Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 19 / 5

Use the Riemannian curvature two-form conventionand define its four-covariant Riemann curvature tensor by $R(X,Y,Z,W)=g(R(X,Y)W,Z)$. This makes $R(X,Y,X,Y)$ positive on an orthonormal two-plane of the unit sphere. The full algebraic symmetries areTogether with multilinearity, these are antisymmetry in each pair, interchange of pairs, and the first Bianchi identity.

Define the Ricci tensor by the basis-independent contractionfor any orthonormal basis $(e_a)$. The trace is invariant under change of basis, and contraction of the smooth curvature tensor gives a smooth covariant two-tensor. Pair interchange gives $R(e_a,Y,e_a,W)=R(e_a,W,e_a,Y)$, so $\mathrm{Ric}(Y,W)=\mathrm{Ric}(W,Y)$. Thus the Ricci tensor is a symmetric covariant 2-tensor.

The sectional curvature of the two-plane $\sigma =\mathrm{span}(u,v)$ isBoth numerator and denominator multiply by the square of the determinant under a change of basis of $\sigma$, so this is a plane invariant. The directional Ricci curvature of $u\ne 0$ is $\mathrm{Ric}(u,u)/g(u,u)$. For an orthonormal basis beginning with $u/|u|$, it equals the sum of the sectional curvatures of the planes containing that basis vector.

In dimension three, suppose this directional value is $c$ at the point under consideration. For any orthonormal triple $(e_1,e_2,e_3)$, put $K_{ij}=K(\mathrm{span}(e_i,e_j))$. Thenso all three are $c/2$. Every two-plane can be the span of the first two members of such a triple. ConsequentlyThis is a pointwise conclusion, as required; it uses sectional curvatures from Ricci curvature in dimension three and asserts no spatial constancy.

For the product, the canonical identification $T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N$ defines the product Riemannian metricIt is smooth in product charts and positive definite, since a nonzero vector has a nonzero component in at least one factor. Those components are orthogonal. The coordinate connection computation in the subparts below givesTaking the trace in orthonormal bases of the two factors therefore gives ${\mathrm{Ric}}_{M\times N}={\mathrm{Ric}}_M\oplus {\mathrm{Ric}}_N$.

For the final example choose the two round unit spheres. Each has sectional curvature one and Ricci tensor equal to its metric. Hence the product has $\mathrm{Ric}=g$, so every directional Ricci curvature is one. A two-plane tangent to one factor has sectional curvature one, whereas a mixed two-plane has sectional curvature zero by (iii). This gives constant Ricci curvature without constant sectional curvature:Both kinds of plane occur at every point, so the sectional curvatures fail to be constant even at a single point.

Use product coordinates $x^i$ on $M$ and $y^{\alpha }$ on $N$. The lifted vector fields have the form $X={\sum }_i{X}^i(x){\partial }_{x^i}$ and $Y={\sum }_{\alpha }{Y}^{\alpha }(y){\partial }_{y^{\alpha }}$. The coefficients of each depend only on its own coordinate set, and the coordinate partial derivatives commute. The Lie bracket of vector fields formula therefore givesThus $\boxed{[X,Y]=0}$ for fields lifted separately from the two factors. The lift hypothesis matters: arbitrary product-manifold fields can have coefficients depending on both factors.

The product Riemannian metric has block matrixIn the given Christoffel symbols formula, every mixed coefficient vanishes. Off-diagonal metric entries are zero, ${\partial }_{y^{\alpha }}{g}_{ij}=0$, and ${\partial }_{x^i}{g}_{\alpha \beta }=0$. These facts give ${\Gamma }_{i\alpha }^A=0$ for every output index $A$, as well as ${\Gamma }_{ij}^{\alpha }=0$ and ${\Gamma }_{\alpha \beta }^i=0$. The remaining pure blocks are precisely the Christoffel symbols of the respective factor metrics.

For the lifted fields of (i), ${\partial }_{x^i}{Y}^{\alpha }=0$. ConsequentlyThus $\boxed{{\nabla }_{X}Y=0}$. The same calculation gives ${\nabla }_{Y}X=0$; differentiating two fields from one factor gives the lifted factor covariant derivative. This identifies the product Levi-Civita connection with the product affine connection and proves the curvature splitting used in the root solution.

Since $Y$ is lifted from $N$, its covariant derivative ${\nabla }_{Y}Y$ is also lifted from $N$. Part (ii) therefore gives ${\nabla }_X({\nabla }_{Y}Y)=0$. It also gives ${\nabla }_{X}Y=0$, and part (i) gives $[X,Y]=0$. Substituting into the Riemannian curvature two-form definition yieldsWith the four-slot convention fixed above, this provesFor nonzero lifted vectors from different factors they are orthogonal, so the sectional-curvature denominator is positive. Hence the corresponding mixed plane has sectional curvature zero, providing the mixed-plane value in the $S^2\times S^2$ example.