---
abstract: 'We compute the C\*-algebra generated by a group of composition operators acting on certain reproducing kernel Hilbert spaces over the disk, where the symbols belong to a non-elementary Fuchsian group. We show that such a C\*-algebra contains the compact operperators, and its quotient is isomorphic to the crossed product C\*-algebra determined by the action of the group on the boundary circle. In addition we show that the C\*-algebras obtained from composition operators acting on a natural family of Hilbert spaces are in fact isomorphic, and also determine the same *Ext*-class, which can be related to known extensions of the crossed product.'
address: |
    Department of Mathematics\
    University of Florida\
    Gainesville, Florida 32603
author:
- 'Michael T. Jury'
bibliography:
- 'compext.bib'
date: 26 Sep 2005
title: '$C^*$-algebras generated by groups of composition operators'
---

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Introduction
============

The purpose of this paper is to begin a line of investigation suggested by recent work of Moorhouse et al. [@moorhouse-thesis; @jm-tk-preprint; @jm-tk-bm-preprint]: to describe, in as much detail as possible, the $C^*$-algebra generated by a set of composition operators acting on a Hilbert function space. In this paper we consider a class of examples which, while likely the simplest cases from the point of view of composition operators, nonetheless produces $C^*$-algebras which are of great interest both intrinsically and for applications. In fact, the $C^*$-algebras we obtain are objects of current interest among operator algebraists, with applications in the study of hyperbolic dynamics [@emerson-thesis], noncommutative geometry [@MR88k:58149], and even number theory [@MR1931172].

Let $f\in H^2(\mathbb{D})$. For an analytic function $\gamma:\mathbb{D}\to\mathbb{D}$, the *composition operator with symbol $\gamma$* is the linear operator defined by $$(C_\gamma f) (z)=f(\gamma(z))$$ In this paper, we will be concerned with the $C^*$-algebra $${\mathcal{C}_\Gamma}=C^*(\{C_\gamma : \gamma\in\Gamma\})$$ where $\Gamma$ is a discrete group of (analytic) automorphisms of $\mathbb{D}$ (i.e. a *Fuchsian group*). For reasons to be described shortly, we will further restrict ourselves to *non-elementary* Fuchsian groups (i.e. groups $\Gamma$ for which the $\Gamma$-orbit of $0$ in $\mathbb{D}$ accumulates at at least three points of the unit circle $\mathbb{T}$.) Our main theorem shows that ${\mathcal{C}_\Gamma}$ contains the compact operators, and computes the quotient ${\mathcal{C}_\Gamma}/\mathcal{K}$:

Let $\Gamma$ be a non-elementary Fuchsian group, and let ${\mathcal{C}_\Gamma}$ denote the $C^*$-algebra generated by the set of composition operators on $H^2$ with symbols in $\Gamma$. Then there is an exact sequence $$\begin{CD}
0 @>>> \mathcal{K} @>\iota>> {\mathcal{C}_\Gamma}@>\pi >> C(\mathbb{T})\times\Gamma @>>> 0
\end{CD}$$

Here $C(\mathbb{T})\times\Gamma$ is the crossed product $C^*$-algebra obtained from the action $\alpha$ of $\Gamma$ on $C(\mathbb{T})$ given by $$\alpha_\gamma(f)(z)=f(\gamma^{-1}(z))$$ (Since the action of $\Gamma$ on $\partial\mathbb{D}$ is amenable, the full and reduced crossed products coincide; we will discuss this further shortly.) We will recall the relevant definitions and facts we require in the next section.

There is a similar result for the $C^*$-algebras $${\mathcal{C}_\Gamma}^n =C^*\left(\{C_\gamma \in \mathcal{B}(A^2_n)|\gamma\in\Gamma\}\right),$$ acting on the family of reproducing kernel Hilbert spaces $A^2_n$ (defined below), namely there is an extension $$\begin{CD}
0 @>>> \mathcal{K} @>>> {\mathcal{C}_\Gamma}^n @>>> C(\mathbb{T})\times\Gamma @>>> 0
\end{CD}$$ and we will show that each of these extensions represents the same element of the $Ext$ group $Ext(C(\partial\mathbb{D})\times\Gamma,\mathcal{K})$, and we will also prove that ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$ are isomorphic as $C^*$-algebras. In fact we obtain a stronger isomorphism result, namely that *any* unital $C^*$-extension of $C(\partial\mathbb{D})\times\Gamma$ which defines the same $Ext$-class as ${\mathcal{C}_\Gamma}$ is isomorphic to ${\mathcal{C}_\Gamma}$:

Let $x\in \rm{Ext}(C(\partial\mathbb{D})\times\Gamma)$ denote the class of the extension $$\nonumber\begin{CD}
0 @>>> \mathcal{K} @>>> {\mathcal{C}_\Gamma}@>>> C(\partial\mathbb{D})\times\Gamma @>>> 0.
\end{CD}$$ If $e\in \rm{Ext}(C(\partial\mathbb{D})\times\Gamma)$ is a unital extension represented by $$\nonumber\begin{CD}
0 @>>> \mathcal{K} @>>> E  @>>> C(\partial\mathbb{D})\times\Gamma @>>> 0
\end{CD}$$ and $e=x$, then $E\cong {\mathcal{C}_\Gamma}$ as $C^*$-algebras.

Finally, we will compare the extension determined by ${\mathcal{C}_\Gamma}$ to tow other recent constructions of extensions of $C(\partial\mathbb{D})\times\Gamma$. We show that the $Ext$-class of ${\mathcal{C}_\Gamma}$ coincides with the class of the $\Gamma$-equivariant Toeplitz extension of $C(\partial\mathbb{D})$ constructed by J. Lott [@lott-preprint], and differs from the extension of crossed products by cocompact groups constructed by H. Emerson [@emerson-thesis]. Finally we show that this extension in fact gives rise to a $\Gamma$-equivariant $KK_1$-cycle for $C(\partial\mathbb{D})$ which also accords with the construction in [@lott-preprint].

Preliminaries
=============

We will consider C\*-algebras generated by composition operators which act on a family of reproducing kernel Hilbert spaces on the unit disk. Specifically we will consider the spaces of analytic functions $A^2_n$, where $A^2_n$ is the space with reproducing kernel $$k_n(z,w)=(1-z\overline{w})^n$$ When $n=1$ this space is the Hardy space $H^2$, and its norm is given by $$\|f\|^2 =\lim_{r\to 1}\frac{1}{2\pi}\int_0^{2\pi} |f(re^{i\theta})|^2\, d\theta$$ For $n\geq 2$, the norm on $A^2_n$ is equivalent to $$\|f\|^2=\frac{1}{\pi}\int_\mathbb{D}|f(z)|^2 (1-|z|^2)^{n-2}\, dA(z)$$ though the norms are equal only for $n=2$ (the Bergman space).

An analytic function $\gamma:\mathbb{D}\to\mathbb{D}$ defines a *composition operator* $C_\gamma$ on $A^2_n$ by $$(C_\gamma f)(z)=f(\gamma(z))$$ In this paper, we will only consider cases where $\gamma$ is a Möbius transformation; in these cases $C_\gamma$ is easily seen to be bounded, by changing variables in the integrals defining the norms. An elementary calculation shows that if $\gamma :\mathbb{D}\to \mathbb{D}$ is analytic, then $$C_\gamma^* k_w(z) = k_{\gamma(w)}(z)$$ where $k$ is any of the reproducing kernels $k_n$.

We recall here the definitions of the full and reduced crossed product $C^*$-algebras; we refer to [@MR81e:46037] for details. Let a group $\Gamma$ act by homeomorphisms on a compact Hausdorff space $X$. This induces an action of $\Gamma$ on the commutative $C^*$-algebra $C(X)$ via $$(\gamma\cdot f)(x)=f(\gamma^{-1}\cdot x)$$ The *algebraic crossed product* $C(X)\times_{alg} \Gamma$ consists of formal finite sums $f=\sum_{\gamma\in\Gamma}f_\gamma [\gamma]$, where $f_\gamma\in C(\mathbb{T})$ and the $[\gamma]$ are formal symbols. Multiplication is defined in $C(X)\times_{alg}
\Gamma$ by $$\left(\sum_{\gamma\in\Gamma}f_\gamma [\gamma] \right)
\left(\sum_{\gamma\in\Gamma}f_\gamma^\prime [\gamma^\prime] \right) =
\sum_{\delta\in\Gamma}\sum_{\gamma\gamma^\prime=\delta} f_\gamma
(\gamma\cdot f^\prime_{\gamma^\prime})[\delta]$$ For $f=\sum_\gamma f_\gamma [\gamma]$, define $f^*\in
C(X)\times_{alg} \Gamma$ by $$f^* = \sum_{\gamma\in\Gamma} (\gamma \cdot \overline{f_{\gamma^{-1}}})
[\gamma]$$ With this multiplication and involution, $C(X)\times_{alg} \Gamma$ becomes a $*$-algebra, and we may construct a $C^*$-algebra by closing the algebraic crossed product with respect to a $C^*$-norm.

To obtain a $C^*$-norm, one constructs $*$-representations of $C(X)\times_{alg}\Gamma$ on Hilbert space. To do this, we first fix a faithful representation $\pi$ of $C(X)$ on a Hilbert space $\mathcal{H}$. We then construct a representation $\sigma$ of the algebraic crossed product on $\mathcal{H}\otimes\ell^2(\Gamma)=\ell^2(\Gamma, \mathcal{H})$ as follows: define a representation $\tilde{\pi}$ of $C(X)$ by its action on vectors $\xi\in\ell^2(\Gamma, \mathcal{H})$ $$(\tilde{\pi}(f))(\xi)(\gamma)=\pi(f\circ\gamma)\xi(\gamma)$$ Represent $\Gamma$ on $\ell^2(\Gamma,\mathcal{H})$ by left translation: $$(U(\gamma))(\xi)(\eta)=\xi(\gamma^{-1}\eta)$$ The representation $\sigma$ is then given by $$\sigma\left(\sum f_\gamma [\gamma]\right)=\sum \tilde{\pi}(f_\gamma) U(\gamma)$$ The closure of $C(X)\times_{alg}\Gamma$ with respect the norm induced by this representation is called the *reduced crossed product* of $\Gamma$ and $C(X)$, and is denoted $C(X)\times_r\Gamma$. The *full crossed product*, denote $C(X)\times \Gamma$, is obtained by taking the closure of the algebraic crossed product with respect to the maximal $C^*$-norm $$\|f\|=\sup_\pi \|\pi(f)\|$$ where the supremum is taken over *all* $*$-representations $\pi$ of $C(X)\times_{alg}\Gamma$ on Hilbert space. When $\Gamma$ is discrete, $C(X)\times \Gamma$ contains a canonical subalgebra isomorphic to $C(X)$, and there is a natural surjective $*$-homomorphism $\rho:C(X)\times\Gamma \to C(X)\times_r \Gamma$.

The full crossed product is important because of its universality with respect to covariant representations. A *covariant representation* of the pair $(\Gamma, X)$ consists of a faithful representation $\pi$ of $C(X)$ on Hilbert space, together with a unitary representation $u$ of $\Gamma$ on the same space satisfying the covariance condition $$u(\gamma)\pi(f)u(\gamma)^* = f\circ \gamma^{-1}$$ for all $f\in C(X)$ and all $\gamma\in\Gamma$. If $\mathcal{A}$ denotes the C\*-algebra generated by the images of $\pi$ and $u$, then there is a surjective \*-homomorphism from $C(X)\times\Gamma$ to $\mathcal{A}$; equivalently, any C\*-algebra generated by a covariant representation is isomorphic to a quotient of the full crossed product.

We now collect properties of group actions on topological spaces which we will require in what follows. Let a group $\Gamma$ act by homeomorphisms on a locally compact Hausdorff space $X$; The action of $\Gamma$ on $X$ is called *minimal* if the set $\{\gamma\cdot x |
\gamma\in \Gamma\}$ is dense in $X$ for each $x\in X$, and called *topologically free* if, for each $\gamma\in \Gamma$, the set of points fixed by $\gamma$ has empty interior.

Suppose now $\Gamma$ is discrete and $X$ is compact. Let $\text{Prob}(\Gamma)$ denote the set of finitely supported probability measures on $\Gamma$. We say $\Gamma$ *acts amenably on* $X$ if there exists a sequence of weak-\* continuous maps $b_x^n:X\to
\text{Prob}(\Gamma)$ such that for every $\gamma\in \Gamma$, $$\lim_{n\to \infty} \sup_{x\in X} \|\gamma\cdot b_x^n - b_{\gamma\cdot x}^n\|_1
=0$$ where $\Gamma$ acts on the functions $b_x^n$ via $(\gamma\cdot
b_x^n)(z)=b_x^n(\gamma^{-1}\cdot z)$, and $\|\cdot \|_1$ denotes the $l^1$-norm on $\Gamma$.

\[T:as\][@MR94m:46101] Let $\Gamma$ be a discrete group acting on a compact Hausdorff space $X$, and suppose that the action is topologically free. If $\mathcal{J}$ is an ideal in $C(X)\times \Gamma$ such that $C(X)\cap \mathcal{J}=0$, then $\mathcal{J}\subseteq \mathcal{J}_\lambda$, where $\mathcal{J}_\lambda$ is the kernel of the projection of the full crossed product onto the reduced crossed product.

\[T:amenability\][@MR2000g:19004; @MR86j:22014] If the $\Gamma$ acts amenably on $X$, then the full and reduced crossed product $C^*$-algebras coincide, and this crossed product is nuclear.

In this paper we are concerned with the case $X=\mathbb{T}$ and $G=\Gamma$, a non-elementary Fuchsian group. The action of $\Gamma$ on $\mathbb{T}$ is always amenable, so by Theorem \[T:amenability\] the full and reduced crossed products coincide, $C(\partial\mathbb{D})\times\Gamma$ is nuclear.

We also require some of the basic terminology concerning Fuchsian groups. Fix a base point $z_0\in\mathbb{D}$. The *limit set* of $\Gamma$ is the set accumulation points of the orbit $\{\gamma(z_0) : \gamma\in\Gamma\}$ on the boundary circle; it is closed (and does not depend on the choice of $z_0$. The limit set can be one of three types: it is either finite (in which case it consists of at most two points), a totally disconnected perfect set (hence uncountable), or all of the circle. In the latter case we say $\Gamma$ is of the *first kind*, and of the *second kind* otherwise. If the limit set is finite, $\Gamma$ is called *elementary*, and *non-elementary* otherwise.

Though we do not require it in this paper, it is worth recording that if $\Gamma$ is of the first kind, then $C(\partial\mathbb{D})\times\Gamma$ is simple, nuclear, purely infinite, and belongs to the bootstrap category $\mathcal{N}$ (and is clearly separable and unital). Hence, it satisfies the hypotheses of the Kirchberg-Phillips classification theorem, and is classified up to isomorphism by its (unital) $K$-theory [@MR98a:46085; @MR2000k:46093].

Finally, we recall briefly the basic facts about extensions of C\*-algebras and the *Ext* functor. An exact sequence of C\*-algebras $$\begin{CD}
0 @>>> B @>>> E @>>> A @>>> 0
\end{CD}$$ is called an *extension* of $A$ by $B$. (In this paper we consider only extensions for which $B=\mathcal{K}$, the C\*-algebra of compact operators on a separable Hilbert space.) Associated to an extension of $A$ by $\mathcal{K}$ is a $*$-homomorphism $\tau$ from $A$ to the Calkin algebra $\mathcal{Q}(\mathcal{H})=\mathcal{B}(\mathcal{H})/\mathcal{K}(\mathcal{H})$; this $\tau$ is called the *Busby map* associated to the extension. Conversely, given a map $\tau:A\to \mathcal{Q}(\mathcal{H})$ there is a unique extension having Busby map $\tau$; we will thus speak of an extension and its Busby map interchangeably. Two extensions of $A$ by $\mathcal{K}$ with Busby maps $\tau_1:A\to \mathcal{Q}(\mathcal{H}_1)$ and $\tau_2:A\to \mathcal{Q}(\mathcal{H}_2)$ are *strongly unitarily equivalent* if there is a unitary $u:\mathcal{H}_1\to \mathcal{H}_2$ such that $$\label{E:ue}
\pi(u)\tau_1(a)\pi(u)^*=\tau_2(a)$$ for all $a\in A$ (here $\pi$ denotes the quotient map from $\mathcal{B}(\mathcal{H}_1)$ to $\mathcal{Q}(\mathcal{H}_1)$). We say $\tau_1$ and $\tau_2$ are *unitarily equivalent*, written $\tau_1\sim_u \tau_2$, if (\[E:ue\]) holds with $u$ replaced by some $v$ such that $\pi(v)$ is unitary. An extension $\tau$ is *trivial* if it lifts to a $*$-homomorphism, that is there exists a $*$-homomorphism $\rho:A\to \mathcal{B}(\mathcal{H})$ such that $\tau(a)=\pi(\rho(a))$ for all $a\in A$. Two extensions $\tau_1, \tau_2$ are *stably equivalent* if there exist trivial extensions $\sigma_1, \sigma_2$ such that $\tau_1\oplus\sigma_1$ and $\tau_2\oplus \sigma_2$ are strongly unitarily equivalent; stable equivalence is an equivalence relation. If $A$ is separable and nuclear (which will always be the case in this paper) then the stable equivalence classes of extensions of $A$ by $\mathcal{K}$ form an abelian group (where addition is given by direct sum of Busby maps) called $Ext(A,\mathcal{K})$, abbreviated $Ext(A)$. Finally, each element of $Ext(A)$ determines an *index homomorphism* $\partial:K_1(A)\to \mathbb{Z}$ obtained by lifting a unitary in $M_n(A)$ representing the $K_1$ class to a Fredholm operator in $M_n(E)$ and taking its Fredholm index.

The Extensions ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$
==================================================================

The Hardy space
---------------

This section is devoted to the proof of our main theorem, in the case of the Hardy space:

\[T:main\] Let $\Gamma$ be a non-elementary Fuchsian group, and let ${\mathcal{C}_\Gamma}$ denote the $C^*$-algebra generated by the set of composition operators on $H^2$ with symbols in $\Gamma$. Then there is an exact sequence $$\begin{CD}
0 @>>> \mathcal{K} @>\iota>> {\mathcal{C}_\Gamma}@>\pi >> C(\mathbb{T})\times \Gamma @>>> 0
\end{CD}$$

.2in

While the proof in the general case of $A^2_n$ follows similar lines, we prove the $H^2$ case first since it is technically simpler and illustrates the main ideas. The proof splits into three parts: first, we prove that ${\mathcal{C}_\Gamma}$ contains the unilateral shift $S$ and hence the compacts (Proposition \[P:shift\]). We then prove that the quotient ${\mathcal{C}_\Gamma}/\mathcal{K}$ is generated by a covariant representation of the topological dynamical system $(C(\mathbb{T}), \Gamma)$. Finally we prove that the $C^*$-algebra generated by this representation is all of the crossed product $C(\mathbb{T})\times \Gamma$.

We first require two computational lemmas.

\[L:factor\] Let $\gamma$ be an automorphism of $\mathbb{D}$ with $a=\gamma^{-1}(0)$ and let $$f(z)=\frac{1-\overline{a}z}{(1-|a|^2)^{1/2}}$$ Then $$C_\gamma C_\gamma^* = M_f M_f^*$$ where $M_f$ denotes the operator of multiplication by $f$.

[**Proof. **]{}It suffices to show $$\label{E:bilin}
\langle C_\gamma C_\gamma^* k_w, k_z\rangle = \langle M_f M_f^* k_w,
k_z\rangle$$ for all $z,w$ in $\mathbb{D}$. Since $C_\gamma^*k_\lambda =k_{\gamma(\lambda)}$ and $M_f^* k_\lambda = \overline{f(\lambda)}k_\lambda$, (\[E:bilin\]) reduces to the well-known identity $$\label{E:phi_ident}
\frac{1}{1-\gamma(z)\overline{\gamma(w)}}=\frac{(1-\overline{a}z)(1-a\overline{w})}{(1-|a|^2)(1-z\overline{w})}.$$

\[L:commute\] Let $\gamma(z)$ be an automorphism of $\mathbb{D}$ and let $S=M_z$. Then $$M_{\gamma} C_\gamma = C_\gamma S$$

[**Proof. **]{}For all $g\in H^2$, $$\begin{aligned}
\nonumber (C_\gamma S)(g)(z) &= \gamma(z)g(\gamma(z)) \\
\nonumber                  &= (M_\gamma C_\gamma)(g)(z)\end{aligned}$$

\[P:shift\] Let $\Gamma$ be a non-elementary Fuchsian group. Then ${\mathcal{C}_\Gamma}$ contains the unilateral shift $S$.

[**Proof. **]{}Let $\Lambda$ denote the limit set of $\Gamma$. Since $\Gamma$ is non-elementary, there exist three distinct points $\lambda_1$, $\lambda_2$, $\lambda_3\in\Lambda$. For each $\lambda_i$ there exists a sequence $\gamma_{n,i}$ such that $\gamma_{n,i}(0)\to \lambda_i$. Let $a_i^n
=\gamma_{n,i} (0)$. By Lemma \[L:factor\], $$(1-|a_i^n|^2)C_{\gamma^{-1}_{n,i}}C_{\gamma^{-1}_{n,i}}^* = (1-\overline{a_i^n} S)(1-\overline{a_i^n} S)^*$$ As $n\to \infty$, $a_i^n \to \lambda_i$ and the right-hand side converges to $$1-\overline{\lambda}S-\lambda S^* +SS^*$$ in ${\mathcal{C}_\Gamma}$. Taking differences of these operators for different values of $i$, we see that $\Re [\mu_1 S],\ \Re [\mu_2
S]\in {\mathcal{C}_\Gamma}$, with $\mu_1 =\overline{\lambda_1 -\lambda_2}$ and $\mu_2=\overline{\lambda_2 -\lambda_3}$. Note that since $\lambda_1$, $\lambda_2$, $\lambda_3$ are distinct points on the circle, the complex numbers $\mu_1$ and $\mu_2$ are linearly independent over $\mathbb{R}$ when identified with vectors in $\mathbb{R}^2$. We now show there exist scalars $a_1,\ a_2$ such that $$a_1 \Re [\mu_1 S] + a_2 \Re [\mu_2 S] =S$$ which proves the lemma. Such scalars must solve the linear system $$\begin{pmatrix} \mu_1 && \mu_2 \\
                \overline{\mu_1} && \overline{\mu_2} \end{pmatrix}
                \begin{pmatrix} a_1 \\
                                a_2 \end{pmatrix} = \begin{pmatrix} 1 \\
                                                                    0
                                                                    \end{pmatrix}$$ Writing $\mu_1=\alpha_1 +i\beta_1$, $\mu_2=\alpha_2+i\beta_2$, a short calculation shows that $$\det \begin{pmatrix} \mu_1 && \mu_2 \\
                \overline{\mu_1} && \overline{\mu_2} \end{pmatrix}
=-2i \det \begin{pmatrix} \alpha_1 && \alpha_2 \\
                          \beta_1 && \beta_2 \end{pmatrix}$$ The latter determinant is nonzero since $\mu_1$, $\mu_2$ are linearly independent over $\mathbb{R}$, and the system is solvable. $\square$

The above argument does not depend on the discreteness of $\Gamma$; indeed it is a refinement of an argument due to J. Moorhouse [@moorhouse-thesis] that the $C^*$-algebra generated by *all* Möbius transformations contains $S$. The argument is valid for any group which has an orbit with three accumulation points on $\mathbb{T}$; e.g. the conclusion holds for any dense subgroup of $PSU(1,1)$.

.2in

**Proof of Theorem \[T:main\]** For an automorphism $\gamma$ of $\mathbb{D}$ set $$U_\gamma =(C_\gamma C_\gamma^*)^{-1/2} C_\gamma,$$ the unitary appearing in the polar decomposition of $C_\gamma$. By Lemma \[L:factor\], $C_\gamma C_\gamma^*=T_f T_f^*$, so we may write $$U_\gamma = T_{|f|^{-1}} C_\gamma +K$$ for some compact $K$. Now by Lemma \[L:commute\], if $p$ is any analytic polynomial, $$\begin{aligned}
\nonumber U_\gamma T_p &=T_{|f|^{-1}} C_\gamma T_p +K^\prime\\
\nonumber &= T_{|f|^{-1}} T_{p\circ \gamma} C_\gamma +K^\prime\\
\nonumber &= T_{p \circ \gamma}T_{|f|^{-1}} C_\gamma +K^{\prime\prime} \\
\nonumber &= T_{p\circ \gamma}U_\gamma +K^{\prime\prime} \end{aligned}$$ where we have used the fact that $T_{|f|^{-1}}$ and $T_p$ commute modulo $\mathcal{K}$. Taking adjoints and sums shows that $$\label{E:polycovar}
U_\gamma T_q U_\gamma^* =T_{q\circ \gamma} +K$$ for any trigonometric polynomial $q$. We next show that, for Möbius transformations $\gamma, \eta$, $$U_\gamma U_\eta = U_{\eta\circ\gamma} +K$$ for some compact $K.$ To see this, write $$\gamma(z)=\frac{az+b}{cz+d}, \qquad \eta(z)=\frac{ez+f}{gz+h}$$ Then $$U_\gamma = T_{|cz+d|^{-1}} C_\gamma +K_1, \qquad U_\eta = T_{|gz+h|^{-1}}
C_\eta +K_2.$$ Note that $$U_{\eta\circ\gamma} = T_{|g(az+b)+h(cz+d)|^{-1}}C_{\eta\circ\gamma} +K_3.$$ Then $$\begin{aligned}
\nonumber U_\gamma U_\eta &= T_{|cz+d|^{-1}} C_\gamma T_{|gz+h|^{-1}} C_\eta +K\\
\nonumber     &= T_{|cz+d|^{-1}} T_{|g\gamma (z)+h|^{-1}} C_\gamma C_\eta
+K^\prime \\
\nonumber     &=T_{|g(az+b)+h(cz+d)|^{-1}}C_{\eta\circ\gamma}+K^\prime \\
\nonumber     &= U_{\eta\circ\gamma} +K^{\prime\prime}\end{aligned}$$

Observe now that ${\mathcal{C}_\Gamma}$ is equal to the C\*-algebra generated by $S$ and the unitaries $U_\gamma$. We have already shown that ${\mathcal{C}_\Gamma}$ contains $S$ and each $U_\gamma$, for the converse we recall that $C_\gamma=(C_\gamma C_\gamma^*)^{1/2}U_\gamma$, and since $C_\gamma C_\gamma^*)$ lies in $C^*(S)$ by Lemma \[L:factor\], we see that $C_\gamma$ lies in the C\*-algebra generated by $S$ and $U_\gamma$. Letting $\pi$ denote the quotient map $\pi:{\mathcal{C}_\Gamma}\to{\mathcal{C}_\Gamma}/\mathcal{K}$, it follows hat ${\mathcal{C}_\Gamma}/\mathcal{K}$ is generated as a $C^*$-algebra by a copy of $C(\mathbb{T})$ and the unitaries $\pi(U_\gamma)$, and the map $\gamma\to\pi(U_{\gamma^{-1}})$ defines a unitary representation of $\Gamma$. Let $\alpha:\Gamma\to\text{Aut}(C(\mathbb{T}))$ be given by $$\alpha_\gamma (f)(z) = f(\gamma^{-1}(z))$$ Then by (\[E:polycovar\]), $$\label{E:cov}
\pi(U_{\gamma^{-1}})\pi(T_f)\pi(U_{\gamma^{-1}}^*)=\pi(T_{f\circ\gamma^{-1}})=\pi(T_{\alpha_\gamma(f)})$$ for all trigonometric polynomials $f$, and hence for all $f\in
C(\mathbb{T})$ by continuity. Thus, ${\mathcal{C}_\Gamma}/\mathcal{K}$ is generated by $C(\mathbb{T})$ and a unitary representation of $\Gamma$ satisfying the relation (\[E:cov\]). Therefore there is a surjective $*$-homomorphism $\rho:C(\mathbb{T})\times\Gamma \to {\mathcal{C}_\Gamma}/\mathcal{K}$ satisfying $$\rho(f)=\pi(T_f), \qquad \rho(u_\gamma)=\pi(U_{\gamma^{-1}})$$ for all $f\in C(\mathbb{T})$ and all $\gamma\in\Gamma$. Let $\mathcal{J}=\text{ker}\, \rho$. The theorem will be proved once we show $\mathcal{J}=0$. To do this, we use Theorems \[T:as\] and \[T:amenability\]. Indeed, it suffice to show that $C(\mathbb{T})\cap\mathcal{J}=0$: then $\mathcal{J}\subset \mathcal{J}_\lambda$ by Theorem \[T:as\], and $\mathcal{J}_\lambda =0$ by Theorem \[T:amenability\].

To see that $C(\mathbb{T})\cap\mathcal{J}=0$, choose $f\in C(\mathbb{T})\cap \mathcal{J}$. Then $\pi(T_f)=\rho(f)=0$, which means that $T_f$ is compact. But then $f=0$, since nonzero Toeplitz operators are non-compact.

$\square$

Other Hilbert spaces
--------------------

In this section we prove the analogue of Theorem \[T:main\] for composition operators acting on the reproducing kernel Hilbert spaces with kernel given by $$k(z,w)=\frac{1}{{{(1-z\overline{w})}^n}}$$ for integers $n\geq2$.

We let $A^2_n$ denote the Hilbert function space on $\mathbb{D}$ with kernel $k(z,w)={(1-z\overline{w})}^{-n}$. For $n\geq2$, this space consists of those analytic functions in $\mathbb{D}$ for which $$\int_\mathbb{D} |f(w)|^2 (1-|w|^2)^{n-2}dA(w)$$ is finite, and the square root of this quantity is a norm on $A^2_n$ equivalent to the norm determined by the reproducing kernel $k$ (they are equal for $n=2$). We fix $n\geq2$ for the remainder of this section, and let $T$ denote the operator of multiplication by $z$ on $A^2_n$. The operator $T$ is a contractive weighted shift, and essentially normal. Moreover, if $f$ is any bounded analytic function on $\mathbb{D}$, then multiplication by $f$ is bounded on $A^2_n$ and is denoted $M_f$. If $\gamma$ is a Möbius transformation, then $C_\gamma$ is bounded on $A^2_n$. We let ${\mathcal{C}_\Gamma}^n$ denote the C\*-algebra generated by the operators $\{C_\gamma :\gamma\in\Gamma\}$ acting on $A^2_n$. In this section we prove:

\[T:main\_gen\] The C\*-algebra ${\mathcal{C}_\Gamma}^n$ contains $C^*(T)$, and in particular the compact operators $\mathcal{K}(A^2_n)$. The map defined by $\pi(C_\gamma)=|\gamma^\prime|^{n/2} u_\gamma$ extends to a $*$-homomorphism from ${\mathcal{C}_\Gamma}^n$ onto $C(\partial\mathbb{D})\times\Gamma$, and there is an exact sequence $$0\to \mathcal{K}(A^2_n)\to {\mathcal{C}_\Gamma}^n\to C(\partial\mathbb{D})\times\Gamma \to 0$$

We collect the following computations together in a single Lemma, the analogue for $A^2_n$ of the lemmas at the beginning of the previous section.

\[L:cgn\_calc\]For $\gamma\in\Gamma$, with $a=\gamma^{-1}(0)$ and $n$ fixed, let $$f(z)=\frac{{(1-\overline{a}z)}^n}{{(1-{|a|}^2)}^{n/2}}$$ Then $C_\gamma C_\gamma^* =M_fM_f^*$, and $M_\gamma C_\gamma = C_\gamma T$.

[**Proof. **]{}As in the Hardy space, for the first equation it suffices to prove that $$\label{E:bilin_general}
\langle C_\gamma C_\gamma^* k_w, k_z\rangle = \langle T_f T_f^* k_w,
k_z\rangle$$ for all $z,w$ in $\mathbb{D}$. This is the same as the identity $$\label{E:phi_ident_general}
{\left( \frac{1}{1-\gamma(z)\overline{\gamma(w)}}\right)}^n=\frac{{(1-\overline{a}z)}^n{(1-a\overline{w})}^n}{{(1-|a|^2)}^n{(1-z\overline{w})}^n}.$$ which is just (ref) raised to the power $n$. As before, the second equation follows immediately from the definitions. $\square$

The proof of Theorem \[T:main\_gen\] follows the same lines as that for the Hardy space, except that it requires more work to show that ${\mathcal{C}_\Gamma}^n$ contains $T$. We first prove that ${\mathcal{C}_\Gamma}^n$ is irreducible.

\[P:reduce\] The $C^*$-algebra ${\mathcal{C}_\Gamma}^n$ is irreducible.

[**Proof. **]{}We claim that the operator $T^n$ lies in ${\mathcal{C}_\Gamma}^n$; we first show how irreducibility follows form this and then prove the claim.

By [@MR2001h:47044], the only reducing subspaces of $T^n$ are direct sums of subspaces of the form $$X_k=\overline{\text{span}}\{z^{k+mn}|m=0,1,2,\dots\}$$ for $0\leq k\leq n-1$. Thus, since $T^n\in {\mathcal{C}_\Gamma}^n$, these are the only possible reducing subspaces for ${\mathcal{C}_\Gamma}^n$. Suppose, then, that $X$ is a nontrivial direct sum of distinct subspaces $X_k$ (i.e. $X\neq
A^2_n$). Observe that, for each $k$, either $X$ contains $X_k$ as a summand, or $X$ is orthogonal to $X_k$. In the latter case (which we assume holds for some $k$), the $k^{th}$ Taylor coefficient of every function in $X$ vanishes.

It is now easy to see that $X$ cannot be reducing (or even invariant) for ${\mathcal{C}_\Gamma}^n$. Indeed, if $X$ does not contain $X_0$ as a summand, then every function $f\in X$ vanishes at the origin, but if $\gamma\in \Gamma$ does not fix the origin then $C_\gamma f\notin X$. On the other hand, if $X$ contains the scalars, then consider the operator $F(\lambda)=(1-\overline{\lambda}T)^n(1-\lambda T^*)^n$, for $\lambda\in\Lambda$. Since $T^*$ annihilates the scalars, we have $F(\lambda)1=(1-\overline{\lambda}z)^n=p(z)$. But since $|\lambda|=1$, the $k^{th}$ Taylor coefficient of $p$ does not vanish, so $p\notin X$. But since $F(\lambda)$ belongs to ${\mathcal{C}_\Gamma}^n$, it follows that $X$ is not invariant for ${\mathcal{C}_\Gamma}^n$ and hence ${\mathcal{C}_\Gamma}^n$ is irreducible.

To prove that $T^n\in {\mathcal{C}_\Gamma}^n$, we argue along the lines of Proposition \[P:shift\], though the situation is somewhat more complicated. Using the same reproducing kernel argument as in that Proposition, we see that the operators $$\begin{aligned}
F(\lambda)&=(1-\overline{\lambda}T)^n(1-\lambda T^*)^n \\
&= \sum_{j=0}^n\sum_{k=0}^n (-1)^{(j+k)} \binom{n}{j}\binom{n}{k}\overline{\lambda}^k \lambda^j T^k T^{*j} \\
&=\sum_{d=0}^n \binom{n}{d}^2 T^d T^{*d} +2\Re \sum_{m=1}^n (-1)^m\overline{\lambda}^m \sum_{m\leq k\leq n}\binom{n}{k}\binom{n}{k-m}T^k T^{*{k-m}}   \\
&= \sum_{d=0}^n \binom{n}{d}^2 T^d T^{*d} + \sum_{m=1}^n \overline{\lambda}^m E_m + \lambda^m E_m^*\end{aligned}$$ lie in ${\mathcal{C}_\Gamma}^n$ for all $\lambda$ in the limit set $\Lambda$ of $\Gamma$. Here we have adopted the notation $$E_m = (-1)^m \sum_{m\leq k\leq n} \binom{n}{k}\binom{n}{k-m}T^k T^{* k-m}$$ Note in particular that $E_n=(-1)^n T^n$. Forming differences as $\lambda$ ranges over $\Lambda$ shows that ${\mathcal{C}_\Gamma}^n$ contains all operators of the form $$G(\lambda,\mu)=F(\lambda)-F(\mu) = \sum_{m=1}^n (\overline{\lambda}^m-\overline{\mu}^m)E_m +(\lambda^m-\mu^m)E_m^*$$ for all $\lambda, \mu \in \Lambda$. We wish to obtain $T^n$ as a linear combination of the $G(\lambda,\mu)$; since $E_n$ is a scalar multiple of $T^n$ it suffices to show that there exist $2n$ pairs $(\lambda_j, \mu_j)\in \Lambda\times\Lambda$ and $2n$ scalars $\alpha_j$ such that $$E_n = \sum_{j=1}^{2n} \alpha_j G(\lambda_j, \mu_j)$$ Let $L$ be the $2n\times 2n$ matrix whose $j^{th}$ column is $$\begin{pmatrix} \overline{\lambda_j}-\overline{\mu_j} \\
                \lambda_j-\mu_j \\
                \overline{\lambda_j}^2-\overline{\mu_j}^2 \\
                \lambda_j^2-\mu_j^2 \\
                \vdots \\
                \overline{\lambda_j}^n-\overline{\mu_j}^n \\
                \lambda_j^n-\mu_j^n \\
\end{pmatrix}$$ We must therefore solve the linear system $$L \begin{pmatrix} \alpha_1 \\ 
                   \vdots \\
                   \alpha_{2n-1}\\
                   \alpha_{2n}\\
  \end{pmatrix}
 = \begin{pmatrix} 0 \\
                   \vdots \\
                   1 \\
                   0 \end{pmatrix}$$ for which it suffices to show that the matrix $L$ is nonsingular for some choice of the $\lambda_j$ and $\mu_j$ in $\Lambda$. To do this, we fix $2n+1$ distinct points $z_0, z_1,\dots
z_{2n}$ in $\Lambda$ and set $\lambda_j=z_0$ for all $j$ and $\mu_j=z_j$ for $j=1,\dots 2n$. The matrix $L$ then becomes the matrix whose $j^{th}$ column is $$\begin{pmatrix} \overline{z_0}-\overline{z_j} \\
                z_0-z_j \\
                \overline{z_0}^2-\overline{z_j}^2 \\
                z_0^2-{z_j}^2 \\
                \vdots \\
                \overline{z_0}^n-\overline{z_j}^n \\
                z_0^n-{z_j}^n \\
\end{pmatrix}$$ To prove that $L$ is nonsingular, we prove that its rows are linearly independent. To this end, let $c_j$ and $d_j$, $j=1,\dots n$ be scalars such that for each $k=1, \dots 2n$, $$\label{E:roots}
\sum_{j=1}^n c_j (z_0^j-z_k^j)+\sum_{j=1}^n d_j (\overline{z_0}^j-\overline{z_k}^j) = 0$$ To see that all of the $c_j$ and $d_j$ must be $0$, consider the harmonic polynomial $$P(z,\overline{z})=\sum_{j=1}^n c_j (z_0^j-z^j)+\sum_{j=1}^n d_j (\overline{z_0}^j-\overline{z}^j)$$ By (\[E:roots\]), $P$ has $2n+1$ distinct zeroes on the unit circle, namely the points $z_0, \dots z_{2n}$. But this means that the rational function $$Q(z)=\sum_{j=1}^n c_j (z_0^j-z^j)+\sum_{j=1}^n d_j (\overline{z_0}^j-\frac{1}{z^j})$$ also has $2n+1$ zeroes on the circle. But since the degree of $Q$ is at most $2n$, it follows that $Q$ must be the zero polynomial, and hence all the $c_j$ and $d_j$ are $0$.

\[P:bergshift\] ${\mathcal{C}_\Gamma}^n$ contains $T$.

[**Proof. **]{} We have established that for each $\lambda$ in the limit set, the operators $$(1-\overline{\lambda}T)^n(1-\overline{\lambda}T)^{*n}$$ lie in ${\mathcal{C}_\Gamma}^n$. Now, since $T$ is essentially normal, the difference $$(1-\overline{\lambda}T)^n(1-\overline{\lambda}T)^{*n} -
[(1-\overline{\lambda}T)(1-\overline{\lambda}T)^*]^n$$ is compact. Given two positive operators which differ by a compact, their (unique) positive $n^{th}$ roots also differ by a compact. The positive square root of the left-hand term above is thus equal to $$(1-\overline{\lambda}T)(1-\overline{\lambda}T)^*+K$$ for some compact operator $K$ (depending on $\lambda$), and lies in ${\mathcal{C}_\Gamma}^n$. Forming linear combinations as in Proposition \[P:shift\], we conclude that $T+K$ lies in ${\mathcal{C}_\Gamma}^n$ for some compact $K$. Now, as ${\mathcal{C}_\Gamma}^n$ is irreducible and contains a nonzero compact operator (namely, the self-commutator of $T+K$), it contains all the compacts, and hence $T$. $\square$

We now prove the analogue of Theorem \[T:main\] for ${\mathcal{C}_\Gamma}^n$:

\[T:berg\_main\] Let $\Gamma$ be a non-elementary Fuchsian group, and let ${\mathcal{C}_\Gamma}^n$ denote the $C^*$-algebra generated by the set of composition operators on $A^2$ with symbols in $\Gamma$. Then there is an exact sequence $$\begin{CD}
0 @>>> \mathcal{K} @>\iota>> {\mathcal{C}_\Gamma}^n @>\pi >> C(\mathbb{T})\times \Gamma @>>> 0
\end{CD}$$

[**Proof. **]{}The proof is similar to that of Theorem \[T:main\]; we define the unitary operators $U_\gamma$ in the same way (using the polar decomposition of $C_\gamma$), and check that he map that sends $\gamma$ to $U_{\gamma^{-1}}$ is a unitary representation of $\Gamma$ on $A^2_n$, modulo $\mathcal{K}$. The computation to check the covariance condition is essentially the same. As ${\mathcal{C}_\Gamma}^n$ is generated by the $T$ and the unitaries $U_\gamma$, the quotient ${\mathcal{C}_\Gamma}^n/\mathcal{K}$ is generated by a copy of $C(\partial\mathbb{D})$ (since $\partial\mathbb{D}$ is the essential spectrum of $T$) and a representation of $\Gamma$ which satisfy the covariance condition; the rest of the proof proceeds exactly as for $H^2$, up until the final step. At the final step in that proof, we have a Toeplitz operator $T_f$ which is compact; this implies that $f=0$ in the $H^2$ case but not for the spaces $A^2_n$. However the symbol of a compact Toeplitz operator on $A^2_n$ must vanish at the boundary of $\mathbb{D}$, so we may still conclude that $f=0$ on $\partial\mathbb{D}$ and the proof is complete.$\square$

$K$-Homology of $C(\partial\mathbb{D})\times\Gamma$
===================================================

$Ext$ classes of ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$
--------------------------------------------------------------------

In this section we prove that the extensions of $C(\partial\mathbb{D})\times
\Gamma$ determined by ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$ determine the same element of the group $\text{Ext}(C(\partial\mathbb{D})\times\Gamma)$, and we prove that ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$ are isomorphic as $C^*$-algebras. (In general neither of these statements implies the other.) We borrow several ideas from \[Rørdam\] concerning the classification of extensions by the associated six-term exact sequence in $K$-theory. The reader is referred to [@MR2001g:46001] for an introduction to $K$-theory and [@MR88g:46082 Chapter 15] for the basic definitions and theorems concerning extensions of $C^*$-algebras and the $Ext$ group.

\[T:extclasses\] The extensions of $C(\partial\mathbb{D})\times\Gamma$ determined by ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$ define the same element of the group $\text{Ext}(C(\partial\mathbb{D})\times\Gamma)$.

We will show that the Busby maps of the extensions determined by ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$ are strongly unitarily equivalent. To do this we introduce an integral operator $V\in\mathcal{B}(A^2_n,A^2_{n-1})$ and prove the following lemma, which describes the properties of $V$ needed in the proof of the theorem.

\[L:v\_operator\]Consider the integral operator defined on analytic polynomials $f(w)$ by $$(Vf)(z)=\int_{\mathbb{D}}\frac{f(w)}{(1-\overline{w}z)}(1-|w|^2)^{-1/2}\, dA(w)$$ where $dA$ is normalized Lebesgue area measure on $\mathbb{D}$. Then:

1.  For each $n\geq0$, $V$ extends to a bounded diagonal operator from $A^2_{n+1}$ to $A^2_n$, and there exists a compact operator $K_n$ such that $V+K_n$ is unitary.

2.  If $T_{n+1}$ and $T_n$ denote multiplication by $z$ on $A^2_{n+1}$ and $A^2_n$ respectively, then $VT_{n+1}-T_nV$ is compact.

3.  If $g$ is continuous on $\overline{\mathbb{D}}$ and analytic in $\mathbb{D}$, then the integral operator $$(V_{\overline{g}}f)(z)=\int_\mathbb{D}\frac{f(w)\overline{g(w)}}{1-\overline{w}z}(1-|w|^2)^{-1/2}dA(w)$$ is bounded from $A^2_{n+1}$ to $A^2_n$, and is a compact perturbation of $VM_g^*$.

[**Proof. **]{}If $f(w)=w^k$, then the integrand is absolutely convergent for each $z\in\mathbb{D}$, and we calculate $$\begin{aligned}
(Vf)(z) &=\int_\mathbb{D}\frac{w^k}{1-\overline{w}z}(1-|w|^2)^{-1/2}dA(w) \\
&= \sum_{j=0}^\infty\int_\mathbb{D}w^k\overline{w}^j z^j (1-|w|^2)^{-1/2}dA(w)\\
&= \left(\int_\mathbb{D}|w|^{2k} (1-|w|^2)^{-1/2}dA(w)\right) z^k \\
&= \alpha_k z^k \end{aligned}$$ It is known that the sequence $(\alpha_k)$ defined by the above integral is asymptotically $(k+1)^{-1/2}$. Each space $A^2_n$ has an orthonormal basis of the form $\beta_k z^k$, where the sequence $\beta_k$ is asymptotically $(k+1)^{(n-1)/2}$. It follows that $V$ intertwines orthonormal bases for $A^2_{n+1}$ and $A^2_n$ modulo a compact diagonal operator, which establishes the first statement. Moreover, the second statement also follows, by observing that the operators $T_{n+1}$ and $T_n$ are weighted shifts with weight sequences asymptotic to $1$.

To prove the last statement, first suppose $g(w)=w$, and let $\beta_k z^k$ be an orthonormal basis for $A^2_{n+1}$. Then a direct computation shows that $V_{\overline{g}}$ is a weighted backward shift with weight sequence $\alpha_{j+1} \beta_j /\beta_{j+1}$, $j=0, 1, \dots$. Thus with respect to this basis, $V^*V_{\overline{z}}$ is a weighted shift with weight sequence $\alpha_{j+1}\beta_j /\alpha_j\beta_{j+1}$. Since $\lim_{j \to\infty}\alpha_{j+1}/\alpha_j=1$, it follows that $V^*V_{\overline{z}}$ is a compact perturbation of $M_z^*$. By linearity, the lemma holds for polynomial $g$. Finally, if $p_n$ is a sequence of polynomials such that $M_{p_n}$ converges to $M_g$ in the operator norm (which is equal to the supremum norm of the symbol), then $M_{p_n}$ converges to $M_g$ in the essential norm, and $p_n$ converges to $g$ uniformly, so that $V_{\overline{p_n}}\to V_{\overline{g}}$ in norm and the result follows.

**Proof of Theorem \[T:extclasses\]**

We will prove that the Busby maps of the extensions determined by ${\mathcal{C}_\Gamma}$ and ${\mathcal{C}_\Gamma}^n$ are strongly unitarily equivalent. It suffices to prove, for each fixed $n$, the equivalence between ${\mathcal{C}_\Gamma}^n$ and ${\mathcal{C}_\Gamma}^{n+1}$. We first establish some notation: for a function $g$ in the disk algebra, we let $M_g$ denote the multiplication operator with symbol $g$ acting on $A^2_{n+1}$, and $\widetilde{M}_g$ the corresponding operator acting on $A^2_n$. Similarly, for the composition operators $C_\gamma$ and the unitaries $U_\gamma$ on $A^2_{n+1}$, a tilde denotes the corresponding operator on $A^2_n$. Now, since each of the algebras ${\mathcal{C}_\Gamma}^n, {\mathcal{C}_\Gamma}^{n+1}$ is generated by the operators $M_z$ (resp. $\widetilde{M}_z$) and the unitaries $U_\gamma$ (resp. $\widetilde{U}_\gamma$) on the respective Hilbert spaces, it suffices to produce a unitary $U$ (or indeed a compact perturbation of a unitary) from $A^2_{n+1}$ to $A^2_n$ such that the operators $UM_z-\widetilde{M}_zU$ and $UU_\gamma-\widetilde{U}_\gamma U$ are compact for all $\gamma\in\Gamma$. In fact we will prove that the operator $V$ of the previous lemma does the job.

To prove this, we will first calculate the operator $\widetilde{C}_\gamma VC_{\gamma^{-1}}$. Written as an integral operator, $$(\widetilde{C}_\gamma VC_{\gamma^{-1}}
f)(z)=\int_\mathbb{D}\frac{f(\gamma^{-1}(w))}{1-\overline{w}\gamma(z)}(1-|w|^2)^{-1/2}\,dA(w)$$ Applying the change of variables $w\to\gamma(w)$ to this integral, we obtain $$(\widetilde{C}_\gamma VC_{\gamma^{-1}}
f)(z)=\int_\mathbb{D}\frac{f(w)}{1-\overline{\gamma(w)}\gamma(z)}(1-|\gamma(w)|^2)^{-1/2}|\gamma^\prime(w)|^2\,dA(w)$$ A little algebra shows that $$1-|\gamma(w)|^2 = (1-|w|^2)|\gamma^\prime(w)|$$ Furthermore, multiplying and dividing the identity (\[E:phi\_ident\]) by $1-\overline{a}w$, we obtain $$\frac{1}{1-\overline{\gamma(w)}\gamma(z)}=\frac{1-\overline{a}z}{(1-\overline{a}w)(1-\overline{w}z)}|\gamma^\prime(w)|^{-1}$$ Thus the integral may be transformed into $$(\widetilde{C}_\gamma VC_{\gamma^{-1}} f)(z)=\int_\mathbb{D}\frac{f(w)}{1-\overline{w}z}\frac{1-\overline{a}z}{1-\overline{a}w}|\gamma^\prime(w)|^{1/2}(1-|w|^2)^{-1/2}\,dA(w)$$ Since $\gamma^\prime$ is analytic and non-vanishing in a neighborhood of the closed disk, there exists a function $\psi$ in the disk algebra such that $|\psi|^2=|\gamma^\prime|^{1/2}$ (choose $\psi$ to be a branch of $(\gamma^\prime)^{1/4}$). Thus we may rewrite this integral as $$\begin{aligned}
(\widetilde{C}_\gamma VC_{\gamma^{-1}} f)(z) &=\int_\mathbb{D}\frac{f(w)}{1-\overline{w}z}\frac{1-\overline{a}z}{1-\overline{a}w}\psi(w)\overline{\psi(w)}(1-|w|^2)^{-1/2}\,dA(w)\\
&= M_{1-\overline{a}z}V_{\overline{\psi}}M_{(1-\overline{a}z)^{-1}\psi(z)}\end{aligned}$$ Now, applying the lemma to $V_{\overline{\psi}}$ and using the fact that $V$ intertwines the multiplier algebras of $A^2_{n+1}$ and $A^2_n$ modulo the compacts, we have $$\begin{aligned}
\widetilde{C}_\gamma VC_{\gamma^{-1}} &= M_{1-\overline{a}z}V_{\overline{\psi}}M_{(1-\overline{a}z)^{-1}\psi(z)}\\
&= M_{1-\overline{a}z} VM_\psi^*M_\psi M_{(1-\overline{a}z)^{-1}} +K\\
&= VM_\psi^*M_\psi +K\end{aligned}$$ modulo the compacts. Multiplying on the right by $C_{\gamma^{-1}}$ and on the left by $V^*$ we get $$\label{E:VCV}
V^*\widetilde{C}_\gamma V =M_\psi^* M_\psi C_\gamma +K$$ Since $\psi^2(z)=(1-|a|^2)^{1/2}(1-\overline{a}z)^{-1}$, the calculations in the proof of \[T:main\] show that $$(C_\gamma C_\gamma^*)^{-1/2}=M_{\psi^{n+1}}M_{\psi^{n+1}}^* \quad\text{and}\quad (\widetilde{C}_\gamma \widetilde{C}_\gamma^*)^{-1/2}=\widetilde{M}_{\psi^{n}}\widetilde{M}_{\psi^{n}}^*$$ Using the fact that $V$ intertwines the C\*-algebras generated by $M_z$ and $\widetilde{M}_z$ modulo $\mathcal{K}$, and that these algebras are commutative modulo $\mathcal{K}$, we conclude after multiplying both sides of (\[E:VCV\]) on the left by $M_{\psi^n}M_{\psi^n}^*$ that $$V^*\widetilde{U}_\gamma V = V^*\widetilde{M}_{\psi^{n}}\widetilde{M}_{\psi^{n}}^* \widetilde{C}_\gamma V+K=M_{\psi^{n+1}}M_{\psi^{n+1}}^* C_\gamma +K =U_\gamma +K$$ which completes the proof.

$C^*$-algebras isomorphic to ${\mathcal{C}_\Gamma}$
---------------------------------------------------

In this section we prove that all unital extensions of $C(\partial\mathbb{D})\times\Gamma$ which determine the same *Ext* class as $\mathcal{C}_\Gamma$ have isomorphic pull-backs, i.e. the middle term in the exact sequence is isomorphic to $\mathcal{C}_\Gamma$. TO prove this, we apply Voiculescu’s theorem to obtain a stable isomorphism and use an analysis of the $K_0$-classes of finite projections in ${\mathcal{C}_\Gamma}$ to obtain a unital isomorphism.

We recall that if $A$ is a unital, nuclear C\*-algebra, an extension of $A$ by $\mathcal{K}$ is called *unital* if the Busby map $\tau:A\to \mathcal{Q}(\mathcal{H})$ is unital. A trivial extension is called *strongly unital* $\tau$ lifts to a unital homomorphism $\rho:A\to \mathcal{B}(\mathcal{H})$. (Not all unital trivial extensions are strongly unital.) We begin with the following lemma, which is a special case of [@MR99b:46108 Proposition 5.1].

Let $A$ be a unital $C^*$-algebra, and let $\tau_0=\pi\circ \alpha$, $\tau_1:A\to \mathcal{Q}(\mathcal{H})$ be unital extensions, with $\tau_0$ trivial. Then there is an isometry $v\in \mathcal{B}(\mathcal{H})$ such that $\alpha(1_A)=vv^*$. Set $\alpha^\prime(a)=v^*\alpha(a)v$ and set $\tau^\prime_0 = \pi\circ \alpha^\prime$, so that $\tau_0^\prime$ is a strongly unital trivial extension. Let $$\nonumber\begin{CD}
e:\quad  0 @>>> \mathcal{K} @>>> E @>\psi >> A @>>> 0
\end{CD}$$ $$\nonumber\begin{CD}
e^\prime:\quad  0 @>>> \mathcal{K} @>>> E^\prime @>\psi^\prime >> A @>>> 0
\end{CD}$$ be the extensions with Busby maps $\tau_1 \oplus \tau_0$, respectively, $\tau_1 \oplus \tau_0^\prime$. Let $s_1, s_2$ be isometries in $\mathcal{B}(\mathcal{H})$ with $s_1s_1^* + s_2s_2^* =1$ and set $w=s_1s_1^* +s_2 vs_2^*$, $p=ww^*$. Then $\beta(b)=w^*bw$ and $\eta(x)=w^*xw$ define an isomorphism $$\nonumber\begin{CD}
0 @>>> p\mathcal{K} p @>>> pEp @>\psi >> A @>>> 0 \\
& & @V\beta VV @VV\eta V @|   \\
0 @>>> \mathcal{K} @>>> E^\prime @>>\psi^\prime > A @>>> 0 
\end{CD}$$

[**Proof. **]{}[@MR99b:46108]

\[T:classify\] Let $x\in \rm{Ext}(C(\partial\mathbb{D})\times\Gamma)$ denote the class of the extension $$\nonumber\begin{CD}
0 @>>> \mathcal{K} @>>> {\mathcal{C}_\Gamma}@>>> C(\partial\mathbb{D})\times\Gamma @>>> 0.
\end{CD}$$ If $e\in \rm{Ext}(C(\partial\mathbb{D})\times\Gamma)$ is a unital extension represented by $$\nonumber\begin{CD}
0 @>>> \mathcal{K} @>>> E  @>>> C(\partial\mathbb{D})\times\Gamma @>>> 0
\end{CD}$$ and $e=x$, then $E\cong {\mathcal{C}_\Gamma}$ as $C^*$-algebras.

[**Proof. **]{}We first prove that the index homomorphism $\partial$ from $K_1(C(\partial\mathbb{D})\times\Gamma)$ to $\mathbb{Z}$ determined by the extension $x$ is surjective (and hence so is the homomorphism determined by $e$). First, consider the function $f(z)=z$; this function is a unitary in $C(\partial\mathbb{D})\times\Gamma$ lying in the canonical subalgebra isomorphic to $C(\partial\mathbb{D})$. By construction, this unitary lifts to the unilateral shift $S$ on $H^2$, which is a Fredholm isometry of index $-1$. Thus $\partial[z]_1 =-1$, and since $-1$ generates $\mathbb{Z}$, $\partial$ is surjective.

Since $\partial$ is surjective, the six-term exact sequence in $K$-theory associated to $E$ $$\begin{CD}
&0 @>>> &K_1(E) @>>> &K_1(C(\partial\mathbb{D})\times\Gamma) \\
&@AAA & & & & @VV\partial V \\
&K_0(C(\partial\mathbb{D})\times\Gamma) @<<< &K_0(E) @<<< &\mathbb{Z}  
\end{CD}$$ splits into the two sequences $$\begin{CD}
0 @>>> K_1(E) @>>> K_1(C(\partial\mathbb{D})\times\Gamma) @>\partial >> \mathbb{Z} @>>> 0
\end{CD}$$ $$\begin{CD}
0 @>>> K_0(E) @>>> K_0(C(\partial\mathbb{D})\times\Gamma) @>>>  0
\end{CD}$$ Thus $K_0(E)\cong K_0(C(\partial\mathbb{D})\times\Gamma)$, and as the isomorphism is induced by the quotient map which annihilates the compacts, it follows that $[p]_0=0$ for any finite projection $p\in E$.

Now let $\tau_1$ and $\tau_2$ denote the Busby maps associated to ${\mathcal{C}_\Gamma}$ and $E$ respectively. Since ${\mathcal{C}_\Gamma}$ and $E$ determine the same element in $\text{Ext}(C(\partial\mathbb{D})\times\Gamma)$, there exist unital trivial extensions $\sigma_1$ and $\sigma_2$ such that $\tau_1 \oplus \sigma_1
\sim_u \tau_2 \oplus \sigma_2$. Let $E_j$ denote the extension with Busby map $\tau_j \oplus \sigma_j$; we have $E_1\cong E_2$. Using $\sigma_1$ and $\sigma_2$, we may choose isometries $v_1$ and $v_2$ in $\mathcal{B}(\mathcal{H})$ as in the lemma to obtain strongly unital extensions $\sigma_1^\prime$ and $\sigma_2^\prime$. Let $E_j^\prime$ be the extension of $C(\partial\mathbb{D})\times\Gamma$ with Busby map $\tau_j \oplus
\sigma_j^\prime$, $j=1,2$. By Voiculescu’s theorem, $\tau_j \sim_u
\tau_j \oplus \sigma_j^\prime$, and it follows (since unitarily equivalent Busby maps determine isomorphic extensions) that $E_1^\prime \cong {\mathcal{C}_\Gamma}$ and $E_2^\prime \cong E$. Applying the lemma to $E_1$ and $E_2$, we obtain projections $p_j\in E_j$ such that $p_j
E_j p_j \cong E_j^\prime$. We claim that $[p_j]_0 = [1_{E_j}]_0$; from this the theorem follows, since we then have $p_j E_j p_j \cong
E_j$.

To prove the claim, observe that for the isometries $v_j$, the projections $1-v_jv_j^*$ are finite, and hence so are the projections $1-p_j$: $$\begin{aligned}
\nonumber
1-p = 1-ww^* &= 1-[s_1s_1^* + s_2vv^*s_2^*] \\
\nonumber             &= s_2s_2^* - s_2vv^*s_2^* \\
\nonumber             &= s_2[1-vv^*]s_2^* \end{aligned}$$ which is finite. Thus $1-p_j$ lies in $E_j$ (since $E_j$ contains the compacts) and $p_j\in Ej$ since $E_j$ is unital. Finally, $[1-p_j]_0
=0$, and $[p_j]_0 =[1_{E_j}]_0$. $\square$

Emerson’s construction
----------------------

When $\Gamma$ is cocompact, there is another extension of $C(\partial\mathbb{D})\times\Gamma$ which was constructed by Emerson [@emerson-thesis], motivated by work of Kaminker and Putnam [@MR98f:46056]; we will show that the $Ext$-class of ${\mathcal{C}_\Gamma}$ differs from the $Ext$-class of this extension.

The extension is constructed as follows: since $\Gamma$ is cocompact, we may identify $\mathbb{T}$ with the Gromov boundary $\partial\Gamma$. Let $f$ be a continuous function on $\partial\Gamma$, extend it arbitrarily to $\Gamma$ by the Tietze extension theorem, denote this extended function by $\tilde{f}$. Let $e_x, x\in\Gamma$ be the standard orthonormal basis for $\ell^2(\Gamma)$, and let $u_\gamma$, $\gamma\in\Gamma$ denote the unitary operator of left translation on $\Gamma$. Define a map $\tau:C(\partial\mathbb{D})\times\Gamma\to \mathcal{Q}(\ell^2(\Gamma))$ by $$\tau(f)e_x = \tilde{f}(x)e_x$$ and $$\tau(\gamma)e_x =u_\gamma e_x = e_{\gamma x},$$ it can be shown that these expressions are well defined modulo the compact operators and determine a $*$-homomorphism from $C(\partial\mathbb{D})\times\Gamma$ to the Calkin algebra of $\ell^2(\Gamma)$, i.e. an extension of $C(\partial\mathbb{D})\times\Gamma$ by the compacts. Let $\pi$ denote the quotient map $\pi:\mathcal{B}(\ell^2(\Gamma))\to \mathcal{Q}(\ell^2(\Gamma))$, and consider the pull-back $C^*$-algebra $E$: $$\begin{CD}
E @>\tilde{\tau}>> C(\partial\mathbb{D})\times\Gamma \\
@VVV   @VV\tau V \\
\mathcal{B}(\ell^2(\Gamma)) @>>\pi> \mathcal{Q}(\ell^2(\Gamma))
\end{CD}$$

We will show this extension is distinct from ${\mathcal{C}_\Gamma}$ by showing that it induces a different homomorphism from $K_1(C(\partial\mathbb{D})\times\Gamma)$ to $\mathbb{Z}$. Indeed, consider the class $[z]_1$ of the unitary $f(z)=z$ in $K_1(C(\partial\mathbb{D})\times\Gamma)$; the extension belonging to ${\mathcal{C}_\Gamma}$ sends this class to $-1$. On the other hand, for the extension $\tau$ we claim the function $z$ lifts to a unitary in $E$; it follows that $\partial ([z]_1) =0$ in this case. To verify the claim, note that $z$ lifts to a diagonal operator on $\ell^2(\Gamma)$, and if $(x_n)$ is a subsequence in $\Gamma$ tending to the boundary point $\lambda\in\mathbb{T}$, then $\tilde{f}(x_n)\to f(\lambda)=\lambda$. We may thus choose all the $\tilde{f}(x)$ nonzero, and by dividing $\tilde{f}(x)$ by its modulus, we may assume they are all unimodular. Thus $z$ lifts to the unitary $diag(\tilde{f}(x))\oplus z$ in $E$.

Lott’s construction
-------------------

In this section we relate the extension $\tau$ given by $\mathcal{C}_\Gamma$ to an extension recently constructed by J. Lott [@lott-preprint]. In particular it will follow that (up to tensoring with $\mathbb{Q}$) $\tau$ lies in the range of the Baum-Douglas-Taylor boundary map $$\partial: K^0 (C(\overline{\mathbb{D}})\times\Gamma, C(\partial\mathbb{D})\times\Gamma )\to K^1(C(\partial\mathbb{D})\times\Gamma)$$ We first describe a construction of the extension $\sigma_+$ of [@lott-preprint]. Let $\mathcal{D}$ denote the Hilbert space of analytic functions on $\mathbb{D}$ with finite Dirichlet integral $$D(f)=\int_\mathbb{D}|f^\prime(z)|^2dA(z)$$ equipped with the norm $$\|f\|^2=|f(0)|^2 + \int_\mathbb{D}|f^\prime(z)|^2dA(z)=|f(0)|^2+D(f)$$ If $f$ is represented in $\mathbb{D}$ by the Taylor series $\sum_{n=0}^\infty a_n z^n$, then this norm is given by $$\|f\|^2 = |a_0|^2 + \sum_{n=1}^\infty n |a_n|^2$$ The operator of multiplication by $z$ on $\mathcal{D}$, denoted $M_z$, is a weighted shift with weight sequence asymptotic to $1$, and hence is unitarily equivalent to a compact perturbation of the unilateral shift on $H^2$. It follows that there is a \*-homomorphism $\rho:C(\partial\mathbb{D})\to \mathcal{Q}(\mathcal{D})$ with $\rho(z)=\pi(M_z)$. Now, by changing variables one checks that if $\gamma$ is a Möbius transformation, $D(f\circ\gamma)=D(f)$. Let $\mathcal{D}_0$ denote the subspace of $\mathcal{D}$ consisting of those functions which vanish at the origin. It then follows from the definition of the norm in $\mathcal{D}$ that the operators $$u_\gamma(f)(z)=f(\gamma(z))-f(\gamma(0))$$ are unitary on $\mathcal{D}_0$, and form a unitary representation of $\Gamma$. We extend this representation to all of $\mathcal{D}$ by letting $\Gamma$ act trivially on the scalars. Moreover, it is simple to verify (by noting that $u_\gamma$ is a compact perturbation of the composition operator $C_\gamma$) that for all $\gamma\in\Gamma$, $$u_\gamma M_z u_\gamma ^* = M_{\gamma(z)}$$ modulo compact operators. Arguing as in the proof of Theorem \[T:main\], we conclude that the pair $(\rho(f), \pi(u_\gamma))$ determines an injective \*-homomorphism from $C(\partial\mathbb{D})\times\Gamma$ to $\mathcal{Q}(\mathcal{D})$, which is unitarily equivalent to the Busby map $\sigma_+$ of [@lott-preprint].

We may now state the main theorem of this subsection:

\[T:lott\] The Busby maps $\tau$ and $\sigma_+$ are unitarily equivalent.

[**Proof. **]{}We first show that the Busby map $\tau: \mathcal{C}_\Gamma\to \mathcal{Q}(H^2)$ lifts to a completely positive map $\eta:C(\partial\mathbb{D})\times\Gamma \to\mathcal{B}(H^2)$. Define a unitary representation of $\Gamma$ on $L^2(\partial\mathbb{D})$ by $$U(\gamma^{-1})=M_{|\gamma^\prime|^{1/2}}C_\gamma$$ Together with the usual representation of $C(\partial\mathbb{D})$ as multiplication operators on $L^2$, we obtain a covariant representation of $(\Gamma, \partial\mathbb{D})$ which in turn determines a representation $\rho:C(\partial\mathbb{D})\times\Gamma
\to \mathcal{B}(L^2)$. Letting $P$ denote the Riesz projection $P:L^2\to H^2$, we next claim that the commutator $[\rho(a),P]$ is compact for all $a\in
C(\partial\mathbb{D})\times\Gamma$, and so the pair $(\rho, P)$ is an abstract Toeplitz extension of $C(\partial\mathbb{D})\times\Gamma$ by $\mathcal{K}$. To see this, it suffices to prove the compactness of the commutators $$[\pi(f), P] \quad \text{and} \quad [U(\gamma),P].$$ Now, it is well known that $[M_f, P]$ is compact, and as $U(\gamma)$ has the form $M_g C_{\gamma^{-1}}$ it suffices to check that $[C_\gamma,
P]$ is compact. It is easily checked that this latter commutator is rank one. Indeed, the range of $P$ is invariant for $C_\gamma$, so $[C_\gamma, P]= PC_\gamma P^\bot$. If we expand $h\in
L^2(\partial\mathbb{D})$ in a Fourier series $$h \sim \sum_{n\in\mathbb{Z}} \hat{h}(n)e^{in\theta}$$ then a short calculation shows that $$PC_\gamma P^\bot h \sim (\sum_{n<0} \hat{h}(n)\overline{\gamma(0)}^{|n|})
\cdot 1$$ so $PC_\gamma P^\bot$ is rank one.

Thus, we have established that the completely positive map $\eta:C(\partial\mathbb{D})\times\Gamma \to\mathcal{B}(H^2)$ given by $$\eta(a)=P\rho(a)P$$ is a homomorphism modulo compacts, and the calculations in the proof of Theorem \[T:main\] show that the map $$\tau(a)=\pi(P\rho(a)P)$$ coincides with the Busby map associated to ${\mathcal{C}_\Gamma}$. Thus, $\eta$ is a completely positive lifting of $\tau$ as claimed.

With the lifting $\eta$ in hand, to prove the unitary equivalence of $\tau$ and $\sigma_+$ it will therefore suffice to exhibit an operator $V:H^2\to \mathcal{D}$ such that $V$ is a compact perturbation of a unitary, and such that $$\pi(V\eta(a)V^*)=\sigma_+(a)$$ for all $a\in C(\partial\mathbb{D})\times\Gamma$. Since the crossed product is generated by the function $f(z)=z$ and the formal symbols $[\gamma]$, it suffices to establish the above equality on these generators. In fact, we may use the operator $V$ of Lemma \[L:v\_operator\]. Since the map $z^n \to n^{-1/2}z^n$ is essentially unitary from $H^2$ to $\mathcal{D}$, the proofs of statements (1) and (2) of Lemma \[L:v\_operator\] are still valid. The conclusion of statement (3) holds provided the hypothesis on $g$ is strengthened, by requiring that $g$ by analytic in a neighborhood of $\overline{\mathbb{D}}$. Moreover, the arguments of Theorem \[T:extclasses\] still apply, since the proof applies statement (3) of Lemma \[L:v\_operator\] only to the function $\psi$, which is indeed analytic across the boundary of $\mathbb{D}$. Thus, the arguments in the proof of Theorem \[T:extclasses\] prove that $V$ intertwines (modulo compacts) multiplication by $z$ on $H^2$ and $\mathcal{H}$, and also intertwines (modulo compacts) $C_\gamma$ acting on $\mathcal{H}$ with $U_\gamma$ on $H^2$. Since $u_\gamma$ is a compact perturbation of $C_\gamma$ on $\mathcal{H}$, it follows that $V$ intertwines $U_\gamma$ and $u_\gamma$.

Now, the Busby map $\sigma_+$ takes the function $z$ to the image of $M_z$ in the Calkin algebra $\mathcal{Q}(\mathcal{H})$. Since $\eta(z)=M_z\in\mathcal{B}(H^2)$, the intertwining property of $V$ (modulo compacts) may be written as $$\pi(V\eta(z)V^*)=\sigma_+(z)$$ Similarly, since $\eta$ applied to the formal symbol $[\gamma]$ (viewed as a generator of $C(\partial\mathbb{D})\times\Gamma$) is $U_\gamma$, the intertwining property for $V$ with respect to $U_\gamma$ and $u_\gamma$ reads $$\pi(V\eta([\gamma])V^*)=\sigma_+([\gamma])$$ Thus the equivalence of $\tau$ and $\sigma_+$ on generators is established, which proves the theorem. $\square$

We conclude by observing that the covariant representation on $L^2$ described in the proof of the previous theorem gives rise to an equivariant $KK_1$-cycle for $C(\partial\mathbb{D})$. Indeed, such a cycle consists of a triple $(U, \pi, F)$ where $(U, \pi)$ is a covariant representation on a Hilbert space $\mathcal{H}$ and $F$ is a bounded operator on $\mathcal{H}$ such that the operators $$F^2-I,\  F-F^*,\  [U(\gamma),F], \text{ and } [\pi(f),F]$$ are compact for all $\gamma\in\Gamma$ and $f\in C(\partial\mathbb{D})$. The computations in the previous proof show that the triple $(U, \pi, 2P-I)$ satisfies all of these conditions, and essentially the same unitary equivalence argument shows that this cycle represents (up to a scalar multiple) the class of [@lott-preprint Section 9.1].

[^1]: \*Research partially supported by NSF VIGRE grant DMS-9983601

[^2]: 2000 [*Mathematics Subject Classification.*]{} 20H10, 46L55, 46L80, 47B33, 47L80
