Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 4

The solder form on $X$ is the $TX$-valued one-formIn a coordinate frame it is $\theta =d{x}^i\otimes {\partial }_i$. The torsion form of $\nabla$ isEquivalently, for vector fields $U,V$,Writingand using $[{\partial }_i,{\partial }_j]=0$ givesHence $\nabla$ is torsion-free exactly whenfor every $i,j,k$.

The connection is orthogonal, or metric-compatible, when $\nabla g=0$; equivalently, its parallel transport preserves the Riemannian metric. Compatibility of the induced connections with tensor contraction givesfor all vector fields $U,V$. Taking $U={\partial }_j$ and $V={\partial }_k$ yieldsConversely, this coordinate identity makes every component of $\nabla g$ vanish, so it is equivalent to orthogonality.

For forms of the same degree, define the Hodge inner product byOn a $p$-form $\beta$, define the codifferentialequivalently $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ in dimension $n$. If $\alpha$ has degree $p-1$, Stokes theorem on the compact boundaryless manifold givesRearranging and using the definition of $\delta$ givesThus $\delta$ is the formal adjoint of $d$.

A form is harmonic whenAdjointness givesTherefore $\Delta \alpha =0$ implies $d\alpha =0$ and $\delta \alpha =0$; the converse follows immediately from the definition of $\Delta$.

The Hodge decomposition theorem states in particular that every de Rham cohomology class has a unique harmonic representative. Hence

For the stated flat metric and orientation,Because the metric coefficients and the coordinate one-forms are constant, the Hodge Laplacian acts coefficientwise:Thus a harmonic one-form $\alpha ={\sum }_i{\alpha }_{i}d{x}^i$ has harmonic coefficient functions. Every harmonic function on the compact connected torus is constant by the maximum principle for harmonic functions. HenceThe Hodge decomposition theorem now gives