0
$\begingroup$

I'm having a brain freeze.

Let $B$ be the space of complex valued measurable functions on the unit disk in the complex plane with essential supremum less than 1. Then, the universal Teichmuller space $T$ can be though of as a quotient space of $B$.

I vaguely recall reading somewhere that for every point $p \in T$ has a unique representative of the form

$\mu(z) = k \frac{\bar{q}}{q}$, where $q$ is a holomorphic function on the unit disk.

First of all, am I crazy and am remembering something that's completely off?

Secondly, if there is a statement like that, where should I be looking?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

  1. Strebel discovered (in 1962) that Teichmüller's existence theorem fails for quasiconformal maps of the unit disk. Uniqueness fails too for points in $T(D^2)$ which do not have Teichmuller representatives. See Strebel's examples in the book by Gardiner and Lakic, "Quasiconformal Teichmüller Theory," p. 177-178.

  2. On the other hand, the uniqueness theorem in certain sense does hold (Theorem 5, Chapter 4 of the same book). More precisely, if a quasiconformal map $f: D^2\to D^2$ is a Teichmüller map, i.e., it has Beltrami differential of the Teichmüller form $t \bar{\phi}/|\phi|$ (where $\phi$ is holomorphic), then $f$ is the unique extremal quasiconformal map in its equivalence class $[f]\in T(D^2)$. In particular, $[f]$ contains no other Teichmüller maps.

  3. On third hand, Strebel proved in 1976 ("On the existence of extremal Teichmueller mappings") that for $[f]\in T(D^2)$ given by, say, smooth boundary values, the Teichmüller existence theorem does hold.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you for the reference. I'm actually looking at a beltrami differential of a very particular form: mu(z) = k \bar{q}/q, where q is the Koebe function precomposed with the map exp(2pi i z). (So it is a beltrami differential on a punctured disk.) Theorem 5 doesn't really apply in my case, but Chapter 9 of the reference you gave me (which contains the pages you gave) seems to be the right place to start. Again, thank you very much for your answer. $\endgroup$
    – BrainDead
    Commented Apr 3, 2012 at 2:32
  • $\begingroup$ @BrainDead: You are welcome. Judging by your description, your $q$ is holomorphic on the upper half-plane (composition of two holomorphic functions), so Theorem 5 would apply. Maybe the problem is that your quadratic differential is unbounded: I forgot if Theorem 5 has boundedness requirement on $q$. $\endgroup$
    – Misha
    Commented Apr 3, 2012 at 3:44
  • $\begingroup$ Right, by punctured disk, I meant the strip [0,1] x [0,infinity] in the upperhalf plane. And the problem is that $\phi = q^2$ does not have a finite $L^1$ norm. Theorem 3 in Chapter 9.3 (pg.183) seems to be exactly what I need. $\endgroup$
    – BrainDead
    Commented Apr 3, 2012 at 4:59
0
$\begingroup$

I think you probably referred to this fact: you can find for your point $p \in T$ a representative in the form of a Beltrami differential $\mu$ expressed as $\mu=k\frac{\bar{\phi}}{\vert \phi \vert}$, with $\phi$ a quadratic differential. But I doubt about the uniqueness, as you can always multiply $\phi$ by some positive constant...

Hubbard's book on Teichm\üller theory contains certainly all of that. Also, I think that Lehto's book has an entire chapter on Universal Teichm\üller space.

Improve this answer
$\endgroup$
2
  • $\begingroup$ @Sylvain: The question is about uniqueness of $\mu$ rather than uniqueness of $\phi$. Once you know $\mu$, the holomorphic quadratic differential $\phi$ is determined uniquely up to a constant positive multiple. This is the usual ambiguity in Teichmüller theory which is usually remedied by assuming that $\phi$ has unit norm (for some choice of the norm on the space of quadratic differentials). $\endgroup$
    – Misha
    Commented Mar 31, 2012 at 1:22
  • $\begingroup$ Thanks for the comment! together with your answers, the whole thing is much clearer now... $\endgroup$
    – Sylvain Bonnot
    Commented Apr 1, 2012 at 16:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .