Shift-Type Properties of Commuting, Completely Non Doubly Commuting Pairs of Isometries

Pairs (V, V′) of commuting, completely non doubly commuting isometries are studied. We show, that the space of the minimal unitary extension of V (denoted by U) is a closed linear span of subspaces reducing U to bilateral shifts. Moreover, the restriction of V′ to the maximal subspace reducing V to a unitary operator is a unilateral shift. We also get a new hyperreducing decomposition of a single isometry with respect to its wandering vectors which strongly corresponds with Lebesgue decomposition.


Introduction and Preliminaries
Let L(H) denote the algebra of all bounded linear operators acting on a complex Hilbert space H. For an operator T ∈ L(H) by its negative power T n we understand T * |n| . Recall that a subspace L ⊂ H reduces T ∈ L(H) if and only if T commutes with the orthogonal projection P L onto L. By the span of E ⊂ H, we always mean the minimal closed linear subspace containing E.
Recall the classical result of von Neumann-Wold [18]: For a given isometry V ∈ L(H) by H u , H s we always mean the subspaces in the decomposition (1.1). The restrictions V | Hu , V | Hs are referred as the unitary part and the shift part of the considered isometry. A natural question arises about generalizations for pairs or families of operators. The most natural generalization, which following [7] is proposed to be called a multiple canonical von Neumann-Wold decomposition, has been achieved only in some special cases ( [4,16]). In the general case various von Neumann-Wold type decompositions or models were established ( [1][2][3]5,9,10,14,17]). Recall that operators T 1 , T 2 ∈ L(H) doubly commute if they commute and T * 1 T 2 = T 2 T * 1 . Consider a pair of isometries (V 1 , V 2 ) on H. One can find a unique maximal subspace reducing it to a doubly commuting pair. In [16] a multiple von Neumann-Wold decomposition in the case of doubly commuting pairs is constructed along with a model for pairs of doubly commuting unilateral shifts. Therefore, we consider only completely non doubly commuting pairs (i.e. such that the only subspace of H reducing (V 1 , V 2 ) to a doubly commuting pair is {0}). Examples of such pairs are: non doubly commuting unilateral shifts or the so called modified bi-shifts (see [14]). Note that if operators commute and one of them is unitary, then they doubly commute. Thus, in completely non doubly commuting pairs both of the isometries have nontrivial unilateral shift parts and restrictions to any nontrivial subspace reducing both operators also have a nontrivial unilateral shift part. However, the unitary part may be, but need not to be trivial.
The property of being bilateral shift is not hereditary (i.e. the restriction of a bilateral shift to some reducing subspace may be not a bilateral shift). Therefore, usually there cannot be found the largest subspace reducing a given isometry to a bilateral shift or a span of bilateral shifts (see Definition 4.8). This means that usually it cannot be constructed a canonical decomposition of an isometry into a bilateral shift (or a span of bilateral shifts) and a completely non bilateral shift operator. However, in Theorem 3.10 we are able to construct a decomposition of a single isometry with respect to its wandering vectors. One of the summands of the constructed decomposition contains all the bilateral shifts.
There are known examples of undecomposable pairs where the unitary part of any isometry is a bilateral shift (the aforementioned modified bishift). We describe the unitary part and the minimal unitary extension of an isometry which commutes but completely non doubly commutes with some other isometry. One of our main results contained in Theorem 4.5 and Corollary 4.9 says that the unitary extension of any member of a completely non doubly commuting pair is a span of bilateral shifts. Since being a span of bilateral shifts is not a hereditary property, it does not mean that the unitary part of a considered isometry is a span of bilateral shifts. In Sect. 5 we show that the unitary part of an isometry may not contain any subspace reducing it to a bilateral shift but its unitary extension can be a span of bilateral shifts

Multiple von Neumann-Wold Decomposition for Pairs of Isometries
Let us recall the notion of multiple canonical von Neumann-Wold decompositions introduced in [7] in the general case, here taking a simplified form in the case of a pair of commuting isometries: By [4,16] there are multiple canonical von Neumann-Wold decompositions in either of the two cases: for doubly commuting pairs of isometries and for pairs satisfying the conditions dim(ker V * 1 ) < ∞ and dim(ker V * 2 ) < ∞. Moreover, in the case of doubly commuting isometries, the subspace ker V * 1 ∩ ker V * 2 is wandering for the semigroup generated by V 1 , V 2 . However in the general case we have only a weaker result. Recall a definition from [14].
The following general decomposition of pairs of commuting isometries obtained in [14] is not necessarily a canonical one.
We are going to focus on the weak bi-shift part. Precisely, we consider a pair of isometries whose decomposition (2.1), trivializes to the weak bi-shift subspace. In such a case the subspace reducing our isometries to a doubly commuting pair is either trivial or reduces the isometries to a pair of unilateral shifts. Indeed, in the other case the decomposition of the restriction to a doubly commuting pair of isometries would give a non trivial subspace orthogonal to H ws . By [14] there can be found a maximal subspace of H ws which reduces these isometries to a doubly commuting pair of unilateral shifts. Their model can be found in [16]. Therefore we reduce our attention to completely non doubly commuting pairs of isometries. Such pairs are a special case of a weak bi-shift class whose finer, but not fully satisfying decomposition has been described in [5].

Decomposition for Single Isometries
A unitary part of an isometry in a modified bi-shift is a bilateral shift. Our aim in the present section is to construct a decomposition with respect to a property which is close to the property of being a span of all the shifts (bilateral or unilateral). The following example shows that "being a bilateral shift" is not a hereditary property.
where m denotes the normalized Lebesgue measure on the unit circle T. Let V be the operator of multiplication by the independent variable, (V f)(z) = zf (z) for f ∈ L 2 (m). Then its spectral measure F satisfies F (α)f = χ α f for f ∈ L 2 (m) and all Borel subsets α of T. Let α be a proper subarc of T. Let V α be the restriction of V to its reducing subspace F (α)H. Since α is not of total measure in the circle, the operator V α is not a bilateral shift.
Note that a canonical decomposition of an operator T ∈ L(H) with respect to some property is in fact a construction of a unique maximal reducing subspace H p ⊂ H such that the restriction T | Hp has the considered property. Since being a bilateral shift is not a hereditary property, there is a problem with construction of a maximal subspace reducing operator to a bilateral shift. Example 5.2 in the last section will show that such a maximal subspace is not unique. The construction of any maximal bilateral shift subspace can be done by considering a maximal wandering subspace. We follow the idea of wandering vectors from [5]. Let G be a semigroup and {T g } g∈G be a semigroup of isometries on H. The vector x ∈ H is called a wandering vector (for a given semigroup of isometries) if for any g 1 = g 2 we have T g1 x, T g2 x = 0. For a semigroup generated by a single isometry we obtain the following definition of a wandering vector.
Note that for any wandering vector x the vector x+V x is not wandering.
However for x ∈ H s this is not so clear. On the other hand, every vector in the set n≥0 V n (ker V * ) is wandering, fulfills the orthogonality condition also for the negative powers and generates the whole H s . Therefore, despite Definition 3.2 is "weaker" it seems to be sufficient.

Theorem 3.3. For any isometry V ∈ L(H) there is a unique decomposition:
Note that for w wandering we have w, P Hs w ∈ H w and consequently also P Hu w = w − P Hs w ∈ H w . Note also that P Hu is wandering as well. Moreover, P Hu V * w = V * P Hu w = P Huw . On the other hand, sincew is wandering, by the previous argumentation we have P Huw ∈ H w . Consequently, P Hu V * w ∈ H w . Since w was an arbitrary wandering vector and H w is spanned by wandering vectors, we get V * P Hu H w ⊂ H w . For H 2 , the Hardy space on the unit circle, the following result is well known (see [11] p. 53).

Assume that the logarithm of the Radon-Nikodym derivative dμ dm is integrable on σ.
Then the operator M μ of the multiplication by independent variable z on the space L 2 (σ, μ), is unitary equivalent to M z -the operator of multiplication by independent variable z on the space L 2 (σ, m).
Proof. Put h := dμ dm + χ T\σ . Obviously, h is positive almost everywhere on the unite circle T. Since log h(z) = 0 for z ∈ T\σ and log dμ dm is Lebesgue integrable on σ then log h is Lebesgue integrable on T.
Consequently, by Proposition 3.6 we have h = |f | 2 for some f ∈ H 2 . Consider the following operator: where u is extended to the whole T as usual by taking its value 0 on T\σ. Since |f | 2 = dμ dm on σ and uf = 0 on T\σ then operator U f preserves the scalar product (it is an isometry). It follows also that uf belongs to L 2 (T, m) if and only if u ∈ L 2 (σ, μ). The range space of U f can be understand as a subspace of L 2 (T, m). We will show that R(U f ) = L 2 (σ, m) and consequently that U f is a unitary mapping onto its range. Since dμ dm is positive almost everywhere on σ then f is not equal 0 almost everywhere on σ. Thus for u ∈ L 2 (σ, m) we can define u f . Since u ∈ L 2 (σ, m) then σ |u| 2 dm exists and is finite. On the other hand, and integers exists simultaneously. Thus u ∈ L 2 (σ, m) implies u f ∈ L 2 (σ, μ).
Consequently, U f gives the unitary equivalence between M μ and M z .
Any isometry V acting on a Hilbert space H has Lebesgue decomposition which combined with von Neumann-Wold decomposition gives us the equality:  (supp μ, μ), where μ is a measure absolutely continuous with respect to the Lebesgue measure on the unit circle T. Since the measure μ is different from 0, we can find > 0 such that the set σ := {z ∈ T : dμ dm (z) > } has positive measure. Then log dμ dm is integrable on σ. By Corollary 3.7 the operator V | L 2 (σ,μ) is unitarily equivalent to multiplication by the independent variable on L 2 (σ, m). Since L 2 (σ, m) ⊂ H x is orthogonal to H w , it can not be a bilateral shift. Therefore T\σ has positive Lebesgue measure. Take h(z) = 1 2 for z ∈ σ and h(z) = 1 for z ∈ T\σ. By Proposition 3.6 there is f ∈ H 2 such that |f | 2 = h. Since we assumed isometry V to be non unitary, there is a subspace of H s reducing V which is identified with multiplication by z on H 2 . In other words we can assume that f ∈ H s . Take g = χσ √ 2 . Note that |f | 2 + |g| 2 = 1. (To be precise function g ∈ L 2 (σ, m) ⊂ L 2 (T, m) can be understand as defined on the unit circle). Since f ∈ H s ⊂ H w it is orthogonal to g. One can check that Similarly, for positive n. It follows that f + g is a non trivial wandering vector. Since g was orthogonal to H w we have m(σ) Thus σ is a set of measure 0 which contradicts the hypothesis x = 0. Consequently, the orthogonal complement of H w is trivial.
where H w is a linear subspace generated by wandering vectors and V | Hac is the whole absolutely continuous part of V in decomposition (3.1).

Proof.
Let v ∈ H w be a nonzero wandering vector. Denote The operator V | L is a unilateral shift and V | K is a non unitary isometry. Thus by Theorem 3.8 we get K w = L ⊕ K ac , where the subspaces are denoted as in Theorem 3.8. Similarly, for any K n := V * n L⊕M we have K n w = V * n L⊕K ac . Since every wandering vector is wandering for operator extended to some superspace then K n w ⊂ H w for every n ≥ 0. Finally, Lebesgue decomposition is hyperreducing (see Thm. 2.1 [13] or Thm. 2.2 [12]). It means that the subspaces H s ⊕ H ac and H sing in (3.1) are hyperreducing. Summing up Theorems 3.3, 3.8, Corollary 3.9 and Remark 3.5, we get where H w is a span of all wandering vectors. Moreover:

Decomposition for Pairs of Isometries
In this section we take the advantage of the decomposition obtained in Theorem 3.3 and construct a decomposition for pairs of isometries. Despite being spanned by wandering vectors is not a hereditary property it has a multiple canonical decomposition.  Let us consider the space Let V 1 be the multiplication by independent variable on K. Let V 2 be the multiplication by independent variable on L 2 (T, m) and be a unilateral shift on ∞ n=1 K n such that the wandering space is K 1 . The isometry V 1 is unitary absolutely continuous and contains a bilateral shift, hence, by Corollary 3.9, its wandering vectors span K. The isometry V 2 is an orthogonal sum of a unilateral shift and unitary absolutely continuous operator, hence its wandering vectors also span K (see Theorem 3.10). Thus we have H ww = K.
Unfortunately the decomposition is not unique. For example take H 00 = L 2 (T − , m) and H 0w := L 2 (T + , m) ⊕ ∞ n=1 K n . Recall from [6], that a pair of commuting contractions (T 1 , T 2 ) is called strongly completely non unitary if there is no proper subspace reducing T 1 , T 2 and at least one of them to a unitary operator. Moreover, there is a decomposition theorem ( [6], Thm. 2.1): ) is a strongly completely non unitary pair of contractions. IEOT The above theorem in the case of pairs of commuting isometries appears also in [2] or more general in [9]. Can be found also in [14]. Note that by the decomposition in the last theorem the subspace H S reduces V 1 , V 2 to such a pair, that there is no subspace reducing both isometries and at least one of them to a unitary operator. As we recalled earlier such a pair is called strongly completely non unitary. Using more general language of [2] such a pair is {1} -pure and {2} -pure. An immediate consequence is the following: By the proof of Theorem 3.3, the projection of a wandering vector onto the unitary part subspace of the isometry is in the span of wandering vectors but may not be wandering.
Remark 4.7. Let V ∈ L(H) be an isometry with a wandering vector w. Note that for every wandering vector in H u the equality V n w, V m w = 0 holds true for every n, m ∈ Z, n = m. Thus the minimal V -reducing subspace generated by w is n∈Z V n (Cw). In other words, the minimal V -reducing subspace generated by a wandering vector in H u reduces V to a bilateral shift.
Let us introduce a definition of the following class of operators. For the reverse implication note that every V wandering vector w ∈ H is U wandering. On the other hand, U is unitary. According to Remark 4.7 the subspace L w := n∈Z V n (Cw) is the minimal U -reducing subspace generated by w. Since H is spanned by wandering vectors, H is contained in the span of {L w : w ∈ W } taken over the set W of all V -wandering vectors. Since L w ⊂ K for w ∈ W , by minimality of U as a unitary extension, K equals to the latter span. On the other hand, U | Lw is a bilateral shift. This finishes the proof.
As a corollary of Proposition 4.6 and Corollary 4.9 we obtain the following result. The following result shows some geometry of completely non-doubly commuting pairs of isometries.
Proof. We make the proof for i = 1. By (1.1) we conclude that the subspace H u1 is hyperinvariant. Thus we can consider the operators Since the pair (V 1 , V 2 ) is completely non doubly commuting, it is a weak bi-shift. Consequently, the product V 1 V 2 is a unilateral shift. It implies that U 1 V 2 is a unilateral shift and By a similar argument applied for the operators U n 1 , V n 2 we obtain ker(U n 1 V n 2 ) * = ker V * n 2 for any n. Note that Consequently Note also, that ker V * 2 is U * 1 -invariant. Since U 1 commutes with V 2 and is unitary, they doubly commute. Consequently, is U 1 -reducing for every n. We have showed the theorem with W 1 = ker V * 2 .
A similar result is known for a normal operator and a unilateral shift (see [15] Proposition 9). Consequently we have: Proof. Any pair of commuting isometries can be decomposed into a doubly commuting pair and a completely non doubly commuting pair. In the case of doubly commuting pair we need to consider only pairs consisting of a unitary operator and a unilateral shift for which the result is trivial. It can be deduced from the mentioned result ([15] Proposition 9) or from the model in [16]. Theorem 4.11 can be deduced also from [2]. By von Neumann-Wold decomposition for the operator V 1 and by Theorem 4.11 we get In the corollary above we get an orthogonal decomposition of H into two orthogonal sums of subspaces wandering for V 1 and V 2 respectively. In Proposition 4.6 for each of the isometries V 1 , V 2 we can find a collection of wandering vectors spanning H.
Summing up, Corollary 4.13 is stronger with respect to the orthogonality of wandering subspaces. On the other hand, Proposition 4.6 is stronger with respect to the fact that the space H is spanned by wandering vectors of a single arbitrarily chosen isometry.
Since v ∈ H s , the sequence V * n v converges to zero. Consequently we obtain a contradiction 1 = f 2 = lim n→∞ − v, V * n v = 0. Thus H can not be spanned by Vwandering vectors.
The next is an example of a span of bilateral shifts which is not a bilateral shift. We would like to thank Professor László Kérchy for this example. Denote by T the unit circle in the complex plane. 4 3 π]}. Then α ∪ α 2 = T. Let H = L 2 (α) ⊕ L 2 (α 2 ) ⊕ L 2 (α) and U ∈ L(H) be multiplying by z . Then H 0 = {0}. If the operator would be unitarily equivalent to some bilateral shift then their spectral multiplicities would be equal. However, the spectral multiplicity of a bilateral shift is constant, while in our example it is not.
It is clear that wandering vectors of an isometry V ∈ L(H) span the whole subspace H s . On the other hand by Remark 4.7 any wandering vector in a subspace H u fulfills the orthogonality V n w ⊥ V m w for every distinct integer powers. The natural question is what will be changed if we make a definition of wandering vectors stronger in the following sense. We call a wandering vector w strongly wandering if it fulfills the condition V n w ⊥ V m w for every n, m ∈ Z, n = m. Denote by H ws the minimal subspace spanned by strongly wandering vectors and by H w the subspace spanned by wandering vectors. Obviously, H ws ⊂ H w and both subspaces are reducing for our isometry V . As we know, the subspace H w is invariant for every isometry commuting with V , but H ws does not need to be. We have the following lemma: Thus x u is a wandering vector and by V n x u , V m x u = − V n x s , V m x s , also x s is wandering. For the second part, note that V n (ker V * ) for every n ≥ 0 is a set of V -strongly wandering vectors. Thus H s ⊂ W . Since H u reduces V , every vector wandering for V | Hu is wandering for V . Thus W u ⊂ W . By the first part of the lemma, the reverse inclusion W ⊂ H s ⊕ W u follows.
We want to show that unitary extension can be a span of bilateral shifts and Hilbert space H is not spanned by strongly wandering vectors. This follows from the next example: Example 5.4. Consider Example 5.2, put K = L 2 (α) ⊕ L 2 (α 2 ) ⊕ L 2 (α) and denote by U the operator of multiplication by z . Find a wandering subspace W for the bilateral shift in L 2 (α 2 ) ⊕ L 2 (α) such that L 2 (α 2 ) ⊕ L 2 (α) = n∈Z U n W . Then take H = L 2 (α) ⊕ n≥0 U n (W ). The restriction U | H is an isometry with its unitary part acting on L 2 (α), and U is its minimal unitary extension. Since σ(T | Hu ) does not contain the unit circle, there is no subspace reducing it to a bilateral shift. Consequently, H u does not contain any wandering vector and H ws = H s = n≥0 U n W . On the other hand, the unitary extension U of U | H is a span of bilateral shifts.
We follow the same idea to show an example of an isometry having unitary part of H 0 type, but whose unitary extension is a bilateral shift. Let L 2 (T), L 2 (T + ), L 2 (T − ) be the subspaces of functions on T, T + , T − respectively, which moduli are square summable with respect to Lebesgue measure and let H 2 (T) be the Hardy subspace of L 2 (T). Consider H = L 2 (T − ) ⊕ n≥0 H 2 (T) and the isometric operator M z of multiplication by z . Then the unitary part of M z acts on L 2 (T − ). Since the spectrum of M z restricted to L 2 (T − ) does not contain T it is not a span of bilateral shifts. On the other hand M z extends to a unitary operator of multiplication by z on the space K = L 2 (T − ) ⊕ n≥0 L 2 (T). Since we have the decomposition L 2 (T) = L 2 (T + ) ⊕ L 2 (T − ) we obtain K = L 2 (T − ) ⊕ L 2 (T + ) ⊕ L 2 (T − ) ⊕ L 2 (T + ) ⊕ L 2 (T − ) ⊕ · · · = n≥0 L 2 (T). Thus the unitary extension is a bilateral shift.