1
$\begingroup$

The Riemann mapping theorem (cf e.g. http://en.wikipedia.org/wiki/Riemann_mapping_theorem) essentially guarantees the existence of a biholomorphic mapping of a simply connected, open subset of the complex plane onto the unit disk.

Questions:

  • Are there any results known about length-preserving mappings from simply connected, open subsets of the complex plane to non-congruent target-subsets of the complex plane?

  • What are the necessary and sufficient conditions, under which such mappings exist?

  • Are any analogues of the Schwarz-Christoffel mapping (cf e.g. http://en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping) known for such length-preserving mappings?

Edit:

Question under relaxed conditions:

  • are there non-isometric mappings between two simply connected regions with equal perimeter of finite length, which preserve the length of a finite collection of coordinate lines?
    A different way of stating the relaxed problem would be that the source-region is partitioned into a finite collection of simply connected sub-regions, whose interiors do not intersect and the question is, whether non-isometric mappings exist, for which the length of the boundary of every sub-region's image equals the length of the boundary of the corresponding original region.
$\endgroup$
  • 1
    $\begingroup$ It's not very clear to me what are you asking for. A length-preserving mapping is a congruence, so for instance, the answer to the second question is that they exist iff the domain is congruent to a subset of the target. What kind of nontrivial information about congruences do you expect to get? $\endgroup$ – Emil Jeřábek Jun 27 '14 at 9:35
  • $\begingroup$ @EmilJeřábek you are right; the current formulation is not clear. I will edit the question later today. $\endgroup$ – Manfred Weis Jun 27 '14 at 9:58
4
$\begingroup$

Any length preserving map from one plane domain to another must be also conformal. Because the angle of a small triangle can be found if you know the sides. This observation solves all your questions, because if a conformal map is also length preserving, than the derivative must have constant absolute value one, thus it is of the form $z\mapsto \lambda z+c$ where $|\lambda|=1$, so it is a "roto-translation".

$\endgroup$
  • 1
    $\begingroup$ This trivial observation is already pointed out in the comments, and the OP suggested he’s going to clarify the question. $\endgroup$ – Emil Jeřábek Jun 27 '14 at 13:21

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.