The Lemniscate Sine — Torsion Points

4 Torsion Points

The duplication formulae above allow us to consider the (1+i)n(1 + i)^ n-torsion points of RR, i.e. the points such that acting by (1+i)n(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 sl\operatorname{sl}, which is the composition C/ΛRπP1\mathbb {C}/\Lambda \to R \overset {\pi }{\to } \mathbb {P}^1. If sl(z)=0\operatorname{sl}(z) = 0, then the corresponding point in RR need not be the origin. It could be the other point (0,1)(0, -1). Fortunately, we already know exactly which point gets mapped to (0,1)(0, -1), namely Ω\Omega , which is the unique (1+i)(1 + i)-torsion point of C/Λ\mathbb {C}/\Lambda .

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

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

and the fact that sl(Ω2)=1\operatorname{sl}(\frac{\Omega }{2}) = 1, we know that the two points at infinity are 1±i2Ω\frac{1 \pm i}{2} \Omega , which are the remaining 22-torsion points. These let us conclude


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

This lets us compute the higher torsion points:


(1+i)n(1 + i)^ n-torsion points


(0,1)(0, 1)


(0,1)(0, -1)


,\infty , \infty


(±1,0),(±i,0)(\pm 1, 0), (\pm i, 0)


(ζ8k,±2)(\zeta _8^ k, \pm \sqrt{2}) k=1,3,5,7k = 1, 3, 5, 7


(ik3±a224,±b2a22)(i^ k \sqrt[4]{3 \pm _ a 2 \sqrt{2}}, \pm _ b \sqrt{2 \mp _ a 2\sqrt{2}})  k=0,1,2,3k = 0, 1, 2, 3

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

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

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

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

  3. The ray class field modulo (1+i)5(1 + i)^5 is Q(i,2+22)\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 Q(i)\mathbb {Q}(i) has class number 11, we can decompose

AQ(i)×=Q(i)×μ4×C××vOv×, \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

CQ(i)=C××vOv×μ4. C_{\mathbb {Q}(i)} = \frac{\mathbb {C}^\times \times \prod _{v \nmid \infty } \mathcal{O}_ v^\times }{\mu _4}.

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

Clm=O(1+i)×μ4×(1+(1+i)kO(1+i))=(Z[i](1+i)k)×/μ4. \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 Z[i]/(1+i)k\mathbb {Z}[i]/(1 + i)^ k are exactly those (1+i)(1 + i) does not divide, i.e. those with odd norm. Thus, we see that

Cl(1+i)3=0,Cl(1+i)4=C2,Cl(1+i)5=C2×C2, \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 Q(i,2+22)\mathbb {Q}(i, \sqrt{2 + 2\sqrt{2}})).