Restricting a quasi-conformal homeomorphism of the disc to the boundary gives a surjective homomorphism from $QC(D^2)$ (quasi-conformal homeos of $D^2$) to $QS(S^1)$ (quasi-symmetric homeos of the circle). Surjetivity here follows, for example, from the existence of natural extensions of QS homeos like Douady-Earle.  

The same restriction-to-the-boundary map from the group of *area preserving smooth diffeos* (symplectomorphisms if you will) to Diff$(S^1)$ is also surjective -- one way to see this is to construct a by-hand extension defined in a collar neighborhood and then use Moser's theorem [(here)](http://www.ams.org/journals/tran/1965-120-02/S0002-9947-1965-0182927-5/S0002-9947-1965-0182927-5.pdf), and details of perhaps another approach have been written up [here](http://link.springer.com/article/10.1007%2FBF01076419).  

My question is: what are the possible "boundary values" of area preserving QC homeos of the disc? My guess is that the map from area preserving QC homeos to $QS(S^1)$ is not surjective... perhaps someone with a good understanding of symplectic homeos has an idea of what the image is. 



quasi conformal, area preserving homomorphisms of the disc
Claim. A QS homeomorphism $f$ of the circle extends to a QC area preserving map of the disk if and only if $f$ is BL (bi-Lipschitz). 


Proof. 


1. Suppose that $f: S^1\to S^1$ is a QS map which admits an area-preserving quasiconformal extension $F: D^2\to D^2$. Then it immediately follows from the definition of quasiconformality that $F$ has to be (globally) bi-Lipschitz. In particular, $f$ was bi-Lipschitz to begin with.
2. Suppose now that $f$ is a BL homeomorphism of the unit circle. I claim that it extends to a BL area preserving homeomorphism $F$ of the disk. By looking at the BL flow from the identity to $f$, we see that the problem reduces to the case of Lipschitz vector fields $v$ tangent to the circle which we have to extend to Lipschitz divergence-free vector fields $V$ on the disk. By removing a point on the circle, we reduce the question to the case when $v=u'$, where $u$ is a $C^{1,1}$ function on the circle (maybe minus a point). Now, extend $u$ to a harmonic function $U$ on the disk. Then the gradient $V$ of $U$ has zero divergence and is Lipschitz. Furthermore, $V$ will be Lipschitz at the boundary (except maybe for one point), to it will extend to the remaining point as well. Qed


I was assuming here that $f$ is orientation-preserving. However, by composing with a symmetry of the disk, the general case reduces to this one, provided that you allow QC maps to reverse orientation. 


