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Let me also mention that the norms are continuous in the limits $q \to 0$ and $q \to 1$. When $q = 1$, the relations define an abelian $C^\ast$-algebra with self-adjoint $a$ and an element $b$ s.t. $|b|^2 = a(1-a)$. It is not hard to check that this is simply the algebra of continuous functions on the unit sphere, with $a = \frac{1+z}{2}$ and $b = \frac{x+iy}{2}$ where $(x,y,z)$ are Cartesian coordinates functions. Thus, we have $\|a\| = 1$ and $\|b\| = \frac{1}{2}$. In $A_q$, when $q \geq \sqrt{\frac{1}{2}}$, recall that we have,

Let me also mention that the norms are continuous in the limits $q \to 0$ and $q \to 1$. When $q = 1$, the relations define an abelian $C^\ast$-algebra with self-adjoint $a$ and an element $b$ s.t. $|b|^2 = a(1-a)$. It is not hard to check that this is simply the algebra of continuous functions on the unit sphere, with $a = \frac{1+z}{2}$ and $b = \frac{x+iy}{2}$ where $(x,y,z)$ are Cartesian coordinate functions (for a detailed proof of this fact, see my answer here). Thus, we have $\|a\| = 1$ and $\|b\| = \frac{1}{2}$. In $A_q$, when $q \geq \sqrt{\frac{1}{2}}$, recall that we have,

Here is a proof that any representation of $A_q$ on a separable Hilbert space is a direct sum of copies of the representation in the first part of my answer (which, for clarity's sake, we shall call the standard representation) and possibly a zero part (i.e., a part on which $a, b$ both act as $0$. The identity operator still acts as the identity).

Here is a proof that any representation of $A_q$ on a separable Hilbert space is a direct sum of copies of the representation in the first part of my answer (which, for clarity's sake, we shall call the standard representation) and possibly a zero part (i.e., a part on which $a, b$ both act as $0$. The identity operator still acts as the identity). Despite the representation not being assumed faithful, I’ll abuse the notations slightly and still just use $a$, $b$ and so on to denote the operators corresponding to these elements of $A_q$ under the given representation.

Here is a proof that any faithful representation of $A_q$ on a separable Hilbert space is a direct sum of copies of the representation in the first part of my answer (which, for clarity's sake, we shall call the standard representation) and possibly a zero part (i.e., a part on which $a, b$ both act as $0$. The identity operator still acts as the identity).

Here is a proof that any representation of $A_q$ on a separable Hilbert space is a direct sum of copies of the representation in the first part of my answer (which, for clarity's sake, we shall call the standard representation) and possibly a zero part (i.e., a part on which $a, b$ both act as $0$. The identity operator still acts as the identity).

Edit: I realized that the argument never assumes the representation is faithful, so any representation on a separable Hilbert space is a direct sum of copies of the standard representation - which is faithful and irreducible, and a zero part (again, this means $a, b$ both act as $0$, but the identity still acts as the identity). So there are exactly two proper closed ideals of $A_q$, namely the zero ideal, which is primitive, and the ideal generated by $a$ and $b$, which is maximal - in fact, $A_q/\langle a, b \rangle = \mathbb{C}$. Actually, one easily verifies that $C^\ast(A, B)$ in the standard representation is just $\mathbb{K}(\ell^2)$, so $A_q$, for any $0 < q < 1$, is simply $\mathbb{K}(\ell^2) \oplus \mathbb{C}$ (the unitization of the algebra of compact operators). In particular, they are all isomorphic to each other.