V ′ ( u )) ⊥ E ∆ n ( u ) we must have ( a ′′ ( u ) V ( u ) − d n ′′ ( u ) V ( u )) ⊥ E ∆ n ( u ) .Then there exist m ( u ) and m ( u ) such that(2.9) a ′′ ( u ) V ( u ) − d n ′′ ( u ) V ( u ) = a ′ ( u ) m ( u ) − d ′ ( u ) m ( u ) , since the left hand side belongs to the plane of a ′ ( u ) and d n ′ ( u ) . One can show thatin fact m ( z ) + im ( z ) is analytic on Ω .Writing the equation s ′ = ( a ′ + id n ′ ) ˜ V + i ˜˜ d = ( a ′′ + id n ′′ ) V + ( a ′ + id n ′ ) V ′ + i ˜ d inits real and imaginary parts, on I we obtain two equations: a ′ ( u )( m ( u ) + V ′ ( u )) − d n ′ ( u )( m ( u ) + V ′ ( u )) = a ′ ( u ) ˜ V ( u ) − d n ′ ( u ) ˜ V ( u ) ,a ′′ ( u ) V ( u ) + d n ′′ ( u ) V ( u ) + a ′ ( u ) V ′ ( u ) + d n ′ ( u ) V ′ ( u ) + ˜ d ′ = a ′ ( u ) ˜ V ( u ) + d n ′ ( u ) ˜ V ( u ) + ˜˜ d ( u ) . Let V = R wu ( ˜ V − m )( ˜ w ) d ˜ w and V = R wu ( ˜ V − m )( ˜ w ) d ˜ w and V = V + iV .Then V is a complex analytic function on Ω since ˜ V and m = m + im are so.Let ˜ d ( u ) = − R uu ( Im ( F ( u )) − ˜˜ d ( u )) du where F ( z ) = ( a ′′ + id n ′′ )( z ) V ( z ) − ( a ′ + id n ′ )( z ) m ( z ) . Note that by Equation(2.9), Re F ( u ) = 0 on I .Thus ˜ d ( u ) = − R uu − i ( Re F ( u ) + i Im F ( u )) + R uu ˜˜ d ( u ) du Then ˜ d ( w ) = − R wu ( − iF − ˜˜ d )( z ) dz belongs to W .Then one checks that s = ( a ′ + id n ′ ) V + i ˜ d . Also, V ( u ) = 0 and ˜ d ( u ) = 0 implies V ∈ C and ˜ d ∈ W . Moreover, Re (( a ′ ( u ) + id n ′ ( u )) V ( u ) + i ˜ d ( u )) ′ ⊥ E ∆ n on I since Re ( τ ) = Re ( s ′ ) ⊥ E ∆ n on I . Since V ( u ) = ˜ d ( u ) = 0 , ( V, ˜ d ) is unique.Thus ( V, ˜ d ) ∈ T is the inverse of s ∈ J n . RUKMINI DEY AND RAHUL KUMAR SINGH (cid:3)
Proposition 2.5. At p = ( γ , d n ) , H − ( J n ) has a Banach manifold structure.Proof. Consider H − ( J n ) ⊂ C ω (Ω , Ω) × C ω (Ω , C ) in subspace topology.Recall T = { ( V, ˜ d ) ∈ C × W such that on I , Re (( a ′ ( u ) + id n ′ ( u )) V ( u ) + i ˜ d ( u )) ′ ⊥ E ∆ n ( u ) } .We show that there is a smooth bijection from a open subset U of p in H − ( J n ) to an open neighbourhood W of T . This would prove that at the point p , H − ( J n ) is modelled on a Banach space T Let ˜ T = { ( V , d ) ⊂ C ω (Ω , C ) × W| c = ( a + id ) ◦ γ ∈ J n where γ = γ + V , d = d n + d } . One can show that this is nonempty and contains (0 , because c = ( a + id n ) ◦ γ ∈ J n .Let V be such that if ( V , d ) ∈ V ⊂ ˜ T then ( γ + tV , d n + td ) ∈ H − ( J l ) and γ t = γ + tV takes I to I forall t ∈ (0 , . It can be shown that V is open.Let ( V , d ) ∈ V and γ = γ + V and d = d n + d . Let us name this map Ψ − . It issmooth, invertible and inverse is also smooth given by V = γ − γ and d = d − d n .Then Ψ :
U ⊂ H − ( J n ) → V where U = Ψ − ( V ) , an open set.Let ( γ, d ) ∈ U ⊂ H − ( J n ) and ( V , d ) ∈ V ⊂ ˜ T be its image under Ψ .Let c = ( a + id ) ◦ γ and c = a + id n .Let F ( t ) = ( a + i ( d n + td )) ◦ ( γ + tV ) .Then F (0) = c and F (1) = ( a + i ( d n + d )) ◦ ( γ + V ) .There exists s a real number < s < such that F (1) − F (0) = ˙ F ( s ) by MeanValue Theorem where · stands for derivative w.r.t. t . Let s be the least suchnumber. It is easy to see that since F is not locally constant, s = 0 .Also, it is easy to check that(2.10) ˙ F ( s ) = ( a ′ + id n ′ )( γ s ) · V + is d ′ ( γ s ) · V + id ( γ s ) Now note that F (1)( γ − s ) = c ( γ − s ) belongs to J n since c ∈ J n and γ s = γ + s V .This is because Re ( c ( γ − s ) ′ ) · γ ′ s is ⊥ E ∆ n as a consequence of the fact that γ − s maps I to I . Similarly c ( γ − s ) ∈ J n .Then c ( γ − s ) − c ( γ − s ) ∈ J n .By invertibility of H n in Proposition (2.4) there is an unique solution ( V χ , d χ ) ofthe equation(2.11) c ( γ − s ) − c ( γ − s ) = (( a ′ + id n ′ ) · V χ + id χ ) . From equations (2.10) and (2.11) one can deduce that F (1)( γ − s ) − F (0)( γ − s ) = ˙ F ( s )( γ − s )= ( a ′ + id n ′ ) · V ( γ − s ) + is d ′ · V ( γ − s ) + id = (( a ′ + id n ′ ) · V χ + id χ ) . By uniqueness of ( V χ , d χ ) we must have ( V χ , d χ ) = ( V ( γ − s ) , s d ′ · V ( γ − s ) + d ) . NTERPOLATION 9
Since ( V χ , d χ ) ∈ T we have a map Φ :
V ⊂ ˜ T → W ⊂ T which maps ( V , d ) → ( V χ , d χ ) given by the above formula.One can show that Φ − also exists from the open set W ⊂ T to V ⊂ ˜ T . We givea sketch of the proof. Take any s ∈ (0 , . Suppose we have ( V χ , d χ ) ∈ W . We wishto solve ( V s , d s ) such that ( V s ( γ − s ) , sd s ′ · V ( γ − s ) + d s ) = ( V χ , d χ ) . First we solve V χ = V s ( γ − s ) . The equation is written as V χ ( γ s ) = V s where γ s = γ + sV s . Solvingthe ODE: V ′ χ ( γ s ) · (1 + sV s ′ ) = V s ′ and using V s ( u ) = 0 we have a unique solution.Then we have γ s and one can determine d s from the ODE: sd s ′ · V s ( γ − s ) + d s = d χ and d s ( u ) = 0 . We can repeat the process for all s ∈ (0 , . Denote by A s themap which takes ( V χ , d χ ) → ( V s , d s ) . Choose s to be such that A s ◦ Φ = id and Φ ◦ A s = id , i.e. ( V s ( γ − s ) , s d s ′ · V s ( γ − s ) + d s ) = ( V χ , d χ ) as in the defintion of Φ . Then Φ − = A s .One can show that Φ and Φ − are smooth (in the sense of Kriegl and Michor [12]).Then Φ ◦ Ψ :
U ⊂ H − ( J l ) → W ⊂ T . We showed this map is locally invertible.Hence Ψ − ◦ Φ − which maps ( V, d ) ∈ W → ( γ, d ) ∈ U is well defined.Thus there is a bijection Φ ◦ Ψ from a open neighbourhood U of p in H − ( J n ) to an open subset of T . One can show that the bijection is smooth ( in the senseof Kriegl and Michor [12]). Thus at the point p , H − ( J n ) is modelled on a Banachspace T . (cid:3) Proposition 2.6.
Let p = ( γ , d n ) as before. T p H − ( J n ) ⊂ C × W consists of all ( V, ˜ d ) ∈ C × W such that on I , Re (( a ′ ( u ) + id n ′ ( u )) V ( u ) + i ˜ d ( u )) ′ ⊥ E ∆ n ( u ) . Inother words, T p H − ( J n ) = T .Proof. We claim that T p H − ( J n ) ⊂ C × W consists of all ( V, ˜ d ) ∈ C × W such thaton I , Re (( a ′ ( u ) + id n ′ ( u )) V ( u ) + i ˜ d ( u )) ′ ⊥ E ∆ n ( u ) .This is because one can take a curve c t = ( γ t , d t ) ∈ H − ( J n ) such that H ( c t ) =( a + id t ) ◦ γ t ∈ J n . We have Re ( c ′ t ( u )) ⊥ E ∆ n ( u ) on I .A calculation as in Proposition (2.3) shows that dc t dt | t =0 = ( a ′ + id l ′ ) V + i ˜ d withthe additional constraint that Re ( dc ′ t dt | t =0 ) = Re ( dc t dt | t =0 ) ′ ⊥ E ∆ n on I . Thus we aredone. (cid:3) Let p = ( γ , d n ) . Then an easy computation shows that DH n | p : T p H − ( J n ) → J n is defined to be ( V, ˜ d ) → ( a ′ + id n ′ ) V + i ˜ d with Re (( a ′ ( u )+ id n ′ ( u )) V ( u )+ i ˜ d ) ′ ( u ) ⊥ E ∆ n ( u ) on I since T p H − ( J n ) = T . Proposition 2.7.
Fix γ ∈ C ω (Ω , Ω) to be γ ( z ) = z and d n ∈ W as before. Let p = ( γ , d n ) . The map DH | p : T p ( H − ( J n )) ⊂ C × W → J n is invertible.Proof. We know that DH | ( γ ,d n ) ( V, ˜ d ) is given by the map ( a ′ + id n ′ ) V + i ˜ d . Thereis an additional constraint that Re (( a ′ + id n ′ ) V + i ˜ d )) ′ ⊥ E ∆ n on I .We also have T = T p ( H − ( J n )) and H n = DH | ( γ ,d n ) . Then DH | p is invertible since H n is invertible, by Proposition (2.4). (cid:3) Theorem 2.2.
Let a and Ω as in Notation (2.1). Let p = a + id ( where d ∈ W ).Then there exist an open neighbourhood of the function p = a + id , namely V Ω ⊂ J (open in n -topology), and open sets U Ω ⊂ C ω (Ω , Ω) and W ⊂ W such that for every c ∈ V Ω , there exists γ ∈ U Ω and d ∈ W such that ( a + id ) ◦ γ = c .Proof. Note that p ∈ J n for all n ∈ N . We wish to invert H n in a neighbourhood of p = ( a + id ) ◦ γ . It is a local problem and we can use Inverse Function Theorem forBanach spaces. Since DH l | p is invertible, by Inverse Function Theorem for Banachspaces [2], there exists a neighbourhood of p = ( a + id ) ◦ γ , namely V n ⊂ J n , suchthat if c ∈ V n , the equation H − n ( c ) = ( γ, d ) has a solution, i.e. ( a + id ) ◦ γ = c has asolution ( γ, d ) , where γ ∈ U nγ ⊂ C ω (Ω , Ω) a neighbourhood of γ and d ∈ W nd ⊂ W of d , W nd an open subset of d .Let V Ω = ∪ n V n . This is open in J in n -topology and contains p . If c ∈ V Ω then c ∈ V n for some n and hence there exists ( γ, d ) ∈ U nγ × W nd such that c = ( a + id ) ◦ γ .Let U Ω = ∪ n U nγ and W = ∪ n W nd . Then the result follows. (cid:3) Interpolation by maximal surface:
Let a = a ( u ) and c = c ( u ) be tworeal analytic curves in S p . Suppose there exists a maximal surface X = X ( u, v ) = Re f ( z ) , (i.e. < f ′ , f ′ > L = 0 and X is a spacelike immersion) and in addition a ( u ) = Re f ( γ ( u )) and c ( u ) = Re ( f ( γ )( u )) where γ and γ are automorphisms ofthe domain Ω . Then we say that a = a ( u ) and c = c ( u ) are interpolated by themaximal surface X = X ( u, v ) .In general, an isotropic curve in L -inner product is defined as follows, analogousto that in E -inner product.Any holomorphic map f : Ω ⊂ C → C is said to be an L-isotropic curve if itsatisfies h f ′ ( z ) , f ′ ( z ) i L = 0 , z ∈ Ω ; where ′ denotes the complex derivative withrespect to z and h , i L is the extension of the inner product in L to C .Let f : Ω ⊂ C → C be a holomorphic map defined by f ( z ) = a ( z ) + id ( z ) such that h a ′ ( z ) , a ′ ( z ) i L = h d ′ ( z ) , d ′ ( z ) i L and h a ′ ( z ) , d ′ ( z ) i L = 0 . Then f satisfies h f ′ ( z ) , f ′ ( z ) i L = 0 for all z ∈ Ω , and hence defines an isotropic curve.Let a ∈ S p ⊂ L .We prove a proposition which will be useful for us.Let n ∈ N . Let us consider d = d n where d n ′ = n × L a ′ and d ( u ) = 0 .Let p = a + id . Note that p ∈ J n for all n ∈ N since Re ( p ′ )( u ) = a ′ ( u ) ⊥ E ∆ n for all n ∈ N .Let V Ω be an open set in J in the n -topology such that p ∈ V Ω = ∪ n ∈ N V n .Since p ∈ ( ∩ J n ) ∩ V Ω , p ∈ V n for all n ∈ N . Then there is a sup-norm-open ballneighbourhood of p , namely B sǫ n ( p ) open in V n since V n is an open set of J n insupnorm subspace topology. (Here s is B sǫ n ( p ) stands for “slice”). NTERPOLATION 11
Let S = { n ∈ N s.t. || n || ≤ M, M >> fixed . } be a closed and bounded subsetof N . This is compact (by direct application of (1 . , page 32, Rudin [21]).We define B = ∪ n ∈ S B sǫ n ( p ) and take ǫ = inf S ǫ n . We first show that ǫ = 0 .Suppose ǫ = 0 . Let us take a minimizing sequence n i ∈ S such that ǫ n i → .Since S is compact, n i → n τ and since S is compact, n τ ∈ S . Consider V n τ . Let B sǫ nτ ( p ) ⊂ V n τ be the “slice” ball about p in V n τ . Since ǫ = ǫ n τ > , we have acontradiction. This shows that ǫ = 0 .Let us take any < ǫ < ǫ where ǫ is the infimum. Let B ǫ ( p ) be the ǫ -ball about p in sup-norm topology.Let v ∈ B ǫ ( p ) ⊂ C ω (Ω , C ) , any ǫ < ǫ . We claim that v ∈ J n for some n ∈ N .We show that in the spacelike plane passing through Re ( v ′ ( u )) and a ′ ( u ) there is avector d n ′ ( u ) of the form d n ′ = n × L a ′ on I where n ∈ N . This can be seenas follows. Let Re ( v ′ ( u )) × E a ′ ( u ) = mN = m ( N , N , N ) , normal to the plane ofRe ( v ′ ( u )) and a ′ ( u ) . Since N is timelike, we can choose m such that < N, N > L = − . We take n = ( N , N , − N ) and solve the equation ( n × L a ′ ) × E a ′ = f N .One can check that ( n × L a ′ ) × E a ′ = < a ′ , a ′ > L N . Then the equation implies < a ′ , a ′ > L = f . We have n + n − n = − since < N, N > L = − . Since < N, a ′ > E = 0 we have < n , a ′ > L = 0 . It is clear that Re ( v ′ ( u )) ⊥ E ∆ n where ∆ n = d n ′ × E a ′ . Thus v ∈ J n .Let us denote n = n v . It is easy to check if v is bounded, || n v || is bounded usingthe holomorphicity of v . In fact one can show that there exists a τ > such that if || v − p || < τ , || n v || < M and in this case, n v ∈ S .We have the following proposition. (Note that the proof here gives a proof ofProposition 2.2 too).Let V Ω be as in Theorem (2.2). Proposition 2.8.
Let p = a + id ∈ V Ω . Then there is an open ball B ǫ ( p ) ⊂ C ω (Ω , C ) open in sup-norm topology such that B ǫ ( p ) ⊂ V Ω .Proof. Let us take < ǫ < ǫ < ǫ n for all n ∈ S and such that ǫ < τ . Thus for every v ∈ B ǫ ( p ) , || n v || < M .If v ∈ B ǫ ( p ) implies that || v − p || < ǫ < ǫ n for all n ∈ S . We have by theprevious argument that v ∈ J n v for some n v ∈ S ⊂ N . Then v ∈ B sǫ nv ( p ) for this n v since || v − p || < ǫ < ǫ n v . Thus v ∈ V Ω . (cid:3) Let a ∈ S p . Let d ∈ W such that d ′ = n × L a ′ , n ∈ N . Thus d ′ ( u ) ⊥ L a ′ ( u ) on I and < d ′ ( u ) , d ′ ( u ) > L = < a ′ ( u ) , a ′ ( u ) > L on I and < d ′ , a ′ > L = 0 . We alsorequire that X ( u, v ) = Re ( a + id ) is an space-like immersion.Since ( a + id ) is isotropic in L -norm and a space-like immersion it is a maximalsurface. It can be written as ( a + id ) = ( R wu Φ , R wu Φ , R wu Φ ) where Φ + Φ − Φ ≡ , Φ i are analytic. This is the standard Weierstrass-Enneper representationof the maximal surface. Since Re ( a + id ) is a spacelike immersion, we have | Φ | + | Φ | − | Φ | = 0 . In fact there exists ζ such that | Φ | + | Φ | − | Φ | > ζ > . Let η > be such that sup Ω ( | Φ i | ) η + η < ζ / and < sup Ω ( | Φ i | ) η − η < ζ / .The second inequality implies η − sup Ω | Φ i | η > − ζ / for each i = 1 , , . Lemma 2.3.
For the above choice of η , for any analytic function F i , i = 1 , , . (2.12) || F i | − | Φ i || < η imples || F i | − | Φ i | | < ζ / And thus | F | + | F | − | F | > .Proof. Then | Φ i | − η < | F i | < | Φ i | + η imples (by squaring) that | F i | < | Φ i | + η + 2 | Φ i | η < | Φ i | + η + 2 sup Ω ( | Φ i | ) η < | Φ i | + ζ / .Similarly, | F i | > | Φ i | + η − | Φ i | η > | Φ i | + η − sup Ω | Φ i | η > | Φ i | − ζ / . | F | + | F | − | F | > | Φ | + | Φ | − | Φ | − ζ / − ζ / > (cid:3) Now we come to the main theorem of this section.
Local interpolation by maximal surface:Theorem 2.4.
Let Ω be as in Notations (2.1) and let d ∈ W be fixed such that a + id is L -isotropic and X = Re ( a + id ) is a spacelike immersion.Then there is an η > and η > , such that for all curves l ∈ S p which satisfy(in sup-norm in C ) k l − a k C < η , k l ′ − a ′ k C < η and k d l − d k C < η , k d l ′ − d ′ k C < η , there exists a maximal surface X : Ω → R and γ : I → Ω such that X ( u,
0) = a ( u ) and ( X ◦ γ )( u ) = l ( u ) where u ∈ I ⊂ Ω and d l ′ = ( a ′ × L l ′ ) × L l ′ | a ′ × L l ′ | ∈ W .Proof. Let n l = ( a ′ × L l ′ ) | a ′ × L l ′ | ∈ N . As before, p = ( a + id ) ∈ J n l ⊂ J for all n l ∈ N since a ′ ( u ) ⊥ E ∆ n l ( u ) on I where ∆ n l ( u ) = d n l ′ ( u ) × E a ′ ( u ) .Let V Ω ⊂ J be an open neighbourhood (in l -topology) of p as in Theorem (2.2).Let ǫ > be such that B ǫ ( p ) ⊂ V Ω ⊂ J as in Proposition (2.8).Let η , η be such that η + η < ǫ and η + η < η where η is as in Lemma (2.3).Let || · || denote the sup-norm in C .Let V = { s ∈ L||| s − a || < η , || s ′ − a ′ || < η on Ω } Let s ∈ L and d s ∈ W such that d s ′ = n s × L a ′ where n s = ( a ′ × L s ′ ) | a ′ × L s ′ | . Note we writethe indexing set as s and not n s to keep notations simple.Note that s + id s ∈ J n s and is L - isotropic, which can be checked by a computation.Let V = { s ∈ L||| d s − d || < η , || d s ′ − d ′ || < η on Ω } Then S p ∩ V ∩ V is an open set in L since S p , V and V are open in L .We consider l ∈ S p ∩ V ∩ V ⊂ L . Note that if l ∈ V , d l ∈ W is such that it is η -close to d and d l ′ is η -close to d ′ where d l = R wu ( a ′ × l ′ ) × L l ′ | a ′ × L l ′ | ( ˜ w ) d ˜ w NTERPOLATION 13 (2.13) C ( w ) = l ( w ) + id l ( w ) = ( C , C , C )( w ) , where w = u + iv ∈ Ω . Then C ( w ) defines an L -isotropic curve in C . This isbecause d l ′ ( u ) ⊥ L l ′ ( u ) and < d l ′ ( u ) , d l ′ ( u ) > L = < l ′ ( u ) , l ′ ( u ) > L for u ∈ I and hence < C ′ , C ′ > L ( u ) = 0 on I . By analyticity of < C ′ , C ′ > L it is zero in Ω .For u ∈ I , let Re C ( u ) = l ( u ) . We name C as the L -isotropic extension of l = l ( u ) .Now we claim that C ( w ) ∈ B ǫ ( p ) ⊂ V Ω , where p = a + id . C ( w ) − ( a ( w ) + id ( w )) = ( l ( w ) − a ( w )) + i ( d l ( w ) − d ( w )) .Thus || C ( w ) − ( a + id )( w ) || < || l ( w ) − a ( w ) || + || d l ( w ) − d ( w ) || < η + η < ǫ since l ∈ V ∩ V .Thus C ( w ) ∈ B ǫ ( p ) ⊂ V Ω ⊂ J and hence we can apply Theorem ( 2.2) to showthat C ( w ) = ( a + id ) ◦ γ . Since C ( w ) is L -isotropic, it is easy to see that ( a + id ) isalso L -isotropic.We have to show that X ( u, v ) = Re (( a + id ) ◦ γ ) = Re ( C ( w )) is a space likeimmersion, i.e. we would have a maximal surface which contains l ( u ) = Re ( C ( u )) = X ( u, . Notice ( X ◦ γ − )( u, v ) = ( a + id )( u, v ) contains a ( u ) when we put v = 0 .Thus X ( u,
0) = l ( u ) and ( X ◦ γ − )( u ) = a ( u ) , hence X interpolates between l and a . Since ( l + id l ) is isotropic (in L -norm) , it can be written as l + id l = ( R wu Φ , R wu Φ , R wu Φ ) where Φ i , i = 1 , , are analytic and such that Φ + Φ − Φ ≡ .Then Re ( l + id l ) = Re ( R wu Φ , R wu Φ , R wu Φ ) where Φ i , i = 1 , , are such that Φ + Φ − Φ ≡ .Recall that since ( a + id ) is isotropic in L -norm, ( a + id ) = ( R wu Φ , R wu Φ , R wu Φ ) where Φ + Φ − Φ ≡ . Since Re ( a + id ) is a spacelike immersion, we have | Φ | + | Φ | − | Φ | = 0 . We have | Φ | + | Φ | − | Φ | > ζ > .Since || ( l + id l ) ′ − ( a + id ) ′ || < η + η < η , we have || Φ i − Φ i || < η for i = 1 , , . ||| Φ i | − | Φ i ||| < || Φ i − Φ i || < η , or by choice of η and by Lemma(2.3) , we have | Φ | + | Φ | − | Φ | > .Thus X ( u, v ) = Re ( l + id l )( u, v ) is a spacelike immersion.Thus X is a space-like immersion and also the real part of an isotropic curve (in L -inner product and hence a maximal surface. (cid:3) Interpolation by Minimal Surfaces
Our starting point is the Björling problem for minimal surfaces. Let a : I → E ,where E := ( R , < · , · > E ) and I an open interval, be a real analytic curve with a ′ ( t ) = 0 , t ∈ I and n : I → E be a real analytic map such that < n ( t ) , a ′ ( t ) > E = 0 a ′ ( t ) can be zero only at isolated points t ∈ I . and | n ( t ) | = 1 , t ∈ I and n is called a normal vector field along a . The Björlingproblem (see page 120, [5]) is to find a minimal surface X : Ω → E with I ⊂ Ω ⊂ C , Ω simply connected, such that the following are satisfied: X ( u,
0) = a ( u ) , N ( u,
0) = n ( u ) , ∀ u ∈ I , where N : Ω → E is a normal to the surface X . The solution to theBjörling’s problem is given by X ( u, v ) = Re ( a(w) − i Z wu n( ˜w) × E a ′ ( ˜w)d ˜w ) ; w = u + iv , u ∈ I , where n ( w ) and a ( w ) are the complex analytic extensions of n ( u ) and a ( u ) over Ω and × E denotes the Euclidean cross-product. Observe that the map, w a ( w ) − i R wu n ( ˜ w ) × E a ′ ( ˜ w ) d ˜ w , from Ω to C is an isotropic curve , i.e. < ( a − id ) ′ , ( a − id ) ′ > E = 0 where d = R wu n ( ˜ w ) × E a ′ ( ˜ w ) d ˜ w . Proposition 3.1.
Given a , the map G : C ω (Ω , Ω) × C ω (Ω , C ) → C ω (Ω , C ) definedby G ( γ, d ) = (( a − id ) ◦ γ ) is a smooth map. Let γ ∈ C ω (Ω , Ω) , γ ( z ) = z and d ∈ C ω (Ω , C ) . Then the derivative of G at the point ( γ , d ) , DG ( γ ,d ) : C ω (Ω , C ) × C ω (Ω , C ) → C ω (Ω , C ) is given by DG ( γ ,d ) ( V, d ) = (( a ′ − id ′ ) V − id ) . Proof.
Proof is very similar to (2.3) with ( a + id ) ◦ γ replaced by ( a − id ) ◦ γ . (cid:3) Notation:
Let us denote by N to be all those normals n ∈ C ω (Ω , C such that n ( u ) = ( n ( u ) , n ( u ) , n ( u )) is real on I and such that n + n + n = 1 . This set iscompact, [21]. Let d n ′ = n × E a ′ and d ( u ) = 0 . Let γ ( z ) = z . Let ∆ n = d n ′ × E a ′ the normal to the plane defined by d n ′ and a ′ . Note that ∆ n is a multiple of n .Let J = ∪ n ∈N J n where J n = { c ∈ C ω (Ω , C ) s.t. Re ( c ′ ( u )) ⊥ E ∆ n ( u ) ∀ u ∈ I } .Let us endow it with the n -topology.Recall ˜ C ⊂ C ω (Ω , Ω) such that its tangent space at γ such that γ ( z ) = z is C ⊂ C ω (Ω , C ) consisting of all V such that V ( u ) = 0 .Let G n be G with the domain restricted to ˜ C × W and range restricted to J n . Inother words, G n : G − ( J n ) → J n defined by G n ( γ, d ) = ( a − id ) ◦ γ . G − ( J n ) isnonempty. This is because for all u ∈ I , Re [(( a − id n ) ◦ γ )( u )] ′ = a ′ ( u ) is ⊥ E to ∆ n ( u ) since ∆ n ( u ) = d n ′ ( u ) × E a ′ ( u ) . Thus ( γ , d n ) ∈ G − ( J n ) .Let T = { ( V, ˜ d ) ∈ C × W such that on I , Re (( a ′ ( u ) − id n ′ ( u )) V ( u ) − i ˜ d ( u )) ′ ⊥ E ∆ n ( u ) } .Let G n : T → J n be a map defined by G n ( V, ˜ d ) = ( a ′ − id n ′ ) V − i ˜ d .We show that G n is invertible. Proposition 3.2.
Fix γ ∈ C ω (Ω , Ω) to be γ ( z ) = z and d n ∈ W as before. Let p = ( γ , d n ) . Then the map G n : T ⊂ C × W → J n is invertible. NTERPOLATION 15
Proof.
The proof is similar to Proposition (2.4) with some minor differences. (cid:3)
Proposition 3.3. At p = ( γ , d n ) , G − ( J n ) has a Banach manifold structure.Proof. The proof runs similar to Proposition (2.5). (cid:3)
Proposition 3.4.
Let p = ( γ , d n ) . T p G − ( J n ) ⊂ C × W consists of all ( V, ˜ d ) ∈C × W such that on I , Re (( a ′ ( u ) − id n ′ ( u )) V ( u ) − i ˜ d ( u )) ′ ⊥ E ∆ n ( u ) . In other words, T p G − ( J n ) = T .Proof. Proof is similar to Proposition ( 2.6). (cid:3)
Let p = ( γ , d n ) . Then an easy computation shows that DG n | p : T p G − ( J n ) → J n is defined to be ( V, ˜ d ) → ( a ′ − id n ′ ) V − i ˜ d with two constraints: V ( u ) = 0 andRe (( a ′ ( u ) − id n ′ ( u )) V ( u ) − i ˜ d ) ′ ( u ) ⊥ E ∆ n ( u ) on I since ( V, ˜ d ) belong to T p G − ( J n ) . Proposition 3.5.
Fix γ ∈ C ω (Ω , Ω) to be γ ( z ) = z and d l ∈ W as before. Let p = ( γ , d n ) . The map DG | p : T p ( G − ( J n )) ⊂ C × W → J n is invertible.Proof. The proof is similar to Proposition (2.7). (cid:3)
Theorem 3.1.
Let a and Ω as in Notations 2.1. Let p = a + id . Then thereexist open neighbourhood V Ω ⊂ J of p (open in n -topology) and U Ω ⊂ C ω (Ω , Ω) and W ⊂ W such that for every c ∈ V Ω , there exists γ ∈ U Ω and d ∈ W such that ( a − id ) ◦ γ = c .Proof. The proof is similar to Theorem ( 2.2). (cid:3)
Interpolation by minimal surfaces.
By interpolation by minimal surface oftwo real analytic curves we mean the following. Let a = a ( u ) and c = c ( u ) be two realanayltic curves in R . Suppose there exists a minimal surface X = X ( u, v ) = Re f ( z ) such that < f ′ , f ′ > = 0 and such that a ( u ) = Re f ( γ ( u )) and c ( u ) = Re ( f ( γ )( u )) where γ and γ are automorphisms of the domain Ω . Then we say that a = a ( u ) and c = c ( u ) are interpolated by the minimal surface X = X ( u, v ) .Recall that an isotropic curve is defined as follows.Any holomorphic map f : Ω ⊂ C → C is said to be an isotropic curve if itsatisfies < f ′ ( z ) , f ′ ( z ) > E = 0 , z ∈ Ω ; where ′ denotes the complex derivative withrespect to z and < · , · > E is the extension of the dot product in E to C .Let f : Ω ⊂ C → C be a holomorphic map defined by f ( z ) = a ( z ) − id ( z ) such that h a ′ ( z ) , a ′ ( z ) i E = h d ′ ( z ) , d ′ ( z ) i E and h a ′ ( z ) , d ′ ( z ) i E = 0 . Then f satisfies h f ′ ( z ) , f ′ ( z ) i E = 0 for all z ∈ Ω , and hence defines an isotropic curve. Local Interpolation by minimal surfaces:
Let l : I → E which havedifferentiable extension l : Ω → C and analytic on Ω . The space of all such l isdenoted by L .We have the following proposition. Proposition 3.6.
Let p = a − id and V Ω be as a neighbourhood of p as above, openin n -topology. Then there is a open ball B ǫ ( p ) in sup-norm topology in C ω (Ω , C ) such that B ǫ ( p ) ⊂ V Ω .Proof. The proof runs similar to Proposition (2.8). (cid:3)
Theorem 3.2.
Let a , Ω be as in Notations (2.1) . Let d ∈ W be fixed such that p = ( a − id ) is isotropic, i.e. < a ′ − id ′ , a ′ − id ′ > E = 0 and X = Re ( a − id ) animmersion.Then there is an η > and η > , such that for all curves l ∈ L which satisfy k l − a k C < η , k l ′ − a ′ k C < η and k d l − d k C < η , k d l ′ − d ′ k C < η , there exists a minimal surface X : Ω → R and γ : I → Ω such that X ( u,
0) = a ( u ) and ( X ◦ γ )( u ) = l ( u ) where u ∈ I ⊂ Ω and d l ′ = ( a ′ × E l ′ ) × E l ′ | a ′ × E l ′ | ∈ W .Proof. The proof runs similar to Theorem ( 2.4). All inner products and cross-products are Euclidean. (cid:3)
Future Directions:
Our methods can be extended to interpolation of tworeal analytic curves by Born Infeld solitons [23] and also by timelike minimal sur-faces [11], [14], [17]. 4.
Acknowledgement
The authors are grateful to Dr. Pradip Kumar, S.N.U., U.P., India, for the usefuldiscussions.This research was supported in part by the International Centre for Theoreti-cal Sciences (ICTS)- Bangalore, for participating in the program - Geometry andTopology for Lecturers (Code: ICTS/gtl2018/06). Rukmini Dey would like to ac-knowledge support of the Department of Atomic Energy, Government of India, underproject no. RTI4001.
References [1] L. J. Alías, R. M. B. Chaves, and P. Mira,
Björling problem for maximal surfaces in Lorentz-Minkowski space , Math. Proc. Cambridge Philos. Soc, 134, No.2, 2003, 289-316.[2] M. S. Berger:
Nonlinearity and Functional Analysis
Academic Press, Inc, 1977.[3] R. Dey, P. Kumar, R. K.Singh:
Existence of maximal surface containing given curve andspecial singularity , J. Ramanujan Math. Soc. (No.4), 455–471 (2018). NTERPOLATION 17 [4] R. Dey, P. Kumar, R. K. Singh:
Local interpolation for minimal surfaces , arXiv:1907.10780[5] U. Dierkes, S. Hildebrandt, A. Küster, O. Wohlrab:
Minimal Surfaces I , Springer Verlag,1991.[6] Jesse Douglas,
Solution of the problem of Plateau , Trans. Amer. Math. Soc. 33, no. 1, 1931,263–321.[7] J. Douglas:
The Problem of Plateau for two contours
J. Math. Phys. 10, 1931, 315–359.[8] S. Fujimori, W. Rossman, M. Umehara, S-D. Yang, K. Yamada :
New maximal surfaces inMinkowski 3-space with arbitrary genus and their cousins in de Sitter 3-space ; Result. Math.56, 2009, 41-82.[9] R. Hamilton:
The Inverse Function Theorem of Nash and Moser , Bulletin of the Am. Math.Soc., Volume 7, Number 1, July 1982, 65-222.[10] R. Hardt and L. Simon:
Boundary regularity and embedded solutions for the oriented Plateauproblem , Ann. of Math. (2) 110, no. 3, 1979, 439—486.[11] Y.M. Kim, S.E. Koh, H. Shin, S.D. Yang:
Spacelike maximal surfaces, timelike minimalsurfaces, and Björling representation formulae , Journal of the Korean Mathematical Society48.5, 2011, 1083-1100.[12] A. Kriegl and P. W. Michor.
The Convenient Setting of Global Analysis , Mathematical Sur-veys, and Monographs. American Mathematical Society, 1997.[13] F. Labourie, J. Toulisse, M. Wolf:
Plateau Problems for Maximal Surfaces in Pseudo-Hyperbolic Spaces , arXiv 2006.12190.[14] S. Lee:
Weierstrass representation for timelike minimal surfaces in Minkowski 3-space , Com-mun. Math. Anal., 2008, Conference 1, 11–19.[15] R. López:
On the existence of spacelike constant mean curvature surfaces spanning two circularcontours in Minkowski space , J. Geom. Phys., 57, No. 11, 2007, 2178-2186.[16] R. López:
Differential Geometry of Curves and Surfaces in Lorentz-Minkowski space , Inter-national Electronic Journal of Geometry, no. 7, 2014, 44-107.[17] M. A. Magid:
Timelike surfaces in Lorentz 3-space with prescribed mean curvature and Gaussmap , Hokkaido Mathematical Journal, 20.3, 1991, 447-464.[18] J.C.C. Nitsche:
Lectures on Minimal surfaces (English edition), Cambridge University Press,1989.[19] R. Osserman:
Survey of minimal surfaces , Dover Publications, New York, 1986.[20] T. Rado,
On Plateau’s problem , Ann. of Math. (2) 31, no. 3, 1930, 457-469.[21] W. Rudin:
Functional Analysis , Tata McGraw Hill, 1973.[22] I. N. Vekua:
A certain functional equation of minimal surface theory. (Russian) Dokl. Akad.Nauk SSSR 217 (1974), 997–1000.[23] G.B. Whitham:
Linear and Nonlinear Waves , John Wiley and Sons, Inc. 1999.
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