On an oriented Euclidean four-space with its metric volume form, the <Hodge star operator> is defined by
$$
\alpha\wedge{}^\star\beta
=\langle\alpha,\beta\rangle\operatorname{vol}
$$
for forms $\alpha$ and $\beta$ of the same degree. In $n$ Euclidean dimensions it satisfies
$$
{}^\star({}^\star\alpha)=(-1)^{p(n-p)}\alpha
\qquad(\alpha\in\Lambda^p).
$$
Hence on two-forms in four dimensions,
$$
\boxed{{}^\star{}^\star=1\quad\text{on }\Lambda^2(\mathbb R^4)}.
$$
The normalization of $\operatorname{vol}$ as the metric volume form is essential; reversing its orientation reverses a single Hodge star but leaves its square unchanged.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 313 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-313/3/a/solution