The Lemniscate SineSine

# 1 Sine

The sine function, as we learnt in high school, is closely related to the arc length of the circle. We will describe this in a slightly funny way. Consider the circle of unit diameter with center $(0, \frac{1}{2})$: By some elementary geometry, we find that $r = \sin \theta$. The arc length $s$ is given by

$(\mathrm{d}s)^2 = (\mathrm{d}r)^2 + (r \; \mathrm{d}\theta )^2.$

We have

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

So we can write the line element as

$\mathrm{d}s = \frac{1}{\sqrt{1 - r^2}}\; \mathrm{d}r$

Thus, the arc length function is given

$s(r_0) = \int _0^{r_0} \frac{1}{\sqrt{1 - r^2}}\; \mathrm{d}r.$

This is, of course, the familiar arcsine function, inverse to the even more familiar sine function.

Note that here we have a square root sitting inside the integrand. For the arc length function, we simply always take the positive square root. However, if we want to extend arcsine and sine to complex functions by replacing the integral with a contour integral, then we cannot always stick with the above choice.

The solution is to instead think about the Riemann surface of the function $\frac{1}{\sqrt{1 - r^2}}$, which is given by the circle $r^2 + t^2 = 1$: We can then write the integral as

$s(z) = \int _0^z \frac{\mathrm{d}r}{t},$

where $z$ is now viewed as a point in $R = \{ (r, t) \in \mathbb {C}^2: r^2 + t^2 = 1\}$. This integral depends not only on $z$, but also on the path taken from $0$ to $z$. In general, it is well-defined only up to a period, namely the integral of $\frac{\mathrm{d}r}{t}$ around a closed loop. Thus, the inverse function, namely sine, is a singly-periodic function.

It is important to note that this circle is different from the previous circle. When we discuss the Lemniscate sine soon, we will work with two rather different shapes.

Before we end the section, note that to study the sine function, which is a function defined on $R$, it is often convenient to pick an isomorphism between $R$ and $\mathbb {P}^1$ given by stereographic projection. This is given by the formulae

$r = \frac{2u}{1 + u^2},\quad t = \frac{1 - u^2}{1 + u^2}$

Substituting in, we find

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

which is the integral of a nice, rational function.