The <Hodge star operator> is defined by
$$
\alpha\wedge *_g\beta
=\langle\alpha,\beta\rangle_g\,\operatorname{vol}_g
$$
for forms of the same degree. In four dimensions, under $\widehat g=e^{2\psi}g$, the inner product on $p$-forms scales by $e^{-2p\psi}$ and the volume form scales by $e^{4\psi}$. Hence
$$
*_{\widehat g}\big|_{\Omega^p}
=e^{(4-2p)\psi}*_g.
$$
For $p=2$ the exponent vanishes, so the Hodge star on two-forms is conformally invariant.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 309 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-309/4/b/solution