From this mathstack page and in particular Qmechanic's answer:
- There exists an n-dimensional generalization δn(f(x)) = f(x(0))=0∑x(0)1|det∂f(x)∂x|δn(x−x(0))
of the substitution formula for the Dirac delta distribution under pertinent assumptions, such as e.g., that the function f:Ω⊆Rn→Rn has isolated zeros. Here the sum on the rhs. of eq. (1) extends to all the zeros x(0) of the function f.
Also from this page on the Faddeev-Popov procedure they say:
For ordinary functions, a property of the Dirac delta function gives: δ(x−x0)=|df(x)dx|x=x0δ(f(x)) assuming f(x) has only one zero at x=x0 and is differentiable there. Integrating both sides gives :1=|df(x)dx|x=x0∫dxδ(f(x))
. Extending over n variables, suppose f(xi)=0 for some xi0. Then, replacing δ(x−x0) with ∏niδi(xi−xi0) :1=(∏i|∂f(xi)∂xi|)∫(∏idxi)δ(f(xi)). Recognizing the first factor as the determinant of the diagonal matrix ∂f(xi)∂xiδij (no summation implied), we can generalize to the functional version of the identity: :1=det|δGδΩ|G=0∫DΩδ[Ga(ϕΩ)], where ΔF[ϕ]≡det|δFδg|F=0 is the Faddeev-Popov determinant.
What I don't understand is that it seems their function f seems to be f:Ω⊆Rn→R. How does the generalized Dirac formula (1) work in this case? I don't really understad their notation in:
1=(∏i|∂f(xi)∂xi|)∫(∏idxi)δ(f(xi))
What does ∂f(xi)∂xi
Answer
The notation ∂fi∂xi
I found this very confusing a few weeks ago so. Here is the proof I wrote up for the identity based on the response I received to an earlier question of mine here:
Recall that if f(x) has one zero at x0 then, ∫dx|df(x)dx|x=x0δ(f(x))=1
We write the continuum generalization of this equation as, ∫Dα(x)δ(G(Aα))det(δG(Aα)δα)=1
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