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 \circ \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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    $\begingroup$ Apologies in advance, but I can't seem to prevent myself from asking if your name is intentionally an anagram of "smurf liver." $\endgroup$ Sep 27, 2011 at 17:55
  • $\begingroup$ Haha no, it is an abbreviation of " I love Riemann Surface " . $\endgroup$ Sep 27, 2011 at 18:19
  • $\begingroup$ A Riemann surfaces lover misspelling Ahlfors's name? I edited that (but kept other misspellings). $\endgroup$ Sep 27, 2011 at 23:30
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    $\begingroup$ 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. $\endgroup$ Sep 28, 2011 at 1:50
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    $\begingroup$ According to The Chicago Manual of Style, Ahlfors's is the proper way to spell: "The possessive of most singular nouns is formed by adding an apostrophe and an s. The possessive of plural nouns (except for a few irregular plurals, like children, that do not end in s) is formed by adding an apostrophe only." and "The general rule extends to proper nouns, including names ending in s, x, or z, in both their singular and plural forms, as well as letters and numbers." $\endgroup$ Feb 17, 2015 at 19:41

1 Answer 1

  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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