Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 115 / 4 / Solution

The Levi-Civita connection of a Riemannian manifold $(M,g)$ is the unique connection on a vector bundle on $TM$ that is torsion-free and compatible with the Riemannian metric. Any such connection must satisfy the Koszul formulaNondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely from the right-hand side. Conversely, define $\nabla$ by this formula. Direct substitution shows that it is $C^{\infty }$-linear in $X$, satisfies the Leibniz rule in $Y$, preserves $g$, and obeys ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. It is therefore a torsion-free metric connection. This proves the Existence and uniqueness of the Levi-Civita connection.

For the metric vector bundle $E$, choose the stated orthonormal local frame $(e_1,\dots ,e_m)$ and write $d_A{e}_j=A^i{}_j{e}_i$. Since $⟨e_i,e_j⟩={\delta }_{ij}$, metric compatibility givesHence the connection matrix in an orthonormal frame is skew-symmetric:

On an oriented Riemannian $d$-manifold, the Hodge star operator is defined byIt is an orthogonal map $\ast :{\Lambda }^r{T}_x^{\ast }M\to {\Lambda }^{d-r}{T}_x^{\ast }M$ and satisfies ${\ast }^2=(-1{)}^{r(d-r)}$. In dimension $d=2n$ on middle-degree forms, the Adjoint of the Hodge star on middle-degree forms isThus it is self-adjoint when $n$ is even, but it is not always self-adjoint. For $n=1$ on the oriented Euclidean plane,so its matrix in the orthonormal basis $(dx,dy)$ is skew-adjoint.

The Laplace-Beltrami operator on differential forms is the Hodge Laplacianwhere the codifferential $\delta$ is the formal $L^2$ adjoint of the exterior derivative. On a compact manifold without boundary, if $\Delta \omega =\lambda \omega$ for a nonzero differential form $\omega$, then integration by parts givesTherefore the nonnegativity of the Hodge Laplacian yields