Past exam of the geometry course of the University of Oxford / 2011 / 5

Choose $s\in \mathbb{C}$ with $s^2=\lambda$. Since $\lambda \ne 0$, the invertible diagonal projective transformationfixes $[0:1:0]$. Writing its image as $[X:Y:Z]$, direct substitution givesThus it maps $C_{\lambda }$ onto $C_{1/\lambda }$. Either choice of square root works. We choose $s=i$ at $\lambda =-1$ and $s=e^{\pi i/6}$ at $\lambda =e^{\pi i/3}$ for the later order calculations.

An appropriate projective transformation isIt fixes $[0:1:0]$ and has an invertible representing matrix. For $X=z-x$, $Y=iy$, $Z=z$,Hence its restriction is an isomorphism $C_{\lambda }\to C_{1-\lambda }$.

Compose the preceding transformations, applying ${\sigma }_{\lambda }$ first and then ${\tau }_{1/\lambda }$. With the square-root convention already specified,This invertible projective transformation fixes $[0:1:0]$ and carries $C_{\lambda }$ to $C_{1-1/\lambda }$. It is important to retain the sign of the chosen lift on the $y$ coordinate when computing its order later.

On the parameter, the two transformations are $S(\lambda )=1/\lambda$ and $T(\lambda )=1-\lambda$. Their compositions yieldEvery step lifts to an invertible projective transformation of the corresponding cubic curves. Therefore all six parameter values give isomorphic curves. Some values coincide for special $\lambda$; this does not affect the conclusion.

At $\lambda =-1$ and $s=i$, the chosen map isThe square is not the identity on the cubic, since a generic point has $y\ne 0$. Thus $\boxed{\mathrm{ord}(\sigma )=4}$.

Choose the cubic's point $[0:1:0]$ as the origin in its complex torus description. Since $\sigma$ fixes it, its torus form is $z+\Lambda \mapsto az+\Lambda$, with $a\Lambda =\Lambda$. Exact order four implies $a=\pm i$; $a^2=-1$. Therefore $\Lambda$ is stable under multiplication by $i$.

To identify its shape without an unproved uniqueness assertion, choose a shortest nonzero period $\omega$. For any ${\lambda }_0\in \Lambda$, approximate ${\lambda }_0/\omega$ by a Gaussian integer $m+ni$ within distance $1/\sqrt{2}<1$. The difference ${\lambda }_0-(m+ni)\omega$ belongs to $\Lambda$ and is shorter than $\omega$, so it must vanish. HenceThe lattice is determined up to a nonzero complex scaling: it is square.

Take $\lambda =e^{\pi i/3}$ and the chosen $s=e^{\pi i/6}$. Then $\lambda s=i$, so $i/(\lambda s)=1$, and $-1/\lambda ={\lambda }^2$. In affine coordinates the chosen lift becomesSince ${\lambda }^2$ is a primitive cube root of unity, $1+{\lambda }^2+{\lambda }^4=0$. Three iterations fix $x$ and $y$, whereas one does not. ThusThe other square-root choice replaces this map by $(x,y)\mapsto (1+{\lambda }^{2}x,-y)$. Its cube is $(x,y)\mapsto (x,-y)$, so that lift has order six. This sign dependence is why a lift was specified explicitly.

Both fix the point at infinity. In the complex torus model their linear multipliers have order three or six, so the lattice is stable under a primitive cube root $\zeta =e^{2\pi i/3}$ and hence under $e^{\pi i/3}=1+\zeta$. Choose a shortest nonzero period $\omega$. The triangular lattice $\mathbb{Z}+\zeta \mathbb{Z}$ has covering radius $1/\sqrt{3}<1$: its equilateral triangles of side one each have circumradius $1/\sqrt{3}$. Approximating any period divided by $\omega$ by this lattice leaves a strictly shorter period, which must be zero. ThereforeThus the cubic has the hexagonal lattice shape, independently of which of the two lifts is chosen.