$ for all polynomials $X$ in the variables $\a_i$ and $\a_j\1*$. This rather non-trivial result has been obtained in \tref{\BoSpei,\Fivel}. Two cases of the \qrel\ can be studied in complete detail. The first is the case of a single relation ($d=1$). Most of the physics literature treats either this case, or else commuting operators, each of which satisfies the \qrel\ separately. It is easy to see that, for $d=1$, the Fock representation is faithful, and contains only one proper two-sided ideal, which is generated by the one-dimensional projection onto the vacuum vector $\Omega$, and is isomorphic to the compact operators. The algebras $\Eq(1)$ are isomorphic, for all $q$, to the universal C*-algebra $\En(1)$ generated by an isometry. $\En(1)$ is the Toeplitz algebra \tref{\Douglas,\Coburn}. It is an extension \tref\BDF\ of $\C({\rm S}\11)$ by the algebra $\compacts$ of compact operators, i.e.\ there is an exact sequence $0\to\compacts\to\En(1)\to\C({\rm S}\11)\to0$. Under the isomorphisms $\En(1)\cong\Eq(1)$ the generators for different $q$ are related by multiplication with a suitable function of the number operator in Fock space, i.e.\ they are ``weighted shifts''\tref\Conway. It has been noted by several authors \tref{\Daskaloy,\Odaka} that such results are not specific to the \qrel, but also hold for more general relations like $\a\a\1*=\CR(\a\1*\a)$ for suitable functions $\CR$. The second case which has been fully analyzed is the case $q=0$, $d<\infty$. Each $\a_i\1*$ is then isometric, and the relations are essentially those studied by \tref\Cuntz. Again we have that the Fock representation is faithful, and contains a single closed two-sided ideal, which is generated by $\idty-\sum_i\a_i\1*\a_i$, and is isomorphic to the compact operators $\compacts$. The quotient of $\En(d)$ by this ideal is a simple C*-algebra known as the Cuntz algebra, and is usually denoted by ${\cal O}_d$. Again $0\to\compacts\to\En(d)\to{\cal O}_d\to0$ is exact. By analogy with the case $d=1$, we will call $\En(d)$ the \CuTa. Combining these two well-known cases one is led for general $q\in(-1,1)$ to the questions (1) whether $\Eq(d)$ is an extension of a simple C*-algebra by the compact operators; (2) whether the Fock representation of $\Eq(d)$ is faithful; (3) whether $\Eq(d)$ and $\En(d)$ are isomorphic for all $q$, and (4) whether such an isomorphism can be obtained by multiplying the generators of $\En(d)$ with a fixed $q$-dependent element of $\En(d)$. Questions (3) and (4) make sense also for the more general ``\Crel'' of the form given in equation (2). For a universal C*-algebra analogous to $\Eq(d)$ to exist, the matrix of functions $\CR_{ij}$ has to satisfy an upper bound guaranteeing a uniform upper bound on any solution of equation (2). Assuming, in addition, a Lipshitz continuity estimate for $\CR_{ij}$, and, most importantly, a lower bound making the matrix $\CR(a_1,\ldots,a_d)$ invertible for all admissible arguments, we show in Theorem 9 that the universal solution of the \Crel\ (2) is of the form $a_i=v_i\rho$, where the $v_i$ are the generators in $\En(d)$, and $\rho\in\En(d)$ is a positive element computed from $\CR$. Hence (3) and (4) hold true, and the \CuTa\ $\En(d)$ is the universal C*-algebra for all relations covered by our Theorem. We emphasize that our stability result does not fall within the range of analytic deformation theory (see Gerstenhaber's article in \tref\Gerstenh, and references cited there), since the $\CR_{ij}$ need not be analytic functions. It is also not a case of isomorphism of ``close C*-algebras'' in the sense of $\tref\Christensen$, since we make explicit use of the fact that the ``perturbed'' algebra is given in terms of generators and relations of a specific type. For the special case of the \qrel, the bounds needed in Theorem 9 are satisfied for $\absq<\sqrt2\,-1$. Thus the above conjectures (1) to (4) are valid for these values of $q$. Moreover, all information that has been accumulated about the Cuntz algebra \tref{\Evans,\Roberts} becomes immediately relevant to the representations of the \qrel. Our paper is organized as follows: In section 1 we construct the universal algebra $\Eq(d)$ for the \qrel, and state some of its elementary properties. In section 2 we prove the stability result for the \CuTa. In section 3 give some applications of this result. Apart from the \qrel, we study in some detail the relations $\a_i\a_j\1*-q\a_i\1*\a_j= \delta_{ij}\idty$ which differ from the \qrel\ in the position of the indices in the term containing $q$. For these modified relations we also compute the spectrum of $\sum_i\a_i\1*\a_i$ in all possible bounded representations, and show that representations fail to exist for some negative values of $q$. \beginsection 2. The C*-algebra $\Eq(\H)$ In this section, we use a slightly different version of the \qrel, which makes the basic symmetries of these relations more transparent. Let $\H$ be a vector space with sesquilinear form $\bra\cdot,\cdot>$, and let $q$ be a number with $-1\idty \qquad g,f\in\H \quad.\eqno(3)$$ Any map $\a$ with these properties will be called a representation, or a realization of the \qrel. The connection with the relations described in the introduction is then, that for some basis $\set{e_i}\subset\H$ which is orthonormal with respect to $\bra\cdot,\cdot>$, we can write $\a_i=\a(e_i)$ and $\a(\sum_if_ie_i)=\sum_if_i\a_i$. First, we need a result about the case of a single generator $\a=\a_1$ where $\H$ is one-dimensional. Further details about this case will be given below in Example 2. \iproclaim Proposition 1. Let $\Alg$ be a C*-algebra with unit, $q\in(-1,1)$, $c\in\Rl$, and $\a\in\Alg$ a non-zero element satisfying $$ \a\a\1*-q\a\1*\a=c\idty \quad.$$ Then $c>0$, and either $\a\a\1*=\a\1*\a=c/(1-q)\idty$, or the spectra of $\a\a\1*$ and $\a\1*\a$ are equal to the closure of the sequence $$ c{1-q\1n\over 1-q} \quad,$$ where $n\in\Nl$, and $n\geq0$ for $\a\1*\a$ and $n\geq1$ for $\a\a\1*$. \eproclaim \proof: \def\aas{\a\a\1*} \def\asa{\a\1*\a}% We may assume $q\neq0$. Let $\spec x$ denote the spectrum of an element $x\in\Alg$. Then $\spec\aas=c+q\spec\asa$, and $\spec\aas\setminus\set0=\spec\asa\setminus\set0$. Hence with $f(x)=c+qx$ we have for $x\neq0$ that $x\in\spec\aas$ if and only if $f(x)\in\spec\aas$. The sequence $f\1{-n}(x)$, $n\geq0$ is unbounded unless $x=x_\infty=c(1-q)\1{-1}$. Thus if $x_\infty\neq x\in\spec\aas$, the backwards iteration from $x$ must terminate in $0$. It follows that $x=f\1n(0)$ for some $n\geq1$. From the explicit solution of the forward iteration given in the Proposition, one sees that the forward iterates are all non-zero, hence the entire sequence must be in $\spec\aas$. The exceptional value $x_\infty$ is in the closure of the sequence. Since $\aas\geq0$ we get $c=(1-q)x_\infty\geq0$. It remains to exclude the possibility $c=0$. Since $\a$ was assumed to be non-zero there must be some non-zero $x\in\spec\aas$. Hence with $c=0$ the whole sequence $q\1{-n}x$ would be in $\spec\aas$. This contradicts the boundedness of $\a$. \QED \iproclaim Corollary 2. Suppose $\a:\H\to\Alg$ satisfies the \qrel. Then the sesquilinear form $\bra\cdot,\cdot>$ on $\H$ is positive semidefinite, and $$\eqalign{ \norm{\a(f)}=c(q)\bra f,f>\1{1/2} \quad,\quad\hbox{where}\qquad\qquad c(q)=\cases{ {1\over\sqrt{1-q}} & for $0\leq q<1$\cr 1 & for $-1< q\leq0$ } \quad.}$$ \eproclaim \noindent It follows that the map $\a$ is continuous for the topology on $\H$ given by the seminorm $\norm{f}=\bra f,f>\1{1/2}$ and the norm topology on $\Alg$. Therefore it extends by continuity to the completion of $\H$, and this extension again satisfies the \qrel. Hence there is no loss of generality in assuming that $\H$ is a Hilbert space, and we will make this assumption from now on. An important consequence of Corollary 2 is that the norm bound on $\a(f)$ is given by a constant independent of the particular realization of $\a$. Only this fact is needed to construct the universal C*-algebra associated with the \qrel, which is given in the following Proposition. \iproclaim Proposition 3. \item{(1)} For a given Hilbert space $\H$, and any $q\in(-1,1)$, there exists a C*-algebra, denoted by $\Eq(\H)$ with a map $\a:\H\to\Eq(\H)$ satisfying the \qrel\ with the following universal property: for any map $\tilde\a:\H\to\Alg$ satisfying the \qrel\ there is a unique *-homomorphism $\tilde\pi:\Eq(\H)\to\Alg$ such that $\tilde\pi(\a(f))=\tilde\a(f)$ for all $f\in\H$. \item{(2)} $\Eq(\H)$ is determined up to C*-isomorphism. For any isometry $V:\H_1\to\H_2$ between Hilbert spaces, there is a unique *-homomorphism $\Eq(V):\Eq(\H_1)\to\Eq(\H_2)$ such that $\Eq(V)\bigl(\a(f)\bigr)=\a(Vf)$. Thus $\Eq$ is a covariant functor from the category of Hilbert spaces with isometries into the category of C*-algebras with *-homomorphisms. \item{(3)} $V\mapsto\Eq(V)$ is continuous on the unitary group of $\H$ in the sense that strong convergence $V_\alpha\to V$ implies $\Eq(V_\alpha)\bigl(X\bigr)\to\Eq(V)\bigl(X\bigr)$ in norm for all $X\in\Eq(\H)$. \item{(4)} $\Eq(\H)$ carries a natural $\Ir$-grading $$ \Eq(\H)=\bigoplus_{m\in\Ir}\Eq(\H)_m \quad,$$ where $X\in\Eq(\H)_m$ iff $\Eq(\eta\idty)(X)=\eta\1m\,X$ for any $\eta\in\Cx$ with $\abs\eta=1$. The projections $\Pr_m:\Eq(\H)\to\Eq(\H)_m$ onto the summands are given explicitly by $$ \Pr_m(X)=\int\!\! d\eta\ \eta\1{-m}\Eq(\eta\idty)(X) \quad,$$ where $d\eta$ denotes the normalized Haar measure on the circle. \eproclaim \proof: (1),(2) Let $\Eqa(\H)$ denote the quotient of the *-algebra of non-commuting polynomials in the generators $\a(f)$ by the ideal generated by the \qrel\ and the relations arising from the linearity of $\a$. It is straightforward to check that $\Eqa(\H)$ satisfies the analogoues of (1) and (2) in the category of *-algebras: if $\tilde\a$ satisfies the relations, the homomorphism $\tilde\pi$ is defined simply by substituting the given $\tilde\a(f)$ into any polynomial in $\Eqa(\H)$. In order to get a universal {\bf C*}-algebra with the analogous property we have to define a C*-seminorm $\norm{\cdot}$ on $\Eqa(\H)$ with the property that for any {\it bounded} representation of the relations the substitution homomorphism is continuous, and hence extends to the completion $\Eq(\H)$ of $\Eqa(\H)$ with respect to that seminorm. This is to say that for any polynomial $X\in\Eqa(\H)$, and any bounded realization $\tilde\a$ we must have $\norm{\tilde\pi(X)}\leq\norm{X}$. Hence we want to define $$ \norm{X}=\sup_{\tilde\a}\norm{\tilde\pi(X)} \quad, $$ where the $\sup$ is over all bounded realizations $\tilde\a$ of the relations, and $\tilde\pi$ denotes the associated substitution homomorphism. There are three problems with this definition. The first is the technical point that the bounded realizations of the relations do not form a set, but a proper class. However, since $\norm{\tilde\pi(X)}$ may be computed in the C*-algebra gene\-rated by the $\tilde\a(f)$, we may restrict the supremum to realizations in C*-algebras with at most a certain cardinal number of elements depending on the number of generators via the dimension of $\H$. We now pick a Hilbert space of sufficiently high dimension such that all the universal representations of all these algebras can be realized in it. It is then clear that the supremum can be restricted to the set of realizations of the \qrel\ in bounded operators on this big Hilbert space. The second problem is whether the supremum is possibly over the empty set, i.e. whether any bounded realizations of the relations exist at all. For the case at hand this question has been settled by \tref\BoSpei, who explicitly construct the so-called Fock representation for all $q\in(-1,1)$ (See Example 1 below). The third problem is that the supremum might turn out to be infinite for some $X$. Since the $X\in\Eqa(\H)$ are polynomials in $\tilde\a(f)$ this can be ruled out by proving an upper bound on $\norm{\tilde\a(f)}$, which is uniform with respect to all realizations $\tilde\a$ of the \qrel. For the case at hand a bound of this kind was established in Corollary 2. Hence $\Eq$ exists and has the universal property (1). The uniqueness and the further properties stated in (2) follow as usual from the universal property. (3) It is clear from Corollary 2 that $V_\alpha\to V$ strongly implies $\Eq(V_\alpha)\bigl(\a(f)\bigr)\to\Eq(V)\bigl(\a(f)\bigr)$ for all $f\in\H$. Hence we also have convergence for all polynomials $X$ in the variables $\a(f),\a(g)\1*$, and by a straigthforward estimate also the convergence for $X$ in the norm closure $\Eq(\H)$ of the polynomials. (4) The integral for $\Pr_m$ is well-defined and in $\Eq(\H)$ because by (3) the integrand is continuous in norm. On monomials in $\a(f)$ and $\a(g)\1*$ we can compute the degree $m$ by counting the number of factors $\a(f)$ and $\a(f)\1*$. Hence on polynomials, the Fourier series decomposition $X=\bigoplus_m\Pr_m(X)$ is evident, and carries over by continuity to all $X\in\Eq(\H)$. \QED % We remark that in constructions of this type it might happen that, % for some non-zero $X\in\Eqa(\H)$, we get $\norm{X}=0$. Such elements % will be mapped to $0\in\Eq(\H)$ under the completion map. They % correspond exactly to the additional algebraic conditions which % follow from the boundedness of the representation. By (2), $\Eq(\H)$ depends only on the dimension of $\H$. Since we will later be mostly interested in the case $\dim\H<\infty$, we will write $\Eq(d)$ for $\Eq(\Cx\1d)$. \examp 1. (The Fock representation): Consider a representation of the \qrel\ by bounded operators in a Hilbert space $\K$. Let $$ \N=\set{\xi\in\K \stt \forall_{f\in\H}\ \a(f)\xi=0} \quad.\eqno(4)$$ We call $\N$ the set of Fock vectors in this representation. Note that by applying the \qrel\ we can arrange any polynomial $X$ in the generators ``in standard form'', i.e. such that all $\a(f)$ are to the right of all $\a(f)\1*$. In this form we have for all $\xi,\eta\in\N$ $$\bra\xi,X\eta>=\fock(X)\bra\xi,\eta> \quad,\eqno(5)$$ where $\fock(X)$ denotes the constant term in the standard form. For each $\xi\in\N\setminus\set0$ we obtain a cyclic representation of $\Eq(\H)$, which is the GNS representation associated with the Fock state $\fock$. This is called the Fock representation. Since the cyclic subspaces generated from orthogonal Fock vectors are orthogonal, we can choose an orthonormal basis in $\N$, and thus obtain a direct sum of isomorphic copies of the Fock representation. The subspace on which this direct sum lives can be represented naturally as $\K_\fock\otimes\N$, where $\K_\fock$ denotes the GNS-Hilbert space of the Fock representation, and $\a(f)\rstr\K_\fock\otimes\N\cong \a_\omega(f)\otimes\idty$. The orthogonal complement of this invariant subspace is characterized by the property that it contains no non-zero Fock vectors. Note that the argument so far does not imply that there is any representation with $\N\neq\set0$. This will be the case if and only if the constant term in the standard form is indeed a positive functional on $\Eqa(\H)$. Equivalently, one has to check that the unique sesquilinear form on the vector space spanned by vectors of the form $\a(f_1)\cdots\a(f_n)\Omega$ which is computed by using the \qrel\ together with the equation $\a(f)\Omega\equiv0$, and the normalization condition $\norm{\Omega}=1$, is positive definite. Bozejko and Speicher \tref\BoSpei\ have shown this by reducing it to the positive definiteness of the function $\pi\mapsto q\1{i(\pi)}$ on the permutation group of $n$ elements, where $i(\pi)=\abs{\set{(i,j)\stt i\pi(j)}}$ denotes the number of inversions in $\pi$. \examp 2. (The case of a single relation): By Proposition 1 a representation of the relation $\a\a\1*-q\a\1*\a=\idty$ containing no Fock vectors has the property that $\a=(1-q)\1{-1/2}u$, with a unitary $u$. Equivalently, in such representations we have the equation $\a\a\1*=\a\1*\a$. By Example 1 every representation is a direct sum of such an abelian representation and a multiple of the Fock representation. We can obtain an explicit picture of the Fock representation by taking the polar decomposition $a=v(\a\1*\a)\1{1/2}$. Since by Proposition 1 $\a\a\1*$ is boundedly invertible it follows that $v\1*=\a\1*(\a\a\1*)\1{-1/2}$ is an isometry, which lies in the C*-algebra $\Eq(\Cx)$. Similarly, the projection $\idty-vv\1*=\supp(\a\1*a)$ onto the set $\N$ of Fock vectors lies in $\Eq$. For studying the Fock representation we may thus take this projection to be one-dimensional in the representation space. Denoting by $\ket0$ the Fock vector, and $\ket n=(v\1*)\1n\ket0$ we get $vt\12v\1*=\idty+qt\12$. Thus $v$ is a one-sided shift implementing the iteration on the spectrum used in the proof of Proposition 1, and we get $\langle n\vert t\12\ket m = \delta_{nm} (1-q\1n)/(1-q)$. The algebra generated by $v$ thus contains $t\12$, and hence $\Eq(\Cx)$ is generated by the single isometry $v\1*$. The ideal generated by $\idty-vv\1*$ is isomorphic to the compact operators, and the quotient by this ideal gives back the abelian representations. Thus all irreducible representations of $\Eq(\Cx)$ are quotients of the the Fock representation, which therefore is faithful. Moreover, and the algebras $\Eq(\Cx)$ are isomorphic for all $q\in(-1,1)$ to the well-known C*-algebra $\En(\Cx)$ \tref\Coburn\ generated by a single isometry. The following Proposition shows that in a representation with several generators each single generator occurs in the Fock representation. \iproclaim Proposition 4. For $\dim\H\geq2$, any bounded representation of the \qrel, and every $0\neq f\in\H$, $0$ is in the spectrum of $\a(f)\1*\a(f)$. \eproclaim \proof: Since $\a(f)$ generates a homomorphic image of $\Eq(\Cx)$ we only have to show that this homomorphism is faithful, i.e. that $u=\norm{f}\1{-1}\sqrt{1-q}\,\a(f)$ is not unitary. Suppose it were. Then from the commutation relations we would get for $g\perp f$: $U\a(g)\1*U=q\a(g)\1*$. Hence $\norm{\a(g)}=\absq\norm{\a(g)}$, Because $\absq<1$ this implies $\a(g)=0$, which implies $g=0$. This implies that $\H$ is one-dimensional. \QED The following is a typical feature of infinite C*-algebras \tref\Cuntz, i.e.\ algebras containing elements with $x\1*x=\idty$, but $xx\1*\neq\idty$, of which the Cuntz algebra is a prototype. \iproclaim Corollary 5. For $-1 0$, and $x,y\geq\epsilon\idty$. Then $$ \norm{\sqrt x-\sqrt y}\leq{1\over2\sqrt\epsilon}\norm{x-y} \quad.$$ \eproclaim \proof: \def\hx{\hat x}% \def\hy{\hat y}% \def\he{\hat\epsilon}% By multiplying $x$ and $y$ with the same factor we may assume that $x=\idty+\hx$, $y=\idty+\hy$ with $\norm{\hx},\norm{\hy}\leq\he=1-\epsilon$. Then $$\eqalign{ \norm{\sqrt{\idty+\hx}-\sqrt{\idty+\hy}} &\leq\sum_{n=0}\1\infty \abs{{1/2\choose n}} \norm{\hat x\1n-\hat y\1n} \leq\norm{\hx-\hy}\sum_{n=0}\1\infty \abs{{1/2\choose n}}n\he\1{n-1} \cr &={1\over2}\norm{x-y}\sum_{n=1}\1\infty \abs{{-1/2\choose n-1}}(-\he)\1{n-1} ={1\over2} (1-\he)\1{-1/2}\norm{x-y} \quad.}$$ \QED For stating our main Theorem it is convenient to introduce the norm $$\norm{a}\12=\Bigl\Vert\sum_{i=1}\1d a_i\1*a_i \Bigr\Vert \eqno(11)$$ on $d$-tuples $a=(a_1,\ldots,a_d)\in\Alg\1d$. Note that by the proof of Proposition 7 we have $\norm{\dyad aa}=\norm{\hat a\hat a\1*} =\norm{\hat a\1*\hat a}=\norm{a}\12$. \iproclaim Theorem 9. Let $\CR_{ij}$, $i,j=1,\ldots,d$ be in the C*-functional calculus of $d$ variables such that $\CR(a)\in\M_d(\Alg)$ is hermitian for all C*-algebras $\Alg$ and all $a\in\Alg\1d$. Suppose there are constants $\mu,\epsilon>0,\lambda<2\sqrt\epsilon$ such that \item{(1)} the \Crel\ $\a_i\a_j\1*= \CR_{ij}(\a)$ imply $\norm{\a}\12\leq\mu$. \item{(2)} For arbitrary $a,b\in\Alg\1d$ with $\norm{a}\12,\norm{b}\12\leq\mu$ we have the estimates $$\eqalignno{ \norm{\CR(a)}&\leq \mu &(a)\cr \CR(a)&\geq\epsilon\idty &(b) \cr \norm{\CR(a)-\CR(b)} &\leq\lambda\norm{a-b} \quad.&(c)}$$ \noindent Then there is a unique $\rho\in\En(d)$ such that $\rho\geq0$, $\rho\sum_iv_i\1*v_i=\rho$, and that $\a_i=v_i\rho$ satisfies the \Crel, where the $v_i$ are the generators in $\En(d)$. \hfill\break Moreover, this solution has the following universal property: if $\tilde\a_1,\ldots,\tilde\a_d\in\Alg$ satisfies the \Crel\ in any C*-algebra $\Alg$, there is a unique *-homomorphism $\tilde\pi:\En(d)\to\Alg$ such that $\tilde\pi(v_i\rho)=\tilde\a_i$ for all $i$. \eproclaim \proof: \def\va{\tilde v}% \def\aa{\tilde a}% \def\ra{\tilde\rho}% \def\tp{\tilde\pi}% Let $\Alg$ be any C*-algebra containing elements $\va_i$, $i=1,\ldots,d$ satisfying the \nullrel. We will iterate $$ f(x)=\Bigl( \sum_{ij}\va_i\1*\CR_{ij}(\va x) \va_j \Bigr)\1{{1\over2}} $$ on the set $$ X=\set{x\in\Alg\stt 0\leq x\12 \leq \mu\, \Fpr } \quad,$$ where $(\va x)_i=\va_ix$, and $\Fpr=\sum_i\va_i\1*\va_i$. Then by equation (10) the fixed points of $f$ are precisely those $x\in\Alg$ with $x\geq0$ and $x\Fpr=x$ such that $\va x$ satisfies the \Crel. The set $X$ is invariant, because $\norm{\va x}\12\leq\norm{x}\12$, so $x\12\leq \mu\Fpr$ implies $\norm{f(x)\12}\leq\norm{\CR(\va x)}\leq\mu$ by (a), and $f(x)\12=\Fpr f(x)\12\Fpr\leq\mu\Fpr$ by the \nullrel. By assumption (b) $\CR(\va x)\geq\epsilon\idty$ for $x\in X$, hence $f(x)\12\geq\epsilon\Fpr$. The square root is thus well defined, and by Lemma 8 (applied in the algebra $\Fpr\Alg\Fpr$ with unit $\Fpr$) we get from (c) the estimate $$ \norm{f(x)-f(y)} \leq {1\over2\sqrt\epsilon}\norm{f(x)\12-f(y)\12} \leq {\lambda\over2\sqrt\epsilon}\norm{\va x-\va y} \leq {\lambda\over2\sqrt\epsilon}\norm{x-y} \quad.$$ Thus $f$ is contractive, and has a unique fixed point $\ra$. We apply this firstly in the case $\Alg=\En(d)$ with $\va=v$ to obtain $\rho\in\En(d)$ as claimed in the Theorem. Now let $\aa\in\Alg\1d$ satsify the \Crel. Then by assumption (1) of the Theorem, $\norm{\aa}\12\leq\mu$, which implies that $\dyad\aa\aa=\CR(\aa)\geq\epsilon\idty$ is invertible. Hence, by Proposition 7, $\aa=\va\ra$, and $\ra$ must be the unique fixed point of the iteration in $\Alg$. On the other hand, by the universal property of $\En(d)$ for the \nullrel\ there is a unique homomorphism $\tp:\En(d)\to\Alg$ such that $\tp(v_i)=\va_i$. By the homomorphism property (7) of the functional calculus we have $$\eqalign{ \CR_{ij}\bigl(\va_1\tp(\rho),\ldots,\va_d\tp(\rho)\bigr) &=\CR_{ij}\bigl(\tp(v_1\rho),\ldots,\tp(v_d\rho)\bigr) =\tp\Bigl(\CR_{ij}(v_1\rho,\ldots,v_d\rho)\Bigr) \cr &=\tp(v_i\rho\12v_j\1*)=\va_i\tp(\rho)\12\va_j \quad.}$$ Hence $\tp(\rho)$ is also a fixed point of the iteration in $\Alg$, and we must have $\tp(\rho)=\ra$, and $\tp(a_i)=\va_i\ra=\aa_i$. \QED \beginsection 4. Applications \vskip-20pt\noindent {\bf 4.1. The \qrel } \noindent In equation (8) we have introduced the dyads $\dyad ab$ for $a,b\in\Alg\1d$. Here we need a second kind of dyad, which is equal to $\dyad\cdot\cdot$ when $a_i$ and $b_i$ commute. We set $$ \dayd ab_{ij}=b_j\1*a_i \qquad,\qquad \dyad ab_{ij}=a_ib_j\1* \quad.\eqno(12)$$ Using these dyads we can write the \qrel\ in the compact form $$ \dyad aa=\idty+q\,\dayd aa \quad.\eqno(13)$$ The norm bounds needed for the application of Theorem 9 are given in the next Lemma. \iproclaim Lemma 10. Let $a,b\in\Alg\1d$. Then $\norm{\dyad aa}=\norm{\sum a_i\1*a_i}\equiv\norm{a}\12$, and $$\norm{\dayd ab}\leq\norm{a}\norm{b} \qquad,\qquad \norm{\dyad ab}\leq\norm{a}\norm{b} \quad.$$ Moreover, $\norm{a\1*}\12\leq d\norm{a}\12$. \eproclaim \proof: The equation for $\norm{\dyad aa}$ was already noted before Theorem 9. $$ \bigl(\dyad ab\1*\dyad ab\bigl)_{ij} =\bigl(\dyad ba\dyad ab\bigl)_{ij} =b_i \bigl(\sum_k a_k\1*a_k\bigr)b_j\1* \quad.$$ The last expression can be written as $\hat b(\sum_k a_k\1*a_k)\hat b\1*$ with a $d\times1$-matrix $\hat b$. Estimating the bracket by its norm we get $\dyad ab\1*\dyad ab\leq\norm{a}\12\dyad bb$, and $\norm{\dyad ab}\12\leq\norm{a}\12\norm{\dyad bb}$. Similarly, we get $$ \bigl(\dayd ab\1*\dayd ab\bigl)_{ij} =\bigl(\dayd ba\dayd ab\bigl)_{ij} =\sum_k a_k\1* b_ib_j\1* a_k =\Phi(b_ib_j\1*) $$ with the completely positive map $\Phi(x)=\sum_k a_k\1* x a_k$. Hence $\dayd ab\1*\dayd ab\leq\norm{\dyad bb}\Phi(\idty)$, and $\norm{\dayd ab}\12\leq\norm{b}\12\norm{a}\12$. Finally we get $\norm{a\1*}=\norm{\sum a_ka_k\1*} \leq\sum_{k=1}\1d\norm{\a_k}\12 \leq d\sum\norm{\sum a_k\1*a_k}$. \QED \proof{ of Proposition 6:} With $\CR(a)=\idty+q\dayd aa$ we get $$ \norm{\CR(a)}=1+\absq\norm{\dayd aa}\leq 1+\absq\norm{a}\12 \quad.$$ Hence the \qrel\ imply $\norm{a}\12\leq 1+\absq\norm{a}\12$, and $\norm{a}\12\leq (1-\absq)\1{-1}\equiv\mu$. For $\norm{a}\12\leq\mu$ we get $\norm{\CR(a)}\leq\mu$ as required in condition (2a) of Theorem 9. Furthermore, $$\CR(a)\geq \idty(1-\absq\norm{\dayd aa} \geq \idty(1-\absq\mu)=\epsilon\idty $$ with $\epsilon=(1-2\absq)/(1-\absq)\idty$. Hence for $\CR(a)$ to be bounded away from zero as required in condition (2b) we need $\absq<1/2$. Using the bilinearity of $\dayd ab$ we get the Lipshitz bound $$\norm{\CR(a)-\CR(b)}=\absq\norm{\dayd aa-\dayd bb} \leq \absq\norm{a-b}(\norm{a}+\norm{b}) \leq 2\absq\sqrt\mu \norm{a-b} \quad.$$ With $\lambda=2\absq\sqrt\mu$ the contractivity condition then becomes $\absq<\sqrt{1-2\absq}$, or $\absq<\sqrt2-1$. \QED For the extension of Proposition 6 to a larger interval, possibly even to the whole range $q\in(-1,1)$, it is crucial to improve the bounds $$ {1-2\absq\over1-\absq}\idty\quad \leq \dyad aa \quad \leq{1\over1-\absq}\idty \eqno(14)$$ for $a$ satisfying the \qrel, which were obtained in the above proof. The upper bound is fairly good. In fact, for $q\geq0$ the upper bound is an equality since by Proposition 1 each diagonal element in $\dyad aa$ has norm $(1-q)\1{-1}$. As a corollary of this observation we find that in every representation with $d+1$ generators there are many vectors on which $\sum_i\1d\a_i\1*\a_i$ is small (namely those on which $a_{d+1}\1*\a_{d+1}$ nearly attains its maximum). The existence of such ``almost Fock'' vectors for a subset of generators can be seen as a generalization of Proposition 4, and can be extended also to some negative values of $q$. For $q<0$ the upper bound in (14) is the best possible $d$-independent bound. (The vectors in Fock space arising from (anti-)symmetrization of $\a_1\1*\cdots\a_n\1*\ket0$ are eigenvectors of $T$ with eigenvalue $(1-(\pm q)\1n)/(1\mp q)$). However, for given finite $d$ the bound can be improved. For example, for $d=2$ we get $$\eqalign{ \dyad aa&=\pmatrix{\a_1\a_1\1* &0\cr 0& \a_2\a_2\1*} +q\pmatrix{0& \a_2\1*\a_1\cr \a_1\1*\a_1} \cr \pmatrix{0& \a_2\1*\a_1\cr\a_1\1*\a_1}\1{\textstyle2} &= \pmatrix{\a_2\1*\a_1\a_1\1*\a_2 &0\cr 0&\a_1\1*\a_2\a_2\1*\a_1} \leq \pmatrix{\a_2\1*\a_2 &0\cr 0& \a_1\1*\a_1} \leq\idty \quad,}$$ where we have used the bound $\a\a\1*\leq1$ from Corollary 2. Hence $$ \norm{\dyad aa}\leq 1+\absq \qquad\hbox{for $d=2$.}\eqno(15)$$ We can insert this into the bound $\dyad aa\geq(1-\absq\norm{\dyad aa})\idty$ to get an improvement of the lower bound in (14): $$ \dyad aa\geq 1-\absq-\absq\12 \qquad\hbox{for $d=2$.}\eqno(16)$$ Hence $\dyad aa$ is boundedly invertible for $q>-\golden\1{-1}=-{1\over2}(\sqrt5-1)$. It is not clear what the best lower bounds on $\dyad aa$ are. Numerical evidence from Fock space suggests that $\dyad aa$ might be bounded away from zero for all $-10$ and $q<0$. For $q>0$ we apply Lemma 10 to get $\norm{\CR(a)}\leq 1+q\,\norm{\dyad{a\1*}{a\1*}} \leq 1+qd\norm{a}\12$, and set set $\mu=(1-qd)\1{-1}$ in Theorem 9. Clearly, we can set $\epsilon=1$. With $\lambda=2qd\sqrt\mu$ the contractivity estimate is $qd<\sqrt{1-qd}$, or $qd<\golden\1{-1}$. For $q<0$ we can take $\mu=1$. Then $$\CR(a)\geq \idty(1-\absq\norm{\dyad{a\1*}{a\1*}})\geq\epsilon\idty $$ with $\epsilon=(1-\absqd)$. The Lipshitz bound is $\lambda=2\absqd$, so contractivity holds once more for $\absqd<\sqrt{1-\absqd}$. \QED The fact that the interval for $q$ for which our technique works now depends on $d$ came in through the estimate $\norm{a\1*}\12\leq d\norm{a}\12$ in Lemma 10, which may appear to be exceedingly crude. However, this $d$-dependence is typical for the relations (17). We will show this now by using the idea of Proposition 1 to obtain rather detailed necessary conditions for such representations. The core of the idea is contained in the following Proposition: \iproclaim Proposition 12. Let $\a_i$, $i=1,\ldots,d$ be bounded operators satisfying the relations (17). Let $T=\sum_{i=1}\1d\a_i\1*a_i$, and consider the function $f(x)=1+qd+q\12x$. Then $$ x\in\spec T \iff f(x)\in\spec T \quad,$$ provided that both $f(x)\neq1$ and $f(x)\neq0$. \eproclaim \proof: We consider two $d\times1$-matrices with entries in $\Alg$, namely $A_i=\a_i$, and $\Ah_i=a_i\1*$. This somewhat redundant notation is necessary to keep apart the different meaning of adjoints. We can then write $$\matrix{ \hfil \dyad aa &=AA\1* \hfil\qquad& \hfil \dyad {a\1*}{a\1*}&=\Ah\Ah\1* \hfil\qquad& \hfil AA\1*&=\idty+q\,\Ah\Ah\1* \cr \hfil T:=\sum_ia_i\1*a_i&=A\1*A \hfil\qquad& \hfil \Th:=\sum_ia_ia_i\1*&=\Ah\1*\Ah \hfil\qquad& \hfil \Th&=d\idty+qT \cr}$$ Now suppose that $x$ satisfies the two conditions in the Proposition. Then $x\in\spec{T}\iff d+qx\in\spec\Th$. Since we have assumed $d+qx=(1/q)(f(x)-1)\neq0$ this is in turn equivalent to $d+qx\in\spec{\Ah\1*\Ah}$, and $d+qx\in\spec{\Ah\Ah\1*}$. This is equivalent to $(1+q(d+qx))=f(x)\in\spec{\A\A\1*}$. Since we have assumed that $f(x)\neq0$ this is equivalent to $f(x)\in\spec\T$. \QED \iproclaim Proposition 13. Let $a_1,\ldots,a_d$ with $d>1$ be bounded operators satisfying the relations (17). Let $T=\sum_{i=1}\1d\a_i\1*a_i$, and $x_\infty=(1+qd)/(1-q\12)$. For $n\in\Nl$, $n\geq0$ let $\lambda_n=x_\infty(1-q\1{2n})$ and $\mu_n=x_\infty+(1-x_\infty)q\1{2n}$. Then the spectrum of $T$ is contained in $$ \set{\lambda_n\stt n\geq0} \cup\set{\mu_n\stt n\geq0} \cup\set{x_\infty} \quad.$$ Moreover, one has \item{(1)} For $q\geq-d\1{-1}$, all $\mu_n$ are in the spectrum, and either all or no $\lambda_n$. Moreover, for $q\neq0$ $\lambda_n\neq\mu_m$ for all $n,m\geq0$. \item{(2)} For $q<-d\1{-1}$, the spectrum of $\T$ is of the form $$\set{\mu_0,\mu_1,...\mu_{N-1}}$$ with $\mu_N=0$ for some finite $N$. Thus representations can exist only for the discrete set of values $q$ for which the equation $\mu_N=0$, or equivalently $$ q\1{2N}={1+qd\over q(d+q)}$$ has a solution. \eproclaim \proof: If $x\in\spec T$, $x\neq x_\infty$, the iteration of $f\1{-1}$, with $f$ as in Proposition 12 yields an unbounded sequence. Thus by Proposition 12 the iterates must either hit $0$ or $1$, which means that either $x=\lambda_n$ or $x=\mu_n$ for some $n\in\Nl$. Next observe that Proposition 1 and the argument in Proposition 4 show that $0\in\spec{a_1\1*a_1}$. Thus by Proposition 1 we can choose a non-zero $\xi\in\ker(a_1\1*a_1)=\ker(a_1)$. Now the vector $(\xi,0,\ldots,0)$ is in $\ker(\Ah\Ah\1*)$. This implies $1\in\spec{AA\1*}$, and $\mu_0\in\spec T$. If $q>-1/d$ then $x_\infty$ is positive, and the sequence $\lambda_n$ is strictly increasing towards $x_\infty$. 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