CLASSIFICATION OF REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS ON THE BERGMAN SPACE VIA THE HARDY SPACE OF THE BIDISK

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1 CLASSIFICATION OF REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS ON THE BERGMAN SPACE VIA THE HARDY SPACE OF THE BIDISK SHUNHUA SUN, DECHAO ZHENG AND CHANGYONG ZHONG ABSTRACT. In this paper we obtain a complete description of nontrivial minimal reducing subspaces of the multiplication operator by a Blaschke product with four zeros on the Bergman space of the unit disk via the Hardy space of the bidisk. Let D be the open unit disk in C. Let da denote Lebesgue area measure on the unit disk D, normalized so that the measure of D equals 1. The Bergman space L 2 a is the Hilbert space consisting of the analytic functions on D that are also in the space L 2 (D, da) of square integrable functions on D. For a bounded analytic function φ on the unit disk, the multiplication operator M φ with symbol φ is defined on the Bergman space L 2 a given by M φ h = φh for h L 2 a. On the basis {e n } n=0, where e n is equal to n + 1z n, the multiplication operator M z by z is a weighted shift operator, said to be the Bergman shift: n + 1 M z e n = n + 2 e n+1. A reducing subspace M for an operator T on a Hilbert space H is a subspace M of H such that T M M T M M. A reducing subspace M of T is called minimal if M does not have any nontrivial subspaces which are reducing subspaces. The goal of this paper is to classify reducing subspaces of M φ for the Blaschke product φ with four zeros by identifying its minimal reducing subspaces. Our main idea is to lift the Bergman shift up as a compression of a commuting pair of isometries on a nice subspace of the Hardy space of the bidisk. This idea was used in studying the Hilbert modules by R. Douglas V. Paulsen [5], operator theory in the Hardy space over the bidisk by R. Dougals R. Yang [6], [18], [19] [20]; the higher-order Hankel forms by S. Ferguson R. Rochberg [7] [8] the lattice of the invariant subspaces of the Bergman shift by S. Richter [12]. On the Hardy space of the unit disk, for an inner function φ, the multiplication operator by φ is a pure isometry. So its reducing subspaces are in one-to-one correspondence with the closed subspaces of H 2 φh 2 [4], [10]. Therefore, it has infinitely many reducing subspaces provided that φ is any inner function other than a Möbius function. Many people have studied the problem of determining reducing subspaces of a multiplication operator on the Hardy space of the unit circle [1], [2] [11]. The multiplication operators on the Bergman space possess a very rich structure theory. Even the lattice of the invariant subspaces of the Bergman shift M z is huge [3]. But the lattice of reducing subspaces of the multiplication operator by a finite Blaschke on the The first author was supported in part by the National Natural Science Foundation of China. The second author was partially supported by the National Science Foundation. 1

2 2 SUN, ZHENG AND ZHONG Bergman space seems to be simple. On the Bergman space, Zhu [21] showed that for a Blaschke product φ with two zeros, the multiplication operator M φ has exact two nontrivial reducing subspaces M 0 M 0. In fact, the restriction of the multiplication operator on M 0 is unitarily equivalent to the Bergman shift. Using the Hardy space of bidisk in [9], we show that the multiplication operator with a finite Blaschke product φ has a unique reducing subspace M 0 (φ), on which the restriction of M φ is unitarily equivalent to the Bergman shift if a multiplication operator has a such reducing subspace, then its symbol must be a finite Blaschke product. The space M 0 (φ) is called the distinguished reducing subspace of M φ is equal to {φ φ n : n = 0, 1,, m, } if φ vanishes at 0 in [15], i.e, φ(z) = cz n k=1 z α k 1 α k z, for some points {α k } in the unit disk a unimodular constant c. The space has played an important role in classifying reducing subspaces of M φ. In [9], we have shown that for a Blaschke product φ of the third order, except for a scalar multiple of the third power of a Möbius transform, M φ has exactly two nontrivial minimal reducing subspaces M 0 (φ) M 0 (φ). This paper continues our study on reducing subspaces of the multiplication operators M φ on the Bergman space in [9] by using the Hardy space of the bidisk. We will obtain a complete description of nontrivial minimal reducing subspaces of M φ for the fourth order Blaschke product φ. This paper is organized as follows. In Section 1 we introduce some notation to lift the Bergman shift as the compression of some isometry on a subspace of the Hardy space of the bidisk state some theorems in [9] which will be used later. In Section 2 we state the main result present its proof. Since the proof is long, two difficult cases in the proof are considered in the last two sections. 1. BERGMAN SPACE VIA HARDY SPACE Let T denote the unit circle. The torus T 2 is the Cartesian product T T. Let dσ be the rotation invariant Lebesgue measure on T 2. The Hardy space H 2 (T 2 ) is the subspace of L 2 (T 2, dσ), where functions in H 2 (T 2 ) can be identified with the boundary value of the function holomorphic in the bidisc D 2 with the square summable Fourier coefficients. The Toeplitz operator on H 2 (T 2 ) with symbol f in L (T 2, dσ) is defined by T f (h) = P (fh), for h H 2 (T 2 ) where P is the orthogonal projection from L 2 (T 2, dσ) onto H 2 (T 2 ). For each integer n 0, let n p n (z, w) = z i w n i. Let H be the subspace of H 2 (T 2 ) spanned by functions {p n } n=0. Thus Let i=0 H 2 (T 2 ) = H cl{(z w)h 2 (T 2 )}. B = P H T z H = P H T w H

3 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 3 where P H is the orthogonal projection from L 2 (T 2, dσ) onto H. So B is unitarily equivalent to the Bergman shift M z on the Bergman space L 2 a via the following unitary operator U : L 2 a(d) H, Uz n = p n(z, w) n + 1. This implies that the Bergman shift is lifted up as the compression of an isometry on a nice subspace of H 2 (T 2 ). Indeed, for each finite Blaschke product φ(z), the multiplication operator M φ on the Bergman space is unitarily equivalent to φ(b) on H. Let L 0 be kert φ(z) kert φ(w) H. In [9], for each e L 0, we construct functions {d k e} d 0 e such that for each l 1, l 1 p l (φ(z), φ(w))e + p k (φ(z), φ(w))d l k e k=0 H p l (φ(z), φ(w))e + p l 1 (φ(z), φ(w))d 0 e H. On one h, we have a precise formula of d 0 e: d 0 e(z, w) = we(0, w)e 0 (z, w) wφ 0 (w)e(z, w), (1.1) where e 0 is the function φ(z) φ(w). On the other h, d k z w e is orthogonal to for a reducing subspace M e M, kert φ(z) kert φ(w) H, l 1 p l (φ(z), φ(w))e + p k (φ(z), φ(w))d l k e M. k=0 Moreover, the relation between d 1 e d 0 e is given by Theorem 1 in [9] as follows: Theorem 1.1. If M is a reducing subspace of φ(b) orthogonal to the distinguished reducing subspace M 0, for each e M L 0, then there is an element ẽ M L 0 a number λ such that d 1 e = d 0 e + ẽ + λe 0. (1.2) Since for Blaschke products with smaller order, it is not difficult to calculate ẽ λ precisely, we are able to classify minimal reducing subspaces of a multiplication operator by a Blaschke product of the fourth order. Main ideas in the proof of Theorems are that by complicated computations we use (1.2) to derive conditions on zeros of the Blaschke product of the fourth order. In this paper we often use Theorem 1.1 Theorems 1 25 in [9] stated as follows. Theorem 1.2. There is a unique reducing subspace M 0 for φ(b) such that φ(b) M0 is unitarily equivalent to the Bergman shift. In fact, M 0 = l 0{p l (φ(z), φ(w))e 0 }, { p l(φ(z),φ(w))e 0 l+1 e0 } 0 form an orthonormal basis of M 0.

4 4 SUN, ZHENG AND ZHONG We call M 0 to be the distinguished reducing subspace for φ(b). M 0 is unitarily equivalent to a reducing subspace of M φ contained in the Bergman space, denoted by M 0 (φ). The space plays an important role in classifying the minimal reducing subspaces of M φ in Theorem 2.1. In [9] we showed that for a nontrivial minimal reducing subspace Ω for φ(b), either Ω equals M 0 or Ω is a subspace of M 0. The condition in the following theorem is natural. Theorem 1.3. Suppose that Ω, M N are three distinct nontrivial minimal reducing subspaces for φ(b) Ω M N. If they are contained in M 0, then there is a unitary operator U : M N such that U commutes with φ(b) φ(b). 2. MAIN RESULT Let φ be a Blaschke product with four zeros. In this section we will obtain a complete description of minimal reducing subspaces of the multiplication operator M φ. First observe that the multiplication operator M z 4 is a weighted shift with multiplicity 4: n + 1 M z 4e n = n + 5 e n+4 where e n equals n + 1z n. By Theorem B [14], M z 4 has exact four nontrivial minimal reducing subspaces: M j = {z n : n j mod 4} for j = 1, 2, 3, 4. Before stating the main result of this paper we need some notation. It is not difficult to see that the set of finite Blaschke products forms a semigroup under composition of two functions. For a finite Blaschke product φ we say that φ is decomposable if there are two Blaschke products ψ 1 ψ 2 with orders greater than 1 such that φ(z) = ψ 1 ψ 2 (z). For each λ in D, let φ λ denote the Möbius transform: φ λ (z) = λ z 1 λz. Define the operator U λ on the Bergman space as follows: U λ f = f φ λ k λ for f in L 2 a where k λ is the normalized reproducing kernel (1 λ 2 ) (1 λz) 2. Clearly, U λ is a selfadjoint unitary operator on the Bergman space. Using the unitary operator U λ we have M 0 (φ) = U λ M 0 (φ φ λ ) where λ is a zero of the finite Blaschke product φ. This easily follows from that φ φ λ vanishes at 0 U λm φ U λ = M φ φλ. We say that two Blaschke products φ 1 φ 2 are equivalent if there is a complex number λ in D such that φ 1 = φ λ φ 2.

5 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 5 For two equivalent Blaschke products φ 1 φ 2, M φ1 M φ2 are mutually analytic function calculus of each other hence share reducing subspaces. The following main result of this paper gives a complete description of minimal reducing subspaces. Theorem 2.1. Let φ be a Blaschke product with four zeros. One of the following holds. (1) If φ is equivalent to z 4, i.e., φ is a scalar multiple of the fourth power φ 4 c of the Möbius transform φ c for some complex number c in the unit disk, M φ has exact four nontrivial minimal reducing subspaces {U c M 1, U c M 2, U c M 3, U c M 4 }. (2) If φ is is decomposable but not equivalent to z 4, i.e, φ = ψ 1 ψ 2 for two Blaschke products ψ 1 ψ 2 with orders 2 but not both of ψ 1 ψ 2 are a scalar multiple of z 2, then M φ has exact three nontrivial minimal reducing subspaces {M 0 (φ), M 0 (ψ 2 ) M 0 (φ), M 0 (ψ 2 ) }. (3) If φ is not decomposable, then M φ has exact two nontrivial minimal reducing subspaces {M 0 (φ), M 0 (φ) }. To prove the above theorem we need the following two lemmas which tell us when a Blaschke product with order 4 is decomposable. Lemma 2.2. If a Blaschke product φ with order four is decomposable, then the numerator of the rational function φ(z) φ(w) has at least three irreducible factors. Proof. Suppose that φ is the Blaschke product with order four. Let f(z, w) be the numerator of the rational function φ(z) φ(w). If φ is decomposable, then φ = ψ 1 ψ 2 for two Blaschke products ψ 1 ψ 2 with order two. Let g(z, w) be the numerator of the rational function ψ 1 (z) ψ 1 (w). Clearly, z w is a factor of g(z, w). Thus we can write g(z, w) = (z w)p(z, w) for some polynomial p(z, w) of z w to get On the other h, we also have g(ψ 2 (z), ψ 2 (w)) = (ψ 2 (z) ψ 2 (w))p(ψ 2 (z), ψ 2 (w)). ψ 2 (z) ψ 2 (w) = (z w)p 2(z, w) q 2 (z, w) for two polynomials p 2 (z, w) q 2 (z, w) which p 2 (z, w) q 2 (z, w) do not have common factor. In fact, q 2 (z, w) the numerator of the rational function p(ψ 2 (z), ψ 2 (w)) do not have common factor also. So we obtain g(ψ 2 (z), ψ 2 (w)) = (z w)p 2(z, w) p(ψ 2 (z), ψ 2 (w)). q 2 (z, w) Since f(z, w) is the numerator of the rational function g(ψ 2 (z), ψ 2 (w)), this gives that f(z, w) has at least three factors. This completes the proof. For α, β D, define f α,β (w, z) = w 2 (w α)(w β)(1 ᾱz)(1 βz) z 2 (z α)(z β)(1 ᾱw)(1 βw).

6 6 SUN, ZHENG AND ZHONG It is easy to see that f α,β (w, z) is the numerator of z 2 φ α (z)φ β (z) w 2 φ α (w)φ β (w). The following lemma gives a criteria when the Blaschke product z 2 φ α (z)φ β (z) is decomposable. Lemma 2.3. For α β in D, one of the following holds. (1) If both α β equal zero, then f α,β (w, z) = (w z)(w + z)(w iz)(w + iz). (2) If α does not equal either β or β, then f α,β (w, z) = (w z)p(w, z) for some irreducible polynomial p(w, z). (3) If α equals either β or β but it does not equal zero, then f α,β (w, z) = (w z)p(w, z)q(w, z) for two irreducible distinct polynomials p(w, z) q(w, z). Proof. Clearly, (1) holds. To prove (2), by the example on page 6 of [13] we may assume that none of α β equals 0. First observe that (w z) is a factor of the polynomial f α,β (w, z). Taking a long division gives where f α,β (w, z) = (w z)g α,β (w, z) g α,β (w, z) = (1 ᾱz)(1 βz)w 3 + (z (α + β))(1 ᾱz)(1 βz)w 2 +(z α)(z β)(1 (ᾱ + β)z)w + z(z α)(z β). Next we will show that g α,β (w, z) is irreducible. To do this, we assume that g α,β (w, z) is reducible to derive a contradiction. Assuming that g α,β (w, z) is reducible, we can factor g α,β (w, z) as the product of two polynomials p(w, z) q(w, z) of z w with degree of w greater than or equal one. Write p(w, z) = a 1 (z)w + a 0 (z) q(w, z) = b 2 (z)w 2 + b 1 (z)w + b 0 (z) where a j (z) b j (z) are polynomials of z. Since g α,β (w, z) equals the product of p(w, z) q(w, z), taking the product comparing coefficients of w k give a 1 (z)b 2 (z) = (1 ᾱz)(1 βz), (2.1) a 1 (z)b 1 (z) + a 0 (z)b 2 (z) = (z (α + β))(1 ᾱz)(1 βz), (2.2) a 1 (z)b 0 (z) + a 0 (z)b 1 (z) = (z α)(z β)(1 (ᾱ + β)z), (2.3) Equation (2.1) gives that either a 0 (z)b 0 (z) = z(z α)(z β). (2.4) a 1 (z) = (1 ᾱz) or a 1 (z) = (1 ᾱz)(1 βz) or a 1 (z) = 1.

7 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 7 In the first case that a 1 (z) = (1 ᾱz), (2.1) gives b 2 (z) = (1 βz). Thus by Equation (2.2), we have a 0 (z)(1 βz) = (1 ᾱz)[(z (α + β))(1 βz) b 1 (z)], to get that (1 ᾱz) is a factor of a 0 (z), hence is also a factor of a factor z(z α)(z β) by (2.4). This implies that α must equal 0. It is a contradiction. In the second case that a 1 (z) = (1 ᾱz)(1 βz), we have that b 2 (z) = 1 to get that either the degree of b 1 (z) or the degree of b 0 (z) must be one while the degrees of b 1 (z) b 0 (z) are at most one. So the degree of a 0 (z) is at most two. Also a 0 (z) does not equal zero. Equation (2.2) gives Thus (1 ᾱz)(1 βz)b 1 (z) + a 0 (z) = (z (α + β))(1 ᾱz)(1 βz). a 0 (z) = c 1 (1 ᾱz)(1 βz) for some constant c 1. But Equation (2.4) gives c 1 (1 ᾱz)(1 βz)b 0 (z) = z(z α)(z β). Either c 1 = 0 or (1 ᾱz)(1 βz) is a factor of z(z α)(z β). This is impossible. In the third case that a 1 (z) = 1, then b 2 (z) = (1 ᾱz)(1 βz). Since the root w of f α,β (w, z) is a nonconstant function of z, the degree of a 0 (z) must be one. Thus the degrees of b 1 (z) b 0 (z) are at most two. By Equation (2.2) we have to get (1 ᾱz)(1 βz)a 0 (z) + b 1 (z) = (z (α + β))(1 ᾱz)(1 βz), b 1 (z) = (1 ᾱz)(1 βz)[(z (α + β)) a 0 (z)]. Since the degree of b 1 (z) is at most two, we have Equations (2.4) (2.3) give a 0 (z) = (z (α + β)) c 0 ; b 1 (z) = c 0 (1 ᾱz)(1 βz). [(z (α + β)) c 0 ]b 0 (z) = z(z α)(z β) b 1 (z)[(z (α + β)) c 0 ] + b 0 (z) = (z α)(z β)(1 (ᾱ + β)z). Multiplying the both sides of the last equality by [(z (α + β)) c 0 ] gives This leads to b 1 (z)[(z (α + β)) c 0 ] 2 + z(z α)(z β) = [(z (α + β)) c 0 ](z α)(z β)(1 (ᾱ + β)z). c 0 (1 ᾱz)(1 βz)[(z (α + β)) c 0 ] 2 + z(z α)(z β) = [(z (α + β)) c 0 ](z α)(z β)(1 (ᾱ + β)z). If c 0 0, then the above equality gives that (z α)(z β) is a factor of [(z (α + β)) c 0 ] 2. This is impossible.

8 8 SUN, ZHENG AND ZHONG If c 0 = 0, then we have z(z α)(z β) = [(z (α + β))](z α)(z β)(1 (ᾱ + β)z). to get ᾱ + β = 0 hence α = β. It is also a contradiction. This completes the proof that g α,β (w, z) is irreducible. To prove (3), we note that if α equals β, an easy computation gives If α = β, we also have f α,β (w, z) = (w z)[(1 ᾱz)w + (z α)] [w(w α)(1 ᾱz) + z(z α)(1 ᾱw)]. f α,β (w, z) = (w z)(w + z)[(1 ᾱ 2 z 2 )w 2 + (z 2 α 2 )]. This completes the proof. Proof of Theorem 2.1. Assume that φ is a Blaschke product with the fourth order. By the Bochner Theorem [17], φ has a critical point c in the unit disk. Let λ = φ(c) be the critical value of φ. Then there are two points α β in the unit disk such that φ λ φ φ c (z) = ηz 2 φ α φ β where η is a unimodule constant. Let ψ be z 2 φ α φ β. Since φ φ c ψ are mutually analytic function calculus of each other, both M φ φc M ψ share reducing subspaces. (1) If φ is equivalent to z 4, then ψ must equal a scalar multiple of z 4. By Theorem B in [14], M ψ has exact four nontrivial minimal reducing subspaces where {M 1, M 2, M 3, M 4 } M j = {z n : n j mod 4} for j = 1, 2, 3, 4. The four spaces above are also reducing subspaces for M φ φc. Noting U c M φ φc U c = M φ, we have that M φ has exact four nontrivial minimal reducing subspaces {U c M 1, U c M 2, U c M 3, U c M 4 }. (2) If φ is decomposable but not equivalent to z 4, i.e, φ = ψ 1 ψ 2 for two Blaschke products ψ 1 ψ 2 with degrees two not both ψ 1 ψ 2 are scalar multiples of z 2, by Lemmas , then α equals either β or β but does not equal 0. By Theorem 1.2, the restriction of M ψ2 on M 0 (ψ 2 ) is unitarily equivalent to the Bergman shift. Thus M 0 (ψ 2 ) is also a reducing subspace of M φ the restriction of M φ = M ψ1 ψ 2 on M 0 (ψ 2 ) is unitarily equivalent to M ψ1 on the Bergman space. By Theorem 1.2 again, there is a unique reducing subspace M 0 (ψ 1 ) on which the restriction M ψ1 is unitarily equivalent to the Bergman shift. Thus there is a subspace of M 0 (ψ 2 ) on which the restriction of M φ is unitarily equivalent to the Bergman shift. Theorem 1.2 implies that M 0 (φ) is contained in M 0 (ψ 2 ). Therefore M 0 (ψ 2 ) M 0 (φ) is also a minimal reducing subspace of M φ L 2 a = M 0 (φ) [M 0 (ψ 2 ) M 0 (φ)] [M 0 (ψ 2 )]. By Theorems 3.1 in [16], {M 0 (φ), [M 0 (ψ 2 ) M 0 (φ)], [M 0 (ψ 2 )] } are nontrivial minimal reducing subspaces of M φ. We will show that they are exact nontrivial minimal reducing

9 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 9 subspaces of M φ. If this is not true, then there is another minimal reducing subspace Ω of M φ. By Theorem 38 [9], we have By Theorem 1.3, there is a unitary operator which commutes with both M φ M φ. But Ω [M 0 (ψ 2 ) M 0 (φ)] [M 0 (ψ 2 )]. U : [M 0 (ψ 2 ) M 0 (φ)] [M 0 (ψ 2 )] dimkerm φ [M 0 (ψ 2 ) M 0 (φ)] = 1 dimkerm φ [M 0 (ψ 2 )] = 2. This is a contradiction. Thus {M 0 (φ), [M 0 (ψ 2 ) M 0 (φ)], [M 0 (ψ 2 )] } are exact nontrivial minimal reducing subspaces of M φ. (3) If φ is not decomposable, by Lemma 2.3, then φ equals z 3 φ α or z 2 φ α φ β for two nonzero points α β in D α does not equal β or β. The difficult cases will be dealt with in Sections 3 4. By Theorems , M φ has exact two nontrivial minimal reducing subspaces {M 0 (φ), M 0 (φ) }. 3. REDUCING SUBSPACES OF M z 3 φ α In this section we will study reducing subspaces of M z 3 φ α for a nonzero point α D. Recall that M 0 is the distinguished reducing subspace of φ(b) as in Theorem 1.2. Theorem 3.1. Let φ = z 3 φ α for a nonzero point α D. Then φ(b) has exact two nontrivial reducing subspaces {M 0, M 0 }. Proof. Let M 0 be the distinguished reducing subspace of φ(b) as in Theorem 1.2. By Theorem 1.3, we only need to show that M 0 is a minimal reducing subspace for φ(b). Assume that M 0 is not a minimal reducing subspace for φ(b). Then by Theorem 3.1 in [16] we may assume 2 H = i=0 such that each M i is a nontrivial reducing subspace for φ(b), M 0 = M 0 is the distinguished reducing subspace for φ(b) Recall that We further assume that M i M 0 = M 1 M 2. φ 0 = z 2 φ α, L 0 = span{1, p 1, p 2, k α (z)k α (w)}, L 0 = (L 0 M 0 ) (L 0 M 1 ) (L 0 M 2 ). dim(m 1 L 0 ) = 1 dim(m 2 L 0 ) = 2.

10 10 SUN, ZHENG AND ZHONG Take 0 e 1 M 1 L 0, e 2, e 3 M 2 L 0 such that {e 2, e 3 } are a basis for M 2 L 0, then By (1.1), we have d 0 e j direct computations show that for 0 k 2, L 0 = span{e 0, e 1, e 2, e 3 } = we j (0, w)e 0 φ(w)e j d 0 e j, p k = we j (0, w)e 0 φ(w)e j, p k = we j (0, w)e 0, p k (by T φ(w)p k = 0) = we j (0, w)e 0 (w, w), p k (0, w) = we j (0, w)φ (w), w k = w 3 e j (0, w)(wφ α(w) + 3φ α (w)), w k = w 3 k e j (0, w)(wφ α(w) + 3φ α (w)), 1 = 0 d 0 e j, k α (z)k α (w) = αe j (0, α)e 0 (α, α) α 3 = αe j (0, α) 1 α. 2 This implies that those functions d 0 e j are orthogonal to {1, p 1, p 2 }. Simple calculations give e 0, p k = 0 for 0 k 1, e 0, p 2 = e 0 (0, w), p 2 (w, w) = 3 2 φ 0(0) = 3α 0 e 0, k α (z)k α (w) = e 0 (α, α) By Theorem 1.1, there are numbers µ, λ j such that = φ (α) α 3 = 1 α 0 2 d 1 e 1 = d 0 e 1 + µe 1 + λ 1 e 0 d 1 e 2 = d 0 e 2 + ẽ 2 + λ 2 e 0 d 1 e 3 = d 0 e 3 + ẽ 3 + λ 3 e 0 where ẽ 2, ẽ 3 M 2 L 0. Now we consider two cases. In each case we will derive a contradiction. Case 1. µ 0. In this case, we get that e 1 is orthogonal to {1, p 1 }. So {1, p 1, e 0, e 1 } form an orthogonal basis for L 0.

11 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 11 First we show that ẽ 2 = 0. If ẽ 2 0, then we get that {1, p 1, e 0, ẽ 2 } are also an orthogonal basis for L 0. Thus ẽ 2 = ce 1 for some nonzero number c. However, ẽ 2 is orthogonal to e 1 since ẽ 2 M 2 e 1 M 1. This is a contradiction. Thus d 1 e 2 = d 0 e 2 + λ 2 e 0. Since both d 1 e 2 d 0 e 2 are orthogonal to p 2 e 0, p 2 = 3α 0, we have that λ 2 = 0 to get that d 0 e 2 = d 1 e 2 is orthogonal to L 0. On the other h, α 3 d 0 e 2, k α (z)k α (w) = αe 2 (0, α) 1 α. 2 Thus e 2 (0, α) = 0. Similarly we get that e 3 (0, α) = 0. Moreover, since e 2 e 3 are orthogonal to {e 0, e 1 }, write e 2 = c 11 + c 12 p 1, e 3 = c 21 + c 22 p 1. Thus we have e 2 (0, α) = c 11 + c 12 α = 0, e 3 (0, α) = c 21 + c 22 α = 0, to get that e 2 e 3 are linearly dependent. This leads to a contradiction in this case. Case 2. µ = 0. In this case we have d 1 e 1 = d 0 e 1 + λ 1 e 0. Similarly to the proof in Case 1 we get that λ 1 = 0, d 1 e 1 = d 0 e 1 L 0 (3.1) e 1 (0, α) = 0. Theorem 2.2 in [16] gives that at least one ẽ j, say ẽ 2 does not equal 0. Assume that ẽ 2 0, write ẽ 2 = d 1 e 2 d 0 e 2 λ 2 e 0. Note that we have shown above that both d 0 e 2 e 0 are orthogonal to both 1 p 1. Thus ẽ 2 {1, p 1 } L 0 = span{1, p 1, e 0, ẽ 2 }. Since e 1 is orthogonal to {e 0, ẽ 2 } we have Noting that e 1 (0, α) = c 1 + c 2 α = 0 we get e 1 = c 1 + c 2 p 1. e 1 = c 2 ( α + p 1 ).

12 12 SUN, ZHENG AND ZHONG Without loss of generality we assume that e 1 = α + p 1. (3.2) Letting e be in M 2 L 0 such that e is a nonzero function orthogonal to ẽ 2, we have that e is orthogonal to {e 0, ẽ 2 }. Thus e must be in the subspace span{1, p 1 }. So there are two constants b 1 b 2 such that Noting we have e = b 1 + b 2 p 1. 0 = e, e 1 = b 1 ᾱ + 2b 2 Hence we may assume that e = b 1 2 (2 + ᾱp 1). e = 2 + ᾱp 1. (3.3) By Theorem 1.1 we have for some number λ ẽ M 2 L 0. Thus 0 = d 1 e 1, d 1 e d 1 e = d 0 e + ẽ + λe 0 = d 1 e 1, d 0 e + ẽ + λe 0 = d 1 e 1, d 0 e However, a simple computation gives =, d 0 e (by (3.1))., d 0 e =, we(0, w)e 0 φ(w)e =, we(0, w)e 0 (by T φ(w)d 0 e 1 = 0) = we 1 (0, w)e 0 φ(w)e 1, we(0, w)e 0 = we 1 (0, w)e 0, we(0, w)e 0 φ(w)e 1, we(0, w)e 0. We need to calculate two terms in the right h of the above equality. By (3.2) (3.3), the first term becomes we 1 (0, w)e 0, we(0, w)e 0 = w( α + w)e 0, w(2 + ᾱw)e 0 = ( α + w)e 0, (2 + ᾱw)e 0 = αe 0, 2e 0 + we 0, 2e 0 + αe 0, ᾱwe 0 + we 0, ᾱwe 0 = α e 0, e we 0, e 0 α 2 e 0, we 0.

13 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 13 The first term in right h of the last equality is The last equality follows from Similarly, we have This gives e 0, e 0 = e 0 (w, w), e 0 (0, w) = wφ 0 + φ 0, φ 0 = w(2wφ α + w 2 φ α), w 2 φ α + φ 0, φ 0. = 2 + wφ α, φ α + 1 = 4. φ α = 1 ᾱ + 1 ᾱ α 1 ᾱw = 1 ᾱ + ( 1 ᾱ α)k α(w). we 0, e 0 = we 0 (w, w), e 0 (0, w) = w(wφ 0 + φ 0 ), φ 0 = α. we 1 (0, w)e 0, we(0, w)e 0 = e 1 (0, w)e 0, e(0, w)e 0 = ( α + w)e 0, (2 + ᾱw)e 0 = 2α e 0, e 0 α 2 e 0, we 0 +2 we 0, e 0 + α we 0, we 0 = 8α α α 2 + 2α + 4α = 2α α α 2 A simple calculation gives that the second term becomes Thus we conclude φ(w)e 1, we(0, w)e 0 = φ 0 (w)e 1, (2 + ᾱw)e 0 = φ 0 (w)e 1, 2e 0 + φ 0 (w)e 1, ᾱwe 0 = 2 φ 0 (w)e 1 (w, w), e 0 (0, w) + α φ 0 (w)e 1 (w, w), we 0 (0, w) = 2 e 1 (w, w), 1 + α e 1 (w, w), w = 2 α + 2w, 1 + α α + 2w, w = 2α + 2α = 0., d 0 e = we 1 (0, w)e 0, we(0, w)e 0 φ(w)e 1, we(0, w)e 0 = 2α α α 2 = α(2 + α 2 ) 0 to get a contradiction in this case. This completes the proof.

14 14 SUN, ZHENG AND ZHONG 4. REDUCING SUBSPACES FOR M z 2 φ αφ β In this section we will classify minimal reducing subspaces of M z 2 φ αφ β for two nonzero points α β in D with α β. Theorem 4.1. Let φ be the Blaschke product z 2 φ α φ β for two nonzero points α β in D. If α does not equal either β or β, then φ(b) has exact two nontrivial reducing subspaces {M 0, M 0 }. Proof. By Theorem 27 in [9], if N is a nontrivial minimal reducing subspace of φ(b) which is not equal to M 0 then N is a subspace of M 0, so we only need to show that M 0 is a minimal reducing subspace for φ(b) unless α = β. Assume that M 0 is not a minimal reducing subspace for φ(b). By Theorem 3.1 in [16], we may assume 2 H = i=0 such that each M i is a reducing subspace for φ(b), M 0 = M 0 is the distinguished reducing subspace for φ(b) M i M 1 M 2 = M 0. Recall that φ 0 = zφ α φ β, L 0 = span{1, p 1, e α, e β }, with e α = k α (z)k α (w), e β = k β (z)k β (w) L 0 = (L 0 M 0 ) (L 0 M 1 ) (L 0 M 2 ). So we further assume that the dimension of M 1 L 0 is one the dimension of M 2 L 0 is two. Take a nonzero element e 1 in M 1 L 0, then by Theorem 1.1, there are numbers µ 1, λ 1 such that d 1 e 1 = d 0 e 1 + µ 1 e 1 + λ 1 e 0. (4.1) We only need to consider two possibilities, µ 1 is zero or nonzero. If µ 1 is zero, then (4.1) becomes In this case, simple calculations give d 1 e 1 = d 0 e 1 + λ 1 e 0. (4.2), p 1 = we 1 (0, w)e 0 (z, w) wφ 0 (w)e 1 (z, w), p 1 (z, w) = we 1 (0, w)e 0 (w, w) wφ 0 (w)e 1 (w, w), p 1 (z, w) = we 1 (0, w)e 0 (w, w) wφ 0 (w)e 1 (w, w), p 1 (0, w) = we 1 (0, w)e 0 (w, w) wφ 0 (w)e 1 (w, w), w = e 1 (0, w)e 0 (w, w) φ 0 (w)e 1 (w, w), 1 = e 1 (0, 0)e 0 (0, 0) φ 0 (0)e 1 (0, 0) = 0,

15 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 15 e 0, p 1 = e 0 (z, w), p 1 (z, w) = e 0 (z, w), p 1 (w, w) = e 0 (0, w), 2w = φ 0 (w), 2w = 2 wφ α (w)φ β (w), w = 2φ α (0)φ β (0) = 2αβ 0. Noting that d 1 e 1 is orthogonal to L 0, by (4.2) we have that λ 1 = 0, hence d 0 e 1 = d 1 e 1 L 0. So On the other h,, e α = 0 =, e β., e α = αe 1 (0, α)e 0 (α, α) αφ 0 (α)e 1 (α, α) = αe 1 (0, α)e 0 (α, α), e β = βe 1 (0, β)e 0 (β, β) βφ 0 (β)e 1 (β, β) = βe 1 (0, β)e 0 (β, β). Consequently e 1 (0, α) = e 1 (0, β) = 0. (4.3) Observe that e 0, e 1 1 are linearly independent. If this is not so, then 1 = ae 0 + be 1 for some numbers a, b. But e 1 (0, α) = 0 e 0 (0, α) = 0. This forces that 1 = 0 leads to a contradiction. By Theorem 1.1, we can take an element e M 2 L 0 such that d 1 e = d 0 e + e 2 + µe 0 with e 2 0 e 2 M 2 L 0. Thus we have that e 2 is orthogonal to 1 so e 2 is in {1, e 0, e 1 } {1, e 0, e 1, e 2 } form a basis for L 0. Moreover for any f M 2 L 0, d 1 f = d 0 f + g + λe 0 for some number λ g M 2 L 0. If g does not equal 0 then g is orthogonal to 1. Thus g is in {1, e 0, e 1 } hence g = ce 2 for some number c. Therefore taking a nonzero element e 3 M 2 L 0 which is orthogonal to e 2, we have d 1 e 2 = d 0 e 2 + µ 2 e 2 + λ 2 e 0, d 1 e 3 = d 0 e 3 + µ 3 e 2 + λ 3 e 0, {e 0, e 1, e 2, e 3 } is an orthogonal basis for L 0.

16 16 SUN, ZHENG AND ZHONG If µ 2 = 0, then by the same reason as before we get So using we have λ 2 = 0, d 0 e 2 = d 1 e 2 L 0 e 2 (0, α) = e 2 (0, β) = 0. p 1 L 0 = span{1, e 0, e 1, e 2 } α = p 1 (0, α) = p 1 (0, β) = β, which contradicts our assumption that α β. Hence µ 2 0. Observe that 1 is in L 0 = span{e 0, e 1, e 2, e 3 } orthogonal to both e 0 e 2. Thus for some numbers c 1 c 3. So By (4.3), we have to obtain that c = c 1 e 1 + c 3 e 3 1 = c 1 e 1 (0, α) + c 3 e 3 (0, α) = c 1 e 1 (0, β) + c 3 e 3 (0, β). 1 = c 3 e 3 (0, α) = c 3 e 3 (0, β), e 3 (0, α) = e 3 (0, β) = 1/c 3. If µ 3 = 0, then by the same reason as before we get e 3 (0, α) = e 3 (0, β) = 0. Hence µ 3 0. Now by the linearality of d 1 ( ) d0 ( ) we have By the same reason as before we get therefore So we get Hence This implies that d 1 µ 3 e 2 µ 2 e 3 = d 0 µ 3 e 2 µ 2 e 3 + (µ 3 λ 2 µ 2 λ 3 )e 0. µ 3 λ 2 µ 2 λ 3 = 0 d 0 µ 3 e 2 µ 2 e 3 = d 1 µ 3 e 2 µ 2 e 3 L 0 µ 3 e 2 (0, α) µ 2 e 3 (0, α) = µ 3 e 2 (0, β) µ 2 e 3 (0, β) = 0. e 2 (0, α) = µ 2 /µ 3 c 3 = e 2 (0, β). p 1 L 0 = span{1, e 0, e 1, e 2 }. α = p 1 (0, α) = p 1 (0, β) = β which again contradicts our assumption that α β.

17 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 17 Another case is that µ 1 is not equal to 0. In this case, (4.1) can be rewritten as e 1 = 1 µ 1 d 1 e 1 1 µ 1 d 0 e 1 λ 1 µ 1 e 0, we have that e 1 is orthogonal to 1 since d 1 e 1, d 0 e 1 e 0 are orthogonal to 1. Thus 1 is in M 2 L 0. By Theorem 1.1, there is an element e M 2 L 0 a number λ 0 such that If e = 0 then λ 0 = 0, hence d 0 1 L 0 d 1 1 = d e + λ 0 e 0. (4.4) 1 = 1(0, α) = 1(0, β). So e 0. Since d 1 1 is in L 0, d 1 1 is orthogonal to 1. Noting that d 0 1 e 0 are orthogonal to 1, we have that e 1. Hence we get an orthogonal basis {e 0, e 1, 1, e} of L 0. Claim. e(0, α) e(0, β) = 0. Proof of the claim. Using Theorem 1.1 again, we have that d 1 e = d 0 e + g + λe 0 for some g L 0 M 2. If g 0, we have that g 1 since d 1 e, d 0 e, e 0 are orthogonal to 1. Thus we have that g = µe for some number µ to obtain d 1 e = d 0 e + µe + λe 0. Furthermore by the linearality of d 1 ( ) d0 ( ) we have that d 1 e µ1 = d 0 e µ1 + (λ µλ 0 )e 0. By the same reason (namely d 1 e µ1 L 0, d 0 e µ1 1 e 0, 1 0) we have that Hence we have λ µλ 0 = 0, d 0 e µ1 = d 1 e µ1 L 0 (e µ1)(0, α) = (e µ1)(0, β) = 0. e(0, α) e(0, β) = µ µ = 0, to complete the proof of the claim. Let us find the value of λ 0 in (4.4) which will be used to make the coefficients symmetric with respect to α β. To do this, we first state a technical lemma which will be used in several other places in the sequel. Lemma 4.2. If g is in H 2 (T), then wgφ 0, φ 0 = g(0) + g(α) + g(β).

18 18 SUN, ZHENG AND ZHONG Proof. Since φ 0 equals zφ α φ β, simple calculations give wgφ 0, φ 0 = wg(wφ α φ β ), wφ α φ β = g(wφ α φ β ), φ α φ β = g(φ α φ β + wφ αφ β + wφ α φ β), φ α φ β = g, 1 + wgφ α, φ α + wgφ β, φ β = g(0) + wgφ α, φ α + wgφ β, φ β Writing φ α as we have φ α = 1 ᾱ + 1 ᾱ α 1 ᾱw = 1 ᾱ + 1 α 2 k α (w), ᾱ wgφ α, φ α = 1 α 2 (wgφ α α)(α) = g(α). The first equality follows from wgφ α, 1 equals 0 the second equality follows from φ α(α) = 1 1 α 2. By the symmetry of α β, similar computations lead to wgφ β, φ β = g(β) the proof is finished. We state the values of λ 0 e 0, e 0 as a lemma. Lemma 4.3. λ 0 = α + β 4 (4.5) e 0, e 0 = 4 (4.6) Proof. Since d 1 1 is orthogonal to L 0, e 0 is in L 0, e is orthogonal to e 0, (4.4) gives 0 = d 1 1, e 0 = d e + λ 0 e 0, e 0 = d 0 1, e 0 + λ 0 e 0, e 0.

19 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 19 We need to compute d 0 1, e 0 e 0, e 0 respectively. d 0 1, e 0 = φ(w) + we 0, e 0 = we 0, e 0 = we 0 (w, w), e 0 (0, w) = w(wφ 0 + φ 0 ), φ 0 = w 2 φ 0, φ 0 + wφ 0, φ 0 = w 2 φ 0, φ 0 = α + β. The last equality follows from Lemma 4.2 with g = w. e 0, e 0 = e 0 (w, w), e 0 (0, w) = wφ 0 + φ 0, φ 0 = wφ 0, φ 0 + φ 0, φ 0 = wφ 0, φ = 4, where the last equality follows from Lemma 4.2 with g = 1. Hence α + β + 4λ 0 = 0 λ 0 = α + β. 4 Let P L0 denote the projection of H 2 (T 2 ) onto L 0. The element P L0 (k α (w) k β (w)) has the property that for any g L 0, g, P L0 (k α (w) k β (w)) = g, k α (w) k β (w) = g(0, α) g(0, β). Thus P L0 (k α (w) k β (w)) is orthogonal to g for g L 0 with g(0, α) = g(0, β). So P L0 (k α (w) k β (w)) is orthogonal to e 0, 1, e. On the other h, p 1, P L0 (k α (w) k β (w)) = α β 0. This gives that the element P L0 (k α (w) k β (w)) is a nonzero element. Therefore there exists a nonzero number b such that Without loss of generality we assume that P L0 (k α (w) k β (w)) = be 1. e 1 = P L0 (k α (w) k β (w)).

20 20 SUN, ZHENG AND ZHONG Observe that p 1 (φ(z), φ(w))e 1 + d 1 e 1 M 1, p 1 (φ(z), φ(w)) + d 1 1 M 2, M 1 M 2, to get p 1 (φ(z), φ(w))e 1 + d 1 e 1, p 1 (φ(z), φ(w)) + d 1 1 = 0. Thus we have 0 = p 1 (φ(z), φ(w))e 1 + d 1 e 1, p 1 (φ(z), φ(w)) + d 1 1 = (φ(z) + φ(w))e 1, φ(z) + φ(w) + d 1 e 1, d 1 1 = d 1 e 1, d 1 1. (4.7) The second equality follows from d 1 e 1, d 1 1 kertφ(z) kertφ(z). The last equality follows from e 1 1 e 1, 1 kert φ(z) kert φ(z). Substituting (4.4) into Equation (4.7), we have 0 = d 1 e 1, d e + λ 0 e 0 = d 1 e 1, d 0 1 = d 1 e 1, φ(w) + we 0 = d 1 e 1, we 0 = + µ 1 e 1 + λ 1 e 0, we 0 =, we 0 + µ 1 e 1, we 0 + λ 1 e 0, we 0. The second equation comes from that d 1 e 1 is orthogonal to L 0 both e e 0 are in L 0. The third equation follows from the definition of d 0 1 the forth equation follows from that d 1 e 1 is in kert φ(z) kert φ(w). We need to calculate d0 e 1, we 0, e 1, we 0, e 0, we 0 separately. To get, we 0, by the definition of d 0 e 1, we have, we 0 = φ(w)e 1 + we 1 (0, w)e 0, we 0 = φ(w)e 1, we 0 + we 1 (0, w)e 0, we 0 Thus we need to compute φ(w)e 1, we 0 we 1 (0, w)e 0, we 0 one by one. The equality φ(w)e 1, we 0 = 0

21 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 21 follows from the following computations. φ(w)e 1, we 0 = wφ 0 (w)e 1, we 0 = φ 0 (w)e 1, e 0 = φ 0 (w)e 1 (w, w), e 0 (0, w) = φ 0 (w)e 1 (w, w), φ 0 (w) = e 1 (w, w), 1 = e 1, 1 = 0. To get we 1 (0, w)e 0, we 0, we continue as follows. we 1 (0, w)e 0, we 0 = e 1 (0, w)e 0, e 0 = e 1 (0, w)e 0 (w, w), e 0 (0, w) = e 1 (0, w)e 0 (w, w), φ 0 (w) = e 1 (0, w)(φ 0 (w) + wφ 0(w)), φ 0 (w) = e 1 (0, w)φ 0 (w), φ 0 (w) + e 1 (0, w)wφ 0(w), φ 0 (w) = e 1 (0, w), 1 + e 1 (0, w)wφ 0(w), φ 0 (w) = e 1 (0, 0) + e 1 (0, w)wφ 0(w), φ 0 (w) = e 1, 1 + e 1 (0, w)wφ 0(w), φ 0 (w) = e 1 (0, w)wφ 0(w), φ 0 (w) = e 1 (0, α) + e 1 (0, β). The last equality follows from Lemma 4.2 e 1 (0, 0) = e 1, 1 = 0. Hence, we 0 = e 1 (0, α) + e 1 (0, β) Recall that d 1 1 = d e + λ 0 e 0 is orthogonal to L 0 e 1 is orthogonal to both e, e 0. Thus 0 = e 1, d e + λ 0 e 0 = e 1, φ(w) + we 0 = e 1, we 0. From the computation of d 0 1, e 0 in the proof of Lemma 4.3 we have showed that we 0, e 0 = α + β. Therefore we have that e 1 (0, α) + e 1 (0, β) + λ 1 (ᾱ + β) = 0. (4.8)

22 22 SUN, ZHENG AND ZHONG On the other h, 0 = d 1 e 1, e 0 = + µ 1 e 1 + λ 1 e 0, e 0 =, e 0 + 4λ 1, e 0 = φ(w)e 1 + we 1 (0, w)e 0, e 0 = we 1 (0, w)e 0, e 0 = we 1 (0, w)e 0 (w, w), e 0 (0, w) = we 1 (0, w)(φ 0 (w) + wφ 0), φ 0 (w) = w 2 e 1 (0, w)φ 0, φ 0 (w) = αe 1 (0, α) + βe 1 (0, β). The last equality follows from Lemma 4.2 with g = we 1 (0, w). Thus αe 1 (0, α) + βe 1 (0, β) + 4λ 1 = 0. So λ 1 = α 4 e 1(0, α) β 4 e 1(0, β). (4.9) Substituting (4.9) into (4.8), we have Recall that to get [1 α(ᾱ + β) ]e 1 (0, α) + [1 4 λ 0 = α + β, 4 β(ᾱ + β) ]e 1 (0, β) = 0. 4 (1 + λ 0 α)e 1 (0, α) + (1 + λ 0 β)e 1 (0, β) = 0. (4.10) We are going to draw another equation about e 1 (0, α) e 1 (0, β) from the property that d 1 e 1 is orthogonal to L 0. To do this, recall that e 1 = P L0 (k α (w) k β (w)) M 1 L 0, d 1 e 1 = d 0 e 1 + µ 1 e 1 + λ 1 e 0 L 0, L 0 = span{1, p 1, e α, e β }, Thus d 1 e 1 is orthogonal to p 1, e α e β. Since d 1 e 1 is orthogonal to p 1 we have e α = k α (z)k α (w), e β = k β (z)k β (w)., p 1 + µ 1 e 1, p 1 + λ 1 e 0, p 1 = 0.

23 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 23 Noting, p 1 = φ(w)e 1 + we 1 (0, w)e 0, p 1 = we 1 (0, w)e 0, p 1 = we 1 (0, w)e 0 (w, w), w = e 1 (0, w)e 0 (w, w), 1 = 0, we have to obtain Since d 1 e 1 e α, we have to get e 1, p 1 = P L0 (K α (w) K β (w)), p 1 = K α (w) K β (w), p 1 = ᾱ β, e 0, p 1 = e 0 (0, w), p 1 (w, w) = φ 0 (w), 2w = wφ α φ β, 2w = 2 φ α φ β, 1 = 2φ α (0)φ β (0) = 2αβ, (ᾱ β)µ 1 + 2αβλ 1 = 0, λ 1 = µ 1 ᾱ β 2αβ. (4.11), e α + µ 1 e 1, e α + λ 1 e 0, e α = 0,, e α + µ 1 e 1, e α µ 1 ᾱ β 2αβ e 0, e α = 0. (4.12) We need to calculate, e α, e 1, e α e 0, e α. Simple calculations show that, e α = φ(w)e 1 + we 1 (0, w)e 0, e α = we 1 (0, w)e 0, e α = αe 1 (0, α)e 0 (α, α), e 1, e α = e 1 (α, α) = P L0 (k α (w) k β (w)), e α = k α (w) k β (w), e α 1 = 1 α α β

24 24 SUN, ZHENG AND ZHONG = α(ᾱ β), (4.13) (1 α 2 )(1 α β) e 0, e α = e 0 (α, α) = αφ 0(α) + φ 0 (α) Thus (4.13) (4.14) give = α 2 1 α β. (4.14) 1 α 2 1 α β e 1 (α, α) e 0 (α, α) = ᾱ β α(α β). Substituting the above equality in Equation (4.12) leads to αe 1 (0, α)e 0 (α, α) + µ 1 e 1 (α, α) µ 1 ᾱ β 2αβ e 0(α, α) = 0. Dividing the both sides of the above equality by e 0 (α, α) gives Hence we have to obtain e 1 (α, α) αe 1 (0, α) + µ 1 e 0 (α, α) µ ᾱ β 1 2αβ = 0. ᾱ αe 1 (0, α) + µ β 1 α(α β) µ ᾱ β 1 2αβ = 0, αe 1 (0, α) + (β + λ 0 ) 2µ 1(ᾱ β) αβ(α β) Similarly, since d 1 e 1 is orthogonal to e β, we have to obtain, e β + µ 1 e 1, e β + λ 1 e 0, e β = 0, = 0. (4.15), e β + µ 1 e 1, e β µ 1 ᾱ β 2αβ e 0, e β = 0. (4.16) We need to calculate, e β, e 1, e β e 0, e β. Simple calculations as above show that, e β = φ(w)e 1 + we 1 (0, w)e 0, e β = we 1 (0, w)e 0, e β = βe 1 (0, β)e 0 (β, β), e 1, e β = e 1 (β, β) = P L0 (k α (w) k β (w)), e β = k α (w) k β (w), e β 1 = 1 ᾱβ 1 1 β 2

25 REDUCING SUBSPACES OF A CLASS OF MULTIPLICATION OPERATORS 25 = β(ᾱ β) (1 ᾱβ)(1 β 2 ) (4.17) e 0, e β = e 0 (β, β) = βφ 0(β) + φ 0 (β) Combining (4.17) with (4.18) gives = β 2 β α 1 (4.18) 1 ᾱβ 1 β 2 e 1 (β, β) e 0 (β, β) = ᾱ β β(α β). Substituting the above equality in (4.16) gives βe 1 (0, β)e 0 (β, β) + µ 1 e 1 (β, β) µ 1 ᾱ β 2αβ e 0(β, β) = 0. Dividing both sides of the above equality by e 0 (β, β) gives Hence we have to get Eliminating 2µ 1(ᾱ β) αβ(α β) e 1 (β, β) βe 1 (0, β) + µ 1 e 0 (β, β) µ ᾱ β 1 2αβ = 0 ᾱ βe 1 (0, β) µ β 1 β(α β) µ ᾱ β 1 2αβ = 0, βe 1 (0, β) (α + λ 0 ) 2µ 1(ᾱ β) αβ(α β) from (4.15) (4.19) gives = 0. (4.19) α(α + λ 0 )e 1 (0, α) + β(β + λ 0 )e 1 (0, β) = 0. (4.20) Now combining (4.10) (4.20), we have the following linear system of equations about e 1 (0, α) e 1 (0, β) If (1 + λ 0 α)e 1 (0, α) + (1 + λ 0 β)e 1 (0, β) = 0 α(α + λ 0 )e 1 (0, α) + β(β + λ 0 )e 1 (0, β) = 0. (4.21) e 1 (0, α) = e 1 (0, β) = 0, then p 1 is in L 0 = span{e 0, e 1, 1, e}. But noting we have e 0 (0, α) = e 0 (0, β) e(0, α) = e(0, β) p 1 (0, α) = p 1 (0, β),

26 26 SUN, ZHENG AND ZHONG which contradicts the assumption that α β. So at least one of e 1 (0, α) e 1 (0, β) is nonzero. Then the determinant of the coefficient matrix of System (4.21) has to be zero. This implies 1 + λ 0 α 1 + λ 0 β α(α + λ 0 ) β(β + λ 0 ) = 0 Making elementary row reductions on the above the determinant, we get (α β) λ λ 0 β (α β)(α + β + λ 0 ) β(β + λ 0 ) = 0. Since we have Exping this determinant we have α + β = 4λ 0 α β 0, λ λ 0 β 3λ 0 β(β + λ 0 ) = 0. 0 = λ 0 (β 2 + βλ 0 ) + 3λ 0 (1 + λ 0 β) = λ 0 (β 2 + βλ 0 + 3βλ 0 ) + 3λ 0 = λ 0 (β 2 + 4βλ 0 ) + 3λ 0 = λ 0 ( αβ) + 3λ 0 Taking absolute value on both sides of the above equation, we have to get This implies to complete the proof. 0 = λ 0 ( αβ) + 3λ 0 λ 0 (3 αβ ) 2 λ 0, λ 0 = 0. α + β = 0, REFERENCES [1] I. Baker, J. Deddens J. Ullman, A theorem on entire functions with applications to Toeplitz operators, Duke Math. J. 41(1974), [2] J. Ball, Hardy space expectation operators reducing subspaces, Proc. Amer. Math. Soc. 47(1975), [3] H. Bercovici, C. Foias C. Pearcy, Dual algebra with applications to invariant subspace dilation theory, CBMS, Vol. 56. [4] A. Brown, On a class of operators, Proc. Amer. Math. Soc. 4 (1953) [5] R. Douglas V. Paulsen, Hilbert Modules over function algebras, Pitman Research Notes in Mathematics, Series 217, Lonman Group UK Limited, [6] R. Douglas, R. Yang, Operator theory in the Hardy space over the bidisk. I. Integral Equations Operator Theory 38 (2000),

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