Icosahedral tiling of the sphere

## On Klein's icosahedral solution of the quintic

### PDFs available

Those interested in the below may wish to consult either:

### A solution of the quintic

Since the work of Ruffini (1799) and Abel (1823) it has been known that it is not possible to solve the quintic in radicals. In 1858, Hermite presented the first solution of the quintic; the non-radical functions he added to his toolbox for the purpose were certain modular functions. This solution received a lot of attention and shortly afterwards Kronecker and Brioschi also published solutions. Eventually in about 1878, the great geometer Klein (drawing on work of Gordon and crucially of Galois) showed how to unify the different approaches and also demonstrated a new, extremely beautiful solution using icosahedral functions. Here is (a slightly simplified version of) that solution:

Given the quintic equation: $$y^5 + 5y + \gamma = 0$$ set: $$\nabla = \sqrt{\gamma^4 + 256}\\ Z = \frac{1}{2\cdot 1728}[2\cdot 1728 + 207\gamma^4 + \gamma^8 - \gamma^2 (81 + \gamma^4)\nabla]\\ z = \frac{{}_2F_1(\frac{31}{60}, \frac{11}{60}; \frac{6}{5}; Z^{-1})} {(1728Z)^{1/5}{}_2F_1(\frac{19}{60}, -\frac{1}{60}; \frac{4}{5}; Z^{-1})}$$ and: $$f(z) = z(z^{10} + 11z^5 - 1)\\ H(z) = -(z^{20} + 1) + 228(z^{15} - z^5) - 494z^{10}\\ T(z) = (z^{30} + 1) + 522(z^{25} - z^5) - 10005(z^{20} + z^{10})\\ B(z) = -1 - z - 7(z^2 - z^3 + z^5 + z^6) + z^7 - z^8\\ D(z) = -1 + 2z + 5z^2 + 5z^4 - 2z^5 - z^6$$ Then: $$y = -\gamma\cdot\frac{f(z)}{H(z)/B(z)} - \frac{7\gamma^2 + 9\nabla}{2\gamma(\gamma^4 + 648)} \cdot\frac{D(z)T(z)}{f(z)^2H(z)/B(z)}$$ is a root!

Replacing $$z$$ with $$e^{2\pi\nu i/5}z$$ for $$\nu=1, 2, 3, 4$$ provides all the other roots.

The above might look a little messy but it's really quite neat from the right point of view. Given this solution, a strong hint at the geometric connection is the five auxiliary polynomials $$f, H, T, B, D$$. The roots of the first three $$f, H, T$$ are, respectively, the locations of the projection of the vertices, face centres and edge midpoints of a regular icosahedron onto its circumsphere (once this circumsphere has been identified with the extended complex plane by stereographic projection). The roots of the last two polynomials, $$B, D$$ are, respectively, the locations of the vertices and face centres of a regular cube inscribed in the icosahedron. Indeed the vertices of this cube are the vertices of a pair of dual tetrahedra inscribed in the icosahedron, one of which I show below:

Icosahedron with inscribed tetrahedron

The even explains the grouping $$H/B$$ above: $$B$$ is a factor of $$H$$ because the vertices of a cube can be placed at the face centres of 8 of the 20 icosahedral faces.

Another important detail to notice above is that the non-radical functions employed are certain hypergeometric functions. Although they may look somewhat arbitrary at first sight, these are extremely special hypergeometric functions which belong to Schwarz's list of algebraic hypergeometric functions (those with finite monodromy). The quotient of the pair used in the above equation inverts the 60-fold branched covering of the complex projective line with $$A_5$$ monodromy. The picture at the top of this post is a picture of that branched covering as well as being a picture of the icosahedral tiling of the sphere.

Also important in the above is the invariant $$Z$$ since it has a neat geometric interpretation. The vector of roots of a quintic defines an $$S_5$$-orbit in complex projective 4-space. If the quintic has no degree 4 or 3 terms, this orbit lies on the quadric surface and in fact the $$S_5$$-action on the quadric is induced from a natural action on the lines in this quadric. If we adjoin a square root $$\nabla$$ of the discriminant then the $$S_5$$-action becomes an $$A_5$$-action and each of the two families of lines in the ruled surface carries and $$A_5$$-action. These families are parameterised by complex projective lines and the quotient by $$A_5$$ defines the icosahedral invariants. Evidently, it is this quotient that we invert using the hypergeometric functions mentioned above.

### More general solution

Using a radical transformation that requires solving only one auxiliary quadratic, any quintic can be put in the form: $$y^5 + 5\alpha y^2 + 5\beta y + \gamma = 0$$ As we have said, in this case because there are no degree 4 or 3 terms, the vector ordered roots lies on a doubly-ruled quadric surface. This is all that is required for Klein's geometric approach to proceed. The notes I linked provide formulae (similar to those above) for the case of a quintic in this form.

### Three roles for $$A_5$$

From an abstract point of view, one reason for the connection between the quintic and the icosahedron is that the group $$A_5$$ can be made to play three roles:
• As the Galois group of a general quintic (together with a distinguished square root of its discriminant)
• As the group of rotations of the icosahedron
• As the monodromy group of the hypergeometric differential equation with appropriate parameters (from Schwarz's list)

### Background

My own interest in this problem dates from when I was at high school and read E.T. Bell's (slightly whimsical) Men of Mathematics books. I remembered that Bell had mentioned that Klein had established a connection between the rotations of the icosahedron and the solution of the quintic equation. Searching through the books now I see that the relevant passage is from Bell's discussion of Cauchy (ironically Bell doesn't discuss Klein):
To give but one instance, the set of all rotations which twirl a regular icosahedron (twenty-sided regular solid) about its axes of symmetry, so that after any rotation of the set the volume of the solid occupies the same space as before, forms a group, and this group of rotations, when expressed abstractly, is the same group as that which appears, under permutations of the roots, when we attempt to solve the general equation of the fifth degree. [...] This beautiful unification was the work of Felix Klein (1849-1925) in his book on the icosahedron (1884).

Last year (2011) I could no longer stand to live in ignorance so I bought a copy of Klein's "Lectures on the Icosahedron and the Solution of the Fifth Degree" and started reading it and various other sources. I made a few notes for myself and as they took shape I thought it might be worth making them available online. Part of my motivation for doing so was this Mathoverflow question.

### Implementing the solution

This repo contains a script that implements the solution in Bring-Jerrard (also shown below) as well as in the more general form.

import mpmath, numpy as N

def nabla(cc):
return N.exp(N.log(cc**4 + 256 + 0j)/2)

def Z(cc):
return (2*1728 + cc**4*(207 + cc**4) - cc*cc*(81 + cc**4)*nabla(cc))/(2*1728)

def z(ZZ):
mpmath.mp.dps = 20
return N.exp(-N.log(1728*ZZ) / 5) * mpmath.hyp2f1(31.0/60, 11.0/60, 6.0/5, 1.0/ZZ) /\
mpmath.hyp2f1(19.0/60, -1.0/60, 4.0/5, 1.0/ZZ)

def HB(zz):
return (zz**4 - 3*zz**3 - zz**2 + 3*zz + 1) *\
(zz**8 + 4*zz**7 + 7*zz**6 + 2*zz**5 + 15*zz**4 - 2*zz**3 + 7*zz**2 - 4*zz + 1)

def D(zz):
return -1 + 2*zz + 5*zz**2 + 5*zz**4 - 2*zz**5 - zz**6

def f(zz):
u = zz**5
return zz*(-1 + u * (11 + u))

def T(zz):
u = zz**5
return 1 + u * (-522 + u * (-10005 + u * u * (-10005 + u * (522 + u))))

def y(cc):
ys = []
eps = N.exp(2*N.pi*1j / 5)
zz = z(Z(cc))
for n in range(5):
ys.append(-cc * f(zz) / HB(zz) -\
(7*cc*cc + 9*nabla(cc))/(2*cc**5 + 2*648*cc)*D(zz)*T(zz)/(HB(zz)*f(zz)**2))
zz *= eps
return [(yy, abs(yy**5 + 5*yy + cc)) for yy in ys]

for (z, err) in y(-2.8234): # Randomly chosen value for quintic (see xkcd #221).
print '%.6f + %.6fi   (%.10f)' % (z.real, z.imag, err)