Delta distribution as an integral...

2010-05-13T11:08:08Z

Hi, I'm actually physics student and I've not been able to understand how the following integration has been performed:

[...]

The fields e and A have support on these discrete structures. The $su(2)$ -valued 1-form field $e$ is represented by the assignment of an $e \in su(2)$ to each 1-cell in $\Delta$ . The connection field A is represented by the assignment of group elements $g_e \in SU(2)$ to each edge in $\mathcal{J}_\Delta$

The partition function is defined by

${\cal Z}(\Delta)=\int \prod_{f \in {\cal J}_{\Delta}} de_f\prod_{e \in {\cal J}_{\Delta}} dg_e e^{i {\rm Tr}\left[e_f U_f\right]}$

where $de_f$ is the regular Lebesgue measure on $R^3$, $dg_e$ is the Haar measure on $SU(2)$, and $U_f$ denotes the holonomy around faces, i.e., $U_f=g^1_e\dots g^{N}_e$ for $N$ being the number of edges bounding the corresponding face. Since $U_f \in SU(2)$ we can write it as $U_f=u^0_f\ {{1}} + F_f$ where $u^0_f\in C$ and $F_f \in su(2)$. $F_f$ is interpreted as the discrete curvature around the face $f$. Clearly ${\rm Tr}[e_f U_f]={\rm Tr}[e_f F_f]$. An arbitrary orientation is assigned to faces when computing $U_f$. We use the fact that faces in ${\cal J}_{\Delta}$ are in one-to-one correspondence with $1$-cells in $\Delta$ and label $e_f$ with a face subindex.

Integrating over $e_f$, we obtain

${\cal Z}(\Delta)=\int \ \prod_{e \in {\cal J}_{\Delta}} dg_e \prod_{f \in {\cal J}_{\Delta}}{\delta}(g^1_e\dots g^{N}_e),$

where $\delta$ corresponds to the delta distribution defined on ${\cal L}^2(SU(2))$.

[...]

The details are not important, what I need to understand is how the integration over the $e_f$'s has been performed

thanks

Answer by userN for Delta distribution as an integral...

2010-05-13T12:41:56Z

Whoever performed that integration is using the following fact from Fourier analysis:

The "delta function supported at the position 0" is the Fourier transform of the constant function 1.

$\delta(x) = \frac{1}{2\pi}\int_{\mathbb{R}} 1 e^{ikx}dk$

You can prove this by approximating the delta-function with a sequence of Gaussian bump functions. Fourier transforming Gaussians inverts the variance, so as the Gaussians approach a "delta function spike", their Fourier transforms approach a constant.

Delta distribution as an integral... - MathOverflow

Delta distribution as an integral...

Answer by userN for Delta distribution as an integral...