 Icosahedral tiling of the sphere

## On Klein's icosahedral solution of the quintic

### Other resources

• My original notes (now published in Expo. Math.)
• This GitHub repo where the solution is implemented in python

### A solution of the quintic

Klein famously showed that certain non-radical functions naturally associated to the icosahedron permit us to solve the quintic equation. Here is (a slightly simplified version of) that solution:

Using only radical transformations, the general quintic may be put in the form: $$y^5 + 5y + \gamma = 0$$ Given such a quintic, 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 an $$A_5$$-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. Both the notes and python repo linked above provide formulae (similar to those above) for the case of a quintic in this form.