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Let $X$ be the space obtained by gluing together two congruent equilateral triangles along corresponding edges.

Note that $X$ has the structure of a Riemannian manifold except at the three cone points. In particular, $X$ is a Riemannian orbifold.

Is there an isometric embedding of $X$ into $\mathbb{R}^3$?

Jim Belk
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2 Answers2

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If you want the image to be convex, then no. Otherwise, my guess is that Kuiper's theorem gives you an embedding...

Igor Rivin
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  • Sorry, which Kuiper's theorem? (It's fine if the image isn't convex.) – Jim Belk Jan 08 '14 at 21:15
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    Kuiper's theorem which says that a Riemannian manifold admits a $C^1$ isometric embedding in its topological embedding dimension. I conjecture is that the proof can be modified for the cone-manifold case. Of course, is $C^1$ enough for you? – Igor Rivin Jan 08 '14 at 21:17
  • I see, thanks. I'll have to look into that. By the way, how do you prove that it can't be embedded convexly? – Jim Belk Jan 08 '14 at 22:49
  • @JimBelk I think the cauchy uniqueness theorem applies to this degenerate case, so if there is a "doubled triangle" convex realization (which there is...), there is no other. – Igor Rivin Jan 08 '14 at 23:28
  • @IgorRivin, which Cauchy uniqueness theorem are you referring to? The theorem on convex imbeddings that I am familiar with applies in the smooth case. – Mikhail Katz Jan 19 '14 at 19:31
  • @user72694 the one Cauchy proved was for polyhedra. The smooth uniqueness theorems are due to a number of people, but the definitive result is due to Pogorelov, and covers all convex surfaces regardless of regularity. – Igor Rivin Jan 19 '14 at 22:32
  • @IgorRivin, thanks. But here we don't really have a polyhedron, or perhaps only a degenerate one. Does Cauchy's theorem cover such a case? – Mikhail Katz Jan 20 '14 at 12:26
  • @user72694 Yes, it does. – Igor Rivin Jan 21 '14 at 05:48
  • @IgorRivin, thanks for the information. Can you elaborate a bit more on this? Surely Cauchy himself did not have the notion of an abstract orbifold, so he could not have envisioned a convex polyhedron that does not imbed. Are you saying that Cauchy's proof applies also in this case? Where is this discussed? – Mikhail Katz Jan 21 '14 at 11:57
  • @user72694 Cauchy could certainly envision a doubled triangle, which is what we are talking about here, but he certainly did not consider this particularly interesting. You can just go through Cauchy's proof, and see that it works (well, Cauchy's proof is actually incorrect, but the corrected version). You can find proofs in a number of places, of which my favorite is Alexandrov's "convex polyhedra". – Igor Rivin Jan 21 '14 at 16:39
  • @IgorRivin, it's hard to get into Cauchy's mind, but I would conjecture that it is equally likely that he could imagine a doubled triangle and that he gave an epsilon, delta definition of continuity :-) – Mikhail Katz Jan 22 '14 at 16:33
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The common boundary of the two triangles is a curve in the plane whose interior is isometric to each of the 2-simplices making up your orbifold $X$. By the solution of Is an isometric embedding of a disk determined by the boundary?, the imbedding of the boundary extends in a unique way to an isometric imbedding of its interior. Hence the images of both 2-simplices must be identical. In particular, any isometric map $X\to\mathbb{R}^3$ cannot be an imbedding.

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Mikhail Katz
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