Trying to piece together an integral addition theorem

Question

If we have a curve $C:\{ P(x,y) = 0 \}$ and define $\omega=\frac{\mathrm{d}x}{y}$ then is

$$\int_0^A \omega + \int_0^B \omega = \int_0^{A \oplus B} \omega$$

(with $\oplus$ being addition on a group on the curve) a theorem?

I'm quite sure this is what's happening when C is an elliptic curve, but I have failed to make this work out when C is defined by $P(x,y) = x^2 + y^2 - 1$ (unit circle) and the group law is defined by firing a ray through $(3/5,4/5)$ (chosen arbitrarily) parallel to $AB$ and taking it's intersection with the circle as $A \oplus B$.

Answer 1

$0$, the lower limit for the integrals, should be whatever point on the curve is $0$ for the group law.

Then $dx/y$ should be rotation-invariant for any circle. It will not be true for a rotated ellipse; there is an invariant differential but it is not $dx/y$.

For an elliptic curve, $dx/y$ being the invariant differential relies on the curve being written as $y^2=f(x)$ with $\deg(f)=3$. If you move the curve the invariant differential will be a different one.

For higher genus the curve does not have an addition law.

Answer 2

muad,

You're right! This is what's happening on elliptic curves. And, if you look at higher genus, as T.. said, there is no group law on the curve. But something silly like there not being a group law on curves of high genus wasn't enough to stop the 19th century mathematicians, they discovered that it's the group law on an abelian variety related to the curve that matters, and the curve and the abelian variety happen to be the same for cubics!

You can find all this worked out in any good book that talks about elliptic integrals or abelian integrals, and I'm personally a big fan of the exposition in Stillwell's "Mathematics and its History"