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Let $H$ be the Heisenberg group with left invariant sub-Riemannian metric and $\varepsilon>0$ is small. Let us denote by $|x-y|_H$ the distance from $x$ to $y$ in $H$.

I have a bi-Lipschitz homeomorphism $f\colon H\to H$ with displacement at most $\varepsilon$; that is $|f(x)-x|_H\le \varepsilon$ for any $x\in H$.

Given a point $p\in H$, I need to construct a bi-Lipschitz map $g\colon H\to H$ which coincide with $f$ in $B_1(p)$ and coinsides with identity in $H\backslash B_2(p)$.

Please help.

Comments.

  • The space $H$ is nice; it is homogenous and scale invariant like $\mathbb R^3$.
  • For $\mathbb R^3$ Sullivan's gluing theorem produces the needed map $g$. His construction is using a discrete cocompact group action on open ball $B^3\subset\mathbb R^3$ by conformal transformations. Maybe his proof will work if there would an open subset in $H$ which admits a discrete cocompact group action by conformal transformations.
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  • $\begingroup$ Anton: I think you should ask somebody like Eliashberg this question in the smooth setting (when you use smooth contactomorphisms instead of BL maps for CC-metrics). $\endgroup$
    – Misha
    Nov 24, 2015 at 18:25

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More of a comment than an answer. This is probably a very hard question in general.

One approach is a technique I learned from Jeremy Tyson, based on the Koranyi-Reimann "Foundations" paper: http://www.ams.org/mathscinet-getitem?mr=1317384 . If you have a specific bi-Lipschitz mapping in mind, you could see if it can be embedded in a contact flow, i.e., is it the time-1 map associated to a vector field of the form $V_{\phi}=\phi T + (Y\phi X - X \phi Y)/4$ where $\phi \colon H \to \mathbb{R}$ is sufficiently smooth. Here X,Y,T are the standard left-invariant vector fields on $H$. Multiplying the potential $\phi$ by a cut-off function $\chi$ and integrating the resulting vector field $V_{\phi \chi}$ would yield a new mapping as you desire - the bilipschitz property for the new mapping can be quantitatively checked by verifying that all second derivatives of $\phi \chi$ are uniformly bounded (just checking that $Z\overline{Z} \phi \chi$ is bounded would yield that the corresponding map is quasiconformal).

Of course, it's very hard to check that a given map can be embedded in a flow. There are plenty of (hard) open questions along these lines.

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