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Section 2.6 Belyi's theorem (Maria)

Definition 2.6.2. Belyi functions.

A meromorphic function with less than 4 branching values is a Belyi function.

Remark 2.6.3.
  1. Branching values can be taken to be in \(\{0,1,\infty\}\text{.}\)
  2. If \(S \ne \PP^1\text{,}\) then \(f \colon S \to \PP^1\) has at least 3 branching values
Definition 2.6.4. Belyi polynomials.

Let \(m,n \in \NN\text{,}\) \(\lambda = m/(m+n)\text{,}\) define

\begin{equation*} P_\lambda(x) = P_{m,n}(x) = \frac{(m+1)^{m+n}}{m^mn^n} x^m(1-x)^n \end{equation*}

Belyi polynomials.

Example 2.6.6.
\begin{equation*} S_\lambda : y^2 = x(x-1)(x-\lambda) \end{equation*}

with \(\lambda = m/(m+n)\text{.}\) From ex. 1.32

\begin{equation*} x\colon S_\lambda \to \PP^1 \end{equation*}
\begin{equation*} (x,y) \mapsto x \end{equation*}
\begin{equation*} \infty \mapsto \infty \end{equation*}

ramifies over \(0,1,\lambda,\infty\text{.}\) Then \(f = P_\lambda \circ x \colon S_\lambda \to \PP^1\) ramifies exactly at \((0,0), (1,0), (\lambda,0), \infty\text{.}\) With branching values \(0,0,1,\infty\) so that \(f\) is a Belyi function.

Subsection 2.6.1 Proof of a) implies b)

Note 2.6.7.

Its enough to show \(\exists f\colon S\to \PP^1\) ramified over \(\{0,1,\infty, \lambda_1, \ldots, \lambda_n\} \subseteq \QQ \cup \{\infty\}\text{.}\) Given this we can repeatedly use Belyi polynomials to obtain \(g\colon S \to \PP^1\) ramified over \(\{0,1, \infty\}\text{.}\)

Write \(S = S_F\)

\begin{equation*} F(x,y) = p_0(x)y^n + \cdots + p_n(x) \end{equation*}

defined over \(\overline \QQ\lb x,y\rb\text{.}\) Let \(B_0 = \{\mu_1, \ldots, \mu_s\}\) be the branching values of \(x\colon S_F\to \PP^1\text{.}\)

Theorem 1.86 says that the each \(\mu_i\) is \(\infty\text{,}\) a root of \(p_0(x)\) or a common root of \(F, F_y\) which implies by lemma 1.84 that \(B_0 \subseteq \overline \QQ \cup\{\infty\}\) . If \(B_0\subseteq \QQ\cup\{\infty\}\) we are done otherwise let \(m_1(T) \in \QQ\lb T \rb\) be the minimal polynomial of \(\{\mu_1, \ldots, \mu_s\}\text{.}\) Let \(\{\beta_1, \ldots, \beta_d\}\) be the roots of \(m_1' (T) \) and \(p'(T)\) their min. poly. Note : \(\deg P(t) \lt \deg m'_1(T)\)

Note: \(\operatorname{Branch}(g\circ f) = \operatorname{Branch}(g) \cup g(\operatorname{Branch}(f))\) branching values.

So \(B_1 \operatorname{Branch}(m_1 \circ x) = m_1(\{\text{roots of }m_1'\}) \cup\{0,\infty\}\text{.}\)

\begin{equation*} S\xrightarrow x \PP^1 \xrightarrow{m_1} \PP^1 \end{equation*}

If \(B_1 \subseteq \QQ \cup \{\infty\}\) done. Otherwise let \(m_2(T) \) be the minimal polynomial \(/\QQ\) of \(\{m_1(\beta_1) ,\ldots, m_1(\beta_d) \}\text{,}\) \(B_2 = \operatorname{Branch}(m_2\circ m_2 \circ x)\text{.}\) Fact: \(\deg(m(t)) \lt \deg(m_1(T))\text{.}\)

Repeat inductively until \(B_k \subseteq \QQ \cup \{\infty\}\) which is guaranteed by the decreasing degrees.

Subsection 2.6.2 Algebraic characterization of morphisms

Remark 2.6.9.
\begin{equation*} \xymatrix{ S_f \ar[r]^f \ar[dr]_m & S_G \ar[d]_h \\ & S_D } \end{equation*}

The fact that this diagram commutes can be expressed by polynomial identities.

Subsection 2.6.3 Galois action

Let \(\operatorname{Gal}(\CC) = \Gal{\CC}{\QQ}\text{.}\)

Definition 2.6.10.

For \(\sigma \in \operatorname{Gal}(\CC)\text{,}\) \(a\in \CC\) denote \(a^\sigma = \sigma(a)\text{,}\)

  1. If \(P = \sum a_{ij} x^iy^j \in \CC\lb x,y\rb\) set
    \begin{equation*} P^\sigma = \sum a^\sigma_{ij} x^iy^j \in \CC\lb x,y\rb \end{equation*}
    if \(R = P/Q\) set \(R^\sigma = P^\sigma / Q^\sigma\text{.}\)
  2. If \(S \simeq S_F\text{,}\) \(S^\sigma = S_{F^\sigma}\text{.}\)
  3. If \(\Psi = (R_1, R_2)\) \(S_F\to S_G\) is a morphism, set \(\Psi^ \sigma = (R_1^\sigma , R_2 ^\sigma) \colon S_{F^\sigma} \to S_{G^\sigma}\text{.}\)
  4. For an equivalence class \((S,f) = (S_F, R(x,y))\) of ramified covers of \(\PP^1\) set \((S,f)^\sigma = (S^\sigma, f^\sigma) = (S_{F^\sigma}, R^\sigma (x,y))\text{.}\)

Verify this Galois action is well-defined (lemma 3.12).

Recall: \(S_F\) is constructed from a noncompact Riemann surface \(S_F^\times \subseteq \CC^2\) by adding finitely many points, (theorem 1.86). If \(P= (a,b) \in S_F^\times\) then \(P^\sigma = (a^\sigma, b^\sigma)\text{.}\) What about the other points?

Subsection 2.6.4 Points and valuations

Definition 2.6.12.

Let \(\mathcal M\) be a function field. A (discrete) valuation of \(\mathcal M\) is \(v \colon \mathcal M^* \to \ZZ\) s.t.

  1. \(\displaystyle v(\phi\psi) = v(\phi) + v(\psi)\)
  2. \(\displaystyle v(\phi\pm\psi) \ge \min\{ v(\phi) , v(\psi)\}\)
  3. \(v(\phi) = 0 \) if \(\phi \in \CC^*\)
  4. \(v\) is nontrivial \(\exists \phi : v(\phi)\ne 0\)

set \(v(0) = \infty\text{.}\)

Facts:

\begin{equation*} A_v = \{\phi \in \mathcal M : v(\phi) \ge 0 \} \subseteq \mathcal M \end{equation*}

is a subring that is a local ring with a maximal ideal

\begin{equation*} M_v = \{ \phi\in \mathcal M : v(\phi) \gt 0\} = (\phi) \end{equation*}

for some \(\phi\) a uniformizer.

If \(v(\phi) = 1\) \(v\) is normalised.

Easy exercise.

Sketch: First prove it for \(S = \PP^1\text{.}\)

Inductively meromorphic functions separate points.

Surjectivity study behaviour of valuations in finite extensions of fields and use a nonconstant morphism \(f\colon S \to \PP^1\) to reduce to the case of \(\PP^1\text{.}\)

Galois action on points.
Definition 2.6.15.
  1. Given a valuation \(v\) on \(\mathcal M(S)\) define a valuation \(v^\sigma\) on \(\mathcal M(S^\sigma)\) by \(v^\sigma = v\circ \sigma^{-1}\) i.e. \(v^\sigma(\psi^\sigma) = v(\psi)\) for all \(\psi \in \mathcal M(S)\text{.}\)
  2. For \(P \in S\) define \(P^\sigma \in S^\sigma\) as the unique point in \(S^\sigma\) s.t. \(v_{P^\sigma} = (v_P)^\sigma\text{.}\)

Sketch

  1. \(a \mapsto a^{\sigma^{-1}}\text{.}\)
  2. Follows as in proof of 3.22
  3. Obvious for \(a \in \QQ\text{,}\) for \(\infty\text{:}\)
    \begin{equation*} (v_\infty)^\sigma (x-1) = v_\infty(x- a^{\sigma^{-1}}) = 1 = v_\infty(x-1) \end{equation*}
    for all \(a\in \CC\) implies \((v_\infty)^{\sigma^{-1}} = v_\infty\) implies \(\infty^\sigma = \infty\text{.}\)

Subsection 2.6.5 Elementary invariants of the action of \(\operatorname{Gal}(\CC)\text{.}\)

Remark 2.6.17.

The bijection \(S \leftrightarrow S^\sigma\) is not holomorphic. In general \(S\) and \(S^\sigma\) are not isomorphic.

We will use properties 1 and 4 at least.

1 implies 2: \(S =S_F\text{,}\) \(F = K\lb x,y\rb\) for \(K\) a number field then

\begin{equation*} | \{F^\sigma \}_{\sigma \in \operatorname{Gal}(\CC)} | \le [K : \QQ] \end{equation*}

2 implies 1 is section 3.7.

Proof of b implies a in Belyi's theorem (3.61).

Suppose \(f \colon S\to \PP^1\) is a morphism of degree \(d\) with branching values \(\{0,1,\infty\}\text{.}\) By theorem 3.28 \(\forall \sigma \in \operatorname{Gal}(\CC)\)

\begin{equation*} f^\sigma \colon S^\sigma \to \PP^1 \end{equation*}

is a morphism of degree \(d\) and branching values are

\begin{equation*} \{\sigma(0), \sigma(1), \sigma(\infty)\} = \{0,1,\infty\}\text{.} \end{equation*}

So \(\{f^\sigma\}_{\sigma \in\operatorname{Gal}(\CC)}\) gives rise to only finitely many monodromy homomorphisms.

\begin{equation*} F_{f^\sigma}\colon \pi_1(\PP^1 \smallsetminus \{0,1,\infty\}) \to \Sigma_d \end{equation*}

the fundamental group is free on two generators so there are only finitely many such maps. Theorem 2.61 implies \(\{S^\sigma\}_{\sigma \in \operatorname{Gal}(\CC)}\) contains only finitely many equivalence classes so by the criterion \(S\) is defined over \(\overline\QQ\text{.}\)

Subsection 2.6.6 The field of definition of Belyi functions (3.8)

Use the same methods as in 3.7.