This paper investigates $\Gamma_n$-contractions, providing sharp estimates, existence and uniqueness results for operator equations, and constructing explicit dilations and models for these contractions, advancing the understanding of their structure.
Contribution
It introduces new estimates, solves operator equations with unique solutions, and constructs explicit dilations and models for $\Gamma_n$-contractions, enhancing their theoretical framework.
Findings
01
Sharp estimates for elementary symmetric polynomials on $\mathbb D^n$
02
Existence and uniqueness of solutions to specific operator equations
03
Construction of explicit dilations and functional models for $\Gamma_n$-contractions
Abstract
We prove some estimates for elementary symmetric polynomials on Dn. We show that these estimates are sharp which allow us to study the properties of closed symmetrized polydisc Γn. Furthermore, we show the existence and uniqueness of solutions to the operator equations Si−Sn−i∗Sn=DSnXiDSnandSn−i−Si∗Sn=DSnXn−iDSn, where Xi,Xn−i∈B(DSn),foralli=1,…,(n−1), with numerical radius not greater than 1, for a Γn-contraction (S1,…,Sn). We construct a conditional dilation of various classes of Γn-contractions. Various properties of a Γn-contraction and its explicit dilation allow us to construct a concrete functional model for a Γn-contraction. We describe the structure and additional characterization of Γn-unitaries and…
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TopicsAdvanced Topics in Algebra · Holomorphic and Operator Theory · Algebraic and Geometric Analysis
Full text
On Γn-contractions and their Conditional Dilations
Avijit Pal
Department of Mathematics , Indian Institute of Technology, Bhilai
We prove some estimates for elementary symmetric polynomials on Dn. We show that these estimates are sharp which allow us to study the properties of closed symmetrized polydisc Γn. Furthermore, we show the existence and uniqueness of solutions to the operator equations
[TABLE]
where Xi,Xn−i∈B(DSn),foralli=1,…,(n−1), with numerical radius not greater than 1, for a Γn-contraction (S1,…,Sn). We construct a conditional dilation of various classes of Γn-contractions. Various properties of a Γn-contraction and its explicit dilation allow us to construct a concrete functional model for a Γn-contraction. We describe the structure and additional characterization of Γn-unitaries and Γn-isometries in detail.
Key words and phrases:
Symmetrized polydisc, Spectral set, Complete spectral set, Wold decomposition, Pure isometry, Conditional dilation, Functional model
2010 Mathematics Subject Classification:
32A60, 32C15, 47A13, 47A15, 47A20, 47A25,
47A45.
The work of A. Pal was supported through the NBHM Post-doctoral Fellowship.
1. Introduction
For n≥2, let s:Cn→Cn be the symmetrization map given by the formula
[TABLE]
where si(z)=∑1≤k1<…<ki≤nzk1…zki and s0=1. The image Γn:=s(Dˉn) under the map s of the unit polydisc is known as closed symmetrized polydisc. The map s is a proper holomorphic map [35]. The set Γn is polynomially convex but not convex [39]. The open symmetrized polydisc is defined to be the set Gn:=s(Dn). Also, the distinguished boundary bΓn of symmetrized polydisc is known to be s(Tn), the image of the n-torus Tn under the map s [20].
A commuting n-tuple of bounded operators (S1,…,Sn) on a Hilbert space H having Γn as a spectral set will be called a Γn-contraction.
Let f=((fij)) be a matrix valued polynomial defined on Γn.
Let
∥f∥=sup{∥((fij(z)))∥op:z∈Γn}.Γn is said
to be complete spectral set for (S1,…,Sn) or (S1,…,Sn) to be a complete Γn-contraction if ∥f(S1,…,Sn)∥≤∥f∥∞,Γn for all f∈A⊗Mk(C),k≥1.
Let (S1,…,Sn) be a commuting n-tuple of bounded operators defined on a Hilbert space H whose joint spectrum lies in Γn, we say that (S1,…,Sn) admits a bΓn normal dilation if there is a Hilbert space K containing H as a subspace and a normal operator N=(N1,…,Nn) whose joint spectrum lies in bΓn such that Sij=PHNij∣H,1≤i≤n,j≥0, where PH is the orthogonal projection of K onto H.
In [1, 2] it was shown that the existence of a
bΓn normal dilation of a commuting tuple (S1,…,Sn) of operators whose joint spectrum lies in Γn is equivalent to (S1,…,Sn) being a complete Γn-contraction.
We recall the Γn-unitary, Γn-isometry and pure Γn-isometry from [14].
Definition 1.1**.**
Let (S1,…,Sn) be a commuting n-tuple of operators on a Hilbert space H. We say that (S1,…,Sn) is
(1)
a Γn-unitary if S1,…,Sn are normal operators and the joint spectrum σ(S1,…,Sn) of (S1,…,Sn) is contained in the distinguished boundary of Γn.
2. (2)
a Γn-isometry if there exists a Hilbert space K⊇H and a Γn-unitary (S~1,…,S~n) on K such that H is a common invariant subspace for S~1,…,S~n and that Si=S~i∣H for i=1,…,n.
3. (3)
a Γn-co-isometry if (S1∗,…,Sn∗) is a Γn-isometry.
4. (4)
a pure Γn-isometry if (S1,…,Sn) is a Γn-isometry and Sn is also a pure isometry.
For a commuting n-tuple of operators (S1,…,Sn) with ∥Sn∥≤1, we consider the following operator equations
[TABLE]
where DSn=(I−Sn∗Sn)21 and DSn=RanDSn. This equations are called the fundamental equations for (S1,…,Sn). The operators Ei for i=1,…,(n−1) are called the fundamental operators. These operators Ei,s play an important role in constructing the conditional dilation of Γn.
The only dilation theorems, in the
multi-variable context are proposed by Agler and Young
[6, Theorem 1.1] and Ando [32, Chapter 5, Theorem 5.5].
Under some additional hypotheses, we prove that Γn is a spectral set for a commuting tuple of operators (S1,…,Sn) if and only if Γn is a complete spectral set.
In Section 2, we prove some estimates for elementary symmetric polynomial on Dn which are important in their own right. These estimates play a pivotal role for studying the characterizations of Γn. We also show that these estimates are sharp.
In Section 3, the operator pencils Φ1(i) and Φ2(i) are defined for i=1,…,(n−1). These operator pencils play a significant role in characterizing the domain Γn. In subsequent section, we find that these operator pencils are also important in studying Γn-contractions.
Section 4 is devoted to Γn-contraction and their fundamental operators. One can easily verify that if (S1,…,Sn) is a Γn-contraction, then ∥Si∥≤(in−1)+(n−in−1) for i=1,…,(n−1) and ∥Sn∥≤1. In Proposition 4.2, we prove that if (S1,…,Sn) is a Γn-contraction, then
[TABLE]
for all α∈D with k(i)=(in−1)+(n−in−1) for i=1,…,(n−1). However, it is not clear whether the converse holds. Using this result, we show the existence and uniqueness theorem for (n−1)-tuple of fundamental operators Ei for i=1,…,(n−1).
In Section 5, we prove the various properties of Γn-unitary and Γn-isometry.
We show that if (S1,…,Sn) is a Γn-contraction and Sn is a unitary, then (S1,…,Sn) is a Γn-unitary (see Theorem 5.1). We prove that if (S1,…,Sn) is a Γn-isometry, then
[TABLE]
for all α∈D with k(i)=(in−1)+(n−in−1) and vice-versa. Similar to the case of Γn-unitary, we further show that if (S1,…,Sn) is a Γn-contraction and Sn is a isometry, then (S1,…,Sn) is also a Γn-isometry (see Theorem 5.2).
In Section 6, we find the necessary conditions for the existence of rational dilation in terms of its fundamental operators ( Proposition 6.5).
In Section 7, we prove that along with the necessary conditions, if we assume some conditions on the fundamental operators of (S1∗,…,Sn∗), where (S1,…,Sn) is a Γn-contraction, then (S1,…,Sn) possesses a Γn-unitary dilation. We produce explicit Γn-isometric dilations of these Γn-contractions.
In Section 8, we construct an explicit functional model for a class of Γn-contractions assuming similar conditions on the fundamental operators of (S1,…,Sn) and (S1∗,…,Sn∗) respectively, where (S1,…,Sn) is a Γn-contraction.
We found some discrepancy in S. Pal’s proof [30, Page 6, Proposition 2.5][30, Page 8,Proposition 2.6],[30, Page 9,Theorem 3.1], [29, Page 10, Theorem 3.5],[29, Page 13, Theorem 3.7],[29, Page 15, Theorem 4.2],[29, Page 19, Theorem 4.3][31, Page-13, Proposition 2.9],[31, Page 16, Theorem 3.2], where the author used the following assertion: if a point si∈Γn then ∥si∥≤n for all i. We believe that this statement is incorrect. We now give a counter-example that supports our claim. For n≥4,i=1,n and i=(n−1), we show that there exists a point si∈Γn such that ∥si∥>n. Choose ε>0 is arbitrary small such that zi=(1−ε) for all i=1,…,n. Then by definition we have si∈Gn and ∥si∥=(1−ε)(in). Since n−1>i, for arbitrary small ε>0 ensure that
∥si∥>n. From above discussion, it follows that for arbitrary small ε>0, there are lots of points si∈Γn such that ∥si∥>n.
2. A sharp estimate for the elementary symmetric polynomial on Dn
Only for this section we use the notation si(m)(z) in place of si(z) to emphasize the fact that the ambient space is Cm. For z∈Dm+1 and i′=m+2−i, let us consider hij(m+1)(z)=∣si(m+1)(z)∣2−∣sm+2+j−i(m+1)(z)∣2 and hi′j(m+1)(z)=∣sm+2−i′(m+1)(z)∣2−∣si′+j(m+1)(z)∣2. Then we see that hi′j(m+1)(z)=−hij(m+1)(z). Our main goal is to prove the following estimate.
Theorem 2.1**.**
For z∈Dm+1 and i′=m+2−i. Then we have the following estimates
for all i with i∈{1,…,m+1} and j=i−m−2,…,(i−1) and j=2i−m−2
[TABLE]
*However, if j=2i−m−2, then hi′j(m+1)(z)=0=hij(m+1)(z).
*
We prove Theorem 2.1 by means of several lemmas and propositions. For i∈{1,…,m} and 0≤l≤i, let Fil(m),Gi′l(m):{i+l−m−2,…,i−2}→R be two functions defined by
[TABLE]
[TABLE]
[TABLE]
where i′=m+2−i.
For positive integer i,l,m, we have either l≥2i−m or l<2i−m. Depending on i,l,j,m and l≥2i−m the following proposition determines the sign of Fil(m) and Gi′l(m) which will be useful to prove Theorem 2.1.
Proposition 2.2**.**
Let i,l,j be integers such that i∈{1,…,m},0≤l≤i,l≥2i−m and
j=i+l−m−2,…,i−2. Then the following conditions hold:
(a)
Fil(m)(j)=Fil(m)(2i−m−4+l−j)* for all j with i+l−m−2≤j≤i−2. In particular, Fil(m)(l−2)=Fil(m)(2i−m−2)=0.*
2. (b)
for all j with 2i−m−2≤j≤l−2,Fil(m)(j)≤0.
3. (c)
for all j with i+l−m−2≤j<2i−m−2,Fil(m)(j)>0
4. (d)
Fil(m)(j)>0* for all j with i−2≥j>l−2.*
5. (e)
Fil(m)([i−2m−l+5]+1)=mini+l−m−2≤j≤i−2Fil(m)(j),* that is, Fil(m) attains minimum value at [i−2m−l+5]+1.Fil(m)([i−2m−l+5]+1)≤0.Fil(m) is decreasing for all j with i+l−m−2≤j≤[i−2m−l+5]+1 and Fil(m) is increasing for all j with [i−2m−l+5]+1≤j≤i−2.*
Proof.
(a) For fixed but arbitrary j, from equation (2.5) we have
[TABLE]
which implies Fil(m)(j)=Fil(m)(2i−m−4+l−j). This shows that Fil(m)(j)=Fil(m)(2i−m−4+l−j) for all j with i+l−m−2≤j≤i−2.
(b) The condition Fil(m)(j)<0 is equivalent to (m−im)(m−i+lm)(m−i+j+2m)(m+2+j−i−lm)>1. Since 2i−m−2<j<l−2, we note that i−j−2<m−i,i−j−2>i−l and m−i+j+2>i. Thus, we have
[TABLE]
Since 2i−m−2<j<l−2, we observe that (m−i+j−2)<m−i+l,…,(i+l−j−2)>(i+1) and so on. Therefore, from (2) we conclude that (m−im)(m−i+lm)(m−i+j+2m)(m+2+j−i−lm)>1. This completes the proof of (b).
(c)
To prove (c), we need to verify (m−im)(m−i+lm)(m−i+j+2m)(m+2+j−i−lm)<1 for all j with i+l−m−2≤j<2i−m−2. Since j<2i−m−2, we get m−i+2+j<i and i+l−j−2>m−i+l. By observing the above facts, note that
[TABLE]
Since l>2i−m and j<2i−m−2, we have i+l−j−2>i,…,m−i+l+1>m−i+3+j and so on. Therefore, from (2.8) we conclude that (m−im)(m−i+lm)(m−i+j+2m)(m+2+j−i−lm)<1. This completes the proof of (c).
(d)
One can prove (d) by simply observing the fact that Fil(m)(j)=Fil(m)(2i−m−4+l−j) for all j with i+l−m−2≤j≤i−2.
(e)
In order to prove (e), we show that (m−i+j+2m)(m+2+j−i−lm) attains maximum value at [i−2m+l−5]+1, that is, find out the maximum value of j such that (m−i+j+3m)(m−i−l+j+3m)(m−i+j+2m)(m+2+j−i−lm)<1.
Now, we have
[TABLE]
where i−j−2=x.
The condition (m−i+j+3m)(m−i−l+j+3m)(m−i+j+2m)(m+2+j−i−lm)<1 is equivalent to x(x+l)(m+1−x)(m+1−x−l)<1, which gives 2j<2i−m+l−5. Similarly, one can also show that (m−i+j+3m)(m−i−l+j+3m)(m−i+j+2m)(m+2+j−i−lm)>1 implies that 2j>2i−m+l−5. The above two conditions together imply that (m−i+j+2m)(m+2+j−i−lm) attains maximum value at [i−2m−l+5]+1.
Now, we show that 2i−m−2≤[i−2m−l+5]+1. If not, we have 2i−m−2>[i−2m−l+5]+1, which implies that l<2i−m+1. This gives a contradiction to our assumption l>2i−m. Similarly, one can also show that [i−2m−l+5]+1≤l−2. By observing the above facts, one can prove that Fil(m)([i−2m−l+5]+1)≤0. From above, it is also clear that Fil(m) is decreasing for all j with i+l−m−2≤j≤[i−2m−l+5]+1 and Fil(m) is increasing for all j with [i−2m−l+5]+1≤j≤i−2. This completes the proof.
∎
As a consequence, we prove the following corollary.
Corollary 2.3**.**
Let i′,l,j be integers such that i∈{1,…,m},0≤l≤m+2−i′,l>m−2i′+4 and
j=l−i′,…,m−i′. Then the following conditions hold:
(a)
Gi′l(m)(j)=Gi′l(m)(m−2i′+l−j)* for all j with l−i′≤j≤m−i′. In particular, Gi′l(m)(l−2)=Gi′l(m)(m−2i′+2)=0.*
2. (b)
for all j with m−2i′+2≤j≤l−2,Gi′l(m)(j)≥0.
3. (c)
for all j with l−i′≤j<m−2i′+2,Gi′l(m)(j)<0.
4. (d)
Gi′l(m)(j)<0* for all j with m−i′≥j>l−2.*
5. (e)
Gi′l(m)([2m+l−1−i′]+1)=maxl−i′≤j≤m−i′Gi′l(m)(j),* that is, Gi′l(m) attains maximum value at [2m+l−1−i′]+1.Gi′l(m)([2m+l−1−i′]+1)≥0.Gi′l(m) is increasing for all j with l−i′≤j≤[2m+l−1−i′]+1 and Gi′l(m) is decreasing for all j with [2m+l−1−i′]+1≤j≤m−i′.*
Proof.
Putting i′=m+2−i in (2.6) and using Proposition 2.2, one can easily show the required inequality.
∎
The following proposition also determines the signs of Fil(m) and Gi′l(m) whenever l<2i−m. As the proof of this proposition is similar to that of Proposition 2.2, we skip this proof.
Proposition 2.4**.**
Let i,l,j be integers such that i∈{1,…,m},0≤l≤i,l<2i−m and
j=i+l−m−2,…,i−2. Then the following conditions hold:
(a)
Fil(m)(j)=Fil(m)(2i−m−4+l−j)* for all j with i+l−m−2≤j≤i−2. In particular, Fil(m)(l−2)=Fil(m)(2i−m−2)=0.*
2. (b)
for all j with l−2≤j≤2i−m−2,Fil(m)(j)≤0.
3. (c)
for all j with i+l−m−2≤j<l−2,Fil(m)(j)>0
4. (d)
Fil(m)(j)>0* for all j with i−2≥j>2i−m−2.*
5. (e)
Fil(m)([i−2m−l+5]+1)=mini+l−m−2≤j≤i−2Fil(m)(j),* that is, Fil(m) attains minimum value at [i−2m−l+5]+1.Fil(m)([i−2m−l+5]+1)≤0.Fil(m) is decreasing for all j with i+l−m−2≤j≤[i−2m−l+5]+1 and Fil(m) is increasing for all j with [i−2m−l+5]+1≤j≤i−2.*
As a consequence, we will prove the following corollary and the proof of this corollary is similar to the Corollary 2.3, we skip the proof.
Corollary 2.5**.**
Let i′,l,j be integers such that i∈{1,…,m},0≤l≤m+2−i′,l<m−2i′+4 and
j=l−i′,…,m−i′. Then the following conditions hold:
(a)
Gi′l(m)(j)=Gi′l(m)(m−2i′+l−j)* for all j with l−i′≤j≤m−i′. In particular, Gi′l(m)(l−2)=Gi′l(m)(m−2i′+2)=0.*
2. (b)
for all j with l−2≤j≤m−2i′+2,Gi′l(m)(j)≥0.
3. (c)
for all j with l−i′≤j<l−2,Gi′l(m)(j)<0.
4. (d)
Gi′l(m)(j)<0* for all j with m−i′≥j>m−2i′+2.*
5. (e)
Gi′l(m)([2m+l−1−i′]+1)=maxl−i′≤j≤m−i′Gi′l(m)(j),* that is, Gi′l(m) attains maximum value at [2m+l−1−i′]+1.Gi′l(m)([2m+l−1−i′]+1)≥0.Gi′l(m) is increasing for all j with l−i′≤j≤[2m+l−1−i′]+1 and Gi′l(m) is decreasing for all j with [2m+l−1−i′]+1≤j≤m−i′.*
Remark 2.6**.**
If l=2i−m, then from Proposition 2.2, one can conclude that Fil(m) attains minimum value at [i−2m−l+5]=2i−m−2 and mini+l−m−2≤j<i−2Fil(m)(j)=0. Suppose l<2i−m. Then between l−2 and 2i−m−2 there are even or odd number of points. If there are even number of points between l−2 and 2i−m−2, then by observing the fact Fil(m)(j)=Fil(m)(2i−m−4+l−j) for all j with 0≤j≤2i−m−4+l, one can conclude that Fil(m)([i−2m+l−5]+1)=Fil(m)([i−2m+l−5]). If it contains odd number of points, then Fil(m)([i−2m+l−5]+1)=mini+l−m−2≤j<i−2Fil(m)(j). We can also arrive the same conclusion when l<2i−m.
For all z∈Dm and w∈D and i∈{1,…,m}, let gijl(m) and gi′lj(m) be the functions on Dm+1 defined by
[TABLE]
[TABLE]
[TABLE]
where i′=m+2−i. Let A:={(z,w):gilj(m)(z,w)>0}andB:={(z,w):gi′lj(m)(z,w)≥0} be two subsets of Dm+1. The following proposition gives the relation between two sets A and B.
Proposition 2.7**.**
Let A and B be as above. Then B=Dm+1∖A.
Proof.
Consider (z,w)∈Dm+1∖A. Then by definition of A, we get gilj(m)(z,w)≤0 which is equivalent to gi′lj(m)(z,w)≥0.
Hence, (z,w)∈B. This shows that Dm+1∖A⊂B. Using similar technique which is described above, one can verify that B⊂Dm+1∖A. Thus, we have Dm+1∖A=B.
∎
From above Proposition, we observe that A and B are complement of each other. Therefore, for each (z,w)∈Dm+1, we have either gilj(m)(z,w)>0 or gi′lj(m)(z,w)≥0. For i≥1,l≥1 and i′≥1, every gilj(m)(z,w) and gi′lj(m)(z,w) can be written as
[TABLE]
[TABLE]
[TABLE]
where z′∈D(m−1).
The following lemma gives the estimate for the norm of gilj(m)(z,w) and gi′lj(m)(z,w). It is expected that this result may be important for proving some elementary properties of Γm.
Lemma 2.8**.**
Let z∈Dm and i′=m+2−i. Then
for all w∈D and for all i with i∈{1,…,m},0≤l≤i and
j=(i+l−m−2),…,(i−2) and j=2i−m−2
[TABLE]
However, if j=2i−m−2, then gilj(m)(z,w)=0=gi′lj(m)(z,w).
Proof.
We prove it by the method of induction on m.
m=1:
For m=1, we get
[TABLE]
[TABLE]
[TABLE]
By Proposition 2.7, we have either g1lj(1)(z,w)>0 or g2lj(1)(z,w)≥0. Suppose g1lj(1)(z,w)>0. For i=1, we have to consider two values of l, that is, l=0,1. For l=0, we can take two values of j, that is, j=−2,−1. If m=1,i=1,l=0 and j=−2, then the fourth case of the inequality (2.16) is satisfied. For m=1,i=1,l=0 and j=−1, we have g10(−1)(1)(z,w)=0. Now, for m=1,i=1 and l=1, the value of j is j=−1. Clearly, g11(−1)(1)(z,w)=0. Similarly, one can prove the required estimates whenever g2lj(1)(z,w)≥0.
This shows that the result is true for m=1.
As the hypothesis of the induction, we assume that m>3 and the result holds for all integers up to (m−1), that is,
(1)
for all z′∈Dm−1,w∈D and for all i with i∈{1,…,(m−1)},0≤l≤i,j=(i+l−m−1),…,(i−2),j=2i−m−1 and i′=m+1−i,
For each (z,w)∈Dm+1, we have either gilj(m)(z,w)>0 or gi′lj(m)(z,w)≥0( See Proposition 2.7). We first present a general outline of the method used in the flowchart below.
(\mathbb{z},w)\in\mathbb{D}^{m+1}$$g^{(m)}_{ilj}(\mathbb{z},w)>0$$l\geq 2i-mStep-1: j=l−2Observation-A: j<l−2Case-1:g(i−1)l(j−1)(m−1)(z′,w)>0Case-2:g(i−1)l(j−1)(m−1)(z′,w)<0Observation-B:j>l−2,Proof is similar to the other counter partStep-2: j=l−2Case-1:g(i−1)l(l−3)(m−1)(z′,w)>0Subcase-1:l−1>2i−mSubcase-2:l−1=2i−mCase-2:g_{(i-1)l(l-3)}^{(m-1)}(\mathbb{z}^{\prime},w)<0$$l<2i-mProof is similar to the other counterpartgi′lj(m)(z,w)≥0,Proof is similar to the other counter part
We prove the required estimate for gilj(m)(z,w)>0, where the proof of the other required estimate for gi′lj(m)(z,w)≥0 is similar. Also, for positive integers i,l,m, we have two possibilities: either l≥2i−m or l<2i−m. To get the above estimates, we consider the case l≥2i−m, while the proof of the other case holds similarly. We prove the above estimates for gilj(m)(z,w)>0 and l≥2i−m in two steps. In the first step we assume j=l−2 whereas in the second we consider the case j=l−2.
Step-1 {j=l−2}: In order to prove the estimate for j=l−2 we need to consider the following observations either j>l−2 or j<l−2.
Observations-A:{ j<l−2}:
For j<l−2, we need to consider the following cases
Case-1: {g(i−1)l(j−1)(m−1)(z′,w)>0}:
From (2), we get
[TABLE]
If any one real part of (2.17) is negative, then we can dominate it by zero and proceed similarly. Therefore, we consider only the subcase when
gil(j+1)(m−1)(z′,w)>0,gi(l+1)(j+1)(m−1)(z′,w)>0,g(i−1)(l−1)(j−1)(m−1)(z′,w)>0.
By mathematical induction, one can easily verify that g_{ilj}^{(m-1)}(\mathbb{z}^{\prime},w)\leq\big{|}F_{il}^{(m-1)}(j)\big{|},g_{(i-1)lj}^{(m-1)}(\mathbb{z}^{\prime},w)\leq\big{|}F_{(i-1)l}^{(m-1)}(j)\big{|},g_{(i(l+1)(j+1)}^{(m-1)}(\mathbb{z}^{\prime},w)\leq\big{|}F_{i(l+1)}^{(m-1)}(j+1)\big{|} and g_{(i-1)(l-1)(j-1)}^{(m-1)}(\mathbb{z}^{\prime},w)\leq\big{|}F_{(i-1)(l-1)}^{(m-1)}(j-1)\big{|}.
Sign of F(i−1)l(m−1)(j),Fil(m−1)(j),Fi(l+1)(m−1)(j+1) and F(i−1)(l−1)(m−1)(j−1) can be determined by Proposition 2.2 depending on i,l,j,m.
By Proposition 2.2, we note that F(i−1)l(m−1)(j),Fil(m−1)(j),Fi(l+1)(m−1)(j+1) and F(i−1)(l−1)(m−1)(j−1) are positive for all j with i+l−m−2≤j<2i−m−3 and F(i−1)l(m−1)(j),Fil(m−1)(j),Fi(l+1)(m−1)(j+1) and F(i−1)(l−1)(m−1)(j−1) are negative for all j with 2i−m−2≤j<l−2. Also, we have Fil(m−1)(2i−m−1)=0 and F(i−1)l(m−1)(2i−m−3)=0. Hence, if i+l−m−2≤j<2i−m−3, then from (2.17) we observe that
[TABLE]
Similarly, we can show that gilj(m)(z,w)≤∣Fil(m)(j)∣ whenever 2i−m≤j<l−2. Also, using Proposition 2.2, we get gil(2i−m−2)(m)(z,w)=0. So, we now have to investigate the case j=2i−m−1 and j=2i−m−3 only. Since F(i−1)l(m−1)(2i−m−3)>0,Fil(m−1)(2i−m−3)>0,Fi(l+1)(m−1)(2i−m−2)>0 and F(i−1)l(m−1)(2i−m−3)=0, from (2.17), we conclude that
[TABLE]
By using similar procedure, one can show that gil(2i−m−1)(m)(z,w)≤∣Fil(m)(2i−m−1)∣. This completes the proof of this subcase-1.
Case-2: {g(i−1)l(j−1)(m−1)(z′,w)<0}:
From (2), we note that
[TABLE]
Similar to the Case-1, if any one real part of (2.18) is negative, then we can dominate it by zero and proceed similarly. Therefore, for this case also we consider only the subcase when gil(j+1)(m−1)(z′,w)>0,g(i−1)(l−1)(j−1)(m−1)(z′,w)>0,gi(l+1)(j+1)(m−1)(z′,w)>0.
We verify this subcase only when j=l−3, because the proof of this subcase for other value of j are similar to the Case-1. It follows from induction hypothesis that
[TABLE]
[TABLE]
Using similar technique which is described in Case-1, from (2.18) we conclude that g_{il(l-3)}^{(m)}(\mathbb{z},w)\leq\big{|}F_{il}^{(m-1)}(j)\big{|}.
Observations-B:{ j<l−2}:
The proof of the observation-B is similar to the Observation-A.
Step-2:{j=l−2}:
In this step, as before we need to consider two cases.
Case-1: {g(i−1)l(l−3)(m−1)(z′,w)>0}: Because l>2i−m, we need to consider two subcases
Subcase-1: {l−1>2i−m}: By Proposition 2.2, we note that F(i−1)l(m−1)(l−2),Fil(m−1)(l−2),Fi(l+1)(m−1)(l−1) and F(i−1)(l−1)(m−1)(l−3) are positive. Using similar technique which is described in Step-1 of Case-1, one can show the required estimate.
Subcase-2: {l−1=2i−m}: By Proposition 2.2, we see that F(i−1)l(m−1)(l−2)=0 and Fil(m−1)(l−2),Fi(l+1)(m−1)(l−1) and F(i−1)(l−1)(m−1)(l−3) are positive. Arguing similarly as in Step-1 of Case-1, one can prove the required estimate.
Case-2: {g(i−1)l(l−3)(m−1)(z′,w)<0}: In this case, by Proposition 2.2, it follows that Fil(m−1)(l−1),Fi(l+1)(m−1)(l−1) and F(i−1)(l−1)(m−1)(l−3) are positive. Using similar argument as in Step-1 of Case-2, one can get the desired estimate. This completes the proof.
∎
The previous lemma will be useful to prove Theorem 2.1.
: We prove it by induction on (m+1). For m=1, the result follows from Lemma 2.8.
By mathematical induction, we assume that m>1 and the result holds for all integers up to m, that is,
(1)
z′∈Dm and for all i with i∈{1,…,m},j=i−m−1,…,(i−1),j=2i−m−1 and i′=m+1−i,
[TABLE]
2. (2)
However, if j=2i−m−1, then ∣hi′j(m)(z′)=∣hij(m)(z′)=0.
Note that
[TABLE]
By Proposition 2.7, we have either hij(m+1)(z)>0 or hi′j(m+1)(z)≥0. We verify the required estimate for hij(m+1)(z)>0, as the proof of the other required estimate for hi′j(m+1)(z)≥0 is similar. As before, for positive integer i,m, we have either 2i>m or 2i<m. We consider the case when 2i>m, as the proof of the other case hold similarly. Thus, we wish to verify our required estimate for hij(m+1)(z)>0 and 2i>m in two steps.
Step-1:{j=−1}:
In this step, we have two observations.
Observation-A:{j<−1}
In this observation, we need to consider two cases.
Case-1:{∣si−1(m)(z′)∣2>∣sm+1+j−i(m)(z′)∣2}: From (2), we have
[TABLE]
If any one real part of (2.24) is negative, then we can dominate it by zero and proceed similarly. Therefore, we verify only the subcase
when ∣si(m)(z′)∣2>∣sm+2+j−i(m)(z′)∣2,Re(sˉi(m)(z′)si−1(m)(z′)−sˉm+2+j−i(m)(z′)sm+1+j−i(m)(z′))>0. By induction hypothesis and using Lemma 2.8, we deduce the following estimates:
By using similar argument which is described in Lemma 2.8, one can easily show the required estimate.
Case-2:{∣si−1(m)(z′)∣2<∣sm+1+j−i(m)(z′)∣2}: From (2), we have
[TABLE]
Similar to the Lemma 2.8, one can verify the required estimate.
Step-2:{j=−1}: Applying similar technique which in Lemma 2.8, one can easily prove the required estimate.
This completes the proof.
∎
Remark 2.9**.**
For j=l−2, we notice that |g_{ilj}^{(m)}((1,\ldots,1),w)|=\big{|}F_{il}^{(m)}(j)\big{|}=|g_{i^{\prime}lj}^{(m)}((1,\ldots,1),w)|.
Therefore, for j=l−2, by using Lemma 2.8, we deduce that \sup_{(\mathbb{z},w)\in\mathbb{D}^{m+1}}|g_{ilj}^{(m)}(\mathbb{z},w)|=\big{|}F_{il}^{(m)}(j)\big{|}=\sup_{(\mathbb{z},w)\in\mathbb{D}^{m+1}}|g_{i^{\prime}lj}^{(m)}(\mathbb{z},w)|. This shows that for j=l−2 the above estimates are sharp. Similarly, for j=−1, from Theorem 2.1, we also have \sup_{\mathbb{z}\in\mathbb{D}^{m+1}}\big{|}h^{(m+1)}_{ij}(\mathbb{z})\big{|}=\big{|}\binom{m+1}{i}^{2}-\binom{m+1}{m+2+j-i}^{2}\big{|}=\sup_{\mathbb{z}\in\mathbb{D}^{m+1}}\big{|}h^{(m+1)}_{i^{\prime}j}(\mathbb{z})\big{|}.
3. Schur’s criterion and some properties of Symmetrized polydisc
It is well known that a point (s1,…,sn−1,sn)∈Gn is equivalent to the polynomial P(z)=zn−s1zn−1+…+(−1)nsn has all roots in D. Schur’s criterion [33, 37] is a standard tool for deciding whether the zeros of the polynomial lie in D.
Let k(i)=(in−1)+(n−in−1) for i=1,…,n−1. For i=1,…,(n−1), we define operator pencils Φ1(i) and Φ2(i) for a commuting n tuple of bounded operators (S1,…,Sn) as
[TABLE]
[TABLE]
[TABLE]
We show that these operator pencils play an important role in determining the structure of Γn-contractions.
In particular, if S1,…,Sn are scalar, then Φ1(i) and Φ2(i) are of the following form
[TABLE]
[TABLE]
[TABLE]
The following propositions are characterization of bΓn and Γn.
Suppose (s1,…,sn−1,sn)∈Cn. Then the following conditions are equivalent:
(1)
(s1,…,sn−1,sn)∈Γn;**
2. (2)
∣sn∣≤1* and there exist (c1,…,cn−1)∈Γn−1 such that si=ci+cˉn−isn and sn−i=cn−i+cˉisn for i=1,…,(n−1).*
For closed symmetrized polydisc Γn, we have the following characterization.
Theorem 3.3**.**
Suppose (s1,…,sn−1,sn)∈Γn. Then for i=1,…,(n−1)
[TABLE]
for all α∈D with k(i)=(in−1)+(n−in−1).
Proof.
Let (s1,…,sn) be a point in Γn. Then (αs1,…,αnsn)∈Gn for all α∈D. Consider the polynomial
[TABLE]
Suppose z1,…,zn are the roots of the polynomial p(z)=0. Then αisi=∑1≤k1<…<ki≤nzk1…zki. Let S be the n×n “Shift matrix” which is of the form S=0⋮001⋮000⋮000⋮00…⋮……0⋮10. Then by a simple computation one can show that
[TABLE]
[TABLE]
[TABLE]
where q(z)=znp(z1). Let L=q(S)∗q(S)−p(S)∗p(S),a=1−∣αnsn∣2 and c=(−1)n−1αˉn−1(sˉn−1−∣α∣2s1sˉn). Then
[TABLE]
where L(i)=a+(∣αs1∣2−∣αn−1sn−1∣2)+…+(∣αisi∣2−∣αn−isn−i∣2)⋮(−1)n−i−1αn−i−1(sn−i−1−∣α∣2(i+1)sˉi+1sn)…⋮…(−1)n−i−1αˉn−i−1(sˉn−i−1−∣α∣2(i+1)si+1sˉn)⋮a,A(i) and B(i) are (n−i)×(n−i),i×i and i×(n−i) matrices respectively and L(0)=L for i=1,…,n.
By Schur’s criterion [33], the matrix L is positive definite. It is well known that if the matrix A is positive definite, then all leading principal minors of A are positive definite[24]. Since L is positive definite, L(i) and S(1,n−i) are positive definite for i=1,…,n, where
[TABLE]
The positive definiteness of S(1,n−i) is equivalent to detS(1,n−i)>0 and a>0 which gives
[TABLE]
In order to prove the required inequality, we need to consider two cases, namely n is odd or n is even. As the proof of the other case is similar, we only consider the case when n is odd,.
Case-1: {n=(2m+1): } For this case, we first prove the following inequality for i=1,…,m,
[TABLE]
[TABLE]
[TABLE]
Let m(i)=∣αisi∣2−∣αn−isn−i∣2. To prove the above required inequality, we need to show the following inequalities:
[TABLE]
To show this, we apply Proposition 3.2, which implies that there exists (ci,…,cn−1)∈Γn−1 such that
We prove the inequality (3.5) by mathematical induction.
For i=1, the result follows from the positivity of S(1,n)>0. Assuming the truth of inequality (3.5) for a fixed value of i, we wish to show that it is true with i replaced by i+1. Let di be the i th diagonal entries of L. Also we observe that
[TABLE]
[TABLE]
[TABLE]
Thus, by observing above fact and the positive definiteness of S(1,n−i)>0, we conclude that
[TABLE]
This completes the induction step.
We now show the inequality (3.6). We prove it by induction on m. We can write the matrix L as
[TABLE]
Since L is positive definite, then the 2×2 principal minor D2,2 which is obtained from L is positive definite, where
[TABLE]
The positive definiteness of D2,2 is equivalent to ∣α(s1−∣α∣2n−2sˉn−1sn)∣2<(n−1)2a2, which shows that the result is true for i=1.
Assume that the inequality (3.6) is true for fixed values of i. Now, we wish to show that it is true with i replaced by i+1. Let
[TABLE]
[TABLE]
[TABLE]
The positive definiteness of L implies that Di+2,i+2 and S(1,i+2) are positive definite. Therefore, we have
[TABLE]
This completes the induction step.
Our next goal is to show that for k(i)=(in−1)+(n−in−1)
[TABLE]
[TABLE]
[TABLE]
Note that
[TABLE]
If m(i) and m(n−i) are negative, then the inequalities in (3.9) and (3.10) are obvious. Therefore, we assume that
m(i)≥0 and m(n−i)≥0. By using (3.5), (3.6),(3.7) and (3.8), we see that for k(i)=(in−1)+(n−in−1)
[TABLE]
[TABLE]
[TABLE]
Therefore, for all ω∈T and i=1,…,m, from (3.13) we have
[TABLE]
Choosing ω=1 and substituting the value of a,m(i) in (3.14), for i=1,…,m, we get
[TABLE]
For i=1,…,m, by continuity, we deduce that Φ2(i)(αisi,αn−isn−i,αnsn)≥0 for all α∈D. Similarly, for i=1,…,m, we can prove that Φ1(i)(αn−isn−i,αisi,αnsn)≥0, for all α∈D. This completes the proof.
∎
Remark 3.4**.**
For i=1,…,(n−1), the positivity of two functions Φ1(i) and Φ2(i) are equivalent to the conditions ∣k(i)αnsn−αn−isn−i∣≤∣k(i)−αisi∣ and ∣k(i)αnsn−αisi∣≤∣k(i)−αn−isn−i∣ with k(i)=(in−1)+(n−in−1). One can easily check that the positivity of two functions Φ1(i) and Φ2(i) imply that for i=1,…,(n−1)
[TABLE]
If we further assume that ∣sn∣=1
and (γ1s1,…,γn−1sn−1)∈Γn−1, then from (3.15), it follows immediately that (s1,…,sn)∈bΓn, where γi=nn−i. For ∣sn∣<1, however it is not clear whether there exists (c1,…,cn−1)∈Γn−1 such that si=ci+cˉn−isn,sn−i=cn−i+cˉisn for i=1,…,(n−1).
4. Γn-contractions and their fundamental operators
Let (S1,…,Sn) be the commuting n-tuples of bounded operators defined on a Hilbert space H with ∥Sn∥≤1. Given a contraction Sn, denote DSn=(I−Sn∗Sn)21 and DSn=RanDSn. In this section, we want to find the solutions Xi and Xn−i of the fundamental equations which are defined in the following:
[TABLE]
The following proposition gives an estimate for the norm of Si, for i=1,…,(n−1). As the proposition follows immediately from the definition of Γn-contraction, we skip the proof.
Proposition 4.1**.**
If (S1,…,Sn) is a Γn-contraction, then ∥Si∥≤(in−1)+(n−in−1) and ∥Sn∥≤1.
Under some suitable condition the following proposition says that the operator pencils Φ1(i) and Φ2(i) are positive.
Proposition 4.2**.**
For n≥2, let (S1,…,Sn) be a Γn-contraction. Then for all i=1,…,(n−1)
[TABLE]
for all α∈D with k(i)=(in−1)+(n−in−1).
Proof.
Since (S1,…,Sn) is a Γn-contraction, σT(S1,…,Sn−1,Sn)⊆Γn. Let f be a holomorphic function in a neighbourhood of Γn. Since Γn is a polynomially convex, by Oka-Weil theorem [22, Theorem 5.1], there exists a sequence of polynomials {pn} that converges uniformly to f on Γn.
Thus, we have
[TABLE]
For fix α∈D, let fi(s1,…,sn)=k(i)−αisik(i)αnsn−αn−isn−iforalli=1,…,(n−1). Since k(i)=(in−1)+(n−in−1),f is well defined holomorphic function in a neighbourhood of Γn. Therefore, we note that
[TABLE]
which gives
[TABLE]
So by definition of Φ1(i), we get that Φ1(i)(αiSi,αn−iSn−i,αnSn)≥0 for all α∈D. By continuity we have Φ1(i)(αiSi,αn−iSn−i,αnSn)≥0 for all α∈Dˉ. Similarly, one can also prove that Φ2(i)(αiSi,αn−iSn−i,αnSn)≥0 for all α∈D. This completes the proof.
∎
Recall that the numerical radius of a bounded operator A on a Hilbert space H is defined to be
[TABLE]
It is well known that r(A)≤ω(A)≤∥A∥, where r(A) is the spectral radius of A. Also if the numerical radius of a bounded operator A is not greater than n then ReαA≤nI for all complex number α with ∣α∣=1 and vice-versa. In [29], it is shown that if two bounded operators A1,A2 with ω(A1+zA2)≤n for all z∈T, then ω(A1+zA2∗)≤n and ω(A1∗+zA2)≤n for all z∈T. We start with pivotal theorem which says that if (S1,…,Sn) is a Γn-contraction then the fundamental equations have unique solutions. The proof of the theorem needs the following lemma. We skip the proof as it is easy to check.
Lemma 4.3**.**
Let A1,…,An−1 be bounded operators on a Hilbert space H. Then the following are equivalent:
(1)
[Ai,Aj]=0* and [Ai,An−j∗]=[Aj,An−i∗] for 1≤i,j≤n−1, where [P,Q]=PQ−QP for two operators P,Q;*
2. (2)
Ai∗+An−iz* and Aj∗+An−jz commute for every z with ∣z∣=1 and for 1≤i,j≤n−1;*
3. (3)
Ai∗+An−iz* is a normal operator for every z with ∣z∣=1 and for 1≤i≤n−1;*
4. (4)
An−i∗+Aiz* is a normal operator for every z with ∣z∣=1 and for 1≤i≤n−1.*
Now here is the theorem which gives the existence and uniqueness of solution to the fundamental equations of a Γn-contraction.
Theorem 4.4**.**
(Existence and Uniqueness)
For n≥2, let (S1,…,Sn) be a Γn-contraction on a Hilbert space H. Then there are unique operators E1,…,En−1∈B(DSn) such that Si−Sn−i∗Sn=DSnEiDSn and Sn−i−Si∗Sn=DSnEn−iDSn,Ei,En−i∈B(DSn), for i=1,…,(n−1). Moreover, ω(Ei+En−iz)≤(in−1)+(n−in−1) for all z∈T.
Proof.
For i=1,…,(n−1), applying Proposition 4.2, we have
[TABLE]
for all α∈D with k(i)=(in−1)+(n−in−1).
In particular, for all α,β∈T and i=1,…,(n−1) , in view of (4.1), this implies that
[TABLE]
and
[TABLE]
respectively.
For all i=1,…,(n−1), set
[TABLE]
Choose β such that βn−i=α2i. Hence we deduce from (4.2), (4.3) and (4.4) that η(i)(α)≥0, for all α∈T and i=1,…,(n−1). Therefore, for all i=1,…,(n−1) and α∈T, by operator Fejer-Reisz Theorem [19, Theorem 1.2], there is a polynomial of degree 2i say P(i)(α)=X0(i)+X1(i)αi+X2(i)α2i such that
[TABLE]
Comparing (4.4) and (4), for all i=1,…,(n−1), it follows that
which, by Douglas’s lemma [18, Lemma 2.1], implies that there are contractions Z0(i),Z1(i),Z2(i) such that
[TABLE]
For all i=1,…,(n−1), putting the value of X0(i)∗,X1(i)∗ and X2(i)∗ in (4.7), we deduce that
[TABLE]
For all i=1,…,(n−1), let
[TABLE]
Then from above it is clear that Ei,En−i are the solutions to the equations Si−Sn−i∗Sn=DSnXiDSn and Sn−i−Si∗Sn=DSnXn−iDSn respectively, for all i=1,…,(n−1).
Uniqueness: Let Ei,Gi be two solutions of the equation Si−Sn−i∗Sn=DSnXiDSn for all i=1,…,(n−1). Then DSn(Ei−Gi)DSn=0, which implies that Ei=Gi on DSn for all i=1,…,(n−1).
Also, for all i=1,…,(n−1), we have that
[TABLE]
where Fi(α)=k(i)1(Ei+αiEn−i). This implies that IDSn−ReαiFi(α)≥0, because Fi(α) is defined on DSn for all i=1,…,(n−1), which gives ω(Fi(α))≤1. Thus, we have ω(Ei+zEn−i)≤(in−1)+(n−in−1) for all z∈T and i=1,…,(n−1). This completes the proof.
∎
Remark 4.5**.**
From Theorem 4.4, it is clear that Si−Sn−i∗Sn is equal to zero on the orthogonal complement of DSn in H for all i=1,…,(n−1). Also for the case of Γn-isometry or Γn-unitary of (S1,…,Sn), the (n−1)-tuple of fundamental operators Ei for i=1,…,(n−1), are zero, because for this case DSn={0}.
The following proposition says that two fundamental operators are invariant under unitary equivalence. We skip the proof because it is easy to verify.
Proposition 4.6**.**
If two Γn-contractions are unitarily equivalent then their fundamental operators are also unitarily equivalent.
5. Γn-Unitaries and Γn-Isometries
We recall that a Γn-unitary is a commuting n-tuple of normal operators (S1,…,Sn)
whose Taylor joint spectrum contained in the distinguished boundary of Γn and a
Γn-isometry is the restriction of a Γn-unitary to a joint invariant subspace
of S1,…Sn. In this section we would like to discuss several properties of Γn-unitaries and Γn-isometries. In order to describe several properties of Γn-unitaries and Γn-isometries, we need a known fact from [12], which says that if T is a bounded operator on a Hilbert space H with ReβT≤0 for all β∈T, then T=0. Parts of the following theorem tells the new characterization of Γn-unitary which were obtained in [14, Theorem-4.2]. Parts (4) and (5) are new. This theorem plays an important role for proving the conditional dilation on Γn which we will see later.
Theorem 5.1**.**
Let (S1,…,Sn) be a commuting n-tuple of operator defined on a Hilbert space H. Then the following are equivalent:
(1)
(S1,…,Sn)* is a Γn-unitary;*
2. (2)
There exists commuting unitary operators U1,…,Un on H such that Si=∑1≤k1<…<ki≤nUk1…Uki for i=1,…,n−1;
3. (3)
Sn* is unitary, (γ1S1,…,γn−1Sn−1)∈Γn−1 and Si=Sn−i∗Sn for i=1,…,n−1;*
4. (4)
(S1,…,Sn)* is a Γn-contraction and Sn is an unitary ;*
5. (5)
Sn* is unitary and there exist Γn−1-unitary (C1,…,Cn−1) on H such that C1,…,Cn−1,Sn commute and
Si=Ci+Cn−i∗Sn,Sn−i=Cn−i+Ci∗Sn for i=1,…,n−1.*
Proof.
We will prove that (1)⇔(2)⇔(3),(2)⇒(4)⇒(3) and (2)⇔(5).
As the equivalence of (1),(2) and (3) is due to [14, Theorem 4.2], we skip the proof. It is also easy to verify (2) implies (4).
We will now show that (4) implies (3). To show (4) implies (3), let (S1,…,Sn) be a Γn-contraction and Sn be a unitary. Then by Proposition 4.2, we see that for i=1,…,(n−1) and for all β∈T with k(i)=(in−1)+(n−in−1)
[TABLE]
which implies
[TABLE]
[TABLE]
[TABLE]
Since Sn is unitary, from (5.1) and (5.2) it follows that for i=1,…,(n−1) and for all β∈T with k(i)=(in−1)+(n−in−1)
[TABLE]
[TABLE]
[TABLE]
Putting βi=1 and βi=−1 respectively in (5.3) and adding them, we obtain for i=1,…,n−1,
[TABLE]
Again, putting βn−i=1 and βn−i=−1 respectively in (5.4) and adding them, we have
[TABLE]
Thus, (5.5) and (5.6) together imply that Si∗Si=Sn−i∗Sn−i for i=1,…,(n−1). From (5.3) we deduce that Reβi(Si−Sn−i∗Sn)≤0 for all β∈T and for i=1,…,(n−1), which, by above fact, gives Si=Sn−i∗Sn. Since (S1,…,Sn) is a Γn-contraction, it is easy to see that (γ1S1,…,γn−1Sn−1) is also a Γn−1-contraction.
We will prove (2) implies (5). Suppose (2) holds. Then, for i=1,…,(n−1),
[TABLE]
where Ci=∑1≤l1<…<li≤n−1Ul1…Uli. Clearly, (C1,…,Cn−1) are Γn−1-unitary.
Now we will show that (5) implies (2). Let Ci=∑1≤l1<…<li≤n−1Ul1…Uli for some commuting unitary unitary U1,…,Un−1 on H. We choose Un=Cn−1∗Sn. Clearly, Un is unitary on H. Thus, we have
[TABLE]
and Sn=Πi=1nUi. Hence (5) implies (2). This completes the proof.
∎
Parts of the following theorem gives the new characterization of Γn-isometry which were obtained in [14, Theorem 4.12]. Parts (4),(5) and (6) are new. The proof of the following theorem works along the lines of [12].
Theorem 5.2**.**
Let S1,…,Sn be commuting operators on a Hilbert space H.
Then the following are equivalent:
(1)
(S1,…,Sn)* is a Γn-isometry ;*
2. (2)
Sn* is a isometry, Si=Sn−i∗Sn and (γ1S1,…,γn−1Sn−1) is Γn−1-contraction;*
3. (3)
( Wold-Decomposition ): there is an orthogonal decomposition H=H1⊕H2 into common invariant subspaces of S1,…,Sn such
that (S1∣H1,…,Sn∣H1) is a Γn-unitary and (S1∣H2,…,Sn∣H2) is a pure Γn-isometry ;
4. (4)
(S1,…,Sn)* is a Γn-contraction and Sn is a isometry;*
5. (5)
(γ1S1,…,γn−1Sn−1)* is a Γn−1-contraction and for all β∈T with k(i)=(in−1)+(n−in−1)*
[TABLE]
Moreover, if r(Si)<(in−1)+(n−in−1) for i=1,…,(n−1), then all of the above are equivalent to :
6. (6)
(γ1S1,…,γn−1Sn−1)* is a Γn−1-contraction and (k(i)βnSn−Sn−i)(k(i)I−βiSi)−1 and (k(i)βnSn−Si)(k(i)I−βn−iSn−i)−1 are isometry for all β∈T and for i=1,…,(n−1) with k(i)=(in−1)+(n−in−1).*
Proof.
As the equivalence of (1),(2) and (3) is due to [14, Theorem 4.12], we skip the proof. It is also easy to verify (1) implies (4).
Suppose (4) holds. Then, by Proposition 4.2 for all β∈T with k(i)=(in−1)+(n−in−1) and for i=1,…,(n−1), we have
[TABLE]
which, together with Sn is isometry, implies that
[TABLE]
[TABLE]
[TABLE]
Using similar argument which is described in Theorem 5.1, one can easily verify that Si∗Si=Sn−i∗Sn−i,Si=Sn−i∗Sn and Sn−i=Si∗Sn for i=1,…,(n−1).
This gives for all β∈T with k(i)=(in−1)+(n−in−1)
[TABLE]
Since (S1,…,Sn−1,Sn) is a Γn-contraction, one can easily show that (γ1S1,…,γn−1Sn−1) is a Γn−1-contraction. Thus, (4) implies (5).
Suppose (5) holds. Then for all β∈T with k(i)=(in−1)+(n−in−1) and for i=1,…,(n−1), we see that
[TABLE]
[TABLE]
[TABLE]
Putting βi=1 and βi=−1 respectively in (5.11) and βn−i=1 and βn−i=−1 respectively in (5.12) and adding them, we get Sn∗Sn=I from which it follows by the same argument as above Si=Sn−i∗Sn and Sn−i=Si∗Sn for i=1,…,(n−1). Thus, we get (5) implies (2).
Now we will prove that (5) is equivalent to (6). Suppose (5) holds. Then for all β∈T with k(i)=(in−1)+(n−in−1) and for i=1,…,(n−1), we obtain
[TABLE]
[TABLE]
[TABLE]
Since r(Si)<(in−1)+(n−in−1), the operator (k(i)I−βiSi) and (k(i)I−βn−iSn−i) are invertible for i=1,…,(n−1). Therefore from (5.11) and (5.12), we conclude that (k(i)βnSn−Sn−i)(k(i)−βiSi)−1 and (k(i)βnSn−Si)(k(i)−βn−iSn−i)−1 are isometry for all β∈T and for i=1,…,(n−1).
Clearly, (6) implies (5). Combining all we conclude that all the above conditions are equivalent. This completes the proof.
∎
6. A Necessary condition for the existence of Dilation
In this section we find out the necessary conditions for the existence of rational dilation. First, we define Γn-isometric dilation of Γn-contraction.
Also, we would like to discuss several properties
of Γn-isometric dilation of Γn-contractions.
Definition 6.1**.**
Let (S1,…,Sn) be a Γn-contraction on a Hilbert space H. A commuting n-tuple of operators (T1,…,Tn−1,V) defined on a Hilbert space K containing H as subspace, is said to be Γn-isometric dilation of (S1,…,Sn) if it satifies the following properties:
•
(T1,…,Tn−1,V) is Γn-isometric;
•
PHT1m1…Tn−1mn−1Vn∣H=S1m1…Snn, for all non-negative integers m1,…,mn−1,n.
If K=span{T1m1…Tn−1mn−1Vnh:h∈Handm1,…,mn−1,n∈N∪{0}}, then we call the n-tuple a minimal Γn-isometric dilation of (S1,…,Sn). Similarly we can define Γn-unitary dilation of a Γn-contraction.
Proposition 6.2**.**
Let (S1,…,Sn) be a Γn-contraction defined on a Hilbert space H. Also assume that (S1,…,Sn) has a Γn-isometric dilation. Then (S1,…,Sn) has a minimal Γn-isometric dilation.
Proof.
Let (T1,…,Tn−1,V) defined on a Hilbert space K containing H as a subspace, be a Γn-isometric dilation of (S1,…,Sn). Let K0 be the space defined by
[TABLE]
One can easily verify that K0 is invariant under T1m1,…,Tn−1mn−1,Vn, for any non-negative integers m1,…,mn−1,n. If we denote T11=T1∣K0,…,T1(n−1)=Tn−1∣K0 and V1=V∣K0, then we get
[TABLE]
Therefore for any any non-negative integers m1,…,mn−1,n we have
[TABLE]
(T11,…,T1(n−1),V1) is a Γn-contraction, because it is the restriction of a Γn-contraction (T1,…,V) to a common invariant subspace K0. Again, V1 is also isometry, because it is the restriction of an isometry to a common invariant subspace K0. Therefore, by Theorem 5.2, (T11,…,T1(n−1),V1) is a Γn-isometry. Hence (T11,…,T1(n−1),V1) is a minimal Γn-isometry of (S1,…,Sn).
∎
Proposition 6.3**.**
Let (T1,…,Tn−1,V) be n-tuple of operators defined on a Hilbert space K containing H as a subspace. Then (T1,…,Tn−1,V) is a minimal Γn-isometric dilation of a Γn-contraction (S1,…,Sn) if and only if (T1∗,…,Tn−1∗,V∗) is a Γn-co-isometric extension of (S1∗,…,Sn∗).
Proof.
We first prove that SiPH=PHTi for i=1,…,(n−1) and SnPH=PHV, where PH:K→H is orthogonal projection onto H. It is also evident that
[TABLE]
Also, for h∈H, we have
[TABLE]
which gives SiPH=PHTi for i=1,…,(n−1). Similarly we can also prove that SnPH=PHV.
Now for any h∈H and for k∈K, we get
[TABLE]
Thus, we have Si∗=Ti∗∣H for i=1,…,(n−1). Similarly, we can also show that Sn∗=V∗∣H.
The converse part is very easy to verify. This completes the proof.
∎
The following proposition, describes a model for pure Γn-isometry which will be used to prove the next proposition.
Proposition 6.4**.**
[14, Theorem 4.10]**
Let S1,…,Sn be commuting operators on a Hilbert space H. Then (S1,…,Sn) is a pure Γn-isometry if and only if there exist a separable Hilbert space Σ and a unitary operator U:H→H2(Σ) and function φ1,…,φn−1 in H∞(B(Σ)) and operators Ei∈B(Σ),i=1,…,(n−1) such that
(1)
Si=U∗MφiU,i=1,…,(n−1),Sn=U∗MzU;**
2. (2)
φi(z)=Ei+En−i∗z* for i=1,…,(n−1);*
3. (3)
∥Ei+En−i∗z∥≤(in−1)+(n−in−1)* for i=1,…,(n−1);*
4. (4)
[Ei,Ej]=0* and [Ei,En−j∗]=[Ej,En−i∗] for i,j=1,…,(n−1).*
Proposition 6.5**.**
Let H1 be a Hilbert space and let (S1,…,Sn) be a Γn-contraction on H with (n−1)-tuple of fundamental operators (E1,…,En−1) and Sn is such that
Sn(DSn)={0} and Snker(DSn)⊆DSn. Also assume that (S1∗,…,Sn∗) has Γn-isometric dilation. Then the (n−1)-tuple of fundamental operators (E1,…,En−1) satisfies the following conditions:
(1)
[Ei,Ej]=0,* where [P,Q]=PQ−QP;*
2. (2)
[Ei,En−j∗]=[Ej,En−i∗]* for 1≤i,j≤(n−1).*
Proof.
Let (T1,…,Tn−1,V) on a Hilbert space K containing H as a subspace, be a Γn-isometric dilation of (S1∗,…,Sn∗). From Proposition 6.2, it follows that (T1,…,Tn−1,V) is a minimal Γn-isometric dilation. By Proposition 6.3, we conclude that
(T1∗,…,Tn−1∗,V∗) is a Γn-co-isometric extension of (S1,…,Sn). Since (T1,…,Tn−1,V) is a Γn-isometriy on K, by Theorem 5.2, we have
[TABLE]
where (T1(1),…,Tn−1(1),U(1)) on K1 is a Γn-unitary and (T1(2),…,Tn−1(2),V(2)) on K2 is a pure Γn-isometry. Since (T1(2),…,Tn−1(2),V(2)) on K2 is a pure Γn-isometry, by Proposition 6.4, K2 can be identified with H2(E)≅H2(D)⊗E and T1(2),…,Tn−1(2),V(2) can be identified with the multiplication operators Mφ1,…,Mφn−1,Mz on H2(E) for some φi∈H∞(B(E)), where φi(z)=Ai+An−i∗z for i=1,…,(n−1),z∈D,E=DV(2)∗ and (A1∗,…,An−1∗) is the (n−1)-tuples of fundamental operators of (T1(2)∗,…,Tn−1(2)∗,V(2)∗). Since H2(E)≅H2(D)⊗E, therefore we have
[TABLE]
We first prove DSn⊆E⊕{0}⊕…⊆H2(E)=K2. To prove this, let h=h1⊕h2∈DSn⊆H, where h1∈K1 and (c0,c1,…)T∈H2(E). Then, together with Sn(DSn)={0} and U(1) is unitary, we get
[TABLE]
which, implies that h1=0 and c1=c2=…=0.
Now we prove that ker(DSn)⊆{0}⊕E⊕{0}⊕…⊆H2(E).
To prove this, take k=k1⊕k2∈ker(DSn)⊆H, where k1∈K1 and k2=(g0,g1,…)T∈H2(E). Notice that
[TABLE]
If DSn2k=0, then, from (6), we have k1⊕(g0,g1,…)T=PH(k1⊕(0,g1,…)T from which it follows that g0=0.
Since Snker(DSn)⊆DSn, we get for every k=k1⊕(0,g1,…)T∈ker(DSn),
[TABLE]
Also by Sn(DSn)={0}, from above we conclude that k1=0 and g2=g3=…=0. Hence, ker(DSn)⊆{0}⊕E⊕{0}⊕…⊆H2(E).
Since H=DSn⊕ker(DSn), we have H⊆E⊕E⊕{0}⊕…⊆H2(E). Therefore, (Mφ1∗,…,Mφn−1∗,Mz∗) on H2(E) is a Γn-co-isometric extension of (S1,…,Sn). We now compute the fundamental operator pairs of (Mφ1∗,…,Mφn−1∗,Mz∗). Note that
[TABLE]
Similarly one can also prove that
[TABLE]
It is also easy to verify that
[TABLE]
Thus, we have DMz∗=E⊕{0}⊕… and DMz∗2=DMz∗=Id on E⊕{0}⊕…. If we consider
[TABLE]
then
we have
[TABLE]
Therefore, from (6.3), we conclude that E^1,…,E^n−1 are the (n−1)-tuples of fundamental operators of (Mφ1∗,…,Mz∗).
Our next claims are E^iDMz∗∣DSn⊆DSn and E^i∗DMz∗∣DSn⊆DSn for i=i,…,(n−1). To prove this, let h=(c0,0,…)T∈DSn. Then we get E^iMz∗h=(Ai∗c0,0,…)T=Mφi∗h, which, together with Mφi∗h∈H and Mφi∗∣H=Si, implies that (Ai∗c0,0,…)T∈DSn. Hence, E^iDMz∗∣DSn⊆DSn for i=1,…,(n−1).
We now compute the adjoint of Sn. Let (c0,c1,0,…)T and (d0,d1,0,…)T be two arbitrary elements in H, where (c0,0,0,…)T,(d0,0,0,…)T∈DSn and (0,c1,0,…)T,(0,d1,0,…)T∈ker(DSn). Note that
[TABLE]
This shows that Sn∗(c0,c1,0,…)T=(0,c0,c1,0,…)T. Therefore, for h0=(c0,0,0,…)T∈DSn, we have Sn∗h0=(0,c0,0,…)T∈H and E^i∗DMz∗h0=(Aic0,0,…)T=Mφn−i∗(0,c0,0,…)T for i=1,…,(n−1). As (Aic0,0,…)T∈DSn, we deduce that E^i∗DMz∗∣DSn⊆DSn for i=i,…,(n−1). Our next aim is to show
E^i∣DSn=Ei and E^i∗∣DSn=Ei∗ for i=i,…,(n−1).
Since Mz∗∣H=Sn,DSn⊆Mz∗=E⊕{0}⊕… and DMz∗ is the projection onto DMz∗, we see that DMz∗∣H=DMz∗2∣DSn=DSn2. Therefore, from (6.3), we get
[TABLE]
Since (Mφ1∗,…,Mz∗) on H2(E) is a Γn-co-isometric extension of (S1,…,Sn), from (6.4), we have
[TABLE]
As (E1,…,En−1) are (n−1)-tuple of fundamental operators of (S1,…,Sn−1,Sn), we get
[TABLE]
Since Si−Sn−i∗Sn is zero on the orthogonal complement of DSn,DMz∗∣DSn=DSn and E^iDMz∗∣DSn⊆DSn, from (6.5), we see that
[TABLE]
By uniqueness of Ei, we have E^i∣DSn=Ei for i=1,…,(n−1). Similarly, one can prove also E^i∗∣DSn=Ei∗ for i=i,…,(n−1).
Since Mφi∗’s are commute, we have
[TABLE]
The above condition is equivalent to for i,j=1,…,(n−1),
[TABLE]
Equivalently, for i,j=1,…,n−1,
[TABLE]
Comparing both sides we get
[An−i∗,An−j∗]=0and[Ai,An−j∗]=[Aj,An−i∗]fori,j=1,…,(n−1).
Also, from (6.2) we obtain
[E^n−i,E^n−j]=0and[E^i,E^n−j∗]=[E^j,E^n−i∗]fori,j=1,…,(n−1). Taking restriction to the subspace DSn, we have
[En−i,En−j]=0and[Ei,En−j∗]=[Ej,En−i∗]fori,j=1,…,(n−1).
This completes the proof.
∎
7. Conditional Dilation
In this section, we want to show that if the commuting (n−1)-tuple of fundamental operators E1,…,En−1 and F1,…,Fn−1 of a Γn-contractions of (S1,…,Sn) and (S1∗,…,Sn∗) respectively, satisfy the conditions
ElEn−k∗−EkEn−l∗=En−k∗El−En−l∗Ek and Fl∗Fn−k−Fk∗Fn−l=Fn−kFl∗−Fn−lFk∗, then (S1,…,Sn) possesses a Γn-unitary dilation. The two conditions of (n−1)-tuple of fundamental operators of (S1,…,Sn) and of (S1∗,…,Sn∗), mentioned above, are the sufficient to have a such Γn-unitary dilation.
Throughout this section, we use the definitions of (n−1)-tuple of fundamental operators as
[TABLE]
The following proposition whose proof we skip because it is routine, gives a relation between Sn,DSn,DSn∗.
Proposition 7.1**.**
[11, Chapter- I]**
Suppose Sn is any contraction on a Hilbert space H. Then SnDSn=DSn∗Sn.
We begin with the following lemma which will be used later to prove the conditional dilation of Γn.
Lemma 7.2**.**
Let (S1,…,Sn) be a Γn-contraction defined on a Hilbert space H. Let (E1,…,En−1) and (F1,…,Fn−1) be the (n−1)-tuple of fundamental operators of (S1,…,Sn) and (S1∗,…,Sn∗) respectively. Then the following properties hold:
(1)
SnEi=Fi∗Sn∣DSn* and Sn∗Fi=Ei∗Sn∗∣DSn∗ for i=1,…,(n−1),*
2. (2)
DSnSi=EiDSn+En−i∗DSnSn* and DSnSn−i=En−iDSn+Ei∗DSnSn for i=1,…,(n−1),*
3. (3)
SiDSn∗=DSn∗Fi∗+SnDSn∗Fn−i* and Sn−iDSn∗=DSn∗Fn−i∗+SnDSn∗Fi for i=1,…,(n−1),*
4. (4)
Sn−j∗Si−Sn−i∗Sj=DSn(En−j∗Ei−En−i∗Ej)DSn,* when [Ei,Ej]=0, for i,j=1,…,(n−1),*
5. (5)
SiSn−j∗−SjSn−i∗=DSn∗(Fi∗Fn−j−Fj∗Fn−i)DSn∗,* when [Fi,Fj]=0, for i,j=1,…,(n−1),*
6. (6)
ω(En−i+Ei∗z)≤(in−1)+(n−in−1)* and ω(Fn−i∗+Eiz)≤(in−1)+(n−in−1) for i=1,…,(n−1).*
Proof.
(1)
For DSnh∈DSn and DSn∗h′∈DSn∗, we have
[TABLE]
which implies first part of (1).
Similarly, it is also easy to verify the other equality.
2. (2)
For h,h′∈H, we see that
[TABLE]
which gives the first part of (1). The other part is also easy to verify.
3. (3)
The proof (3) is similar to (2).
4. (4)
Note that
[TABLE]
5. (5)
The proof of (5) is same as that of (4).
6. (6)
By Theorem 4.4, we have for all z∈T,ω(Ei+En−iz)≤(in−1)+(n−in−1) for i=1,…,n−1. The condition ω(Ei+En−iz)≤(in−1)+(n−in−1), implies that ω(Ei∗+En−iz)≤(in−1)+(n−in−1) for all z∈T and i=1,…,n−1. Similarly, one can also show that ω(Fn−i∗+Eiz)≤(in−1)+(n−in−1) for i=1,…,(n−1).
∎
The following theorem allows us to construct the conditional dilation of Γn of a Γn-contraction.
Theorem 7.3**.**
Suppose (S1,…,Sn) is a Γn-contraction defined on a Hilbert space H such that the commuting (n−1)-tuple of fundamental operators E1,…,En−1 and F1,…,Fn−1 of (S1,…,Sn) and (S1∗,…,Sn∗) respectively satisfy the conditions
ElEn−k∗−EkEn−l∗=En−k∗El−En−l∗Ek and Fl∗Fn−k−Fk∗Fn−l=Fn−kFl∗−Fn−lFk∗. Let K=…⊕DSn⊕DSn⊕DSn⊕H⊕DSn∗⊕DSn∗⊕DSn∗⊕… and let (R1,…,Rn−1,U) be a n-tuple of operators defined on K by
[TABLE]
and
[TABLE]
Also we assume that (R1,…,Rn−1,U) is a Γn-contraction. Then (R1,…,Rn−1,U) is a minimal Γn-unitary dilation of (S1,…,Sn).
Proof.
It is immediate from Sz.-Nazy-Foias dilation [11, Chapter- I] that U is the minimal unitary dilation of Sn. The minimality of Γn-unitary dilation follows from the fact that K and U are the minimal dilation space and minimal unitary dilation of Sn respectively. Since U is unitary, in order to show that (R1,…,Rn−1,U) is a a minimal Γn-unitary dilation of (S1,…,Sn) one has to verify the following steps:
(1)
RiRj=RjRi for i,j=1,…,(n−1),
2. (2)
RiU=URi for i=1,…,(n−1),
3. (3)
Ri=Rn−i∗U for i=1,…,(n−1),
4. (4)
RiRi∗=Ri∗Ri for i=1,…,(n−1).
Step 1:
[TABLE]
[TABLE]
[TABLE]
First we verify the equality of the entities (−1,0),(0,1),(−1,1) in the matrices of RiRj and RjRi. To show this, we need to check the following operator identities.
By above given conditions, all other entries of RiRj and RjRi are equal. Hence, RiRj=RjRi.
Step 2: We now show that RiU=URi. Notice that
[TABLE]
[TABLE]
[TABLE]
Using Lemma 7.2, one can verify that the entities of the positions (−1,2),(−1,0) and (0,1) of RiU and URi are equal. To complete the above equality we need to verify
[TABLE]
Since the left hand side and right hand side of (7.4) are defined on DSn∗ to DSn, we have
[TABLE]
which gives RiU=URi.
Step 3: We now show that Ri=Rn−i∗U. Note that
[TABLE]
To prove Ri=Rn−i∗U, we need to verify the following operator identities:
(a)
−DSnEiSn∗+Sn−i∗DSn∗=DSn∗Fn−i,
2. (b)
SnEiSn∗+Fi∗DSn∗2=Fi∗.
(a) As the LHS and RHS are defined on DSn∗, we need only to show that
The other equality follow from Lemma 7.2 and (7.1). Hence, we have Ri=Rn−i∗U.
Step 4: First we will prove that Ri is normal. Clearly, Rn−i=Ri∗U. Since RiU=URi, using Fuglede’s theorem [35], we conclude that Ri∗U=URi∗. Therefore, we have
[TABLE]
Since Ri’s are normal for i=1,…,(n−1) and (R1,…,Rn−1,U) is a Γn-contraction with U is a unitary, by Theorem 5.2, we conclude that (R1,…,U) is a Γn-unitary. This completes the proof.
∎
Corollary 7.4**.**
Let N⊆K be defined as N=H⊕l2(DSn) and (R1,…,Rn−1,U) be a Γn-contraction. Then N is a common invariant subspace of R1,…,Rn−1,U and (T1,…,Tn−1,V)=(R1∣N,…,Rn−1∣N,U∣N) is a minimal Γn-isometric dilation of (S1,…,Sn).
Proof.
Clearly, N is a common invariant subspace of R1,…,Rn−1,U. Therefore, by definition of Γn -isometry, the restriction of (R1,…,Rn−1,U) to the common invariant subspace N, that is, (T1,…,Tn−1,V) is a Γn-isometry. The matrices of T1,…,Tn−1,V with respect to the decomposition H⊕DSn⊕… of N are as follows:
[TABLE]
One can easily check that the adjoint of (T1,…,Tn−1,V) is a Γn-co-isometry extension of (S1∗,…,Sn∗).
Hence, by Proposition 6.3, we conclude that (T1,…,Tn−1,V) is a Γn-isometric dilation of (S1,…,Sn). Since N and V are the minimal isometric space and minimal isometric dilation of Sn, the minimality of this Γn-isometric dilation follows from this fact. This completes the proof.
∎
Remark 7.5**.**
From Theorem 7.3 and Corollary 7.4, we conclude that (R1,…,Rn−1,U) is a minimal Γn-unitary extension of (T1,…,Tn−1,V).
As an easy consequence the following theorem gives a sufficient condition for an n-tuple of operators (S1,…,Sn) to become a complete spectral set for Γn.
Theorem 7.6**.**
Let (S1,…,Sn) be a commuting n-tuple of operators on a Hilbert space H having Γn is a spectral set. Let E1,…,En−1 be a commuting (n−1)-tuple of bounded operators on DSn satisfying the following conditions:
(1)
Si−Sn−i∗Sn=DSnEiDSn,Sn−i−Si∗Sn=DSnEn−iDSn,* for i=1,…,n−1;*
2. (2)
[Ei∗,Ej∗]=0* for i,j=1,…,n−1;*
3. (3)
[Ei,En−j∗]=[Ej,En−i∗]* for i,j=1,…,n−1.*
Then Γn is a complete spectral set for (S1,…,Sn).
8. A functional model for a class of Γn-contractions
In this section, we want to construct a concrete and explicit functional model for a
class of Γn-contractions (S1,…,Sn) for
which the adjoint (S1∗,…,Sn∗) admits an (n−1)-tuple of
fundamental operators (F1,…,Fn−1) satisfying the
following conditions:
(1)
[Fn−i,Fn−j]=0 for i,j=1,…,n−1;
2. (2)
[Fi,Fn−j∗]=[Fj,Fn−i∗] for i,j=1,…,n−1.
The following proposition plays an important role for proving the
model theorem.
Proposition 8.1**.**
[29, Proposition 10.1]**
Suppose T is a contraction and V is the minimal isometric
dilation of T. Then T∗ and V∗ have defect spaces of same
dimension.
Theorem 8.2**.**
Suppose (S1,…,Sn) is a Γn-contraction on a
Hilbert space H such that the adjoint
(S1∗,…,Sn∗) has commuting (n−1)-tuples of
fundamental operators (F1,…,Fn−1) with
[Fi,Fn−j∗]=[Fj,Fn−i∗] for i,j=1,…,n−1. Let
(T^1,…,T^n−1,V^) on N∗=H⊕DSn∗⊕DSn∗⊕… be defined as
[TABLE]
Then
(1)
(T^1,…,T^n−1,V^)* is a Γn-co-isometry, H is a
common invariant subspace of
T^1,…,T^n−1,V^ and
T^1∣H=S1,…,T^n−1∣H=Sn−1,V^∣H=Sn;*
2. (2)
there is a orthogonal decomposition N∗=N1⊕N2 into reducing subspace of
T^1,…,T^n−1,V^ such that
T^1∣N1,…,T^n−1∣N1,V^∣N1 is a Γn unitary and
T^1∣N2,…,T^n−1∣N2,V^∣N2 is a pure Γn-co-isometry;
3. (3)
N2* can be identified with H2(DV^), where DV^ has the same dimension
as of DSn. The operators T^1∣N2,…,T^n−1∣N2,V^∣N2 are unitarily equivalent to
TA1+An−1∗zˉ,…,TAn−1+A1∗zˉ,Tzˉ
on the vectorial Hardy space H2(DV^), where
A1,…,An−1 are the (n−1)-tuples of fundamental operators
of (T^1,…,T^n−1,V^).*
Proof.
By Corollary 7.4, we conclude that
(T^1∗,…,T^n−1∗,V^∗) is minimal
Γn-isometric dilation of of (S1∗,…,Sn∗),
where V^∗ is the minimal isometric dilation of Sn∗. By
Proposition 6.3, it follows immediately that
(T^1,…,T^n−1,V^) is the
Γn-co-isometric extension of (S1,…,Sn).
Therefore, we conclude that H is a common invariant
subspace of T^1,…,T^n−1,V^ and
T^1∣H=S1,…,T^n−1∣H=Sn−1,V^∣H=Sn. Since
(T^1∗,…,T^n−1∗,V^∗) is a Γn-isometry, by Wold decomposition theorem, there is a orthogonal
decomposition N∗=N1⊕N2 into
reducing subspace of T^1,…,T^n−1,V^ such
that T^1∣N1,…,T^n−1∣N1,V^∣N1 is a Γn-unitary and T^1∣N2,…,T^n−1∣N2,V^∣N2 is a pure Γn-co-isometry. Therefore, we can write
(T^1,…,T^n−1,V^) with respect to the
orthogonal decomposition as
[TABLE]
for i=1,…,n−1. Since
DV^=DV2, one can easily show that
(T^1,…,T^n−1,V^) and
(T21,…,T2(n−1),V2) have the same (n−1)-tuples of
fundamental operators. Also, by applying Proposition 6.4, we have the
following:
(1)
N2 can be identified with H2(DV^)
2. (2)
The operators T^1∣N2,…,T^n−1∣N2,V^∣N2 are unitarily equivalent to
TA1+An−1∗zˉ,…,TAn−1+A1∗zˉ,Tzˉ
on the vectorial Hardy space H2(DV^), where
A1,…,An−1 are the (n−1)-tuples of fundamental operators
of (T^1,…,T^n−1,V^).
Since V^∗ is the minimal isometric dilation of Sn∗, using
Proposition 8.1 one can easily verify that DV^ has the same dimension as of DSn. This
completes the proof.
∎
Acknowledgements:
The author gratefully acknowledge the help that was received from Dr. Sourav Pal, Dr. Shibananda Biswas, Dr. Subrata Shyam Roy, Dr. Sayan Bagchi and Prof. Gadadhar Misra.
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