REDUCING SUBSPACES FOR A CLASS OF TOEPLITZ OPERATORS ...

Bull. Korean Math. Soc. 52 (2015), No. 5, pp. 1649–1660 http://dx.doi.org/10.4134/BKMS.2015.52.5.1649

REDUCING SUBSPACES FOR A CLASS OF TOEPLITZ OPERATORS ON THE BERGMAN SPACE OF THE BIDISK Mohammed Albaseer, Yufeng Lu, and Yanyue Shi Abstract. In this paper, we completely characterize the nontrivial reducing subspaces of the Toeplitz operator Tz N zM on the Bergman space 1

A2 (D2 ), where N and M are positive integers.

2

1. Introduction Let D be the open unit disk in the complex plane C. For −1 < α < ∞, let L2 (D, dAα ) be the Hilbert space of square integrable functions on D with the inner product Z f (z)g(z)dAα (z), f, g ∈ A2α (D), hf, giα = D

where

dAα (z) = (α + 1)(1 − |z|2 )α dA(z), and dA is the normalized area measure on D. The weighted Bergman space A2α (D) is the subspace of L2 (D, dAα ) consisting of all the analytic functions in D. We denote s n!Γ(2 + α) n γn = kz kα = Γ(n + α + 2) for n = 0, 1, 2, . . . . Therefore, kf k2α =

+∞ X

γn2 |an |2 < ∞,

n=0

where f (z) =

P+∞

A20 (D). In this

2 n=0 an z ∈ qAα (D). 1 case, γn = n+1 . n

Especially when α = 0, we write A2 (D) =

Received November 3, 2014; Revised February 19, 2015. 2010 Mathematics Subject Classification. Primary 47B35, 46E22. Key words and phrases. reducing subspace, Toeplitz operator, polydisk. This research is supported by NSFC (11271059,11201438), Research Fund for the Doctoral Program of Higher Education of China and Shandong Province Young Scientist Research Award Fund (BS2012SF031). c

2015 Korean Mathematical Society

1649

1650

M. ALBASEER, Y. LU, AND Y. SHI

Denote by D2 = D × D the bidisk. The Bergman space A2 (D2 ) is the space of all holomorphic functions in L2 (D2 , dµ) where dµ(z) = dA(z1 )dA(z2 ). For multi-index β = (β1 , β2 ), denote z β = z1β1 z2β2 and eβ =

zβ . γβ 1 γβ 2

Then {eβ }β0 (β 0 means that β1 ≥ 0 and β2 ≥ 0) is an orthogonal basis in A2 (D2 ). For a bounded measurable function f ∈ L∞ (D2 ), the Toeplitz operator with symbol f is defined by Tf h = P (f h) for every h ∈ A2 (D2 ), where P is the Bergman orthogonal projection from L2 (D2 , dµ) onto A2 (D2 ). Recall that for a bounded linear operator T on a Hilbert space H, a closed subspace M is called a reducing subspace of the operator T , if T (M) ⊂ M and T ∗ (M) ⊂ M. A reducing subspace M is said to be minimal if there is no nonzero reducing subspace N such that N is properly contained in M. On the Bergman space over D, it is proved that TB has just two non-trivial reducing subspaces [13, 16], where B is the product of two Blaschke factors. In [12], M. Stessin and K. Zhu gave a complete description of the reducing subspaces of weighted unilateral shift operators of finite multiplicity. In particular, Tzn has n distinct minimal reducing subspaces. If B is a finite Blaschke product (order n ≥ 2), the number of nontrivial minimal reducing subspaces of TB equals the number of connected components of the Riemann surface of B −1 ◦ B over D (see [2, 3, 4, 8, 9, 14] for details). Further, if B is an infinite Blaschke product or a covering map, the relative research can be founded in [5, 6, 7]. On the Bergman space of bidisk, Y. Lu and X. Zhou [10] characterized the reducing subspaces of Tz1N z2N , Tz1N and Tz2N , respectively. The reducing subspaces of Tz1N z2M on the weighted Bergman space A2α (D2 ) have been completely described in [11]. For p = αz k + βwl , the minimal reducing subspaces of Tp on A2 (D2 ) and the commutant algebra V ∗ (p) = {Tp , Tp∗ }′ was described in [1, 15]. In this paper, we mainly consider the reducing subspaces for the Toeplitz on the Bergman space A2 (D2 ), where N and M are positive operator Tz1N zM 2 integers.

2. Main results In this section, we will give a complete characterization of the reducing subspaces of Tz1N z M . To state our results, we need some notations and lemmas. 2 , where N and M are positive Through out this paper, denote T = Tz1N zM 2 integers. Denote by [f ] the reducing subspace of T generated by f ∈ A2 (D2 ). Let N be the set of all the nonnegative integers.

REDUCING SUBSPACES FOR A CLASS OF TOEPLITZ OPERATORS

By direct calculation, we know that ( γl2 z1k+hN z2l−hM , 2 h k l γ l−hM T (z1 z2 ) = 0, ( γk2 k−hN l+hM z1 z2 , 2 γk−hN T ∗h (z1k z2l ) = 0,

if

l ≥ hM

if

l < hM

if

k ≥ hN

if

k < hN

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;

for k, l, h ∈ N. Set E0 = {(k, l) ∈ N × N : 0 ≤ k < N, 0 ≤ l < M }, E1 = {(k, l) ∈ N × N : k ≥ 2N }, E2 = {(k, l) ∈ N × N : l ≥ 2M, 0 ≤ k < 2N }, E3 = {(k, l) ∈ N × N : N ≤ k < 2N, M ≤ l < 2M }, E4 = {(k, l) ∈ N × N : 0 ≤ k < N, M ≤ l < 2M }, E5 = {(k, l) ∈ N × N : 0 ≤ l < M, N ≤ k < 2N }. Clearly, A2 (D2 ) =

5 M

span{z1p z2q : (p, q) ∈ Ei }.

i=0

Notice that M0 = span{z1p z2q : (p, q) ∈ E0 } is a reducing subspace of T . To find other reducing subspaces, we first study the orthogonal decomposition of z1k z2l with respect to M. Lemma 2.1. Suppose M ⊂ M⊥ 0 is a reducing subspace of T . Let PM be the orthogonal projection from A2 (D2 ) onto M. (i) If (k, l) ∈ E1 ∪ E2 ∪ E3 , then PM z1k z2l = λz1k z2l with some λ ∈ C. (ii) If (k, l) ∈ E4 , then PM z1k z2l ∈ span{z1n z2m : (n, m) ∈ E4 }. (iii) If (k, l) ∈ E5 , then PM z1k z2l ∈ span{z1n z2m : (n, m) ∈ E5 }. Proof. Let k, l ∈ N. Since M⊥M0 , hPM (z1k z2l ), z1p z2q i = 0 for (p, q) ∈ E0 . In the following, we consider the inner product hPM (z1k z2l ), z1p z2q i for (p, q) ∈ 5 S Ei . i=1

For every nonnegative integer h satisfying l ≥ hM , T h∗ T h (z1k z2l ) =

(1)

2 γl2 γk+hN zkzl . 2 γl−hM γk2 1 2

By computation, 2 γl2 γk+hN hPM (z1k z2l ), z1p z2q i = hPM T h∗ T h (z1k z2l ), z1p z2q i 2 γl−hM γk2

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= hPM (z1k z2l ), T h∗ T h (z1p z2q )i ( 2 2 γq γp+hN hPM (z1k z2l ), z1p z2q i, 2 γq−hM γp2 = 0,

q ≥ hM q < hM.

Recall that [s] = max{n ∈ Z : n ≤ s} for real number s. By above equality, we get that if hPM (z1k z2l ), z1p z2q i = 6 0, then 2 2 γq2 γp+hN γl2 γk+hN = 2 2 γl−hM γk2 γq−hM γp2

(2)

for 0 ≤ h ≤ [ Ml ], q ≥ [ Ml ]M . Equivalently, (k + 1 + hN )(q + 1 − hM ) (k + 1)(q + 1) = (3) (p + 1)(l + 1) (p + 1 + hN )(l + 1 − hM ) for 0 ≤ h ≤ [ Ml ], q ≥ [ Ml ]M . (i) If (k, l) ∈ E1 ∪ E2 ∪ E3 , we will show that the equality (2) holds if and only if p = k and q = l. Case one: l ≥ 2M . Let g1 (λ) = (k + 1)(q + 1)(p + 1 + λN )(l + 1 − λM ), g2 (λ) = (p + 1)(l + 1)(k + 1 + λN )(q + 1 − λM ) and g(λ) = g1 (λ) − g2 (λ). Since l ≥ 2M , we have g(0) = g(1) = g(2) = 0. Considering g(λ) is a quadratic polynomial, we have g(λ) ≡ 0 on C. Therefore, g1 and g2 have the same zeros, i.e.,   (k + 1)(q + 1)N M = (p + 1)(l + 1)N M k+1 (k + 1)(q + 1) p+1 N = (p + 1)(l + 1) N  q+1 l+1 (k + 1)(q + 1) M = (p + 1)(l + 1) M .

It follows that p = k and q = l. Case two: k ≥ 2N . Replacing T ∗ T by T T ∗ in Case one, we can get the desire result. The details are listed as follows. Since γ2γ2 k T h T h∗ (z1k z2l ) = k2 l+hM2 z1k z2l , ∀0 ≤ h ≤ [ ], γk−hN γl N we know that 2 γk2 γl+hM hPM (z1k z2l ), z1p z2q i = hPM T h T h∗ (z1k z2l ), z1p z2q i 2 γk−hN γl2 = hPM (z1k z2l ), T h T h∗ (z1p z2q )i ( 2 2 γp γq+hM hPM (z1k z2l ), z1p z2q i if 2 γp−hN γq2 = 0 if Therefore, (4)

hPM (z1k z2l ), z1p z2q i

6= 0 will give that

2 γk2 γl+hM 2 γk−hN γl2

=

2 γp2 γq+hM 2 γp−hN γq2

p ≥ hN p < hN.


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k k for 0 ≤ h ≤ [ N ] and p ≥ [ N ]N . Equivalently,

(5)

(k + 1 − hN )(q + 1 + hM ) (k + 1)(q + 1) = (p + 1)(l + 1) (p + 1 − hN )(l + 1 + hM )

k k for 0 ≤ h ≤ [ N ] and p ≥ [ N ]N . So when k ≥ 2N , the above equality follows for h = 0, 1, 2. In this case we will get p = k and q = l by the same arguments as the case l ≥ 2M has done. Case three: (k, l) ∈ E3 = {(n, m) ∈ N2 : N ≤ n < 2N, M ≤ m < 2M }. k In this case, [ N ] ≥ 1 and [ Ml ] ≥ 1. Then equalities (3) and (5) hold for h = 0, 1. Recall that g(λ) = g1 (λ) − g2 (λ), where g1 (λ) = (k + 1)(q + 1)(p + 1 + λN )(l + 1 − λM ) and g2 (λ) = (p + 1)(l + 1)(k + 1 + λN )(q + 1 − λM ). We get g(0) = g(1) = g(−1) = 0. Therefore, we obtain that p = k and q = l. (ii) Suppose that (k, l) ∈ E4 . We need only prove that

PM (z1k z2l ) ⊥ span{z1n z2m : (n, m) ∈ (

3 [

i=1

Ei )

[

E5 }.

If (n, m) ∈ E1 ∪ E2 ∪ E3 , the conclusion (i) implies that PM z1n z2m = λz1n z2m for some λ ∈ C. Thus hPM z1k z2l , z1n z2m i = hz1k z2l , PM z1n z2m i = λhz1k z2l , z1n z2m i = 0. That is, PM z1k z2l ⊥ span{z1p z2q : (p, q) ∈ E1 ∪ E2 ∪ E3 }. If (n, m) ∈ E5 = {(k, l) ∈ N × N : 0 ≤ l < M, N ≤ k < 2N }, hPM z1k z2l , z1n z2m i = =

2 γl−M γk2 hPM T ∗ T z1k z2l , z1n z2m i 2 2 γl γk+N 2 γl−M γk2 hT PM z1k z2l , T z1nz2m i = 0, 2 γl2 γk+N

where the last equality comes from span{z1p z2q : (p, q) ∈ E5 } ⊆ KerT . Thus PM z1k z2l ⊥ span{z1p z2q : (p, q) ∈ E5 }. (iii) Replacing T ∗ T by T T ∗ in (ii), we get the desired result. Remark 2.1. Let M ⊂ M⊥ 0 is a nonzero reducing subspace of T . In (i) of Lemma 2.1, we indeed get that λ = 0 or 1, that is z1k z2l ∈ M or z1k z2l ∈ M⊥ for each (k, l) ∈ E1 ∪ E2 ∪ E3 . If z1k z2l ∈ M, then (6)

[z1k z2l ] = span{z1k−hN z2l+hM : k − hN ≥ 0, l + hM ≥ 0, h ∈ Z}

is a minimal reducing subspace of T , containing in M. Moreover, if z1k z2l , z1p z2q ∈ M and (k, l), (p, q) ∈ E1 ∪ E2 ∪ E3 , then it’s clear that eitherP [z1k z2l ]⊥[z1p z2q ] or p q k l [z1 z2 ] = [z1 z2 ]. So for any non-zero function f (z) = ak,l z1k z2l , (k,l)∈E1 ∪E2 ∪E3

[f ] is the direct sum of some minimal reducing subspace as (6). We define two equivalences on E4 and E5 respectively by:

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(i) for (p, q), (k, l) ∈ E4 , (p, q) ∼1 (k, l) ⇔ (ii) for (p, q), (k, l) ∈ E5 , (p, q) ∼2 (k, l) ⇔

(k+1)(q+1) (p+1)(l+1) (k+1)(q+1) (p+1)(l+1)

= =

(k+1+N )(q+1−M) (p+1+N )(l+1−M) ; (k+1−N )(q+1+M) (p+1−N )(l+1+M) .

It is easy to check that (i) (p, q) ∈ E4 ⇔ (p + N, q − M ) ∈ E5 ; (ii) for (p, q), (k, l) ∈ E4 , (p, q) ∼1 (k, l) ⇔ (p+N, q−M ) ∼2 (k+N, l−M ); (iii) for (p, q), (k, l) ∈ E5 , (p, q) ∼2 (k, l) ⇔ (p−N, q+M ) ∼1 (k−N, l+M ). For (n, m) ∈ E4 and (k, l) ∈ E5 , let Pn,m : A2 (D2 ) → span{z1p z2q : (p, q) ∼1 (n, m), (p, q) ∈ E4 }, Qk,l : A2 (D2 ) → span{z1k z2l : (p, q) ∼2 (k, l), (p, q)) ∈ E5 } be two orthogonal projections. For f ∈ A2 (D2 ) and Pn,m f 6= 0, we have [Pn,m f ] = span{Pn,m f, T Pn,m f },

(7)

since T ∗ Pn,m f = 0, T 2 Pn,m f = 0 and T ∗ T Pn,m f = if f ∈ M and Qk,l f 6= 0, then

2 2 γm γn+N 2 2 Pn,m f . γm−M γn

Similarly,

[Qk,l f ] = span{Qk,l f, T ∗ Qk,l f }.

(8)

Lemma 2.2. Let M ⊂ M⊥ 0 be a reducing subspace of T and (n, m) ∈ E4 . Then the following statements hold. If f ∈ M, then [Pn,m f ] ⊂ M and [Qn+N,m−M f ] ⊂ M. If f1 , f2 ∈ Pn,m M and f1 ⊥f2 , then [f1 ]⊥[f2 ]. Pn,m T ∗ f = T ∗ Qn+N,m−M f and T Pn,m f = Qn+N,m−M T f , ∀f ∈ M. If f ∈ M, then [Pn,m f ] = [Qn+N,m−M T f ] and [Qn+N,m−M f ] = [Pn,m T ∗ f ]. (e) Pn,m M ⊕ Qn+N,m−M M ⊂ M is a reducing subspace of T .

(a) (b) (c) (d)

Proof. (a) For every f ∈ M, we know that PM Pn,m f = Pn,m f , since PM Pn,m = Pn,m PM , which obtained by the following simple facts: (i) if (k, l) ∈ E4 , then PM z1k z2l ∈ span{z1p z2q : (p, q) ∈ E4 }; (ii) if (k, l) ∈ / E4 , then PM z1k z2l ⊥span{z1p z2q : (p, q) ∈ E4 }. So Pn,m f ∈ M, which implies that [Pn,m f ] ⊂ M. Similarly, we have PM Qn+N,m−M f = Qn+N,m−M f , which shows that Qn+N,m−M f ∈ M. Thus [Qn+N,m−M f ] ⊂ M. (b) It is clear that T f1 , T f2 ∈ span{z1k z2l : (k, l) ∈ E5 } and hT f1 , T f2 i = hT ∗ T f1 , f2 i =

2 2 γn+N γm hf1 , f2 i = 0. 2 γn2 γm−M

Equality (7) shows that [f1 ] = span{f1 , T f1 }, [f2 ] = span{f2 , T f2}. So [f1 ]⊥[f2 ].


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(c) For every (n, m) ∈ E4 , let Mn,m = span{z1k z2l : (k, l) ∼1 (n, m), (k, l) ∈ E4 }, Mn+N,m−M = span{z1k z2l : (k, l) ∼2 (n + N, m − M ), (k, l) ∈ E5 }. Then Mn,m and Mn+N,m−M are finite dimension, and the following statements hold: (i) T Mn,m = Mn+N,m−M and T ∗ Mn+N,m−M = Mn,m ; ⊥ ∗ ⊥ ⊥ (ii) T (M⊥ n,m ) ⊂ Mn+N,m−M and T (Mn+N,m−M ) ⊂ Mn,m . Therefore, T Pn,m f = Qn+N,m−M T f and Pn,m T ∗ f = T ∗ Qn+N,m−M f for any f ∈ M. (d) By equality (7), conclusion (c) and (9)

T ∗ T Pn,m f =

2 2 γn+N γm Pn,m f, 2 γn2 γm−M

we have [Qn+N,m−M T f ] = span{Qn+N,m−M T f, T ∗ Qn+N,m−M T f } = span{T Pn,mf } ⊕ span{T ∗T Pn,m f } = span{T Pn,mf } ⊕ span{Pn,m f } = [Pn,m f ]. Similarly, [Qn+N,m−M f ] = [Pn,m T ∗ f ] comes from equality (8), conclusion (c) and (10)

T T ∗Qn+n,m−M f =

2 2 γn+N γm Qn+N,m−M f. 2 γn2 γm−M

(e) By equalities (9), (10) and conclusion (c), we have (11)

Qn+N,m−M M = T T ∗(Qn+N,m−M M) = T Pn,m T ∗ M, Pn,m M = T ∗ T (Pn,m M) = T ∗ Qn+N,m−M T M.

Therefore, we only need to show that Pn,m M ⊕ Qn+N,m−M M is an invariant subspace of T and T ∗ . In fact, T (Pn,m M ⊕ Qn+N,m−M M) = T Pn,m M = Qn+N,m−M M, where the last equality comes from T Pn,m f = Qn+N,m−M T f ∈ Qn+N,m−M M and Qn+N,m−M f ∈ T Pn,m T ∗ M ⊂ T Pn,m M for all f ∈ M. Therefore, T (Pn,m M ⊕ Qn+N,m−M M) ⊂ Pn,m M ⊕ Qn+N,m−M M. Similarly, we can prove that T ∗ (Pn,m M ⊕ Qn+N,m−M M) = T ∗ Qn+N,m−M M = Pn,m M. So we finish the proof.

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Remark 2.2. In the prove of (e), we also get that [Pn,m M] = Pn,m M ⊕ Qn+N,m−M M = [Qn+N,m−M M], where [Pn,m M] and [Qn+N,m−M M] are the reducing subspaces generated by Pn,m M and Qn+N,m−M M, respectively. Theorem 2.1. Let M ⊂ M⊥ 0 be a non-zero reducing subspace of T on the bidisk. Then M = M1 ⊕ M2 , where (i) M1 is a direct sum of minimal reducing subspace [z1p z2q ] with z1p z2q ∈ M for some (p, q) ∈ E1 ∪ E2 ∪ E3 ; (ii) M2 is a direct sum of minimal reducing subspace [f ] with f ∈ Pn,m M for some (n, m) ∈ E4 . Proof. Firstly, we prove that M M M (12) M = M1 (Pn,m M Qn+N,m−M M), (n,m)∈E

with Λ = {(p, q) ∈ E1 ∪E2 ∪E3 : z1p z2q ∈ M}, and E is L the partition of E4 by the equivalence ∼1 . Set Hn,m =Pn,m M Qn+N,m−M M. L L On the one hand, M1 Hn,m ⊂ M, since M1 ⊂ M is a reducing (n,m)∈E L subspace of T , and conclusion (e) in Lemma 2.2 implies that Hn,m ⊂

where M1 =

[z1p z2q ] (p,q)∈Λ L

(n,m)∈E

M. On the other hand, for g = g1 + g2 ∈ M with X X ap,q z1p z2q , g2 (z) = (13) g1 (z) = (p,q)∈E1 ∪E2 ∪E3

ap,q z1p z2q .

(p,q)∈E4 ∪E5

Remark 2.1 shows that P g1 ∈ M1 ⊂ M, which implies that L g2 = g − g1 ∈ M. Therefore, g2 = (Pn,m g2 + Qn+N,m−M g2 ) ∈ Hn,m . It follows (n,m)∈E

(n,m)∈E

that M is in the direct sum of M1 and {Hn,m } with (n, m) ∈ E. So we have equality (12) holds. Secondly, for each (n, m) ∈ E4 , we prove that Hn,m is the direct sum of minimal reducing subspaces as [f ] = span{f, T f } with f ∈ Pn,m M. There are some steps in the proof. Step 1. Take 0 6= f1 ∈ Pn,m M. Then [f1 ] = span{f1 , T f1 } ⊂ Hn,m . Step 2. If Pn,m M 6= Cf1 , take 0 6= f2 ∈ Pn,m M Cf1 . Then [f2 ] = span{f2 , T f2 } ⊂ Hn,m [f1 ]. Step 3. If Pn,m M 6= span{f1 , f2 }, take 0 6= f3 ∈ Pn,m M span{f1 , f2 }. Then [f3 ] = span{f3 , T f3 } ⊂ Hn,m [f1 ] [f2 ]. If Pn,m M 6= span{f1 , f2 , f3 }, continue this process. This process will stop in finite steps, since the dimension of Hn,m is finite. Thus, we finish the proof.


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Remark 2.3. In particular, if M is a reducing subspace generated by g = g1 + g2 ∈ A2 (D2 ) as in (13), then [g] = [g1 ] ⊕ [g2 ] and M [Pn,m g, Qn+N,m−M g], [g2 ] = (n,m)∈E

where [Pn,m g, Qn+N,m−M g] is the reducing subspace generated by Pn,m g and Qn+N,m−M g. By conclusions (a) and (d) in Lemma 2.2 and equalities in (11), we get [Pn,m g, Qn+N,m−M g] = [Pn,m g, Pn,m T ∗ g] = span{Pn,m g, Pn,m T ∗ g} ⊕ span{Qn+N,m−M g, Qn+N,m−M T g}. Notice that span{Pn,m g, Pn,m T ∗ g} has an orthonormal basis {e1 , . . . , ek }, since the dimension of span{Pn,m g, Pn,m T ∗ g} is finite. Conclusion (b) in Lemma 2.2 shows that [ei ]⊥[ej ] for i 6= j. Then we get [Pn,m g, Pn,m T ∗ g] =

k k M M span{ej , T ej }. [ej ] = j=1

j=1

Similarly, we can prove that M [Qn+N,m−M g, Qn+N,m−M T g], [g2 ] = (n,m)∈E

and [Qn+N,m−M g, Qn+N,m−M T g] =

l l M M span{hj , T ∗ hj }, [hj ] = j=1

j=1

where {h1 , . . . , hl } is an orthonormal basis of span{Qn+N,m−M g, Qn+N,m−M T g}. In the last part of this paper, we give some examples of the reducing subspaces of Tz1N zM for the case that N = M and N 6= M , respectively. 2 Example 2.1. Fix a, b, c, d, e ∈ C with e 6= 0. Let f (z1 , z2 ) = az19 z214 + bz17 z215 + cz15 z217 + dz14 z219 + ez111 z212 , generated by f . Then and [f ] be the reducing subspace of Tz110 z10 2 [f ] = span{f1 , f2 } ⊕ span{z111+10h z212−10h : h = −1, 0, 1}, where f1 (z1 , z2 ) = az19 z214 + bz17 z215 + cz15 z217 + dz14 z219 , f2 (z1 , z2 ) = a3 z119 z24 +

3b 17 5 8 z1 z2

+

4c 15 7 9 z1 z2

+ d2 z114 z29 .

Proof. Notice that (11, 12) ∈ E3 and (9, 14) ∈ E4 . A direct computation shows that (9, 14) ∼1 (7, 15) ∼1 (5, 17) ∼1 (4, 19). Remark 2.1 implies that f1 = P4,19 f and z111 z212 are in M. As in Remark 2.3, there is span{P4,19 f, P4,19 T ∗ f } = [f1 ] = span{f1 , f2 }. Therefore we get the desired result.

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Example 2.2. Let f (z1 , z2 ) = z14 z214 + z17 z27 + z13 z215 and [f ] be the reducing subspace of Tz15 z10 generated by f . Then 2 [f ] = span{z14 z214 + z13 z215 , 31 z19 z24 + 83 z18 z25 } ⊕ span{z17 z27 , z12 z217 }. Proof. Notice that (7, 7) ∈ E5 , (4, 14), (3, 15) ∈ E4 and (4, 14) ∼1 (3, 15). Let f1 = P4,14 f = z14 z214 + z13 z215 and f2 = Q7,7 f = z17 z27 . Then [P4,14 f, P4,14 T ∗ f ] = [f1 ] = span{z14 z214 + z13 z215 , 31 z19 z24 + 38 z18 z25 }, [P2,17 f, P2,17 T ∗ f ] = [Q7,7 f, Q7,7 T f ] = [f2 ] = span{z17 z27 , z12 z217 }. Then we finish the proof. Example 2.3. Let f (z1 , z2 ) = z13 z28 + z17 z23 , and [f ] be the reducing subspace of Tz14 z 52 generated by f . Then [f ] = span{z13 z28 , z17 z23 }. Proof. Notice that (3, 8) ∈ E4 , (7, 3) ∈ E5 . It is easy to check that Tz14 z52 z13 z28 = 4 7 3 1 3 8 ∗ 3 8 7 3 3 8 7 3 7 3 9 z1 z2 and Tz14 z 52 z1 z2 = 2 z1 z2 . So [z1 z2 ] = [z1 z2 ] = span{z1 z2 , z1 z2 }. It means 3 8 7 3 that [f ] = span{z1 z2 , z1 z2 }. Example 2.4. Let f (z1 , z2 ) = z12 z217 + z14 z214 + z19 z24 + z13 z215 + z18 z25 and [f ] be the reducing subspace of Tz15 z10 generated by f . Then 2 [f ] = [z12 z217 ] ⊕ [z14 z214 ] ⊕ [z13 z215 ] = [z12 z217 ] ⊕ [z14 z214 + z13 z215 ] ⊕ [z14 z214 − = [z17 z27 ] ⊕ [z19 z24 + z18 z25 ] ⊕ [z19 z24 −

64 3 15 75 z1 z2 ] 27 8 5 25 z1 z2 ].

Proof. Notice that (2, 17), (4, 14), (3, 15) ∈ E4 , (9, 4), (8, 5) ∈ E5 and (4, 14) ∼1 (3, 15), (9, 4) ∼2 (8, 5). (i) Since P4,14 T f = T ∗ (z19 z24 + z18 z25 ) = 21 z14 z214 + 49 z13 z215 , we have ∗

span{P4,14 f, P4,14 T ∗ f } = span{z14 z214 , z13 z215 }. Therefore, [f ] = [z12 z217 ] ⊕ [z14 z214 ] ⊕ [z13 z215 ] = span{z12 z217 , z17 z27 } ⊕ span{z14 z214 , z19 z24 } ⊕ span{z13 z215 , z18 z25 }. 64 3 15 4 14 3 15 75 z1 z2 , z1 z2 + z1 z2 i = 0 and 3 15 span{z14 z214 + z13 z215 , z14 z214 − 64 75 z1 z2 }.

(ii) It is easy to check that hz14 z214 − span{P4,14 f, P4,14 T ∗ f } =

So [f ] = [z14 z214 + z13 z215 ] ⊕ [z14 z214 − (iii) Notice that

64 3 15 75 z1 z2 ]

⊕ [z12 z217 ].

span{Q9,4 f, Q9,4 T f } = span{z19 z24 + z18 z25 , 13 z19 z24 + 83 z18 z25 } = span{z19 z24 + z18 z25 , z19 z24 − where z19 z24 −

27 8 5 25 z1 z2 ⊥Q9,4 f . Then [f ] = [z17 z27 ] ⊕ [z19 z24

+ z18 z25 ] ⊕ [z19 z24 −

27 8 5 25 z1 z2 },

27 8 5 25 z1 z2 ].


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Remark 2.4. In Example 2.4, since T ∗ (z19 z24 + z18 z25 ) = 21 z14 z214 + 49 z13 z215 and 12 3 15 27 8 5 z1 z2 ) = 12 z14 z214 − 25 z1 z2 , conclusion (d) in Lemma 2.2 implies T ∗ (z19 z24 − 25 1 4 14 4 3 15 2 17 3 15 that [f ] = [z1 z2 ] ⊕ [ 2 z1 z2 + 9 z1 z2 ] ⊕ [ 21 z14 z214 − 12 25 z1 z2 ]. and g = z14 z214 + z19 z24 + z13 z215 , then [g] = [g + Moreover, let T = Tz15 z10 2 8 5 4 14 3 15 az1 z2 ] = [z1 z2 ] ⊕ [z1 z2 ] for a 6= 89 . In fact, span{P4,14 (g + az18 z25 ), P4,14 T ∗ (g + az18 z25 )} = span{z14 z214 , z13 z215 }, since T ∗ (z19 z24 + az18 z25 ) and z14 z214 + z13 z215 are linearly independent. For the case that a = 89 , we have [g + 98 z18 z25 ] = span{z14 z214 + z13 z215 , z19 z24 + 89 z18 z25 } = [z14 z214 + z19 z24 ] since T ∗ (g + 89 z18 z25 ) = 21 P4,14 g. Acknowledgments. The authors thank the reviewer very much for his helpful suggestions which led to the present version of this paper. References [1] H. Dan, and H. Huang, Multiplication operators defined by a class of polynomials on L2α (D2 ), Integral Equations Operator Theory 80 (2014), no. 4, 581–601. [2] R. G. Douglas, M. Putinar, and K. Wang, Reducing subspaces for analytic multipliers of the Bergman space, J. Funct. Anal. 263 (2012), no. 6, 1744–1765. [3] R. G. Douglas, S. Sun, and D. Zheng, Multiplication operators on the Bergman space via analytic continuation, Adv. Math. 226 (2011), no. 1, 541–583. [4] K. Guo and H. Huang, On multiplication operators on the Bergman space: Similarity, unitary equivalence and reducing subspaces, J. Operator Theory 65 (2011), no. 2, 355– 378. , Multiplication operators defined by covering maps on the Bergman space: the [5] connection between operator theory and von Neumann algebras, J. Funct. Anal. 260 (2011), no. 4, 1219–1255. , Geometric constructions of thin Blaschke products and reducing subspace prob[6] lem, Proc. Lond. Math. Soc. 109 (2014), no. 4, 1050–1091. [7] , Multiplication Operators on the Bergman Space, Lecture Notes in Mathematics 2145, Springer-Verlag Berlin Heidelberg 2015. [8] K. Guo, S. Sun, D. Zheng, and C. Zhong, Multiplication operators on the Bergman space via the Hardy space of the bidisk, J. Reine Angew. Math. 628 (2009), 129–168. [9] J. Hu, S. Sun, X. Xu, and D. Yu, Reducing subspace of analytic Toeplitz operators on the Bergman space, Integral Equations Operator Theory 49 (2004), no. 3, 387–395. [10] Y. Lu and X. Zhou, Invariant subspaces and reducing subspaces of weighted Bergman space over bidisk, J. Math. Soc. Japan 62 (2010), no. 3, 745–765. [11] Y. Shi and Y. Lu, Reducing subspaces for Toeplitz operators on the polydisk, Bull. Korean Math. Soc. 50 (2013), no. 2, 687–696. [12] M. Stessin and K. Zhu, Reducing subspaces of weighted shift operators, Proc. Amer. Math. Soc. 130 (2002), no. 9, 2631–2639. [13] S. L. Sun and Y. Wang, Reducing subspaces of certain analytic Toeplitz operators on the Bergman space, Northeast. Math. J. 14 (1998), no. 2, 147–158. [14] S. Sun, D. Zheng, and C. Zhong, Classification of reducing subspaces of a class of multiplication operators on the Bergman space via the Hardy space of the bidisk, Canad. J. Math. 62 (2010), no. 2, 415–438.

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[15] X. Wang, H. Dan, and H. Huang, Reducing subspaces of multiplication operators with the symbol αz k + βω l on L2a (D2 ), Sci. China Math. 58 (2015), doi:10.1007/s11425-0154973-9. [16] K. Zhu, Reducing subspaces for a class of multiplication operators, J. Lond. Math. Soc. 62 (2000), no. 2, 553–568. Mohammed Albaseer School of Mathematical Sciences Dalian University of Technology Dalian 116024, P. R. China E-mail address: [email protected] Yufeng Lu School of Mathematical Sciences Dalian University of Technology Dalian 116024, P. R. China E-mail address: [email protected] Yanyue Shi School of Mathematical Sciences Ocean University of China Qingdao 266100, P. R. China E-mail address: [email protected], [email protected]