MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The noncommutative torus $A_\theta$ is a $C^*$-algebra corresponding to an irrational foliation of the torus $\mathbb{T}^2$ by lines of slope $\theta \notin \mathbb{Q}$.

As far as I am reading it is generated by two bounded linear operators $U,V$ acting on $L^2(\mathbb{T})$. $$ U(f)(z) = z f(z) \hspace{0.5in} V(f)(z) = f(e^{-2\pi i \theta}z) $$

This is cool... but what quantum mechanical system does this correspond to?
In particular,

  • can we find a hamiltonian $\mathcal{H}$ corresponding to this system?
  • Or is this a quantization of a different Poisson structure than the free particle?

In order to address this question, I tried writing the two formulas in bra-ket notation:

$$ \langle f |U | z \rangle = z \langle f| z\rangle \hspace{0.5in} \langle f |V| z \rangle = \langle f| e^{-2\pi i \theta} z\rangle $$

Here $\langle f | \in L^2 (\mathbb{T})$ and $|z\rangle \in \mathbb{T}$. In a sense it didn't matter which function $f$ we picked we can just write:

$$ U | z \rangle = z | z\rangle \hspace{0.5in} V| z \rangle = | e^{-2\pi i \theta} z\rangle $$

These look like the position and momentum operators and we get an representation of the Heisenberg algebra, since $UV = e^{2\pi i \theta} VU$.

Basically... $C^\ast$ algebras supposed to be the "same" as quantum mechanical systems. So what is the quantum mechanical system corresponding to the non-commutative torus $A_\theta$ ?

The article says define Poisson bracket on the phase space $C^\infty(\mathbb{T}^2)$ by $$\{ f,g \} = \theta \left( \frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_2} - \frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_1} \right)$$

Then build this deformed multiplication $\ast_\hbar$ such that $$f \ast_\hbar g - g \ast_\hbar f = i\hbar\{ f,g\} + O(\hbar^2) $$

Then quantizing we get $[ x_1, x_2] = i\theta\hbar$, which looks like $\theta$ could be absorbed into $\hbar$ ? This might be semiclassical analysis or something.

I don't understand this:

When $n=2$ the skew-symmetric matrix $\theta$ is just determined by a real number, again denoted $\theta$, and $A_\theta$ will be isomorphic to the crossed-product $C^\ast$-algebra for the action of $\mathbb{Z}$ on the circle $\mathbb{T}$ coming from rotation by angle $2\pi \theta$. For this reason the algebras are often called rotation algebras or irrational rotation algebras if $\theta \notin \mathbb{Q}$.

share|cite|improve this question
Seems to be the phase space of the 1D free particle on the circle. Notes by Marc Reiffel explain the deformation of the Poisson structure... – john mangual May 29 '14 at 19:09
$A_\theta$ can be obtained as the $C^*$-algebra generated by two unitaries on $\ell^2(\mathbb{Z}^2)$ associated with the formulation of the problem of the Bloch electron (describe the motion of a free electron on the square lattice submitted to a uniform magnetic field orthogonal to the lattice). See a famous paper by D. Hofstadter:… – Alain Valette May 29 '14 at 20:10
@AlainValette There is a formula for the deformed product for the convolution in $\ell^2(\mathbb{Z}^2)$ $$(\phi \ast_\hbar \psi)(p) = \sum_q \phi(q)\psi(p-q) e^{\pi i \hbar \langle q | \theta | p-q \rangle}$$ Can $\hbar$ be absorbed into $\theta$ ? – john mangual May 29 '14 at 21:52
@johnmangual --- a free particle on a circle would have phase space a cylinder, not a torus. – Nik Weaver May 30 '14 at 1:04
@NikWeaver OK - the best we can do is say there are two variables $x_1, x_2 \in S^1$ and a pairing $\{\cdot, \cdot\}$ on the torus which makes a Poisson manifold. – john mangual May 30 '14 at 1:08

I am not completely sure what the question is here but I'all attempt an answer anyway, hopefully with some reasonable informations.

Point 1 - Dependence on $\theta$ How does the non commutative algebra $A_\theta$ depends on $\theta$? Much. The first obvious difference is between rational and irrational, If $\theta=\frac p q $ something similar to what happens on quantum groups at roots of unity shows up: a big center, in fact $Z(A_{\frac p q})={\cal C}(\mathbb T^2)$ and $A_{\frac p q}$ turns out to be isomorphic to the algebra of continous sections of a vector bundle over $\mathbb T^2$ whose fibers are $q\times q$ matrix algebras. What if $\theta$ is irrational? Well, well, things are not so easy in that case as well. Consider, for example, that if $\theta,\theta'\in ]0,\frac 1 2 [$ are two irrationals then $A_\theta$ and $A_{\theta'}$ are not isomorphic as $C^*$--algebras. But we know that isomorphism as $C^*$--algebras is pretty difficult to obtain, right? Let us relax things and ask for a weaker condtion, Morita equivalence. Then you can prove that two quantum torus algebras for two different values of the parameter are Morita equivalent if and only if

$$ \exists\, A=\left(\begin{array}{cc}a&b\\ c& d\end{array}\right)\in GL(2;\mathbb Z)\, : \theta'=\frac{a\theta+b}{c\theta+d} $$

thus showing that dependence on the parameter is quite subtle. In this sense it is not easily "rescaled"... (best reference on this, imho, chapter 12 of Elements of NC Geometry, Gracia-Bondia, Varilly, Figueroa).

Point 2 - Relation with Poisson The formula as a $\star_\hbar=\star_{\hbar,\theta}$--product shows that there is a relation with a specific Poisson structure on the torus. In fact that $\theta$--depending Poisson brackets you wrote exhausts all possible left invariant Poisson structures on $\mathbb T^2$: they are all symplectic. From each fixed $A_\theta$ you can recover the corresponding Poisson bracket as a semiclassical limit, basically as $$ \{f,g\}_\theta=\frac{f\star_{\hbar,\theta}g-g\star_{\hbar,\theta}f}{\hbar}\, \mathrm{mod}\,\hbar $$ which can be made more precise in the context of continous fields of $C^*$--algebras. Remark that the semiclassical limit requires an identification of ${\cal C}^\infty(\mathbb T^2)$ with a subalgebra of $A_\theta$ and this identification is $\theta$--depending. For this part papers by Rieffel keep being the best possible source of information. Let me add this reference as source of very interesting consideration on the relation between classification of Poisson structures (as Dirac ones) and Morita equivalence of $C^*$--algebras. Always keep in mind that quantization, here, is not a functor...

Point 3 - Poisson geometry+irrational rotation algebra which seems to puzzle you somehow. Either you take it as a fact: life is hard and NC torus are isomorphic as $C^*$--algebras to certain crossed products, or you try to give it a geometric meaning which, as often happens, brings in a certain amount of complication... One possible way to quantize Poisson manifolds is the following: if the Poisson manifold is integrable, integrate it to a symplectic groupoid and then perform geometric quantization on this symplectic groupoid. Here the symplectic groupoid, being $\mathbb T^2$ symplectic itself, is $T^*\mathbb T^2$. You have two possible reasonable choices of polarization: one is the vertical one, like in every vector bundle: since fibers are planes you have no Bohr-Sommerfeld conditions and you basically end up with an identification of ${\cal C}^\infty(\mathbb T^2)$ with a subalgebra of $A_\theta$ when you identify functions on the base with polarized sections. Or you can take a cylindrical polarization in which leaves are of the form $\mathbb T\times\mathbb R$; then you have a Bohr-Sommerfeld conditions on the first factor so that your space of quantizable leaves is of the form $\mathbb Z\times\mathbb R$ and when you carry through the whole procedure you end up with the irrational rotation algebra. There are too many details to be explained here. This is described originally in Weinstein, A., Symplectic groupoids, geometric quantization, and irrational rotation algebras, in Symplectic geometry, groupoids, and integrable systems, Seminaire sud-Rhodanien de geometrie a Berkeley (1989), P. Dazord and A. Weinstein, eds., Springer-MSRI Series (1991), 281–290; but you can also profit by much later work of Eli Hawkins in

Hope this answers some of your questions...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.