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Hello,

I am afraid that my main question might be a bit too elementary, but still I ask :

In short, my question is "what is the version of Montel's theorem for a family of holomorphic maps from an open subset $U$ to a hyperbolic Riemann surface $X$ whose range does not lie in a single co-ordinate chart of X ?" Here is how I got to think about the version :

I was reading up the proof of jenkins-Strebel quadratic differential of prescribed heights $b_i$ on a Riemann surface $X$ from kurt Strebel's book " Quadratic Differentials ", Chapter VI, section 21, page 108. Let us, for simplicity, use the version of the theorem in the case of one single Jordan curve $\gamma$. So, in the beginning of the proof, they consider a sequence of ring domains $R_n \subset X $ of homotopy type $\gamma $ with moduli $M_n$ such that $M_n \to M = sup_{n \ge 1 }M_n, 0 < M < \infty $. Then they consider a sequence of conformal maps $g_n : A_n \to R_n $, where $ A_n= { z \in \mathbb{C} : r_n < |z| < 1 } $, where $r_n \to 0 $.

Then they claim that : $ g_n : A_n \to R_n\subset X $ form a normal family. I understand that somehow they are trying to use Montels theorem from complex analysis : a family of holomorphic functions on a domain $U$ in $\mathbb{C}$ which omits two points is normal. But how exactly does this theorem apply to our present case, where the codomain of $g_n$ is a Riemann surface, not a subset of $\mathbb{C}$, so that we can NOT even talk about the absolute value symbol $ | g_n - g_m | $ , so what is the meaning of uniform convergence of a family of holomorphic maps which take values into a Riemann surface $X$ ? And what is the meaning of normality ?

I was thinking of ideas like lifting the maps $g_n $ first to the universal cover $D$ of $X$ with the normalization $g_n(p) =0 \forall n $ for some fixed $p \in A_1 $. Then we can apply Momtels's theorem to the lifted maps since all of them are missing three points, the co-domain being $D$, but then I needed to prove that the projection map $\pi: D\to X $ sends normal family to the normal family, which I was unable to prove.

This is my main question. A detailed explanations would be appreciated ! Thanks !

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The assumption is that the modulus $M$ is bounded, so what I think Strebel is doing is applying the theorem that a uniformly bounded family of analytic functions is normal ($=$ has a uniformly convergent subsequence). The Riemann surface is endowed with a natural metrics and analytic structure. If you regard the composition of the conformal mapping $g_n$ with a local analytic chart $\varphi$, it becomes a complex-valued analytic function $ \varphi \circ g_n$, so you can apply to it standard theorems from Complex Analysis. Also the notion $|g_n-g|$ makes a perfect sense.

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  • $\begingroup$ Yes, this shows how to apply the right theorem. There are two necessary conditions for normality of families of holomorphic maps that were proved by Montel and both are referred to as "Montel's theorem". One says that a uniformly bounded family of holomorphic functions on an open subset of the complex plane is normal. The other (which can be obtained as a corollary) says that a family of holomorphic functions all of which omit the same two values $a,b \in \mathbb{C}$ is normal. $\endgroup$
    – Margaret Friedland
    Commented Jan 31, 2012 at 15:42
  • $\begingroup$ ...But I am not assuming that the range of all $g_n$ lie in a single co-ordinate chart of $X$ ( in which case your argument goes through ), and, if I don't assume that, the composition with $\phi$ just gives me that the family/sequence is locally uniformly bounded, which is not the hypothesis for getting a locally uniformly convergent subsequence . $\endgroup$
    – Analysis Now
    Commented Jan 31, 2012 at 15:42
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    $\begingroup$ @Hypgeo1: I don't think you need the chart. There's a general version of Montel's theorem for maps between Riemann surfaces. See the Section on Montel's Theorem in Milnor's "Dynamics in one complex variable" or Theorem 3.4 here: bit.ly/xjXYps $\endgroup$
    – Autumn Kent
    Commented Jan 31, 2012 at 16:46
  • $\begingroup$ @Richard Kent : that really does help, thank you ! $\endgroup$
    – Analysis Now
    Commented Jan 31, 2012 at 18:49
  • $\begingroup$ @Hypgeo1: No problem. $\endgroup$
    – Autumn Kent
    Commented Jan 31, 2012 at 20:34

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