Past exam of the geometry course of the University of Oxford / 2008 / B

For a projective frame, choose representative vectors $v_1,\dots ,v_{n+1}$ forming a basis. Express the final point by $\sum c_i{v}_i$. Every $c_i$ is nonzero, since otherwise some $n+1$ of the given representatives would be dependent. Rescale the first representatives to $c_i{v}_i$, so the last point is represented by their sum. Do the same for the target frame, with basis $w_i$. The linear map taking each rescaled source basis vector to the corresponding target vector induces the required projective transformation. For uniqueness, a transformation fixing the first $n+1$ coordinate points is represented by a diagonal matrix. Fixing their sum forces every diagonal entry to agree. It is therefore scalar, giving a unique projective transformation.

If three intersection points of the conics lay on a line, Bézout's theorem would force that line to be a component of either nonsingular conic, impossible. Thus the four points form a projective frame and can be moved to the three coordinate points and $[1:1:1]$. Vanishing at the coordinate points removes the square terms from each conic equation; vanishing at the last point gives $a+b+c=0$.

Substitution gives, by direct expansion,The substitution matrixhas determinant two, hence is invertible over $\mathbb{C}$. If the initial frame transformation is represented in the direction old coordinates equal $Q$ times new coordinates, take $P=QS$. Both quadratic forms then have no mixed terms, proving $P^{T}AP$ and $P^{T}BP$ are simultaneously diagonal. The same argument applies when an individual conic is reducible but the intersection consists of exactly four distinct points: a line containing three would be a common component, contradicting finiteness of the intersection.

An inflection point is a smooth point where the tangent line has intersection multiplicity at least three with the curve. For a nonsingular plane curve of degree $n\ge 3$, these are the intersections with its Hessian curve, whose degree is $3(n-2)$. The original curve is not a component of its Hessian: otherwise in each graph chart the tangent direction would have zero derivative identically, making the curve locally a line and hence a line globally. Bézout therefore gives the upper bound $3n(n-2)$, counting multiplicity. A nonsingular conic has no inflection points; lines are excluded from this finite-contact definition.

Take $F=y^{2}z-x^3+x{z}^2$. Its Hessian determinant isAt infinity the curve has only $O=[0:1:0]$, which is an inflection point: its tangent is $z=0$ and the restricted equation is $-x^3=0$. In the chart $z=1$, substitute $y^2=x^3-x$ into the Hessian equation to obtainThus the full answer isThe four complex values of $x$ are distinct and none is $0,1,-1$, so each gives two distinct nonzero values of $y$. Together with $O$ these are all nine inflection points, exhausting the Bézout bound.

For a degree-$d$ nonconstant holomorphic map from a compact connected Riemann surface of genus $g$ to the projective line, the Riemann-Hurwitz formula iswhere $e_p$ is the local degree. A meromorphic function defines a map to the sphere by giving its ordinary finite values away from poles and value infinity at each pole. The reciprocal function is holomorphic in a neighborhood of a pole, proving holomorphicity there. For use in Riemann-Hurwitz the function must be nonconstant; a nonzero constant still defines a holomorphic map, but of degree zero.

Let $X=\mathbb{C}/\Lambda$. The double pole of $\wp$ gives a degree-two map to the sphere. The torus has genus one, so total ramification is four. One simple ramification point is zero, above infinity. At the three nonzero half-period classes $h_i$, evenness and periodicity give ${\wp }^{\prime }(h_i)=0$. Each is a ramification point of local degree two, and these account for all remaining ramification. Put $e_i=\wp (h_i)$. They are distinct, since a degree-two fiber cannot contain two points each of local degree two.

At each $h_i$, $\wp -e_i$ has a double zero and ${\wp }^{\prime }$ has a simple zero. Consequently the elliptic quotienthas removable singularities there and no zeros or poles elsewhere except possibly zero. Near zero, $\wp =z^{-2}+O(z^2)$ and ${\wp }^{\prime }=-2z^{-3}+O(z)$, so $R\to 4$. Thus $R$ is holomorphic on the compact torus and constant, provingThe printed additional requirement that all three $e_i$ be nonzero is false. For the square lattice $\Lambda =\mathbb{Z}+i\mathbb{Z}$, invariance of the defining series under multiplication by $i$ gives $\wp (iz)=-\wp (z)$. At $h=(1+i)/2$, one has $ih=h-1$, so periodicity gives $\wp (h)=\wp (ih)=-\wp (h)$ and hence $\wp (h)=0$. This is one of the three branch values. Distinctness, which is all the cubic construction needs, remains valid.

DefineThe image lies on $Y^{2}Z=4(X-e_{1}Z)(X-e_{2}Z)(X-e_{3}Z)$. Distinct roots make its affine part smooth, and the point at infinity is smooth because the derivative of its equation with respect to $Z$ is nonzero there. Near zero the projective coordinates $X/Y=\wp /{\wp }^{\prime }\sim -z/2$ and $Z/Y=1/{\wp }^{\prime }\sim -z^3/2$ extend holomorphically. Every finite $X/Z$ value has two preimages under $\wp$ except at a branch value. Evenness pairs those preimages as $z,-z$, and oddness of ${\wp }^{\prime }$ distinguishes them unless they are the same half-period class. Therefore the two coordinates together give injectivity and reach both signs of $Y$, proving surjectivity onto the cubic. Thus $\Phi$ is a bijective holomorphic map to a nonsingular cubic, including when one branch value is zero.

A divisor on an algebraic curve is a finite formal sum $D=\sum n_{p}p$ with integer coefficients. Its degree is $\mathrm{deg}D=\sum n_p$. The principal divisor of a nonzero meromorphic function is $(f)=\sum {\mathrm{ord}}_p(f)p$. For a nonconstant $f$, zeros and poles are fibers over zero and infinity of the same map to the sphere, each of total multiplicity equal to its degree; hence $\mathrm{deg}(f)=0$. A nonzero constant has zero divisor.

Write $L(D)=\{f:(f)+D\ge 0\}\cup \{0\}$ and $\ell (D)=\dim L(D)$. The Riemann-Roch theorem stateswhere $K_C$ is a canonical divisor. Taking $D=p_1+\cdots +p_{g+1}$ gives $\ell (D)=2+\ell (K_C-D)\ge 2$. Constants span only one dimension, so there is a nonconstant $f\in L(D)$. It has no poles outside these points and at most a simple pole at each. Thus the required nonconstant function exists, with actual poles at some nonempty subset because a holomorphic function on a compact connected curve is constant.

Since every $p_i$ is fixed, $f\circ \phi$ has poles only among the first $g+1$ points, with order at most one: a biholomorphic change of coordinate preserves pole order. Thus $h=f-f\circ \phi$ also has at most $g+1$ poles, counted with multiplicity. At every $p_i$ with $i>g+1$, both terms are finite and equal, so $h(p_i)=0$.

If $h$ were not zero, its principal divisor would have degree zero. It would therefore have at most $g+1$ zeros counted with multiplicity. But it has at least $n-(g+1)$ distinct zeros. When $n>2g+2$ this number exceeds $g+1$, a contradiction. Consequently

Choose an actual pole $p_i$ of $f$; one exists by nonconstancy and compactness. In a local coordinate $z$ centered there, its simple pole has expansion $f(z)=a_{-1}/z+O(1)$ with $a_{-1}\ne 0$. HenceThe holomorphic inverse function theorem makes $1/f$ a local coordinate. Here the source notation $f^{-1}$ denotes the reciprocal, rather than an inverse map of the meromorphic function.

The chosen pole is among the fixed points, so $\phi$ maps a sufficiently small neighborhood of it into the coordinate neighborhood. If $f=f\circ \phi$, then $w=w\circ \phi$ there. Injectivity of this coordinate forces $\phi (x)=x$ throughout a nonempty open set. The stated identity principle then gives $\phi$ equal to the identity everywhere.

If a nonidentity holomorphic automorphism had more than $2g+2$ fixed points, choose that many distinct ones and apply the function constructed above. The zero-and-pole count would give $f=f\circ \phi$, and the local reciprocal coordinate would force $\phi$ to be the identity, a contradiction. Therefore