-3
$\begingroup$

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

$\endgroup$
4
  • 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

7
$\begingroup$

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 $B$.

$\endgroup$
3
  • $\begingroup$ 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

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.