also skew, there exists a unique projective transformation of 3-space that fixes the line o and every plane through o, while taking the lines a and b to a0 and b0. Proof Choose three distinct planes through o, sayπ1,π2, andπ3. In each plane πi, there is a unique common transversal of the three lines o, a, and b. The three lines(o,a,b)together with those three common transversals form a grid.
Repeat this construction, starting with the three lines(o,a0,b0)and using the same three planesπ1,π2, andπ3through o. The result is a second grid.
The unique projective transformation that maps the first grid to the second fixes all three of the planes(πi), so it fixes all planes through o. tu
Exercise 9.3-5 Recall from Section 4.5 what it means to be a ‘configuration’ in the narrow sense of that term. A grid is obviously a configuration of points and lines in 3-space of type(92,63). Show that a grid can also be viewed as a configuration of points and planes of type(95,95).
[Answer: Each intersecting pair of lines spans a plane that passes through five points, and each point lies on five such planes.]
9.4 The Projection Theorem
Once we have formulas, in terms of some parameters, for all twelve of the points in a generic representation of B2,1,1, the cubic case of the Projection Theorem re-duces to algebra: showing that the determinant of a certain matrix is zero. Indeed, we could probably have proved the Projection Theorem for B2,1,1 based on the P-first construction and the formulas in terms of the parameters [u1,u2,u3] and [v1, v2, v3] in the proof of the Bm,1,1Representation Theorem. But we can get more symmetric formulas from the P-last construction, in terms of some new parame-ters(a1,a2,a3,a4;b1,b2,b3,b4); so we shall do that instead.
Theorem 9.4-1 (Projection Theorem, cubic case) Given any representation
2 1 1
P1 A1 B1
P2 A2 B2 P3 A3 B3 P4 A4 B4
of the matroid B2,1,1in some 3-space and given a generic line o in that 3-space, the twelve planes joining o to those four triples of points form a 3-dependent block.
Proof We restrict the line o to be generic only so that we don’t have to consider such degenerate cases as o = a, where projecting any one of the four A-points from the line o yields the indeterminate plane.
122 CHAPTER 9. ON B2,1,1AND 3-DEPENDENCY We proceed by analytic geometry, choosing our coordinate system on points [w,x,y,z] with some care. To make Lemma 8.5-1 applicable, we want the line a to be the line y = z = 0, the line b to be the line w = x = 0, and the plane π to be the planew = z, which has homogeneous coefficientsh1,0,0,−1i. We also want the center line o of the projection to have simple coordinates — in fact, to be the line y− w = z − x = 0, whose points have the form [u, v,u, v]. To verify that we have enough freedom to achieve all of those goals at once, we apply Corollary 9.3-3.
Since the twelve given points represent the matroid B2,1,1, the lines a and b are skew, and neither of them lies in the planeπ. Hence, there is a unique common transversal of the lines a and b that lies in the planeπ. Since the line o is generic, we can assume that o is skew to that common transversal. It follows that no com-mon transversal of a, b, and o lies in the planeπ.
Those same geometric properties hold of the lines and plane with the coordi-nates that we want a, b, o, andπ to have. The three lines y= z =0,w = x =0, and y−w =z−x =0 are indeed skew, since combining the equations for any two of them yields only the indeterminate solutionw = x = y = z = 0. The plane w= z does not contain any common transversal of those three lines, since the three points [0,1,0,0], [0,0,1,0], and [1,1,1,1] at which the three lines y = z = 0, w= x = 0, and y−w =z−x =0 cut the planew =z are not collinear.
It follows from Corollary 9.3-3 that there is a unique coordinate system for 3-space in which the three lines(a,b,o)and the planeπ have the coordinates that we want them to have. Where are the twelve points of the given representation in that coordinate system? The point Aπ := a∩π has the homogeneous coordi-nates [0,1,0,0]. The point Ai, for i in [1. .4], lies on the line a but is distinct from Aπ; so we must have Ai = [1,a1,0,0], for some finite scalar ai. Similarly, we have Bπ = [0,0,1,0] and Bi = [0,0,bi,1], for some finite scalar bi. We are going to use the scalars(a1,a2,a3,a4;b1,b2,b3,b4)as our parameters. Note that these parameters involve eight degrees of freedom, while the homogeneous parameter vectors [u1,u2,u3] and [v1, v2, v3] that we used in analyzing the P-first construction involve only four. We need an additional four degrees of freedom in this analysis because we have constrained our choice of coordinate system to put the center line o of the projection in a fixed, simple place.
Lemma 8.5-1 now tells us the coordinates of the P-points. When{i,j,k,l} = that swapping any two of the three indices j , k, and l would negate all four of the coordinates; but that doesn’t matter, because they are homogeneous.
9.4. THE PROJECTION THEOREM 123 Given some point [w,x,y,z], what is the slope of the plane joining that point to the center line o? It is easy to check that the homogeneous coefficients of the joining plane arehz−x, w−y,x−z,y−wi: The point [w,x,y,z] lies on that plane because(z−x)w+(w−y)x+(x−z)y+(y−w)z=0, while every point [u, v,u, v] on the line o lies on that plane because(z−x)u+(w−y)v+(x−z)u+(y−w)v = 0. As for what slope to assign to that joining plane, we showed in Lemma 2.5-1 that the notion of n-dependence is projectively invariant. Hence, we can take any three distinct planes we like, in the pencil of planes through o, and assign to those planes the slopes 0, 1, and∞. To be definite, let’s measure the slope of the plane hz−x, w−y,x−z,y−wiby using the ratio z−x : y−wof the first homogeneous coefficient to the last.
Measuring slopes in this way, the plane joining the line o to the point Ai = [1,ai,0,0] has slope−ai : −1, which is the same as ai : 1. The plane joining o to Bi =[0,0,bi,1] has slope 1 : bi. And the plane joining o to Pi = [wi,xi,yi,zi] has slope zi − xi : yi −wi. To build the ith row of the 4-by-4 matrix that tests 3-dependence, we take those three ratios and combine them as in the elementary symmetric polynomials — that is, by the process that takes the three simple ratios t↑ : t↓, u↑ : u↓, andv↑ :v↓into the compound ratio
t↑u↑v↑ : t↑u↑v↓+t↑u↓v↑+t↓u↑v↑ : t↑u↓v↓+t↓u↑v↓+t↓u↓v↑ : t↓u↓v↓. Doing so for each of the four rows in turn, we conclude that the twelve slopes are 3-dependent just if this determinant is zero:
That determinant is zero for the simple reason that the first and third rows of the matrix have the same sum as the second and fourth. To see this, start by considering the term a1z1 in the upper-left entry. The determinant
is zero, since its first and third columns are equal; if we expand by cofactors down the first column, we find that a1z1−a2z2+a3z3−a4z4 =0. A similar argument
124 CHAPTER 9. ON B2,1,1AND 3-DEPENDENCY shows that a1x1−a2x2+a3x3−a4x4 =0 as well; so the alternating sums of each of the two terms in the first column is zero. In the second column, of the six terms in each entry, similar arguments handle four, leaving the two terms aibizi +aiyi. The alternating sums of those two terms are not separately zero, but they cancel each other out. Using cofactors down the first column again, the alternating sum a1b1z1−a2b2z2+a3b3z3−a4b4z4is the determinant
Those two determinants are negatives of each other because we can convert one matrix into the other by swapping the first and third columns. tu
Exercise 9.4-2 In the P-last construction, suppose that the cross ratios A(1,2,3,4) and B(1,2,3,4)are distinct, which means that the determinant of the matrix
m :=
is nonzero, and suppose that the planeπ is generic. In analyzing the P-last con-struction in Section 9.2, we showed that the four P-points given by Pi := π∗i will then form a projective frame for the planeπ. Verify this directly and alge-braically, using the formulas from Lemma 8.5-1 for the homogeneous coordinates [wi,xi,yi,zi] of the point Pi. In particular, whenever {i,j,k,l} = {1,2,3,4},
[Hint: Given any n-by-n matrix s, let’s borrow a term from the theory of deter-minants [34] and refer to the matrix of cofactors of s as the adjugate1of s, and let’s
1Artin [2] uses the name ‘adjoint’ for the transpose of the adjugate, but that conflicts with the Hermitian adjoint, which is the transpose of the complex conjugate.
9.5. THE WITNESS THEOREM 125