# Noncommutative torus as a von Neumann algebra

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.

How to prove that $$\mathfrak{A}''$$ is a type $$II_1$$ factor (correct me if it isn't true)?

• In context, ‘jest’ must be ‘is’. – Branimir Ćaćić Aug 3 '20 at 21:33
• Hi @JonBannon : I'm not too familiar with crossed products, is it clear that the "natural" vN completion of the abstract crossed product $C(S^1)\rtimes {\mathbb Z}$ coincides with the concrete representation in the question? Does this work by showing that the representation in the question is the GNS representation for the unique tracial state on ${\mathfrak A}$? – Yemon Choi Aug 3 '20 at 21:59
• Thank you: it seems to me that this solves my problem-as one can construct the faithful tracial state on $A_{\theta}$ (and $A_{\theta}$ is infinite dimensional) then this factor has to be of type $II_1$. Or am I wrong? – truebaran Aug 3 '20 at 22:00
• No. This is irreducible. The commutant of $U$ is $L^\infty({\mathbb T})$ and its intersection with $V$-commutant is just ${\mathbb C}1$. – Narutaka OZAWA Aug 3 '20 at 23:06
• @JonBannon it's fine, especially that your idea looked very plausible – truebaran Aug 4 '20 at 1:19

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

To support Ruy's answer: in my opinion the most natural representation of the quantum torus C*-algebra is the GNS representation coming from its tracial state. This can be explicitly described as the action on $$l^2(\mathbb{Z}^2)$$ given by $$Ue_{m,n} = e^{-i\hbar n/2}e_{m+1,n}$$ and $$Ve_{m,n} = e^{i\hbar m/2}e_{m,n+1}.$$ The von Neumann algebra they generate is indeed a $$II_1$$ factor.

I would even say this is the "quantum torus von Neumann algebra". There's more in Section 6.6 of my book Mathematical Quantization.

• I wonder if people know what are the indices of finite index subfactors of this vN algebra. I was ready to guess that these should involve the rotation angle $\theta$, or $\hbar$ in your answer, but I now see that this is wrong. If I am reading it correctly you state in the 2nd paragraph of page 146 of your book that these algebras are all isomorphic to each other for all positive values of $\hbar$ (presumably just the irrational ones). Do you have a reference for this result? – Ruy Aug 14 '20 at 14:35
• I was wondering that too! It's been 20 years since I wrote that comment. I think the reason is because they're all hyperfinite (and there is only one hyperfinite $II_1$ factor). But I really don't remember how we know this ... I will think about it. – Nik Weaver Aug 14 '20 at 15:41
• I guess it follows from Theorem 4 in "Elliott, George A.; Evans, David E., The structure of the irrational rotation C*-algebra, Ann. Math. (2) 138, No. 3, 477-501 (1993)". – Ruy Aug 14 '20 at 17:58
• Yes, that's it. Thanks! – Nik Weaver Aug 14 '20 at 18:40

As I was reading Yemon Choi's comment above it occurred to me that the situation of the crossed product $$C(S^1)\times_\theta\mathbb{Z}$$ is in fact a bit peculiar since the most standard representation of $$C(S^1)$$ one usually thinks of, namely as multiplication operators on $$L^2(S^1)$$, already comes equipped with a unitary representation of $$\mathbb{Z}$$ implementing the action by rotation.

This is not always the case for a general crossed product $$A\times\mathbb{Z}$$, so one usually starts with one's favorite representation of $$A$$ on some Hilbert space $$H$$ and builds the "regular representation" of the crossed product on the Hilbert space $$H\otimes \ell^2(\mathbb{Z})$$.

Even though that was not the representation the OP had in mind it is interesting to observe that, if the irrational rotation C*-algebra is completed in this other representation, one does indeed get a type $$II_1$$ factor, partly because the standard trace is a vector state in this representation and hence duly extends to a normal state on the weak closure.

PS: It was my original intention to reply to a comment by Yemon Choi, but I could not fit all of this within the 600 charactes size limitation. I therefore hope to be excused for shamelessly attempting to sidestep the rules and I am ready to delete this post should anyone complain!