for inclusion as chapter 7 in the volume:

The Geometry of Rigid Structures version:'2o Jan 1987

Bipartite Frameworks Henry Crapo

Introduction There is one class of bar-and-joint structures for which

virtually all the essential statical and mechanical properties are known. These are those structures whose underlying graphs are complete bipartite. The complete bipartite framework K(A, B) consists of two sets A, B of nodes; for every node a e A and for

every node b e B, there is a bar eab joining a to b. If the set A has cardinality m, and B has cardinality n, we write K(A, B)= Km,n· Such structures first attracted interest because it was rather a surprise to fmd rigid structures with no triangles. It had been known since [Crofton 1878] that the complete bipartite structure · K3,3 is generically isostatic, but becomes infinitesimally flexible (and simultaneously dependent) whenever its six nbdes lie.on a quadric curve in the plane. This result, once remarkable for its isolation from the rest of the developing theory of structural rigidity, was frequently the subject of discussion in the Structural · Topology group during the 1970's. It seemed that quadric surfaces should play a major role in rigidity theory, as defming loci of special position, if for no other reason than that the conditions imposed by bars on the locations of nodes are quadratic in the coordinates of those nodes. But at that stage, the K 3,3 result had not been extended to any larger family of structures. The most tempting family to consider as possible generalizations of K3,3 were the complete bipartite structures. At a 1978 meeting of the AMS in Syracuse, Whiteley conjectured that K4,6 (with 24 = 3v-6 bars on v = 10 nodes) should be generically isostatic in 3-dimensional space. (One might expect a failure of infmitesimal rigidity when the vertices lie on a quadric surface, by analogy with the case ofK3,3 in the plane.) He further

conjectured that Ks,s (with 25 = 3v-5 bars on v = 10 nodes) should be generically a circuit in 3-space.

The first breakthrough came in 1980, through work of Bolker and Roth [1]. By separating the structure equations for a general bipartite graph, they showed that the problem of determining the stresses in a bipartite graph could be broken down into two parts. The fust part could be solved with reference to a tensor product of vector spaces of projective dependencies; the second part could be solved by seeing how many quadric swfaces pass through a certain subset of the set of nodes. This was a very big step forward, but the resulting count of stresses still seemed rather formidable. In a review of the work by Bolker and Roth, [Whiteley 1979] simplified matters by finding a direct link between quadric surfaces and motions of bipartite frameworks. The present chapter will summarize these results, and . will investigate the possible application of these rather sophisticated methods to arbitrary bar-and-joint structures.

Splitting the problem into two parts Consider two abstract sets A, B of "points". Let R d be real

Euclidean space of dimension d = k-1. To every point a e A, be B we assign position vectors Pa' Pb in Rd. Furthermore, let

• (1) a= (pa, 1), b = (pb, 1),

indicated by bold-face letters, be vectors in Rk fonneP. by appending a fmal coordinate 1 to the position vectors Pa• Pb· These vectors a, b are standard projective coordinates of points · in real projective space pd of dimension d (rank k = d+l). The bold-face letters A, B will denote the sets of projective coordinate vectors of points a (for a e A), b (forb e B). The hyperplane with equation Xk = 0 we call the hyperplane at infinity. (Note that the points in the sets A, B have been assigned finite positions.)

Consider the complete bipartite graph K(A, B) with vertices a e A, b e B, in positions Pa, Pb· respectively. An infinitesimal motion of K(A, B) is thus an assigtlii!ent of vectors va, vb in Rd such that for all a e A, be B,

(2)

A stress in K(A, B) is any assignment of scalars Aab, for a e A, b

e B, to the edges, such that

Let

(3)

for all a e A, Lb e B (Pa - Pb) = 0, and

for all b e B, La e A (Pa - Pb) = 0.

be the row and columns sums of the rectangular matrix A of

coefficients Aab. Then the conditions that A be a stress can be rewritten:

(4) for all a e A, Lb e B A.ab Pb = PaPa,

for all b E B, La e A Aab Pa = lbPb·

(Recall that infinitesimal ·motions and stresses are the vector spaces orthogonal to the row space and column space, respectively, of the rigidity matrix, a matrix whose entry in row ab, column c, is Pa- Pb if c =a, Pb- Pa if c = b, and is otherwise equal to 0.)

Let Q (A, B) denote the vector space of stresses in a complete bipartite framework K(A, B). The defming properties (3) and (4) of a stress can be written in simplified form, in tenns of standard projective cordina.tes (1), as follows:

(5) for all a e A, Lb e B Aab b = Paa,

for all b e B, La e A Aab a = "fbb. ..

Represent each stress A. by an m by n matrix A whose en~es are

the coefficients Aab· Let p be the vector (Pa, ... ) for a e A, y be

the vector ('Yb , ••• ) forb e B. Consider the linear mapping

't: O.(A, B)~ Rm+n

defined by 't(l) = {p, y), the ve~tor whose m+n entries are the row

and column sums of the matrix A.. This linear mapping 't has a

kernel which we shall call O.o(A, B), consisting of all those stresses whose matrices have row and column sums all equal to zero.

Our aim is to give an accurate description of both the kernel no and of the image of the mapping 't. Standard results from linear algebra will then provide a complete description of the space 0. of Stresses.

•

The kernel For any set A ofn points a in positions Pain Rd, ann-tuple

of scalars A.a, a e A provides a linear dependence among the

associated projective points a= <Pa, 1) if and only if~a A.a a= 0.

Let D(A) ~ R m be the vector space of such linear dependencies. The rank of D(A) is the nullity n(A) of theset A in the linear space <A> spanned by A, where n(A) = IAI - rank(A). The tensor product

D(A) ® D(B)

is easily representable as the vector space of those m by n matrices, all of whose rows are in D(B), all of whose columns are in D(A). We give a concrete example.

Example 7.1 Let A= {a, b, c, d} and B = {e, f, g, h} be two four-point

lines, the points having standard projective coordinates

a b c d e f g h

o 1 2 3 o 2 3 s· 1 1 1 1 1 1 1 1

• Then D(A) and D(B) are spanned by the vectors

a b c d e f g h

a 1 -2 1 0 ~ 1 -3 2 0 a' 0 1 -2 1 ~· 0 2 -3 1

respectively. The tensor product space D(A) ® D(B) has as basis

the four matrices a®~, a®~', a'®~, a.'®~', these being decomposable tensor products of the form

a®~= a.a.J3b, for a e A, be B.

Recall that calculations involving decomposable tensors are carried out using the algebraic rules of bilinearity. For any scalar

k and vectors a, ~' 'Y ,

k(a®~) = ka.® ~=a® k~,

(a+ y) ® ~ = a ® ~ + g ® ~ ,

a ® (~ + y) = a ® ~ + a ® ~ .

A general element of the tensor product space is uniquely dete~ined by its values in a submatrix with index set A 1 X B 1'

where At is a basis for the linear space <A> spanned by A, and Br is a basis for the linear space <B> spanned by B. In the case at hand, the matrix

1 1 -4 2 l= 0 ~ 9 ~

-3 '9 -6 0 2 -4 1 1

is in the tensor product, because each column is a vector in D(A),

each row is in D(B). This matrix A. is not decomposable ; it is not itself a tensor product of vectors in D(A), D(B ). But it is expressible as a sum of such products of vectors:

l - (1, 0, -3, 2) ® (1, 1, -4, 2) +

0 0 0 0 0 -6 9 -3 0 12 -18 6 0 -6 9 -3

= (1, 0, -3, 2) ® (1, 1, -4, 2) + (0, 3, -6, 3) ® (0, ·2, 3, -1).

Theorem 7.2 The space !lo{A, B) (consisting of those stresses in a complete bipartite framework K{A, B) for which all roV{ .and column sums

Pa = Lbe B OOab and 'Yb = Lae A Aab

are equal to zero) is equal to the tensor product

D(A) ® 1D(B)

of the spaces of linear dependencies among the projective representations a, b of the points of A, B, respectively.

Proof: If a. = ( CXa, ••• ) and ~ = (~b' . . . ) are dependencies among the projective representations of the points in sets A, B,

respectively, then a.®~ e 0o(A, B) because

La (a.® ~)ab a= La O.a ~b a= ~b La O.a a= 0,

Lb <a. ® ~)ab b = Lb a.a ~b b = a.a Lb Pb b = o.

- Consequently, D(A) ® D(B), the vector space spanned by all

products a® b, a e D(A), be D(B), is a subspace of flo( A, B).

Conversely, if A. e i'lo(A, B), then

La Aab a =· 0, Lb Aab b = 0,

so the rows and columns of A. are dependencies of B and A

respectively, and A. e D(A) ® D(B). + The sum of the dimensions of the kernel and of the image

of. the linear transformation -c: Q(A, B)~ Rm+n is equal to the

dimension of the domain n. So

(6) dim O(A, B)'= null A null B +dim (Image -c),

where null A is the nullity of the set A, the linear dimension of the space of dependencies among the projective representations of the points in the set A (similarly for rank B). Our task is to see what vectors in Rm+n can occur as row-and-column sums -c(A.) of

stresses A. in K(A, B).

The image Select bases At~ A, B1 s;; B for the linear subspaces <A>,·

<B> spanned by the projective coordinate vectors a e A, b e B, respectively. Let x, s be variable points in the sets A 1, A 2 = A \A 1, and y, t be variable points in the sets B b B2 = B\B 1,

respectively. (This notation will permit us to rewrite any, sum La as Lx + L5 , any sum Lb as Ly +Lt.)

Theorem 7.3 Assume A, B (as usual) are sets of coordinate vectors of points in projective space of dimension d, that A. is a

matrix A.: Ax B ~ R, and that p, yare vectors p : A~ R, y: B __,

R. Let A1 ~A, B1 ~ B be bases for the spaces <A>, <B>, respectively. Assume further that

for all s e A2 = A\A~t Lb Asb b = Ps s, and

for all b e B, La Aab a = 'Yb b.

Then the following are equivalent:

(7) for all x e A 1 , Lb fvxb b = Px x

(8) La Pa a ® a = Lb 'Yb b ® b.

Proof: (7) => (8) ...

La Pa a ® a = La a ® Pa a = La a ® (Lb Aab b)

= L~b Aab a ® b = Lb 'Yb b ® b .

(8) => (7) . ..

Lx.Px X® X= -Ls Ps S ® S + Lb Pb b ® b

= -Ls,b Asb s ® b + L~b Aab a ® b

= Lx,b Axb X ® b .

Thus Lx x ® (pxx - Lb A. xb b) = 0. Since the vectors x e A 1 are independent, all the vectors PxX - Lb Axb b must be zero. (The

reasoning is as follows: If I. i pi ® qi = 0 for a sum of tensor

products of vectors, then in the jth coordinate position of the second factor,

Li pi qij = o. If the vectors pi are independent, the jth coordinates of all the vectors qi are zero, and this, for all j.) +

•

Although, by Theorem 7 .3, any stress A. has a matrix A. with row and column sums Pa, 'Yb satisfying ..

La Pa a ® a = Lb 'Yb b ® b,

the converse is not true. There may exist dependencies among the tensor squares which do not arise as row and column sums of stresses.

Example 7.4 Let A = {a, b, c} and B = { d, e, f} be two three­point lines in the real plane,

a b c d e f

0 0 0 1 1 1 0 1 2 0 1 2 1 1 1 1 . 1 1.

Then

a®a-2b®b+c®c=

0 0 0 0 2 0

0 0 0

There is no stress A. with

Pa = 1, Pb = -2, Pc = 1, Pd = 1, Pe = -2, Pf = 1

because no element of A is in the span <B> of B, and no element of B is in the span <A> of A. None of the defming equations (5) of a stress can be satisfied using these values of row and column sums.

Define asetC ~ Au B by

C = (A n <B>) u (<A> n B), the. set of projective points in each set which are in the linear span of the other set. It is clear from the defining equations (5) that every stress will have row and column sums which are zero outside this set C . With this added restriction, the image of the mapping 't: Q(A, B) ~ Rm+n is easily determined, as follows.

Theorem 7.5 A pair of vectors (pa, a e A), ("fb, be B) is the set of row and column sums of a stress on the complete bipartite framework K(A, B) if and only if

(8) La Pa a·® a = Lb 'Yb b ®b.

(9) Pa = 0 for a e: <B>, 'Yb = 0 forb e: <A> .. Proof: If A. is a stress, then its row and column sums Pa, 'Yb satisfy equation (7), therefore equation (8) of theorem 7.3. Whenever a point a e A is not in the closure <B> ofB, then the coefficient Pa

in equation (5) must be equal to zero. (Similarly, for b e B\<A>.)

Assume that equations (8) and (9) hold for the scalars Pa, a

e A, 'Yb, be B. Let At and B1 be, as in·the statement of Theorem

7.3, bases for <A> and <B>. Then for every points e A2 we

may choose scalars Ash, forb e B, to fonn a row such that

PsS= LAs b.

Furthetmore, since Ar is a basis for <A>, spanning all points in

A2 = A\A1 and all points 'Yb b (these being zero ifb E <A>), we

may choose scalars Axb for x e At, be B such that

'Yb b = Lx Axb X + Ls Ash S, and thus

'Yb b = La Aab a.

By Theorem 7.3 (in the form (8) => (7)) we conclude that

.Px X = Lb Axb b

for all points x e A 1, so A. is a stress with the given row and

column sums. +

Corollary 7.6 The dimension of the space O(A, B) of stresses in a complete bipartite structure K(A, B) is given by the formula

(10) dim Q(A, B)= null (A) null (B)+ null (C2)

where C =(An <B>) u (<A> n B), and C2 = {c®c; c eC} .

Proof: Replace dim(Image 't) by null(C2) in formula 6. +

Example 7. 7 Assume that the sets A and B span planes P and Q, respectively, in Euclidean 3-space, and that none of the points lie on the intersection P n Q of the two planes. See Figure la. If lA I = m, IBI = n, the~ null(A) = m-3, null(B) = n-3, and C = 0. So ·

the stress space Q(A, B) has dim~nsion (m-3)(n-3). Since there are mn bars, with (m-3)(n-3) independent linear relations among the constraints supplied by those mn bars, the space M(A, B) of infmitesimal motions of the structure has dimension

dim M(A, B)= 3(m+n)- mn + (m-3)(n-3) = 9.

Leaving aside 6 degrees of freedom for rigid motions of the structure, there remain 3 internal degrees of freedom, a constant value, independent of the cardinalities of the sets A, B.

(Figure 1)

Exercise 7.8 Calculate the dimension of the space of stresses in a bipartite structure in which the sets A and B of vertices span, respectively, skew lines L, M in 3-space. Show that these structures have space of infinitesimal motions .of dimension m+n+4, if IAI = m, IBI = n. Give a simple geometric description of these motions.

Now asswne that

A= (-1, 0, 0) (0, 0, 0) (1, 0, 0)

B = (0, -1, 1) (0, 0, 1) (0~ 1, 1)

(Euclidean coordinates). Calculate the stress (unique up to one scalar multiple) in the bipartite structure K(A, B).

Quadric surfaces In this section, we discuss the . coordinatization of

geometric points relative to function spaces, and show that the coordinatization by tensor squares and that by quadratic functions yield the same higher dimensional geometric structures.

Since we have represented Euclidean points a, b in positions Pa' Pb by standard projective coordinates

a= (pa, 1), b = (pb, 1),

any linear dependency

ra Pa a ® a = Lb 'Yb b ® b.

among the tensor squares yields (in the last column of the associated matrix equation) a linear dependency

La Pa a = Lb 'Yb b

among the projective coordinates for the points themselves. The converse is not true. For instance, if A = {a, b, c, d} is the set of four collinear points used· in example 7.1, there is a 2- ·

· dimensional space D(A) of linear dependencies among the four vectors: ..

aa + ( -2a+~)b + (a-2~)c + ~d = 0, for all a, ~'

but only a !-dimensional space of linear dependencies among the tensor squares:

"(cl® a- 3 yb® b + 3yc® c- yd® d = 0, for all 'Y·

(Take a = -~ as 'Yin the preceding equation.)

This decrease of dependence, as one passes from projective coordinates, a, to tensor squares, a® a, of those coordinate vectors, is accounted for by the observation that the coordinatization by tensor squares is a higher dimensional coordinatization of the same set of abstract poirlts. Projection (onto the last column, in the matrix representation of those tensors) is a linear transfonnation which preserves, but also introduces, dependencies.

A natural way to vary the coordinatization of a set of points in a space S, without actually "moving" the points, is to coordinatize them with respect to different function spaces. Let 1F be a (real) function space, that is, any subspace of the vector space of all real-vaalued functions f: S ~ R. For each point a e A, the operator a, "evaluation at a"'

: a(f) = f(a)

is a linear functional a: lF ~ R, and is thus an element of the dual

space 1F*. This assignment a ~ a is the coordinatization of the set A relative to the function space lF. The standard example is as follows.

Example 7.9 Let L be the space of linear functions defined on Euclidean space Rd. This space L is (d+l)-dimensional, being spanned by one constant function (generating a !-dimensional subspace) together with the d different coordinate projection functions. The function space coordinatization of points in Euclidean space as elements of L * is isomorphic to their representation by standard projective coordinates. To see this, form the ( d+ 1) by IAI matrix M whose columns are the standard projective coordinates of the points in A. The the ith row of this matrix M lists the values of a typical constant function. Since the columns of the matrix M list the values of the linear functionals a . on a basis of the space L, the columns of M and the linear

functionals a, a e A, provide equivalent coordinatizations ~{ the set A, equivalent under a linear isomorphism.

For any function space lF of real-valued functions on Rd, and for any subset A !:: Rd, let lF(A) be the vector space of restrictions of functions in lF to the set A. It is the dimension of this vector space lF(A) which is equal to the rank of the set A in the function space coordinatization. For example, if A consists of four collinear points in R3, the space L of linear functions on R3 is 4.-dimensional, but the space L(A) of restrictions to the set A is only 2-dimensional.

Quadratic functions in d variables can be made homogeneous quadratic in k = d+ 1 variables, simply by multiplying all constant and linear monomial terms by xk2, xk, respectively. Let q be a quadratic polynomial in d variables, and q its corresponding homogeneous polynomial. Then for any point a e R d, with standard projective coordinates a,

q(a) = 0 if and only if q(a) = 0.

Thus, for any subset A Rd, the vector spaces

. { q quadratic, q: Rd--? R; q(a) = 0 for all a e A}

and

(q homogeneous quadratic, q: Rk--? R; q(a) = 0 for all a e A}

are linearly isomorphic. . .

Theorem 7.10 For any set A of points in Rd, a linear relation

:Ea A.a a®a = 0

holds among the tensor squares of the standard projective coordinates for those points if and only if the same linear relation

IaA.aa= o holds among the points a e A as coordinatized by the space Q of homogeneous quadratic functions in k = d+l variables (or, equally well, by the space of quadratic polynomial funmctions in d variables).

Proof: Since the ijth coordinate of a®a is equal to the product

aiaj, the relation l:a A.a a®a = 0 holds if and only if for each pair i, j of coordinate places,

1~, 1Sj~, l:a Aa aiaj ==- 0 .

But aiaj = a(xixj), the value at a of a quadratic monomial. So

!a A.a a®a = 0 if and only if

(l:a Aa a)(XiXj) = 0 for all i, j, if and only if

l:a"-aa =0, this final equivalence being due to the fact that the monomials XiXj

span the space Q of homogeneous quadratic functions. •

Corollary 7.11 The dimension of the space !l(A; B) of stresses in a complete bipartite structure K(A, B) is given by the formula

dim !l(A, B) = null(A)null(B) + nnllQ(C)

where C =(A t1 <B>) u (<A> t1 B), and nnllQ(C) is the nullity

of the set C of points, coorclinatized relative to the space Q of homogeneous quadratic functions.

Let P be the set of all points in the Euclidean (or projective) plane, and let PQ be the same set of points,

coordinatized relative to the function space Q of quadratic (homogeneous quadratic) functions, respectively. It is an easy matter to predict the rank rQ(A) and the nullity

nllliQ(A) = IAI - rQ(A)

of point sets A ~ P, by reference to the concept of Q-closed sets. For each set A of points in the plane, let <A>Q be the Q-closure

of A,

<A>Q = { q e P; for all f e Q, f(A) = 0 implies f(q) = 0},

that is, the set of points q in the plane at which f( q) = 0 for every quadratic function f such that f(p) = 0 for all p e A. A subset A ~ P is Q-closed if and only if A = <A>Q. The Q-rank of any subset

A ~ P is equal to the Q-rank of its Q-closure, and there are relatively few types of Q-closed sets, as shown in Figure ti.

rank 6 rank 5 rank 4

rank 3

rank2

rank 1

rank 0

..

Figure2

Q-closed subsets of the plane

the entire plane HI all points on one conic )-; .. ( 0 V ,;;:-<'

four points, no three collinear, or o o

• • a line, plus one point not on that line ~

three non-collinear points, or 0

0 0

all points on a line ------two distinct points 0 •

one point 0

the empty set ¢

The space Q of homogeneous quadratic functions in k = d+ 1 variables is of dimension (k+ lz) (printer: use normal notation for

binomial coefficient), and has the set { XiXj; 1 <i<1j~} of monomials as a

basis. Consider the matrix M(C) whose columns are indexed by the points_ in the set

C =(A r1 <B>) u (<A> r1 B),

whose rows are indexed by the (k+l2) monomials XiXj, and whose entries provide a table of the values of those monomial functions at those points. Observe. that for d=2, a point a with projective coordinates (at, a2, a3) gives rise to a column of M with entries

a= (at2, ata2, ala3, a22., a2a3, a32).

The matrix M has a certain rank, measurable either as its row rank or column rank, and satisfying

# rows - row nullity = rank = # columns - column nullity,

so

# rows + colwnn nullity = # columns + row nullity.

We now know the column nullity of the matrix M: it is equal to nuJ.lQC =null C2, the nullity of the set C of points,

coordinatized relative to quadratic functions, or coordinatized as tensor squares. The row nullity is equal to the dimension of the

. space of (k+l2)-vectors q such that q•a = 0 for all points a e C.

Since q•a is simply the value at a of the homogeneous quadratic polynomial q with coefficients given by the vector q, the row · nullity is equal to the dimension of the space

AnnihQ(C), .. the space of homogeneous quadratic polynomials which are equal to zero at all projective points a, for a e C, that is, which annihilate C.

Corollary 7.12 The dimension of the space Q(A, B) of stresses in a complete bipartite structure K(A, B) in a space of dimension d = k-1 is given by the fonnula

dim Q(A, B)= null(A)null(B) + ICI- (k+l2) +dim (AnnihQ(C)),

where C =(A r1 <B>) u (<A> r1 B).

Proof: Since

# rows + colunm nullity = # columns + row nullity,

we know

(k+12) + Nul.IQ(C) = ICI +dim (AnnihQ(C)).

make this substitution for NuJ.lQ(C) in the formula of Corollary

7.11..

Problem 7.13 Given six points a, ... , fin the plane, let A= {a, b, c}, B = { d, e, f}. Then the bipartite structure K(A, B) is generically isostatic, but will have a one-dimensional space of stresses (and thus a one-dimensional space of internal infmitesimal motions) if all six points lie on a single quadratic curve. Use the formula of Corollary 7.11 to calculate the dimension dim Q(A, B) of the stress space in the following special cases:

(a) a, ... , fare distinct points on a non-degenerate conic

(b) a, b, c collinear, d, e, f collinear (on distinct lines)

(c) a, b, d collinear, c, e, f collinear (on distinct lines)

(d) . a, b, c, d collinear, e, f off that line.

Observe that the same result is computed in quite different ways: sometimes the contribution is from null(A)null(B), sometimes from nuJ.lQ(C). But the cases (b), (c), (d) are specializations of

the non-degenerate situation (a). It would seem that some other way of counting stresses would yield numerical results varying more smoothly with respect to specialization. A first guess would be that

dimQ(A, B) = nuJ.lQ(A, B),

which is true in all of the above cases. Show this formula fails·if a, b, c, d are collinear, e, f, g in general position in the plane, with A = (a, b, c}, B = { d, e, f, g}. Here the stress space has rank 2, while there is only a !-dimensional space of dependencies among the quadratically represented points (a dependence supported on the set of four collinear points a, b, c, d). As a special case of this example, let a, . . . , g be the seven points indicated in Figure ti. Calculate the stresses in the bipartite framework K(A, B), the space D(A) ® D(B), and the space DQ(A u B)= lDQ(a, b, c, d).

Hint: the stress space is spanned by stresses with scalars given by matrices a,~: ·

d e f g_

a 1 3 -3 0

ex: b -3 -6 6 0

c 3 3 -3 0

l a... ' c e. C\

0 ' l 0 \ 2. 3 3 0 0 0 0 l l 'l...

I ' ' J • I 1

I~

I

o_5

I

~ } c., 0

c c:. c:.

" .. 'o ~ c. J

d e f g

a 0 1 -3 1

J3: b 1 -2 6 -2

c -2 1 -3 1

Exercise 7.14 Show that K4,6 on the moment curve (t, t2, t3) is dependent and has infinitesimal motions.

Exercise 7.15 Show that K4,5, four points coplanar, is dependent. What is its space of motions? (This structure is 1-underbraced, yet requires only one projective condition for special position: the K4,6 had a repeated factor!)

Exercise 7.16 Calculate the stresses and motions of a structure K4,4 with its vertices on two planes in 3-space.

Problem 7.17 Given a single edge eab in a complete bipartite

structure K(A, B), how many of the stresses A. in .Q(A, B) have 0

as the scalar lab? Such stresses are in 1-1 correspondance with arbitrary stresses on the bipartite structure K(A, B)\ eab· An answer to this question will perhaps suggest a solution to the problem of determining the ~tresses on general (not necessarily complete) bipartite structures. ·

.. Motions

A complete bipartite framework K(A, B) with IAI = m, IBI = n, in d-space has a space of infinitesimal motions of dimension

(10) d(m+n) - mn +dim Q(A, B),

equal to the dimension d(m+n) of the space of unrestricted motions of the separate vertices, less the number of constraints on motion furnished by the bars. That is, d(m+n) less the number mn of bars, plus the dimension dim Q(A, B) of the stress space, the space of linear dependencies among the bars, as constraints on motion.

Theorem 7.18 The dimension of the space· M(A, B) of infinitesimal motions of a complete bipartite framework K(A, B) in d-space is given by the fonnula

dim M(A, B) = d(m+n) -mn + Null(A)Null(B) +Null Q(C),

where IAI = m, IBI = n, and C =(<A> n B) u (An <B>).

Proof: Use the formula of Corollary 7.7 to substitute for W(A, B) in line (10). • .

Corollary 7.19 [Whiteley 1984]

If both vertex sets A, B span the d·dimensional space R d, then the space MintCA, B) of internal motions of the complete bipartite framework K(A, B) has dimension

dim Mint (A, B) =dim (AnnihQ(A u B).

Proof: If A and B both span Rd, then Null A = m·k, Null B = n-k, where k = d+1 and C =Au B. To count internal motions, we must subtract the(~) rigid motions of d-space. By Theorem 7.18 and Corollary 7.12,

dim Mint (A, B) =dim M(A, B) - (~)

= (k·l)(m+n) - mn • CkU + Null(A)Null(B) + NlillQ(A u B)

=(k-l)(m+n)- mn- {~) + (m-k)(n-k) + (m+n)

- (k+l2) +dim (AnnihQ(A u B)

= dim (AnnihQ(A u B)

This remarkably simple formula for the number of internal motions of reasonably general bipartite frameworks · suggests an intimate connection between

internal motions

and

quadratic functions which are zero at all nodes of the framework.

To see that this is the case, let Q be a symmetric matrix of size k by k. Then, indicating points p by their standard projective coordinates p = (p, 1),

(11) pQp = 0

is the equation of a quadric surface. Furthermore, any (inhomogeneous) quadratic polynomial equation can be represented in this way. For instance, the equation

3pl2 - 5PlP3 + P22 - Pl + 6 = 0 -

(for d-3) is given by equation (11) when

3 0 -5/2 -1

Q= 0 1 0 0

-5/2 0 0 0

-1 0 0 6 Theorem . 7.20 [Whiteley 1984] If all nodes of a complete bipartite framework K(A, B) lie on a quadric surlace given by the formula pQp = 0, then

Va = (aQ)r, ... , d for a e A

Vb = (-bQ)t, ... , d forb e B

is the set of Euclidean velocity assignments of an inimitesimal motion of the framework . . <Note that we take all components except the kth (final) component of aQ·and of -bQ). If there is at least one pair a, a' of points in A (or one pair b, b' of points in-B) which do n9t lie together on a ruling of the quadric surface pQp = 0, the n the motion described above is an internal motion.

Proof: Consider any bar ab, for a e A, b e B. Then

(va- vb) • (a- b)

= (aQ + bQ) • (a - b)

= aQa - aQb + bQa - bQb

= -aQb + bQa = -aQb + aQb = 0

because aQa = bQb = 0 on the quadric surface, and aQb = bQa by s~etry of Q.

Now consider a pair a·, a' of points in A. If the distance · beween a and a' is preserved by the motion Va, Va•, then

0 = (va- Va•) • (a- a')

= aQa- aQa' - a'Qa + a'Qa'

= -2 aQa'.

..

The equation of the hyperplane Ha tangent to the surpace pQp = 0 at the point a is

(aQ) • p = 0,

so a' lies in the hyperplane Ha. Since the line aa' is tangent to the surface at a, and has an additional point a' :~: a also in the surface, it follows from the fact that the surface is quzdratic that the entire line aa' must lie within the surlace pQp = 0, and is thus a ruling of the swface. Conversely, if there is any pair a, a' of points in A (or points b, b' in B) which are not together on a ruling of the surface pQp = 0, then the distance aa' (bb', respectively) is not preserved, and the motion is not a rigid motion of the framework. +

.. ·."" ...

I

\ i

··. ··· . . . ·· .. .. ..... ~-...

' ' . , .

...

'·~ -... ..

'

.. ... .. .......

'\ ~---·-­---

' '

. .

' ' ' .,...

........ ~;. .. ...

' ' ' '

• • ' . '

I \

I I

{ I

)

Exercise 7.21 Ve~ that the velocity vector Va, for a A, is the gradient of the (inhomogeneous) quadratic polynomial function given by the formula pQp = 0.

In the simplest case~ where the sets A and B both span Rd~ all inf"mitesimal motions are obtained by combining the internal motion

Va = aQ~ vb = ·bQ

with arbitrary rigid motions. Figure ti shows a typical infinitesimal motion of the parabola y = x 2 + 1, together with its "gradient" motion. It doesn't matter what size sets we take for A and B, so long as there are at least three points in each, and all lie on the quadratic curve. In fact, we might as well take

A = B = {p; pQp = 0}

to be the entire curve. This infinite bipartite framework still has one internal motion, for a total of four infmitesimal degrees of freedom.

Probl-em 7.22 It seems that in any infmitesimal motion of a sufficiently large bipartite structure, the vertices of which lie on a quadric surface, the "tips" of the velocity vectors of the A-points (of the B-points, respectively) generate a quadric surface. Verify.

Figure ti illustrates the "gradient" motion of a sphere (here, the circle). All A-points move radially outward, alfB-. points move radially inward. All bars ab, for a e A, be B, have length preserved in this motion because the inward and outward normals at points on a sphere have the same projection on· the chord which joins them.

Extensions It would be strange if the Balker-Roth analysis of stresses,

which produced the formula (10):

dim O(A, B) = null (A) null (B) + null (C2)

would not have at least some implications in a broader context. We could hope to apply a similar analysis either to bipartite frameworks which are not necessarily complete bipartite~ or else even to frameworks which are not bipartite. Rather surprisingly, it is in the latter case that the Bolker-Roth analysis seems to give more useful information. In the following paragraphs, we use their ideas to analyze stresses in a complete graph. Such an analysis seems potentially useful, in view of the fact that the stresses on an arbitrary strucrure with a set A of n nodes and edge

set E ~ ( A:z) are simply those stresses on the complete graph Kn which happen to be zero on all edges in the complementary set (A:z)\E.

Consider a single abstract set A of n "points", with positions Pa' ae A, in Euclidean space of dimension d = k -1. Let a = (pa, 1), ae A, be the associated standard projective coordinates.

A stress on the structure with graph K n will be any assignment A.

of scalars "-ab, for a, b e A such that

l:b; a;eb Aab (b-a) = 0

that is,

(12)

where

and where

a = (a, 1), b = (b, 1)

are standard projective coordinates for the points. We see from fonnula (12) that for any point a e A, the vector

(J.la, ~b' Aac, • • • )

is the set of scalars for a linear dependence among the projective .. coordinates of the points in the set A. Fonn the matrix A(A.):

J.la

A('J....) Aab

Theorem 7.23 An n by n matrix of real numbers is the matrix A(A.) of a self-stress A. in the complete-graph framework K0 if and only if ·

(1) A is symmetric

(2) The rows of A are dependencies among the vectors a, for a e A. That is, L is in the symmetric tensor square li)2(A). •

Let {ex, ~, . . . } be a basis for the space D(A) of dependencies arriong the standard projective coordinates a, for a A. Then

ex®ex, ex®~ + ~®ex, ex®y + y®ex, ...

form a basis for the symmetric tensor square D2(A). If the point set A has rank k (spanning a projective space of dimension k-1), then there are n-k vectors in a basis for D(A), and the space of all tensors D ® D, of rank (n-k) 2, is the sum of the space D v D of antisymmetric tensors, rank (n-k2), and the space 02 of symmetric tensors, rank (n-k+ 12).

Example 7.24 Let A consist of four collinear points with projective coordinates

a b c d

0 1 2 3

1 1 1 1

Then

ex= 1 -2 1 0

{3= 0 1 -2 1

form a basis for D(A), and

1 -2 1 0 0 0 0 0

a®a= -2 4 -2 0 {3®~ = 0 1 -2 1

1 -2 1 0 0 -2 4 2

0 0 0 0 0 1 -2 1

0 0 0 0 0 1 -2 1

ex®f3 1 -2 1 0 + 0 -2 4 2

+~®ex= -2 4 -2 0 0 1 -2 1

1 -2 1 0 0 0 0 0

0 1 -2 1

- 1 -4 5 -2

-2 5 -4 1

1 -2 1 0

fonn a basis for the space

{A(A.); A. is a self stress}

of a K4 constructed on a line.

Any set of n points spanning a projective space of dimension n-2 (rank n-1) have a .!-dimensional space of dependencies. Thus all dependencies are a multiple of

d-1: a[bc] - b[ac] t c[ab] = 0

d=2: a[bcd] - b[acd] + c[abd] - d[abc] = 0

d=3:. a[bcde]- b[acde] + c[abde]- d[abce] + e[abcd] = 0

Theorem 7.25 The Wlique stress (up to one scalar multiple) on the complete graph structure Kn in a space of dimension d = n-2

has a matrix A which is the tensor square of the unique dependence on those n points, and thus has scalars

A.(ah aj) = (-l)i+j [al ... ai ... an][al ... aj .•. anl·

Example 7.26 For the K4 structure in the plane, the unique ·

self-stress A has matrix

a b c d .•

a [bcd]2 -[bcd][acd] [bcd][abd] -[bed] [abc]

b -[aqi][bcd] [acd]2 -[ acd] [ abd] [acd][abc]

c [abd][bcd] -[abd][acd] [abd]2 -[abd][abc]

d -[abc] [bed] [abc][acd] -[abc][abd] [abc]2

Exercise 7.27 Consider the complete graph Ks as a framework

with all five points collinear. Prove that the vector ("fa, a e A} can be any 5-vector, and that once this 5-vector has been chosen, there remains a 2-dimensional space of self-stres$eS with those selected "sums" 'Ya·

Research problems

1. Given a single edge eab in a co.11_1plete bipartite graph K(A, B),

how many of the stresses A. in K(A, B) have 0 as the scalar "-ab? Such stresses are in 1-1 correspondence with arbitrary stresses on the bipartite graph K(A, B)\ e ab· An answer to this question will perhaps suggest a solution to the problem of determining the stresses on general (not necessarily complete) bipartite structUres.

Changes already suggested (V{alter)

Generic rigidity is also directly accessible, using Henneberg type insertions and the motions for quadrics. The complex pattern of stresses could come after the initial results, which would motivate the implicit geometry.

The exercise~ could include deducing the pure condition of some basic examples, such as Kd+l, d(d+l)/2· These are important examples of repeated factors, multiple stresses for the corresponding condition, and consequent finite motions (realizing a repeated factor guarantees thet the infinitesimal motion becomes finite!).

. .

Bibliography

[Bolker & Roth 1980]

Ethan D. Bolker and Ben Roth, When is a bipartite graph a rigid framework?, Pacific Journal of Mathematics, 90 (1980), 27-43.

[Crofton 1878] ti

[Whiteley 1979]

Walter Whiteley, Motions of a Bipartite Framework, Structural Topology, 3 (1979), 62-63.

[Whiteley 1982]

Walter Whiteley, Motions of Trusses and Bipartite Frameworks, Structural Topology, 7 (1982), 61-68.

[Whiteley 1984]

Walter Whiteley, Infinitesimal motions of a bipartite framework, Pacific Journal of Mathematics, 110 (1984), 233-255. . .