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Take a convex polyhedron $P$ in $\mathbb R^3$ and remove all the faces, i.e. leave only the edges. Call this graph $E$. Let us now try to continuously deform $E$ in $\mathbb R^3$ so that all the edges of $E$ keep their length and remain straight (like metal sticks), but allow the change of angles between the edges.

Question. Is it true that $E$ admits non-trivial deformations (not globally isomteric) if at least one face of $P$ is not a triangle?

Note that if all the faces of $P$ are triangles it is not deformable by Cauchy's rigidity theorem: https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(geometry)

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5 Answers 5

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Yes, this is true.

One strategy is to use the naïve approach of counting degrees of freedom and constraints. For triangulated polyhedra one can easily show with the Euler characteristic that the expected dimension (number of variables minus number of equations) of the realization space modulo Euclidean isometries is zero, and Cauchy would then imply that this is the case for a convex embedding. Then we can get to any polyhedron with a nontriangular face by removing some number of edges from a triangulated one and thus the dimension of the realization space is positive. However, this alone doesn't guarantee that the set of real points is positive dimensional. This point is made in Pak's lectures, e.g. Remark 26.3 and footnote 55 on page 248.

To deal with this issue, we appeal to a theorem of Asimow and Roth (Section 3 of "The rigidity of graphs", L. Asimow and B. Roth, Trans. Amer. Math. Soc. 245 (1978), 279-289).

This theorem states: Let $G$ be a graph with $v$ vertices, $e$ edges and let $p\in\mathbb{R}^{nv}$ be a regular point of the length-squared function $f_G:\mathbb{R}^{nv}\rightarrow \mathbb{R}^e$ (which maps a point in $\mathbb{R}^{nv}$, to the vector of squared-edge lengths). Suppose that the affine span of $p$ viewed as a configuration of $v$ points in $\mathbb{R}^n$ is all of $\mathbb{R}^n$. Then $(G,p)$ is (locally) rigid (every $q$ sufficiently close to $p$ with $f_G(q)=f_G(p)$ is induced by an isometry of $\mathbb{R}^n$) if and only if the rank of $df_G(p)=nv-n(n+1)/2$.

Now let $P'$ be a triangulation of your polyhedron $P$ and let $E'$ be the 1-skeleton of $P'$. Suppose that $E$ (and hence $E'$) both have $v$ vertices with placement $p\in\mathbb{R}^{3v}$. Since $E'$ is the graph of a triangulated polyhedron, $E'$ has $e'=3v-6$ edges and by assumption $E$ has $e<3v-6$ edges. Cauchy's rigidity theorem guarantees that $E'$ is rigid, hence the rank of $df_{E'}(p)=3v-6=e'$. Note that the matrix for $df_{E}(p)$ consists of a strict subset of rows of $df_{E'}(p)$ and so by basic linear algebra its rank is $e<3v-6$. Asimow-Roth thus implies that $(E,p)$ cannot be rigid.


Ivan Izmestiev has pointed out the following doesn't work: see the comments.

This follows from Theorem 30.1 (section 30.9, page 283) in Igor Pak's lecture notes, which is proved in section 37.3 (page 332). To see this, consider any triangulation $P'$ of $P$. Then let $F_1,F_2$ be a pair of adjacent triangles in $P'$ which are part of a triangulation of a non-triangular face of $P$. Theorem 30.1 implies that $P''=P'\setminus (F_1\cup F_2)$ is a flexible polyhedral surface, and the flexes of $P''$ give rise to a desired flex of $E$, the 1-skeleton of $P$.

The proof relies on the Alexandrov existence theorem, which states that "every intrinsically convex 2-dimensional surface homeomorphic to a sphere is isometric to the surface of a convex polytope in $\mathbb{R}^3$ or a doubly covered polygon".

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  • $\begingroup$ Triangulate a cube, but then replace the top face by a square (not two triangles). I believe this is rigid, which would imply the answer to the posed question is No. What am I missing? $\endgroup$ Oct 18, 2017 at 22:32
  • $\begingroup$ @JosephO'Rourke I believe the asker is asking about the rigidity of the underlying graph, not the rigidity of the polyhedron with the face included. $\endgroup$
    – j.c.
    Oct 18, 2017 at 22:41
  • $\begingroup$ Thanks j.c.! Indeed this follows from Alexandrov existence theorem. $\endgroup$
    – aglearner
    Oct 18, 2017 at 22:56
  • $\begingroup$ In fact, the proof of Theorem 30.1 (given in Section 37.3) does not work in this situation, because Theorem deals with removing the edge between two non-coplanar triangular faces. $\endgroup$ Oct 19, 2017 at 4:58
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    $\begingroup$ On the other hand, the original argument can be slightly modified: keep the intrinsic metric of all faces except one non-triangular face. This gives a continuous family of metrics with cone angles $<2\pi$. By Alexandrov's theorem, there is a convex polyhedral realization, besides it depends continuously on the metric. We don't pay attention how does the combinatorics change, the non-triangular faces can bend along different diagonals, changing with the time. This answers the original question, but not the strenghened version of Joseph O'Rourke. $\endgroup$ Oct 19, 2017 at 11:51
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I am not seeing the nontrivial deformation of the $E$ below, a triangulation of a cube except for the top face. Can someone explain? Thanks!


          RigidCube


Update. The deformation of this cardboard model is essentially what Nik Weaver described in a comment.


          CubeRhombus


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    $\begingroup$ It shouldn't be too hard to write down explicitly the (17 x 24) rigidity matrix for this example (I'd do it with my code but my Mathematica license expired recently). We can add 6 rows to this matrix to kill the Euclidean isometries, e.g. pin down vertex 1, pin down the xz coordinates of vertex 2 and pin down the z coordinate of vertex 3 (assuming that edge 12 is in the y direction and edge 23 is in the x direction). The resulting (23 x 24) matrix will have a 1-dimensional null space which is the space of infinitesimal flexes. Maybe plotting a picture of such an infinitesimal flex will help? $\endgroup$
    – j.c.
    Oct 19, 2017 at 10:31
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    $\begingroup$ @j.c. Thanks. The top square must squeeze to a rhombus and likely all the dihedral angles along the diagonal bars change simultaneously in concert. Not easy to visualize. $\endgroup$ Oct 19, 2017 at 10:35
  • $\begingroup$ It should be possible to find numerical solutions to the full system of quadratic length equations by adding a small multiple of an infinitesimal flex to the initial positions to get a starting guess. I've been able to make movies of the 1-parameter families of motions in similarly-sized linkages in Mathematica this way. If you're not able to do it, I will try to whenever my institution finally approves my new license. $\endgroup$
    – j.c.
    Oct 19, 2017 at 10:42
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    $\begingroup$ This would be a fun exercise with toothpicks and marshmallows. $\endgroup$
    – Neal
    Oct 20, 2017 at 13:35
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    $\begingroup$ A picture of this case being deformed appears in this old article: jstor.org/stable/24936915 $\endgroup$
    – supergra
    Oct 9, 2021 at 3:37
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Let me show how to reduce the problem to Alexandrov existence theorem; that is, the original idea of j.c. works.

Let us modify the polyhedron by adding a low pyramid over each non-triangular face. The edges which we add this way will be called new.

This graph is rigid by Cauchy theorem. Moreover if we slightly change the length of a new edge, then by Alexandrov existence we get one parameter family of rigid graphs.

Evidently the old part of graph moves, hence we get the required deformation.

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As Anton's answer suggests, it is not possible to isometrically flex the edge graph while keeping all the coplanar edges in the same plane. This follows quickly from Cauchy's proof of his rigidity theorem. Indeed if all the faces remain flat, then we can compare the dihedral angles around each vertex of the new polyhedron $P'$ with those of the original polyhedron $P$. By Cauchy's lemma there must be at least 4 sign changes around each vertex, if there are any sign changes at all, when we consider the difference between the corresponding dihedral angles of $P$ and $P'$. So it follows from Euler's formula that all dihedral angles must be equal. It remains only to observe that if two combinatorially isomorphic convex polyhedra have equal edge lengths and dihedral angles then they are congruent.

In short the answer would have been No if there had been an additional requirement to preserve the combinatorial structure of the polyhedron.

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I would like to share one picture which shows how flexibility of polyhedrons with faces removed can be tested. This is an illustration for the polytope proposed by Joseph:

enter image description here

The whole construction is made of plastic tubes that are hold by rubber bands.

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  • $\begingroup$ Cool model! However, note that because the tubes in the photograph have finite thickness and do not end right at the vertices, the joints depicted behave rather differently from the ideal mathematical model in the question. It may be OK for small deformations of generically placed bars but you may get very different behavior for configurations with special geometry / symmetry. $\endgroup$
    – j.c.
    Dec 24, 2017 at 15:21

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