/ / Article

Research Article | Open Access

Volume 2015 |Article ID 209307 | 12 pages | https://doi.org/10.1155/2015/209307

# Reducing Subspaces of Some Multiplication Operators on the Bergman Space over Polydisk

Revised14 Feb 2015
Accepted09 Mar 2015
Published01 Apr 2015

#### Abstract

We consider the reducing subspaces of on , where , , and for . We prove that each reducing subspace of is a direct sum of some minimal reducing subspaces. We also characterize the minimal reducing subspaces in the cases that and , respectively. Finally, we give a complete description of minimal reducing subspaces of on with .

#### 1. Introduction

Denote by the open unit disk in the complex plane and the normalized area measure on . For , denote . For a positive integer , the weighted Bergman space is the space of all holomorphic functions on which are square integrable with respect to the measure . is the Hilbert space with inner product and . In particular, if , then is the weighted Bergman space on . Denote by the set of all the nonnegative integers. For a -dimension multi-index ( means that for any ), write and . Then , where . Obviously, is an orthogonal basis of .

For every bounded analytic function on , the multiplication operator is defined by

Recall that, in a Hilbert space , a (closed) subspace is called reducing subspace of an operator if and . Moreover, is called minimal if does not contain any proper reducing subspaces other than .

Although the definition of multiplication operator seems simple, the invariant subspace lattice is very complicated. Even on the Bergman space , the characterization of invariant subspaces for the Bergman shift remains a very fascinating open problem in operator theory. To get some deeper information about , much effort has been devoted to studying the structure of the reducing subspaces of on (see  and its references).

Firstly, it is proved that the multiplication operator , where is the product of two Blaschke factors, has exactly two nontrivial reducing subspaces by Sun and Wang  and Zhu  independently. On the weighted sequence space , Stessin and Zhu  gave a complete description of the reducing subspaces of weighted unilateral shift operators. In particular, they show that has distinct minimal reducing subspaces on . For finite Blaschke product , Hu et al.  obtained that has at least a reducing subspace on which the restriction of is unitary equivalent to . Later on, Xu and Yan  generalized this result to the weighted Bergman space with . In 2009, Guo et al.  proved that if is a Blaschke product of degree 3, then the number of minimal reducing subspaces of is at most . For finite Blaschke product , they also raised a conjecture that the number of nontrivial minimal reducing subspaces of equals the number of connected components of the Riemann surface of over . By different techniques, some partial results are obtained in . Finally, an affirmative answer to the conjecture is given by Douglas et al. . Furthermore, when is an infinite Blaschke product, some relative results are obtained by Guo and Huang in [12, 13].

On , known results about the reducing subspaces of are quite few. If is a monomial, the reducing subspaces of are characterized in . If , Dan and Huang  described the minimal reducing subspaces of and the commutant algebra .

Let be a nonzero reducing subspace of on with . Suppose satisfies . By , we know that every must be in . However, on the unweighted Bergman space , it is not true. For example, is a reducing subspace of . But does not belong to .

To know more about how influences the structure of reducing subspaces, we consider the reducing subspaces of over for .

Fix integer and distinct positive integers for . Denote for . In Section 2, we prove that each reducing subspace of is a direct sum of some minimal reducing subspaces. To classify the minimal reducing subspaces, we consider three cases: (i) is irrational; (ii) ; (iii) is rational and . For cases (i) and (ii), we describe the minimal reducing subspaces of . For case (iii), we find that the minimal reducing subspaces of are varied. In Section 3, we give a complete characterization of the reducing subspaces of when the dimension .

#### 2. Reducing Subspaces on

The aim of this section is to give a complete description of the reducing subspaces of on . Denote Define an equivalence on by Write for . For , letClearly, and , where is the partition of by the equivalence . Let be the orthogonal projection from onto .

Theorem 1. Let be a nonzero reducing subspace of on . Then, contains a minimal reducing subspacewhere and with coefficients .

Proof. Let be the orthogonal projection from onto . For abbreviation, we denote .
Firstly, we show that for every . Let . We only need to prove that if , then . If , then ; that is, for . Therefore, If , we find that for . Then,Thus, we get that if , thenSince , we have . Therefore, which implies .
Thus . We also obtain that for any .
Next, we claim that there is a nonzero function in for some positive integer .
Choose a nonzero function in . Let be the minimal integer such that , where is the orthogonal projection from onto . Namely, there exists such that . Then, we can prove that . In fact,(i)if , thenwhere the second equality comes from and the last equality comes from ;(ii)if is out of , then .
Therefore, we get , where is the reducing subspace of induced by . Notice thatHence, we conclude that is a minimal reducing subspace of .

In the following, we will prove that each nonzero reducing subspace of is the orthogonal sum of some minimal reducing subspaces.

Theorem 2. Let be a nonzero reducing subspace of on . Then, where is the reducing subspace of induced by . If , then where is the orthogonal basis of and .

Proof. Denote . Firstly, we know that , since(i);(ii);(iii), for ;(iv), for .
Secondly, we prove that . On the one hand, in the proof of Theorem 1, we get . Then, . On the other hand, if , choose a nonzero function in . Theorem 1 shows that there are and such that . However, for , which is a contradiction.
Finally, we prove that if , then . Choose an orthogonal basis () of the subspace . Theorem 1 shows that . We have that for , sinceLet . Clearly, . Assume that . Take a nonzero function . As in Theorem 1, there is an integer such that and . Since , we have , which is in contradiction with . So we finish the proof.

From this theorem, we know that the reducing subspaces of are determined by the sets . There arises the following question: what are the elements in the set exactly? We begin the research with the case that is irrational.

Lemma 3. If is irrational, then for every .

Proof. Suppose ; that is, , for all . Then, we haveThis is equivalent toWriteClearly, is a polynomial over and for any . Fundamental theorem of algebra shows that , for all . Denote Since is irrational, . Then, implies . So we get and .
Without loss of generality, we may assume . Then there exist nonnegative integers and making If , then , which is in contradiction with the assumption. So .
Equality (21) implies . Then, . Hence, we get and .
Therefore,LetWithout loss of generality, assume . As above, it is easy to get . Applying this process again, we can prove that for .

By Theorems 1 and 2 and Lemma 3, we obtain the following theorem.

Theorem 4. If is irrational, then each reducing subspace of on is a direct sum of some minimal reducing subspaces of the form where .

Proof. Lemma 3 shows that . In light of Theorem 1, we have . For , Theorem 2 implies that , . Thus, .

Next, we consider the case that . Denote by the permutation group of the set . Let for .

Lemma 5. If , then

Proof. Suppose . By the definition of , we haveLet . We have , since is a polynomial on with infinitely many roots. Therefore,For each , there is only one integer such thatthat is, . LetHence .
Conversely, for every , . By definition of , we have Therefore, equality (27) holds, implying . Therefore, .

From this result, we find .

Example 6. LetDenote by the reducing subspace of on induced by . Let(i) for ;(ii) for ;(iii) for .
Then, .

Proof. Let and let . It is easy to check that That is,Similarly,By Lemma 5, we get and . Therefore, . Notice thatwhere , , and . So

If is a nonzero rational number, the structure of minimal reducing subspace turns to be more complicated. In particular, we will study the reducing subspaces of on in the next section.

#### 3. Reducing Subspaces on

Let be rational. We consider the reducing subspaces of on . Recall and ; that is, if and only if for . For every , if , we assume that for . Otherwise, if there exists such that , we can prove that for as in . Since and are decreasing as is increasing, there exist and satisfying and .

This section is organized as follows. Firstly, we consider under the assumption that , , and . Let , , and . We give a description of in the cases that is in , and , respectively. Secondly, we get all the possible cases by symmetry (see Corollaries 11 and 13). Finally, we obtain and Theorem 14.

Lemma 7. Let and let . If satisfies , , and , then one of the following statements holds:(1);(2).

Proof. Let . By definition of , as in Lemma 3, we havefor any . Denotefor . Then,where denotes the disjoint union. Since is not an integer, for .
It is easy to see that Since for and , we haveWithout loss of generality, assumeFirstly, we prove that by contradiction. Otherwise, if , thenSince , we have andTherefore, . Since , it holds that We will find the contradictions under the assumptions (a) and (b) , respectively.
(a) If , then . So andSince , we have By (43) and (48), we get and . It follows thatThus, andBy and equality (48), we conclude thatEqualities (46) and (51) imply . Thus, , which is in contradiction with the assumption.
(b) Suppose . Notice that .
If , then equality (40) implies . Equivalently, and . Hence, , which is impossible.
If , then . It follows that . Since , equality (43) implies . Therefore, and . Then, , which is a contradiction.
Summing up, we must have .
Next, we prove that .
If , then ; that is, In this case, . Otherwise, and , which is in contradiction with .
If , then . It follows that In this case, and . Or else, and , which is in contradiction with . So we get the desired results.

Lemma 8. Fix and . If satisfies , , and , then one of the following statements holds:(1);(2).

Proof. Let for . Then, for . Let for . Then, and for . As in Lemma 7, we assume equality (43) holds. Then, we can prove that or .
Since , we have and . It means for .
If , then statement (1) or statement (2) holds.
If , then and . Equalities (43) and (53) imply thatSince , and are distinct, equality (56) shows thatThen, we have and