In this post, I will describe the solution of Jacobi sextic in terms of Weierstrass elliptic function. But before that, we need to transform the coefficients of the Jacobi sextic (21), namely the polyhedral invariants, into the elliptic invariants :

First, let us write the Brioschi normal form (22) like this

\(\displaystyle t^5 + \frac{10}{\Delta} + \frac{45}{\Delta^2} t - \frac{216 g_3}{\Delta^3} = 0 \;\; (23)\)

Where \(\displaystyle g_2, g_3\) are elliptic invariants and \(\displaystyle \Delta = g_2^3 - 27 g_3^2\) is the elliptic discriminant. However, (23) implies

\(\displaystyle f = - \frac{1}{\Delta} \;\, H = \frac{12g_2}{\Delta^2} \;\, T = \frac{216 g_3}{\Delta^3}\)

Hence, the Jacobi sextic in the elliptic form would be :

\(\displaystyle s^6 + \frac{10}{\Delta} s^3 - \frac{12 g_2}{\Delta^2} s - \frac{5}{\Delta^2} = 0 \;\; (24)\)

Now we can begin our work. First let me introduce an interesting property of the Weierstrass elliptic function :

\(\displaystyle \frac{\sigma(5 u)}{\sigma^{(25)}(u)} = \frac{1}{82944} \left | \begin{matrix} \wp' & \wp'' & \wp''' & \wp^{\text{IV}} \\ \wp'' & \wp''' & \wp^{\text{IV}} & \wp^{\text{V}} \\ \wp''' & \wp^{\text{IV}} & \wp^{\text{V}} & \wp^{\text{VI}} \\ \wp^{\text{IV}} & \wp^{\text{V}} & \wp^{\text{VI}} & \wp^{\text{VII}} \end{matrix} \right | \;\; (25)\)

(

NOTE : The sigma function is the Weierstrass sigma) Note that the determinant vanishes at the zeros of \(\displaystyle \sigma\), explicitly, at

\(\displaystyle u_{m, n} = \frac{2m \omega + 2n \omega'}{5} \;\; (26)\)

Where \(\displaystyle \omega, \omega'\) are the two periods of \(\displaystyle \wp\) and m,n are integers mod 5 not both zero.

Expanding the determinant (25) and setting it to zero, we get the 12-th degree polynomial :

\(\displaystyle \( (\wp'')^2 - 12 \wp (\wp')^2 \)^3 - 16 (\wp')^4 \wp'' \( (\wp'')^2 - 12 \wp (\wp')^2 \) - 64 (\wp')^8 = 0 \;\; (27)\)

The 24 roots of (26) relates to the 12 roots of (24) by the relation \(\displaystyle \wp_{m, n} = \wp_{-m,-n}\) due to the mod 5 arithmetic of m,n.

We will now set a variable defined as

\(\displaystyle y_{m, n} = \wp_{2m, 2n} - \wp_{m, n} = \wp\(\frac{4m \omega + 4n \omega}{5}\) - \wp\(\frac{2m \omega + 2n \omega}{5}\)\)

Using the formula \(\displaystyle \wp(u) - \wp(2u) = \frac{\wp'(u) \wp'''(u) = \wp''(u)^2}{4 \wp'(u)^2}\) and some simplification on the derivatives we reckon,

\(\displaystyle y = y_{m, n} = \frac{\wp''_{m,n}^2 - 12 \wp_{m,n} \wp'_{m,n}^2}{4 \wp'_{m,n}} \;\; (28)\)

Substituting this into (27) and dividing by \(\displaystyle 64 (\wp')^6\) gives

\(\displaystyle y^3 - \wp'' y - (\wp')^2 = 0 \;\; (29)\)

Also, by solving for \(\displaystyle (\wp'')^2\) in (28),

\(\displaystyle (\wp'')^2 = (4y + 12 \wp)(\wp')^2 = (4y + 12 \wp)(y^3 - \wp'' y) \;\; (30)\)

By the closed form expression of the derivatives of \(\displaystyle \wp\), we have two equations now

\(\displaystyle y^3 - (6 \wp^2 - g_2/2) y - (4 \wp^3 - g_2 \wp - g_3) = 0 \;\; (31) \\ (6 \wp^2 - g_2/2 )^2 = (4y + 12 \wp)(y^3 - 6 \wp^2 y + g_2 y /2) \;\; (32)\)

Let us define an auxiliary variable z defiend as

\(\displaystyle z = 2 \wp + y\)

so that the equations (30) and (31) becomes (respectively) :

\(\displaystyle z^3 - (3y^2 + g_2) z + 2 g_3 = 0 \;\; (32) \\ 9z^4 - 6(5y^2 + g_2)z^2 + (5y^4 - 2 g_2 y^2 + g_2^2) = 0 \;\; (33)\)

Using quadratic formula for (33), we get

\(\displaystyle 3 z^2 = 5 y^2 + g_2 + 2y \sqrt{5 y^2 + 3 g_2} \;\; (34)\)

Now let us write (32) as

\(\displaystyle z(z^2 - 3y^2 - g_2) = 2g_2 \;\; (35)\)

Squaring (35) and substituting (34) gives

\(\displaystyle \(5y^2 + g_2 + 2y \sqrt{5y^2 + 3g_2}\)\(-4y^2 - 2g_2 + 2y\sqrt{5y^2+3g_2}\)^2 = 108 g_3^2\)

Multiplying out the factors and dividing both sides by 4 gives the sextic

\(\displaystyle 5 y^6 - 2 y^5 \sqrt{5y^2 + 3 g_2} + g_2^3 = 27 g_2^3\)

Using the definition of the elliptic discriminant, this is equivalent to

\(\displaystyle 5 y^6 + \Delta = 2 y^5 \sqrt{5y^2 + 3 g_2}\)

Squaring both sides, rearranging the equation by the powers of y gives

\(\displaystyle 5y^{12} - 12 g_2 y^{10} + 10 \Delta y^6 + \Delta^2 = 0\)

If we let \(\displaystyle y = \frac{1}{s^2}\) and multiply everything by \(\displaystyle \frac{s^6}{\Delta^2}\), we get the Jacobi sextic (24)

\(\displaystyle s^6 + \frac{10}{\Delta} y^3 - \frac{12 g_2}{\Delta^2} s - \frac{5}{\Delta^2} = 0\)

Hence, we reckon that the roots of (24) are actually of the form :

\(\displaystyle \sqrt{s_{m,n}} = \frac{1}{\wp\(\frac{4m\omega + 4n \omega'}{5}\) - \wp\(\frac{2m\omega + 2n\omega'}{5}\)}\)

After going through all the 12 roots, we see that the correct ones are :

\(\displaystyle \sqrt{s_5} = \frac{1}{\wp\(\frac{2\omega}{5}\) - \wp\(\frac{4\omega}{5}\)}\)

\(\displaystyle \sqrt{s_k} = \frac{1}{\wp\(\frac{2\omega' + 48k\omega}{5}\) - \wp\(\frac{4\omega' + 96 k \omega}{5}\)} \;\;\; 0 \le k \le 4\)

Hence, we have the roots of the Jacobi sextic, undoing all the transformation will give the equivalent roots of general quintic equation, and this way, the Kiepert's algorithm ends.

- [1] R. B. King, Beyond the quartic equation

- [2] S. Bessels, One step beyond the solvable equation