Conjugations in $L^2$ and their invariants

Abstract

Conjugations in space $L^2$ of the unit circle commuting with multiplication by z or intertwining multiplications by z and ${{\bar{z}}}$ are characterized. We also study their behaviour with respect to the Hardy space, subspaces invariant for the unilateral shift and model spaces.

1 Introduction

Let ${\mathcal {H}}$ be a complex Hilbert space and denote by $B({{\mathcal {H}}})$ the algebra of all bounded linear operators on ${\mathcal {H}}$. A conjugationC in ${\mathcal {H}}$ is an antilinear isometric involution, i.e., $C^2=id_{{\mathcal {H}}}$ and

$$\begin{aligned} \langle Cg,Ch \rangle =\langle h,g\rangle \quad \text { for } g,h\in {\mathcal {H}}. \end{aligned}$$

(1.1)

Conjugations have recently been intensively studied and the roots of this subject comes from physics. An operator $A\in B({{\mathcal {H}}})$ is called C–symmetric if $CAC=A^*$ (or equivalently $AC=CA^*$). A strong motivation to study conjugations comes from the study of complex symmetric operators, i.e., those operators that are C–symmetric with respect to some conjugation C. For references see for instance [2, 3, 7,8,9,10]. Hence obtaining the full description of conjugations with certain properties is of great interest.

Let ${\mathbb {T}}$ denote the unit circle, and let m be the normalized Lebesgue measure on ${\mathbb {T}}$. Consider the spaces $L^2=L^2(\mathbb {T,}m)$, $L^{\infty }=L^\infty ({\mathbb {T}},m)$, the classical Hardy space $H^2$ on the unit disc ${\mathbb {D}}$ identified with a subspace of $L^2$, and the Hardy space $H^\infty$ of all analytic and bounded functions in ${\mathbb {D}}$ identified with a subspace of $L^\infty$. Denote by $M_\varphi$ the operator defined on $L^2$ of multiplication by a function $\varphi \in L^\infty$.

The most natural conjugation in $L^2$ is J defined by $Jf={{\bar{f}}}$, for $f\in L^2$. This conjugation has two natural properties: the operator $M_z$ is J–symmetric, i.e., $M_z J=J M_{{{\bar{z}}}}$, and J maps an analytic function into a co-analytic one, i.e., $J H^2=\overline{H^2}$.

Another natural conjugation in $L^2$ is $J^{\star }f=f^{\#}$ with $f^{\#}(z)\overset{df}{=}\overline{ f({{\bar{z}}})}.$ The conjugation $J^{\star }$ has a different behaviour: it commutes with multiplication by z ($M_z J^\star =J^\star M_{ z}$) and leaves analytic functions invariant, $J^\star H^2\subset {H^2}$. The map $J^{\star }$ appears for example in connection with Hankel operators (see [11, pp. 146–147]). Its connection with model spaces was studied in [4] (Lemma 4.4, see also [1, p. 37]). Hence a natural question is to characterize conjugations with respect to these properties. The first step was done in [2] where all conjugations in $L^2$ with respect to which the operator $M_z$ is C–symmetric were characterized, see Theorem 2.2. In Sect. 2 we give a characterization of all conjugations which commute with $M_z$, Theorem 2.4. In Sect. 3, using the above characterizations we show that there are no conjugations in $L^2$ leaving $H^2$ invariant, with respect to which the operator $M_z$ is C–symmetric. We also show that $J^{\star }$ is the only conjugation commuting with $M_z$ and leaving $H^2$ invariant.

Beurling’s theorem makes subspaces of $H^2$ of the type $\theta H^2$ ($\theta$ inner function, i.e., $\theta \in H^\infty$, $|\theta |=1$ a.e. on ${\mathbb {T}}$) exceptionally interesting, as the only invariant subspaces for the unilateral shift S, $Sf(z)=zf(z)$ for $f\in H^2$. On the other hand, model spaces (subspaces of the type $K_\theta =H^2\ominus \theta H^2$), which are invariant for the adjoint of the unilateral shift, are important in model theory, [12].

In [2] all conjugations C with respect to which the operator $M_z$ is C–symmetric and mapping a model space $K_\alpha$ into another model space $K_\theta$ were characterized, with the assumption that $\alpha$ divides $\theta$ ($\alpha \leqslant \theta$). Recall that for $\alpha$ and $\theta$ inner, $\alpha \leqslant \theta$ means that $\theta /\alpha$ is also an inner function. In what follows we will show that the result holds without the assumption $\alpha \leqslant \theta$.

In Sect. 4 conjugations commuting with $M_z$ and preserving model spaces are described. Section 5 is devoted to conjugations between S–invariant subspaces (i.e., subspaces of the form $\theta H^2$ with $\theta$ an inner function). In the last section we deal with conjugations commuting with the truncated shift $A^\theta _z$ ($A^\theta _z=P_{\theta }M_{z|K_{\theta }}$ where $P_{\theta }$ is the orthogonal projection from $L^2$ onto $K_{\theta }$) or conjugations such that $A^\theta _z$ is C–symmetric with respect to them.

2 $M_z$ and $M_z$–commuting conjugations in $L^2$

Denote by J the conjugation in $L^2$ defined as $Jf={{\bar{f}}}$, for $f\in L^2$. This conjugation has the following obvious properties:

Proposition 2.1

• (1) $M_z J=J M_{{{\bar{z}}}}$;

• (2) $M_\varphi J=JM_{{{\bar{\varphi }}}}$ for all $\varphi \in L^\infty$;

• (3) $J H^2=\overline{H^2}$.

Let us consider all conjugations C in $L^2$ satisfying the condition

$$\begin{aligned} M_z C=C M_{{{\bar{z}}}}. \end{aligned}$$

(2.1)

Such conjugations were studied in [2] and are called $M_z$–conjugations. The following theorem characterizes all $M_z$–conjugations in $L^2$.

Theorem 2.2

[2] Let C be a conjugation in $L^2$. Then the following are equivalent:

• (1) $M_z C=C M_{{{\bar{z}}}}$,

• (2) $M_\varphi C=C M_{{{\bar{\varphi }}}}$ for all $\varphi \in L^\infty$,

• (3) there is $\psi \in L^\infty$, with $|\psi |=1$, such that $C=M_\psi J$,

• (4) there is $\psi ^\prime \in L^\infty$, with $|\psi ^\prime |=1$, such that $C=JM_{\psi ^\prime }$.

Another natural conjugation in $L^2$ is defined as

$$\begin{aligned} J^{\star }f=f^{\#}\ \text { with }\ f^{\#}(z)\overset{df}{=}\overline{ f({{\bar{z}}})}. \end{aligned}$$

(2.2)

The basic properties of $J^{\star }$ are the following:

Proposition 2.3

 

• (1) $M_z J^{\star }=J^{\star } M_{ z}$;

• (2) $M_{{{\bar{z}}}} J^{\star }=J^{\star } M_{{{\bar{z}}}}$;

• (3) $M_\varphi J^{\star }=J^{\star }M_{\varphi ^{\#}}$ for all $\varphi \in L^\infty$;

• (4) $J^\star H^2=H^2$.

In the context of the Proposition 2.3 it seems natural to consider all conjugations C in $L^2$ commuting with $M_z$, i.e.,

$$\begin{aligned} M_z C=C M_{ z}. \end{aligned}$$

(2.3)

Such conjugations will be called $M_z$–commuting. In what follows we will often deal with functions $f\in L^2$ such that $f(z)=f({{\bar{z}}})$ a.e. on ${\mathbb {T}}$, which will be called symmetric. Observe that if f is symmetric and $f\in H^2$, then we also have $f\in \overline{H^2}$ and so it is a constant function.

Theorem 2.4

Let C be a conjugation in $L^2$. Then the following are equivalent:

• (1) $M_z C=C M_{z}$ ,

• (2) $M_\varphi C=C M_{\varphi ^{\#}}$ for all $\varphi \in L^\infty$,

• (3) there is a symmetric unimodular function $\psi \in L^\infty$ such that $C=M_\psi J^{\star }$,

• (4) there is a symmetric unimodular function $\psi ^\prime \in L^\infty$ such that $C=J^{\star }\,M_{\psi ^\prime }$.

Proof

We will show that $(1)\!\Rightarrow \!(3)$. The other implications are straightforward. Assume that $M_{ z}C =CM_z$. Then $M_z CJ^{\star }=CM_{ z}J^{\star }=CJ^{\star }M_z$. It follows that the linear operator $CJ^{\star }$ commutes with $M_z$. By [13, Theorem 3.2], there is $\psi \in L^\infty$ such that $CJ^{\star }=M_{\psi }$. Hence $C=M_{\psi }J^{\star }=J^{\star }M_{\psi ^{\#}}$.

By (1.1) for any $f,g\in L^2$ we have

$$\begin{aligned} \int _{\mathbb {T}} g\bar{f} dm(z)&=\langle g, f\rangle = \langle Cf,Cg\rangle \\&=\langle \psi f^{\#},\psi g^{\#}\rangle =\int _{\mathbb {T}}|\psi (z)|^2\,\overline{f({{\bar{z}}})}g(\bar{z})\,dm(z)\\&=\int _{\mathbb {T}}|\psi ({{\bar{z}}})|^2\,\overline{f( z)}g( z)\,dm(z). \end{aligned}$$

Hence $|\psi |=1$ a.e. on ${\mathbb {T}}$. On the other hand, since $C^2=I_{L^2}$, for all $f\in L^2$ we have

$$\begin{aligned} f=C^2 f=M_\psi J^{\star } M_\psi J^{\star } f=M_\psi J^{\star }({\psi f^{\#}})=\psi \psi ^{\#} f, \end{aligned}$$

which implies that $\psi \psi ^{\#}=1$ a.e. on ${\mathbb {T}}$. Therefore $\psi$ is symmetric, i.e., $\psi (z)=\psi ({{\bar{z}}})$ a.e. on ${\mathbb {T}}$. $\square$

3 Conjugations preserving $H^2$

In the previous section all $M_z$–conjugations C in $L^2$, i.e., such that

$$\begin{aligned} M_{ z} C=C M_{{{\bar{z}}}} \end{aligned}$$

or $M_z$–commuting conjugations, i.e., such that

$$\begin{aligned} M_{z} C=C M_{ z} \end{aligned}$$

were characterized. Let us now consider the question which of them preserve $H^2$. Clearly, if C is a conjugation in $L^2$ and $C(H^2)\subset H^2$, then $C(H^2)= H^2$. Since $J^\star$ preserves $H^2$, it can be considered as a conjugation in $H^2$. The following result shows that $J^\star$ is in that sense unique.

Corollary 3.1

Let C be an $M_z$–commuting conjugation in $L^2$. If $C(H^2)\subset H^2$, then $C=\lambda J^{\star }$ for some $\lambda \in {\mathbb {T}}$.

Proof

By Theorem 2.4 we have that $C=M_{\psi }J^{\star }$ for some $\psi \in L^{\infty }$ with $|\psi |=1$ and $\psi (z)=\psi ({\overline{z}})$ a.e. on ${\mathbb {T}}$. Since C preserves $H^2$, we have

$$\begin{aligned}\psi =M_{\psi }J^{\star }(1)=C(1)\in L^{\infty }\cap H^2=H^{\infty }.\end{aligned}$$

Thus $\psi$ is analytic. Since it is symmetric, it is also co-analytic. Hence $\psi$ must be a constant function, so $\psi =\lambda \in {\mathbb {C}}$ and $|\lambda |=|\psi |=1$. $\square$

Corollary 3.2

There are no $M_z$–conjugations in $L^2$ which preserve $H^2$.

Proof

If C is an $M_z$–conjugation in $L^2$, then by Theorem 2.2 it follows that $C=M_{\psi }J$ for some $\psi \in L^{\infty }$ with $|\psi |=1$. As in the proof of Corollary 3.1 the assumption $C(H^2)\subset H^2$ implies that $\psi \in H^{\infty }$, which in turn means that $\psi$ is an inner function. Moreover, for $n=0,1,2,\dots$ we have

$$\begin{aligned} 0=\langle C z^{n+1},{\overline{z}}\rangle = \langle \psi {\overline{z}}^{n+1},{\overline{z}}\rangle =\langle \psi ,z^n\rangle =0. \end{aligned}$$

So $\psi =0$ which is a contradiction. $\square$

The following example shows that not all conjugations in $L^2$ satisfy either (2.1) or (2.3).

Example 3.3

There is a set of naturally defined conjugations. For $k,l\in {\mathbb {Z}}$, $k<l$, define $C_{k,l}\ :\ L^2\rightarrow L^2$ by

$$\begin{aligned} C_{k,l}\Big (\sum _{n\in {\mathbb {Z}}}a_n z^n\Big )={\overline{a}}_l z^k+{\overline{a}}_k z^l+\sum _{n\notin \{k,l\}}{\overline{a}}_n z^n, \end{aligned}$$

(3.1)

where $\{z^n\}$ is the standard basis in $L^2$. Then (2.1) and (2.3) are not satisfied since

$$\begin{aligned} M_{ z} C_{k,l}(z^k)=M_z(z^l)=z^{l+1},\quad C_{k,l} M_{\bar{z}}(z^k)=C_{k,l} (z^{k-1})=z^{k-1} \end{aligned}$$

and

$$\begin{aligned} C_{k,l} M_{z}(z^k)=C_{k,l} (z^{k+1})={\left\{ \begin{array}{ll}z^{k+1}&{}\text {if}\ k+1\ne l,\\ z^k&{}\text {if}\ k+1=l.\end{array}\right. } \end{aligned}$$

Note that, on the other hand, $C_{k,l}$ preserves $H^2$ whenever $k\geqslant 0$ or $l<0$.

4 Conjugations preserving model spaces

There is another class of conjugations in $L^2$ which appear naturally in connection with model spaces. For a nonconstant inner function $\theta$, denote by $K_\theta$ the so called model space of the form $H^2\ominus \theta H^2$. The conjugation $C_\theta$ defined in $L^2$ by

$$\begin{aligned} C_\theta f=\theta {{\bar{z}}}{{\bar{f}}} \end{aligned}$$

has the important property that it preserves the model space $K_\theta$, i.e., $C_\theta K_\theta =K_\theta$. Thus $C_\theta$ can be considered as a conjugation in $K_\theta$. Such conjugations are important in connection with truncated Toeplitz operators (see for instance [6]). Here we present several simple properties of such conjugations, which we will use later.

Proposition 4.1

Let $\alpha ,\beta ,\gamma$ be nonconstant inner functions. Then

• (1) $C_\beta C_\alpha =M_{\beta {{\bar{\alpha }}}}$,

• (2) $M_{\gamma } C_\alpha M_{{{\bar{\gamma }}}}$ is a conjugation in $L^2$,

• (3) $C_\beta \, M_\gamma = M_{{{\bar{\gamma }}}}\,C_\beta$.

Now we will consider relations between $M_z$–conjugations and model spaces. The theorem below was proved in [2, Theorem 4.2] with the additional assumption that $\alpha \leqslant \theta$. As we prove here, this assumption is not necessary.

Theorem 4.2

Let $\alpha , \gamma , \theta$ be inner functions ($\alpha ,\theta$ nonconstant). Let C be a conjugation in $L^2$ such that $M_z C=C M_{{{\bar{z}}}}$. Assume that $C(\gamma K_\alpha )\subset K_\theta$. Then there is an inner function $\beta$ such that $C=C_\beta$, with $\gamma \alpha \leqslant \beta \leqslant \gamma \theta$ and $\alpha \leqslant \theta$.

Proof

Recall the standard notation for the reproducing kernel function at 0 in $K_\alpha$, namely, $k_0^\alpha =1-\overline{\alpha (0)}{\alpha }$ and its conjugate ${{\tilde{k}}}_0^\alpha =C_\alpha k_0^\alpha =\bar{z}(\alpha -\alpha (0))$. By Theorem 2.2 we know that $C=M_\psi J$ for some function $\psi \in L^\infty$, $|\psi |=1$. Hence

$$\begin{aligned} K_\theta \ni C (\gamma {{\tilde{k}}}_0^\alpha )= M_\psi J (\gamma \tilde{k}_0^\alpha )=\psi \overline{\gamma \bar{z}(\alpha -\alpha (0))}={{\bar{\gamma }}}{{\bar{\alpha }}} z\psi (1-\overline{\alpha (0)}\alpha ). \end{aligned}$$

Thus there is $h\in K_\theta$ such that $h={{\bar{\gamma }}}{{\bar{\alpha }}} z\psi (1-\overline{\alpha (0)}\alpha )$. Since $(1-\overline{\alpha (0)}\alpha )^{-1}$ is a bounded analytic function, we have

$$\begin{aligned} {{\bar{\gamma }}}{{\bar{\alpha }}} z\psi =h(1-\overline{\alpha (0)}\alpha )^{-1}\in H^2. \end{aligned}$$

Since ${{\bar{\gamma }}}{{\bar{\alpha }}} z\psi \in H^2$ and $|\bar{\gamma }{{\bar{\alpha }}} z\psi |=1$ a.e. on ${\mathbb {T}}$, it has to be an inner function. Moreover $\beta =z\psi$ has to be inner and divisible by $\gamma \alpha$, i.e., $\gamma \alpha \leqslant \beta$.

On the other hand, we have similarly

$$\begin{aligned} K_\theta \ni C_\theta C (\gamma k_0^\alpha ) =C_\theta (\psi \overline{\gamma (1-\overline{\alpha (0)}\alpha )}=\theta \gamma \bar{z}{{\bar{\psi }}} (1-\overline{\alpha (0)}\alpha ), \end{aligned}$$

and $\theta \gamma {{\bar{\beta }}}={\theta }{\gamma } {{\bar{z}}}{{\bar{\psi }}}\in H^2$. Hence $\beta$ divides ${\theta }{\gamma }$, i.e., $\beta \leqslant \gamma \theta$. It is clear that $C=C_\beta$. Finally, we have $\alpha \leqslant \theta$ as a consequence of $\gamma \alpha \leqslant \beta \leqslant \gamma \theta$. $\square$

Note that if $\alpha \leqslant \theta$ and $C=C_\beta$ for some inner $\beta$ with $\gamma \alpha \leqslant \beta \leqslant \gamma \theta$, then $K_{\alpha }\subset K_{\tfrac{\beta }{\gamma }}\subset K_{\theta }$, $C_{\beta }M_{\gamma }=C_{\tfrac{\beta }{\gamma }}$ and

$$\begin{aligned} C(\gamma K_{\alpha })=C_{\beta }M_{\gamma }(K_{\alpha })=C_{\tfrac{\beta }{\gamma }}(K_{\alpha }) \subset K_{\tfrac{\beta }{\gamma }}\subset K_{\theta }. \end{aligned}$$

Hence the implication in Theorem 4.2 is actually an equivalence.

The corollary bellow strengthens [2, Proposition 4.5].

Corollary 4.3

Let $\alpha , \theta$ be nonconstant inner functions, and let C be a conjugation in $L^2$ such that $M_z C=C M_{{{\bar{z}}}}$. Assume that $C(K_\alpha )\subset K_\theta$. Then $\alpha \leqslant \theta$ and there is an inner function $\beta$ such that $C=C_\beta$, with $\alpha \leqslant \beta \leqslant \theta$.

Let us turn to discussing the relations between $M_z$–commuting conjugations and model spaces. The following proposition describes some more properties of $J^\star$.

Proposition 4.4

Let $\alpha$ be an inner function. Then

• (1) $J^\star (\alpha H^2)=\alpha ^{\#}H^2$;

• (2) $J^\star (K_\alpha )=K_{\alpha ^{\#}}$;

• (3) $J^{\star }C_\alpha =C_{\alpha ^{\#}}J^{\star }$.

Proof

The condition (1) is clear, (2) and (3) were proved in [4, Lemma 4.4]. $\square$

Hence the conjugation $J^{\star }$ has a nice behaviour in connection with model spaces, namely $J^{\star }(K_\alpha )=K_{\alpha ^{\#}}$. Theorem 4.6 below says that the conjugation $J^{\star }$ is, in some sense the only $M_z$–commuting conjugation with this property.

We start with the following:

Proposition 4.5

Let $\alpha , \gamma , \theta$ be inner functions ($\alpha ,\theta$ nonconstant). Let C be an $M_z$–commuting conjugation in $L^2$. Assume that $C(\gamma K_\alpha )\subset K_\theta$. Then $\alpha \leqslant \theta ^{\#}$ and there is an inner function $\beta$ with $\gamma \alpha \leqslant \beta \leqslant \gamma \theta ^{\#}$ such that $C=J^{\star } M_{\frac{\beta }{\gamma \alpha } {{\bar{\gamma }}}}$.

Proof

Observe that since C is an $M_z$–commuting conjugation, taking antilinear adjoints and applying [2, Proposition 2.1] we get $M_{{{\bar{z}}}} C=C M_{{{\bar{z}}}}$. Since by Proposition 4.1 the antilinear operator ${M_{\gamma }} C_\alpha {M_{{\overline{\gamma }}}}$ is a conjugation, then $J^{\star }C{M_{\gamma }} C_\alpha {M_{{\overline{\gamma }}}}$ is also a conjugation. Note also that ${M_{{\gamma }}} C_\alpha {M_{{\overline{\gamma }}}}M_z = M_{\bar{z}}{M_{{\gamma }}} C_\alpha {M_{{\overline{\gamma }}}}$. Hence

$$\begin{aligned} J^{\star }C {M_{{\gamma }}} C_\alpha {M_{{\overline{\gamma }}}} M_z= M_{{{\bar{z}}}}J^{\star }C{M_{{\gamma }}}C_\alpha {M_{{\overline{\gamma }}}}. \end{aligned}$$

(4.1)

On the other hand,

$$\begin{aligned} J^{\star }C{M_{{\gamma }}} C_\alpha {M_{{\overline{\gamma }}}}(\gamma K_\alpha )&\subset J^{\star }C{M_{{\gamma }}} C_\alpha (K_\alpha )\\&\subset J^{\star }C(\gamma K_\alpha )\subset J^{\star }(K_{\theta })\subset K_{\theta ^{\#}}. \end{aligned}$$

By Theorem 4.2 there is an inner function $\beta$ such that $J^{\star }C {M_{{\gamma }}}C_\alpha {M_{{\overline{\gamma }}}}=C_\beta$, with $\gamma \alpha \leqslant \beta \leqslant \gamma \theta ^{\#}$ and $\alpha \leqslant \theta ^{\#}$. Hence $C=J^{\star } C_\beta {M_{{\gamma }}} C_\alpha {M_{{\overline{\gamma }}}}$. Therefore

$$\begin{aligned} C=J^{\star } M_{\frac{\beta }{\gamma \alpha } {{\bar{\gamma }}}} \end{aligned}$$

$\square$

As in Theorem 4.2 the implication in Proposition 4.5 can be reversed. Indeed, if $\alpha \leqslant \theta ^{\#}$ and $C=J^{\star } M_{\frac{\beta }{\gamma \alpha } {\overline{\gamma }}}$ for some inner function $\beta$ with $\gamma \alpha \leqslant \beta \leqslant \gamma \theta ^{\#}$, then $K_{\alpha }\subset K_{\tfrac{\beta }{\gamma }}\subset K_{\theta ^{\#}}$ and

$$\begin{aligned} C(\gamma K_{\alpha })=J^{\star } M_{\frac{\beta }{\gamma } {\overline{\alpha }}}(K_{\alpha })=J^{\star } C_{\frac{\beta }{\gamma }}C_{\alpha }(K_{\alpha })\subset J^{\star }(K_{\tfrac{\beta }{\gamma }})\subset J^{\star }( K_{\theta ^{\#}})= K_{\theta }. \end{aligned}$$

Theorem 4.6

Let $\alpha , \theta$ be nonconstant inner functions, and let C be an $M_z$–commuting conjugation in $L^2$, i.e., $M_z C=C M_{ z}$. Assume that $C(K_\alpha )\subset K_\theta$. Then $\alpha \leqslant \theta ^{\#}$ and $C=\lambda J^{\star }$ with $\lambda \in {\mathbb {T}}$.

Corollary 4.7

Let C be an $M_z$–commuting conjugation in $L^2$. Assume that there is some nonconstant inner function $\theta$ such that $C(K_\theta )\subset K_{\theta ^{\#}}$. Then $C=\lambda J^{\star }$ with $\lambda \in {\mathbb {T}}$.

Proof of Theorem 4.6

By Proposition 4.5 there is an inner function $\beta$ with $\alpha \leqslant \beta \leqslant \theta ^{\#}$ such that $C=J^{\star } M_{\frac{\beta }{ \alpha } }$. The function $\frac{\beta }{ \alpha }$ is inner and by Theorem 2.4 it is symmetric. As observed before it follows that it is constant. Hence $C=J^{\star }$ up to multiplication by a constant of modulus 1. $\square$

5 Conjugations preserving S-invariant subspaces of $H^2$

Beurling’s theorem says that all invariant subspaces for the unilateral shift S are of the form $\theta H^2$ with $\theta$ inner. We will now investigate conjugations in $L^2$ which preserve subspaces of this form. Since $C_{\theta }$ transforms $\theta H^2$ onto $\overline{z H^2}$, the operator

$$\begin{aligned} C_{\theta }J^{\star }C_{\theta }=M_{\theta }J^{\star }M_{{\overline{\theta }}} \end{aligned}$$

is an example of such a conjugation. Note that

$$\begin{aligned} (C_{\theta }J^{\star }C_{\theta })M_z=M_z(C_{\theta }J^{\star }C_{\theta }). \end{aligned}$$

Let $\alpha$, $\theta$ be two inner functions. Then the operator $C_{\theta }J^{\star }C_{\alpha }\ :\ L^2\rightarrow L^2$ is an antilinear isometry which maps $\alpha H^2$ onto $\theta H^2$ and commutes with $M_z$. This operator however does not have to be an involution.

Lemma 5.1

Let $\alpha$, $\theta$ be two inner functions. The operator $C_{\theta }J^{\star }C_{\alpha }$ is an involution (and hence a conjugation in $L^2$) if and only if the function $\theta \overline{\alpha ^{\#}}$ is symmetric (or equivalently $\alpha \alpha ^{\#}=\theta \theta ^{\#}$).

Proof

Note that by Proposition 4.1,

$$\begin{aligned} (C_{\theta }J^{\star }C_{\alpha })(C_{\theta }J^{\star }C_{\alpha })=C_{\theta }C_{\alpha ^{\#}}C_{\theta ^{\#}}C_{\alpha }=M_{\theta \overline{\alpha ^{\#}}}M_{\theta ^{\#}{\overline{\alpha }}}=M_{\theta \overline{\alpha ^{\#}}\theta ^{\#}{\overline{\alpha }}}. \end{aligned}$$

Therefore $C_{\theta }J^{\star }C_{\alpha }$ is an involution if and only if

$$\begin{aligned} \theta \overline{\alpha ^{\#}}\theta ^{\#}{\overline{\alpha }}=1\quad \text {a.e. on }{\mathbb {T}}, \text { i.e., } \theta \theta ^{\#}=\alpha \alpha ^{\#}, \end{aligned}$$

which means that

$$\begin{aligned} (\theta \overline{\alpha ^{\#}})(z) =(\overline{\theta ^{\#}}\alpha )(z)=\theta ({\overline{z}})\overline{\alpha ^{\#}(\bar{z})}=(\theta \overline{\alpha ^{\#}})({\overline{z}})\quad \text {a.e. on }{\mathbb {T}}. \end{aligned}$$

$\square$

The theorem bellow characterizes all $M_z$–commuting conjugations mapping one S–invariant subspace into another S–invariant subspace.

Theorem 5.2

Let $\theta$ and $\alpha$ be two inner functions and let C be a conjugation in $L^2$ such that $CM_z=M_zC$. Then $C(\alpha H^2)\subset \theta H^2$ if and only if $\theta \theta ^{\#}\leqslant \alpha \alpha ^{\#}$ and $C=C_{\beta }J^{\star }C_{\alpha }$, where $\beta$ is an inner function such that $\theta \leqslant \beta$, $\beta \beta ^{\#}=\alpha {\alpha ^{\#}}$. Moreover, in that case $C(\alpha H^2)=\beta H^2$.

Let $\alpha$ be a fixed inner function. By Lemma 5.1, for each inner function $\beta$ with $\beta \beta ^{\#}=\alpha \alpha ^{\#}$ there exists an $M_z$–commuting conjugation C which maps $\alpha H^2$ onto $\beta H^2$, namely $C=C_{\beta }J^{\star }C_{\alpha }$. On the other hand, if $\beta$ is an inner function and there exists an $M_z$–commuting conjugation C which maps $\alpha H^2$ onto $\beta H^2$, then by Theorem 5.2, $\beta \beta ^{\#}\leqslant \alpha {\alpha ^{\#}}$ and $C=C_{\gamma }J^{\star }C_{\alpha }$ for some inner function $\gamma$ such that $\beta \leqslant \gamma$, $\gamma \gamma ^{\#}=\alpha \alpha ^{\#}$. In particular, $C(\alpha H^2)=\gamma H^2=\beta H^2$ and so $\gamma$ is a constant multiple of $\beta$, $\beta \beta ^{\#}=\alpha \alpha ^{\#}$.

It follows from the above that Lemma 5.3 characterizes all possible spaces of type $\beta H^2$ such that for a given S–invariant subspace $\alpha H^2$ there is an $M_z$–commuting conjugation mapping $\alpha H^2$ onto $\beta H^2$.

Lemma 5.3

Let $\alpha$ be a nonconstant inner function. Then

$$\begin{aligned}&\{\beta : \beta \text { is inner, }\alpha \alpha ^{\#}= \beta \beta ^{\#} \}\nonumber \\&\quad = \{\lambda \, uv^{\#}: u, v \text { are inner, } \alpha =uv, \lambda \in {\mathbb {T}}\}. \end{aligned}$$

(5.1)

For two inner functions $\alpha$ and $\beta$ denote by $\alpha \wedge \beta$ the greatest common divisor of $\alpha$ and $\beta$. We will write $\alpha \wedge \beta =1$ if the only common divisor of $\alpha$ and $\beta$ is a constant function.

Proof

Note that for $\alpha =uv$ and $\beta =\lambda uv^{\#}$ we have $\alpha \alpha ^{\#}=\beta \beta ^{\#}$, hence one inclusion is proved. For the other inclusion let $u=\alpha \wedge \beta$ and we can write $\alpha =uv$ and $\beta =u v_1$. From the condition $\alpha \alpha ^{\#}=\beta \beta ^{\#}$ it follows that

$$\begin{aligned} uvu^{\#}v^{\#}=uv_1u^{\#}v_1^{\#}. \end{aligned}$$

Hence $vv^{\#}=v_1v_1^{\#}$. Since $v\wedge v_1=1$, we have that v divides $v_1^{\#}$ and $v_1$ divides $v^{\#}$ and vice–versa. Thus we can take $v_1=\lambda v^{\#}$ with $\lambda \in {\mathbb {T}}$, and so $\beta = \lambda uv^{\#}$. $\square$

Proof of Theorem 5.2

Assume firstly that $CM_z=M_zC$ and $C(\alpha H^2)\subset \theta H^2$. By Theorem 2.4, $C=M_{\psi }J^{\star }$ for some unimodular symmetric function $\psi \in L^{\infty }$. In particular,

$$\begin{aligned} \psi \alpha ^{\#}=M_{\psi }J^{\star }(\alpha )=C(\alpha )\in \theta H^2, \end{aligned}$$

and there exists $u\in H^2$ such that $\psi \alpha ^{\#}=\theta u$. Note that u must be inner and so $\psi =\beta \overline{\alpha ^{\#}}$ with $\beta =\theta u$, $\theta \leqslant \beta$. Clearly $\beta \overline{\alpha ^{\#}}$ is symmetric, i.e., $\beta \beta ^{\#}=\alpha \alpha ^{\#}$. Hence $\theta \theta ^{\#}\leqslant \alpha \alpha ^{\#}$.

Assume now that $\theta \theta ^{\#}\leqslant \alpha \alpha ^{\#}$, and let $\alpha = \alpha _1\cdot (\alpha \wedge \theta )$ and $\theta =\theta _1\cdot (\alpha \wedge \theta )$. Since $(\alpha \wedge \theta )^{\#}=\alpha ^{\#}\wedge \theta ^{\#}$, we get $\alpha ^{\#}= \alpha _1^{\#}\cdot (\alpha ^{\#}\wedge \theta ^{\#})$ and $\theta ^{\#}=\theta _1^{\#}\cdot (\alpha ^{\#}\wedge \theta ^{\#})$. Note also that $\theta _1\theta _1^{\#}\leqslant \alpha _1\alpha _1^{\#}$ and $\theta _1\wedge \alpha _1=1$, so $\theta _1\leqslant \alpha _1^{\#}$. Thus

$$\begin{aligned} \frac{\alpha \alpha ^{\#}}{\theta \theta ^{\#}}=\frac{\alpha _1\alpha _1^{\#}}{\theta _1 \theta _1^{\#}}=\frac{\alpha _1^{\#}}{\theta _1}\frac{\alpha _1}{\theta _1^{\#}}=u u^{\#}, \end{aligned}$$

where $u=\frac{\alpha _1^{\#}}{\theta _1}$ is an inner function. Now we may take $\beta =\theta u$. Since $\theta \leqslant \beta$ and $\beta \beta ^{\#}=\alpha {\alpha ^{\#}}$, by Lemma 5.1 and by Proposition 4.1, $C=M_{\beta \overline{\alpha ^{\#}}}J^{\star }=C_{\beta }J^{\star }C_{\alpha }$ is a conjugation which maps $\alpha H^2$ onto $\beta H^2\subset \theta H^2$. $\square$

Corollary 5.4

Let $\theta$ be an inner function and let C be an $M_z$–commuting conjugation in $L^2$. Then

• (1) $C(\theta H^2)\subset \theta H^2$ if and only if $C=\lambda C_{\theta }J^{\star }C_{\theta }$ with $\lambda \in {\mathbb {T}}$;

• (2) $C(\theta H^2)\subset \theta ^{\#} H^2$ if and only if $C=\lambda J^{\star }$ with $\lambda \in {\mathbb {T}}$.

Proof

By Theorem 5.2, $C(\theta H^2)\subset \theta H^2$ if and only if there exists an inner function $\beta$ such that $\theta \leqslant \beta$ and $\beta \beta ^{\#}=\theta {\theta ^{\#}}$. This is only possible if $\beta$ is constant multiple of $\theta$ and (1) is proved. The proof of (2) is similar. $\square$

Note that by Theorem 5.2 (Lemma 5.1, actually) if $\theta \overline{\alpha ^{\#}}$ is symmetric, then there exists an $M_z$–commuting conjugation from $\alpha H^2$ into $\theta H^2$. The following example shows that in that case there may be no such conjugation between the corresponding model spaces $K_{\alpha }$ and $K_{\theta }$.

Example 5.5

Fix $a,b\in {\mathbb {D}}$ such that $a\ne b$, $a\ne {\overline{a}}$ and $b\ne {\overline{b}}$, and put

$$\begin{aligned} \alpha (z)=\tfrac{a-z}{1-{\overline{a}}z}\ \tfrac{b-z}{1-{\overline{b}}z}\qquad \text {and}\qquad \theta (z)=\tfrac{a-z}{1-{\overline{a}}z}\ \tfrac{{\overline{b}}-z}{1-{b}z}. \end{aligned}$$

Then

$$\begin{aligned} \alpha ^{\#}(z)=\tfrac{{\overline{a}}-z}{1-{a}z}\ \tfrac{{\overline{b}}-z}{1-{b}z}\qquad \text {and}\qquad \theta ^{\#}(z)=\tfrac{{\overline{a}}-z}{1-{a}z}\ \tfrac{{b}-z}{1-{\overline{b}}z} \end{aligned}$$

and so $\alpha \alpha ^{\#}=\theta \theta ^{\#}$. Thus there exists an $M_z$–commuting conjugation from $\alpha H^2$ onto $\theta H^2$. In this case however neither $\alpha \leqslant \theta ^{\#}$ nor $\theta \leqslant \alpha ^{\#}$, so by Theorem 4.6 no $M_z$–commuting conjugation between $K_{\alpha }$ and $K_{\theta }$ exists. Here also neither $\alpha \leqslant \theta$ nor $\theta \leqslant \alpha$, and so by Theorem 4.2 no $M_z$–conjugation between $K_{\alpha }$ and $K_{\theta }$ exists.

Finally, consider $M_z$–conjugations preserving S–invariant subspaces.

Proposition 5.6

Let $\theta$ and $\alpha$ be two inner functions. There are no $M_z$–conjugations in $L^2$ which map $\alpha H^2$ into $\theta H^2$.

Proof

If C was such a conjugation, then by Theorem 2.2, $C=M_{\psi }J$ for some unimodular function $\psi \in L^{\infty }$ and, in particular,

$$\begin{aligned} C(\alpha )=\psi {\overline{\alpha }}=\theta g\quad \text {for some }g\in H^2. \end{aligned}$$

Clearly g must be an inner function and $\psi =\alpha \theta g$. Then, for every $h\in H^2$,

$$\begin{aligned} C(\alpha h)=\alpha \theta g\overline{\alpha h}=\theta g\overline{ h}\in \theta H^2, \end{aligned}$$

and so $g{\overline{h}}\in H^2$. It follows that $g=0$ and $C(\alpha )=0$ which is a contradiction. $\square$

6 Conjugations and truncated Toeplitz operators

For $\varphi \in L^2$ define the truncated Toeplitz operator$A^\theta _\varphi$ by

$$\begin{aligned} A^\theta _\varphi f=P_\theta (\varphi f), \text { for } f\in H^\infty \cap K_\theta , \end{aligned}$$

were $P_\theta :L^2\rightarrow K_\theta$ is the orthogonal projection (see [14]). The operator $A^\theta _\varphi$ is closed and densely defined, and if it is bounded, it admits a unique bounded extension to $K_\theta$. The set of all bounded truncated Toeplitz operators on $K_\theta$ is denoted by ${\mathcal {T}}(\theta )$. Note that $A^\theta _\varphi \in {\mathcal {T}}(\theta )$ for $\varphi \in L^\infty$. It is known that every operator from ${\mathcal {T}}(\theta )$ is $C_{\theta }$–symmetric (see [14, Lemma 2.1]).

Observe that if $k\geqslant 0$, then the conjugation $C_{k,l}$ defined by (3.1) satisfies neither $M_zC_{k,l}=C_{k,l}M_z$ nor $SC_{k,l}=C_{k,l}S$. However, for $0\leqslant n<k$ and $\theta (z)=z^n$,

$$\begin{aligned} C_{k,l}(K_\theta )=K_\theta \quad \text {and}\quad A_z^{\theta }C_{k,l}=C_{k,l}A_z^{\theta } \end{aligned}$$

(since here $C_{k,l|K_{\theta }}=J^{\star }_{|K_{\theta }}$ and $\theta ^{\#}=\theta$).

Theorem below characterizes conjugations intertwining truncated shifts $A_z^{\theta }$ and $A_z^{\theta ^{\#}}$.

Theorem 6.1

Let $\theta$ be a nonconstant inner function and let C be a conjugation in $L^2$ such that $C(K_\theta )\subset K_{\theta ^{\#}}$. Then the following are equivalent:

• (1) $A_{\varphi ^{\#}}^{\theta ^{\#}} C=C A_{\varphi }^{\theta }$ on $K_{\theta }$ for all $\varphi \in H^\infty$,

• (2) $A_z^{\theta ^{\#}} C=C A_z^{\theta }$ on $K_{\theta }$,

• (3) there is a function $\psi \in H^{\infty }$ such that $C_{|K_{\theta }}= J^{\star }A_{\psi }^{\theta }$ and $A_{\psi }^{\theta }$ is an isometry,

• (4) there is a function $\psi '\in H^{\infty }$ such that $C_{|K_{\theta }}= A_{\psi '}^{\theta ^{\#}}J^{\star }_{|K_{\theta }}$ and $A_{\psi '}^{\theta ^{\#}}$ is an isometry.

Proof

We will only prove that $(2)\Rightarrow (3)$. Since $J^{\star }(K_{\theta ^{\#}})=K_\theta$ and $J^{\star }A_{\varphi ^{\#}}^{\theta ^{\#}} = A_{\varphi }^{\theta }J^{\star }$ for all $\varphi \in H^{\infty }$ (see [4, Lemma 4.5]), we have

$$\begin{aligned} J^{\star }CA_z^{\theta }=J^{\star }A_z^{\theta ^{\#}}C=A_z^{\theta }J^{\star }C \end{aligned}$$

on $K_{\theta }$ and so $J^{\star }C_{|K_{\theta }}=A_{\psi }^{\theta }$ for some $\psi \in H^{\infty }$ ([5, Theorem 14.38]). Hence $A_{\psi }^{\theta }$ is an isometry and

$$\begin{aligned} C_{|K_{\theta }}=J^{\star }A_{\psi }^{\theta }. \end{aligned}$$

$\square$

It is much more restrictive if $\theta ^{\#}=\theta$.

Proposition 6.2

Let $\theta$ be an inner function such that $\theta ^{\#}=\theta$ and let C be a conjugation in $K_\theta$. Then the following are equivalent:

• (1) $A_{\varphi ^{\#}}^{\theta }C=CA_{\varphi }^{\theta }$ for all $\varphi \in H^\infty$,

• (2) $A_z^{\theta } C= CA_z^{\theta }$,

• (3) $C=\lambda J^{\star }_{|K_{\theta }}$ with $\lambda \in {\mathbb {T}}$.

Proof

Implications $(1)\Rightarrow (2)$ and $(3)\Rightarrow (1)$ are clear. To prove $(2)\Rightarrow (3)$ apply Theorem 6.1 to the conjugation ${\tilde{C}}$ in $L^2$ defined by

$$\begin{aligned}{\tilde{C}}=C\oplus C_{\theta }\ :\ K_{\theta }\oplus (K_{\theta })^{\perp }\rightarrow K_{\theta }\oplus (K_{\theta })^{\perp }.\end{aligned}$$

It follows that ${\tilde{C}}_{|K_{\theta }}=C=J^{\star }A_{\psi }^{\theta }$ for $\psi \in H^{\infty }$ such that $A_{\psi }^{\theta }$ is an isometry. Since $C(K_{\theta })=K_{\theta }$ and $J^{\star }(K_{\theta })=K_{\theta ^{\#}}=K_{\theta }$, we see that $A_{\psi }^{\theta }=J^{\star }C$ maps $K_{\theta }$ onto $K_{\theta }$ and is in fact unitary. Thus we have

$$\begin{aligned} A_{{\overline{\psi }}}^{\theta }A_{\psi }^{\theta }=A_{\psi }^{\theta }A_{{\overline{\psi }}}^{\theta }=I_{K_{\theta }}. \end{aligned}$$

On the other hand, $C^2=I_{K_{\theta }}$ so

$$\begin{aligned} C^2=J^{\star }A_{\psi }^{\theta }J^{\star }A_{\psi }^{\theta }=A_{\psi ^{\#}}^{\theta }A_{\psi }^{\theta }=A_{\psi }^{\theta }A_{\psi ^{\#}}^{\theta }=I_{K_{\theta }}. \end{aligned}$$

Hence $A_{{\overline{\psi }}}^{\theta }=A_{\psi ^{\#}}^{\theta }$ and $A_{{\overline{\psi }}-\psi ^{\#}}^{\theta }=0$, which gives ${\overline{\psi }}-\psi ^{\#}\in \overline{\theta H^2}+\theta H^2$ (see [14]). In other words, ${\overline{\psi }}-\psi ^{\#}=\overline{\theta h_1}+\theta h_2$ for some functions $h_1, h_2\in H^2$. Thus there exists a constant $\lambda$ such that

$$\begin{aligned}{\psi }-\theta h_1=\overline{\psi ^{\#}+\theta h_2}={\overline{\lambda }}.\end{aligned}$$

We now have

$$\begin{aligned} A_{\psi }^{\theta }=A_{\theta h_1+{\overline{\lambda }}}^{\theta }={\overline{\lambda }} I_{K_{\theta }}. \end{aligned}$$

Moreover $\lambda \in {\mathbb {T}}$, since $A_{\psi }^{\theta }$ is unitary. Hence

$$\begin{aligned} C=J^{\star }A_{\psi }^{\theta }={\lambda } J^{\star }_{|K_{\theta }}. \end{aligned}$$

$\square$

Now we characterize conjugations intertwining the truncated shifts $A_z^{\theta }$ and $A_{{{\bar{z}}}}^{\theta }$.

Theorem 6.3

Let $\theta$ be an inner function and let C be a conjugation in $K_\theta$. Then the following are equivalent:

• (1) $A_{\varphi }^{\theta } C=C A_{{\overline{\varphi }}}^{{\theta }}$ for all $\varphi \in H^\infty$,

• (2) $A_z^{\theta } C=C A_{{{\bar{z}}}}^{\theta }$ ,

• (3) there is a function $\psi \in H^\infty$ such that $C=A_\psi ^{\theta }C_{\theta }$ and $A_\psi ^{\theta }$ is unitary,

• (4) there is a function $\psi ^\prime \in H^\infty$ such that $C=C_{\theta }A_{\overline{\psi ^\prime }}^{\theta }$ and $A_{\psi ^\prime }^{\theta }$ is unitary.

Proof

Let us start with $(2)\Rightarrow (3)$. Since $A_z^{\theta }$ is $C_{\theta }$–symmetric,

$$\begin{aligned} A_z^{\theta } CC_{\theta }=CA_{\bar{z}}^{\theta }C_{\theta }=CC_{\theta }A_z^{\theta }. \end{aligned}$$

Hence, by [1, Proposition 1.21], $CC_{\theta }=A_\psi ^{\theta }$ for some $\psi \in H^\infty$. Clearly, $A_\psi ^{\theta }$ is unitary and

$$\begin{aligned} C=A_\psi ^{\theta }C_{\theta }=C_{\theta }(A_\psi ^{\theta })^{*}=C_{\theta }A_{{\overline{\psi }}}^{\theta }. \end{aligned}$$

To prove that $(4)\Rightarrow (1)$ note that, since $A_{\overline{\psi ^\prime }}^{\theta }$ and $A_{{\overline{\varphi }}}^{\theta }$ commute, we have

$$\begin{aligned} A_{\varphi }^{\theta }C=A_{\varphi }^{\theta }C_{\theta }A_{\overline{\psi ^\prime }}^{\theta } =C_{\theta }A_{{\overline{\varphi }}}^{\theta }A_{\overline{\psi ^\prime }}^{\theta }=C_{\theta }A_{\overline{\psi ^\prime }}^{\theta }A_{{\overline{\varphi }}}^{\theta }=CA_{{\overline{\varphi }}}^{\theta }=C(A_{{\varphi }}^{\theta })^{*}. \end{aligned}$$

All other implications are straightforward. $\square$

Corollary 6.4

If C is a conjugation in $K_\theta$ and every $A\in {\mathcal {T}}(\theta )$ is C–symmetric, then $C=A_\psi ^{\theta }C_{\theta }$ for some $\psi \in H^\infty$ such that $A_\psi ^{\theta }$ is unitary.

For a complete description of unitary operators from ${\mathcal {T}}(\theta )$ see [15, Proposition 6.5].