In Ahlfors' complex analysis text, page 281 it says

By reflection the region $\Omega'$ that is symmetric to $\Omega$ with respect to the imaginary axis is mapped onto the lower half plane, and thus both regions together correspond to the whole plane, except for the points $0$ and $1$.

Here $\Omega$ is the region bounded by the lines $\Re \tau=0,\Re \tau=1$ and the circular arc of $|\tau-1/2|=1/2$ in the upper half plane. By "correspondence" he means under the modular $\lambda$ function.

Now, both regions together is IMHO the region $\Omega \cup \Omega'$ along with the positive imaginary axis. I don't think that real values are taken on this region except those in the open interval $(0,1)$. Thus, in my opinion both regions together should correspond to $$\mathbb C \setminus \left[ (-\infty,0] \cup [1, +\infty) \right] $$ rather than $$\mathbb C \setminus \{0,1 \} $$ as the author claims.

Am I right? If not, please tell me why.



You are right, but Ahlfors is also right.

It's caused by a not perfectly clear formulation. You are (naturally) thinking of the "region" $\Omega \cup \Omega' \cup i\cdot (0,\infty)$, understanding "region" as "connected open set". In this part of the theory, however, Ahlfors considers the "fundamental region" of $\lambda$, which is the set $\overline{\Omega} \cup \Omega'$, as witnessed by

Theorem 8. Every point $\tau$ in the upper half plane is equivalent under the congruence subgroup $\mod 2$ to exactly one point in $\overline{\Omega}\cup \Omega'$.

That fundamental region (to add further confusion, the closure $\overline{\Omega}$ should be understood as the closure in the upper half plane, so $0,1 \notin\overline{\Omega}$) is mapped bijectively to $\mathbb{C}\setminus \{0,1\}$ by $\lambda$.

In that view, "both regions together correspond to the whole plane, except for the points $0$ and $1$."

  • $\begingroup$ as always, thanks Daniel. $\endgroup$ – user1337 Nov 24 '13 at 21:13

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