Thursday, December 22, 2011

Outline of Commutative Geometry

So, preparing for my discussion of noncommutative geometry, I need to discuss "commutative geometry".

What's going on here? Well, lets begin with the simplest notion of a space: a topological space.

I have discussed in my notebk [] the notion of a topological space and continuous functions.

However, the algebra of continuous real-valued (or, more generally, complex-valued) functions encode the topology.

So I need to write notes reconstructing the topological data for $X$ from the ring structure and properties which $C(X)$ satisfy.

Vector Bundles

The next sort of space we can work with is a vector bundle. What's this guy?

Well, it's really a fibre bundle whose fibre forms a vector space. What's a fibre bundle?

It's a generalization of the product space where we fix one of the spaces.

Where does this occur? In vector calculus!

We are working with $\mathbb{R}^{3}$. A vector field assigns to each point in $\mathbb{R}^{3}$ a vector. But vectors live in "linear spaces" (or vector spaces).

So secretly we have $X=\mathbb{R}^{3}$ be the underlying space, and the total space be $E=F\times X$ consisting of "tangent vectors" (an ordered pair consisting of the vector assigned by the vector field, and its base point).

The fibre here is a vector space. Moreover, it is $F=\mathbb{R}^{3}$ as a vector space.

This is the simplest example of a vector bundle. So what?

Well, vector fields can be represented through ordered triples. That is, three smooth functions represent each component of the vector field (the x part, y part, and z part).

So algebraically we have $C^{\infty}(\mathbb{R}^{3})\times C^{\infty}(\mathbb{R}^{3})\times C^{\infty}(\mathbb{R}^{3})$ represent all possible vector fields on our space.

This is a free module over $C^{\infty}(\mathbb{R}^{3})$. So are vector bundles represented by free modules?

Not really, we use projective modules (which is more general).

There are a few other things to discuss on this matter, e.g., global sections, and so forth.

Spinor Bundles

This should be discussed in some detail, as there are few good references on the subject.

Even nLab's entry on Spinor bundles is lacking, alas!

However, we need to encode this data in a spectral triple. See the nLab's entry, it is quite good.

See also Alain Connes' "On the spectral characterization of manifolds" (arXiv:0810.2088) for details.

This would take some time to write up.

Noncommutative Rejoinder

I suspect that by taking an arbitrary ring, instead of the ring of smooth functions (or continuous functions, or...), we begin working with noncommutative geometry.

Is this algebraic geometry? No, not really. Algebraic geometers use polynomials to encode their geometric objects.

On the other hand, what we are doing here is considering the structure of rings and modules over our rings to encode geometric properties and data.

Of course, I may be misinformed on what algebraic geometers do...I frankly never understood it well enough to satisfy myself.

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