Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 / 2

The wedge product of differential forms is the alternating tensor productFor nonzero $\eta \in {\Lambda }^{n-1}(({\mathbb{R}}^n{)}^{\ast })$, choose a volume form $\mu$. There is a nonzero vector $v$ with $\eta ={\iota }_v\mu$. Extend $v$ to a basis and use its dual coframe; then $\eta$ is a scalar multiple of $e^2\wedge \cdots \wedge e^n$, hence equals $\xi \wedge {\eta }_0$. The zero form is immediate.

To integrate a top form on a compact oriented $n$-manifold, choose a finite oriented atlas and a subordinate partition of unity; integrate each compactly supported coordinate expression and sum. A smooth map pulls forms back byIf $F^{\ast }\omega =\omega$ for a nowhere-zero top form, $F$ preserves its orientation. The change-of-variables theorem gives

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The metric on covectors is induced by the inverse matrix $g^{-1}$, and on $p$-forms by the determinant pairingThe Riemannian volume form is the unique positive top form taking value one on every oriented orthonormal frame. The Hodge star operator is uniquely determined byNondegeneracy of the wedge pairing proves existence and uniqueness pointwise, and the smooth metric dependence makes $\ast$ a well-defined smooth bundle map.

On compactly supported forms, Stokes theorem and the graded Leibniz rule givewhere $m=\dim M$; this is the formal $L^2$ adjoint of $d$. The Hodge Laplace-Beltrami operator is

For $g_f=e^{2f}g$, the covector metric scales by $e^{-2f}$, the $p$-form metric by $e^{-2pf}$, and the volume form by $e^{mf}$. ThereforeIf $f$ is constant, the two star factors in the codifferential contribute $e^{-2f}$, so ${\delta }_f=e^{-2f}\delta$. Since $d$ is metric-independent,

Solved by gpt-5.6-sol high.