# 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 \begin{equation} 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 \end{equation} After a few steps we encounter the matrix, whose determinant we're interested in.

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

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,

\begin{equation} F_{xa,yb}[\phi_\Lambda] = \int J_{xa, zc} [\phi, \Lambda] R^{zc}_{yb}[\Lambda] d^4z\tag3 \end{equation} 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

## 1 Answer

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 In Weinberg, the discrete index has formally been replaced by a double index $$A~\leftrightarrow~ (x,a),\qquad B~\leftrightarrow~ (y,b),\qquad C~\leftrightarrow ~(z,c), \tag{B}$$ which consists of a continuous spacetime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.

• But are we taking the determinant over the infinite number of indices or is it only on the two color indices? Kindly could you post the link which answer has the discussion you mentioned.? – levitt Mar 14 '17 at 10:43
• It's over all indices. – Qmechanic Mar 14 '17 at 10:55