from which it is easy to calculate that(adg(adg s))=(det s)n−2s. The coordinates (xi), (yi), and(zi) appear, up to sign, as entries in the adjugate matrix(adg m), which means that the determinant that we want to evaluate is, up to sign, an entry in the matrix(adg(adg m)).]
9.5 The Witness Theorem
The place where the P-last construction really shines is in proving the cubic case of the Witness Theorem.
Theorem 9.5-1 (Witness Theorem, cubic case) Let {(ρi, αi, βi)}i∈[1..4] be some 3-dependent, ordered block of planes in 3-space, all passing through a common line o.2 In generic cases, there exists a representation{(Pi,Ai,Bi)}i∈[1..4]of the budget matroid B2,1,1each of whose twelve vertices lies on the corresponding plane and none of whose vertices lies on the line o itself — and which hence witnesses to the 3-dependence of the planes. Furthermore, this representation is unique in the sense that any witnessing representation can be carried to any other by a projective transformation of 3-space that fixes the line o and fixes every plane through o.
Proof To construct a witnessing representation, choose the lines a and b to be any two lines that are skew to each other and skew to o. Let Ai denote the point Ai :=αi∩a and Bidenote Bi :=βi∩b, for i in [1. .4]. In generic cases, the points Ai and Bi will be distinct and the two cross ratios A(1,2,3,4) and B(1,2,3,4) will also be distinct. The P-last construction tells us how to finish up building a represen-tation of B2,1,1: We choose some planeπ and, whenever{i, j,k,l} = {1,2,3,4}, we define the point Pi := π∗i to be the pole of the plane π in the null system Ni := N((Aj,Bj), (Ak,Bk), (Al,Bl))determined by the skew-Pappian hexagon AjBkAlBjAkBl.
How should we choose the plane π? Note that the point Pi = π∗i will lie on the planeρi just when the planeπpasses through the point Ri :=ρi∗i. So we want π to pass through all four of the points(Ri). The assumed 3-dependence of the block of planes had better imply that the four points(Ri)are coplanar.
To see that it does, choose any three of them — say R1, R2, and R3 — and chooseπ to be the planeπ := R1R2R3 that those three points determine. The resulting configuration of twelve points{(Pi,Ai,Bi)}will, in generic cases, be a representation of B2,1,1, and it will have all of its vertices on the appropriate input planes, except that P4might not lie onρ4. Letρ40 be the unique plane through o and P4. It follows from the Projection Theorem that the slopes of the twelve planes {(ρ1, α1, β1), (ρ2, α2, β2), (ρ3, α3, β3), (ρ40, α4, β4)}will be 3-dependent. But, in
2We refer to the first plane in the ithtriple asρi, rather than asπi, in order to avoid confusion with the planeπ :=Span(P1,P2,P3,P4)below.
126 CHAPTER 9. ON B2,1,1AND 3-DEPENDENCY generic cases once again, there is only one slope that the twelfth plane can have, given the other eleven slopes, that makes the twelve slopes 3-dependent. So we must haveρ4 =ρ40, and we are done with the construction.
As for uniqueness, the only free choices that we made were the choices of the skew lines a and b. By Corollary 9.3-4, there is a unique projective transformation of 3-space that fixes the line o, fixes every plane through o, and takes the two lines a and b, skew to each other and to o, to any other two such lines a0 and b0. Thus, any witnessing representation differs from any other only by a projective transfor-mation. tu
From the proof of the Witness Theorem, we can extract a flat-side geometric construction that solves the following problem: Given eleven planes through a line o — sayαi andβi for i in [1. .4] andρi for i in [1. .3] — construct the unique twelfth planeρ4 through o that makes the block {(ρi, αi, βi)}i∈[1..4] 3-dependent.
We choose two lines a and b that are skew to each other and skew to o, and we let Ai be the point where the line a cuts the planeαi, and similarly for Bi. For i in [1. .3], we construct the point Ri := ρi∗i, the pole of the planeρi in the null system determined by the skew-Pappian hexagon AjBkAlBjAkBl. Finally, letting πdenote the planeπ := R1R2R3, we construct the point P4 :=π∗4. The planeρ4
that achieves 3-dependence is the unique plane through o that passes also through P4. Note that, if the plane ρ4 is our only goal, there is no need to construct the vertices P1, P2, and P3 of the witnessing B2,1,1representation.
Exercise 9.5-2 (Cubic harmonic conjugacy) Here is an ordered block of scalars that is 3-dependent:
Choose a line o in 3-space, find the twelve planes through o with those slopes, and carry out the construction for a witnessing representation of B2,1,1. Is what results a valid representation, or are these slopes a degenerate case? What is special about the resulting configuration and its relationship with the center line o?
[Answer: The result is a valid representation of B2,1,1, but it has the special property that the first three row planes P1A1B1, P2A2B2, and P3A3B3 intersect, not in a point, as is typically the case, but in an entire line. In addition, the center line o of the projection coincides with that line of intersection.
Commentary: This situation is the cubic analog of harmonic conjugacy. Recall that two scalars u andvare quadratic harmonic conjugates of two distinct scalars
p and q just when the block
9.5. THE WITNESS THEOREM 127 is 2-dependent. In a similar way, we say that three scalars u,v, andw are cubic harmonic conjugates of three distinct scalars p, q, and r just when the block
is 3-dependent. In the example above, the scalars(−5,−2,1)are cubic harmonic conjugates of the distinct scalars(−1,4,5).
The new wrinkle that arises in the cubic case is that only those representations of B2,1,1that have a special property can serve as witnesses to the harmonic conju-gacy of two triples. In the quadratic case, given any representation of B2,1 — that is, given any complete quadrilateral — there are three places where we can put the center point O so as to end up, as in Figure 2.4, with two pairs of lines through O that are harmonic conjugates. In the cubic case, on the other hand, in order to end up with two triples of planes that are harmonic conjugates, we must start with a representation of B2,1,1in which three of the four row planes lie in a common pen-cil; and then we must choose the center line o of the projection to coincide with the axis of that pencil.
For any degree n, harmonic conjugacy gives a binary relation on unordered n-tuples of distinct scalars, and that relation is always symmetric. For example, in the cubic case, suppose that both of the triples{p,q,r}and{u, v, w}consist of distinct scalars. Then, when either of the blocks
But the relation of n-ic harmonic conjugacy is reflexive only when the degree n is odd. For example, the 1-block
p p
is always 1-dependent and the 3-block
is always 3-dependent. Indeed, if we interpret the 3-dependence of that latter block geometrically, we can demonstrate a property of twisted cubic curves that we men-tioned back in Section 2.9. The cubic polynomial(X − p)(X−q)(X −r)is the intersection of the osculating planes to the twisted cubic curve Ft := (X −t)3 at the three points Fp, Fq, and Fr. We mentioned in Section 2.9 that the intersection
128 CHAPTER 9. ON B2,1,1AND 3-DEPENDENCY point(X− p)(X−q)(X −r)always lies in the plane FpFqFr determined by the three points of osculation.
When the degree n is even, on the other hand, the relation of n-ic harmonic conjugacy is not reflexive. For example, the 2-block
is 2-dependent only when p and q are equal. We can interpret that fact geomet-rically also. The quadratic polynomial(X − p)(X −q)is the intersection of the tangent lines to the conic curve Gt :=(X−t)2at the points Gp and Gq. But, for
p and q distinct, that intersection point never lies on the chord GpGq.]