The Lemniscate Sine — Complex Multiplication

3 Complex Multiplication

We have already observed above that

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

We can think of this as follows — we fix O=(0,1)O = (0, 1) as our basepoint of RR, which corresponds to the point 0C/Λ0 \in \mathbb {C}/\Lambda . Then multiplication by ii is an automorphism of C/Λ\mathbb {C}/\Lambda , which should then descend to an automorphism of RR. This formula gives a very concrete description of what the induced map on RR is — it is simply multiplication by ii in the xx coordinates!

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

sin(z+w)=sinz1sin2w+sinw1sin2z \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=2u1+u2x = \frac{2u}{1 + u^2} that turned things into rational functions. We might try a similar substitution. If we want 1x41 - x^4 to become a perfect square and we can get rid of the radical, we should set

x2=2u21+u4. x^2 = \frac{2u^2}{1 + u^4}.

Doing the substitution yields

dx1x4=1+u41u44u(1u4)(1+u4)2du2x=2  du1+u4 \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=ζ8vu = \zeta _8 v, and then we get

dx1x4=(1+i)  dv1+v4. \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((1+i)v01v04)=(1+i)s(v0). s\left(\frac{(1 + i) v_0}{\sqrt{1 - v_0^4}}\right) = (1 + i) s(v_0).

In other words, we have

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}}.

Taking complex conjugates, we get

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

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

sl(2z)=2sl(z)1sl(z)41+sl(z)4. \operatorname{sl}(2z) = \frac{2\operatorname{sl}(z)\sqrt{1 - \operatorname{sl}(z)^4}}{1 + \operatorname{sl}(z)^4}.

The presence of these 1sl(z)4\sqrt{1 - \operatorname{sl}(z)^4} may be a bit off-putting, but there is no reason to fear them — they are simply the yy coordinates of RR, which we should think of as the lemniscate arcsine of zz. The ambiguity in sign is fixed by what happens in a small neighbourhood of 22, where we choose 1sl(z)4\sqrt{1 - \operatorname{sl}(z)^4} to be positive whenever zz is small and positive.

It is an elementary exercise to iterate these formulas to obtain

sl(2(1+i)z)=2(1+i)sl(z)1sl(z)4(1+sl(z)4)16sl(z)4+sl(z)8sl(4z)=4sl(z)1sl(z)4(1+sl(z)4)(16sl(z)4+sl(z)8)1+20sl(z)426sl(z)8+20sl(z)12+sl(z)16 \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 sl(2z)\operatorname{sl}(2z), it is reasonable to guess that we might have an addition formula of the form

sl(z+w)=sl(z)1sl(w)4+sl(w)1sl(z)41+sl(z)2sl(w)2, \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 sl(iz)=isl(z)\operatorname{sl}(iz) = i \operatorname{sl}(z).

A poor man's way of proving the formula would be to set w=czw = c - z for some constant cc, 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 C/Λ\mathbb {C}/\Lambda to give an addition rule on RR.