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I have a question about Ahlfors's proof of modular function being a covering space of the twice punctured plane .See Ahlfors' complex analysis, second edition, page 272. You can either explain or suggest a better reference.

Let $\Omega$ be defined by the open domain in $\mathbb{H} $ bounded by the lines $\Re(\tau)=0,\Re(\tau)=1, $ and the circle $|z-1/2|=1/2 $. ( $\Re$ means real part. )

In Ahlfors's complex analysis, second edition, page 272, Ahlfors proves that the modular function $\lambda $ maps $\Omega $ ( which is the open right half of the fundamental domain of the congruence subgroup modulo 2 group $\Gamma(2)$ ) conformally onto the upper half plane $\mathbb{H} $ ( which, while combined with the fact $\lambda o \phi = \lambda \forall \phi \in \Gamma(2) $ and that $\Omega \cup \Omega^* $$\cup $ {positive y-axis} is a fundamental domain for $\Gamma(2)$,( $\Omega^*$ is the reflection of $\Omega $ in the positive y- axis ) and that $\lambda$ is surjective,proves that $\lambda$ is a covering space for $\mathbb{C}\backslash{0,1}$.

I have some questions regarding the proof of covering space :

  1. How exactly do we prove that $\lambda : \mathbb{H} \to \mathbb{C}\backslash \{0,1} $ is surjective ? I think this should follow from the my queston # 2.

  2. I am also unable to follow Ahlfors's argument on the first paragraph of P. 273, apparently which seems rather sketchy to me ( see P. 272 ) that $ \forall w_0\in \mathbb{H}, \frac{1}{2\pi i}\int_\Gamma \frac{\lambda'(\tau)}{\lambda(\tau)- \ w_0}d\tau = 1 $ and $ \forall w_0\in \mathbb{H^*}, $ ( the lower half plane )$ \frac{1}{2\pi i}\int_\Gamma \frac{\lambda'(\tau)}{\lambda(\tau) -\ w_0}d\tau = 0 $ , where $\Gamma$ is obtained by taking the boundary of the truncated region bounded by $\Re(\tau)= t_0 > 0,\Im (\tau) =0, \Im(\tau)=1$ where $t_0$ is sufficiently large, and two sufficiently small circles tangent to the x-axis at 0 and 1 respectively. This will prove, by argument principle, that $\lambda$ takes each $ w_0 \in \mathbb{H} $ exactly once in $\Omega$. But why are the integrals 1 and 0 in the two above cases ? Could you please explain that in more detail ? Or suggest a better reference that is easier to follow ?

Thanks very much !

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5  
Apologies in advance, but I can't seem to prevent myself from asking if your name is intentionally an anagram of "smurf liver." –  Cam McLeman Sep 27 '11 at 17:55
    
Haha no, it is an abbreviation of " I love Riemann Surface " . –  Analysis Now Sep 27 '11 at 18:19
    
A Riemann surfaces lover misspelling Ahlfors's name? I edited that (but kept other misspellings). –  Igor Belegradek Sep 27 '11 at 23:30
3  
I guess except for the first one ( which was a mistake ), the other spellings were correct if we follow the grammatical convention of not adding an extra s after the apostrophe (') when the name already ends with s. For example, I would write "Weyl's lemma " but not Ahlfors's lemma, rather Ahlfors' lemma, I would agree people might have different ways of writing the same thing. –  Analysis Now Sep 28 '11 at 1:50

1 Answer 1

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  1. Ahlfors explains that $\lambda(\Omega) = \mathbb H$, $\lambda(\Omega^\ast) = \mathbb H^\ast$ and $\lambda(\overline{\Omega} \backslash \Omega) = \mathbb R\backslash \{0,1\}$ (here $\overline{\Omega}$ is the closure of $\Omega$ in $\mathbb H$). Thus $\lambda$ maps $\overline{\Omega} \cup \Omega^\ast$, which is a fundamental domain for $\Gamma(2)$, onto $\mathbb C \backslash \{0,1\}$.

  2. The point is this: $\lambda$ maps $\overline{\Omega}$ into $\{y \geq 0\}$. So the image of the truncated region (i.e. what you call $\Gamma$) will roughly look like a semicircle in $\mathbb H$ with its base on the real line but with two semicircular "dents". (These "dents" will be above the points $0$ and $1$.) Thus, if we choose $t_0$ appropriately, $\lambda(\Gamma)$ will contain a nonreal $w_0$ in its interior iff $w_0 \in \mathbb H$, and in this case $\lambda(\Gamma)$ will wind around $w_0$ once, as you can verify.

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