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Cauchy's rigidity theorem is usually cites briefly as

Any two (convex, 3-dimensional) polyhedra with pairwise congruent faces are themselves congruent.

As a more formal generalization to general dimensions, one often reads something like this (also known as Alexandrov's theorem, see also here):

Theorem (naive). Given two combinatorially equivalent polytopes $P_i\subset\Bbb R^{d_i},i\in\{1,2\},d_i\ge 3$, and let $\phi:\mathcal F(P_1)\to\mathcal F(P_2)$ be a corresponding face-lattice isomorphism. Then, if each facet $\sigma\in\mathcal F(P_1)$ is congruent to the corresponding facet $\phi(\sigma)\in\mathcal F(P_2)$, then $P_1$ and $P_2$ are congruent.

Except that this formulation is incorrect, especially in dimension three! For example, consider the following two combinatorially equivalent polyhedra found by Matt F. (over here) and depicted by მამუკა ჯიბლაძე (over here):

Either polyhedron consists of two stacked twisted antiprisms, which are chiral, and depending on the orientation of each antiprism, the resulting polyhedron is metrically distinct, even though corresponding faces (aka. facets) are congruent.

To deal with this contradiction in my knowledge, I looked up the common proofs of Cauchy's rigidity theorem for polyhedra. And it appears as if each proof uses an ingredient that is often not explicitly stated: not only corresponing faces must be congruent, but also corresponding interior angles of faces must be equal. For me, this was a surprise, because I always considered Cauchy's theorem as especially elegant for its relatively weak assumptions. This subtlety however, makes it unclear to me what the right generalization to higher dimensions should be.

Apparently, one can formulate the theorem also with matching edge-lengths rather than matching interior angles, which brings me to the following guess on what a generalization might look like:

Theorem (better). [...] If each (proper) face $\sigma\in\mathcal F(P_1)$ is congruent to the corresponding face $\phi(\sigma)\in\mathcal F(P_2)$, then $P_1$ and $P_2$ are congruent.

I highlighted the relevant changes: we ask for all (proper) faces to be pair-wise congruent, rather than just the facets.


Question 1: Is this a correct generalization?

Question 2: Do we need this modification in dimensiona $d\ge 4$, or is the "naive" version still valid except for dimension three?


Here is another observation, which was quite surprising to me. Instead of making polytopes rigid by requiring congruent facets, it apparently suffices to require congruent 2-faces (and maybe edges):

Theorem. [...] If each edge and each 2-face $\sigma\in\mathcal F(P_1)$ is congruent to its corresponding image $\phi(\sigma)\in\mathcal F(P_2)$, then $P_1$ and $P_2$ are congruent.

Because if all edges and all 2-faces are pair-wise congruent, then so are the 3-faces (by classical Cauchy), and so are then the 4-faces (by generalized Cauchy), and the 5-faces, ..., and the facets.

Maybe again, in $d\ge 4$ it suffices to require only 2-faces. This feels quite opposite to the usual Cauchy theorem, but is essentially the same.

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  • $\begingroup$ The paper "Cauchy’s Theorem and Edge Lengths of Convex Polyhedra" claims that "the dihedral angles of a convex polyhedron are completely determined by the facial angles and the combinatorial structure." Springer link. $\endgroup$ – Joseph O'Rourke Oct 11 at 14:20
  • $\begingroup$ I find the statement natural. We consider labeled polytopes, that is, every vertex has distinct label. Then the congruence must respect this labeling, and if so, the theorem holds true. $\endgroup$ – Joseph Gordon Oct 11 at 21:41
  • $\begingroup$ @JosephGordon Do you mean that the original version talks about labeled polytopes and that the congruences are implicitly label preserving? Don't you think these are quite some assumptions that should be stated explicitly? $\endgroup$ – M. Winter Oct 14 at 12:30
  • $\begingroup$ Yes, I think that this is what is implicitly meant in this theorem. In my comment I was referring to this: I always considered Cauchy's theorem as especially elegant for its relatively weak assumptions. While I do agree that the condition should be explicitly stated, I can't think of it as inelegant as it is quite a natural thing to assume in this kind of setting. $\endgroup$ – Joseph Gordon Oct 15 at 2:33

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