Multivariable Dirac Delta and Faddeev-Popov Determinant

From this mathstack page and in particular Qmechanic's answer:

1. There exists an $n$-dimensional generalization $$\tag{1} \delta^n({\bf f}({\bf x})) ~=~\sum_{{\bf x}_{(0)}}^{{\bf f}({\bf x}_{(0)})=0}\frac{1}{|\det\frac{\partial {\bf f}({\bf x})}{\partial {\bf x}} |}\delta^n({\bf x}-{\bf x}_{(0)})$$ of the substitution formula for the Dirac delta distribution under pertinent assumptions, such as e.g., that the function ${\bf f}:\Omega \subseteq \mathbb{R}^n \to \mathbb{R}^n$ has isolated zeros. Here the sum on the rhs. of eq. (1) extends to all the zeros ${\bf x}_{(0)}$ of the function ${\bf f}$.

For ordinary functions, a property of the Dirac delta function gives: $\delta(x-x_0) = \left|\frac{df(x)}{dx}\right|_{x=x_0}\delta(f(x))\,$ assuming $f(x)\,$ has only one zero at $x=x_0\,$ and is differentiable there. Integrating both sides gives :$$1 = \left|\frac{df(x)}{dx}\right|_{x=x_0}\int\!dx\,\delta(f(x))\,$$. Extending over $n$ variables, suppose $f(x^i) = 0\,$ for some $x^i_0\,$. Then, replacing $\delta(x-x_0)\,$ with $\prod_i^n \delta^i(x^i-x^i_0)\,$ :$$1 = \left(\prod_i \left|\frac{\partial f(x^i)}{\partial x^i}\right|\right) \int\!\left(\prod_i dx^i\right)\,\delta(f(x^i))\,$$. Recognizing the first factor as the determinant of the diagonal matrix $\frac{\partial f(x^i)}{\partial x^i}\delta^{ij}\,$ (no summation implied), we can generalize to the functional version of the identity: :$$1 = \det\left|\frac{\delta G}{\delta \Omega}\right|_{G=0} \int\!\mathcal{D}\Omega\,\delta[G_a(\phi^\Omega)]\,$$, where $\Delta_F[\phi] \equiv \det\left|\frac{\delta F}{\delta g}\right|_{F=0}\,$ is the Faddeev-Popov determinant.

What I don't understand is that it seems their function $f$ seems to be $f:\Omega \subseteq \mathbb{R}^n \to \mathbb{R}$. How does the generalized Dirac formula $(1)$ work in this case? I don't really understad their notation in:

$$1 = \left(\prod_i \left|\frac{\partial f(x^i)}{\partial x^i}\right|\right) \int\!\left(\prod_i dx^i\right)\,\delta(f(x^i))\,$$

What does $$\frac{\partial f(x^i)}{\partial x^i}$$ mean here?

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The notation $$\frac{ \partial f_i}{ \partial x ^i }$$ means the diagonal elements of the matrix: $$J _{ ij} = \frac{ \partial f _i }{ \partial x ^j }$$ where $f_i$ is the component of the vector $\vec{f} (x)$.

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 $x _0$ then, $$\int d x \left| \frac{ df (x) }{ d x } \right| _{ x = x _0 } \delta \left( f (x) \right) = 1$$ We want to generalize this to instead of having $f (x)$ we have, ${\mathbf{g}} ( {\mathbf{a}} )$ for vectors of arbitrary size. To do this consider the Taylor expansion of ${\mathbf{g}}$ around its root (we assume it only has one root, ${\mathbf{a}} _0$): $$g _i ( {\mathbf{a}} ) = \overbrace{g _i ( {\mathbf{a}} _0 )}^0 + \sum _{ j} \frac{ \partial g _i }{ \partial a _j } \bigg|_{ a _0 } ( a _j - a _{ 0,j }) + ...$$ We want to insert this into a delta function, $\delta ^{ ( n ) } ( {\mathbf{g}} ( {\mathbf{a}} ) )$. This will only be nonzero near ${\mathbf{a}} = {\mathbf{a}} _0$. Thus we have, \begin{align} \delta \left( {\mathbf{g}} ( {\mathbf{a}} ) \right) & = \prod _i \delta \left( g _i ( {\mathbf{a}} ) \right) \\ & = \prod _i \delta \big( \sum _j J _{ ij} ( a _j - a _{ 0,j} ) \big) \end{align} where $J _{ ij}$ is the Jacobian matrix defined by $J _{ ij} \equiv \frac{ \partial g _{ i} }{ \partial a _j } \big|_{ a _0 }$. We have, \begin{align} \delta \left( {\mathbf{g}} ( {\mathbf{a}} ) \right) & = \delta \big( \sum _j J _{ 1j} ( a _j - a _{ 0,j} ) \big) \delta \big( \sum _j J _{ 2j} ( a _j - a _{ 0,j} ) \big) ... \end{align} We now use the identity, $$\delta ( \alpha ( a - a _0 ) ) = \frac{ \delta ( a - a _0 ) }{ \left| \alpha \right| }$$ We choose to isolate each delta function in the equation above for a different $a _j$: \begin{align} \delta \big( {\mathbf{g}} ( {\mathbf{a}} ) \big) & = \frac{ \delta ( a _1 - a _{ 0,1 } ) }{ \left| J _{ 1,1 } \right| } \frac{ \delta ( a _2 - a _{ 0,2 } ) }{ \left| J _{ 2,2 } \right| } ... \end{align} If we take the Jacobian matrix to be greater then zero then we have the product: $$( J _{ 1,1 } J _{ 2,2} .. ) ^{-1} = \frac{1}{ \det J }$$ where we have used the fact that the determinant of $J$ is independent of a unitary transformation. So we finally have, \begin{align} \left( \int \prod _{ i} d a _i \right) \delta ^{ ( n ) } \big( {\mathbf{g}} ( {\mathbf{a}} ) \big) \det \big( \frac{ \partial g _i }{ \partial a _j } \big) & = 1 \end{align} where it is understood that the Jacobian matrix is evaluated at the root of ${\mathbf{g}}$.

We write the continuum generalization of this equation as, $$\int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1$$

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Hi, thanks for the answer, I have some question though. The notation $J _{ ij} \equiv \frac{ \partial g _{ i} }{ \partial a _j } \big|_{ a _0 }$ makes sense because $g_i$ is a component of the vector function $\bf g$. But in the first paragraph you state that $$J _{ ij} = \frac{ \partial f ( x ^i ) }{ \partial x ^j }$$ which doesn't make sense or at least does not agree with the one involving g. I get confused by the notation. Your derivation seems correct and I get that but the first part is a little unclear, which was my original question. Thanks. – Faq Feb 28 '14 at 14:07
Whoops, got to exciting to post an answer... My mistake. I updated it now. It is indeed a vector. – JeffDror Feb 28 '14 at 14:30
Makes much more sense to me now :) thanks for your great answer. – Faq Feb 28 '14 at 14:35
I think you have to be more careful here since the function we are considering is $f:\mathbb{R}^n \to \mathbb{R}$ and not $f:\mathbb{R}^n \to \mathbb{R}^m$ then $\frac{\partial f(x^i)}{\partial x^i}$ is simply the normal partial derivative $\frac{\partial f}{\partial x^i}$ and keeping the argument is just misleading. This is different from $\frac{\partial f_i}{\partial x^i}$, which indeed would be a diagonal matrix to the matrix of partial derivatives of all the functions $f_j$ which form an $f:\mathbb{R}^m \to \mathbb{R}^n$ function as $f=(f_1(x^i),f_2(x^i),\cdots,f_m(x^i))$. – Julio Parra Feb 28 '14 at 16:47
Nice answer but it seems only to be valid for diagonal Jacobians. In the step after the statement "We choose to isolate each delta function in the equation above for a different $a_j$" you quickly choose $J_{1,1}$ to be the prefactor to the $a_{1}-a_{0,1}$ zero but you could just as well have chosen $J_{2,1}$. For non-diagonal Jacobians this doesn't hold up. Another indication of this problem is in the identification $J_{1,1}J_{2,2}... = \det J$ which only works for diagonal J. – AltLHC Oct 8 '15 at 17:23

Again assuming it only has a zero $x^i=x_0^i$ what you have is $$\delta(f(x^i)) = \frac{\delta(x^1-x_0^1)}{\left|\frac{\partial f}{\partial x^1}\right|_{x^i=x_0^i}} \frac{\delta(x^2-x_0^2)}{\left|\frac{\partial f}{\partial x^2}\right|_{x^i=x_0^i}}\cdots \frac{\delta(x^n-x_0^n)}{\left|\frac{\partial f}{\partial x^n}\right|_{x^i=x_0^i}} = \prod_{j=1}^n \frac{\delta(x^j-x_0^j)}{\left|\frac{\partial f}{\partial x^j}\right|_{x^i=x_0^i}} = \frac{\prod_{j=1}^n\delta(x^j-x_0^j)}{\prod_{j=1}^n\left|\frac{\partial f}{\partial x^j}\right|_{x^i=x_0^i}}$$ then $$\prod_{j=1}^n\left|\frac{\partial f}{\partial x^j}\right|_{x^i=x_0^i} \delta(f(x)) = \prod_{j=1}^n\delta(x^j-x_0^j)$$ and integrating at both sides in all the variables $\int\left(\prod_{j=1}^n dx^j\right)$ you get $$\int\left(\prod_{j=1}^n dx^j\right)\prod_{j=1}^n\delta(x^j-x_0^j)=\int\prod_{j=1}^n dx^j\delta(x^j-x_0^j)=1$$ and since the derivatives are evaluated in the zero and are just numbers $$\int\left(\prod_{j=1}^n dx^j\right)\prod_{j=1}^n\left|\frac{\partial f}{\partial x^j}\right|_{x^i=x_0^i} \delta(f(x)) = \prod_{j=1}^n\left|\frac{\partial f}{\partial x^j}\right|_{x^i=x_0^i}\int\left(\prod_{j=1}^n dx^j\right) \delta(f(x^i))$$ so finally we get $$1== \prod_{j=1}^n\left|\frac{\partial f}{\partial x^j}\right|_{x^i=x_0^i}\int\left(\prod_{j=1}^n dx^j\right) \delta(f(x^i))$$

The notation is a bit clumsy but I believe this is what you were looking for.

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+1 Nice proof! Sorry, my answer is essentially identical to yours... I didn't see you already posted. – JeffDror Feb 28 '14 at 12:39
If I am not mistaken the original delta function has to be $\delta^{(n)}(f(x^{i}))$ for the initial equality to be valid? Otherwise only one variable would be constrained on the left side whereas all are constrained on the right. – AltLHC Oct 8 '15 at 17:10

What confused me was the explanation from the tangentbundle homepage (second yellow box in OP). The generalization is straightforward, for simple zeros we have:

$$\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] = \delta(x-x_0)$$ integrate

$$\int \mathrm{d}x\,\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] = 1$$ generalize $$\int \mathrm{d}{\bf x\,\bigg\vert \mathrm{det}\frac{\partial f(x)}{\partial x}\bigg\vert_{x_0}\delta[f(x)]} = 1$$ generalize $$\int \mathcal{D}\alpha(x)\,\bigg\vert \mathrm{Det}\frac{\delta G(A^\alpha)}{\delta \alpha}\bigg\vert_{A^\alpha_0}\delta[G(A^\alpha)] = 1,$$ where it is to be understood that $A^\alpha_0$ is such that $G(A^\alpha_0)=0.$

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