# Trying to piece together an integral addition theorem

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 $A \oplus B$.

-
Not all curves are groups in a sensible way... –  Mariano Suárez-Alvarez Aug 6 '10 at 4:53

$0$, 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.

-
Thanks! I tried my calculation again today with that in mind and it worked out. Now I just need to find out how to prove this theorem and if it applies to anything more than conics and cubics. –  anon Aug 6 '10 at 12:27