Why Der($A_{\theta}$) is spanned by two elements only? In the work of Connes and Marcolli, on page 20, it state that:

Just as in the classical case of a (commutative) manifold, what ensures that the derivations
considered are enough to span the whole tangent space is the condition of ellipticity for the
Laplacian...

The derivations stated are given by (6.15), which are only two elements.
First thing I don't understand is how to prove this. Second thing I don't understand is, if we consider $\delta(U)=V^{*}-U^{2}Ve^{2\pi i\theta}$, $\delta(V)=0$, then it is also a derivation, and does not belongs to module spanned by $\delta_{1},\delta_{2}$. Am I missing something?
 A: Bratteli–Elliott–Jorgensen prove a range of classification results for unbounded derivations on a totally irrational noncommutative torus $C(\mathbb{T}^n_\theta)$, which basically say that any reasonable $\ast$-derivation will be the sum of a $\mathbb{R}$-linear combination of the infinitesimal generators $\delta_1,\dotsc,\delta_n$ of the translation action of $\mathbb{T}^n$ and an (approximately) inner derivation.
Here’s a typical special case, which follows from a combination of results in Bratteli–Elliott–Jorgensen. Let $\theta \in \mathbb{R}$ be irrational, let $U$ and $V$ be the usual unitary generators of the continuous noncommutative $2$-torus $C(\mathbb{T}^2_\theta)$, and let $\mathcal{O}(\mathbb{T}^2_\theta)$ be the dense $\ast$-subalgebra of Fourier polynomials in $U$, $V$. Then any $\ast$-derivation $\delta : \mathcal{O}(\mathbb{T}^2_\theta) \to \mathcal{O}(\mathbb{T}^2_\theta)$ has a unique decomposition $$\delta = c_1 \delta_1 + c_2 \delta_2 + \tilde{\delta},$$ where $c_1,c_2 \in \mathbb{R}$ and $\tilde{\delta}$ is inner, i.e., there exists self-adjoint $H \in \mathcal{O}(\mathbb{T}^2_\theta)$, such that $\tilde{\delta}(a) = i[H,a]$ for all $a \in \mathcal{O}(\mathbb{T}^2_\theta)$.
Please note that your would-be derivation $\delta$ is ill-defined, since you’d have
$$
 \delta(VU) = 1 - (e^{2\pi i\theta})^3U^2 V^2, \quad \delta(e^{2\pi i \theta} UV) = e^{2\pi i \theta} - (e^{2\pi i \theta})^2 U^2V^2.
$$
