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Almost all circle polyhedra are rigid

John C. Bowers Department of Computer Science, James Madison University, Harrisonburg VA 22807 bowersjc@jmu.edu Philip L. Bowers Department of Mathematics, The Florida State University, Tallahassee FL 32306 bowers@math.fsu.edu  and  Kevin Pratt Computer Science Department, Carnegie Mellon University, Pittsburgh PA 15213 kpratt@andrew.cmu.edu
(Date: August 3, 2025)
Abstract.

We verify the infinitesimal inversive rigidity of almost all triangulated circle polyhedra in the Euclidean plane 𝔼2\mathbb{E}^{2}, as well as the infinitesimal inversive rigidity of tangency circle packings on the 22-sphere 𝕊2\mathbb{S}^{2}. From this the rigidity of almost all triangulated circle polyhedra follows. The proof adapts Gluck’s proof in [7] of the rigidity of almost all Euclidean polyhedra to the setting of circle polyhedra, where inversive distances replace Euclidean distances and Möbius transformations replace rigid Euclidean motions.

Introduction

The infinitesimal rigidity theory of bar-and-joint frameworks traces its origins to the investigations of James Clerk Maxwell in 1864. Before Maxwell’s investigations, Legendre and Cauchy investigated the global rigidity of polyhedra, culminating in the famous Cauchy Rigidity Theorem [4] of 1813 that avers that convex, bounded polyhedra in Euclidean 33-space 𝔼3\mathbb{E}^{3} are globally rigid. In 1916 Max Dehn [6] generalized in one direction the Cauchy Theorem by proving the infinitesimal rigidity of strictly convex polyhedra in 𝔼3\mathbb{E}^{3}, and in the decade of the 1950’s, A.D. Alexandrov [1] greatly extended Dehn’s insights in his articulation of a rather broad theory of rigidity for 33-dimensional polyhedra. Nonetheless, the question of whether or not all polyhedra, even non-convex ones, in 𝔼3\mathbb{E}^{3} were rigid remained open. In 1975 Herman Gluck [7], in the vein of Maxwell and Dehn, proved that almost all of them are infinitesimally rigid, and therefore rigid. At that time, many believed that all polyhedra in 𝔼3\mathbb{E}^{3} were rigid, but in 1977 Robert Connelly [5] surprised the community by constructing a flexible polyhedron in 𝔼3\mathbb{E}^{3}, necessarily non-convex, and by Gluck’s result, a rather rare example among polyhedra.

In [2], the authors of the present work began a study of the rigidity theory of circle frameworks, and in particular of circle polyhedra. There we showed that an analogue of the Cauchy Rigidity Theorem remains true when vertices in 𝔼3\mathbb{E}^{3} are replaced by circles in the 22-sphere 𝕊2\mathbb{S}^{2} that are placed in the pattern of a Euclidean polyhedron. There global rigidity adheres when the circle polyhedron is convex and proper, and an example of Ma and Schlenker [9] shows that this fails for non-convex ones, which mirrors the classical case for Euclidean polyhedra. Of course for circle polyhedra, Euclidean distance is replaced by inversive distance and rigidity is understood with respect to Möbius transformations. In the present paper, we show that an analogue of Gluck’s Theorem on the infinitesimal rigidity as well as the rigidity of almost all polyhedra holds in the setting of circle polyhedra in the plane 𝔼2\mathbb{E}^{2}.

The proof, not surprisingly, mirrors the plan developed over the last couple of centuries by Cauchy [4], Dehn [6], Alexandrov [1], and Gluck [7] in their investigations of rigid Euclidean polyhedra, with Gluck’s paper in particular serving as a valuable guide for our development. The first task is identifying appropriate rigidity matrices and stress matrices in the setting of circle frameworks, where the trivial motions are by Möbius transformations and inversive distance between circles is the desired preserved parameter. This is accomplished in Sections 1 and 2. With the right notions of these matrices in place, the argument proceeds as in Gluck [7]. The infinitesimal rigidity of planar circle polyhedra is related to inversive stresses in Section 3. In the classical Euclidean development, Dehn’s Theorem plays the role of showing that the collection of infinitesimally rigid polyhedra is non-empty, from which Gluck argues that the collection is dense in the space of all polyhedra. The role played by Dehn’s Theorem in our development is of independent interest and appears in Section 4 where we show that univalent tangency circle packings of the 22-sphere are infinitesimally rigid. The four preceding sections come together in the proof of the infinitesimal rigidity of almost all circle polyhedra in the final section, Section 5, as well as their rigidity.

Because of the well-known, intimate connection between the hyperbolic geometry of the unit ball in 𝔼3\mathbb{E}^{3} (the Beltrami-Klein model), the inversive geometry of its 22-sphere ideal boundary 𝕊2\mathbb{S}^{2}, and the projective geometry of 3\mathbb{RP}^{3} in which these models reside, (infinitesimal) rigidity results for circle polyhedra and frameworks have implications for the (infinitesimal) rigidity of hyperbolic polyhedra. These have been articulated in works of Thurston, Rivin, Hogdsen-Rivin, Bao-Bonahon, and our previous paper [2]. We do not take the time to translate the results of this paper to the setting of, for example, strictly hyperideal hyperbolic polyhedra in 3\mathbb{H}^{3} (see the final section of [2]) as we are content with this mere mention of the connection.

1. Infinitesimal Inversive Rigidity of Circle Frameworks

1.1. Circle-frameworks and motions

The general definition of a circle framework in the 22-sphere allows for the case where the disks bounded by the circles of the framework cover the whole of the 22-sphere.111For example, non-univalent tangency circle packings are circle frameworks, and their corresponding disks can cover the 22-sphere multiple times. Our study will restrict this generality by considering only those circle frameworks that have at least one intersticial region, an open set not covered by the disks that the circles of the framework bound. This allows us to project stereographically to the plane and define the inversive rigidity matrix using the very simple Euclidean formulæ for the inversive distances among the adjacent circles of the framework. Our study, then, is of circle frameworks in the plane. Nonetheless, our setting is general enough to cover all univalent circle frameworks in the 22-sphere, the ones the interiors of whose companion disks are pairwise disjoint, as well as many with overlapping adjacent circles.

We parameterize the planar circles by their centers and radii and write C=(x,y,r)C=(x,y,r) when CC is the circle with center (x,y)(x,y) and radius rr. A motion of CC is a continuously differentiable path C(t)=(x(t),y(t),r(t))C(t)=(x(t),y(t),r(t)) such that C(0)=CC(0)=C. Let GG be a connected graph on nn vertices labeled by the integers 1,,n1,\dots,n and with mm unoriented edges, each labeled by its pair of vertices. The set 𝒞={C1,,Cn}\mathcal{C}=\{C_{1},\cdots,C_{n}\} of circles in the plane indexed by the vertex set V(G)={1,,n}V(G)=\{1,\dots,n\} is called a circle framework with adjacency graph GG, or a c-framework for short, and is denoted as G(𝒞)G(\mathcal{C}). Two circles CiC_{i} and CjC_{j} of the c-framework G(𝒞)G(\mathcal{C}) are adjacent provided ijij is an edge of GG. If all adjacent circles are tangent in G(𝒞)G(\mathcal{C}), we call it a tangency circle framework, or a tc-framework. If the closed disks in the collection 𝒟={D1,,Dn}\mathcal{D}=\{D_{1},\dots,D_{n}\}, where Di=Ci\partial D_{i}=C_{i} for i=1,,ni=1,\dots,n, have pairwise disjoint interiors, then the circle framework G(𝒞)G(\mathcal{C}) is said to be univalent.

A motion of a framework G(𝒞)G(\mathcal{C}) is given by a motion of each of the circles in 𝒞\mathcal{C} that at each time tt preserves inversive distances among adjacent circles: for all tt and each edge ijE(G)ij\in E(G), the inversive distance between Ci(t)C_{i}(t) and Cj(t)C_{j}(t) remains constant. If there exists a smooth 11-parameter family MtM_{t} of Möbius transformations such that M0=id2M_{0}=\text{id}_{\mathbb{R}^{2}} and Mt(Ci)=Ci(t)M_{t}(C_{i})=C_{i}(t) for all iV(G)i\in V(G), then the motion is trivial; otherwise, the motion is a flex of G(𝒞)G(\mathcal{C}). If there do not exist any flexes, then G(𝒞)G(\mathcal{C}) is rigid, otherwise flexible.

1.2. Inversive rigidity matrix

Let a motion of G(𝒞)G(\mathcal{C}) be given as

𝒞(t)={C1(t)=(x1(t),y1(t),r1(t)),,Cn(t)=(xn(t),yn(t),rn(t))}\mathcal{C}(t)=\{C_{1}(t)=(x_{1}(t),y_{1}(t),r_{1}(t)),\cdots,C_{n}(t)=(x_{n}(t),y_{n}(t),r_{n}(t))\}

and consider an edge ijij. Differentiating the expression

Inv(Ci(t),Cj(t))=(xi(t)xj(t))2+(yi(t)yj(t))2ri(t)2rj(t)22ri(t)rj(t)=constant,\text{Inv}(C_{i}(t),C_{j}(t))=\frac{(x_{i}(t)-x_{j}(t))^{2}+(y_{i}(t)-y_{j}(t))^{2}-r_{i}(t)^{2}-r_{j}(t)^{2}}{2r_{i}(t)r_{j}(t)}=\text{constant},

multiplying by ri2rj2-r_{i}^{2}r_{j}^{2}, letting dij2=(xixj)2+(yiyj)2=dji2d_{ij}^{2}=(x_{i}-x_{j})^{2}+(y_{i}-y_{j})^{2}=d_{ji}^{2}, and evaluating at t=0t=0 gives

(1.1) rirj(xjxi)xi+rjri(xixj)xj+rirj(yjyi)yi+rjri(yiyj)yj+(1/2)rj(ri2+dij2rj2)ri+(1/2)ri(rj2+dji2ri2)rj=0.\begin{split}&r_{i}r_{j}(x_{j}-x_{i})x_{i}^{\prime}+r_{j}r_{i}(x_{i}-x_{j})x_{j}^{\prime}+\\ &r_{i}r_{j}(y_{j}-y_{i})y_{i}^{\prime}+r_{j}r_{i}(y_{i}-y_{j})y_{j}^{\prime}+\\ &(1/2)r_{j}(r_{i}^{2}+d_{ij}^{2}-r_{j}^{2})r_{i}^{\prime}+\\ &(1/2)r_{i}(r_{j}^{2}+d_{ji}^{2}-r_{i}^{2})r_{j}^{\prime}=0.\end{split}

Equation 1.1 is linear in the derivatives xix_{i}^{\prime}, yiy_{i}^{\prime}, rir_{i}^{\prime}, xjx_{j}^{\prime}, yjy_{j}^{\prime}, and rjr_{j}^{\prime}. The system of linear equations over all edges ijij can be represented as a matrix equation Rc=𝟎Rc=\mathbf{0}, where RR is an m×3nm\times 3n matrix corresponding to the coefficients of the linear system defined by Eq. 1.1 and c=(x1,y1,r1,,xn,yn,rn)Tc=(x_{1}^{\prime},y_{1}^{\prime},r_{1}^{\prime},\cdots,x_{n}^{\prime},y_{n}^{\prime},r_{n}^{\prime})^{T} is the column vector of derivatives. The rows are indexed by the mm edges and the columns by the 3n3n coordinates of the nn circles of the framework. When ijij is an edge of GG, the row in RR corresponding to that edge has the entries

(1.2) rirj(xjxi),rirj(yjyi),(1/2)rj(ri2+dij2rj2)r_{i}r_{j}(x_{j}-x_{i}),\quad r_{i}r_{j}(y_{j}-y_{i}),\quad(1/2)r_{j}(r_{i}^{2}+d_{ij}^{2}-r_{j}^{2})

in the three columns corresponding to the three coordinates of the circle CiC_{i}, and similarly for CjC_{j} with ii and jj exchanged, and with all other entries of that row equal to zero. We call the matrix R=RG(𝒞)R=R_{G(\mathcal{C})} the inversive rigidity matrix for G(𝒞)G(\mathcal{C}).

1.3. Infinitesimal rigidity

The derivative of the motion 𝒞(t)\mathcal{C}(t) at t=0t=0 gives a solution vector cc to Rc=𝟎Rc=\mathbf{0}. We call any vector cc satisfying Rc=𝟎Rc=\mathbf{0} an infinitesimal motion of G(𝒞)G(\mathcal{C}). Such an infinitesimal motion is trivial if it is the derivative of a trivial motion of G(𝒞)G(\mathcal{C}), one by a smooth 11-parameter family of Möbius transformations. A non-trivial infinitesimal motion is called an infinitesimal flex of G(𝒞)G(\mathcal{C}). If no infinitesimal flex exists for G(𝒞)G(\mathcal{C}), we say G(𝒞)G(\mathcal{C}) is infinitesimally rigid; otherwise, infinitesimally flexible.

The space of infinitesimal motions of G(𝒞)G(\mathcal{C}) is the kernel of RR. Since trivial infinitesimal motions are derivatives of Möbius transformations, a 66-dimensional Lie group, describing them requires six parameters. Thus, the dimension of the kernel is at least six and equals six if and only if the framework is infinitesimally rigid. By the rank-nullity theorem, we have the following result.

Lemma 1.1.

G(𝒞)G(\mathcal{C}) is infinitesimally rigid if and only if the rank of the inversive rigidity matrix RR is equal to 3n63n-6.

2. Inversive Stress of a c-Framework

For the graph GG with vertex set V(G)={1,,n}V(G)=\{1,\dots,n\}, the set 𝔭={p1,,pn}\mathfrak{p}=\{p_{1},\dots,p_{n}\} of points in d\mathbb{R}^{d} indexed by the vertex set V(G)={1,,n}V(G)=\{1,\dots,n\} is called a Euclidean framework and denoted as G(𝔭)G(\mathfrak{p}). A key concept that has found many uses in the rigidity theory of Euclidean frameworks is that of a stress on a framework. Informally, a stress is a real number ωij\omega_{ij} attached to each edge ijij of a framework and is used to compute force vectors Fij=FjiF_{ij}=-F_{ji} acting on ii and jj so that a positive stress attracts ii and jj towards each other, a negative stress repels, and a zero stress results in Fij=Fji=0F_{ij}=F_{ji}=0. For each vertex ii a force FiF_{i} is computed by the sum Fi=j:ijE(G)FijF_{i}=\sum_{j:ij\in E(G)}F_{ij}. If for all ii, Fi=0F_{i}=0, the stress is said to be an equilibrium stress. An equilibrium stress is non-trivial whenever there exists at least one edge ijij such that Fij0F_{ij}\neq 0. Alexandrov [1] proved that the only equilibrium stress on the edge-framework of a convex polyhedron in 3\mathbb{R}^{3} is the trivial one if and only if the framework is infinitesimally rigid. Our aim in this section is to develop a similar notion of stress for c-frameworks.

2.1. An inversive distance edge-vector

We want to define stresses in such a way that an edge of a c-framework with a non-zero stress generates an attractive or repellent force between its adjacent circles. The question is in what direction should the two circles move and how should their radii change? In the Euclidean case, a stressed edge attracts or repels its endpoints along the edge vector. One way to think of an edge vector in Euclidean space is as the direction in which the distance between two points increases or decreases most rapidly. The edge vector from pip_{i} to pjp_{j} in 𝔼3\mathbb{E}^{3} is pjpip_{j}-p_{i}. Define the function fif_{i} as fi(p)=ppif_{i}(p)=||p-p_{i}||, the distance function from all points p𝔼3p\in\mathbb{E}^{3} to pip_{i}. Then the gradient of fif_{i} at pjp_{j} is equal to the normalized edge vector fi(pj)=(pjpi)/pjpi\nabla f_{i}(p_{j})=(p_{j}-p_{i})/||p_{j}-p_{i}||. In physics, forces are usually the negative of the gradient of a potential. When thinking about the edge vector as representing a force, a more physical description might be that one should think of the point pjp_{j} as exerting a force on the point pip_{i} determined by the potential function fjf_{j}, the distance from point pjp_{j}, so that the force is fj(pi)-\nabla f_{j}(p_{i}). Of course it is straightforward to see that fj(pi)=fi(pj)-\nabla f_{j}(p_{i})=\nabla f_{i}(p_{j}) so it doesn’t matter which you use. It seems then that there are two natural candidates for the inversive distance edge-vector between two circles Ci=(xi,yi,ri)C_{i}=(x_{i},y_{i},r_{i}) and Cj=(xj,yj,rj)C_{j}=(x_{j},y_{j},r_{j}). Fixing the circle CkC_{k} and denoting the inversive distance from CkC_{k} to all other circles C=(x,y,r)C=(x,y,r) as

fk(x,y,r)=Inv(C,Ck)=dk2r2rk22rrk,f_{k}(x,y,r)=\text{Inv}(C,C_{k})=\frac{d_{k}^{2}-r^{2}-r_{k}^{2}}{2rr_{k}},

where dk2=(xxk)2+(yyk)2d_{k}^{2}=(x-x_{k})^{2}+(y-y_{k})^{2}, we might define the inversive distance edge-vector from vertex ii to vertex jj as either fi(xj,yj,rj)\nabla f_{i}(x_{j},y_{j},r_{j}) or as fj(xi,yi,ri)-\nabla f_{j}(x_{i},y_{i},r_{i}). These in general fail to be equal, so we have a choice to make. It turns out that using the more physically inspired expression works and we define the inversive distance edge-vector from circle CiC_{i} to circle CjC_{j} as Vij=fj(xi,yi,ri)V_{ij}=-\nabla f_{j}(x_{i},y_{i},r_{i}).

The partial derivatives of fjf_{j} with respect to xx, yy, and rr are

(2.1) fjx=xxjrrj,fjy=yyjrrj,fjr=rj2dj2r22r2rj.\frac{\partial f_{j}}{\partial x}=\frac{x-x_{j}}{rr_{j}},\quad\frac{\partial f_{j}}{\partial y}=\frac{y-y_{j}}{rr_{j}},\quad\frac{\partial f_{j}}{\partial r}=\frac{r_{j}^{2}-d_{j}^{2}-r^{2}}{2r^{2}r_{j}}.

It follows that

Vij=fj(xi,yi,ri)=(xjxirirj,yjyirirj,ri2+dij2rj22ri2rj)T.V_{ij}=-\nabla f_{j}(x_{i},y_{i},r_{i})=\left(\frac{x_{j}-x_{i}}{r_{i}r_{j}},\frac{y_{j}-y_{i}}{r_{i}r_{j}},\frac{r_{i}^{2}+d_{ij}^{2}-r_{j}^{2}}{2r_{i}^{2}r_{j}}\right)^{T}.

Notice that unlike the standard case of Euclidean frameworks, the edge vectors are not negatives of one another since, in general, VijVjiV_{ij}\neq-V_{ji}, this because of the asymmetry between ii and jj in the denominators of the third coordinates.

2.2. Inversive stresses

An inversive stress ω={ωij:ijE(G)}\omega=\{\omega_{ij}:ij\in E(G)\} on a c-framework G(𝒞)G(\mathcal{C}) is an assignment of real numbers ωij=ωji\omega_{ij}=\omega_{ji} to each edge ijE(G)ij\in E(G). For a fixed vertex ii, each edge ijij adjacent to ii exerts a force ωijVij\omega_{ij}V_{ij} on vertex ii. If the sum of these forces is 0, i.e., if

(2.2) j:ijE(G)ωijVij=0,\sum_{j:ij\in E(G)}\omega_{ij}V_{ij}=0,

we say that the stress is in equilibrium at vertex ii. If the stress is in equilibrium at all the vertices of a framework, then ω\omega is an equilibrium inversive stress on G(𝒞)G(\mathcal{C}). An inversive stress is non-trivial if there exists at least one edge ijij for which ωij0\omega_{ij}\neq 0; otherwise it is trivial.

The set of linear equations given by Eq. 2.2 for i=1,,ni=1,\dots,n gives a linear system Vω=0V\omega=0. The matrix VV has dimension 3n×m3n\times m and its ijij column is given as follows: its (3i2,ij)(3i-2,ij), (3i1,ij)(3i-1,ij), and (3i,ij)(3i,ij) entries are the components of VijV_{ij}, its (3j2,ij)(3j-2,ij), (3j1,ij)(3j-1,ij), and (3j,ij)(3j,ij) entries are the components of VjiV_{ji}, and the remaining entries in the column are zero. Notice that there are exactly six possible non-zero entries in the ijijth column, which are the components of the edge vectors VijV_{ij} and VjiV_{ji} in the appropriate rows.

It follows that a non-trivial equilibrium inversive stress exists on G(𝒞)G(\mathcal{C}) if and only if dim(kerV)0\mathrm{dim}(\mathrm{ker}V)\neq 0. We call VV the inversive stress matrix for G(𝒞)G(\mathcal{C}).

Remark.

We may scale each vector VijV_{ij} by an arbitrary non-zero real number λij=λji\lambda_{ij}=\lambda_{ji} without affecting the existence of a non-trivial equilibrium stress. To see this, suppose we have an inversive stress ω\omega on G(𝒞)G(\mathcal{C}). Choose any set of non-zero numbers Λ={λij:ijE(G)}\Lambda=\{\lambda_{ij}:ij\in E(G)\}, one for each edge ijij. Now scale each vector VijV_{ij} by its corresponding number λij\lambda_{ij} to obtain the scaled edge vector V^ij=λijVij\widehat{V}_{ij}=\lambda_{ij}V_{ij}. Let ω^ij=ωij/λij\widehat{\omega}_{ij}=\omega_{ij}/\lambda_{ij}. Then it follows immediately that

(2.3) j:ijE(G)ωijVij=j:ijE(G)ω^ijV^ij,\sum_{j:ij\in E(G)}\omega_{ij}V_{ij}=\sum_{j:ij\in E(G)}\widehat{\omega}_{ij}\widehat{V}_{ij},

and ωij=0\omega_{ij}=0 if and only if ω^ij=0\widehat{\omega}_{ij}=0. In other words, the existence of a non-trivial equilibrium inversive stress on a framework G(𝒞)G(\mathcal{C}) is independent of any choice of scale on the edge vectors VijV_{ij}. Eq. 2.2 with the scaled edge vectors may be written as j:ijE(G)ω^ijV^ij=0\sum_{j:ij\in E(G)}\widehat{\omega}_{ij}\widehat{V}_{ij}=0, which gives the linear system V^ω^=0\widehat{V}\widehat{\omega}=0. We call the matrix V^\widehat{V} a scaled inversive stress matrix.

We can state this rather nicely if we let Λ=diag[ωij]\Lambda=\text{diag}[\omega_{ij}] be the diagonal m×mm\times m matrix of rescale values rather than just the collection of rescale values. Then V^=VΛ\widehat{V}=V\Lambda and ω^=Λ1ω\widehat{\omega}=\Lambda^{-1}\omega, and Eq. 2.3 becomes Vω=VΛΛ1ω=V^ω^V\omega=V\Lambda\Lambda^{-1}\omega=\widehat{V}\widehat{\omega}. This implies in particular that the kernels kerV\ker V and kerV^\ker\widehat{V} are isomorphic via the invertible matrix Λ\Lambda. The next lemma summarizes this discussion.

Lemma 2.1.

Let VV be the inversive stress matrix for G(𝒞)G(\mathcal{C}) and let V^=VΛ\widehat{V}=V\Lambda be any scaled inversive stress matrix. Then dim(kerV)=dim(kerV^)\dim(\ker V)=\dim(\ker\widehat{V}). In particular, dim(kerV)=0\mathrm{dim}(\mathrm{ker}\,V)=0 if and only if dim(kerV^)=0\mathrm{dim}(\mathrm{ker}\,\widehat{V})=0 so that G(𝒞)G(\mathcal{C}) has a non-trivial inversive stress if and only if dim(kerV)0\dim(\ker V)\neq 0 if and only if dim(kerV^)0\dim(\ker\widehat{V})\neq 0.

We use this fact with two different scalings in the next two sections. First, we choose one set of λij\lambda_{ij}-values to connect the concepts of inversive stresses and infinitesimal rigidity, and then use a different set of λij\lambda_{ij}-values to prove that all tc-frameworks are infinitesimally rigid.

3. Non-trivial Equilibrium Inversive Stresses and Infinitesimal Rigidity of Triangulated c-Polyhedra

Before specializing to circle polyhedra, which are the c-frameworks of interest in the remainder of the paper, we make some observations connecting the inversive rigidity matrix RR to the inversive stress matrix VV. For each edge ijij of GG, let λij=ri2rj2\lambda_{ij}=r_{i}^{2}r_{j}^{2} and let V^\widehat{V} be the scaled inversive stress matrix scaled by the non-zero constants Λ={λij:ijE(V)}\Lambda=\{\lambda_{ij}:ij\in E(V)\}. A moment’s inspection reveals that

(3.1) V^T=R,\widehat{V}^{T}=R,

the transpose of the scaled inversive stress matrix is precisely the inversive rigidity matrix. In particular, the ranks of the matrices V^\widehat{V} and RR agree, and by Lemma 2.1 so too does the rank of VV. The next lemma now follows from Lemma 1.1.

Lemma 3.1.

G(𝒞)G(\mathcal{C}) is infinitesimally rigid if and only if the rank of the inversive stress matrix VV is equal to 3n63n-6.

We now specialize to those c-frameworks whose graphs are polyhedral graphs. A c-framework P(𝒞)P(\mathcal{C}) is a c-polyhedron if the graph PP is the 1-skeleton of an abstract triangulated polyhedron, equivalently, the 11-skeleton of a simplicial triangulation of the 22-sphere.222This is more general than the development of c-polyhedra in [2] since we do not require the existence of ortho-circles for each face, as required there, and at the same time less general in that we do require the polyhedra to be triangulated, unlike there. The importance of the triangulation assumption is seen in the lemma following.

Lemma 3.2.

A c-polyhedron P(𝒞)P(\mathcal{C}) in the plane is infinitesimally rigid if and only if it has no non-trivial equilibrium inversive stress.

Proof.

Let mm be the number of edges and nn the number of vertices of PP. Since PP is a triangulated polyhedron, the Euler characteristic gives m=3n6m=3n-6 so that the stress matrix VV for PP has dimension 3n×(3n6)3n\times(3n-6) and the rigidity matrix RR has dimension (3n6)×3n(3n-6)\times 3n. The result now is immediate. Indeed, P(𝒞)P(\mathcal{C}) is infinitesimally rigid \iff the rank of RR is 3n63n-6 \iff the rank of RT=V^R^{T}=\widehat{V} is 3n63n-6 \iff the m=3n6m=3n-6 columns of V^\widehat{V} are linearly independent \iff dim(kerV^)=0\dim(\ker\widehat{V})=0 \iff dim(kerV)=0\dim(\ker{V})=0 \iff there is no non-trivial equilibrium inversive stress. ∎

Note that this proof works precisely because the number of edges mm is equal to 3n63n-6 and the Möbius group is a 66-dimensional Lie group.

4. Univalent Tangency packings of the Sphere are Infinitesimally Rigid

A tc-polyhedron is a tc-framework that is also a c-polyhedron, so a configuration of circles in the pattern of a triangulated polyhedron whose adjacent circles are tangent. Recall that a c-framework is univalent if the open disks bounded by the circles are pairwise disjoint.

Lemma 4.1.

Let P(𝒞)P(\mathcal{C}) be a univalent tc-polyhedron in the plane. Then P(𝒞)P(\mathcal{C}) is infinitesimally rigid.

Proof.

In our argument we use Cauchy’s Combinatorial Lemma, which Cauchy used in the proof if his celebrated rigidity theorem for convex Euclidean polyhedra.

Cauchy’s Combinatorial Lemma.

Let GG be a graph that triangulates the 22-sphere. Label some edges with ++ sign, some with - sign, and some not at all. Consider a vertex vv of the graph and a traversal of adjacent edges, say in counter-clockwise order. Ignoring the unlabeled edges, count the number of times a sign change occurs, from - to ++ or vice versa, in traversing in order the edges adjacent to vv. It is easy to see that this count must be even. If there is a vertex of GG with at least two sign changes, then there must exist a vertex with exactly two sign changes.

To verify the lemma, it suffices by Lemma 3.2 to show that PP has no non-trivial equilibrium inversive stresses. We argue by contradiction.

Assume then that PP admits a non-trivial equilibrium stress ω\omega. Then Vω=0V\omega=0 and ω\omega has a non-zero component for at least one edge. Let V^\widehat{V} be the scaled inversive stress matrix obtained from VV by applying the scale λij=rirj\lambda_{ij}=r_{i}r_{j}. Then,

V^ij=λijVij=(xjxi,yjyi,rj3+dij2rj+ri2rj2rirj)T.\widehat{V}_{ij}=\lambda_{ij}V_{ij}=\left(x_{j}-x_{i}\ ,\ y_{j}-y_{i}\ ,\ \frac{-r_{j}^{3}+d_{ij}^{2}r_{j}+r_{i}^{2}r_{j}}{2r_{i}r_{j}}\right)^{T}.

Since this is a tangency packing, dij2=(ri+rj)2d_{ij}^{2}=(r_{i}+r_{j})^{2}, and a quick calculation shows that the third component of V^\widehat{V} simplifies to ri+rjr_{i}+r_{j} so that V^ij=(xjxi,yjyi,ri+rj)T\widehat{V}_{ij}=\left(x_{j}-x_{i}\,,y_{j}-y_{i}\,,r_{i}+r_{j}\right)^{T}. Because Vω=0V\omega=0, V^ω^=0\widehat{V}\widehat{\omega}=0 by Lemma 2.1, and because ω\omega has at least one non-zero entry, so too does ω^\widehat{\omega}.

For any edge ijE(V)ij\in E(V), label ijij with a ++ sign if ω^ij>0\widehat{\omega}_{ij}>0 and a - sign if ω^ij<0\widehat{\omega}_{ij}<0. We argue that there is no vertex ii that is incident to a vertex labeled with a ++ or a - sign such that all labeled edges incident to ii have the same label. To see this, treat V^\widehat{V} as a vector in 3\mathbb{R}^{3} and note that since all radii ri>0r_{i}>0, the zz-component of each vector V^ij\widehat{V}_{ij} is positive. Suppose that for a vertex ii some of the edges ijij incident to ii have a ++ label, but no incident edge is labeled with a - sign. Then all of the ω^ij\widehat{\omega}_{ij} values are strictly positive. From this it follows that the zz-component of the sum jAdj(i)ω^V^ij\sum_{j\in Adj(i)}\widehat{\omega}\widehat{V}_{ij} is strictly positive, which contradicts that V^ω^=0\widehat{V}\widehat{\omega}=0. We conclude that any vertex ii that is adjacent to an edge labeled with a ++ or a - sign must give rise to at least two sign changes as one traverses in order the edges adjacent to ii. Since ω^\widehat{\omega} has at least one non-zero entry, Cauchy’s combinatorial lemma applies, and there exists a vertex ii with exactly two sign changes.

Let v1,,vkv_{1},\cdots,v_{k} denote the vertices adjacent to ii given in a counter-clockwise rotation. Without loss of generality assume that all the ++ signs occur at the smaller indices, starting with iv1iv_{1}. Let ivsiv_{s} be the last edge adjacent to ii labeled with a ++ sign so that any of the remaining edges ivs+1,,ivkiv_{s+1},\dots,iv_{k}, if labeled, are labeled with a - sign. Consider the corresponding edge vectors V^iv1,,V^ivk\widehat{V}_{iv_{1}},\dots,\widehat{V}_{iv_{k}}. Note first that each vector V^ivj\widehat{V}_{iv_{j}} lies on the 4545^{\circ} cone L:z2=x2+y2L:z^{2}=x^{2}+y^{2}. This follows from the facts that circle CiC_{i} is tangent to circle CvjC_{v_{j}} for j=1,,kj=1,\dots,k, that (xvjxi)2+(yvjyi)2(x_{v_{j}}-x_{i})^{2}+(y_{v_{j}}-y_{i})^{2} is the squared-distance from the center of CiC_{i} to that of CvjC_{v_{j}}, and that V^ivj=(xvjxi,yvjyi,ri+rvj)T\widehat{V}_{iv_{j}}=\left(x_{v_{j}}-x_{i}\,,y_{v_{j}}-y_{i}\,,r_{i}+r_{v_{j}}\right)^{T}.

Assuming that circle CiC_{i} is centered at the origin of the xyxy-plane, the orthogonal projection of the vector V^ivj\widehat{V}_{iv_{j}} is merely the position vector of the center of circle CvjC_{v_{j}}. Since the tc-polyhedron P(𝒞)P(\mathcal{C}) is a univalent tangency circle packing, the centers of the circles CvjC_{v_{j}} appear in the order 1,,k1,\dots,k as one walks counterclockwise about the origin. It follows that there are two rays in the xyxy-plane, R1R_{1} and R2R_{2}, such that one open sector of the plane bounded by the union R1R2R_{1}\cup R_{2} contains the centers of Cv1,,CvsC_{v_{1}},\dots,C_{v_{s}} and the complementary open sector contains the centers of the remaining circles Cvs+1,CvkC_{v_{s+1}},\dots C_{v_{k}}. Let R^1\widehat{R}_{1} and R^2\widehat{R}_{2} be the orthogonal, vertical lifts of the respective rays R1R_{1} and R2R_{2} to the 4545^{\circ} cone LL, and let Π\Pi be the plane in 𝔼3\mathbb{E}^{3} containing these lifted rays. Then it is easy to see that the vectors V^iv1,,V^ivs\widehat{V}_{iv_{1}},\dots,\widehat{V}_{iv_{s}} lie in one open half-space bounded by Π\Pi and the vectors V^ivs+1,,V^ivk\widehat{V}_{iv_{s+1}},\dots,\widehat{V}_{iv_{k}} lie in the complementary open half-space. Since ++ signs may occur only at the edges iv1,,ivsiv_{1},\dots,iv_{s} and - signs at edges ivs+1,,ivkiv_{s+1},\dots,iv_{k} among the edges incident with vertex ii, all the non-zero vectors in the list ω^iv1V^iv1,,ω^ivkV^ivk\widehat{\omega}_{iv_{1}}\widehat{V}_{iv_{1}},\dots,\widehat{\omega}_{iv_{k}}\widehat{V}_{iv_{k}} lie in the same open half-space bounded by Π\Pi. It follows that the sum jAdj(i)ω^ijV^ij\sum_{j\in Adj(i)}\widehat{\omega}_{ij}\widehat{V}_{ij} is non-zero since there are exactly two sign changes about vertex ii, which implies that not all the vectors ω^iv1V^iv1,,ω^ivkV^ivk\widehat{\omega}_{iv_{1}}\widehat{V}_{iv_{1}},\dots,\widehat{\omega}_{iv_{k}}\widehat{V}_{iv_{k}} are zero. This contradicts the fact that V^ω^=0\widehat{V}\widehat{\omega}=0 and finishes the proof. ∎

5. Almost All c-Polyhedra are (Infinitesimally) Rigid

The arguments of Gluck [7], adapted to the setting of c-polyhedra, now apply to verify our main theorem. Here are the details.

Let PP be the 11-skeleton of a simplicial triangulation of the 22-sphere with vertices labeled as 1,,n1,\dots,n. Identify the c-polyhedron P(𝒞)P(\mathcal{C}) where 𝒞\mathcal{C} is the circle collection {C1,,Cn}\{C_{1},\dots,C_{n}\} with the point p3np\in\mathbb{R}^{3n} whose coordinates are p3i2=xip_{3i-2}=x_{i}, p3i1=yip_{3i-1}=y_{i}, and p3i=rip_{3i}=r_{i} when Ci=(xi,yi,ri)C_{i}=(x_{i},y_{i},r_{i}) under our parameterization of circles. The collection of all c-polyhedra in the pattern of PP is then parameterized by the points of the open subspace 𝕆\mathbb{O} of 3n\mathbb{R}^{3n} determined by the inequalities ri>0r_{i}>0 for i=1,,ni=1,\dots,n.

By Lemma 1.1, the c-polyhedron P(𝒞)P(\mathcal{C}) is infinitesimally flexible precisely when the rank of the inversive rigidity matrix R=RP(𝒞)R=R_{P(\mathcal{C})} is less than 3n63n-6. Since the rank of a matrix is the greatest integer dd such that some d×dd\times d sub-matrix has non-zero determinant, this occurs when every (3n6)×(3n6)(3n-6)\times(3n-6) sub-matrix of the rigidity matrix RR has zero determinant. By Equations 1.2, the coefficients of the rigidity matrix RR are polynomials in the coordinates of the parameter point p3np\in\mathbb{R}^{3n} that represents P(𝒞)P(\mathcal{C}), and this implies that the determinant of any (3n6)×(3n6)(3n-6)\times(3n-6) sub-matrix of RR is a polynomial in the coordinates of pp, i.e., in the variables xi,yi,rix_{i},y_{i},r_{i}, i=1,,ni=1,\dots,n. It follows that the c-polyhedron P(𝒞)P(\mathcal{C}) is infinitesimally flexible if and only if the point pp representing P(𝒞)P(\mathcal{C}) lies in the real algebraic variety 𝕍\mathbb{V} of 3n\mathbb{R}^{3n} determined by the polynomials detD=0\det D=0, as DD ranges over the (3n6)×(3n6)(3n-6)\times(3n-6) sub-matrices of RR.

Main Theorem.

The space 𝕆𝕍\mathbb{O}\setminus\mathbb{V} of parameter points corresponding to the infinitesimally rigid c-polyhedra in the pattern of PP is open and dense in 𝕆\mathbb{O}, and contains those parameter points corresponding to the rigid c-polyhedra in the pattern of PP.

Proof.

To prove the first statement, that the space 𝕆𝕍\mathbb{O}\setminus\mathbb{V} is open and dense in 𝕆\mathbb{O}, it suffices to show that 𝕍\mathbb{V} is a proper subvariety of 3n\mathbb{R}^{3n}. For this we need but demonstrate the existence of a single c-polyhedron P(𝒞)P(\mathcal{C}) that is infinitesimally rigid. The Koebe Circle Packing Theorem333This was proved first in Koebe [8], rediscovered by Thurston [10], and now is a part of what is known as the Koebe-Andre’ev-Thurston Theorem. There are many proofs in the literature. See Bowers [3] for a fairly quick proof and a survey of results relating to the theorem, as well as a bibliography of literature surrounding the theorem. implies the existence of a univalent, tangency circle packing of the 22-sphere in the pattern of PP, which then stereographically projects to give a univalent tc-polyhedron P(𝒞)P(\mathcal{C}) in the plane. Lemma 4.1 implies that P(𝒞)P(\mathcal{C}) is infinitesimally rigid.

For the second part, that the infinitesimal rigidity of a c-polyhedron implies its rigidity, we follow almost exactly the proof of Theorem 4.1 of Gluck [7]. It deserves to be separated out as its own theorem, whose proof will finish off the proof of this Main Theorem. ∎

Theorem 5.1.

If the c-polyhedron P(𝒞)P(\mathcal{C}) is infinitesimally rigid, it is rigid.

Proof.

As in Gluck [7], the proof is an application of the implicit function theorem. It works in the present setting because the number of edges mm of PP is 3n63n-6 by an Euler characteristic argument, since PP is the 11-skeleton of a triangulation of 𝕊2\mathbb{S}^{2}, and the Möbius group is a 66-dimensional Lie group. Here are the details.

The map f:𝕆m=3n6f:\mathbb{O}\to\mathbb{R}^{m=3n-6} that assigns the inversive distances between adjacent circles in the c-polyhedron P(𝒞)P(\mathcal{C}) corresponding to p𝕆p\in\mathbb{O} via

f(p)ij=Inv((Ci,Cj))whenijE(P)f(p)_{ij}=\text{Inv}((C_{i},C_{j}))\quad\text{when}\quad ij\in E(P)

has derivative

dfp=ApRP(𝒞),df_{p}=A_{p}R_{P(\mathcal{C})},

where pp corresponds to P(𝒞)P(\mathcal{C}) and ApA_{p} is the diagonal matrix whose ijij diagonal entry is 1/ri2rj2-1/r_{i}^{2}r_{j}^{2} when ijE(P)ij\in E(P). As ApA_{p} is invertible, we have P(𝒞)P(\mathcal{C}) is infinitesimally rigid \iff R=RP(𝒞)R=R_{P(\mathcal{C})} has rank 3n63n-6 \iff RR is surjective \iff dfpdf_{p} is surjective \iff pp is a regular value of ff. By the implicit function theorem, f1f(p)f^{-1}f(p) is a 66-dimensional manifold near pp. A moment’s consideration shows that f1f(p)f^{-1}f(p) parameterizes the set of c-polyhedra in the pattern of PP with the same inversive distances between adjacent circles as those of P(𝒞)P(\mathcal{C}). Let Möb(p)\text{M\"{o}b}(p) denote the set of parameter points qq that correspond to the c-polyhedra in the orbit of P(𝒞)P(\mathcal{C}) under the action of the Möbius group on the extended plane. Since the Lie group of Möbius transformations is 66-dimensional, Möb(p)\text{M\"{o}b}(p) is a 66-dimensional manifold near pp. Since Möb(p)f1f(p)\text{M\"{o}b}(p)\subset f^{-1}f(p) and both are 66-dimensional manifolds near pp, they coincide in a neighborhood of pp. But this says precisely that any motion of the c-polyhedron P(𝒞)P(\mathcal{C}) is in fact trivial. Hence P(𝒞)P(\mathcal{C}) is rigid. ∎

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