# Weierstrass elliptic function identity

For a lattice $$\Lambda = [\lambda_1, \lambda_2] \subset \mathbb C$$, the Weierstrass $$\wp$$-function defined as $$$$\wp(z) = \frac{1}{z^2} + \sum_{\lambda \in \Lambda \setminus \{0\}} \left( \frac{1}{(z - \lambda)^2} - \frac{1}{\lambda^2} \right)$$$$

and the theta function $$\sigma$$ given by $$$$\sigma(z) = z \prod_{\lambda \in \Lambda \setminus \{0\}} \left( 1 - \frac{z}{\lambda} \right) e^{z/\lambda + \frac{1}{2} z^2/\lambda^2},$$$$

I want to show that, for $$a \notin \Lambda$$, $$$$\wp(z) - \wp(a) = - \frac{\sigma(z - a)\sigma(z + a)}{\sigma(a)^2 \sigma(z)^2},$$$$

c.f. exercise 2.12 in http://pub.math.leidenuniv.nl/~luijkrmvan/elliptic/2011/ec.pdf.

I have shown both sides share zeros and poles (counting multiplicities), and now I want to show they are proportional by a factor 1 to prove the identity. To do so I consider the expression near the pole at $$z = 0$$. Then $$$$\wp(z) - \wp(a) \sim \frac{1}{z^2} \quad \text{and} \quad - \frac{\sigma(z - a)\sigma(z + a)}{\sigma(a)^2 \sigma(z)^2} \sim \frac{1}{z^2} \prod_{\lambda \in \Lambda \setminus \{0\}} \frac{\lambda^2 - a^2}{\lambda^2(\lambda - a)^2}.$$$$

However, I cannot find a way to deduce $$$$\prod_{\lambda \in \Lambda \setminus \{0\}} \frac{\lambda^2 - a^2}{\lambda^2(\lambda - a)^2} = -1,$$$$

and I am also not sure whether I have made a mistake. Any help would be greatly appreciated!

• The function $\sigma$ is usually called Weierstrass sigma, that might help you find more information. Also, do you insist on finding the result in this manner, or are you happy with another solution as well? Mar 28, 2020 at 18:04
• I also believe your formula for the residue is incorrect, check for example what happens in the exponent. Mar 28, 2020 at 18:13
• Just commenting that this is also an exercise in Silverman's AEC, VI.6.3 (a). Sep 27, 2023 at 14:45

Define $$f(z):=\wp(z)-\wp(a)$$. Clearly $$f$$ is an elliptic function with order 2. Looking at poles and zeros actually shows that $$f(z)=C\frac{\sigma(z-a)\sigma(z+a)}{\sigma(z)^2}$$ for a constant $$C$$. We now need to compute such constant. One the one hand, since $$\wp$$ has a pole of order $$2$$ at $$0$$, we find $$\lim_{z\to 0}z^2f(z)=\lim_{z \to 0}z^2 \wp(z)-\lim_{z \to 0}z^2 \wp(a)=\lim_{z \to 0}z^2 \wp(z)-0=1$$ On the other hand, using the definition of $$\sigma$$ we get $$\lim_{z \to 0}\frac{\sigma(z)}{z}=1$$ and therefore $$\lim_{z\to 0}z^2f(z)=C \lim_{z \to 0}\frac{\sigma(z-a)\sigma(z+a)}{\frac{\sigma(z)^2}{z^2}}=-C\sigma(a)^2$$ Thus, $$C=\frac{-1}{\sigma(a)^2}$$ and the desired result now follows.