1 Introduction
One of the most fruitful directions of research in order to develop a suitable spectral theory for nonselfadjoint operators was opened by the theorem of Sz.-Nagy [Reference Sz.-Nagy56] on the existence of a unitary dilation for every contraction operator on a Hilbert space. The matrix construction for such dilations, proposed by Schäffer in [Reference Schäffer52], was the starting point to obtain their precise geometrical structure (cf. [Reference Halperin29, Reference Sz.-Nagy58, Reference Sz.-Nagy and Foiaş61]).
The problem of finding isometric or unitary dilations for families of commuting contractions was proposed by Sz.-Nagy and solved in the case when the family in discussion is double commuting (cf. [Reference Sz.-Nagy57, Reference Sz.-Nagy60]). Ando [Reference Ando2] proved that every pair of commuting contractions has isometric dilation. Unfortunately, Ando’s result cannot be extended for arbitrary families of more than two contractions, according to the example given by Parrot [Reference Parrott46].
Later developments show that the problem of finding a unitary dilation for a family
$T=(T_\omega )_{\omega \in \Omega }$
of commuting contractions on a Hilbert space
$\mathfrak {H}$
can be reduced, by the Naimark theorem [Reference Naimark45], to the possibility of extending the function
$$ \begin{align*} \mathbb{Z}_+^\Omega\ni n\mapsto T^n\in\mathcal{B}(\mathfrak{H}) \end{align*} $$
to a positive definite one on
$\mathbb {Z}^\Omega .$
It was the idea of Brehmer [Reference Brehmer9] to consider the regular extension
$$ \begin{align*} \mathbb{Z}^\Omega\ni n\mapsto(T^{n^-})^*{T}^{n^+}\in\mathcal{B}(\mathfrak{H}) \end{align*} $$
and to obtain the so-called regular unitary dilations. Their systematic study was completed and simplified later by Sz.-Nagy [Reference Sz.-Nagy59] and Halperin [Reference Halperin30, Reference Halperin31]. In the multi-variable case, even supposing that the existence is assured, minimal isometric or unitary dilations are, in general, not unique. However, the minimality condition ensures the uniqueness of a regular isometric or unitary dilation. Regular dilation results have been used to provide models for commuting multi-operators [Reference Curto and Vasilescu13, Reference Popovici47, Reference Popovici49, Reference Timotin63], in connection with intertwining liftings [Reference Gaşpar, Suciu and Crăciunescu26], von Neumann inequalities [Reference Barik and Das6], [Reference Gaşpar and Suciu25], operator moment problems [Reference Popovici and Sebestyén50], Markov processes [Reference Loubaton41], completely contractive representations of product systems of correspondences [Reference Solel55], or in the context of right LCK semigroups [Reference Li37].
The large class of applications involving dilation theory, for example, in operator interpolation problems, optimization, control, and systems theory (excellent references are given by the survey of Shalit [Reference Shalit, Bastos, Castro and Karlovich53] or by the books of Foiaş–Frazho [Reference Foiaş and Frazho22], Foiaş–Frazho–Gohberg–Kaashoek [Reference Foiaş, Frazho, Gohberg and Kaashoek23] and Rosenblum–Rovnyak [Reference Rosenblum and Rovnyak51]), but also in prediction theory [Reference Kakihara35] motivate our work.
It is natural to assume that such a theory on spaces with indefinite metric (in particular on Kreĭn or Pontryagin spaces) will provide at least a similar set of applications. We should mention in this context that operators on Kreĭn have been used recently, for example, in machine learning [Reference Liu, Huang, Chen, Suykens, Banerjee and Fukumizu39, Reference Liu, Shi, Huang, Yang and Suykens40] or frame theory [Reference Dong and Li17, Reference Jahan and Johnson34, Reference Li and Dong38].
The following section (Section 2) is devoted to some preliminary facts concerning Kreĭn spaces, their Kreĭn subspaces, and bounded operators on such objects. Basic facts on the theory of Kreĭn spaces and operators on them are given in [Reference Ando3, Reference Bognár8, Reference Iokhvidov, Kreĭn and Langer32, Reference Kreĭn36]; to see also the excellent monograph [Reference Gheondea27].
One variable dilation theory on Kreĭn spaces is the subject of Section 3. The indefinite case started with the theorem of Davis [Reference Davis14] proving that every bounded operator on a Hilbert space
$\mathfrak {H}$
has a unitary dilation on a Kreĭn space
$\mathfrak {K}$
containing
$\mathfrak {H}$
as a regular subspace. The result holds true even if we suppose that
$\mathfrak {H}$
is a (more general) Kreĭn space, as showed by Dijskma–Langer–de Snoo [Reference Dijskma, Langer and de Snoo16] using Carathéodory-type representations for holomorphic operator functions, or by Constantinescu–Gheondea [Reference Constantinescu and Gheondea10] following a Schäffer-type matrix construction. Geometric structure results are obtained for any minimal isometric or unitary dilation (Theorem 3.1). In such a generality, a minimal isometric dilation of a bounded operator T is unique (up to a unitary equivalence) if and only if T is either contractive or expansive [Reference Gheondea and Popescu28]. We prove that T has a unique minimal unitary dilation if and only if T and
$T^*$
have unique minimal isometric dilations.
The main result of Section 4, the existence of a minimal unitary extension for every commuting pair of bounded isometries on a Kreĭn space, is based on a matrix construction similar to the one given by the author in [Reference Popovici48, Theorem 3.3.1]. The Hilbert space case was obtained by Itô [Reference Itô33] (cf. also Brehmer [Reference Brehmer9] and Douglas [Reference Douglas18]). The problem of finding conditions for the uniqueness of a minimal unitary extension reduces to the unidimensional case by the observation in [Reference Berger, Coburn and Lebow7] (on Hilbert spaces), and extended here, that
$U_1U_2$
is a minimal unitary extension for
$V_1V_2$
if
$U=(U_1, U_2)$
is a minimal unitary extension for
$V=(V_1, V_2).$
Two-variable dilation theory on Kreĭn spaces has also been considered earlier. The first result in this generalized context has been obtained by Azizov, Barsukov, and Dijksma in [Reference Azizov, Barsukov and Dijksma4]. We should also remark that, in the Hilbert space case, Ando’s theorem [Reference Ando2] was proved to be equivalent with the commutant lifting theorem given by Sz.-Nagy and Foiaş [Reference Sz.-Nagy, Foiaș, Bercovici and Kérchy62]. In the indefinite case, several versions of this last mentioned theorem have been obtained by Alpay [Reference Alpay1], Baidiuk and Hassi [Reference Baidiuk and Hassi5], Constantinescu and Gheondea [Reference Constantinescu and Gheondea10, Reference Constantinescu and Gheondea11], Dritschel [Reference Dritschel19], Dritschel and Rovnyak [Reference Dritschel and Rovnyak21], or Dijksma, Dritschel, Marcantognini, and de Snoo [Reference Dijskma, Dritschel, Marcantognini and de Snoo15]. Some of these proofs could lead to different Ando-type dilations. An excellent survey on this topic is presented in the paper [Reference Dritschel and Alpay20] of Dritschel. There are attempts to a several (more than two) variable dilation theory on Kreĭn spaces (see, e.g., [Reference Marcantognini42]). The theory of unitary extensions for pairs of Kreĭn space isometries has been initiated in the paper of Marcantognini and Moran [Reference Marcantognini and Moran43].
The last section contains structure results for the minimal (regular) isometric dilation provided that such a dilation exists. The geometric structure given by Theorem 5.3 is the indefinite correspondent of some Hilbert space results appeared in [Reference Słociński54] (for double commuting contractions) or, more generally, in [Reference Gaşpar and Suciu24] (for commuting contractive pairs having regular dilation). The most important result of the article is the existence of a minimal regular isometric dilation for every commuting pair
$T=(T_1, T_2)$
of bounded operators on a Kreĭn space such that
$T_1,\ T_2$
are contractions and T is a bidisc contraction or
$T_1,\ T_2$
are expansions and T is a bidisc expansion (Theorem 5.6). If the conditions above are satisfied, a minimal regular isometric dilation is unique up to a unitary equivalence (Theorem 5.8). The unitary extension (obtained in Section 4) for a regular isometric dilation provides a regular unitary dilation (Corollary 5.7).
We remark that similar results also hold true for finite families of more than two commuting operators. These topics will be treated elsewhere.
2 Preliminaries on Kreĭn spaces
2.1 Kreĭn spaces, regular subspaces, and operators
A Kreĭn space is a complex linear space
$\mathfrak {K}$
equipped with a Hermitian sesquilinear form
$\langle \cdot , \cdot \rangle _{\mathfrak {K}}$
and having a decomposition
$$ \begin{align} \mathfrak{K}=\mathfrak{K}_+\oplus\mathfrak{K}_-\ , \end{align} $$
where (
$\mathfrak {K}_\pm , \pm \langle \cdot , \cdot \rangle _{\mathfrak {K}}$
) are Hilbert spaces (“
$\oplus $
” denotes an orthogonal direct sum). Decomposition (2.1) is said to be a fundamental decomposition of the Kreĭn space
$\mathfrak {K}$
and, in general, it is not unique. It induces on
$\mathfrak {K}$
a Hilbert space structure: if
$P_\pm $
are the orthogonal projections onto
$\mathfrak {K}_\pm $
and
$J=P_+-P_-$
(called a fundamental symmetry or signature operator) then
$\mathfrak {K}$
becomes a Hilbert space (denoted
$\mathfrak {K}_J$
) when equipped with the inner product
$$ \begin{align*} \mathfrak{K}_J\times\mathfrak{K}_J\ni(x, y)\mapsto[x, y]_J:=\langle Jx, y\rangle_{\mathfrak{K}}\in\mathbb{C}. \end{align*} $$
The strong topology of this Hilbert space is independent of the choice of a fundamental decomposition and is usually called the Mackey topology of
$\mathfrak {K}$
. All topological notions on a Kreĭn space are to be understood with respect to this strong topology.
The cardinal numbers
$$ \begin{align*} \kappa^\pm(\mathfrak{K})=\dim_{alg}(\mathfrak{K}_\pm) \end{align*} $$
are the positive (respectively, negative) indices of
$\mathfrak {K}$
and are also independent of the choice of a fundamental decomposition. The rank of indefiniteness of
$\mathfrak {K}$
is
$\kappa (\mathfrak {K})=\min \kappa ^\pm (\mathfrak {K}).$
A subspace
$\mathfrak {H}$
of a Kreĭn space
$\mathfrak {K}$
is a closed linear manifold of
$\mathfrak {K}.$
It is positive (respectively, negative) if
$$ \begin{align*} \langle h, h\rangle_{\mathfrak{K}}\ge 0\qquad (\text{respectively, } \langle h, h\rangle_{\mathfrak{K}}\le 0), \end{align*} $$
for every
$h\in \mathfrak {K}.$
A positive (respectively, negative) subspace is said to be maximal positive (respectively, maximal negative) if it is not contained in a larger positive (respectively, negative) subspace. It is called uniformly positive (respectively, uniformly negative) if, for a certain
$\delta _J>0,$
$$ \begin{align*} \langle h, h\rangle_{\mathfrak{K}}\ge\delta_J\lVert h\rVert_J^2\quad (\text{respectively, } \langle h, h\rangle_{\mathfrak{K}}\le -\delta_J\lVert h\rVert_J^2),\quad h\in\mathfrak{H}. \end{align*} $$
Similarly, one defines maximal uniformly positive (respectively, maximal uniformly negative) subspaces.
The orthogonal subspace of
$\mathfrak {H}$
is
$\mathfrak {H}^\bot =\{ k\in \mathfrak {K} \mid \langle h, k\rangle =0,\ h\in \mathfrak {H}\}.$
For each pair
$(\mathfrak {M}, \mathfrak N)$
of subspaces in
$\mathfrak {K},$
we use the notation
$\mathfrak {M}\bot \mathfrak N$
if
$\mathfrak {M}\subset \mathfrak N^\bot $
, and
$\mathfrak {M}\oplus \mathfrak N$
if the sum
$\mathfrak {M}+\mathfrak N$
is closed, orthogonal, and direct.
$\mathfrak {H}$
is said to be regular (or ortho-complemented) if
$\mathfrak {K}=\mathfrak {H}\oplus \mathfrak {H}^\bot .$
Proposition 2.1 The following conditions are equivalent:
(i)
$\mathfrak {H}$
is regular.
(ii)
$\mathfrak {H}$
is a Kreĭn space in the inner product inherited from
$\mathfrak {K}$
.
(iii) There exists a fundamental symmetry J on
$\mathfrak {K}$
such that
$J\mathfrak {H}\subset \mathfrak {H}$
(hence
$J\mathfrak {H}=\mathfrak {H}$
).
Condition (ii) justifies the use of the term Kreĭn subspace for any regular subspace.
If
$T:\mathfrak {D}(T)\subset \mathfrak {K}_1\to \mathfrak {K}_2$
is a densely defined linear operator between Kreĭn spaces
$\mathfrak {K}_1$
and
$\mathfrak {K}_2$
, then its Kreĭn adjoint
$T^*:\mathfrak {D}(T^*)\subset \mathfrak {K}_2\to \mathfrak {K}_1$
is uniquely determined by the relation
$$ \begin{align*} \langle x, T^*y\rangle_{\mathfrak{K}_1}=\langle Tx, y\rangle_{\mathfrak{K}_2},\quad x\in\mathfrak{D}(T),\ y\in\mathfrak{D}(T^*). \end{align*} $$
The Kreĭn adjoint
$T^*$
and the Hilbert adjoint
$T^\times $
computed relative to fundamental symmetries
$J_1$
(on
$\mathfrak {K}_1$
) and
$J_2$
(on
$\mathfrak {K}_2$
) are related by
$T^*=J_1T^\times J_2.$
If T belongs to
$\mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
(the set of all bounded linear operators between
$\mathfrak {K}_1$
and
$\mathfrak {K}_2$
), then
${T^*\in \mathcal B(\mathfrak {K}_2, \mathfrak {K}_1).}$
Note that any fundamental symmetry J on a Kreĭn space
$\mathfrak {K}$
belongs to
$\mathcal B(\mathfrak {K})$
and
$J^*=J^\times =J^{-1}=J.$
A linear operator
$V:\mathfrak {D}(V)\subset \mathfrak {K}_1\to \mathfrak {K}_2$
is isometric if
$\langle Vx, Vy\rangle _{\mathfrak {K}_2}=\langle x, y\rangle _{\mathfrak {K}_1}, x, y\in \mathfrak {K}_1.$
An isometric operator U between Kreĭn spaces
$\mathfrak {K}_1$
and
$\mathfrak {K}_2$
is said to be unitary if
$\mathfrak {D}(U)=\mathfrak {K}_1$
and
$\mathfrak R(U)=\mathfrak {K}_2.$
An everywhere defined isometry is bounded if and only if its range is closed. Then, the range is a regular subspace. We deduce that any unitary operator is continuous. However, a densely defined Kreĭn space isometry may fail to have a continuous extension.
Two subspaces
$\mathfrak {M}_1$
(of
$\mathfrak {K}_1$
) and
$\mathfrak {M}_2$
(of
$\mathfrak {K}_2$
) are said to be isometrically isomorphic if there exists a boundedly invertible isometric operator
$U:\mathfrak {M}_1\to \mathfrak {M}_2.$
In this situation,
$\mathfrak {M}_1$
is regular if and only if
$\mathfrak {M}_2$
is regular. Note that two regular subspaces are isometrically isomorphic if and only if they have the same positive (respectively, negative) indices. Let us finally remark that definite Kreĭn spaces (their rank of indefiniteness is null)
$\mathfrak {K}$
can be characterized by the fact that every unitary operator U on
$\mathfrak {K}$
is power bounded (
$\sup _n\lVert U^n\rVert <\infty $
).
2.2 Hardy-type Kreĭn spaces and operator extensions
If
$\mathfrak {M}_1, \mathfrak {M}_2, \dots , \mathfrak {M}_n$
is a finite family of mutually orthogonal regular subspaces of a Kreĭn space
$\mathfrak {K}$
, then the subspace
$\mathfrak {M}_1\oplus \mathfrak {M}_2\oplus \dots \oplus \mathfrak {M}_n$
is clearly regular. The result remains no longer true if the family is infinite. For a finite or infinite family
$\{\mathfrak {K}_n\}_{n\ge 0}$
of Kreĭn spaces we can, however, compute their external orthogonal direct sum as the set
$\bigoplus _{n\ge 0}\mathfrak {K}_n$
of all sequences
$k=\{k_n\}_{n\ge 0}$
with
$\sum _{n\ge 0}\lVert k_n\rVert ^2_n<\infty $
(here, for each n, the norm
$\lVert \cdot \rVert _n$
is computed relative to a given fundamental decomposition of
$\mathfrak {K}_n$
). It becomes a Kreĭn space relative to the indefinite inner product
$\langle \{h_n\}_n, \{k_n\}_n\rangle :=\sum _{n\ge 0}\langle h_n, k_n\rangle _{\mathfrak {K}_n},\ \{h_n\}_n, \{k_n\}_n\in \bigoplus _{n\ge 0}\mathfrak {K}_n.$
If
$\mathfrak {K}$
is a Kreĭn space, then the external orthogonal direct sum of a family of identical copies of
$\mathfrak {K}$
can be obviously identified with the Hardy-type Kreĭn space
$H^2_{\mathfrak {K}}(\mathbb {T}),$
of functions on the torus
$$ \begin{align*} z\mapsto f(z)=\sum_{n\ge 0}z^nk_n, \ \text{ with } \sum_{n\ge 0}\lVert k_n\rVert^2<\infty. \end{align*} $$
If our family is doubly indexed, we obtain similarly
$H^2_{\mathfrak {K}}(\mathbb {T}^2).$
The following set of bounded operators will be frequently used in our constructions.
Let
$\mathfrak {K}, \mathfrak {K}_1, \mathfrak {K}_2$
be given Kreĭn spaces:
-
• the multiplication by the independent variable z on the Hardy-type space
$H^2_{\mathfrak {K}}(\mathbb {T}):$
has the adjoint
$$ \begin{align*} (T_zf)(z):=zf(z),\quad z\in\mathbb{T},\ f\in H^2_{\mathfrak{K}}(\mathbb{T}), \end{align*} $$
$T^*_z$
given by
$$ \begin{align*} (T^*_zf)(z):=\bar z(f(z)-f(0)),\quad z\in\mathbb{T},\ f\in H^2_{\mathfrak{K}}(\mathbb{T}); \end{align*} $$
-
• the pair
$(T_{z_1}, T_{z_2})$
of multiplications by coordinate functions
$z_1$
and
$z_2$
on
$H^2_{\mathfrak {K}}(\mathbb {T}^2)$
– defined similarly; -
• any
$T\in \mathcal B(\mathfrak {K})$
can be extended to a bounded operator
$[T]$
on
$H^2_{\mathfrak {K}}(\mathbb {T})$
by
$$ \begin{align*} ([T]f)(z):=T(f(z)),\quad z\in\mathbb{T},\ f\in H^2_{\mathfrak{K}}(\mathbb{T}); \end{align*} $$
-
• any
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
can be extended to
$[T]_0\in \mathcal B(\mathfrak {K}_1, H^2_{\mathfrak {K}_2}(\mathbb {T}))$
by its adjoint is given by
$$ \begin{align*} ([T]_0k_1)(z):=z^0Tk_1,\quad z\in\mathbb{T},\ k_1\in\mathfrak{K}_1; \end{align*} $$
$$ \begin{align*} [T]^*_0f=T^*(f(0)),\quad f\in H^2_{\mathfrak{K}_2}(\mathbb{T}); \end{align*} $$
-
• finally, any
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
can be extended to a bounded operator
$[T]_i\in \mathcal B(H^2_{\mathfrak {K}_1}(\mathbb {T}), H^2_{\mathfrak {K}_2}(\mathbb {T}^2))$
by Moreover,
$$ \begin{align*} ([T]_if)(z_1, z_2):=T(f(z_i)),\quad z_i\in\mathbb{T},\ f\in H^2_{\mathfrak{K}_1}(\mathbb{T}),\ i=1, 2. \end{align*} $$
(here
$$ \begin{align*} ([T]^*_if)(z)=T^*(f(ze_i)),\quad z\in\mathbb{T},\ f\in H^2_{\mathfrak{K}_2}(\mathbb{T}^2),\ i=1, 2 \end{align*} $$
$e_1:=(1,0)$
and
$e_2:=(0,1)$
).
Their joint properties are mentioned in the following.
Proposition 2.2 Let
$\mathfrak {K}, \mathfrak {K}_1, \mathfrak {K}_2, \mathfrak {K}^{\prime }_1, \mathfrak {K}^{\prime }_2$
be given Kreĭn spaces. Then,
-
(i) the map
is a
$$ \begin{align*} \mathcal B(\mathfrak{K})\ni T\mapsto [T]\in \mathcal B(H^2_{\mathfrak{K}}(\mathbb{T})) \end{align*} $$
$*$
-algebra homomorphism;
-
(ii) the map
is linear; moreover, for
$$ \begin{align*} \mathcal B(\mathfrak{K}_1, \mathfrak{K}_2)\ni T\mapsto [T]_0\in \mathcal B(\mathfrak{K}_1, H^2_{\mathfrak{K}_2}(\mathbb{T})) \end{align*} $$
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
,
$[S]_0T=[ST]_0$
(when
$S\in \mathcal B(\mathfrak {K}_2, \mathfrak {K}^{\prime }_2)$
),
$[S]^*_0[T]_0=S^*T$
(when
$S\in \mathcal B(\mathfrak {K}^{\prime }_1, \mathfrak {K}_2)$
), and (when
$$ \begin{align*} [S]_0[T]^*_0f=ST^*(f(0))z^0,\ f\in H^2_{\mathfrak{K}_2}(\mathbb{T}) \end{align*} $$
$S\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}^{\prime }_2)$
);
-
(iii) the map
is linear; moreover, for
$$ \begin{align*} \mathcal B(\mathfrak{K}_1, \mathfrak{K}_2)\ni T\mapsto [T]_i\in \mathcal B(H^2_{\mathfrak{K}_1}(\mathbb{T}), H^2_{\mathfrak{K}_2}(\mathbb{T}^2)) \end{align*} $$
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
,
$[S]^*_i[T]_i=[S^*T]$
(when
$S\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
) and (when
$$ \begin{align*} ([S]_i[T]_i^*f)(z_1, z_2)=ST^*(f(z_ie_i)), \quad z_i\in\mathbb{T},\ f\in H^2_{\mathfrak{K}_2}(\mathbb{T}^2),\ i=1, 2 \end{align*} $$
$S\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}^{\prime }_2)$
);
-
(iv) let
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
. Then,-
–
$[T]T_z=T_z[T]$
and
$T_z[T]^*=[T]^*T_z$
(when
$\mathfrak {K}_1=\mathfrak {K}_2$
); -
–
$[T]^*_0T_z=0;$
-
–
$[T]_iT_z=T_{z_i}[T]_i$
,
$T_z[T]^*_i=[T]_i^*T_{z_i}$
and
$[T]_i^*T_{z_{3-i}}=0,\ i=1, 2;$
-
-
(v) let
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2)$
and
$S\in \mathcal B(\mathfrak {K}^{\prime }_1, \mathfrak {K}^{\prime }_2)$
. Then,-
–
$[S][T]_0=[ST]_0$
(when
$\mathfrak {K}^{\prime }_1=\mathfrak {K}^{\prime }_2=\mathfrak {K}_2$
); -
–
$[S]_i[T]=[ST]_i,\ i=1, 2$
(when
$\mathfrak {K}_1=\mathfrak {K}_2=\mathfrak {K}^{\prime }_1$
); -
–
$[S]_1^*[T]_2=[S]^*_2[T]_1=[S^*]_0[T^*]^*_0$
(when
$\mathfrak {K}^{\prime }_2=\mathfrak {K}_2$
); -
–
$[S]_1[T]_0=[S]_2[T]_0$
(when
$\mathfrak {K}^{\prime }_1=\mathfrak {K}_2$
); -
–
$[T]_i[S]_0k_1=z_1^0z_2^0TSk_1,\quad k_1\in \mathfrak {K}_1,\ i=1,2$
(when
$\mathfrak {K}_1=\mathfrak {K}^{\prime }_2$
).
-
3 Isometric and unitary dilations for bounded operators
It is well known that any bounded self-adjoint operator A on a Kreĭn space
$\mathfrak A$
can be factorized into the form
$$ \begin{align*} A=BB^*, \end{align*} $$
for a certain operator
$B\in \mathcal B(\mathfrak B, \mathfrak A)$
with zero kernel on a Kreĭn space
$\mathfrak B.$
By a defect operator for
$T\in \mathcal B(\mathfrak {K}_1, \mathfrak {K}_2),\ \mathfrak {K}_1$
and
$\mathfrak {K}_2$
Kreĭn spaces, we mean an operator
$D_T\in \mathcal B(\mathfrak {D}_T, \mathfrak {K}_1)$
with zero kernel on a Kreĭn space
$\mathfrak {D}_T$
(called its defect space) such that
$$ \begin{align} I-T^*T=D_TD_T^*. \end{align} $$
Let
$T\in \mathcal B(\mathfrak {H}),\ \mathfrak {H}$
Kreĭn space. An isometric (respectively, unitary) dilation of T is a bounded isometric (respectively, unitary) operator U on a Kreĭn space
$\mathfrak {K}\supset \mathfrak {H}$
satisfying
$$ \begin{align} \langle T^nh, h'\rangle_{\mathfrak{H}}=\langle U^nh, h'\rangle_{\mathfrak{K}},\quad h, h'\in\mathfrak{H},\ n\in\mathbb{Z}_+. \end{align} $$
An isometric (respectively, unitary) dilation
$U\in \mathcal B(\mathfrak {K})$
of
$T\in \mathcal B(\mathfrak {H})$
is said to be minimal if
$\mathfrak {K}=\bigvee _{n\ge 0}U^n\mathfrak {H}$
(respectively,
$\mathfrak {K}=\bigvee _{n=-\infty }^\infty U^n\mathfrak {H}$
).
For bounded operators
$T\in \mathcal B(\mathfrak {H})$
, minimal isometric (respectively, unitary) dilations always exist [Reference Constantinescu and Gheondea10, Reference Dijskma, Langer and de Snoo16]. A Schäffer-like matrix construction is still possible in these generalized settings [Reference Constantinescu and Gheondea10].
Define
$\mathfrak {K}_+=\mathfrak {H}\oplus H^2_{\mathfrak {D}_T}(\mathbb {T}).$
Then, an isometric dilation V of T on
$\mathfrak {K}_+$
is given by the representation
$$ \begin{align} V= \begin{pmatrix} T && 0 \\ [D^{*}_{T}]_{0} && T_z \end{pmatrix}. \end{align} $$
A minimal unitary dilation U of T on the Kreĭn space
$\mathfrak {K}=H^2_{\mathfrak {D}_{T^*}}(\mathbb {T})\oplus \mathfrak {H}\oplus H^2_{\mathfrak {D}_T}(\mathbb {T})$
can be built in terms of a Julia operator or elementary rotation (to see [Reference Constantinescu and Gheondea11, Reference Ćurgus, Dijksma, Langer and de Snoo12]) for T, i.e., a unitary operator of the form
$$ \begin{align*} \begin{pmatrix} T && D_{T^*}\\ D^*_T && L \end{pmatrix}\in\mathcal B(\mathfrak{H}\oplus\mathfrak{D}_{T^*}, \mathfrak{H}\oplus\mathfrak{D}_T). \end{align*} $$
More precisely,
$$ \begin{align} U= \begin{pmatrix} T_z^* &&& 0 &&& 0\\ [D^*_{T^*}]^*_0 &&& T &&& 0\\ [L]_0[I_{\mathfrak{D}_{T^*}}]^*_0 &&& [D^*_T]_0 &&& T_z \end{pmatrix}. \end{align} $$
As regarding the geometry of minimal dilations, we could mention the following theorem.
Theorem 3.1 Let
$V\in \mathcal B(\mathfrak {K}_+)$
(respectively,
$U\in \mathcal B(\mathfrak {K})$
) be any minimal isometric (respectively, unitary) dilation of
$T\in \mathcal B(\mathfrak {H}).$
-
(a)
-
(i)
$\mathfrak {L}=\overline {(V-T)\mathfrak {H}}$
is wandering for V (i.e.,
$V^n\mathfrak {L}\bot V^m\mathfrak {L},\ n,m\ge 0,\ n\ne m$
), regular, and isometrically isomorphic with
$\mathfrak {D}_T$
; -
(ii)
$M_+(\mathfrak {L})=\bigvee _{n\ge 0} V^n\mathfrak {L}$
is regular and (3.5)
$$ \begin{align} \mathfrak{K}_+=\mathfrak{H}\oplus M_+(\mathfrak{L}); \end{align} $$
-
-
(b)
-
(i)
$\mathfrak {L}=\overline {(U-T)\mathfrak {H}}$
and
$\mathfrak {L}^*=\overline {(U^*-T^*)\mathfrak {H}}$
are wandering for U, regular, and isometrically isomorphic, respectively, with
$\mathfrak {D}_T$
and
$\mathfrak {D}_{T^*}$
; -
(ii)
$M_+(\mathfrak {L})=\bigvee _{n\ge 0} U^n\mathfrak {L}$
and
$M_-(\mathfrak {L}^*)=\bigvee _{n\le 0}U^n\mathfrak {L}^*$
are regular and (3.6)
$$ \begin{align} \mathfrak{K}=M_-(\mathfrak{L}^*)\oplus\mathfrak{H}\oplus M_+(\mathfrak{L}). \end{align} $$
-
Proof All the orthogonality properties involved here can be easily checked following the definition of an isometric or unitary dilation and, therefore, we shall omit the details.
Observe firstly that a successive application of the formula
$$ \begin{align*} Vh=Th+(V-T)h,\quad h\in\mathfrak{H} \end{align*} $$
will lead to
$$ \begin{align} V^nh=T^nh+\sum_{k=0}^{n-1}V^k(V-T)T^{n-k-1}h,\quad h\in\mathfrak{H},\ n\in\mathbb{N}. \end{align} $$
Hence, (3.5) holds. Consequently,
$M_+(\mathfrak {L})$
is regular and, since
$M_+(\mathfrak {L})=\mathfrak {L}\oplus VM_+(\mathfrak {L}),\ \mathfrak {L}$
is also regular. Moreover,
$$ \begin{align*} \langle(V-T)h, (V-T)h'\rangle_{\mathfrak{K}_+} &=\langle h, h'\rangle_{\mathfrak{H}}-\langle Th, Th'\rangle_{\mathfrak{H}}\\&=\langle(I-T^*T)h, h'\rangle_{\mathfrak{H}}\\&=\langle D^*_Th, D^*_Th'\rangle_{\mathfrak{D}_T},\quad h, h'\in\mathfrak{H} \end{align*} $$
shows that the map
$$ \begin{align*} \mathfrak{L}\ni(V-T)h\mapsto D^*_Th\in\mathfrak{D}_T \end{align*} $$
is well defined (
$\mathfrak {D}_T$
is a Kreĭn space), isometric, densely defined, and with dense range. It is also injective (since
$\mathfrak {L}$
is regular) and, therefore,
$\kappa ^\pm (\mathfrak {L})=\kappa ^\pm (\mathfrak {D}_T).$
Deduce that
$\mathfrak {L}$
and
$\mathfrak {D}_T$
are isometrically isomorphic and the proof of
$(a)$
is complete.
To obtain (3.6), we apply (3.7) for
$(U, T)$
and then for
$(U^*, T^*)$
instead of
$(V, T).$
Since
$M_-(\mathfrak {L}^*)$
and
$M_+(\mathfrak {L})$
are regular we show, as before, that
$\mathfrak {L}$
and
$\mathfrak {L}^*$
are regular. A similar argument as for the minimal isometric dilation allows us to conclude that
$\mathfrak {L}$
and
$\mathfrak {L}^*$
are isometrically isomorphic, respectively, with
$\mathfrak {D}_T$
and
$\mathfrak {D}_{T^*}.$
Corollary 3.2 Let
$U\in \mathcal B(\mathfrak {K})$
be any minimal unitary dilation of
$T\in \mathcal B(\mathfrak {H}).$
Then,
$$ \begin{align*} \mathfrak{K}_+=\bigvee_{n\ge 0}U^n\mathfrak{H} \end{align*} $$
is a regular subspace of
$\mathfrak {K}$
, invariant to U, and
$V_+=U|_{\mathfrak {K}_+}$
is a minimal isometric dilation of
$T.$
Similarly,
$$ \begin{align*} \mathfrak{K}_-=\bigvee_{n\le 0} U^n\mathfrak{H} \end{align*} $$
is a regular subspace of
$\mathfrak {K}$
, invariant to
$U^*$
, and
$V_-=U^*|_{\mathfrak {K}_-}$
is a minimal isometric dilation of
$T^*.$
Remark 3.3
$\bullet $
Proposition 2.1 gives another condition, usually used as an axiom in the isometric or unitary dilations definition: if
$U\in \mathcal B(\mathfrak {K})$
is any minimal isometric or unitary dilation of
$T\in \mathcal B(\mathfrak {H}),$
then
$\mathfrak {H}$
is a regular subspace of
$\mathfrak {K}.$
Theorem above indicates the structure of its orthogonal complement. Therefore, (3.2) can be re-written as
$$ \begin{align*} T^n=P_{\mathfrak{H}} U^n|_{\mathfrak{H}},\quad n\ge 0, \end{align*} $$
where
$P_{\mathfrak {H}}$
is the orthogonal projection onto
$\mathfrak {H}.$
Another consequence of (3.2) is
$$ \begin{align*} T^{*n}=P_{\mathfrak{H}} U^{*n}|_{\mathfrak{H}},\quad n\ge 0. \end{align*} $$
$\bullet $
Let
$V\in \mathcal B(\mathfrak {K}_+)$
be a minimal isometric dilation of
$T\in \mathcal B(\mathfrak {H}).$
Then, for any
$h, h'\in \mathfrak {H}$
and
$n\ge 0,$
$$ \begin{align*} \langle V^*h-T^*h, h'\rangle=\langle h, Vh'\rangle-\langle h, Th'\rangle=0 \end{align*} $$
and
$$ \begin{align*} \langle V^*h-T^*h, V^n(V-T)h'\rangle &=\langle h, V^{n+2}h'\rangle-\langle h, V^{n+1}Th'\rangle\\&\quad -\langle T^*h, V^{n+1}h'\rangle+\langle T^*h, V^nTh'\rangle=0. \end{align*} $$
By the geometrical structure of
$\mathfrak {K}_+$
given by (3.5), we deduce that
$\mathfrak {H}$
is invariant to
$V^*$
and
$V^*|_{\mathfrak {H}}=T^*.$
$\bullet $
Let
$U\in \mathcal B(\mathfrak {K})$
be a minimal unitary dilation of
$T\in \mathcal B(\mathfrak {H})$
. It is clear that
$M_+(\mathfrak {L})$
is invariant to U and
$M_-(\mathfrak {L}^*)$
to
$U^*.$
Since
$M_+(\mathfrak {L})$
and
$M_-(\mathfrak {L})$
are regular,
$U|_{M_+(\mathfrak {L})}$
and
$U|_{M_-(\mathfrak {L}^*)}$
are unilateral shifts (according to the definition in [Reference McEnnis44]).
An operator
$T\in \mathcal B(\mathfrak {H}, \mathfrak {K}),\ \mathfrak {H}, \mathfrak {K}$
Kreĭn spaces, is a contraction (respectively, expansion) if
$$ \begin{align*} \langle Th, Th\rangle_{\mathfrak{K}}\le \langle h, h\rangle_{\mathfrak{H}}\quad (\text{respectively, }\langle Th, Th\rangle_{\mathfrak{K}}\ge \langle h, h\rangle_{\mathfrak{H}}),\quad h\in\mathfrak{H} \end{align*} $$
or, equivalently,
$$ \begin{align*} I-T^*T\ge 0\quad (\text{respectively, } I-T^*T\le 0). \end{align*} $$
Contractions (respectively, expansions) can be characterized by the fact that their defect spaces are Hilbert (respectively, anti-Hilbert) spaces.
Let
$U\in \mathcal B(\mathfrak {K}),\ U'\in \mathcal B(\mathfrak {K}')$
be two minimal unitary dilations of
$T\in \mathcal B(\mathfrak {H})$
and
$V=U|_{\mathfrak {K}_+}, V'=U'|_{\mathfrak {K}^{\prime }_+}$
, where
$\mathfrak {K}_+=\bigvee _{n\ge 0}U^n\mathfrak {H}$
,
$\mathfrak {K}^{\prime }_+=\bigvee _{n\ge 0}U^{\prime n}\mathfrak {H}$
the corresponding minimal isometric dilations (according to Corollary 3.2).
$V\in \mathcal B(\mathfrak {K}_+)$
and
$V'\in \mathcal B(\mathfrak {K}^{\prime }_+)$
are unitarily equivalent if there exists a unitary operator
$\Phi :\mathfrak {K}_+\to \mathfrak {K}^{\prime }_+$
which intertwines V and
$V'$
(i.e.,
$\Phi V=V'\Phi $
) and
$\Phi |_{\mathfrak {H}}=I_{\mathfrak {H}}$
. If, moreover,
$\Phi $
can be extended to a unitary operator on
$\mathfrak {K}$
onto
$\mathfrak {K}'$
which intertwines U and
$U'$
, then
$U\in \mathcal B(\mathfrak {K})$
and
$U'\in \mathcal B(\mathfrak {K}')$
are said to be unitarily equivalent.
A result by Gheondea and Popescu shows that minimal isometric dilations are, in general, not unique (up to a unitary equivalence).
Theorem 3.4 [Reference Gheondea and Popescu28]
A bounded operator T on a Kreĭn space
$\mathfrak {H}$
has a unique minimal isometric dilation if and only if T is either contractive or expansive.
The same kind of conditions holds for minimal unitary dilations.
Theorem 3.5
$T\in \mathcal B(\mathfrak {H})$
has a unique minimal unitary dilation if and only if T is either contractive or expansive and
$T^*$
is either contractive or expansive.
Proof Assume that, for example, T is contractive and
$T^*$
is expansive. Then,
$\mathfrak {D}_T$
(and hence also
$H^2_{\mathfrak {D}_T}(\mathbb {T})$
) is a Hilbert space and
$\mathfrak {D}_{T^*}$
(and hence also
$H^2_{\mathfrak {D}_{T^*}}(\mathbb {T})$
) is an anti-Hilbert space.
Let U be the minimal unitary dilation of T given by (3.4) and acting on the Kreĭn space
$\mathfrak {K}=H^2_{\mathfrak {D}_{T^*}}(\mathbb {T})\oplus \mathfrak {H}\oplus H^2_{\mathfrak {D}_{T}}(\mathbb {T}).$
If
$U'\in \mathcal B(\mathfrak {K}')$
is any other minimal unitary dilation of T then, according to (3.6),
$\mathfrak {K}'$
has an orthogonal decomposition of the form
$\mathfrak {K}'=M_-(\mathfrak {L}^{\prime *})\oplus \mathfrak {H}\oplus M_+(\mathfrak {L}')$
, with
$\mathfrak {L}'=\overline {(U'-T)\mathfrak {H}}$
and
$\mathfrak {L}^{\prime *}=\overline {(U^{\prime *}-T^*)\mathfrak {H}}.$
Since, for arbitrary finite sequences
$\{h_n\}_n,\ \{g_n\}_n$
of vectors in
$\mathfrak {H}$
, we have
$$ \begin{align*} \langle\sum_{n\ge 0}U^{\prime n}(U'-T)h_n, & \sum_{m\ge 0}U^{\prime m}(U'-T)g_m\rangle_{\mathfrak{K}'}\\ & \quad =\sum_{n\ge 0}\langle(U'-T)h_n, (U'-T)g_n\rangle_{\mathfrak{K}'}\\ & \quad =\sum_{n\ge 0}\langle(I-T^*T)h_n, g_n\rangle_{\mathfrak{H}}\\ & \quad =\langle\sum_{n\ge 0}z^nD^*_Th_n, \sum_{m\ge 0}z^mD^*_Tg_m\rangle_{H^2_{\mathfrak{D}_T}(\mathbb{T})},\ \qquad\quad\qquad \end{align*} $$
the mapping
$$ \begin{align*} M_+(\mathfrak{L}')\ni\sum_{n\ge 0} U^{\prime n}(U'-T)h_n\stackrel{\Phi_+}{\longmapsto}\sum_{n\ge 0}z^nD^*_Th_n\in H^2_{\mathfrak{D}_T}(\mathbb{T}) \end{align*} $$
is well defined, and the linear operator
$\Phi _+$
is isometric, densely defined, and with dense range. It can be uniquely extended to a unitary operator
$\Phi _+\in \mathcal B(M_+(\mathfrak {L}'), H^2_{\mathfrak {D}_T}(\mathbb {T}))$
(since
$H^2_{\mathfrak {D}_T}(\mathbb {T}))$
is a Hilbert space). Similarly, we can define a unitary operator
$\Phi _-\in \mathcal B(M_-(\mathfrak {L}^{\prime *}), H^2_{\mathfrak {D}_{T^*}}(\mathbb {T})).$
Then,
$\Phi =\Phi _-\oplus I_{\mathfrak {H}}\oplus \Phi _+\in \mathcal B(\mathfrak {K}', \mathfrak {K})$
is unitary,
$\Phi |_{\mathfrak {H}}=I_{\mathfrak {H}}$
and
$\Phi U'=U\Phi .$
Moreover,
$\Phi \mathfrak {K}^{\prime }_+=\mathfrak {K}_+$
and, therefore, U and
$U'$
are unitarily equivalent.
Conversely, suppose that, for example, T is neither contractive nor expansive or, equivalently, the Kreĭn space
$\mathfrak {D}_T$
is indefinite. If
$Z\in \mathcal B(\mathfrak {D}_T)$
is a unitary operator with
$\lVert Z^n\rVert \stackrel {n}{\longrightarrow }\infty $
, then the matrix
$$ \begin{align*} U'=\begin{pmatrix} T^*_z &&& 0 &&& 0\\ [D^*_{T^*}]^*_0 &&& T &&& 0\\ [L]_0[I_{\mathfrak{D}_{T^*}}]^*_0 &&& [D^*_T]_0 &&& T_z[Z] \end{pmatrix} \end{align*} $$
defines a minimal unitary dilation of T on
$\mathfrak {K}=H^2_{\mathfrak {D}_{T^*}}(\mathbb {T})\oplus \mathfrak {H}\oplus H^2_{\mathfrak {D}_T}(\mathbb {T}).$
If T would have a unique minimal unitary dilation, then
$U'$
and the minimal unitary dilation U on
$\mathfrak {K}$
given by (3.4) would be unitarily equivalent via a unitary operator
$\Phi \in \mathcal B(\mathfrak {K})$
which leaves invariant
$H^2_{\mathfrak {D}_T}(\mathbb {T}).$
We get a contradiction since
$\lim _{n\to \infty }\lVert (\Phi ^*|_{H^2_{\mathfrak {D}_T(\mathbb {T})}}T_z \Phi |_{H^2_{\mathfrak {D}_T(\mathbb {T})}})^n\rVert =\lim _{n\to \infty }\lVert \Phi ^*|_{H^2_{\mathfrak {D}_T(\mathbb {T})}}T^n_z \Phi |_{H^2_{\mathfrak {D}_T(\mathbb {T})}}\rVert <\infty ,$
while
$\lim _{n\to \infty }\lVert (T_z[Z])^n\rVert =\lim _{n\to \infty }\lVert Z^n\rVert =\infty .$
Corollary 3.6
T has a unique minimal unitary dilation if and only if T and
$T^*$
have unique minimal isometric dilations.
4 Commuting isometric pairs and their unitary extensions
Using dilation theory, we can easily deduce that every bounded isometry V on a Kreĭn space
$\mathfrak {H}$
has a unitary extension U on a Kreĭn space
$\mathfrak {K}\text{ containing } \mathfrak {H}$
as a regular subspace which is minimal in the sense that
$$ \begin{align*} \mathfrak{K}=\bigvee_{n\in\mathbb{Z}} U^n\mathfrak{H}. \end{align*} $$
We just have to take the minimal isometric dilation of
$V^*$
given by (3.3) and observe that it is a unitary operator
$U^*$
on
$\mathfrak {K}=\mathfrak {H}\oplus H^2_{\ker V^*}(\mathbb {T}).$
More precisely, the linear operator given by the matrix
$$ \begin{align} U=\begin{pmatrix} V &&& [P^{\mathfrak{H}}_{\ker{V^*}}]^*_0\\ 0 &&& T^*_z \end{pmatrix} \end{align} $$
is a minimal unitary extension of
$V.$
Remark 4.1
$\bullet $
$U\in \mathcal B(\mathfrak {K})$
is a minimal unitary extension of
$V\in \mathcal B(\mathfrak {H})$
if and only if
$U^*$
is a minimal isometric dilation of
$V^*.$
$\bullet $
If
$V\in \mathcal B(\mathfrak {K}_+)$
is any minimal isometric dilation of
$T\in \mathcal B(\mathfrak {H})$
and
$U\in \mathcal B(\mathfrak {K})$
is any minimal unitary extension of V, then U is a minimal unitary dilation of
$T.$
$\bullet $
Suppose that V is the minimal isometric dilation of T given by (3.3) on the Kreĭn space
$\mathfrak {K}_+=\mathfrak {H}\oplus H^2_{\mathfrak {D}_T}(\mathbb T).$
The minimal unitary extension U given by (4.1) on the Kreĭn space
$\mathfrak {K}=\mathfrak {H}\oplus H^2_{\mathfrak {D}_T}(\mathbb T)\oplus H^2_{\ker {V^*}}(\mathbb T)$
is a minimal unitary dilation of T and its matrix representation depends only on T and on its corresponding defect operators (does not require the construction of a Julia operator or of an elementary rotation for T).
The following theorem explains the geometrical structure of minimal unitary extensions.
Theorem 4.2 Let V be a bounded isometry on a Kreĭn space
$\mathfrak {H}$
and
$U\in \mathcal B(\mathfrak {K})$
be any minimal unitary extension of
$V.$
Then,
-
(i)
$\mathfrak {L}^*=\overline {(U^*-V^*)\mathfrak {H}}$
is wandering for U, regular, and isometrically isomorphic with
$\ker V^*;$
-
(ii)
$M_-(\mathfrak {L}^*)=\bigvee _{n\le 0}U^n\mathfrak {L}^*$
is a regular subspace of
$\mathfrak {K}$
and
$$ \begin{align*} \mathfrak{K}=\mathfrak{H}\oplus M_-(\mathfrak{L}^*); \end{align*} $$
-
(iii)
$M_-(\mathfrak {L}^*)$
is invariant to
$U^*$
and
$U^*|_{M_-(\mathfrak {L}^*)}$
is a unilateral shift.
Proof The proof follows Theorem 3.1
$(a)$
with the observation that
$U^*$
is a minimal isometric dilation of
$V^*.$
Following Remark 4.1 and Theorem 3.4, we can characterize the uniqueness of a minimal unitary extension.
Theorem 4.3 Let
$V\in \mathcal B(\mathfrak {H}),\ \mathfrak {H}$
Kreĭn space, be isometric. The following conditions are equivalent:
-
(i) V has a unique minimal unitary extension;
-
(ii)
$V^*$
is either contractive or expansive; -
(iii)
$\ker V^*$
is either uniformly positive or uniformly negative; -
(iv)
$\ker V^*$
is either Hilbert or anti-Hilbert space.
If
$U\in \mathcal B(\mathfrak {K})$
is a minimal unitary extension of V, then
$\mathfrak {K}\ominus \mathfrak {H}$
is a Hilbert or an anti-Hilbert space.
A pair
$T=(T_1,T_2)$
of bounded linear operators on a Kreĭn space
$\mathfrak {H}$
is said to be commuting if
$T_1T_2=T_2T_1$
. T is called double commuting if
$T_1$
commutes not only with
$T_2$
, but also with its adjoint
$T_2^*$
. By the end of this article, any pair of bounded operators acting on a Kreĭn space will be considered a commutative pair. In case the components
$T_1$
and
$T_2$
are clear from the context or they are not needed in the corresponding discussion, in order to avoid repetitions, we simply use the notation
$T\in \mathcal B(\mathfrak {H})^2$
instead of
$T=(T_1,T_2)\in \mathcal B(\mathfrak {H})^2$
. If
$n=(n_1, n_2)\in \mathbb {Z}^2$
and
$T=(T_1, T_2)\in \mathcal B(\mathfrak {H})^2$
, the notation
$T^n=T_1^{n_1}T_2^{n_2}$
will be frequently used whenever the computations
$T_1^{n_1}$
and
$T_2^{n_2}$
make sense.
Definition 4.1 Let
$V=(V_1, V_2)$
be a pair of commuting isometries in
$\mathcal B(\mathfrak {H}),\ \mathfrak {H}$
a Kreĭn space. A unitary extension of V is a commuting pair
$U=(U_1,U_2)$
of unitary operators on a Kreĭn space
$\mathfrak {K}\ \mathrm{containing}\ \mathfrak {H}$
as a regular subspace such that
$U_1, U_2$
extend, respectively,
$V_1, V_2$
. U is said to be minimal if, in addition,
$$ \begin{align*} \mathfrak{K}=\bigvee_{n\in\mathbb{Z}^2}U^n\mathfrak{H}. \end{align*} $$
Just to give an example, observe that the pair
$(T_{z_1}, T_{z_2})$
of multiplications by coordinate functions
$z_1$
and
$z_2$
on a certain Hardy-type Kreĭn space
$H^2_{\mathfrak {H}}(\mathbb T^2)$
can be extended by the commuting unitary pair
$(M_{z_1}, M_{z_2})$
,
$$ \begin{align*} f\stackrel{M_{z_i}}{\longmapsto}z_i f \end{align*} $$
on the
$L^2$
-type Kreĭn space
$L^2_{\mathfrak {H}}(\mathbb T^2)$
introduced in an obvious manner. This unitary extension is minimal.
Theorem 4.4 Let
$V=(V_1, V_2)$
be a commuting pair of isometric operators in
$\mathcal B(\mathfrak {H})$
. Then, the pair
$U=(U_1, U_2)$
given by the matrix representation
$$ \begin{align} U_1=\begin{pmatrix} V_1 && [P^{\mathfrak{H}}_{\ker(V_1V_2)^*}V_2|_{\ker(V_1V_2)^*}]^*_0\\ 0 && [V_1(I-V_2V_2^*)|_{\ker(V_1V_2)^*}]+[V^*_2|_{\ker(V_1V_2)^*}]T_z^* \end{pmatrix} \end{align} $$
and
$$ \begin{align} U_2=\begin{pmatrix} V_2 && [P^{\mathfrak{H}}_{\ker(V_1V_2)^*}V_1|_{\ker(V_1V_2)^*}]^*_0\\ 0 && [V_2(I-V_1V_1^*)|_{\ker(V_1V_2)^*}]+[V^*_1|_{\ker(V_1V_2)^*}]T_z^* \end{pmatrix} \end{align} $$
is a minimal unitary extension of V on the Kreĭn space
$$ \begin{align*} \mathfrak{K}=\mathfrak{H}\oplus H^2_{\ker(V_1V_2)^*}(\mathbb T). \end{align*} $$
Proof Standard computations with operator matrices (in view of Proposition 2.2) show that, in fact, formulas (4.2) and (4.3) define commuting unitary operators on
$\mathfrak {K}=\mathfrak {H}\oplus H^2_{\ker (V_1V_2)^*}(\mathbb T).$
Obviously, U is a unitary extension of
$V.$
It remains to show the minimality. We will actually prove an apparently stronger result, namely,
$$ \begin{align*} \mathfrak{K}=\bigvee_{n\le 0} (U_1U_2)^n\mathfrak{H}. \end{align*} $$
To this end, let
$h\in \mathfrak {H}$
and
$n\in \mathbb {N}$
be arbitrary. We see that
$$ \begin{align*} (U_1U_2)^{*n}\left[ (U_1U_2)^*\begin{pmatrix}h\\ 0\end{pmatrix} - \begin{pmatrix}(V_1V_2)^*h\\ 0\end{pmatrix}\right]=\begin{pmatrix}0\\ z^nP^{\mathfrak{H}}_{\ker(V_1V_2)^*}h\end{pmatrix}, \end{align*} $$
hence,
$$ \begin{align*} H^2_{\ker(V_1V_2)^*}(\mathbb T)=\bigvee\Bigl\{(U_1U_2)^{*(n+1)}\left(\begin{smallmatrix}h\\ 0\end{smallmatrix}\right) - (U_1U_2)^{*n} \left(\begin{smallmatrix}(V_1V_2)^{*}h\\ 0\end{smallmatrix}\right) \mid h\in\mathfrak{H}, n\ge 0\Bigr\}, \end{align*} $$
which proves our claim.
Remark 4.5 The conclusion of the previous theorem also holds for arbitrary finite families
$V=(V_1, V_2, \dots , V_n)$
of commuting bounded isometries on a Kreĭn space
$\mathfrak {H}.$
More precisely, the family
$U=(U_1, U_2, \dots , U_n)$
given by
$$ \begin{align*} U_i=\begin{pmatrix} V_i &&&& [P^{\mathfrak{H}}_{\ker(V_1\dots V_n)^*}W_i|_{\ker(V_1\dots V_n)^*}]^*_0\\ 0 &&&& [V_i(I-W_iW_i^*)|_{\ker(V_1\dots V_n)^*}]+[W_i^*|_{\ker(V_1\dots V_n)^*}]T^*_z \end{pmatrix}, \end{align*} $$
where
$W_i=\prod _{j\ne i}V_j,\ i=1,2, \dots , n,$
is a minimal unitary extension of V on the Kreĭn space
$$ \begin{align*} \mathfrak{K}=\mathfrak{H}\oplus H^2_{\ker(V_1\dots V_n)^*}(\mathbb T). \end{align*} $$
For the Hilbert space case, we refer to [Reference Popovici48, Theorem 3.3.1].
Proposition 4.6 If
$U=(U_1, U_2)\in \mathcal B(\mathfrak {K})^2$
is a minimal unitary extension of the commuting isometric pair
$V=(V_1, V_2)\in \mathcal B(\mathfrak {H})^2$
, then
$U_1U_2$
is a minimal unitary extension of
$V_1V_2.$
Proof It is clear that, under the given hypothesis,
$U_1U_2$
is a unitary extension of
$V_1V_2.$
Since, for
$m\le n\le 0,\ U_1^mU_2^n\mathfrak {H}\subset (U_1U_2)^m\mathfrak {H},$
we observe that
$$ \begin{align*} \mathfrak{K}=\bigvee_{m, n\le 0}U_1^mU_2^n\mathfrak{H}\subset \bigvee_{m\le 0} (U_1U_2)^m\mathfrak{H}\subset\mathfrak{K}, \end{align*} $$
hence,
$\mathfrak {K}=\bigvee _{m\le 0} (U_1U_2)^m\mathfrak {H},$
and the minimality of the unitary extension
$U_1U_2$
is proved.
In view of Theorem 4.2, we can deduce a geometrical structure for the minimal unitary extension.
Corollary 4.7 Let
$U=(U_1, U_2)\in \mathcal B(\mathfrak {K})^2$
be any minimal unitary extension of the commuting isometric pair
$V=(V_1, V_2)\in \mathcal B(\mathfrak {H})^2.$
Then,
-
(i)
$\mathfrak {L}^*=\overline {((U_1U_2)^*-(V_1V_2)^*)\mathfrak {H}}$
is wandering for
$U_1U_2$
, regular, and isometrically isomorphic with
$\ker (V_1V_2)^*;$
-
(ii)
$M_-(\mathfrak {L}^*)=\bigvee _{n\le 0}(U_1U_2)^n\mathfrak {L}^*$
is a regular subspace of
$\mathfrak {K}$
and (4.4)
$$ \begin{align} \mathfrak{K}=\mathfrak{H}\oplus M_-(\mathfrak{L}^*); \end{align} $$
-
(iii)
$M_-(\mathfrak {L}^*)$
is invariant to
$(U_1U_2)^*$
and
$(U_1U_2)^*|_{M_-(\mathfrak {L}^*)}$
is a unilateral shift.
Let
$U=(U_1, U_2)\in \mathcal B(\mathfrak {K})^2,\ U'=(U^{\prime }_1, U^{\prime }_2)\in \mathcal B(\mathfrak {K}')^2$
be two minimal unitary extensions of the pair
$V=(V_1, V_2)$
of commuting bounded isometries on
$\mathfrak {H}$
. U and
$U'$
are said to be unitarily equivalent if there exists a unitary operator
$\Phi :\mathfrak {K}\to \mathfrak {K}'$
which intertwines
$U_1$
and
$U^{\prime }_1$
, respectively,
$U_2$
and
$U^{\prime }_2$
and such that
$\Phi |_{\mathfrak {H}}=I_{\mathfrak {H}}.$
Theorem 4.8 Let
$V=(V_1, V_2)\in \mathcal B(\mathfrak {H})^2$
be a commuting isometric pair such that
$(V_1V_2)^*$
is either contractive or expansive. Then, V has a unique minimal unitary extension.
Proof Assume that, for example,
$(V_1V_2)^*$
is contractive. Then,
$\ker (V_1V_2)^*$
(and, hence, also
$H^2_{\ker (V_1V_2)^*}$
) is a Hilbert space.
Let
$U=(U_1, U_2)$
be the minimal unitary extension of V given by (4.2) and (4.3) and acting on the Kreĭn space
$\mathfrak {K}=\mathfrak {H}\oplus H^2_{\ker (V_1V_2)^*}(\mathbb T).$
If
$U'=(U^{\prime }_1, U^{\prime }_2)\in \mathcal B(\mathfrak {K}')^2$
is any other minimal unitary extension of V, then, according to (4.4),
$\mathfrak {K}'$
has an orthogonal decomposition of the form
$\mathfrak {K}'=\mathfrak {H}\oplus M_-(\mathfrak {L}^{\prime *}),$
with
$\mathfrak {L}^{\prime *}=\overline {((U^{\prime }_1U^{\prime }_2)^*-(V_1V_2)^*)\mathfrak {H}}.$
Since, for arbitrary finite sequences
$\{h_n\}_n,\ \{g_n\}_n$
of vectors in
$\mathfrak {H}$
, we have
$$ \begin{align*} & \langle\sum_{n\ge 0}(U^{\prime}_1U^{\prime}_2)^{*n}((U^{\prime}_1U^{\prime}_2)^*-(V_1V_2)^*)h_n, \sum_{m\ge 0}(U^{\prime}_1U^{\prime}_2)^{*m}((U^{\prime}_1U^{\prime}_2)^*-(V_1V_2)^*)g_m\rangle_{\mathfrak{K}'}\\ & \qquad =\langle\sum_{n\ge 0}z^n(I-V_1V_2(V_1V_2)^*)h_n, \sum_{m\ge 0}z^m(I-V_1V_2(V_1V_2)^*)g_m\rangle_{H^2_{\ker(V_1V_2)^*}(\mathbb T)}, \end{align*} $$
the mapping
$$ \begin{align*} M_-(\mathfrak{L}^{\prime *})\ni\sum_{n\ge 0}(U^{\prime}_1U^{\prime}_2)^{*n}((U^{\prime}_1U^{\prime}_2)^*- & (V_1V_2)^*)h_n\stackrel{\phi_+}{\longmapsto} \\ & \sum_{n\ge 0}z^n(I-V_1V_2(V_1V_2)^*)h_n\in H^2_{\ker(V_1V_2)^*}(\mathbb T) \end{align*} $$
is well defined, and the linear operator
$\Phi $
is isometric, densely defined, and with dense range. It can be uniquely extended to a unitary operator
$\Phi \in \mathcal B(M_-(\mathfrak {L}^{'*}), H^2_{\ker (V_1V_2)^*}(\mathbb T))$
(since
$H^2_{\ker (V_1V_2)^*}(\mathbb T)$
is a Hilbert space).
Then,
$\Phi =I_{\mathfrak {H}}\oplus \Phi _+\in \mathcal B(\mathfrak {K}', \mathfrak {K})$
is unitary and
$\Phi |_{\mathfrak {H}}=I_{\mathfrak {H}}.$
Moreover, for any
$h\in \mathfrak {H},$
$$ \begin{align*} \Phi U^{\prime}_1h=\Phi V_1h=V_1h=V_1\Phi h=U_1\Phi h, \end{align*} $$
$$ \begin{align*} \Phi U^{\prime}_1((U^{\prime}_1U^{\prime}_2)^*- & (V^{\prime}_1V^{\prime}_2)^*)h \\& = \Phi[(I-V_1V_1^*)V_2^*h+((U^{\prime}_1U^{\prime}_2)^*-(V_1V_2)^*)V_1(I-V_2V_2^*)h] \\& = z^0(I-V_1V_1^*)V_2^*h+zV_1(I-V_2V_2^*)h\ \\& = U_1(z^0(I-V_1V_2V_1^*V_2^*)h)\ \\& = U_1\Phi((U^{\prime}_1U^{\prime}_2)^*-(V_1V_2)^*)h, \end{align*} $$
and, by a similar argument,
$$ \begin{align*} \Phi U^{\prime}_1(U^{\prime}_1U^{\prime}_2)^{*n}((U^{\prime}_1U^{\prime}_2)^*-(V_1V_2)^*)h= U_1\Phi(U^{\prime}_1U^{\prime}_2)^{*n}((U^{\prime}_1U^{\prime}_2)^*-(V_1V_2)^*)h, \end{align*} $$
for any positive integer
$n.$
Conclude that
$\Phi U^{\prime }_1=U_1\Phi $
since they are continuous and coincide on a dense subset. By symmetry, we also obtain
$\Phi U^{\prime }_2=U_2\Phi $
and, therefore, U and
$U'$
are unitarily equivalent.
Corollary 4.9 Let
$V=(V_1, V_2)$
be a commuting isometric pair in
$\mathcal B(\mathfrak {H})^2$
. If
$V_1V_2$
has a unique minimal unitary extension, then V has a unique minimal unitary extension.
5 Regular dilations for commuting pairs
Definition 5.1 Let
$T=(T_1, T_2)$
be a commuting pair of bounded operators on a Kreĭn space
$\mathfrak {H}$
.
$\bullet $
An isometric (respectively, unitary) dilation of T is a pair
$U=(U_1, U_2)$
of bounded commuting isometric (respectively, unitary) operators on a Kreĭn space
$\mathfrak {K}$
containing
$\mathfrak {H}$
as a Kreĭn subspace and satisfying
$$ \begin{align} T^n=P_{\mathfrak{H}} U^n|_{\mathfrak{H}},\quad n\in\mathbb{Z}^2_+. \end{align} $$
$\bullet $
An isometric or unitary dilation
$U=(U_1,U_2)\in \mathcal B(\mathfrak {K})^2$
of
$T=(T_1, T_2)\in \mathcal B(\mathfrak {H})^2$
is said to be regular if
$$ \begin{align} ({T^{n^-}})^*{T^{n^+}}=P_{\mathfrak{H}}({U^{n^-}})^*{U^{n^+}}|_{\mathfrak{H}},\quad n\in\mathbb{Z}^2. \end{align} $$
Here, for
$n=(n_1,n_2)\in \mathbb {Z}^2$
, the usual notations
$n^+:=(\max \{n_1,0\},\max \{n_2,0\})$
and
$n^-:=(-\min \{n_1,0\},-\min \{n_2,0\})$
are used. Formula (5.2) is consistent with the dilation definition (5.1) which can be obtained for
$n\in \mathbb {Z}^2_+$
(in this case
$n^-=(0,0)$
and
$n^+=n$
). If (5.1) holds true, then (5.2) is actually equivalent with
$$ \begin{align*} T^{*m}_1T^n_2=P_{\mathfrak{H}} U^{*m}_1U^n_2|_{\mathfrak{H}},\quad m, n\ge 0. \end{align*} $$
$\bullet $
An isometric (respectively, unitary) dilation
$U\in \mathcal B(\mathfrak {K})^2$
of
$T\in \mathcal B(\mathfrak {H})^2$
is called minimal if
$\mathfrak {K}=\bigvee _{n\in \mathbb {Z}^2_+}U^n\mathfrak {H}$
(respectively,
$\mathfrak {K}=\bigvee _{n\in \mathbb {Z}^2}U^n\mathfrak {H}$
).
Remark 5.1 Suppose that the commuting pair
$T\in \mathcal B(\mathfrak {H})^2$
has a minimal isometric (respectively, minimal regular isometric) dilation
$V\in \mathcal B(\mathfrak {K}_+)^2$
. Let
$U\in \mathcal B(\mathfrak {K})^2$
be the minimal unitary extension of
$V\in \mathcal B(\mathfrak {K}_+)^2$
as constructed in Theorem 4.4. Then, U is a minimal unitary (respectively, minimal regular unitary) dilation of T.
As in the one-dimensional case, a minimal isometric dilation for a commuting pair T ensures the existence of a co-isometric extension for
$T^*$
.
Proposition 5.2 Let
$V=(V_1, V_2)\in \mathcal B(\mathfrak {K}_+)^2$
be a minimal isometric dilation of the commuting pair
$T=(T_1, T_2)\in \mathcal B(\mathfrak {H})^2$
. Then,
$\mathfrak {H}$
is invariant to
$V^*=(V^*_1, V^*_2),$
$$ \begin{align*} T^n P_{\mathfrak{H}}=P_{\mathfrak{H}} V^n,\ n\in\mathbb{Z}^2_+ \quad\text{and}\quad T^*=V^*|_{\mathfrak{H}}. \end{align*} $$
Proof For every
$m, n\in \mathbb {Z}^2_+$
and
$h\in \mathfrak {H}$
, the dilation definition shows that
$$ \begin{align*} T^n P_{\mathfrak{H}}(V^m h)=T^{n+m}h=P_{\mathfrak{H}} V^{n+m} h=P_{\mathfrak{H}} V^n(V^m h). \end{align*} $$
Since
$\{ V^m h \mid m\in \mathbb {Z}^2_+, h\in \mathfrak {H}\}$
is dense in
$\mathfrak {K}_+$
and
$P_{\mathfrak {H}}, T, V$
are bounded, we actually deduce that
$$ \begin{align*} T^n P_{\mathfrak{H}}=P_{\mathfrak{H}} V^n. \end{align*} $$
We will prove that
${V^n}^*h={T^n}^*h,$
for every
$n\in \mathbb {Z}^2_+$
and
$h\in \mathfrak {H}.$
To this end, let
${m\in \mathbb {Z}^2_+}$
and
$h'\in \mathfrak {H}.$
Then,
$$ \begin{align*} \langle{V^n}^*h-{T^n}^*h, V^m h'\rangle_{\mathfrak{K}_+}=\langle h, V^{n+m}h'\rangle_{\mathfrak{K}_+}-\langle{T^n}^*h, T^m h'\rangle=0. \end{align*} $$
Use again the minimality of V to obtain that
${V^n}^*h={T^n}^*h.$
Consequently,
$\mathfrak {H}$
is invariant to
$V^*$
and
$T^*=V^*|_{\mathfrak {H}}.$
Let
$T=(T_1, T_2)$
be a pair of commuting bounded operators on a Kreĭn space
$\mathfrak {H}.$
By a defect operator for
$T,$
we mean an operator
$D_T\in \mathcal B(\mathfrak {D}_T, \mathfrak {H})$
with zero kernel on a Kreĭn space
$\mathfrak {D}_T$
(called its defect space) such that
$$ \begin{align*} I-T^*_1T_1-T^*_2T_2+T^*_1T^*_2T_1T_2=D_T D^*_T. \end{align*} $$
T is said to be a bidisc contraction, respectively, bidisc expansion if
$$ \begin{align*} \langle T_1h, T_1h\rangle_{\mathfrak{H}}+\langle T_2h, T_2h\rangle_{\mathfrak{H}}\le\langle h, h\rangle_{\mathfrak{H}}+\langle T_1T_2h, T_1T_2h\rangle_{\mathfrak{H}},\quad h\in\mathfrak{H}, \end{align*} $$
respectively,
$$ \begin{align*} \langle T_1h, T_1h\rangle_{\mathfrak{H}}+\langle T_2h, T_2h\rangle_{\mathfrak{H}}\ge\langle h, h\rangle_{\mathfrak{H}}+\langle T_1T_2h, T_1T_2h\rangle_{\mathfrak{H}},\quad h\in\mathfrak{H} \end{align*} $$
or, equivalently, the defect space of T is a Hilbert, respectively, an anti-Hilbert space.
In what follows, we shall use the notations
$D_1=D_{T_1}, D_2=D_{T_2}, D=D_T$
for the defect operators and
$\mathfrak {D}_1=\mathfrak {D}_{T_1}, \mathfrak {D}_2=\mathfrak {D}_{T_2}, \mathfrak {D}=\mathfrak {D}_T$
for the corresponding defect spaces.
As in the one-dimensional case, recall that a subspace
$\mathfrak {L}$
is said to be wandering for a commuting isometric pair V if
$V^n\mathfrak {L}\bot V^m\mathfrak {L}$
for all
$n,m\in \mathbb {Z}_+^2, n\ne m$
.
Regarding the geometrical structure of a minimal regular isometric dilation, we could mention the following theorem.
Theorem 5.3 Let
$V=(V_1, V_2)\in \mathcal B(\mathfrak {K}_+)^2$
be a minimal regular isometric dilation of a commuting pair
$T=(T_1, T_2)\in \mathcal B(\mathfrak {H})^2.$
Then,
-
(i)
$\mathfrak {L}_1=\overline {(V_1-T_1)\mathfrak {H}}, \mathfrak {L}_2=\overline {(V_2-T_2)\mathfrak {H}}, \mathfrak {L}=\overline {(V_1V_2-V_1T_2-V_2T_1+T_1T_2)\mathfrak {H}}$
are regular, wandering, respectively, for
$V_1, V_2, V$
and isometrically isomorphic, respectively, with
$\mathfrak {D}_1, \mathfrak {D}_2, \mathfrak {D};$
-
(ii)
$M^1_+(\mathfrak {L}_1)=\bigvee _{m\ge 0}V_1^m\mathfrak {L}_1, M^2_+(\mathfrak {L}_2)=\bigvee _{n\ge 0}V_2^n\mathfrak {L}_2, M_+(\mathfrak {L})=\bigvee _{p\in \mathbb {Z}^2_+}V^p\mathfrak {L}$
are regular and (5.3)
$$ \begin{align} \mathfrak{K}_+=\mathfrak{H}\oplus M_+(\mathfrak{L})\oplus M^1_+(\mathfrak{L}_1)\oplus M^2_+(\mathfrak{L}_2); \end{align} $$
-
(iii)
$M^i_+(\mathfrak {L}_i)$
is invariant to
$V_i$
and
$V_i|_{M^i_+(\mathfrak {L}_i)}$
is a unilateral shift,
$i=1, 2;$
-
(iv)
$M_+(\mathfrak {L})$
is invariant to V and
$V|_{M_+(\mathfrak {L})}$
is a pair of double commuting unilateral shifts.
Proof As in the Hilbert space case [Reference Gaşpar and Suciu24], it is not hard to check that
$\mathfrak {L}_1, \mathfrak {L}_2, \mathfrak {L}$
are wandering, respectively, for
$V_1, V_2, V$
and
$\mathfrak {H}, M^1_+(\mathfrak {L}_1), M^2_+(\mathfrak {L}_2),$
$M_+(\mathfrak {L})$
are pairwise orthogonal. Therefore, we prefer to omit the details.
Proceed similarly as in the proof of Theorem 3.1 to obtain
$$ \begin{align} V_1^mh=T_1^mh+\sum_{k=0}^{m-1}V_1^k(V_1-T_1)T_1^{m-k-1}h,\quad h\in\mathfrak{H},\ m\ge 1. \end{align} $$
We use (5.4) in conjunction with the formulas
$$ \begin{align} V_2h=T_2h+(V_2-T_2)h,\quad h\in\mathfrak{H} \end{align} $$
and
$$ \begin{align} V_2(V_1-T_1)h=(V_1V_2-V_1T_2-V_2T_1+T_1T_2)h+(V_1-T_1)T_2h,\quad h\in\mathfrak{H}, \end{align} $$
applied successively, to finally get
$$ \begin{align*} V_1^mV_2^nh = &\ T_1^mT_2^nh\\& +\sum_{0\le p\le(m-1, n-1)} V^p(V_1V_2-V_1T_2-V_2T_1+T_1T_2)T^{(m-1, n-1)-p}h\\& +\sum_{i=0}^{m-1}V_1^i(V_1-T_1)T_1^{m-1-i}T_2^nh+ \sum_{j=0}^{n-1}V_2^j(V_2-T_2)T_1^{m}T_2^{n-1-j}h,\\& \qquad\qquad\qquad h\in\mathfrak{H},\ m, n\in\mathbb{N}^*. \end{align*} $$
More precisely, we firstly apply
$V_2$
to (5.4) and then use (5.5) for
$T_1^mh$
and (5.6) for
$T_1^{m-k-1}h, k\in \{0,1,\dots ,m-1\}$
, instead of h. We obtain that
$$ \begin{align*} V_1^mV_2h&=T_1^mT_2h+(V_2-T_2)T_1^mh+ \sum_{k=0}^{m-1}V_1^k(V_1V_2-V_1T_2-V_2T_1+T_1T_2)T_1^{m-k-1}h\\& \quad+\sum_{k=0}^{m-1}V_1^k(V_1-T_1)T_1^{m-k-1}T_2h. \end{align*} $$
Following again (5.5) and (5.6) for the computation of the vectors
$V_2T_2T_1^mh$
and, respectively,
$V_2(V_1-T_1)T_1^{m-k-1}T_2h, k\in \{0,1,\dots ,m-1\}$
, another application of
$V_2$
shows that
$$ \begin{align*} V_1^mV_2^2h&= T_1^mT_2^2h+(V_2-T_2)T_1^mT_2h+V_2(V_2-T_2)T_1^mh\\& \quad +\sum_{k=0}^{m-1}V_1^kV_2(V_1V_2-V_1T_2-V_2T_1+T_1T_2)T_1^{m-k-1}h\\& \quad +\sum_{k=0}^{m-1}V_1^k(V_1V_2-V_1T_2-V_2T_1+T_1T_2)T_1^{m-k-1}T_2h\\& \quad +\sum_{k=0}^{m-1}V_1^k(V_1-T_1)T_1^{m-k-1}T_2^2h. \end{align*} $$
This iterative procedure is repeated n times.
Since, obviously,
$\mathfrak {K}_+$
contains
$\mathfrak {H}, M_+(\mathfrak {L}), M_+^1(\mathfrak {L}_1),$
and
$M_+^2(\mathfrak {L}_2)$
, we deduce that
$$ \begin{align*} \mathfrak{K}_+=\mathfrak{H}\vee M_+(\mathfrak{L})\vee M^1_+(\mathfrak{L}_1)\vee M^2_+(\mathfrak{L}_2) \end{align*} $$
and, by orthogonality, (5.3) also holds.
It is then clear that
$\mathfrak {H}, M_+(\mathfrak {L}), M_+^1(\mathfrak {L}_1), M_+^2(\mathfrak {L}_2)$
are all regular and, since
$M_+(\mathfrak {L}) = \mathfrak {L}\oplus V_1V_2M_+(\mathfrak {L})\oplus V_1(V_2M_+(\mathfrak {L}))^\bot \oplus V_2(V_1M_+(\mathfrak {L}))^\bot , M_+^1(\mathfrak {L}_1)=\mathfrak {L}_1\oplus V_1M_+^1(\mathfrak {L}_1), M_+^2(\mathfrak {L}_2)=\mathfrak {L}_2\oplus V_2M_+^2(\mathfrak {L}_2),$
we obtain that
$\mathfrak {L}, \mathfrak {L}_1, \mathfrak {L}_2$
are also regular.
By a similar argument as in the proof of Theorem 3.1, we can deduce that
$\mathfrak {L}_1, \mathfrak {L}_2, \mathfrak {L}$
are isometrically isomorphic, respectively, with
$\mathfrak {D}_1, \mathfrak {D}_2, \mathfrak {D}.$
It is obvious that
$M_+(\mathfrak {L}), M_+^1(\mathfrak {L}_1), M_+^2(\mathfrak {L}_2)$
are invariant to
$V_1, V_2$
and, respectively, V and that
$V_1|_{M_+^1(\mathfrak {L}_1)}$
,
$V_2|_{M_+^2(\mathfrak {L}_2)}$
are unilateral shifts. Since
$M_+(\mathfrak {L})=M_+^1(M_+^2(\mathfrak {L})) =M_+^2(M_+^1(\mathfrak {L})),$
we obtain that
$V|_{M_+(\mathfrak {L})}$
is a pair of commuting unilateral shifts which, moreover, doubly commute.
To this aim, we firstly note that it is only necessary to prove that
$(V_1|_{M_+(\mathfrak {L})})^*$
and
$V_2|_{M_+(\mathfrak {L})}$
commute on the set
$\{V_1^mV_2^nl\mid m,n\ge 0, l\in \mathfrak {L}\},$
which generates
$M_+(\mathfrak {L})$
.
Indeed, for
$n\ge 0$
,
$$ \begin{align*} \bigl((V_1|_{M_+(\mathfrak{L})})^*V_2\bigr)V_2^nl=(V_1|_{M_+(\mathfrak{L})})^*V_2^{n+1}l=0=\bigl(V_2(V_1|_{M_+(\mathfrak{L})})^*\bigr)V_2^nl, \end{align*} $$
since
$V_2^nl\in M_+^2(\mathfrak {L})=\ker (V_1|_{M_+(\mathfrak {L})})^*$
. Also, in view of the fact that
$V_1|_{M_+(\mathfrak {L})}$
is isometric (i.e.,
$(V_1|_{M_+(\mathfrak {L})})^*V_1|_{M_+(\mathfrak {L})}=I_{M_+(\mathfrak {L})}$
), the following equalities
$$ \begin{align*} \bigl((V_1|_{M_+(\mathfrak{L})})^*V_2\bigr)V_1^mV_2^nl=(V_1|_{M_+(\mathfrak{L})})^*V_1V_1^{m-1}V_2^{n+1}l=V_1^{m-1}V_2^{n+1}l \end{align*} $$
and
$$ \begin{align*} \bigl(V_2(V_1|_{M_+(\mathfrak{L})})^*\bigr)V_1^mV_2^nl=V_2(V_1|_{M_+(\mathfrak{L})})^*V_1V_1^{m-1}V_2^nl=V_1^{m-1}V_2^{n+1}l \end{align*} $$
hold true for every
$m>0$
and
$n\ge 0$
.
Corollary 5.4 Let
$V=(V_1, V_2)\in \mathcal B(\mathfrak {K})^2$
be a minimal regular isometric dilation of the commuting pair
$T=(T_1, T_2)\in \mathcal B(\mathfrak {H})^2.$
The following conditions are equivalent:
-
(i)
$M_+^1(\mathfrak {L}_1)$
is invariant to
$V_2$
; -
(ii)
$M_+^2(\mathfrak {L}_2)$
is invariant to
$V_1$
; -
(iii) T is a bidisc isometry (i.e.,
$I-T_1^*T_1-T_2^*T_2+T_1^*T_2^*T_1T_2=0$
).
Proof The conclusion follows from the geometrical structure of
$\mathfrak {K}_+$
given by the theorem above since
$$ \begin{align*} V_{3-i}V^n_i(V_i-T_i)h=V_i^n(V_1V_2-V_1T_2-V_2T_1+T_1T_2)h + & V_i^n(V_i-T_i)T_{3-i}h,\\ & \quad h\in\mathfrak{H},\ i=1, 2,\ n\in\mathbb{N} \end{align*} $$
and
$(V_1V_2-V_1T_2-V_2T_1+T_1T_2)h=0$
, for all
$h\in \mathfrak {H}$
, if and only if
$I-T_1^*T_1-T_2^*T_2+T_1^*T_2^*T_1T_2=0$
.
For the Hilbert space case, we refer to [Reference Gaşpar and Suciu24].
For the rest of the article, we shall suppose that
$T=(T_1, T_2)$
is a pair of commuting bounded operators on a Kreĭn space
$\mathfrak {H}$
such that
$T_1, T_2$
are both contractive and T is a bidisc contraction or
$T_1, T_2$
are both expansive and T is a bidisc expansion. Equivalently, the defect spaces
$\mathfrak {D}_1, \mathfrak {D}_2$
, and
$\mathfrak {D}$
are either Hilbert or anti-Hilbert spaces. Denote by
$\lVert \cdot \rVert _1, \lVert \cdot \rVert _2, \lVert \cdot \rVert $
the Hilbert space norms, respectively, on
$\mathfrak {D}_1, \mathfrak {D}_2, \mathfrak {D}.$
Remark 5.5 (i) Observe firstly that
$$ \begin{align} I-T_1^*T_1-T_2^*T_2 & +T_1^*T_2^*T_1T_2 \nonumber \\ & =D_1D_1^*-T_2^*(I-T_1^*T_1)T_2 \nonumber \\ & =D_1D_1^*-(T_2^*D_1)(T_2^*D_1)^* \nonumber \\ & =D_2D_2^*-(T_1^*D_2)(T_1^*D_2)^*. \end{align} $$
Use the inequality
$$ \begin{align*} \lVert D_1^*T_2h\rVert_1\le\lVert D_1^*h\rVert_1\quad (\text{respectively, } \lVert D_2^*T_1h\rVert_2\le\lVert D_2^*h\rVert_2),\quad h\in\mathfrak{H} \end{align*} $$
to introduce a densely defined Hilbert space contraction on
$\mathfrak {D}_1$
(respectively,
$\mathfrak {D}_2$
) by
$$ \begin{align} R_2D_1^*h=D_1^*T_2h\quad (\text{respectively, } R_1D_2^*h=D_2^*T_1h),\quad h\in\mathfrak{H}, \end{align} $$
which can be extended, by continuity, to the whole space. In fact, the maps above are (under our Kreĭn space terminology) contractions if T is a bidisc contraction, respectively, expansions if T is a bidisc expansion.
(ii) Taking into account the operators
$R_1$
and
$R_2$
(defined by (5.8)), formula (5.7) can be re-written as
$$ \begin{align*} DD^*=(D_1D_{R_2})(D_1D_{R_2})^*=(D_2D_{R_1})(D_2D_{R_1})^* \end{align*} $$
or, equivalently, as
$$ \begin{align*} \lVert D^*h\rVert=\lVert D_{R_2}^*D_1^*h\rVert_{\mathfrak{D}_{R_2}}= \lVert D_{R_1}^*D_2^*h\rVert_{\mathfrak{D}_{R_1}},\quad h\in\mathfrak{H}. \end{align*} $$
Hence, the linear operators
$U_1: \mathfrak {D}\to \mathfrak {D}_{R_1}$
and
$U_2: \mathfrak {D}\to \mathfrak {D}_{R_2}$
given by
$$ \begin{align} U_1D^*h=D_{R_1}^*D_2^*h\quad \text{and}\quad U_2D^*h=D_{R_2}^*D_1^*h,\quad h\in\mathfrak{H} \end{align} $$
are well defined unitary operators.
The next construction of a regular isometric dilation is the main result of this section.
Theorem 5.6 Let
$T=(T_1, T_2)$
be a pair of commuting bounded operators on a Kreĭn space
$\mathfrak {H}$
such that
$T_1, T_2$
are both contractive and T is a bidisc contraction or
$T_1, T_2$
are both expansive and T is a bidisc expansion. The pair
$V=(V_1, V_2)\in \mathcal B(\mathfrak {K}_+)^2$
given by the matrix representation
$$ \begin{align} V_1=\begin{pmatrix} T_1 &&& 0 &&& 0 &&& 0\\ 0 &&& T_{z_1} &&& 0 &&& [U_1^*D_{R_1}^*]_2\\ [D_1^*]_0 &&& 0 &&& T_{z} &&& 0\\ 0 &&& 0 &&& 0 &&& [R_1] \end{pmatrix} \end{align} $$
and
$$ \begin{align} V_2=\begin{pmatrix} T_2 &&& 0 &&& 0 &&& 0\\ 0 &&& T_{z_2} &&& [U_2^*D_{R_2}^*]_1 &&& 0\\ 0 &&& 0 &&& [R_2] &&& 0\\ [D_2^*]_0 &&& 0 &&& 0 &&& T_{z} \end{pmatrix} \end{align} $$
is a minimal regular isometric dilation of T on the Kreĭn space
$$ \begin{align*} \mathfrak{K}_+=\mathfrak{H}\oplus H^2_{\mathfrak{D}}(\mathbb T^2)\oplus H^2_{\mathfrak{D}_1}(\mathbb T)\oplus H^2_{\mathfrak{D}_2}(\mathbb T). \end{align*} $$
Proof Direct computations with matrices show that, for
$i=1,2$
,
$V_i$
is an isometric operator on
$\mathfrak {K}_+$
if and only if
$[D_i^*]_0^*T_z=0, [U_i^*D_{R_i}^*]^*_{3-i}T_{z_i}=0, T_i^*T_i+[D_i^*]_0^*[D_i^*]_0=I_{\mathfrak {H}}$
and
$[U_i^*D_{R_i}^*]^*_{3-i}[U_i^*D_{R_i}^*]_{3-i}+[R_i]^*[R_i]=I_{H^2_{\mathfrak {D}_{3-i}}(\mathbb {T})}$
.
While the first two equalities hold true by Proposition 2.2 (iv), the last two are consequences of the conditions (ii), respectively, (iii) of the same proposition. Indeed,
$[D_i^*]^*_0[D_i^*]_0=D_iD_i^*$
and, hence,
$T_i^*T_i+D_iD_i^*=I_{\mathfrak {H}}$
, by (3.1). Also,
$$ \begin{align*} & [U_i^*D_{R_i}^*]^*_{3-i}[U_i^*D_{R_i}^*]_{3-i}+[R_i]^*[R_i]& \\ & \qquad\qquad=[(U_i^*D_{R_i}^*)^*U_i^*D_{R_i}^*+R_i^*R_i]& (\text{by Proposition }2.2 \text{ (i) and (ii)})\\ & \qquad\qquad=[D_{R_i}D_{R_i}^*+R_i^*R_i]& (\text{since } U_i \text{ is unitary})\\ & \qquad\qquad=[I_{\mathfrak{D}_{3-i}}]=I_{H^2_{\mathfrak{D}_{3-i}}(\mathbb{T}).}& (\text{by }(3.1)) \end{align*} $$
Similarly,
$V_1V_2=V_2V_1$
if and only if
$[U_1^*D_{R_1}^*]_2[D_2^*]_0=[U_2^*D_{R_2}^*]_1[D_1^*]_0$
,
$[U_i^*D_{R_i}^*]_{3-i}T_z=T_{z_{3-i}}[U_i^*D_{R_i}^*]_2$
,
$T_z[R_i]=[R_i]T_z$
and
$[D_i^*]_0T_{3-i}=[R_{3-i}][D_i^*]_0$
,
$i=1,2$
. The first condition follows by Proposition 2.2 (v) and (5.9):
$$ \begin{align*} [U_i^*D_{R_i}^*]_{3-i}[D_{3-i}^*]_0h=z_1^0z_2^0U_i^*D_{R_i}^*D_{3-i}^*h=z_1^0z_2^0D^*h,\quad h\in\mathfrak{H}, i=1,2. \end{align*} $$
The following two conditions are consequences of Proposition 2.2 (iv). The last equality uses Proposition 2.2 (v) and formula (5.8):
$$ \begin{align*} [D_i^*]_0T_{3-i}=[D_i^*T_{3-i}]_0=[R_{3-i}D_i^*]_0=[R_{3-i}][D_i^*]_0. \end{align*} $$
Moreover, by an inductive method,
$V_2^{*n}V_1^m$
has the form
$$ \begin{align*} V_2^{*n}V_1^m=\begin{pmatrix} T_2^{*n}T_1^m & 0 & 0 & *\\ 0 & (T_{z_2})^{*n}(T_{z_1})^m & 0 & *\\ * & * & [R_2^{*n}](T_z)^m & *\\ 0 & 0 & 0 & (T_{z})^{*n}[R_1^m] \end{pmatrix}, \end{align*} $$
which proves that
$$ \begin{align*} T_2^{*n}T_1^m=P_{\mathfrak{H}} V_2^{*n}V_1^m|_{\mathfrak{H}},\quad m, n\ge 0. \end{align*} $$
We can also obtain, by a similar argument, that
$$ \begin{align*} T_1^{m}T_2^n=P_{\mathfrak{H}} V_1^{m}V_2^n|_{\mathfrak{H}},\quad m, n\ge 0. \end{align*} $$
Hence, V is a regular isometric dilation of
$T.$
It remains to prove the minimality. To this end, take
$h\in \mathfrak {H}$
and observe that
$$ \begin{align*} (V_1-T_1)h=(0, 0, [D_1^*]_0h, 0). \end{align*} $$
Proceed inductively to show that
$$ \begin{align*} V_1^m(V_1-T_1)h=(0, 0, (T_z)^n[D_1^*]_0h, 0),\quad m\ge 0, \end{align*} $$
that is,
$$ \begin{align} H^2_{\mathfrak{D}_1}(\mathbb T)=\bigvee_{m\ge 0}V_1^m(V_1-T_1)\mathfrak{H}. \end{align} $$
By symmetry, it also holds
$$ \begin{align} H^2_{\mathfrak{D}_2}(\mathbb T)=\bigvee_{n\ge 0}V_2^n(V_2-T_2)\mathfrak{H}. \end{align} $$
Now, the relation
$$ \begin{align*} (V_1V_2-V_1T_2-V_2T_1+T_1T_2)h=(0, z_1^0z_2^0D^*h, 0, 0) \end{align*} $$
applied successively gives
$$ \begin{align*} V_1^mV_2^n(V_1V_2-V_1T_2-V_2T_1+T_1T_2)h=(0, z_1^mz_2^nD^*h, 0, 0), \quad m, n\ge 0, \end{align*} $$
that is,
$$ \begin{align} H^2_{\mathfrak{D}}(\mathbb T^2)=\bigvee_{n\in\mathbb{Z}_+^2}V^n(V_1V_2-V_1T_2-V_2T_1+T_1T_2)\mathfrak{H}. \end{align} $$
Equations (5.12–5.14) show that the regular isometric dilation given by (5.10) and (5.11) is minimal.
Use Theorem 4.4, Remark 5.1, and Theorem 5.6 to obtain the following.
Corollary 5.7 Let
$T\in \mathcal B(\mathfrak {H})^2$
be as in Theorem 5.6 and
$V\in \mathcal B(\mathfrak {K}_+)^2$
be the minimal regular isometric dilation of T given by (5.10) and (5.11). Then, T has a minimal regular unitary dilation
$U\in \mathcal B(\mathfrak {K})^2$
given by (4.2) and (4.3) on the Kreĭn space
$$ \begin{align*} \mathfrak{K}=\mathfrak{H}\oplus H^2_{\mathfrak{D}}(\mathbb T^2)\oplus H^2_{\mathfrak{D}_1}(\mathbb T)\oplus H^2_{\mathfrak{D}_2}(\mathbb T)\oplus H^2_{\ker(V_1V_2)^*}(\mathbb T). \end{align*} $$
Let
$V=(V_1, V_2)\in \mathcal B(\mathfrak {K}_+)^2$
and
$V'=(V^{\prime }_1, V^{\prime }_2)\in \mathcal B(\mathfrak {K}^{\prime }_+)^2$
be two minimal regular isometric dilations of
$T\in \mathcal B(\mathfrak {H})^2.$
V and
$V'$
are said to be unitarily equivalent if there exists a unitary operator
$\Phi : \mathfrak {K}_+\to \mathfrak {K}^{\prime }_+$
which intertwines
$V_1$
and
$V^{\prime }_1$
, respectively,
$V_2$
and
$V^{\prime }_2$
and such that
$\Phi |_{\mathfrak {H}}=I_{\mathfrak {H}}$
.
Theorem 5.8 Let
$T=(T_1, T_2)$
be a pair of commuting bounded operators on a Kreĭn space
$\mathfrak {H}$
such that
$T_1, T_2$
are both contractive and T is a bidisc contraction or
$T_1, T_2$
are both expansive and T is a bidisc expansion. Then, T has a unique minimal regular isometric dilation (up to a unitary equivalence).
Proof Let
$V\in \mathcal B(\mathfrak {K}_+)^2$
be the minimal regular isometric dilation of
$T\in \mathcal B(\mathfrak {H})^2$
given by (5.10) and (5.11) on
$\mathfrak {K}_+=\mathfrak {H}\oplus H^2_{\mathfrak {D}}(\mathbb T^2)\oplus H^2_{\mathfrak {D}_1}(\mathbb T)\oplus H^2_{\mathfrak {D}_2}(\mathbb T).$
If
$V'=(V^{\prime }_1, V^{\prime }_2)\in \mathcal B(\mathfrak {K}^{\prime }_+)^2$
is any other minimal regular isometric dilation of T then, according to (5.3),
$\mathfrak {K}'$
has an orthogonal decomposition of the form
$$ \begin{align*} \mathfrak{K}^{\prime}_+=\mathfrak{H}\oplus M_+(\mathfrak{L}')\oplus M_+^1(\mathfrak{L}^{\prime}_1)\oplus M_+^2(\mathfrak{L}^{\prime}_2), \end{align*} $$
with
$\mathfrak {L}'=\overline {(V^{\prime }_1V^{\prime }_2-V^{\prime }_1T_2-V^{\prime }_2T_1+T_1T_2)\mathfrak {H}}$
and
$\mathfrak {L}^{\prime }_i=\overline {(V^{\prime }_i-T_i)\mathfrak {H}}, i=1,2$
.
The maps
$$ \begin{align*} M_+(\mathfrak{L}')\ni V_1^{'m}V_2^{'n}(V^{\prime}_1V^{\prime}_2-V^{\prime}_1T_2-V^{\prime}_2T_1+T_1T_2)h\stackrel{\Phi}{\longmapsto} z_1^mz^n_2D^*h\in H^2_{\mathfrak{D}}(\mathbb T^2), \end{align*} $$
$$ \begin{align*} M_+(\mathfrak{L}^{\prime}_1)\ni V_1^{'m}(V^{\prime}_1-T_1)h\stackrel{\Phi_1}{\longmapsto} z^mD_1^*h\in H^2_{\mathfrak{D}_1}(\mathbb T) \end{align*} $$
and
$$ \begin{align*} M_+(\mathfrak{L}^{\prime}_2)\ni V_2^{'n}(V^{\prime}_1-T_2)h\stackrel{\Phi_2}{\longmapsto} z^nD_2^*h\in H^2_{\mathfrak{D}_2}(\mathbb T) \end{align*} $$
are well defined and can be extended by linearity to densely defined isometries with dense ranges. Since
$H^2_{\mathfrak {D}}(\mathbb T^2), H^2_{\mathfrak {D}_1}(\mathbb T),$
and
$H^2_{\mathfrak {D}_2}(\mathbb T)$
are either Hilbert or anti-Hilbert spaces, the applications above can be extended to unitary operators.
A routine check shows that
$I_{\mathfrak {H}}\oplus \Phi \oplus \Phi _1\oplus \Phi _2:\mathfrak {K}^{\prime }_+\to \mathfrak {K}_+$
is unitary and intertwines
$V_1$
and
$V^{\prime }_1$
, respectively,
$V_2$
and
$V^{\prime }_2.$
Hence, V and
$V'$
are unitarily equivalent.
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