Past exam of the geometry course of the University of Oxford / 2010 / B

First consider the unlettered projective frame assertion. A list of $n+2$ points in ${\mathbb{CP}}^n$ is in general position when every $n+1$ of their representative vectors are linearly independent, equivalently no $n+1$ of the points lie in a projective hyperplane. Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points as a basis. Write $v_{n+2}=\sum c_i{v}_i$. Each $c_i$ is nonzero, since otherwise $v_{n+2}$ and the other $n$ basis vectors would be dependent. Rescaling the representatives makes $v_{n+2}=v_1+\cdots +v_{n+1}$.

Do the same for the second frame with representatives $w_i$. The invertible linear map $v_i\mapsto w_i$ induces a projective transformation carrying all $n+2$ points as required. A transformation fixing a normalized frame is represented by a diagonal matrix, since it fixes the first $n+1$ points. Fixing the sum point forces every diagonal entry to coincide, so the matrix is scalar and the projective transformation is the identity. This proves existence and uniqueness.

Return to the four chosen points on the two lines. Since the six points are distinct, at most one is $K\cap L$; put its index third if necessary. No three of $p_1,p_2,q_1,q_2$ are collinear: every triple contains two on one line and a point on the other line away from the intersection. Thus they form a projective frame in ${\mathbb{CP}}^2$. The four target points also have no collinear triple, as their three-by-three determinants are nonzero. The frame result gives a unique projective transformation with the prescribed four images.

The normalized lines are $K=\{z=0\}$ and $L=\{y=0\}$. A point of $K$ with first coordinate zero is $p_1$, so $p_3\ne p_1$ permits unique normalization $p_3=[1,\alpha ,0]$. The inequality $p_3\ne p_2$ gives $\alpha \ne 1$. Similarly, $q_3=[1,0,\beta ]$ uniquely with $\beta \ne 1$.

The lines $p_1{q}_2$ and $p_2{q}_1$ are $x-z=0$ and $x-y=0$, giving $r_{12}=[1,1,1]$. The lines $p_1{q}_3$ and $p_3{q}_1$ are $\beta x-z=0$ and $\alpha x-y=0$, giving $r_{13}=[1,\alpha ,\beta ]$. Finally, the two lines defining $r_{23}$ have equationsTaking the cross product of their coefficient vectors givesThe last vector cannot vanish: if $\alpha \beta =1$, both parameters are nonzero and vanishing of its other coordinates would force $\alpha =\beta =1$, which is excluded.

Let $u=(1,1,1)$ and $v=(1,\alpha ,\beta )$. The vector representing $r_{23}$ satisfiesConsequently the three representative vectors are linearly dependent, so $r_{12},r_{13},r_{23}$ are collinear in the projective plane. Equivalently their three-by-three determinant vanishes. This is the two-line configuration of Pappus's hexagon theorem.

The strong Bézout theorem says that if two projective plane algebraic curves $C,D$ over $\mathbb{C}$ have degrees $m,n$ and no common irreducible component, their intersection is finite andEach intersection multiplicity is a positive integer. Scheme-theoretic multiplicities, rather than just distinct intersection points, are essential: tangencies and singularities can make individual terms larger than one.

For curves with no common local component at $p$,In local affine coordinates, their defining equations have linearly independent gradients exactly in this case. These independent linear terms form local coordinates, so the local intersection quotient has dimension one. Conversely, if either equation lacks a nonzero linear term or the two linear terms are dependent, the local intersection quotient cannot have dimension one. This is transverse intersection of plane curves of two smooth plane curves.

Write the irreducible curve as $F=0$, with $F$ homogeneous of degree $n$. For $n=1$ there are no singular points. Otherwise choose a nonzero partial derivative, say $F_x$, which has degree $n-1$. Such a derivative exists in characteristic zero. Irreducibility and its smaller degree show that $F$ and $F_x$ have no common component.

At every singular point $p$ of $C$, $F_x(p)=0$ and the curve multiplicity of $F$ is at least two. The standard local lower bound for intersection multiplicity gives $I_p(F,F_x)\ge 2$. Applying the Bézout theorem,ThereforeThis is the polar bound for singular points of a plane curve.

Take the union of $n$ distinct projective lines in general position, with no three concurrent. Its defining polynomial is the product of their linear equations, so its degree is $n$. Exactly two lines meet at each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise intersections, producing an ordinary double point; elsewhere just one line passes through and the curve is smooth. HenceFor $n=1$, the line itself gives zero singular points, as required.

The curves have no common component, since their intersection is finite. Let $R=0$ define the irreducible degree-$m$ curve $E$. Choose $p\in E$ outside $C\cap D$, and select $(\lambda ,\mu )\ne (0,0)$ with $\lambda P(p)+\mu Q(p)=0$. The polynomial $H=\lambda P+\mu Q$ is not identically zero, since $P,Q$ are not proportional.

The degree-$n$ curve $H=0$ meets $E$ at the prescribed $nm$ base points and also at $p$. If $R$ did not divide $H$, irreducibility of $R$ and the Bézout theorem would bound the number of distinct intersections by $nm$, contradicting these $nm+1$ points. ThusAt every remaining common zero of $P,Q$ lying off $E$, $R$ is nonzero and hence $S=0$. ThereforeThis is residual intersection through a curve pencil. The existence of a point $p$ outside the finite base locus is automatic on the projective curve $E$.

Let the six distinct conic points be $p_1,\dots ,p_6$, and write ${\ell }_{ij}$ for their joining line. Form two reducible cubicsNo three distinct points of an irreducible conic are collinear, by the Bézout theorem applied to a line and the conic. Thus all six side lines are distinct, the cubics have no common component, and six of their intersections are precisely the vertices. The remaining pairings giveThese three points are off the conic and distinct. For example, equality of $q_1$ and $q_2$ would put that point on both ${\ell }_{34}$ and ${\ell }_{45}$, making it $p_4$ and then placing $p_4$ on ${\ell }_{12}$, impossible. The other cases are identical. Hence the two cubics meet in nine distinct points.

Apply residual intersection through a curve pencil with $n=3$, $m=2$ and $E$ the conic: six intersections lie on $E$, so the other three lie on a curve of degree at most one. Therefore $q_1,q_2,q_3$ are collinear, proving Pascal's theorem.

First record the unlettered definitions. A divisor on an algebraic curve on a compact Riemann surface is a finite formal sum $D={\sum }_p{n}_p[p]$ with $n_p\in \mathbb{Z}$. Its degree is $\mathrm{deg}D={\sum }_p{n}_p$. A principal divisor is $(f)={\sum }_p{\mathrm{ord}}_p(f)[p]$ for a nonzero meromorphic function, with positive order at a zero and negative order at a pole. Two divisors are linearly equivalent if their difference is principal. Principal divisors form a subgroup because $(fg)=(f)+(g)$ and $(1/f)=-(f)$.

For the doubly periodic function, put $h=f^{\prime }/f$. It is periodic with the same period lattice, and its residue at $p_i$ equals $n_i$. Opposite sides of the fundamental parallelogram are translates by a period with opposite directions, so their integrals of $h$ cancel. The residue theorem givesThusIn particular an elliptic function has equally many zeros and poles counted with multiplicity in a period cell.

Order the period lattice basis so the indicated contour is positively oriented; reversing its orientation only changes the signs of the resulting integers. SetPairing the translated sides while retaining the factor $w$ givesFor example, the right side is the translate by ${\lambda }_1$ of the left side with reversed orientation; their combined contribution is ${\lambda }_{1}B$. The bottom and top similarly give $-{\lambda }_{2}A$.

Along either path there is a continuously continued logarithm of the nonzero finite function $f$. Since its endpoint values agree by periodicity, each logarithmic increment is an integer multiple of $2\pi i$. Therefore $A=2\pi ir$, $B=2\pi is$ for integers $r,s$. The residue theorem applied to $wh$ now yieldsThis is the elliptic divisor-sum identity.

For a divisor on an algebraic curve $D={\sum }_j{n}_j[p_j+\Lambda ]$, defineChanging the representatives $p_j$ changes the second coordinate only by an element of $\Lambda$. The two preceding contour arguments show that a principal divisor has degree zero and period lattice sum zero. Thus the formula is unchanged by linear equivalence of divisors and is a well-defined group homomorphism on the quotient.

It is nontrivial, since $[0]$ maps to $(1,0+\Lambda )$. In fact it is surjective: for any $d\in \mathbb{Z}$ and $z+\Lambda \in \mathbb{C}/\Lambda$, the divisor on an algebraic curve $(d-1)[0]+[z+\Lambda ]$ maps to $(d,z+\Lambda )$. Only well-definedness and a nontrivial homomorphism are required here; identifying its entire kernel would be an additional theorem.

Set $g(w)=a\wp (w)+b{\wp }^{\prime }(w)+c$. The Weierstrass elliptic function is doubly periodic, with a double pole at the origin modulo $\Lambda$ and no other poles. Its derivative has leading term $-2w^{-3}$. Since $b\ne 0$, no lower-order term cancels this, so $g$ has a pole of order exactly three at zero and no others.

The point $\Phi (0+\Lambda )=[0,1,0]$ does not lie on the line, because its line equation evaluates to $b\ne 0$. At every other point of the image, lying on the line is equivalent to $g(w)=0$. The three distinct image points therefore yield three distinct zeros modulo $\Lambda$. By the zero-pole count, these are all the zeros and each is simple. Hence its principal divisor on the torus isApplying the elliptic divisor-sum identity givesThis is the period lattice form of the chord-and-tangent group law on an elliptic cubic.