Posted by: hilbertthm90 | July 10, 2008

NCG 4

So this is going slower than I thought it would. I’m on 4 and I’ve basically given a definition that was incomplete. Maybe I’ll quit this endeavor if this one doesn’t go any better than the previous. I’m now going to make the assumption that if you could read NCG 2, then you were bored and so to cover more ground I’m going to make more demands on the reader.

A good example of the overview in NCG 2 is in electromagentism. Let V be a manifold. Let \mathcal{A} be the algebra of functions on V. It turns out the group of unitary elements of \mathcal{A} is the local gauge group of electromagnetism. Let D be the covariant derivative associated to the potential.

We can then express D by \mathcal{H}\stackrel{D}\to \Omega^1(V)\otimes_\mathcal{A} \mathcal{H}. Where \mathcal{H} is a \mathcal{A}-module and D as a derivation satisfies the Leibniz rule: D(f\psi)=df\otimes \psi + fD\psi where f\in\mathcal{A} and \psi\in\mathcal{H}. Now we know that each component of the potential couples equally with the Dirac spinors.

This is tending to be a little physics-like, so I’ll take a little diversion here. If you look at the Dirac equation it requires a collection of four objects satisfying: \gamma_0^2=I, \gamma_1^2=\gamma_2^2=\gamma_3^2=-I, and \gamma_i\gamma_j=-\gamma_j\gamma_i when i\neq j. These \gamma have 4 by 4 matrix representation that I won’t type (too cumbersome, but you can look up anywhere if you are terribly curious). Using Einstein summation convention, we get a nice compact form of the Dirac equation: i\hbar\gamma^i\partial_i\psi = mc\psi where \partial_i=\frac{\partial}{\partial x_i}.

So we have an interaction with the EM-field F^{ij} through the potential (A^0, A^1, A^2, A^3)=(\frac{1}{c}V, A_x, A_y, A_z). Just replace i\hbar \partial^i \to i\hbar\partial^i-eA^i to get the new Dirac equation \gamma_i(i\hbar\partial^i-eA^i)\psi=mc\psi where the state function \psi is a column spinor (\psi_i)_1^4\in\mathbb{C}^4.

OK. So back to what I was driving at. This means we can identify \mathcal{H} with \mathcal{A}. So if we define D by the rule D1=A\otimes 1=A along with the Leibniz rule. This leads to the operator D behaving exactly as I derived (no that wasn’t for nothing) above. After redefining some constants we get D\psi is the left side of the “new” Dirac equation.

And so now we have an electromagnetic theory for noncommutative geometry. Which is actually pretty important since like we said before, if we want the four forces in a unified theory we need them to work in a quantum/noncommutative framework. Hmm…sorry, since I feel like the vote that went for NCG was not a vote for physics examples.

Posted in math, physics | Tags: , , , , , , ,

«
»

Leave a response






Your response:

Categories