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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)?

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    $\begingroup$ In context, ‘jest’ must be ‘is’. $\endgroup$ – Branimir Ćaćić Aug 3 at 21:33
  • $\begingroup$ 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 ${\mathfrak A}$? $\endgroup$ – Yemon Choi Aug 3 at 21:59
  • $\begingroup$ 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? $\endgroup$ – truebaran Aug 3 at 22:00
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    $\begingroup$ No. This is irreducible. The commutant of $U$ is $L^\infty({\mathbb T})$ and its intersection with $V$-commutant is just ${\mathbb C}1$. $\endgroup$ – Narutaka OZAWA Aug 3 at 23:06
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    $\begingroup$ @JonBannon it's fine, especially that your idea looked very plausible $\endgroup$ – truebaran Aug 4 at 1:19
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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

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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.

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  • $\begingroup$ 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? $\endgroup$ – Ruy Aug 14 at 14:35
  • $\begingroup$ 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. $\endgroup$ – Nik Weaver Aug 14 at 15:41
  • $\begingroup$ 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)". $\endgroup$ – Ruy Aug 14 at 17:58
  • $\begingroup$ Yes, that's it. Thanks! $\endgroup$ – Nik Weaver Aug 14 at 18:40
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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!

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