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



Delta distribution as an integral...
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.


