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.


  • 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?


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.
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    $\begingroup$ 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? $\endgroup$ Jun 27, 2014 at 9:35
  • $\begingroup$ @EmilJeřábek you are right; the current formulation is not clear. I will edit the question later today. $\endgroup$ Jun 27, 2014 at 9:58

1 Answer 1


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".

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    $\begingroup$ This trivial observation is already pointed out in the comments, and the OP suggested he’s going to clarify the question. $\endgroup$ Jun 27, 2014 at 13:21

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