Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 16 / 3

Let $\tau >0$ be the first conjugate time and choose a nonzero Jacobi field $J$ with $J(0)=J(\tau )=0$. It is normal by part 1(b), and $D_{t}J(\tau )\ne 0$, since zero value and derivative there would force $J=0$. Fix any $T>\tau$ and extend $J$ by zero to a continuous piecewise smooth vector field $V$ on $[0,T]$.

For endpoint-vanishing fields, the Riemannian index form isIntegration by parts on $[0,\tau ]$ and the Jacobi field equation give $I(V,V)=0$ andChoose a smooth normal field $W$, zero at $0,T$, with $W(\tau )=-D_{t}J(\tau )$. Thenfor sufficiently small positive $\epsilon$. The corner in $V$ causes no difficulty: smooth endpoint-vanishing fields approximate $V+\epsilon W$ in the Sobolev space $H^1$, and the Riemannian index form is continuous in this norm on the compact segment. A smooth approximant therefore still has negative index form.

Realize this smooth field as a fixed-endpoint curve variation using the Riemannian exponential map. The allowed second variation of geodesic energy gives $E^{\prime }(0)=0$ and $E^{\prime \prime }(0)<0$, so some nearby curve has energy less than $T/2$, the energy of the unit-speed reference geodesic. The Cauchy-Schwarz inequality bounds its length byIt is shorter than the reference segment. Thus a geodesic cannot minimize beyond its first conjugate time. This argument deliberately uses $T>\tau$; it does not rule out minimization up to the conjugate point itself.

Let $t_0$ be the finite cut time and put $q=\gamma (t_0)$. Continuity of Riemannian distance shows that the segment to $q$ still minimizes and $d(p,q)=t_0$. Choose $t_j\downarrow t_0$ with $t_j>t_0$. By the Hopf-Rinow theorem, there is a minimizing geodesic from $p$ to $q_j=\gamma (t_j)$ with length ${\ell }_j=d(p,q_j)<t_j$ and initial unit vector $v_j$. Continuity gives ${\ell }_j\to t_0$.

The unit tangent sphere is compact, so after a subsequence $v_j\to v$. Smooth dependence of the Riemannian exponential map gives ${\exp }_p(t_{0}v)=q$. The geodesic with initial vector $v$ has length $t_0=d(p,q)$ up to $q$, hence minimizes. If $v\ne \dot{\gamma }(0)$, this is the required second minimizing geodesic.

Suppose instead $v=\dot{\gamma }(0)$ and $q$ is not a conjugate point. Then the differential of ${\exp }_p$ at $t_0\dot{\gamma }(0)$ is nonsingular. The inverse function theorem makes ${\exp }_p$ injective on a neighborhood of this vector. For large $j$, both ${\ell }_j{v}_j$ and $t_j\dot{\gamma }(0)$ belong to that neighborhood and exponentiate to $q_j$. They must be equal, implying ${\ell }_j=t_j$, a contradiction. Thus either a distinct minimizing geodesic exists or $q$ is conjugate. In the latter case it is the first conjugate point, since part (a) excludes any earlier one on a minimizing segment. We have proved the cut-point dichotomy for complete Riemannian manifolds.

Here the relevant concept is a Riemannian cut point, not the topological cut point obtained by disconnecting a space through removal of a point.

Take unit-curvature Real projective space ${\mathbb{RP}}^n=S^n/\{\pm 1\}$ with $n\ge 2$, and any point $p=[v]$. It is complete with positive sectional curvature. A unit-speed geodesic through $p$ lifts toSince the two representatives of an endpoint are antipodal, Riemannian distance in the quotient is the smaller of the two spherical distances:This equals $t$ for $0\le t\le \pi /2$ and is strictly smaller just afterwards. Thus the Riemannian cut point occurs at time $\pi /2$ in every direction.

Along a curvature-one geodesic, every normal Jacobi field with initial value zero is $J(t)=\sin t{P}_{t}a$, where $P_t$ is parallel transport and $a\perp \dot{\gamma }(0)$. Its first further zero is $t=\pi$. The quotient map is a local isometry, so the Jacobi field equation and conjugacy times are unchanged. HenceThis is cut and conjugate times in round real projective space.