Essentially normal Hilbert modules and K- homology III: Homogenous quotient modules of Hardy modules on the bidisk

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1 Science in China Series A: Mathematics Mar., 2007, Vol. 50, No. 3, Essentially normal Hilbert modules and K- homology III: Homogenous quotient modules of Hardy modules on the bidisk Kun-yu GUO & Peng-hui WANG School of Mathematics, Fudan University, Shanghai , China ( Abstract In this paper, we study the homogenous quotient modules of the Hardy module on the bidisk. The essential normality of the homogenous quotient modules is completely characterized. We also describe the essential spectrum for a general quotient module. The paper also considers K- homology invariant defined in the case of the homogenous quotient modules on the bidisk. Keywords: MSC(2000): Hardy modules, bidisk, essential normality, K-homology. 47A13, 47A20, 46H25, 46C99 1 Introduction Let T = (T 1,..., T n ) be a tuple of commuting operators acting on a Hilbert space H. Using Douglas-Paulsen s Hilbert module language [1], we endow H with a C[z 1,..., z n ]-module structure by p x = p(t 1,..., T n )x, p C[z 1,..., z n ], x H. In Arveson s language [2 4], a Hilbert module is called essentially normal if the self-commutators Tk T j T j Tk of its canonical operators are all compact (one can also refer to such Hilbert modules as essentially reductive, see [1, 5, 6]). Much work has been done along this line [1 11]. In the course of Arveson s studies [2 4], he considered the essential normality of quotients of standard Hilbert modules. For a d-shift Hilbert module with finite-multiplicity, he established a p-essential normality in case the submodule is generated by monomials [2]. That result on monomial submodules was generalized by Douglas to the case in which the d-shift is replaced by more general weighted shifts [5]. In the dimension d = 2, Guo [7] obtained trace estimates, which implies that the homogeneous quotient modules of 2-shift Hilbert module are essentially normal. In the recent work of Guo and Wang [8], these results were generalized to cases in which the 2-shift is replaced by U-invariant Hilbert modules with finite-multiplicity over 2-dimensional unit ball. In [9], the essential normality and the a K-homology of quasi-homogeneous quotient modules over a 2-dimensional unit ball were characterized by a hard analysis, and the method Received July 3, 2006; accepted November 21, 2006 DOI: /s Corresponding author This work is partially supported by the National Natural Science Foundation of China (Grant No ), the Young Teacher Fund, the National Key Basic Research Project of China (Grant No. 2006CB805905) and the Specialized Research for the Doctoral Program

2 388 Kun-yu GUO & Peng-hui WANG is different from the case of homogeneous quotient modules. The paper will be devoted to studying the essential normality of quotient modules of the Hardy module over the bidisk. At this point it is completely different from the Hilbert modules on the unit ball. It is well known that neither the Hardy module H 2 (D 2 ) nor its nonzero submodules are essentially normal. Let H 2 (D 2 ) be the Hardy module on the bidisk with the module action defined by multiplication of coordinate functions. The structure of submodules of H 2 (D 2 ) is much more complex than that of submodules of H 2 (D), and for more work on this subject, see [1, 6, 12 19] and references therein. Let M be a submodule of H 2 (D 2 ), set N = H 2 (D 2 ) M. Then N can be endowed with a C[z, w]-module structure by p f = p(s z, S w )f, p C[z, w], f N, where S z = P N M z N and S w = P N M w N. Some efforts have been done to describe the structure of some certain quotient modules over the bidisk, see [20 23]. For any nonzero submodule M of H 2 (D 2 ), using the fact that the coordinate functions define a pair of isometries, both of infinite multiplicity, M is not essentially normal. However, in [24], Douglas and Misra showed that some quotient modules of H 2 (D 2 ) are essentially normal and some are not. Some other related work has been done in [11, 25]. If a quotient module N of H 2 (D 2 ) is essentially normal, then the C -algebra C (N) generated by {S z, S w } is essentially commutative. A result of Yang [18,Theorem 4.4] implies that for any quotient module N of H 2 (D 2 ), the C -algebra C (N) is irreducible. Hence, we have a C - extension 0 K C (N) C(σ e (S z, S w )) 0. This C -extension gives an element of K 1 (σ e (S z, S w )), which is an invariant for the Hilbert modules [1,26,27]. In this paper, we mainly concern the homogenous quotient modules. Let I be an ideal of C[z, w], and [I] be the submodule of H 2 (D 2 ) generated by I. If I is homogeneous, then the submodule [I] also is homogeneous, and in this case, the quotient module H 2 (D 2 )/[I] is homogenous. Since the polynomial ring is Noetherian [28], the ideal I is generated by finitely many polynomials. This implies that I has a greatest common divisor p, and so, I can be uniquely written as I = pl, which is called the Beurling form of I (cf. [29]). In dimension d = 2, by a lemma of Yang [15], dim C[z, w]/l < and hence [p] [I] is of finite dimension, where [p] is the submodule of H 2 (D 2 ) generated by p. This means that H 2 (D 2 )/[p] and H 2 (D 2 )/[I] have the same essential normality. Notice if I is homogeneous, then both p and L are homogeneous. This paper is organized as follows. In sec. 2, we will consider the compactness of the commutators [P M, M z ] and [P M, M w ] for homogenous submodules M. For a homogenous submodule M = [p], both [P M, M z ] and [P M, M w ] are compact if and only if Z(p) D 2 T 2. In sec. 3, we will develop some properties of the asymptotic orthogonality. It plays an important role in the paper. Sec. 4 and sec. 5 are devoted to characterizing the essential normality of homogenous quotient modules of H 2 (D 2 ). Let p be a nonzero homogenous polynomial in C[z, w], then p

3 Homogenous quotient modules of Hardy modules on the bidisk 389 can be factorized as p(z, w) = (α i z β i w) ni. Set p 1 (z, w) = (α i z β i w) ni, p 2 (z, w) = (α i z β i w) ni. α i = β i α i β i Then p = p 1 p 2, (1.1) with Z(p 1 ) D 2 T 2 and Z(p 2 ) D 2 (D T) (T D). We state our main theorem as follows: Theorem 1.1. Let p be a nonzero homogenous polynomial in C[z, w], and p = p 1 p 2 as in (1.1), then the quotient module H 2 (D 2 )/[p] is essentially normal if and only if p 2 has one of the following forms : (1) p 2 = c, with c 0; (2) p 2 = αz + βw, with α β ; (3) p 2 = c(z αw)(w βz), with α < 1, β < 1 and c 0. In sec. 6, we will describe the essential spectrum of quotient modules. For any quotient module N of H 2 (D 2 ), the essential spectrum σ e (N) is defined by σ e (S z, S w ). It is shown that for a homogenous polynomial p, σ e (H 2 (D 2 )/[p]) = Z(p) D 2. Therefore, if a homogenous quotient module H 2 (D 2 )/[p] is essentially normal, then we have a C -extension 0 K C (S z, S w ) C(Z(p) D 2 ) 0. It is shown that this extension yields a nontrivial K-homology element in K 1 (Z(p) D 2 ). Moreover, we also describe the essential spectrum of a general quotient module. Using the result in this paper, one can construct a quotient module N of H 2 (D 2 ), such that σ e (N) = D 2. 2 Compactness of [P M, M z ] and [P M, M w ] Let M be a submodule of H 2 (D 2 ), and N = H 2 (D 2 ) M be the corresponding quotient module. The quotient module N is called essentially normal, if both [S z, Sz ] and [S w, Sw ] are compact, where for two operators A and B, the operator [A, B] = AB BA is the commutator of A and B. If both S z and S w are essentially normal, then by Fuglede-Putnam Theorem, the commutator [S z, S w] is compact. Since N is an invariant subspace of M z and M w, the quotient module N is essentially normal if and only if both M z N and M w N are essentially normal. In [24], Douglas and Misra established by a direct calculation that the quotient module H 2 (D 2 ) [(z w) 2 ] is essentially normal and H 2 (D 2 ) [z 2 ] is not. They also showed that H 2 (D 2 ) [z n w m ] is essentially normal, and this result was generalized by Clark in [25]. Recently, Izuchi and Yang [11] showed that if ϕ(w) H 2 (D) is an inner function, then H 2 (D 2 ) [z ϕ(w)] is essentially normal if and only if ϕ is a finite Blaschke product. In this section, the homogenous quotient modules H 2 (D 2 )/[p] with will be concerned. Z(p) D 2 T 2 Let p be a homogenous polynomial, and M = [p] be the submodule of H 2 (D 2 ) generated by p. We will prove that both [P M, M z ] and [P M, M w ] are compact if and only if Z(p) D 2 T 2, where P M denotes the orthogonal projection from H 2 (D 2 ) onto M.

4 390 Kun-yu GUO & Peng-hui WANG As a corollary, we will prove that if Z(p) D 2 T 2, then the quotient module H 2 (D 2 )/[p] is essentially normal. At first, we need some results on the evaluation operator and the difference quotient operator, which are studied by Yang [16,17,19]. The evaluation operators L(0) and R(0) at z = 0 and w = 0 respectively are defined by L(0)f(z, w) = f(0, w), R(0)f(z, w) = f(z, 0), f H 2 (D 2 ). In fact, L(0) = 1 M z M z and R(0) = 1 M w M w are projections. Let M be a submodule of H 2 (D 2 ), and N = H 2 (D 2 ) M be the corresponding quotient module. Write then we can rewrite D z = P M M z N, D w = P M M w N, D z = P M M z (1 P M ) = [P M, M z ], D w = P M M w (1 P M ) = [P M, M w ]. The operators D z and D w are given by the difference quotient operators in [19], D zf(z, w) = f(z, w) f(0, w), D z wf(z, w) = f(z, w) f(z, 0). w The following proposition comes from [16], which gives the relationship between the evaluation operators and the difference quotient operators. Proposition 2.1 quotient module, then If M is a submodule of H 2 (D 2 ), let N = H 2 (D 2 ) M be the corresponding (1) S z S z + D zd z = I N, S ws w + D wd w = I N ; (2) S z S z + P NL(0)P N N = I N, S w S w + P NR(0)P N N = I N. Let M be a submodule of H 2 (D 2 ), and N = H 2 (D 2 ) M. Proposition 2.1 implies that, if both [P M, M z ] and L(0)P N are compact, then S z is essentially normal. Similarly, if both [P M, M w ] and R(0)P N are compact, then S w is essentially normal. We have the following Lemma 2.2 Let p be a homogeneous polynomial with the factorization n p(z, w) = λ (w α i z), with α i 0. Set M = [p] and N = H 2 (D 2 ) M, then [P M, M z ] is compact if and only if L(0)P N is compact. Similarly, the commutator [P M, M w ] is compact if and only if R(0)P N is compact. Proof. p as Notice that Firstly, we will prove that both ker S z and ker S z p(z, w) = λ n (w α i z) = λw n + zp 1 (z, w). ker S z = ker M z [p] = H 2 w [p], are of finite dimension. We rewrite where H 2 w = span{wn, n = 0, 1, 2,...}. If f ker S z, then f = k=0 a kw k and f [p]. For any m 0, λa m+n = λa m+n + M z f, w m p 1 = f, λw m+n + f, zw m p 1 = f, w m p = 0.

5 Homogenous quotient modules of Hardy modules on the bidisk 391 It follows that dim(ker Sz ) n + 1. Now we will show that ker S z = {0}. Notice that ker S z = {f f [p], zf [p]}. (2.1) For any f ker S z, let f = i=0 F i(z, w) be the homogeneous expansion of f. Since p is homogeneous and zf [p], zf i [p]. By the fact that F i is a polynomial, it follows that there is a polynomial q such that zf i = pq. Noting that GCD(z, p) = 1, this means that q/z is a polynomial, and hence F i = pq/z [p]. So f [p]. By (2.1), f = 0. Therefore ker S z = 0. By Proposition 2.1, if one of [P M, M z ] and L(0)P N is compact, then S z is Fredholm, and hence Sz is the essential inverse of S z. This insures that the other of [P M, M z ] and L(0)P N is compact. The same reason gives the case of [P M, M w ] and R(0)P N. We state the main theorem in this section as follows. Theorem 2.3. Let p be a homogenous polynomial in C[z, w]. Set M = [p], then both [P M, M z ] and [P M, M w ] are compact if and only if Z(p) D 2 T 2. Combining Theorem 2.3, Lemma 2.2 with Proposition 2.1, we have the next corollary. Corollary 2.4. Let p be a homogenous polynomial in C[z, w] satisfying Z(p) D 2 T 2, then the quotient module H 2 (D 2 ) [p] is essentially normal. Proof of the necessity of Theorem 2.3. one of [P M, M z ] and [P M, M w ] is not compact. We will prove that if Z(p) D 2 T 2, then at least As mentioned in the introduction, for any homogenous polynomial p, p can be factorized as p = λz k w l n (w α i z), with α i 0. Below, the necessity of Theorem 2.3 will be proved in two cases. Case 1. Both k and l are zero. Since Z(p) D 2 T 2, there is an i 0 such that α i0 1. Set β = α i0. Suppose now that 0 < β < 1. We rewrite p as p = (z βw)p, where p is a homogenous polynomial. Set N 1 = [z βw] and N 2 = [z βw] [p], then M = N 1 N 2. Since {1} {e j = ( βz) j +( βz) j 1 w+ +w j β 2j + β 2j } j=1 is an orthonormal basis of N 1, L(0)P N1 e j = 1 β 2j + β 2j wj. Then lim j L(0)P N1 e j = 1 β 2 0. It follows that L(0)P N1 is not compact, and hence L(0)P M is not compact. Lemma 2.2 implies that [P M, M z ] is compact if and only if L(0)P M is compact. Thus [P M, M z ] is not compact. The same reason implies that if β > 1, then [P M, M w ] is not compact. Case 2. At least one of k and l is not zero. We suppose now k 0, then p = zp, where p is a homogenous polynomial. It follows that M = [z] ([z] M), and hence {w j } j=1 M. Let m = deg p, then p = m i=0 a iz i w m i. Now taking a i0 0, we claim that P M M z i 0 M is not compact.

6 392 Kun-yu GUO & Peng-hui WANG Since p is homogenous, {f j = w j p/ p } j=1 is an orthonormal set of M. By Bessel s inequality, for sufficiently large j, P M M z i 0 M w j 2 = P M z i0 w j 2 ( ) z i0 w j 2, f i f i = z i0 w j, f j m+i0 2 = a i0 2 / p 2 0. Since w j converges to zero weakly, [P M, M z i 0 ] = P M M z i 0 M is not compact. Since {P M } e = {A B(H 2 (D 2 )) [P M, A] is compact} is a C -algebra, the compactness of [P M, M z ] implies that [P M, M z i 0 ] is compact. And hence [P M, M z ] is not compact. The same reason implies that if l 0, then [P M, M w ] is not compact, thus completing the proof. The proof of the sufficiency of Theorem 2.3 is long. The remaining part of this section will be devoted to the proof of the sufficiency of Theorem 2.3. Let H n = {f n C[z, w] f n is homogenous with degree n}, then H 2 (D 2 ) can be decomposed as H 2 (D 2 ) = n=0 H n. It is easy to see that dim H n = n + 1. Given a homogenous polynomial p of degree m, then for sufficiently large n, and hence dim([p] H n ) = m. dim([p] H n ) = n m + 1, To prove the sufficiency of Theorem 2.3, we need the next lemma. In the case n = 1, the next lemma is considered in [16]. Lemma 2.5. For n 1, let M 0 = [(z w) n ], then both [P M0, M z ] and [P M0, M w ] are compact. Proof. then Let N k i Set N 0 = H 2 (D 2 ) M 0. For i = 1, 2,..., n, set N i = [(z w) i 1 ] [(z w) i ], N 0 = N 1 N n. = N i H k, k = 0, 1, 2,..., then N i has a homogeneous decomposition N i = Ni k. k=0 It is easy to see that for sufficiently large k, N k i is of dimension 1. Let e k i (z, w) be in N k i with e k i = 1. Define a unitary operator V : H 2 (D 2 ) H 2 (D 2 ) by V f(z, w) = f(w, z), f H 2 (D 2 ). Obviously, every N i is a reduced subspace of V, and hence every Ni k is a reduced subspace of V. It follows that there are complex numbers λ (k) i with λ (k) i = 1, such that e k i (w, z) = V e k i (z, w) = λ (k) i e k i (z, w).

7 Homogenous quotient modules of Hardy modules on the bidisk 393 This implies that z k, e k i = V z k, V e k i = w k, λ (k) i e k i = w k, e k i. (2.2) Below, we will prove that L(0)P Ni L(0) is compact for i = 1,..., n, by induction. {1, zk +z k 1 w+ +w k k+1, k = 1, 2,...} is an orthonormal basis of N 1, Since L(0)P N1 L(0)w k = wk k + 1, and this insures that L(0)P N1 L(0) is compact. Suppose now that L(0)P Ni L(0) is compact for all i < m. Since for sufficiently large k, L(0)P Ni L(0)w k = L(0)P N k i L(0)w k = w k, e k i ek i, wk w k, i = 1,..., m 1, we have lim k w k, e k i 2 = 0 for i = 1,..., m 1, and hence by (2.2) lim k m 1 z k, e k i e k i, w k m 1 lim k w k, e k i 2 = 0. (2.3) By [16, Lemma 4.1.2], for any quotient module N, the operator L(0)P N R(0) is Hilbert-Schmidt. Now take N = H 2 (D 2 ) [(z w) m ], then N = m N i. Since P Ni z k = P N k i z k = z k, e k i ek i, m m L(0)P N R(0)z k = L(0)P Ni R(0)z k = z k, e k i e k i, w k w k. Let L(0)P N R(0) H.S denote the Hilbert-Schmidt norm of L(0)P N R(0), then m 2 z k, e k i e k i, w k = L(0)P N R(0)z k 2 L(0)P N R(0) 2 H.s < +. Thus k=m By (2.3) and (2.4), we have It follows that k=m lim k m z k, e k i ek i, wk = 0. (2.4) lim k zk, e k m ek m, wk = 0. L(0)P Nm L(0)w k = w k, e k m ek m, wk = z k, e k m ek m, wk 0. So L(0)P Nm L(0) is compact, and hence for 1 i n, L(0)P Ni L(0) is compact. This insures that ( n L(0)P N0 L(0) = L(0) P Ni )L(0) is compact, and hence L(0)P N0 is compact. By Lemma 2.2, [P M0, M z ] is compact. The same reason implies that [P M0, M w ] is compact. Let p i C[z, w], i = 1, 2, setting r = GCD(p 1, p 2 ), the greatest common divisor of p 1 and p 2, then p 1 = r p 1 and p 2 = r p 2. If Z(r) D 2 =, then by [12, Proposition ], [r] = H 2 (D 2 ), and hence [p i ] = [ p i ]. To prove the sufficiency of Theorem 2.3, we still need several lemmas.

8 394 Kun-yu GUO & Peng-hui WANG Lemma 2.6. Given p 1, p 2 C[z, w], if Z(p 1 ) Z(p 2 ) D 2 =, then [p 1 ] + [p 2 ] = [p 1 p 2 ]. Proof. At first, we claim that [p 1 ] + [p 2 ] is closed. Since Z(p 1 ) Z(p 2 ) D 2 =, as mentioned above, we may assume GCD(p 1, p 2 ) = 1. By [15, Lemma 6.1], it follows that [p 1 ] [p 2 ] = ([p 1 ] + [p 2 ]) is of finite dimension. Let M = [p 1 ] ([p 1 ] [p 2 ] ) and N = [p 2 ] ([p 1 ] [p 2 ] ), then [p 1 ] + [p 2 ] is closed if and only if M + N is closed. Now suppose that M + N is not closed, then we can take {f n f n = 1} n=1 M and {g n g n = 1} n=1 N, such that both f n and g n converge weakly to 0, and f n + g n tends to 0 as n. Since both Mp 1 f n = 0 and M p 2 g n = 0, we have lim ( (M p 1 Mp 1 + M p2 Mp 2 )f n, f n + (M p1 Mp 1 + M p2 Mp 2 )g n, g n ) = lim ( M p 2 M p 2 f n, f n + M p1 Mp 1 g n, g n + M p1 Mp 1 f n, f n + g n + M p1 Mp 1 g n, f n + M p2 Mp 2 g n, f n + g n + M p2 Mp 2 f n, g n ) = lim (M p 1 M p 1 + M p2 M p 2 )(f n + g n ), f n + g n lim M p 1 M p 1 + M p2 M p 2 f n + g n 2 = 0. Since Z(p 1 ) Z(p 2 ) D 2 =, ker(m p1 M p 1 + M p2 M 2 ) = [p 1] [p 2 ] is of finite dimension. By [30, Theorem 4.3] and [30, Lemma 4.5], the operator M p1 Mp 1 + M p2 Mp 2 has a closed range, and hence it is a positive Fredholm operator. This implies that there exist a positive invertible operator A and a compact operator K, such that M p1 M p 1 + M p2 M p 2 = A + K. Since both f n and g n converge weakly to zero as n, we have lim Kf n = 0 and lim Kg n = 0. Since A is positive and invertible, there are some c i > 0, i = 1, 2, such that and lim (M p 1 M p 1 + M p2 Mp 2 )f n, f n = lim (A + K)f n, f n lim c 1 f n 2 = c 1, lim (M p 1 M p 1 + M p2 M p 2 )g n, g n = lim (A + K)g n, g n lim c 2 g n 2 = c 2.

9 Homogenous quotient modules of Hardy modules on the bidisk 395 This contradiction implies that M + N is closed, and hence [p 1 ] + [p 2 ] is closed, the claim is proved. Using the characteristic space theory for polynomials in [12, Chapter 2], we have [p 1 ] [p 2 ] = [p 1 p 2 ], and hence [p 1 ] + [p 2 ] = ([p 1 ] [p 2 ]) = [p 1 p 2 ], thus completing the proof. Lemma 2.7. Let p i C[z, w], i = 1, 2, such that Z(p 1 ) Z(p 2 ) D 2 =. Write P i = P [pi] and set M = [p 1 p 2 ]. If both [P i, M z ] and [P i, M w ] are compact for i = 1, 2, then so are [P M, M z ] and [P M, M w ]. Proof. Since both [P i, M z ] and [P i, M w ] are compact for i = 1, 2, a simple reason shows that for any polynomial p, the commutator [P i, M p ] are compact for i = 1, 2. Since Z(p 1 ) Z(p 2 ) D 2 =, by Lemma 2.6, M = [p 1 ] + [p 2 ]. It is easy to check that Mp 1 M [p 2 ], and hence Mp 1 P M = P2 Mp 1 P M, where P2 = I P 2. Moreover, notice that M = [p 1 p 2 ] [p 2 ], then for any polynomial p, we have P M M p M p1 Mp 1 P M = P M (P 2 M p M p1 P2 )M p 1 P M = P M [P 2, M pp1 ]Mp 1 P M. This implies that the operator P M M p M p1 Mp 1 P M is compact. Similarly, the operator P M M p M p2 M p 2 P M is compact. Therefore, the operator is compact. P M M p (M p1 M p 1 + M p2 M p 2 )P M Since Z(p 1 ) Z(p 2 ) D 2 =, the same argument as in the proof of Lemma 2.6 implies that the operator M p1 Mp 1 + M p2 Mp 2 is a positive Fredholm operator, and hence there exist a finite rank operator F and a positive invertible operator X, such that M p1 Mp 1 + M p2 Mp 2 = X + F. Hence P M M p XP M = P M M p (M p1 M p 1 + M p2 M p 2 )P M P M M p F P M (2.5) is compact. Considering the matrix representation of X, X = X 11 X 12 X12 X 22 M M, and taking p = 1 in the left side of (2.5), we have that X 12 = P M XP M X 22 is Fredholm. is compact. Hence Noting that M is an invariant subspace, the operator M p has a matrix representation ( ) Sp D p M, and hence 0 R p M M p X = S px 11 + D p X12 S p X 12 + D p X 22 R p X12 R p X 22 M M.

10 396 Kun-yu GUO & Peng-hui WANG By (2.5) again, S p X 12 + D p X 22 is compact. Since X 12 is compact, the operator D p X 22 is compact. The Fredholmness of X 22 implies that D p is compact. Now taking p = z and p = w, we have that both D z and D w are compact, that is, both [P M, M z ] and [P M, M w ] are compact, as desired. Proof of the sufficiency of Theorem 2.3. The hypothesis Z(p) D 2 T 2 implies that p can be factorized as m p(z, w) = λ (z α i w) ni, with α i = 1 and λ 0. Without loss of generality, we may assume λ = 1. Below, we will prove the theorem by induction. When m = 1 and α = 1, define a unitary operator U : H 2 (D 2 ) H 2 (D 2 ) by then Uf(z, w) = f(z, αw), for f H 2 (D 2 ), M z U = UM z, αm w U = UM w. Let M 0 = [(z w) n ] and M = [(z αw) n ], then P M0 = U P M U, P M 0 = U P M U. Notice that P M M z P M = UP M0 U M z UP M 0 U = UP M0 M z P M 0 U = U[P M0, M z ]U. Lemma 2.5 implies that [P M0, M z ] is compact, and hence [P M, M z ] = P M M z P M Similarly, [P M, M w ] is compact. Now we suppose that the theorem holds for m = k, that is, for is compact. k p k (z, w) = (z α i w) ni, with α i = 1 both [P k, M z ] and [P k, M w ] are compact, where P k = P [pk ]. For any β = 1 and β α i, i = 1,..., k, setting p k+1 = (z βw) n k+1 p k, write P k+1 = P [pk+1 ]. It suffices to show that both [P k+1, M z ] and [P k+1, M w ] are compact. Since β α i, Z(z βw) Z(p k ) = {0}, and hence by Lemma 2.7, both [P k+1, M z ] and [P k+1, M w ] are compact. This completes the proof. 3 Asymptotic orthogonality In this section, we will develop some properties of the asymptotic orthogonality for closed subspaces. It plays an important role in this paper.

11 Homogenous quotient modules of Hardy modules on the bidisk 397 Let H be a Hilbert space, N 1 and N 2 be two closed subspaces of H. The subspaces N 1 and N 2 are called asymptotically orthogonal, denoted by N 1 N 2 if P N1 P N2 is compact. Proposition 3.1. N 1 N 2, then N 1 + N 2 is closed. Let H be a Hilbert space, N 1 and N 2 be two closed subspaces of H. If Proof. Suppose N 1 + N 2 is not closed. Then there are {f n f n = 1} n=1 N 1 and {g n g n = 1} n=1 N 2 satisfying f n w 0 and gn w 0, such that fn + g n tends to zero as n. Since P N2 P N1 is compact, and since f n converges weakly to zero, lim P N 2 f n = lim P N 2 P N1 f n = 0. Similarly, lim P N 1 g n = 0. Now, on the one hand, lim (P N 2 + P N1 )(f n + g n ), (f n + g n ) lim P N 2 + P N1 f n + g n 2 = 0, on the other hand, lim (P N 2 + P N1 )(f n + g n ), (f n + g n ) = lim [ P N 1 f n, f n + P N2 g n, g n + P N1 g n, f n + g n + P N2 f n, f n + g n + P N1 f n, g n + P N2 g n, f n ] = lim ( f n 2 + g n 2 ) = 2. This contradiction shows that N 1 + N 2 is closed. The closeness of N 1 + N 2 is not enough to insure N 1 N 2. The following proposition helps understanding the asymptotic orthogonality. Proposition 3.2. Let H be a Hilbert space, N 1 and N 2 be two closed subspaces of H. Then N 1 and N 2 are asymptotically orthogonal if and only if N 1 + N 2 is closed and P N1+N 2 (P N1 + P N2 ) is compact. Proof. Suppose first that N 1 + N 2 is closed and P N1+N 2 (P N1 + P N2 ) is compact. Since P N1 P N2 + P N2 P N1 = (P N1 + P N2 ) 2 (P N1 + P N2 ) = (P N1 + P N2 ) 2 P N1+N 2 + P N1+N 2 (P N1 + P N2 ) = [(P N1 + P N2 P N1+N 2 )(P N1 + P N2 + P N1+N 2 )] +[P N1+N 2 (P N1 + P N2 )], we have that P N1 P N2 + P N2 P N1 is compact. It follows that P N1 P N2 P N1 P N2 = 1 2 P N 1 (P N1 P N2 + P N2 P N1 )P N1 P N2 is compact. Since (P N2 P N1 P N2 ) (P N2 P N1 P N2 ) = P N2 P N1 P N2 P N1 P N2, and hence P N2 P N1 P N2 is compact. From the equality P N2 P N1 P N2 = (P N1 P N2 ) (P N1 P N2 ),

12 398 Kun-yu GUO & Peng-hui WANG we see that P N1 P N2 is compact. Suppose now that P N1 P N2 is compact. By Proposition 3.1, N 1 + N 2 is closed. Notice that the compactness of P N1 P N2 implies N 1 N 2 being of finite dimension, and hence we can assume without loss of generality that N 1 N 2 = {0}. Now, we claim that Q = P N1 + P N2, viewed as an operator on N 1 + N 2, is Fredholm. At first, for f 1 N 1, f 2 N 2, if Q(f 1 + f 2 ) = 0, then by N 1 N 2 = {0}, f 1 + P N1 f 2 = 0 and f 2 + P N2 f 1 = 0, and hence f 2 = P N2 P N1 f 2. Since N 1 N 2 = {0}, we have f 2 = 0. Thus f 1 = 0. Therefore ker Q = {0}. To prove that Q is Fredholm, it suffices to show that Q has a closed range. Let N 1 N 2 = {(f 1, f 2 ) f i N i, (f 1, f 2 ) = f 1 + f 2 }, then N 1 N 2 is a Banach space. Define an operator T : N 1 N 2 N 1 +N 2 by T (f 1, f 2 ) = f 1 +f 2, then T is bounded and invertible. It follows that there are c 0 > 0 and c 1 > 0 such that c 0 f 1 + f 2 f 1 + f 2 c 1 f 1 + f 2, for f 1 N 1, f 2 N 2. Notice that P N1 P N2 P N1 can be regarded as an operator on N 1, and it is compact. By the Fredholm Alternative Theorem [31,Theorem 5.22], for any ε > 0, there is a finitely codimensional subspace N1 ε of N 1 such that, for any f N1 ε, P N2 f = (P N1 P N2 P N1 ) 1 2 f < ε f. Similarly, there is a finitely codimensional subspace N2 ε of N 2, such that for any f N2 ε, P N1 f = (P N2 P N1 P N2 ) 1 2 f < ε f. Hence for f 1 N1 ε and f 2 N2 ε, (P N1 + P N2 )(f 1 + f 2 ) f 1 + f 2 ( P N1 f 2 + P N2 f 1 ) = f 1 + f 2 ( (P N2 P N1 P N2 ) 1 2 f2 + (P N1 P N2 P N1 ) 1 2 f1 ) f 1 + f 2 ε( f 1 + f 2 ) c 0 f 1 + f 2 εc 1 f 1 + f 2 = (c 0 εc 1 ) f 1 + f 2. We can take ε to be enough small, such that c 0 εc 1 > 0. Now, fix ε, since Ni ε is a finitely codimensional subspace of N i, the space N ε = N1 ε + N2 ε is a finitely codimensional subspace of N 1 + N 2. The above reason implies that Q N ε is bounded below, and hence Q has a closed range. Therefore, the operator Q is Fredholm, and this means that 0 / σ e (Q). Now, since P N1 P N2 is compact, Q 2 Q = (P N1 + P N2 ) 2 (P N1 + P N2 ) = P N1 P N2 + P N2 P N1 is compact, and hence σ e (Q) {0, 1}. It follows that σ e (Q) = {1}. Since Q is self-adjoint, there is an exact sequence, 0 K C (Q) + K π C 0,

13 Homogenous quotient modules of Hardy modules on the bidisk 399 where K is the ideal of all compact operators on N 1 +N 2 and C (Q) is the C -algebra generated by Q and the identity operator I on N 1 + N 2. Since Q is Fredholm, π(q) 0. Notice that π(q) 2 = π(q), and we have π(q) = 1. It follows that I Q is compact. This insures that P N1+N 2 (P N1 + P N2 ) is compact, as desired. Let H be an infinitely dimensional Hilbert space, T B(H). Suppose N 1 and N 2 are invariant subspaces of T satisfying N 1 N 2, by Proposition 3.1, N = N 1 + N 2 is closed. It is easy to see that N is an invariant subspace of T. Let T 1 = T N1 and T 2 = T N2 be the restrictions of T to N 1 and N 2 respectively. The following theorem maybe is familiar to some readers, we sketch it here for convenience. Theorem 3.3 Under the above assumption, if both T 1 and T 2 are essentially normal, then T N, the restriction of T to N, is essentially normal. Proof. At first, we consider the matrix of T N relative to the decomposition N = N 1 (N N 1 ) to obtain T N = P N 1 T P N1 P N1 T P N N1 P N N1 T P N1 P N N1 T P N N1 Since N 1 is an invariant subspace of T, we have T P N1 = P N1 T P N1, and hence P N N1 T P N1 = P N N1 P N1 T P N1 = 0. Since N 2 is also an invariant subspace of T, we have T P N2 = P N2 T P N2, and hence. P N1 T P N N1 = P N1 T (P N N1 P N2 ) + P N1 T P N2 = P N1 T (P N N1 P N2 ) + P N1 P N2 T P N2. Since N 1 N 2, by Proposition 3.2, both P N N1 P N2 and P N1 P N2 are compact, and hence P N1 T P N N1 is compact. A simple reason implies that T N is essentially normal if and only if both P N1 T P N1 and P N N1 T P N N1 are essentially normal. Since T 1 is essentially normal, P N1 T P N1 = T 1 is essentially normal. Now noticing that P N N1 P N2 is compact, we have P N N1 T P N N1 = P N2 T P N2 + a compact operator. Since P N2 T P N2 = T 2 is essentially normal, this implies that P N N1 T P N N1 is essentially normal, and hence T N is essentially normal, thus completing the proof. Now, let us turn to quotient modules of H 2 (D 2 ). Let M 1, M 2 be two submodules of H 2 (D 2 ), and N 1 = H 2 (D 2 ) M 1, N 2 = H 2 (D 2 ) M 2 be the corresponding quotient modules. If N 1 N 2, then by Proposition 3.1, N = N 1 + N 2 is closed, and hence N = H 2 (D 2 ) (M 1 M 2 ). As an immediate application of Theorem 3.3, we have Corollary 3.4. Under the above assumption, if both N 1 and N 2 are essentially normal, then N is essentially normal. 4 The sufficiency of Theorem 1.1 In this section, we use some results of the asymptotic orthogonality to prove the sufficiency of Theorem 1.1. To this end, we still need several facts. Proposition 4.1. Let p 1 and p 2 be two homogenous polynomials in C[z, w], setting N 1 = H 2 (D 2 ) [p 1 ] and N 2 = H 2 (D 2 ) [p 2 ], then the followings are equivalent :

14 400 Kun-yu GUO & Peng-hui WANG (i) N 1 N 2, (ii) for any f n N 1 H n, g n N 2 H n satisfying f n = 1 and g n = 1, lim f n, g n = 0, (iii) for f n N 1 H n with f n = 1, there exist q n [p 2 ] H n such that lim f n q n = 0. Proof. (i)= (iii): Since H 2 (D 2 ) = N 2 [p 2 ], there are f n (1) N 2 H n and f n (2) [p 2 ] H n such that f n = f n (1) + f n (2). Since P N2 P N1 is compact and f n converges weakly to zero, lim f n f (2) n = lim (1) f n = lim P N 2 P N1 f n = 0. (iii)= (ii): By (iii), for any f n N 1 H n with f n = 1, there is q n [p 2 ] H n such that lim f n q n = 0. It follows that for any g n N 2 H n with g n = 1, we have lim f n, g n lim ( f n q n, g n + q n, g n ) lim f n q n g n = 0. (ii)= (i): Set k 1 = deg p 1 and k 2 = deg p 2. Then, as mentioned in sec. 2, there is a positive integer N, such that for any n N, dim(n 1 H n ) = k 1 and dim(n 2 H n ) = k 2. Let {e i n }k1 and {f j n }k2 j=1 be orthonormal bases of N 1 H n and N 2 H n respectively, then there is a finite rank operator F, so that ( P N1 P N2 = F + = F + k 1 n=n ( k1 n=n e i n e i n )( k 2 m=n j=1 k 2 ) fn, j e i n e i n fn j. j=1 f j m f j m Since k 1 k2 j=1 f n j, ei n ei n f n j can be viewed as an operator from N 2 H n to N 1 H n, we have ( k1 k 2 ) ( k1 k 2 ) fn, j e i n e i n fn j = fn, j e i n e i n fn j. n=n j=1 Since for any i, j, lim f j n, e i n = 0, this gives lim n=n j=1 k 1 k 2 fn, j e i n e i n fn j = 0. j=1 Therefore, P N1 P N2 is compact. From Proposition 4.1, the next corollary follows easily. Corollary 4.2. Let α 1 < 1 and α 2 < 1, then [z α 1 w] [w α 2 z]. )

15 Homogenous quotient modules of Hardy modules on the bidisk 401 Proof. and Let e k = (ᾱ 1z) k + (ᾱ 1 z) k 1 w + w k α1 2k + α 1 2k , f k = (ᾱ 2w) k + (ᾱ 2 w) k 1 z + z k α2 2k + α 2 2k , then e k [z α 1 w] H k and f k [w α 2 z] H k, with e k = 1 and f k = 1. Note that dim([z α 1 w] H k ) = 1 and dim([w α 2 z] H k ) = 1. It is easy to verify lim e k, f k = 0, by Proposition 4.1, Proposition 4.3. [z α 1 w] [w α 2 z]. Let p i, i = 1, 2, 3, be the homogenous polynomials in C[z, w] satisfying GCD(p 2, p 3 ) = 1. If both [p 1 ] [p 2 ] and [p 1 ] [p 3 ], then [p 1 ] [p 2 p 3 ]. Proof. Since p 2 and p 3 are homogenous and GCD(p 2, p 3 ) = 1, Z(p 2 ) Z(p 3 ) = {0}. By Lemma 2.6, [p 2 p 3 ] = [p 2 ] + [p 3 ]. Below, we will prove [p 1 ] ([p 2 ] + [p 3 ] ). By [15, Lemma 6.1], [p 2 ] [p 3 ] is of finite dimension, and hence for sufficiently large n, ([p 2 ] H n ) ([p 3 ] H n ) = {0}. Using the argument in the proof of Lemma 2.6, there are constants c i > 0, i = 1, 2, such that for sufficiently large n and f n [p 2 ] H n, g n [p 3 ] H n, f n c 1 f n + g n and g n c 2 f n + g n. Suppose now f n +g n = 1, then f n c 1 and g n c 2. Since [p 1 ] [p 2 ] and [p 1 ] [p 3 ], by Proposition 4.1, for any h n [p 1 ] H n satisfying h n = 1, lim (f n + g n ), h n = lim f n, h n + lim g n, h n = 0. By Proposition 4.1 again, [p 1 ] ([p 2 ] + [p 3 ] ), thus completing the proof. Lemma 4.4. For α 1 and β = 1, the quotient modules [z αw] [(z βw) n ]. Proof. Without loss of generality, assume α < 1. Setting α = α β, we claim that [z α w] [(z w) n ]. Since α < 1, it follows from Corollary 4.2 that [z] [w ᾱ z]. Now, let N i = [(z w) i 1 ] [(z w) i ], and e k i N i H n with e k i = 1. The argument in the proof of Lemma 2.5 implies that for i = 1, 2,..., n, lim k w k, e k i = 0. By Proposition 4.1, P [z] P N is i k compact, and hence P [z] P [(z w) n ] is compact, which means [z] [(z w) n ]. Since α < 1, GCD((ᾱ z w), (z w) n ) = 1, and hence by Proposition 4.3, the quotient modules [z]

16 402 Kun-yu GUO & Peng-hui WANG [(ᾱ z w)(z w) n ]. By Proposition 4.1, it follows that there are polynomials p k H k n, such that lim k wk+1 (ᾱ z w)(z w) n p k = 0. Now, setting f k 1 = k 1 i=0 (ᾱ z) i w k 1 i, we have (ᾱ z) k w k = (ᾱ z w)f k 1, and hence 0 lim (ᾱ z w)f k (ᾱ z w)(z w) n p k lim (ᾱ z) k+1 + lim wk+1 (ᾱ z w)(z w) n p k = 0. Since for any f H 2 (D 2 ), (ᾱ z w)f(z, w) 2 = (ᾱ z w)f(z, w) 2 dm 2 (1 α ) 2 f 2, T 2 we have 0 lim k f k (z w) n p k 1 1 α Noticing that lim k f k = (1 α 2 ) 1 2, we have lim k (ᾱ z w)f k (ᾱ z w)(z w) n p k = 0. lim f k (z w) n p k / f k = 0. k Taking q k = p k f k, then q k [(z w) n ]. Since f k f k [z α w] H k, by Proposition 4.1, the quotient modules [z α w] [(z w) n ]. The claim is proved. For β = 1, define a unitary operator U : H 2 (D 2 ) H 2 (D 2 ) by Then it is easy to see that Uf(z, w) = f(z, βw) for f H 2 (D 2 ). [z αw] = U[z α w], [(z βw) n ] = U[(z w) n ]. This implies that [z αw] [(z βw) n ], thus completing the proof. For a homogenous polynomial p satisfying Z(p) D 2 T 2, p can be factorized as p(z, w) = λ m (z α i w) ni, with α i = 1 and λ 0. Combining Lemma 4.4 with Proposition 4.3, we have the following Lemma 4.5. Let p be a homogenous polynomial in C[z, w] satisfying then for α 1, [z αw] [p]. Z(p) D 2 T 2, With the above preparations, we are able to prove the sufficiency of Theorem 1.1. As mentioned in the introduction, let p be a nonzero homogenous polynomial in C[z, w], then p can be factorized as p = p 1 p 2,

17 Homogenous quotient modules of Hardy modules on the bidisk 403 with Z(p 1 ) D 2 T 2 and Z(p 2 ) D 2 (D T) (T D). We restate the sufficiency of Theorem 1.1 as follows. Theorem 4.6. Let p be a nonzero homogenous polynomial in C[z, w], and p = p 1 p 2 as in (1.1), if p 2 has one of the following forms: (1) p 2 = c, with c 0; (2) p 2 = αz + βw, with α β ; (3) p 2 = c(z αw)(w βz), with α < 1, β < 1 and c 0, then the quotient module H 2 (D 2 )/[p] is essentially normal. Proof. (1) Suppose p 2 = c, then [p] = [p 1 ]. Corollary 2.4 implies that the quotient module H 2 (D 2 )/[p] is essentially normal. (2) Suppose p 2 = αz + βw, with α β. Without loss of generality, we may assume that α 0. Notice that Z(p 2 ) Z(p 1 ) D 2 =. By Lemma 2.6, [p] = [p 1 ] + [p 2 ]. Since α 0 and α β, [p 2 ] = [ z β α w ], β with 1. α Notice that Z(p 1 ) D 2 T 2. Lemma 4.5 insures [p 2 ] [p 1 ]. To show that [p] is essentially normal, by Corollary 3.4, it suffices to show that both [p 1 ] and [p 2 ] are essentially normal. Since Z(p 1 ) D 2 T 2, the essential normality of [p 1 ] comes from Corollary 2.4. Now, set α = β α, then [p 2] = [z α w]. It follows that e k = (ᾱ z) k + (ᾱ z) k 1 w + + w k α 2k + α 2k is an orthonormal basis of [p 2 ]. So S z e k = ze k, e k+1 e k+1, and hence it is easy to verify that S z is essentially normal. Similarly, S w is essentially normal. This implies that [p 2 ] is essentially normal. (3) Suppose p 2 = c(z αw)(w βz), with α < 1, β < 1 and c 0, then [p] = [(z αw)(w βz)p 1 ]. Since by Lemma 2.6, we have Z(z αw) Z((w βz)p 1 ) = {0}, [p] = [z αw] + [(w βz)p 1 ]. As done in proof (2), the same reason shows that both [z αw] and [(w βz)p 1 ] are essentially normal. To show that [p] is essentially normal, by Corollary 3.4, it suffices to show [z αw] [(w βz)p 1 ]. Since α < 1, β < 1, Corollary 4.2 insures [z αw] [w βz]. Since α < 1 and Z(p 1 ) D 2 T 2, Lemma 4.5 implies [z αw] [p 1 ].

18 404 Kun-yu GUO & Peng-hui WANG Notice that Z(w βz) Z(p 1 ) D 2 =, then by Proposition 4.3, [z αw] [(w βz)p 1 ], as desired, thus completing the proof. 5 The necessity of Theorem 1.1 In this section, we will prove the necessity of Theorem 1.1. In fact, we will prove the following result, from which the necessity of Theorem 1.1 easily follows. Theorem 5.1. Let M be a submodule of H 2 (D 2 ), and α i < 1, i = 1, 2, then the quotient module [(z α 1 w)(z α 2 w)m ] is not essentially normal. Similarly, the quotient module [(w α 1 z)(w α 2 z)m ] is not essentially normal. Let p be a homogenous polynomial. Factorizing p = p 1 p 2 as in Theorem 1.1, if p 2 has not one of the forms in Theorem 1.1, then p can be factorized as p = (z α 1 w)(z α 2 w)p, or p = (w α 1 z)(w α 2 z)p, with α i < 1, i = 1, 2. Now taking M = [p ] in Theorem 5.1, the necessity of Theorem 1.1 immediately comes from Theorem 5.1. Let H be an infinitely dimensional Hilbert space, and T be a bounded linear operator on H. For an invariant space M of T, T has a matrix representation, T = P M T M P M T M P M T M M M. To prove Theorem 5.1, the following lemma is needed, whose proof is similar to the proofs of [3, Theorem 1] and [5, Theorem 4.3]. Lemma 5.2. Under the above assumption, if P M T M is essentially normal, then the followings are equivalent: (1) T is essentially normal, (2) P M T M is compact, and P M T M is essentially normal. To continue, the following lemma is needed. Lemma 5.3. Given α i < 1, i = 1, 2, write M = [(z α 1 w)(z α 2 w)] and N = H 2 (D 2 ) M. Then the quotient module N is not essentially normal. Proof. We will show that S z is not essentially normal. Let V 1 = [z α 1 w] and V 2 = [z α 1 w] M, then N = V 1 V 2. For any f V 1 and g V 2, it is easy to see Sz f, g = f, zg = 0, that is, V 1 is an invariant subspace of Sz. It follows that S z has the matrix representation, S z = A z B z C z V 1 V 2.

19 Homogenous quotient modules of Hardy modules on the bidisk 405 Since V 1 = [z αw], Theorem 4.6 implies that A z = P V1 M z V1 is essentially normal. To show that S z is not essentially normal, by Lemma 5.2, it suffices to show that B z is not compact. Notice that for n 2, dim V i H n = 1, i = 1, 2. Let e n = (ᾱ 1z) n + (ᾱ 1 z) n 1 w + + w n α1 2n + α 1 2(n 1) + + 1, then e n V 1 H n and e n = 1. For g n V 2 H n satisfying g n = 1, we have B z e n = ze n, g n+1 g n+1. Hence B z is compact if and only if ze n, g n+1 0 as n. We will show that B z is not compact in the following three cases. Case 1. α 1 α 2. Since Z(z α 1 w) Z(z α 2 w) = {0}, by Lemma 2.4, N = M = [z α 1 w] + [z α 2 w]. Set and e n = (ᾱ 1z) n + (ᾱ 1 z) n 1 w + + w n α1 2n + α 1 2(n 1) + + 1, f n = (ᾱ 2z) n + (ᾱ 2 z) n 1 w + + w n α2 2n + α 2 2(n 1) Since α 1 α 2, by the Cauchy inequality, e n, f n n α 1 α 2 j/( n n ) 1 α 1 2j α 2 2j 2 < 1. j=0 It follows that for n 2, {e n, f n } is a linear basis of N H n. orthogonalization, j=0 j=0 g n = ( f n f n, e n e n )/( 1 en, f n 2) 1 2. Since α i < 1, i = 1, 2, by direct calculations, we have Using the Gram-Schmidt and lim e n+1, f n+1 = lim ze n, f n+1 = α 2 (1 α1 2 )(1 α 2 2 ), 1 ᾱ 1 α 2 (1 α1 2 )(1 α 2 2 ), 1 ᾱ 1 α 2 lim ze n, e n+1 = α 1. Set c 0 = 1 ᾱ1α2 (1 α 1 2 )(1 α 2 2 ) (1 ᾱ 1α 2) α 1 α 2, then we have lim ze ze n, f n+1 e n+1, f n+1 ze n, e n+1 n, g n+1 = lim (1 e n, f n 2 ) 1 2 = c 0 (α 2 α 1 ).

20 406 Kun-yu GUO & Peng-hui WANG In the case α 1 α 2, we have c 0 0, and hence lim ze n, g n+1 0. This implies that in this case, B z is not compact. Case 2. α 1 = α 2 = α 0. Let e n = (ᾱz)n + (ᾱz) n 1 w + + w n α 2n + α 2n Now it is easy to verify that for any nonnegative integers j 1, j 2, (z αw) 2 z j1 w j2, (n + 1)(ᾱz) n + n(ᾱz) n 1 w + + w n = 0, for n 2. It follows that (n + 1)(ᾱz) n + n(ᾱz) n 1 w + + w n N H n. Write h n = (n + 1)(ᾱz)n + n(ᾱz) n 1 w + + w n (n + 1)2 α 2n + n 2 α 2n , then h n = 1. For n 2, by the Cauchy inequality, ( n )/( n n ) 1 e n, h n = (j + 1) α 2j (j + 1) 2 α 2j α 2j 2 < 1. j=0 Since dim N H n = 2 for n 2, {e n, h n } is a linear basis of N H n. Using the Gram-Schmidt orthogonalization again, j=0 j=0 g n = (h n h n, e n e n ) / (1 e n, h n 2 ) 1 2 V2 H n, and g n = 1. Now, we will show that Since e n+1, g n+1 = 0 and α 0, (5.1) holds if and only if lim ze n, g n+1 0. (5.1) lim ᾱze n be n+1, g n+1 0 for any b R. Taking we have b = ( n+1 / n ) 1 α 2j α 2j 2, j=1 j=1 ᾱze n be n+1 = w n+1 /( n j=1 α 2j ) 1 2. Therefore, it suffices to show that 1 α 2 lim wn+1, g n+1 = lim Since 0 < α < 1, by direct calculations, we have w n+1 /( n j=1 α 2j ) 1 2, gn+1 0. lim wn, e n = 1 α 2,

21 Homogenous quotient modules of Hardy modules on the bidisk 407 and / lim wn, h n = (1 α 2 ) 3 2 (1 + α 2 ) 1 2, lim e 1 n, h n =. 1 + α 2 Therefore, lim wn, g n = ( lim w n, h n e n, h n w n, e n ) = 1 lim 1 en, h n 2 α Therefore, in this case, B z is not compact. Case 3. ( ) (1 α 2 ) 3 2 (1 α 2 ) α 1 = α 2 = 0. This case is considered in [24], we sketch the proof here for convenience. M = N 1 N 2, where N 1 = H 2 (D 2 ) zh 2 (D 2 ) and N 2 = zh 2 (D 2 ) z 2 H 2 (D 2 ). It is easy to see that N 1 = span{w n n = 0, 1,...}, N 2 = span{zw n n = 0, 1,...}. Since S z can be decomposed as S z w n = zw n and S z zw n = 0, S z = 0 0 U 0 N 1 N 2, where Uw n = zw n. That is, B z = U is not compact. The above reason shows that S z is not essentially normal, thus completing the proof. The proof of Theorem 5.1. submodule M of H 2 (D 2 ), the quotient module is not essentially normal. Let We will only prove that, if α i < 1 for i = 1, 2, then for any N = H 2 (D 2 ) [(z α 1 w)(z α 2 w)m ] M 1 = [z α 1 w], M 2 = [(z α 1 w)(z α 2 w)] and M 3 = [(z α 1 w)(z α 2 w)m ], then N can be decomposed as N = M 1 (M 1 M 2 ) (M 2 M 3 ). Relative to this decomposition, S z has the matrix representation, S z = A z B z C z E z F z G z M 1 M 1 M 2. M 2 M 3

22 408 Kun-yu GUO & Peng-hui WANG Since M 1 = [z α 1 w], Theorem 4.6 implies that A z = P M 1 M z M 1 is essentially normal. The ( ) argument in the proof of Lemma 5.3 implies that B z is not compact, and hence Bz is not E z compact. By Lemma 5.2, S z is not essentially normal, and hence Theorem 5.1 is proved. 6 Essential spectrum of the quotient modules and K-homology In this section, we will describe the essential spectrum of quotient modules of H 2 (D 2 ). Some similar techniques can be seen in [10]. The K-homology will also been considered. Let M be submodule of H 2 (D 2 ), Z(M) = {λ D 2 f(λ) = 0, f M}, and Z (M) is defined by { } Z (M) = λ D 2 there are λ n Z(M), such that lim λ n = λ. Theorem 6.1. σ e (N). For any submodule M of H 2 (D 2 ), set N = H 2 (D 2 ) M. Then Z (M) Proof. The proof is routine. Some similar techniques appear in [10]. For the reader s convenience, we give the detail of the proof. Given λ Z (M), by the definition of Z (M), there is µ n Z(M) such that lim µ n = λ. If the pair (λ 1 S z, λ 2 S w ) is Fredholm, by a result of Curto [32, Corollary 3.11], the operator A = λ 1 S z λ 2 S w (λ 2 S w ) (λ 1 S z ) : N N N N is Fredholm. Set T 1 = λ 1 S z and T 2 = λ 2 S w. It follows that AA = T 1T1 + T 2 T2 0 0 T1 T 1 + T2 T 2 is Fredholm. Hence (λ 1 S z )(λ 1 S z ) + (λ 2 S w )(λ 2 S w ) = T 1 T1 + T 2T2 is Fredholm. This implies that there exist a positive invertible operator B and a compact operator K, such that (λ 1 S z )(λ 1 S z ) + (λ 2 S w )(λ 2 S w ) = B + K. Now, let k µn be the normalized reproducing kernel of H 2 (D 2 ) at µ n. Since k µn converges weakly to zero, and since B is positive and invertible, there is a positive constant c such that lim [(λ 1 S z )(λ 1 S z ) + (λ 2 S w )(λ 2 S w ) ]k µn, k µn = lim (B + K)k µ n, k µn = lim Bk µ n, k µn c. However, since µ n = (µ (1) n, µ (2) n ) Z(M), for any f M, f, k µn = 0, and then k µn N. It follows that lim [(λ 1 S z )(λ 1 S z ) + (λ 2 S w )(λ 2 S w ) ]k µn, k µn = lim ( λ 1 µ (1) n 2 + λ 2 µ (2) n 2 ) = 0.

23 Homogenous quotient modules of Hardy modules on the bidisk 409 This contradiction implies Z (M) σ e (N), as desired. Example. Let ᾱ n α n z φ(z) = α n 1 ᾱ n z, ϕ(w) = β n β n w β n 1 β n w n=1 be two infinite Blaschke products. It is well-known that there exist infinite Blaschke products φ and ϕ, such that T Z(φ) and T Z(ϕ). For detailed information of this kind of inner functions, one can see [33, sec. 3]. Now, let M = [φϕ], then Z (M) = D 2. By Theorem 6.1, D 2 σ e (S z, S w ). By [19, Theorem 4.3], σ e (S z, S w ) = D 2. Theorem 6.2. Given a homogenous polynomial p. Set N = H 2 (D 2 ) [p], then σ e (N) = Z(p) D 2. Proof. On the one hand, since p is a homogenous polynomial, Z ([p]) = Z([p]) D 2 = Z(p) D 2. By Theorem 6.1, Z(p) D 2 σ e (N). On the other hand, by [19, Theorem 4.3], σ e (N) D 2. Since p(s z, S w ) = 0, by the Spectral Mapping Theorem [34,Theorem 4.8], σ e (N) Z(p) D 2. This completes the proof. Let p be a homogenous polynomial. If the quotient module [p] is essentially normal, then we get an extension 0 K C ([p] ) C(Z(p) D 2 ) 0. This extension yields a K-homology element in K 1 (Z(p) D 2 ), which is denoted by e p. By Theorem 1.1, the polynomial p can be factorized as p = (w α 1 z) n1 (z α 2 w) n2 m i=3 n=1 (z α i w) ni, with α i < 1, n i 1 for i = 1, 2, and α i = 1 for i 3. Set p 1 = (w α 1 z) n1, and p i = (z α i w) ni for i 2. Without loss of generality, assume that Z(p i ) Z(p j ) = {0} for i j. The same argument as [8, Proposition 4.2] implies that e p = e p1 e p2 e pm. (6.1) Below, we will show that if deg p i 0, then e pi are nontrivial K-homology elements in K 1 (Z(p i ) D 2 ) for i = 1,..., m. In fact, we will show that the corresponding extension is not split. Proposition 6.3. Let α be a complex number with α < 1, then the extension 0 K C ([z αw] ) C(Z(z αw) D 2 ) 0 is not split.

24 410 Kun-yu GUO & Peng-hui WANG Proof. By Theorem 6.2, σ e ([z αw] ) = Z(z αw) D 2 = {(αw, w) w = 1}. By the Spectral Mapping Theorem [34,Theorem 4.8], σ e (S w ) = T. This implies that S w is Fredholm. It is easy to see that the Fredholm index Ind(S w ) = 1. By [8, Proposition 4.6], the extension 0 K C ([z αw] ) C(Z(z αw) D 2 ) 0 is not split, as desired. Proposition 6.4. Let α be a complex number with α = 1, then the extension 0 K C ([(z αw) n ] ) C(Z(z αw) D 2 ) 0 is not split. Proof. The same reason as in the proof of Proposition 6.3 shows that S z is Fredholm. The argument in the proof of Lemma 2.2 implies that ker S z = {0}. Now since 1 ker Sz, the Fredholm index Ind(S z ) 0. By [8, Proposition 4.6] again, the extension 0 K C ([(z αw) n ] ) C(Z(z αw) D 2 ) 0 is not split, thus completing the proof. Combining Proposition 6.3, Proposition 6.4 with (6.1), we have the following theorem. Theorem 6.5. Let p be a homogenous polynomial. If [p] is essentially normal, then the short exact sequence 0 K C ([p] ) C(Z(p) D 2 ) 0 is not split. References [1] Douglas R, Paulsen V. Hilbert modules over function algebras. Pitman Research Notes in Mathematics Series, 217, 1989 [2] Arveson W. p-summable commutators in dimension d. J Oper Theory, 54: (2005) [3] Arveson W. Quotients of standard Hilbert modules. Trans AMS, to appear [4] Arveson W. The dirac operator of a commuting d-tuple. J Funct Anal, 189: (2002) [5] Douglas R. Essentially reductive Hilbert modules. J Oper Theory, 55: (2006) [6] Douglas R. Invariants for Hilbert Modules. Proceedings of Symposia in Pure Mathematics, Vol. 51, Part 1, 1990, [7] Guo K. Defect operator for submodules of Hd 2. J Reine Angew Math, 573: (2004) [8] Guo K, Wang K. Essentially normal Hilbert modules and K-homology. Preprint [9] Guo K, Wang K. Essentially normal Hilbert modules and K-homology II: Quasi-homogeneous Hilbert modules over two dimensional unit ball. Preprint [10] Guo K, Duan Y. Spectrum property of the submodule of the Hardy space over B d. Studia Math, to appear [11] Izuchi K, Yang R. N ϕ-type quotient modules on the torus. Preprint [12] Chen X, Guo K. Analytic Hilbert modules. π-chapman & Hall/CRC Research Notes in Math, 433, 2003 [13] Curto R, Muhly P, Yan K. The C -algebra of an homogeneous ideal in two variables is type I. Current Topics in Operator Algebras (Nara, 1990). River Edge, NJ: World Sci. Publishing,

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