An addition property of Weierstrass $\wp$

Question

I want to show

$$ \left( \begin{array}{ccccc} &1 &\wp(v) &\wp'(v) \\ &1 &\wp(w) &\wp'(w) \\ &1 &\wp(v+w) &-\wp'(v+w) \end{array} \right)=0 $$

where $\wp$ denotes the Weierstrass elliptic function.

Answer 1

As already mentioned in David's comment and Bruno's answer, this determinant is merely saying that the triangle formed by the three points $(\wp(v),\wp^\prime(v))$, $(\wp(w),\wp^\prime(w))$, and $(-\wp(v+w),-\wp^\prime(v+w))$ has area zero (i.e. they are collinear). (I presume the determinantal formula for the area of a triangle was taught to you in your geometry classes, no?)

In particular, you'll want to use the relation $\wp^\prime(z)^2=4\wp(z)^3-g_2 \wp(z)-g_3$. Your determinant is now equivalent to saying that a line in general position must intersect the elliptic curve $y^2=4x^3-g_2 x-g_3$ (or parametrically, $x=\wp(u)\quad y=\wp^\prime(u)$) in three points. I presume you already know how to show that a line and a cubic will have three intersection points at most. If you do the computations, you can obtain an expression for the third intersection point of a line going through the points $(\wp(v),\wp^\prime(v))$ and $(\wp(w),\wp^\prime(w))$ with the elliptic curve in terms of the coordinates of those two given points.

To get the equivalent expression in terms of $\wp(v+w)$ and $\wp^\prime(v+w)$, as Bruno mentions in his answer, one assembles the linear combination $\wp(u)+a\wp^\prime(u)+b$ and note that it has a triple pole at the origin (since $\wp^\prime(u)$ itself has a triple pole from the $-\frac2{u^3}$ term in its lattice series expansion). One can then find appropriate values of $a$ and $b$ such that $v$ and $w$ are zeroes of $\wp(u)+a\wp^\prime(u)+b$, and then find that $u=-v-w$ is a zero as well due to the properties of the Weierstrass functions within their fundamental period parallelogram. Make use of the symmetry of the two Weierstrass functions ($\wp$ is even; $\wp^\prime$ is odd), and you're golden.

Answer 2

This is the last of a series of exercises in Ahlfors' Complex Analysis (Pages 276-277). Those exercises are arranged in a logical sequence. I will copy and paste those that lead to this last result.

1. $$\wp(z)-\wp(u) = -\frac{\sigma(z-u)\sigma(z+u)}{\sigma(z)^2\sigma(u)^2} \tag{16} $$

(Use (14) to show that the right-hand member is a periodic function of $z$. Find the multiplicative constant by comparing the Laurent developments.)

2. $$\frac{\wp'(z)}{\wp(z)-\wp(u)} = \zeta(z-u) + \zeta(z+u) - 2\zeta(z)\tag{17}$$

(Follows from (16) by taking logarithmic derivatives.)

3. $$\zeta(z+u) = \zeta(z) + \zeta(u) + \frac{1}{2}\frac{\wp'(z)-\wp'(u)}{\wp(z)-\wp(u)} \tag{18}$$

(This is a symmetrized version of (17).)

4. The addition theorem for the $\wp$-function: $$\wp(z+u)=-\wp(z)-\wp(u)+\frac{1}{4}\left(\frac{\wp'(z)-\wp'(u)}{\wp(z)-\wp(u)}\right)^2 \tag{19}$$

Differentiation of (18) leads to a formula which contains $\wp''(z)$. It can be eliminated by (15) which gives $\wp" = 6\wp^2 - \frac{1}{2}g_2$. Symmetrization yields (19). Observe that this is an algebraic addition theorem, for $\wp'(z)$ and $\wp'(u)$ can be expressed algebraically through $\wp(z)$ and $\wp(u)$.)

I omit exercises 5 and 6 because they are not needed.

7. Prove that $$\begin{vmatrix} \wp(z) & \wp(u) & 1 \\ \wp(u) & \wp'(u) & 1 \\ \wp(u+z) & -\wp'(u+z) & 1 \end{vmatrix} = 0.$$

(14) and (15) are as follows.

$$\sigma(z+\omega_1)=-\sigma(z)e^{\eta_1(z+\omega_1/2)} \\ \sigma(z+\omega_2)=-\sigma(z)e^{\eta_2(z+\omega_2/2)} \tag{14}$$

$$\wp'(z)^2=4\wp(z)^3-g_2\wp(z)-g_3 \tag{15}$$

Below is how to derive the final result.

Differentiating (19), using $\wp''=6\wp^2-\frac{1}{2}g_2$, and symmetrising the result, we have

$$\wp’(z+u) = -\frac{1}{2} \left(\wp’(z)+\wp’(u)\right) + \frac{3}{2} \frac{\left(\wp’(z)-\wp’(u)\right)\left(\wp(z)+\wp(u)\right)}{\wp(z)-\wp(u)} - \frac{1}{4} \left(\frac{\wp’(z)-\wp’(u)}{\wp(z)-\wp(u)}\right)^3.$$

Substituting (19) and this into the determinant, we can easily show it vanishes.

(Denote the three rows of the determinant as (1), (2), (3). We replace (3) by (1)+(2)-2(3). It is then quite easy to calculate.)