Feeds:
Posts
Geometric quantization for three-dimensional configuration spaces $M$ can be described in the following way when it embeds into $S^4(1/h)$.  First, the tangent bundle and its dual cotangent bundle are both topologically trivial in three dimensions isomorphic to $M\times\mathbf{R}^3$ and so are its powers which describe $N$-particle phase spaces in classical mechanics.  Now prequantization requires a line bundle $L\rightarrow T^*M=M\times\mathbf{R}^3$.  The line bundle can be identified with a circle bundle topologically.  We can map the fibers $L_x$ of the circle bundle choosing a fixed base section to $M\subset S^4(1/h)$.  This allows us to map the fibers to the normal circle at every point from the embedding $M\subset S^4(1/h)$.  Now we have a surjective map $L\rightarrow S^4(1/h)\times\mathbf{R}^3$ that factors through the bundle projection $L\rightarrow M\times\mathbf{R}^3$ and the embedding map $M\rightarrow S^4(1/h)$.  Now a connection on the circle-bundle $L$ is equivalent to a 1-form defined on the total space.  We should be able to produce a 1-form on $S^4(1/h)$ which pulls back to $L$.  Geometric quantization prescribes a connection on $L$ whose curvature is $i\omega$ where $\omega$ is the symplectic form.
In geometric quantization one picks a polarization, a maximal distribution $T$ on $T^*M$ such that $\omega(v,v')=0$ and the quantum algebra consists of functions $f$ such that $[X_f,T]\subset T$.
Geometric quantization maps functions $f\in C^{\infty}(T^*M)$ to the operators with values in the prequantization line bundle $f\cdot + i\hbar^{1/2} \mathcal{L}_{X_f}$.  So we can assume a geometric quantization and use the map $L\rightarrow S^4(1/h)$ to make a trivial change where the values are in Slatex S^4(1/h)\$.  But this is most likely the tip of the iceberg for most likely the entire geometric quantization programme is hiding an actual structure of the actual universe which is being approached by geometric quantization in a roundabout manner.