Lets have a look at the two simply connected domains $D,G \subset \mathbb{C}$ and a biholomorphic function $f:D \rightarrow G$ which maps $D$ conformal onto $G$.

For some $n \in \mathbb{N}$ there are points $b_k \in B_n := \lbrace b_d \in D \mid d = 1,...,n \rbrace$ such that the delaunay triangulation $\Delta_n(B_n)$ approximates the domain $D$.

I know that the delaunay triangulation is not unique but lets map all those points $B_n$ onto $G$:

Let $f(B_n)$ bei the set of the correspondenting points in $G$. Lets connect all the points in $f(B_n)$ the same way we did in $\Delta_n(B_n)$. We get a triangulation $\Delta_n(f(B_n)) for $G$. I ask myself if this triangulation is a valid delaunay triangulation again.

In other words: keeps the validation of a delaunay triangulation stable under conformal mappings?


1 Answer 1


It's very unlikely that a Delaunay triangulation is stable under a conformal mapping. Try for instance the map $z \mapsto 1/z$ on a disc not containing the origin. Let $B=4$ be the vertices of a triangle and its barycenter.

One chance of preserving Delaunay triangulation would be if the mapping preserves circles. Perhaps it would be stable under a Möbius transformation, though I doubt it.

  • $\begingroup$ But $z \mapsto z^2$ is not a conformal mapping. $\endgroup$
    – Arji
    Nov 10, 2018 at 18:34
  • $\begingroup$ Yes. I should have mentioned that. Im considering conformal mappings to be biholomorphic funtions. $\endgroup$
    – Arji
    Nov 10, 2018 at 18:55

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