# Minimal data required to determine a convex polytope

Let $$P\subset \Bbb R^d$$ be a convex polytope. Suppose that I know

1. its combinatorial type (aka. the face-lattice),
2. the length $$\ell_i$$ of each edge, and
3. the distance $$r_i$$ of each vertex from the origin.

Question: Does this already determine $$P$$ (up to orthogonal transformation)?

This is the case if all $$\ell_i$$ are the same, and all $$r_i$$ are the same (see this question). But what if they are not the same? What if I do not know the combinatorial type but only the edge-graph?

Update

I am not sure whether the formulation of my question was too vague, so below I added a second equivalent version of what I am asking:

Given two combinatorailly equivalent polytopes $$P_1,P_2\subset\Bbb R^d$$, and a corresponding face-lattice isomorphism $$\phi:\mathcal F(P_1)\to\mathcal F(P_2)$$. Now suppose that each edge $$e\in\mathcal F_1(P_2)$$ has the same length as $$\phi(e)\in\mathcal F_1(P_2)$$, and that each vertex $$v\in\mathcal F_0(P_1)$$ has the same distance from the origin as $$\phi(v)\in\mathcal F_0(P_2)$$. Is it then true that $$P_1$$ and $$P_2$$ are congruent (related by an orthogonal transformation)?

• Do you assume the origin is inside of the polytope? – Sam Hopkins Sep 19 '20 at 23:16
• Why doesn't the same proof as in the question you mentioned apply? Base for $d\le 2$ is still true: in any dimension the $2$-face and the origin generate $3$-dimensional pyramid with fixed edge lengths. It should be rigid by Cauchy lemma about sign changes. – Joseph Gordon Sep 19 '20 at 23:22
• @SamHopkins Does this make a difference? But yes, we can assume that it is inside. – M. Winter Sep 20 '20 at 7:51
• @JosephGordon I can imagine that the same technique applies if all the $r_i$ are the same, because then you know that the facets are inscribed and you know all edge lengths. But if the $r_i$ of $P$ are distinct, then "the $r_i$ of a facet" are not predetermined uniquely (up to some common factor or so), at least not obviously. So the shape of a facet might not be uniquely determined. – M. Winter Sep 20 '20 at 8:20
• @JosephGordon What is Cauchy's lemma about sign changes? I know Cauchy's rigidity theorem which needs the shape of facets rather than 2-faces. – M. Winter Sep 20 '20 at 8:37

Multiple polytopes can have the same data, as pictured below.

Take a pyramidal frustum, and twist it slightly clockwise or slightly counterclockwise. Make one polytope by gluing two identical versions, and make another polytope by gluing two opposite versions.

These will have the same combinatorial type, the same edge lengths, and the same distances from the origin to the edges, but they are not orthogonally equivalent.

The images show polytopes with vertices at \begin{align} &(\cos (k+\frac15)\alpha,&\sin (k+\frac15)\alpha, &\ \ \ \ +1)\\ &(\ \ \ \ 3\cos k\alpha,&3\sin k\alpha,\ \ \ \ &\ \ \ \ \ \ \ \ \ 0)\\ &(\cos (k\pm\frac15)\alpha,&\sin (k\pm\frac15)\alpha, &\ \ \ \ -1) \end{align} with $$\alpha=\pi/2$$, and $$+$$ for one polytope, $$-$$ for the other.

• But wait, why isn't this also a counterexample to Cauchy's rigidity theorem? – M. Winter Sep 22 '20 at 6:03
• Okay, this is not a counterexample to Cauchy's theorem, but it points to a subtlety that is apparently often overlooked when stating it: it does not suffice to state that the two polytopes in question have congruent faces, but the faces must also have identical interior angles at the respective vertices. Neither Wikipedia, nor "Proofs from the Book" states this. It is briefly mentioned in Igor Pak's book before the theorem. Moreover, in this light, my previous question seems unanswered again. – M. Winter Sep 22 '20 at 8:04
• @M.Winter Does one need that much for this example? I believe in it already edge lengths do not match: say, you have a triangle $ABC$ in the first polytope whose corresponding triangle $A'B'C'$ in the second one has $|AC|=|A'C'|$, $|AB|=|B'C'|$ and $|BC|=|A'B'|$ but not $|AB|=|A'B'|$ and not $|BC|=|B'C'|$ as it should be. – მამუკა ჯიბლაძე Sep 22 '20 at 12:52
• @მამუკაჯიბლაძე Yes, non-matching edge lengths are also enough to see that Cauchy's theorem cannot be applied, but this condition too is rarely mentioned. I mentioned interior angles because those where primarily used in those proofs of Cauchy's theorem that I have seen. – M. Winter Sep 22 '20 at 13:27
• I just realized the following, and correct me if I am wrong: there is no face-lattice isomorphism $\phi:\mathcal F(P_1)\to\mathcal F(P_2)$ between your two polytopes that maps edges in $P_1$ to edges in $P_2$ of the same length. So actually this example fails to match the requirement that I know the length of each specific edge in the combinatorial type (maybe my question was to imprecise in this regard). – M. Winter Oct 11 '20 at 13:39

Not an answer, just an additional illustration for the existing answer by @MattF. (which I find exhaustive).

A 2-dimensional counterexample

In the image below, the white dot represents the origin and is located outside the polygon. And it has to be: if the origin were inside, the shape would be unique, as shown here.

One can imagine to build higher dimensional counterexamples from this, e.g. prisms over these shapes.

Proof of a special case

Suppose that for each 2-face $$\sigma\in\mathcal F(P)$$ the (perpendicular) projection of the origin onto $$\mathrm{aff}(\sigma)$$ ends up in the relative interior of $$\sigma$$. Then the polytope is uniquely determined by its edge-lengths and vertex-origin-distances.

Proof.

Let $$P$$ be a $$d$$-polytope.

Each 2-face of $$P$$ and the origin form a (possibly degenerate) pyramid, in which all edge-length are known, and the apex projects to the interior of the base. This case is discussed in this question where it is also proven that the base face of the pyramid is uniquely determined.

If $$d=2$$, we are done. If $$d\ge 3$$, we can apply Cauchy's rigidity theorem in the 2-face version, that is, the last version mentioned over here, to obtain that $$P$$ is uniquely determined.

$$\square$$