2 The Lemniscate Sine

The lemniscate sine is an example of an elliptic integral. Elliptic integrals first arose when people studied problems analogous to the above but for ellipses, and integrals that looked similar were called elliptic integrals. We shall not be interested in ellipses here, because they are less interesting. Fix two foci $F_{\pm } = (\pm a, 0) \in \mathbb {R}^2$, and consider the locus of all points $P = (x, y)$ such that

$$\| P - F_+\| \cdot \| P - F_-\| = \| 0 - F_+\| \cdot \| 0 - F_-\| .$$

Explicitly, this is given by

$$((x - a)^2 + y^2)((x + a)^2 + y^2) = a^4.$$

Equivalently, this is

$$(x^2 + y^2)^2 = 2a^2 (x^2 - y^2).$$

It is convenient to set $a = \frac{1}{\sqrt{2}}$, and using polar coordinates, this equation becomes

$$r^2 = \cos 2\theta .$$

We then have

$$\mathrm{d}\theta = \frac{-r}{\sin 2\theta }\; \mathrm{d}r = \frac{-r}{\sqrt{1 - r^4}}\; \mathrm{d}r.$$

Thus, the line element is

$$\mathrm{d}s = \frac{1}{\sqrt{1 - r^4}}\; \mathrm{d}r,$$

and we have a lemniscate arcsine

$$s(r_0) = \int _0^{r_0} \frac{1}{\sqrt{1 - r^4}}\; \mathrm{d}r$$

with inverse $\operatorname{sl}(s)$. Again, we want to promote this to a complex function. From now on, we will replace $r$ with $x$, since explicit coordinates for the lemniscate itself will not be of much use. Then the associated Riemann surface is given by

$$R = \{ (x, y) \in \mathbb {C}^2: y^2 = 1 - x^4\} .$$

Then

$$s = \int \frac{\mathrm{d}x}{y}.$$

Observe that this surface is singular at infinity. Which is bad. However, we can blow this up at infinity to resolve the singularity, and since we are working with curves, any rational map is automatically a morphism, and so we don't have to think about infinity much.

A key observation is that projection onto the $x$ coordinate exhibits $R$ as a double cover of $\mathbb {P}^1$ branched at four points (it cannot be branched at infinity since there is always an even number of branch points), and so by Riemann-Hurwitz, $R$ is a torus. This means $R$ admits a holomorphic and non-vanishing differential, and one can check $\frac{\mathrm{d}x}{y}$ is one.

Since $R$ is topologically a torus, its homology is generated by two loops, and $s$ will be defined up to the integrals of those loops, which we expect to be an integer lattice $\Lambda$ in $\mathbb {C}$. We will explicitly identify this lattice soon. What this means is that $s$ actually gives an explicit identification $R \overset {\sim }{\to } \mathbb {C}/\Lambda$. Surjectivity is automatic, and injectivity is given by the Abel–Jacobi theorem, since if $s(p) = s(p')$, then $p - p'$ is killed by the Abel–Jacobi map, and hence $p \sim p'$. This is possible only if $p = p'$ as $R$ is not $\mathbb {P}^1$.

To understand the lattice, we pick two generated loops as follows:

The horizontal loop has a very geometric interpretation, since the path lies entirely within the real axis. Indeed, this just corresponds to integrating the line element along the whole lemniscate, and the integral is just the total arc length of the lemniscate, $2\Omega$. It is written this way so that the length of half of the lemniscate is $\Omega$, which is exactly the contribution we get if we go around the branch point at $1$ and then return to the “origin” (of course, on the Riemann surface, this is a distinct point that the origin, as it is on the other sheet).

It remains to understand what happens when we go around the branch point at $i$. This follows from the simple observation that the integrand is invariant under $x \mapsto ix$, while $\mathrm{d}x$ gets multiplied by $i$ under this operation. So we find that

$$s(iz) = i s(z).$$

So it follows that the other homology class has an integral of $(1 + i)\Omega$. Thus,

$$\Lambda = \langle 2\Omega , (1 + i)\Omega \rangle .$$

The inverse function $\operatorname{sl}$ is then a double periodic function with period lattice $\Lambda$ with zeroes given by $\Lambda \cup (\Omega + \Lambda )$.