§23.22 Methods of Computation

Suppose that the lattice $𝕃$ is given. Then a pair of generators $2{\omega }_1$ and $2{\omega }_3$ can be chosen in an almost canonical way as follows. For $2{\omega }_1$ choose a nonzero point of $𝕃$ of smallest absolute value. (There will be $2$, $4$, or $6$ possible choices.) For $2{\omega }_3$ choose a nonzero point that is not a multiple of $2{\omega }_1$ and is such that $\mathrm{\Im }\tau >0$ and $|\tau |$ is as small as possible, where $\tau ={\omega }_3/{\omega }_1$. (There will be either $1$ or $2$ possible choices.) This yields a pair of generators that satisfy $\mathrm{\Im }\tau >0$, $|\mathrm{\Re }\tau |\le \frac{1}{2}$, $|\tau |>1$. In consequence, $q={\mathrm{e}}^{\mathrm{i}\pi {\omega }_3/{\omega }_1}$ satisfies $|q|\le {\mathrm{e}}^{-\pi \sqrt{3}/2}=0.0658\mathrm{\dots }$. The corresponding values of $e_1$, $e_2$, $e_3$ are calculated from (23.6.2)–(23.6.4), then $g_2$ and $g_3$ are obtained from (23.3.6) and (23.3.7).

Suppose that the invariants $g_2=c$, $g_3=d$, are given, for example in the differential equation (23.3.10) or via coefficients of an elliptic curve (§23.20(ii)). The determination of suitable generators $2{\omega }_1$ and $2{\omega }_3$ is the classical inversion problem (Whittaker and Watson (1927, §21.73), McKean and Moll (1999, §2.12); see also §20.9(i) and McKean and Moll (1999, §2.16)). This problem is solvable as follows:

1. (a)

   In the general case, given by $cd\ne 0$, we compute the roots $\alpha$, $\beta$, $\gamma$, say, of the cubic equation $4t^3-ct-d=0$; see §1.11(iii). These roots are necessarily distinct and represent $e_1$, $e_2$, $e_3$ in some order.

   If $c$ and $d$ are real, and the discriminant is positive, that is $c^3-27d^2>0$, then $e_1$, $e_2$, $e_3$ can be identified via (23.5.1), and $k^2$, $k_{}^{\prime }{}_{}{}^2$ obtained from (23.6.16).

   If , or $c$ and $d$ are not both real, then we label $\alpha$, $\beta$, $\gamma$ so that the triangle with vertices $\alpha$, $\beta$, $\gamma$ is positively oriented and $[\alpha ,\gamma ]$ is its longest side (chosen arbitrarily if there is more than one). In particular, if $\alpha$, $\beta$, $\gamma$ are collinear, then we label them so that $\beta$ is on the line segment $(\alpha ,\gamma )$. In consequence, $k^2=(\beta -\gamma )/(\alpha -\gamma )$, $k_{}^{\prime }{}_{}{}^2=(\alpha -\beta )/(\alpha -\gamma )$ satisfy $\mathrm{\Im }{k}^2\ge 0\ge \mathrm{\Im }k_{}^{\prime }{}_{}{}^2$ (with strict inequality unless $\alpha$, $\beta$, $\gamma$ are collinear); also $|k^2|$, $|k_{}^{\prime }{}_{}{}^2|\le 1$.

   Finally, on taking the principal square roots of $k^2$ and $k_{}^{\prime }{}_{}{}^2$ we obtain values for $k$ and $k^{\prime }$ that lie in the 1st and 4th quadrants, respectively, and $2{\omega }_1$, $2{\omega }_3$ are given by

   23.22.1		$$2{\omega }_{1}M(1,k^{\prime })=-2\mathrm{i}{\omega }_{3}M(1,k)=\frac{\pi }{3}\sqrt{\frac{c(2+k^{2}k_{}^{\prime }{}_{}{}^2)(k_{}^{\prime }{}_{}{}^2-k^2)}{d(1-k^{2}k_{}^{\prime }{}_{}{}^2)}},$$
   ⓘ Symbols: $M(a,g)$: arithmetic-geometric mean, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{i}$: imaginary unit, ${\omega }_1$, ${\omega }_3$, ${\omega }_2=-{\omega }_1-{\omega }_3$: lattice generators, $k$: modulus and $k^{\prime }$: complementary modulus Proof sketch: Combine (23.6.16), (23.6.17), (22.20.6), (23.3.5)–(23.3.7), and use analytical continuation. Referenced by: §23.22(ii) Permalink: http://dlmf.nist.gov/23.22.E1 Encodings: TeX, pMML, png See also: Annotations for §23.22(ii), §23.22(ii), §23.22 and Ch.23

   where $M$ denotes the arithmetic-geometric mean (see §§19.8(i) and 22.20(ii)). This process yields 2 possible pairs ($2{\omega }_1$, $2{\omega }_3$), corresponding to the 2 possible choices of the square root.

2. (b)

   If $d=0$, then

   23.22.2		$$2{\omega }_1=-2\mathrm{i}{\omega }_3=\frac{{\left(\mathrm{\Gamma }\left(\frac{1}{4}\right)\right)}^2}{2\sqrt{\pi }{c}^{1/4}}.$$
   ⓘ Symbols: $\mathrm{\Gamma }\left(z\right)$: gamma function, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{i}$: imaginary unit and ${\omega }_1$, ${\omega }_3$, ${\omega }_2=-{\omega }_1-{\omega }_3$: lattice generators Referenced by: §22.20(iv), §23.22(ii) Permalink: http://dlmf.nist.gov/23.22.E2 Encodings: TeX, pMML, png See also: Annotations for §23.22(ii), §23.22(ii), §23.22 and Ch.23

   There are 4 possible pairs ($2{\omega }_1$, $2{\omega }_3$), corresponding to the 4 rotations of a square lattice. The lemniscatic case occurs when $c>0$ and ${\omega }_1>0$.

3. (c)

   If $c=0$, then

   23.22.3		$$2{\omega }_1=2{\mathrm{e}}^{-\pi \mathrm{i}/3}{\omega }_3=\frac{{\left(\mathrm{\Gamma }\left(\frac{1}{3}\right)\right)}^3}{2\pi d^{1/6}}.$$
   ⓘ Symbols: $\mathrm{\Gamma }\left(z\right)$: gamma function, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit and ${\omega }_1$, ${\omega }_3$, ${\omega }_2=-{\omega }_1-{\omega }_3$: lattice generators Referenced by: §23.22(ii) Permalink: http://dlmf.nist.gov/23.22.E3 Encodings: TeX, pMML, png See also: Annotations for §23.22(ii), §23.22(ii), §23.22 and Ch.23

   There are 6 possible pairs ($2{\omega }_1$, $2{\omega }_3$), corresponding to the 6 rotations of a lattice of equilateral triangles. The equianharmonic case occurs when $d>0$ and ${\omega }_1>0$.

Assume $c=g_2=-4(3-2\mathrm{i})$ and $d=g_3=4(4-2\mathrm{i})$. Then $\alpha =-1-2\mathrm{i}$, $\beta =1$, $\gamma =2\mathrm{i}$; $k^2=(7+6\mathrm{i})/17$, and $k_{}^{\prime }{}_{}{}^2=(10-6\mathrm{i})/17$. Working to 6 decimal places we obtain