The Icosahedral Equation

By now, it should be clear that — given an arbitrary quintic equation — it’s possible to massage this quintic into a form that can be used to easily determine the roots: first one passes from the general quintic to a corresponding “principal quintic” with roots given by

where $\lambda_\pm$, $\mu_\pm$ are determined in terms of the coefficients of the principal quintic. Once these roots are computed, the corresponding roots of the general quintic are found by solving one last quadratic equation:

for some $A(p_i)$, $B_\pm(p_i)$ which determine the principal quintic, given the coefficients $p_i$ of the original quintic.

The Icosahedral Inverse and the Schwarzian¶

We wish to formally invert the equation

such that a value $z_0$ can be computed from a known icosahedral parameter $I_0$. In other words, we seek a function $s(x)$ such that $s(I_0) = z_0$ for any $I_0$ away from the singular points $(0,1,\infty)$.

It has been shown previously that the function $I(z)$ maps vertices, edge midpoints, and face centres to the singular points $(\infty, 1,0)$, respectively. $I(z)$ also satisfies the symmetries of the icosahedron. As such, we expect that — once we invert it on a small triangle inscribed by the points

near the top of the icosahedron (where $z\to\infty$) — we’ll find that we can uniquely define a particular $z_0$ for every $I_0$ (up to some conditions, which we’ll specify). (Recall the parameters $z_n$, $j$, $k$ from earlier sections, which made easy the computation of the icosahedral-invariant polynomials; they are defined to be

in terms of the angle $\alpha := \cos^{-1}\left[1/(2\,\sin(\pi/5))\right]$ subtended by the two-fold symmetry axis and the vertical.

We only expect the function $I(z)$ to have an inverse on this region and its complex conjugate, since the icosahedral symmetries will map out every other $z_0$ found to reproduce a particular $I_0$ (and so the inverse would be multi-valued outside of this wedge). These singular points allow us to uniquely determine the Schwarzian of $s(x)$, as we’ll see next.

Consider the series expansion of $I(z)$ about the three singular points $z \in \{s_0,s_1,s_\infty\}$:

Note that, of these three expansions, only the one about $z=s_\infty$ embodies the corresponding $(2,3,5)$-fold symmetry about that point. The reason for this is that the origin coincides with the five-fold axis of symmetry, since the antipodes along this axis are $\{0,\infty\}$.

Given this observation, we can perform a Möbius transformation which maps $(v_1, v_{20})$, $(w_3, w_{28})$ to $(0,\infty)$ in each case:

Before expanding about $w=0$, let’s first consider the icosahedral parameter $I(z(w))$ by substituting $z = [w(z_n -j) - z_n -j]/(w-1)$ (found by inverting the above and substituting $v_1=z_n+j$, $v_{20}=z_n-j$):

Since both $H(w)$ and $f(w)$ are functions of $w^2$, the two-fold symmetry is now explicit. Performing the series expansion about $p:=w^2 = 0$, we find

Rewriting in terms of $z$, we have the resummed expansion

Similarly, if we instead transform $z$ such that $(w_3, w_{28})$ go to $(0, \infty)$, i.e.

then we find that $I(z(w))$ is instead a rational polynomial of $q:=w^3$:

Rewriting in terms of $z$, we have the resummed expansion at $z=s_0$:

Lastly, for completeness, we can apply the same reasoning to expand at $z\to\infty$ by transforming to $w = -1/z$ (swapping antipodes) and expanding as before as a function of $r := w^5$:

Moreover, the resummed expressions can be easily inverted to give the corresponding expansions of the inverse about each singular point: since $I\Big|_{s=s_i}$ has the general form

for $s_i$ with $i\in\{0,1\}$ (we denote the antipode to $s_i$ as ${\hat s}_i$) then the corresponding inverse is defined implicitly to be

(from which $z = s(I)$).

The above gives us insight into the structure of singular points of $s(x)$, as we’ll now see. Consider the Schwarzian, $S[s(x)]$ given by

Any Möbius transformation has zero Schwarzian (by definition), so applying the Schwarzian to the above power series gives

Therefore, the leading and subleading terms in each expansion give all the singular behaviour in the Schwarzian.

Near $x=I_0 = 0$, we that — although the expressions for $A_0$ and $B_0$ are quite complex (involving square roots of square roots via $k$) — we find that the ratio

gives a rational number.

Similarly, we find near $x=I_1=1$ that

gives a rational number as well.

We’ll see that the reason for this later on, but, to preview: the vertices $v_0$, $v_1$, and $v_\infty$ form a Schwarz triangle on the Riemann sphere with internal angles $(\pi/\nu_0,\pi/\nu_1,\pi/\nu_\infty)$, where

The properties of analytic functions on a Schwarz triangle are well-known, but it’s sufficient to know that the singularities near $s = s_0$, $s = s_1$ — together with the asymptotic expansion near $s = s_\infty$ — are sufficient to determine the Schwarzian of $s(x)$. In particular, we’ll find that

where $\alpha_i = (1-1/\nu_i^2)/2$, $\beta_0 = -\beta_1 = (1+1/\nu_\infty^2 -1/\nu_0^2-1/\nu_1^2)/2$, and $\gamma=0$.

As $s \to s_\infty$, we find instead that the inverted series has the form

where $A_\infty = 1/1728$ and $B_\infty = 739/1728$. We can show this by inverting the series for $\tilde I(r(w)) := 1728\,I(z(w))$, i.e.

to determine $r(x)$ (for which $r(\tilde I(r)) = 1$), and then taking the fifth root on the principal branch (since $r(z) = -1/z^5$) to determine $s(x)$:

Since we find

upon inverting, the inverse $s(x)\big|_{x \to s_\infty}$ has the asymptotic series

Computing the Schwarzian of the asymptotic expansion above, we find

Combining these with the expansions about $x=0,1$, namely

we find that, indeed, these asymptotic expressions can be found by expanding the full expression

in the appropriate limits.

As one last check of the expression above for $S[s(x)]$, one can take the demonstrated leading behaviour of $S[s(x)]$ at $x=0,1$ and show that, in order for

as well, the singular terms must have exactly this form. To see this, take the expression

and expand:

(We could instead consider $\gamma$ to be a regular function of $x$, but Schwarz reflection across the real line would then impose that it be constant.) At each order, the conditions are $\gamma = 0$ at ${\cal O}(x^0)$, $\beta_0 = -\beta_1$ at ${\cal O}(x^{-1})$, and $\beta_1 = (1+1/\nu_\infty^2 - 1/\nu_0^2-1/\nu_1^2)/2$ at ${\cal O}(x^{-2})$.

Solving the Schwarzian¶

It can be shown that the general solution to the the Schwarzian equation $S[s(x)] = f(x)$ is given by

where $g_i(x)$, $i\in\{1,2\}$ are two independent solutions (i.e. with Wronskian $g_1(x)g'_2(x) - g_2(x)\,g'_1(x)\neq 0$) to the second-order equation

Since the singular points of $s(x)$ are at 0, 1, and $\infty$, we expect the hypergeometric function

to play a prominent role. It’s well understood to be a solution of the differential equation

which can be rewritten as

by substituting $h(x) = x^{-c/2}(1-x)^{(c-1-a-b)/2} g(x)$. (The prefactor has been engineered to cancel the first-derivative term, but is nevertheless unimportant once the ratio $s(x) = g_2(x) / g_1(x)$ is computed.)

Comparing the bracketed term above to

we find that

Denoting the sign choices as $\sigma_0$, $\sigma_1$, $\sigma_\infty$ (where $\sigma_i = \pm 1$), we can invert and find

Since we’d like to fix $s(x)$ such that $s(\infty) \to \infty$, we’ll choose the expansion point for the hypergeometric solutions to be $1/x$. As such, the linearly independent solutions in this case are

To fix the overall scale of $s(x)$, as well as the sign $\sigma_\infty$, we examine the asymptotic expansion of the ratio as $x\to\infty$ and determine that

Since we expect the expansion at $x\to\infty$ to be local, it’s expected to not depend on the sign choices $\sigma_0$, $\sigma_1$. However, $\sigma_\infty$ must be chosen to be positive such that series expansion agrees with the inverted series found previously.

Similarly, if take the expansion point to be $x=0$, we find that the ratio

(with $\sigma_0 = +1$) gives the appropriate expansion near $x=0$.

Lastly, at $x=1$, we find that

(with $\sigma_0 = +1$) gives a series expansion appropriate to that singularity.