Let $\mathbb T\_\theta^2$ be quantum tori generated by two unitary operators $u,v$. can $u,v$ be finite dimensional?



A question on quantum tori
Let's fix conventions, so that $vu = e^{2\pi i \theta}uv$. It follows from a result of [Slawny](https://projecteuclid.org/euclid.cmp/1103857742) 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](https://eudml.org/doc/152389) 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 $M\_q(\mathbb{C})$.




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**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 $C(\mathbb{T}^2\_\theta)$.


