I have been wondering this for a while; I get how elliptic curves of the Weierstrass form $y^2=4x^3-g_2x-g_3$ have the lowest degree and are the simplest way to study a genus $1$ curve. But why aren’t there (or at least that I could find) elliptic functions that parametrize the Edwards Curve and derived modular forms? They are highly symmetrical, like the addition theorems on the curve $$C_d:=\{(x,y)\in\mathbb C^2\mid y^2+x^2=1+dx^2y^2\}$$ $$ (x_1,y_1) + (x_2,y_2) = \left( \frac{x_1 y_2 + x_2 y_1}{1 + dx_1 x_2 y_1 y_2}, \frac{y_1 y_2 - x_1 x_2}{1 - dx_1 x_2 y_1 y_2} \right) $$ And by the morphism $$ \psi : C_d \longrightarrow E\qquad \psi(X,Y) =\left( \frac{1+y}{\,1-y\,} \;-\; \frac{2(d+1)}{3} \;,\; \frac{4(1+y)}{\,x(1-y)\,}\right) $$ we get that $$[d(\tau)^2+14d(\tau)+1]^3\left[g_2(\tau)^3 - 27 g_3(\tau)^2\right] \;-\; 108\,d(\tau)[1-d(\tau)]^4 g_2(\tau)^3 = 0. $$ So $d(\tau)$ could behave as a modular form in a subgroup of $\mathrm{SL(2,\mathbb Z)}$. Also the $j$-invariant $$ j(\tau) = 16\,\frac{(d^2+14d+1)^3}{\,d(1-d)^4\,}.$$
1 Answer
You are correct that the Edwards curve is highly symmetric, and the missing functions you are looking for do exist. They are the Jacobi elliptic functions, but applied with a specific coordinate rotation.
While the Weierstrass function is unique to the lattice, the Edwards curve $x^2 + y^2 = 1 + d x^2 y^2$ corresponds to the Jacobi quartic.
If we set the curve parameter $d = k^2$ (where $k$ is the elliptic modulus), the analytic parametrization for the Edwards curve is:
$$ x(u) = \operatorname{sn}(u, k), \qquad y(u) = \operatorname{cd}(u, k) \equiv \frac{\operatorname{cn}(u, k)}{\operatorname{dn}(u, k)} $$
We can verify this satisfies the Edwards equation using standard Jacobi identities: $$ \operatorname{sn}^2(u) + \frac{\operatorname{cn}^2(u)}{\operatorname{dn}^2(u)} = \frac{1 - k^2 \operatorname{sn}^4(u)}{\operatorname{dn}^2(u)} = 1 + k^2 \operatorname{sn}^2(u) \frac{\operatorname{cn}^2(u)}{\operatorname{dn}^2(u)} $$
The use of Edwards coordinate isn't standard because the preference for the Weierstrass form is geometric. When using Weierstrass, the coordinate function $\wp(z)$ has a pole at $z=0$, which corresponds to the group identity (the point at infinity). This aligns the analytic pole with the algebraic identity. In the Edwards parametrization, the functions $x(u)$ and $y(u)$ are holomorphic at the group identity ($u=0$), taking values $0$ and $1$ respectively. The poles of the parametrization are located away from the origin in the complex plane. Don't quote me on this but I think this regularity at the identity is why Edwards curves are preferred for computational arithmetic, but it makes them less canonical for defining the curve via Laurent series at the origin.
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2$\begingroup$ Ohhh nice! Never considered Jacobi’s elliptic curves. Thanks lot! $\endgroup$Roccooi– Roccooi2025-11-29 18:19:55 +00:00Commented yesterday
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2$\begingroup$ @Roccooi You're very welcome. $\endgroup$saver_of_light– saver_of_light2025-11-29 18:21:07 +00:00Commented yesterday