Download Notes on Rigidity Theory James Cruickshank

Notes on Rigidity Theory
James Cruickshank
Contents
Chapter 0. Introduction
5
Chapter 1. Cauchy’s Rigidity Theorem
1. Polygons and Polyhedra
2. Cauchy’s Theorem
3. Flexible Polyhedra
7
7
12
15
Chapter 2. Bar and Joint Frameworks
1. Basic Definitions
2. Infinitesimal Rigidity
3. The Rigidity Matrix
4. The Maxwell Counting Conditions
17
17
17
21
22
Chapter 3. Laman’s Theorem
1. Generic Placements and the Statement of Laman’s Theorem
2. Henneberg Extensions
3. Proof of Laman’s Theorem
4. The Pebble Game Algorithm
25
25
27
30
32
Chapter 4. Three Dimensional Rigidity
1. The Double Banana
2. Gluck’s Theorem
33
33
34
3
CHAPTER 0
Introduction
These notes are meant to be used as source material for a course
on the rigidity theory of bar and joint frameworks.
5
CHAPTER 1
Cauchy’s Rigidity Theorem
If we join two metal plates together with a hinge, the resulting
structure is not rigid - it is possible to flex the structure by changing the ’dihedral angle’ of the hinge. See Figure 1 for an illustration.
Now, suppose that you construct a polyhedron by joining appropriately shaped metal polygons together with hinges. Will the resulting
structure flex? The answer to this question could conceivably depend
on many different aspects of the structure. It could also depend on
precisely what you mean by the word ’flex’. So to give a correct and
precise answer to the question, we must begin by precisely defining our
terms.
1. Polygons and Polyhedra
A set is convex if the line segment joining any two of its points is
contained entirely within the set (we are being deliberately vague about
the setting here). The convex hull of a set U is the intesection of all
E
I
B
A
J
F
H
C
D
Figure 1. Some polyhedral surfaces. The one on the
right is rigid whereas the one on the left is not.
7
K
8
1. CAUCHY’S RIGIDITY THEOREM
E
F
D
G
C
A
B
Figure 2. A combinatorial 7-gon.
the convex sets that contain U . A convex polygon is the convex hull of
a finite set of points in the Euclidean plane. A convex polyhedron is the
convex hull of a finite set of non coplanar points in three dimensional
Euclidean space. Observe that the boundary of a polygon is topologically equivalent to a circle and that the boundary of a polyhedron is
topologically equivalent to a sphere.
1.1. Combinatorial Polygons and Polyhedra. Note that the
vertices and edges of a convex poly(gon/hedron) form a graph. We
want to characterise the graphs that arise in this way. Thus we make
the following definitions. A combinatorial k-gon or a cycle is a graph
of the form
({v1 , · · · , vk }, {v1 v2 , v2 v3 , · · · , vk−1 vk , vk v1 }).
Combinatorial polyhedra a little more difficult to characterise. Nevertheless, we make the following definition. Recall that a graph is 3connected if there is no pair of vertices whose removal leaves a disconnected graph. Now, a polyhedral graph (or combinatorial polyhedron)
is a 3-connected, planar graph with at least four vertices. Why is this
the correct definiton?
Theorem 1 (Steinitz). A graph is the graph of a three dimensional
convex polyhedron if and only if it is 3-connected, planar and has at
least four vertices.
Proof. Omitted.
Exercise 1. Let G be the graph obtained by deleting a single edge
from K5 . Make a sketch of a polyhedron whose graph is G.
1. POLYGONS AND POLYHEDRA
F
9
D
A
C
B
E
Figure 3. A self-intersecting 6-gon.
Now a polygon is a combinatorial polygon together with an embedding of the vertex set into some plane. Note that polygons may be self
intersecting.
A face of a polyhedral graph is a non-separating cycle without repeated vertices in the graph. A polyhedron is a polyhedral graph together with an embedding of the vertex set into some three dimensional
real vector space such that each face of the polyhedral graph is embedded into a plane.
It is important to note that these definitions are not standard.
There are many different definitions of these terms in the literature
some of which may not be equivalent to the ones given here.
1.2. Euler’s Polyhedral Formula. A plane embedding of a graph
is a embedding of the vertices and edges in the plane without any ’edge
crossings’. A face of such a plane embedding is a connected component
of its complement in the plane.
Lemma 2. If G is a polyhedral graph then there is a one to one
correspondence between faces of the G and faces of any plane embedding
of G.
Proof. The only thing that we need to check is that the boundary
cycle of any face of the plane graph contains no repeated vertices. This
follows from the fact that G is 3-connected and so there is no vertex
whose removal disconnects the graph.
Theorem 3. Let G be a connected plane graph with f faces, e edges
and v vertices. Then
f −e+v =2
10
1. CAUCHY’S RIGIDITY THEOREM
B
C
E
A
H
F
D
I
Figure 4. A (non-convex) polyhedron. ABCDE is a
pentagonal face of this polyhedron.
Proof. We proceed by induction on e. If e = 0 then f = 1 and
v = 1 since G is connected. Now suppose that e ≥ 1 and suppose
that the conclusion is true for all for all connected plane graphs with
fewer than e edges. for some k ≥ 1. Choose an edge uv of G and let
G0 − G − e and let f 0 , e0 respectively v 0 be the number of faces, edges
respectively vertices of G0 . There are two cases to consider.
Case 1: G0 is connected. So f 0 − e0 + v 0 = 2. Moreover, in this case,
0
f = f − 1, e0 = e − 1 and v 0 = v. Therefore,
f − e + v = f 0 + 1 − (e0 + 1) + v 0 = f 0 − e0 + v 0 = 2
Case 2: G0 is not connected. Let G1 and G2 be the components
of G0 . The induction hypothesis applies to each of these components.
Therefore f1 − e1 + v1 = f2 − e2 + v2 = 2 (with the obvious notation).
Moreover, e1 + e2 = e − 1, f1 + f2 = f + 1 and v1 + v2 = v. It follows
immediately that
f −e+v =2
as required
Combining Lemma 2 and Theorem 3 we obtain . . .
Theorem 4 (Euler). Let G be a polyhedral graph with f faces, e
edges and v vertices. Then
f −e+v =2
1. POLYGONS AND POLYHEDRA
11
Exercise 2. Suppose that G is a polyhedral graph in which every
face is a triangle. Prove that e = 3v − 6.
Exercise 3. Prove that there exists a plane graph G, with v vertices, such that every face of G is a pentagon if and only if v ≡ 2
mod 3 and v ≥ 5.
1.3. Congruence. An isometry of Rd is a function f : Rd →
Rd that preserves distance. We say that subset A and B of Rd are
congruent if there is an isometry f of Rd such that F restricts to a
bijection A → B.
As we have observed in previous sections, polygons and polyhedra
are not just subsets of R2 or R3 . They also come equipped with a
combinatorial structure - namely the graph of vertices and edges. So
congrunces of polygons or polyhedra should respect this extra structure. Thus, suppose that P1 and P2 are poly(gons/hedra) in Rd with
the same underlying graph G and with embeddings ρ1 and ρ2 of the
vertex set. We say that P1 and P2 are congruent if there is an isometry
f of Rd such that f ◦ ρ1 = ρ2 .
Note that the Jordan curve theorem implies that if a polygon is
non self intersecting then it divides the plane into precisely two distinct regions, one of which is bounded and one of which is unbounded.
Thus in the case of non self intersecting polygons we can unambiguously refer to the internal angles of the polygon. Similarly a non self
intersecting polyhedron divides R3 into two distinct regions and we can
unambiguously identify the internal dihedral angles of the polyhedron.
We cite some facts from elementary geometry here concerning congruence.
Proposition 5. Combinatorially equivalent non self intersecting
polygons P1 and P2 are congruent if and only if corresponding sides
have the same length and corresponding internal angles have the same
measure.
Proof. The ’only if’ direction is clear since congruences preserve
lengths and angles. For the converse see your lecture notes.
Exercise 4. Fill in the missing details of the proof of the above
proposition.
The analgous proposition for polyhedra is the following.
Proposition 6. Combinatorially equivalent non self intersecting
polyhedra P1 and P2 are congruent if and only if corresponding faces
are congruent and corresponding internal dihedral angles have the same
measure.
12
1. CAUCHY’S RIGIDITY THEOREM
Proof. Choose a congruence between a pair of corresponding faces
and show that this congruence extends to a congruence of the polyhedra. See lecture notes.
2. Cauchy’s Theorem
Suppose that we build a polyhedron from rigid metal plates that
are joined along the edges by hinges. A natural question to consider is
whether or not such a structure will be rigid. What would a non trivial
’flex’ look like? Intuitively, such a flex would allow us to continuously
deform the structure so that the faces preserve their shape but so that
at least one of the dihedral angles between a pair of faces changes.
Cauchy’s rigidity theorem (Theorem 7 below) demonstrates that no
such flex can arise - in fact it is a stronger statement than this.
2.1. Statement of the theorem and some examples.
Theorem 7 (Cauchy). Let P1 and P2 be convex polyhedra in R3
that are combinatorially equivalent (ie they have the same graph of
vertices and edges). If all pairs of corresponding faces of P1 and P2 are
congruent then P1 and P2 are congruent.
Observe that if we could continuously ’flex’ a convex polyhedron
in a non trivial way then for a sufficiently small such flex, the resulting polyhedron would also be convex but would not be congruent to
the original polyhedron. Thus such a continuous flex is ruled out by
Cauchy’s Theorem.
To prove Theorem 7 we first need to prove an elementary fact concerning flexes of ’arms’.
2.2. The Arm Lemma.
Lemma 8 (The Arm Lemma). Suppose that P and Q are convex
k-gons in the plane with vertices p1 , · · · , pk and q1 , · · · , qk respectively.
Suppose also that kpi − pi+1 k = kqi − qi+1 k for i = 1, · · · , k − 1. Finally
suppose that the internal angle of Q at qi is greater than or equal to the
internal angle of P at pi for i = 2, · · · , k−1. Then kq1 −qk k ≥ kp1 −pk k
with equality if and only if all the corresponding internal angles are
equal.
Proof.
Exercise 5. Given an example to show that the conclusion of
Lemma 8 can be false if one of P1 or P2 is allowed to be non-convex.
2. CAUCHY’S THEOREM
13
For the proof of Cauchy’s rigidity theorem we need to prove the arm
lemma for polygons on the sphere. In the case of a convex spherical
polygon, the internal angle at a vertex is the angles that is less than π.
More precisely,
Lemma 9 (The Spherical Arm Lemma). Suppose that P and Q
are convex k-gons on a sphere with vertices p1 , · · · , pk and q1 , · · · , qk
respectively. Suppose also that d(pi , pi+1 ) = d(qi , qi+1 ) for i = 1, · · · , k−
1. Finally suppose that the internal angle of Q at qi is greater than or
equal to the internal angle of P at pi for i = 2, · · · , k − 1. Then
d(q1 , qk ) ≥ d(p1 , pk ) with equality if and only if all the corresponding
internal angles are equal.
Proof. The proof is essentially the same as the planar version
above.
Exercise 6. Use the spherical cosine rule to verify the base case
of the induction argument in the proof of Lemma 9.
To see the relevance of Lemma 9, observe that, given a vertex p
of a convex polyhedron, the intersection of a small sphere centred at
v with the polyhedron is a spherical polyhedron, whose side lengths
are determined by the face angles and whose internal angles are the
dihedral angles of the polyhedron.
2.3. Proof of Cauchy’s Theorem. Suppose that G is a plane
graph (ie a planar graph together with a planar embedding in the
plane). Suppose also that each of the edges of G is coloured either black
or white. For each vertex we can count the number of colour changes as
we traverse the edges around the vertex. That is the number of times
that there is an edge of one color followed immediately by an edge of
the other colour as we traverse the edges around the vertex once. Let
c(v) be the number of colour chages around vertex v. Observe that
c(v) is a nonnegative even number for every vertex v.
Exercise 7. Prove that c(v) is even.
Exercise 8. Suppose that G is an i-gon and that each of the edges
is coloured either black or white. Prove that there are at most 2 2i
vertices of G that are incident to both a white edge and a black edge.
Proposition 10. Given a bicolouring of the edges of a simple plane
graph G, as above, there is some vertex v for which c(v) ≤ 2.
Proof. Let fi denote the number of i-gonal faces of G. Note that
f2 = 0 since G is a simple graph. Observe that there are at most 2 2i
14
1. CAUCHY’S RIGIDITY THEOREM
colour changes as we circumnavigate the edges of a bicoloured i-gon.
Therefore we see that
X
X
i
(1)
c(v) ≤
2fi
2
i≥3
v∈V
Now suppose that c(v) ≥ 4 for all v ∈ V . Then
X
(2)
c(v) ≥ 4|V |.
v∈V
Now Euler’s formula (Theorem 4) tells us that
X
(3)
|V | = 2 + |E| −
fi .
i≥3
Moreover, it is clear that
(4)
X
ifi = 2|E|
i≥3
Combining Equations (3) and (4) we obtain
X
(5)
4|V | = 8 +
(2i − 4)fi .
i≥3
Now, combining (5), (2) and (1) we see that
X
i
(6)
2i − 4 − 2
fi ≤ −8.
2
i≥3
However, this last inequality must be false since 2i − 4 − 2 2i ≥ 0 for
i ≥ 3 and fi ≥ 0 for all i. Thus we have arrived at a contradiction. So
our assumption that c(v) ≥ 4 for all v must be false.
Proof of Theorem 7. Suppose that P and Q are convex polyhedra as in the statement of the theorem. Let G be the graph of P
(and of Q). For an edge of G, colour it white if the dihedral angle in
P is strictly greater than that in Q and colour it black if the dihedral
angle in P is strictly less than that in Q. Do not assign any colour if
the dihedral angles are equal. Let G be the subgraph of G consisting
only of the coloured edges and only the vertices incident to at least one
coloured edge. Suppose that G is not empty (ie it has at least one vertex). Then G is a simple plane graph with a bicolouring of the edges.
By Proposition 10, there is at least one vertex with at most two color
changes. Let P, respectively Q, be the spherical polygon obtained by
intersecting a suitably small sphere centred at the vertex with the faces
of the polyhedron P , respectively Q, that are incident at the vertex.
Color the vertices of P and Q black, white, or not at all, according
3. FLEXIBLE POLYHEDRA
15
to whether the corresponding edges of G are coloured black, white or
not all. Now corresponding edges of P and Q are equal in length since
the edge lengths of these polygons are essentially determined by the
face angles of the faces in the polyhedra. Suppose that there are at
most two colour changes as we traverse around the vertices of P. That
means that there are vertices u and v in P such that there are no black
vertices contained in one of the arcs from u to v and no white vertices
contained in the other arc from u to v. Applying the spherical Arm
Lemma (Lemma 9) to the polygonal arc without white vertices, we
conclude that the distance between the points in Q that correspond
to u and v is less than or equal to the distance between u and v. On
the other hand, applying the Arm Lemma to the polygonal arc with
no black vertices we conclude that the distance between the points in
Q that correspond to u and v is greater than or equal to the distance
between u and v. Thus we must have equality between these two distances and the Arm Lemma entails that none of the vertices of P (or
Q) are in fact coloured black or white and the vertex does not belong
to G, which contradicts our assumption. Thus we conclude that in fact
G is the empty graph - so all of the dihedral angles of Q are equal to
the corresponding dihedral angles of P. Therefore, by Proposition 6,
we conclude that P and Q are congruent.
3. Flexible Polyhedra
Cauchy published a proof of his rigidity theorem in 1813. Although
some errors were later discovered in his proof, it was essentially the
same as the (hopefully correct!) proof given above. It is such a beautiful theorem that it is natural to seek the most general version of it.
While it is fairly clear that the convexity assumption is necessary if one
is to allow ’discrete flexes’ (see the example ???), it is at first tempting
to suspect that there should be no ’continuous flex’ of any polyhedron,
convex or not.
In 1897, Bricard discovered some examples of continuously flexible
structures that we would describe in modern language as self intersecting octahedra. The problem of the continuous rigidity/flexibility
of actual non self intersecting polyhedra remained open for another 80
years however. The suspicion that convexity was not necessary was
further strenghtened when, in 1974, Gluck showed that almost all (in
a certain precise sense) polyhedra are continuously rigid. Thus, it was
a rather surprising when, in 1977, Connelly constructed an example of
a continuously flexible non convex non self intersecting polyhedron.
3.1. Connelly’s Flexor.
CHAPTER 2
Bar and Joint Frameworks
1. Basic Definitions
A simple graph G is a pair (V, E) where E is a subset of the set of
unordered pairs of distinct elements of V . A d-dimensional framework
is a pair (G, ρ), where ρ : V → Rd . The function ρ is called a placement
of the vertex set in Rd . Frameworks (G, ρ1 ) and (G, ρ2 ) are said to be
congruent if there is an isometry f : Rd → Rd such that ρ2 = f ◦ ρ1 .
We say that (G, ρ2 ) is a global flex of (G, ρ1 ) if kρ1 (u) − ρ1 (v)k =
kρ2 (u) − ρ2 (v)k for all uv ∈ E. We say that (G, ρ) is globally rigid if
every global flex of (G, ρ) is congruent to (G, ρ). A continuous flex of
(G, ρ) is a family of frameworks (G, ρt ) where t ∈ [0, ε) for some ε > 0
and t 7→ ρt (v) is a continuous function for all v ∈ V , and ρ0 = ρ. We
say that (G, ρ) is continuously rigid if given any continuous flex (G, ρt )
of (G, ρ), (G, ρt ) is congruent to (G, ρ) for all t.
2. Infinitesimal Rigidity
Cauchy’s Rigidity Theorem can be viewed as a global rigidity theorem for frameworks that arise from convex polyhedra.
Exercise 9. State Cauchy’s Theorem in terms of frameworks.
However, global rigidity is a notoriously hard property to work with
in a more general setting. Some progress has been made recently but
there are very few statements that one can make about global rigidity
that hold for arbitrary graphs. Indeed continuous rigidity is already
very difficult to work with. In order to make significant progress we
take the step, very common in applied mathematics, of linearising the
problem.
Suppose that (G, ρt ) is a continuous flex of a framework. Now further suppose that ρt is a actually a smooth (ie differentiable) function
of t. What can we say about the derivative of ρt with respect to t?
Proposition 11. Suppose that ρt is a smooth flex as described
above. Then for all ij ∈ E,
(ρt (i) − ρt (j)).(ρt (i)0 − ρt (j)0 ) = 0
17
18
2. BAR AND JOINT FRAMEWORKS
where ρt (i) =
d
(ρ (i)).
dt t
Proof. By assumption, for each ij ∈ E, there is some constant cij
such that
(ρt (i) − ρt (j)).(ρt (i) − ρt (j)) = cij
for all t. Therefore
d
((ρt (i) − ρt (j)).(ρt (i) − ρt (j))) = 0
dt
Now apply the product rule to the left hand side of this equation to
obtain the desired conclusion.
Thus we arrive at the following definition. An infinitesimal flex of
a framework (G, ρ) is a function u : V → Rd satisfying
(7)
(ρ(i) − ρ(j)).(u(i) − u(j)) = 0
for all ij ∈ E.
Proposition 11 tells us that any smooth flex of a framework induces
an infinitesimal flex of that framework (note however that it does not
automatically follow that the derivative of a nontrivial smooth flex is a
nontrivial infinitesimal). It is natural to wonder if the converse is true:
does every infinitesimal flex arise in this way. For arbitrary frameworks,
the answer is no.
Exercise 10. Let G = K3 with vertex set {1, 2, 3} and define a
two dimensional placement, ρ, as follows: ρ(1) = (0, 0), ρ(2) = (1, 0)
and ρ(3) = (2, 0). Show that there is an infinitesimal flex of (G, ρ) that
does not arise as the derivative of any smooth flex.
Later we will introduce a special, but very important, class of placements for which this type of pathology cannot occur.
Let F(G, ρ) denote the set of all infinitesimal flexes of (G, ρ). By
definition
Proposition 12. The set F(G, ρ) forms a real vector space under
the operations of pointwise addition and scalar multiplication.
Proof. By definition, F(G, ρ) is the set of solutions of a homogeneous system of linear equations.
Exercise 11. Find a basis for F(G, ρ) where
(1) G is the complete graph on vertex set {1, 2, 3} and ρ(1) =
(0, 0), ρ(2) = (1, 0) and ρ(3) = (0, 1).
(2) G is the complete graph on vertex set {1, 2, 3} and ρ(1) =
(0, 0), ρ(2) = (1, 0) and ρ(3) = (2, 0).
2. INFINITESIMAL RIGIDITY
19
(3) G is a quadrilateral with vertex set {1, 2, 3, 4} and ρ(1) =
(0, 0), ρ(2) = (1, 0), ρ(3) = (1, 1) and ρ(4) = (0, 1).
In the next section we identify a canonical subspace of F that is
always present regardless of the particular frammework.
2.1. Infinitesimal Isometries of Rd . An infinitesimal isometry
of Rd is a vector field x →
7 ξx satisfying
(x − y).(ξx − ξy ) = 0
(8)
for all x, y ∈ Rd .
Observe that the set of all infinitesimal isometries of Rd is a vector
space under the natural operations of pointwise additions and scalar
multiplication. Also, it is clear that if ξ is an infinitesimal isometry
of Rd and if (G, ρ) is a framework in Rd then the function i 7→ ξρ(i) is
an infinitesimal flex of (G, ρ). We call such flexes trivial flexes of the
framework.
We would like to understand the structure of space of infinitesimal
isometries of Rd , and in particular, to compute its dimension.
Proposition 13. Let A be a d × d skew symmetric matrix (ie
A = −A. Then x 7→ Ax is an infinitesimal isometry of Rd .
T
Proof. Observe that
(x − y).(Ax − Ay) =
=
=
=
(x − y)T A(x − y)
(x − y)T AT (x − y)
−(x − y)T A(x − y)
−(x − y).(Ax − Ay)
Therefore (x − y).(Ax − Ay) = 0 as required.
Conversely,
Proposition 14. Let ξ be an infinitesimal isometry of Rd and
suppose that ξ0 = 0. Then there is a unique d × d skew symmetric
matrix A such that
ξx = Ax
d
for all x ∈ R .
Proof. We can prove this by induction on the dimension d. First
observe that if ξ is an infinitesimal isometry and ξ0 = 0, then x.ξx = 0
for all x ∈ Rd . Now let v = ξed . The matrix veTd − ed v T is clearly skew
symmetric (check it!). Therefore x 7→ (veTd − ed v T )x is an infinitesimal
isometry of Rd (by Proposition 13) which we will denote by ν. Now
20
2. BAR AND JOINT FRAMEWORKS
we calculate that νed = v. Therefore if ξ = ξ − ν we have ξ ed = 0.
Moreover for any y ∈ Rd ,
eTd ξ y = eTd ξy − eTd (veTd − ed v T )y
= eTd ξy + v T y
= −eTd ξed + eTd ξy + v T y − y T ξy
= −(ed − y)(ξed − ξy )
= 0
∼ d−1 .
It follows that ξ restricts to an infinitesimal isometry of the e⊥
d = R
By induction it follows that there is some d × d skew symmetric matrix
A such that ξ x = Ax and such that the last row and the last column
of A are zero. Now let A = A + (ved AT − ed v t ).
Proposition 15. Let ξ be an infinitesimal isometry of Rd . There
is a unique vector b ∈ Rd and a unique d × d skew symmetric matrix A
such that
ξx = Ax + b
for all x ∈ Rd .
Proof. Let b = ξ0 . . .
Corollary 16. The set of all infinitesimal isometries of a d.
dimensional Euclidean space is a vector space of dimension d(d+1)
2
Pm of points p1 , · · · , pm is a linear combination
PmAn affine combination
λ
p
such
that
i=1 λi = 1. The affine span (or affine hull) of
i=1 i i
p1 , · · · , pm is the set of all affine combinations of those points.
Proposition 17. Suppose that p1 , · · · , pm affinely span Rd . Then
any infinitesimal isometry of Rd is determined by its values at p1 , · · · , pm
Proof. This follows immediately from the observation that any
affine function Rd → Rd is determined by its values on any set that
affinely spans Rd .
Corollary 18. Let (G, ρ) be a framework in Rd and suppose that
the affine span of ρ(V ) is Rd . The set of trivial infinitesimal flexes of
(G, ρ) is a real vector space of dimension d(d+1)
.
2
Proof. By Proposition 17 we see that there is a bijection between
the set of trivial flexes and the set of infinitesimal isometries of Rd . In
fact it is easy to see that this bijection is in fact a linear isomorphism.
The result now follows from Corollary 16.
3. THE RIGIDITY MATRIX
21
Exercise 12. Let ξ be an infinitesimal isometry of R2 . Prove that
either
• there is some x ∈ R2 such that ξx = 0, or,
• there is some constant b such that ξy = b for all y ∈ R2 .
Exercise 13. Give an example of a non constant, nowhere zero
infnitesimal isometry of R3 .
Exercise 14. Recall that an orthogonal matrix (over the real numbers) is a d × d matrix U that satisfies U T U = Id (i.e. the transpose of
U is the inverse of U ). Let O(d) denote the set of all orthogonal d × d
matrices and let α : R → O(d) be a differentiable function such that
α(0) = Id . Prove that α0 (0) is a skew symmetric matrix.
Exercise 15. Let O(d) be the group of orthogonal d × d matrices
with real entries. Observe that O(d) is a smooth submanifold of Rd×d .
Let T be the tangent space to O(d) at the identity element. Prove that
there is a canonical linear isomorphism between T and the space of d×d
skew symmetric matrices with real entries. In other words prove that
there is a unique skew symmetric matrix corresponding to each tangent
vector to O(d) at the identity and prove that this correspondence is a
linear isomorphism of vector spaces.
3. The Rigidity Matrix
Given a framework (G, ρ) in Rd , we define its rigidity matrix, R(G, ρ),
to be a matrix whose rows are indexed by the edges of G and that has
d|V | columns. For an edge ij, the ijth row has zeroes everywhere except possibly in columns (i − 1)d + 1 to id and (j − 1)d + 1 to jd. In
columns (i − 1)d + 1 to id the entries are given by (ρ(i) − ρ(j))T and
in columns (j − 1)d + 1 to jd the entries are given by (ρ(j) − ρ(i))T
This description is accurate if perhaps somewhat opaque. A more
insightful viewpoint is perhaps to consider the rigidity matrix as the
coefficient matrix of the system of equations in d|V | variables (denoted
by u : V → Rd ) defined by Equation (7) as ij varies over all edges of G.
In particular the rigidity matrix is a |E| × d|V | matrix whose (right)
kernel consists precisely of the infinitesimal flexes of the framework
(indeed the kernel is this by construction essentially).
Proposition 19. Suppose that ρ is a placement of G in Rd and
that the affine span of ρ(V ) is Rd . Then the rank of R(G, ρ) is at most
d|V | − d(d+1)
. Moreover, the rank of R(G, ρ) is equal to d|V | − d(d+1)
2
2
if and only if (G, ρ) is infinitesimally rigid.
22
2. BAR AND JOINT FRAMEWORKS
Proof. The rank-nullity theorem tells us that
(9)
rank(R(G, ρ)) + dim(ker(R(G, ρ)) = d|V |.
Now, let F be the space of infinitesimal flexes of (G, ρ) and let T be the
space of trivial infinitesimal flexes of (G, ρ). We know that T ≤ F with
equality if and only if (G, ρ) is infinitesimally rigid (this is the definition
of infinitesimal rigidity). Corollary 18 tells us that dim(T ) = d(d+1)
.
2
∼
Also, as we have observed above F = ker(R(G, ρ)). Putting all of
,
these observations together, we see that dim(ker(R(G, ρ))) ≥ d(d+1)
2
with equality if and only if (G, ρ) is infinitesimally rigid. The required
conclusions follow immediately from Equation (9).
The following two corollaries demonstrate the utility of these elementary linear algebraic results in extracting purely combinatorial
information about the graph of a rigid framework. We will pursue this
theme in much more depth as we proceed.
Corollary 20. Suppose that G is a graph with v vertices (v ≥ 3)
and that (G, ρ) is an infinitesimally rigid framework in the plane. Then
G has at least 2v − 3 edges.
Corollary 21. Suppose that G is a graph with v vertices (v ≥ 4)
and that (G, ρ) is an infinitesimally rigid framework in R3 . Then G
has at least 3v − 6 edges.
Exercise 16. Suppose that (G, ρ) is an infinitesimally rigid framework in Rd and that G has at least d + 1 vertices. Prove that ρ(V )
affinely spans Rd . (Hint: Show that if all of the vertices are contained
in some proper affine subspace then any function V → the perp of that
affine subspace is a flex of the framework.)
Exercise 17.
(1) Make a sketch of the framwork (G, ρ), where
G = ({1, 2, 3, 4}, {12, 23, 34, 14}) and where ρ : V → R2 is
defined by ρ(1) = (0, 0), ρ(2) = (1, 1), ρ(3) = (1, 2) and
ρ(4) = (0, 2).
(2) Write down the rigidity matrix R(G, ρ) for this framework and
calculate its rank. What can you conclude about the framework?
(3) Give an example of a non-trivial infinitesimal flex of this framework and add arrows to the sketch from part (a) to indicate this
infnitesimal flex.
4. The Maxwell Counting Conditions
Corollaries 20 and 21 provide necessary conditions for a framework
to be infinitesimally rigid in two and three dimensions. However these
4. THE MAXWELL COUNTING CONDITIONS
23
conditions are clearly not sufficient to ensure rigidity as the following
examples demonstrate. This suggests two obvious questions
(1) Can we find a stronger necessary condition?
(2) Can we find a nice necessary and sufficient condition?
The answer to the first question is unequivocally yes and this section
is devoted to explaining that answer. For the second question, we can
answer in the affirmative in the plane but in three dimensions this
question remains open and indeed is one of the major open problems
in contemporary research in rigidity theory.
We say that a framework is minimally infinitesimally rigid (or isostatic) if it is infinitesimally rigid and if the removal of any edge results
in a non rigid framework.
Some graph theoretic notation: suppose that G = (V, E) is a graph
and that V 0 ⊂ V . Then E(V 0 ) denotes the subset of E consisting of
edges both of whose vertices lie in V 0 - ie the set of edges of G spanned
by V 0 .
Theorem 22. Suppose that G = (V, E) has at least d + 1 vertices
and that (G, ρ) is a minimally infinitesimally rigid framework in Rd .
Then
(1) |E| = d|V | − d(d+1)
.
2
0
(2) For any subset V ⊂ V such that |V 0 | ≥ d, |E(V 0 )| ≤ d|V 0 | −
d(d+1)
.
2
Proof. By Exercise 16 we know that the affine span of ρ(V ) is Rd .
Therefore by Proposition 19 we see that rank(R(G, ρ)) = d|V | − d(d+1)
.
2
Since (G, ρ) is minimally infinitesimally rigid, Proposition 19 also tells
us that the rank of the matrix obtained by deleting any row of R(G, ρ)
is strictly less than d|V | − d(d+1)
. This means that the rows of R(G, ρ)
2
are linearly independent. Therefore rank(R(G, ρ)) = |E| which proves
(1).
Now suppose that |E(V 0 )| > d|V 0 | − d(d+1)
for some V 0 ⊂ V with
2
at least d + 1 elements. Consider the submatrix of R(G, ρ) consisting
of those rows corresponding to the edges in E(V 0 ). Call this submatrix S. It is clear that S is the rigidity matrix of the framework
(G0 , ρ0 ) where G0 = (V 0 , E(V 0 )) and ρ0 = ρ|0V augmented with some
columns of zeroes (that correspond to the vertices in V − V 0 ). Now
adding columns of zeroes has no effect on the rank of a matrix. There- the last inequality
fore rank(S) = rank(R(G0 , ρ0 )) ≤ d|V 0 | − d(d+1)
2
coming from Proposition 19. But S has at least d|V 0 | − d(d+1)
+1
2
24
2. BAR AND JOINT FRAMEWORKS
rows by assumption. Therefore the rows of S are not linearly independent. Therefore the rows of R(G, ρ) are not linearly independent and
rank(R(G, ρ)) < |E| = d|V |− d(d+1)
. By Proposition 19 this means that
2
(G, ρ) is not infinitesimally rigid, contradicting our hypothesis.
Statements (1) and (2) from Theorem 22 are sometimes collectively
referred to as the Maxwell counts (or Maxwell counting conditions).
The planar case is worth particular mention at this point. In this case
the Maxwell counts are
(1) |E| = 2|V | − 3.
(2) |E(V 0 )| ≤ 2|V 0 | − 3 for any V 0 ⊂ V with at least two elements.
Note that we need only require that |V 0 | ≥ 2 in statement (2) since the
inequality is automatically also satisfied for doubletons in this case (ie
the planar case). A graph satisfying condition (2) is known as a (2, 3)sparse graph. Sparse graphs have been the subject of much independent
interest in purely combinatorial graph theory and the Maxwell counting conditions demonstrate the existence of an intriguing connection
between the apparently geometric problems of rigidity theory and the
purely combintorial study of sparse graphs. Indeed in the next chapter,
we will see that in the planar case the connection is even stronger than
we might suspect at this point.
CHAPTER 3
Laman’s Theorem
1. Generic Placements and the Statement of Laman’s
Theorem
A placement ρ is said to be generic if the multiset consisting of all of
the coordinates of all of ρ(v), v ∈ V is algebraically independent over
the rational numbers. In other words there is no nontrivial rational
polynomial relationship between the coordinates of ρ(V ).
A placement ρ : V → Rd is said to be general or in general position
if, for i = 1, · · · , d + 1 there is no i-element subset of V whose image
under ρ lies in an 1 − 2-dimensional affine subspace of Rd .
Exercise 18. Prove that if |V | ≥ d + 1 then ρ is general if and
only if the image of every |V | + 1-element subset of V affinely spans
Rd .
Proposition 23. Generic placements are general.
Proof. Suppose that ρ is a placement of k points in Rd . If k ≤ d
then the image of ρ lies in some k − 1 dimensional affine subspace so
we may assume wlog that k ≥ d + 1.
Suppose that the affine span of ρ(1), · · · , ρ(d + 1) is contained in
some d-dimensional affine hyperplane. Then


1
ρ(1)
1
ρ(2) 
=0
det 
..
 ...

.
1 ρ(d + 1)
(Why?) Therefore ρ is not generic.
Exercise 19. Give an example of a general placement that is not
generic.
Exercise 20. Find a nontrivial polynomial C in eight variables
with rational coefficients such that
C(x1 , y1 , x2 , y2 , x3 , y3 , x4 , y4 ) = 0
if (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) and (x4 , y4 ) lie on a common circle. (Hint:
Ptolemy’s Theorem)
25
26
3. LAMAN’S THEOREM
Theorem 24. Let ρ be a generic placement of a graph G. Then
(G, ρ) is continuously rigid if and only if it is infinitesimally rigid.
Proof. Possibly omit the proof of this theorem.
Proposition 25. Suppose that (G, π) is an infinitesimally rigid
framework in Rd . Then (G, ρ) is infinitesimally rigid for any generic
placement ρ in Rd .
Proof. We begin by recalling a basic characterisaton of the rank
of a matrix. A submatrix of a matrix A is a matrix obtained by deleting some rows and columns of A. One can show that rank(A) is the
largest k for which there exists a k × k submatrix of A with nonzero determinant. We apply this observation to the rigidity matrices R(G, π)
and R(G, ρ). Let B be any square submatrix of R(G, π). The entries
of B are algebraic combinations of the coordinates of π(V ). Therefore
the determinant of B is a polynomial function of the coordinates of
π(V ). Suppose that this determinant is nonzero. That means that
the polynomial in question is nontrivial (it has a non zero value). Let
B 0 be the corresponding submatrix of R(G, ρ). Since ρ is generic and
the polynomial corresponding to the determinant of B 0 is a nontrivial
polynomial, we conclude that det(B 0 ) 6= 0. Therefore rank(R(G, ρ)) ≥
rank(R(G, π)). Now, since (G, π) is infinitesimally rigid, Proposition
19 of Chapter 2 tells us that rank(R(G, π)) = d|V | − d(d+1)
. There2
d(d+1)
fore rank(R(G, π)) = d|V | − 2 also, and (G, ρ) is infinitesimally
rigid.
Theorem 26. Suppose that ρ1 and ρ2 are generic placements of G.
Then (G, ρ1 ) is rigid if and only if (G, ρ2 ) is rigid.
Proof. This follows immediately from Proposition 25.
A graph G is said to be generically rigid in Rd , or generically d-rigid
if there is some generic placement ρ for which (G, ρ) is infinitesimally
rigid. Note that we do not need to say generically infinitesimally rigid
or generically continuously rigid, by dint of Theorem 24.
Generic d-rigidity is a graph theoretic property. So it makes sense
to look for purely graph theoretic characterisations of this property for
various values of d. For example,
Exercise 21. Prove that G is generically 1-rigid if and only if G
is a connected graph.
It is worth noting that there are various inequivalent definitions of
generic in the literature. The one given here seems to be the most
commonly used in the recent literature on rigidity theory. However,
2. HENNEBERG EXTENSIONS
27
the reader should be aware of this lack of standardisation and should
take care to ensure that she is fully aware of the particular definition
that is being used at any time.
1.1. Statement of Laman’s Theorem. The following theorem
provides the desired characterisation of generic rigidity in the plane (ie
generic 2-rigidity).
Theorem 27 (Laman). Suppose that G = (V, E) satisfies the following two conditions:
(1) |E| = 2|V | − 3.
(2) |E(V 0 )| ≤ 2|V 0 | − 3 for all V 0 ⊂ V such that |V 0 | ≥ 2.
Then G is generically rigid in the plane.
2. Henneberg Extensions
Let G = (V, E) be a graph. A Henneberg extension of G is a graph
formed by deleting some edges of G and then adding a new vertex
together with some edges between the new vertex and the existing
vertices of G. We will also require that the set of neighbours of the
new vertex contains all of the vertices that were incident to the deleted
edges. If G0 is obtained from G by deleting k edges and adjoining a
new vertex of degree k +d then we say that G0 is a Henneberg extension
of G of type (d, k) or alternatively, G0 is a d-dimensional Henneberg kextension of G. In some of the literature, different terminology is used.
For example, what we will call a Henneberg 0-extension is sometimes
called a vertex addition (usually the context will make the value of
d clear in this case). What we will call a Henneberg 1-extension is
sometimes called an edge split.
Exercise 22. Why is a 1-extension called an edge split?
The basic question that we want to investigate in this section is
this: Under what circumstances do d-dimensional Henneberg extensions preserve generic rigidity in Rd ? The answer to this question is
only known for certain special cases that we will describe below.
Proposition 28. Suppose that G0 is a 2 dimensional 0-extension
of G. Then G is generically rigid in the plane if and only if G0 is
generically rigid in the plane.
Proof. Let ρ0 be a generic placement of V 0 in the plane. It is
clear that ρ0 restricts to a generic placement of ρ of V in the plane.
So it suffices to prove that (G, ρ) is rigid if and only if (G0 , ρ0 ) is rigid.
28
3. LAMAN’S THEOREM
We suppose that V = {1, 2, · · · , n}, V 0 = {0, 1, 2, · · · , n} and that
E 0 = E ∪ {01, 02}.
Now suppose that (G, ρ) is rigid and let u0 be an infinitesimal flex
of G0 . If u is the restriction of u0 to V , then there is some infinitesimal
isometry f of R2 such that
(10)
u(i) = f (ρ(i)), i = 1, · · · , n
(since every infinitesimal flex of rigid framework is a trivial flex). Now
consider the flex u0 of (G0 , ρ0 ) given by u0 (i) = u0 (i) − f (ρ0 (i)). By
Equation (10) we see that u0 (i) = 0 for i = 1, · · · , n. Also we know
that
(11)
(ρ0 (0) − ρ0 (i)).(u0 (0) − u0 (i)) = 0
for i = 1, 2 since u0 is a flex of (G0 , ρ0 ). Therefore
(ρ0 (0) − ρ0 (1)).u0 (0) = 0
(ρ0 (0) − ρ0 (2)).u0 (0) = 0
But ρ0 (0) − ρ0 (1) and ρ0 (0) − ρ0 (2) are linearly independent since ρ0 is
a generic placement. Therefore u0 (0) = 0. Therefore u0 (0) = f (ρ0 (0))
and we conclude that u0 is a trivial flex. Therefore (G0 , ρ0 ) is rigid.
On the other hand suppose that (G0 , ρ0 ) is rigid and let u be a flex
of (G, ρ). Since ρ0 (0) − ρ0 (1) and ρ0 (0) − ρ0 (2) are linearly independent
as above, we see that the system of equations (in variable v)
(ρ0 (0) − ρ0 (1)).v = (ρ0 (0) − ρ0 (1)).u(1)
(ρ0 (0) − ρ0 (2)).v = (ρ0 (0) − ρ0 (1)).u(2)
has a unique solution. Set u0 (0) equal to this solution and set u0 (i) =
u(i) for i = 1, · · · , n. It is clear that u0 is a flex of (G0 , ρ0 ) and so must
be a trivial flex. Therefore u is a also a trivial flex and we conclude
that (G, ρ) is rigid.
Exercise 23. Prove that 0-extensions preserve generic rigidity in
all dimensions. (Hint: the proof is essentially as easy generalisation of
the proof of Proposition 28)
Now we consider 1-extensions. Again in this situation we have a
strong result.
Proposition 29. Suppose that G0 is a 2 dimensional 1-extension
of G. If G is generically rigid in the plane then G0 is generically rigid
in the plane.
2. HENNEBERG EXTENSIONS
29
Proof. We will assume that V = {1, · · · , n}, V 0 = {0, · · · , n} and
that E 0 = E − {12} ∪ {01, 02, 03}. Suppose first that G is generically rigid in the plane and let ρ be a generic placement of G in R2 .
By Proposition 25 it suffices to find just one (not necessarily generic)
placement ρ0 of G0 for which (G0 , ρ0 ) is infinitesimally rigid. We define
ρ0 as follows. For i ≥ 1, ρ0 (i) = ρ(i) and ρ0 (0) is defined to be any point
on the line joining ρ(1) and ρ(2) that is distinct from both ρ(1) and
ρ(2) (note that this line is unique since ρ is generic).
Now suppose that u0 is an infinitesimal flex of (G0 , ρ0 ). Observe
that ρ0 (0) − ρ0 (1) = λ(ρ0 (1) − ρ0 (2)) for some λ 6= 0 and ρ0 (0) − ρ0 (2) =
µ(ρ0 (1) − ρ0 (2) for some µ 6= 0. Therefore
(ρ(1) − ρ(2)).(u0 (1) − u0 (2)) = (ρ0 (1) − ρ0 (2)).(u0 (1) − u0 (0))
+(ρ0 (1) − ρ0 (2)).(u0 (0) − u0 (2))
1 0
=
(ρ (0) − ρ0 (1)).(u0 (1) − u0 (0))
λ
1
+ (ρ0 (0) − ρ0 (2)).(u0 (0) − u0 (2))
µ
= 0+0
= 0.
In other words, (G0 , ρ0 ) flexes as though 12 is an edge of G0 even though
it is not. Therefore u0 in fact restricts to a flex of (G, ρ) which is rigid.
Thus we can in fact assume that u0 restricts to the zero flex of (G, ρ)
(just subtract an appropriate trivial flex as necessary - as in the proof
of Proposition 28). So we have shown that u0 (i) = 0 for i = 1, · · · , n.
It remains to show that u0 (0) = 0. However, ρ0 (0), ρ0 (1) and ρ(3) are
not collinear (why?). Moreover u0 (1) = u0 (3) = 0. Now, it follows (as
in the proof of Proposition 28 that u0 (0) = 0 also. Therefore u0 is the
zero flex. Therefore (G0 , ρ0 ) is infinitesimally rigid as required.
Exercise 24. Give an example that shows that the converse to
Proposition 29 is not in general true.
Exercise 25. Generalise Proposition 29 to higher dimensions.
One might hope that there are similar results for k-extensions where
k ≥ 2. However only much weaker results are known in this case. We
will return to this topic when we discuss three dimensional rigidity in
greater depth later on.
Exercise 26. Find example that demonstrates that two dimensional 2-extensions need not preserve generic rigidity in the plane.
30
3. LAMAN’S THEOREM
3. Proof of Laman’s Theorem
Proposition 30. Suppose that G = (V, E) is a graph that has the
following properties.
(1) |E| = 2|V | − 3.
(2) |E(V 0 )| ≤ 2|V 0 | − 3 for all V 0 ⊂ V such that |V 0 | ≥ 2.
Then there is a sequence of graphs
K3 = G1 , G2 , · · · , Gm = G
where Gi+1 is a two dimensional k-extension of Gi for k = 0, 1.
Proof. The proof that we present here is essentially the same as
the original proof of Laman. We prove the theorem by induction on |V |.
If |V | = 3, then G = K3 and there is nothing to prove. Now suppose
that |V | ≥ 4. First we will show that G has no vertices of degree
1. Suppose that i is such a vertex. Then let V 0 = V − {i}. Clearly
|E(V 0 )| = |E| − 1 = 2|V | − 4 = 2|V 0 | − 2 > 2|V 0 | − 3 contradicting our
hypothesis. Now suppose that G has no vertices of degree 2 or 3. Then
4|V | ≤ 2|E| which contradicts the hypothesis that |E| = 2|V | − 3. So
we conclude that if G has no vertices of degree 2 then it must have
some vertices of degree 3.
Now suppose that G has a vertex of degree 2. Let H be the graph
obtained by deleting this vertex and its incident edges. Clearly G is
a 2-dimensional 0-extension of H. Moreover, it is easy to see that H
satisfies the hypothesis of the proposition (exercise!) and since it has
one fewer vertices we can apply the induction hypothesis to H and we
see that the conclusion of the proposition is valid for G.
Now suppose that G has no vertex of degree 2. Then, as above, we
know that G must have a vertex of degree 3. Let V = {0, 1, 2, 3, · · · , n}
and suppose (wlog) that the set of neighbours of 0 is {1, 2, 3}.
First observe that least one of the edges 12, 23 and 13 is not present
in G. If they were all present, then the vertex set {0, 1, 2, 3} would
violate condition (2) of the hypotheses.
Now let us make the following supposition:
(S) There are subsets Vi ⊂ V for i = 1, 2, 3 such that
({1, 2, 3} − {i}) ⊂ Vi and such that |E(Vi )| = 2|Vi | − 3
for i = 1, 2, 3.
Observe that i ∈
/ Vi , for if it was, then Vi ∪ {0} would violate the
Maxwell-Laman counts for G.
Now we will show that V1 ∩ V2 = {3}. Suppose that there is some
a ∈ V1 ∩ V2 , a 6= 3. Now consider the vertex set W = V1 ∪ V2 ∪ {0}.
3. PROOF OF LAMAN’S THEOREM
31
Clearly {1, 2, 3} ⊂ W . Therefore
|E(W )| =
≥
=
≥
=
=
=
>
E(V1 ∪ V2 ) + 3
|E(V1 )| + |E(V2 )| − |E(V1 ) ∩ E(V2 )| + 3
2|V1 | − 3 + 2|V2 | − 3 − |E(V1 ∩ V2 )| + 3
2|V1 | − 3 + 2|V2 | − 3 − 2|V1 ∩ V2 | + 3 + 3
2(|V1 | + |V2 | − |V1 ∩ V2 |)
2|V1 ∪ V2 |
2|W | − 2
2|W | − 3
which contradicts hypothesis (2). Similarly V2 ∩V3 = {1} and V1 ∩V3 =
{2}.
Now let U = V1 ∪ V2 ∪ V3 ∪ {0}. Then, by the principle of inclusion
and exclusion
|U | = |V1 |+|V2 |+|V3 |−|V1 ∩V2 |−|V2 ∩V3 |−|V1 ∩V3 |+|V1 ∩V2 ∩V3 |+1.
Therefore,
|U | = |V1 | + |V2 | + |V3 | − 2.
Now
|E(U )| = |E(V1 ∪ V2 ∪ V3 )| + 3
≥ |E(V1 ) ∪ E(V2 ) ∪ E(V3 )| + 3
= |E(V1 )| + |E(V2 )| + |E(V3 )|
−|E(V1 ) ∩ E(V2 )| − |E(V2 ) ∩ E(V3 )| − |E(V1 ) ∩ E(V3 )|
+|E(V1 ) ∩ E(V2 ) ∩ E(V3 )| + 3
= |E(V1 )| + |E(V2 )| + |E(V3 )|
−|E(V1 ∩ V2 )| − |E(V2 ∩ V3 )| − |E(V1 ∩ V3 )|
+|E(V1 ∩ V2 ∩ V3 )| + 3
= |E(V1 )| + |E(V2 )| + |E(V3 )| + 3
= 2|V1 | − 3 + 2|V2 | − 3 + 2|V3 | − 3 + 3
= 2(|V1 | + |V2 | + |V3 |) − 6
= 2(|U | + 2) − 6
= 2|U | − 2
> 2|U | − 3
which contradicts hypothesis (2).
Thus, we have shown that (S) must be false. Therefore for some
i ∈ {1, 2, 3} the following is true:
32
3. LAMAN’S THEOREM
(T) For any V 0 ⊂ V such that {1, 2, 3} − {i} ⊂ V 0 , we
have |E(V 0 )| ≤ 2|V 0 | − 4.
Let us assume (wlog) that (T) is true for i = 1. In particular we
note that (T) implies that 23 is not an edge of G (just apply (T) with
V 0 = {2, 3}). Now let G0 be the graph obtained by deleting 0 and
its incident edges from G and inserting the edge 23. It is clear that
(T) implies that G0 satisfies the hypothesis of the proposition, so the
inductive hypothesis ensure that the conclusion of the proposition holds
true for G0 . It is also clear that G is a 1-extension of G0 . Therefore, by
induction, the conclusion holds for G.
Proof of Theorem 27 (Laman’s Theorem). By Proposition
30 there is a sequence of graphs K3 , G2 , · · · , Gm = G where Gi+1 is a
two dimensional 0 or 1-extension of Gi . Now K3 is generically rigid
in the plane (why?). Therefore by Propositions 28 and 29 we see that
Gi is generically rigid in the plane for i = 2, 3, · · · , m. Therefore G is
generically rigid in the plane.
Exercise 27. Find a sequence of Henneberg 0 or 1-extensions, as
in Proposition 30 that starts with K3 and terminates with K3,3 .
Exercise 28. Let G be the graph obtained by removing a 6-cycle of
distinct edges from K6 . Prove that G is generically rigid in the plane.
Exercise 29. Let G be a 2-connected planar graph which has exactly one face that is not a triangle and such that every vertex of the
graph belongs to the non triangular face. Prove that G is generically
rigid in the plane. (Hint: Draw some examples of graphs that satisfy
the hypotheses, now try to prove that every such graph has a vertex of
degree 2.)
4. The Pebble Game Algorithm
Theorem 27 is a beautifully clean characterisation of graphs that
are generically rigid in the plane. However, from an algorithmic point
of view it is not ideal. There are 2|V | − |V | − 1 subsets of V with at
least two elements, so the naive idea of checking the Maxwell counts
directly does not lead to a very efficient algorithm for checking rigidity.
It turns out, however, that there is a nice polynomial time algorithm.
CHAPTER 4
Three Dimensional Rigidity
1. The Double Banana
Let G be the graph shown in Figure 1 It is clear that G is not
generically 3-rigid. However, it is also clear that G satisfies the three
dimensional Maxwell counts. That is |E| = 3|V | − 6 and |E(V 0 )| ≤
3|V 0 | − 6 for all V 0 ⊂ V , |V | ≥ 4. So the obvious generalisation of
Laman’s Theorem to three dimensions is false. One might hope that
if we insist that G is also 3-connected, then the three dimensional
Maxwell counts will be sufficient to ensure rigidity. However . . .
Exercise 30. Show that there is a 0-extension of the double banana
that is 3-connected, satisfies the three dimensional Maxwell counts and
is not generically 3-rigid.
1.1. The Graver-Tay-Whiteley Conjecture. We know that 0
and 1-extensions preserve generic rigidity in all dimensions. However,
it is possible to find graphs satisfying the three dimensional three dimensional Maxwell counts that have no vertices of degree 2, 3 or 4.
Exercise 31.
(1) Suppose that G is a polyhedral graph and
that every face of G is a triangle. Prove that G satisfies the
three dimensional Maxwell counts.
(2) Suppose that G is a polyhedral graph with only triangular faces
and with no vertices of degree less than 5. Use Euler’s formula
to find a lower bound for the number of vertices.
(3) Prove that this lower bound is sharp (by finding an example).
This means that we need to understand three dimensional Henneberg 2-extensions if we hope to generalise the inductive arguments
that we used to prove Laman’s Theorem.
Example 31. Let G be K5 with one edge removed. It is clear that
G is generically 3-isostatic (prove it!). Let e and f be two edges of
G that share a common vertex and whose common vertex has degree
Figure 1. The double banana.
33
34
4. THREE DIMENSIONAL RIGIDITY
Figure 2. G0 is a non generically rigid 2-extension of
the generically rigid G.
Figure 3. G0 is a non generically rigid 2-extension of
the generically rigid G.
3 in G. Let G0 be the 2-extension of G obtained by deleting e and f
and adjoining a vertex of degree 5. It is clear that G0 is not generically
3-rigid (it is not even 3-connected). See Figure 2
Is it possible to find other essentially different examples of 2-extensions
that break rigidity? One might consider an example such as depicted
in Figure 3 as being different. However, we observe that in this case,
the graph obtained by deleting the two edges e and f flexes as if all of
the possible edges between 1, 2, 3 and 4 are present, even though some
of them are not.
Let G be a graph and let i and j be vertices of G. We say that
ij is a generically 3-implied edge of G is given a flex u of some generic
framework (G, ρ),
(ρ(i) − ρ(j)).(u(i) − u(j)) = 0.
In other words, infintesimal flexes of (G, ρ) behave as if the edge ij is
present, regardless of whether or not that edge is actually in G. Clearly,
of course, every edge of G is also an implied edge.
Exercise 32. Prove that e is a generically 3-implied edge of G if
and only if, given a generic placement ρ : V → R3 , the rank of R(G, ρ)
is equal to the rank of R(G ∪ e, ρ).
Conjecture 32 (Graver, Tay and Whiteley). Suppose that G is a
generically 3-rigid graph and that G0 is a three dimensional 2-extension
of G that is not generically 3-rigid. Let N be the set of neighbours of
the new vertex of G and let e and f be the edges of G that are not in G0 .
Then e and f share a common vertex v ∈ N and, for all i, j ∈ N −{v},
ij is a generically 3-implied edge of G − {e, f }.
2. Gluck’s Theorem