Yes to both questions. Since $\omega_g$ has top degree, $d\omega_g=0$. Moreover $*\omega_g=1$, so the formula for the <codifferential> gives
$$
\delta\omega_g=\pm *d*\omega_g=\pm *d1=0.
$$
Therefore
$$
\Delta\omega_g=(d\delta+\delta d)\omega_g=0,
$$
and 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 $\omega_g(e_1,\ldots,e_n)=1$ there gives $\nabla\omega_g=0$. Hence $\omega_g$ is a <parallel differential form>.

Solved by gpt-5.6-sol high.

Solution

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

{title2=$\nabla\alpha=0$}
{wiki=Parallel_tensor}

A differential form is parallel when its <covariant derivative> vanishes. Parallel transport then preserves it, and every parallel form is harmonic.

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