Past exam of the geometry of surfaces course of the University of Oxford / 2023 / 1

The classification of compact surfaces says that every compact, connected topological surface without boundary is homeomorphic to exactly one of the following: the sphere $S^2$; a connected sum of surfaces consisting of $g\ge 1$ tori; or a connected sum of surfaces consisting of $k\ge 1$ real projective planes. The first two cases are orientable; the last case is a nonorientable surface. Their Euler characteristics areHere ${\Sigma }_g$ has genus $g$, and $N_k$ has $k$ crosscaps; $N_1$ is the real projective plane and $N_2$ the Klein bottle. Thus orientability together with the Euler characteristic determines the closed surface up to homeomorphism.

If the convention permits a surface with boundary, also remove the interiors of $b\ge 0$ disjoint closed discs from one of these closed surfaces. The Euler characteristic decreases by $b$, giving $2-2g-b$ in the orientable case and $2-k-b$ in the nonorientable surface case. The number of boundary components is then an additional invariant.

Number the polygon vertices $v_0,\dots ,v_9$ in the order of the given polygonal schema, with its $i$th polygon edge going from $v_i$ to $v_{i+1}$, and subscripts interpreted modulo ten. Equal labels identify equal arrow endpoints. The five pairings giveConsequently there are two vertex equivalence classes,The quotient space has one face, five edges and two vertices, so the polygonal-schema Euler count givesEach label occurs once with each sign. Hence every pairing reverses the direction induced by traversing the boundary of the polygon, so an orientation of a surface on its interior extends consistently across every glued edge. Thus $X$ is orientable. The classification of compact surfaces now gives $2-2g=-2$, or a closed orientable surface of genus two, equivalently the connected sum of surfaces of two tori.

Write ${\theta }_i$ for the internal angle at $v_i$, using the numbering in part (b), and use the hyperbolic plane with Gaussian curvature $-1$. The necessary and sufficient conditions are equal lengths for each pair of equally labelled polygon sides, and total angle $2\pi$ at each vertex equivalence class. Explicitly, the two angle conditions areNecessity follows because the identification must preserve arc length along each seam, and a smooth interior point of a Riemannian surface has total angle $2\pi$. For sufficiency, equally long hyperbolic geodesic segments can be identified by isometries. Across their interiors, two half-disc neighbourhoods join to a hyperbolic disc. At an identified vertex, the incident sectors form a cyclic disc with total angle $2\pi$, so no cone singularity of a surface metric remains. These local isometries therefore extend the Riemannian metric smoothly everywhere. This is the smooth hyperbolic polygon gluing criterion.

The ten internal angles sum to $4\pi$. The area of a hyperbolic geodesic polygon consequently givesEquivalently, the Gauss-Bonnet theorem gives $-\mathrm{Area}(X)=2\pi \chi (X)=-4\pi$.

The uniformization theorem says that every simply connected Riemann surface is biholomorphic to the Riemann sphere, the complex plane, or the unit disc. Its consequence for a compact Riemann surface of genus at least two is that the universal cover is the unit disc, equivalently the hyperbolic plane.

To explain the restriction, a spherical universal cover would make the fundamental group finite. Since a finite covering map multiplies Euler characteristic, $2=m(2-2g)$ is impossible for $g\ge 2$. If the universal cover were the complex plane, its deck transformations would be biholomorphisms $z\mapsto az+b$. A nonidentity such map with $a\ne 1$ has a fixed point, impossible for a deck transformation. Thus the deck transformation group would consist of translations. A discrete subgroup of translations with compact quotient space is a rank-two Euclidean lattice, whose quotient space is a torus, again incompatible with $g\ge 2$.

On the unit disc, the Poincare metric on a Riemann surfacehas Gaussian curvature $-1$ and is invariant under every biholomorphism of the unit disc. It therefore descends through the deck transformation group to a hyperbolic metric on $X$. Its local expression is a positive multiple of $|dz{|}^2$ in every holomorphic coordinate chart, so it is compatible with the given complex structure. In fact, this hyperbolic metric is unique within that conformal class.

Take a hexagonal polygonal schema with boundary wordIdentify equally labelled polygon edges in the indicated directions. The two consecutive occurrences of $a$ identify the first three vertices; the two $b$ edges extend this class to the next two vertices, and the $c$ edges include the remaining vertex. Thus all six polygon vertices become one, while the six edges become three and the interior is one face. Its Euler characteristic isThis is the usual polygonal schema for the connected sum of surfaces of three real projective planes. The same-sign edge pairs force a nonorientable surface, consistently with $\chi (N_3)=2-3=-1$. A closed orientable surface cannot have odd Euler characteristic.