Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 25G (Differential Geometry)

For a positively oriented local parametrization $X(u,v)$ with first fundamental formthe area element of a surface isUnder an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation byor equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If $T$ is its unit tangent and $N$ is the chosen unit normal to the surface, its signed geodesic curvature iswhere $D_s$ is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners iswhere $K$ is the Gaussian curvature.

Because $S_1$ and $S_2$ are isometric,The two induced orientations on their common boundary $\gamma$ are opposite. Ifis computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the Gauss-Bonnet theorem to the two halves givesThe equalities above imply $B=0$. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry ${\varphi }_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv 0$. A curve has zero geodesic curvature exactly when it is a geodesic, so $\gamma$ is geodesic.

No. On the round sphere, take a latitude circle other than the equator. Rotations about the polar axis preserve the circle and act transitively on its points, so all the isometries ${\varphi }_{x,y}$ exist. The latitude separates the sphere into two caps of unequal area, so those caps are not isometric. It is not a great circle and therefore is not a geodesic; equivalently, its geodesic curvature is nonzero. This is a homogeneous nongeodesic latitude.

Again, no. On the unit round sphere and for a sufficiently small nonzero $\epsilon$, considerThis is a smooth embedded closed curve and may be reparametrized by arc length. It obeysso the antipodal isometry preserves the curve and exchanges the two complementary discs. Its restriction therefore gives an isometry $S_1\to S_2$.

The curve is not contained in any plane through the origin: its third-harmonic height cannot satisfy a nontrivial linear relation with $\cos \theta$ and $\sin \theta$. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles, $\gamma$ is not geodesic. This antipodally symmetric nongeodesic spherical curve shows that the point-transitivity assumption cannot be omitted.