Past exam of the mathematics course of the University of Cambridge 2005 ib Paper 2 12A Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 115 4 Solution Created 2026-10-03 Updated 2026-10-05
Use the energy of a curve normalizationFor a variation put and . The Levi-Civita connection is torsion free, so . Differentiating energy once gives . Along a geodesic, . Differentiating again, commuting covariant derivatives, and integrating by parts gives the second variation of geodesic energy:The curvature convention is , consistent with the specified positive sectional curvature. Indeed , and . The endpoint term vanishes for fixed endpoints or for periodic variations of a closed geodesic. The integral is the Riemannian index form .
Let . Parallel transport around the closed geodesic preserves the metric and orientation, so lies in the special orthogonal group. It fixes the nonzero tangent , and its restriction to is an orientation-preserving orthogonal map of odd dimension . An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues occur in conjugate pairs, while an odd-dimensional real orthogonal map with determinant one must have an eigenvalue . Choose a nonzero fixed vector normal to and parallel-transport it along the curve. It gives a nonzero smooth periodic normal field with .
For the exponential variation , the endpoints match periodically. Strictly positive sectional curvature givesThus for small nonzero . By the Cauchy-Schwarz inequality,where the final equality uses the geodesic's constant speed. This proves the instability of a closed geodesic in positive even-dimensional curvature.
For an embedded closed geodesic, sufficiently small variations remain embeddings, hence give a smooth isotopy with strictly shorter curves. For a nonembedded closed geodesic the construction gives a smooth deformation through immersions; an isotopy class of embeddings is not literally defined for such a curve. The stated isotopy conclusion therefore uses the usual embedded-curve interpretation, while the shorter-loop variation holds without it.
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 3 12E Solution Created 2026-09-24 Updated 2026-09-25
For a smooth curve on a Riemannian surface, its energy of a curve isChoose a local parameterization and write . With coefficients of the first fundamental formthe Lagrangian isThe Euler-Lagrange equations are equivalentlyAfter multiplying by the inverse metric these become the geodesic equationwhere the Christoffel symbols are determined by .
If a straight line segment lies in the surface, parameterize it byThen , so its acceleration has zero tangential component. The two displayed equations hold directly, and the segment is a geodesic.
For the one-sheeted hyperboloidputTwo distinct ruling lines through areIndeed, the direction satisfiesso substitution shows that every point of the line lies in . These are geodesics by the straight-line argument.
A third geodesic is the meridian through . Choose with ; thenIts acceleration is , which is normal to , so it is a geodesic. If , these give the required three distinct subsets.
If , there is also the equatorial circle. Writing ,Its acceleration is normal to along , so it is a fourth geodesic distinct from the meridian and the two rulings.
Finally, write . Clairaut's relation gives the conserved quantityChoose initial data in withSince , every point of the resulting geodesic satisfies , and thereforeContinuity keeps the geodesic in the component , and the stated completeness assumption defines it for every real time. The continuum of choices of supplies infinitely many such geodesics.