I have found several references where the elliptic curve $y^2=x^3-1$ is called the equianharmonic curve, and, more often, where the half-period of this curve $$ \omega_1 = \frac{\Gamma(1/3)^3}{4\pi} $$ is called the equianharmonic constant, the corresponding semi-period for the elliptic curve $y^2=x^3-x$ is called the Lemniscate constant.

I think that the anharmonic ratio of four points is another name for the cross-ratio, if we look the action of $S_4$ in the cross ratio we see that it is fixed by $K$, Klein's four-group, and $S_4/K \simeq S_3$ is called the anharmonic group, that can be viewed as the group of Mobius transformations $$ \lambda, \frac{1}\lambda, 1-\lambda, \frac{1}{1-\lambda}, \frac{1-\lambda}{\lambda}, \frac{\lambda}{1-\lambda} $$ This group has three especial orbits with less than 6 elements: $\{0,1,\infty\}$, $\{-1,\tfrac{1}{2}, 2\}$ called the harmonic case, and $\{\pm e^{2\pi i/6}\}$ which might be called the equianharmonic case but I am bot sure.

My question is why is it called equianharmonic curve/constant? I expect it to have a relationship with the group but I can't figure out which. If so, does the harmonic case have a similar relationship with some other elliptic curve ($y^2=x^3-x$ for instance)?

Thank you and sorry if it is a silly question.

  • $\begingroup$ Maybe the idea is that it's not just "anharmonic" (not harmonic) but as far as possible from the "harmonic" case $-$ and the distance is the same (thus "equi-") in each of the three directions that you could deform the tetrahedrally symmetric curve to reach the dihedrally symmetric "harmonic" curve. $\endgroup$ Feb 27, 2021 at 18:33

1 Answer 1


The name refers to the concept of an anharmonic ratio, or cross-ratio. Four points $A,B,C,D$ are called equianharmonic if their cross-ratio is a cube root of 1. In that case the 6 cross-ratios obtained by combining the four points in all possible ways are the same ("equal anharmonic ratio" = "equianharmonic").

The curve $y^2=x^3-1=(x-p_1)(x-p_2)(x-p_3)$ has $p_1=1$, $p_2=e^{i2\pi/3}$, $p_3=e^{i4\pi/3}$ equal to the three cube roots of 1. The four intersection points of a line with the four tangents, drawn from any point on the curve, are equianharmonic.

If you read German, you might want to know how Wiener described this when he explained the concept in 1901:

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  • $\begingroup$ Thank you very much, that what was exactly the kind of description I was hoping for. I don't understand German but I will do my best with the help of Google :-) $\endgroup$ Feb 28, 2021 at 10:05

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