Lecture 7: Wednesday, September 4, 2013

Let $K$ denote either $\mathbb Q$, $\mathbb R$, or $\mathbb C$. The collection of rational functions $\gamma(z) \in K(z)$ which have $\deg(\gamma) = 1$ are called Möbius Transformations. That is, they are in the form $\gamma(z) = \dfrac {a \, z + b}{c \, z + d}$ where $\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \in GL_2(K) = \left \{ \gamma \in \text{Mat}_{2\times 2}(K) \ \biggl| \ a \, d - b \, c \neq 0 \right \}$. We denote this collection by $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$. We found in the previous lecture that $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ forms a group under composition. Moreover, $\text{Aut} \bigl( \mathbb P^1(K) \bigr)$ acts on $K(z)$ via $\gamma^\ast \, f(z) = f \bigl( \gamma^{-1}(z) \bigr)$. For any rational function $f(z) \in K(z)$, we define the Galois group of $f$ as the subgroup
$\text{Gal}(f) = \left \{ \gamma \in \text{Aut} \bigl( \mathbb P^1(K) \bigr) \ \biggl| \ f \bigl( \gamma(z) \bigr) = f(z) \right \}.$
Which finite groups $G$ can appear as the Galois group of some rational function $f(z)$? We discuss this question in detail today.

Inverse Galois Problem

We begin with a motivating example. Consider the general quadratic polynomial equation $a \, z^2 + 2 \, b \, z + c = 0$. Its roots are $z_0 = \dfrac {-b \pm \sqrt{b^2 - a \, c}}{a}$. They are permuted by the Galois automorphism $r: \dfrac {-b + \sqrt{b^2 - a \, c}}{a} \mapsto \dfrac {-b - \sqrt{b^2 - a \, c}}{a}$. In fact, we can think of the Galois group of this polynomial as the the cyclic group $Z_2 = \left \langle r \, \bigl| \, r^2 = 1 \right \rangle$.

Let’s think of this in a slightly different way. Consider instead the rational function $f(z) = \dfrac {a \, z^2 + 2 \, b \, z + c}{a \, z + b}$ with $\deg(f) = 2$. It is easy to see that $f(z_0) = 0$. The Möbius transformation $r(z) = - \dfrac {b \, (a \, z + b) + (b^2 - a \, c)}{a \, (a \, z + b)}$ has a couple of nice properties: $f \bigl( r(z) \bigr) = f(z)$ while $r^2 = 1$ is the identity transformation. In fact, $f \bigl( r(z_0) \bigr) = f(z_0) = 0$, so that $r$ acts as the same Galois automorphism as before: it permutes these two roots. This Möbius transformation gives an embedding $Z_2 \hookrightarrow \text{Aut} \bigl( \mathbb P^1(K) \bigr)$.

There is a larger question at play here:

Inverse Galois Problem.
Which finite groups $G$ appear as the permutations of the roots of some polynomial $F(z)$ with coefficients in $K$? If one of these groups appears once for some polynomial, does it appear infinitely many times for infinitely many polynomials?

Here is one approach to this problem. Say that we can find a rational function $f(z) = p(z) / q(z)$ in $K(z)$ such that $\text{Gal}(f) \simeq G$. Then the extension $K(z) / K \bigl( f(z) \bigr)$ has Galois group $G$. For each $w_0 \in K$, consider the roots $z_0 \in \mathbb C$ to the equation $f(z_0) = w_0$. Since $f \bigl( \gamma(z_0) \bigr) = f(z_0) = w_0$ for all $\gamma \in G$, the field $L = K(z_0)$ is an extension of $K = K(w_0)$ with Galois group $G \simeq \text{Gal}(L/K)$. As we range over all $w_0 \in K$, we find infinitely many polynomials $F(z) = p(z) - w_0 \cdot q(z)$ such that $G$ appears as the permutations of its roots. Of course, this approach is much more stringent than the Inverse Galois Problem, but it is an approach nonetheless.

Which Groups do We Know?

For the sake of brevity, assume for the rest of the lecture that $K = \mathbb C$. Now let $G$ be a finite group, and say that we have a faithful representation $\phi: G \hookrightarrow GL_2(\mathbb C) \twoheadrightarrow \text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr)$. Then $G \simeq \phi(G)$ is a subgroup of the group of Möbius Transformations. There are several examples we can construct in this way:

\begin{aligned} Z_n & = \left \langle r \, \bigl| \, r^n = 1 \right \rangle : & r(z) & = \zeta_n \, z \\[5pt] D_n & = \left \langle r, \, s \, \bigl| \, s^2 = r^n = (s \, r)^2 = 1 \right \rangle: & \qquad r(z) & = \zeta_n \, z & \qquad s(z) & = \dfrac {1}{z} \\[5pt] S_3 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^2 = 1 \right \rangle: & r(z) & = \dfrac {z-1}{z} & s(z) & = \dfrac {z}{z-1} \\[5pt] A_4 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^3 = 1 \right \rangle: & r(z) & = \zeta_3 \, z & s(z) & = \dfrac {1-z}{2 \, z + 1} \\[5pt] S_4 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^4 = 1 \right \rangle: & r(z) & = \dfrac {\zeta_4 + z}{\zeta_4 - z} & s(z) & = \dfrac {1 - z}{1 + z} \\[5pt] A_5 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^5 = 1 \right \rangle: & r(z) & = \dfrac {z + {\zeta_5}^3 \, \varphi}{\varphi \, z - {\zeta_5}^3} & s(z) & = \dfrac {z + \varphi}{\varphi \, z - 1} \end{aligned}

where $\zeta_n = e^{2 \pi i/n}$ is a root of unity and $\varphi = - \bigl( {\zeta_5}^2 + {\zeta_5}^3 \bigr) = \dfrac {1 + \sqrt{5}}{2}$ is the golden ratio. Observe that we need $K = \mathbb C$ in order to contain the roots of unity, although we can work with other fields.

Given a faithful representation $\phi: G \hookrightarrow \text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr)$ of a finite group, the subgroup $G$ acts on $\mathbb C(z)$ because $\text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr)$ acts on $\mathbb C(z)$. We wish to compute the set $\mathbb C(z)^G = \left \{ g(z) \in \mathbb C(z) \ \biggl| \ g\bigl( \gamma(z) \bigr) = g(z) \ \text{for all} \ \gamma(z) \in \phi(G) \right \}$. We say that $\mathbb C(z)^G$ is the field fixed by $G$. We can write $\mathbb C(z)^G = \mathbb C(J)$ for some rational function $J = f(z) \in \mathbb C(z)$; we will show examples below. We may think of $G$ as the Galois group of the extension $\mathbb C(z)/\mathbb C(J)$. Here are several examples:

\begin{aligned} Z_n & = \left \langle r \, \bigl| \, r^n = 1 \right \rangle : & f(z) & = z^n \\[5pt] D_n & = \left \langle r, \, s \, \bigl| \, s^2 = r^n = (s \, r)^2 = 1 \right \rangle: & f(z) & = \dfrac {4 \, z^n}{( z^n + 1)^2} \\[5pt] S_3 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^2 = 1 \right \rangle: & f(z) & = \frac{4}{27} \, \dfrac {( z^2 - z + 1 )^3}{z^2 \, (z-1)^2} \\[5pt] A_4 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^3 = 1 \right \rangle: & f(z) & = -\dfrac {1}{64} \, \dfrac {( 8 \, z^3 + 1)^3}{z^3 \, (z^3 - 1)^3} \\[5pt] S_4 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^4 = 1 \right \rangle: & f(z) & = \frac{1}{108} \, \dfrac {( z^8 + 14 \, z^4 + 1 )^3}{z^4 \, (z^4 - 1)^4} \\[5pt] A_5 & = \left \langle r, \, s \, \bigl| \, s^2 = r^3 = (s \, r)^5 = 1 \right \rangle: & f(z) & = -\frac{1}{1728} \, \dfrac {( z^{20} - 228 \, z^{15} +494 \, z^{10} + 228 \, z^5 + 1)^3}{z^5 \, (z^{10} + 11 \, z^5 - 1)^5} \end{aligned}

We will see in a future lecture that these formulas were all discovered by Felix Klein! Here are a few observations:

• There may be more than one faithful representation $\phi: G \hookrightarrow \text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr)$, and hence more than one choice of rational function $J = f(z)$. For example, $S_3 \simeq D_3$, the unique non-abelian group of order 6, admits two fundamentally different such rational functions.

• The rational functions $J = f(z) \in \mathbb C(z)$ satisfy $\deg(f) = |G|$ for each of the groups listed. This is because $\text{Aut} \bigl( \mathbb C(z) / \mathbb C(J) \bigr) \simeq G$ has size $\deg(f) = \bigl[ \mathbb C(z) : \mathbb C(J) \bigr] = |G|$.

• Each of the groups has a presentation $G \simeq \left \langle r, \, s \, \bigl| \, s^m = r^n = (s \, r)^k = 1 \right \rangle$. These are examples of triangle groups. It is no coincidence that $\dfrac 1m + \dfrac 1n + \dfrac 1k > 1$. We will return to this later.

Are There Others?

Felix Klein proved a remarkable theorem which classifies all such finite subgroups of $\text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr)$.

Theorem.
If $G \hookrightarrow \text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr)$ is a finite group, then $G \simeq Z_n, \, D_n, \, S_3, \, A_4, \, S_4, \, A_5$. In particular, $G \simeq \text{Gal}(f) = \left \{ \gamma \in \text{Aut} \bigl( \mathbb P^1(\mathbb C) \bigr) \ \biggl| \ f \bigl( \gamma(z) \bigr) = f(z) \right \}$ if and only if $G$ is one of the previously discussed examples.

We will discuss this more in the next lecture. Let me end the lecture with a thought: this result does not say the Inverse Galois Problem is limited to these five families of groups $G$. Indeed, the approach we outlined earlier is restrictive to the projective line $\mathbb P^1(K)$. What if we replace this projective variety with another one?

If we consider elliptic curves
$E(K) = \left \{ (\tau_1 : \tau_2 : \tau_0) \in \mathbb P^2(K) \, \biggl| \, \tau_2^2 \, \tau_0 + a_1 \, \tau_1 \, \tau_2 \, \tau_0 + a_3 \, \tau_2 \, \tau_0^2 = \tau_1^3 + a_2 \, \tau_1^2 \, \tau_0 + a_4 \, \tau_1 \, \tau_0^2 + a_6 \, \tau_0^3 \right \}$
then the group $\text{Aut} \bigl( E(K) \bigr)$ is much more robust. We replace $K(z)$ with the quotient field $K(E)$ of the coordinate ring $K[E] = K[x,y] / (y^2 + a_1 \, x \, y + a_3 \, y - x^3 - a_2 \, x^2 - a_4 \, x - a_6)$. Any finite group $G \simeq Z_n \times Z_m$ can appear as the kernel of an isogeny $\phi: E' \to E$, so that the extension $K(E) / \phi^\ast \, K(E')$ has Galois group $G$.

We will return to this line of reasoning later.
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