0
$\begingroup$

The Riemann mapping theorem states that given any two simply connected open domains $A$ and $B$ of $\mathbb C$ that are neither empty nor equal to $\mathbb C$, there exists a unique (up to normalization) conformal mapping from one to the other. In higher dimensions of space Liouville's theorem states that the conformal mappings are only the Möbius transforms. One can loosen the requirement of being conformal but the exemple of Whitehead manifold shows that the topological properties shared by $A$ and $B$ would have to be extended beyond contractibility. These thoughts lead me to the following question

For $n\gt2$, is there a class of mapping $\mathscr M$ such that for any two homeomorphic simply connected open subsets $A$ and $B$ of $\mathbb R^n$ that are neither empty nor equal to $\mathbb R^n$ there is a unique map $f\in\mathscr M$ verifying $f(A)=B$ ?

Note. This question actually came up while I was considering $A(f)=\left\{(x,y,z)\in\mathbb R^3, z\gt f(x,y)\right\}$ and $B=\{(x,y,z)\in\mathbb R^3, z\gt0\}$.

$\endgroup$
  • 2
    $\begingroup$ The result that you mention in the first sentence is due to Riemann, and the map is not unique, unless normalized correctly. $\endgroup$ – Malik Younsi Jun 3 '15 at 9:39
  • $\begingroup$ What do you mean by "homeomorphic ... and share topological invariants"? $\endgroup$ – Alex Degtyarev Jun 3 '15 at 9:47
  • $\begingroup$ @AlexDegtyarev. You are right, this is quite redundant. I will edit my question. Thanks. $\endgroup$ – Tom-Tom Jun 3 '15 at 9:59
  • $\begingroup$ @MalikYounsi. I have edited the question according to your remarks. Thanks $\endgroup$ – Tom-Tom Jun 3 '15 at 10:01
  • $\begingroup$ The uniqueness part of your question makes it very unlikely to have a reasonable answer. Taking for $A=B$ a ball, you see that the class in question should not contain any rigid motion other than identity. But I can hardly imagine a reasonable class of maps that exclude rigid motions. $\endgroup$ – Alex Degtyarev Jun 3 '15 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.