I have a right-angled square pyramid, $A$, whose height and base-length are $l$. By 'right-angled', I mean that the apex of $A$ lies vertically above one of the vertices in its base. Now supposed I form a new polyhedron, $B$, by gluing a cube with side-length $l$ to the base of $A$ ($B$ now has a base-length $l$ and height $2l$). Additionally, suppose that I have a new pyramid $A^{'}$ that is similar to $A$, but with a height and base-length of $2l$.

Does anyone know how to derive a conformal map from the interior of $B$ to that of $A^{'}$?

P.s. I have also cross-posted this on MathStackExchange: https://math.stackexchange.com/questions/3503921/conformal-map-from-a-7-sided-polyhedron-to-a-square-pyramid

  • 2
    $\begingroup$ Duplication of math.stackexchange.com/questions/3503921/… $\endgroup$
    – user64494
    Jan 11, 2020 at 9:45
  • $\begingroup$ I figured I was going to get a different calibre of responses on MO as opposed to MSE. Is it not common to post the same question on both? Sorry I'm relatively new to SE. $\endgroup$
    – niran90
    Jan 11, 2020 at 9:50
  • 1
    $\begingroup$ Your conformal map is just the similarity. $\endgroup$ Jan 11, 2020 at 13:46
  • $\begingroup$ niran90, I would delete the duplicate at math.se since it's been answered here. Then, if I were you, I would post the smooth bijection question over there first, and wait some days for a response. If there is none, you can try asking here (although it might easily be closed as off-topic for this site). $\endgroup$
    – Todd Trimble
    Jan 11, 2020 at 14:16

1 Answer 1


No such conformal map exists.

Conformal mapping in dimensions above 2 is very different from conformal mapping in dimension 2. In dimensions above 2, any conformal mapping is a (finite) composition of rigid motions, dilations, and inversions. In particular, such a mapping carries planes and spheres to planes and spheres and preserves the intersection angles between them. Your pyramid has 5 boundary planes, and so any conformal image of it will have 5 faces that are parts of either planes or spheres. In particular, the image cannot look like the pyramid+box that you describe as your polyhedron $B$.

  • $\begingroup$ Okay I see. Suppose I relax the requirement for conformality, and simply require that the mapping is smooth and bijectve, would that be feasible? I imagine there are infinitely many such mappings. But what's the most straightforward way to derive one of them? Actually for my specific application, conformality is not crucial. I just thought it would be a sufficient condition for a smooth bijective map. $\endgroup$
    – niran90
    Jan 11, 2020 at 9:41
  • $\begingroup$ Actually, I assumed that you wanted to map the interior of $B$ to the interior of $A'$, and that's what my answer addressed. Did you instead want to map the actual polyhedrons (i.e., the surfaces) conformally? $\endgroup$ Jan 11, 2020 at 19:15
  • $\begingroup$ No, you assumed correctly - I want to map the interiors to each other. I've now clarified the ambiguity in the post. Thanks. $\endgroup$
    – niran90
    Jan 12, 2020 at 5:27

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