Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 38E (General Relativity)

For a covariant rank-two tensor,Impose metric compatibility $\nabla g=0$, write this equation and its two cyclic permutations in $\alpha ,\mu ,\nu$, and use the assumed torsion-free symmetry of the connection. Solving gives the unique Levi--Civita connection

Set $U_{\mu }={\partial }_{\mu }\varphi$. Torsion-freeness and commutation of ordinary partial derivatives imply that its covariant Hessian is symmetric:Take the cyclic sum of the three commutators acting respectively on $U_{\gamma },U_{\alpha },U_{\beta }$. Their left-hand sides cancel pairwise by this Hessian symmetry, leavingAt a chosen point, a scalar field can be selected whose gradient has any prescribed covector value. The coefficient of $U_{\nu }$ must therefore vanish. Relabelling indices gives the first Bianchi identity

Using the Leibniz rule for the connection, or first checking decomposable tensors and extending linearly, the commutator acts on each covariant index:Choose $W_{\mu \nu }=g_{\mu \nu }$. Metric compatibility makes the left side zero. Applying the formula with commutator indices $\beta ,\gamma$ givesHence

Define the Ricci tensor by contracting the first and third Riemann indices:Contract the first Bianchi identity in its upper index and third lower index. The middle term vanishes by antisymmetry in the first pair, while antisymmetry in the final pair turns the last term into $-R_{\gamma \alpha }$. Thusso the Ricci tensor is symmetric.

For the stated constant-curvature form,The contracted Bianchi identity isUsing $\nabla g=0$ givesorIn four dimensions,More generally, $K$ is constant for every $n>2$. In dimension $n=2$, the contracted Bianchi identity gives no such conclusion, and the Gaussian curvature may vary from point to point.