MULTIPLICATION OPERATORS ON THE BERGMAN SPACE VIA THE HARDY SPACE OF THE BIDISK

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1 MULTIPLICATION OPERATORS ON THE BERGMAN SPACE VIA THE HARDY SPACE OF THE BIDISK KUNYU GUO, SHUNHUA SUN, DECHAO ZHENG AND CHANGYONG ZHONG Abstract. In this paper, we develop a machinery to study multiplication operators on the Bergman space via the Hardy space of the bidisk. Using the machinery we study the structure of reducing subspaces of a multiplication operator on the Bergman space. As a consequence, we completely classify reducing subspaces of the multiplication operator by a Blaschke product φ with order three on the Bergman space to solve a conjecture of Zhu [40]. 1. Introduction Let D be the open unit disk in C. Let da denote the 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. Our main objective is to study multiplication operators on L 2 a by bounded analytic functions on the unit disk D via the Hardy space of bidisk. The theme is to use the theory of multivariable operators to study a single operator. Our main idea is to lift the Bergman shift up as the compression of a commuting pair of isometries on a nice subspace of the Hardy space of bidisk. This idea was used in studying the Hilbert modules by R. Douglas V. Paulsen [12], operator theory in the Hardy space over the bidisk by R. Dougals R. Yang [13], [37], [38] [39]; the higher-order Hankel forms by S. Ferguson R. Rochberg [10] [11] the lattice of the invariant subspaces of the Bergman shift by S. Richter [22]. For a bounded analytic function φ on the unit disk, the multiplication operator M φ is defined on the Bergman space L 2 a given by M φ h = φh for h L 2 a. Let e n = n + 1z n. Then {e n } 0 form an orthonormal basis of the Bergman space L 2 a. On the basis {e n }, the multiplication operator n+1 M z by z is a weighted shift operator, called the Bergman shift: M z e n = n+2 e n+1. 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 [4]. The Bergman shift M z has a universal property [4]: for any strict contraction S on a Hilbert space H, there always exist a pair of invariant subspaces of M z, say M N in LatM z ( the invariant subspace lattice of M z is the set of subspaces M of L 2 a with M z M M), such that S = P M N {M z M N }, where P M N denotes the orthogonal projection of L2 a(d) onto M N. This indicates that existence of the invariant subspace problem for Hilbert space operator is equivalent to whether LatM z is saturated, i.e., for any M, N LatM z, with M N dim(m N)=, whether there always exists some Ω LatM z such that N Ω M. Let T denote the unit circle. The torus T 2 is the Cartesian product T T. The Hardy space H 2 (T 2 ) over the bidisk is H 2 (T) H 2 (T). Let P be the orthogonal projection from the space L 2 (T 2 ) of the Lebesgue square integrable functions on the torus T 2 onto H 2 (T 2 ). The Toeplitz operator on H 2 (T 2 ) with symbol f in L (T 2 ) is defined by T f (h) = P (fh), for h H 2 (T 2 ). The first author the second author were supported in part by the National Natural Science Foundation of China. The third author was partially supported by the National Science Foundation. 1

2 2 GUO, SUN, ZHENG AND ZHONG Clearly, T z T w are a pair of doubly commuting pure isometries on H 2 (T 2 ). For each integer n 0, let n p n (z, w) = z i w n i = zn+1 w n+1. z w i=0 Let H be the subspace of H 2 (T 2 ) spanned by functions {p n } n=0. The orthogonal complement of H in H 2 (T 2 ) is [z w] = closure H2 (T 2 ){(z w)h 2 (T 2 )}. Then [z w] is an invariant subspace of analytic Toeplitz operators T f for f H (T 2 ). Let P H be the orthogonal projection from L 2 (T 2 ) onto H. It is easy to check that P H T z H = P H T w H. Let B denote the operator above. It was shown explicitly in [29] implicitly in [12] that 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. Clearly, for each f(z, w) H, (U f)(z) = f(z, z), for z D. The simple observation that p n (z, w) = zn+1 w n+1 z w gives that for each f(z, w) H, there is a function f(z) in the Dirichlet space D such that f(z, w) = f(z) f(w). z w Thus, for each Blaschke product φ(z) with finite order, the multiplication operator M φ on the Bergman space is unitarily equivalent to φ(b) on H. In this paper we will study the operator φ(b) on the Hardy space of the bidisk to shed light on properties of the multiplication operator M φ. This method seems to be effective in dealing with the multiplication operators on the Bergman space because functions in the Hardy space of the bidisk behave slightly better than the functions in the Bergman space. The difficulty to study B on H is to get better understing the projection P H. The price that we pay is that we will get a lot of mileage from developing a heavy machinery on the Hardy space of the bidisk how to get rid of P H in the expression φ(b) n f = 1 n + 1 P H (p n(φ(z), φ(w))f), for f H. To do this, letting L 0 denote the space kert φ(z) kert φ(w) H, for each e L 0, we construct functions {d k e} in Section 3.1 d 0 e in Section 3.2 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. On the other h, d k e is orthogonal to L 0. These constructions are useful in studying the reducing subspaces of φ(b). 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 only reducing subspaces contained in M are M {0}. As in [16], a subspace N of H is a wering subspace of T if N is orthogonal to T n N for each n = 1, 2,. If M is an invariant subspace of T, it is clear that M T M is a wering subspace of T, we will refer this subspace as the wering subspace of M.

3 OPERATORS ON THE BERGMAN SPACE 3 In fact, for a reducing subspace M of φ(b), e in the wering subspace of M l 1 p l (φ(z), φ(w))e + p k (φ(z), φ(w))d l k e M. k=0 Although for a Blaschke product φ of finite order, M φ is not an isometry, using the machinery on the Hardy space of bidisk we will show that there exists a unique reducing subspace M 0, the so called distinguished subspace, on which the restriction of M φ is unitarily equivalent to the Bergman shift, which will play an important role in classifying reducing subspaces of M φ. The functions d 1 e d 0 e have the following relation. Theorem 1. If M is a reducing subspace of φ(b) orthogonal to the distinguished reducing subspace M 0, for each e in the wering subspace for M, then there is an element ẽ in the wering subspace for M a number λ such that d 1 e = d 0 e + ẽ + λe 0. (1) To underst the structure of minimal reducing subspaces of φ(b) we lift a reducing subspace of φ(b) as a reducing subspace of the pair of doubly commuting isometries T φ(z) T φ(w). For a given reducing subspace M of φ(b), define the lifting M of M M = span{φ(z) l φ(w) k M, l, k 0}. Since M is a reducing subspace of φ(b) M is a reducing subspace of the pair of doubly commuting isometries T φ(z) T φ(w), by the Wold decomposition of the pair of isometries on M, we have M = φ(z) l φ(w) k L M, where L is the wering subspace M L M = kert φ(z) kert φ(w) M. The following theorem gives a complete description of the wering subspace L M. Theorem 2. Suppose that M is a reducing subspace of φ(b) orthogonal to M 0. If { 1,, } is a basis of the wering subspace of M, then L M = span{e(m) 1,, ; d 1,, d 1 }, 1 diml M = 2. To prove Theorem 2, first we use the Wold decomposition of the pair of doubly commuting isometries T φ(z) T φ(w) on the lifting K φ (= H) of H to get the dimension of the wering subspace L φ (= L ). By means of the Fredholm theory in [8], we are able to show that the H dimension of L φ equals 2N 1, where N is the order of the Blaschke product φ. Then by means of the finite dimension of the wering subspace of these isometries on the reducing subspace we will be able to obtain some structure theorems on reducing subspaces of the multiplication operators by finite Blaschke products on the Bergman space. Theorem 3. Suppose that Ω, M N are three distinct nontrivial minimal reducing subspaces contained in M 0 for φ(b). If Ω M N, then there is a unitary operator U : M N such that U commutes with φ(b) φ(b).

4 4 GUO, SUN, ZHENG AND ZHONG The machinery on the Hardy space of the bidisk is not only useful in classifying the reducing subspaces of multiplication operators on the Bergman space, but also it is helpful in understing the lattice of invariant subspaces of the Bergman shift as in [1] hence the invariant subspace problem. One of our goals is to develop the Bergman function theory [15], [17] via the Hardy space of the bidisk. The multiplication operators on the Bergman space is completely different from that in the Hardy space. By the famous Beurling Theorem [9], the lattice of the invariant subspaces of the multiplication operator by z on the Hardy space is completely determined by inner functions. A Beurling s theorem was recently obtained for the Bergman space [1]. On one h, on the Hardy space, for an inner function φ, the multiplication operator by φ is a pure isometry hence unilateral shift (with arbitrary multiplicity). So its reducing subspaces are in one-to-one correspondence with the closed subspaces of H 2 φh 2 [5], [16]. 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 [2], [3] [20]. For an inner function φ, the multiplication operator by φ on the Bergman space is a contraction but not an isometry. On the other h, surprisingly, on the Bergman space, it was shown in [28] [40] that for a Blaschke product φ with two zeros, the multiplication operator M φ has only two nontrivial reducing subspaces. Zhu [40] conjectured that for a Blaschke product φ with N zeros, the lattice of reducing subspaces of the operator M φ is generated by N elements. In other words, M φ has exactly N nontrivial minimal reducing subspaces. Applying the machinery developed in the paper, we will be able to disproves Zhu s conjecture in the following theorem. For a holomorphic function φ on the unit disk a point c in the unit disk, we say that c is a critical point of φ if its derivative vanishes at c. Theorem 4. Let φ be a Blaschke product with three zeros. If φ(z) has a multiple critical point in the unit disk, then M φ has three nontrivial minimal reducing subspaces. If φ does not have any multiple critical point in the unit disk, then M φ has only two nontrivial minimal reducing subspaces. Bochner s theorem [35], [36] says that every Blaschke product with N zeros has exactly N 1 critical points in the unit disk D. Theorem 4 gives a classifcation of reducing subspaces for M φ for a Blaschke product φ with three zeros. Critical points of φ have important geometric meaning about the self-mapping of the unit disk. The work of Stephonson [24], [25], [26] suggests that the geometric version of the above theorem should be in terms of the Riemann surfaces. A finite Blaschke product φ with N zeros is an N to 1 conformal map of D onto D. Bochner s theorem [35], [36] says that φ has exactly N 1 critical points in the unit disk D none on the unit circle. Let C denote the set of the critical points of φ in D F = φ 1 φ(c). Then F is a finite set, φ 1 φ is an N-branched analytic function defined arbitrarily continuable in D/F. 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. The Riemann surface for φ 1 φ over D is an N-sheeted cover of D at most N(N 1) branch points, it is not connected. The geometric version of Theorem 4 is the following theorem. Theorem 5. Let φ be a Blaschke product with three zeros. Then 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. We would like to point out that there are many essential differences in analysis geometry between Blaschke products with order three Blaschke products with order two. On one h, for Blaschke products φ with order two, φ 1 φ contains two analytic functions on the unit disk hence the Riemann surface for φ 1 φ over D is just two copies of the unit disk. On the other h, for the most Blaschke products with order three, φ 1 φ has three multivalue functions

5 OPERATORS ON THE BERGMAN SPACE 5 on the unit disk the Riemann surface for φ 1 φ over D has two connected components. This phenomenon makes it difficult for us to classify the reducing subspaces of a multiplication operator with the Blaschke product of order highter than two. It seems that the machinery developed in the paper is inevitable in classifying the reducing subspaces of the multiplication operator by a Blaschke product of higher order. The problem of determining reducing subspaces of a multiplication operator is equivalent to finding projections in the commutant of the operator which is the set of operators commuting with the multiplication operators. Every von Nuemann algebra is generated by its projections. Theorem 4 says that every von Nuemann algebra contained in the commutant of mulitplication operator by the Blaschke product with third order is commutative. A lot of nice deep work on the commutant of a multiplication operator has been done on the Hardy space [6], [33], [34] while Blaschke products with finite zeros play an important role. Indeed Cowen proved that for f H, if the inner factor of f f(α) is a Blaschke product φ with finite order for some α D, then the commutant of the multiplication operator by f equals the commutant of the multiplication operator by the finite Blaschke product φ [6]. Thus the structure of lattice of reducing subspaces of the multiplication operator by a Blaschke product with finite order is useful in studying the general multiplication operators on the Bergman space. One of applications of the machinery on the Hardy space of the bidsk is that it was proved in [32] that the multiplication operator on the Bergman space is unitarily equivalent to a weighted unilateral shift operator of finite multiplicity if only if its symbol is a constant multiple of the N-th power of a Mobius transform. Another one is that we have obtained a complete description of nontrivial minimal reducing subspaces of the multiplication operator by φ on the Bergman space of the unit disk for the fourth order Blaschke product φ [31]. Using Theorems 1 3, for a finite Blaschke product φ, we are able to show that for two distinct nontrivial minimal reducing subspaces of φ(b), either they are orthogonal or φ(b) has two distinct unitarily equivalent reducing subspaces has also infinitely many minimal reducing subspaces (Theorem 31). Thus either φ(b) has infinitely many minimal reducing subspaces or the number of nontrivial minimal reducing subspaces of φ(b) is less than or equal to the order of φ (Theorem 32). We say that two reducing subspaces M N of φ(b) are unitarily equivalent if there is a unitary operator U : M N such that U commutes with φ(b) φ(b). The adjoint of the multiplication operator by a finite Blaschke product is in a Cowen-Douglas class [7]. The theory of Cowen-Douglas classes will be useful in studying the multiplication operators on the Bergman space. On the other h, we would like to see some applications of the results obtained in the paper to the study of the Cowen-Dougals classes. We thank R. Douglas for his insightful comments on the relations between multiplication operators Cowen-Dougals classes, R. Rochberg for his drawing our attention to his papers [10] [11] with S. Ferguson, K. Stephenson for his drawing our attention to his papers [24], [25], [26] K. Zhu for his useful comments on his conjecture. 2. The wering subspace of the lifting of the Bergman space As pointed out before, we can identify the Bergman space with H. First we introduce notations show some properties of functions in H. Then we compute the dimension of the wering space for the lifting H of H. The dimension is useful for us to find the wering space for the lifting M of a reducing subspace M of φ(b). For α D, let k α be the reproducing kernel of the Hardy space H 2 (T) at α. That is, for each function f in H 2 (T), f(α) = f, k α. In fact, k α = 1 (1 ᾱz). For φ in H (T), let ˆT φ denote the analytic Toeplitz operator with symbol φ on H 2 (T), given by ˆT φ h = φh.

6 6 GUO, SUN, ZHENG AND ZHONG Thus it is easy to check that For an integer s 0, let Lemma 6. For each f H (T), ˆT f k s α = ˆT φk α = φ(α)k α. (2) k s α(z) = s l=0 s!z s (1 ᾱz) s+1. s! l!(s l)! f (l) (α)k s l α. The proof of the above lemma is left for readers. Lemma 6 gives that the kernel of the Toeplitz operator ˆT φ on the Hardy space of the unit circle is spanned by {{ks k α k } sk =0,,n k } k=0,,k. Recall that H is the subspace of H 2 (T 2 ) spanned by functions {p n } n=0. The following two lemmas give some properties for functions in H or H. Lemma 7. If f is in H 2 (T 2 ) continuous on the closed bidisk e is in H, then f(z, w), e(z, w) = f(z, z), e(z, 0) = f(w, w), e(0, w). The proof of Lemma 7 is left for readers. Lemma 8. For h(z, w) H 2 (T 2 ), h is in H iff h(z, z) = 0, for z D. Proof. As pointed out before, H = cl{(z w)h 2 (T 2 )}. Let z be in D. For each function f(z, w) (z w)h 2 (T 2 ), f(z, z) = 0. Thus h(z, z) = 0 for each h H. Conversely, assume that for a function h H 2 (T 2 ), h(z, z) = 0, for z D. For 0 < r < 1, define h r (z, w) = h(rz, rw). Then for each fixed 0 < r < 1, h r (z, z) = 0, h r (z, w) is continuous on the closed bidisk in H 2 (T 2 ). By Lemma 7, for each e(z, w) in H, On the other h, by Theorem in [23], h r (z, w), e(z, w) = h r (z, z), e(z, 0) = 0. h(z, w), e(z, w) = lim r 1 h r (z, w), e(z, w) = 0. Hence we conclude that h is in H. The Dirichlet space D consists of analytic functions on the unit disk whose derivative is in the Bergman space L 2 a. Theorem 9. For each f(z, w) in H 2 (T 2 ), f is in H if only if there is a function f(z) in D such that f(z, w) = f(z) f(w), z w for two distinct points z w in the unit disk. This immediately gives the following three lemmas, which proofs are left for readers. Lemma 10. Suppose that e(z, w) is in H. e(z, w) = 0 for (z, w) on the torus. If e(z, z) = 0 for each z in the unit disk, then

7 OPERATORS ON THE BERGMAN SPACE 7 Lemma 11. If e(z, w) is in H, then e(z, w) = e(w, z). Lemma 12. Suppose f(z, w) is in H. Let F (z) = f(z, 0). Then for each λ D. f(λ, λ) = λf (λ) + F (λ), For an operator T on a Hilbert space, let kert denote the kernel of T. Then kert = H T H. Given a pure isometry U on a Hilbert space H, the classical Wold decomposition theorem [19] states that H = n 0 U n E, where E = H UH is the wering subspace for U equals kert. For a function φ in H (D), we can view φ(z) φ(w) as functions on the torus T 2. While M φ is not an isometry on the Bergman space of the unit disk, the analytic Toeplitz operators T φ(z) T φ(w) are a pair of doubly commuting pure isometries on the Hardy space H 2 (T 2 ) of torus. Since for n 1 T z p n = T wp n = p n 1 T z p 0 = T wp 0 = 0, H is an invariant subspace for both Tz Tw. So H is also an invariant subspace for both Tφ(z) T φ(w). Recall the lifting K φ of H: K φ = span{φ l (z)φ k (w)h; l, k 0}. Then K φ is a reducing subspace for both T φ(z) T φ(w), so T φ(z) T φ(w) are also a pair of doubly commuting isometries on K φ. We consider the Wold decompositions for the pair on both K φ K φ (H 2 (T 2 ) K φ ). Introduce wering subspaces L φ = kert φ(z) kert φ(w) K φ, L φ = kert φ(z) kert φ(w) K φ. To get the dimension of the wering subspaces L φ L φ, we will identify the wering subspace L φ for the Blaschke product φ with distinct zeros. The following lemma follows from the remark after Lemma 6 about ker ˆT φ. Lemma 13. If φ(z) is a Blaschke product with distinct zeros {α i } N i=1, then intersection of the kernel of Tφ(z) T φ(w) is spanned by {k α i (z)k αj (w)} N i,j=1. The following lemma is implicit in the proof of Theorem 3 [29]. But we give a complete proof of the lemma. Lemma 14. Suppose that φ(z) is a Blaschke product with distinct zeros {α i } N i=1. Then the wering space L φ is equal to the space spanned by {k αi (z)k αj (w) k αj (z)k αi (w) : 1 i < j N} {(Tz w[k αl (z)k αl+1 (w)+k αl+1 (z)k αq (w)+k αq (z)k αl (w)] : 2 l +1 < q N}. Moreover, the dimension of L φ equals (N 1) 2.

8 8 GUO, SUN, ZHENG AND ZHONG Proof. First we show Since H K φ, L φ = kert φ(z) kert φ(w) H. L φ kert φ(z) kert φ(w) H. Conversely, if f is in kertφ(z) kert φ(w) H, then f is in kertφ(z) kert φ(w) orthogonal to H. Thus for each g(z, w) = φ(z)l φ(w) k h kl K φ where h kl H, we have f, g = f, φ(z) l φ(w) k h kl So f is also in L φ. Hence we have k,l 0 = [Tφ(z) ]l [Tφ(w) ]k f, h lk = 0. k,l 0 L φ = kert φ(z) kert φ(w) H. We are to prove that the dimension of L φ is (N 1) 2. Without loss of generality, we assume that α 1 = 0. By Lemma 13, the N 2 dimensional space kertφ(z) kert φ(w) is spanned by {k αi (z)k αj (w)} N i,j=1. So it follows from Lemma 8 that L φ consists of the elements h in kertφ(z) kertφ(w) which satisfy h(z, z) = 0. That is, N N N N L φ = {h = c ij k αi (z)k αj (w) : h(z, z) = c ij k αi (z)k αj (z) = 0}. i=1 j=1 i=1 j=1 For any h L φ, taking the limit at infinity testing the multiplicity at its poles 1/ᾱ j of the function h(z, z), we immediately have that h(z, z) = 0 implies c jj = 0, j = 1, 2..., N. That is, L φ = {h = N i j,i=1 j=1 N c ij k αi (z)k αj (w) : h(z, z) = N i j,i=1 j=1 N c ij k αi (z)k αj (z) = 0}. Observe that k αi (z)k αj (z) = a ij k αi (z) + b ij k αj (z) where a ij = āi ā i ā j b ij = āj ā i ā j, k α2 (z),..., k αn (z) are linearly independent. Write h(z, z) as linear combination of k αj (z), j = 2,..., N, then all the coefficients of k αj (z) must be zero. So we have a system of another N 1 linear equations governing c ij, i j, i, j = 1,..., N. It is easy to check that the rank of the coefficient matrix of the system is N 1. Hence the dimension of L φ (as the solution space of N 2 N unknown variables governed by N 1 linearly independent equations) equals N 2 N (N 1). The proof is finished. We are ready to prove our main result in the section. Theorem 15. Let φ be a Blaschke product with N zeros in the unit disk. Then K φ = φ l (z)φ k (w)l φ, H 2 (T 2 ) K φ = φ l (z)φ k (w) L φ. The dimension of L φ equals (N 1) 2 the dimension of L φ equals 2N 1. Proof. As pointed out early in this section, T φ(z) T φ(w) are a pair of doubly commuting isometries on both K φ H 2 (T 2 ) K φ. Consider the Wold decomposition of T φ(z) on K φ to get K φ = l φ(z) l E

9 OPERATORS ON THE BERGMAN SPACE 9 where E is the wering subspace for T φ(z) given by E = K φ [T φ(z) K φ ] = ker[t φ(z) K φ ] = kert φ(z) K φ. Since T φ(z) T φ(w) are doubly commuting, E is a reducing subspace of T φ(w). Thus T φ(w) E is still an isometry. The Wold decomposition theorem again gives E = k 0 φ(w) k E 1 where E 1 is the wering subspace for T φ(w) E given by This gives E 1 = E T φ(w) E = kert φ(w) E = kert φ(z) T φ(w) K φ. K φ = φ l (z)φ k (w)l φ. Considering the Wold decompositions of T φ(z) T φ(w) on H 2 (T 2 ) K φ, similarly we obtain Noting we have H 2 (T 2 ) K φ = φ l (z)φ k (w) L φ. kert φ(z) kert φ(w) = L φ L φ dim[kert φ(z) kert φ(w) ] = dim[l φ] + dim[ L φ ]. By Lemma 13, the dimension of kert φ(z) kert φ(w) equals N 2. Hence dim[l φ ] = N 2 dim[ L φ ]. To finish the proof, it suffices to show that the dimension of L φ equals (N 1) 2. By Lemma 14, for a Blaschke product φ(z) with distinct zeros, the dimension of L φ equals (N 1) 2. We need to show that this is still true for a Blaschke product B with N zeros which perhaps contains some repeated zeros. To do so, for a given λ D, let φ λ (z) be the Möbius transform z λ 1 λz. Then φ λ φ(z) is a Blaschke product with N zeros in the unit disk Thus K φ = K φλ φ, so L φλ φ The last equality follows from that We have the fact that T φλ φ(z) = (T φ(z) λi)(i λt φ(z) ) 1. = kert φ λ φ(z) kert φ λ φ(w) [H2 (T 2 ) K φλ φ] = kert φ(z) λ kert φ(w) λ [H2 (T 2 ) K φ ]. kert φ(z) λ = kert φ λ φ(z), kert φ(w) λ = kert φ λ φ(w). dim L φλ φ = index(t φ(z) λ, T φ(w) λ ), where index(tφ(z) λ, T φ(w) λ ) is the Fredholm index of the pair (T φ(z) λ, T φ(w) λ ), which was first introduced in [8]. The proof of the fact is left for readers. It was shown [8] that the Fredholm index of the pair (Tφ(z) λ, T φ(w) λ ) is a continuous mapping from the set of the Fredholm tuples to the set of integers. Thus for a sufficiently small λ, index(t φ(z) λ, T φ(w) λ ) = index(t φ(z), T φ(w) ).

10 10 GUO, SUN, ZHENG AND ZHONG If λ is not in the critical values set C = {µ D : µ = φ(z) φ (z) = 0 for some z D} of φ, then φ λ φ(z) is a Blaschke product with N distinct zeros in D. In fact, Bochner s theorem implies that there are at most N 1 points in C. In this case, by Lemma 14, dim L φλ φ = (N 1) 2. Since the set C has zero area, we conclude that the dimension of L φ equals (N 1) Basic constructions In this section we will construct a family {d k e} of functions a function d 0 e in L φ for each e kertφ(z) kert φ(w) H, which have properties in Theorem 1 in Section 1 to present the proof of Theorem 1 that gives a relation between d 1 e d 0 e. The relation is very useful for us to underst the structure of the minimal reducing subspaces in the rest of the paper First Construction. Let L 0 be kertφ(z) kert φ(w) H. It is easy to check that the dimension of L 0 equals the order of the Blaschke product. First we will show that for a given reducing subspace M for φ(b), for each e M L 0 each integer l 1, there are a family of functions {d k e} l k=1 such that l 1 p l (φ(z), φ(w))e + p k (φ(z), φ(w))d l k e M. k=0 These functions are useful in studying the structure of the multiplication operator M φ on the Bergman space. The following lemma shows that for each reducing subspace M of φ(b), the intersection of M L 0 is nontrivial. Lemma 16. If M is a nontrivial reducing subspace for φ(b), then the wering subspace of M is contained in L 0. Proof. Let M be a nontrivial reducing subspace for φ(b). For each f in H, P M f is in M. Thus for each e in the wering subspace M φ(b)m of M, 0 = e, φ(b)p M f = e, P M φ(b)f = e, φ(b)f = Tφ(z) e, f. The second equality follows from that M is a reducing subspace the last equality follows from the fact that for each f H, φ(b) f = T φ(z) f = T φ(w) f. So T φ(z) e = 0. Similarily, we also have that T φ(w) e = 0. This gives that e is in L 0 to complete the proof. Lemma 17. If M is a reducing subspace for φ(b), then φ(b) M = M. Proof. First note that for a Blaschke product φ(z) with finite order, φ(b) is Fredholm the kernel of φ(b) contains only zero. Thus φ(b) H = H. Suppose that M is a reducing subspace for φ(b). Let N = M. Then φ(b) = φ(b) M φ(b) N under the decomposition H = M N. Since φ(b) is subjective, This completes the proof. φ(b) M M = M.

11 OPERATORS ON THE BERGMAN SPACE 11 Theorem 18. Suppose that M is a reducing subspace for φ(b). For a given e in the wering subspace of M, there are a unique family of functions {d k e} L φ L 0 such that (1) p l (φ(z), φ(w))e + l 1 k=0 p k(φ(z), φ(w))d l k e is in M, for each l 1. (2) P H [p l (φ(z), φ(w))d k e] is in M for each k 1, l 0. Proof. For a given e in the wering subspace of M, first we will use mathematical induction to construct a family of functions {d k e}. By Lemma 16, e is in L 0. A simple calculation gives T φ(z) [(φ(z)+φ(w))e] = e, T φ(w) [(φ(z)+ φ(w))e] = e. By Lemma 17, there is a unique function ẽ M L 0 such that This gives T φ(z)ẽ = T φ(w)ẽ = e. Tφ(z) [ẽ (φ(z) + φ(w))e] = e e = 0, Tφ(w) [ẽ (φ(z) + φ(w))e] = e e = 0, to get that letting d 1 e = ẽ (φ(z) + φ(w))e, d 1 e is in kert φ(z) kert φ(w), p 1 (φ(z), φ(w))e + d 1 e = (φ(z) + φ(w))e + d 1 e M. Because both ẽ e are in M, we have that d 1 e is in K φ, hence d 1 e is in L φ. Next we show that d 1 e is orthogonal to L 0. To do so, let f L 0. A simple calculation gives d 1 e, f = ẽ (φ(z) + φ(w))e, f = ẽ, f (φ(z) + φ(w))e, f = 0 e, Tφ(z) f + T φ(w) f = 0. The third equality follows from that ẽ is in M L 0. This gives that d 1 e is in L φ L 0. Assume that for n < l there are a family of functions {d k e} n k=1 L φ L 0 such that n 1 p n (φ(z), φ(w))e + p k (φ(z), φ(w))d n k e M. k=0 Let E = p n (φ(z), φ(w))e + n 1 k=0 p k(φ(z), φ(w))d n k e. By Lemma 17 again, there is a unique function Ẽ in M L 0 such that T φ(z)ẽ = T φ(w)ẽ = E. Let F = p n+1 (φ(z), φ(w))e + n k=1 p k(φ(z), φ(w))d n+1 k e. Since T φ(z) [p k(φ(z), φ(w))f] = T φ(w) [p k(φ(z), φ(w))f] = p k 1 (φ(z), φ(w))f, for each f in L φ k 1, simple calculations give Thus T φ(z) F = T φ(w) F = E. T φ(z) (Ẽ F ) = T φ(w) (Ẽ F ) = E E = 0.

12 12 GUO, SUN, ZHENG AND ZHONG So letting d n+1 e = Ẽ F, dn+1 e that for each f L 0, to get that d n+1 e d n+1 e, f = Ẽ, f F, f is in kert φ(z) kert φ(w). Noting Ẽ is orthogonal to L 0, we have = [ p n+1 (φ(z), φ(w))e, f + = 0, is in L φ L 0. Hence p n+1 (φ(z), φ(w))e + n k=1 n p k (φ(z), φ(w))d n+1 k, f ] k=1 p k (φ(z), φ(w))d n+1 k e + d n+1 e M. This gives a family of functions {d k e} L φ L 0 satifying Property (1). To finish the proof we need only to show that Property (2) holds. A simple calculation gives This implies 2φ(B)e = P H (p 1 (φ(z), φ(w))e) = P H (p 1 (φ(z), φ(w))e + d 1 e) P H (d 1 e) = p 1 (φ(z), φ(w))e + d 1 e P H (d 1 e). P H (d 1 e) = [p 1 (φ(z), φ(w))e + d 1 e] 2φ(B)e M. Noting that (d 1 e P H d 1 e) is in H = [z w] [z w] is an invariant subspace for analytic Toeplitz operators, we have that so to get [p l 1 (φ(z), φ(w))(d 1 e P H d 1 e)] H, P H [p l 1 (φ(z), φ(w))(d 1 e P H d 1 e)] = 0, P H [p l 1 (φ(z), φ(w))(d 1 e)] = P H {p l 1 (φ(z), φ(w))[p H d 1 e]} M. Assume that P H [p l (φ(z), φ(w))d k e] M for k n any l 0. To finish the proof by induction we need only to show that for any l 0. A simple calculation gives Thus P H [p l (φ(z), φ(w))d n+1 e )] M (n + 2)φ(B) n+1 e = P H [p n+1 (φ(z), φ(w))e + {P H [d n+1 e ] + P H [ n k=1 P H [d n+1 e ] = P H [p n+1 (φ(z), φ(w))e + {(n + 2)φ(B) n+1 e + P H [ n k=1 n k=0 e p k (φ(z), φ(w))d n+1 k e ] p k (φ(z), φ(w))d n+1 k e ]}. n k=0 p k (φ(z), φ(w))d n+1 k e ] p k (φ(z), φ(w))d n+1 k e ]}.

13 OPERATORS ON THE BERGMAN SPACE 13 Property (1) gives that the first term in the last equality is M, the induction hypothesis gives that the last term is in M the second term belongs to M since e M M is a reducing subspace for φ(b). So P H [d n+1 e ] is in M. Therefore we conclude P H [p l (φ(z), φ(w))d n+1 e ] = P H [(p l (φ(z), φ(w))(p H d n+1 e )] M, to complete the proof. In the special case for H, as H is a reducing subspace for φ(b), Theorem 18 immediately gives the following theorem. Theorem 19. For a given e L 0 there are a unique family of functions {d k e} L φ L 0 such that l 1 p l (φ(z), φ(w))e + p k (φ(z), φ(w))d l k e H, for each l 1. k= Second Construction. Next for a given e L 0, we will show that the function d 0 e(z, w) given by d 0 e(z, w) = we(0, w)e 0 (z, w) wφ 0 (w)e(z, w) is in L φ satisfies p l (φ(z), φ(w))e + p l 1 (φ(z), φ(w))d 0 e H for each l 1. Recall that φ is a Blaschke product with zeros {α k } K 0 α k repeats n k + 1 times, φ(z) = zφ 0 (z) where φ 0 is a Blaschke product with N 1 zeros. Let e 0 = φ(z) φ(w) z w. Theorem 9 gives that e 0 is in H since φ is a Blaschke product with finite order. This also gives that e 0 (z, 0) = φ 0 (z). Theorem 20. Let f be a nonzero function f in H. p l (φ(z), φ(w))f H, for some l 1 if only if f = λe 0 for some constant λ The proof of Theorem 20 is left for readers. Theorem 20 gives that M 0 = span l 0 {p l (φ(z), φ(w))e 0 } is a reducing subspace of φ(b). We will study the space in next section. For each e(z, w) in L 0, let d 0 e(z, w) = we(0, w)e 0 (z, w) wφ 0 (w)e(z, w). Theorem 21. For each e(z, w) in L 0, d 0 e(z, w) is a function in L φ such that for l 1. p l (φ(z), φ(w))e + p l 1 (φ(z), φ(w))d 0 e H, (3) Proof. First we show that the function d 0 e(z, w) is in kertφ(z) kert φ(w). To do this, by Theorem 9, write e(z, w) = φ e(z) φ e (w) (4) z w for some function φ e in the Dirichlet space D with φ e (0) = 0. Letting w = 0 in the above equality gives that e(z, 0) = e(0, z) = zφ e (z). d 0 e(z, w) = we(0, w)e 0 (z, w) wφ 0 (w)e(z, w) = φ(z) φ(w) φ e (w)[ z w = φ e(w)φ(z) φ(w)φ e (z). z w ] φ(w)[ φ e(z) φ e (w) ] z w

14 14 GUO, SUN, ZHENG AND ZHONG This gives that d 0 e(z, w) is a symmetric function of z w. symmetric functions of z w in L 0, we have T φ(z) [d0 e(z, w)] = T φ(z) [we(0, w)e 0(z, w) wφ 0 (w)e(z, w)] Since e 0 (z, w) e(z, w) are = we(0, w)tφ(z) [e 0(z, w)] wφ 0 (w)tφ(z) e(z, w) = 0, T φ(w) [d0 e(z, w)] = T φ(w) [d0 e(w, z)] = T φ(w) [ze(0, z)e 0(w, z) zφ 0 (z)e(w, z)] = ze(0, z)tφ(w) [e 0(z, w)] zφ 0 (z)tφ(w) e(z, w) = 0, to get that d 0 e(z, w) is in kert φ(z) kert φ(w). Next we show that d 0 e(z, w) satisfies (3). To do this, let We show that E l = p l (φ(z), φ(w))e + p l 1 (φ(z), φ(w))d 0 e. E l = φ e(z)φ l (z) φ e (w)φ l (w). z w By Theorem 9, this gives that E l is in H. Simple calculations give p l (φ(z), φ(w))e = [ φl+1 (z) φ l+1 (w) φ(z) φ(w) ][ φ e(z) φ e (w) ] z w = φl+1 (z)φ e (z) φ l+1 (z)φ e (w) φ l+1 (w)φ e (z) + φ l+1 (z)φ e (w) (φ(z) φ(w))(z w) p l 1 (φ(z), φ(w))d 0 e = [ φl (z) φ l (w) φ(z) φ(w) ][φ e(w)φ(z) φ(w)φ e (z) ] z w Thus = φl+1 (z)φ e (w) φ l (z)φ(w)φ e (z) φ l (w)φ(z)φ e (w) + φ l+1 (w)φ e (z). (φ(z) φ(w))(z w) E l = p l (φ(z), φ(w))e + p l 1 (φ(z), φ(w))d 0 e = φl (z)φ e (z)(φ(z) φ(w)) φ l (w)φ e (w)(φ(z) φ(w)) (φ(z) φ(w))(z w) = φ e(z)φ l (z) φ e (w)φ l (w). z w Since p 1 (φ(z), φ(w))e + d 0 e is in H p 1 (φ(z), φ(w))e is in K φ, we conclude that d 0 e is in K φ. Hence d 0 e is in L φ. This completes the proof. Now we are ready to prove Theorem 1. Proof of Theorem 1. Since M is orthogonal to M 0, we have H = M 0 M 0 = M 0 M [M 0 M ]. Thus L 0 = Ce 0 [M L 0 ] [M 0 M L 0 ]. So e is orthogonal to e 0, L 0 e 0 = [M (L 0 e 0 )] [M 0 M (L 0 e 0 )]. Let P 0 denote the orthogonal projection from H 2 (T 2 ) onto the space Ce 0. Let d e = d 0 e P 0 d 0 e. Then d e is orthogonal to e 0. Theorems give p l (φ(z), φ(w))e + p l 1 (φ(z), φ(w))d e H, (5)

15 OPERATORS ON THE BERGMAN SPACE 15 for l 1. By Theorem 19, there is a function d k e L φ L 0 such that for each l 1. Thus l 1 p l (φ(z), φ(w))e + p k (φ(z), φ(w))d l k e M, k=0 So d e d 1 e is in L 0 e 0. Write d e d 1 e = p 1 (φ(z), φ(w))e + d e (p 1 (φ(z), φ(w))e + d 1 e) H. d e d 1 e = e + e for e M (L 0 e 0 ) e M (L 0 e 0 ). Thus (5) gives that the following function is in H: p 2 (φ(z), φ(w))e + p 1 (φ(z), φ(w))d e = [p 2 (φ(z), φ(w))e + p 1 (φ(z), φ(w))d 1 e + d 2 e] + [p 1 (φ(z), φ(w))e + d 1 e ] +[p 1 (φ(z), φ(w))e + d 1 e ] (d2 e + d 1 e + d1 e ). Theorem 18 gives that the first term the second term in the right h side are in M the third term is in M. Thus the last term must be in H hence By Theorem 18 again, we have d 2 e + d 1 e + d1 e H kert φ(z) T φ(w) = L 0. d 2 e + d 1 e + d1 e L φ L 0, to get This gives d 2 e + d 1 e + d1 e = 0. P H d 1 e = (P H d1 e + P H d2 e). On the other h, Theorem 18 gives P H d 1 e + P H d2 e is in M P H d 1 e P H d 1 e = 0, so simple calculations give is in M. Thus d 1 e 2 = d 1 e, d1 e = d 1 e, p 1(φ(z), φ(w))e + d 1 e = d 1 e, P H [p 1(φ(z), φ(w))e + d 1 e ] = P H (d 1 e ), p 1(φ(z), φ(w))e + d 1 e = 0. Hence we have that d 1 e = 0, to get p1(φ(z), φ(w))e H. Theorem 20 gives that e = λe 0, for some constant λ. Since e M (L 0 e 0 ) we conclude that e = 0 to get d e = d 1 e + e. Letting ẽ = e M, we obtain d 1 e = d e + ẽ = d 0 e P 0 d 0 e + ẽ = d 0 e + ẽ + λe 0, as desired. The last equality follows from that P 0 d 0 e = λe 0 for some constant. This completes the proof.

16 16 GUO, SUN, ZHENG AND ZHONG 4. The distinguished reducing subspace Theorems 1 19 are useful in studying reducing subspaces of φ(b). In this section we will use them to show that there always exists a unique reducing subspace M 0 for φ(b) such that the restriction of φ(b) on M 0 is unitarily equivalent to the Bergman shift. The existence of such reducing subspace is the main result in [18]. Moreover, we will show that such reducing space is unique. We call M 0 to be the distinguished reducing subspace for φ(b). In fact, M 0 is unitarily equivalent the subspace span{φ φ m : m = 0,, n, } of the Bergman space [27] if φ(0) = 0. Furthermore we will show that only the multiplication operator by a finite Blashcke product has such nice property. Assume that φ be a Blaschke product of order N with φ(0) = 0. Recall e 0 (z, w) = φ(z) φ(w) z w. The following lemmas will be used in the proofs of Theorems The proofs of those lemmas are left for readers. Lemma 22. Let f be a function in H 2 (T 2 ). Then P H [φ(z)p n (φ(z), φ(w))f] = n + 1 n + 2 P H [p n+1(φ(z), φ(w))f]. Lemma 23. Let φ(z) be an inner function satisfying φ(z) φ(w) z w φ(z) φ(w) φ(z)h 2 (T 2 ). z w H 2 (T 2 ), then Lemma 24. For an inner function φ(z), φ(z) φ(w) z w is in H 2 (T 2 ) iff φ(z) is a finite Blaschke product. Moreover, for a Blaschke product φ of order N, e 0 2 = N. Now we are ready to prove the first main result in this section. Theorem 25. Let φ be a Blaschke product of order N. 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 = span l 0 {p l (φ(z), φ(w))e 0 }, { p l(φ(z),φ(w))e 0 l+1 N } 0 form an orthonormal basis of M 0. Proof. First we show that there exists a reducing subspace M 0 of φ(b) such that φ(b) M0 is unitarily equivalent to the Bergman shift. Let M 0 = span l 0 {p l (φ(z), φ(w))e 0 }. As pointed out before, Theorem 20 gives that M 0 is a reducing subspace of φ(b). Here e 0 (z, w) = φ(z) φ(w) z w. A simple calculation gives p l (φ(z), φ(w))e = (l + 1) e 0 2 2, p l (φ(z), φ(w))e 0, p n (φ(z), φ(w))e 0 = 0, for n l. Let E n = pn(φ(z),φ(w))e0. Thus {E n } are an orthonormal basis of M 0. By Lemma 22 (n+1) e0 2 we have φ(b)[p n (φ(z), φ(w))e 0 ] = P H [φ(z)p n (φ(z), φ(w))e 0 ] = P H [ n + 1 n + 2 p n+1(φ(z), φ(w))e 0 ] = n + 1 n + 2 p n+1(φ(z), φ(w))e 0,

17 OPERATORS ON THE BERGMAN SPACE 17 to obtain φ(b)e n = φ(b)[p n(φ(z), φ(w))e 0 ] (n + 1) e0 2 = n + 1 p n+1 (φ(z), φ(w))e 0 n + 1 = n + 2 (n + 1) e0 2 n + 2 E n+1. Clearly, φ(b) E 0 = 0. This implies that φ(b) M0 is unitarily equivalent to the Bergman shift. Suppose that M 1 is a reducing subspace of φ(b) φ(m) M1 is unitarily equivalent to the Bergman shift, i.e., there is an orthonormal basis {F n } of M 1 such that n + 1 φ(b)f n = n + 2 F n+1. Next we will show that M 1 = M 0. Observe P H [(φ(z) + φ(w))f 0 ] = 2φ(B)F 0 = 2 F 1. 2 Thus Since a simple calculation gives Thus we obtain because P H [(φ(z) + φ(w))f 0 ] 2 = 2. T φ(z) F 0 = φ(b) F 0 = 0, (φ(z) + φ(w))f 0 2 = (φ(z) + φ(w))f 0, (φ(z) + φ(w))f 0 = φ(z)f 0, φ(z)f 0 + φ(w)f 0, φ(w)f 0 = 2 F 0, F 0 = 2. + φ(z)f 0, φ(w)f 0 + φ(w)f 0, φ(z)f 0 P H [(φ(z) + φ(w))f 0 ] = 0 (φ(z) + φ(w))f 0 2 = P H [(φ(z) + φ(w))f 0 ] 2 + P H [(φ(z) + φ(w))f 0 ] 2. So p 1 (φ(z), φ(w))f 0 = (φ(z) + φ(w))f 0 is in H. Theorem 20 gives that F 0 = λe 0 for some constant λ. Thus M 0 is a subspace of M 1 but M 0 is a reducing subspace of φ(b) M1, which is unitarily equivalent to the Bergman shift. But the Bergman shift is irreducible. So we conclude that M 1 = M 0, to complete the proof. For φ(z) H (D), let S φ denote P H M φ H. Then U S φ U = M φ, where M φ is the multiplication operator on L 2 a(d). Indeed, for each g H anyz D, we have (U S φ g)(z) = (S φ g)(z, z) = (P H φg)(z, z) = (φg P H φg)(z, z) = φ(z)g(z, z) = (M φ U g)(z). The last equality follows from Lemma 8. This gives that U S φ = M φ U. Thus U S φ U = M φ. Theorem 25 tells us that for each finite Blaschke product φ, M φ has a unique the distinguished reducing subspace. The following theorem shows that only a multiplication operator by a finite Blaschke product has such property.

18 18 GUO, SUN, ZHENG AND ZHONG Theorem 26. Let φ H (D). Then M φ acting on L 2 a(d) has the distinguished reducing subspace iff φ is a finite Blaschke product. Proof. We only need to prove that if M φ has the distinguished reducing subspace, then φ is a finite Blaschke product. Now, assume M φ has the distinguished reducing subspace M such that M φ M is unitarily equivalent to the Bergman shift M z, that is, there exists a unitary operator U : M L 2 a(d) such that U M z U = M φ M. Let K M λ be the reproducing kernel of M for λ D. Clearly, K M λ 0, except for at most a countable set. Thus we have M φ K M λ, K M λ = φ(λ) K M λ 2 = M z UK M λ, UK M λ M z UK M λ 2 K M λ 2, to get that φ(λ) 1 except for at most a countable set. So φ 1. Since S φ acting on H = H 2 (T 2 ) [z w] is unitarily equivalent to M φ acting on L 2 a(d), this means that S φ, restricted on its corresponding reducing subspace N, is unitarily equivalent to M z acting on L 2 a(d), that is, there exists a unitary operator V : N L 2 a(d) such that V M z V = S φ N. Set e n = V e n, where e n = n + 1z n for n = 0, 1,. Then S φ e 0 = 0, hence M φ(z) e 0 = 0 M φ(w) e 0 = 0, where M φ(z) M φ(w) are the operators acting on H 2 (T 2 ). Noticing S φ(z) = S φ(w), we have to obtain Thus Since we have V S (φ(z)+φ(w)) e 0 2 = z + z 2 = 2, φ(z)e 0, φ(w)e 0 = M φ(w) e 0, M φ(z) e 0 = 0. (φ(z) + φ(w))e 0 2 = φ(z)e φ(w)e = V S (φ(z)+φ(w)) e 0 2 = V P H (φ(z) + φ(w))e 0 2 = P H (φ(z) + φ(w))e 0 2, (φ(z) + φ(w))e 0 H, to obtain φ(z) φ(w) e 0 = c z w for some constant c. This follows from Theorem 20. On the other h, (φ(z) + φ(w))e 0 2 = φ(z)e φ(w)e 0 2 = 2. As showed above, φ 1. We have that φ(z)e 0 2 = 1 to get T 2 ( φ(z) 2 1) e 0 2 dm 2 = 0. Thus φ(z) = 1 almost all on the unit circle so φ is an inner function. Lemma 24 gives that φ is a finite Blaschke product. This completes the proof. 5. Structure of minimal reducing subspaces In this section we will first show that every nontrivial minimal reducing subspace of φ(b) is orthogonal to the distinguished subspace M 0 if it is other than M 0. We will show the proof of Theorem 3 in the section. Theorem 27. Suppose that Ω is a nontrivial minimal reducing subspace for φ(b). If Ω does not equal M 0 then Ω is a subspace of M 0.

19 OPERATORS ON THE BERGMAN SPACE 19 Proof. By Lemma 16, there is a function e in Ω L 0 such that e = λe 0 + e 1 for some constant λ a function e 1 in M 0 L 0. By Theorem 18 p 1 (φ(z), φ(w))e + d 1 e Ω. Here d 1 e is the function constructed in Theorem 18. Let Since p 1 (φ(z), φ(w))e 0 is in H, we obtain E = φ(b) [φ(b)e] 1 2 e. Simple calculations give φ(b)e 0 = p 1(φ(z), φ(w))e 0. 2 E = φ(b) {φ(b)[λe 0 + e 1 ]]} 1 2 [λe 0 + e 1 ] = 1 2 φ(b) P H d 1 e 1. The sixth equality holds because that p 1 (φ(z), φ(w))e 1 + d 1 e 1 H. The eighth equality follows from that d 1 e 1 is in L φ. We claim that E 0. If this is not true, we would have This gives that P H d 1 e 1 is in L 0. And hence 1 2 φ(b) P H d 1 e 1 = 0. 0 = P H d 1 e 1, d 1 e 1 = d 1 e 1 2. This gives that d 1 e 1 = 0. Thus we obtain that p 1 (φ(z), φ(w))e 1 H. By Theorem 20, we get that e 1 is linearly dependent on e 0. This contradicts that e 1 M 0. By Theorem 18, P H d 1 e 1 is in M so is E = 1 2 φ(b) P H d 1 e 1. This implies that E is in Ω M 0. We conclude that Ω M 0 = Ω since Ω is minimal to complete the proof. Lemma 28. If M N are two mutually orthogonal reducing subspaces of φ(b), then M is orthogonal to Ñ. Proof. Let f = φ(z)l φ(w) k m lk g = φ(z)l φ(w) k n lk for finite numbers of elements m lk M n lk N. Then f, g = φ(z) l φ(w) k m lk, φ(z) l φ(w) k n lk = l 1,k 1 0 φ(z) l l1 φ(w) k k1 m lk, n l1k 1. Since M is orthogonal to N both M N are invariant subspaces of T φ(z) T φ(w), the above inner product f, g must be zero. Thus we conclude that M is orthogonal to Ñ to complete the proof. Proof of Theorem 2. Suppose that { 1,, } form a basis of M L 0. First we show span{ 1,, ; d 1,, d 1 } L 1 M. Note that { 1,, ; d 1,, d 1 } are contained in L 1 φ. It suffices to show { 1,, ; d 1,, d 1 } M. 1

20 20 GUO, SUN, ZHENG AND ZHONG Since M L 0 contains { 1,, }, for each l, k 0, φ(z) l φ k (w) i is in M for 1 i. Thus p 1 (φ(z), φ(w)) i is in M. By Theorem 18, we have So we have that d 1 i M, to obtain p 1 (φ(z), φ(w)) i Next we will show that { 1,, that for some constants λ i µ i, q Thus + d 1 i M. span{ 1,, ; d 1,, d 1 } L 1 M. i=1 q i=1 λ i i + λ i i ; d 1,, d 1 } are linearly independent. Suppose 1 q i=1 = µ i d 1 i q i=1 = 0. µ i d 1. i The right h side of the above equality is in L 0 but the left h side of the equality is orthogonal to L 0. So we have q λ i i = 0, i=1 q i=1 µ i d 1 i = 0. The first equality gives that λ i = 0 the second equality gives d 1 q i=1 µie(m) i = 0. Because M is orthogonal to M 0, by Theorem 20, we have q µ i i = 0. i=1 This gives that µ i = 0. Hence { 1,, ; d 1,, d 1 } are linearly independent. So far, 1 we have obtained diml M 2. To finish the proof, we need only to show that To do so, we consider the decomposition of H, Then diml M 2. H = M 0 M [M 0 M ], L 0 = [M 0 L 0 ] [M L 0 ] {[M 0 M ] L 0 }. dim{[m 0 M ] L 0 } = diml 0 dim[m 0 L 0 ] dim[m L 0 ] Letting N = [M 0 M ], Lemma 28 gives = N 1. K φ = M 0 M Ñ,

21 OPERATORS ON THE BERGMAN SPACE 21 L φ = L M 0 L M LÑ. Replacing M by N in the above argument gives By Theorem 15, so we have Hence This completes the proof. dim[l M ] dimlñ 2(N 1 ). 2N 1 = 1 + dim[l M ] + dim[lñ]. = 2N 2 dim[lñ] 2N 2 2(N 1 ) = 2. Lemma 29. Suppose that M, N, Ω are three distinct nontrivial minimal reducing subspaces of φ(b) such that Ω M N. If M, N, Ω are orthogonal to M 0, then M Ω = Ñ Ω = {0}. Proof. Since the intersection M Ω is also a reducing subspace of the pair of isometries T φ(z) T φ(w), the Wold decomposition of the pair of isometries on M Ω gives where L is the wering space given by M Ω M Ω = φ(z) l φ(w) k L M Ω, L M Ω = kert φ(z) T φ(w) M Ω = [kert φ(z) T φ(w) M] [kert φ(z) T φ(w) Ω] = L M L Ω. To prove that M Ω = {0}, it suffices to show L M L Ω = {0}. To do this, let q L M L Ω. By Theorem 2, there are functions e M, ẽ M M L 0 e Ω, ẽ Ω Ω L 0 such that The above two equalities give q = e M + d 1 ẽ M = e Ω + d 1 ẽ Ω. e M e Ω = d 1 ẽ M ẽ Ω. On the other h, d 1 ẽ M ẽ Ω is orthogonal to L 0. Thus d 1 ẽ M ẽ Ω = e M e Ω = 0. This gives e M = e Ω But e M is in M e Ω is in Ω hence both e M e Ω are zero. Since d 1 ẽ M ẽ Ω = 0, Theorem 20 implies that ẽ M ẽ Ω linearly depends on e 0. Since both M Ω are orthogonal to M 0, we have that ẽ M = ẽ Ω. Thus we obtain ẽ M = 0 to conclude that q = 0, as desired. So M Ω = {0}.

22 22 GUO, SUN, ZHENG AND ZHONG Similarly we obtain Ñ Ω = {0}. Lemma 30. Suppose that M, N, Ω are three distinct nontrivial minimal reducing subspaces of φ(b) such that Ω M N. If M, N, Ω are orthogonal to M 0, then P M L Ω = L M, PÑL Ω = LÑ, where P M denotes the orthogonal projection from H2 (T 2 ) onto M. Proof. Since M is orthogonal to N, Lemma 28 gives that M is orthogonal to Ñ We will show that P M L Ω = L M. Since Ω M N, we have Ω M Ñ. Ω L 0 [M L 0 ] [N L 0 ]. For each e (Ω) Ω L 0, there are two functions M L 0 e (N) N L 0 such that e (Ω) = + e (N) d 1 e (Ω) = d 1 + d 1 e (N). By Theorem 2, d 1 is in M d 1 e (N) is in Ñ. Since M, N, Ω are orthogonal to M 0, the above decompositions are unique. Thus P M e(ω) =, P M d1 e (Ω) = d 1. So for each f = e (Ω) + d 1 ẽ L Ω, where e (Ω) ẽ (Ω), we have (Ω) P M f = e(m) + d 1 ẽ (M) is in L M to obtain P M L Ω L. M To prove that P M L Ω = L, it suffices to show that M P M : L Ω L M is subjective. If this is not so, by Theorem 2, there are two functions e, ẽ M L 0 such that 0 e + d 1 ẽ is orthogonal to P M L Ω. Assume that {e 1,, e qω } are a basis of Ω L 0. Then P M L Ω = span{ 1,, q Ω ; d 1,, d 1 }. 1 q Ω If e 0, then e, i = 0, for 1 i q Ω. Thus 0 = e, i = e, i + e (N) i = e, e i, e, d 1 e i = 0,

23 OPERATORS ON THE BERGMAN SPACE 23 for each 1 i q Ω. So e is orthogonal to L Ω = span{e 1,, e qω ; d 1 e 1,, d 1 e qω }. Noting e is in L 0, we see that e is orthogonal to φ(z) l φ(w) k L Ω, for each l > 0 or k > 0. This gives that e is orthogonal to Ω hence orthogonal to Ω. Since e is in M, e must be orthogonal to the closure of P M Ω M, which is also a reducing subspace of φ(b). Therefore e is orthogonal to M, which is a contradiction. If e = 0, then d 1 ẽ 0 0 = d 1 ẽ, d 1 i = d 1 ẽ, P M d1 e i = d 1 ẽ, d 1 e i, d 1 ẽ, e i = 0, for each 1 i q Ω. This gives that d 1 ẽ is orthogonal to L Ω. But d 1 ẽ is also in L φ. We have that for any f L Ω, d 1 ẽ, φ(z) l φ(w) k f = 0, for l > 0 or k > 0. We have that d 1 ẽ is orthogonal to Ω hence orthogonal to Ω to obtain that P H d 1 ẽ is orthogonal to Ω. On the other h, by Theorem 18, P H d1 ẽ is in M. Thus P H d1 ẽ is orthogonal to the closure of P M Ω so P H d 1 ẽ must be zero because the closure of P M Ω equals M. Therefore, 0 = P H d 1 ẽ, p 1 (φ(z), φ(w))ẽ + d 1 ẽ = d 1 ẽ, p 1 (φ(z), φ(w))ẽ + d 1 ẽ = d 1 ẽ, d 1 ẽ = d 1 ẽ 2. The second equality follows from that p 1 (φ(z), φ(w))ẽ + d 1 ẽ is in H the third equality follows that d 1 ẽ is orthogonal to p 1(φ(z), φ(w))ẽ. This gives that d 1 ẽ = 0, which is a contradiction. We have obtained that P M : L Ω L is subjective hence M Similarly we obtain This completes the proof. Now we are ready to prove Theorem 3. Proof of Theorem 3. First we will show P M L Ω = L M. PÑL Ω = LÑ. P M = P H P M. Let N 1 denote the orthogonal complementary of M N in H. Write Lemma 28 gives For each function f in H 2 (T 2 ), write H = M N N 1. H = M Ñ Ñ1. f = f H f 2 = f M fñ fñ1 f 2, where f 2 is orthogonal to H, f H H, f M M, fñ Ñ, fñ1 Ñ1. Since M contains M, we write f M = f M f 3,

24 24 GUO, SUN, ZHENG AND ZHONG for two functions f M M f 3 M M. Thus f 3 is orthogonal to both Ñ Ñ1 hence orthogonal to both N N 1. So f 3 is orthogonal to This gives that P H f 3 = 0. We have to get P H P M f H = M N N 1. = P H f M = P H f M + P H f 3 = P H f M = f M, P M f = f M, P M = P H P M. Next we will show that P M is subjective from Ω onto M. For each q M, by Lemma 30, there are functions q lk L Ω such that q = φ(z) l φ(w) k m lk, q 2 = m lk 2 <, where m lk = P M q lk. Since L Ω L M are finite dimension spaces, there are two positive constants c 1 c 2 such that c 1 q lk m lk c 2 q lk. Define Thus q = φ(z) k φ(w) l q lk. q 2 = q lk 2 So we obtain that q is in Ω, q = φ(z) l φ(w) k q lk c 2 m lk 2 <. = φ(z) l φ(w) k [P M q lk + PÑqlk ] = φ(z) l φ(w) k m lk + φ(z) l φ(w) k [PÑqlk ] = q + q N, where q N = φ(z)k φ(w) l [PÑqlk ] is in Ñ. Hence P M q = q. We have to obtain P H P M q = P H q = q, P M q = P H P M q = q.