According to Wikipedia, one can construct noncommutative tori as follows: Letting $\theta \in \mathbb{R}$ be a parameter, consider the hilbert space $H = L^2(\mathbb{T})$, where $\mathbb{T}$ is the 1-torus, i.e. unit circle. Let $U$ and $V$ be the unitary operators on $H$ given by
$$(Uf)(z) = zf(z)$$ $$(Vf)(z) = f(e^{-2 \pi i \theta }z)$$
Then the noncommutative torus $A_{\theta}$ is $C^*(U,V)$, the unitary $C^*$-subalgebra of $B(H)$ generated by $U$ and $V$.
Wikipedia also claims that with $\theta = 0$, we recover the ordinary algebra of continuous functions on the 2-torus.
However, setting $\theta = 0$, $V$ becomes the identity operator on $H$, so $C^*(U,V) = C^*(U)$. Then by Gelfand duality, $C^*(U) \cong \mathcal{C}(\Omega(C^*(U))) \cong \mathcal{C}(\sigma(U))$, where $\Omega(B)$ denotes the character space of an algebra $B$. I claim that $\sigma(U)$ is homeomorphic to $\mathbb{T}$: The inverse of $U-\lambda I$, if it exists, must be $$(U-\lambda I)^{-1}f(z) = \frac{f(z)}{z-\lambda}$$ but this exists for all $f$ if and only if $| \lambda | \neq 1$, in which case the operator is square-integrable.
But this shows that $C^*(U) \cong \mathcal{C}(\mathbb{T})$ which must be wrong, because we should get the 2-torus and not the 1-torus. So obviously I have done some mistake here, and it would be greatly appreciated if someone could point me in the right direction.
The specific unitaries appearing in the OP are a representation of the relations defining the rotation algebra $A_\theta$ that is faithful when $\theta$ is irrational (for otherwise $V^q=I$ for some integer $q$). The more general definition of the rotation algebra $A_\theta$ is as the universal C*-algebra generated by two unitaries $u,v$ that satisfy the relation
$$uv=e^{i2\pi\theta}vu.$$
When $\theta=0$, these two unitaries commute and they give rise to a commutative C*-algebra which is isomorphic to the algebra of continuous functions on the joint spectrum of $u$ and $v$, i.e. the 2-torus $S^1\times S^1$.