Past exam of the geometry course of the University of Oxford / 2007 / B

Choose representative vectors $x_1,\dots ,x_{n+1}$ of the first $n+1$ points; they form a basis. Write a representative of the last point as ${\sum }_i{a}_i{x}_i$. General position forces each $a_i\ne 0$, since otherwise $n+1$ of the points would lie in a hyperplane. Rescale the basis vectors to $a_i{x}_i$. The tuple is then projectively equivalent to the standard frameApply the same normalization to the second tuple and compose the two transformations to obtain existence. An automorphism fixing the coordinate points is represented by a diagonal matrix. Fixing their sum forces all its diagonal entries to agree, so its projective action is the identity. This proves existence and uniqueness. At $n=1$ it says the Möbius group acts sharply triply transitively on the projective line: any ordered triple of distinct points can be sent to any other in exactly one way.

By the preceding frame result take the four points to be $[1:0:0],[0:1:0],[0:0:1],[1:1:1]$. The conics through them have equationsCoefficient vectors differing by a common nonzero scalar define the same conic. Their parameter space is the projectivization of a two-dimensional vector space, hence a projective line.

Each partition of the four points into two unordered pairs gives a reducible conic formed by the two joining lines. There are three pairings. Conversely a reducible conic is a union of two lines, possibly repeated. General position prevents any one line from carrying three of the points, so the lines must be distinct and carry exactly one of these pairings. Thus there are exactly three reducible members. In these coordinates they are the parameters $[0:1:-1]$, $[1:0:-1]$, $[1:-1:0]$, giving respectively $x(z-y)=0$, $y(z-x)=0$, and $z(y-x)=0$.

Write $P_5=[u:v:w]$ in the same normalized coordinates. Its coordinates are all nonzero, because the coordinate lines join pairs of the first three points. Evaluation at $P_5$ gives the additional linear equation $avw+bwu+cuv=0$ on the two-dimensional coefficient plane $a+b+c=0$. It is not identically zero there: that would imply $vw=wu=uv$, hence $u=v=w$ and $P_5=P_4$, contradicting the hypothesis. Its kernel is therefore one-dimensional. Thus there is exactly one projective conic through all five points. If it were reducible, $P_5$ would lie on a line joining two of the other points, contrary to no three being collinear. Hence the unique conic is irreducible.

For a plane affine curve given by a reduced equation $f(x,y)=0$, a point is singular when $f=f_x=f_y=0$ there. For a projective curve given by a homogeneous equation $F(x,y,z)=0$, a point is singular when $F=F_x=F_y=F_z=0$ on a nonzero representative. These conditions are invariant under projective rescaling. In characteristic zero, Euler's identity $dF=x{F}_x+y{F}_y+z{F}_z$ makes the equation $F=0$ follow from vanishing of the partials when $d\ne 0$ in the field. Over an arbitrary field one should retain the equation itself, particularly if the characteristic divides the degree.

Bézout's theorem says that two projective plane curves over $\mathbb{C}$ of degrees $d,e$ with no common irreducible component have total intersection multiplicity $de$, so at most $de$ distinct intersection points. Let the irreducible curve have reduced homogeneous equation $F$ of positive degree $d$. In characteristic zero some partial derivative is nonzero. For $d>1$ this derivative has degree $d-1$ and cannot share a component with the irreducible degree-$d$ polynomial $F$. Every singular point belongs to their intersection. Bézout bounds their number by $d(d-1)$, proving there are finitely many singular points. For $d=1$ a partial is a nonzero constant and there are none.

Put $F=y^{2}z-x^3+3x{z}^2-2z^3$. Its partials areAt infinity $z=0$, the curve equation forces $x=0$, giving $[0:1:0]$, where $F_z\ne 0$. In the chart $z=1$, the derivative equations give $y=0$ and $x=\pm 1$. Only $x=1$ lies on the curve and also makes $F_z=0$. Thus the only singular point is $[1:0:1]$. As a check, the affine equation factors as $y^2=(x-1{)}^2(x+2)$, whose quadratic tangent cone there is $y^2-3(x-1{)}^2$, giving two distinct tangent lines.

For $f=y^2-x^3-17$, the partials are $f_x=-3x^2$ and $f_y=2y$. If $p\notin \{2,3\}$ their simultaneous vanishing forces $x=y=0$, and that point lies on the curve exactly when $p=17$. In characteristic two, the equations force $x=0$ and then $y^2=1$, so $y=1$. In characteristic three they force $y=0$ and then $x^3=1$, so $x=1$. Consequentlyand there are no singular points for any other prime.

With local indices $e_p$, the required specialization isOnly the finitely many points of index greater than one contribute. Their images are the branch points on the target.

For $d>1$, the map $z\mapsto z^d$ has local index $d$ at zero and at infinity; at infinity use coordinates $w=1/z$ in source and target, giving again $w\mapsto w^d$. At every nonzero finite point its derivative is nonzero, so there are no further ramification points. Thus the ramification sum is $2(d-1)$ and the formula gives $-2=-2d+2(d-1)$, as required for the sphere. For $d=1$ the map is the identity and has no ramification, giving the same equality separately.

For a smooth complex projective plane curve of degree $d$,Project generically from a point outside the curve to a projective line. Its degree is $d$, by intersection with a generic line through the center. Ramification occurs when such a line is tangent. In coordinates for projection $[x:y:z]\mapsto [x:z]$, these points satisfy $F=F_y=0$. The polar has degree $d-1$, and Bézout gives total ramification $d(d-1)$; generic projection makes the ramification simple, or the intersection multiplicities can be read as the corresponding ramification orders. The Riemann-Hurwitz formula yields $2g-2=-2d+d(d-1)$ and hence the formula. The smoothness qualification is essential for geometric genus: singular plane curves have the same arithmetic degree-genus expression, while the genus of their normalization can be smaller.

The ramification sum equals $2g-2+2d$. Since $g\ge 0$ and $d>1$, it is at least $2d-2>0$. Thus some point has analytic ramification index greater than one.

For a degree-one map, the sum of local degrees in every fiber is one. Therefore each fiber contains exactly one point and its local degree is one. There is no ramification, and the Riemann-Hurwitz formula gives $2g-2=-2$, so $g=0$. The map is a continuous bijection from compact $C$ onto the Hausdorff sphere and hence a homeomorphism. In fact its local degree-one coordinate inverses are holomorphic, so it is biholomorphic as well.

For degree two every ramification point has index two, so contributes exactly one. The formula gives $2g+2$ ramification points. No two can share an image: that would give a fiber of total local degree at least four. Hence there are also $2g+2$ distinct branch points on the sphere.

A Riemann surface is a connected Hausdorff second-countable topological surface equipped with coordinate charts to open subsets of $\mathbb{C}$ whose transition maps are holomorphic. This is a complex manifold of complex dimension one. The atlas makes holomorphic functions, meromorphic functions and holomorphic differentials coordinate-independent notions.

Choose $r>0$ so the closed disk centered at $a$ of radius $r$ lies in $U$. The Cauchy integral formula givesParametrize $z=a+r{e}^{i\theta }$; then $dz=ir{e}^{i\theta }d\theta$ and the denominator cancels. Thus

Choose a disk on which $|f(z)|\le M=|f(a)|$. If $M=0$ the assertion is immediate. Otherwise multiply $f$ by a fixed unit scalar so that its value at $a$ is the positive real number $M$. The mean-value identity shows the average of $\mathrm{Re}f$ on every sufficiently small circle is $M$. Since $\mathrm{Re}f\le |f|\le M$ there, continuity forces equality at every point on each circle. Equality of real part to $M$ together with modulus at most $M$ forces the imaginary part to vanish. Varying the radius covers a neighborhood of $a$. Thus both $|f|$ and $f$ are constant near $a$, proving the local maximum-modulus assertion.

Compactness makes the continuous function $|f|$ attain a maximum. Apply the preceding argument in a chart at a maximizing point to make $f$ constant on a neighborhood. The permitted identity theorem and connectedness then make it constant on all of $S$. Hence every holomorphic function from a compact connected Riemann surface to $\mathbb{C}$ is constant.

A divisor on an algebraic curve is a finite integer sum $D={\sum }_p{n}_{p}p$; its degree is ${\sum }_p{n}_p$. For a meromorphic function its principal divisor records its zero and pole orders, andLet $K$ be the divisor of a nonzero meromorphic differential, called a canonical divisor. The Riemann-Roch theorem statesMultiplication by that differential identifies $L(K)$ with the space of holomorphic differentials. By the preceding result $L(0)$ consists exactly of constants, so $\ell (0)=1$. Applying the formula at $D=0$ gives $1-\ell (K)=1-g$, hence $\ell (K)=g$.

Apply Riemann-Roch at $D=K$ and use $\ell (K)=g$ and $\ell (0)=1$:Rearranging gives $\mathrm{deg}K=2g-2$. This holds for every canonical divisor, since any two differ by a principal divisor of degree zero.