The <Hodge star operator> is defined by
$$
\alpha\wedge\star\beta=\langle\alpha,\beta\rangle_\eta\,\mathrm{vol}.
$$
On two-forms in oriented Euclidean four-space, $\star^2=1$ and the wedge product is symmetric. Therefore
$$
\star\omega\wedge\star\omega
=\omega\wedge\star^2\omega
=\boxed{\omega\wedge\omega}.
$$
For an anti-self-dual curvature $F=-\star F$, the <Yang-Mills instanton> action becomes purely topological. With
$$
c_2=-\frac1{8\pi^2}\int_{\mathbb R^4}\operatorname{Tr}(F\wedge F),
$$
the standard positive-action convention gives
$$
\boxed{S_{\rm YM}=8\pi^2c_2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 313 / 2 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-313/2/i/solution