Past exam of the geometry course of the University of Oxford / 2011 (B3, DMAB 2642, DMPB 2642)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

For a finite CW complex with $V$ vertices, $E$ edges and $F$ faces,This Euler characteristic is unchanged by subdividing an edge, which adds one vertex and one edge, or subdividing a face by an edge, which adds one edge and one face. Common refinements therefore give the same value for different subdivisions of the same surface.

Form the dual graph of the finite triangulation: its vertices are the triangles, and two vertices are joined when their triangles share an edge. Connectedness of the surface makes this graph connected. Choose a spanning tree and begin with one disjoint triangle. Attach the remaining triangles along the tree edges, one at a time, always joining a new triangle along one boundary edge. At every step the result is a topological disk, so the final union is a polygonal disk.

The edges not used for these attachments still occur twice on the boundary, once for each incident triangle. Identify each pair as in the original triangulation. The result recovers the original surface. This proves a compact connected closed surface is a polygon with its boundary edges identified in pairs. Closedness is the convention here; a boundary would leave some edges unpaired.

Count incidences of triangular faces with edges. Every triangle contributes three and every edge of a closed surface occurs in two triangles, hence $3F=2E$. Substituting $F=2E/3$ into the Euler characteristic gives

The simplicial intersection conditions prohibit loops and multiple edges between the same pair of vertices, so the one-skeleton is a simple graph andUsing the preceding identity gives $V^2-7V+6\chi \ge 0$. A closed surface has $\chi \le 2$ and a simplicial triangulation has $V\ge 4$. The smaller root is at most three, so this inequality forces $V$ to lie above the larger root:For completeness, fewer than four vertices could give at most one triangle under the intersection rules, which cannot form a closed surface.

Here the surface is the real projective plane, with Euler characteristic one. The bound gives $V\ge 6$. The following ten triangles on vertices $1,\dots ,6$ attain it:Take these as the maximal simplices of an abstract simplicial complex, identifying edges with the same vertex labels. Every one of the 15 pairs of vertices occurs as an edge in exactly two triangles. The links of the six vertices are single five-cycles; listed cyclically they areThus every point has a disk neighborhood, and the realization is a connected closed surface. Its Euler characteristic is $6-15+10=1$, so the classification of compact surfaces identifies it with the real projective plane. This is an explicit triangulation, proving $\boxed{V_{\min }=6}$.

For a nonconstant holomorphic map $f:X\to Y$ between compact connected Riemann surfaces, let $n$ be its degree and $e_p$ its local degree at $p$. The Riemann-Hurwitz formula isEquivalently, $2g_X-2=n(2g_Y-2)+{\sum }_p(e_p-1)$. Only finitely many terms are nonzero; their sum is the total ramification. In local coordinates the map has the form $w=z^{e_p}$ after suitable coordinate changes.

Let $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ and $T=\mathbb{C}/\Lambda$. Its fundamental parallelogram with opposite sides identified is a torus, so $\chi (T)=0$. The Weierstrass elliptic function is even and doubly periodic. One way to verify the latter from its series is to differentiate: the locally normally convergent derivative series iswhich is absolutely convergent away from the lattice and periodic by reindexing. Thus $\wp (z+{\omega }_j)-\wp (z)$ is constant. Evaluating at $z=-{\omega }_j/2$ and using evenness shows that this constant is zero for each generating period.

Therefore $[z]\mapsto \wp (z)$ descends to a meromorphic function on $T$, with one double pole at $[0]$. Viewed as a map to the Riemann sphere, a pole is a holomorphic point in the target coordinate $1/w$. The map is consequently holomorphic everywhere and

The Riemann-Hurwitz formula gives total ramification four because $0=2\cdot 2-R$. The double pole at $[0]$ has local degree two and contributes one. Since ${\wp }^{\prime }$ is odd and periodic, it vanishes at each nonzero point satisfying $[h]=[-h]$. These areThey are three distinct finite points, each contributing at least one to ramification. Together with $[0]$ they exhaust the total of four. Each therefore has local degree exactly two, meaning its derivative has a simple zero. There are no other zeros of ${\wp }^{\prime }$, and their translates by $\Lambda$ give the full list in the complex plane.

Multiplication by two on the complex torus is a holomorphic unramified coveringIts kernel is $(\frac{1}{2}\Lambda )/\Lambda$, of order four; its derivative is the nonzero constant two. Therefore $\mathrm{deg}[2]=4$. Composition with the degree-two Weierstrass elliptic function givesIt is well-defined because changing $z$ by $\omega \in \Lambda$ changes $2z$ by the period $2\omega$.

A pole occurs exactly when $2z\in \Lambda$. Thus the poles on $T$ areEach is a double pole, since the multiplication-by-two map is locally unramified and $\wp$ has a double pole at $[0]$. There are four distinct poles, each of order two, consistent with the degree eight.

Put $R(z)=\frac{1}{4}({\wp }^{\prime \prime }/{\wp }^{\prime }{)}^2-2\wp (z)$. This is an elliptic function. Its only possible poles modulo $\Lambda$ are $0$ and the three simple zeros $h$ of ${\wp }^{\prime }$. At such a nonzero half-period, periodicity and evenness give $\wp (h+t)=\wp (h-t)$, soIt follows that $R(h+t)=1/(4t^2)+O(1)$ with no simple-pole term. This is exactly the principal part of $\wp (2h+2t)=\wp (2t)$.

At zero the source expansion yieldsConsequently $R(z)=1/(4z^2)+O(z^2)$, again matching $\wp (2z)$. Thus $R(z)-\wp (2z)$ is a holomorphic function on a compact connected Riemann surface, hence constant. Its limit at zero is zero, provingThe equality extends over the apparent singularities as an identity of meromorphic functions.

For a regular parametrization $\mathbf{r}(u,v)$, the first fundamental form is the induced Riemannian metricIt measures squared lengths of tangent vectors and therefore lengths of curves. In the plane put $x=\rho \cos \theta$, $y=\rho \sin \theta$. The derivatives are orthogonal, of lengths one and $\rho$, soThese polar coordinates are regular for $\rho >0$.

Two regular coordinate patches are locally isometric under a coordinate correspondence exactly when it preserves their first fundamental forms. More explicitly, a diffeomorphism $\Phi :U\to \tilde{U}$ of parameter domains induces a local isometry if and only ifIn common coordinates this means equality of the three coefficient functions $E,F,G$. The criterion follows because the first fundamental form gives the inner product of every pair of tangent vectors. Equality therefore preserves lengths of all curves; conversely a smooth isometry preserves tangent-vector lengths and their inner products by polarization.

Using the two first fundamental forms calculated below, chooseThe surface of revolution is then a catenoid, with metric ${\cosh }^{2}s(d{s}^2+d{v}^2)$. On the helicoid, take $u=\sinh s/a$, assuming $a\ne 0$ so that its parametrization is regular. Substitution givesThus the coordinate mapis a local isometry on sufficiently small patches. The constant $a$ only changes the radial parametrization of the given helicoid, whose pitch is fixed at one. For $a=0$ the displayed helicoid parametrization is not a regular surface.

Differentiate to getTheir squared lengths are $a^2$ and $a^2{u}^2+1$, and their inner product is zero. Hence the helicoid has first fundamental form

For the surface of revolution,The mixed inner product vanishes, givingTaking $f(u)=\cosh u$ makes both coefficients ${\cosh }^{2}u$, as used in the parent calculation.

Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points. The general-position condition says that these vectors form a basis of ${\mathbb{C}}^{n+1}$. Write a representative of the last point as $v_{n+2}={\sum }_i{c}_i{v}_i$. Every $c_i$ is nonzero; otherwise the last point and the other $n$ basis points would lie in a hyperplane. Rescaling $v_i$ by $c_i$ gives representatives with $v_{n+2}={\sum }_i{v}_i$. Normalize representatives $w_i$ of the second frame in the same way.

There is a unique invertible linear map $L$ with $L(v_i)=w_i$ for $i\le n+1$. It also maps $v_{n+2}$ to $w_{n+2}$, so its projectivization has the required effect. For uniqueness, the quotient of two candidate projective transformations fixes the normalized frame. A representing matrix must be diagonal in the $v_i$ basis, with entries $a_i$. Fixing $[{\sum }_i{v}_i]$ forces all $a_i$ to be equal. Such a scalar represents the identity projective transformation. Thus the ordered projective frame determines the transformation uniquely.

Represent each point by a nonzero column $v_i$. The source's cross-ratio isRescaling representatives cancels between numerator and denominator. If the projective transformation is represented by $A\in {\mathrm{GL}}_2(\mathbb{C})$, then $\det (A{v}_i,A{v}_j)=\det (A)\det (v_i,v_j)$. Two factors of $\det (A)$ occur in both numerator and denominator and cancel. Hence the cross-ratio is invariant under every projective transformation.

Injectivity makes the images of any three distinct points distinct. By the ordered-frame result, there is a projective transformation $L$ agreeing with the given map on three chosen points. Composing by $L^{-1}$ reduces to an injective map fixing those points and preserving every cross-ratio. Change coordinates so that the three points are $\infty ,0,1$.

For any other point with affine coordinate $z$, the convention in the source givesPreservation therefore forces the image of $z$ to have the same coordinate $z$. The normalized map fixes every point and is the identity. Undoing the normalization proves the original map is exactly $L$, hence projective. No prior continuity or surjectivity assumption is needed.

Choose $s\in \mathbb{C}$ with $s^2=\lambda$. Since $\lambda \ne 0$, the invertible diagonal projective transformationfixes $[0:1:0]$. Writing its image as $[X:Y:Z]$, direct substitution givesThus it maps $C_{\lambda }$ onto $C_{1/\lambda }$. Either choice of square root works. We choose $s=i$ at $\lambda =-1$ and $s=e^{\pi i/6}$ at $\lambda =e^{\pi i/3}$ for the later order calculations.

An appropriate projective transformation isIt fixes $[0:1:0]$ and has an invertible representing matrix. For $X=z-x$, $Y=iy$, $Z=z$,Hence its restriction is an isomorphism $C_{\lambda }\to C_{1-\lambda }$.

Compose the preceding transformations, applying ${\sigma }_{\lambda }$ first and then ${\tau }_{1/\lambda }$. With the square-root convention already specified,This invertible projective transformation fixes $[0:1:0]$ and carries $C_{\lambda }$ to $C_{1-1/\lambda }$. It is important to retain the sign of the chosen lift on the $y$ coordinate when computing its order later.

On the parameter, the two transformations are $S(\lambda )=1/\lambda$ and $T(\lambda )=1-\lambda$. Their compositions yieldEvery step lifts to an invertible projective transformation of the corresponding cubic curves. Therefore all six parameter values give isomorphic curves. Some values coincide for special $\lambda$; this does not affect the conclusion.

At $\lambda =-1$ and $s=i$, the chosen map isThe square is not the identity on the cubic, since a generic point has $y\ne 0$. Thus $\boxed{\mathrm{ord}(\sigma )=4}$.

Choose the cubic's point $[0:1:0]$ as the origin in its complex torus description. Since $\sigma$ fixes it, its torus form is $z+\Lambda \mapsto az+\Lambda$, with $a\Lambda =\Lambda$. Exact order four implies $a=\pm i$; $a^2=-1$. Therefore $\Lambda$ is stable under multiplication by $i$.

To identify its shape without an unproved uniqueness assertion, choose a shortest nonzero period $\omega$. For any ${\lambda }_0\in \Lambda$, approximate ${\lambda }_0/\omega$ by a Gaussian integer $m+ni$ within distance $1/\sqrt{2}<1$. The difference ${\lambda }_0-(m+ni)\omega$ belongs to $\Lambda$ and is shorter than $\omega$, so it must vanish. HenceThe lattice is determined up to a nonzero complex scaling: it is square.

Take $\lambda =e^{\pi i/3}$ and the chosen $s=e^{\pi i/6}$. Then $\lambda s=i$, so $i/(\lambda s)=1$, and $-1/\lambda ={\lambda }^2$. In affine coordinates the chosen lift becomesSince ${\lambda }^2$ is a primitive cube root of unity, $1+{\lambda }^2+{\lambda }^4=0$. Three iterations fix $x$ and $y$, whereas one does not. ThusThe other square-root choice replaces this map by $(x,y)\mapsto (1+{\lambda }^{2}x,-y)$. Its cube is $(x,y)\mapsto (x,-y)$, so that lift has order six. This sign dependence is why a lift was specified explicitly.

Both fix the point at infinity. In the complex torus model their linear multipliers have order three or six, so the lattice is stable under a primitive cube root $\zeta =e^{2\pi i/3}$ and hence under $e^{\pi i/3}=1+\zeta$. Choose a shortest nonzero period $\omega$. The triangular lattice $\mathbb{Z}+\zeta \mathbb{Z}$ has covering radius $1/\sqrt{3}<1$: its equilateral triangles of side one each have circumradius $1/\sqrt{3}$. Approximating any period divided by $\omega$ by this lattice leaves a strictly shorter period, which must be zero. ThereforeThus the cubic has the hexagonal lattice shape, independently of which of the two lifts is chosen.

The genus-degree formula for a nonsingular plane curve givesA divisor is a finite formal sum $D={\sum }_p{n}_p[p]$ with integer coefficients; its degree is $\mathrm{deg}D={\sum }_p{n}_p$. An effective divisor has all $n_p\ge 0$. A hyperplane divisor is the intersection with a projective line not containing the curve, counted with local intersection multiplicities. Its degree is $d$.

Define $L(D)=\{0\}\cup \{f\text{ meromorphic}:(f)+D\ge 0\}$ and $\ell (D)={\dim }_{\mathbb{C}}L(D)$. For a canonical divisor $K$, the Riemann-Roch theorem statesHere $(f)$ records zero orders positively and pole orders negatively, and $\mathrm{deg}K=2g-2$.

A meromorphic function on a compact Riemann surface has finitely many poles, say $p_j$ of orders $r_j$. For each $p_j$, choose a projective line through it that does not contain $C$. Its hyperplane divisor contains $p_j$ with multiplicity at least one. Include $r_j$ copies of this divisor. The sum then dominates the pole divisor of $f$.

Add further arbitrary hyperplane divisors until their number $m$ satisfies $m\ge d$ and $md>2g-2$. Adding effective divisors cannot destroy the pole bound, so for $D=H_1+\cdots +H_m$ we haveFor a constant function, including zero, no initial poles need to be covered; simply add sufficiently many lines.

The divisor $D={\sum }_i{H}_i$ has degree $md$. The hypothesis gives $\mathrm{deg}(K-D)=2g-2-md<0$. A divisor of negative degree has no nonzero section: if $h\in L(K-D)\setminus \{0\}$, then $(h)+K-D$ would be effective, but its degree would be negative because the divisor of a meromorphic function has degree zero. Thus $\ell (K-D)=0$. The Riemann-Roch theorem now gives

The monomials $x^i{y}^j{z}^k$ with nonnegative integers $i+j+k=m$ form a basis for the vector space of homogeneous polynomials of degree $m$. For each $i=0,\dots ,m$, there are $m-i+1$ choices of $(j,k)$. Hence

Write $L_i=a_{i}x+b_{i}y+c_{i}z$ and $S={\prod }_i{L}_i$. The numerator and denominator have the same homogeneous degree, so their ratio is well-defined on projective points away from its poles and extends as a meromorphic function on $C$. Locally the pole order cannot exceed the zero order of $S{|}_C$, whose divisor is $D$. Thus $f_Q\in L(D)$.

Equality of two ratios is equivalent to $Q-Q^{\prime }$ vanishing on $C$. A nonsingular projective plane curve is irreducible: distinct positive-degree components would intersect by Bézout's theorem, producing a singular point. Its homogeneous vanishing ideal is therefore generated by its irreducible defining polynomial $P$. One can also see this using Bézout's theorem: a homogeneous polynomial not divisible by $P$ has only finitely many intersections with $C$, so cannot vanish on the whole curve. ConsequentlyMultiplication by the nonzero polynomial $P$ is injective. The linear map $V_m\to L(D)$ therefore has kernel $P{V}_{m-d}$ and image dimensionThe assumption $m\ge d$ makes both polynomial-space dimensions applicable directly.

Given $f$, choose the hyperplane divisors as in (a). The ratios $f_Q$ form a linear subspace of $L(D)$ by (d), and its dimension equals $md+1-g=\ell (D)$ by (b). A linear subspace of a finite-dimensional vector space with the same dimension is the whole space. Hence $f=f_Q$ for some homogeneous polynomial $Q$ of degree $m$, givingThe zero function corresponds to $Q=0$. Thus every allowed meromorphic function is a ratio of homogeneous polynomials of equal degree, with a denominator that can be chosen as a product of linear forms.