Checking the incidences

It remains to show that the A-points and B-points, chosen as above, do represent the budget matroid Bm,n.

Did we achieve every required incidence? The A-points lie in the m-flatαand the B-points lie in the n-flatβ. Also, for every partition(I;J)of [1. .m+n] into

an m-set I and an n-set J , the perfect set of points{Ai | i ∈ I} ∪ {Bj | j ∈ J}is

84 CHAPTER 6. REPRESENTING THE MATROID Bm,n

To show that we have avoided every forbidden incidence, it suffices to check that we have mapped every base of the matroid Bm,nto a mutually skew set of m+n

points — a set that spans all of S. There are three types of bases. A base of the first type has the form{Ai |i ∈ I}∪{Bj | j ∈ J}where I is any(m−1)-set and J is any (n+1)-set. We haveβJ =β, whileαI is some(m−2)-flat inα. Note that adding one new element to the set I to produce an m-set I0 ⊃ I gives us a hyperplaneαI0

inαthat cuts the line c at the point CI0. Since the point CI0 varies, depending upon

which m-set I0 ⊃ I we choose, the(m−2)-flatαI must be skew to c. It follows that Span(αI ∪βJ)= S. The second type of base is symmetric, with m+1 points in the A column and n−1 in the B column. A base of the third and final type has the form{Ai |i ∈ I} ∪ {Bj | j ∈ J0}where I is an m-set and J0is an n-set, but the

partitions(I;J)and(I0;J0)of [1. .m+n] are distinct. The span of such a base

includes the flatsαI andβJ0, so it contains the two distinct points C_I and C_I0 along

c, so it includes the entire line c, so it includes bothαandβ, so it coincides with

S. And that, at long last, finishes the proof of the Bm,n Representation Theorem.

Exercise 6.2-3 (For matroid mavens) We have finally finished proving that the budget matroid Bm,nis representable over the rationals. If it were also binary (that

is, representable over the Galois field of order 2), it would follow from standard results that it was regular (that is, representable over all fields). And indeed, the budget matroids B1,1 and B2,1 are both binary and regular. But show that those two are the only budget matroids Bb1,...,bk that are binary.

[Hint: Recall from Exercise 4.2-4 what the circuits of a budget matroid are. It is a standard result that a matroid is binary if and only if the symmetric difference

C4D of any two distinct circuits C and D always includes a circuit [41]. Let X be

the subset of Bb1,...,bkof size b that contains, for each j , precisely the top bj elements

in the jth column. Show that one can typically find two distinct elements f and g so that C := X ∪ {f}and D := X ∪ {g}are circuits (of the type called ambient circuits in Exercise 4.2-4), and note that the symmetric difference C4D = {f,g}

Chapter 7

Representing the matroid B

_m,1,1

As the number of parts in the partition b=b1+ · · · +bk increases and as the parts

themselves increase, it rapidly becomes quite unclear whether or not the budget matroid Bb1,...,bk is representable. But for the simplest of the budget matroids with

three columns, those of the form Bm,1,1, we can still get a general representability result over the rational numbers. Already in this case, however, avoiding forbidden incidences is much harder than it was in the case of the budget matroids Bm,n, with

only two columns.

7.1

The case m

=

2 of the matroid B

Let’s begin by considering the matroid B2,1,1— the one that we are eventually go- ing to show characterizes the dependence of four cubic polynomials. Recall that a representation of B2,1,1consists of twelve points

    2 1 1 P0 A0 B0 P1 A1 B1 P2 A2 B2 P3 A3 B3    

in 3-space. The four A-points lie on a common line a, the four B-points lie on a common line b, and the four P-points lie on a common planeπ. Furthermore, each of the twelve perfect sets{Pi,Pj,Ak,Bl}for{i, j,k,l} = {0,1,2,3}must be

coplanar. Note that, in this chapter, it is convenient to index the four rows starting at 0, rather than at 1.

Suppose that we choose any planeπin 3-space and four points(Pi)in the plane π with no three collinear. We choose a line a whose intersection withπ — call it

A_π :=a∩π— is not on any line joining two of the P-points. And we choose four distinct points(Ai)on the line a, none coinciding with Aπ. Whenever{i,j,k,l} =

{0,1,2,3}, the point Bl is restricted to lie in the three planes PiPjAk, PiPkAj, and

86 CHAPTER 7. REPRESENTING THE MATROID Bm,1,1

PkPjAi. By Lemma 6.2-1, those three planes intersect at a unique point Bl, which

does not lie in the planeπ. But are the four points(Bl)so determined collinear?

It turns out that they are; in fact, much more is true.

Note that the A-points did not have to be collinear, in order to enable the con- struction of the B-points. Given any sequence of four A-points, none lying inπ, we get a sequence of four B-points, none lying inπ; and the relationship between the sequences(Ai)and(Bi)is symmetric. It turns out that the spans of the two se-

quences(Ai)and(Bi)always have the same dimension. Furthermore, this result

generalizes from 3-space to projective space of any dimension.