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.

  • $\begingroup$ For the record, I deleted an answer as I misunderstood the question. $\endgroup$
    – Hao Chen
    Aug 27, 2021 at 17:19

1 Answer 1


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.


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