Multiplication Operators, Riemann Surfaces and Analytic continuation

Multiplication Operators, Riemann Surfaces and Analytic continuation Dechao Zheng Vanderbilt University This is a joint work with Ronald G. Douglas and Shunhua Sun.

Bergman space Let D be the open unit disk in C:

Bergman space Let D be the open unit disk in C: Let da denote normalized Lebesgue area measure on D.

Bergman space Let D be the open unit disk in C: Let da denote normalized Lebesgue area measure on D. L 2 (D, da) = {f : f (z) 2 da(z) < }. D

Bergman space Let D be the open unit disk in C: Let da denote normalized Lebesgue area measure on D. L 2 (D, da) = {f : f (z) 2 da(z) < }. The Bergman space D L 2 a = {f L 2 (D, da) : f is analytic in D}

Reproducing kernel Inner product f, g = = D n=0 f (z)g(z)da(z) a n b n n + 1 if f (z) = n=0 a nz n and g(z) = n=0 b nz n.

Reproducing kernel Inner product f, g = = D n=0 f (z)g(z)da(z) a n b n n + 1 if f (z) = n=0 a nz n and g(z) = n=0 b nz n. For each z D, f f (z) is a bounded linear functional on L 2 a.

Reproducing kernel Inner product f, g = = D n=0 f (z)g(z)da(z) a n b n n + 1 if f (z) = n=0 a nz n and g(z) = n=0 b nz n. For each z D, f f (z) is a bounded linear functional on L 2 a. There is a unique function k z L 2 a such that f (z) = f, k z.

Multiplication Operators For φ H (D), the multiplication operator M φ on L 2 a is defined by M φ h = φh for h L 2 a.

Multiplication Operators For φ H (D), the multiplication operator M φ on L 2 a is defined by M φ h = φh for h L 2 a. Let e n = n + 1z n. Then {e n } n=0 of L 2 a. form an orthonormal basis

Multiplication Operators For φ H (D), the multiplication operator M φ on L 2 a is defined by M φ h = φh for h L 2 a. Let e n = n + 1z n. Then {e n } n=0 of L 2 a. M z is the Bergman shift form an orthonormal basis M z e n = n + 1 n + 2 e n+1.

Multiplication Operators M αφ+βψ = αm φ + βm ψ. M φ M ψ = M φψ.

Multiplication Operators M αφ+βψ = αm φ + βm ψ. M φ M ψ = M φψ. for each α D. kerm φ φ(α) = φ(β)=φ(α) k β,

Multiplication Operators M αφ+βψ = αm φ + βm ψ. M φ M ψ = M φψ. for each α D. kerm φ φ(α) = φ(β)=φ(α) k β, σ(m φ ) = φ(d).

Bergman spaces?

Bergman spaces? On the Hardy Space Beurling Theorem Every invariant subspace M of the multiplication operator T z by z is equal to θh 2 for some inner function θ.

Bergman spaces? On the Hardy Space Beurling Theorem Every invariant subspace M of the multiplication operator T z by z is equal to θh 2 for some inner function θ. 1. M = [M T z M] = [θ]. 2. dim[m T z M] = 1. On the Bergman Space Aleman-Richter-Sundberg Theorem Every invariant subspace M of the multiplication operator M z by z is generated by its wandering subspace M M z M., i.e., M = [M M z M]. Index Theorem For each n equal to positive integer or +, there is an invariant subspace M of M z such that dim [M M z M] = n.

Multiplication operators? On the Hardy Space On the Bergman Space Lattice T z : {θh 2 : inner θ}.

Multiplication operators? On the Hardy Space On the Bergman Space Lattice M z? Lattice T z : {θh 2 : inner θ}. Apostol-Bercovici-Foias-Pearcy Invariant subspace problem Does every bounded operator on a Hilbert space have an invariant subspace?

Multiplication Operators via Complex Geometry Cowen-Douglas classes If φ is a rational function with poles outside of the closed unit disk, Mφ B n(ω) for some n and open set Ω C.

Multiplication Operators via Complex Geometry Cowen-Douglas classes If φ is a rational function with poles outside of the closed unit disk, Mφ B n(ω) for some n and open set Ω C. T B n (Ω) on a Hilbert space H if Ω σ(t ). Ran(T ω) = H for each ω Ω. ω Ω ker(t ω) = H. dimker(t ω) = n for each n Ω.

Multiplication Operators via Theory of Subnormal Operators Subnormal operators M φ is subnormal.

Multiplication Operators via Theory of Subnormal Operators Subnormal operators M φ is subnormal. An operator S on a Hilbert space H is subnormal if there are a Hilbert space K containing H and a normal operator N on K (i.e., NN = N N) such that H is an invariant subspace of N and S = N H. Suppose S 1 and S 2 are two subnormal operators on a Hilbert space H and N 1 and N 2 are the minimal normal extensions of S 1 and S 2 on K respectively. If S 1 and S 2 are unitarily equivalent, i.e., W S 1 W = S 2 for some unitary operator W on H, then there is a unitary operator W on K such that W = W H, W N 1 W = N 2.

Minimal Normal Extension of M φ For φ in H, let M φ denote the multiplication operator by φ on L 2 (D, da) given by M φ g = φg for each g in L 2 (D, da). The minimal normal extension of the Bergman shift M z is the operator M z on L 2 (D, da). For each finite Blaschke product φ, the minimal normal extension of M φ is the operator M φ on L 2 (D, da). Thus for each unitary operator W commuting with M φ, there is a unitary operator W on L 2 (D, da) such that W = W L 2 a, W M φ W = M φ.

Multiplication Operators via H 2 (D 2 ) Let [z w] denote the subspace of H 2 (T 2 ) spanned by (z w)h 2 (T 2 ).

Multiplication Operators via H 2 (D 2 ) Let [z w] denote the subspace of H 2 (T 2 ) spanned by (z w)h 2 (T 2 ). Let H = [z w].

Multiplication Operators via H 2 (D 2 ) Let [z w] denote the subspace of H 2 (T 2 ) spanned by (z w)h 2 (T 2 ). Let H = [z w]. For each integer n 0, let p n (z, w) = n z i w n i. i=0 Then H is the subspace of H 2 (T 2 ) spanned by functions {p n } n=0.

Lifiting the Bergman shift Let P H be the orthogonal projection from L 2 (T 2, dσ) onto H.

Lifiting the Bergman shift Let P H be the orthogonal projection from L 2 (T 2, dσ) onto H. B def = P H T z H = P H T w H.

Lifiting the Bergman shift Let P H be the orthogonal projection from L 2 (T 2, dσ) onto H. B def = P H T z H = P H T w H. The Bergman shift M z = B via the unitary operator U.

Lifiting the Bergman shift Let P H be the orthogonal projection from L 2 (T 2, dσ) onto H. B def = P H T z H = P H T w H. The Bergman shift M z = B via the unitary operator U. Thus the Bergman shift is lifted up as the compression of an isometry on a nice subspace of H 2 (T 2 ).

Problem 1 Classify reducing subspaces of M φ? Finite Blaschke product: φ(z) = n j=1 where α i are points in the unit disk D. z α i 1 ᾱ i z,

Problem 1 Classify reducing subspaces of M φ? Finite Blaschke product: φ(z) = n j=1 where α i are points in the unit disk D. z α i 1 ᾱ i z, An invariant subspace M for an operator T on a Hilbert space H is a subspace M of H such that T M M. A reducing subspace M for an operator T on a Hilbert space H is a subspace M of H such that T M M and T M M.

Commutant For a bounded operator S, the commutant {S} = {T B(L 2 a) : ST = TS} Let A φ = {M φ } {Mφ }. Then A φ = {T : TM φ = M φ T, TMφ = M φ T }. So A φ is a von Neumann algebra.

Commutant For a bounded operator S, the commutant Let Then {S} = {T B(L 2 a) : ST = TS} A φ = {M φ } {M φ }. A φ = {T : TM φ = M φ T, TM φ = M φ T }. So A φ is a von Neumann algebra. If M is a reducing subspace of M φ and P M is the orthogonal projection from the Bergman space onto M, then P M M φ = M φ P M, P M M φ = M φ P M. Problem 2 Structure of the von Neumann algebra A φ?

Why finite Blaschke product φ? Theorem (Cowen-Thomson) For each rational function f with poles outside of the closed unit disk, there is a finite Blaschke product φ such that {M f } = {M φ }.

Why finite Blaschke product φ? Theorem (Cowen-Thomson) For each rational function f with poles outside of the closed unit disk, there is a finite Blaschke product φ such that {M f } = {M φ }. The local inverses {ρ j } n j=1 of an n-th order Blaschke product φ have group-like property under composition. The Riemann surface φ 1 φ over D is completely determined by the polynomial P(w)Q(z) P(z)Q(w) where P(z) an Q(z) are coprime polynomials of two complex variables z and w such that φ(z) = P(z) Q(z).

Why finite Blaschke product φ? Theorem There exists a unique reducing subspace M 0 (φ) for φ(b) such that φ(b) M0 (φ) is unitarily equivalent to the Bergman shift. In fact, M 0 (φ) = l 0{p l (φ(z), φ(w))e 0 } H, and { p l (φ(z),φ(w))e l+1 e0 0 } 0 form an orthonormal basis of M 0(φ). Here e 0 (z, w) = φ(z) φ(w) z w. Theorem For a function φ in H, if M φ on the Bergman space has the distinguished reducing subspace on which the restriction of M φ is unitarily equivalent to the Bergman shift, then φ must be a finite Blaschke product.

Distinguished Reducing Subspace M 0 (φ) Theorem (Stessin-Zhu) If φ is a finite Blaschke product and φ(0) = 0, M 0 (φ) φ n φ. n=0

Second Order Blaschke products A reducing subspace M of T is called minimal if only reducing subspaces contained in M are M and {0}.

Second Order Blaschke products A reducing subspace M of T is called minimal if only reducing subspaces contained in M are M and {0}. M z 2 has only two nontrivial minimal reducing subspaces

Second Order Blaschke products A reducing subspace M of T is called minimal if only reducing subspaces contained in M are M and {0}. M z 2 has only two nontrivial minimal reducing subspaces M even = span k=0 {z2k } M odd = span k=0 {z2k+1 } Theorem (Sun-Wang-Zhu) Let φ(z) = 2 j=1 z α i 1 ᾱ i z. Then M φ has only two nontrivial minimal reducing subspaces {M 0 (φ), M 0 (φ) }.

Third Order Blaschke products via H 2 (D 2 ) Theorem 1 (Guo-Sun-Zheng-Zhong) Let φ be a Blaschke product with three zeros. If φ(z) has a multiple critical point in the unit disk, (i.e., φ z 3 ) then the lattice of reducing subspaces of M φ the lattice generated by {M 1, M 2, M 3 }, where M j = k=0 {z3k+j } for j = 1, 2, 3. If φ does not have any multiple critical point in the unit disk, then the lattice of reducing subspaces of M φ is generated by {M 0 (φ), M 0 (φ) }. By Bochner Theorem that every Blaschke product with n zeros has exactly n 1 critical points in the unit disk D, we obtain the classification of reducing subspaces of M φ for a Blaschke product φ with the third order.

Fourth order Blaschke products Theorem (Sun-Zheng-Zhong) Let φ be a fourth order Blaschke product. (1) If φ z 4, the lattice of reducing subspaces of M φ the lattice generated by {M 1, M 2, M 3, M 4 }, where M j = k=0 {z4k+j } for j = 1, 2, 3, 4. (2) If φ z 4 but φ is decomposable, i.e., φ = ψ 1 ψ 2 for two Blaschke products ψ 1 and ψ 2 with orders 2 then the lattice of reducing subspaces of M φ is generated by {M 0 (φ), M 0 (ψ 2 ) M 0 (φ), M 0 (ψ 2 ) }. (3) If φ is not decomposable, then the lattice of reducing subspaces of M φ is generated by {M 0 (φ), M 0 (φ) }.

Reducing Subspaces vs Geometry Theorem 1 (Guo-Sun-Zheng-Zhong) Let φ be a Blaschke product with three zeros. If φ(z) has a multiple critical point in the unit disk, (i.e., φ z 3 ) then the lattice of reducing subspaces of M φ the lattice generated by {M 1, M 2, M 3 }, where M j = k=0 {z3k+j } for j = 1, 2, 3. If φ does not have any multiple critical point in the unit disk, then the lattice of reducing subspaces of M φ is generated by {M 0 (φ), M 0 (φ) }. Theorem 1 suggests that the lattice of reducing subspaces of M φ is related to the geometry of φ.

Reducing Subspaces vs Geometry Conjecture For a Blaschke product φ of finite order, the number of nontrivial minimal reducing subspaces of M φ equals the number of connected components of the Riemann surface of φ 1 φ over D.

Main Result Theorem Let φ be a finite Blaschke product. The von Neumann algebra A φ is generated by E 1,, E q and has dimension q. Here q is the number of connected components of the Riemann surface φ 1 φ over the unit disk and E k f (z) = ρ G ik ρ (z)f (ρ(z)). The Riemann surface φ 1 φ over the unit disk. G ik is the collection of local inverses of φ which are mutually analytically continuable. Local inverses Analytical continuation

Abelian C -algebras Conjecture For a Blaschke product φ of finite order, the number of nontrivial minimal reducing subspaces of M φ equals the number of connected components of the Riemann surface of φ 1 φ over D. Our main result implies that Conjecture is equivalent to that the C -algebra A φ is abelian. Theorem Let φ be a finite Blaschke product with order less than or equal to 8. Then A φ is commutative and hence, in these cases, the number of minimal reducing subspaces of M φ equals the number of connected components of the Riemann surface φ 1 φ over the unit disk.

Analytic continuation An analytic function element is a pair (f, U), which consists of an open disk U and an analytic function f defined on this disk. A finite sequence U = {(f j, U j )} m j=1 is a continuation sequence if U j U j+1 is not empty for j = 1,, m 1 and f j f j+1 on U j U j+1, for j = 1,, m 1.

Analytic continuation Let γ be an arc with parametrization γ(t), γ(t) being a continuous function on an interval [0, 1]. A sequence {U 1,, U m } is admissible or a covering chain for γ if each U j is an open disk, and if there exist increasing numbers t 1, t m in [0, 1] such that γ(t j ) U j for j = 1,, m and γ(t) U 1, 0 t t 1 U j U j+1, t j t t j+1 U m, t m t 1.

A continuation sequence U = {(f j, U j )} m j=1 is an analytic continuation along the arc γ if the sequence U 1,, U m is admissible for γ. Each of {(f j, U j )} m j=1 is an analytic continuation of the other along the curve γ. We say that the analytic function f 1 on U 1 admits a continuation in U m. Theorem (Riemann Monodromy Theorem) Suppose Ω C is a simply connected open set. If an analytic element, (f, U) can be analytically continued along any path inside Ω, then this analytic function element can be extended to be a single-valued holomorphic function defined on the whole of Ω.

Local Inverses Bochner s theorem says that φ has exactly n 1 critical points in the unit disk D and none on the unit circle.

Local Inverses Bochner s theorem says that φ has exactly n 1 critical points in the unit disk D and none on the unit circle. Let C denote the set of the critical points of φ in D and the set of branch points: Let E = D/F. For an open set V D, we define a local inverse of φ in V to be a function f analytic in V with f (V ) D such that φ(f (z)) = φ(z) for every z in V. That is, f is a branch of φ 1 φ defined in V. F = φ 1 φ(c) = {z 1,, z m }.

Not all of the branches of φ 1 φ can be continued to a different branch, for example z is a single valued branch of φ 1 φ. A local inverse (f, V ) admits an analytic continuation along the curve γ in E if there is a continuation sequence U = {(f j, U j )} m j=1 admissible for γ and (f 1, U 1 ) equals (f, V ). A local inverse in V E is admissible for φ if it admits unrestricted continuation in E. As φ : E D is an n-to-one mapping, for each z E, φ 1 φ(z) contains n distinct points {z 1,, z n } in E. For each z E, φ(ρ(z)) = φ(z) has n solutions φ 1 φ = {ρ k (z)} n k=1 on an open neighborhood D z of z (ρ j, D z ) is an analytic element and it is locally analytic and arbitrarily continuable in E. {ρ j } n j=1 is the family of admissible local inverses in some invertible open disc V D.

For a point z 0 V, label those local inverses as {ρ j (z)} n i=1 on V. If there is a loop γ in E at z 0 such that ρ j and ρ j in {ρ i (z)} n i=1 are mutually analytically continuable along γ, we can then write ρ j ρ j, and it is easy to check that is an equivalence relation. Using the equivalence relation, we divide {ρ i (z)} n i=1 into equivalence classes {G i1, G i2,, G iq } where i 1 = 1 < i 2 < i 3 < < i q n. Each element in G ik extends analytically to the other in G ik, but can not extend to any element in G il if i k i l.

{ρ j } n j=1 form a group under composition. The set of admissible local inverses has the useful property that is closed under composition ρ ˆρ(z) = ρ(ˆρ(z)).

{ρ j } n j=1 form a group under composition. The set of admissible local inverses has the useful property that is closed under composition ρ ˆρ(z) = ρ(ˆρ(z)). φ(ρ ˆρ(z)) = φ(ρ(ˆρ(z))) = φ(ρ(z)) = φ(z). Is ρ ˆρ arbitrarily continuable in E?

Let f and g be admissible local inverses in open discs V and W centered at a and f (a), respectively, with f (V ) W. Let γ be a curve in E with initial point a. f can be analytically continued along γ. There is an obvious image curve γ of γ under this analytic continuation along γ. g can be analytically continued along γ. By refining the covering chain of γ, we can assume that if Ṽ is a covering disc of γ and ( f, Ṽ ) the corresponding function element, then f (Ṽ ) is contained in one of the covering discs of γ. We now compose corresponding function elements in the analytic continuations along γ and γ to obtain an analytic continuation for (g f, V ) along γ.

Main Result Theorem Let φ be a finite Blaschke product. The von Neumann algebra A φ is generated by E 1,, E q and has dimension q. Here q is the number of connected components of the Riemann surface φ 1 φ over the unit disk and E k f (z) = ρ G ik ρ (z)f (ρ(z)). Russo-Dye Theorem Every element in a von Neumann algebra A can be written as a finite linear combination of unitary operators in A. Representation of unitary operators For each unitary operator W in the von Neumann algebra A φ, we hope to get a nice representation of W on the Bergman reproducing kernel k α.

Local Representation Theorem Let φ be a finite Blaschke product. Let U be an invertible open set of E. Then for each T in {M φ }, there are analytic functions {s i (α)} n i=1 on U such that for each h in the Bergman space L2 a, Th(α) = T k α = n s i (α)h(ρ i (α)), i=1 n s i (α)k ρi (α) i=1 for each α in U. Moreover, those functions {s i (α)} n i=1 admit unrestricted continuation in E.

Riemann surfaces of φ 1 φ over D A finite Blaschke product φ with n zeros is an n to 1 conformal map of D onto D. Let φ = P(z) Q(z) be a Blaschke product of order n where P(z) and Q(z) are two coprime polynomials of degree n. Let f (w, z) = P(w)Q(z) P(z)Q(w). Then f (w, z) is a polynomial of w with degree n and of z. For each z D, f (w, z) = 0 has exactly n solutions in D. The Riemann surface S φ equals the locus S f of solutions of the equation f (w, z) = 0 in D 2 except for a few branch points. The Riemann surface S φ for φ 1 φ over D is an n-sheeted cover of D with at most n(n 1) branch points, and it is not connected.

By unique factorization theorem for the ring C[z, w] of polynomials of z and w, we can factor f (w, z) = q j=1 p j(w, z) n j where p 1 (w, z),, p q (w, z) are irreducible polynomials. Bochner s Theorem says that φ has finite critical points in the unit disk D. Thus we have f (w, z) = q j=1 p j(w, z). Theorem Let φ(z) be an n-th order Blaschke product and f (w, z) = q j=1 p j(w, z). Suppose that p(w, z) is one of factors of f (w, z). Then the Riemann surface S p is connected if and only if p(w, z) is irreducible. Hence q equals the number of connected components of the Riemann surface S φ = S f.

Visualizing Riemann Surfaces Visualization of Riemann surfaces is complicated by the fact that they are embedded in C 2, a four-dimensional real space. One aid to constructing and visualizing them is a method known as cut and paste. We begin with n copies of the unit disk D, called sheets. The sheets are labeled D 1,, D n and stacked up over D.

Visualizing Riemann Surfaces Visualization of Riemann surfaces is complicated by the fact that they are embedded in C 2, a four-dimensional real space. One aid to constructing and visualizing them is a method known as cut and paste. Let {z 1,, z m } be the branch points. Suppose a curve Γ drawn through those branch points and a fixed point on the unit circle so that D/Γ is a simply connected region.

Visualizing Riemann Surfaces Visualization of Riemann surfaces is complicated by the fact that they are embedded in C 2, a four-dimensional real space. One aid to constructing and visualizing them is a method known as cut and paste. Various sheets are glued to others along opposite edges of cuts. With the point in the k-th sheet over a value z in D/Γ we associate the pair of values (ρ k (z), z). In this way a one-to-one correspondence is set up between the points in S f over D/Γ and the pair of points on the n sheets over D/Γ.

Example If φ(z) = 2 j=1 z α i 1 ᾱ i z, the Riemann surface of φ 1 φ over D has two connected components..

Example If φ(z) = 2 j=1 z α i 1 ᾱ i z, the Riemann surface of φ 1 φ over D has two connected components.. Example If φ(z) = z 3, then the Riemann surface of φ 1 φ over D has three connected components.

Example (Stephenson) If φ(z) = z 2 z α 1 ᾱz for α 0, the Riemann surface of φ 1 φ over D has only two connected components.

We need to order {ρ j } n j=1 globally over a simply connected subset of E. To do this, take an invertible small open set U of E such that the intersection of ρ j (U) and ρ k (U) is empty for j k. We always can do so by shrinking U sufficiently. Fix the small invertible open set U and label {ρ j (z)} n j=1 as and assume ρ 1 (z) = z. Take a curve Γ through the finite set F and connecting a point on the unit circle so that E\Γ is simply connected and disjoint from the set n j=1 ρ j(u). {ρ 1 (z), ρ 2 (z),, ρ n (z)}

Label {{ρ j (z)} n j=1, U} The Riemann Monodromy theorem gives that each of {ρ j (z)} n j=1 has a uniquely analytic continuation on E\Γ and hence we can view each of {ρ j (z)} n j=1 as an analytic function on E\Γ satisfying φ(ρ j (z)) = φ(z), We still order {ρ j (z)} n j=1 as {ρ 1(z), ρ 2 (z),, ρ n (z)} at for z E\Γ and j = 1,, n. every point in E\Γ. The label is denoted by {{ρ j (z)} n j=1, U}.

Global representation Theorem Let φ be a finite Blaschke product. Let U be a small invertible open set of E. Let {ρ i (z)} n j=1 a complete collection of local inverses and E\Γ with the label {{ρ j (z)} n j=1, U}. Then for each T in {M φ }, there are analytic functions {s i (α)} n i=1 on E\Γ such that for each h in the Bergman space L 2 a, Th(α) T k α = n i=1 s i(α)h(ρ i (α)), = n i=1 s i(α)k ρi (α) for each α in E\Γ.

Representation of unitary operators If W is a unitary operator in {M φ }, then W is also in {M φ }.

Representation of unitary operators If W is a unitary operator in {M φ }, then W is also in {M φ }. Theorem Let φ be a finite Blaschke product. Let U be a small invertible open set of E. Let {ρ i (z)} n j=1 be a complete collection of local inverses and E\Γ with the label {{ρ j (z)} n j=1, U}. Then for each unitary operator W in {M φ }, there is a unit vector {r i } n i=1 in C n such that for each h in the Bergman space L 2 a, Wh(α) = W k α = n r i ρ i(α)h(ρ i (α)), i=1 n r i ρ i (α)k ρ i (α) i=1 for each α in E\Γ.

Theorem Let U be a small invertible open set of E. Let {ρ i (z)} n j=1 be a complete collection of local inverses. Then for each α in U, W k α = q ˆr k k=1 ρ G ik ρ (α)k ρ(α), where ˆr k = r ρ for some ρ in G ik. Take a curve γ through a point z 0 in E. The analytic continuation along γ of W k α is still W k α, but for each ρ G ik, the analytic continuation of ρ (α)k ρ(α) will turn to ˆρ (α)kˆρ(α) for some ˆρ G ik

Theorem Let U be a small invertible open set of E. Let {ρ i (z)} n j=1 be a complete collection of local inverses. Then for each α in U, W k α = q ˆr k k=1 ρ G ik ρ (α)k ρ(α), where ˆr k = r ρ for some ρ in G ik. For each 1 k q, define a bounded linear operator E k : L 2 a L 2 a by E k f (z) = ρ G ik ρ (z)f (ρ(z)) for z E and each f L 2 a. W = q ˆr k Ek. k=1

Proof of Main Result Russo-Dye Theorem Every element in a von Neumann algebra A can be written as a finite linear combination of unitary operators in A. For each unitary operator W A φ, W = q ˆr k E k. k=1 The von Neumann algebra A φ is generated by E 1,, E q and its dimension q if each E k is in A φ. The dimension of A φ equals q if E 1,, E q are linearly independent.

Problems E k f (z) = ρ G ik ρ (z)f (ρ(z)). Is E k well-defined? E k A φ? What is E k? Are E 1,, E q linearly independent?

E k is well-defined and commutes with M φ For each polynomial f, ρ G ik ρ (z)f (ρ(z)) extends analytically on E and for each ρ. E k (M φ f ) = ρ G ik ρ (z)φ(ρ(z))f (ρ(z)) = φ(z) ρ G ik ρ (z)f (ρ(z)) = M φ E k (f ). Thus E k M φ = M φ E k.

Change Variable Formula To get Ek, we need the following change variable formula. Lemma For each ρ {ρ j } n j=1 and f L2 a, f (ρ(z)) 2 ρ (z) 2 da(z) = E\Γ D f (w) 2 da(w). This follows from that for each ρ {ρ j } n j=1, ρ(e\γ) contains D\ˆΓ where ˆΓ denotes the set Γ n j=1{w E\Γ : ρ j (w) Γ}, which consists of finitely many curves on D. Since ρ maps E into E and is locally analytic and injective on E, ˆΓ is a closed set which area measure equals zero.

E k = E k Since ρ 1 is also in {ρ i } n i=1 for each ρ {ρ i} n i=1, let G i k the subset of {ρ i } n i=1 : denote {ρ : ρ 1 G ik }. We need the following lemma to get E k. Lemma For each i k, there is an integer k with 1 k q such that Theorem G i k = G ik. For each integer k with 1 k q, there is an integer k with 1 k q such that E k = E k.

E 1,, E q are linearly independent. Assume that there are constants c 1,, c q such that Thus for each α in E, we have For each i, define An easy calculation gives c 1 E 1 + + c q E q = 0. [c 1 E 1 + + c q E q ] k α = 0. P i (α, z) = n (z ρ j (α)). j i P ik (α,.), [c 1 E 1 + + c q E q ] k α = c k ρ i k (α)p ik (α, ρ ik (α)). Since P ik (α, ρ ik (α)) 0 and ρ i k (z) vanishes only on a countable subset of D, we have that c k must be zero for each k. We conclude that E 1, E 2,, E q are linearly independent

Abelian C -algebras Theorem Let φ be a finite Blaschke product with order less than or equal to 8. Then A φ is commutative. Let φ be a finite Blaschke product. Let G be those local inverses of φ which extend analytically to only themselves in D\F. Lemma G is an elementary subgroup of Aut(D) consisting of elliptic Möbius transforms and identity ρ 1. Moreover, there are a point α in D, a unimodular constant λ and an integer n G such that G = { α 2 λ 1 α 2 λ φ α(1 λ) (z) : λ n G = 1}. 1 α 2 λ