The Lemniscate Sine — Torsion Points

# 4 Torsion Points

The duplication formulae above allow us to consider the $(1 + i)^ n$-torsion points of $R$, i.e. the points such that acting by $(1 + i)^ n$ sends the point back to the origin. There is a slight subtlety here, due do the fact that we are working with $\operatorname{sl}$, which is the composition $\mathbb {C}/\Lambda \to R \overset {\pi }{\to } \mathbb {P}^1$. If $\operatorname{sl}(z) = 0$, then the corresponding point in $R$ need not be the origin. It could be the other point $(0, -1)$. Fortunately, we already know exactly which point gets mapped to $(0, -1)$, namely $\Omega$, which is the unique $(1 + i)$-torsion point of $\mathbb {C}/\Lambda$.

We might also worry about the two points at infinity, which is not adequately captured by our formula. Observing that

$\operatorname{sl}((1 + i)z) = \frac{(1 + i)\operatorname{sl}(z)}{\sqrt{1 - \operatorname{sl}(z)^4}},$

and the fact that $\operatorname{sl}(\frac{\Omega }{2}) = 1$, we know that the two points at infinity are $\frac{1 \pm i}{2} \Omega$, which are the remaining $2$-torsion points. These let us conclude

Proposition

$\operatorname{sl}((1 + i)^ n z) = \infty$ iff $z$ is a $(1 + i)^{n + 2}$-torsion point but not an $(1 + i)^{n + 1}$-torsion point.

This lets us compute the higher torsion points:

 $n$ $(1 + i)^ n$-torsion points 0 $(0, 1)$ 1 $(0, -1)$ 2 $\infty , \infty$ 3 $(\pm 1, 0), (\pm i, 0)$ 4 $(\zeta _8^ k, \pm \sqrt{2})$ $k = 1, 3, 5, 7$ 5 $(i^ k \sqrt{3 \pm _ a 2 \sqrt{2}}, \pm _ b \sqrt{2 \mp _ a 2\sqrt{2}})$  $k = 0, 1, 2, 3$

Finding higher-order torsion points will require advanced WolframAlpha techniques.

Recall that we can find ray class fields of $\mathbb {Q}(i)$ by adjoining the torsion points of the curve with CM by $\mathbb {Q}(i)$. We need to first identify a Weber function, which is given by quotienting out $R$ by the automorphism group. The automorphism group multiplies by powers of $i$, and so we find that $h(x, y) = y$ is a perfectly good Weber function on $R$. So we find that

1. The ray class field modulo $(1 + i)^3$ is still just $\mathbb {Q}(i)$.

2. The ray class field modulo $(1 + i)^4 = (4)$ is $\mathbb {Q}(i, \sqrt{2})$.

3. The ray class field modulo $(1 + i)^5$ is $\mathbb {Q}(i, \sqrt{2 + 2 \sqrt{2}})$.

We can compute the ray class groups explicitly, using the definition of the ray class group, and see that they match. Observe that since $\mathbb {Q}(i)$ has class number $1$, we can decompose

$\mathbb {A}_{\mathbb {Q}(i)}^\times = \frac{\mathbb {Q}(i)^\times }{\mu _4} \times \mathbb {C}^\times \times \prod _{v \nmid \infty } \mathcal{O}_ v^\times ,$

and so

$C_{\mathbb {Q}(i)} = \frac{\mathbb {C}^\times \times \prod _{v \nmid \infty } \mathcal{O}_ v^\times }{\mu _4}.$

The ray class group modulo $\mathfrak {m} = (1 + i)^ k$ is then given by

$\mathrm{Cl}_\mathfrak {m} = \frac{\mathcal{O}_{(1 + i)}^\times }{\mu _4 \times (1 + (1 + i)^ k \mathcal{O}_{(1 + i)})} = \left(\frac{\mathbb {Z}[i]}{(1 + i)^ k}\right)^\times \Big/ \mu _4.$

The invertible elements in $\mathbb {Z}[i]/(1 + i)^ k$ are exactly those $(1 + i)$ does not divide, i.e. those with odd norm. Thus, we see that

$\mathrm{Cl}_{(1 + i)^3} = 0,\quad \mathrm{Cl}_{(1 + i)^4} = C_2,\quad \mathrm{Cl}_{(1 + i)^5} = C_2 \times C_2,$

in agreement with above (exercise: find the other subfields on $\mathbb {Q}(i, \sqrt{2 + 2\sqrt{2}})$).