from loose to snug, there is some other column of X that was loose and that remains loose.
Case 3: This case is very similar to Case 2. Suppose that X has only one loose column, the mth, but that column is actually baggy. We must have(Y\X)∩Cm 6= ∅, so we add any element of(Y\X)∩Cm to X. The mthcolumn remains loose even after the augmentation, so both the Column and Perfect Rules hold.
In what follows, we can assume that none of the first three cases pertains. Thus, there is precisely one column that is loose in X, say the mth, but this column is not baggy. It follows from this and the constraint|X| < b that, in fact,|X| = b−1,
|Y| =b, and, for all j 6=m, we haveX ∩Cj=bj and(Y \X)∩Cj = ∅. Case 4: Suppose that, in the mth column, Y has at least two elements that X doesn’t have; that is,|(Y \X)∩Cm| ≥2. Since|X| =b−1, we know that X has at least one empty row; choose one of them. Because we have at least two elements of Y \X in the mth column to pick from, we can find one to add to X that does not spoil our chosen empty row.
Case 5: In the remaining case, we can assume that Case 4 also fails to pertain, which means that|(Y \X)∩Cm| =1. Since(Y \X)∩Cj=0 for all j different from m, we deduce that|Y \X| =1. Since we also have|Y| − |X| = 1, it follows that X ⊂Y . We can hence take the unique element of Y\X and add it to X without destroying independence. tu
5.4 Weakening the rules for independence
Given column budgets(bj)and dimensions(dj), the budgetary matroid Bdb1,...,dk
1,...,bk
arises out of the interplay among the three rules that define the notion of indepen-dence: the Ambient Rule, the Column Rule, and the Perfect Rule. In this section, we attempt to justify those three rules by considering what would happen if we omitted one of the three.
What would happen if we kept the Ambient and Perfect Rules, but dropped the Column Rule? Then, every set with less than b elements would be independent, every set with more than b elements would be dependent, and, among the sets with b elements, precisely the perfect sets would be dependent. Note that, if a set X has b−1 elements, there is at most one perfect set that includes X. From this alone, it is straightforward to check that the resulting structure would be a matroid of a standard type, called a paving matroid [39].
If we kept the Ambient and Column Rules, but dropped the Perfect Rule, we would again get a matroid of a standard type: a truncation of a direct sum of uni-form matroids [40].
So what about omitting the Ambient Rule? Since the Ambient Rule was ar-guably the least well motivated of the three rules to begin with, it is particularly interesting to study what happens when we omit that rule — or perhaps just weaken
64 CHAPTER 5. THE BUDGETARY MATROIDS it somewhat. We’ll call a weakened version of the Ambient Rule successful when the structure that results from the Column Rule, the Perfect Rule, and that weak-ened Ambient Rule is still a matroid.
If we weaken the Ambient Rule too far, we lose all hope of being successful. In particular, whenever the total budget b is at least 3, there are some sets of size b+1 that satisfy both the Column Rule and the Perfect Rule but that must be classified as dependent, in order for the resulting structure to be a matroid. To construct such a set, let P be any perfect set in the structure Bdb11,...,d,...,bkkwith b ≥3, and let f , g, and h be any three elements of P with f and g in different columns. Let f0 be the element that is in the same row as f and the same column as g, and let g0 be the fourth vertex of that rectangle. Consider applying the Augmentation Axiom to the sets X := P ∪ {f0} \ {g}and Y := P ∪ {f0,g0} \ {h}. Here are pictures of what since each has an empty row. And both X and Y satisfy the Column Rule, since dj ≥bj for all j . So, if we weaken the Ambient Rule far enough, the set Y will be independent, as well as X. But adding either of the elements of Y \X = {g,g0} to X results in a set that has a perfect subset, and hence violates the Perfect Rule.
If we want to satisfy the Augmentation Axiom, we must preserve enough of the Ambient Rule to classify the set Y as dependent.
When the parameters(bj)and (dj)satisfy bj = dj > 0 for all j — that is, for those budgetary matroids that are budget matroids and all of whose column budgets are positive, which are the cases of primary interest — much more is true:
There is no way to weaken the Ambient Rule at all and still be successful. To see this, recall from Exercise 4.2-4 that the circuits (the minimal dependent sets) of the budget matroid Bb1,...,bk are of three types:
perfect circuits the perfect sets,
column circuits the subsets of column j of size bj +2, and
ambient circuits those sets of size b+1 that do not include, as a subset, any per-fect circuit or column circuit.
Let’s refer to the perfect and column circuits as the key circuits, since they will remain circuits, even if the Ambient Rule is somehow weakened. Suppose that M is some other matroid on the same ground set as the budget matroid Bb1,...,bk, and suppose that all of the key circuits of Bb1,...,bk are circuits also in M. As the next exercise shows, it follows from those assumptions alone that the rank of M
5.4. WEAKENING THE RULES FOR INDEPENDENCE 65 is at most b. Hence, in these cases, weakening the Ambient Rule however slightly would mean that we would no longer have a matroid.
Exercise 5.4-1 Show that the Ambient Rule can’t be successfully weakened when bj =dj >0 for all j . In particular, let M be any matroid whose ground set is the ground matrix of the budget matroid Bb1,...,bk. Show that, if every key circuit of the matroid Bb1,...,bkis a circuit also in M, then the rank of M is at most the total budget b=b1+ · · · +bk.
[Hint: Let P be any perfect set, let j be any column index, let f be an element of the jth column that belongs to P, let g be an element of the jth column that does not belong to P, and let X denote the set X := P ∪ {g} \ {f}. Because
|X| =b, it suffices to show that all of the elements of the ground matrix e lie in the flat Span(X), where that span is computed in the matroid M. Since the perfect set P is a circuit in M, the element f depends on the elements in the set P\{f}, which is a subset of X; hence, f lies in Span(X). The set X ∪ {f} = P ∪ {g}exceeds the dimension for the jth column by precisely 1; it follows that every element of the jth column depends on some subset of X ∪ {f}, and hence lies in Span(X). Show next, by considering new perfect sets, that every element in the same row as f belongs to Span(X). Conclude that, in fact, all of the elements of the ground matrix e must belong to Span(X).]
Exercise 5.4-2 Show that the Ambient Rule can be successfully weakened, even when bj =dj for all j , if at least one of the column budgets is zero. In particular, after reviewing Exercise 4.2-2, consider those matroids M on the same ground set as the budget matroid B2,1,0,0,...,0 and with the property that every key circuit of B2,1,0,0,...,0is a circuit also in M. Find such a matroid M whose rank is 1003.
[Hint: Put the points Z3 through Z1002 in general position with respect to the plane of the complete quadrilateral.]
Exercise 5.4-3 Show that the Ambient Rule can sometimes be successfully weak-ened, even when bj is positive for all j , if the dimensions(dj)aren’t equal to the budgets(bj). In particular, consider the budgetary matroid B2,2,21,1,1, whose key cir-cuits are precisely the six perfect sets. Find a matroid M of rank 4 on the same 3-by-3 ground matrix
[Hint: Let the circuits of M be precisely the perfect sets, the 2-by-2 subma-trices, and the unions of a row and a column. To check that M is a matroid, we can either verify the circuit-based axioms for a matroid or — probably easier —
66 CHAPTER 5. THE BUDGETARY MATROIDS
A1
A2
A3
B3
B1
B2
C2
C3
C1
r1
r2
r3
s1
s2
s3
Figure 5.1: A representation in 3-space of a matroid of rank 4 that has, as circuits, all six of the key circuits of the budgetary matroid B21,,21,,21. (Imagine the points A1 and A3as being above the plane of the paper, while B2and C2 are below it.) construct a representation. To do the latter, let r1, r2, and r3be three skew lines in 3-space and let s1, s2, and s3 be three of their common transversals, as shown in Figure 5.1. (We are going to study this configuration of nine points and six lines in Section 9.3, where we call it a grid.) We set Ai :=ri∩si, Bi :=ri+1∩si−1, and Ci :=ri−1 ∩si+1 for i in [1. .3], where all subscripts are interpreted modulo 3.
By the way, a matroid theorist [42] would think of this matroid M as the dual of the cycle matroid of the complete bipartite graph K3,3. The nine edges of the graph K3,3correspond to the nine points in Figure 5.1, while its six vertices correspond to the three lines(ri)and the three lines(si).]