# Regarding the Weierstrass $\wp$-function of the hexagonal lattice

Playing with the Weierstrass $$\wp$$-function of the hexagonal (or triangular) lattice $$\mathbb{T}$$, $$\wp'(z)^2 = 4 \wp(z)^3 - 1,$$ I noticed that the zeros of $$\wp'(z) + \sqrt{3}$$ are $$\frac{\varpi}{3} + \omega, \quad e^{\pi i/3} \frac{2\varpi}{3} + \omega, \quad e^{2\pi i/3} \frac{\varpi}{3} + \omega, \qquad \omega \in \mathbb{T}$$ (clearly they are all simple), where $$\varpi$$ is the real period of $$\wp$$ given by $$\varpi = \int_{4^{-1/3}}^{\infty} \frac{dx}{\sqrt{x^3 - 1/4}} = \frac{1}{2 \pi}\, \Gamma(1/3)^3$$ (the zeros of $$\wp'(z) - \sqrt{3}$$ can be also determined). Since $$\wp'(z)^2 - 3 = 4 \left[\wp(z)^3 - 1\right]$$ and $$\wp(x)$$ is real for $$0 < x < \varpi$$, we also get $$\wp\left(\pm\frac{\varpi}{3}\right) = 1$$ or, equivalently, $$\int_1^{\infty} \frac{dx}{\sqrt{4x^3 - 1}} = \int_{\wp\left(\frac{\varpi}{3}\right)}^{\infty} \frac{dx}{\sqrt{4x^3 - 1}} = \frac{\varpi}{3} = \frac{1}{6 \pi}\, \Gamma(1/3)^3$$ (the detailed calculations can be found in arXiv:2105.04307).

Since I am not an expert in elliptic functions, my question is whether the above facts are known.

• For the record, I deleted an answer as I misunderstood the question. Aug 27, 2021 at 17:19

There is also a more algebraic perspective.

For this, it is easier (at least for me) to reverse engineer'' things by starting with your points, and then deducing something about the elliptic function values.

I will working on the elliptic curve $$E: y^2 = 4 x^3 - 1$$, which is identified with the quotient $$\mathbb C/\mathbb T$$ via $$z \bmod \mathbb T \mapsto \bigl( \wp(z), \wp'(z) \bigr).$$

Let $$P$$ be the point of $$E$$ which is the image of $$\varpi/3$$. It is a point of order $$3$$ in $$E$$. Now $$E$$ has complex multiplication'' by the ring $$\mathcal O$$ obtained by adjoining $$\zeta = e^{2\pi i/3}$$ to $$\mathbb Z$$ (because the lattice $$\mathbb T$$ is a submodule of $$\mathbb C$$ with respect to this ring). The element $$\zeta$$ sends $$(x,y)$$ to $$(\zeta x, y)$$.

In the ring $$\mathcal O$$, the number $$3$$ is not prime: it factors as $$3 = (1 - \zeta)(1 - \zeta^{-1}) = (1-\zeta)^2 \cdot (-\zeta^{-1}).$$

So among the $$3$$-torsion points, there are some even more special points, namely those killed by multiplication by $$(1-\zeta)$$ (and not just by $$(1-\zeta)^2$$). Looking at the formula for multiplication by $$\zeta,$$ we see these are the points with $$x = 0$$, so the points $$(0,\pm i).$$

There are $$9$$ points of order dividing $$3$$ altogether, namely the origin (i.e. the point at infinity), these $$2$$ points $$(0,\pm i)$$, and then the remaining $$6$$ points $$(x,y)$$ which are killed by $$3$$, but not by $$(1-\zeta)$$.

If $$P$$ is one of these latter points, then we see that $$[1-\zeta] P = (0,\pm i).$$

We can compute the $$x$$-coordinate of $$[1-\zeta] P$$ explicitly, in terms of the $$x$$-coordinate of $$P$$, just using the usual addition formula. The answer is $$x^3 - 1 = 0.$$

So we find that these $$6$$ points satisfy $$x^3 = 1$$, and then (from the equation for $$E$$) $$y^2 = 3$$.

In terms of the lattice, these points are the cosets of $$\varpi/3,$$ $$2 \varpi/3$$, $$\zeta \varpi/3$$, $$\zeta 2 \varpi/3$$, $$\zeta^{-1}\varpi/3,$$ and $$\zeta^{-1} 2 \varpi/3$$. (The remaining cosets of order $$3$$ are exactly $$0,$$ $$(1 + \zeta) \varpi/3$$, and $$2(1+\zeta)\varpi/3$$, and when you multiply these by $$(1-\zeta)$$, you actually land in $$\mathbb T$$.)

So there are $$6$$ lattice cosets, and $$6$$ sets of $$(x,y)$$-coordinates, which match when you apply $$z \bmod \mathbb T \mapsto \bigl(\wp(z),\wp'(z)\bigr).$$

The cosets $$\varpi/3$$, $$\zeta \varpi/3$$, and $$\zeta^{-1} \varpi/3$$ are related by successive multiplications by $$\zeta,$$ and so have the same $$y$$-value. (Either $$\sqrt{3}$$ or $$-\sqrt{3}$$.)

Similarly the cosets $$2\varpi/3$$, $$\zeta 2 \varpi/3,$$ and $$\zeta^{-1} 2 \varpi/3$$ have the same $$y$$-value. (Again, either $$\sqrt{3}$$ or $$-\sqrt{3}$$.)

To see that $$\varpi/3$$ actually maps to $$-\sqrt{3}$$, it it just a matter of checking that $$\wp'(\varpi/3)$$ is negative, which I'd guess one can fairly easily do.

This answer may look elaborate, but I've tried to include details to help non-experts in this perspective. I think it's fair to say that typically all such identities are part of the general theory of complex multiplication, and derivable by this kind of analysis.