Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 12 / 3

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.