Past exam of the geometry course of the University of Oxford / 2009 / A

The classification of compact surfaces states that every compact connected topological surface without boundary is homeomorphic to exactly one of the sphere, a connected sum of surfaces consisting of $g\ge 1$ tori, or a connected sum of $h\ge 1$ real projective planes. The torus family is orientable and the projective-plane family is nonorientable; their respective genera are unique. Including $g=0$ makes the sphere the genus-zero orientable case. If boundaries are allowed, the number of removed open discs is an additional invariant.

A finite subdivision into vertices, edges and polygonal faces defines the Euler characteristicCount cells after all identifications. Subdividing an edge adds one vertex and one edge; splitting a face adds one edge and one face. Each operation leaves $V-E+F$ unchanged. Comparing subdivisions through a common refinement proves independence of the subdivision.

For the connected sum of surfaces, remove an embedded open disc from each surface and identify the resulting boundary circles by a homeomorphism. The resulting closed connected surface is denoted $M\#N$.

Removing a disc decreases the Euler characteristic by one. Gluing the remaining surfaces along their common boundary subtracts the characteristic of a circle, which is zero. ThusThis can be verified by choosing matching subdivisions of the boundary circles and counting the vertices and edges identified in the gluing.

A square model of the torus has one vertex, two edges and one face, so its Euler characteristic is zero. The connected-sum formula givesThe real projective plane has a bigon model with one vertex, one edge and one face, hence characteristic one. Therefore

Take $M=T^2$, $N={\mathbb{RP}}^2\#{\mathbb{RP}}^2$ and $S={\mathbb{RP}}^2$. The first is orientable, while $N$, the Klein bottle, is not, so $M$ and $N$ are not homeomorphic.

Both $M\#S$ and $N\#S$ are nonorientable and have Euler characteristic $-1$. By the classification of compact surfaces, both are homeomorphic to the connected sum of three projective planes. HenceThis is failure of cancellation for connected sums of surfaces.

For a nonconstant holomorphic map of degree $d$ between compact connected Riemann surfaces, the Riemann-Hurwitz formula isHere $e_p$ is the local degree, characterized by local coordinates in which the map is $z\mapsto z^{e_p}$. The sum is finite and its terms are nonnegative; a ramification point has $e_p>1$.

If $g_X=0$, the Riemann-Hurwitz formula gives $2=d(2-2g_Y)-R$, where $R\ge 0$. If $g_Y\ge 1$, the right side is nonpositive, a contradiction. Therefore

The given Weierstrass elliptic function is even, by replacing the lattice index $\omega$ with $-\omega$. Differentiation givesFor any $\lambda \in \Lambda$, reindexing the sum shows ${\wp }^{\prime }(z+\lambda )={\wp }^{\prime }(z)$. Thus $\wp (z+\lambda )-\wp (z)=c_{\lambda }$ is constant off the lattice. Replacing $z$ by $-z-\lambda$ and using evenness changes this difference to its negative, so $c_{\lambda }=-c_{\lambda }$ and $c_{\lambda }=0$. Hence $\wp$ is doubly periodic.

It therefore descends to a meromorphic function on $\mathbb{C}/\Lambda$, with a double pole at zero modulo the lattice and no others. A meromorphic function is a holomorphic map to the Riemann sphere, sending poles to infinity. Locally at zero, $1/\wp (z)=z^2+O(z^4)$ is holomorphic and vanishes there, so the sphere chart at infinity supplies the required extension. The map is nonconstant because it has a pole.

The total pole order in one period cell is two, so the map has degree two. The Riemann-Hurwitz formula for genus one to genus zero gives $0=2\cdot 2-R$, hence $R=4$. Each local degree is at most two, so every ramification contribution is exactly one. ThereforeThey are zero and the three nonzero half-period classes. Zero is ramified because its reciprocal has a double zero. At a nonzero half-period, oddness and periodicity of ${\wp }^{\prime }$ force ${\wp }^{\prime }=0$; the count shows these exhaust the remaining points.

For a degree-two map to the Riemann sphere, the Riemann-Hurwitz formula gives $2-2g=4-R$. Again each ramification index is two, so the number of ramification points equals $R$. Thus

For a regular parametrization $r(u,v)$, the first fundamental form is the induced metricIt measures the lengths and angles of tangent vectors. A geodesic is a curve whose unit tangent has zero covariant derivative along itself. For an arc-length parametrized surface curve in Euclidean three-space, this means its acceleration is normal to the surface, or ${\gamma }^{\prime \prime }\cdot r_u={\gamma }^{\prime \prime }\cdot r_v=0$.

Since ${\gamma }^{\prime }=r_u{u}^{\prime }+r_v{v}^{\prime }$, we have ${\gamma }^{\prime }\cdot r_u=E{u}^{\prime }+F{v}^{\prime }$. The normal-acceleration characterization of a geodesic givesExpanding and differentiating the metric coefficients yields $r_u\cdot r_{uu}=E_u/2$, $r_v\cdot r_{uu}+r_u\cdot r_{uv}=F_u$, and $r_v\cdot r_{uv}=G_u/2$. ThereforeApplying the same calculation to ${\gamma }^{\prime }\cdot r_v=F{u}^{\prime }+G{v}^{\prime }$ givesThese are also the Euler-Lagrange equations of the quadratic metric energy.

For the surface of revolution, the arc-length condition on its profile gives $E=1$, $F=0$, $G=f(u{)}^2$. Thus the geodesic equations reduce toConsequentlythe Clairaut first integral for a surface of revolution.

On a meridian, take $v=b$ and $u=\pm s+u_0$. Both equations hold, so every meridian is a geodesic. On a cross-section $u=c$, arc length requires $v^{\prime }=\pm 1/f(c)$. The first equation then holds exactly when $f^{\prime }(c)=0$, while the second holds automatically. Hence