A question on quantum tori Let $\mathbb T_\theta^2$ be quantum tori generated by two unitary operators $u,v$. can $u,v$ be finite dimensional?
 A: Let's fix conventions, so that $vu = e^{2\pi i \theta}uv$. It follows from a result of Slawny that $C(\mathbb{T}^2_\theta)$ is simple iff $\theta$ is irrational, so that $C(\mathbb{T}^2_\theta)$ has a finite-dimensional representation only if $\theta$ is rational.
Now, suppose that $\theta = p/q$ for $p \in \mathbb{Z}$, $q \in \mathbb{N}$. Define $U_{q}, V_{q} \in B(\ell^2(\mathbb{Z}/q\mathbb{Z})) \cong M_q(\mathbb{C})$ by
$$
 \forall m \in \mathbb{Z}/q\mathbb{Z}, \quad U_{q}\delta_m := e^{2\pi i m/q}\delta_m, \quad V_{q}\delta_m := \delta_{m+1},
$$
which are just the clock and shift operators on $\ell^2(\mathbb{Z}/q\mathbb{Z}) \cong \mathbb{C}^q$. Then $U_q^p$ and $V_q$ are unitaries satisfying the commutation relation $V_q U_q^p = e^{2\pi i \theta} U_q^p V_q$, and hence
$$
 \pi_{p,q} : C(\mathbb{T}^2_\theta) \to B(\ell^2(\mathbb{Z}/q\mathbb{Z})), \quad \pi_{p,q}(u) := U_q^p, \quad \pi_{p,q}(v) := V_q
$$
gives you a $q$-dimensional unitary representation, whose image is called a “fuzzy [noncommutative] torus” in the mathematical physics literature. In fact, it follows from a result of Høegh-Krohn–Skjelbred that if $\theta = p/q$ in least terms, then $C(\mathbb{T}^2_\theta)$ can be identified with the $C^\ast$-algebra of global sections of a locally trivial bundle of $C^\ast$-algebras over $\mathbb{T}^2$ with fibre $M_q(\mathbb{C})$.

ADDENDUM
The $C^\ast$-algebra $C(\mathbb{T}^2_\theta)$, which is the universal $C^\ast$-algebra generated by unitaries $u$ and $v$ satisfying $vu = e^{2\pi i \theta}uv$, is necessarily infinite-dimensional, and what the above (likely sub-optimal) argument shows is that it admits a finite-dimensional $\ast$-representation iff $\theta$ is irrational.
One way to construct $C(\mathbb{T}^2_\theta)$ is as a strict deformation quantisation of $C(\mathbb{T}^2)$. This means that it is a certain $C^\ast$-completion of the $\ast$-algebra of trigonometric polynomials on $\mathbb{T}^2$ together with the deformed multiplication $$z_1^{m_1}z_2^{m_2} \star_\theta z_1^{n_1} z_2^{n_2} := e^{-\pi i \theta(n_1m_2-m_2 n_1)} z_1^{m_1+n_1}z_2^{m_2+n_2},$$
so that $u := z_1$ and $v := z_2$ with $v \star_\theta u = e^{2\pi i \theta}u \star_\theta v$. Since the vector space of trigonometric polynomials on $\mathbb{T}^2$ is literally untouched by this procedure, the infinite set $$\{z_1^{m_1}z_2^{m_2}\}_{(m_1,m_2) \in \mathbb{Z}^2} = \{e^{\pi i \theta m_1m_2}u^{m_1} \star_\theta v^{m_2}\}_{(m_1,m_2) \in \mathbb{Z}^2}$$ remains linearly independent in the $C^\ast$-algebra $C(\mathbb{T}^2_\theta)$.
