Vector Delta Function Identity I'm trying to prove the the vector extension of the identity 
\begin{equation} 
1 = \int \left|\sum_i\frac{ \partial  g  }{ \partial a }\big| _{a =a _i} \right| \delta ( g ( a ) ) da
\end{equation} 
where the sum is over all the zeros of $g$. The vector extension is:
\begin{equation} 
1 = \left( \prod _{i =  1 } ^n  \int d a _i \right) \delta ^{ ( n ) } \left( {\mathbf{g}} ( {\mathbf{a}} )  \right) \det \left( \frac{ \partial g _i }{ \partial a _j } \right) 
\end{equation}
where ${\mathbf{g}}$ and $ {\mathbf{a}} $ are $n$ dimensional vectors. This identity is used in Quantum Field Theory by Peskin and Schroeder (pg. 295).
I'm a physics graduate student and I don't know much formal mathematics.  Any help on how to prove this would be greatly appreciated.
 A: Here is a partial answer, in the case of a single root $\mathbf{a}_1$.
For each number $h>0$ you can define a function
$d(\mathbf{x})$ to be zero outside the sphere $|\mathbf{x}|=h$, and
1/volume inside.
When $h$ is small, $d$ is close to $\delta$ as
distributions. Also, you can use an ellipsoid rather
than a sphere.
$\mathbf{g}(\mathbf{a})$ is in such a disc if $\mathbf{a}$
is close enough
to a zero of $\mathbf{g}$. Specifically,
$$
     \mathbf{g}(\mathbf{a})
       = D\mathbf{g}(\mathbf{a}_1)(\mathbf{a}-\mathbf{a}_1)
$$
nearly, when $\mathbf{a}$ is near $\mathbf{a}_1$, where
$D\mathbf{g}(\mathbf{a}_1)$ means the
Jacobian matrix
$\left(\frac{\partial g _i}{\partial a _j}\right)$
 evaluated at the root.
The matrix maps any small sphere centered at $\mathbf{a}_1$ to
some ellipsoid centered at $\mathbf{0}$, and the ellipsoid volume is
 $|\det(D\mathbf{g}(\mathbf{a}_1))|$ times the sphere volume.
So,
$$
\delta^{(n)} \left(
           \mathbf{g}(\mathbf{a})
             \right)
$$
is well approximated by
$$
  \frac{1}{|\det(D\mathbf{g}(a_1))|({\rm sphere vol.})}
$$
 when $\mathbf{a}$ is in any small sphere centered at $\mathbf{a}_1$
and zero outside the sphere.
Then
$$
\delta^{(n)} \left(
           \mathbf{g}(\mathbf{a})
             \right)
        |\det(D\mathbf{g}(\mathbf{a}_1))|
$$
is nearly 1/sphere volume when $\mathbf{a}$ is inside,
and 0 if outside. That is, it is nearly $\delta(\mathbf{x}-\mathbf{a}_1)$.
I don't see how to handle more than one root and I don't see why
your authors don't have absolute values on the derivative.
A: The proof below uses only the chain rule (to change integration variables). It's what I, as a physics student, would write. 
But first, I believe that you have an error in your original posting. I believe that the 1D identity should be:
\begin{align}
\int \Big( \sum_{\substack{\textrm{roots $a_i$}\\ \textrm{ of $g$}}}  \frac{1}{g'(a)}  \Big)^{-1}  \delta(g(a)) da = 1.
\end{align} 
I know that you asked about the multidimensional case. However, let's review the one-dimensional case first. They are extremely similar.
One-dimensional Case
\begin{align}
\int_{\substack{\textrm{small $a$-region} \\ \textrm{ containing root $a_i$}}} \delta (g(a)) da = &\int_{\substack{\textrm{small $g$-region} \\ \textrm{ corresponding to} \\ \textrm{ root $a_i$}}} \delta(g) \frac{1}{g'(a)} dg
\end{align}
We knew $g$ as a function $a$ (i.e., $g(a)$), but we can also think of $a$ as a function of $g$. This works because in the small region near the root, $g(a)$ is invertible. So we can think of $\frac{1}{g'(a)}$ as a function of $g$. The delta function extracts the value of this function near $g=0$. But in the small region we care about, $g=0$ means $a=a_i$. Thus:
\begin{align}
\int_{\substack{\textrm{small $a$-region} \\ \textrm{ containing root $a_i$}}} \delta (g(a)) da = \frac{1}{g'(a_i)}
 \end{align}
Integrating over all space instead of a small region gives:
\begin{align}
\int \delta(g(a)) da =  \sum_{\textrm{roots $a_i$ of $g$}} \frac{1}{g'(a_i)}
\end{align}
which is a special case of an identity you can find at the Wikipedia article on delta functions
We can rewrite this to obtain:
\begin{align}
\int \Big( \sum_{\substack{\textrm{roots $a_i$}\\ \textrm{ of $g$}}}  \frac{1}{g'(a)}  \Big)^{-1}  \delta(g(a)) da = 1.
\end{align} 
Which is  (except for the inversion) the 1D identity that you had. You can bring the factor $\left( \sum_{\textrm{roots $a_i$ of $g$}} \right)^{-1}$ inside the sum because it's just a constant. 
Multidimensional case
The only difference in the multidimensional case is that when you change variables you need to use the Jacobian determinant (which Peskin and Schroeder call $\frac{\partial g_i }{\partial a_j}$).  So, instead of $da = \frac{1}{g'(a)} dg$, we have
\begin{align}
\prod_k d a_k = \frac{1}{\det{ \frac{\partial g_i }{\partial a_j} } } \prod d g_k 
\end{align}
Note that $\frac{\partial g_i }{\partial a_j}$ really stands for ``the matrix whose $ij$ entry is $\frac{\partial g_i }{\partial a_j}$''. 
Hence the final identity from before becomes:
\begin{align}
\left( \prod_k \int d a_k \right) \Big( \sum_{\substack{\textrm{roots $\vec{a}_i$}\\ \textrm{ of $g$}}} \frac{1}{\det{ \frac{\partial g_i }{\partial a_j} } } \Big)^{-1}  \delta^{(n)}(\vec{g}(\vec{a}))  = 1.
\end{align} 
where $\vec{a}_i$ are the roots of $\vec{g}(\vec{a})$ and each determinant is evaluated at the root $\vec{a}_i$.
If there's only one root
, then the sum only has one term, and so inverting the terms is undone by inverting the sum:
\begin{align}
\left( \prod_k \int d a_k \right) \det{ \frac{\partial g_i }{\partial a_j} }  \delta^{(n)}(\vec{g}(\vec{a}))  = 1.
\end{align} 
[Which is the identity on p295 of Peskin and Schroeder.] The fact that these two inversions cancel is nice, but I think it is what led you to the error in your original post.
