Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 15 / 2

For fixed $p$, the covariant derivative is linear in its tangent-vector input, so $L_p:T_{p}M\to T_{p}M$, $v\mapsto ({\nabla }_{v}X)(p)$, is a well-defined linear map. Its trace defines the divergence of a Riemannian vector field. In the given Riemannian orthonormal frame, write $X={\sum }_j{f}_j{E}_j$. The Levi-Civita connection product rule givesAt $p$, the second sum vanishes by the normal orthonormal frame hypothesis. Thus the $i$th diagonal coefficient of $L_p$ is $E_i(f_i)(p)$, andEquivalently one can take ${\sum }_{i}g({\nabla }_{E_i}X,E_i)$; the trace does not depend on the orthonormal basis chosen.

Since the Riemannian orthonormal frame is positively oriented, its dual coframe satisfies ${\omega }_g={\omega }_1\wedge \cdots \wedge {\omega }_n$. The form $\nu$ is the interior product ${\iota }_X{\omega }_g$. The contraction rule for an exterior product givesOne can also obtain the sign by moving $E_i$ from the first argument into slot $i$ of the alternating form, using $i-1$ transpositions. Linearity in $X={\sum }_i{f}_i{E}_i$ proves the interior contraction of the Riemannian volume form formula:The exponents $i-1$ and $i+1$ have the same parity.

Fix any point $p$ and use a normal orthonormal frame there. Such frames exist: in geodesic normal coordinates, $g_{ij}(p)={\delta }_{ij}$, ${\partial }_k{g}_{ij}(p)=0$ and the Christoffel symbols vanish at $p$. Applying the Gram-Schmidt process smoothly to the coordinate fields gives identity coefficients at $p$ and zero first derivatives there, so it produces the required frame. Reverse one frame vector if necessary to match the orientation.

Because the Levi-Civita connection is torsion-free, $[E_j,E_k](p)=({\nabla }_{E_j}{E}_k-{\nabla }_{E_k}{E}_j)(p)=0$. The supplied formula for the exterior derivative of a one-form givesThus $d{\omega }_i(p)=0$, and the graded Leibniz rule implies $d{\theta }_i(p)=0$. Differentiate the formula in part (b):Since $d{f}_i={\sum }_j{E}_j(f_i){\omega }_j$, every summand with $j\ne i$ vanishes because ${\omega }_j$ already occurs in ${\theta }_i$. The remaining term has ${\omega }_i\wedge {\theta }_i=(-1{)}^{i-1}{\omega }_g$, so both signs cancel. HenceThe point was arbitrary, proving the global exterior derivative of contracted volume form identity

Use the standard convention that these compact smooth manifolds have no boundary. The interior product identity for a one-form gives $df\wedge {\iota }_X{\omega }_g=df(X){\omega }_g=X(f){\omega }_g$. For example, in the coframe from part (b), only the matching index survives in this wedge product, with the same pair of cancelling signs as in part (c). Thus the graded Leibniz rule and part (c) giveIntegrate and apply the Generalized Stokes theorem. Since $\partial M=\varnothing$, the integral of $d(f\nu )$ is zero. Rearranging proves integration by parts for Riemannian divergence:If a boundary were allowed, the exact general formula would instead have the additional term ${\int }_{\partial M}f\nu$ on the right; compactness alone does not remove that boundary contribution.