Le $\theta$ be irrational. One can define the noncommutative torus $A\_{\theta}$ as a universal algebra generated by two unitaries $u,v$ satisfying the relation $vu=e^{2 \pi i \theta} uv$. This is an abstract defnition: however one can show that this algebra is simple and can be concretely represented as a $C^\*$-subalgebra of $B(L^2(\mathbb{T}))$ generated by $U$ and $V$ where $Uf(x)=e^{2\pi i x}f(x)$ and $Vf(x)=f(x+\theta)$. Denote this concrete algebra as $\mathfrak{A}$ and consider $\mathfrak{A}''$ which is von Neumann algebra.



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> How to prove that $\mathfrak{A}''$ is a type $II\_1$ factor (correct me if it isn't true)?
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Noncommutative torus as a von Neumann algebra
No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant:
$$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$
By the way, the invariant subspace problem for the Bishop operator $f(x)\mapsto xf(x+\theta)$ is still open in full generality.
<https://mathscinet.ams.org/mathscinet-getitem?mr=353015>


