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

Authors: Yanyue Shi; Na Zhou
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

We consider the reducing subspaces ofM
z
N on A
α
2
(
D
k
), where k
≥
3, z
N
=
z
1
N
1
⋯
z
k
N
k, and N
i
≠
N
j for i
≠
j. We prove that each reducing subspace of M
z
N is a direct sum of some minimal reducing subspaces. We also characterize the minimal reducing subspaces in the cases that α
=
0 and α
∈
(
-
1
,
+
∞
)
∖
Q, respectively. Finally, we give a complete description of minimal reducing subspaces of M
z
N on A
α
2
(
D
3
) with α
>
-
1.


---

## Body

## 1. Introduction

Denote byD the open unit disk in the complex plane and d
A the normalized area measure on D. For -
1
<
α
<
∞, denote d
A
α
(
z
)
=
(
α
+
1
)
(
1
-
|
z
|
2
)
α
d
A
(
z
). For a positive integer k, the weighted Bergman space A
α
2
(
D
k
) is the space of all holomorphic functions on D
k which are square integrable with respect to the measure d
v
α
(
z
)
=
d
A
α
(
z
1
)
⋯
d
A
α
(
z
k
). A
α
2
(
D
k
,
d
v
α
) is the Hilbert space with inner product (1)
f
,
g
α
=
∫
D
k
‍
f
(
z
)
g
z
¯
d
v
α
,and f
α
2
=
〈
f
,
f
〉
α. In particular, if k
=
1, then A
α
2
(
D
) is the weighted Bergman space on D. Denote by N
0 the set of all the nonnegative integers. For a k-dimension multi-index β
=
(
β
1
,
…
,
β
k
)
∈
N
0
k (β
⪰
0 means that β
i
≥
0 for any i
=
1,2
,
…
,
k), write z
β
=
z
1
β
1
⋯
z
k
β
k and γ
β
=
z
β
α
2. Then γ
β
=
ω
β
1
⋯
ω
β
k, where ω
β
i
=
z
i
β
i
α
2
=
β
i
!
Γ
(
2
+
α
)
/
Γ
(
2
+
α
+
β
i
). Obviously, {
z
β
/
γ
β
}
β
∈
N
0
k is an orthogonal basis of A
α
2
(
D
k
).For every bounded analytic functionφ on D
k, the multiplication operator M
φ is defined by (2)
M
φ
(
h
)
=
φ
h
,
∀
h
∈
A
α
2
(
D
k
)
.Recall that, in a Hilbert spaceH, a (closed) subspace M is called reducing subspace of an operator T if T
(
M
)
⊂
M and T
∗
(
M
)
⊂
M. Moreover, M is called minimal if M does not contain any proper reducing subspaces other than {
0
}.Although the definition of multiplication operatorM
φ seems simple, the invariant subspace lattice Lat
M
φ is very complicated. Even on the Bergman space A
2
(
D
), the characterization of invariant subspaces for the Bergman shift M
z remains a very fascinating open problem in operator theory. To get some deeper information about Lat
M
z, much effort has been devoted to studying the structure of the reducing subspaces of M
φ on A
2
(
D
) (see [1] and its references).Firstly, it is proved that the multiplication operatorM
B, where B is the product of two Blaschke factors, has exactly two nontrivial reducing subspaces by Sun and Wang [2] and Zhu [3] independently. On the weighted sequence space H
ω
2, Stessin and Zhu [4] gave a complete description of the reducing subspaces of weighted unilateral shift operators. In particular, they show that M
z
n has n distinct minimal reducing subspaces on A
2
(
D
). For finite Blaschke product B, Hu et al. [5] obtained that M
B has at least a reducing subspace on which the restriction of M
B is unitary equivalent to M
z. Later on, Xu and Yan [6] generalized this result to the weighted Bergman space A
α
2
(
D
) with α
∈
N
0. In 2009, Guo et al. [7] proved that if B is a Blaschke product of degree 3, then the number of minimal reducing subspaces of M
B is at most 3. For finite Blaschke product B, they also raised a conjecture that the number of nontrivial minimal reducing subspaces of M
B equals the number of connected components of the Riemann surface of B
-
1
∘
B over D. By different techniques, some partial results are obtained in [8–10]. Finally, an affirmative answer to the conjecture is given by Douglas et al. [11]. Furthermore, when B is an infinite Blaschke product, some relative results are obtained by Guo and Huang in [12, 13].OnA
α
2
(
D
2
), known results about the reducing subspaces of M
φ are quite few. If φ is a monomial, the reducing subspaces of M
φ are characterized in [14–17]. If φ
=
z
1
m
+
z
2
n, Dan and Huang [18] described the minimal reducing subspaces of M
φ and the commutant algebra {
M
φ
,
M
φ
∗
}
′.LetM be a nonzero reducing subspace of T
z
1
m
z
2
n on A
α
2
(
D
2
) with α
≠
0. Suppose f
∈
M satisfies 〈
f
,
z
1
p
z
2
q
〉
≠
0. By [16], we know that every z
1
p
z
2
q must be in M. However, on the unweighted Bergman space A
2
(
D
2
), it is not true. For example, M
=
span
⁡
¯
{
(
z
1
z
2
5
+
z
1
3
z
2
2
)
z
1
2
h
z
2
3
h
:
h
=
0,1
,
…
} is a reducing subspace of M
z
1
2
z
2
3. But z
1
z
2
5 does not belong to M.To know more about howα influences the structure of reducing subspaces, we consider the reducing subspaces of M
z
N over D
k for k
≥
3.Fix integerk
≥
3 and distinct positive integers N
i for i
=
1
,
…
,
k. Denote M
z
N
=
M
z
1
N
1
z
2
N
2
⋯
z
n
N
k for N
=
(
N
1
,
…
,
N
k
). In Section 2, we prove that each reducing subspace of M
z
N is a direct sum of some minimal reducing subspaces. To classify the minimal reducing subspaces, we consider three cases: (i) α is irrational; (ii) α
=
0; (iii) α is rational and α
≠
0. For cases (i) and (ii), we describe the minimal reducing subspaces of M
z
N. For case (iii), we find that the minimal reducing subspaces of M
z
N are varied. In Section 3, we give a complete characterization of the reducing subspaces of M
z
N when the dimension k
=
3.


## 2. Reducing Subspaces onA
α
2
(
D
k
)

The aim of this section is to give a complete description of the reducing subspaces ofM
z
N on A
α
2
(
D
k
). Denote (3)
Ω
=
n
=
n
1
,
…
,
n
k
∈
N
0
k
:
0
≤
n
i
<
N
i
for
some
i
.Define an equivalence on Ω by (4)
q
~
n
⟺
γ
q
+
h
N
=
γ
n
+
h
N
,
∀
h
∈
N
0
.Write γ
n
+
h
N
=
∏
i
=
1
k
‍
ω
n
i
+
h
N
i
=
∏
i
=
1
k
‍
(
(
n
i
+
h
N
i
)
!
Γ
(
2
+
α
)
/
Γ
(
2
+
α
+
n
i
+
h
N
i
)
) for h
∈
N
0. For n
∈
Ω, let(5)
J
n
∶
=
q
∈
Ω
:
q
~
n
,
H
n
∶
=
span
⁡
¯
z
J
:
J
∈
J
n
.Clearly, ⋃
n
∈
F
‍
J
n
=
Ω and ⨁
n
∈
F
H
n
=
span
⁡
¯
{
z
J
:
J
∈
Ω
}, where F is the partition of Ω by the equivalence ~. Let P
n be the orthogonal projection from A
α
2
(
D
k
) onto H
n.Theorem 1.
LetM be a nonzero reducing subspace of M
z
N on A
α
2
(
D
k
). Then, M contains a minimal reducing subspace(6)
f
=
span
⁡
¯
f
z
h
N
:
h
∈
N
0
,where n
∈
Ω and f
=
∑
J
∈
J
n
‍
b
J
z
J with coefficients b
J
∈
C.Proof.
LetP
M be the orthogonal projection from A
α
2
(
D
k
) onto M. For abbreviation, we denote M
=
M
z
N
=
M
z
1
N
1
z
2
N
2
⋯
z
k
N
k.
Firstly, we show thatP
M
(
z
m
)
∈
H
n for every m
∈
J
n. Let l
=
(
l
1
,
…
,
l
k
)
∈
N
0
k. We only need to prove that if 〈
P
M
z
m
,
z
l
〉
≠
0, then l
∈
J
n. If l
∉
Ω, then l
⪰
N; that is, l
i
≥
N
i for 1
≤
i
≤
k. Therefore, (7)
P
M
z
m
,
z
l
=
P
M
z
m
,
M
z
l
-
N
=
P
M
M
∗
z
m
,
z
l
-
N
=
0
.If l
∈
Ω, we find that (8)
M
∗
h
M
h
z
J
=
∏
i
=
1
k
‍
ω
j
i
+
h
N
i
ω
j
i
z
j
i
=
γ
J
+
h
N
γ
J
z
Jfor J
=
(
j
1
,
…
,
j
k
)
∈
N
0
k. Then,(9)
γ
m
+
h
N
γ
m
P
M
z
m
,
z
l
=
P
M
M
∗
h
M
h
z
m
,
z
l
=
M
∗
h
M
h
P
M
z
m
,
z
l
=
P
M
z
m
,
M
∗
h
M
h
z
l
=
γ
l
+
h
N
γ
l
P
M
z
m
,
z
l
.Thus, we get that if 〈
P
M
z
m
,
z
l
〉
≠
0, then(10)
∏
i
=
1
k
‍
ω
m
i
+
h
N
i
ω
m
i
=
∏
i
=
1
k
‍
ω
l
i
+
h
N
i
ω
l
i
,
∀
h
∈
N
0
.Since lim
h
→
+
∞
⁡
(
ω
m
i
+
h
N
i
/
ω
l
i
+
h
N
i
)
=
1, we have ∏
i
=
1
k
‍
(
ω
l
i
/
ω
m
i
)
=
1. Therefore, (11)
∏
i
=
1
k
‍
ω
l
i
+
h
N
i
=
∏
i
=
1
k
‍
ω
m
i
+
h
N
i
,
∀
h
∈
N
0
,which implies l
∈
J
m
=
J
n.
ThusP
M
(
z
m
)
∈
H
n. We also obtain that P
M
(
z
l
)
⊥
H
n for any l
∉
J
n.
Next, we claim that there is a nonzero functionf
0 in H
n
0
∩
M for some positive integer n
0.
Choose a nonzero functionf in M. Let h
0 be the minimal integer such that P
Ω
M
∗
h
0
(
f
)
≠
0, where P
Ω is the orthogonal projection from A
α
2
(
D
k
) onto span
⁡
¯
{
z
J
:
J
∈
Ω
}. Namely, there exists n
0
∈
Ω such that f
0
=
P
n
0
M
∗
h
0
f
=
P
n
0
P
Ω
M
∗
h
0
f
=
∑
J
∈
J
n
0
‍
b
J
z
J
≠
0. Then, we can prove that f
0
=
P
M
f
0
∈
M. In fact,(i)
ifm
∈
J
n
0, then(12)
P
M
f
0
,
z
m
=
P
M
P
n
0
M
∗
h
0
f
,
z
m
=
M
∗
h
0
f
,
P
M
P
n
0
z
m
=
P
n
0
P
M
M
∗
h
0
f
,
z
m
=
f
0
,
z
m
,
where the second equality comes fromz
m
,
P
M
(
z
m
)
∈
H
n and the last equality comes from M
∗
h
0
f
∈
M;
(ii)
ifm is out of J
n
0, then 〈
P
M
f
0
,
z
m
〉
=
0
=
〈
f
0
,
z
m
〉.
Therefore, we get[
f
0
]
⊂
M, where [
f
0
] is the reducing subspace of M induced by f
0. Notice that(13)
a
M
q
f
0
z
h
N
=
f
0
z
h
+
q
N
for
h
,
q
≥
0
;
b
M
∗
q
f
0
z
h
N
=
γ
n
0
+
h
N
γ
n
0
+
(
h
-
q
)
N
f
0
z
(
h
-
q
)
N
,
if
h
≥
q
≥
1
0
,
if
0
≤
h
<
q
;
c
f
0
z
h
1
N
⊥
f
0
z
h
2
N
with
h
1
≠
h
2
,
since
f
0
z
h
1
N
,
f
0
z
h
2
N
=
M
h
1
f
0
,
M
h
2
f
0
=
γ
n
0
+
h
2
N
γ
n
0
f
0
z
h
1
-
h
2
N
,
f
0
,
if
h
1
>
h
2
≥
0
γ
n
0
+
h
1
N
γ
n
0
f
0
,
f
0
z
h
2
-
h
1
N
,
if
h
2
>
h
1
≥
0
.Hence, we conclude that [
f
0
]
=
span
⁡
¯
{
f
0
z
h
N
:
h
∈
N
0
}
=
⨁
h
=
0
+
∞
span
⁡
{
f
0
z
h
N
}
⊂
M is a minimal reducing subspace of M.In the following, we will prove that each nonzero reducing subspace ofM
z
N is the orthogonal sum of some minimal reducing subspaces.Theorem 2.
LetM be a nonzero reducing subspace of M
z
N on A
α
2
(
D
k
). Then, (14)
M
=
⨁
n
∈
F
P
n
M
,where [
P
n
M
] is the reducing subspace of M
z
N induced by P
n
M. If P
n
M
≠
{
0
}, then (15)
P
n
M
=
⨁
h
=
0
+
∞
z
h
N
P
n
M
=
⨁
j
=
1
q
e
n
,
j
,where {
e
n
,
j
}
j
=
1
q is the orthogonal basis of P
n
M and 1
≤
q
≤
+
∞.Proof.
DenoteM
=
M
z
N. Firstly, we know that [
P
n
M
]
=
⨁
h
=
0
+
∞
z
h
N
P
n
M, since(i)
z
h
1
N
P
n
M
⊥
z
h
2
N
P
n
M;
(ii)
M
(
z
h
N
P
n
M
)
=
z
(
h
+
1
)
N
P
n
M;
(iii)
M
∗
(
P
n
M
)
=
{
0
}, M
∗
(
z
h
N
P
n
M
)
=
z
(
h
-
1
)
N
P
n
M for h
≥
1;
(iv)
M
∗
M
(
z
h
N
P
n
M
)
=
z
h
N
P
n
M, M
M
∗
(
z
h
N
P
n
M
)
=
z
h
N
P
n
M for h
≥
1.
Secondly, we prove thatM
=
⨁
n
∈
F
[
P
n
M
]. On the one hand, in the proof of Theorem 1, we get P
n
M
⊂
M. Then, [
P
n
M
]
⊂
M. On the other hand, if M
≠
⨁
n
∈
F
[
P
n
M
], choose a nonzero function f
⊥
⨁
n
∈
F
[
P
n
M
] in M. Theorem 1 shows that there are n
0 and h
0 such that 0
≠
f
0
=
P
n
0
M
∗
h
0
f
∈
M. However, 〈
f
0
,
g
〉
=
〈
f
,
M
h
0
P
n
0
g
〉
=
0 for g
∈
M, which is a contradiction.
Finally, we prove that ifP
n
M
≠
{
0
}, then [
P
n
M
]
=
⨁
j
=
1
q
[
e
n
,
j
]. Choose an orthogonal basis {
e
n
,
j
}
j
=
1
q (q
≤
+
∞) of the subspace P
n
M. Theorem 1 shows that [
e
n
,
j
]
=
span
⁡
¯
{
e
n
,
j
z
h
N
:
h
∈
N
0
}
⊂
M. We have that [
e
n
,
p
1
]
⊥
[
e
m
,
p
2
] for n
≠
m, since(16)
e
n
,
p
1
z
h
1
N
,
e
m
,
p
2
z
h
2
N
=
M
h
1
e
n
,
p
1
,
M
h
2
e
m
,
p
2
=
γ
m
+
h
2
N
γ
m
e
n
,
p
1
z
h
1
-
h
2
N
,
e
m
,
p
2
,
if
h
1
≥
h
2
≥
0
γ
n
+
h
1
N
γ
n
e
n
,
p
1
,
e
m
,
p
2
z
h
2
-
h
1
N
,
if
h
2
>
h
1
≥
0
.Let M
n
=
⨁
j
=
1
q
[
e
n
,
j
]. Clearly, M
n
⊂
[
P
n
M
]. Assume that M
n
≠
[
P
n
M
]. Take a nonzero function g
∈
[
P
n
M
]
⊖
M
n. As in Theorem 1, there is an integer h
0 such that g
0
∶
=
P
n
M
∗
h
0
g
∈
P
n
M and g
0
≠
0. Since g
⊥
M
n, we have 〈
g
0
,
e
n
,
j
〉
=
〈
g
,
M
h
0
P
n
e
n
,
j
〉
=
0, which is in contradiction with g
0
≠
0. So we finish the proof.From this theorem, we know that the reducing subspaces ofM
z
N are determined by the sets {
J
n
}
n
∈
Ω. There arises the following question: what are the elements in the set J
n exactly? We begin the research with the case that α is irrational.Lemma 3.
Ifα is irrational, then J
n
=
{
n
} for every n
∈
Ω.Proof.
Supposem
∈
J
n; that is, γ
n
+
h
N
=
γ
m
+
h
N, for all h
∈
N
0. Then, we have(17)
γ
n
+
h
N
γ
n
+
(
h
+
1
)
N
=
γ
m
+
h
N
γ
m
+
(
h
+
1
)
N
,
∀
h
∈
N
0
.This is equivalent to(18)
∏
i
=
1
k
‍
∏
j
=
1
N
i
‍
n
i
+
h
N
i
+
α
+
1
+
j
n
i
+
h
N
i
+
j
=
∏
i
=
1
k
‍
∏
j
=
1
N
i
‍
m
i
+
h
N
i
+
α
+
1
+
j
m
i
+
h
N
i
+
j
,
∀
h
∈
N
0
.Write(19)
g
λ
=
∏
i
=
1
k
‍
∏
j
=
1
N
i
‍
n
i
+
λ
N
i
+
α
+
1
+
j
m
i
+
λ
N
i
+
j
-
∏
i
=
1
k
‍
∏
j
=
1
N
i
‍
m
i
+
λ
N
i
+
α
+
1
+
j
n
i
+
λ
N
i
+
j
.Clearly, g is a polynomial over C and g
(
h
)
=
0 for any h
∈
N
0. Fundamental theorem of algebra shows that g
(
λ
)
≡
0, for all λ
∈
C. Denote (20)
E
1
=
n
i
+
α
+
1
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
,
E
2
=
n
i
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
,
F
1
=
m
i
+
α
+
1
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
,
F
2
=
m
i
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
.Since α is irrational, E
1
∩
E
2
=
F
1
∩
F
2
=
∅. Then, g
(
λ
)
≡
0 implies E
1
∪
F
2
=
E
2
∪
F
1. So we get E
1
=
F
1 and E
2
=
F
2.
Without loss of generality, we may assumemax
⁡
⁡
F
2
=
(
m
k
+
N
k
)
/
N
k. Then there exist nonnegative integers i and j making (21)
m
i
+
α
+
1
+
j
N
i
=
n
k
+
α
+
2
N
k
.
Ifi
≠
k, then α
=
(
(
n
k
+
2
)
N
i
-
(
m
i
+
1
+
j
)
N
k
)
/
(
N
k
-
N
i
)
∈
Q, which is in contradiction with the assumption. So i
=
k.
Equality (21) implies (
m
k
+
j
)
/
N
k
=
(
n
k
+
1
)
/
N
k. Then, max
⁡
⁡
E
2
≥
(
n
k
+
N
k
)
/
N
k
=
(
m
k
+
N
k
+
j
-
1
)
/
N
k
≥
max
⁡
⁡
F
2. Hence, we get j
=
1 and m
k
=
n
k.
Therefore,(22)
∏
i
=
1
k
-
1
‍
∏
j
=
1
N
i
‍
n
i
+
h
N
i
+
α
+
1
+
j
n
i
+
h
N
i
+
j
=
∏
i
=
1
k
-
1
‍
∏
j
=
1
N
i
‍
m
i
+
h
N
i
+
α
+
1
+
j
m
i
+
h
N
i
+
j
,
∀
h
∈
N
0
.
Let(23)
E
~
1
=
n
i
+
α
+
1
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
-
1
,
E
~
2
=
n
i
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
-
1
,
F
~
1
=
m
i
+
α
+
1
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
-
1
,
F
~
2
=
m
i
+
j
N
i
:
j
=
1
,
…
,
N
i
,
i
=
1
,
…
,
k
-
1
.Without loss of generality, assume max
⁡
⁡
F
~
2
=
(
m
k
-
1
+
N
k
-
1
)
/
N
k
-
1. As above, it is easy to get m
k
-
1
=
n
k
-
1. Applying this process again, we can prove that m
i
=
n
i for i
=
1
,
…
,
k.By Theorems1 and 2 and Lemma 3, we obtain the following theorem.Theorem 4.
Ifα is irrational, then each reducing subspace M of M
z
N on A
α
2
(
D
k
) is a direct sum of some minimal reducing subspaces of the form (24)
span
⁡
¯
z
n
+
h
N
:
h
∈
N
0
,where n
∈
A
=
{
n
∈
Ω
:
z
n
∈
M
}.Proof.
Lemma3 shows that J
n
=
{
n
}. In light of Theorem 1, we have A
≠
∅. For n
∈
A, Theorem 2 implies that P
n
M
=
span
⁡
{
z
n
}, [
z
n
]
=
span
⁡
¯
{
z
n
+
h
N
:
h
∈
N
0
}. Thus, M
=
⨁
n
∈
A
[
z
n
].Next, we consider the case thatα
=
0. Denote by S
k the permutation group of the set {
1,2
,
…
,
k
}. Let ρ
i
j
(
x
)
=
(
x
+
1
)
N
i
/
N
j
-
1 for x
∈
R.Lemma 5.
Ifα
=
0, then(25)
J
n
=
ρ
1
σ
1
n
σ
1
,
ρ
2
σ
2
n
σ
2
,
…
,
ρ
k
σ
k
n
σ
k
:
n
σ
k
σ
∈
S
k
.Proof.
Supposem
∈
J
n. By the definition of J
n, we have(26)
∏
i
=
1
k
‍
h
N
i
+
m
i
+
1
=
∏
i
=
1
k
‍
h
N
i
+
n
i
+
1
,
∀
h
∈
N
0
.Let g
(
λ
)
=
∏
i
=
1
k
‍
(
λ
+
(
m
i
+
1
)
/
N
i
)
-
∏
i
=
1
k
‍
(
λ
+
(
n
i
+
1
)
/
N
i
). We have g
(
λ
)
≡
0, since g is a polynomial on C with infinitely many roots. Therefore,(27)
n
i
+
1
N
i
:
i
=
1
,
…
,
k
=
m
i
+
1
N
i
:
i
=
1
,
…
,
k
.For each j
∈
{
1,2
,
…
,
k
}, there is only one integer i
∈
{
1,2
,
…
,
k
} such that(28)
n
j
+
1
N
j
=
m
i
+
1
N
i
;that is, m
i
=
(
n
j
+
1
)
N
i
/
N
j
-
1
=
ρ
i
j
(
n
j
). Let(29)
E
=
ρ
1
σ
1
n
σ
1
,
ρ
2
σ
2
n
σ
2
,
…
,
ρ
k
σ
k
n
σ
k
:
n
σ
k
σ
∈
S
k
.Hence m
∈
E.
Conversely, for everym
∈
E, (
m
i
+
1
)
/
N
i
=
(
ρ
i
σ
(
i
)
(
n
σ
(
i
)
)
+
1
)
/
N
i. By definition of ρ
i
j
(
n
j
), we have (30)
ρ
i
σ
(
i
)
(
n
σ
(
i
)
)
+
1
N
i
=
n
σ
(
i
)
+
1
N
σ
(
i
)
.Therefore, equality (27) holds, implying m
∈
J
n. Therefore, J
n
=
E.From this result, we findCard
(
J
n
)
≤
n
!.Example 6.
Let(31)
f
z
=
a
1
z
1
z
3
+
a
2
z
1
2
z
2
z
3
3
+
a
3
z
1
2
z
2
3
z
3
+
a
4
z
1
z
2
z
3
5
+
a
5
z
1
z
3
11
+
a
6
z
1
5
z
3
3
+
a
7
z
1
11
z
3
.Denote by M
=
[
f
] the reducing subspace of M
z
1
3
z
2
z
3
6 on A
2
(
D
3
) induced by f. Let(i)
M
1
=
span
⁡
¯
{
f
1
(
z
)
z
1
3
h
z
2
h
z
3
6
h
:
h
=
0,1
,
2
,
…
} for f
1
(
z
)
=
z
1
z
3;
(ii)
M
2
=
span
⁡
¯
{
f
2
(
z
)
z
1
3
h
z
2
h
z
3
6
h
:
h
=
0,1
,
2
,
…
} for f
2
(
z
)
=
a
2
z
1
2
z
2
z
3
3
+
a
4
z
1
z
2
z
3
5
+
a
5
z
1
z
3
11
+
a
6
z
1
5
z
3
3;
(iii)
M
3
=
span
⁡
¯
{
f
3
(
z
)
z
1
3
h
z
2
h
z
3
6
h
:
h
=
0,1
,
2
,
…
} for f
3
(
z
)
=
a
3
z
1
2
z
2
3
z
3
+
a
7
z
1
11
z
3.
Then,M
=
⨁
i
=
1
3
M
i.Proof.
Letn
=
(
n
1
,
n
2
,
n
3
)
=
(
2,1
,
3
) and let m
=
(
m
1
,
m
2
,
m
3
)
=
(
11,0
,
1
). It is easy to check that (32)
1,1
,
5
=
ρ
13
n
3
,
ρ
22
n
2
,
ρ
31
n
1
,
5,0
,
3
=
ρ
12
n
2
,
ρ
21
n
1
,
ρ
33
n
3
,
1,0
,
11
=
ρ
13
n
3
,
ρ
21
n
1
,
ρ
32
n
2
.That is,(33)
J
n
=
2,1
,
3
,
1,1
,
5
,
1,0
,
11
,
5,0
,
3
.Similarly,(34)
J
m
=
11,0
,
1
,
2,3
,
1
,
0,3
,
5
,
0,0
,
23
.By Lemma 5, we get f
2
(
z
)
=
a
2
z
1
2
z
2
z
3
3
+
a
4
z
1
z
2
z
3
5
+
a
5
z
1
z
3
11
+
a
6
z
1
5
z
3
3
∈
[
f
] and f
3
(
z
)
=
a
3
z
1
2
z
2
3
z
3
+
a
7
z
1
11
z
3
∈
[
f
]. Therefore, a
1
z
1
z
3
∈
[
f
]. Notice that(35)
M
∗
h
M
q
f
(
z
)
=
∑
i
=
1
3
‍
μ
i
f
i
(
z
)
z
(
q
-
h
)
N
,
if
0
≤
h
≤
q
0
,
if
0
≤
q
<
h
,where μ
1
=
γ
(
1,0
,
1
)
+
q
N
/
γ
(
1,0
,
1
)
+
(
q
-
h
)
N, μ
2
=
γ
(
2,3
,
1
)
+
q
N
/
γ
(
2,3
,
1
)
+
(
q
-
h
)
N, and μ
3
=
γ
(
11,0
,
1
)
+
q
N
/
γ
(
11,0
,
1
)
+
(
q
-
h
)
N. So(36)
f
=
⨁
i
=
1
3
[
f
i
]
=
⨁
i
=
1
3
M
i
.Ifα 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 M
z
N on A
α
2
(
D
3
) in the next section.


## 3. Reducing Subspaces onA
α
2
(
D
3
)

Letα
≠
0 be rational. We consider the reducing subspaces of M
z
N on A
α
2
(
D
3
). Recall (37)
Ω
=
n
=
n
1
,
n
2
,
n
3
∈
N
0
3
:
0
≤
n
i
<
N
i
for
some
i
,and J
n
=
{
q
∈
Ω
:
q
~
n
}; that is, m
∈
J
n if and only if γ
m
+
h
N
=
γ
n
+
h
N for h
∈
N
0. For every n
∈
Ω, if m
∈
J
n, we assume that m
i
≠
n
i for i
=
1,2
,
3. Otherwise, if there exists j such that m
j
=
n
j, we can prove that m
i
=
n
i for i
=
1,2
,
3 as in [16]. Since γ
β
=
∏
i
=
1
3
‍
ω
β
i
=
∏
i
=
1
3
‍
(
β
i
!
Γ
(
2
+
α
)
/
Γ
(
2
+
α
+
β
i
)
) and ω
β
i are decreasing as β
i is increasing, there exist i and j satisfying n
i
>
m
i and n
j
<
m
j.This section is organized as follows. Firstly, we considerm
∈
J
n under the assumption that n
1
>
m
1, n
2
>
m
2, and m
3
>
n
3. Let Q
1
=
{
1,2
,
…
,
n
,
…
}, Q
2
=
{
α
∈
Q
∖
Q
1
:
α
>
0
}, and Q
3
=
(
-
1,0
)
∩
Q. We give a description of m
∈
J
n in the cases that α is in Q
1
,
Q
2, and Q
3, respectively. Secondly, we get all the possible cases by symmetry (see Corollaries 11 and 13). Finally, we obtain Card
(
J
n
)
≤
2 and Theorem 14.Lemma 7.
Letα
∈
Q
2 and let n
∈
Ω. If m
∈
J
n satisfies n
1
>
m
1, n
2
>
m
2, and m
3
>
n
3, then one of the following statements holds:(1)
m
=
n
-
(
1,1
-
1
);
(2)
m
=
n
-
(
1,1
,
-
2
).Proof.
Letm
∈
J
n. By definition of J
n, as in Lemma 3, we have(38)
∏
i
=
1
2
‍
∏
j
=
1
n
i
-
m
i
‍
n
i
+
α
+
2
-
j
+
λ
N
i
m
i
+
j
+
h
N
i
=
∏
j
=
1
m
3
-
n
3
‍
m
3
+
α
+
2
-
j
+
λ
N
3
n
3
+
j
+
h
N
3for any λ
∈
C. Denote(39)
E
i
=
n
i
+
α
+
1
N
i
,
n
i
+
α
N
i
,
…
,
m
i
+
α
+
2
N
i
,
E
3
=
m
3
N
3
,
m
3
-
1
N
3
,
…
,
n
3
+
1
N
3
,
F
i
=
n
i
N
i
,
n
i
-
1
N
i
,
…
,
m
i
+
1
N
i
,
F
3
=
m
3
+
α
+
1
N
3
,
m
3
+
α
N
3
,
…
,
n
3
+
α
+
2
N
3
,for i
=
1,2. Then,(40)
⨆
i
=
1
3
E
i
=
⨆
i
=
1
3
F
i
,where ⨆ denotes the disjoint union. Since α
>
0 is not an integer, E
i
∩
F
i
=
∅ for i
=
1,2
,
3.
It is easy to see that(41)
max
⁡
⁡
n
1
+
α
+
1
N
1
,
n
2
+
α
+
1
N
2
,
m
3
N
3
=
max
⁡
⁡
n
1
N
1
,
n
2
N
2
,
m
3
+
α
+
1
N
3
.Since (
n
i
+
α
+
1
)
/
N
i
>
n
i
/
N
i for i
=
1,2 and (
m
3
+
α
+
1
)
/
N
3
>
m
3
/
N
3, we have(42)
m
3
+
α
+
1
N
3
∈
n
1
+
α
+
1
N
1
,
n
2
+
α
+
1
N
2
.Without loss of generality, assume(43)
n
1
+
α
+
1
N
1
=
m
3
+
α
+
1
N
3
.
Firstly, we prove thatn
1
-
m
1
=
1 by contradiction. Otherwise, if n
1
-
m
1
≥
2, then(44)
max
⁡
⁡
⨆
i
=
1
3
E
i
∖
n
1
+
α
+
1
N
1
=
max
⁡
⁡
⨆
i
=
1
3
F
i
∖
m
3
+
α
+
1
N
3
.Since max
⁡
{
(
n
1
+
α
)
/
N
1
,
(
n
2
+
α
+
1
)
/
N
2
}
>
max
⁡
{
n
1
/
N
1
,
n
2
/
N
2
}, we have m
3
-
n
3
≥
2 and(45)
max
⁡
n
1
+
α
N
1
,
n
2
+
α
+
1
N
2
,
m
3
N
3
=
max
⁡
n
1
N
1
,
n
2
N
2
,
m
3
+
α
N
3
=
m
3
+
α
N
3
>
m
3
N
3
.Therefore, (
m
3
+
α
)
/
N
3
∈
{
(
n
1
+
α
)
/
N
1
,
(
n
2
+
α
+
1
)
/
N
2
}. Since N
1
≠
N
3, it holds that (46)
m
3
+
α
N
3
=
n
2
+
α
+
1
N
2
.
We will find the contradictions under the assumptions (a)n
2
-
m
2
≥
2 and (b) n
2
-
m
2
=
1, respectively.
(a) Ifn
2
-
m
2
≥
2, then max
⁡
{
(
n
1
+
α
)
/
N
1
,
(
n
2
+
α
)
/
N
2
,
m
3
/
N
3
}
>
max
⁡
{
n
1
/
N
1
,
n
2
/
N
2
}. So m
3
-
n
3
≥
3 and(47)
max
⁡
n
1
+
α
N
1
,
n
2
+
α
N
2
,
m
3
N
3
=
max
⁡
n
1
N
1
,
n
2
N
2
,
m
3
+
α
-
1
N
3
.Since N
2
≠
N
3, we have (48)
m
3
+
α
-
1
N
3
=
n
1
+
α
N
1
.
By (43) and (48), we get 1
/
N
1
=
2
/
N
3 and m
3
/
N
3
>
n
1
/
N
1. It follows that(49)
max
⁡
n
1
+
α
-
1
N
1
,
n
2
+
α
N
2
,
m
3
N
3
≥
max
⁡
n
2
+
α
N
2
,
m
3
N
3
>
max
⁡
n
1
N
1
,
n
2
N
2
.Thus, m
3
-
n
3
≥
4 and(50)
m
3
+
α
-
2
N
3
=
max
⁡
n
1
N
1
,
n
2
N
2
,
m
3
+
α
-
2
N
3
.By N
1
≠
N
3 and equality (48), we conclude that(51)
m
3
+
α
-
2
N
3
=
n
2
+
α
N
2
.
Equalities (46) and (51) imply 2
/
N
3
=
1
/
N
2. Thus, N
2
=
N
1, which is in contradiction with the assumption.
(b) Supposen
2
-
m
2
=
1. Notice that E
2
=
{
(
n
2
+
α
+
1
)
/
N
2
}
⊂
F
3.
IfF
2
=
{
n
2
/
N
2
}
⊂
E
1, then equality (40) implies F
1
=
E
3. Equivalently, n
1
/
N
1
=
m
3
/
N
3 and (
n
1
-
1
)
/
N
1
=
(
m
3
-
1
)
/
N
3. Hence, N
1
=
N
3, which is impossible.
IfF
2
=
{
n
2
/
N
2
}
⊂
E
3, then F
1
∪
F
2
=
E
3. It follows that max
⁡
{
n
1
/
N
1
,
n
2
/
N
2
}
=
m
3
/
N
3. Since N
1
≠
N
3, equality (43) implies m
3
/
N
3
=
n
2
/
N
2. Therefore, n
1
/
N
1
=
(
m
3
-
1
)
/
N
3 and (
n
1
-
1
)
/
N
1
=
(
m
3
-
2
)
/
N
3. Then, N
3
=
N
1, which is a contradiction.
Summing up, we must haven
1
-
m
1
=
1.
Next, we prove thatn
2
-
m
2
=
1.
IfF
1
⊂
E
2, then F
2
=
E
3; that is, (52)
n
2
N
2
=
m
3
N
3
.In this case, n
2
-
m
2
=
m
3
-
n
3
=
1. Otherwise, n
2
/
N
2
=
m
3
/
N
3 and (
n
2
-
1
)
/
N
2
=
(
m
3
-
1
)
/
N
3, which is in contradiction with N
2
≠
N
3.
IfF
1
⊂
E
3, then E
2
=
F
3
∖
{
(
m
3
+
α
+
1
)
/
N
3
}. It follows that (53)
n
2
+
α
+
1
N
2
=
m
3
+
α
N
3
.In this case, n
2
-
m
2
=
1 and m
3
-
n
3
=
2. Or else, (
n
2
+
α
+
1
)
/
N
2
=
(
m
3
+
α
)
/
N
3 and (
n
2
+
α
)
/
N
2
=
(
m
3
+
α
-
1
)
/
N
3, which is in contradiction with N
2
≠
N
3. So we get the desired results.Lemma 8.
Fixα
∈
Q
1 and n
∈
Ω. If m
∈
J
n satisfies n
1
>
m
1, n
2
>
m
2, and m
3
>
n
3, then one of the following statements holds:(1)
m
=
n
-
(
1,1
-
1
);
(2)
m
=
n
-
(
1,1
,
-
2
).Proof.
Letk
i
=
min
⁡
{
α
+
1
,
|
n
i
-
m
i
|
} for i
=
1,2
,
3. Then, (54)
∏
i
=
1
2
‍
∏
j
=
1
k
i
‍
n
i
+
α
+
2
-
j
+
λ
N
i
m
i
+
j
+
λ
N
i
=
∏
j
=
1
k
3
‍
m
3
+
α
+
2
-
j
+
λ
N
3
n
3
+
j
+
λ
N
3for λ
∈
C. Let (55)
E
~
i
=
n
i
+
α
+
1
N
i
,
n
i
+
α
N
i
,
…
,
n
i
+
α
+
2
-
k
i
N
i
;
E
~
3
=
n
3
+
1
N
3
,
n
3
+
2
N
3
,
…
,
n
3
+
k
3
N
3
;
F
~
i
=
m
i
+
1
N
i
,
m
i
+
2
N
i
,
…
,
m
i
+
k
i
N
i
;
F
~
3
=
m
3
+
α
+
1
N
3
,
m
3
+
α
N
3
,
…
,
m
3
+
α
+
2
-
k
3
N
3for i
=
1,2. Then, (56)
⨆
i
=
1
3
E
~
i
=
⨆
i
=
1
3
F
~
i
,and E
~
i
∩
F
~
i
=
∅ for i
=
1,2
,
3. As in Lemma 7, we assume equality (43) holds. Then, we can prove that (
k
1
,
k
2
,
k
3
)
=
(
1,1
,
2
) or (
k
1
,
k
2
,
k
3
)
=
(
1,1
,
1
).
Sinceα
+
1
>
1, we have k
1
=
n
1
-
m
1
=
1 and k
2
=
n
2
-
m
2
=
1. It means Card
(
E
~
i
)
=
Card
(
F
~
i
)
=
1 for i
=
1,2.
Ifk
3
=
m
3
-
n
3, then statement (1) or statement (2) holds.
Ifk
3
=
α
+
1, then Card
(
E
~
3
)
=
Card
(
F
~
3
)
=
k
3
=
2 and α
=
1. Equalities (43) and (53) imply that(57)
n
1
+
2
N
1
=
m
3
+
2
N
3
,
n
2
+
2
N
2
=
m
3
+
1
N
3
.Since N
1
,
N
2, and N
3 are distinct, equality (56) shows that(58)
n
2
N
2
=
n
3
+
2
N
3
,
n
1
N
1
=
n
3
+
1
N
3
.
Then, we haveN
1
:
N
2
:
N
3
=
2
:
6
:
3 and (
n
1
,
n
2
,
m
3
,
n
3
)
=
r
(
N
1
,
N
2
,
N
3
,
N
3
)
+
(
0,2
,
1
,
-
1
). In this case, m
3
-
n
3
=
2
=
k
3. Hence, statement (2) holds.Lemma 9.
Fix a rational numberα
>
0 and a vector n
∈
Ω. If m
=
n
-
(
1,1
,
-
1
)
∈
J
n, then one of the following statements holds:(1)
1
/
N
1
+
1
/
N
2
-
1
/
N
3
=
0 and (59)
n
∈
r
N
1
,
N
2
,
N
3

+

α

+

2
N
1
N
3

-

α
-

1
,
N
2
N
3
,
0
:
r
∈
R
,
(2)
1
/
N
1
+
1
/
N
2
-
1
/
N
3
=
0 and (60)
n
∈
r
N
1
,
N
2
,
N
3

+

N
1
N
3
,
α

+

2
N
2
N
3
-

α

-

1,0
:
r
∈
R
.Proof.
As in Lemma7, we have (
m
3
+
α
+
1
)
/
N
3
∈
{
(
n
1
+
α
+
1
)
/
N
1
,
(
n
2
+
α
+
1
)
/
N
2
}.
If(
n
1
+
α
+
1
)
/
N
1
=
(
m
3
+
α
+
1
)
/
N
3, Lemmas 7 and 8 show that(61)
n
1
+
α
+
1
N
1
=
m
3
+
α
+
1
N
3
,
m
3
N
3
=
n
2
N
2
,
n
1
N
1
=
n
2
+
α
+
1
N
2
.Therefore,(62)
n
1
,
n
2
,
m
3
=
r
(
N
1
,
N
2
,
N
3
)
+
α
+
2
N
1
N
3
-
α
-
1
,
N
2
N
3
,
1
,where r
∈
R and 1
/
N
1
+
1
/
N
2
-
1
/
N
3
=
0. Statement (1) holds.
If(
n
2
+
α
+
1
)
/
N
2
=
(
m
3
+
α
+
1
)
/
N
3, then we have(63)
n
2
+
α
+
1
N
2
=
m
3
+
α
+
1
N
3
,
m
3
N
3
=
n
1
N
1
,
n
2
N
2
=
n
1
+
α
+
1
N
1
.In this case,(64)
n
1
,
n
2
,
m
3
=
r
N
1
,
N
2
,
N
3
+
N
1
N
3
,
α
+
2
N
2
N
3
-
α
-
1,1
,where r
∈
R and 1
/
N
1
+
1
/
N
2
-
1
/
N
3
=
0. So statement (2) holds.Lemma 10.
Fix a rational numberα
>
0 and a vector n
∈
Ω. If m
=
n
-
(
1,1
,
-
2
)
∈
J
n, then one of the following statements holds:(1)
N
1
:
N
2
:
N
3
=
α
(
α
+
1
)
:
(
α
+
1
)
(
α
+
2
)
:
α
(
α
+
2
) and (65)
n
∈
r
N
1
,
N
2
,
N
3
+
0
,
N
2
N
3
,
-
1
:
r
∈
R
,
(2)
N
1
:
N
2
:
N
3
=
(
α
+
1
)
(
α
+
2
)
:
α
(
α
+
1
)
:
α
(
α
+
2
) and (66)
n
∈
r
N
1
,
N
2
,
N
3
+
N
1
N
3
,
0
,
-
1
:
r
∈
R
.In this case, 1
/
N
1
+
1
/
N
2
-
2
/
N
3
=
0.Proof.
As in Lemma8, if (
n
1
+
α
+
1
)
/
N
1
=
(
m
3
+
α
+
1
)
/
N
3, then (
n
2
+
α
+
1
)
/
N
2
=
(
m
3
+
α
)
/
N
3. Since N
1
≠
N
3, we have n
1
/
N
1
≠
m
3
/
N
3. Similarly, equality (53) implies that n
2
/
N
2
≠
(
m
3
-
1
)
/
N
3. Therefore,(67)
n
1
+
α
+
1
N
1
=
m
3
+
α
+
1
N
3
,
n
2
+
α
+
1
N
2
=
m
3
+
α
N
3
,
n
2
N
2
=
m
3
N
3
,
n
1
N
1
=
m
3
-
1
N
3
.Hence,(68)
n
1
,
n
2
,
m
3
=
r
N
1
,
N
2
,
N
3
+
0
,
N
2
N
3
,
1
,where r
∈
R and N
1
:
N
2
:
N
3
=
α
(
α
+
1
)
:
(
α
+
1
)
(
α
+
2
)
:
α
(
α
+
2
). In this case, 1
/
N
1
+
1
/
N
2
-
2
/
N
3
=
0. Thus, we get (1).
If(
n
2
+
α
+
1
)
/
N
2
=
(
m
3
+
α
+
1
)
/
N
3, then it is easy to check that (2) holds.By symmetry, we can get the following corollary.Corollary 11.
Fix a rational numberα
>
0 and n
∈
Ω. If m
∈
J
n
∖
{
n
}, then one of the following statements holds.(1)
m
σ
(
1
)
=
n
σ
(
1
)
-
1
,
m
σ
(
2
)
=
n
σ
(
2
)
-
1
,
m
σ
(
3
)
=
n
σ
(
3
)
+
1, where σ
∈
S
k. In this case, 1
/
N
σ
(
1
)
+
1
/
N
σ
(
2
)
-
1
/
N
σ
(
3
)
=
0 and(69)
n
∈
r
(
N
1
,
N
2
,
N
3
)
+
(
k
1
,
k
2
,
k
3
)
:
α
+
2
N
σ
1
N
σ
3
N
σ
2
N
σ
3
k
σ
(
3
)
=
0
,
k
σ
1
=
α
+
2
N
σ
1
N
σ
3
-
α
-
1
k
σ
2
=
N
σ
2
N
σ
3
k
σ
3
=
0
,
r
∈
R
.
(2)
m
σ
(
1
)
=
n
σ
(
1
)
+
1
,
m
σ
(
2
)
=
n
σ
(
2
)
+
1
,
m
σ
(
3
)
=
n
σ
(
3
)
-
1, where σ
∈
S
k. In this case, 1
/
N
σ
(
1
)
+
1
/
N
σ
(
2
)
-
1
/
N
σ
(
3
)
=
0 and(70)
n
∈
r
(
N
1
,
N
2
,
N
3
)
+
(
k
1
,
k
2
,
k
3
)
:
α
+
2
N
σ
1
N
σ
3
N
σ
2
N
σ
3
-
1
k
σ
(
3
)
=
1
,
k
σ
(
1
)
=
α
+
2
N
σ
1
N
σ
3
-
α
-
2
k
σ
(
2
)
=
N
σ
2
N
σ
3
-
1
k
σ
(
3
)
=
1
,
r
∈
R
.
(3)
m
σ
(
1
)
=
n
σ
(
1
)
-
1
,
m
σ
(
2
)
=
n
σ
(
2
)
-
1
,
m
σ
(
3
)
=
n
σ
(
3
)
+
2, where σ
∈
S
k. In this case, N
σ
(
1
)
:
N
σ
(
2
)
:
N
σ
(
3
)
=
α
(
α
+
1
)
:
(
α
+
1
)
(
α
+
2
)
:
α
(
α
+
2
) and(71)
n
∈
r
(
N
1
,
N
2
,
N
3
)
+
(
k
1
,
k
2
,
k
3
)
:
k
σ
(
1
)
N
σ
2
N
σ
3
k
σ
(
3
)
,
k
σ
(
1
)
=
0
k
σ
(
2
)
=
N
σ
2
N
σ
3
k
σ
(
3
)
=
-
1
,
r
∈
R
.
(4)
m
σ
(
1
)
=
n
σ
(
1
)
+
1
,
m
σ
(
2
)
=
n
σ
(
2
)
+
1
,
m
σ
(
3
)
=
n
σ
(
3
)
-
2, where σ
∈
S
k. In this case, N
σ
(
1
)
:
N
σ
(
2
)
:
N
σ
(
3
)
=
α
(
α
+
1
)
:
(
α
+
1
)
(
α
+
2
)
:
α
(
α
+
2
) and (72)
n
∈
r
N
1
,
N
2
,
N
3
+
k
1
,
k
2
,
k
3
:
k
σ
(
1
)
=
-
1
N
σ
2
N
σ
3
-
1
k
σ
(
3
)
=
1
,
k
σ
(
1
)
=
-
1
k
σ
(
2
)
=
N
σ
2
N
σ
3
-
1
k
σ
(
3
)
=
1
,
r
∈
R
.Lemma 12.
Fix a rational numberα
∈
(
-
1,0
) and n
∈
Ω. If there exists m
∈
J
n such that n
1
>
m
1, n
2
>
m
2, and m
3
>
n
3, then one of the following statements holds: (1)
m
=
n
-
(
1,1
,
-
1
). In this case, 1
/
N
1
+
1
/
N
2
+
1
/
N
3
=
0 and(73)
n
∈
r
(
N
1
,
N
2
,
N
3
)
N
1
N
3
-
α
-
1
,
N
2
N
3
+
α
+
2
N
1
N
3
-
α
-
1
,
N
2
N
3
,
0
:
r
∈
R
,
or(74)
n
∈
r
(
N
1
,
N
2
,
N
3
)
N
1
N
3
,
N
2
N
3
+
N
1
N
3
,
α
+
2
N
2
N
3
-
α
-
1,0
:
r
∈
R
,
(2)
m
=
n
-
(
2,1
,
-
1
). In this case, N
1
:
N
2
:
N
3
=
-
α
(
α
+
2
)
:
(
α
+
1
)
(
α
+
2
)
:
-
α
(
α
+
1
) and (75)
n
∈
r
N
1
,
N
2
,
N
3
+
1
,
α
+
1
N
2
N
1
,
-
1
:
r
∈
R
.
(3)
m
=
n
-
(
1,2
,
-
1
). In this case, N
1
:
N
2
:
N
3
=
(
α
+
1
)
(
α
+
2
)
:
-
α
(
α
+
2
)
:
-
α
(
α
+
1
) and (76)
n
∈
r
N
1
,
N
2
,
N
3
+
α
+
1
N
1
N
2
,
1
,
-
1
:
r
∈
R
.Proof.
DefineE
i and F
i as in the proof of Lemma 7. Then, ⨆
i
=
1
3
E
i
=
⨆
i
=
1
3
F
i and E
i
∩
F
i
=
∅ for i
=
1,2
,
3. Assume that (77)
n
1
+
α
+
1
N
1
=
m
3
+
α
+
1
N
3
∈
F
3
.Then,(78)
max
⁡
⨆
i
=
1
3
E
i
∖
n
1
+
α
+
1
N
1
=
max
⁡
⨆
i
=
1
3
F
i
∖
m
3
+
α
+
1
N
3
.
Since(
n
2
+
α
+
1
)
/
N
2
>
n
2
/
N
2, (
n
1
+
α
)
/
N
1
<
n
1
/
N
1, and (
m
3
+
α
)
/
N
3
<
m
3
/
N
3, we have (79)
max
⁡
n
2
+
α
+
1
N
2
,
m
3
N
3
=
n
1
N
1
.By (77) and N
1
≠
N
3, we get (80)
n
1
N
1
=
n
2
+
α
+
1
N
2
∈
E
2
.
(a) Ifn
1
-
m
1
=
1, then F
2
=
E
3. Therefore, n
2
-
m
2
=
m
3
-
n
3
=
1. Otherwise, n
2
/
N
2
=
m
3
/
N
3 and (
n
2
-
1
)
/
N
2
=
(
m
3
-
1
)
/
N
3, which is in contradiction with N
2
≠
N
3.
In this case, we have(81)
n
1
+
α
+
1
N
1
=
m
3
+
α
+
1
N
3
,
n
1
N
1
=
n
2
+
α
+
1
N
2
,
n
2
N
2
=
m
3
N
3
.Hence, (82)
(
n
1
,
n
2
,
m
3
)
=
r
N
1
,
N
2
,
N
3

+

α

+

2
N
1
N
3

-

α

-

1
,
N
2
N
3
,
1
,where r
∈
R and (83)
1
N
1
+
1
N
2
=
1
N
3or, equivalently, (84)
n
=
r
(
N
1
,
N
2
,
N
3
)
+
α
+
2
N
1
N
3
-
α
-
1
,
N
2
N
3
,
0
;
m
=
n
1
-
1
,
n
2
-
1
,
n
3
+
1
.So (i) holds.
(b) Assumen
1
-
m
1
≥
2.
Since(
n
1
-
1
)
/
N
1
<
(
n
1
+
α
)
/
N
1, (
n
2
+
α
)
/
N
2
<
n
2
/
N
2, and (
m
3
+
α
)
/
N
3
<
m
3
/
N
3, we get(85)
max
⁡
n
1
+
α
N
1
,
n
2
+
α
N
2
,
m
3
N
3
=
max
⁡
n
1
+
α
N
1
,
m
3
N
3
=
n
2
N
2
=
max
⁡
n
1
-
1
N
1
,
n
2
N
2
,
m
3
+
α
N
3
.
We prove that(
n
1
+
α
)
/
N
1
=
n
2
/
N
2 by contradiction. Otherwise, if m
3
/
N
3
=
n
2
/
N
2, then equalities (77) and (80) imply equality (83). Therefore, 1
/
N
3
>
1
/
N
1 and (
n
1
+
α
)
/
N
1
>
(
m
3
+
α
)
/
N
3. Since -
1
<
α
<
0, we have (86)
max
⁡
n
1
+
α
N
1
,
m
3
-
1
N
3
>
max
⁡
n
1
-
1
N
1
,
m
3
+
α
N
3
,
max
⁡
n
1
+
α
N
1
,
n
2
+
α
N
2
,
m
3
-
1
N
3
>
max
⁡
n
1
-
1
N
1
,
n
2
-
1
N
2
,
m
3
+
α
N
3
,which contradicts ⨆
i
=
1
3
E
i
=
⨆
i
=
1
3
F
i. Thus, (87)
n
1
+
α
N
1
=
n
2
N
2
.
Combining equality (80), we get (88)
-
α
N
1
=
α
+
1
N
2
.
In the following, we prove thatn
2
-
m
2
=
1 by contradiction. Assume n
2
-
m
2
≥
2. By (
n
1
-
1
)
/
N
1
>
(
n
1
+
α
-
1
)
/
N
1, (
n
2
+
α
)
/
N
2
>
(
n
2
-
1
)
/
N
2, and m
3
/
N
3
>
(
m
3
+
α
)
/
N
3, we get (89)
max
⁡
n
2
+
α
N
2
,
m
3
N
3
=
max
⁡
n
1
-
1
N
1
,
n
2
-
1
N
2
=
n
1
-
1
N
1
.
If(
n
1
-
1
)
/
N
1
=
(
n
2
+
α
)
/
N
2, combining the equalities (87) and (88), we get N
1
=
N
2, which is a contradiction.
If(
n
1
-
1
)
/
N
1
=
m
3
/
N
3, equality (77) shows that N
1
:
N
3
=
(
α
+
2
)
:
(
α
+
1
). We also have m
3
-
n
3
≥
2, since (90)
max
⁡
n
1
+
α
-
1
N
1
,
n
2
+
α
N
2
>
max
⁡
n
1
-
2
N
1
,
n
2
-
1
N
2
.Therefore,(91)
max
⁡
n
1
+
α
-
1
N
1
,
n
2
+
α
N
2
,
m
3
-
1
N
3
=
m
3
+
α
N
3
=
max
⁡
n
1
-
2
N
1
,
n
2
-
1
N
2
,
m
3
+
α
N
3
.We conclude that(92)
m
3
+
α
N
3
=
max
⁡
n
2
+
α
N
2
,
n
1
+
α
-
1
N
1
.
If(
m
3
+
α
)
/
N
3
=
(
n
1
+
α
-
1
)
/
N
1, equality (77) shows that (93)
N
1
:
N
3
=
2
:
1
.So (
α
+
2
)
:
(
α
+
1
)
=
2
:
1; that is, α
=
0 which is in contradiction with α
<
0.
If(
m
3
+
α
)
/
N
3
=
(
n
2
+
α
)
/
N
2, the fact(94)
max
⁡
n
1
+
α
-
1
N
1
,
n
2
+
α
-
1
N
2
,
m
3
-
1
N
3
=
max
⁡
n
1
-
2
N
1
,
n
2
-
1
N
2
,
m
3
+
α
-
1
N
3
,implies that (95)
max
⁡
n
1
+
α
-
1
N
1
,
m
3
-
1
N
3
=
n
2
-
1
N
2
.However, (
n
2
-
1
)
/
N
2
≠
(
m
3
-
1
)
/
N
3 because of (
m
3
+
α
)
/
N
3
=
(
n
2
+
α
)
/
N
2 and N
2
≠
N
3 and (
n
2
-
1
)
/
N
2
≠
(
n
1
+
α
-
1
)
/
N
1, because of equality (87) and N
1
≠
N
2. We also get a contradiction. Hence, n
2
-
m
2
=
1.
Therefore,E
2
=
{
(
n
2
+
α
+
1
)
/
N
2
}
⊂
F
1 and F
2
=
{
n
2
/
N
2
}
⊂
E
1. Further, F
1
∖
{
n
1
/
N
1
}
=
E
3. It follows that m
3
-
n
3
=
1, n
1
-
m
1
=
2, and m
3
/
N
3
=
(
n
1
-
1
)
/
N
1. Summing up, we have(96)
n
1
+
α
+
1
N
1
=
m
3
+
α
+
1
N
3
,
n
1
N
1
=
n
2
+
α
+
1
N
2
,
n
2
N
2
=
n
1
+
α
N
1
,
m
3
N
3
=
n
1
-
1
N
1
.Hence, (
n
1
,
n
2
,
m
3
)
=
r
(
N
1
,
N
2
,
N
3
)
+
(
1
,
(
α
+
1
)
N
2
/
N
1
,
0
), where r
∈
R and(97)
N
1
:
N
2
:
N
3
=
-
α
(
α
+
2
)
:
(
α
+
1
)
(
α
+
2
)
:
-
α
(
α
+
1
)
.Thus, (2) holds. In this case, 2
/
N
1
+
1
/
N
2
-
1
/
N
3
=
0.
Similarly, if(
n
2
+
α
+
1
)
/
N
2
=
(
m
3
+
α
+
1
)
/
N
3, by symmetry, we can get (3) and another part of (1).By symmetry, we also have the following corollary.Corollary 13.
Letα
∈
(
-
1,0
)
∩
Q and let n
∈
Ω. If there exists m
∈
J
n such that n
≠
m, then there is a permutation σ
∈
S
k such that one of the following statements holds.(1)
1
/
N
σ
(
1
)
+
1
/
N
σ
(
2
)
-
1
/
N
σ
(
3
)
=
0 and(98)
n
∈
r
(
N
1
,
N
2
,
N
3
)
+
(
k
1
,
k
2
,
k
3
)
:
α
+
2
N
σ
1
N
σ
3
N
σ
2
N
σ
3
k
σ
(
3
)
,
k
σ
(
1
)
=
α
+
2
N
σ
1
N
σ
3
-
α
-
1
k
σ
(
2
)
=
N
σ
2
N
σ
3
k
σ
(
3
)
=
0
,
r
∈
R
.
In this case,m
σ
(
1
)
=
n
σ
(
1
)
-
1, m
σ
(
2
)
=
n
σ
(
2
)
-
1, and m
σ
(
3
)
=
n
σ
(
3
)
+
1.
(2)
1
/
N
σ
(
1
)
+
1
/
N
σ
(
2
)
-
1
/
N
σ
(
3
)
=
0 and (99)
n
∈
r
N
1
,
N
2
,
N
3
+
k
1
,
k
2
,
k
3
:
α
+
2
N
σ
1
N
σ
3
-
α
-
2
N
σ
2
N
σ
3
-
1
k
σ
(
3
)
=
1
,
k
σ
(
1
)
=
α
+
2
N
σ
1
N
σ
3
-
α
-
2
k
σ
(
2
)
=
N
σ
2
N
σ
3
-
1
k
σ
(
3
)
=
1
,
r
∈
R
.
In this case,m
σ
(
1
)
=
n
σ
(
1
)
+
1, m
σ
(
2
)
=
n
σ
(
2
)
+
1, and m
σ
(
3
)
=
n
σ
(
3
)
-
1.
(3)
N
σ
(
1
)
:
N
σ
(
2
)
:
N
σ
(
3
)
=
-
α
(
α
+
2
)
:
(
α
+
1
)
(
α
+
2
)
:
-
α
(
α
+
1
) and (100)
n
∈
r
(
N
1
,
N
2
,
N
3
)
+
(
k
1
,
k
2
,
k
3
)
:
k
σ
(
1
)
α
+
1
N
σ
2
N
σ
1
k
σ
(
3
)
=
,
k
σ
(
1
)
=
1
k
σ
(
2
)
=
α
+
1
N
σ
2
N
σ
1
k
σ
(
3
)
=
-
1
,
r
∈
R
.
In this case,m
σ
(
1
)
=
n
σ
(
1
)
-
2, m
σ
(
2
)
=
n
σ
(
2
)
-
1, and m
σ
(
3
)
=
n
σ
(
3
)
+
1.
(4)
N
σ
(
1
)
:
N
σ
(
2
)
:
N
σ
(
3
)
=
-
α
(
α
+
2
)
:
(
α
+
1
)
(
α
+
2
)
:
-
α
(
α
+
1
) and (101)
n
∈
r
N
1
,
N
2
,
N
3
+
k
1
,
k
2
,
k
3
:
k
σ
(
1
)
=
-
1
α
+
1
N
σ
2
N
σ
1
k
σ
(
3
)
=
0
,
k
σ
(
1
)
=
-
1
k
σ
(
2
)
=
α
+
1
N
σ
2
N
σ
1
-
1
k
σ
(
3
)
=
0
,
r
∈
R
.
In this case,m
σ
(
1
)
=
n
σ
(
1
)
+
2, m
σ
(
2
)
=
n
σ
(
2
)
+
1, and m
σ
(
3
)
=
n
σ
(
3
)
-
1.By careful computation, we find that each choice ofn and N cannot simultaneously satisfy two of the statements in Corollary 11 or Corollary 13. So Card
(
J
n
)
≤
2. Moreover, there are finite numbers of n
∈
Ω such that Card
(
J
n
)
=
2. Denote(i)
M
n
=
span
⁡
¯
{
z
n
+
h
N
:
h
∈
N
0
};
(ii)
H
n
=
span
⁡
¯
{
(
a
z
n
+
b
z
n
-
(
i
,
j
,
k
)
)
z
h
N
:
h
∈
N
0
}, (
i
,
j
,
k
) is a permutation of {
1,1
,
-
1
} or {
1,1
,
-
2
};
(iii)
N
n
=
span
⁡
¯
{
(
a
z
n
+
b
z
n
-
(
i
,
j
,
k
)
)
z
h
N
:
h
∈
N
0
}, (
i
,
j
,
k
) is a permutation of {
1,1
,
-
1
} or {
2,1
,
-
1
}.Theorem 2 implies the following statements.(a)
Ifα
>
0, then each reducing subspace of M
z
N is the direct sum of some minimal reducing as in (i) and (ii), where the number of reducing subspaces as (ii) is finite.
(b)
If-
1
<
α
<
0, then each reducing subspace of M
z
N is the direct sum of some minimal reducing as in (i) and (iii), where the number of reducing subspaces as (iii) is finite.Finally, we consider the reducing subspaces from the viewpoint of von Neumann algebras. Denote byW
∗
(
z
N
) the von Neumann algebra generated by M
z
N and ν
∗
(
z
N
) the commutant of W
∗
(
z
N
). Then v
∗
(
z
N
) is a von Neumann algebra, and it is generated by its self-adjoint projections. For each reducing subspace M of M
z
N, denote by P
M the orthogonal projection from A
α
2
(
D
3
) onto M. It is known that P
M is a self-adjoint projection in ν
∗
(
z
N
). Conversely, if P is a self-adjoint projection in ν
∗
(
z
N
), then the range of P is a reducing subspace of M
z
N. So our results can be written in the following form.Theorem 14.
The von Neumann algebraν
∗
(
z
N
) is ∗
- isomorphic to (102)
⨁
n
=
1
m
M
2
C
⨁
⨁
n
=
1
+
∞
C
,where 0
≤
m
<
+
∞.



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*Source: 209307-2015-04-01.xml*