/sci/ - Science & Math

>>11533750
Note that for any group [math]G[/math], [math]P(G) \cong 2^G[/math] is the space of functions [math]G\rightarrow\mathbb{Z}_2[/math]; in other words, it is a group ring [math]\mathbb{Z}_2[G][/math]. Now we know that the complex group algebra [math]\mathbb{C}[G] = \widehat{G}[/math] is the Pontrjagyn dual, which is spanned by the space of complex irreducible characters [math]\chi[/math]. Define the map [math]\operatorname{det}:\mathbb{C} \rightarrow \mathbb{Z}_2: z\mapsto \operatorname{det}z = \operatorname{sgn}\operatorname{arg}z[/math], where [math]\operatorname{sgn}\theta = \begin{cases}1 &; \theta\in [0,\pi] \\ -1 &; \theta\in [\pi,2\pi]\end{cases}[/math], we can send [math]\mathbb{C}[G]\xrightarrow{\operatorname{det}} \mathbb{Z}_2[G][/math]. By projecting the space of characters [math]\{\chi\}\xrightarrow{\operatorname{det}} \{\chi\}_{\mathbb{Z}_2}[/math], we obtain a span of [math]\mathbb{Z}_2[G] \cong 2^G[/math]. For discrete Abelian [math]G[/math], we can find the [math]\chi[/math]'s by Fourier transform.

>>11434528
GNS construction.

>>11009280
J_____

>>10922819
First let us identify what "[math]p[/math]" means. Under holomorphic quantization, we promote [math]f\in C^\infty(M)[/math] to linear operators [math]\hat{f} \in \operatorname{End}B[/math] on the prequantum Hermitian line bundle [math]B\rightarrow M[/math]. Now with the polarization [math]\{X_p\mid \qquad p = (0,p) \in M\}\subset TM[/math], we polarize out the momentum dependence on the operators [math]\hat{f}(p,q) \rightarrow \hat{f}(q)[/math]. In this way, the function [math]f(p,q) = p[/math] is promoted to the vector field [math]\hat{f}(q) = \partial_q[/math].
Now the Hermitian inner product [math]\langle \cdot,\cdot\rangle: B\times B \rightarrow \mathbb{C}[/math] allows us to form [math]\langle \cdot, \hat{f}\cdot\rangle: B \times (\operatorname{End}B \times B) \rightarrow B\times B \rightarrow \mathbb{C}[/math] on which [math]\operatorname{End}B[/math] acts on [math]B[/math] as usual. Hence given a section [math]\psi: M\rightarrow B[/math], we can form [math]\langle\psi_q,\hat{f}(q)\psi_q\rangle \in \mathbb{C}[/math] for each [math]q \in M[/math].
Let us denote the polarized manifold again by [math]M[/math]. Treat [math]\omega(q) = \langle \psi_q,\hat{f}(q)\psi_q\rangle d\operatorname{vol}\in\Omega^n(M)[/math] as a top form in [math]M[/math], where [math]d\operatorname{vol}[/math] is thevolume form and [math]\operatorname{dim}M = n[/math]. Then we may use the non-degenerate de Rham pairing to form
[eqn]\int_M \omega= \int_M \langle\psi_q,\hat{f}(q)\psi_q\rangle d\operatorname{vol} \equiv \langle \hat{f}\rangle.[/eqn] Put [math]\hat{f} = \hat{p}[/math] yields the result.

>>10720117
Existence of non-measurable sets.