# 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 ${\bf f}$.

Also from this page on the Faddeev-Popov procedure they say:

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?

## 3 Answers

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, $$\int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1$$

• 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. Commented Feb 28, 2014 at 14:07
• Whoops, got to exciting to post an answer... My mistake. I updated it now. It is indeed a vector. Commented Feb 28, 2014 at 14:30
• Makes much more sense to me now :) thanks for your great answer. Commented Feb 28, 2014 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 $f=(f_1(x^i),f_2(x^i),\cdots,f_m(x^i))$.
– jpm
Commented Feb 28, 2014 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. Commented Oct 8, 2015 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.

• +1 Nice proof! Sorry, my answer is essentially identical to yours... I didn't see you already posted. Commented Feb 28, 2014 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. Commented Oct 8, 2015 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 $G(A^\alpha_0)=0.$