MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

I find myself needing to construct some noncommutative geometries. I want to take various (algeba-) geometric objects and look at their noncommutative analogs. Is there a constructive way to do this? It seems hard to find a topological space's algebra of continuous functions. I do see examples of $\mathbb{C}^n$ and $\mathbb{P}^n$ and how to proceed if you have a variety. But even for $\mathbb{P}^2$ or $\mathbb{P}^3$, I have not been able to find WHY those* algebras are appropriate. I have also found the noncommutative torus, but I can't find any explanation as how to derive this from the torus. So to sum up, I guess my question is if I have an arbitrary geometric object, say a complex manifold, then is there any general way to create a noncommutative analog?

*$\mathbb{P}^3 \cong T(z_1, z_2, z_3,z_4)/ \langle[z_3, z_i] = [z_4, z_i] = 0, [z_1, z_2] = 2hz_3z_4\rangle$ where $T$ is the tensor algebra.

Why do we make $z_3$ and $z_4$ commute with everything? Why is the commutator of $z_1$ and $z_2$ proportional to $z_3z_4$ and not, say, $z_3^2$?

share|cite|improve this question
What you're probably looking for is deformation quantization, for which there are several methods appearing in the literature. In a specifically operator-algebraic context, what you might want to use is Rieffel's strict deformation quantisation: see, for instance, these survey articles by Rieffel himself: For instance, the noncommutative torus can be very nicely obtained from the usual torus through strict deformation quantisation. What context are you working in, anyway? – Branimir Ćaćić Feb 2 '14 at 21:35
I'm working with K3 surfaces in general, but for the immediate context, I'm trying to get my hands on a "noncommutative Kummer surface". – user46348 Feb 3 '14 at 19:45

There is not a unique or canonical noncommutative analogue of the algebra of functions on a classical space.

In the context of deformation quantizations non commutative algebras are constructed starting from a Poisson structure on the classical space; thus you have as many NC deformations as Poisson brackets on the classical algebra of functions. In your proposed example you are quantizing the Poisson bracket $$ \{z_3,z_i\}=\{z_4,z_i\}=0\,, \qquad \{z_1,z_2\}=z_3z_4 $$ a quadratic Poisson bracket which is zero on the subvariety $z_3=0=z_4$ and has rank two everywhere else. In the NC torus case you are quantizing invariant symplectic structures on the torus.

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.