Addition After Projective Embedding

On the complex torus, the group operation is simple: \(z_1+z_2 \quad \text{mod } \Lambda .\)

The projective embedding is made from three theta sections: \(\phi(z)=[s_0(z):s_1(z):s_2(z)].\)

Let \(P=\phi(z_1),\qquad Q=\phi(z_2).\)

A line in $\mathbb P^2$ has equation \(aX_0+bX_1+cX_2=0.\)

Pull this line back through the embedding. Then we get a theta section \(F(z)=as_0(z)+bs_1(z)+cs_2(z).\)

So saying that the line passes through $P$ and $Q$ means \(F(z_1)=0,\qquad F(z_2)=0.\)

Since $F$ is a section of a degree $3$ line bundle, it has three zeros on the torus: \(z_1,\quad z_2,\quad z_3.\)

Why does the sum of zeros appear? Here $\vartheta$ is an auxiliary degree $1$ theta factor, not one of the degree $3$ basis sections $s_0,s_1,s_2$. It satisfies \(\vartheta(z+\tau) = -e^{-\pi i\tau-2\pi iz}\vartheta(z).\)

If a degree $3$ theta section has zeros $z_1,z_2,z_3$, then locally it behaves like \(F(z)\sim \vartheta(z-z_1)\vartheta(z-z_2)\vartheta(z-z_3).\)

After translating by $\tau$, this product picks up the factor \(e^{-3\pi i\tau-6\pi iz} \cdot e^{2\pi i(z_1+z_2+z_3)}.\)

But the line bundle already fixes the allowed transformation factor of a degree $3$ section. The basis sections themselves satisfy \(s_k(z+\tau)=e^{-3\pi i\tau-6\pi iz}s_k(z), \qquad k=0,1,2.\) So any linear combination \(F(z)=a_0s_0(z)+a_1s_1(z)+a_2s_2(z)\) must have the same factor.

Therefore the extra term \(e^{2\pi i(z_1+z_2+z_3)}\) must be fixed.

After choosing the origin used in the embedding, this fixed value is normalized to $1$. So the three zeros satisfy \(z_1+z_2+z_3=0 \quad \text{in } \mathbb C/\Lambda\)

Therefore \(z_3=-(z_1+z_2).\)

Geometrically, $\phi(z_3)$ is the third intersection point of the line with the cubic.

Thus the embedded group law is: \(\phi(z_1+z_2) = \text{inverse of the third intersection}.\)

In Weierstrass coordinates, inverse means \((x,y)\mapsto (x,-y).\)