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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, $z_3^2$?

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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: math.berkeley.edu/~rieffel/papers/deformation.pdf math.berkeley.edu/~rieffel/papers/quantization.pdf 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 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 at 19:45

1 Answer 1

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.

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