Einstein manifold 2026-10-07
An Einstein manifold has Ricci curvature equal to a constant multiple of its metric. In the Riemannian setting, the round unit sphere has . This condition is weaker than constant sectional curvature in higher dimensions. The definition can also be applied to pseudo-Riemannian metrics, but positivity is assumed in the completeness theorems discussed here.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 15 4 Solution Created 2026-10-03 Updated 2026-10-07
Fix the curvature conventionTo prove tensoriality, use and the connection rules. The two terms cancel, giving . Antisymmetry in gives linearity over smooth functions in the second input. Expanding the third input givesIt is therefore a smooth tensor of type . Lowering the output with gives the type Riemann curvature tensor .
For an independent pair , define sectional curvature byMetric compatibility gives by applying to . Together with antisymmetry in , this shows that replacing the pair by multiplies both numerator and denominator by . Thus the value depends only on the plane. Define Ricci curvature by for any orthonormal basis; a trace is independent of the orthonormal basis.
For the curvature of the round unit sphere, the outward unit normal is the position vector . The tangential projection of ambient differentiation is torsion-free and has metric compatibility, so uniqueness identifies it with . Ambient differentiation satisfies andsince differentiating gives its normal component. The ambient curvature is zero. Take tangential components of to obtainHenceEvery two-plane has sectional curvature one. Tracing the first formula gives , and consequentlyThus the sphere is an Einstein manifold. When there are no tangent two-planes, the curvature tensor is zero and the same Ricci formula gives zero. The declared slot convention fixes all signs.