Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You might find this thread useful. – t.b. Mar 18 '12 at 10:22
In particular, look at Bruno's answer. Your determinant vanishing is equivalent to the fact that $(P(v), P'(v))$, $(P(w), P'(w))$ and $(-P(v+w), -P'(v+w))$ are colinear, which is shown in the last paragraph of Bruno's answer. – David Speyer Mar 18 '12 at 15:41

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.