Similarity of analytic Toeplitz operators on the Bergman spaces

Transcription

1 Journal of Functional Analysis 258 (200) wwwelseviercom/locate/jfa Similarity of analytic Toeplitz operators on the Bergman spaces Chunlan Jiang a, echao Zheng b, a epartment of Mathematics, Hebei Normal University, Shijianzhuang 05006, PR China b epartment of Mathematics, Vanderbilt University, Nashville, TN 37240, USA Received 3 July 2009; accepted 6 September 2009 Available online 3 October 2009 Communicated by N Kalton Abstract In this paper we give a function theoretic similarity classification for Toeplitz operators on weighted Bergman spaces with symbol analytic on the closure of the unit disk 2009 Elsevier Inc All rights reserved Keywords: Bergman spaces; Toeplitz operator; Similarity; Blaschke product; Strongly irreducible Introduction Let da denote Lebesgue area measure on the unit disk, normalized so that the measure of equals For α>, the weighted Bergman space A 2 α is the space of analytic functions on which are square-integrable with respect to the measure da α (z) = (α + )( z 2 ) α da(z) For u L (,da),thetoeplitz operator T u with symbol u is the operator on A 2 α defined by T u f = P(uf); here P is the orthogonal projection from L 2 (,da α ) onto A 2 α T g is called to be the analytic Toeplitz operator if g H (the set of bounded analytic functions on ) In this case, T g is just the operator of multiplication by g on A 2 α In this paper we study the similarity of Toeplitz operators with symbol analytic on the closure of the unit disk on A 2 α OntheHardy The first author was partially supported by NSFC and the second author was partially supported by NSF * Corresponding author addresses: cljiang@hebtueducn (C Jiang), dechaozheng@vanderbiltedu ( Zheng) /$ see front matter 2009 Elsevier Inc All rights reserved doi:006/jjfa

2 2962 C Jiang, Zheng / Journal of Functional Analysis 258 (200) space, Cowen showed that two Toeplitz operators with symbol analytic on the closure of the unit disk are similar if and only if they are unitarily equivalent [3] However this is not true on the Bergman space In [7], Sun showed that for two functions f and g analytic on the closure of the unit disk, the analytic Toeplitz operator T f on the Bergman space A 2 0 is unitarily equivalent to T g if and only if there is an inner function χ of order one such that g = f χ Also in [8], Sun and Yu showed that if the direct sum of two analytic Toeplitz operators is unitarily equivalent to an analytic Toeplitz operator, then they must be constants So the Bergman space is rigid But in [2], Jiang and Li showed that if f is a finite Blaschke product, then the analytic Toeplitz operator T f is similar to the direct sum of finite copies of the Bergman shift T z on the unweighted Bergman space A 2 0 In this paper, we will completely determine when two Toeplitz operators with symbol analytic on the closure of the unit disk are similar on the weighted Bergman spaces in terms of symbols, which is analogous to the result on the Hardy space [3] While the Beurling theorem plays an important role on the Hardy space [3] and the Beurling theorem does not hold on the weighted Bergman spaces [8], we apply the general results [9 ] on similarity classification of the Cowen ouglas classes to analytic Toeplitz operators It was shown in [9,] that two strongly irreducible members of Cowen ouglas operator class B n (Ω) [4] are similar if and only if the respective commutant algebras have isomorphic K 0 groups and strongly irreducible decomposition operators give the similarity classification for Cowen ouglas operator classes As the adjoint of Toeplitz operators with symbol analytic on the closure of the unit disk is in the Cowen ouglas operator classes, our main ideas are to identify the commutant of analytic Toeplitz operators as the commutant of Toeplitz operators with some finite Blaschke products and to use the strongly irreducible decomposition of analytic Toeplitz operators and K 0 -groups of the commutants The following theorem is our main result It gives a function theoretic similarity classification of Toeplitz operators with symbol analytic on the closure of the unit disk Theorem Suppose that f and g are analytic on the closure of the unit disk T f is similar to T g on the weighted Bergman spaces A 2 α if and only if there are two finite Blaschke products B and B with the same order and a function h analytic on the closure of the unit disk such that f = h B and g = h B 2 Toeplitz operators with symbol as a finite Blashcke product A finite Blashcke product B is given by B = n k= z a k a k z for n numbers {a k } n k= in the unit disk and some positive integer n Heren is said to be the order of the Blaschke product B In this section we will show that two Toeplitz operators with symbols as finite Blaschke products are similar on the weighted Bergman spaces if and only if their symbols have the same order This was conjectured in [6] and proved in [2] on the unweighted Bergman space For another proof, see [7] Even for a very special Toeplitz operator T z n on the weighted Bergman space A 2 α, the following result was established in [4] recently

3 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Theorem 2 Let B be a Blaschke product with order n Then T B on the weighted Bergman space A 2 α is similar to the direct sum n j= T z on n j= A 2 α We define the composition operator C B on A 2 α by C B (f )(z) = f ( B(z) ) for f A 2 α Since B is a finite Blaschke product, the Nevanlinna counting function of B is equivalent to ( z 2 ) on the unit disk Thus C B is bounded below and hence has the closed range For each k, define M k = { } f(b) a k z : f A2 α Since T is invertible, M k is a closed subspace of A 2 α a k z Assume that the n-th order Blaschke B has n distinct zeros In [6], Stessin and Zhu showed that the Bergman space A 2 0 is spanned by {M,,M n } In [2], Jiang and Li showed that the Bergman space A 2 0 is the Banach direct sum M M 2 M n The following theorem extends Jiang and He s result to the weighted Bergman spaces, which immediately gives Theorem 2 Theorem 22 Let B be an n-th order Blaschke product with distinct zeros {a k } n k= in the unit disk Then the weighted Bergman space A 2 α is the Banach direct sum of M,,M n, ie, A α = M M 2 M n Before going to the proof of the above theorems we need the following simple lemma Lemma 23 Suppose that {a k } n k= following system are n distinct nonzero numbers in the unit disk Then the has only the trivial solution a 2 a 2 a a a 2 a 2 2 a a 3 a 2 a 3 a 3 a a 3 a 2 a 3 2 a a n a 2 a n a 3 a n a n a a n a 2 a n a 3 a n 2 c c 2 c 3 c n 0 0 = 0 0 (2) Proof Using row reductions and induction we obtain that the determinant of the coefficient matrix of system (2) equals

4 2964 C Jiang, Zheng / Journal of Functional Analysis 258 (200) a 2 a 2 a a a 2 a 2 2 a 3 a a 3 a 2 a a 3 a 2 a 3 a 3 2 a a n a 2 a n a 3 a n = a n a a n a 2 a n a 3 a n 2 a 2 a 2 a a (a a 2 ) a 2 (a a 2 ) a 3 (a a 2 ) ( a a 2 )( a 2 ) ( a 2 2 )( a 2 a ) a (a a 3 ) a 2 (a a 3 ) a 3 (a a 3 ) ( a a 3 )( a 2 ) ( a 2 a 3 )( a 2 a ) ( a 3 2 )( a 3 a ) a (a a n ) ( a a n )( a 2 ) a 2 (a a n ) ( a 2 a n )( a 2 a ) a 3 a ( a 3 a 2 )( a 3 a ) a 3 (a a n ) ( a 3 a n )( a 3 a ) a n a a n (a a 2 ) ( a n a 2 )( a n a ) a n (a a 3 ) ( a n a 3 )( a n a ) a n (a a n ) ( a n 2 )( a n a ) = = [ a a j a 2 a a j j> [ a a j a 2 a a j j> ] a a 2 a 3 ( a 3 a 2 ) a n ( a a 2 ) a ( a a 3 ) a ( a a n ) = ( ) n [ a a j 2 a 2 ( a a j ) 2 = ( ) n(n ) 2 j> nj= ( a j 2 ) ( a 2 2 ) a 2 ( a 2 a 3 ) a 2 ( a 2 a n ) a 3 ( a 3 2 ) a 3 ( a 3 a n ) ( a n a 2 ) a n ( a n a 3 ) a n ( a n 2 ) (a 0 a 2 ) (a a 3 ) ] ( a 2 2 )( a a 2 ) ( a 3 a 2 )( a a 2 ) (a 0 a 2 ) (a a 3 ) ( a 2 a 3 )( a a 3 ) ( a 3 2 )( a a 3 ) (a 0 a 2 ) (a a 3 ) ( a 2 a n )( a a n ) ( a 3 a n )( a a n ) ] a 2 2 a 2 a 3 a 3 a 2 a 3 2 a 2 a n a 3 a n n [ aj a k 2 ] ( a j a k ) 2 0 j= j<k a n a 2 a n a 3 a n 2 (a a n ) ( a n a 2 )( a a 2 ) (a a n ) ( a n a 3 )( a a 3 ) (a a n ) ( a n 2 )( a a n ) This gives that system (2) has only the trivial solution and hence c = c 2 ==c n = 0

5 C Jiang, Zheng / Journal of Functional Analysis 258 (200) From the above proof we immediately see a b a 2 b a b 2 a 2 b 2 a b 3 a 2 b 3 a 3 b a 3 b 2 a 3 b 3 a b n a 2 b n a 3 b n = ( ) n(n ) 2 nj= ( a j b j ) a n b a n b 2 a n b 3 a n b n n [ (aj a k )(b j b k ) j= j<k ( a j b k ) 2 ] (22) The above formula will be used in the proof of Theorem 22 Proof of Theorem 22 Without loss of generality we may assume that B has n distinct zeros {a k } n k= with a k 0 for each k We will show that the sum n k= M k is closed Suppose that {G m } is a Cauchy sequence in the sum n k= M k and converges to a function G in A 2 α There are functions f km in A 2 α such that G m (z) = f m(b(z)) a z + f 2m(B(z)) a 2 z ++ f nm(b(z)) a n z (23) To show that G is in n j= M j, we need only to show that for each j, {f jm } is Cauchy in A 2 α Since B is a finite Blaschke product, the Bochner theorem [2] gives that critical points of B in the closed unit disk are contained in a compact subset of the open unit disk So we may assume that for each point z on the unit circle, there is an open neighborhood U(z)of z such that n (local) inverses ρ j : U(z) ρ j (U(z)) of B on U(z) (we have to emphasize that {ρ j } n j= depend on U(z)) satisfying the following conditions: () B(ρ j (w)) = w for w U(z); (2) each ρ j is analytic on U(z); (3) ρ j (w) ρ k (w) for w in the closure of U(z) if j k; (4) each ρ j (w) does not vanish on the closure of U(z); and (5) ρ j (U(z) ) for j =,,n Substituting ρ j into both sides of equality (23) by ρ j and condition () give G m ( ρj (w) ) = f m(b(ρ j (w))) a ρ j (w) + f 2m(B(ρ j (w))) a 2 ρ j (w) ++ f nm(b(ρ j (w))) a n ρ j (w) = f m(w) a ρ j (w) + f 2m(w) a 2 ρ j (w) ++ f nm(w) a n ρ j (w)

6 2966 C Jiang, Zheng / Journal of Functional Analysis 258 (200) for w U(z) and each j =,,n We obtain the following system of equations: a ρ (w) a ρ 2 (w) a ρ 3 (w) a ρ n (w) a 2 ρ (w) a 2 ρ 2 (w) a 2 ρ 3 (w) a 2 ρ n (w) a 3 ρ (w) a 3 ρ 2 (w) a 3 ρ 3 (w) a 3 ρ n (w) a n ρ (w) a n ρ 2 (w) a n ρ 3 (w) a n ρ n (w) f m (w) G m (ρ (w)) f 2m (w) G f 3m (w) = m (ρ 2 (w)) G m (ρ 3 (w)) f nm (w) G m (ρ n (w)) Using Cramer s rule to solve the above system of equations, by conditions () and (5) and equality (22), we have that there are uniformly bounded functions F jk (w) on U(z) such that f jm (w) = n ( F jk (w)g m ρk (w) ) k= Therefore for some positive constants C and C z and any positive integers m and m, f jm (w) f jm (w) 2 da α U(z) C C [ n j= U(z) [ n CC z [ j= ρ j (U(z) ) ( G m ρj (w) ) ( G m ρj (w) ) ] 2 da α G m (λ) G m (λ) [ 2 ρ ρ j (ρ j (λ)) 2 Gm (w) G m (w) 2 da α ] The last inequality follows from condition (4) and This comes from condition () and the fact that since B(λ) 2 = ( λ 2)[ a 2 w 2 ρ j (w) 2 λ 2 B(λ) 2 a λ 2 + φa (λ) 2 a 2 2 a 2 λ 2 ++ ( n j= j (λ) 2 λ 2 ] α da α ] φaj (λ) ) ] 2 a n 2 a n λ 2

7 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Here φ a (z) is the Mobius transform φ a (z) = z a az Since the unit circle is compact, there are finitely many {U(z k )} l k= covering the unit circle Thus there is 0 <r< such that {w : r< w < } is contained in l k= U(z j ) and so for any integers m and m, r< w < f jm (w) f jm (w) 2 da α l k= U(z k ) ( lc )[ max C z i i l f jm (w) f jm (w) 2 da α G m (w) G m (w) 2 da α ] Since χ {r< w <} da α is a reversed Carleson measure for A 2 α, by [5], there is a positive constant C 2 such that for all f A 2 α, f(w) 2 daα C 2 f(w) 2 daα r< w < Thus for any integers m and m, f jm (w) f jm (w) 2 da α C 2 f jm (w) f jm (w) 2 da α r< w < ( )[ lcc 2 max C z i i l G m (w) G m (w) 2 da α ] This implies that for each j, {f jm } is Cauchy in A 2 α since {G m} is Cauchy in A 2 α Wemay assume that f jm converges to a function f j in A 2 α and hence converges pointwise to f j Taking the pointwise limit on both sides of (23) gives G(z) = f (B(z)) a z + f 2(B(z)) a 2 z ++ f n(b(z)) a n z Therefore the sum n k= M k is a closed subspace of A 2 α Next we will show that the sum n k= M k is a Banach direct sum on the weighted Bergman space A 2 α M M 2 M n

8 2968 C Jiang, Zheng / Journal of Functional Analysis 258 (200) To do so, let f k(b) a k z be functions in M k such that n k= f k (B) a k z = 0 for some functions f k = m=0 c km z m in the weighted Bergman space A 2 α On the other hand, easy calculations give n k= f k (B) n a k z = a k z = k= n m=0 k= c km B m m=0 c km a k z Bm Taking the inner product of both sides of the above equality with each reproducing kernel k aj of the weighted Bergman space A 2 α at a j and noting that the powers of B vanish at each a j,we have the following system of equations a 2 a 2 a a a 2 a 2 2 a a 3 a 2 a 3 a 3 a a 3 a 2 a 3 2 a a n a 2 a n a 3 a n a n a a n a 2 a n a 3 a n 2 c 0 c 20 c 30 c n0 0 0 = 0 0 Lemma 23 gives that the above system has only the trivial solution and hence Let c 0 = c 20 ==c n0 = 0 g k = c kj z j j= Then {g k } n k= are functions in A2 α such that f k = zg k and 0 = = = B n k= n k= f k (B) a k z Bg k (B) a k z n k= g k (B) a k z

9 C Jiang, Zheng / Journal of Functional Analysis 258 (200) So we obtain n k= g k (B) a k z = 0 Repeating the above argument gives c = c 2 ==c n = 0 By the induction we have that c ij = 0 for any i, j to get f k (B) a k z = 0 This implies that 0 in the sum n k= M k has the unique decomposition 0 = n {}}{ Therefore the sum n j= M k is a Banach direct sum M M 2 M n Next we will show that M M 2 M n is dense in the weighted Bergman space A 2 α and hence equals A 2 α Suppose that f is a function in the weighted Bergman space A 2 α but orthogonal to each M k Then f(z) da α = 0 za k for each k We claim f(z) ( za k ) m da α = 0 for each k and all m 0 Assuming the claim, we have f(z) ( z 2) α da = 0 (24) ( za k ) m for all m This implies that the Bergman projection P [f(z)( z 2 ) α ] of the function f(z)( z 2 ) α into the unweighted Bergman space A 2 0 also equals zero at a k and all derivatives of P [f(z)( z 2 ) α ] equal zero at a k and so P [ f(z) ( z 2) α] = 0

10 2970 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Here P is the Bergman projection from L 2 (,da) onto the unweighted Bergman space A 2 0 Taking derivatives of P [f(z)( z 2 ) α ] and evaluating at 0 give f(z) ( z 2) α z m da = 0 for all m 0 Thus f(z) ( z 2) α p(z)da = 0 for any polynomials p(z) and hence f(z) 2( z 2 ) α da = 0 as polynomials are dense in A 2 α This gives that f equals 0 We are going to prove our claim by induction We warn the reader that although having (24) for one point a k is enough, one needs it for all a k for the induction step Clearly, our claim is true for m = Assume that the claim is true for m K, ie, f(z) ( za k ) m da α = 0 for each k and all m KForm = K +, we note that for each k, the rational function a BK k z can be written as a sum of partial fractions with the highest power of ( a k z) is K + and the highest power of the rest term ( a j z) is K for j k Thus the term is a linear combination of terms { ( a j z) m } j=,,n; m=,,k and a BK k z So f(z) ( za k ) K+ da α = 0 ( a k z) K+ This gives the claim For two bounded operators S and T, we say that T is similar to S if there is an invertible operator X such that T = XSX We use T S to denote that T is similar to S Now we turn to the proof of Theorem 2 Proof of Theorem 2 Since for most λ in the unit disk, is the n-th order Blaschke product λb with n distinct zeros in the unit disk, without loss of generality we may assume that B has n B λ

11 C Jiang, Zheng / Journal of Functional Analysis 258 (200) distinct zeros {a k } n k= with a k 0 for each k Now we use the same notation as in the above proof First we will show that on each M k, T B is similar to T z on the weighted Bergman space To do this, let Y k = T a k z C B : A 2 α M k Clearly, Y k is a bounded invertible operator from A 2 α onto M k Moreover, Y k T z f = Y k (zf ) Bf (B) = a k z ( ) f(b) = T B a k z = T B Y k f for each f A 2 α By Theorem 22, the weighed Bergman spaces Thus we conclude that A 2 α = M M 2 M n n T B T z k= 3 Some notation and lemmas In this section we introduce some notation and give a few lemmas which will be used in the proof of our main result in the last section Let {T f } denote the commutant of T f,thesetof bounded operators commuting with T f on the Bergman space L 2 a We need the following theorem to study the similarity of analytic Toeplitz operators The version of the following theorem on the Hardy space was obtained in [20] For more general results on the Hardy space, see [2] and [9] The method in [20] also works on the weighted Bergman spaces For the details of the proof of the following theorem, see the proof of theorem in [20, pp ] by replacing the reproducing kernel on the Hardy space by one on weighted Bergman spaces A 2 α Theorem 3 If f is analytic on the closure of the unit disk, then there are a finite Blaschke product B and a function h analytic on the closure of unit disk such that and f = h B {T f } ={T B } Let T be a bounded operator on a Hilbert space H An idempotent Q in the commutant {T } is said to be minimal if for every idempotent R in {T }, Q = R, whenever Ran R Ran QHere Ran R denotes the range of the operator R

12 2972 C Jiang, Zheng / Journal of Functional Analysis 258 (200) A bounded operator T on a Hilbert space is said to be strongly irreducible if there is no nontrivial idempotent operator in its commutant In the other words, T is strongly irreducible if and only if XT X is irreducible for each invertible operator X Lemma 32 If Q is a minimal idempotent in {T }, then T Ran Q is strongly irreducible Proof Let A = T Ran Q LetR be an idempotent in {A} In other words, and We write H as a Banach direct sum and extend R to H by AR = RA, Ran R Ran Q H = Ran Q Ran(I Q) R(x + y) = Rx if x Ran Q and y Ran(I Q) Clearly, R is an idempotent For each x in Ran Q, TRx= T Ran Q Rx = ARx = RAx = RT Ran Q x = RT x For y Ran(I Q), write y = (I Q)z and then and This gives RTy = RT (I Q)z = R(I Q)T z = 0 T Ry = 0 T Ry = RTy

13 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Thus we obtain RT = T R So R is in {T } Since Q is a minimal idempotent in {T }, we conclude that R = Q and hence R = Q Lemma 33 If f is analytic on the closure of the unit disk and B is a Blaschke product B with order n such that {T f } ={T B }, then there is a function h bounded and analytic on the unit disk such that and T h is strongly irreducible T f n T h, {T h } ={T z } j= Proof First we get a decomposition of T f as a Banach direct sum of strongly irreducible operators By Theorem 2, there is an invertible operator X from A 2 α to n j= A 2 α such that XT B X = n T z j= Let P i be the projection from n j= A 2 α onto the i-th component Clearly, [ n ] T z = T z Ran Pi j= Since {T z } ={T g : g H } is isomorphic to the Banach algebra H and H does not contain any nontrivial idempotents, T z is strongly irreducible Thus P i is a minimal idempotent in the commutant { n j= T z } LetQ i = X P i XSo{Q i } n i= are n minimal idempotents in {T f } with This follows from the hypothesis: n Q j = I j= {T f } ={T B } and the fact that {T B } contains exactly n minimal idempotents

14 2974 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Let S i = T f Ran Qi and T i = XS i X Wehave XT f X = X = n S i X i= n T i The above direct sum follows from the fact that {P i } n i= are orthogonal projections on n j= A 2 α Since we have i= { n } T z = { ( T F : F M n H )}, j= T i T z = T z T i It is well known that there are functions f,,f n in H such that T i = T fi for each i This gives the decomposition of T f as we desired: on n j= A 2 α Next we need to show that T f T f2 0 0 XT f X = 0 0 T f T fn The above equalities follow from f = f 2 ==f n T f = T f2 ==T fn To simplify the proof, we are going to show only that T f = T f2 since the same argument gives that T fi = T fj

15 C Jiang, Zheng / Journal of Functional Analysis 258 (200) To do so, let 0 I 0 0 I V = 0 0 I I on n j= A 2 α Clearly, V is in { n j= T z } Easy computations give and 0 I 0 0 I I I T f T f T f3 0 = T fn T f I T f2 0 0 I T f I 0 = T fn I Since we have [ n ] [ n V T fj = T fj ]V, j= j= 0 T f2 0 0 T f T f T fn 0 T f 0 0 T f T f T fn to obtain T f = T f2 XT f X = n j T h where h = f = f 2 ==f n Moreover, since P j is a minimal idempotent, by Lemma 32, T h is strongly irreducible Finally we show that {T h } ={T z }

16 2976 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Letting A be in {T h }, we define Thus U is in the commutant A A 0 0 U = 0 0 A A { n } T h = { XT f X } and so A is an analytic Toeplitz operator Hence j {T h } = { T g : g H } ={T z } The decomposition of T f in the above lemma is the so-called strongly irreducible decomposition of T f In fact the decomposition for the above Toeplitz operator is unique up to the similarity To state the result in [] and [3], we need some notation Let B be a Banach algebra and Proj(B) be the set of all idempotents in B Murray von Neumann equivalence a is introduced in Proj(B) Lete and ẽ be in Proj(B) We say that e a ẽ if there are two elements x,y B such that xy = e, yx =ẽ Let Proj(B) denote the equivalence classes of Proj(B) under Murray von Neumann equivalence a Let M (B) = M n (B) n= where M n (B) is the algebra of n n matrices with entries in BLet (B) = Proj ( M (B) ) The direct sum of two matrices gives a natural addition in M (B) and hence induces an addition + in Proj(M (B)) by [p]+[q]=[p q] where [p] denotes the equivalence class of p ( (B), +) forms a semigroup and depends on B only up to stable isomorphism K 0 (B) is the Grothendieck group of (B) Let A be a bounded operator on a Hilbert space H WeuseH (n) to denote the direct sum of n copies of H and A (n) to denote the direct sum of n copies of A acting on H (n)

17 C Jiang, Zheng / Journal of Functional Analysis 258 (200) Let T be a bounded operator on a Hilbert space H and Q ={Q j : j l} be a family of minimal idempotents such that and l Q j = I j= Q j Q i = 0 if j i We say that Q is a strongly irreducible decomposition of T In fact, let T j = T Ran Qj Then each T j is strongly irreducible and T is similar to l Tj j= Under similarity, {T,,T n } is classified as equivalence classes {[T j ],,[T jk ]}Letn i be the number of elements in [T ji ] Then T is similar to k i= T (n i ) j i If for any two strongly irreducible decompositions Q ={Q j : j l } and Q 2 = { Q j : j l 2 } of T, we have that l = l 2 and there are a permutation π of {, 2,,l } and invertible bounded operators X j from Ran Q j onto Ran Q π(j) such that X j T Ran Qj = T Ran Q π(j) X j, then we say that T has unique strongly irreducible decomposition up to similarity We need the following theorem [] and [3] Theorem 34 Let T be a bounded operator on a Hilbert space H The following are equivalent: (a) T is similar to k i= A (n i ) i under the decomposition of the space H = k i= H (n i ) i, where k and n i are finite, A i is strongly irreducible and A i is not similar to A j if i j Moreover T (n) has a unique strongly irreducible decomposition up to similarity (b) The semigroup ({T } ) is isomorphic to the semigroup N (k) where N ={0,, 2,} and the isomorphism φ sends

18 2978 C Jiang, Zheng / Journal of Functional Analysis 258 (200) [I] n e + n 2 e 2 ++n k e k where {e i } k i= are the generators of N(k) and n i 0 Remark The above theorem immediately gives that if T T (n ) T (n 2), both T and T 2 are strongly irreducible and ({T } ) is isomorphic to the semigroup N, then T is similar to T 2 4 Proof of main result In this section we give the proof of Theorem For a Fredholm operator T on a Hilbert space, we use index T to denote the Fredholm index: index T = dim ker T dim ker T Clearly, the product formula of the Fredholm index [5] gives if T = S (k) and both T and S are Fredholm index T = k index S Proof of Theorem Suppose that there are two finite Blaschke products B and B with the same order n and a function h analytic on the closure of the unit disk such that f = h B, g = h B By Theorem 2, T f is similar to T h(z n ) and T g is similar to T h(z n ) Thus T f is similar to T g Conversely, suppose that T f is similar to T g Since f and g are analytic functions on the closure of the unit disk, by Theorem 3, there are two finite Blaschke products B and B and two functions h and h analytic on the closure of the unit disk such that f = h B, g = h B, and {T f } ={T B }, {T g } ={T B } Let n be the order of B and n the order of B By Lemma 33, T f has the following strongly irreducible decomposition T f n j= T h = T (n) h

19 C Jiang, Zheng / Journal of Functional Analysis 258 (200) and T g has the following strongly irreducible decomposition First we show that n T g T h = T (n ) h j= To do so, let T = T f T g Then T is similar to T h T h T (n) h T (n ) h and hence it is similar to T (2n) h This means that there is an invertible operator Y such that This implies YTY = T (2n) h Y {T } Y = { T (2n) } { ( h = TF : F M 2n H )} Thus the Banach algebra {T } is isomorphic to M 2n (H ) By Theorem 6 in [3, p 203] or Lemma 29 in [, p 248], we have ( M2n ( H )) = ( H ) = N So ( {T } ) = ( M2n ( H )) = N The first equality follows from the fact that the semigroup (B) is invariant for Banach algebra isomorphisms By the remark after Theorem 34, we get that T h is similar to T h Next we show that n = n Since we just proved that T h is similar to T h and T f is similar to T g,wehave T f T (n ) h Noting that T f λ and T h λ are Fredholm with nonzero index for some λ, by the Fredholm index product formula, we have and index T f λ = index [ T (n) h λ] = n index Th λ

20 2980 C Jiang, Zheng / Journal of Functional Analysis 258 (200) index T f λ = index [ T (n ) h λ ] = n index T h λ to get that n = n Thus B and B have the same order Finally we need to show that there is a Mobius transform χ such that h = h χ Noting that {T h } ={T z } and {T h } ={T z },ast h is similar to T h, we have that there is an invertible operator Z such that and Thus By the fact that we have ZT h Z = T h Z{T h } Z ={T h } Z{T z } Z ={T z } {T z } = { T G : G H }, Z { T G : G H } Z = { T G : G H } In particular, there is a function χ in H such that ZT z Z = T χ The spectral picture of T z forces χ to be a Mobius transform Since h is analytic on the closure of the unit disk, we can write h as with for some r> Thus h = a n z n n=0 a n r n < j=0

21 in the norm topology So Since ZT h Z = T h,wehave to obtain C Jiang, Zheng / Journal of Functional Analysis 258 (200) n=0 ZT h Z = h = = = a n T n z = T h ( a n ZTz Z ) n n=0 a n (T χ ) n n=0 a n T χ n n=0 = T n=0 a n χ n a n χ n = h χ, n=0 f = h B, g = h B = h (χ B) = h B where B = χ B is a Blaschke product with order n Acknowledgment We thank the referee for his many useful suggestions and pointing out Theorem 22 which is contained in the proof of Theorem 2 in the first version of this paper References [] Y Cao, J Fang, C Jiang, K-groups of Banach algebras and strongly irreducible decompositions of operators, J Operator Theory 48 (2002) [2] C Cowen, The commutant of an analytic Toeplitz operator, Trans Amer Math Soc 239 (978) 3 [3] C Cowen, On equivalence of Toeplitz operators, J Operator Theory 7 (982) [4] M Cowen, R ouglas, Complex geometry and operator theory, Acta Math 4 (978) [5] R ouglas, Banach Algebra Techniques in Operator Theory, Academic Press, New York, London, 972 [6] R ouglas, Operator theory and complex geometry, arxiv:070880v [mathfa] [7] K Guo, H Huang, On multiplication operators of the Bergman space: Similarity, unitary equivalence and reducing subspaces, J Operator Theory, in press [8] H Hedenmalm, B Korenblum, K Zhu, Theory of Bergman Spaces, Grad Texts in Math, vol 99, Springer-Verlag, New York, 2000 [9] C Jiang, Similarity classification of Cowen ouglas operators, Canad J Math 56 (2004) [0] C Jiang, X Guo, K Ji, K-group and similarity classification of operators, J Funct Anal 225 (2005) 67 92

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