Cross commutators on backward shift invariant subspaces over the bidisk II

http://dx.doi.org/10.4134/JKMS.2012.49.1.139

CROSS COMMUTATORS ON BACKWARD SHIFT INVARIANT SUBSPACES OVER THE BIDISK II

Kei Ji Izuchi and Kou Hei Izuchi

Abstract. In the previous paper, we gave a characterization of backward shift invariant subspaces of the Hardy space over the bidisk on which [Szⁿ, S^∗_w] = 0 for a positive integer n≥ 2. In this case, it holds that Szⁿ= cI for some c∈ C. In this paper, it is proved that if [Sφ, S^∗_w] = 0 and φ∈ H^∞(Γz), then Sφ= cI for some c∈ C.

1. Introduction

Let Γ

be the 2-dimensional unit torus. We write (z, w) = (e

, e

) for variables in Γ

= Γ

× Γ

. Let L

= L

(Γ

) be the usual Lebesgue space on Γ

with the norm

∥f∥

= ( ∫

0

∫

|f(e

, e

) |

dsdt (2π)

)

.

With the usual inner product, L

is a Hilbert space. Let H

= H

(Γ

) be the Hardy space over Γ

. We denote by H

(Γ

) and H

(Γ

) the Hardy spaces on the unit circle Γ in variables z and w, respectively. We think of H

(Γ

) and H

(Γ

) as closed subspaces H

. For each f ∈ H

, we can write f as

f =

∑

⊕f

(w)z

, f

(w) ∈ H

(Γ

).

Let P be the orthogonal projection from L

onto H

. For a closed subspace M of L

, we denote by P

the orthogonal projection from L

onto M . For a function ψ ∈ L

, the Toeplitz operator T

on H

is defined by T

f = P (ψf ) for f ∈ H

. It is well known that T

= T

, and T

T

= T

T

for every φ(z) ∈ H

(Γ

) and ψ(w) ∈ H

(Γ

). A function f ∈ H

is called inner if |f| = 1 on Γ

almost everywhere. A nonzero closed subspace M of H

is

Received September 20, 2010; Revised December 17, 2010.

2010 Mathematics Subject Classification. Primary 47A15, 32A35; Secondary 47B35.

Key words and phrases. backward shift invariant subspace, invariant subspace, Hardy space, cross commutator.

The first author is partially supported by Grant-in-Aid for Scientific Research (No.16340037), Japan Society for the Promotion of Science.

⃝2012 The Korean Mathematical Societyc 139

called invariant if zM ⊂ M and wM ⊂ M. In one variable case, the well known Beurling theorem [2] says that an invariant subspace M of H

(Γ

) has a form M = q(z)H

(Γ

), where q(z) is an inner function. In two variable case, the structure of invariant subspaces of H

is extremely complicated, see [3, 10].

Let M be an invariant subspace of H

with M ̸= {0} and M ̸= H

. Then T

(H

⊖ M) ⊂ H

⊖ M and T

(H

⊖ M) ⊂ H

⊖ M. In this paper, we write

N = H

⊖ M.

Usually, N is called a backward shift invariant subspace of H

. See [1, 9] for studies of backward shift invariant subspaces over the unit circle Γ.

For a function ψ ∈ L

, we denote by R

the operator on M defined by R

f = P

(ψf ) for f ∈ M. It holds R

= R

and R

= T

|

. We denote by [R

, R

] the cross commutator of R

and R

, that is, [R

, R

] = R

R

− R

R

. In [8], Mandrekar proved that [R

, R

] = 0 if and only if M is Beurling type, that is, M = qH

for some inner function q on Γ

. This is a nice characterization of Beurling type invariant subspaces of H

. More generally, in [4] the authors proved that [R

, R

] = 0 if and only if [R

, R

2(w)

] = 0 for nonconstant functions ψ

(z), ψ

(w) ∈ H

(Γ).

We define the operator S

on N by S

f = P

(ψf ) for f ∈ N. Then we have S

= S

and S

= T

|

. In [6], it is proved that [S

, S

] = 0 if and only if N has one of the following forms;

· N = H

⊖ q

(z)H

,

· N = H

⊖ q

(w)H

,

· N = (H

⊖ q

(z)H

) ∩ (H

⊖ q

(w)H

)

for nonconstant one variable inner functions q

(z) and q

(w). In [7], it is shown that the condition [S

, S

] = 0 does not imply [S

, S

] = 0. In [5], the authors proved that for n ≥ 2, [S

, S

] = 0 if and only if one of the following conditions holds;

(i) [S

, S

] = 0, (ii) S

S

= 0,

(iii) there exists a Blaschke product b(z) with b(z) =

∏

z − α

1 − α

z , 0 < |α

| < 1,

where α

̸= α

for every i, j with i ̸= j and α

= α

= · · · = α

such that N ⊂ H

⊖ b(z)H

.

In [7, Theorem 2.2], it is proved that (ii) holds if and only if either N ⊂ H

(Γ

) or N ⊂ H

⊖ z

H

. If N ⊂ H

(Γ

), then we have [S

, S

] = 0. Moreover, in [5] it is proved that if [S

, S

] = 0 and [S

, S

] ̸= 0, then M ∩ H

(Γ

) = θ(z)H

(Γ

) for an inner function θ(z), and z

∈ C + θ(z)H

(Γ

). In this case, we have S

= cI for some c ∈ C.

The purpose of this paper is to generalize the above phenomeron. Let

φ(z) ∈ H

(Γ

) be a nonconstant function. Suppose that [S

, S

] = 0 and

[S

, S

] ̸= 0. In Section 2, we prove that M ∩H

(Γ

) ̸= {0} and M ∩H

(Γ

) ̸=

H

(Γ

). Hence by the Beurling theorem, M ∩ H

(Γ

) = θ(z)H

(Γ

) for a non- constant inner function θ(z). Thus we get θ(z)H

⊂ M. Write

M

= M ⊖ θ(z)H

.

We prove that M

̸= {0} and T

M

⊂ M

. In another word, φ(z)N ⊂ N ⊕ θ(z)H

holds. In Section 3, we study on the one variable Hardy space H

(Γ

). Let N

, N

be backward shift invariant subspaces of H

(Γ

) satisfying {0} ̸= N

⫋ N

̸= H

(Γ

). It is proved that φ(z)N

⊂ N

⊕ (H

(Γ

) ⊖ N

) if and only if φ(z) ∈ C+(H

(Γ

) ⊖N

). As applications of these facts, in Section 4 we prove that φ(z) ∈ C + θ(z)H

(Γ

) and S

= cI for some c ∈ C.

2. Equivalent conditions for [S

, S

] = 0

Let N be a backward shift invariant subspace of H

with N ̸= {0} and N ̸=

H

, and let φ(z) ∈ H

(Γ

) be a nonconstant function. We write operators T

and T

on H

= M ⊕ N in the matrix forms as

T

=

( ∗ P

T

|

0 S

)

, T

=

( ∗ 0

P

T

|

S

)

on H

=



 M

⊕ N



 .

Let

A = P

T

|

and B = P

T

|

. Since T

T

= T

T

on H

, we have

S

S

= BA + S

S

. Hence we get the following.

Lemma 2.1. [S

, S

] = 0 if and only if BA = 0.

It is not difficult to see that

ker B = {f ∈ M : T

f ∈ M}

= {f ∈ M ⊖ wM : T

f = 0 } ⊕ wM

= (

M ∩ H

(Γ

) )

⊕ wM and

range A = M ⊖ ker A

= M ⊖ {f ∈ M : T

f ∈ M}.

Then by Lemma 2.1, we have the following.

Lemma 2.2. [S

, S

] = 0 if and only if M ⊖ {f ∈ M : T

f ∈ M} ⊂ (

M ∩ H

(Γ

) )

⊕ wM.

Lemma 2.3. If [S

, S

] = 0 and [S

, S

] ̸= 0, then M ∩H

(Γ

) is a nontrivial

invariant subspace of H

(Γ

).

Proof. Since M ̸= H

, trivially M ∩ H

(Γ

) ̸= H

(Γ

) holds. Suppose that M ∩ H

(Γ

) = {0}. By Lemma 2.2,

M ⊖ {f ∈ M : T

f ∈ M} ⊂ wM.

Hence

M ⊖ wM ⊂ {f ∈ M : T

f ∈ M}.

Since T

T

= T

T

on H

, if f ∈ M and T

f ∈ M, then T

(w

f ) = w

T

f ∈ M for every n ≥ 0, so that by the above we get

w

(M ⊖ wM) ⊂ {f ∈ M : T

f ∈ M}.

Therefore

M =

∑

⊕w

(M ⊖ wM) ⊂ {f ∈ M : T

f ∈ M}.

Thus we get T

M ⊂ M. This shows that φ(z)N ⊂ N.

Let

A = {ψ(z) ∈ H

(Γ

) : ψN ⊂ N}.

Then both functions 1 and φ(z) are contained in A. For ψ ∈ A and h ∈ N, we have

N ∋ T

(ψh) = (T

ψ)h + ψ(0)T

h.

Hence (T

ψ)N ⊂ N, so that T

A ⊂ A. It is easy to see that A is a weak-

closed subalgebra of H

(Γ

). Let

L = {

f (z) ∈ H

(Γ

) :

∫

f (e

)ψ(e

) dθ

2π = 0 for every ψ(z) ∈ A } . Then L is a closed subspace of H

(Γ

). Since T

A ⊂ A and 1 ∈ A, we have zL ⊂ L.

Suppose that L ̸= {0}. By the Beurling theorem, L = q(z)H

(Γ

) for an inner function q(z). Since 1 ∈ A, q(0) = 0. Hence zq(z) ∈ H

(Γ

). Since φ(z)

∈ A for n ≥ 1,

∫

e

q(e

)φ(e

)

e

h(e

) dθ 2π =

∫

q(e

)h(e

)φ(e

)

dθ 2π = 0 for every h(z) ∈ H

(Γ

). Hence zq(z)φ(z)

∈ H

(Γ

) for every n ≥ 1. By the Schneider theorem [11], we have φ(z) ∈ H

(Γ

). This shows that φ(z) is constant. Since we assumed that φ(z) is nonconstant, this is a contradiction.

Therefore L = {0}. Hence A = H

(Γ

). Especially, we have z ∈ A and zN ⊂ N. Then T

|

= S

. Since T

|

= S

and T

T

= T

T

on H

, we have S

S

= S

S

. This is a desired contradiction. □ In the rest of this section, we assume that M ∩H

(Γ

) ̸= {0}. Since M ̸= H

, M ∩ H

(Γ

) ̸= H

(Γ

). By the Beurling theorem,

M ∩ H

(Γ

) = θ(z)H

(Γ

)

for some nonconstant inner function θ(z). Hence θ(z)H

⊂ M. Write M

= M ⊖ θ(z)H

.

Then

M = M

⊕ θ(z)H

and H

⊖ θ(z)H

= M

⊕ N.

By the definition of M

, we have wM

⊂ M

and M

∩ H

(Γ

) = {0}. Note that if [S

, S

] = 0 and [S

, S

] ̸= 0, then M

̸= {0}. For, if M

= {0}, then M = θ(z)H

and N = H

⊖ θ(z)H

. Then we have [S

, S

] = 0, see [6], and this is a contradiction.

Lemma 2.4. Let f ∈ M

. Then T

f ∈ M

if and only if f ∈ wM

. Proof. Suppose that T

f ∈ M

. Then

f − f(z, 0) ∈ wM

⊂ M

.

Since f ∈ M

, f (z, 0) ∈ M

. Since M

∩ H

(Γ

) = {0}, f(z, 0) = 0. Hence

f ∈ wM

. The converse is trivial. □

Let P

be the orthogonal projection from H

onto H

⊖ θ(z)H

, and Q

be the operator on H

⊖ θ(z)H

defined by Q

f = P

(φf ) for f ∈ H

⊖ θ(z)H

. We can write both operators Q

and T

|

as

Q

=

( ∗ P

T

|

0 S

)

on H

⊖ θ(z)H

=



 M

⊕ N





and

T

|

=

( ∗ 0

P

T

|

S

)

on H

⊖ θ(z)H

=



 M

⊕ N



 .

Let

A

= P

T

|

and B

= P

T

|

. Lemma 2.5. [S

, S

] = 0 if and only if B

A

= 0.

Proof. Let f ∈ H

⊖θ(z)H

= M

⊕N. We have T

(φ(z)f ) = φ(z)T

f . Write φ(z)f = Q

f ⊕ f

∈ (M

⊕ N) ⊕ θ(z)H

.

Since T

f

∈ θ(z)H

and T

(Q

f ) ⊥ θ(z)H

, we get T

Q

f = Q

T

f . Thus Q

T

= T

Q

on M

⊕ N. Similarly as Lemma 2.1, we can prove the

assertion. □

The following is a slight generalization of [7, Theorem 4.4].

Theorem 2.6. The following conditions are equivalent;

(i) [S

, S

] = 0,

(ii) M

⊖ {f ∈ M

: T

f ∈ M

} ⊂ wM

,

(iii) T

M

⊂ M

,

(iv) φ(z)N ⊂ N ⊕ θ(z)H

. Proof. By Lemma 2.4,

ker B

= {f ∈ M

: T

f ∈ M

} = wM

. Also we have

range A

= M

⊖ ker A

= M

⊖ {f ∈ M

: T

f ∈ M

}.

Hence by Lemma 2.5, we get (i) ⇔ (ii).

If (ii) holds, then

M

⊖ wM

⊂ {f ∈ M

: T

f ∈ M

}.

Hence for each n ≥ 0, we have

T

w

(M

⊖ wM

) = w

T

(M

⊖ wM

) ⊂ w

M

⊂ M

. Since

M

=

∑

⊕w

(M

⊖ wM

), we have T

M

⊂ M

. Thus we get (iii).

(iii) ⇒ (ii) is trivial.

It is not difficult to see that (iii) ⇔ (iv). □

Suppose that [S

, S

] = 0 and [S

, S

] ̸= 0. Then we proved that θ(z)H

⫋ M and φ(z)(H

⊖ M) ⊂ (H

⊖ M) ⊕ θ(z)H

.

Note that θ(z)H

and M are invariant subspaces of H

. Now we fix an inner function θ(z). Here we have a question for which φ(z) ∈ H

(Γ

) satisfies the above condition. In the next section, we study a similar question in the one variable Hardy space H

(Γ

). In Section 4, we revisit on this question.

3. A theorem on the unit circle In this section, we prove the following theorem.

Theorem 3.1. Let N

, N

be backward shift invariant subspaces of H

(Γ

) with 0 ̸= N

⫋ N

̸= H

(Γ

), and φ(z) ∈ N

. Then

φ(N

∩ H

(Γ

)) ⊂ N

⊕ (H

(Γ

) ⊖ N

)

if and only if φ(z) = cP

1 for some c ∈ C. In this case, if we define the operator S

on N

by S

f = P

(φf ) for f ∈ N

, then S

= cI.

To prove the theorem, we need two lemmas which are not difficult to show.

Lemma 3.2. Let N be a backward shift invariant subspace of H

(Γ

). Then N ∩ H

(Γ

) is dense in N .

Lemma 3.3. Let N be a backward shift invariant subspace of H

(Γ

) with N ̸= {0} and N ̸= H

(Γ

). If φ ∈ H

(Γ

) is a nonconstant function, then φ (

N ∩ H

(Γ

) )

̸⊂ N.

Proof of Theorem 3.1. By the Beurling theorem, H

(Γ

) ⊖ N

= θH

(Γ

) for some nonconstant inner function θ.

First, suppose that

φ(N

∩ H

(Γ

)) ⊂ N

⊕ (H

(Γ

) ⊖ N

).

Since N

̸= {0}, by Lemma 3.2 there exists h

∈ N

∩ H

(Γ

) with h

(0) = 1.

Write

(3.1) φh

= f

⊕ θg

∈ N

⊕ (

H

(Γ

) ⊖ N

) = N

⊕ θH

(Γ

).

Also for each h ∈ N

∩ H

(Γ

), we can write

(3.2) φh = f ⊕ θg ∈ N

⊕ θH

(Γ

).

When h(0) = 0, we shall prove that

(3.3) g(0) = 0.

Since

T

(φh) = φT

h + h(0)T

φ = φT

h, by (3.2) we have

φT

h = T

(φh) = T

(f + θg)

= T

f + θT

g + g(0)T

θ

= (T

f + g(0)T

θ) + θT

g.

Note that T

h ∈ N

∩H

(Γ

) and T

f + g(0)T

θ ⊥ θH

(Γ

). By the assump- tion, φT

h ∈ N

⊕ θH

(Γ

). Hence

T

f + g(0)T

θ ∈ N

.

Since T

f ∈ N

, g(0)T

θ ∈ N

. To prove (3.3), suppose that g(0) ̸= 0. Then T

θ ∈ N

. Let N be a backward shift invariant subspace generated by T

θ.

Since N

= H

(Γ

) ⊖ θH

(Γ

), we have N = N

. Since T

θ ∈ N

, N ⊂ N

. This contradicts N

⫋ N

. Therefore g(0) = 0. Thus we get (3.3).

By (3.1) and (3.2),

φ(h − h(0)h

) = (f − h(0)f

) ⊕ θ(g − h(0)g

) ∈ N

⊕ θH

(Γ

).

Since (h − h(0)h

)(0) = 0, by (3.3) we get

(3.4) g(0) = h(0)g

(0).

By (3.2) again,

φT

h + h(0)T

φ = T

(φh) = (T

f + g(0)T

θ) + θT

g, so that

φT

h = (

− h(0)T

φ + T

f + g(0)T

θ )

⊕ θT

g.

Since T

h ∈ N

∩ H

(Γ

) and φ ⊥ θH

(Γ

), by the assumption we have

−h(0)T

φ + T

f + g(0)T

θ ∈ N

.

Similarly we have φT

h =

( − (T

h)(0)T

φ − h(0)T

φ + T

f + g(0)T

θ

+(T

g)(0)T

θ

) ⊕ θT

g.

Repeating the same argument, we get φT

h =

[ − (

∑

j=0

( T

h )

(0)T

φ )

+ T

f

+ (

∑

j=0

(T

g)(0)T

θ

)] ⊕ θT

g.

Since h ∈ N

∩ H

(Γ

), T

h ∈ N

∩ H

(Γ

). Hence by (3.2) and (3.4), (T

g)(0) = (T

h)(0)g

(0)

for every n ≥ 0. This shows that g = g

(0)h. By (3.2), we obtain (φ − g

(0)θ)h = f ∈ N

for every h ∈ N

∩ H

(Γ

). By Lemma 3.3, φ − g

(0)θ is constant. Write φ − g

(0)θ = c. Since φ ∈ N

, we have φ = cP

1.

Next, suppose that φ = cP

1. Then

φ = cP

1 = c(1 − θ(0)θ).

Hence for f ∈ N

∩ H

(Γ

), we have

φf = cf − cθ(0)θf ∈ N

⊕ θH

(Γ

).

Thus we get φ (

N

∩ H

(Γ

) )

⊂ N

⊕ (H

(Γ

) ⊖ N

). □ Corollary 3.4. Let N

, N

be backward shift invariant subspaces of H

(Γ

) with {0} ̸= N

⫋ N

̸= H

(Γ

), and φ(z) ∈ L

(Γ

). Define the operator S

on N

by S

h = P

(φh) for h ∈ N

. Then S

N

⊂ N

if and only if φ ∈ C + H

(Γ

)

+ (H

(Γ

) ⊖ N

) = H

(Γ

) + (H

(Γ

) ⊖ N

).

Proof. Write H

(Γ

) ⊖ N

= θH

(Γ

) for some inner function θ. Let φ = φ

⊕ φ

⊕ θφ

∈ H

(Γ

)

⊕ N

⊕ θH

(Γ

).

It is easy to see that P

( φ

(N

∩ H

(Γ

)) )

⊂ N

and

P

( θφ

(N

∩ H

(Γ

)) )

= {0}.

Hence S

N

⊂ N

if and only if P

(

φ

(N

∩ H

(Γ

)) )

⊂ N

. By Theorem 3.1, S

N

⊂ N

if and only if

φ = φ

+ cP

1 + θφ

= φ

+ c(1 − θ(0)θ) + θφ

= φ

+ c + θ(φ

− cθ(0)).

This completes the proof. □

The following corollaries follow from Corollary 3.4 directly.

Corollary 3.5. Let N

, N

be backward shift invariant subspaces of H

(Γ

) with {0} ̸= N

⫋ N

̸= H

(Γ

), and φ(z) ∈ H

(Γ

). Then φN

⊂ N

⊕ (H

(Γ

) ⊖ N

) if and only if φ ∈ C + (H

(Γ

) ⊖ N

).

Corollary 3.6. Let N

, N

be backward shift invariant subspaces of H

(Γ

) with {0} ̸= N

⊂ N

̸= H

(Γ

), and φ(z) ∈ H

(Γ

). If φN

⊂ N

⊕(H

(Γ

) ⊖ N

), then N

= N

if and only if φ / ∈ C + (H

(Γ

) ⊖ N

).

Corollary 3.7. Let M

, M

be invariant subspaces of H

(Γ

) with {0} ̸= M

⫋ M

̸= H

(Γ

), and φ(z) ∈ H

(Γ

). Then T

(M

⊖ M

) ⊂ M

⊖ M

if and only if φ ∈ C + M

.

Corollary 3.8. Let M

, M

be invariant subspaces of H

(Γ

) with {0} ̸= M

⫋ M

⊂ H

(Γ

), and φ(z) ∈ H

(Γ

). If T

(M

⊖ M

) ⊂ M

⊖ M

, then φ / ∈ C + M

if and only if M

= H

(Γ

).

4. The main theorem

As applications of the results in Sections 2 and 3, we prove the following.

Theorem 4.1. Let N be a backward shift invariant subspace of H

with N ̸=

{0} and N ̸= H

. Let φ(z) ∈ H

(Γ

) be a nonconstant function. If [S

, S

] = 0 and [S

, S

] ̸= 0, then φ(z) − c ∈ M ∩ H

(Γ

) for some c ∈ C and S

= cI.

Proof. By Lemma 2.3, M ∩ H

(Γ

) = θ(z)H

(Γ

) for a nonconstant inner function θ(z). Since θ(z)H

⊂ M, as in Section 2 we write

(4.1) M

= M ⊖ θ(z)H

.

Since [S

, S

] ̸= 0, we have M

̸= {0}. By Theorem 2.6,

(4.2) φ(z)N ⊂ N ⊕ θ(z)H

and

(4.3) T

M

⊂ M

.

To prove the assertion, we assume that

(4.4) φ(z) − c /∈ θ(z)H

(Γ

)

for every c ∈ C. We shall prove that [S

, S

] = 0. This will be a desired

contradiction. We consider two cases θ(0) = 0 and θ(0) ̸= 0 separately.

Case 1. Suppose that θ(0) = 0. If θ(z) = cz for some constant c with |c| = 1, then it is easy to see that

M = θ(z)H

+ q(w)H

for either a nonconstant inner function q(w) or q(w) = 0. In this case, by [6]

we have [S

, S

] = 0. So, we may assume that θ(z) = zθ

(z) for a nonconstant inner function θ

(z). Then

(4.5) H

⊖ θ(z)H

= H

(Γ

) ⊕ z (

H

⊖ θ

(z)H

) . We divide the proof into two subcases.

Subcase 1.1. Assume that θ

(z)M

⊂ θ(z)H

. Then M

⊂ zH

. Hence H

(Γ

) ⊂ N. For each nonnegative integer n, let

L

= {

f (z) ∈ H

(Γ

) ⊖ θ(z)H

(Γ

) : w

f (z) ∈ N } .

Then 1 ∈ L

, L

is a nonzero closed subspace of H

(Γ

) ⊖ θ(z)H

(Γ

), and T

L

⊂ L

. By (4.2),

w

φ(z)L

⊂ φ(z)N ⊂ N ⊕ θ(z)H

, so we have

φ(z)L

⊂ L

⊕ θ(z)H

(Γ

).

By (4.4) and Corollary 3.6, L

= H

(Γ

) ⊖ θ(z)H

(Γ

). Hence w

(

H

(Γ

) ⊖ θ(z)H

(Γ

) )

⊂ N for every n ≥ 0. Therefore

H

⊖ θ(z)H

=

∑

⊕w

(

H

(Γ

) ⊖ θ(z)H

(Γ

) )

⊂ N.

By (4.1), H

⊖θ(z)H

= M

⊕N, so that M

= {0}. This contradicts [S

, S

] ̸=

0.

Subcase 1.2. Assume that θ

(z)M

̸⊂ θ(z)H

. By (4.5), for every g ∈ M

we can write

(4.6) g = f

(w) ⊕ zh

(z, w),

where f

∈ H

(Γ

) and h

∈ H

⊖ θ

(z)H

. Since θ

(z)M

⊂ M, we have θ

(z)g = θ

(z)f

(w) ⊕ zθ

(z)h

(z, w) ∈ M = M

⊕ θ(z)H

,

so that θ

(z)f

(w) ∈ M

. Since θ

(z)M

̸⊂ θ(z)H

, f

(w) ̸= 0 for some g ∈ M

. Then {f

(w) : g ∈ M

} ̸= {0}. Since wM

⊂ M

, by (4.6) {f

(w) : g ∈ M

} is a nonzero T

-invariant subspace of H

(Γ

). Hence there is a one variable inner function q(w) such that

(4.7) q(w)H

(Γ

) = {f

(w) : g ∈ M

}.

Since θ

(z) {f

(w) : g ∈ M

} ⊂ M

, we have (4.8) θ

(z)q(w)H

(Γ

) ⊂ M

. If q(w) is constant, then θ

(z) ∈ M

and

θ(z)H

(Γ

) ⫋ C · θ

(z) + θ(z)H

(Γ

) ⊂ M ∩ H

(Γ

),

so that θ(z)H

(Γ

) ̸= M ∩ H

(Γ

). This is a contradiction. Hence q(w) is nonconstant. By (4.6) and (4.7), we get

(4.9) (

H

(Γ

) ⊖ q(w)H

(Γ

) )

⊥ M

. For each nonnegative integer n, let

L

= {

f (z) ∈ H

(Γ

) ⊖ θ(z)H

(Γ

) : f (z)w

q(w) ∈ M

} .

By (4.8), θ

(z) ∈ L

. Since zM

⊂ M

⊕ θ(z)H

, L

⊕ θ(z)H

(Γ

) is an invariant subspace of H

(Γ

). By (4.3), we have T

L

⊂ L

. By (4.4) and Corollary 3.8, L

= H

(Γ

) ⊖ θ(z)H

(Γ

). Hence

w

q(w) (

H

(Γ

) ⊖ θ(z)H

(Γ

) )

⊂ M

for every n ≥ 0. Thus we get

(4.10) q(w) (

H

⊖ θ(z)H

)

⊂ M

.

By (4.9), H

(Γ

) ⊖q(w)H

(Γ

) ⊂N. For each ψ(w)∈H

(Γ

) ⊖q(w)H

(Γ

), let

L

= {

f (z) ∈ H

(Γ

) ⊖ θ(z)H

(Γ

) : f (z)ψ(w) ∈ N } .

Then 1 ∈ L

, and in the same way as Subcase 1.1, L

is a nonzero closed subspace of H

(Γ

) ⊖ θ(z)H

(Γ

) such that T

L

⊂ L

and φ(z)L

⊂ L

⊕ θ(z)H

(Γ

). Hence by (4.4) and Corollary 3.6, L

= H

(Γ

) ⊖ θ(z)H

(Γ

).

Therefore

ψ(w) (

H

(Γ

) ⊖ θ(z)H

(Γ

) )

⊂ N for every ψ(w) ∈ H

(Γ

) ⊖ q(w)H

(Γ

), and hence

(4.11) (

H

⊖ θ(z)H

)

⊖ q(w) (

H

⊖ θ(z)H

)

⊂ N.

Since H

⊖ θ(z)H

= M

⊕ N, by (4.10) and (4.11) we get N = (

H

⊖ θ(z)H

)

⊖ q(w) (

H

⊖ θ(z)H

) . By [6], this shows that [S

, S

] = 0.

Case 2. Suppose that θ(0) ̸= 0. Let φ

(z) = φ(z) − ⟨φ, θ⟩θ(z). Then S

= S

, so that we may assume that φ ⊥ θ. Write

(4.12) φ(z) = φ

(z) + θ(z)zφ

(z),

where φ

∈ H

(Γ

) ⊖ θH

(Γ

) and φ

∈ H

(Γ

). By (4.4), φ

(z) ̸= 0. Since θ(0) ̸= 0, T

φ

(z) ̸= 0. For each h ∈ N, by (4.2) we can write

φh = f

+ θg

∈ N ⊕ θH

.

Applying T

for the both side of the above, we have

φT

h + h(0, w)T

φ = T

f

+ g(0, w)T

θ + θT

g

. Hence by (4.12),

φT

h = −h(0, w)T

φ + T

f

+ g

(0, w)T

θ + θT

g

= −h(0, w)T

φ

+ T

f

+ g

(0, w)T

θ + θ(T

g

− h(0, w)φ

).

Note that

−h(0, w)T

φ

+ T

f

+ g

(0, w)T

θ ⊥ θH

. Since h ∈ N, we have T

h ∈ N, so that by (4.2) we have

−h(0, w)T

φ

+ T

f

+ g

(0, w)T

θ ∈ N.

Since f

∈ N, also we have T

f

∈ N and

(4.13) −h(0, w)T

φ

+ g

(0, w)T

θ ∈ N.

Write

Θ(z) = θ

(z) − θ(0)θ(z).

We have

T

θ^θ−θ(0)_z

T

= T

= T

. Since

T

θ^θ−θ(0)_z

N ⊂ N,

−h(0, w) (

T

φ

+ aT

θ )

+ g(0, w)T

θ ∈ N.

Since φ

∈ N ⊂ H

⊖ θH

, we have T

φ

= 0. Since T

θ = −θ(0), we get aθ(0)h(0, w) − θ(0)g(0, w) ∈ N.

Since θ(0) ̸= 0,

ah(0, w) − g(0, w) ∈ N.

Thus we get

ah(0, w) − g(0, w) ⊥ θ(z)H

.

Because θ(0) ̸= 0, we have ah(0, w) − g(0, w) = 0. Hence by (4.9),

−h(0, w)T

φ

(z) ∈ N.

Note that T

φ

(z) ̸= 0. In the same way as Subcase 1.2, h(0, w) (

H

(Γ

) ⊖ θ(z)H

(Γ

) )

⊂ N ⊂ H

⊖ θ(z)H

for every h ∈ N. Since T

N ⊂ N and N ̸= {0}, {h(0, w) : h ∈ N} is a nontrivial T

-invariant subspace of H

(Γ

), so that

{h(0, w) : h ∈ N} = H

(Γ

) ⊖ q(w)H

(Γ

) for either nontrivial inner function q(w) or q(w) = 0. Hence

(H

⊖ θ(z)H

) ⊖ q(w)(H

⊖ θ(z)H

) ⊂ N.

For every f ∈ N, write

f =

∑

⊕f

(w)z

.

Since T

N ⊂ N, f

(w) ∈ H

(Γ

) ⊖ q(w)H

(Γ

) for every n ≥ 0. Hence N ⊂ (H

⊖ θ(z)H

) ⊖ q(w)(H

⊖ θ(z)H

).

Therefore

N = (H

⊖ θ(z)H

) ⊖ q(w)(H

⊖ θ(z)H

).

This shows that [S

, S

] = 0. This completes the proof. □ References

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[3] X. Chen and K. Guo, Analytic Hilbert Modules, Chapman & Hall/CRC, Boca Raton, FL, 2003.

[4] K. J. Izuchi and K. H. Izuchi, Commutativity in two variable Jordan blocks on the Hardy space, preprint.

[5] , Cross commutators on backward shift invariant subspaces over the bidisk, Acta Sci. Math. (Szeged) 72 (2006), no. 1-2, 251–270.

[6] K. J. Izuchi, T. Nakazi, and M. Seto, Backward shift invariant subspaces in the bidisc.

II, J. Operator Theory 51 (2004), no. 2, 361–376.

[7] , Backward shift invariant subspaces in the bidisc. III, Acta Sci. Math. (Szeged) 70 (2004), no. 3-4, 727–749.

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103 (1988), no. 1, 145–148.

[9] N. Nikol’skiˇi, Treatise on the Shift Operator, Springer-Verlag, Berlin, 1986.

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[11] R. Schneider, Isometries of H^p(Uⁿ), Canad. J. Math. 25 (1973), 92–95.

Kei Ji Izuchi

Department of Mathematics Niigata University

Niigata 950-2181, Japan

E-mail address: [email protected]

Kou Hei Izuchi Faculty of Education Yamaguchi University Yamaguchi 753-8513, Japan

E-mail address: [email protected]