Why they’re called elliptic curves

Divisors 

Everyone who works with elliptic curves has probably heard that elliptic curves are called "elliptic curves" because they're connected to ellipses in some way, but not many people have actually taken the time to understand exactly what the connection really is. I just came across some notes that I made on this very topic several years ago and thought that the material might make a good blog post. It's tedious (perhaps even very tedious) stuff, but it's something that it's good to at least see once.

Let's see if I can write this up without introducing too many errors into it.

First, remember that elliptic curves in the Weierstrass form can be parameterized by the Weierstrass ℘-function. In particular, the ℘-function satisfies the differential equation

℘'(z)2 = 4 ℘(z)3g2 ℘(z) – g3

and we have that

zz0 = ℘(z0)℘(z) (4 w3g2 wg3)-1/2

so that the ℘-function is the inverse of a particular elliptic integral.

But why is that particular integral called an "elliptic integral?"

Suppose that we have an ellipse defined by

(x / a)2 + (y / b)2 = 1

which we can parameterize by

x(t) = a cos t

y(t) = b sin t

To get the length of this ellipse between t1 and t2 we need to calculate

t1t2 ( (dx/dt)2 + (dy/dt)2 )1/2 dt

The length of the entire ellipse is four times the length of just one quadrant of it, which we can find by calculating

L = 4 0π/2 (a2 cos2t + b2 sin2t)1/2 dt

Now

a2 cos2t + b2 sin2t = a2 cos2t + a2 sin2ta2 sin2t + b2 sin2t

= a2 (cos2t + sin2t) – (a2 + b2) sin2t

= a2 – (a2 + b2) sin2t

= a2 (1 – k2 sin2t)

where

k = (a2b2) / a2

Now let

cos t = (1 – u2)-1/2

and

du = cos t dt

to get

L = 4 a 01 (1 – k2u2) ((1 – u2) (1 – k2u2))-1/2 du

= 4 a 01 ((1 – u2) (1 – k2u2))-1/2 du

– 4 a k2 01 u2 ((1 – u2) (1 – k2u2))-1/2 du

So a reasonable first question to ask is how to evaluate the first of those two integrals, or how to find

∫ ((1 – u2) (1 – k2u2))-1/2 du

To do this, let's write

v2 = (1 – u2) (1 – k2u2) = (u – α) (β) (u – γ) (u – δ)

Dividing by (u – α)4 we get that

(v / (u – α)2)2 = (1 + (α – β) / (u – α)) (1 + (α – γ) / (u – α)) (1 + (α – δ) / (u – α))

Changing variables to

x = 1 / (u – α)

y = v / (uα)2

we get

y2 = (1 + (α – β) x) (1 + (α – γ) x) (1 + (α – δ) x)

= x3A x2B x + C

so that we find that we're interested in an integral that looks like

∫ (x3A x2B x + C)-1/2 dx

That's just one more change of variable (and if you've read this far it's one that I'm sure that you can easily figure out) away from being

∫ (4 x3g2 x – g3)-1/2 dx

which is exactly the sort of integral that's the inverse of the Weierstrass ℘-function.

So there you have it – there really is a good reason for elliptic curves being called "elliptic curves." It's because they're connected to arc length integrals on an ellipse, and that's what the connection is.

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