Compatible pairs of commuting isometries

Linear Algebra and its Applications 479 (2015) 216–259

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Linear Algebra and its Applications www.elsevier.com/locate/laa

Compatible pairs of commuting isometries Zbigniew Burdak a,∗ , Marek Kosiek b , Marek Słociński b a

Katedra Zastosowań Matematyki, Uniwersytet Rolniczy, Balicka 253c, 30-198, Kraków, Poland b Instytut Matematyki, Uniwersytet Jagielloński, Łojasiewicza 6, 30-348, Kraków, Poland

a r t i c l e

i n f o

Article history: Received 9 November 2013 Accepted 8 April 2015 Available online 2 May 2015 Submitted by P. Semrl

a b s t r a c t The paper is devoted to pairs of commuting isometries. A unique decomposition of such pairs into a compatible and a completely non-compatible part is constructed. The compatible part is fully described. © 2015 Elsevier Inc. All rights reserved.

MSC: primary 47B20 secondary 47A13 Keywords: Isometries Wold decomposition Compatible isometries

1. Preliminaries

In the present section we introduce the notation and definitions that will be used throughout.

* Corresponding author. E-mail addresses: [email protected] (Z. Burdak), [email protected] (M. Kosiek), [email protected] (M. Słociński). http://dx.doi.org/10.1016/j.laa.2015.04.011 0024-3795/© 2015 Elsevier Inc. All rights reserved.

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We let Z+ denote the nonnegative integers. Suppose V1 and V2 are commuting isometries on a separable Hilbert space H. Denote L = ker V1∗ V2∗ and Li = ker Vi∗ , the wandering subspace for V1 V2 and Vi respectively, where i = 1, 2. Denote also P1m = V1m V1∗m and P2n = V2n V2∗n , m,n P1,2

=

V1∗m V2n V2∗n V1m

and

n,m P2,1

=

V2∗n V1m V1∗m V2n

(1.1) (1.2)

for m, n ∈ Z+ . If for a set J we have vectors ei ∈ H, i ∈ J, we denote ei : i ∈ J = Span{ei : i ∈ J}. Remark 1.1. The operators Pim are the orthogonal projections onto the range R(Vim ) for i = 1, 2. Consequently, Pim Vim = Vim and Vi∗m Pim = Vi∗m . n,m m,n Remark 1.2. The projections P1m and P2n are particular cases of P2,1 and P1,2 . In fact, 0,m 0,n P1m = P2,1 and P2n = P1,2 .

(1.3)

m,n k m+k,n n,m l n+l,m V1∗k P1,2 V1 = P1,2 and V2∗l P2,1 V2 = P2,1 ,

(1.4)

Moreover,

which for m = 0 in the first case and for n = 0 in the second case is k,n l,m V1∗k P2n V1k = P1,2 and V2∗l P1m V2l = P2,1 .

(1.5)

Now we recall the following Definition 1.3. With the above notation, a pair of commuting isometries V1 , V2 on a Hilbert space H is called compatible if P1m P2n = P2n P1m for every m, n ∈ Z+ (see [1,2] and [5]). Observe that compatibility is preserved by taking powers of commuting isometries. In fact, we have Remark 1.4. If a pair V1 , V2 of commuting isometries is compatible then for all m, n > 0 the pair V1m , V2n is compatible. It is easy to see that Remark 1.5. Any pair of doubly commuting isometries is compatible [1].

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Recall that a subspace is called reducing for an operator if it is invariant for the operator and its adjoint. We introduce the following Definition 1.6. We will call a pair of commuting isometries completely non-compatible if there is no reducing subspace for them on which they are compatible. For any pair of commuting isometries there is a unique maximal subspace reducing them to a compatible pair. Theorem 1.7. For a pair of commuting isometries V1 , V2 on a Hilbert space H there is a unique decomposition H = H1 ⊕ H2 such that H1 and H2 reduce both V1 and V2 and the restriction of the pair to H1 is compatible and to H2 is completely non-compatible. Proof. Denote 1 = H



2 = H  H 1. ker(P1n P2m − P2m P1n ) and H

n,m≥0

Then define H2 =



 2 ⊂ H0 , H0 reduces V1 and V2 } and H1 = H  H2 . {H0 : H

 1 , thus V1 and V2 are compatible on H1 . On the other hand, if there Evidently H1 ⊂ H 1 exists a space K ⊂ H2 such that K reduces V1 and V2 to a compatible pair then K ⊂ H  and H2 ⊂ H2  K. Since H2  K reduces V1 , V2 , by definition of H2 we get H2 ⊂ H2  K. Thus K = {0}, which finishes the proof. 2 It is shown in [1] that there are completely non-compatible isometries, so the second part in the above decomposition may be nontrivial. Our goal is to describe the compatible part. Remark 1.8. Since there is a maximal subspace reducing isometries to a doubly commuting pair, and a model for doubly commuting isometries has been given in [4], we have only to describe compatible isometries which are not doubly commuting. The following proposition gives some properties of compatible isometries. Proposition 1.9. Let V1 , V2 ∈ L(H) be a pair of commuting, compatible isometries. Then for all k, l, m, n ∈ Z+ the following hold: k,l n,m P1,2 , P2,1 are orthogonal projections,

(1.6)

k,l n,m P1,2 and P2,1 commute,

(1.7)

k,l m,n P1,2 and P1,2 commute,

(1.8)

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l,k n,m P2,1 and P2,1 commute,

if for some x ∈ H we have V2∗n V1m x = x then k,l ∗n m k+m,l+n V1∗m V2n P1,2 V2 V1 x = P1,2 x,

for any x ∈ H such that V2∗l V1k x = x and m ≥ k, n ≥ l we have m,n m−k,n−l ∗l k V2∗l V1k P1,2 x = P1,2 V2 V1 x,

for m ≥ k

m,n m−k,n k V1k P1,2 = P1,2 V1 .

219

(1.9) (1.10)

(1.11) (1.12)

Proof. In the proof we use the properties shown in Remarks 1.1 and 1.2 without explicit mention. k,l n,m (1.6): Since P1,2 and P2,1 are self-adjoint, it is enough to show they are idempotents. Indeed, k,l k,l P1,2 P1,2 = V1∗k P2l V1k V1∗k P2l V1k = V1∗k P2l P1k P2l V1k = V1∗k (P2l )2 P1k V1k k,l = V1∗k P2l V1k = P1,2 . n,m The proof for P2,1 is similar. (1.7): Recall that two orthogonal projections commute if their product is self-adjoint. Let us compute k,l n,m P1,2 P2,1 = V1∗k P2l V1k V2∗n P1m V2n = V1∗k P2l V2∗n V2n V1k V2∗n P1m V2n

= V1∗k P2l V2∗n V1k V2n V2∗n P1m V2n = V1∗k P2l V2∗n V1k P2n P1m V2n = V1∗k P2l V2∗n V1k P1m P2n V2n = V1∗k P2l V2∗n V1k P1m V2n = V1∗k V2∗n V2n P2l V2∗n V1k P1m V1∗k V1k V2n = V2∗n V1∗k P2n+l P1m+k V1k V2n . The last operator is self-adjoint by compatibility. (1.8): Without loss of generality we can assume that m ≥ k. Then k,l m,n P1,2 P1,2 = V1∗k P2l V1k V1∗m P2n V1m = V1∗m V1m−k P2l V1∗m−k V1m V1∗m P2n V1m

= V1∗m V1m−k P2l V1∗m−k P1m P2n V1m = V1∗m V1m−k P2l V1∗m−k P2n P1m V1m = V1∗m V1m−k P2l V1∗m−k P2n V1m = V1∗m V2l V1m−k V1∗m−k V2∗l V2n V2∗n V1m  l,m−k m V1 , V1∗m V2l V1m−k V1∗m−k V2∗l V1m = V1∗m P2,1 = ∗m l m−k ∗m−k n−l ∗n−l ∗l m V1 V2 V2 V2 V1 = V1∗m V2l P1m−k P2n−l V2∗l V1m , V 1 V2 V1

l ≥ n, l < n.

For l ≥ n the last expression is evidently self-adjoint. For l < n the last operator is self-adjoint due to commutativity of P1m−k and P2n−l .

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(1.9): Analogous to (1.8). (1.10): The assumption implies that P2n V1m x = V2n V2∗n V1m x = V2∗n V1m x = x = V1m x, and consequently, since P2n is an orthogonal projection, we have P2n V1m x = V1m x. Now we can check k,l ∗n m V1∗m V2n P1,2 V2 V1 x

= V1∗m V2n V1∗k P2l V1k V2∗n V1m x = V1∗m+k V1k V2n V1∗k P2l V1k V2∗n V1m x = V1∗m+k V2n V1k V1∗k P2l V1k V2∗n V1m x = V1∗m+k V2n P1k P2l V1k V2∗n V1m x = V1∗m+k V2n P2l P1k V1k V2∗n V1m x = V1∗m+k V2n P2l V1k V2∗n V1m x = V1∗m+k V2l+n V2∗l V1k V2∗n V1m x = V1∗m+k V2n+l V2∗n+l V2n V1k V2∗n V1m x = V1∗m+k V2n+l V2∗n+l V1k V2n V2∗n V1m x = V1∗m+k V2n+l V2∗n+l V1k P2n V1m x = . . . (since P2n V1m x = V1m x) m+k,n+l x. . . . = V1∗m+k V2n+l V2∗n+l V1m+k x = P1,2 l,k ∗l k (1.11): As in the proof of (1.10), the assumption implies that P2,1 V2 V1 x = V2∗l V1k x. Thus by (1.9) and (1.10) we get m−k,n−l ∗l k m−k,n−l l,k ∗l k l,k m−k,n−l ∗l k P1,2 V2 V1 x = P1,2 P2,1 V2 V1 x = P2,1 P1,2 V2 V1 x m−k,n−l ∗l k m,n V2 V1 x = V2∗l V1k P1,2 x. = V2∗l V1k V1∗k V2l P1,2

(1.12): This follows from (1.11) for l = 0.

2

2. Diagrams The crucial role in a model for a pair of commuting, completely non-doubly commuting pair of compatible isometries is played by diagrams. In fact we will show that diagrams describe pairs of compatible isometries. Let us recall the definition. 2

Definition 2.1. A set J ⊂ Z2 is called a diagram (in Z2 ) if for any g ∈ (Z+ ) and any j ∈ J the element g + j belongs to J [2]. Let us introduce an equivalence relation among diagrams. Later we will show that every equivalence class of diagrams is closely connected to some family of compatible isometries. We will see that for isometries defined by diagrams this connection is one-toone (see Definition 3.1), and for generalized powers a periodic diagram defines exactly one pair of isometries only for a fixed period and a fixed unitary operator (see Definition 7.3).

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Definition 2.2. We say that two diagrams I, J are translation equivalent if there is (i, j) ∈ Z2 such that I = {(x + i, y + j) : (x, y) ∈ J}. This is clearly an equivalence relation in the set of diagrams. For a simpler description of diagrams we introduce the concept of border and shape. Definition 2.3. For a given diagram J the sets JVB = {(i, j) ∈ J : (i − 1, j) ∈ / J} and JHB = {(i, j) ∈ J : (i, j − 1) ∈ / J} are called the vertical and horizontal borders of J respectively; the set JB = JVB ∪ JHB is called the border of J. Note that Z2 is the only diagram having empty borders. Let us consider the more general case of JVB ∩ JHV = ∅. Remark 2.4. If JVB ∩ JHB = ∅ then J is translation equivalent to one of the diagrams Z2 , Z × Z+ , Z+ × Z or Z2 \ Z2− where Z− denotes the negative integers. Indeed, note that for J = Z2 either JVB = ∅ or JHB = ∅. Consider the case JHB = ∅ and take (i, j) ∈ JHB . We will show that (i − 1, j) ∈ JHB . Thus we need to show that (i − 1, j) ∈ J and (i − 1, j − 1) ∈ / J. Since JVB ∩ JHB = ∅, we have (i, j) ∈ / JVB and we get (i − 1, j) ∈ J. On the other hand, if (i − 1, j − 1) were in J then also (i−1+1, j−1) = (i, j−1) ∈ J, which contradicts (i, j) ∈ JHB . Thus indeed (i−1, j) ∈ JHB and by induction it follows that {(k, j) : k ≤ i} ⊂ JHB . On the other hand, if (i1 , j1 ), (i2 , j2 ) ∈ JHB then by the previous result there is an i / J, such that (i, j1 ), (i, j2 ) ∈ JHB . As (i, j2 ) ∈ JHB , for j1 < j2 we would have (i, j1 ) ∈ which is not true. Similarly j2 < j1 is impossible, so j1 = j2 . Consequently, JHB ⊂ Z ×{j} for some j ∈ Z. Summing up, JHB = {(i, j) : i < k} for some j ∈ Z and k ∈ Z ∪ {∞}, and by similar arguments JVB = {(i, j) : j < l} for some i ∈ Z and l ∈ Z ∪ {∞}. If k = ∞ we have JHB = Z × {j}, which means that the diagram is a translation of Z × Z+ . Similarly for l = ∞ the diagram is a translation of Z+ × Z. In the case of k, l finite the diagram is a translation of Z2 \ Z2− . Let us now consider the case JVB ∩ JHB = ∅. Note that for (i, j), (i , j  ) ∈ JVB ∩ JHV we have i < i if and only if j > j  . Thus we can introduce an order in the set JVB ∩ JHB . Moreover, for each (iα , jα ) ∈ JVB ∩JHB there are (mα , nα ) such that (iα +mα , jα ) ∈ JHB , (iα +mα +1, jα ) ∈ / JHB , (iα , jα +nα ) ∈ JVB , (iα , jα +nα +1) ∈ / JVB . If (iα +m, jα ) ∈ JHB for every m > 0, we take mα = ∞, and similarly for nα . The following picture explains this idea:

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The role of the sequence {(mα , nα )}tα=s just considered is to describe the shape of the diagram. Definition 2.5. The shape of a diagram J is a sequence {(mα , nα )}tα=s defined as follows: Let {(iα , jα )}tα=s be the set JHB ∩JVB ordered according to the first coordinate, precisely . . . < iα−1 < iα < iα+1 < . . . and . . . > jα−1 > jα > jα+1 > . . . . Then: • mα = iα+1 − iα − 1 and nα = jα−1 − jα − 1 for s < α < t, • if t is finite, then mt = min{k > 0 : (it + k + 1, jt ) ∈ / JHB }, possibly mt = ∞ if (it + k + 1, jt ) ∈ JHB for every k > 0, • if s is finite, then ns = min{l > 0 : (is , jt + l + 1) ∈ / JVB }, possibly ns = ∞ if (is , js + l) ∈ JVB for every l > 0. If the set JHB ∩ JVB is empty then we say that the diagram has empty shape. Intuitively the shape of a diagram is the number and size of “outer corners.” This explains why the diagrams from Remark 2.4 have empty shapes. In the next section we explain how diagrams define pairs of compatible isometries. The idea of introducing the shape of a diagram is to replace in this role the whole diagram by its shape. This is intuitively correct, because pairs of compatible isometries are uniquely determined by equivalence classes of diagrams. On the other hand, such an equivalence class is determined (under some assumption) by the shape of a diagram. Remark 2.6. Nonempty shape determines a diagram uniquely up to translation, so it uniquely determines the equivalence class of the diagram. In the more general case any diagram is determined by its borders, namely Z2 for empty borders and

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J = {(x + i, y) : (x, y) ∈ JHB and i ≥ 0} ∪ {(x, y + j) : (x, y) ∈ JVB and j ≥ 0} for nonempty border(s). If the border of a diagram is periodic, then the diagram is periodic in the following sense: Definition 2.7. The diagram J is periodic if there exist m, n such that defining J0 := ({0, 1, . . . , m − 1} × Z) ∩ J and Jk := {(i + km, j − kn) : (i, j) ∈ J0 } for k ∈ Z we have  J = k∈Z Jk , and Jk are disjoint for different k. Then the set J0 is called a period of the diagram. The idea of a periodic diagram and its period is explained in the following picture:

Note that the border of the period J0 can be described as a finite sequence (m1 , n1 ), . . . , (mβ , nβ ) where n1 defines the position of J−1 relative to J0 . Then β is a period of the shape of J and (m1 , n1 ), . . . , (mβ , nβ ) are the values in the period of the shape of J. 3. Outline of the model construction The idea of pairs of isometries given by a diagram appeared in [2]. For a given diagram J consider a set of orthonormal vectors ei,j in some Hilbert space where (i, j) ∈ J.  Then define a new Hilbert space H := (i,j)∈J Cei,j and isometries V1 , V2 ∈ L(H) by V1 ei,j = ei+1,j , V2 ei,j = ei,j+1 for all (i, j) ∈ J. This is a simplified example of a pair of isometries given by a diagram. Instead of Cei,j we can take a multidimensional subspace Hi,j .

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Definition 3.1. Let J in Z2 be a diagram and H be a complex Hilbert space. Set 

H=

Hi,j where Hi,j = H.

(i,j)∈J

For x = {xi,j }(i,j)∈J ∈ H, define y = {yi,j }(i,j)∈J ∈ H and z = {zi,j }(i,j)∈J ∈ H by  yi,j =

0, (i − 1, j) ∈ / J, xi−1,j , (i − 1, j) ∈ J,

 and zi,j =

0, xi,j−1 ,

(i, j − 1) ∈ / J, (i, j − 1) ∈ J.

Define isometries V1 and V2 of H by V1 x = y ∈ H,

V2 x = z ∈ H.

We call them the isometries defined by the diagram J and the space H. Remark 3.2. Note that a given diagram and its translation define the same pair of isometries. Thus we need to consider only equivalence classes of diagrams. On the other hand, recall that diagrams in the same equivalence class have the same shapes. Hence the isometries defined by a diagram are uniquely determined by the sequence which is its shape. However, recall that diagrams with empty shape may differ. On the other hand, we have assumed to consider completely non-doubly commuting isometries. Since isometries given by translations of the diagrams Z2 , Z ×Z+ , Z+ ×Z, Z2+ are doubly commuting, they were excluded. According to Remark 2.4 the only completely non-doubly commuting pair of isometries given by a diagram with empty shape is the pair given by Z2 \ Z2− , which is a modified bi-shift known from [3]. Summarizing, sequences that are shapes of diagrams uniquely determine non-doubly commuting pairs of isometries given by diagrams. Every pair of isometries given by a diagram is compatible. On the other hand, there are compatible pairs not given by a diagram. An example is the pair V 2 , V 3 where V is a unilateral shift. We are going to show that also those compatible pairs which are not given by diagrams are in some sense close to pairs given in such a way. We explain the idea on examples. Let us start with two examples of pairs of compatible isometries given by diagrams. Example 3.3 (Pair of shifts). Let J = {(i, j) ∈ Z2 : (i ≥ −4 and j ≥ 2) or (i ≥ −2 and j = 1) or (i ≥ 0 and j = 0) or (i ≥ 1 and j = −1) or (i ≥ 3 and j = −2)}. For an orthonormal basis {ei,j }(i,j)∈J in the space H define V1 ei,j = ei+1,j and V2 ei,j = ei,j+1 for (i, j) ∈ J.

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Then L1 = e−4,j : j ≥ 2, e−2,1 , e0,0 , e1,−1 , e3,−2 , L2 = e−4,2 , e−3,2 , e−2,1 , e−1,1 , e0,0 , e1,−1 , e2,−1 , ei,−2 : i ≥ 3, L1 ∩ L2 = e−4,2 , e−2,1 , e0,0 , e1,−1 , e3,−2 , L = e−4,j : j ≥ 2, e−3,2 , e−2,2 , e−2,1 , e−1,1 , e0,1 , e0,0 , e1,0 , e1,−1 , e2,−1 , e3,−1 , ei,−2 : i ≥ 3. The following picture explains the above:

Example 3.4 (Undecomposable pair of isometries). Let J = {(i, j) ∈ Z2 : (j ≥ 3) or (i ≥ −4 and j = 2) or (i ≥ −2 and j = 1) or (i ≥ 0 and j = 0) or (i ≥ 1 and j = −1) or (i ≥ 3 and j = −2) or (i ≥ 6 and j ≤ −3)}. If {ei,j }(i,j)∈J is an orthonormal basis in the space H then define V1 ei,j = ei+1,j and V2 ei,j = ei,j+1 for (i, j) ∈ J. Consequently, L1 = e−4,2 , e−2,1 , e0,0 , e1,−1 , e3,−2 , e6,j for j ≤ −3, L2 = ei,3 for i ≤ −5, e−4,2 , e−3,2 , e−2,1 , e−1,1 , e0,0 , e1,−1 , e2,−1 , e3,−2 , e4,−2 , e5,−2 , L1 ∩ L2 = e−4,2 , e−2,1 , e0,0 , e1,−1 , e3,−2 , L = ei,3 : i ≤ −4, e−4,2 , e−3,2 , e−2,2 , e−2,1 , e−1,1 , e0,1 , e0,0 , e1,0 , e1,−1 , e2,−1 , e3,−1 , ei,−2 : 3 ≤ i ≤ 6, e6,j : j ≤ −3. A picture for this example looks as follows:

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The next two examples are not given by diagrams. However, their construction also involves diagrams in a weaker way. Example 3.5 (Pair of powers). Let J0 = {(i, j) ∈ Z2 : (i = 0 and j ≥ 1) or (i = 1 and j ≥ 0)}. For {fi,j }(i,j)∈J0 an orthonormal basis in H define  V1 fi,j =

f1,j

for i = 0

f0,j+3 for i = 1

and V2 fi,j = fi,j+1 for (i, j) ∈ J0 .

One can check that V12 = V23 . An easy computation shows that L1 = f0,1 , f0,2 , f1,0 ,

L2 = f0,1 , f1,0 ,

L1 ∩ L2 = f0,1 , f1,0 ,

L = f0,1 , f0,2 , f0,3 , f1,0 , f1,1 .  f1,j−1 for i = 0 then we have V2 = V02 and V1 = V03 . This is f0,j+2 for i = 1 illustrated by the following picture: If we define V0 fi,j =

To get a diagram style picture define J = {(i + 2k, j − 3k) ∈ Z2 : (i, j) ∈ J0 , k ∈ Z} and ei+2k,j−3k = fi,j for (i, j) ∈ J0 , k ∈ Z. The following pictures explain this idea:

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Observe that the vectors fi,j are orthonormal but the vectors ei,j are only normal (i.e., have norm 1 but need not be orthogonal or linearly independent). Moreover, V1 ei,j = ei+1,j and V2 ei,j = ei,j+1 for (i, j) ∈ J. The next example will be a little more complicated. Example 3.6 (Pair of generalized powers). Let J0 = {(i, j) ∈ Z2 : (i = 0 and j ≥ 4) or (i = 1, 2 and j ≥ 3) or (i = 3 and j ≥ 2) or (i = 4 and j ≥ 0)}. For {fi,j }(i,j)∈J0 an orthonormal basis in H define  V1 fi,j =

fi+1,j for i < 4 f0,j+6 for i = 4

and V2 fi,j = fi,j+1 for (i, j) ∈ J0 .

It is easy to check that V15 = V26 . Note that if there were a unilateral shift V0 such that V1 = V06 and V2 = V05 then L1 ∩ L2 = L2 . Such a V0 does not exist since L1 = f0,5 , f0,4 , f1,3 , f3,2 f4,1 , f4,0 , L2 = f0,4 , f1,3 , f2,3 , f3,2 , f4,0 , L1 ∩ L2 = f0,4 , f1,3 , f3,2 , f4,0 , L = f0,6 , f0,5 , f0,4 , f1,4 , f1,3 , f2,3 , f3,3 , f3,2 , f4,2 , f4,1 , f4,0 . As in the previous example, define J = {(i + 5k, j − 6k) ∈ Z2 : (i, j) ∈ J0 , k ∈ Z} and ei+5k,j−6k = fi,j for (i, j) ∈ J0 , k ∈ Z. Then we have a diagram style picture

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As in the previous example we have V1 ei,j = ei+1,j and V2 ei,j = ei,j+1 for (i, j) ∈ J. In the last two examples the isometries are defined using diagrams. However, since the vectors {ei,j }(i,j)∈J are not orthogonal to each other (some of them are equal), the isometries are not given by a diagram J in the sense of Definition 3.1. Let us describe a wider class than in Definition 3.1, of isometries almost given by a diagram. Note some properties of all the above examples. (1) The Hilbert space is spanned by a set of vectors {ei,j }(i,j)∈J where J is a diagram. (2) The vectors {ei,j }(i,j)∈I are orthonormal, where I = J0 for a periodic diagram or I = J for a nonperiodic diagram. (3) The isometries are defined by V1 ei,j = ei+1,j and V2 ei,j = ei,j+1 for (i, j) ∈ J. For nonperiodic diagrams the above situation is equivalent to that in Definition 3.1, but for periodic diagrams it is more general, as can be seen in Examples 3.5 and 3.6. Note

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that the above properties do not define a pair of isometries uniquely (for a fixed diagram) as long the relations among vectors ei,j are not known. For nonperiodic diagrams the pair of isometries fulfilling the above properties is unique — the vectors ei,j have to be orthogonal. If a diagram is periodic, a pair of isometries is not unique. Note that in Examples 3.5, 3.6 we have ei,j = ei+km,j−kn where m, n are as in the definition of a periodic diagram and k ∈ Z is arbitrary. It turns out that one can construct a pair of isometries with properties (1)–(3) above and such that ei,j = U k ei+km,j−kn for a unitary operator U defined on a suitable Hilbert space where U k = U ∗|k| for negative k. In other words, the operator U fixes the relations among the vectors ei,j . Let us explain the meaning of this unitary operator. Remark 3.7. Let U ∈ L(H) be a unitary operator with a ∗-cyclic vector e and let J0 be  a period of a periodic diagram J = k∈Z Jk where the Jk are as in Definition 2.7.  Consider the Hilbert space H := (i,j)∈J0 Hi,j where Hi,j = H for every (i, j) ∈ J0 . We will view Hi,j as subspaces of H. Note that for every (ι, κ) ∈ J there are unique (i, j) ∈ J0 and k ∈ Z such that ι = i + km and κ = j − kn. Denote by eι,κ a vector such that PHi,j eι,κ = U k e if ι = i +km and κ = j −kn, and 0 otherwise. Define V1 eι,κ = eι+1,κ and V2 eι,κ = eι,κ+1 . Note that {ei+km,j−kn : k ∈ Z} ⊂ Hi,j . Moreover, since e is a ∗-cyclic vector, we have ei+km,j−kn : k ∈ Z = Hi,j and consequently H = eι,κ : (ι, κ) ∈ J. Thus V1 , V2 are defined on the whole H. Although the vectors {eι,κ : (ι, κ) ∈ J} may not be linearly independent, the operators V1 , V2 are well defined, which will be proved in Theorem 7.2.   Obviously V1 , V2 are commuting isometries. Note also that each of P1m eι,κ and P2n eι,κ   is either 0 or eι,κ for any m , n ∈ Z+ and (ι, κ) ∈ J. Thus P1m commute with P2n on the generators of H, and consequently V1 , V2 are compatible. In Examples 3.5 and 3.6 the Hilbert space H is one-dimensional and U = I. If we take for U a bilateral shift for which the vector e is wandering, we obtain the orthogonality of all vectors {ei,j } for (i, j) ∈ J. This shows that a periodic diagram with a suitably chosen unitary operator can define a pair of isometries given by a diagram as in Definition 3.1. Definition 3.8. We call a pair of isometries almost given by a diagram if either it is defined by a diagram as in Definition 2.1, or it is defined by a period of a periodic diagram and a unitary operator as in Remark 3.7. Note that in Remark 3.7 the pair of isometries is given by a unitary operator and a period of a diagram, and not by the diagram itself. Indeed, if J0 is a period of a diagram J, then also J0 := J0 ∪ J1 (as in Definition 2.7) is a period of J. However, taking the same unitary operator, the same diagram J but different periods, J0 and J0 , we generally (except when U is a bilateral shift) obtain different examples. On the

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other hand, as in Remark 3.2, we can reduce a period of a diagram to a period of its shape. Remark 3.9. According to Remark 3.2 any diagram is uniquely (up to translation) determined by its shape. Obviously the shape of a periodic diagram is a periodic sequence of pairs of integers. As mentioned at the end of the previous section, any period of a diagram can be uniquely (up to translation) determined by a finite sequence {(mα , nα )}tα=0 of pairs of nonnegative integers. Consequently, the pair of compatible isometries defined in Remark 3.6 is determined by such a finite sequence and a unitary operator. A precise construction is presented in Theorem 7.2. We are going to prove that the model for a pair of compatible isometries is a pair of isometries almost given by a diagram. Following Remarks 3.2 and 3.9 we do not really need to consider the whole diagram but only its shape (in the case of a periodic diagram, just the values of one period). Thus we now interpret a diagram border and its shape in the Hilbert space where the isometries almost given by the diagram act. Note that L1 = {ei,j : (i, j) ∈ JVB },

L2 = {ei,j : (i, j) ∈ JHB }.

Consequently, L1 ∩ L2 = {ei,j : (i, j) ∈ JVB ∩ JHB }. Using the order in JVB ∩ JHB we have L1 ∩ L2 = {eiα ,jα }tα=s , where −∞ ≤ s < t ≤ ∞. The interpretation of a shape {(mα , nα )}tα=s in the Hilbert space is the following: V1mα eiα ,jα ∈ L2 , V1mα +1 eiα ,jα ∈ R(V2 ), V2nα eiα ,jα ∈ L1 , and V2nα +1 eiα ,jα ∈ R(V1 ). For those reasons we consider the subspaces

Lm,n := x ∈ L1 ∩ L2 : V1m x ∈ L2 , V1m+1 x ∈ R(V2 ), V2n x ∈ L1 , V2n+1 x ∈ R(V1 ) . The above subspaces are defined also for indices with one or both coordinates equal to ∞. A precise definition is given in the next section and examples can be seen in Remarks 4.5, 4.6, 4.7, 4.8. Definition 3.10. By a pair of isometries generated by a subspace M ⊂ H we mean the restriction of some pair of isometries V1 , V2 ∈ L(H) to the minimal closed subspace reducing V1 , V2 and containing M . As already mentioned, the aim of this section is to present the main idea of our construction, which is the following: for an arbitrary pair of compatible isometries, using the subspaces Lm,n decompose L1 ∩ L2 into subspaces which generate pairs of isometries almost given by a diagram. As explained above, it is enough to focus on the shape of a diagram. To get a suitable decomposition we will proceed as follows (the assertions below will be proved in the next sections):

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I. L1 ∩ L2 can be decomposed into orthogonal subspaces Lm,n , where (m, n) ∈ Z+ ∪ {∞}. Since we consider only completely non-doubly commuting pairs, the subspace L∞,∞ is always trivial. In other words, we can assume that each index (m, n) can have at most one coordinate equal to ∞. II. For any subspace Lm0 ,n0 with m0 = ∞ one can find a family of maximal orthogonal subspaces L(m0 ,n0 ),(m,n) (for various (m, n)) such that for every x ∈ L(m0 ,n0 ),(m,n) we have x ∈ Lm0 ,n0 and V2∗n+1 V1m0 +1 x ∈ Lm,n . The part of Lm0 ,n0 orthogonal to every L(m0 ,n0 ),(m,n) is denoted by L(m0 ,n0 ),∞ . III. For any subspace L(m0 ,n0 ),(m1 ,n1 ) with n0 = ∞ one can find a family of maximal orthogonal subspaces L(m,n),(m0 ,n0 ),(m1 ,n1 ) (for various (m, n)) inheriting the properties of L(m0 ,n0 ),(m1 ,n1 ) and such that for x ∈ L(m,n),(m0 ,n0 ),(m1 ,n1 ) we have V1∗m+1 V2n0 +1 x ∈ Lm,n . The part of L(m0 ,n0 ),(m1 ,n1 ) orthogonal to every L(m,n),(m0 ,n0 ),(m1 ,n1 ) is denoted by L∞,(m0 ,n0 ),(m1 ,n1 ) . Similarly we can find L(m,n),(m0 ,n0 ),∞ , maximal subspaces of L(m0 ,n0 ),∞ inheriting the properties of L(m0 ,n0 ),∞ and such that for any x ∈ L(m,n),(m0 ,n0 ),∞ we have V1∗m+1 V2n0 +1 x ∈ Lm,n . The orthogonal complement of those subspaces in L(m0 ,n0 ),∞ is denoted by L∞,(m0 ,n0 ),∞ . IV. By induction we show that for any sequence {(mα , nα )}tα=s where −∞ < s ≤ 0 ≤ t < ∞ there is a maximal subspace L{(mα ,nα )}tα=s such that for every 0 ≤ i < t and for every s < j ≤ 0, ∗ni+1 +1

V2

V1mi +1 . . . V2∗n1 +1 V1m0 +1 L{(mα ,nα )}tα=s ⊂ Lmi ,ni ,

∗m +1 n +1 V1 j−1 V2 j

∗m +1 . . . V1 −1 V2n0 +1 L{(mα ,nα )}tα=s

()

⊂ Lmj ,nj ,

where mα , nα are positive integers except for ns and mt which may be ∞. For mt = ∞ one can find maximal subspaces L{(mα ,nα )}t+1 (for various (mt+1 , nt+1 )) α=s fulfilling conditions in () additionally for i = t + 1. For ns = ∞ one can find maximal subspaces L{(mα ,nα )}tα=s−1 fulfilling conditions in () additionally for j = s − 1. We denote the respective orthogonal complements of those subspaces in L{(mα ,nα )}tα=s by L{(mα ,nα )}tα=s ,∞ as in II or L∞,{(mα ,nα )}tα=s as in III. Similarly we can find maximal subspaces of L{(mα ,nα )}tα=s ,∞ denoted by L{(mα ,nα )}tα=s−1 ,∞ (for various (ms−1 , ns−1 )) for ns = ∞ and subspaces L∞,{(mα ,nα )}t+1 of L∞,{(mα ,nα )}tα=s for mt = ∞. The respective orthogonal α=s complements in both subspaces turns out to be equal and are denoted by L∞,{(mα ,nα )}tα=s ,∞ . This way we obtain subspaces of the type: • L∞,{(mα ,nα )}tα=s ,∞ , • L∞,{(mα ,nα )}tα=s where mt = ∞, • L{(mα ,nα )}tα=s ,∞ where ns = ∞, • L{(mα ,nα )}tα=s where mt = ns = ∞.

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It turns out that each of those subspaces generates a pair of isometries almost given by a diagram having the shape {(mα , nα )}tα=s . The idea of the diagrams generated by the above subspaces is presented in the following pictures:

L∞,{(mα ,nα )}2α=−2 ,∞

L{(mα ,nα )}2α=−2 ,∞ where n−2 = ∞

L∞,{(mα ,nα )}2α=−2 where m2 = ∞

L{(mα ,nα )}2α=−2 where n−2 = m2 = ∞

The idea of the shape of a diagram is to describe the number and size of “outer corners” in the diagram. For finite shapes we have two possibilities: the last (or first) element of a shape has one value ∞ which finishes the diagram on the corresponding side, or the last (resp. first) element of a shape has both values finite and the diagram ends with an infinite vertical (resp. horizontal) part on that side. Thus a finite shape determines a diagram uniquely up to translation equivalence. Diagrams with infinite shape (infinite number of elements in the shape) are obtained from the limits of infinite sequences of subspaces of one of the following forms: • Lm0 ,n0 , L{(mα ,nα )}1α=0 , L{(mα ,nα )}1α=−1 , . . . , L{(mα ,nα )}tα=s , . . . for s → −∞, t → ∞, • Lm0 ,n0 , L{(mα ,nα )}1α=0 , L{(mα ,nα )}1α=−1 , . . . , L{(mα ,nα )}tα=s , L{(mα ,nα )}tα=s−1 , L{(mα ,nα )}tα=s−2 , . . . where t is finite fixed and mt = ∞, • Lm0 ,n0 , L{(mα ,nα )}1α=0 , L{(mα ,nα )}1α=−1 , . . . , L{(mα ,nα )}tα=s , L{(mα ,nα )}tα=s ,∞ , L{(mα ,nα )}tα=s−1 ,∞ , L{(mα ,nα )}tα=s−2 ,∞ , . . . where t is finite fixed and mt < ∞, • Lm0 ,n0 , L{(mα ,nα )}1α=0 , L{(mα ,nα )}1α=−1 , . . . , L{(mα ,nα )}tα=s , L{(mα ,nα )}t+1 , α=s L{(mα ,nα )}t+2 , . . . where s is finite fixed and n = ∞, s α=s

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• Lm0 ,n0 , L{(mα ,nα )}1α=0 , L{(mα ,nα )}1α=−1 , . . . , L{(mα ,nα )}tα=s , L∞,{(mα ,nα )}tα=s , L∞,{(mα ,nα )}t+1 , L∞,{(mα ,nα )}t+2 , . . . where s is finite fixed and ns > −∞. α=s−1 α=s−2 Since each of those sequences is decreasing, its limit is the intersection of the subspaces and inherits the properties () for arbitrary small j and/or large i depend, ing on the case. We denote the respective limits of sequences by L{(mα ,nα )}∞ α=−∞ L{(mα ,nα )}tα=−∞ (mt = ∞), L{(mα ,nα )}tα=−∞ ,∞ (mt < ∞), L{(mα ,nα )}∞ (n = ∞), s α=s L∞,{(mα ,nα )}∞ (ns < ∞). It turns out that also each of those subspaces genα=s erates a pair of isometries almost given by a diagram having an infinite shape {(mα , nα )}tα=s where at least one of s, t is infinite according to the case. V. The described construction is performed in such a way that L1 ∩ L2 is decomposed into subspaces L{(mα ,nα )}tα=s for different sequences {(mα , nα )}tα=s . Thus by the previous step, L1 ∩ L2 can be decomposed into subspaces generating pairs of isometries almost given by diagrams. Consequently, a pair of isometries can be decomposed into pairs almost given by diagrams and a pair for which L1 ∩ L2 = {0}. VI. If L1 ∩ L2 = {0} then a pair of isometries which is assumed to be completely nondoubly commuting is given by the diagram Z2 \ Z2− . This is the so-called modified bi-shift known from [3]. VII. If the diagram is periodic and the vectors ei,j are not orthogonal we need additionally to define the relevant unitary operator. In that case there are vectors in L1 ∩ L2 not orthogonal to each other. Recall that the vectors {ei,j ∈ L1 ∩ L2 } are ordered according to the first subscript. Assume ei,j , ei ,j  ∈ L1 ∩ L2 are two closest, in this order, vectors such that (ei,j , ei ,j  ) = 0. The numbers m := |i − i|, n := |j  − j| do not depend on the choice of the closest pair ei,j , ei ,j  . Moreover, U := V2∗n V1m is the unitary operator defining the relations among the vectors ei,j . 4. Geometry of the space for a compatible pair In this section we show a decomposition of L1 ∩ L2 into subspaces Lm,n described in assertion I. First we show a relation between the Wold decomposition of one isometry and the wandering subspace of the other in the case of a compatible pair. Lemma 4.1. Let V1 , V2 be a pair of commuting, compatible isometries on a Hilbert space H. Denote by H = Hui ⊕ Hsi the Wold decomposition of Vi for i = 1, 2. Then m,n n,m and PHu1 , PHs1 , P2,1 PHu2 , PHs2 , P1,2

(4.1)

commute for any n, m ≥ 0, and L1 = PHu2 L1 ⊕ PHs2 L1 , PHu2 L1 = L1 ∩ Hu2 ,

L2 = PHu1 L2 ⊕ PHs1 L2 ,

(4.2)

PHu1 L2 = L2 ∩ Hu1 .

(4.3)

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Proof. We will prove the statement for V2 (H = Hu2 ⊕ Hs2 ). The proof of the second case is similar. (4.1): Observe that P2n (I − P2n+1 ) is an orthogonal projection on V2n L2 for every ∞ n ∈ Z+ and consequently, PHs2 = n=0 P2n (I − P2n+1 ). By (1.6)–(1.8) and Remark 1.2, m,n n,m the projections P2n (I −P2n+1 ) commute with P1,2 and P2,1 for every n, m. Consequently m,n n,m m,n PHs2 commutes with P1,2 and P2,1 . Thus PHu2 = I − PHs2 commutes with P1,2 and n,m P2,1 . Now it is easy to see that PHs1 and PHu1 commute with PHs2 and PHu2 . (4.2): Note that PL1 = I − P11 , which by (4.1) commutes with PHu2 and PHs2 . Thus L1 is an invariant subspace for such projections. (4.3): Note that (4.2) yields PHu2 L1 ⊂ L1 ∩ Hu2 . Since L1 ∩ Hu2 = PHu2 (L1 ∩ Hu2 ) ⊂ PHu2 (L1 ), we have the desired equality. 2 Let us now give a precise definition of the subspaces Lm,n introduced in the previous section. Definition 4.2. For a compatible pair of isometries V1 , V2 and m, n ≥ 0 define

Lm,n := x ∈ L1 ∩ L2 : V1m x ∈ L2 , V1m+1 x ∈ R(V2 ), V2n x ∈ L1 , V2n+1 x ∈ R(V1 ) . If m or n equals ∞ we define

Lm,∞ := x ∈ L1 ∩ L2 : V1m x ∈ L2 , V1m+1 x ∈ R(V2 ), V2n x ∈ L1 for any n ≥ 0 ,

L∞,n := x ∈ L1 ∩ L2 : V1m x ∈ L2 for any m ≥ 0, V2n x ∈ L1 , V2n+1 x ∈ R(V1 ) , L∞,∞ := {x ∈ L1 ∩ L2 : V1m x ∈ L2 for any m ≥ 0, V2n x ∈ L1 for any n ≥ 0} . We will call the above subspaces semi-wandering. In the proof of the decomposition we use the following fact, which is a consequence of compatibility. Proposition 4.3. For the semi-wandering subspaces we have the equality Lm,n = Lm,· ∩ L·,n ,

(4.4)

where m, n ∈ Z+ ∪ {∞} and Lm,· := V1∗m (L2 ∩ V1m L1 )  V1∗m+1 (L2 ∩ V1m+1 L1 )

for m < ∞,

L·,n := V2∗n (L1 ∩ V2n L2 )  V2∗n+1 (L1 ∩ V2n+1 L2 ) for n < ∞,   L∞,· := V1∗m (L2 ∩ V1m L1 ), L·,∞ := V2∗n (L1 ∩ V2n L2 ). m>0

n>0

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Proof. Let us first show that V1∗m (L2 ∩ V1m L1 ) = {x ∈ L1 : V1m x ∈ L2 }. Since L2 is V1∗ -invariant, we have V1∗m (L2 ∩ V1m L1 ) ⊂ L1 ∩ L2 . Since L2 ∩ V1m L1 ⊂ R(V1m ), we have V1m (V1∗m (L2 ∩ V1m L1 )) = P1m (L2 ∩ V1m L1 ) = L2 ∩ V1m L1 ⊂ L2 . Consequently, V1∗m (L2 ∩ V1m L1 ) ⊂ {x ∈ L1 : V1m x ∈ L2 }. For the opposite inclusion take x ∈ L1 such that y = V1m x ∈ L2 . Then y ∈ L2 ∩ V1m L1 and x = V1∗m V1m x = V1∗m y ∈ V1∗m (L2 ∩ V1m L1 ). The next step is to show that Lm,· = {x ∈ L1 : V1m x ∈ L2 , V1m+1 x ∈ R(V2 )} for m < ∞. Recall that P1m (I − P1m+1 ) is the orthogonal projection onto V1m L1 for every m ∈ Z+ . As in the proof of Lemma 4.1, parts (4.2) and (4.3), we can show that L2 is an invariant subspace for such projections and that P1m (I − P1m+1 )L2 = L2 ∩ V1m L1 = P1m (L2 ∩ V1m L1 ) = V1m (V1∗m (L2 ∩ V1m L1 )). Thus PHs1 L2 =

∞ 

P1m (I − P1m+1 )L2 =

m=0

∞ 

V1m (V1∗m (L2 ∩ V1m L1 )).

(4.5)

m=0

Since Lm,· ⊂ V1∗m (L2 ∩ V1m L1 ) = {x ∈ L1 : V1m x ∈ L2 }, we have V1m (Lm,· ) ⊂ L2 . By (4.5), V1m+1 (V1∗m+1 (L2 ∩ V1m+1 L1 )) = PV1m+1 L1 L2 . By the definition, Lm,· is orthogonal to V1∗m+1 (L2 ∩ V1m+1 L1 ). Consequently, V1m+1 Lm,· is orthogonal to PV1m+1 L1 L2 . On the other hand, V1m+1 Lm,· is a subspace of V1m+1 L1 . It follows that V1m+1 (Lm,· ) is orthogonal to the whole L2 . We have V1m+1 Lm,· ⊂ R(V2 ). Next we show that L∞,· = {x ∈ L1 : V1m x ∈ L2 for any m ≥ 0}. By the equality V1∗m (L2 ∩ V1m L1 ) = {x ∈ L1 : V1m x ∈ L2 } it is enough to show that {V1∗m (L2 ∩ V1m L1 )} is a decreasing sequence. Since L2 ∩V1m L1 = P1m (L2 ∩V1m L1 ) for every nonnegative m, we have V1m−1 V1∗m (L2 ∩ V1m L1 ) = V1∗ P1m (L2 ∩ V1m L1 ) = V1∗ (L2 ∩ V1m L1 ) ⊂ L2 ∩ V1m−1 L1 = P1m−1 (L2 ∩ V1m−1 L1 ) = V1m−1 (V1∗m−1 (L2 ∩ V1m−1 L1 )). Since V1m−1 is an isometry, we get V1∗m (L2 ∩ V1m L1 ) ⊂ V1∗m−1 (L2 ∩ V1m−1 L1 ). By the same arguments we can show that L·,n = {x ∈ L2 : V2n x ∈ L1 , V2n+1 x ∈ R(V1 )} for n < ∞ and L·,∞ = {x ∈ L2 : V2n x ∈ L1 for any n ≥ 0}. From the formulas for Lm,· , L·,n the equality Lm,n = Lm,· ∩ L·,n is obvious. 2 As an easy corollary we can refine the decomposition (4.5): Lemma 4.4. Let V1 , V2 be a pair of commuting, compatible isometries on H. Let H = Hui ⊕ Hsi be the Wold decomposition of Vi for i = 1, 2. There are decompositions L1 ∩ L2 =

∞ 

Lm,· =

m=0

∞ 

L·,n

n=0

(including L∞,· and L·,∞ respectively) such that PHs1 L2 =

∞  m  m=0 i=0

V1i (Lm,· ) and PHs2 L1 =

∞  n  n=0 i=0

V2i (L·,n ).

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Proof. It is sufficient to prove the assertion for the decomposition of V1 (and for the subspaces Lm,· ). Recall that the sequence {V1∗m (L2 ∩V1m L1 )}∞ m=0 is decreasing (see the proof of Proposition 4.3). Its first element (m = 0) is equal to L1 ∩ L2 . Thus by the definition of Lm,· ∞ including L∞,· we immediately obtain the decomposition L1 ∩ L2 = m=0 Lm,· . By the proof of Proposition 4.3 we have Lm,· = {x ∈ L1 : V1m x ∈ L2 , V1m+1 x ∈ R(V2 )} for m < ∞. Thus Lm,· , V1 (Lm,· ), . . . , V1m (Lm,· ) are subspaces of L2 and being subspaces of L1 , V1 (L1 ), . . . , V1m (L1 ) respectively, they are mutually orthogonal. Since by definition, Lm,· are mutually orthogonal, we obtain an orthogonal sum ∞ k i and V1k (Lm,· ) ⊂ L2 for k ≤ m, we have k=0 i=0 V2 (Lk,· ). Since Lm,· ⊂ L1 ∩ L2  m ∞ k k i V1 (Lm,· ) ⊂ L2 ∩ V1 (L1 ) ⊂ PHs1 L2 . Hence m=0 i=0 V1 (Lm,· ) ⊂ PHs1 L2 . For the  ∞ reverse inclusion note that V1∗m (L2 ∩ V1m L1 ) = j=m Lj,· . Thus V1m V1∗m (L2 ∩ V1m L1 ) ⊂ ∞ k i k=0 i=0 V1 (Lk,· ) for every m, and (4.5) proves the reverse inclusion. 2 Before we show the decomposition theorem, in the following remarks we calculate the subspaces Lm,n for the previous examples. Remark 4.5. For the pair from Example 3.3 (pair of shifts) we have L0,0 = e0,0 , L1,0 = e−2,1 , e1,−1 , L∞,0 = e3,−2 , L1,∞ = e−4,2 , Li,j = {0} for all other i and j. The pictures below illustrate these spaces:

Remark 4.6. For the pair from Example 3.4 (undecomposable pair of isometries) we have L0,0 = e0,0 , L1,0 = e−4,2 , e−2,1 , e1,−1 , L2,0 = e3,−2 , Li,j = {0} for all other i and j.

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Remark 4.7. For the pair from Example 3.5 (pair of powers) we have L0,0 = f1.0 , L0,1 = f0,1 , Li,j = {0} for all other i and j. The following picture shows these subspaces:

Remark 4.8. For the pair from Example 3.6 (pair of generalized powers) we have L0,0 = f3,2 , L1,0 = f1,3 , L0,1 = f0,4 , f4,0 , Li,j = {0} for all other i and j. The picture in this case looks as follows:

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Now the decomposition of L1 ∩ L2 (assertion I) can be proved. Theorem 4.9. Let V1 , V2 be a pair of commuting, compatible isometries on H. Let H = Hui ⊕ Hsi be the Wold decomposition of Vi for i = 1, 2. Then there is a decomposition ∞ L1 ∩ L2 = m,n=0 Lm,n (including ∞) such that

PHs1 L2 =

∞  m 

 V1i

m=0 i=0

PHs2 L1 =

∞  n  n=0 i=0

 V2i

∞ 

Lm,n

n=0 ∞ 

,

Lm,n

,

m=0

where the subspaces Lm,n are as in Definition 4.2. Proof. Denote by Lm,· , L·,n the subspaces as in Proposition 4.3, which by Lemma 4.4 span PHs2 L1 and PHs1 L2 respectively. Assume that PL·,n Lm,· ⊂ Lm,· for all positive m, n including ∞. Then PL·,n Lm,· = ∞ ∞ Lm,· ∩ L·,n = Lm,n . Since Lm,· ⊂ L1 ∩ L2 = n=0 L·,n , we obtain Lm,· = n=0 Lm,n .

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∞ ∞ Consequently, by Lemma 4.4 we obtain L1 ∩ L2 = Lm,· = m=0 m,n=0 Lm,n and ∞ m ∞ i PHs1 L2 = m=0 i=0 V1 n=0 Lm,n . The other decomposition is similar. Thus it remains to show that indeed PL·,n Lm,· ⊂ Lm,· for all positive m, n including ∞. Recall that Lm,· = {x ∈ L1 : V1m x ∈ L2 , V1m+1 x ∈ R(V2 )} and L·,n = {x ∈ L2 : V2n x ∈ L1 , V2n+1 x ∈ R(V1 )} (see the proof of Proposition 4.3). Let e ∈ Lm,· . Then V1i e ∈ L2 for i = 0, . . . , m and V1m+1 e ∈ R(V2 ). We have either e ∈ L·,∞ or there is a minimal n+1,1 nonnegative number n such that V1∗ V2n+1 e > 0. It is enough to show that P2,1 e is the orthogonal projection of e onto L·,n and belongs to Lm,· . n+1,1 Let us show that P2,1 e ∈ L·,n . By a symmetric case of (1.12) for i ≤ n + 1, we have n+1,1 n+1−i,1 i V2i P2,1 = P2,1 V2 .

(∗)

Since n was chosen minimal, V2i e ∈ L1 for i ≤ n. Consequently, by (∗) we get n+1,1 n+1−i,1 i n+1−i,1 1 i PR(V1 ) V2i P2,1 e = P11 P2,1 V2 e = P2,1 P1 V2 e = 0. Similarly for i = n + 1, n+1,1 n+1 n+1,1 1 n+1 by (∗) we obtain V2 P2,1 e = P1 V2 e ∈ R(V1 ). Thus P2,1 e ∈ L·,n . n+1,1 To prove that P2,1 e is the orthogonal projection of e onto L·,n , it remains to check n+1,1 that e − P2,1 e is orthogonal to L·,n . Note that n+1,1 V1∗ V2n+1 P2,1 e = V1∗ V2n+1 V2∗n+1 V1 V1∗ V2n+1 e

= V1∗ P2n+1 P11 V2n+1 e = V1∗ P11 P2n+1 V2n+1 = V1∗ V2n+1 e, n+1,1 where the last equality follows from Remark 1.1. Therefore, V1∗ V2n+1 (e − P2,1 e) = 0 and for any f ∈ L·,n we have n+1,1 n+1,1 (e − P2,1 e, f ) = (V2n+1 (e − P2,1 e), V2n+1 f ) n+1,1 n+1,1 = (V2n+1 (e − P2,1 e), V1 V1∗ V2n+1 f ) = (V1∗ V2n+1 (e − P2,1 e), V1∗ V2n+1 f ) = 0. n+1,1 Hence P2,1 e is the orthogonal projection of e onto L·,n . n+1,1 n+1,1 Let us show that P2,1 e ∈ Lm,· . The vector P2,1 e belongs to L1 by the commun+1,1 tativity of P11 and P2,1 . For i ≤ m + 1 we have n+1,1 = V1i V2∗n+1 V1 V1∗ V2n+1 = V1i V2∗n+1 V1∗i V1i+1 V1∗i+1 V1i V2n+1 V1i P2,1 n+1,i+1 i n+1,i+1 i i = V1i V1∗i V2∗n+1 V1i+1 V1∗i+1 V2n+1 V1i = P1i P2,1 V1 = P2,1 P 1 V1 n+1,i+1 i = P2,1 V1 .

(∗∗)

n+1,1 For i ≤ m we have V1i e ∈ L2 . Consequently, by (∗∗) we get P21 V1i P2,1 e = n+1,i+1 1 i 1 n+1,i+1 i P2 P2,1 V1 e = P2,1 P2 V1 e = 0. Similarly, in the case of m finite we get n+1,1 n+1,1 V1m+1 P2,1 e ∈ R(V2 ). Indeed, as V1m+1 e ∈ R(V2 ), by (∗∗) we have V1m+1 P2,1 e= n+1,m+2 m+1 n+1,m+2 1 m+1 n+1,m+2 m+1 1 P2,1 V1 e = P2,1 P 2 V1 e = P2 P2,1 V1 e ∈ R(V2 ). 2

We have obtained the decomposition of L1 ∩L2 which is assertion I of the construction. For the next steps the following property of the subspaces Lm,n will be useful.

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m,n n,m Lemma 4.10. The subspaces Li,j are invariant for P1,2 and P2,1 for i, j, m, n ∈ Z+ . m,n Proof. Recall that P1,2 = V1∗m V2n V2∗n V1m (see (1.2)). Note that for m ≤ i we have ∗ m V2 V1 (Li,j ) = {0} by the definition of Li,j . Therefore we can assume m > i. m,n m,n We check that V1i P1,2 (Li,j ) ⊂ L2 and V1i+1 P1,2 (Li,j ) ⊂ R(V2 ). By (1.12) we get m−i,n i m,n i m,n 1 i V1 P1,2 = P1,2 V1 . Consequently, from P2 V1 (Li,j ) = {0} we get P21 V1i P1,2 (Li,j ) = m−i,n 1 i i+1 1 m−i,n i P2 P1,2 V1 (Li,j ) = P1,2 P2 V1 (Li,j ) = {0}. Since V1 (Li,j ) ⊂ R(V2 ) and m,n m−i−1,n i+1 by (1.12), similarly we deduce that V1i+1 P1,2 (Li,j ) = P1,2 V (Li,j ) = m−i−1,n 1 i+1 m−i−1,n 1 i+1 P1,2 P2 V (Li,j ) = P2 P1,2 V (Li,j ) ⊂ R(V2 ). m,n m,n We need to check also that V2j P1,2 (Li,j ) ⊂ L1 and V2j+1 P1,2 (Li,j ) ⊂ R(V1 ). We have m,n V2t P1,2 = V2t V1∗m V2n V2∗n V1m = V2t V1∗m V2∗t V2n+t V2∗n+t V2t V1m m,n+t t m,n+t t t m,n+t t = V2t V2∗t V1∗m V2n+t V2∗n+t V1m V2t = P2t P1,2 V2 = P1,2 P2 V2 = P1,2 V2 .

By the definition of Li,j we have P11 V2j (Li,j ) = {0}. Thus taking t = j in the above equality, by Proposition 1.9 we obtain m,n m,n+j j m,n+j−1 1 j P11 V2j P1,2 (Li,j ) = P11 P1,2 V2 (Li,j ) = P1,2 P1 V2 (Li,j ) = {0}. m,n (Li,j ) ⊂ L1 . For every x ∈ Li,j and for t = j + 1 we have Consequently, V2j P1,2 j+1 m,n m,n+j+1 j+1 m,n+j+1 1 j+1 m,n+j+1 j+1 V2 P1,2 x = P1,2 V2 x = P1,2 P1 V2 x = P11 P1,2 V2 x. Thus j+1 m,n V2 P1,2 x ∈ R(V1 ). n,m the calculations are similar. 2 For P2,1

5. Decomposition theorem The following definition is crucial for the decomposition. t

Definition 5.1. Let V1 , V2 be a pair of compatible isometries on H and {(mα , nα )}α=s be 2 a sequence such that −∞ ≤ s ≤ 0 ≤ t ≤ ∞ and (mα , nα ) ∈ (Z+ ) except for mt and ns which may be ∞. Then we call a sequence {Lα }tα=s of subspaces of L1 ∩ L2 a generating L-sequence t given by {(mα , nα )}α=s if: – – – –

Lα ⊂ Lmα ,nα , ∗n +1 ∗m +1 V2 α+1 V1mα +1 (Lα ) = Lα+1 and V1 α−1 V2nα +1 (Lα ) = Lα−1 for s < α < t, for t < ∞, if mt = ∞ then V1mt +1 Lt ∈ Hu2 , else V1m Lt ⊂ L2 for any m ≥ 0, for s > −∞, if ns = ∞ then V2ns +1 Ls ∈ Hu1 , else V2n Ls ⊂ L1 for any n ≥ 0,

where the subspaces Lm,n are as in Definition 4.2. The subspace L0 and the orthogonal projection of H onto L0 will be called the subspace and the orthogonal projection given

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by this L-sequence. By analogy to diagrams, the sequence (mα , nα ) will be called the shape of the L-sequence. Remark 5.2. Our general assumption is that isometries are compatible and completely non-doubly commuting. Then by Remarks 3.2 and 3.9 we do not need to consider a diagram but only its shape (possibly empty for Z2 \ Z2− ). Thus the definition and considerations below involve sequences and pairs of isometries but are closely connected to diagrams. The L-sequence is supposed to define a pair of isometries almost given by a diagram where the shapes of the L-sequence and of the diagram are equal. It would be much easier to refer to a subspace L0 as given by the shape of a diagram. However, different L-sequences can be chosen for the same shape of the border. To have a well (uniquely) defined L-sequence or a subspace given by the shape of a diagram we will show that for a given pair of compatible isometries there is a maximal L-sequence given by the shape. An order among L-sequences having the same shape can be introduced thanks to the following property. ∗n

+1

Remark 5.3. For a given L-sequence we have V2 α+1 V1mα +1 (Lα ) = Lα+1 and n +1 n +1 ∗n +1 V1∗mα +1 V2 α+1 (Lα+1 ) = Lα . It follows that V1∗mα +1 V2 α+1 V2 α+1 V1mα +1 (Lα ) n +1 ∗n +1 = Lα . Since V1∗mα +1 V2 α+1 V2 α+1 V1mα +1 is a projection, it is the identity on nα+1 +1 ∗nα+1 +1 Lα . It follows that V2 V2 is the identity on V1mα +1 (Lα ), and consequently ∗nα+1 +1 mα +1 V2 V1 is an isometry from Lα onto Lα+1 . The dimensions of Lα are equal for all s ≤ α ≤ t. Obviously, it can happen that L0 = {0}. By Remark 5.3 the order among L-sequences having the same shape is given by the inclusion order among its subspaces L0 . Later we will construct a maximal L-sequence (or equivalently a maximal subspace L0 ) for a given shape. It will be called a maximal L-sequence (subspace) given by the shape of a diagram. Let us point out a property which explains the name of generating L-sequence. Let V1 |H0 , V2 |H0 be a pair generated by L0 . Then H0 is the minimal subspace containing L0 reducing V1 , V2 , and all the subspaces {Lα }tα=s of the whole L-sequence are included in H0 . The conditions in Definition 5.1 describe the kernel of V1∗ |H0 V2∗ |H0 , which is equal to L ∩ H0 . It will be shown that the whole subspace L can be described by L-sequences. In the language of L-sequences, conditions () in assertion IV are as follows: ∗ni+1 +1

V2

∗mj−1 +1

V1

V1mi +1 . . . V2∗n1 +1 V1m0 +1 L0 = Li , nj +1

V2

∗m−1 +1

. . . V1

V2n0 +1 L0 = Lj .

Thus assertion IV claims that any L-sequence generates a pair almost given by a diagram. To describe the pair generated by an L-sequence we use the following:

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Definition 5.4. For x ∈ L1 ∩ L2 take ⎧ i j for V V x ⎪ ⎪ ⎪ 1 2 j ⎪ ⎨ V ∗−i V x for 2 1 xi,j = ⎪ V2∗−j V1i x for ⎪ ⎪ ⎪ ⎩ 0 for

i, j ≥ 0; i < 0, j > 0; i > 0, j < 0; i ≤ 0, j ≤ 0 except when i = j = 0,

and define Hx = xi,j : (i, j) ∈ Jx  where Jx = {(i, j) ∈ Z2 : xi,j = 0}. Note that, since V1∗−i V2j x or V2∗−j V1i x may be zero, the set Jx may be smaller than Z2 \ {Z− ∪{0}}2 ∪{(0, 0)}. We can treat Hx as the subspace generated by the vectors xi,j for all (i, j) ∈ Z2 . In general the subspace Hx does not reduce the operators V1 , V2 , for example V2 V1∗ V2 x need not be in Hx . However, if we consider a normal vector x generating a pair of isometries given by a diagram, then the vectors xi,j defined above either vanish or have norm 1. Before we focus on such vectors, note the following, explaining the idea of L-sequences. Remark 5.5. For the compatible pair from Example 3.3 (pair of shifts) we can choose: s = −2,

t = 2,

(n−2 , m−2 ) = (1, ∞), (n−1 , m−1 ) = (1, 0), (n0 , m0 ) = (0, 0), (n1 , m1 ) = (1, 0), (n2 , m2 ) = (∞, 0) and L−2 = e−4,2  ⊂ L1,∞ , L−1 = e−2,1  ⊂ L1,0 , L0 = e0,0  ⊂ L0,0 , L1 = e1,−1  ⊂ L1,0 , L2 = e3,−2  ⊂ L∞,0 . If we choose x = e0,0 then Jx = J (as defined in Example 3.3) and xi,j = ei,j for (i, j) ∈ Jx . Consequently, Hx = H. Obviously, we could take some other numbering of indices, for example s = −1, t = 3. However, up to the numbering of indices, the sequence {(mα , nα )}tα=s is unique. Remark 5.6. For the pair from Example 3.4 (undecomposable pair of isometries) we can choose s = −2,

t = 2,

(n−2 , m−2 ) = (1, 0), (n−1 , m−1 ) = (1, 0), (n0 , m0 ) = (0, 0), (n1 , m1 ) = (1, 0), (n2 , m2 ) = (2, 0) and

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L−2 = {e−4,2 } ⊂ L1,0 , L−1 = e−2,1  ⊂ L1,0 , L0 = e0,0  ⊂ L0,0 , L1 = e1,−1  ⊂ L1,0 , L2 = e3,−2  ⊂ L2,0 . If we choose x = e0,0 then Jx = J (as defined in Example 3.4), xi,j = ei,j for (i, j) ∈ Jx and consequently Hx = H. Remark 5.7. For the compatible pair from Example 3.5 (pair of powers) we have s = −∞,

t = ∞,

and for any k ∈ Z, (m2k , n2k ) = (0, 1), (m2k+1 , n2k+1 ) = (0, 0) and L2k = e2k,1−3k  = f0,1  = L0,1 , L2k+1 = e2k+1,−3k  = f1,0  = L0,0 . If we choose x = f0,1 then Jx = Jd (as defined in Example 3.5), xi,j = ei,j for (i, j) ∈ Jx and consequently Hx = H. We can see that in this case the xi,j are not in general orthogonal. Remark 5.8. For the compatible pair from Example 3.6 (pair of generalized powers) we have s = −∞,

t = ∞,

and for any k ∈ Z we get (m4k , n4k ) = (0, 1), (m4k+1 , n4k+1 ) = (1, 0), (m4k+2 , n4k+2 ) = (0, 0), (m4k+3 , n4k+3 ) = (0, 1). Consequently, L4k = e5k,4−6k  = f0,4  ⊂ L0,1 , L4k+1 = e5k+1,3−6k  = f1,3  = L1,0 , L4k+2 = e5k+3,2−6k  = f3,2  = L0,0 , L4k+3 = e5k+4,−6k  = f4,0  ⊂ L0,1 . If we choose x = f0,4 then Jx = Jd (as defined in Example 3.6), xi,j = ei,j for (i, j) ∈ Jx , and consequently Hx = H. Also in this case xi,j are not in general orthogonal. The next result shows that for x ∈ L0 (the subspace given by an L-sequence) the subspace Hx reduces the pair V1 , V2 . Moreover, if the vectors xi,j are orthogonal then the pair V1 |Hx , V2 |Hx is given by a diagram Jx . In Section 7 it will be shown that for nonorthogonal vectors the pair V1 |Hx , V2 |Hx is almost given by a diagram (although not

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defined by it). Consequently, any pair generated by an L-sequence turns out to be almost given by a diagram (assertion IV). Lemma 5.9. Let V1 , V2 ∈ L(H) be a pair of commuting, compatible isometries and P an t orthogonal projection given by an L-sequence of shape {(mα , nα )}α=s . Then for every x ∈ P H, the subspace Hx reduces V1 , V2 and V1 xi,j = xi+1,j , V2 xi,j = xi,j+1 for (i, j) ∈ Jx . Proof. Assume that x = 1. Obviously, xi,j  = 1 for i, j > 0. Define induc∗n +1 ∗m +1 tively x0 = x, xα+1 = V2 α+1 V1mα +1 xα , and xα−1 = V1 α−1 V2nα +1 xα for ∗m +1 n +1 s ≤ α ≤ t. By Remark 5.3 we have, xα+1−1 = V1 α+1−1 V2 α+1 xα+1 = ∗mα+1−1 +1 nα+1 +1 ∗nα+1 +1 mα +1 V1 V2 V2 V1 xα = xα . Thus the sequence is well defined and xα  = 1 for s ≤ α ≤ t. For i > 0, j < 0 denote i = m0 + . . . + mβ + β + k where for i < m0 + . . . + mt + t + 1 we choose β and k such that 0 ≤ k < mβ+1 + 1, while for i ≥ m0 + . . . + mt + t + 1 we take β = t and a suitable nonnegative k. One can check that for n = n1 + n2 + . . . + nβ + β by (1.11) we have xβ = V2∗n V1i−k x = V1∗k V2∗n V1i x, and since xβ  = x, it follows that V2∗n V1i x = V1k xβ . Thus for j ≤ n we have V2∗j V1i x = V2n−j V1k xβ . For j > n we obtain V2∗j V1i x = V2∗j−n V1k xβ . The vector V1k xβ is in L2 for i < m0 + . . . + mt + t, and V1k xβ ∈ Hu2 for i ≥ m0 + . . . + mt + t + 1. In the first case (i, j) ∈ / Jx , in the second xi,j  = V2∗j−n V1k xβ  = 1. For i < 0, j > 0 the proof is similar. Thus {xi,j }(i,j)∈Jx are normal vectors. For i, j ≥ 0 obviously V1 xi,j = xi+1,j and V1∗ xi,j = V1∗ V1i V2j x = V1i−1 V2j x = xi−1,j . Recall that the condition (i − 1, j) ∈ / Jx means xi−1,j = 0. For i < 0, j > 0, obviously ∗−(i−1) j ∗ ∗−i j V1 V1 V2 x = V1 V2 x = xi−1,j . For V1 we have 1 = V1 xi,j  = V1 V1∗−i V2j x = ∗−(i+1) j V2 x = P11 xi+1,j  ≤ xi+1,j  = 1. Thus xi+1,j = P11 xi+1,j = V1 xi,j . V1 V1∗ V1 For i > 0, j < 0 we have 1 = V1 xi,j  = V1 V2∗j V1i x = V1 V2∗j V1∗ V1i+1 x = P11 xi+1,j  ≤ xi+1,j  = 1. Hence xi+1,j = P11 xi+1,j = V1 xi,j . For V1∗ we have V1∗ V2∗−j V1i x = V2∗−j V1∗ V1i x = xi−1,j . Note that it may happen that (i − 1, j) ∈ / Jx , which means xi−1,j = 0. For V2 the proof is similar. 2 t

Assume that {(mα , nα )}α=s is the shape of an L-sequence. Consider the sequence in t the next corollary as a finite subsequence of {(mα , nα )}α=s . Corollary 5.10. Let (m0 , n0 ), . . . , (mα+1 , nα+1 ) be a sequence of pairs of nonnegative integers, except for n0 , mα+1 which may be ∞. Let L be a subspace such that: (1) L ⊂ Lm0 ,n0 , ∗n +1 m +1 (2) V2 γ+1 V1 γ . . . V2∗n1 +1 V1m0 +1 (L) ⊂ Lmγ+1 ,nγ+1 for every γ ≤ α, ∗nα+1 +1 mα +1 V1 . . . V2∗n1 +1 V1m0 +1 x = x for x ∈ L. (3) V2 k,l j,i Then P1,2 (L) and P2,1 (L) preserve properties (1)–(3) for i, j, k, l ∈ Z+ . ∗n

Proof. Note that by (3) we have V2 γ+1 V1 γ . . . V2∗n1 +1 V1m0 +1 x = x for any γ ≤ α. Thus if the assumption is fulfilled for α, it will be fulfilled for any shorter +1

m +1

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sequence (m0 , n0 ), . . . , (mγ+1 , nγ+1 ). Consequently, it is enough to show the assertion for α — the inclusions for γ < α will be a consequence of that result applied to shorter sequences. Property (1) is a consequence of Lemma 4.10. Take x ∈ L. Denote for convenience m = m0 + . . . + mα + α + 1 and n = n1 + . . . + nα+1 + α + 1. By (3) we can show that ∗n +1 V2n (V2 α+1 V1mα +1 . . . V2∗n1 +1 V1m0 +1 x) = V1m x and consequently ∗nα+1 +1

V2

V1mα +1 . . . V2∗n1 +1 V1m0 +1 x = V2∗n V1m x.

(5.1)

m,n Since by (3) again x = P1,2 x, by (1.10) we have j,i j,i ∗m n ∗n m n+j,m+i ∗n m V2∗n V1m P2,1 x = V2∗n V1m P2,1 V1 V2 V2 V1 x = P2,1 V2 V1 x. ∗n+j,m+i ∗n m By (2) and Lemma 4.10 we get P2,1 V2 V1 x ∈ Lmα nα . Thus we have shown that j,i P2,1 L preserves property (2). k,l L, we need to consider the following cases: To show (2) for P1,2 k,l • k ≥ m, l ≥ n: Since we know that V2∗n V1m x = x, by (1.11) we have V2∗n V1m P1,2 = ∗k−m,l−n ∗n m ∗k−m,l−n ∗n m V2 V1 x. By (2) and Lemma 4.10 we have P1,2 V2 V1 x ∈ Lmα ,nα , P1,2 which finishes the proof. • k ≥ m, l < n: By V2∗n V1m x = x we have V1m x ∈ R(V2n ) and consequently k,l V1k x ∈ V1k−m (R(V2n )) ⊂ R(V2n ) ⊂ R(V2l ). Thus P1,2 |L = I. ∗n m • k < m, l ≥ n: From V2 V1 L ⊂ L1 ∩ L2 we have V2∗l V1k L = V2∗l V1∗m−k V1m L = V2∗l−n V1∗m−k V2∗n V1m L = {0}. • k < m, l < n: Take the maximal β < α such that for m = m0 + . . . + mβ + β + 1 and n = n1 + . . . + nβ+1 + β + 1 one of the previous cases holds. The second and third cases are obvious. If the first case holds, by the maximality of β we    have m ≤ k < m + mβ+1 + 1, and consequently V1k−m V2∗n V1m x ∈ L2 . Since          V1k−m V2∗n V1m x = x we have V1k−m V2∗n V1m x = V1k−m V2∗n V1∗k−m V1k x =       V1k−m V1∗k−m V2∗n V1k x = V2∗n V1k x. Thus V2∗n V1k x = x and V2∗n V1k x ∈ L2 . k,l Consequently, P1,2 x = V1∗k V2l V2∗l V1k x = x for l = n and V1∗k V2l V2∗l V1k x = 0 for  l>n. j,i i,j or P1,2 . We use induction on α. It remains to show (3). Denote by P any of P2,1 m0 +1,n1 +1 m0 +1,n1 +1 ∗n1 +1 m0 +1 2 2 For α = 0 we have x = V2 V1 x = (P1,2 x, x) = P1,2 x2 . m0 +1,n1 +1 m +1,n +1 1 Thus x = P1,2 x and consequently V2∗n1 +1 V1m0 +1 P x2 = (P1,20 P x, P x) = m0 +1,n1 +1 2 (P P1,2 x, P x) = (P x, P x) = P x . Assume the result holds for α − 1. More m +1 precisely, V2∗nα +1 V1 α−1 . . . V2∗n1 +1 V1m0 +1 P x = P x. In a similar way to (5.1) we m +1 ∗n−nα+1 −1 m−mα −1 get V2∗nα +1 V1 α−1 . . . V2∗n1 +1 V1m0 +1 P x = V2 V1 P x. Consequently, ∗nα+1 +1

V2

V1mα +1 V2∗nα +1 V1

∗nα+1 +1

= V2

mα−1 +1

∗n−nα+1 −1

V1mα +1 V2

. . . V2∗n1 +1 V1m0 +1 P x2

V1m−mα −1 P x2

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n−nα+1 −1

= (V1∗m−mα −1 V2

∗n−nα+1 −1

(since V2

m +1,nα+1 +1

P1,2α

∗n−nα+1 −1

V2

V1m−mα −1 P x, P x) = . . .

V1m−mα −1 P x = P x, (1.10) yields)

m,n m,n P x, P x) = (P P1,2 x, P x) = (P x, P x) = P x2 . . . . = (P1,2

2

The next theorem and the following remark prove assertions II–IV. Theorem 5.11. Let V1 , V2 ∈ L(H) be a pair of commuting, compatible isometries. Denote by H = Huι ⊕ Hsι the von Neumann–Wold decomposition of Vι for ι = 1, 2. Then the subspace L(m0 ,n0 ),...,(mα ,nα ) for a sequence (m0 , n0 ), . . . , (mα , nα ) as in Corollary 5.10 can be chosen maximal. Moreover, if mα = ∞ then L(m0 ,n0 ),...,(mα ,nα ) =  L(m0 ,n0 ),...,(mα ,nα ),∞ ⊕ i,j L(m0 ,n0 ),...,(mα ,nα ),(i,j) including i = ∞ but j only finite, where: • V1mα +1 V2∗nα +1 V1 α−1 . . . V2∗n1 +1 V1m0 +1 (L(m0 ,n0 ),...,(mα ,nα ),∞ ) ⊂ Hu2 , • V2∗j+1 V1mα +1 V2∗nα +1 V1mα −1 . . . V2∗n1 +1 V1m0 +1 (L(m0 ,n0 ),...,(mα ,nα ),(∞,j) ) ⊂ L∞,j for j ≥ 0 finite, • L(m0 ,n0 ),...,(mα ,nα ),(i,j) are subspaces as in Corollary 5.10 for an extended sequence (m0 , n0 ), . . . , (mα , nα ), (i, j). m

+1

Proof. We argue by induction on the length of a sequence.  Base step α = 0: By Theorem 4.9 the subspace L1 ∩ L2 decomposes as i,j Li,j . Take L(i,j) = Li,j . That decomposition immediately yields the maximality of the subspaces L(i,j) . Inductive step: Fix a sequence (m0 , n0 ), . . . , (mα , nα ) and use the shorter notation Lα := L(m0 ,n0 ),...,(mα ,nα ) , Lα,∞ := L(m0 ,n0 ),...,(mα ,nα ),∞ , and Lα,(i,j) := L(m0 ,n0 ),...,(mα ,nα ),(i,j) . Assume the theorem is shown for some α. Thus we have a maximal subspace Lα fulfilling the conditions of Corollary 5.10. Let L be any subspace fulfilling those conditions for the extended sequence (m0 , n0 ), . . . , (mα , nα ), (i, j) for any (i, j). Obviously L fulfills the conditions of Corollary 5.10 for the shorter sequence (m0 , n0 ), . . . , (mα , nα ). Thus, by the maximality of Lα we have L ⊂ Lα . Consequently, if we show the decomposition  Lα = Lα,∞ ⊕ i,j Lα,(i,j) , then the maximality of Lα,(i,j) follows from the maximality of Lα .  The decomposition Lα = Lα,∞ ⊕ i,j Lα,(i,j) will be shown in two steps. First we construct projections, denoted Qj , which decompose Lα to obtain the expected properties of a subspace Lα,(i,j) , according to the second coordinate j. More precisely, since Qj Lα ⊂ Lα , it inherits from Lα the properties described in Corollary 5.10, and additionally V2∗j+1 V1i+1 . . . V2∗n1 +1 V1m0 +1 Qj Lα ⊂ L·,j , as denoted in Proposition 4.3. Then we construct projections Qi,j ≤ Qj , which decompose Lα preserving the properties of Lα,(i,j) according to the second coordinate j obtained for Qj and having the expected properties with respect to the first coordinate i. More precisely,

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V2∗j+1 V1i+1 . . . V2∗n1 +1 V1m0 +1 Qi,j Lα ⊂ Li,j . At the end we show that the subspaces fulfill condition (3) of Corollary 5.10 for the extended sequence. Recall the commutativity described in (1.6)–(1.8) and in Remarks 1.1, 1.2. These properties will be used quite often in the proof, without explicit mention. Denote m = m0 + . . . + mα−1 + α and n = n1 + . . . + nα + α. By property (3) of Corollary 5.10 it can be shown that V2∗nα +1 V1

mα−1 +1

m,n . . . V2∗n1 +1 V1m0 +1 x = V2∗n V1m x and P1,2 x=x

(†)

m,n x = x is equivalent to V1m x ∈ R(V2n ). Since the range for every x ∈ Lα . Note that P1,2 m+mα +1,n space is hyperinvariant, we have V1m+mα +1 x ∈ R(V2n ) or equivalently P1,2 x = x. m+mα +1,n+j+1 ∗mα +m+1 n ∗n m+mα +1 Set Qj := P1,2 = V1 V2 PR(V j+1 ) V2 V1 . By Corollary 5.10 2 we have Qj (Lα ) ⊂ Lα . By property (3) of Corollary 5.10 we have x = V2∗n V1m x = V1mα +1 V2∗n V1m x = V1mα +1 V2∗n V1∗mα +1 V1mα +1 V1m x = P1mα +1 V2∗n V1m+mα +1 x ≤ V2∗n V1m+mα +1 x ≤ x for all x ∈ Lα . It follows that

V1mα +1 V2∗n V1m x = P1mα +1 V2∗n V1m+mα +1 x = V2∗n V1m+mα +1 x.

(‡)

By definition Q0 = V1∗mα +m+1 V2n PR(V2 ) V2∗n V1m+mα +1 . Since V2∗n V1m x ∈ Lmα ,nα , we m,n get V2∗n V1m+mα +1 x = V1mα +1 V2∗n V1m x ∈ R(V2 ) and so Q0 |Lα = P1,2 |Lα = I|Lα . By (1.6) and (1.8) the operator Qj Qj+1 is the orthogonal projection onto the intersection of the range spaces of Qj and Qj+1 . From PR(V j+2 ) ≤ PR(V j+1 ) and (1.6)–(1.8) it is 2 2 easy to check that Qj+1 ≤ Qj . Thus Qj is a decreasing sequence of orthogonal projections whose first term is the identity on Lα and which has a limit Q := limj→∞ Qj .  Note that PHs2 = j≥0 P2j (I − P2j+1 ). However, by (‡) and the inductive assumption, V2∗n V1∗m+mα +1 (Lα ) = V1mα +1 V2∗n V1m (Lα ) ⊂ R(V2 ). Thus in the following calculation the summand for j = 0 is trivial and is omitted: V1∗m+mα +1 V2n PHs2 V2∗n V1m+mα +1 (Lα )  = V1∗m+mα +1 V2n P2j (I − P2j+1 )V2∗n V1m+mα +1 (Lα ). j≥1

By (3) of Corollary 5.10 and (‡) we get V2∗n V1m+1+mα x = x or equivalently = V2∗n V1m+1+mα x, so the above equals

n,m+1+mα ∗n m+1+mα P2,1 V2 V1 x



n,m+mα +1 ∗n m+mα +1 V1∗m+mα +1 V2n P2j (I − P2j+1 )P2,1 V2 V1 (Lα )

j≥1

=



n,m+mα +1 V1∗m+mα +1 V2n P2j P2,1 (I − P2j+1 )V2∗n V1m+mα +1 (Lα )

j≥1

=

 j≥0

Qj (I − Qj+1 )(Lα ) =

 j≥0

Qj (I − Qj+1 )(Lα ),

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and V1∗m+mα +1 V2n PHu2 V2∗n V1m+mα +1 (Lα ) = Q(Lα ). Obviously, Corollary 5.10 can be applied to the operator Qj . It is easy to see that the operator I − Qj also preserves properties (1)–(3) of Corollary 5.10. Thus by the maximality of Lα we have Qj (I − Qj+1 )(Lα ) ⊂ Lα . Finally we obtain the decomposition Lα = V1∗m+mα +1 V2n V2∗n V1m+mα +1 (Lα ) = Q(Lα ) ⊕



Qj (I − Qj+1 )(Lα ).

(♣)

j≥0

Consequently, Lα,∞ := Q(Lα ) ⊂ Lα . By (†), (‡) and Lemma 4.1 we have V1mα +1 V2∗nα +1 V1mα −1 . . . V2∗n1 +1 V1m0 +1 (Lα,∞ ) = V1mα +1 V2∗n V1m (Lα,∞ ) = V2∗n V1m+mα +1 Q(Lα ) n,m+mα +1 = P2,1 PHu2 V2∗n V1m+mα +1 (Lα ) n,m+mα +1 ∗n m+mα +1 V2 V1 (Lα ) ⊂ Hu2 . = PHu2 P2,1

Thus we have shown the required properties of the subspace Lα,∞ . To obtain the subspaces Lα,(i,j) we decompose Qj (I − Qj+1 )(Lα ). Define Qi,j = V1∗m+mα +1 V2n+j+1 V1∗i PR(V2 ) V1i V2∗n+j+1 V1m+mα +1 . Using properties (1.6)–(1.8) and Remarks 1.1 and 1.2 one can check that Qi,j and Qi,j Ql are orthogonal projections for every positive l and Qi,j = Qi,j Qj . Thus Qi,j is a projection commuting with Ql for every l. The proof of these facts is similar to the following proof of the equality Qi,j Qi+1,j = Qi,j . First we show that ι,1 ι+1,1 ι,1 P1,2 P1,2 = P1,2 .

()

Since R(V2 ) is V1 -invariant, we find that P21 V1 P21 = V1 P21 . Consequently, we have ι+1,1 ι,1 P1,2 P1,2 = V1∗ι+1 P21 P1ι+1 V1 P21 V1ι = V1∗ι+1 P1ι+1 P21 V1 P21 V1ι = . . . by Remark 1.2 ι,1 . . . = V1∗ι+1 P21 V1 P21 V1ι = V1∗ι+1 V1 P21 V1ι = V1∗ι P21 V1ι = P1,2 . Taking ι = i we can check that i,1 n+j+1,m+mα +1 i+1,1 ∗n+j+1 m+mα +1 Qi,j Qi+1,j = V1∗m+mα +1 V2n+j+1 P1,2 P2,1 P1,2 V2 V1 i,1 i+1,1 n+j+1,m+mα +1 ∗n+j+1 m+mα +1 = V1∗m+mα +1 V2n+j+1 P1,2 P1,2 P2,1 V2 V1 i,1 ∗n+j+1 m+mα +1 n+j+1 m+mα +1 = V1∗m+mα +1 V2n+j+1 P1,2 V2 P1 P2 V1 i,1 ∗n+j+1 n+j+1 m+mα +1 m+mα +1 = V1∗m+mα +1 V2n+j+1 P1,2 V2 P2 P1 V1 i,1 ∗n+j+1 m+mα +1 = V1∗m+mα +1 V2n+j+1 P1,2 V2 V1 = Qi,j .

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Thus the sequence Qi,j is increasing in i. Consequently, Q∞,j = limi→∞ (I − Qi,j ) is a well defined orthogonal projection. Let Lα,(i,j) := Qi+1,j (I − Qi,j )Qj (I − Qj+1 )(Lα ) and Lα,(∞,j) := Q∞j Qj (I − Qj+1 )(Lα ). Assume i is finite. By (1.10), the restriction Qi,j |Qj (Lα ) is equal to V1∗m+mα +i+1 V2n+j+2 V2∗n+j+2 V1m+mα +i+1 |Qj (Lα ) . Thus Corollary 5.10 can be applied to Qi,j . Note that also I − Qi,j preserves properties (1)–(3) of Corollary 5.10. Thus property (3) is also preserved by Q∞,j . Since the subspaces Li,j are closed, also properties (1), (2) are preserved by Q∞,j . Consequently, the subspaces Lα,(i,j) (including i = ∞) preserve all properties of Lα . By the maximality of Lα we get Lα,(i,j) ⊂ Lα , and by Corollary 5.10 also Qi,j Lα ⊂ Lα . Consequently, since Qi,j commutes with Ql , we obtain Lα,(i,j) = Qj (I − Qj+1 )Qi+1,j (I − Qi,j )(Lα ) ⊂ Qj (I − Qj+1 )(Lα ). Eventually, we obtain the decomposition Qj (I − Qj+1 )(Lα ) =



Lα,(i,j) including i = ∞.

i≥0

 As a consequence, from (♣) we have Lα = Lα,∞ ⊕ i,j Lα,(i,j) . It remains to show that the subspaces Lα,(i,j) fulfill the conditions of Corollary 5.10 for an extended sequence (m0 , n0 ), . . . , (mα , nα ), (i, j). Since Lα,(i,j) ⊂ Lα we need to show the relevant properties with respect to the last element of the sequence, which is (i, j). By (†) and (‡) this is equivalent to the inclusion V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) ⊂ Li,j or the following conditions: (a) V2j V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) ⊂ L1 , (b) V2j+1 V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) ⊂ R(V1 ), (c) V1k V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) ⊂ L2 — in case i = ∞ for every k > 0 (or equivalently for all large k), and in the case of i finite for k = i, (d) V1i+1 V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) ⊂ R(V2 ) — only for i finite. Note that Lα,(i,j) ⊂ R(Qj (I − Qj+1 )). By the definition of Qj we have n,m+mα +1 V2∗n V1m+mα +1 Qj (I − Qj+1 ) = P2,1 (PR(V j+1 ) − PR(V j+2 ) )V2∗n V1m+mα +1 2

= (PR(V j+1 ) − 2

2

n,m+mα +1 ∗n m+mα +1 PR(V j+2 ) )P2,1 V2 V1 . 2

Thus we obtain V2∗n V1m+mα +1 (Lα,(i,j) ) ⊂ R(V2j+1 )  R(V2j+2 ). On the other hand, V1mα V2∗n V1m (Lα,(i,j) ) ⊂ V1mα (Lmα ,nα ) ⊂ L2 . From these two inclusions and (‡) we have V1∗ V2j V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) = V1∗ V2∗n+1 V1m+mα +1 (Lα,(i,j) ) = V2∗ V1mα V2∗n × V1m (Lα,(i,j) ) = {0}, which shows (a). Similarly we have V2j+1 V2∗n+j+1 V1m+mα +1 (Lα,(i,j) ) = V2∗n V1m+mα +1 (Lα,(i,j) ) = V1mα +1 V2∗n V1m (Lα,(i,j) ) ⊂ V1mα +1 (Lmα ,nα ) ⊂ R(V2 ), which shows (b). For (c), note that Lα,(i,j) ⊂ (I − Qi,j )(Lα ). Also Lα,(∞,j) ⊂ (I − Qi,j )(Lα ) for all i. Thus for both i = ∞ and i finite it is enough to show that V1k V2∗n+j+1 V1m+mα +1 (I − m+mα +1,n+j+1 |Lα = I|Lα , we have Qk,j )(Lα ) ⊂ L2 . Since P1,2

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V1k V2∗n+j+1 V1m+mα +1 (I − Qk,j )(Lα ) m+mα +1,n+j+1 = V1k V2∗n+j+1 V1m+mα +1 (P1,2 − Qk,j )(Lα )

= V1k V2∗n+j+1 V1m+mα +1 V1∗m+mα +1 V2n+j+1 (I − V1∗k P21 V1k )V2∗n+j+1 V1m+mα +1 (Lα ) n+j+1,m+mα +1 = V1k P2,1 (I − V1∗k P21 V1k )V2∗n+j+1 V1m+mα +1 (Lα ) n+j+1,m+mα +1 ∗n+j+1 m+mα +1 = V1k (I − V1∗k P21 V1k )P2,1 V2 V1 (Lα ) n+j+1,m+mα +1 ∗n+j+1 m+mα +1 V2 V1 (Lα ) = (V1k − P1k P21 V1k )P2,1 n+j+1,m+mα +1 ∗n+j+1 m+mα +1 V2 V1 (Lα ) = (V1k − P21 P1k V1k )P2,1 n+j+1,m+mα +1 ∗n+j+1 m+mα +1 = (V1k − P21 V1k )P2,1 V2 V1 (Lα ) n+j+1,m+mα +1 ∗n+j+1 m+mα +1 = (I − P21 )V1k P2,1 V2 V1 (Lα ).

Since (I − P21 ) is the projection onto L2 , we have (c). For (d) we use the inclusion Lα,(i,j) ⊂ R(Qi+1,j (I − Qi,j )). We have V2∗n+j+1 V1m+mα +1 Qi+1,j (I − Qi,j ) = V2∗n+j+1 V1m+mα +1 (Qi+1,j − Qi,j ) n+j+1,m+mα +1 i+1,1 i,1 = P2,1 (P1,2 − P1,2 )V2∗n+j+1 V1m+mα +1 i+1,1 i,1 n+j+1,m+mα +1 ∗n+j+1 m+mα +1 = (P1,2 − P1,2 )P2,1 V2 V1 i+1,1 i,1 1,1 = (P1,2 − P1,2 )V2∗n+j+1 V1m+mα +1 = V1∗i (P1,2 − P21 )V1i V2∗n+j+1 V1m+mα +1 1,1 = V1∗i P1,2 (I − P21 )V1i V2∗n+j+1 V1m+mα +1 ,

where the last equality follows from () for ι = 0. Thus V2∗n+j+1 V1m+mα +1 Lα,(i,j) ⊂ 1,1 1,1 1,1 (I − P21 )V1i ). On the other hand, V1i+1 V1∗i P1,2 (I − P21 )V1i = V1 P1i P1,2 (I − R(V1∗i P1,2 1,1 1 i 1 i ∗ 1 1 i i 1 ∗ 1 i P2 )V1 = V1 P1,2 (I − P2 )V1 = V1 V1 P2 V1 (I − P2 )P1 V1 = P2 V1 V1 V1 (I − P2 )V1 . Thus 1,1 V1i+1 (R(V1∗i P1,2 (I − P21 )V1i )) ⊂ R(V2 ), which finishes the proof. 2 By the preceding theorem we can show assertion II and also the corresponding parts of assertion IV. More precisely, having the subspace L(m−k ,n−k ),...,(mk ,nk ) we are able to construct the succeeding subspace L(m−k ,n−k ),...,(mk ,nk ),(mk+1 ,nk+1 ) . This is the case where the index is extended to the right. The construction of L(m−k−1 ,n−k−1 ),(m−k ,n−k ),...,(mk ,nk ),(mk+1 ,nk+1 ) where the index is extended to the left (assertion III and the remaining parts of assertion IV) is possible by the following remark. Remark 5.12. Exchange the roles of V1 and V2 and take V2∗n V1m (Lα ) instead of Lα in Theorem 5.11. Then we have a sequence (nα , mα ), . . . , (n0 , m0 ), which can be extended by any pair (j, i). By Theorem 5.11 we obtain (nα , mα ), . . . , (n0 , m0 ), (j, i); reordered backwards, this is (i, j), (m0 , n0 ), . . . , (mα , nα ).

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We are now able to construct a maximal subspace (L-sequence) given by a fixed shape of a diagram. t

Corollary 5.13. There is a maximal L-sequence having a given shape {(mα , nα )}α=s of the border. Proof. Note that by similar arguments to those in Remark 5.12 we can see that the t maximal subspace given by the shape {(mα , nα )}α=s is equal to the maximal subspace −s for the pair V2 , V1 and the shape {(nα , mα )}α=−t . Therefore without loss of generality we can assume |s| ≥ |t|. Consider the sequence of subsequences {(m0 , n0 )}, {(m0 , n0 ), (m1 , n1 )}, {(m−1 , n−1 ), (m0 , n0 ), (m1 , n1 )}, . . . . Take L1 ∩ L2 and decompose it by means of Theorem 4.9 into subspaces Li,j . Choose L1 := Lm0 ,n0 . Decompose by Theorem 5.11 and choose L2 := L(m0 ,n0 ),(m1 ,n1 ) . Then use Theorem 5.11 backward, as described in Remark 5.12 to obtain L3 := L(m−1 ,n−1 ),(m0 ,n0 ),(m1 ,n1 ) . Proceeding this way we obtain a decreasing sequence of subspaces Lk . For s = −∞, t = ∞ we obtain an infinite, decreasing sequence of subspaces. If t is finite, we have two possibilities: • (mt , nt ) = (∞, j). Since |s| ≥ |t|, at some step of the construction we obtain L2t = L(m−(t−1) ,n−(t−1) ),...,(mt−1 ,nt−1 ),(∞,j) . • (mt , nt ) are finite numbers. We use Theorem 5.11 twice: first to construct L(m−t ,n−t ),...,(mt ,nt ) , and then to get L2t+1 = L(m−t ,n−t ),...,(mt ,nt ),∞ . In both cases we continue the construction but extending the index only at the beginning. Despite the assumption of Theorem 5.11 this can be done even if mt = ∞. Indeed, by the method described in Remark 5.12 the last element of the index is not (mt , nt ) but n−(t−1) , m−(t−1) and it fulfills the assumption of the theorem. If s = −∞ we again obtain an infinite, decreasing sequence of subspaces. If there is a first element (s > −∞), we finish the construction with one of the cases similar to the case of t finite. This way we obtain a finite, decreasing sequence of subspaces. The case when s is finite and t = ∞ is similar.  In each case the required subspace is k Lk . The maximality follows from the fact that after each step of the construction we obtain a maximal subspace having the required properties for a suitable subsequence. 2 The subspace and the L-sequence in Corollary 5.13 will be called respectively the maximal subspace and the maximal L-sequence given by the shape of a diagram. The maximal t subspace given by the shape {(mα , nα )}α=s is the subspace L0 in the L-sequence. Note that the subspace L1 in that L-sequence is the maximal subspace given by the shape t+1 {(mα , nα )}α=s+1 where mα = mα−1 and nα = nα−1 . In other words, the choice of

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a subspace in the L-sequence depends on the position of index 0 in the shape sequence. Therefore each subspace in the maximal L-sequence is the maximal subspace given by some shape of a diagram. The following theorem shows that L1 ∩ L2 can be decomposed into maximal subspaces given by shapes of diagrams. Theorem 5.14. Let V1 , V2 ∈ L(H) be a pair of commuting, compatible isometries. There  is a set S of shapes of diagrams such that L1 ∩ L2 = s∈S Ls , where Ls denotes the maximal subspace given by the shape s. Proof. By the construction of the maximal subspace given by the shape of a diagram, and by the orthogonality of the decomposition in Theorem 5.11, we deduce that for two differ ent shapes of diagrams, their subspaces are orthogonal. Consider H0 = L1 ∩L2  s∈S Ls , where S denotes the set of all possible shapes of diagrams which give a nontrivial maximal subspace Ls . We can decompose H0 into subspaces Li,j . The projection onto Li,j is equal to the identity on the maximal subspace given by the shape of a diagram if (n0 , m0 ) = (i, j), and is equal to 0 if (n0 , m0 ) = (i, j). Therefore every Ls and consequently H0 are invariant subspaces for such projections. Similarly we can show that the projection onto the maximal subspace L(ni ,mi ),...,(nj ,mj ) given by any finite sequence (ni , mi ), . . . , (nj , mj ) is invariant for H0 . Thus we can decompose H0 by applying Theorem 5.11 to the restrictions V1 |H0 , V2 |H0 . If H0 = {0} then at least one L1 := Ln0 ,m0 is not {0}. We decompose it further by using Theorem 5.11, to obtain again at least one nontrivial subspace L2 := L(n0 ,m0 ),(n1 ,m1 ) . Continuing, we obtain a nontrivial maximal subspace given by some sequence which can be interpreted as the shape of a diagram. This is a contradiction, because in H0 there are no such subspaces. 2 Each element of a maximal L-sequence generates the same minimal V1 , V2 reducing subspace. Therefore it is enough to take only one subspace in each L-sequence to obtain a set of vectors generating the whole H. This is described in the following theorem, which is in fact the decomposition theorem. Theorem 5.15. Let V1 , V2 ∈ L(H) be a pair of commuting, compatible isometries. There is a decomposition H = H0 ⊕



Hx ,

x∈X

where X ⊂ L1 ∩ L2 is a set of vectors fulfilling the conditions of Lemma 5.9 and H0 ∩ L1 ∩ L2 = {0}. Note that by Lemma 5.9, each Hx , and hence also H0 , reduces V1 , V2 .  Proof. By Theorem 5.14 we have L1 ∩ L2 = s∈S Ls , where for each s we assume Ls = {0}. By Lemma 5.9 for every x ∈ Ls , the subspace Hx reduces V1 , V2 , and Hx is

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linearly spanned by vectors xi,j as in Definition 5.4. For those xi,j in L1 ∩L2 we have either ∗n +1 m +1 ∗m +1 n +1 xi,j = V2 β V1 β−1 . . . V2∗n1 +1 V1m0 +1 x, or xi,j = V1 β V2 β−1 . . . V1∗m1 +1 V2n0 +1 x for some β, where (m· , n· ) are elements of the shape of the diagram s. Then Hx = Hxi,j and xi,j ∈ Lt where Lt belongs to the maximal L-sequence given by the shape of s. Therefore, for every xi,j and any sequence p ∈ S, either xi,j ∈ Lp , or xi,j is orthogonal to Lp . Consequently, L1p := PHHx Lp = Lp  (Lp ∩ Hx ) for every p ∈ S. Consider V1 |HHx , V2 |HHx and denote the respective subspaces of the restrictions by  1 L1 and L12 . Then we have L11 ∩ L22 = s∈S 1 L1s , where L1s ⊂ Ls . Since we consider only nontrivial L-sequences, if some L1s is {0} we get s ∈ / S 1 . Therefore S 1 ⊂ S. We repeat the whole procedure until L1 ∩ L2 = {0}, where L1 , L2 are the kernels of the adjoints of the respective restrictions. The remaining subspace is H0 . 2 6. Isometries defined by diagrams In this section we describe pairs of isometries given by a diagram. We start by proving that such pairs are compatible. Remark 6.1. For any diagram J and space H the pair of isometries defined by them  m is compatible. Indeed, it is easy to see that P1n H = (i,j)∈J+(n,0) Hi,j and P2 H =  H . Consequently, P1n P2m (or P2m P1n ) is the projection onto P1n P2m H = (i,j)∈J+(0,m) i,j (i,j)∈J+(n,m) Hi,j , which proves the compatibility. Note that for any vector in the maximal subspace given by a shape of a diagram all the assumptions of the following theorem are fulfilled except the orthogonality of the vectors xi,j . Theorem 6.2. Let V1 , V2 ∈ L(H) be a pair of compatible commuting isometries and x ∈ L1 ∩L2 with x = 1. If there are sequences {xα }tα=s ⊂ H and a shape {(mα , nα )}tα=s of a diagram such that {xα } is orthonormal and: (1) (2) (3) (4)

xα ∈ Lmα ,nα , ∗n +1 x0 = x and V2 α+1 V1mα +1 xα = xα+1 for α < t, if t < ∞ then either mt = ∞ or V1mt +1 xt ∈ Hu2 , if −∞ < s then either ns = ∞ or V2ns +1 xs ∈ Hu1 ,

then Hx reduces V1 , V2 to isometries defined by the diagram Jx , and L = x, were Hx , Jx are as in Definition 5.4. Proof. By Lemma 5.9 all vectors xi,j for (i, j) ∈ Jx are normal and Hx reduces V1 , V2 . Note that {xα }tα=s ⊂ {xi,j : i, j ∈ Jx } ∩ L1 ∩ L2 , and by assumption, the xα are orthogonal. Thus there is an increasing sequence {iα }tα=s and a decreasing sequence {jα }tα=s such that (iα , jα ) ∈ J and xα = xiα ,jα .

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To have isometries defined by a diagram we need to show orthogonality of all xi,j for (i, j) ∈ Jx . For all nonnegative p, q, either V1∗p V2q x = 0 or V1∗p V2q x = xp,q  = x and so V1p V1∗p V2q x = V2q x. Similarly, either V2∗q V1p x = 0 or V2q V2∗q V1p x = V1p x. Therefore the scalar product (V1p1 V2q1 x, V1p2 V2q2 x) for any pairs of integers (p1 , q1 ) = (p2 , q2 ) is either 0 or can be simplified to (V1i V2j x, x) or (V1i x, V2j x) where i, j are nonnegative. Since x ∈ L1 ∩ L2 we have (V1i V2j x, x) = 0. For the product (V1i x, V2j x), note that either i > it or there is α such that iα ≤ i < iα+1 . We will show that V1i x = V1k xα for that α and some integer k. By assumption (2) we have iα+1 − iα = mα + 1. Thus i = m0 + . . . + mα−1 + α + k where α is the maximal number ≤ t such that i ≥ m0 + . . . + mα−1 + α and k ≥ 0. Note that k ≤ mα for α < t (by the maximality of α), or mα = ∞ for α = t. However, for α = t and mt finite, it is possible that k > mα . Take m +1 y = V1k V2∗nα +1 V1 α−1 . . . V2∗n1 +1 V1m0 +1 x = V1k xα , which is in L2 for all cases where k ≤ mα , and in Hu2 for k > mα . Since all xα are normal, we have y = x and so V2n1 +...+nα +α y = V1i x. Hence the following cases show that (V1i x, V2j x) = 0: • for y ∈ Hu2 , from x ∈ L2 we have (V2n1 +...+nα +α y, V2j x) = 0, • for x, y ∈ L2 and j = n1 + . . . + nα + α we have (V2n1 +...+nα +α y, V2j x) = 0, • for x, y ∈ L2 and j = n1 + . . . + nα + α we have (V2n1 +...+nα +α y, V2j x) = (y, x), – if k = 0 then y = xα and (xα , x) = 0 by assumption, – if k = 0 then (y, x) = 0 since y ∈ R(V1 ) and x ∈ L1 . 2 The next result proves assertion VI. Denote V p := V ∗|p| for p ∈ Z− . Theorem 6.3. Let V1 , V2 ∈ L(H) be a pair of commuting, compatible, completely nondoubly commuting isometries such that L1 ∩ L2 = {0}. Set N = L  (L1 ⊕ L2 ). Then the subspace 

V1k V2l N

(k,l)∈Z2 \Z2−

reduces the pair (V1 , V2 ). Proof. Let H = Hui ⊕ Hsi be the Wold decomposition of Vi for i = 1, 2. Since I − P11 , I − P21 commute, we have PL1 L2 = (I − P11 )L2 ⊂ L1 ∩ L2 = {0} and so L2 ⊂ R(V1 ). Note that N ⊂ R(V1 ) and V1∗k N ⊂ L2 ⊂ R(V1 ) for k ≥ 1. Hence V1k+1 V1∗k+1 |N = V1k V1 V1∗ V1∗k |N = V1k V1∗k |N = . . . = V1 V1∗ |N = I|N . Thus N ⊂ Hu1 . By similar arguments, N ⊂ Hu2 . Since N ⊂ Hui , we have Vik Vim |N = Vik+m |N for all k, m ∈ Z. Moreover, we are going to show that V1k V2l |N = V2l V1k |N for all k, l ∈ Z. We need only consider the case when the powers have different signs. For k < 0, l > 0 we ∗|k| ∗|k| |k| ∗|k| ∗|k| |k| ∗|k| ∗|k| have V1k V2l |N = V1 V2l |N = V1 V2l V1 V1 |N = V1 V1 V2l V1 |N = V2l V1 |N = V2l V1k |N . The case k > 0, l < 0 is similar.

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k l (k,l)∈Z2 \Z2− V1 V2 N reduces V1 , V2 . We V1k−1 V2l N for every (k, l) ∈ Z2 \ Z2− . It

Now we are ready to show that

255

prove this

for V1 . Note that V1∗ (V1k V2l N ) = that (k − 1, l) ∈ / Z2 \ Z2− , namely for k = 0 and ∗|l|−1 ∗ ∗ ∗ l V1 V 2 N ⊂ V 2 V1 V2 L = {0}. Note that V1k V2l = Similarly V1 V1k restricted to N is equal to V1k+1 for

is possible l < 0. However, for such powers, V2l V1k for every k, l ∈ Z only on N . every k ∈ Z. Such an equality is not obvious in the restriction to V2l (N ) for negative k. In other words, clearly V1 (V1k V2l N ) = V1k+1 V2l N for k ≥ 0, l ∈ Z and for k ∈ Z, l = 0, but for k < 0 a proof is required. If k < 0 then V1k V2l (N ) = {0} for l < 0. Thus we can assume k < 0, l > 0. Then we have V1 (V1k V2l N ) = V1 (V2l V1k N ) = V2l (V1 V1k N ) = V2l V1k+1 N = V1k+1 V2l N . So, it remains to show that V1k V2l N ⊥ V1m V2n N for (k, l) = (m, n) ∈ Z2 \ Z2− . Without loss of generality, we can assume m ≥ 0. Then (V1k V2l x, V1m V2n y) = (V1−m V1k V2l x, V2n y) = (V1k−m V2l x, V2n y) = (V2l V1k−m x, V2n y) = (V1k−m x, V2n−l y) for x, y ∈ N , where the second equality holds because −m ≤ 0 and the last one because y ∈ N ⊂ Hu2 . Depending on the signs of k − m, n − l the last scalar product is equal to one of the following: |k−m| |n−l| |n−l| |k−m| |k−m| |n−l| |k−m| |n−l| (V1 x, V2 y), (V2 x, V1 y), (V1 V2 x, y), (x, V1 V2 y). In other words, it is enough to show that V1k V2l N ⊥ N , V1k N ⊥ V2l N , V1k N ⊥ N , V2l N ⊥ N for positive k, l. By definition, N ⊥ R(V1 V2 ), which shows the first case. For the second, note that V1k N = V1k V2 V2∗ N = V2 V1k V2∗ N ⊂ V2 V1k L1 ⊂ V2 Hs1 . On the other hand, V2l N ⊂ V2l Hu1 ⊂ V2 Hu1 . Since Hs1 ⊥ Hu1 and V2 preserves the scalar product, we have V1k N ⊥ V2l N . The last two cases are similar: V1k N = V2 V1k V2∗ N ⊂ V2 V1k L1 ⊥ V2 L1 ⊃ V2 V2∗ N = N . 2 The pair described in the preceding theorem is in fact a so-called modified bi-shift which appeared in the decomposition obtained in [3]. The following fact is more interesting. Remark 6.4. The isometries as in Theorem 6.3 are a special case of isometries given by a diagram. 7. Generalized powers For a pair V1 , V2 ∈ L(H) defined by a diagram J there is an orthonormal basis {ei,j }(i,j)∈J such that V1 (ei,j ) = ei+1,j and V2 (ei,j ) = ei,j+1 . In fact, the existence of such a basis is equivalent to the pair being defined by a diagram. In the present section we consider analogous pairs of isometries. The important difference is that we assume that the basis {ei,j }(i,j)∈J consists of normal vectors, but not necessarily pairwise orthogonal. We will show later that such pairs can be compatible if the diagram J is periodic. Remark 7.1. Let J0 be a period of a diagram J, and (m1 , n1 ), . . . , (mβ , nβ ) be a period of the shape of the diagram. Denote m = m1 + . . . + mβ + β and n = n1 + . . . + nβ + β. Then

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J = {(i + km, j − kn) : (i, j) ∈ J0 , k ∈ Z}. Moreover, for every pair in J the pair (i, j) ∈ J0 and the number k ∈ Z are uniquely determined. If a sequence (m1 , n1 ), . . . , (mβ , nβ ) is as in the above remark we say that a shape of J is given by the sequence (m1 , n1 ), . . . , (mβ , nβ ). The following theorem gives a generalization of Examples 3.5 and 3.6. Theorem 7.2. Let there be given: (1) a finite sequence {(mα , nα )}βα=1 of pairs of integers, (2) a unitary operator U ∈ L(H) and e ∈ H such that U n e : n ∈ Z = H. Define: (1) (2) (3) (4) (5) (6) (7)

J to be a periodic diagram with period J0 , having a shape given by {(mα , nα )}βα=1 , β β m = α=1 (mα + 1) and n = α=1 (nα + 1),  H := (i,j)∈J0 Hi,j where Hi,j = H, U ∈ L(H) where U(⊕(i,j)∈J0 xi,j ) = ⊕(i,j)∈J0 U (xi,j ), ei,j ∈ H with (i, j) coordinate e and the remaining zero, for (i, j) ∈ J0 , ei+km,j−kn = U k ei,j for (i, j) ∈ J0 and k ∈ Z, V1 (ei,j ) = ei+1,j and V2 (ei,j ) = ei,j+1 .

Then the operators V1 and V2 are well defined on H and form a commuting, compatible pair of isometries satisfying V1m = UV2n . Proof. The subspaces Hi,j can be understood as subspaces of H. Denote the natural embedding of H into Hi,j by pi,j for (i, j) ∈ J0 . More precisely, pi,j : H  X −→ x ∈ Hi,j ⊂ H where  X for (i , j  ) = (i, j), PHi ,j x = 0 otherwise. It follows that pi,j defines a unitary equivalence between H and Hi,j , and hence pi,j p−1 i ,j  establishes a unitary equivalence between Hi,j and Hi ,j  . Moreover, pi,j U = Upi,j . Therefore Hi,j = pi,j (H) = pi,j U n e : n ∈ Z = ei+km,j−kn : k ∈ Z. Consequently, since by Remark 7.1 the vector eι,κ is uniquely determined for every (ι, κ) ∈ J, the set {eι,κ }(ι,κ)∈J generates the whole H. On the other hand, the vectors eι,κ may not be  linearly independent. Every x ∈ H can be uniquely decomposed as (i,j)∈J0 xi,j with xi,j ∈ Hi,j . By the previous argument, each xi,j is linearly spanned by {ei+km,j−kn }k∈Z ,  but not in a unique way. However, for xi,j = k∈Z αk ei+km,j−km we get V2 xi,j =

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αk ei+km,j+1−km . Using the unitary equivalence between Hi,j and Hi,j+1 established by pi,j p−1 i,j+1 we see that V2 xi,j does not depend on the choice of a linear combination equal to xi,j . The proof for V1 is similar. Note that Ueι,κ = eι+m,κ−n . Thus, UV2n (eι,κ ) = U(eι,κ+n ) = eι+m,κ−n+n = eι+m,κ = V1m (eι,κ ). To show that the isometries are compatible note that for every r and (ι, κ) ∈ J, either P1r (eι,κ ) = eι,κ or P1r eι,κ = 0. Similarly for P2q for some q. This implies the compatibility of V1 , V2 . 2 k∈Z

Definition 7.3. The pair of isometries defined in Theorem 7.2 is a pair of generalized powers given by the unitary operator U ∈ L(H) and the finite sequence (m1 , n1 ), . . . , (mβ , nβ ). Remark 7.4. The pairs of isometries defined in Examples 3.5 and 3.6 are pairs of generalized powers (with U equal to the identity). The following theorem characterizes those compatible pairs which are not given by a diagram. Theorem 7.5. Let V1 , V2 ∈ L(H) be a pair of compatible commuting isometries and x ∈ L1 ∩ L2 with x = 1. If there are sequences {xα }tα=s and a shape of a diagram {(mα , nα )}tα=s such that xα are not pairwise orthogonal and: (1) xα ∈ Lmα ,nα , ∗n +1 (2) x0 = x and V2 α+1 V1mα +1 xα = xα+1 for α < t, then: (1) t = ∞, s = −∞, (2) the subspace Hx defined in Definition 5.4 reduces V1 , V2 , (3) the shape of the border is periodic. Denote by (m0 , n0 ), . . . , (mβ−1 , nβ−1 ) the values in a basic period and k = m0 + . . . + mβ−1 + β and l = n0 + . . . + nβ−1 + β. Then V2∗l V1k is unitary and V1 , V2 is a pair of generalized powers defined by the sequence (m0 , n0 ), . . . , (mβ−1 , nβ−1 ) and the respective restriction of V2∗l V1k . Proof. Let us start by showing that the sequence is periodic. By Lemma 5.9 all the vectors xi,j for (i, j) ∈ Jx are normal and Hx reduces V1 , V2 . For convenience, we can assume that x0 is not orthogonal to some xβ . Note that by Lemma 5.9 for any powers i, j and every α, the vector V2∗j V1i xα is either in the kernel of V1 or is orthogonal to it. The same holds for V2 . Since x0 and xβ are not orthogonal, either both

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V2∗j V1i x0 , V2∗j V1i xβ are in the kernel of V1 (or V2 respectively), or both are orthogonal to it. Thus taking pairs (i, j) = (1, 0), . . . , (m0 + 1, 0), (0, 1), . . . , (0, n1 + 1) we can show that Lm0 ,n0 = Lmβ ,nβ . Consequently, m0 = mβ , n0 = nβ . Following the same idea for the pairs (i, j) = (m0 + 1, 1), . . . , (m0 + 1, n1 + 1), we find that n1 = nβ+1 . By assumption we get V2∗n1 +1 V1m0 +1 x0 = x1 , and since m0 = mβ , n1 = nβ+1 , we also get V2∗n1 +1 V1m0 +1 xβ = xβ+1 . Hence V2∗n1 +1 V1m0 +1 is isometric on x0 and xβ . Thus 0 = (x0 , xβ ) = (V2∗m1 V1n0 x0 , V2∗m1 V1n0 xβ ) = (x1 , xβ+1 ). Consequently, x1 and xβ+1 are not orthogonal. In fact, we have shown that if two vectors in the sequence {xα} are not orthogonal, then the indices of their L·,· subspaces are equal and their successors are not orthogonal. Thus by induction we get mβ+k = mk , nβ+k = nk . Therefore, the sequence (mα , nα )α≥0 is periodic. A symmetric argument shows the periodicity of the sequence for negative α. Thus (mα , nα )tα=s is periodic with period β and so s = −∞, t = ∞. The value β may be chosen to be smallest such that (x0 , xβ ) = 0. However, the minimal period of the sequence {(mα , nα )}α∈Z may be smaller. Let β denote the minimal period. Then the sets {mα , mα+1 , . . . , mα+β−1 } and {m0 , . . . , mβ−1 } are equal for every α. Consequently, mα +mα+1 +. . .+mα+β−1 +β = k and similarly nα +nα+1 +. . .+nα+β−1 +β = l ∗n +1 m +1 ∗n +1 for every α. Note that xα+β = V2 α+β+1 V1 α+β . . . V2 α+1 V1mα +1 xα . Since xα+β  = xα , as in (†) in the proof of Theorem 5.11 we can show that ∗nα+β +1

V2l xα+β = V2l V2

mα+β−1 +1

V1

∗nα+1 +1

. . . V2

V1mα +1 xα = V2l V2∗l V1k xα = V1k xα .

Consequently, V1∗k V2l xα+β = xα . Summing up, V1∗k V2l xα+β  = V2∗l V1k xα  = 1. Since α was arbitrary, the operators V1∗k V2l and V2∗l V1k acts isometrically on the sequence {xα }. Hence for every i, j ≥ 0 and any α we get V1∗k V2l (V1i V2j xα ) ≥ V1∗i V2∗j V1∗k V2l V1i V2j xα  = V1∗k V2l xα  = 1. Similarly V2∗l V1k (V1i V2j xα ) = 1. Since {V1i V2j xα : i, j ≥ 0, α ∈ Z} span the whole Hx , the operator V2∗l V1k is unitary on Hx . It remains to show that V1 , V2 is a pair of generalized powers. Note that xα ∈ Lmα ,nα ⊂ L1 ∩ L2 . Thus we can take a periodic diagram J with a period J0 , both given by the sequence (m0 , n0 ), . . . , (mβ−1 , nβ−1 ). Further ei,j := xi,j , U = V2∗l V1k , H = U n x : n ∈ Z and U = U|H . Then the conditions of Theorem 7.2 are fulfilled, which means that V1 , V2 is a pair of generalized powers. 2 8. Decomposition and model for compatible isometries Theorem 5.15 decomposes every pair of compatible, commuting isometries into pairs given by diagrams and generalized powers. Indeed, if the wandering subspaces of both isometries V1 , V2 have a nontrivial intersection L1 ∩ L2 = {0} then we have one of the cases described in the previous sections.

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Remark 8.1. For any x ∈ X as in Theorem 5.15 we have the dichotomy: either the assumption of Theorem 6.2 or of Theorem 7.5 is fulfilled. It remains to describe the restriction of the pair V1 , V2 to the subspace H0 as in Theorem 5.15. Remark 8.2. The pair of isometries V1 |H0 , V2 |H0 , where H0 reduces V1 , V2 and H0 ∩ L1 ∩ L2 = {0}, fulfills the assumption of Theorem 6.3. Hence, by Remark 6.4, the pair V1 |H0 , V2 |H0 is given by a diagram. Consequently, we can formulate the decomposition theorem. Theorem 8.3. Any non-doubly commuting, compatible pair of commuting isometries can be decomposed into a sum of isometries such that each summand is defined by a diagram or is a pair of generalized powers. Proof. This follows immediately from Theorems 5.15, 6.2, 7.5 and Remarks 8.1 and 8.2. 2 Remark 8.4. For doubly commuting isometries the model is described in [4]. Finally, from Remark 8.4 and Theorem 8.3 we get Corollary 8.5. Any compatible pair of commuting isometries has a fully described model. References [1] Karel Horák, Vladimir Müller, On the structure of commuting isometries, Comment. Math. Univ. Carolin. 28 (1) (1987) 165–171. [2] Karel Horák, Vladimir Müller, Functional model for commuting isometries, Czechoslovak Math. J. 39 (2) (1989) 370–379. [3] Dan Popovici, A Wold-type decomposition for commuting isometric pairs, Proc. Amer. Math. Soc. 132 (8) (2004) 2303–2314. [4] Marek Słociński, On the Wold-type decomposition of a pair of commuting isometries, Ann. Polon. Math. XXXVII (1980) 255–262. [5] Marek Słociński, Models of commuting contractions and isometries, in: Report of the 11th Conference on Operator Theory, Bucharest, 1986.