Conjugations of unitary operators, I

If U is a unitary operator on a separable complex Hilbert space \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}$$\end{document}H, an application of the spectral theorem says there is a conjugation C on \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}$$\end{document}H (an antilinear, involutive, isometry on \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {H}$$\end{document}H) for which \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ C U C = U^{*}.$$\end{document}CUC=U∗. In this paper, we fix a unitary operator U and describe all of the conjugations C which satisfy this property. As a consequence of our results, we show that a subspace is hyperinvariant for U if and only if it is invariant for any conjugation C for which \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$CUC = U^{*}$$\end{document}CUC=U∗.


INTRODUCTION
A version of the spectral theorem says that any unitary operator U on a separable complex Hilbert space H is unitarily equivalent to a multiplication operator M ϕ f = ϕf on a Lebesgue space L 2 (µ, X) [33, p. 13].Here X is a compact Hausdorff space, µ is a finite positive Borel measure on X, and ϕ ∈ L ∞ (µ, X) is unimodular µ-almost everywhere.If J is the mapping from L 2 (µ, X) to itself defined by (Jf )(x) = f (x), x ∈ X (the bar denotes complex conjugation), then J is an antilinear, isometric, and involutive map and is called a conjugation.Furthermore, the conjugation J induces the adjoint identity JM ϕ J = M ϕ = M * ϕ .Via unitary equivalence, given any unitary operator U on H, this results in a conjugation C on H for which CU C = U * (see Lemma 2.3).In the parlance of operator theory, one says that U is a C-symmetric operator [18,19].As it turns out, there are many conjugations C on H for which CU C = U * .The goal of this paper is to describe them all.
Towards this goal, let As discussed in the previous paragraph, C s (U ) = ∅ (see also Proposition 2.5).On the other extreme, C s (I), where I is the identity operator on H, is simply the set of all conjugations on H.The subscript s, for "symmetric", in the above definition of C s (U ) might seem superfluous.However, in a follow up paper to this one [28], we will explore the set C c (U ) (notice the subscript c), the conjugations on H that commute with U , i.e., CU C = U .This paper provides several characterizations of C s (U ).One such characterization (Theorem 8.2) involves the spectral theorem.When one is fortunate enough to have a concrete spectral decomposition of a unitary operator, for example when U is an n × n unitary matrix, one can give a very tangible description of C s (U ).In particular, Theorem 4.2 says that if U is an n × n unitary matrix with spectral decomposition where W is an n × n unitary matrix, ξ 1 , . . ., ξ d (1 d n) are the distinct eigenvalues of U , I n j is the n j ×n j identity matrix, and n 1 +n 2 +• • •+n d = n, then any conjugation C on C n for which CU C = U * must take the form where each V j is an n j × n j unitary matrix satisfying V t j = V j (t denotes the transpose of a matrix), and J denotes the conjugation on C n defined by A version of (1.1) can be obtained in the infinite dimensional setting when the eigenvectors for U are complete in H.In particular, we give a concrete description of C s (U ) when U is the Fourier-Plancherel transform on L 2 (R) (Example 4.4) and when U is the Hilbert transform on L 2 (R) (Example 4.6).
The main driver of nearly all the results in this paper is a measure-theoretic decomposition of any C ∈ C s (U ) given in Theorem 3.2.This enables us to examine manageable pieces of C of H on certain reducing subspaces of U .
Another description of C s (U ) relies less on the spectral decomposition of a unitary operator U , which is often quite intangible in the general setting, and more on some special properties of U .For example, if U is a bilateral shift (see Definition 6.1), C s (U ) was described in [5] (see also Theorem 6.4).In particular, when U is the bilateral shift (M ξ f )(ξ) = ξf (ξ) on L 2 (m, T) (m is normalized Lebesgue measure on the unit circle T), then C ∈ C s (U ) if and only if C = M u J, where J is the conjugation on L 2 (m, T) given by Jf = f and u ∈ L ∞ (m, T) is unimodular almost everywhere on T (see Example 6.6).
The unitary multiplication operator M ψ on L 2 (m, T), where ψ is an inner function (a bounded analytic function on D whose radial boundary values on T have modulus one m-almost everywhere [10,23]), turns out to be a bilateral shift [29,Proposition 5.17] and a concrete description of C s (M ξ ) is given in Theorem 7.8.In fact (see Remark 7.1), such M ψ serve as models for all bilateral shifts and so, in a way, Theorem 7.8, is a canonical model for C s (U ) for any bilateral shift U .
As a byproduct of several results in this paper, we can connect conjugations with the hyperinvariant subspaces of a unitary operator U on H (those subspace which are invariant for any bounded operator on H commuting with U ). Certainly the invariant and hyperinvariant subspaces of unitary operators have been discussed before (for example [8,9,31]).Here we connect them with conjugations in the following way.For a (closed) subspace M of H, the following are equivalent: (i) M is hyperinvariant for U ; (ii) CM ⊆ M for every C ∈ C s (U ).This result is obtained, along with several other equivalent conditions covered in this paper, via Theorems 3.2, 3.8, and 8.6 and is officially stated in Theorem 8.8.

BASIC FACTS ABOUT CONJUGATIONS AND SPECTRAL MEASURES
All Hilbert spaces H in this paper will be complex and separable.Let B(H) denote the set of all bounded linear transformations on H and AB(H) denote the set of all bounded antilinear transformations on H.By this we mean that C ∈ AB(H) when C(x + αy) = Cx + αCy for all x, y ∈ H and α ∈ C (C is antilinear) and sup{ Cx : x = 1} is finite (C is bounded).We say that C ∈ AB(H) is a conjugation if C satisfies the two additional conditions Cx = x for all x ∈ H (C is isometric) and C 2 = I (C is involutive).By the polarization identity, a conjugation also satisfies (2.1) Cx, Cy = y, x for all x, y ∈ H.
Example 2.2.Many types of conjugations were outlined in [17,18,19].Below are a few basic ones that are relevant to this paper.
(a) As discussed in the introduction, the mapping Cf = f defines a conjugation on L 2 (µ, X).In particular, the mapping Throughout this paper we will use the notation t to represent the transpose of a matrix.In addition, vectors in C n will be viewed as column vectors since, for an n × n matrix A of complex numbers, we will often consider the linear transformations on C n defined by x → Ax.
(b) One can consider the conjugations (Cf )(t) = f (t) and (Cf )(t) = f (−t) on L 2 (R).These were used in [1,2] to study symmetric operators and their connections to physics.
(c) If u is an inner function, H 2 is the standard Hardy space, and K u := H 2 ⊖uH 2 is the model space associated with u (considering everything as a subspace of L 2 (m, T) via radial boundary values), then (Cf )(ξ) = u(ξ)ξf (ξ) defines a conjugation on K u [15,Ch. 8].
This next lemma enables us to transfer a conjugation on one Hilbert space to a conjugation on another.The (easy) proof is left to the reader.
Lemma 2.3.Suppose H and K are Hilbert spaces and V : For a conjugation C on H and an A ∈ B(H), we say that A is C-symmetric if CAC = A * .A multitude of operators from a variety of settings enjoy this property [17,18,21,22].Here are a few examples that are relevant to this paper.
For a normal operator N ∈ B(H), the spectral theorem says that N is unitarily equivalent to M ϕ on L 2 (µ, X) for some finite positive Borel measure on some compact Hausdorff space X.Now use the discussion above and Lemma 2.3 to see that N is a C-symmetric operator for some conjugation C on H. Though not used in this paper, but certainly in our follow up paper [28], we recall the following result from [24] which shows that any unitary operator can be built from conjugations.We include a short proof (slightly different from the original) for the reader's convenience.
Proposition 2.5.For each unitary operator U on H, there are conjugations J 1 and Proof.For a given unitary U on H, we can use the spectral theorem argument from Example 2.4(a) (and also from the introduction) to produce a conjugation J 1 on H such that J 1 U J 1 = U * .Now observe that J 2 := U * J 1 is a conjugation, J 2 U J 2 = U * , and The following result will be used from time to time in this paper.
Proposition 2.6.Suppose U, V, W are unitary operators on The following result from [20, Lemma 3.2], which will be generalized below, gives an explicit description of all the conjugations on C n .
Proposition 2.7.A mapping C on C n is a conjugation if and only if C = V J, where V is an n × n unitary matrix with V t = V and J is the conjugation on C n defined by Our discussion below needs a generalization of the previous result for C n to an abstract (separable) Hilbert space H, but, of course, we need a substitute for V t since the "transpose" of a linear operator V on H needs a proper definition (as opposed to the adjoint V * which has a clear and well understood definition).For a fixed orthonormal basis B = {u j } j 1 for H and an A ∈ B(H), we use the notation to denote the matrix representation of A with respect to B, considered as a bounded operator on the sequence space (2.8) For this fixed orthonormal basis B, we also define a conjugation J B on H by (2.9) In other words, J B fixes every basis element u j and extends antilinearly to H.Here is a version of Proposition 2.7 for a general, possibly infinite dimensional, separable Hilbert space.
Proposition 2.10.For an orthonormal basis B = {u j } j 1 for a separable Hilbert space H, the mapping C is a conjugation on H if and only if there is a unitary operator Proof.Suppose C is a conjugation on H. Then V = CJ B is a unitary operator on H (since it is linear, isometric, and onto).Moreover, applying (2.9) and then (2.1), we see that Observe that C is antilinear and isometric.We just need to verify that C 2 = I.Since C is antilinear, note that C 2 = VJ B VJ B is a bounded linear operator on H.Moreover, since J B VJ B is also a bounded linear operator on H, we see that Now observe that for all m, n 1, again making use of (2.1) and (2.9), Thus, from our assumption that [V ] t B = [V ] B , we see that (2.12) and so C 2 = I.Thus, C is a conjugation on H. Remark 2.13.If J is any conjugation on H, one can show there exists an orthonormal basis B for H such that J = J B [18, Lemma 1].Thus, there is some freedom in the above analysis, if so desired, to choose a conjugation J first instead of the orthonormal basis B.
A version of the spectral theorem for unitary operators (see [7, defines a positive finite Borel measure on T, sometimes called an elementary measure.
The following proposition from [24] describes the relationship between a unitary operator, its spectral measure, and a conjugation.

DECOMPOSITIONS OF CONJUGATIONS
As we will see in subsequent sections, the set C s (U ) is quite large and so an important step in understanding it is to decompose each C ∈ C s (U ) into more manageable pieces.This decomposition will involve various types of invariant subspaces.Recall that a (closed) subspace M of a Hilbert space and hyperinvariant if T M ⊆ M for every T ∈ B(H) that commutes with A.
We begin with the following lemma which follows from (2.1) and the fact that C 2 = I.
Below is a useful decomposition theorem for a conjugation.For the rest of this paper, M + (T) will denote the set of all finite positive Borel measures on the unit circle T. For µ, σ ∈ M + (T) use use the standard notation µ ≪ σ for µ is absolutely continuous with respect to σ and µ ⊥ σ for µ is singular with respect to σ.
Theorem 3.2.Let U be a unitary operator on H, E(•) be its associated spectral measure, and for each x ∈ H, let µ x denote the associated elementary measure from (2.15).For any µ ∈ M + (T), let Then we have the following.
(a) H µ is a reducing subspace for U .
Proof.The proof of (a) is routine and we omit it (see [25, §65, §66] for the details).
To prove (b), let x ∈ H µ and Ω be a Borel subset of σ(U ) for which µ(Ω) = 0.By Proposition 2.16, C commutes with E(Ω) and thus, via (2.1), This proves that Cx ∈ H µ .Since C is a conjugation, we see that . This proves (i) and (ii).Part (iii) follows from the facts that H µ is a reducing subspace for U and C ∈ C s (U ).
As a note here, other properties of the reducing subspace H µ were explored by Halmos in [25, §65].Observe that if α ∈ T and δ α denotes the point mass at α, one can see that H δα = ker(U − αI).

Corollary 3.3. For a unitary operator
The following corollary will play an important role later on.

Corollary 3.4. If U is a unitary operator on H with
and Remark 3.5.Apply Theorem 3.2 to the case when µ = m (normalized Lebesgue measure on T) and let H ac := H m and H sing := H ⊥ m .The decomposition H = H ac ⊕ H sing was first explored in [25] and considered further in [3,29,30].One can prove the following: Let U be a unitary operator on H A von Neumann-Wold type decomposition surveyed in [29] yields the following decomposition of a unitary operator U on H by reducing subspaces H ac , L, and α∈T ker(U − αI) such that Here one proves the following: Let U be a unitary operator on H and where each summand is a conjugation on the corresponding space.
Though Theorem 3.2 seems relatively simple, it yields interesting results about hyperinvariant subspaces of unitary operators.For a unitary operator, every hyperinvariant subspace is also reducing (but not the converse).Theorem 3.8 below connects hyperinvariant subspaces with conjugations.The culmination of this discussion will be given in Theorems 8.6 and 8.8.

Theorem 3.7. Let U be a unitary operator on H with spectral measure E(•).
Then every hyperinvariant subspace for U is invariant for every C ∈ C s (U ).Moreover, every hyperinvariant subspaces can be written as E(Ω)H, where Ω is a Borel subset of σ(U ).
Proof.From [33, Proposition 6.9] (see also [8]), every hyperinvariant subspace for U can be written as E(Ω)H for some Borel set We now connect H µ with hyperinvariant subspaces.A first step in this direction is this following result -which might be contained somewhere in the literature but we include a proof for the reader's convenience.We will see an extension of this result in Theorems 8. 6 To complete the proof, we will show that E(Ω)H = H µe .For the ⊆ containment, observe that if x ∈ E(Ω)H and ∆ is a Borel subset of Ω, then (3.9) and thus, since µ e is a scalar valued spectral measure for V = U | E(Ω)H , we see that if µ e (∆) = 0, then E(∆) = 0. Thus, by (3.9), µ x ≪ µ e and so x ∈ H µe .This verifies E(Ω)H ⊆ H µe .
We end this section with a discussion of when where denotes the closed linear span, denote the * -cyclic invariant subspace generated by x.For any Recall [25, §48] the standard Boolean operations ∧ and ∨ for µ 1 , µ 2 ∈ M + (T) defined on Borel subsets Ω of T by Proposition 3.10.Let U be a unitary operator on H and µ be any scalar spectral measure for U .For ν 1 , ν 2 ∈ M + (T) the following are equivalent.
For the proof of (a) =⇒ (b), let us assume that ν 1 ∧ µ is not absolutely continuous with respect to ν 2 ∧ µ.Then there is a nonzero measure ν ∈ M + (T) such that ν ≪ ν 1 ∧ µ and ν ⊥ ν 2 ∧ µ.Let e ∈ H be a vector such that µ e is a scalar spectral measure (for example a separating vector for von Neumann algebra generated by U -from the proof of Theorem 3.8).Since µ and µ e are mutually absolutely continuous as scalar spectral measures for U , it follows that ν ≪ µ e .By [7, Ch.IX, Lemma 8.6] there is a nonzero Corollary 3.12.Let U be a unitary operator on H and µ be any scalar spectral measure for U .For ν 1 , ν 2 ∈ M + (T) the following are equivalent.

A MATRIX DESCRIPTION OF C s (U )
Our first description of C s (U ) will involve matrix a representation of U with respect to an orthonormal basis.Though this might seem a bit difficult to apply at first, we will see, through our decomposition theorem from the previous section, that it can often yield a tangible description of C s (U ).Recall our notation from §2, where, for a fixed orthonormal basis B = {u j } j 1 for a (separable) Hilbert space H and for A ∈ B(H), [A] B denotes the matrix representation of A with respect to B, and J B denotes the conjugation on H for which J B u n = u n for all n 1.
Theorem 4.1.Suppose U is a unitary operator on H. Then C ∈ C s (U ) if and only if there is a unitary operator V on H satisfying Proof.Proposition 2.10 says that C is a conjugation on H if and only if Notice the use of (2.12) above.Thus, which completes the proof.
One can use the above discussion to describe C s (U ) when U is an n × n unitary matrix.
Theorem 4.2.Suppose U is an n × n unitary matrix with spectral decomposition where W is an n × n unitary matrix, ξ 1 , . . ., ξ d ∈ T are the distinct eigenvalues of U , I n j is the n j × n j identity matrix, and where, for each 1 j d, V j is an n j ×n j unitary matrix which satisfies V t j = V j , and J is the conjugation on C n defined by Proof.Suppose that C ∈ C s (U ).For the unitary matrix Corollary 3.4 says that any C ∈ C s ( U ) can be written as where (which is the span of appropriate standard basis vectors for C n ) and Theorem 4.1 says that for each 1 j d there is an n j × n j unitary matrix V j such that such that C j = V j J| E j and V t j = V j .Note that J| E j fixes the appropriate standard basis vectors.Since U | E ξ j = ξ j I n j , condition (b) of Theorem 4.1 is automatic.Thus, every Now apply Lemma 2.3 and Proposition 2.6 to see that C takes the form in (4.3).
Conversely suppose that C is the conjugation from (4.3).Then, due to the fact that each of the unitary matrices V j satisfy V t j = V j and JAJ = A for any matrix A, CU C is equal to Thus C ∈ C c (U ), which completes the proof.
In certain circumstances, we can extend Theorem 4.2 to the infinite dimensional setting.We mention two interesting examples of unitary operators from harmonic analysis.
Example 4.4.Let denote the classical Fourier-Plancherel transform on L 2 (R) (defined initially on L 1 (R) ∩ L 2 (R) by the above integral and extended to be a unitary operator on , where H n is the nth Hermite function, i.e., , and h n is the n-th Hermite polynomial.Moreover, B = {H n } n 0 forms an orthonormal basis for L 2 (R) [16,Ch.11].Thus, using our earlier notation from Corollary 3.4, where we can write F in block operator form with respect to the decomposition in (4.5) as Furthermore, any conjugation C on L 2 (R) for which CFC = F * must take the (block) form where, for each k ∈ {1, −i, −1, i}, J k is the conjugation on E k which fixes every element of B k and V k is a unitary operator on E k for which is the Hilbert transform on L 2 (R).Then σ(H ) = σ p (H ) = {i, −i} and, again using the notation from Corollary 3.4, and E −i has orthonormal basis B −i = {g n } n 0 , where See [16,Ch. 12] for the details.Moreover, by a similar discussion as in the previous example, any conjugation C on L 2 (R) for which CH C = H * must take the (block) form where for each k ∈ {i, −i}, J k is the conjugation on E k which fixes every element of B i and V k is a unitary operator on E k for which This section provides a model for conjugations on vector valued Lebesgue spaces.It will be useful for our description of C s (U ) in §8 and also sets up our discussion of models for bilateral shifts, and their associated conjugations, in the next section.
For a Hilbert space H with norm • H and µ ∈ M + (T), consider the set L 0 (µ, H) of all H-valued µ-measurable functions f on T and the Hilbert space One often sees this using tensor notation as L 2 (µ) ⊗ H. Also consider L ∞ (µ, B(H)), the µ-essentially bounded B(H)-valued functions U on T.
For U ∈ L ∞ (µ, B(H)), define the multiplication operator M U on L 2 (µ, H) by (5.1) for f ∈ L 2 (µ, H) and µ-almost every ξ ∈ T. Clearly M U ∈ B(L 2 (µ, H)).If we use the notation U * (ξ) = U(ξ) * , one can verify that to denote the set of all scalar valued µ-essentially bounded functions on T. For ease of notation, ee will write M ϕ , when ϕ ∈ L ∞ (µ), in place of the more cumbersome M ϕI H , that is, for f ∈ L 2 (µ, H) and for µ-almost every ξ ∈ T. Of particular importance is when ϕ(ξ) = ξ which yields the (bilateral) shift M ξ on L 2 (µ, H).As a specific example, we have the bilateral shift Recall from §2 that AB(H) denotes the space of all bounded antilinear operators on H .By L ∞ (µ, AB(H)) we denote the space of all µ-essentially To prove (b), observe that for any f ∈ L 2 (µ, H) we use the fact that JU(ξ)J = U * (ξ) for µ-almost every ξ ∈ T to see that = (Mξ f)(ξ).

CONJUGATIONS AND BILATERAL SHIFTS
Many interesting, and naturally occurring, unitary operators are bilateral shifts.Examples include the Hilbert and Fourier transforms on L 2 (R); the translation operator ; and the special class of multiplication operators U f = ψf on L 2 (m, T) (where ψ is an inner function) which will be the focus of the next section.The fact that the above operators are bilateral shifts were carefully explained in [29].This section gives an initial description of C s (U ) for this class of operators.Another description will be discussed in the next section.We begin with precisely what we mean by the term "bilateral shift".Definition 6.1.A unitary operator U on H is a bilateral shift if there is a subspace M ⊆ H for which (a) U n M ⊥ M for all n ∈ Z \ {0}; In the above, note that U −1 = U * .The subspace M is called an associated wandering subspace for the bilateral shift U .Of course there is the bilateral shift M ξ on L 2 (m, T) defined by (M ξ f )(ξ) = ξf (ξ) where the wandering subspace M is the space of constant functions.
Though the wandering subspace M in Definition 6.1 is not unique, its dimension is [26].The term "bilateral shift " comes from the fact that since This allows us to define a natural unitary operator (6.2) Moreover, W U W * = M ξ , where M ξ is the bilateral shift from (5.2) defined on L 2 (m, M) by (b) Let U : L 2 (R) → L 2 (R) be the unitary dilation operator defined by x 1, 0 otherwise, then (Haar) wavelet theory [6] says that the functions form an orthonormal basis for L 2 (R).For fixed ℓ ∈ Z, define  , where ψ is any inner function whose degree is N .When the inner function ψ is a finite Blaschke product with N zeros (repeated according to multiplicity), the degree of ψ is defined to be N .In all other cases, the degree of ψ is defined to be N = ∞.In the next section, will give a concrete description of C s (M ψ ) written in terms of operators on L 2 .
For a bilateral shift U on H we wish to describe We will see another path to this result in Example 7.10.
Example 6.7.In a similar way as with the previous example, let B = {v k } k 1 be an orthonormal basis for M.Here N = dim M (which might be infinite).Recall the conjugation J B on M defined by almost every ξ ∈ T. Thus, for JU(ξ)J = U(ξ) * it must be the case that [U(ξ) * ] B = [U(ξ)] B .This means that every conjugation C on L 2 (m, M) for which CM ξ C = Mξ must take the form (Cf)(ξ) = U(ξ)J(f(ξ)), where [U(ξ) * ] B = [U(ξ)] B for almost every ξ ∈ T. Note that for any conjugation J on M, we can find an orthonormal basis B for M for which J = J B [18, Lemma 1] .Thus, in a way, the example above is canonical.
The results from [29,Theorem 3.3] show that for any unitary operator U on H we have H = K ⊕ K ′ , where K and K ′ reducing subspaces for U , where U | K is a bilateral shift and U | K ′ has no a bilateral shift part.For a given conjugation C, even assuming that U is C-symmetric, the conjugation C cannot be nesseserily decomposed according to this decomposition.

Define the conjugation
and observe that CU C = U * .As in [29,Example 5.6], the decomposition , is a desired bilateral shift decomposition, but it is not C invariant (CK K).

UNITARY MULTIPLICATION OPERATORS ON L 2 (m, T)
As mentioned in Example 6.3(c), we have a model for any bilateral shift U on H as the multiplication operator M ψ on L 2 = L 2 (m, T), where ψ is an inner function whose degree is that of the (uniquely defined) dimension of any wandering subspace for U .In this section we give a concrete and tangible description of C s (M ψ ).If J is the conjugation Jf = f on L 2 and C ∈ C s (M ψ ), then A = CJ is a unitary operator on L 2 for which AM ψ = M ψ A. Thus, in order to describe C s (M ψ ), we first need to characterize the bounded operators on L 2 that commute with M ψ .Remark 7.1.For an inner function ψ, a known result (see for example [29, Proposition 5.17]), says that where K ψ = H 2 ⊖ ψH 2 is the model space associated with ψ.In other words, K ψ is a wandering subspace for M ψ (recall Definition 6.1).Let us set up some notation to be used below.Let N = dim K ψ ∈ N ∪ {∞}.Observe that N is finite if and only if ψ is a finite Blaschke product with N zeros, repeated according to multiplicity [15,Prop. 5.19].Also define the space The norm of an element f When N = ∞, we need to assume that the sum defining f is finite.Furthermore, the operator 1 j N M ξ (called the inflation of the bilateral shift M ξ on L 2 ) is given by We also define When N = ∞, note that ℓ 2 N is equal to ℓ 2 + which was the space discussed earlier in (2.8).Finally, observe that (7.3) . Theorem 7.4.For an inner function ψ, let {h j } 1 j N be an orthonormal basis for the model space K ψ .Then we have the following.
(a) Every f ∈ L 2 has the unique decomposition where each f j belongs to L 2 and f = ( (b) The operator Proof.By (7.2), every f ∈ L 2 can be written (uniquely) as Moreover, since ψ j K ψ ⊥ ψ j ′ K ψ for all j = j ′ and {h k } 1 k N forms an orthonormal basis for K ψ , we have Rewrite the expression in (7.6) as Finally, by Parseval's theorem and (7.7), note that This verifies statements (a) and (b).To verify statement (c), note that Similarly, one can verify the second equality in (c).
To prove (d) and (e), recall from [33, Chapter III] that the bounded operators on the space 1 j N L 2 that commute with 1 j N M ξ must take the form of multiplication by the matrix function Φ Putting this all together, we see that if A ∈ B(L 2 ) commutes with M ψ , then W AW * commutes with 1 j N M ξ and so W AW * = M Φ , equivalently A = W * M Φ W .This translates to an operator on L 2 by which completes the proof of (d) and (e).
From our earlier discussion, we know that the standard conjugation Jf = f on L 2 induces the standard conjugation on J on Here is our description of C s (M ψ ) when ψ is inner.
, for all f ∈ L 2 with the decomposition from (7.5).
Proof.Suppose C ∈ C s (M ψ ).Note that C := W CW * defines a conjugation on 1 j N L 2 .Since CM ψ C = M ψ , we can use Theorem 7.4(c) to see that N ) (recall (7.3)), Theorem 6.4 yields (with the choice of conjugation J on ℓ 2 N ) that is unitary valued m-almost everywhere, i.e., ΦΦ * = Φ * Φ = I, and such that C = M Φ J.Moreover, since C 2 = I we see that

It remains to verify the formula in (c). Observe that for all
N ) such that conditions (a), (b), and (c) hold.Theorem 7.4(e) and the argument above shows that C ∈ C s (M ψ ).
Remark 7.9.The description of C s (M ψ ) for an inner function ψ depends on knowing an orthonormal basis for the model space K ψ .There are several "natural" bases one could choose.See [15,Prop. 5.25] for some particular examples when dim K ψ is finite.
Example 7.10.Suppose ψ(z) = z.In this case, M ψ becomes the bilateral shift M ξ on L 2 .Furthermore, K ψ = C (the constant functions) and the expansion of an f ∈ L 2 from Theorem 7.4 becomes the classical Fourier expansion.Finally, Theorem 7.4 says that every C ∈ C s (M ξ ) must take the form (Cf )(ξ) = u(ξ)f (ξ) for some u ∈ L ∞ that is unimodular almost everywhere.We observed this by a different method in Example 6.6.
Example 7.11.When ψ(z) = z 2 above, one can check that the functions h 1 (z) ≡ 1 and h 2 (z) = z form an orthonormal basis for K ψ .Furthermore, using the notation from Theorem 7.4, (7.12) Then, one can check that Theorem 7.8 says that every C ∈ C s (M ξ 2 ) takes the form and observe that the conditions above yield Using the notation from (7.12), this means that every C ∈ C s (M ξ 2 ) must take the form where t = Arg(ξ) ∈ (−π, π] and s(t), α(t), β(t) are any 2π-periodic bounded real-valued measurable functions.As a specific nontrivial example we can have the interesting C ∈ C s (M ξ 2 ) defined by where t = Arg(ξ) (s(t) = sin t, α(t) ≡ 0, β(t) ≡ −π) Another interesting example comes from setting s(t) ≡ s, α The main driver of the results of this section is the following.(c) For each j = ∞, 1, 2, . . .there are C j ∈ L ∞ (µ j , AB(H j )) such that C j (ξ) is a conjugation for µ j almost every ξ ∈ T and (d) For each i = ∞, 1, 2, . . .and for any conjugation is unitary and J i -symmetric for µ i -almost every ξ ∈ T and and, for a conjugation C on H define C = ICI * which, by Lemma 2.3, will be a conjugation on Let J i be any a conjugation on H i .Then, as in (5.3), this induces a conjugation J i on L 2 (µ i , H i ) defined by This says that the operator C J commutes with M ξ .The spectral theorem applied to M ξ also yields the commutant [7, p. 307, Theorem 10.20], namely there are operator valued functions Since C J is unitary (being linear, isometric, and onto), it follows that each M U i is unitary and, consequently, U i (ξ) is unitary for µ i -almost every ξ ∈ T. Therefore, Since each C| L 2 (µ i ,H i ) is a conjugation, Proposition 5.6 says that the operator U i (ξ) is J i -symmetric µ i almost everywhere.Thus, we have verified the implication (a) =⇒ (d).
(c) =⇒ (b): For each i = ∞, 1, 2, . .., it is enough to take C i = A C i and prove that ξ A C i .Indeed, for each f i ∈ L 2 (µ i , H i ) we have As an application of the above, we have the following connection between conjugations and hyperinvariant subspaces.Proof.For any fixed conjugation J on H, (5.4) says that J ∈ C s (M ξ ) and thus K is invariant for J.By Theorem 8.4, K is also invariant for all of the conjugations M u J, where u ∈ L ∞ (µ) is unimodular µ-almost everywhere.
Therefore, K is invariant for every M u = M u JJ, where u ∈ L ∞ (µ) is unimodular.From here, one can argue that K is also invariant for M χ Ω for any Borel set Ω ⊆ σ(U ) (indeed let u = 1 on Ω and −1 on T \ Ω and note that χ Ω = (1 + u)/2) and, consequently, for any M v where v ∈ L ∞ (µ).
Now fix any unitary U 0 on H and let U 0 ≡ U 0 ∈ L ∞ (µ, B(H)) denote the operator-valued constant function.By our discussion in the introduction (see also Proposition 2.5), U 0 is J 0 -symmetric for some conjugation J 0 .Let J 0 be the conjugation on L (µ, H) given by (J 0 f)(ξ) = J 0 (f(ξ)).Therefore, K is invariant for J 0 and for U 0 J 0 and thus for U 0 .Similarly, K is invariant for U * 0 .It follows that K is invariant for the von Neumann algebra containing all constant unitary valued functions in B(H).Finally, K is invariant for the von Neumann algebra generated by {M v : v ∈ L ∞ (µ)} and the constant operator-valued functions from L ∞ (µ, B(H)).From here, one can fashion an argument that K is invariant for every element of L ∞ (µ, B(H)).Since L ∞ (µ, B(H)) forms the class of operators M Φ , Φ ∈ L ∞ (µ, B(H)), that commute with M ξ , K is hyperinvariant for M ξ .
From here, we can state our main connection between conjugations and hyperinvariant subspaces.Theorem 8.6.Let U be a unitary operator on a separable Hilbert space H.If M ⊆ H is an invariant subspace for every C ∈ C s (U ), then M is hyperinvariant for U .
Proof.The notation from Theorem 8.2 proves that the invariance of M for all C ∈ C s (U ) implies that IM = K i , where the subspace K i is invariant for all all M U i J i , where U i ∈ L ∞ (µ i , H i ) is unitary valued µ-almost everywhere.Now apply Proposition 8.5 to see that K i is hyperinvariant for each M (i) ξ .Finally, using the description of the commutant of M ξ [7, p. 307, Theorem IX.10.20], we obtain that IM is hyperinvariant for M ξ .This shows that M is a hyperinvariant subspace for U .
The simple example bellow illustrates why requirement in Theorem 8.6 that M is invariant for every C ∈ C s (U ) is an important one.(and of course extend antilinearity to all of C 2 ).One can check that C 1 , C 2 ∈ C s (U ).Note that each C i separately has more invariant subspaces than the symmetric unitary operator.This conjugation also makes the Hilbert transform a Csymmetric unitary operator.(c) The conjugation (Cf )(t) = f (t) from Example 2.2(b) makes the Fourier-Plancherel transform a C-symmetric unitary operator on L 2 (R).

Proposition 2 . 16 .
For a conjugation C and a unitary operator U on H with associated spectral measure E(•), the following are equivalent.(a) C ∈ C s (U ); (b) CE(Ω)C = E(Ω) for every Borel set Ω ⊆ σ(U ).

Example 6 . 3 .
Examples of bilateral shifts come from a variety of places.These were explored in[29] using a Wold-type decomposition.(a)If U : L 2 (R) → L 2 (R) is the unitary translation operator defined by (U f )(x) = f (x − 1)and one sets M = χ [0,1] L 2 (R), then M satisfies conditions (a) and (b) of Definition 6.1 and hence U is unitarily equivalent to M ξ on L 2 (m, M).
and U W ℓ = W ℓ+1 for all ℓ ∈ Z.This means that the subspace M = W 0 satisfies conditions (a) and (b) of Definition 6.1 and hence U is unitarily equivalent to M ξ on L 2 (m, M).
(c) For an inner function ψ (a bounded analytic function on D whose radial boundary values are unimodular for m-almost every ξ ∈ T), the operator M ψ f = ψf is unitary on L 2 = L 2 (m, T) and the model space M = K ψ = H 2 ∩ (ψH 2 ) ⊥ satisfies the hypothesis of conditions (a) and (b) of Definition 6.1 [29, Proposition 5.17].Hence M ψ is unitarily equivalent to M ξ on L 2 (m, M).Observe that if U is any bilateral shift on H with wandering subspace N with dim N = N ∈ N ∪ {∞}, then the above discussion shows that U is unitarily equivalent to M ψ on L 2

Theorem 6 . 4 .Corollary 6 . 5 . 4 . 6 . 6 .
where W : H → L 2 (m, M) is the unitary operator from (6.2), Lemma 2.3 says that C = W CW * is a conjugation on L 2 (m, M) such that CM ξ C = Mξ.The following result from[5, Thm.4.8]  describes C. For a conjugation C on L 2 (m, M), the following are equivalent.(a)C ∈ C s (M ξ ); (b) There is a C ∈ L ∞ (m, AB(M)) such that C = A C and C(ξ) is a conjugation for almost all ξ ∈ T; (c) For any conjugation J on M there is a U ∈ L ∞ (m, B(M)) such that U(ξ) is a J-symmetric unitary operator for almost all ξ ∈ T and C = M U J.This yields a description of C s (U ) when U is a bilateral shift.Recall the unitary operator W : H → L 2 (m, M) from (6.2) associated with a bilateral shift with wandering subspace M. Let U be a bilateral shift on H with an associated wandering subspace M. For a conjugation C on H, the following are equivalent:(a) C ∈ C s (U ); (b) C = W CW * is a conjugation on L 2 (m, M) that satisfies any of the equivalent conditions of Theorem 6.Example For the bilateral shift M ξ on L 2 = L 2 (m, T), we can examine the set C s (M ξ ).Here M = C (the constant functions) and J : C → C is Jz = z and so J :

Theorem 7 . 8 .
Suppose that ψ is an inner function and {h j } 1 j N is an orthonormal basis forK ψ .Then C ∈ C s (M ψ ) if and only if there is a Φ = [ϕ ij ] 1 i,j N ∈ L ∞ (m, ℓ2N ) such that (a) Φ * Φ = I almost everywhere on T;(b) Φ t = Φ almost everywhere on T; and so Φ = Φ * .Combine this with the previous identity to see that Φ t = Φ.So far we have shown that if C ∈ C s (M ψ ) then C = W * (M Φ J)W where Φ satisfies the conditions of (a) and (b).If Φ satisfies the conditions (a) and (b) then a short argument will show that Φ is unitary valued almost everywhere and the condition (b) will show that JΦJ = Φ * almost everywhere.Proposition 5.6 now says that M Φ J ∈ C s (M ξ ) and hence W * (M Φ J)W ∈ C s (M ψ ).So we have shown that C ∈ C s (M ψ ) if and only if C = W * (M Φ J)W , where Φ satisfies the conditions in (a) and (b).

8 .Theorem 8 . 1 (
CONJUGATIONS VIA THE SPECTRAL THEOREM This section describes C s (U ) using the multiplicity version of the spectral theorem.Important applications of this are Theorem 8.6 and Theorem 8.8 below which connect the invariant subspaces of C ∈ C s (U ) with the hyperinvariant subspaces of U .We begin with the following (multiplicity) version of the spectral theorem for unitary operators from [7, Ch.IX, Theorem 10.20].The reader might need a refresher of the notation from §5.Spectral Theorem).For a unitary operator U on a separable Hilbert space H, there are mutually singular measures µ ∞ , µ 1 , µ 2 , . . .∈ M + (T), along with Hilbert spaces H ∞ , H 1 , H 2 , . . .each with corresponding dim H k = k, k = ∞, 1, 2, 3, . . ., and an isometric isomorphism

Theorem 8 . 2 .
Let U be a unitary operator on a separable Hilbert space H with spectral representation as in Theorem 8.1.For a conjugation C on H, the following are equivalent.

=Theorem 8 . 4 .
(b) =⇒ (a): Recalling the notation from (8.3), note that CU C = I C i M I MξI * = U * .The multiplication operator M ξ on L 2 (µ, H) is a special case of Theorem 8.2 -which we record here for what follows.Let µ ∈ M + (T) and C be a conjugation on L 2 (µ, H).Then following are equivalent.(a) C ∈ C s (M ξ ).(b) There are C ∈ L ∞ (µ, AB(H)) such that C(ξ) is a conjugation for µ almost every ξ ∈ T and C = A C .(c)For any conjugation J on H there is a U ∈ L ∞ (µ, B(H)) such that U(ξ) is unitary and J-symmetric for µ almost every ξ ∈ T and C = M U J.