3 Complex Multiplication

We have already observed above that

$$\operatorname{sl}(iz) = i \operatorname{sl}(z).$$

We can think of this as follows — we fix $O = (0, 1)$ as our basepoint of $R$, which corresponds to the point $0 \in \mathbb {C}/\Lambda$. Then multiplication by $i$ is an automorphism of $\mathbb {C}/\Lambda$, which should then descend to an automorphism of $R$. This formula gives a very concrete description of what the induced map on $R$ is — it is simply multiplication by $i$ in the $x$ coordinates!

We might hope that there are more formulas of this sort. For example, an addition formula like

$$\sin (z + w) = \sin z \sqrt{1 - \sin ^2 w} + \sin w \sqrt{1 - \sin ^2 z}$$

would be helpful. To look for such formulae, we may try playing around with some substitutions and see what we get. Previously, for the circle, we had the substitution $x = \frac{2u}{1 + u^2}$ that turned things into rational functions. We might try a similar substitution. If we want $1 - x^4$ to become a perfect square and we can get rid of the radical, we should set

$$x^2 = \frac{2u^2}{1 + u^4}.$$

Doing the substitution yields

$$\frac{\mathrm{d}x}{\sqrt{1 - x^4}} = \frac{1 + u^4}{1 - u^4} \cdot \frac{4u(1 - u^4)}{(1 + u^4)^2} \cdot \frac{\mathrm{d}u}{2x} = \frac{\sqrt{2}\; \mathrm{d}u}{\sqrt{1 + u^4}}$$

This looks quite like what we originally had, except for the change in sign below. Thus, we put $u = \zeta _8 v$, and then we get

$$\frac{\mathrm{d}x}{\sqrt{1 - x^4}} = \frac{(1 + i)\; \mathrm{d}v}{\sqrt{1 + v^4}}.$$

Integrating and putting in the right bounds, we find that

$$s\left(\frac{(1 + i) v_0}{\sqrt{1 - v_0^4}}\right) = (1 + i) s(v_0).$$

In other words, we have

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

Taking complex conjugates, we get

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

Crucially, since $(1 + i)(1 - i) = 2$, we can iterate these two formulas to obtain

$$\operatorname{sl}(2z) = \frac{2\operatorname{sl}(z)\sqrt{1 - \operatorname{sl}(z)^4}}{1 + \operatorname{sl}(z)^4}.$$

The presence of these $\sqrt{1 - \operatorname{sl}(z)^4}$ may be a bit off-putting, but there is no reason to fear them — they are simply the $y$ coordinates of $R$, which we should think of as the lemniscate arcsine of $z$. The ambiguity in sign is fixed by what happens in a small neighbourhood of $2$, where we choose $\sqrt{1 - \operatorname{sl}(z)^4}$ to be positive whenever $z$ is small and positive.

It is an elementary exercise to iterate these formulas to obtain

$$\begin{aligned} \operatorname{sl}(2(1 + i)z) & = \frac{2(1 + i) \operatorname{sl}(z) \sqrt{1 - \operatorname{sl}(z)^4} (1 + \operatorname{sl}(z)^4)}{1 - 6 \operatorname{sl}(z)^4 + \operatorname{sl}(z)^8}\\ \operatorname{sl}(4z) & = \frac{4\operatorname{sl}(z) \sqrt{1 - \operatorname{sl}(z)^4} (1 + \operatorname{sl}(z)^4) (1 - 6 \operatorname{sl}(z)^4 + \operatorname{sl}(z)^8)}{1 + 20 \operatorname{sl}(z)^4 - 26 \operatorname{sl}(z)^8 + 20 \operatorname{sl}(z)^{12} + \operatorname{sl}(z)^{16}} \end{aligned}$$

Given the duplication formula for $\operatorname{sl}(2z)$, it is reasonable to guess that we might have an addition formula of the form

$$\operatorname{sl}(z + w) = \frac{\operatorname{sl}(z) \sqrt{1 - \operatorname{sl}(w)^4} + \operatorname{sl}(w) \sqrt{1 - \operatorname{sl}(z)^4}}{1 + \operatorname{sl}(z)^2 \operatorname{sl}(w)^2},$$

from which we can deduce all the above formulae using that $\operatorname{sl}(iz) = i \operatorname{sl}(z)$.

A poor man's way of proving the formula would be to set $w = c - z$ for some constant $c$, and then differentiate the right-hand side to see it is constant. There are more modern ways of doing so, but is out of the scope of this discussion.

We should think of this formula as transporting the obvious addition structure on $\mathbb {C}/\Lambda$ to give an addition rule on $R$.