# Question on Faddeev-Popov method derivation

In the text book of Weinberg, there is a proof to show that path integral is independent of gauge fixing functional $f_a[\phi; x]$. $\phi_\Lambda$ is the result of gauge transformation on $\phi$ by an arbitrary gauge $\Lambda^a(x)$, then $$I = \int \left [\prod_{n,x} d\phi_{\Lambda,n}(x)\right] G[\phi_\Lambda]B[f[\phi_\Lambda]]\ \mathrm{Det}\, F[\phi_\Lambda]\tag1$$ After a few steps we encounter the matrix, whose determinant we're interested in.

$$F_{xa,yb}[\phi_\Lambda] = \left. \frac{\delta f_a[\phi_\Lambda;x]}{\delta \lambda^b(y)} \right\rvert_{\lambda=0} \tag2$$

Since gauge transformations form a group, gauge transformation with $\Lambda^a(x)$ followed by $\lambda^a(x)$ is a single transformation $\tilde \Lambda^a(x)$, i.e $(\phi_\Lambda)_\lambda = \phi_{\tilde \Lambda(\Lambda, \lambda)}$. Using chain rule,

$$F_{xa,yb}[\phi_\Lambda] = \int J_{xa, zc} [\phi, \Lambda] R^{zc}_{yb}[\Lambda] d^4z\tag3$$ where \begin{align} J_{xa, zc} [\phi, \Lambda] &\equiv \left. \frac{\delta f_a[\phi_{\tilde \Lambda};x ] }{\delta \tilde \Lambda^c(z)} \right \rvert_{\tilde \Lambda=\Lambda} = \left. \frac{\delta f_a[\phi_\Lambda;x]}{\delta \lambda^c(z)} \right\rvert_{\lambda=0} \tag4\\ R^{zc}_{yb} &\equiv \left. \frac{\delta \tilde \Lambda_c[z; \Lambda , \lambda ]}{\delta \lambda^b(y)} \right\rvert_{\lambda=0}\tag5 \end{align}

Now I don't understand how (6) follows from (3)? What happens to the integral on spacetime coordinates when we take determinants on both sides? If I understand correctly, determinant is only on gauge group indices.

$$\mathrm{Det}\, F[\phi_\Lambda] = \mathrm{Det}\, J[\phi, \Lambda]\ \mathrm{Det}\, R[\Lambda] \tag6$$

• $\uparrow$ Which page? – Qmechanic Mar 13 '17 at 19:44
• @AccidentalFourierTransform. Sorry for the confusion. How (6) follows from (3)? We're taking Det on both sides, aren't we? – levitt Mar 14 '17 at 4:12
• @Qmechanic. It is from page 21, vol ii of Weinberg. – levitt Mar 14 '17 at 4:15

It's an infinite-dimensional version of the identity $$F_{AB}~=~ \sum_C J_{AC}R^{}_B\qquad \Rightarrow\qquad \det F~=~ \det J \det R \tag{A}$$ for quadratic matrices. In Weinberg, the discrete index has formally been replaced by a double index $$A~\leftrightarrow~ (x,a),\qquad B~\leftrightarrow~ (y,b),\qquad C~\leftrightarrow ~(z,c), \tag{B}$$ which consists of a continuous spacetime variable and a discrete color-index; and the sum has been replaced by an integral (and an implicitly written sum over a color index) $$\sum_C ~\leftrightarrow~\sum_c\int\! d^4z.\tag{C}$$ See also e.g. my Phys.SE answer here for a similar discussion.