Our next goal is to convert the process that Proposition 10.1-2 uses to produce rep-resentations of the matroid B1,1,1,1into a geometric construction. In order to do that, we need some geometric tool more powerful than a flat-side; essentially, we
10.2. A PROJECTIVE COMPASS 135 need a tool that solves quadratic equations — at least, solves those quadratics that do have solutions. In Euclidean geometry over the real numbers, the standard tool for this job is the compass; so, it seems reasonable to call the tool that we need here a projective compass. There are lots of alternatives for precisely what a projective compass might do, but the different models all have equivalent power:
Model 1 Given three distinct points A, B, and C along a line ` and three more distinct points A0, B0, and C0 along that same line`, the projective compass constructs for us the two fixed points of the unique projectivity from`to` that maps(A,B,C)to(A0,B0,C0).
Model 2 Given two pairs of points(A,B)and(C,D)along a line`, the projective compass constructs for us the two fixed points of the unique involution of` that swaps A with B and swaps C with D.
Model 3 Given five points and one line in the plane, the projective compass con-structs for us the two intersections of the line with the unique conic through the five points.
Model 4 Given four skew lines in 3-space, the projective compass constructs for us their two common transversals.
Of course, if we are working over a field in which some scalars don’t have square roots, then the projective compass may fail on some problem instances; that is true for any of the four models of a projective compass.
Exercise 10.2-1 Show that the four models of a projective compass have equiva-lent power.
[Hint: Implementing Model 2 with Model 1 is trivial.
To implement Model 3 with Model 2, suppose that we are given five points and a line`. Construct some point P on the conic through the five given points and construct the tangent line m to the conic at P, both of which we can do with just a straightedge. Also with a straightedge, we can construct the second tangent m0 to the conic that passes through the point`∩m and the point P0where m0touches the conic. The mapping that takes P to P0 is an involution on the conic, and the fixed points of that involution give us the points where the line`cuts the conic.
To implement Model 1 with Model 3, note first that we can think of a projec-tivity as acting either on the points in a range or on the lines in a pencil. Using the second form, suppose that we are given two triples(a,b,c)and(a0,b0,c0)of lines through a point P and that we want to construct the two fixed lines of the projec-tivity of the P pencil that takes(a,b,c)to(a0,b0,c0). Let Q be some point and let
`be some line. Given any line e through P, let e?be the line that joins Q to e∩`. There is a unique projective correspondence from the P pencil to the Q pencil that takes(a,b,c)to(a0?,b0?,c0?), and the intersections of corresponding pairs of lines
136 CHAPTER 10. THE BUDGET MATROID B1,1,1,1
(e,e0?)trace out a conic. The two points where the line` cuts that conic give us the two fixed lines of the original projectivity.
To implement Model 4 with Model 1, let a, b, c, and d be the four skew lines.
The one-parameter family of all common transversals of a, c, and d determines one projective correspondence between c and d, while the similar family of all com-mon transversals of b, c, and d determines another. If we map from c to d via one of those correspondences and then back from d to c via the other, we get a projec-tivity of the line c whose two fixed points give us the two common transversals.
Finally, to implement Model 1 with Model 4, let P 7→ P0 be the given pro-jectivity of the line`, and choose two lines m and n that are skew to each other and to`. For each point P on`, let P?denote the point on m that lies in the plane Span(P,n). As the point P varies along`, the lines of the form P0P? sweep out a ruled quadric. The lines`and m are generators of that same quadric, from the other family. Let k be a third generator from that other family, skew to`and m.
The common transversals of k,`, m, and n give us the two fixed points of the orig-inal projectivity.]
Which model of the projective compass is most convenient depends upon the application. When constructing a representation of B1,1,1,1in Proposition 10.1-2, we needed to find a common transversal of four skew lines. But there is an impor-tant special case of the same construction where what we need to do, instead, is to find the fixed points of a projectivity.
Suppose that the A-points and B-points in Proposition 10.1-2 are chosen so that the cross ratios A(1,2,3,4)and B(1,2,3,4)are equal. This equality does not hold in general, but it might be desirable to have it hold in certain cases — for example, when trying to construct a representation of B1,1,1,1with lots of geometric symme-tries. When those two cross ratios are equal, the four null systems(Ni)coincide, so the four lines(c∗i)all coincide with the polar c∗ of the line c in that single null system. It follows from Lemma 10.1-1 that we must take d to be the line c∗. But where on the lines c and d =c∗should we put the C-points and the D-points?
One way to figure out where is as follows: Choose any point along the line c, and tentatively call it C1. Using one of the perfect coplanarities{A3,B4,C1,D2} or{A4,B3,C1,D2}, figure out where D2should go along the line d, given that C1
is as we have guessed. In a similar way, compute the points C3, D4, and C1again, each from the previous one. If the final point C1 coincides with our initial guess for C1, we were very lucky, and we have found a representation of B1,1,1,1. Typ-ically, the initial and final values of C1 will not coincide. But the mapping that takes our initial guess for C1 to the resulting final value is some projectivity of the line c. We can characterize that projectivity uniquely by carrying out this four-step tracing process three times, starting with three distinct, initial guesses for C1. We then employ a Model-1 projective compass to construct the two fixed points of that projectivity — those two points being the two places where C1 can go in a valid representation.