Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 19 / 4

There is a normalization issue in the printed constant. Let ${\mathrm{Ric}}_{\mathrm{tr}}$ denote the trace convention in the catalog's Ricci curvature, and write $\overline{\mathrm{Ric}}={\mathrm{Ric}}_{\mathrm{tr}}/(n-1)$ for normalized Ricci curvature, with $n\ge 2$. The claimed $\pi /\sqrt{k}$ bound corresponds to $\overline{\mathrm{Ric}}\ge kg$, or equivalently ${\mathrm{Ric}}_{\mathrm{tr}}\ge (n-1)kg$. We prove the estimate in both conventions.

Take a unit-speed minimizing geodesic of length $L>0$ and perpendicular parallel orthonormal fields $E_1,\dots ,E_{n-1}$. Set $V_i(t)=\sin (\pi t/L)E_i(t)$. The fields vanish at the endpoints. The allowed second variation of geodesic energy makes each Riemannian index form nonnegative. Summing them givesunder the normalized lower bound. Thus $L\le \pi /\sqrt{k}$. Completeness and the Hopf-Rinow theorem supply a minimizing geodesic between any two points, so this bounds the diameter. A closed ball of that radius about any point contains the whole manifold and is compact, again by the Hopf-Rinow theorem.

Hence the intended Bonnet-Myers theorem conclusion isIf the printed $\mathrm{Ric}\ge k$ means trace Ricci curvature instead, the same integral has lower curvature term $kL/2$ and yieldsThe smaller bound is false under that convention when $n>2$: the round unit $S^n$ has trace Ricci tensor $(n-1)g$ and diameter $\pi$. Taking $k=n-1$ contradicts the printed $\pi /\sqrt{k}$ inequality. The compactness assertion remains true.

Pointwise positive Ricci curvature does not imply compactness or a finite diameter. The complete positively curved paraboloid $z=x^2+y^2$ has induced metricand Gaussian curvatureFor example, this follows from the graph formula $K=\det D^{2}z/(1+|Dz{|}^2{)}^2$. The metric dominates the Euclidean plane metric, so it is complete by upward stability of Riemannian completeness. Its diameter is infinite since its distance dominates Euclidean distance. On a surface the Ricci tensor is $Kg$, so it is strictly positive everywhere, but has no uniform positive lower bound.

There is no positive-Ricci metric on $S^2\times S^1$. If one existed, compactness of the unit tangent bundle would give ${\mathrm{Ric}}_{\mathrm{tr}}\ge cg$ for some $c>0$. The lifted metric on the universal cover is complete by Q2 and satisfies the same positive lower bound. Part (a) makes this cover compact. A covering fiber is closed and discrete and therefore finite, implying a finite fundamental group. Butwhich is infinite. This is the finite fundamental group from a uniform positive Ricci bound obstruction; it rules out every possible metric, not just the standard product metric.

Parametrize the nonconstant closed geodesic at unit speed on $[0,L]$. Its parallel transport around the loop fixes the tangent $\dot{\gamma }(0)$. Since the manifold is orientable, it preserves orientation on the full tangent space, and hence has determinant one on the normal space. This normal space has odd dimension because the manifold has even dimension.

An odd-dimensional special orthogonal transformation has a fixed vector. Indeed, nonreal eigenvalues of a real orthogonal transformation occur in conjugate pairs with product one; the real eigenvalues are $\pm 1$, and determinant one makes the number of $-1$ eigenvalues even. Odd dimension therefore requires a $+1$ eigenvalue. Transport a unit fixed normal vector around the geodesic to obtain a periodic parallel field $V$, with $D_{t}V=0$ and $V\perp \dot{\gamma }$.

The variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$ is defined uniformly for small $s$, because the image of the original loop is compact. Periodicity gives closed curves and a homotopy through loops. With the standard positive-sectional-curvature Riemannian index form, the second variation of geodesic energy isThe first variation vanishes for a closed geodesic, since the endpoint terms cancel. Therefore $E({\gamma }_s)<E({\gamma }_0)=L/2$ for sufficiently small nonzero $s$. The Cauchy-Schwarz inequality givesThus the closed geodesic is homotopic to a strictly shorter closed curve. This proves the instability of a closed geodesic in positive even-dimensional curvature; completeness of the manifold is unnecessary for this compact-loop variation. Constant loops are excluded by the usual nonconstant meaning of a closed geodesic in this assertion.