Given the homogeneous coordinates of the vertices of a skew-Pappian hexagon, it will be helpful to be able to compute the associated null system explicitly — that is, to go back and forth between the homogeneous coordinates of a pole point and the
8.5. THE HOMOGENEOUS COORDINATES OF A POLE 111 homogeneous coefficients of its polar plane. Finding formulas that worked in full generality would be a bit clumsy, because of the skew-Pappian condition: The ver-tices of a skew-Pappian hexagon must alternate between two skew lines, and that puts constraints on the coordinates of the vertices. For our purposes, though, it will suffice to deal with a very special case, one in which the skew-Pappian collinear-ities are guaranteed by zero coordinates.
is the pointπ∗ with homogeneous coordinates π∗ =
Proof We verify that the pointπ∗results from the construction in Exercise 8.4-3.
As a check of plausibility, note that the first and last homogeneous coordinates of the pointπ∗are the same, so the pointπ∗does lie on the planeπ.
The points on the line A1B2have the form [u,ua1, vb2, v] for some ratio u :v. Thus, the line A1B2intersects the planeπat the point [1,a1,b2,1]. Similarly, the line A2B1meets the planeπ at the point [1,a2,b1,1]. The construction of Exer-cise 8.4-3 tells us that the pointπ∗ lies on the line in the planeπ that joins those two intersections. This happens precisely if the matrix
Chapter 9
On B 2 , 1 , 1 and 3-dependency
9.1 Constructing the P-points last
Using the theory of null systems, we can construct representations of the matroid B2,1,1 in a new order: Choose the A-points, the B-points, and the planeπ; then construct the P-points. We shall call this the P-last construction, to contrast it with the P-first construction used in proving the Bm,1,1Representation Theorem.
Since the four rows play completely symmetric roles in the P-last construction — there is no need to treat one row specially, as there was in the P-first case — we return to indexing the rows starting with 1:
2 1 1
P1 A1 B1 P2 A2 B2
P3 A3 B3 P4 A4 B4
Proposition 9.1-1 (The P-Last Construction) In any representation of the bud-get matroid B2,1,1, the point P1 is the pole of the planeπ := Span(P1,P2,P3,P4) in the unique null system N1 := N((A2,B2), (A3,B3), (A4,B4)) for which the skew-Pappian hexagon A2B3A4B2A3B4is skew-polar, and analogous results hold for the points P2, P3, and P4. We shall denote these results by writing Pi = π∗i, where the superscript ‘∗i’ means ‘polar in the null system Ni’.
Conversely, suppose that we choose any four collinear A-points in 3-space, any four collinear B-points, and any plane π. For generic choices of those nineteen parameters, the four A-points will be distinct, the four B-points will be distinct, and the lines a and b on which they lie will be skew. For i in [1 . .4], we can then define the null system Ni, and we can construct the point Pi := π∗i. For generic choices of the nineteen parameters, once again, the resulting twelve points {(Pi,Ai,Bi)}i∈[1..4]will represent the matroid B2,1,1.
113
114 CHAPTER 9. ON B2,1,1AND 3-DEPENDENCY
P1
P2
P3
P4
t12
t13
t14
t23
t24
t34
π
Figure 9.1: The complete quadrangle that arises in the planeπ when the P-last construction is used to build a representation of the budget matroid B2,1,1.
Proof Given any representation of B2,1,1, consider what happens in the planeπ. The four P-points lie inπ, and no three of them are collinear. Thus, they determine a complete quadrangle, as shown in Figure 9.1. For{i, j,k,l} = {1,2,3,4}, note that the quadruples{Pi,Pj,Ak,Bl}and{Pi,Pj,Al,Bk}are both coplanar. Thus, the line PiPj is the unique common transversal in the planeπ of the skew lines AkBl and AlBk — call this common transversal tk l. In particular, the point Pi lies on all three of the common transversals tj k, tjl, and tk l. We argued in Exercise 8.4-3 that those three transversals are indeed concurrent and that the point Pi = π∗i where they concur is the pole of the planeπ in the null system Ni determined by the skew-Pappian hexagon AjBkAlBjAkBl.
Conversely, given the A-points, the B-points, and the planeπ, suppose that we construct Pi :=π∗i for i in [1. .4]; will the result be a valid representation?
It is easy to see that every set that should be mutually incident will be. The four P-points are coplanar because the pole of the plane π in any null system always lies inπ. The four points in the perfect set{Pi,Pj,Ak,Bl}are coplanar because Pi is placed at the point of concurrence of the three common transversals tj k, tjl, and tk l, while Pj is placed at the point of concurrence of tik, til, and tk l — so both Pi and Pj lie on the line tk l, which meets the line AkBl.
To show that no forbidden incidences arise in generic cases, it suffices to check that none arise in some particular case, and any representation of the matroid B2,1,1
provides such a case. Readers who made it through Chapter 7 already know that the matroid B2,1,1is representable; the rest can verify for themselves that the twelve points given in Exercise 9.1-2 constitute a representation. tu
9.1. CONSTRUCTING THE P-POINTS LAST 115
P1
P2
P3
P4
A1′ A2′
A3′ A4′
B1′ B2′
B3′ B4′
Figure 9.2: The projections onto the XY -plane of the twelve vertices of a particu-larly symmetric representation of the budget matroid B2,1,1.
Exercise 9.1-2 Verify that the following is a representation of the budget matroid B2,1,1, sitting in Euclidean(X,Y,Z)-space:
P A B
1 (2,4,0) (−10,0,1) (0,−5,−1) 2 (4,−2,0) (−5,0,1) (0,10,−1) 3 (−4,2,0) (5,0,1) (0,−10,−1) 4 (−2,−4,0) (10,0,1) (0,5,−1)
The planeπis the plane Z =0, which we take to be horizontal; the line a is parallel to the X axis, but one unit above it; the line b is parallel to the Y axis, but one unit below it. Figure 9.2 shows the(X,Y)-planeπ, together with the points A0i and Bi0, the vertical projections of Ai and Bi onto the planeπ. Note that this representation is carried to itself by the following symmetry of order 4: Map each point(X,Y,Z) to the point(−Y,X,−Z), permute the row indices through the 4-cycle(1,3,4,2), and swap the A and B columns.
[Hint: For any i and j , the line AiBj meets the planeπ at the midpoint of the segment A0iB0j. To show that the perfect sets are coplanar, check that, whenever i, j , and k are distinct, the point Pk lies on the line joining the midpoint of A0iBj0 to the midpoint of A0jBi0. To show that there are no forbidden incidences, it suffices to verify the absence of the primitive degeneracies listed in Exercise 7.4-4.]
Exercise 9.1-3 Figure 9.3 is a close-up of the rope-and-pole structure found by the
116 CHAPTER 9. ON B2,1,1AND 3-DEPENDENCY
Figure 9.3: A representation of the budget matroid B2,1,1in the Brazilian jungle.
9.2. DEGENERATE CASES IN THE P-LAST CONSTRUCTION 117