Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 24F (Riemann Surfaces)

LetIts complex partial derivatives areThere is no point of $X^{\prime }$ at which both vanish. Indeed, if $F=F_x=F_y=0$, then $F_y=0$ and $F=0$ giveNeither $x$ nor $y$ can be zero because $F_x=0$. Substitution into $F_x=0$ gives $12y^7+1=0$, whereas substitution into $F_y=0$ gives $8y^7+3=0$; these equations are incompatible.

For every $p=(x_0,y_0)\in X^{\prime }$, at least one partial derivative is nonzero. If $F_y(p)\ne 0$, the holomorphic implicit function theorem expresses the curve near $p$ as $y=g(x)$, and the restriction of the $x$-projection is a holomorphic coordinate chart. If $F_x(p)\ne 0$, use the $y$-projection instead. On overlaps, the transition functions are restrictions of holomorphic coordinate functions and are holomorphic. These charts define an atlas making $X^{\prime }$ a one-dimensional complex manifold, hence a Riemann surface.

Homogenizing the affine equation gives the projective closureThis is the Klein quartic. It is smooth: if all three projective partial derivatives vanished, thenMultiplying them gives $28X^3{Y}^3{Z}^3=0$. If one coordinate is zero, the displayed equations force all three to vanish, which is impossible in projective space. Thus $C$ is a smooth compactification of $X^{\prime }$. The smooth compactification compatible with the extended coordinate functions is its normalization, and since $C$ is already smooth it is biholomorphic to $C$.

For a direct genus computation, consider the extended coordinate map ${\pi }_x:X\to {\mathbb{C}}_{\infty }$. A generic $x$ leaves the cubic equationso $\mathrm{deg}{\pi }_x=3$. In the affine part, ramification occurs where $F_y=0$. Solving $F=F_y=0$ gives $(0,0)$ and the seven pointsAt the origin, $x=-y^3+O(y^6)$, so the ramification index of a holomorphic map is three and the contribution is two. At each of the other seven points, $F_x=-7/2\ne 0$, $F_y=0$, and $F_{yy}=6y\ne 0$, so the ramification index is two and each contributes one.

There are two points at infinity. Near $P=[1:0:0]$, in coordinates $u=Y/X$, $v=Z/X$, the equation isand $1/x=v$, so $P$ is unramified. Near $Q=[0:1:0]$, in coordinates $r=X/Y$, $s=Z/Y$, the equation isThus $s=-r^3+O(r^{10})$ and $1/x=s/r=-r^2+O(r^9)$, so $Q$ has ramification index two and contributes one. The total ramification is thereforeThe Riemann-Hurwitz formula for the degree-three map to the sphere givesand henceThis also agrees with the genus of a smooth plane curve of degree four.

Let $\zeta =e^{2\pi i/7}$ and defineEach term in the defining polynomial is multiplied by ${\zeta }^3$:Thus $\varphi$ is a holomorphic automorphism of $X^{\prime }$, with inverse obtained by replacing $\zeta$ by ${\zeta }^{-1}$. By the assumption it extends to a holomorphic automorphism of $X$; in the projective model it isIt has order exactly seven because $\zeta$ is primitive and $y$ is not identically zero. Hence $\mathrm{Aut}(X)$ contains an element of order seven.

The extended coordinate function ${\pi }_x:X\to {\mathbb{C}}_{\infty }$ is a meromorphic function as a holomorphic map to the Riemann sphere. Compose it with the seventh-power map of the sphere and setThis map is nonconstant and holomorphic as a map to ${\mathbb{C}}_{\infty }$. Since ${\pi }_x\circ \varphi ={\zeta }^3{\pi }_x$,