BUNTES Riemann Surfaces and Discrete Groups (Rod)

Section 2.4 Riemann Surfaces and Discrete Groups (Rod)

Welcome to BUGLES (Boston university geometry learning expository seminar), the reason it is called bugles is because bugles are hyperbolic, and today we will see a lot of hyperbolic objects.

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Plan

Uniformization
Fuchsian groups
Automorphisms of Riemann surfaces

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Proposition 2.4.1.

$\begin{equation*} \Aut(\hat \CC) = \{ z\mapsto \frac{az+b }{cz+d}\} \end{equation*}$

$\begin{equation*} \Aut(\CC) = \left\{z \mapsto za+b\right\} \end{equation*}$

$\begin{equation*} \Aut(\HH) = \{ z\mapsto \frac{az+b }{cz+d},a,b,c,d\in \RR \} = \PSL_2(\RR) \end{equation*}$

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Theorem 2.4.2.

$\Sigma$ has a universal cover $\widetilde \Sigma$ with $\pi_1 ( \Sigma) = 1\text{.}$ $\widetilde \Sigma \to \Sigma$ holomorphic. $\Sigma = \widetilde \Sigma /G$ for $G = \pi_1(\Sigma)\text{.}$ $G$ acts freely and properly discontinuously.

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Subsection 2.4.1 Uniformization

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Theorem 2.4.3.

The only simply connected Riemann surfaces are $\hat \CC\text{,}$ $\CC\text{,}$ $\HH\text{.}$

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Theorem 2.4.4.

$\Sigma$ is a Riemann surface then

$\begin{align*} g= 0 : \Sigma \amp\cong \hat \CC\\ g= 1 : \Sigma \amp\cong \CC/ \Lambda\\ g\ge2: \Sigma \amp\cong \HH/ K\text{.} \end{align*}$

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Proof.

$g = 0$ Uniformization.

$g \ge 1$ $\hat \CC$ can't be a cover by Riemann-Hurwitz. $g = 1$ $\pi_1 (\Sigma) = \ZZ \oplus \ZZ$ abelian.

Claim: no subgroup of $\Aut (\HH) $ is isomorphic to $\ZZ \oplus \ZZ$ acting freely and properly discontinuously. So $\widetilde\Sigma = \hat \CC$ $z\mapsto az+b$ free id $a=1$ so $z\mapsto z+\lambda_1$ $z\mapsto z+\lambda_2\text{.}$

$g= 2$ $\pi_1(\Sigma)$ is not abelian but $z\mapsto z + \lambda_1$ is abelian!

\begin{equation*} \Sigma = \HH/K,\, K \subseteq \PSL_2(\RR)\text{.} \end{equation*}

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Goal.

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Understand

$\Sigma$ through

$\widetilde \Sigma$ and

$G\text{.}$

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Fuchsian groups.

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$g \ge 2\text{.}$

$\begin{equation*} \Aut (\HH) = \PSL_2(\RR) = \operatorname{Isom}^+ ( \HH, \frac{|\diff z|^2}{\Im Z}) \end{equation*}$

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hyperbolic

$\HH, \mathbf D$ and

$\PSL_2(\RR)$ acts transitively on geodesics.

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Definition 2.4.5. Fuchsian groups.

A Fuchsian group is a discrete subgroup of $\PSL_2(\RR)\text{.}$

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Remark 2.4.6.

(proof in book) Even if $\Gamma$ doesn't act freely the quotient

$\begin{equation*} \HH \to \HH/\Gamma \end{equation*}$

is still a covering map and $\HH/\Gamma$ is a Riemann surface.

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Reflections on $\HH$ .

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Say

$\mu$ is a geodesic in

$\HH\text{,}$ i.e. a horocycle. There is

$M \in \PSL_2(\RR)$ with

$M\mu$ the imaginary axis. Then

$R = -\bar z$ is the reflection over the imaginary axis. Now

$R_\mu = M\inv\circ R \circ M$ is a reflection over

$\mu\text{.}$

$\begin{equation*} R_\mu = \frac{a\bar z + b}{ c\bar z + d}\not\in \PSL_2(\RR) \end{equation*}$

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this is a a problem for us.

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Triangle groups.

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Given

$n, m, l\in \ZZ \cup \{\infty\}$ then there is a hyperbolic triangle with angles

$\pi/n,\pi/m, \pi/l$ if

$\begin{equation*} \frac 1n + \frac 1m + \frac 1l \lt 1\text{.} \end{equation*}$

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With area

$\pi(1 - \frac 1n - \frac 1m - \frac 1l)\text{.}$

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In the disk model we can start with a wedge of the disk and by adding a choice third geodesic with endpoints on the edge we can adjust the other angles to be what we like.So we can construct hyperbolic triangles with whatever angles we like. Then let

$R_1$ be the reflection over 1 edge,

$R_2$ ,

$R_3$ similarly. By reflecting our original triangle

$T$ with these reflections we can tessellate the disk, colouring alternately the triangles obtained using an odd or even number of reflections.

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The only remaining problem is that

$R_i$ 's are not in

$\PSL_2(\RR)\text{.}$ The solution is to define

$x_1 = R_3 \circ R_1\text{,}$

$x_2 = R_1 \circ R_2\text{,}$

$x_3 = R_2 \circ R_3$ which are all in

$\PSL_2(\RR)$ now. Now we need to take the union of two adjacent triangles before as a fundamental domain, some quadrilateral that still tessellates. So we have formed a Fuchsian group from our triangles.

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A presentation for this group is

$\begin{equation*} \langle x_1, x_2, x_3 | x_1 ^n = x_2 ^ m = x_3 ^l = x_1x_2x_3 = 1\rangle \end{equation*}$

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note

$n,m,l$ can still be

$\infty\text{.}$

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Definition 2.4.7. Triangle groups.

Let $\Gamma_{n,m,l}$ be the triangle group with signature $(1/n, 1/m, 1/l)\text{.}$

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Remark 2.4.8.

$\begin{equation*} \frac 1n + \frac 1m + \frac 1l = 1 \end{equation*}$

$\begin{equation*} \frac 1n + \frac 1m + \frac 1l \gt 1 \end{equation*}$

still work on $\CC$ and $\hat \CC$ respectively.

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Example 2.4.9. $\PSL_2(\ZZ)$ .

Consider $\Gamma_{2,3,\infty}$ angles $\pi/2, \pi/3, 0\text{.}$ We can draw such a triangle in the upper half plane with vertices $i, e^{\pi i/3}, \infty\text{.}$ So a fundamental domain will be the region obtained by reflecting through the imaginary axis, given by $-\frac 12 \le \Re z \le \frac 12\text{,}$ $|z| \ge 1\text{.}$ We have $R_1 = \frac{1}{\bar z}, R_2 = -\bar z +1,R_3 = -\bar z$ so $x_1 = \frac{-1}{z}, x_2 = \frac{1}{-z+1}, x_3 = z+1\text{.}$ Then $\Gamma_{2,3,\infty} \cong \PSL_2(\ZZ)\text{.}$ Sometimes denoted $\Gamma(1)\text{.}$

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Observation 2.4.10.

If $\Gamma_1 \lt \Gamma_2$ and $T$ is a fundamental domain of $\Gamma_2$ then if $\gamma_1, \gamma_2, \ldots, \gamma_n \in \Gamma_2$ are representatives of $\Gamma_1\backslash \Gamma_2$ then

$\begin{equation*} \bigcup \gamma_i (T) \end{equation*}$

is a fundamental domain for $\Gamma_1\text{.}$

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Example 2.4.11. $\Gamma(1)$ .

$\begin{equation*} \Gamma(2) = \{\begin{pmatrix} a\amp b \\ c\amp d\end{pmatrix} = \id \pmod 2 \} \end{equation*}$

then

$\begin{equation*} [\Gamma(1) : \Gamma(2)] = 6 \end{equation*}$

representatives of $\Gamma(2) \backslash \Gamma(1)$ are

$\begin{equation*} x_1 = \id,\,x_2 = \frac{-1}{z-1}, x_3 = \frac{z-1}{z},\, x_4 = \frac{z-1}{z},\,x_5 = \frac{-z}{x-1},\,x_6 = \frac{-1}{z}\text{.} \end{equation*}$

Lets see what these do, for example if $z= e^{i\theta}$

$\begin{equation*} \Re(x_2(z) = \frac{-1}{e^{i\theta} - 1} = \frac{-e^{i\theta} + 1}{2- 2\cos \theta}) = \frac{1-\cos \theta}{ 2- 2\cos \theta} \frac 12 \end{equation*}$

if we plot this we see we get two copies of a 0,0,0 triangle so this corresponds to $\Gamma_{\infty,\infty,\infty}\text{.}$

$\begin{equation*} \langle x_1, x_2, x_3 | x_1x_2x_3 = 1\rangle = \langle x_1,x_2\rangle = \pi_1(\PP^1\smallsetminus\{0,1,\infty\})\text{.} \end{equation*}$

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