For a gauge orbit near a root $\alpha_0$ of $G(A^\alpha)=0$, functional linearization gives
$$
G(A^\alpha)=\left.\frac{\delta G(A^\alpha)}{\delta\alpha}\right|_{\alpha_0}
(\alpha-\alpha_0)+O((\alpha-\alpha_0)^2).
$$
The multidimensional delta-function change-of-variables formula therefore gives
$$
\int\mathcal D\alpha\,\delta[G(A^\alpha)]
=\det\left(\left.\frac{\delta G(A^\alpha)}{\delta\alpha}\right|_{G=0}\right)^{-1}.
$$
Comparison with the defining identity proves
$$
\Delta_{\rm FP}[A]=\det\left(\left.\frac{\delta G(A^\alpha)}{\delta\alpha}\right|_{G=0}\right),
$$
up to the usual field-independent normalization and a choice of determinant sign on the gauge patch.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-304/4/a/solution