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Consider the family of convex simplicial polytopes with vertices in the unit sphere of $\mathbb{R}^n$ which have the origin as an interior point.

My question is the following:

Let $P, P'$ be two non-congruent combinatorially identical polytopes from the above family, with vertices of each polytope labelled so that the face lattices of the two polytopes are identical. Is it possible that $\|a-b\|\leq\|a'-b'\|$ whenever edge $\{a,b\}$ in $P$ corresponds to edge $\{a',b'\}$ in $P'$?

In other words, can you perturb one such polytope to another making all the edges grow?

This is impossible in $\mathbb{R}^2$: Take an inscribed polygon from the family and perturb it so that it remains in the family and has the same combinatorial structure. If all the edges grow, then all the central angles grow, which would make those central angles add up to a value larger than $2\pi$. Therefore in a perturbed polygon, if some edges grow then other edges must contract.

Does this idea transfer to dimension $n=3$, or more generally to $n\geq3$? And, if yes, is there a reference?

Miek Messerschmidt
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  • I think you're right. The title is a bit clumsy. Thank you for the suggestion. I'll make the edit. – Miek Messerschmidt May 11 '22 at 15:47
  • I'm glad you revised the title, and you can see I modified the body so that the short question and the precise question will have the same answer rather than opposite answers. –  May 11 '22 at 16:47
  • ...minor edit to clarify that not all edges need to increase in length. – Miek Messerschmidt May 11 '22 at 17:07
  • Do you know the answer if you consider the areas of the faces instead of the length of the edges? – Del May 11 '22 at 17:19
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    ...I thought about that for n=3. Blowing up the triangular faces to triangles on the sphere, their area has to add up to 4pi. If the areas of the some of the corresponding spherical triangles change in the perturbation, then some must increase and some must decrease in area. Firstly, does non-congruency imply that a spherical triangle has to change area? Further, it is not true in general that decreasing of the area of a (spherical) triangle implies that an edge must decrease in length. Is this forced in this situation? – Miek Messerschmidt May 11 '22 at 17:59
  • Nice question, which might be hard. One thing that seems worth trying to extend the 2D idea is to consider (say in dimension 3) the surface areas of the central projection of the faces of the polytope. These are related to the length of (projection of) edges and their pairwise angles. Using that the angles sum to $2\pi$, it might be possible to prove that if all edges where growing, the sum of the surface areas would itself grow. – Benoît Kloeckner May 14 '22 at 08:41
  • Yes, I thought a bit in this direction too. Since the diheral angles at any vertex must always add to $2\pi$, there must exist a vertex where at least one dihedral angle increases and at least one decreases (else the AAA congruence for spherical triangles would imply that the polyhedra are congruent). Still, the areas the spherical triangle can increase or decrease in this situation... – Miek Messerschmidt May 14 '22 at 13:24
  • ...what I am currently thinking is that this may be a purely combinatoric problem. For $n=3$, consider the set of labels $L:={+,-,0}$. Call "tiles" all the possible ways to label a spherical triangle by placing a label from $L$ on the face, on each of the edges and each of the angles. The labelling is meant to represent whether, in the perturbation of a polyhedron, the "area of the face"/"edge length"/"dihedral angle magnitude" either increases/decreases/remains the same. Some tiles are disqualified, for example, by Girard's theorem, "+" on the face and "-" on all the angles (or vice versa). – Miek Messerschmidt May 14 '22 at 13:25
  • ...There are some rules to assigning tiles to the spherical triangular faces of a polyhedron: If there is a "+" assigned to a face then some other face must have a "-" assigned (since areas always must add to $4\pi$). Similarly, if a dihedral angle at a vertex has a "+" assigned, then another angle at the same vertex must have a "-" assigned (Since dihedrals angles must add to $2\pi$). Matching edges must have the same label ... – Miek Messerschmidt May 14 '22 at 13:25
  • ... We'd be done it can be shown that it is impossible to assign such "admissible tiles" to triangles in a planar triangulated graph, satisfying the stated rules with all the edges labeled with only 0's or +'s. – Miek Messerschmidt May 14 '22 at 13:25
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    Convexity is clearly important, even in dimension 3. Otherwise one can start with two pyramids from North and South poles, connected together by edges in the middle and then rotate one of the pyramids (while keeping the edge structure). – Lev Borisov May 14 '22 at 13:31
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    I asked a very similar (though more general) question some months ago. I also gave a talk on this problem. As you can see from the answers to my question, the more general conjecture (the one that is also in my talk) is false, but I strongly believe that it holds in your case (i.e. you cannot perturb the polytope in this way). I also believe that you do not need to restrict to simplicial polytopes. ... – M. Winter May 14 '22 at 15:10
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    ... I can prove you a result in the same spirit if you allow me to replace "circumradius" by a different measure of polytope size (see slide 11). Proving this requires results from spectral graph theory and convex geometry, which gives an idea for what tools might eventually be necessary. At least I can derive that the answer to your question is affirmative (i.e. such a perturbation does not exist) if the polytopes are centrally symmetric or otherwise very symmetric. Let me know if anything of this is relevant to you. I think this is a very interesting but also hard problem. – M. Winter May 14 '22 at 15:13
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    Are you interested in partial results? I have a proof for simplices, but this seems to be known. – Ilya Bogdanov May 17 '22 at 12:29
  • @ Ilya Bogdanov: I would be interested in the proof. I can show this only if every $n$-simplex that contains the circumcenter has a $n-1$ face that contains its circumcenter, but could not prove this yet. – Karl Fabian May 17 '22 at 16:54
  • @ Ilya Bogdanov: Partial results or their references will definitely be helpful. – Miek Messerschmidt May 18 '22 at 07:14
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    @KarlFabian For every point inside a convex polytope, its projection to the closest hyperplane containing a facet belongs to the facet. However, I will post my proof soon; it is different, as it does not rely on that fact. – Ilya Bogdanov May 18 '22 at 09:27
  • @ Ilya Bogdanov: This is a nice observation. Thanks for the proof. – Karl Fabian May 18 '22 at 18:30
  • We mentioned this problem in a dual and more general form here https://arxiv.org/abs/1505.05263 (the very end of the paper). – Arseniy Akopyan Nov 26 '22 at 20:32

1 Answers1

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OK, let me address the case of a simplex.

In fact, it follows from the `dual Kneser--Poulsen'conjecture, as stated, e.g., in this nice paper. A good thing is that the simplex has only $n+1$ vertices, and in such partial case, according to that paper, the conjecture has been established by Gromov. To reach the desired result, apply the conjecture to the balls of unit radius centered at the vertices of both simplices. (One still needs to check that the intersection strictly changes, but that is doable).

Anyway, here is a direct proof.

Let $u_0,\dots,u_n$ be the vertices of the first simplex (hence unit vectors), and let $v_0,\dots,v_n$ be the vertices of the second one, with $\|u_i-u_j\|\leq\|v_i-v_j\|$ for all $i$ and $j$, where at least one inequality is strict.

Choose the positive $\alpha_i$ such that $$ \sum_i\alpha_i=1 \quad\text{and}\quad \sum_i \alpha_iu_i=0; $$ these are the barycentric coordinates of $0$ in the first simplex. Take the point $p$ with the same barycentric coordinates in the second, i.e., $$ p=\sum_i \alpha_iv_i. $$ Clearly, $p$ lies inside the second simplex.

Notice that $\langle v_i,v_j\rangle\leq\langle u_i,u_j\rangle$. Therefore, for all $j$ we have $$ \langle v_j,v_j-p\rangle =-\sum_{i\neq j}\alpha_i\langle v_j,v_i\rangle +(1-\alpha_j)\langle v_j,v_j\rangle \geq -\sum_{i\neq j}\alpha_i\langle u_j,u_i\rangle +(1-\alpha_j)\langle u_j,u_j\rangle =\langle u_j,u_j\rangle=1, $$ where sometimes an inequality is strict. Hence $\|v_j-p\|\geq 1$, with sometimes strict inequality; in particular, $p\neq 0$.

This shows that all the $v_i$ lie on the unit sphere centered at $0$, but outside the (open) unit ball centered at $p$. This shows that they all (along with $0$( are on the same side of the hyperplane equidistant from $0$ and $p$. Hence the second simplex cannot contain $p$ --- a contradiction.

Ilya Bogdanov
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  • You could also just consider $|p|^2$, which turns out to be less than zero from the same inequalities for scalar products. The difficulty is exactly that not all distances need to grow, though the upside is that with more vertices we have way more freedom in constructing the linear combinations. – fedja May 18 '22 at 17:07
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    Your proof reminds me very much of the one here under "1B The miraculous bit" (which I found because all this reminded me of the Kirszbraun Theorem). If the similarity is not just superficial (there are some differences in the details), then your argument probably applies to 2-neighborly polytopes, i.e. the essential point is not that $P$ is a simplex, but that each pair of vertices forms an edge. – M. Winter May 19 '22 at 00:27
  • @MWinter: You are perfectly right, the similarity is not superficial! In fact, what I prove is a version of the same Lemma, and literally the same works for 2-intersecting polytopes. All you need is to find the linear combination all whose coefficients are positive, which is easy. – Ilya Bogdanov May 19 '22 at 06:48
  • @IlyaBogdanov That is a neat little argument. Thanks for taking the time. – Miek Messerschmidt May 19 '22 at 10:23