# Can you perturb an inscribed polytope so all its edges grow?

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?

• ...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? May 11 at 17:59
• 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). May 14 at 13:31
• 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. ... May 14 at 15:10
• ... 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. May 14 at 15:13
• Are you interested in partial results? I have a proof for simplices, but this seems to be known. May 17 at 12:29

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.

• 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. May 18 at 17:07
• 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. May 19 at 0:27
• @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. May 19 at 6:48
• @IlyaBogdanov That is a neat little argument. Thanks for taking the time. May 19 at 10:23