Zero-sum cycles in flexible polyhedra
ZZero-sum cycles in flexible polyhedra
Matteo Gallet (cid:5)
Georg Grasegger ∗ ,. Jan Legerský ◦ Josef Schicho ∗ , ◦ We show that if a polyhedron in the three-dimensional affine space withtriangular faces is flexible, i. e., can be continuously deformed preservingthe shape of its faces, then there is a cycle of edges whose lengths sum up tozero once suitably weighted by 1 and −
1. We do this via elementary com-binatorial considerations, made possible by a well-known compactificationof the three-dimensional affine space as a quadric in the four-dimensionalprojective space. The compactification is related to the Euclidean met-ric, and allows us to use a simple degeneration technique that reduces theproblem to its one-dimensional analogue, which is trivial to solve.
Introduction
Flexibility of polyhedra — namely, the existence of a continuous deformation preserv-ing the shapes of all faces — is a well-studied topic, a limit case of the theory ofrigidity of surfaces (investigated by, among others, Cohen-Vossen, Nirenberg, Alexan-drov, Gluck, see [Gho] for an overview) that allows for combinatorial and topologicalor algebro-geometric techniques. Although flexible polyhedra have been studied forlong, most of their aspects are mysterious. With the notable exception of Bricardoctahedra (see [Bri97]), very few families are known and classified, and even naturalquestions like the existence of not self-intersecting flexible polyhedra are non-trivial toanswer (see [Con77, Kui79]). Therefore, providing necessary conditions for flexibilityof polyhedra seems a relevant step towards a better understanding of these objects. ∗ Supported by the Austrian Science Fund (FWF): W1214-N15, project DK9. ◦ Supported by the Austrian Science Fund (FWF): P31061. (cid:5)
Supported by the Austrian Science Fund (FWF): Erwin Schrödinger Fellowship J4253. . Supported by the Austrian Science Fund (FWF): P31888. a r X i v : . [ m a t h . M G ] S e p he starting point of the theory of rigidity and flexibility of polyhedra can be consid-ered to be the work by Cauchy [Cau13], in which he proved that convex polyhedra arerigid. Gluck [Glu75] then showed that “almost all” simply connected polyhedra arerigid as well. The research then focused on finding necessary and sufficient conditionsfor the flexibility of polyhedra, and it is still an active research field, as witnessed by,among others, recent works by Gaifullin [Gai18] and Alexandrov [Ale19, Ale20].The goal of this paper is to prove that for every flexible polyhedron in the three-dimensional affine space with triangular faces there is a cycle of edges and a signassignment such that the sum of the signed edge lengths is zero. Notice that we donot ask the polyhedron to be homeomorphic to a sphere, neither to be embedded norimmersed. This is a generalization of the analogous statement that holds for suspen-sions , i. e., simplicial complexes that have the combinatorics of a double pyramid; see[Con, Mik01, AC11]. The first statement of this kind appeared, to our knowledge,in [Leb67] concerning flexible octahedra (see Figure 1). In a previous work [GGLS20]we re-prove this statement in the case of flexible octahedra by means of symbolic com-putation. There, however, our method provides a finer control on which edges takethe positive sign, and which the negative sign. Here, instead, our result is more gen-eral, but gives no information about which edges are counted as positive and which asnegative.Figure 1: Flexible octahedra and the cycles indicated by colors in which the signedlengths of edges sum up to zero. The one on the left is a so-called Type Iflexible octahedron, which is symmetric with respect to a line, while the oneon the right is a so-called Type III flexible octahedron, which admits twoflat positions, one of which is depicted here.This is the main result of our paper (Theorem 2.2): Theorem.
Consider a polyhedron with triangular faces that admits a flex, i. e., a ontinuous deformation preserving the shapes of all faces. Let { w , w } be an edge,and let s and n be the two vertices adjacent to both w and w . If the dihedral anglebetween the faces { w , w , s } and { w , w , n } is not constant along the flex, then thereis an induced cycle of edges containing { w , w } but neither the vertex s nor n andthere is a sign assignment such that the signed sum of lengths of the edges in the cycleis zero. To understand the core argument of the proof, consider a one-dimensional analogue ofthe situation in the theorem: we are given a cycle with vertices mapped to R , namelya sequence x , . . . , x k , x k +1 = x of real numbers. The lengths of the edges of thiscycle are given by | x i − x i +1 | for i ∈ { , . . . , k } . If we multiply each edge length bysign( x i − x i +1 ) we get x i − x i +1 . Therefore, if we sum the edge lengths, each multipliedby sign( x i − x i +1 ), we get a telescoping sum, which yields 0. What we do is to reducethe main result of the paper to this simple statement, and this gives the proof of thetheorem. The way we perform this reduction is by considering “limits” of the flex of thepolyhedron: the key point is that these “limits” are taken in a suitable compactificationof R ; moreover, they are given by points whose coordinates are complex numbers. Thecompactification, which is a quadric hypersurface in the four-dimensional projectivespace, is chosen in such a way that the usual Euclidean distance of R extends to thesepoints “at infinity”. A “limit” of the flex then determines a coloring of the vertices of thepolyhedron. Combinatorial properties of this coloring are obtained from the algebro-geometric features of the compactification of R . These combinatorial properties yieldthe existence of the cycle passing through the edge { w , w } in the statement, suchthat all its vertices lie, in the “limit” situation, on the same tangent space to thequadric hypersurface. A direct computation shows that on such tangent space theextension of the Euclidean distance behaves similarly to a one-dimensional distance,and the argument at the beginning of the paragraph concludes the proof.The paper is structured as follows. Section 1 introduces the compactification of thethree-dimensional affine space we are going to use, and proves the basic relationsbetween the compactification and the standard Euclidean metric. Section 2 formalizesthe notion of flexibility, sets up the combinatorial constructions that are needed forthe main result, and proves it. We introduce a particular compactification of R , which we call the conformal com-pactification (Definition 1.1). This means that we consider R as a subset of a compact3pace (in our case, a projective variety) and so we add to R some “points at infinity”,similarly as it is done in the construction of the projective space. The nice featureof this compactification is that it behaves well with the standard Euclidean distance,in a way that allows extending this distance in a coherent way also to some of thesepoints at infinity (Proposition 1.3). To really use these points at infinity, we need toextend our setting to the complex numbers. This section then includes a series of smallresults regarding various properties of this extension of the Euclidean distance, whichconstitute the algebro-geometric backbone of the main result of the paper. Definition 1.1.
We embed R into P by the map( x, y, z ) ( x : y : z : x + y + z : 1) , which we call the conformal , or Möbius , embedding . Every point ( x : y : z : r : h ) inthe image of R under the conformal embedding lies on the hypersurface M ⊂ P ofthe equation x + y + z − rh = 0 , which is a projective model for conformal geometry, and this is where the name of theembedding comes from. More precisely, each and every real point on M with h = 0corresponds to a unique point in R , and vice versa.Here we make precise the connection between M and the Euclidean distance. Definition 1.2.
The symmetric bilinear form on R × R associated to the equationof M is denoted by h· , ·i M , and is given by h ( x , y , z , r , h ) , ( x , y , z , r , h ) i M = x x + y y + z z −
12 ( r h + r h ) . In the following, we apply this bilinear form to points in P : by this we mean thatwe choose vector representatives of the points for the computation. Whenever needed,we specify explicitly which representative we choose: if p ∈ P , by a slight abuse ofnotation whenever we write p = ( x : y : z : r : h ) we take the vector ( x, y, z, r, h ) asrepresentative of p . Proposition 1.3.
Let u , u ∈ R and let p , p be the corresponding points in M under the conformal embedding, with p i = ( x i : y i : z i : r i : 1) . Then − k u − u k = h p , p i M , where k·k is the Euclidean norm in R . roof. By construction, we have that u i = ( x i , y i , z i ). Then expanding both left andright hand side using the definitions yields the same expression.We now start the exploration of the “points at infinity” of M . Definition 1.4.
The points ( x : y : z : r : h ) in M such that h = 0 are called finite .The other points of M , namely the ones on the quadric M ∞ := M ∩ { h = 0 } , form acone over the plane quadric A := { ( x : y : z ) | x + y + z = 0 } whose vertex is thepoint ω ∞ := (0 : 0 : 0 : 1 : 0). We call the points in M ∞ \ { ω ∞ } simple infinite .From the description of Definition 1.4 we get that the only real point of M ∞ is ω ∞ . Tofully unveil the information contained in M ∞ we then pass to the complex numbers.All the constructions we made so far are algebraic, and so they make sense also over C .Hence from now on, all the points, projective spaces and varieties, and quadratic forms— unless otherwise stated — are considered over the complex numbers. Remark 1.5.
In particular, using the same argument as in Proposition 1.3, for any p , p ∈ M with p i = ( x i : y : z i : r i : 1) — even with complex coordinates — we have − (cid:2) ( x − x ) + ( y − y ) + ( z − z ) (cid:3) = h p , p i M . Due to Proposition 1.3, we can extend the (squared) distance in R to a rationalfunction d : P × P (cid:57)(cid:57)(cid:75) P , defined by d ( p , p ) := ( h p , p i M : h h ) , where p i = ( x i : y i : z i : r i : h i ) for i ∈ { , } . Notice that, however, the value of d ( p , p ) does not depend on the choice of the representatives. If d ( p , p ) = (1 : 0),we write d ( p , p ) = ∞ . From the definition, one derives that d ( p , p ) is not definedif and only if (cid:0) p ∈ M ∞ and p ∈ T p M (cid:1) or (cid:0) p ∈ M ∞ and p ∈ T p M (cid:1) , where T p M is the embedded tangent space of M at p , namely, if p = ( x : y : z : r : h ),then T p M = { ( x : y : z : r : h ) ∈ P | xx + yy + zz ) − ( rh + hr ) = 0 } . In fact, an immediate computation shows that h p , p i M = 0 ⇐⇒ p ∈ T p M ⇐⇒ p ∈ T p M . d , in particular when it isapplied to points in M ∞ .A direct computation shows the first result. Lemma 1.6. If p ∈ M is finite, then d ( p, ω ∞ ) = ∞ . Definition 1.7.
For simple infinite points, we define a mapΨ : M ∞ \ { ω ∞ } −→ A, ( x : y : z : r : 0) ( x : y : z ) . Lemma 1.8. If p , p ∈ M are simple infinite and Ψ( p ) = Ψ( p ) , then d ( p , p ) = ∞ .Proof. Since p , p ∈ M ∞ , either d ( p , p ) is undefined, or d ( p , p ) = ∞ , and thelatter happens precisely when h p , p i M = 0. Write p i = ( x i : y i : z i : 1 : 0). Let B i := { ( x, y, z ) ∈ C | x i x + y i y + z i z = 0 } . From p i ∈ M we get x i + y i + z i = 0,namely, ( x i , y i , z i ) ∈ B i . On the other hand, since Ψ( p ) = Ψ( p ), we have that( x , y , z ) and ( x , y , z ) are linearly independent. Therefore the intersection B ∩ B is one-dimensional and it cannot happen that ( x , y , z ) and ( x , y , z ) are both in B ∩ B , because otherwise they would be linearly dependent. However, the statements( x , y , z ) / ∈ B and ( x , y , z ) / ∈ B are equivalent, so none of the two points belongto B ∩ B . Hence, h p , p i M = x x + y y + z z = 0 . This proves the statement.
Lemma 1.9.
Let q and q be simple infinite points such that Ψ( q ) = Ψ( q ) . If p isfinite, and both d ( q , p ) and d ( q , p ) are not ∞ , then q = q and d ( q i , p ) is undefined.Proof. Write q = ( x : y : z : r : 0). Then q = ( x : y : z : r : 0) = ( αx : αy : αz : r : 0) for some α ∈ C \ { } by the assumption Ψ( q ) = Ψ( q ). We want to showthat r = αr . Let p = ( x : y : z : r : 1). Then d ( p, q i ) = ( h p, q i i M : 0) = ( x i x + y i y + z i z − r i : 0) . Since both the values d ( p, q i ) are not ∞ by assumption, they are both undefined.Hence, r i = 2( x i x + y i y + z i z ) and we get r = 2( x x + y y + z z ) = 2 α ( x x + y y + z z ) = αr , which shows that q = q .We conclude this section with a key technical result, showing that the function d , oncerestricted to particular linear subsets, behaves like a 1-dimensional distance function.6 efinition 1.10. If p is simple infinite, we define Fin p to be the set of all finite points p such that d ( p, p ) = ∞ , namely d ( p, p ) is undefined. Hence Fin p = M ∩ T p M ∩{ h = 0 } . Lemma 1.11.
Let p be a simple infinite point. There exists a function π : Fin p −→ C such that for all q , q ∈ Fin p , we have d ( q , q ) = (cid:16)(cid:0) π ( q ) − π ( q ) (cid:1) : 1 (cid:17) . Proof.
Write p = ( e x : e y : e z : e r : 0). By possibly re-labeling the coordinates, we cansuppose that e x = 0, since p is different from ω ∞ = (0 : 0 : 0 : 1 : 0). The embeddedtangent space T p M has equation2 e x x + 2 e y y + 2 e z z − e r h = 0 . Therefore for any q , q ∈ Fin p we get, once we write q i = ( x i : y i : z i : r i : 1) e r = 2 e x x + 2 e y y + 2 e z z = 2 e x x + 2 e y y + 2 e z z . Since we suppose that e x = 0, we can write x i = e r e x − e y e x y i − e z e x z i . Remark 1.5 shows that − h q , q i M = ( x − x ) + ( y − y ) + ( z − z ) = (cid:18) − e y e x y − e z e x z + e y e x y + e z e x z (cid:19) + ( y − y ) + ( z − z ) = 1 e x h(cid:0)e y ( y − y ) + e z ( z − z ) (cid:1) + e x ( y − y ) + e x ( z − z ) i = 1 e x (cid:2) e y e z ( y − y )( z − z ) + ( e x + e y )( y − y ) + ( e x + e z )( z − z ) (cid:3) = − e x [ e z ( y − y ) − e y ( z − z )] , where in the last step we use that e x + e y + e z = 0. Define the function π : Fin p −→ C by ( x : y : z : r : 1) √ e x ( e zy − e yz ) . Notice that π does not depend on the chosen representative of p . The statement thenfollows, since d ( q , q ) = (cid:0) h q , q i M : 1 (cid:1) = (cid:16)(cid:0) π ( q ) − π ( q ) (cid:1) : 1 (cid:17) . Colorings and zero-sum cycles
This section is devoted to the proof of the main result of the paper (Theorem 2.2).Let us first formalize the notion of flexibility. Recall that a triangular polyhedron isa finite two-dimensional abstract simplicial complex such that every edge belongs toexactly two faces. The of a triangular polyhedron is the graph defined bythe vertices and edges of the polyhedron. We consider only polyhedra whose 1-skeletonis connected.
Definition 2.1. A realization of a triangular polyhedron whose 1-skeleton is G =( V, E ) is a map ρ : V −→ R such that ρ ( u ) = ρ ( v ) for every { u, v } ∈ E . Therealization ρ induces edge lengths λ = ( λ e ) e ∈ E where λ { i,j } := k ρ ( i ) − ρ ( j ) k ∈ R > for { i, j } ∈ E . Here k·k is the standard Euclidean norm. Two realizations ρ and ρ arecalled congruent if there exists an isometry σ of R such that ρ = σ ◦ ρ .A flex of a realization ρ is a continuous map f : [0 , −→ ( R ) V such that . f (0) is the given realization ρ ; . for any t ∈ [0 , f ( t ) and f (0) induce the same edge lengths; . for any two distinct t , t ∈ [0 , f ( t ) and f ( t ) are not con-gruent.To prove the main result, the first step is to convert the information on the existence ofa flex for a triangular polyhedron into a combinatorial object, namely a coloring of thevertices of the polyhedron (Definition 2.6). This is done by considering special “limits”of the images of realizations in the flex under the conformal embedding, which existdue to the fact that M is compact (Definition 2.4). These limits are not realizations,because some of the vertices are sent to points in M that do not correspond to pointsin R ; these points are, actually, not even given by real coordinates. The propertiesof this coloring follow from the results of Section 1. Then, by arguing purely combi-natorially we derive the existence of a monochromatic cycle in the 1-skeleton of thepolyhedron that passes through the edge whose dihedral angle is supposed to changealong the flex (Lemma 2.10). At this point, we notice that the vertices in this cyclesatisfy the hypotheses of Lemma 1.11 (Lemma 2.9). Then the function d behaves likea 1-dimensional distance function, for which the main result is thus trivial.We start by precisely stating our main result. Theorem 2.2.
Let G be the 1-skeleton of a triangular polyhedron with a realizationthat admits a flex. Let λ be the edge lengths induced by the realizations in the flex.Let { w , w } be an edge of G and let s and n be the two opposite vertices of the wo triangles containing { w , w } . If the dihedral angle between the faces { w , w , s } and { w , w , n } is not constant along the flex, then there is an induced cycle in G containing { w , w } but neither the vertex s nor n and there a sign assignment suchthat the signed sum of the edge lengths λ e over the edges e in the cycle is zero. Let us start laying the foundation of the proof. For the rest of this section, we fixthe skeleton G = ( V, E ), the flex, the edge { w , w } , the edge lengths λ and the twovertices s and n satisfying the assumptions of Theorem 2.2. Notice that the realizationsin the flex of the two triangles { w , w , s } and { w , w , n } are non-degenerate – i.e.,the vertices are not collinear — because otherwise their dihedral angle is not defined.The Zariski closure in ( C ) V of the set of real realizations inducing λ is an algebraicset and we denote it by W . By construction, W is the set of maps ρ : V −→ C suchthat for all { v , v } ∈ E with ρ ( v i ) = ( x i , y i , z i ) we have( x − x ) + ( y − y ) + ( z − z ) = λ { v ,v } . ( ∗ )Notice that, as soon as a real realization ρ belongs to W , then all real realizationscongruent to ρ belong to W . To effectively use the hypothesis of having a flex, weneed to get rid of this abundance of “copies” of a single realization. To do so, we takea “slice” of W , which has the effect to kill the action of the group of isometries byfixing a triangle. Let ρ be the realization of the flex at time 0. Consider the subset Z := (cid:8) ρ ∈ W | ρ ( w ) = ρ ( w ) , ρ ( w ) = ρ ( w ) , ρ ( n ) = ρ ( n ) (cid:9) . By construction, no two different elements of Z coming from real realizations maybe congruent by a direct isometry; they may be congruent by the reflection alongthe plane of the triangle with vertices ρ ( w ), ρ ( w ), and ρ ( n ), but this does notinfluence our argument. Now, the realizations in the flex may not be elements of Z ,but for each realization in the flex there is a congruent realization in Z . The imageof Z via the product of conformal embeddings C , → M is a subset of M V . Definition 2.3.
Let Y be the Zariski closure in M V of the image of Z under theproduct of conformal embeddings C , → M .Notice that the elements of Y are maps ρ : V −→ M which need not to correspondto realizations, since the image of some vertices may lie on M ∞ or may have complexcoordinates. Consider the projection Y s of Y on the copy of M indexed by the vertex s . Given a subset V of the vertices V of a graph G , the induced subgraph determined by V is thesubgraph of G with vertices V and all edges of G having both vertices in V . A subgraph is called induced if it is the induced subgraph determined by a subset of vertices.
9y the assumption that the dihedral angle at { w , w } changes during the flex, weget that Y s contains infinitely many points. Hence Y s must intersect the hyperplanesection { h = 0 } ∩ M = M ∞ , since it is a positive-dimensional projective subvarietyof M . Definition 2.4.
From the previous discussion we know that we can pick an element ρ ∞ in Y such that ρ ∞ ( s ) ∈ M ∞ . We fix such an element for the rest of the section.Notice that, by construction of Z , all three points ρ ∞ ( w ), ρ ∞ ( w ), and ρ ∞ ( n ) arefinite. The latter property is crucial for our argument; if we just wanted to achieve ρ ∞ ( s ) ∈ M ∞ we could have used the set W without the need of introducing Z butthis would not be enough to prove the main result of the paper. Lemma 2.5. If v and w are adjacent vertices in G , then d (cid:0) ρ ∞ ( v ) , ρ ∞ ( w ) (cid:1) is eitherfinite or undefined.Proof. The map d (cid:0) · ( v ) , · ( w ) (cid:1) : Y (cid:57)(cid:57)(cid:75) P is a rational map that is constant on thedense subset of Y of finite real points. In fact, these points are indeed realizationsand since v and w are adjacent their distance in R is constant, which implies, dueto Proposition 1.3, that d is constant on them. The constant equals ( − λ { v,w } : 1).Since d ( · ( v ) , · ( w )) is a rational map, it is constant on Y wherever it is defined, and aswe showed this constant is different from ∞ .From Lemma 2.5 we obtain that ρ ∞ ( s ) cannot be ω ∞ , because otherwise we wouldhave d ( ρ ∞ ( s ) , ρ ∞ ( w )) = ∞ due to Lemma 1.6. Definition 2.6.
We color the vertices of G with three colors:a vertex v ∈ V is red if ρ ∞ ( v ) is finite , blue if ρ ∞ ( v ) is simple infinite and Ψ (cid:0) ρ ∞ ( v ) (cid:1) = Ψ (cid:0) ρ ∞ ( s ) (cid:1) , gold otherwise . Lemma 2.7.
A gold vertex cannot be adjacent to a blue and a red vertex simultane-ously. In particular, there is no triangle with three colors.Proof.
For a contradiction, let { u, w } and { v, w } be edges such that u is red, v is blue,and w is gold. There are two cases: . ρ ∞ ( w ) = ω ∞ : since ρ ∞ ( u ) is finite and u is adjacent to w , by Lemma 2.5 andLemma 1.6 we get a contradiction. 10 ρ ∞ ( w ) is simple infinite and Ψ (cid:0) ρ ∞ ( w ) (cid:1) = Ψ (cid:0) ρ ∞ ( s ) (cid:1) : however by assumption wehave Ψ (cid:0) ρ ∞ ( s ) (cid:1) = Ψ (cid:0) ρ ∞ ( v ) (cid:1) , so Ψ (cid:0) ρ ∞ ( w ) (cid:1) = Ψ (cid:0) ρ ∞ ( v ) (cid:1) , and this contradictsLemma 2.5 and Lemma 1.8, since v and w are adjacent.Let T be the set of triangles of G such that the vertices are only red and blue, andboth colors occur. Note that the triangle { w , w , s } belongs to T since w , w arered and s is blue by construction; instead, the triangle { w , w , n } does not belongto T , since all its vertices are red by construction. We define a graph G T with vertexset T , and two triangles are adjacent if they share an edge whose vertices have differentcolors (see Figure 2). By Lemma 2.7 and the fact that every edge is in exactly twotriangles, all vertices of G T have degree two. Hence, the graph G T is a disjoint unionof cycles.Let C be the cycle in G T containing the triangle { w , w , s } . Let T C be the verticesof C (i. e., triangles) and let V := S t ∈ T C t be the set of vertices of G appearing in thetriangles of C . The subgraph of G induced by the red vertices in V defines a closedwalk in G , namely, a triangle in C contributes to the walk by an edge if it has two redvertices, and just by a vertex otherwise. The edge { w , w } is contained in the walk.Similarly, we obtain a blue closed walk, containing s . Figure 2 portraits the situation. Lemma 2.8.
For all vertices v in the blue walk, we have ρ ∞ ( v ) = ρ ∞ ( s ) .Proof. Let v and v be two adjacent vertices of the blue walk. Let r be the thirdvertex of the triangle in T C containing v and v . The vertex r must be red, therefore ρ ∞ ( r ) is finite. For i ∈ { , } , the value d (cid:0) ρ ∞ ( v i ) , ρ ∞ ( r ) (cid:1) cannot be ∞ by Lemma 2.5.Since v and v are blue, Ψ (cid:0) ρ ∞ ( v ) (cid:1) = Ψ (cid:0) ρ ∞ ( v ) (cid:1) . Hence, ρ ∞ ( v ) = ρ ∞ ( v ) byLemma 1.9. The claim follows since the blue walk is connected and contains s . Lemma 2.9.
For all vertices v in the red walk, ρ ∞ ( v ) is contained in Fin ρ ∞ ( s ) .Proof. There exists an adjacent vertex w of v in the blue walk. By Lemma 2.8, ρ ∞ ( w ) = ρ ∞ ( s ). The value d (cid:0) ρ ∞ ( v ) , ρ ∞ ( w ) (cid:1) is not infinity by Lemma 2.5 and ρ ∞ ( v )is finite, therefore the point ρ ∞ ( v ) is in Fin ρ ∞ ( s ) by Definition 1.10. Lemma 2.10.
There is a cycle such that it contains the edge { w , w } , its verticesare in the red walk and it is an induced subgraph of G .Proof. If the red walk, which is closed, contained the edge { w , w } twice, then bothtriangles containing { w , w } would belong to the cycle C . But this is not possible A walk is a finite sequence of vertices such that consecutive vertices are adjacent (i. e., edges orvertices might repeat). Closed means that the first and last vertex are the same. B C DEF G H IJK AB CIJKDE F G H w s w n Figure 2: On the left, a portion of a triangular polyhedron, where we highlight thecolors of the vertices and we label the triangles in T . In the center, thegraph G T . On the right, the red and blue walks. We do not claim that thesesituations do indeed come from flexible polyhedra: this example serves solelyto illustrate the concepts.since { w , w , n } is not an element of T because all its vertices are red. Hence, there isa cycle in G containing { w , w } such that all its vertices are in the red walk. Amongall such cycles, we take one with the minimum number of edges. This guarantees thatthe cycle is an induced subgraph.Now we are ready to prove the main result. Proof of Theorem 2.2.
Let ρ ∞ be the special element of Y such that ρ ∞ ( s ) ∈ M ∞ from Definition 2.4. Let further D := ( v , . . . , v k , v k +1 = v ) be the induced cycleconstructed in Lemma 2.10. Due to Lemma 2.9, for all vertices v j in D the point ρ ∞ ( v j ) is in Fin ρ ∞ ( s ) . Then we know from Lemma 1.11 that there exists a map π : Fin ρ ∞ ( s ) −→ C such that for all j ∈ { , . . . , k } d (cid:0) ρ ∞ ( v j ) , ρ ∞ ( v j +1 ) (cid:1) = (cid:16)(cid:0) π ( ρ ∞ ( v j )) − π ( ρ ∞ ( v j +1 )) (cid:1) : 1 (cid:17) . By construction of the set W , see Equation ( ∗ ), and taking into account Remark 1.5,12or all elements ρ ∈ Y we have λ { v j ,v j +1 } = − h ρ ( v j ) , ρ ( v j +1 ) i M , where we take the representatives for ρ ( v j ) and ρ ( v j +1 ) with h -coordinate equal to 1.Hence in particular λ { v j ,v j +1 } = − h ρ ∞ ( v j ) , ρ ∞ ( v j +1 ) i M = − (cid:0) π ( ρ ∞ ( v j )) − π ( ρ ∞ ( v j +1 )) (cid:1) . We conclude that λ { v j ,v j +1 } = ± ı √ (cid:16) π (cid:0) ρ ∞ ( v j ) (cid:1) − π (cid:0) ρ ∞ ( v j +1 ) (cid:1)(cid:17) , where ı is the imaginary unit. Therefore we can choose integers η j ∈ { , − } such that k X j =1 η j λ { v j ,v j +1 } is a telescoping sum, which yields 0.Clearly, the cycle D does not contain the vertex s as it is blue. If the vertex n wasin D , then the construction in Lemma 2.10 would imply that D = ( w , w , n , w ).But then the condition on the edge lengths of D would imply that the three vertices w , w , n are collinear throughout the flex. This would contradict the assumptionthat the dihedral angle at { w , w } changes. Thus the statement follows.Notice that in our previous works in which we study flexible realizations of graphsin the plane [GLS19] and on the sphere [GGLS19], a flex implies the existence of acombinatorial object — namely, a special edge coloring — called a NAC , resp.
NAP-coloring . On the other hand, the existence of a NAC, resp. NAP-coloring, for a graphprovides a construction of a flex of the graph in the plane, resp. on the sphere, thoughthe obtained flexible realizations might be rather degenerate. Exploiting the ideaof so-called butterfly motions (see Figures 3 and 4) described, among other sources,in [GLS19, Section 6], we obtain analogous results also for flexible polyhedra. Thecombinatorial structure in this case is a separating cycle with a sign assignment to itsedges.
Proposition 2.11.
Let G be the -skeleton of a triangular polyhedron. Let S be acycle in G that separates the graph, namely, removing edges and vertices of S from G yields a disconnected graph. For any sign assignment to the edges of S , not all havingthe same sign, the polyhedron admits a realizations with a flex such that the signed sumof the edge lengths induced by the realization in the cycle S is zero and the dihedralangles at all edges of S vary along the flex. roof. Write S = ( v , . . . , v k , v k +1 = v ). We construct a realization ρ of the polyhe-dron as follows: set ρ ( v j ) = ( x j , , x j < x j +1 if { v j , v j +1 } has the positivesign and x j > x j +1 otherwise. The assumption that not all signs on the edges of S are the same guarantees that we can choose numbers { x j } kj =1 satisfying the previousrequirements. The rest of the vertices of the graph is mapped to arbitrary points. Therealization ρ has a flex since the vertices of different connected components of G \ S can rotate independently around the x -axis. Hence, the statement follows.We remark that if a polyhedron is not homeomorphic to a sphere, then a cycle is notnecessarily separating. On the other hand, if the polyhedron is homeomorphic to asphere, then any induced cycle with at least four edges separates the graph.Figure 3: A butterfly motion of an octahedron. Notice the collinear realizations ofvertices of a separating cycle, highlighted in red.Figure 4: Instances, in a symmetric and a non-symmetric layout, of butterfly motionsof a more complicated polyhedron (on the left). Notice the collinear realiza-tions of vertices of a separating cycle, highlighted in red.14 eferences [AC11] Victor Alexandrov and Robert Connelly. Flexible suspensions with a hexag-onal equator. Illinois J. Math. , 55(1):127–155, 2011. doi:10.1215/ijm/1355927031 .[Ale19] Victor Alexandrov. A sufficient condition for a polyhedron to be rigid.
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Vestnik Moskov. Univ. Ser. I Mat. Mekh. , (3):15–21, 77, 2001. (MG) International School for Advanced Studies/Scuola Internazio-nale Superiore di Studi Avanzati (ISAS/SISSA), Via Bonomea 265, 34136Trieste, Italy
Email address: [email protected] (GG) Johann Radon Institute for Computation and Applied Mathematics(RICAM), Austrian Academy of Sciences
Email address: [email protected] (JL, JS) Johannes Kepler University Linz, Research Institute for Sym-bolic Computation (RISC)
Email address: [email protected] , [email protected] (JL) Department of Applied Mathematics, Faculty of Information Tech-nology, Czech Technical University in Prague(JL) Department of Applied Mathematics, Faculty of Information Tech-nology, Czech Technical University in Prague