# A curious observation on the elliptic curve $y^2=x^3+1$

Here is a calculation regarding the $$2$$-torsion points of the elliptic curve $$y^2=x^3+1$$ which looks really miraculous to me (the motivation comes at the end).

• Take a point of $$y^2=x^3+1$$ and consider its sum with the $$2$$-torsion points. Except for finitely many cases, the $$y$$-coordinates of the resulting four points are distinct. The subsets of size four of the Riemann sphere formed by these $$y$$-coordinates are all conformally equivalent (i.e. they are the same up to automorphisms of the sphere).

This can be verified as follows: The non-identity elements of order two of the elliptic curve $$y^2=x^3+1$$ are given by $$(-1,0)$$, $$(-\omega,0)$$ and $$(-\omega^2,0)$$ where $$\omega:={\rm{e}}^{\frac{2\pi{\rm{i}}}{3}}$$. Given a point $$(a,b)$$ ($$a$$ and $$b$$ complex numbers with $$b^2=a^3+1$$), one can directly check that $$(a,b)\oplus(-\lambda,0)=\left(\left(\frac{b}{a+\lambda}\right)^2-a+\lambda,-\frac{3\lambda^2b}{(a+\lambda)^2}\right).$$ for any third root of unity $$\lambda$$. Thus the $$y$$-coordinates of the points obtained from adding a $$2$$-torsion point to $$(a,b)$$ are $$b,-\frac{3b}{(a+1)^2},-\frac{3\,\omega^2\,b}{(a+\omega)^2},-\frac{3\,\omega\,b}{(a+\omega^2)^2}.$$ These are distinct complex numbers unless $$b=0$$ (i.e. $$(a,b)$$ a $$2$$-torsion itself), $$b=\pm 1$$ (in which case $$a=0$$) or $$b=\pm 3$$ (in which case $$a^3=8$$). Except in these special cases, the cross-ratio of the four points appearing above is independent of $$a$$ and $$b$$: $$T\left(b,-\frac{3b}{(a+1)^2},-\frac{3\,\omega^2\,b}{(a+\omega)^2},-\frac{3\,\omega\,b}{(a+\omega^2)^2}\right) =T\left(1,-\frac{3}{(a+1)^2},-\frac{3\,\omega^2}{(a+\omega)^2},-\frac{3\,\omega}{(a+\omega^2)^2}\right)\\ =T\left(-3,(a+1)^2=a^2+2a+1,\frac{(a+\omega)^2}{\omega^2}=\omega\,a^2+2\,\omega^2\,a+1,\frac{(a+\omega^2)^2}{\omega}=\omega^2\,a^2+2\,\omega\,a+1\right) =\frac{(-a^2-2a-4)((\omega-\omega^2)(a^2-2a))}{(-\omega\,a^2-2\,\omega^2\,a-4)((1-\omega^2)a^2+2(1-\omega)a)}\\ =\omega\,\frac{(a^2+2a+4)(a-2)}{(\omega\,a^2+2\,\omega^2\,a+4)((1+\omega)a+2)}\\ =\omega\,\frac{(a-2\,\omega)(a-2\,\omega^2)(a-2)}{(a-2)(\omega\, a-2)((1+\omega)a+2)} =\omega\,\frac{a-2\,\omega}{(1+\omega)a+2}\,\frac{a-2\,\omega^2}{\omega\, a-2}=\omega\,\frac{1}{1+\omega}\frac{1}{\omega}=\frac{1}{1+\omega}.$$

Question. Is there any explanation for this observation? Is this a special case of a more general phenomenon?

Remark. This property does not hold for all elliptic curves. For instance, the $$2$$-torsion points of $$y^2=x^3-x$$ are $$(0,0)$$, $$(\pm 1,0)$$ and the identity (the point at infinity). One can directly check that for any $$\lambda\in\{0,\pm 1\}$$ the $$y$$-coordinate of $$(a,b)\oplus(\lambda,0)$$ is $$\frac{b\left((3\lambda^2-1)a-2\lambda^3\right)}{(a-\lambda)^3}.$$ Hence the $$y$$-coordinates of the translations of $$(a,b)$$ with $$2$$-torsions are $$b,-\frac{b}{a^2},\frac{2b}{(a-1)^2},\frac{2b}{(a+1)^2}.$$ The cross-ratio of these points (in the order above) is $$\frac{\left(2(1+a^2)\right)(-4a)}{(a^2-2a-1)(3a^2+2a+1)}$$.

Motivation. I came across this observation through studying the rational map $$f(z)=-\frac{z(z-2)^3}{(2z-1)^3}$$. This is a very special rational map: it is Belyi; and, as the diagram below indicates, is induced by an endomorphism of an elliptic curve (it is a rigid Lattès map using the terminology of complex dynamics). $$\require{AMScd}$$ $$\begin{CD} \left\{y^2=x^3+1\right\}@>[-2]>> \left\{y^2=x^3+1\right\}\\ @V(x,y)\mapsto\frac{y+1}{2} V V @VV(x,y)\mapsto\frac{y+1}{2} V\\ \Bbb{CP}^1@>f>>\Bbb{CP}^1 \end{CD}$$ The computation above shows that all regular fibers of $$f$$ are isomorphic as subsets of size four of $$\Bbb{CP}^1$$.

• The 2-torsion is $O,R,\omega R,\omega^2 R$ and $y(P)=y(Q)$ iff $P=\omega^k Q$ from which the zeros and poles of $(y(P),y(P+R);y(P+\omega R),y(P+\omega^2 R))$ cancel. Not sure there is anything easier or deeper going on. Dec 25, 2021 at 11:44