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 (0,12)(0, \frac{1}{2}):

    \draw (-4, 0) -- (4, 0) node [right] {$x$};
    \draw (0, -0.5) -- (0, 3.5) node [above] {$y$};

    \draw [blue, semithick] (0, 1.5) circle [radius=1.5];

    \draw (0, 0) -- (0.75, 2.799) node [pos=0.5, left] {$r$};
    \draw (0.3, 0) arc(0:75:0.3) node [right] {\;\,$\theta$};

By some elementary geometry, we find that r=sinθr = \sin \theta . The arc length ss is given by

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

We have

dθ=drcosθ=dr1sin2θ=dr1r2. \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

ds=11r2  dr \mathrm{d}s = \frac{1}{\sqrt{1 - r^2}}\; \mathrm{d}r

Thus, the arc length function is given

s(r0)=0r011r2  dr. 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 11r2\frac{1}{\sqrt{1 - r^2}}, which is given by the circle r2+t2=1r^2 + t^2 = 1:

    \draw (-2, 0) -- (2, 0) node [right] {$r$};
    \draw (0, -2) -- (0, 2) node [above] {$t$};

    \draw [red, semithick] circle [radius=1.3];

We can then write the integral as

s(z)=0zdrt, s(z) = \int _0^z \frac{\mathrm{d}r}{t},

where zz is now viewed as a point in R={(r,t)C2:r2+t2=1}R = \{ (r, t) \in \mathbb {C}^2: r^2 + t^2 = 1\} . This integral depends not only on zz, but also on the path taken from 00 to zz. In general, it is well-defined only up to a period, namely the integral of drt\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 RR, it is often convenient to pick an isomorphism between RR and P1\mathbb {P}^1 given by stereographic projection.

\begin{tikzpicture} [scale=0.65]
    \draw [red, semithick] circle [radius=2];
    \draw (-5.5, 0) -- (5.5, 0);
    \draw (0, 2) node [circ] {};
    \node at (1.414, 1.414) [anchor = south west] {$(r, t)$};
    \node at (1.414, 1.414) [circ] {};
    \draw (0, 2) -- (4.8259, 0) node [circ] {} node [above] {$u$};
    \draw (0, -3) -- (0, 3);

This is given by the formulae

r=2u1+u2,t=1u21+u2 r = \frac{2u}{1 + u^2},\quad t = \frac{1 - u^2}{1 + u^2}

Substituting in, we find

dr1r2=11(2u1+u2)22(1+u2)4u2(1+u2)2  du=1+u21u22(1u2)(1+u2)2  du=2  du1+u2, \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.