Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 18 / 5

Write the projective plane as $\mathbb{P}(V)$ for a three-dimensional complex vector space $V$. A projective line is the zero set of a nonzero linear functional $\ell \in V^{\ast }$. Two such functionals give the same line exactly when they differ by a nonzero scalar. Hence the parameter space of lines is the dual projective space $\mathbb{P}(V^{\ast })$. A choice of basis identifies $V^{\ast }$ with ${\mathbb{C}}^3$, giving $\boxed{({\mathbb{P}}^2{)}^{\vee }\cong {\mathbb{P}}^2}$. The description as $\mathbb{P}(V^{\ast })$ is intrinsic; the identification with the original $\mathbb{P}(V)$ requires a choice.

Represent the smooth plane conic by $v^{\mathsf{T}}Qv=0$, with $Q$ an invertible symmetric $3\times 3$ matrix. The tangent line at $[v]$ has coefficients $\ell =Qv$, because the differential is $2v^{\mathsf{T}}Q$. Thus its image in the dual projective space satisfiesConversely any nonzero $\ell$ satisfying this equation gives $v=Q^{-1}\ell$, a point of the original smooth plane conic whose tangent is $[\ell ]$. This correspondence is the restriction of an invertible projective linear transformation. The dual conic is therefore the smooth conic

To analyze the genus one tangent incidence curve of two conics, choose projective coordinates in which $C_1$ has equation $XZ-Y^2=0$. Its points and tangent lines are parametrized by $[a:b]\in {\mathbb{P}}^1$ asFor $x=[X:Y:Z]\in C_2$, the incidence equation is a nonzero homogeneous quadratic in $a,b$, so the projection $E\to C_2$ is finite of degree two. Its discriminant vanishes exactly when $Y^2-XZ=0$, that is, at $C_1\cap C_2$. There are four such points; Bézout theorem says their intersection multiplicities sum to four, so all four intersections are transverse.

Locally on $C_2$, after choosing an affine tangent-parameter chart and completing the square, the incidence equation is $w^2=u(s)$, where $s$ is a local parameter and $u$ has a simple zero at each intersection. This is smooth, with ramification index two there; away from those points the roots are distinct and the cover is étale. The analogous chart covers a root at infinity. Thus $E$ is smooth everywhere, and the cover has exactly four ramification points. It is connected: the discriminant has odd valuation at each of its four zeros, so it cannot be a square in the function field of $C_2$. The associated quadratic extension is therefore a field, not two separate sheets.

A smooth plane conic over $\mathbb{C}$ is isomorphic to the projective line. Apply the Riemann-Hurwitz formula to this connected degree-two cover:Consequently $\boxed{g(E)=1}$, and $E$ is a smooth projective genus one curve.

The printed construction is not well defined: the next tangent must pass through $x_1$, rather than $x_0$. If $x_0\ne x_1$ and ${\ell }_1\ne {\ell }_0$, a second line through $x_0$ cannot also pass through $x_1$, because the unique line through both points is ${\ell }_0$. Here is a concrete counterexample satisfying all the conic hypotheses. In the affine chart takewith ${\ell }_0:x=1$, $x_0=(1,3\sqrt{3}/2)$ and $x_1=(1,-3\sqrt{3}/2)$. The second tangent through $x_0$ is $-23x+12\sqrt{3}y=31$. It does not contain $x_1$: its left side there is $-77$. The conics meet transversely at the four complex points with $x^2=32/5$, $y^2=-27/5$. Thus this is a defect in the original PDF, not just in its conversion.

For the corrected construction, let ${\iota }_{\ell }$ exchange the two points of $C_2$ on a fixed tangent line, and let ${\iota }_x$ exchange the two tangents to $C_1$ through a fixed point of $C_2$. Both projections from $E$ are degree-two morphisms to a smooth plane conic: for the line projection this follows from intersecting a line with $C_2$, which has no line component. Since $E$ is smooth and the ground field has characteristic zero, their quadratic function-field extensions define regular involutions on the whole curve. At a ramification point “the second” point is the same point, counted with multiplicity. Each switch is an involution of a degree-two map from a genus one curve. Hence the corrected step is the everywhere-defined automorphism $F={\iota }_x\circ {\iota }_{\ell }$.

Choose an origin $O$ on the genus one curve. The Abel-Jacobi map of a genus-one curve identifies $E$ with ${\mathrm{Pic}}^0(E)$ by $p\mapsto {\mathcal{O}}_E(p-O)$. Fibers of each degree-two projection are linearly equivalent Weil divisors, since they are pullbacks of points of ${\mathbb{P}}^1$. Thus their group sums are constant: for suitable $a,b\in E$,This also holds at the ramification points, where $2p=a$ or $2p=b$. Therefore the corrected step is a translation on an elliptic curve. If $F^n(p_0)=p_0$ for one point, then $n(b-a)=0$. It follows that $F^n(p)=p$ for every $p\in E$. The corrected construction has the Poncelet porism: one periodic orbit implies all orbits are periodic, with the same least period. The literal printed construction fails before this conclusion; the proof establishes the intended, explicitly corrected assertion.