Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 24H (Differential Geometry)

For a compact triangulated surface, the Euler characteristic is $\chi =V-E+F$, independent of the triangulation. For a connected closed orientable surface, the genus $g$ is its number of handles, characterized byA connected compact orientable surface with $b$ boundary components instead has $\chi =2-2g-b$. For multiple connected components the Euler characteristics add, with genus specified for each component. A compact boundaryless surface embedded in ${\mathbb{R}}^3$ is orientable, so the usual closed-surface formula applies to each connected component.

Parametrize a regular curve by arclength $s$, write its unit tangent as $T$, and choose an oriented surface unit normal $N$. Its signed geodesic curvature isIt vanishes exactly when the acceleration of the unit-speed curve is normal to the surface, equivalently when the curve is a geodesic with arclength parametrization. For an arbitrary regular parametrization, this means the curve becomes a geodesic after reparametrization.

The global Gauss-Bonnet theorem for a compact oriented surface with smooth boundary, positively oriented with the surface on the left, isFor piecewise smooth boundary, add the signed exterior turning angles at its corners to the left side.

A simple closed curve on the sphere bounds two topological disks. Equal areas give area $2\pi$ for each, since the unit sphere has area $4\pi$. Apply Gauss-Bonnet theorem to either disk $D$, with $K=1$ and $\chi (D)=1$:A continuous geodesic curvature that were strictly positive everywhere, or strictly negative everywhere, would have a correspondingly strict sign for this integral. Therefore neither everywhere strictly positive nor everywhere strictly negative geodesic curvature is possible. Orientation reversal changes its sign but not this conclusion.

The assertion is false without further global hypotheses. For a concrete non-spherical surface of constant Gaussian curvature one, fix $a=1/2$ and setHere $f>0$, $g^{\prime }>0$ and $f^{\prime 2}+g^{\prime 2}=1$, so this is a connected regular embedded surface of revolution with unit-speed meridian. Its metric is $d{u}^2+f(u{)}^{2}d{v}^2$, whose Gaussian curvature is $-f^{\prime \prime }/f=1$. This formula also follows by multiplying its principal curvatures $-f^{\prime \prime }/g^{\prime }$ and $g^{\prime }/f$.

At $u=0$ these principal curvatures are $a=1/2$ and $1/a=2$. Every point of a radius-one sphere is umbilical with both principal curvature magnitudes one. The displayed surface therefore cannot lie in any such sphere, despite having $K=1$ everywhere.