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Matching univalent functions Pavel Gumenyuk joint research with Erlend Grong and Alexander Vasilev University of Bergen Saratov State University 2 Matching functions and conformal welding 1. Matching functions and conformal welding


  1. Matching univalent functions Pavel Gumenyuk joint research with Erlend Grong and Alexander Vasil’ev University of Bergen Saratov State University

  2. 2 Matching functions and conformal welding 1. Matching functions and conformal welding Definition 1. Suppose that: • f is a conformal mapping of D := { z : | z | < 1 } onto a Jordan domain D ; • ϕ is a conformal mapping of D ∗ := C \ D onto a Jordan domain D ∗ . Then the functions f and ϕ are said to be matching if D and D ∗ are complementary domains, i. e. D ∩ D ∗ = ∅ and Γ := ∂D = ∂D ∗ . University of Bergen Saratov State University

  3. 3 Matching functions and conformal welding Using fractional-linear change of variables, we can assume that: (i) 0 ∈ D and ∞ ∈ D ∗ ; (ii) f (0) = f ′ (1) − 1 = 0; (iii) ϕ ( ∞ ) = ∞ . S := { f : D → C : f is analytic, univalent, and subject to normalization (ii) } . Problem 1. Given f ∈ S s. t. f ( D ) is a Jordan domain, find a univalent meromorphic function ϕ which matches the function f . A pair of matching functions ( f, ϕ ) defines the homeomorphism of the unit circle S 1 , γ = f − 1 ◦ ϕ, γ : S 1 → S 1 . (1) Definition 2. Representation (1) of a homeomorphism γ : S 1 → S 1 by means of matching functions is called the conformal welding . University of Bergen Saratov State University

  4. 4 Matching functions and conformal welding Problem 2. Find the conformal welding for a given orientation preserv- ing homeomorphism γ : S 1 → S 1 , i. e. the pair of matching univalent functions ( f, ϕ ) such that γ = f − 1 ◦ ϕ . Problem 2 has a unique solution for all homeomorphisms γ that are quasisymmetric, i. e. satisfies � e i ( t + h ) � � e it � � � γ − γ � � � � � < C γ < + ∞ , for all t, h ∈ R , 0 < | h | < π. (2) � � � � � � e i ( t − h ) e it − γ γ � � � A. Pfluger, 1960; O. Lehto & K.I. Virtanen, 1960. Also follows from the Ahlfors – Beurling Extension Theorem, A. Beurling & L. Ahlfors, 1956 Existence and uniqueness of the conformal welding for the constant C γ replaced in right-hand side of (2) with ρ ( h ) = O (log h ), O. Lehto, 1970; G.L. Jones, 2000. University of Bergen Saratov State University

  5. 5 Matching functions and conformal welding Denote by: • Lip α ( X, Y ) the class of all functions h : X → Y which are H¨ older- continuous with exponent α ; • S the class of all analytic univalent functions f : D → C such that f (0) = f ′ (0) − 1 = 0; • S qc := { f ∈ S : f can be extended to q. c. homeomorphism of C } ; • S 1 ,α := { f ∈ S : ∂f ( D ) is a C 1 ,α -smooth Jordan curve } , α ∈ (0 , 1); • S ∞ := { f ∈ S : ∂f ( D ) is a C ∞ -smooth Jordan curve } ; • Homeo + qs ( S 1 ) the group of all orientation preserving q. s. homeo- morphisms γ : S 1 → S 1 . Remark 1. The conformal welding establishes one-to-one correspon- dence between S qc and Homeo + qs ( S 1 ) / Rot( S 1 ). • To calculate γ ∈ Homeo + qs ( S 1 ) for given f ∈ S qc one have to solve Problem 1, which is to find the function ϕ matching f . University of Bergen Saratov State University

  6. 6 Matching functions and conformal welding • To determine f ∈ S qc for given γ ∈ Homeo + qs ( S 1 ) one have to solve the Beltrami equation � ¯ if z ∈ D ∗ , ∂u ( z ) /∂u ( z ) , ∂ ˜ ¯ f ( z ) = µ ( z ) ∂ ˜ f ( z ) , µ ( z ) := (3) 0 , otherwise, � � � � ∂ := 1 ∂x − i ∂ ∂ ∂ := 1 ∂x + i ∂ ∂ ¯ , , 2 ∂y 2 ∂y with the normalization f ′ (0) − 1 = 0 , f ( ∞ ) = ∞ and ˜ ˜ f (0) = ˜ (4) where u is any q. c. automorphism of D ∗ such that u | S 1 = γ − 1 . u ( ∞ ) = ∞ and Then f | D ∗ ◦ u − 1 f := ˜ ϕ := ˜ f | D , (5) are matching functions, f ∈ S qc , and γ = f − 1 ◦ ϕ . University of Bergen Saratov State University

  7. 7 Main results 2. Main results The following theorem establishes more explicit relation between f , ϕ , For fixed f ∈ S 1 ,α define the operator and γ for the smooth case. I f : Lip α ( S 1 , R ) → Hol( D ) by the formula � 2 sf ′ ( s ) � I f [ v ]( z ) := − 1 v ( s ) ds � s , z ∈ D . (6) 2 πi S 1 f ( s ) f ( s ) − f ( z ) Suppose f ∈ S 1 ,α and ϕ , ϕ ( ∞ ) = ∞ , are matching Theorem 1. Then the kernel of the operator I f : Lip α ( S 1 , R ) → Hol( D ) functions. is the one-dimensional manifold ker I f = span { v 0 } , where ( ψ ◦ f )( z ) v 0 ( z ) := 1 ψ := ϕ − 1 , z ∈ S 1 . (7) f ′ ( z )( ψ ′ ◦ f )( z ) , z Moreover, the function v 0 is positive on S 1 and satisfies the following condition � 2 π dt v 0 ( e it ) = 2 π. (8) 0 University of Bergen Saratov State University

  8. 8 Kirillov’s manifold Remark 2. Theorem 1 reduces Problem 1 to solution of the equation I f [ v ] = 0 . (9) Indeed, given f and v 0 , one can calculate ψ = ϕ − 1 on the boundary of D ∗ by solving the following differential equation ψ ′ ( u ) = H ( u ) ψ ( u ) , u ∈ ∂D ∗ , 1 H ◦ f − 1 , for z ∈ S 1 . H := ˜ ˜ H ( z ) := (10) zf ′ ( z ) v 0 ( z ) Complex Solutions to I f [ v ] = 0 . Suppose f ∈ S 1 ,α and ϕ , ϕ ( ∞ ) = ∞ , are matching Theorem 2. functions and γ := f − 1 ◦ ϕ . Then the kernel of the operator I f : Lip α ( S 1 , C ) → Hol( D ) is the set of all functions v of the form v ( z ) = v 0 ( z ) · ( h ◦ γ − 1 )( z ) , z ∈ S 1 , (11) where v 0 is defined by (7) and h is an arbitrary holomorphic function in D ∗ admitting Lip α ( S 1 , C ) -extension to S 1 . University of Bergen Saratov State University

  9. 9 Kirillov’s manifold 3. Operator I f and Kirillov’s manifold By Diff + ( S 1 ) denote the Lie – Fr´ echet group of all orientation preserv- ing C ∞ -smooth diffeomorphisms of S 1 . In 1987 A.A. Kirillov proposed to use the 1-to-1 correspondence be- tween S qc and Homeo + qs ( S 1 ) / Rot( S 1 ) established by conformal weld- ing to represent the homogeneous manifold M := Diff + ( S 1 ) / Rot( S 1 ) (Kirillov’s manifold) via univalent functions. The bijection K : M → S ∞ allows to bring the complex structure from S ∞ to M . A.A. Kirillov proved that the (left) action of Diff + ( S 1 ) on M is holomorphic w.r.t. this complex structure. The infinitesimal version of K : M → S ∞ is more explicit and expressed by means of I f [ v ]. Consider the variation of γ ∈ Diff + ( S 1 ) given by γ ε ( ζ ) := γ ( ζ ) δγ ( ζ ) , δγ := exp iε ( v ◦ γ ) , (12) University of Bergen Saratov State University

  10. 10 Kirillov’s manifold where v ∈ C ∞ ( S 1 , R ) is regarded as an element of T id Diff + ( S 1 ). Variation (12) of γ results in the following variation of f ∈ S ∞ f ε := K ( γ ε ) = f + δf, � 2 f 2 ( z ) v ( s ) sf ′ ( s ) � δf ( z ) = ε ds � s = iεf 2 ( z ) I f [ v ]( z ) . (13) 2 π S 1 f ( s ) f ( z ) − f ( s ) Remark 3. A natural consequence of this is that I f [ v ]( z ) = 0 for all z ∈ D if and only if the variation of γ produces no variation of [ γ ] ∈ M (up to higher order terms), which can be reformulated as follows: the element of T γ Diff + ( S 1 ) represented by v ◦ γ is tangent to the one- dimensional manifold γ ◦ Rot( S 1 ) = [ γ ] ⊂ Diff + ( S 1 ) . The latter is equivalent to � � T id Rot( S 1 ) � constant functions on S 1 � v ∈ Ad γ = Ad γ , (14) University of Bergen Saratov State University

  11. 11 Kirillov’s manifold where Ad γ stands for the differential of β �→ γ ◦ β ◦ γ − 1 at β = id. Elementary calculations show that Ad γ u = u ◦ γ − 1 � ′ , β # := π − 1 ◦ β ◦ π � � # , � γ − 1 where π : R → S 1 is the universal covering, π ( x ) = e ix . As a conclusion we get The kernel of I f : C ∞ ( S 1 , R ) → Hol( D ) is a one- Proposition 1. dimensional manifold and coincides with span { 1 / ( γ − 1 ) # } . Remark 4. This Proposition is the special case of Theorem 1 for C ∞ - smooth case. It shows that Problem 2 (of finding conformal welding) is reduced by Theorem 1 to finding solution to I f [1 / ( γ − 1 ) # ] = 0 , (15) regarded as equation w.r.t. f ∈ S ∞ . University of Bergen Saratov State University

  12. 12 An Example 4. An Example of matching functions Given an integer n > 1, let us consider quadratic differentials Ψ( ζ ) dζ 2 := − dζ 2 ζ 2 ; n − 1 W ( w ) dw 2 := − w n − 2 dw 2 w k := e 2 πik/n ; � , P ( w ) := ( w − w k ) , P ( w ) k =0 Z ( z ) dz 2 := − z n − 2 dz 2 , Q ( z ) n − 1 | z k | z k := re 2 πik/n , � Q ( z ) := κ ( z k − z )( z − 1 /z k ) , r ∈ (0 , 1) . z k k =0 where κ > 0 is such that � � Z ( z ) dz = 2 π (16) S 1 for the appropriately chosen branch of the square root. University of Bergen Saratov State University

  13. 13 An Example n = 5 1 1 0.5 0.5 -1 -0.5 0.5 1 -1 -0.5 0.5 1 -0.5 -0.5 -1 -1 The structure of trajectories W ( w ) dw 2 > 0 and Z ( z ) dz 2 > 0. University of Bergen Saratov State University

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