The notation \begin{equation} \frac{ \partial f_i}{ \partial x ^i } \end{equation} means the diagonal elements of the matrix: \begin{equation} J _{ ij} = \frac{ \partial f _i }{ \partial x ^j } \end{equation} 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, \begin{equation} \int d x \left| \frac{ df (x) }{ d x } \right| _{ x = x _0 } \delta \left( f (x) \right) = 1 \end{equation} 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 $): \begin{equation} 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 }) + ... \end{equation} 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, \begin{equation} \delta ( \alpha ( a - a _0 ) ) = \frac{ \delta ( a - a _0 ) }{ \left| \alpha \right| } \end{equation} 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: \begin{equation} ( J _{ 1,1 } J _{ 2,2} .. ) ^{-1} = \frac{1}{ \det J } \end{equation} 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, \begin{equation} \int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1 \end{equation}

The notation \begin{equation} \frac{ \partial f_i}{ \partial x ^i } \end{equation} means the diagonal elements of the matrix: \begin{equation} J _{ ij} = \frac{ \partial f _i }{ \partial x ^j } \end{equation} 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, \begin{equation} \int d x \left| \frac{ df (x) }{ d x } \right| _{ x = x _0 } \delta \left( f (x) \right) = 1 \end{equation} 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 $): \begin{equation} 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 }) + ... \end{equation} 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, \begin{equation} \delta ( \alpha ( a - a _0 ) ) = \frac{ \delta ( a - a _0 ) }{ \left| \alpha \right| } \end{equation} 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: \begin{equation} ( J _{ 1,1 } J _{ 2,2} .. ) ^{-1} = \frac{1}{ \det J } \end{equation} 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, \begin{equation} \int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1 \end{equation}

The notation \begin{equation} \frac{ \partial f ( x ^i ) }{ \partial x ^i } \end{equation} means the diagonal elements of the matrix: \begin{equation} J _{ ij} = \frac{ \partial f ( x ^i ) }{ \partial x ^j } \end{equation}

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, \begin{equation} \int d x \left| \frac{ df (x) }{ d x } \right| _{ x = x _0 } \delta \left( f (x) \right) = 1 \end{equation} 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 $): \begin{equation} 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 }) + ... \end{equation} 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, \begin{equation} \delta ( \alpha ( a - a _0 ) ) = \frac{ \delta ( a - a _0 ) }{ \left| \alpha \right| } \end{equation} 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: \begin{equation} ( J _{ 1,1 } J _{ 2,2} .. ) ^{-1} = \frac{1}{ \det J } \end{equation} 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, \begin{equation} \int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1 \end{equation}

The notation \begin{equation} \frac{ \partial f_i}{ \partial x ^i } \end{equation} means the diagonal elements of the matrix: \begin{equation} J _{ ij} = \frac{ \partial f _i }{ \partial x ^j } \end{equation} 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, \begin{equation} \int d x \left| \frac{ df (x) }{ d x } \right| _{ x = x _0 } \delta \left( f (x) \right) = 1 \end{equation} 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 $): \begin{equation} 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 }) + ... \end{equation} 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, \begin{equation} \delta ( \alpha ( a - a _0 ) ) = \frac{ \delta ( a - a _0 ) }{ \left| \alpha \right| } \end{equation} 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: \begin{equation} ( J _{ 1,1 } J _{ 2,2} .. ) ^{-1} = \frac{1}{ \det J } \end{equation} 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, \begin{equation} \int {\cal D} \alpha (x) \delta \left( G ( A ^\alpha ) \right) \det \left( \frac{ \delta G ( A ^\alpha ) }{ \delta \alpha } \right) = 1 \end{equation}