Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 131 / 3

An orientation selects the positive ordered bases in each tangent space. On an oriented $n$-dimensional Riemannian manifold, the Riemannian volume form ${\omega }_g$ is the unique smooth $n$-form satisfyingfor every positively oriented orthonormal frame. In positively oriented local coordinates,

The metric induces an inner product on the bundle ${\Lambda }^p{T}^{\ast }M$ of $p$-forms. The Hodge star operator is the unique linear mapsuch thatfor all $p$-forms $\alpha ,\beta$. With the codifferential $\delta$, the Laplace-Beltrami operator on differential forms is

The Hodge decomposition theorem says that on a compact oriented Riemannian manifold,an $L^2$-orthogonal direct sum, where ${\mathcal{H}}^p(M)=\ker \Delta$ is the finite-dimensional space of harmonic $p$-forms. Every de Rham cohomology class has exactly one harmonic representative.

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Yes to both questions. Since ${\omega }_g$ has top degree, $d{\omega }_g=0$. Moreover $\ast {\omega }_g=1$, so the formula for the codifferential givesThereforeand the Riemannian volume form is a harmonic differential form.

The Levi-Civita connection preserves both the Riemannian metric and its chosen orientation. At any point, extend a positively oriented orthonormal basis to a local frame whose covariant derivatives vanish at that point. Differentiating ${\omega }_g(e_1,\dots ,e_n)=1$ there gives $\nabla {\omega }_g=0$. Hence ${\omega }_g$ is a parallel differential form.

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On $p$-forms in dimension $n$, the defining identity for the Hodge star operator givesFor $n=4$ and $p=2$, therefore, ${\ast }^2=1$. For every $\alpha \in {\Omega }^2(N)$ defineThen $\ast {\alpha }_{+}={\alpha }_{+}$, $\ast {\alpha }_{-}=-{\alpha }_{-}$, and $\alpha ={\alpha }_{+}+{\alpha }_{-}$. The two eigenspaces of the involution $\ast$ have zero intersection, which proves uniqueness. They are respectively the spaces of self-dual and anti-self-dual two-forms.

Now suppose $N$ is compact and let $\beta$ be an exact three-form, say $\beta =d\theta$. Apply the Hodge decomposition theorem to the two-form $\theta$:Set $a=\delta \psi$. Then $da=d\theta =\beta$ and $\delta a={\delta }^2\psi =0$. For a two-form in dimension four, $\delta =-\ast d\ast$, so $d\ast a=0$. The self-dual formsatisfiesThus every exact three-form is the exterior derivative of a self-dual two-form.

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