# Infinitesimal rigidity of panel-hinge frameworks

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Infinitesimal rigidity of panel-hinge frameworks
Shin-ichi Tanigawa　　 (joint work with Naoki Katoh)　 Kyoto University

Theorem A graph can be realized as an infinitesimally rigid body-hinge framework in R3 if and only if it can be realized as a panel-hinge framework in R3 Originally, posed by Tay and Whiteley in 1984, and called the Molecular conjecture. Implying a combinatorial characterization of generic rigidity of molecular frameworks [Tay&Whiteley84, Whiteley99,04, Jackson&Jordán08]

3-dimensional body-hinge framework (G,p): G = (V,E): a graph,
p is a mapping, (called a hinge-configuration), e ∊ E ↦ a line p(e) in R3 a vertex　⇔　a 3-d body an edge　⇔　a hinge (=a line)

Tay-Whiteley’s theorem (Whiteley 88, Tay 89, 91)
For a generic hinge-configuration p, (G,p) is rigid in R3 if and only if 5G contains six edge-disjoint spanning trees. 5G G

Tay-Whiteley’s theorem (Whiteley 88, Tay 89, 91)
For a generic hinge-configuration p, (G,p) is rigid in R3 if and only if 5G contains six edge-disjoint spanning trees. It cannot be applied to “special” hinge configurations

Infinitesimal motion of a rigid body
a combination of six independent isometric motions three translations + three rotations ⇒ The set of all infinitesimal motions forms a 6-dimensional vector space. each rotation (transformation) can be coordinatized by a so-called 2-extensor (Plűcker coordinate) of the axis-line.

Hinge constraints [Crapo and Whiteley 82]
Two bodies 1 and 2 are connected by a hinge A. Si : a 6-vector assigned to the body i, (representing an infinitesimal motion). SA: a 2-extensor (that is, a 6-vector) of the line A. The constraint by the hinge A: S1 – S 2= t SA for some t ∈ R ⇔ ri・( S1 – S2 ) = 0, i=1,...,5 {r1,r2,...,r5}: a basis of the orthogonal complement of the 1-dimensional vector space spanned by SA.

Rigidity Matrix Def. infinitesimal motion m: V → R6 s.t.
ri(p(e))・(m(u) - m(v)) = 0 　i=1,...,5 , for each e=(u,v) ∊ E {r1(p(e)),...,r5(p(e))} is a basis of the orthogonal complement of the vector space spanned by a 2-extensor of p(e). Rigidity matrix R(G,p): 5|E|×6|V|-matrix 6 columns for u 5 rows for e=uv m(u) m(v) =0 6 columns for v (row space represents the orthogonal complement of a 2-extensor of p(e).)

Def. (G,p) is infinitesimal rigid ⇔ rank R(G,p)=6|V|-6
6 columns for u 5 rows for e=uv m(u) m(v) =0 6 columns for v Def. (G,p) is infinitesimal rigid ⇔ rank R(G,p)=6|V|-6 dim ker R(G,p)≧6 the dimension of the space of all trivial motions is 6 three transformations + three rotations Def. G can be realized as an inf. rigid framework ⇔　∃p s.t. (G,p) is infinitesimal rigid

[Jackson and Jordán09] deficiency of G, def(G):= maxP 6(|P |-1) - 5d(P ) P : partition of V d(P): the number edges connecting different components of P def(G)≧0 (since P ={V}) There exist 6 edge-disjoint spanning trees in 5G ⇔ def(G)= (by Tutte’s tree packing theorem) Theorem(Jackson and Jordán09) The followings are equivalent for G=(V,E): For a generic hinge-configuration p，(G,p) has k d.o.f (i.e., rank R(G,p)=6(|V|-1)-k) def(G)=k The rank of matroid G6(5G) is equal to 6(|V|-1)-k

Panel-hinge frameworks
3-dimentional panel-hinge framework: (G,p) G=(V,E): a graph p: hinge-configuration satisfying “hinge-coplanarity condition” (i.e., all hinges incident to a body lie on a common hyperplane.) vertex　⇔　panel = a hyperplane in R3 edge　⇔　hinge　 Rem. “hinge-coplanarity” is a special geometric relation among hinges; Tay-Whiteley’s characterization may be false...

Theorem G can be realized as an infinitesimally rigid panel-hinge framework if and only if 5G contains six-edge-disjoint spanning trees Def. G is a k-graph if def(G)=k; it is a minimal k-graph if def(G)=k and removal of any edge results in a non k-graph. Theorem: If G is a minimal k-graph, then it can be realized as a panel-hinge framework with k degree of freedom. (Outline) combinatorial part： propose an inductive construction of minimal k-graphs w.r.t. # vertices. algebraic part：provide an explicit construction of a k-dof panel-hinge framework (G,p) following the inductive construction given in combinatorial part.

Rigid subgraphs (combinatorial part)
a subgraph G’=(V’,E’) is called rigid if def(G’)=0 a rigid subgraph is called proper if 1＜|V’|＜|V|. 　　Lemma： For a minimal k-graph G, the graph obtained by contracting a proper rigid subgraph is a minimal k-graph.

Splitting-off operation
A splitting-off at a vertex of degree two: Lemma：Let G be a k-graph with a vertex v of degree 2. Then the graph Gvab obtained by splitting off at v is a k-graph or a (k-1)-graph. Even though G is minimal, Gvab may not be minimal. v a b a b v a b

Splitting-off operation
Lemma：Let G = (V,E) be a minimal k-dof-graph which has no proper rigid subgraph. (i) If k=0, then Gvab is a minimal 0-graph. (ii) If k>0, then Gvab is a minimal (k-1)-graph.

Small degree vertices Lemma：Let G = (V,E) be a minimal k-graph which has no proper rigid subgraph. Then, G has two consecutive vertices of degree two. (sketch of the existence of a degree two vertex) If no proper rigid subgraph exists, there is a basis of G6(5G) that contains 5(E-e) for any edge e ⇒ 5|E-e|≦6(|V|-1) ⇒ 5|E|≦6|V| ⇒ (average degree) = 2|E|/|V| ≦ 2.4 ⇒ there exists a vertex of degree two

Summary of combinatorial part
For a minimal k-graph G=(V,E) if G contains a proper rigid subgraph G’ G/G’ is a minimal k-graph otherwise, G contains a vertex v of degree two if k>0, Gvab is a minimal (k-1)-graph if k=0, Gvab is a minimal 0-graph v a b

Main theorem (algebraic part)
Theorem：Let G=(V,E) be a minimal k-graph. Then G=(V,E) can be realized as a panel-hinge framework (G,p) with rank R(G,p)=6(|V|-1)-k. By induction on |V| |V|=2 |V|＞2 Case 1：G has a proper rigid subgraph G’ Case 2：G has no proper rigid subgraph with k>0 Case 3: G has no proper rigid subgraph with k=0

Case 1: G has a proper rigid subgarph G’=(V’,E’)
G/G’ is a minimal k-graph (by Lemma) (Also, G’ is minimal 0-graph.) By induction, we have panel-hinge realizations (G’,p1) and (G/G’,p2) s.t. (G’,p1) is rigid and (G/G’,p2) has k-d.o.f. Let v* be the vertex obtained by the contraction. Idea: We can consider the rigid framework (G’,p1) as a rigid body. Hence, replacing the panel associated with v* in (G/G’,p2) by a rigid body (G’,p1), the resulting framework has the desired property. v* G’ G G/G’

Case 2: G has no proper rigid subgraph and k>0
G has a vertex v of degree two. For k>0, Gvab is a minimal (k-1)-dof-graph (Gvab,q) (k-1)-d.o.f. Gvab ((k-1)-graph) (G,p1) column operations R(G,p1) R(Gvab,q) *

= 5+6(|V-{v}|-1)-(k-1) = 6(|V|-1)-k
va vb v a b R(p1(va)) -R(p1(va)) R(p1(vb)) -R(p1(vb)) RG, p1[E-va-vb, V-{v}] R(G,p1) = column operations va vb v a b R(p1(va)) R(p1(vb)) -R(p1(vb)) RG, p1[E-va-vb,, V-{v}] (G,p1) va vb a b R(p1(va)) R(q(ab)) -R(q(ab)) RGvab, q[Ev,, V-{v}] v = (Gvab,q) va vb V-{v} R(p1(va)) R(q(ab)) v R(Gvab,q) = rank R(G,p1) ≥ rank R(p1(va)) + rank R(Gvab,q) = 5+6(|V-{v}|-1)-(k-1) = 6(|V|-1)-k

Case 3: G has no proper rigid subgraph (k=0)
G has two vertices v and a of degree two which are adjacent to each other Gvab is a minimal 0-graph (G,p1) (Gvab,q) (G,p2) isomorphism identical (Gavc,q’) (G,p3) *** Show that at least one of (G,p1), (G,p2), and (G,p3) is rigid ***

= 5+6(|V-{v}|-1)=6(|V|-1)-1
va R(p1(va)) (G,p1) vb R(G,p1)= R(p1(vb)) R(Gvab,q) rank R(G,p1) ≥ rank R(p1(va)) + rank R(Gvab,q) = 5+6(|V-{v}|-1)=6(|V|-1)-1 We need to show rank R(G,p1)≥ 6(|V|-1)

Claim: there exists a redundant row in R(Gvab,q) among those associated with ab.
(Sketch) From a combinatorial argument, there exists a redundant edge among 5ab in the combinatorial matroid G6(5Gvab) This redundancy also happen in the rigidity matrix by induction At most 4 edges are used among 5ab v a b a b

q(ab) in (Gvab, q) = p1(vb) in (G,p1)
Analysis of R(G,p1) q(ab) in (Gvab, q) = p1(vb) in (G,p1) R(q(ab)) va vb V-{v} R(p1(va)) v R(Gvab,q) redundant row, say (ab)1 R(G,p1)= va (vb)1* v V-{v} R(p1(va)) R(Gvab,q) - (the row (ab)1) * r1 row operations a linear comb. of rows of R(q(ab)), denoted r1. If is non-singular, we are done!! Note: r1 is nonzero and is determined by R(Gvab,q)

each row of ab in R(Gvab, q) ⇔ each row of va in R(G,p2)
Analysis of R(G,p2) each row of ab in R(Gvab, q) 　⇔ each row of va in R(G,p2) vb va V-{v} R(p2(vb)) R(q(ab)) v R(Gvab,q) the redundant row (ab)1 R(G,p2)= vb (va)1* v V-{v} R(p2(vb)) R(Gvab,q) - (the row (ab)1) * r1 row operations a linear comb. of rows of R(q(ab)), which is equal to r1. If is non-singular, we are done!!

q(ab) in R(Gvab, q) = p3(vb) in R(G,p2)
Analyzing R(G,p3) q(ab) in R(Gvab, q) = p3(vb) in R(G,p2) q(ac) in R(Gvab, q) = p3(va) in R(G,p2) ac va V-{v,a,b,c} p3(ac) q(ac) a v c b -p3(ac) q(ab) -q(ab)) vb R(G,p3)= -q(ac) *** vb ac va V-{a} R(p3(ac)) R(q(ac)) a R(Gvab,q) add columns of a to those of c the redundant row (ab)1

q(ab) in R(Gvab, q) = p3(vb) in R(G,p2)
Analyzing R(G,p3) q(ab) in R(Gvab, q) = p3(vb) in R(G,p2) q(ac) in R(Gvab, q) = p3(va) in R(G,p2) a V-{a} ac R(p3(ac)) the redundant row (ab)1 R(G,p2)= vb va R(q(ac)) R(Gvab,q) ac (vb)1* v V-{v} R(p3(ac)) R(Gvab,q) - (the row (ab)1) * r3 a linear comb. of rows of R(q(ac)), denoted by r3. row operations

Claim: r1 + r3=0 (intuition): R(Gvab,q) =
r1 can be considered as a force applied to the panel Π(a) in (Gvab, q) through the hinge q(ab) r3 can be considered as a force applied to the panel Π(a) in (Gvab,q) through the hinge q(ac) Since the panel Π(a) is incident to only q(ab) and q(ac), these two forces must be in sign-inverse. R(Gvab,q) = ab ac a b c R(q(ab)) -R(q(ab)) R(q(ac)) -R(q(ac)) ***

Analyzing R(G,p3) the redundant row (ab)1 R(Gvab,q) R(Gvab,q)
vb ac va V-{a} R(p3(ac)) R(q(ac)) a R(Gvab,q) the redundant row (ab)1 R(G,p3)= ac (vb)1* v V-{v} R(p3(ac)) R(Gvab,q) - (the row (ab)1) * r3=-r1 row operations

Summary of matrix-transformations
If at least one of is non-singular, we are done. v V-{v} R(p1(va)) R(Gvab,q) - (the row (ab)1) * r1 v V-{v} R(p2(vb)) R(Gvab,q) - (the row (ab)1) * r1 v V-{v} R(p3(ac)) R(Gvab,q) - (the row (ab)1) * r3=-r1

Last step R(p1(va)) r1 R(p2(vb)) r1 R(p3(ac)) r3=-r1 Suppose is singular. Then, r1 is orthogonal to a 2-exntesor of p1(va). Suppose is singular for any choice of p1(va) on Π(a). Then, r1 is orthogonal to every 2-extensors of a line on Π(a). (Similar for the other two matrices.) Suppose all of are singular for every choice of p1(va), p2(vb), p3(ac). Then, r1 is orthogonal to 2-extensors of all lines on Π(a), Π(b), or Π(c). 　 ⇒ These 2-extensors span 6-dimentional space. ⇒ r1=0, a contradiction.

Unsolved Problems Corollary: Let G2 be the square of a graph G. Then, the rank of G2 in generic 3-rigidity matroid is 3|V|-6-def(G) (by Jackson&Jordán 07, 08) Conjecture (Jacobs, Jackson&Jordán07) : Let G be a graph and let u,v ∈V. Then, r(G2+uv) = r(G2) if and only if u and v belong to the same rigid component of G2 (where r is the rank function of 3-rigidity matroid). A rigid component is an inclusionwise-maximal rigid subgraph. Conjecture : Let (G, p) be a panel-hinge framework. Suppose two panels Π(u) and Π(v) are relatively flexible. If connecting between these panels by a bar lying on the intersection of them, then the degree of freedom always decreases. Problem: Efficient computation of the decomposition into redundantly rigid components Open Problem: Provide a simpler proof!!