10.3 Representations with Euclidean symmetries
If all of the column budgets of a budget matroid are equal, the columns play sym-metric roles. It is tempting to try to find representations of such matroids in which as many as possible of the combinatorial symmetries that permute the columns are modeled by geometric symmetries — that is, by projective transformations of the ambient space that map the representation to itself.
Exercise 4.1-1 considered this issue for the Pappus matroid B1,1,1, in which the three columns play symmetric roles. For the representations of B1,1,1discussed in that exercise, all six permutations of the three columns can be achieved by pro-jective transformations — even by Euclidean transformations — of the ambient plane. Furthermore, all six of those transformations leave the rows fixed. In this section and the next, we see how close we can come to that same behavior for the matroid B1,1,1,1in 3-space.
As a warm-up for the projective case, let’s first consider representing the ma-troid B1,1,1,1 in Euclidean 3-space so that as many as possible of the 24 column permutations are achieved by Euclidean transformations. We shan’t do very well with Euclidean transformations — in particular, we shall handle only those 6 of the 24 permutations that leave one of the four columns fixed. But at least the re-sulting representations are easy to visualize, as demonstrated in the accompanying videotape [44].
The basic idea is as follows: Given a hyperboloid of one sheet with circular cross sections, we choose three of the four column lines of the representation — say b, c, and d — to be generating lines of that hyperboloid, so arranged that they are cyclically permuted by a 120◦rotation around the axis of the hyperboloid — which we choose to be the remaining column line, a. One convenient way to set this up in coordinates is to choose, as the hyperboloid, the surface XY+X Z+Y Z+1=0 in Euclidean(X,Y,Z)-space. That surface is rotationally symmetric about the main diagonal a and it includes, as one generating line, the line b parameterized by t 7→
(t,−1,1). Cyclically permuting the three coordinates X, Y , and Z corresponds to rotating this hyperboloid by 120◦about its axis, which also cyclically permutes the generating line b : t 7→ (t,−1,1)and its two cyclic shifts c : t 7→(1,t,−1)and d : t 7→(−1,1,t).
Given any point(X,Y,Z), let us refer to the sum of its coordinates X+Y+Z as its height. Note that the planes of constant height are orthogonal to the main diagonal a, and that the hyperboloid XY+X Z+Y Z+1=0 is symmetric under reflection through the plane of height 0. We envision a representation of B1,1,1,1of the form:
A B C D
1 (q1,q1,q1) (p1,−1,1) (1,p1,−1) (−1,1,p1) 2 (q2,q2,q2) (p2,−1,1) (1,p2,−1) (−1,1,p2) 3 (q3,q3,q3) (p3,−1,1) (1,p3,−1) (−1,1,p3) 4 (q4,q4,q4) (p4,−1,1) (1,p4,−1) (−1,1,p4)
138 CHAPTER 10. THE BUDGET MATROID B1,1,1,1
The parameter pi, for i in [1. .4], is the common height of the points Bi, Ci, and Di, while the point Ai lies at height 3qi along the main diagonal.
For arbitrary values of the parameters(pi), there are unique values for the pa-rameters(qi)that make the 24 perfect coplanarities of the matroid B1,1,1,1hold. To determine the requisite value of qi, we intersect the plane BjCkDl with the main diagonal a, where{i,j,k,l} = {1,2,3,4}.
Exercise 10.3-1 Verify that the plane BjCkDl intersects the main diagonal at the point(qi,qi,qi), where
qi := (pj + pk + pl)+ pjpkpl 9+(pjpk + pjpl + pkpl).
Because of the three-fold rotational symmetry, the formula for qi clearly had to be symmetric under cyclic shifts of the indices(j,k,l). It turns out to be sym-metric under arbitrary permutations of(j,k,l). Thus, if we set Ai := (qi,qi,qi) for each i, where qi is as given in Exercise 10.3-1, the resulting configuration has all of the incidences that are required for a representation of the matroid B1,1,1,1. It may also have forbidden incidences, of course, depending upon our choices for the parameters(pi).
We can double the number of Euclidean symmetries, from three to six, by stip-ulating that p1+ p4and p2+p3are both zero, from which it follows that q1+q4 and q2+q3will also be zero. Under this stipulation, a 180◦rotation about the line X =Y+Z =0 — which maps the point(X,Y,Z)to the point(−X,−Z,−Y)— then maps Ai to A5−i, Bi to B5−i, Ci to D5−i, and Di to C5−i, for all i in [1. .4]. That is, the A and B columns are held fixed while the C and D columns are swapped, but at the cost of uniformly swapping row 1 with row 4 and row 2 with row 3.
There is no way that a Euclidean symmetry of the configuration could avoid fix-ing the axis a of the hyperboloid; so we aren’t gofix-ing to be able to achieve more than 6 of the 24 column permutations with Euclidean symmetries. Hence, we might as well set the two remaining free parameters p1and p2to achieve some lesser goal, such as beauty. For example, we can set p1 :=3 p2, so that the B-points, C-points, and D-points are equally spaced, along their lines. We can then choose the remain-ing parameter p2so that the A-points also end up equally spaced, along the main diagonal.
Exercise 10.3-2 Assuming that we set p1 := 3 p2, p3 := −p2, and p4 := −3 p2, find the positive, real values of p2 that result in the four A-points being equally spaced, in some order, along the main diagonal a.
[Answer: The values p2 = (√
17±2√
2)/3 make the four A-points equally spaced in the order(A1,A3,A2,A4). But the resulting configurations are not rep-resentations of B1,1,1,1, because they have twelve forbidden coplanarities, of which {A1,B1,C4,D1}is an example. Luckily, the values p2=(√
53±2√ 2)/3√
5 also make the A-points equally spaced, this time in the order(A3,A1,A4,A2). The
10.3. REPRESENTATIONS WITH EUCLIDEAN SYMMETRIES 139 configurations that result from these two values have no forbidden incidences, and hence do represent B1,1,1,1. The representation shown in the videotape [44] has
p2 =(√
53−2√ 2)/3√
5=0.6636+.]
This is a convenient time to interject a comment about cross ratios. Recall that there is no representation of the budget matroid B2,1,1in which the two cross ratios A(1,2,3,4) and B(1,2,3,4)along the two column lines are equal. Given that, one might wonder whether there is any constraint on the four cross ratios A(1,2,3,4), B(1,2,3,4),
C(1,2,3,4), and D(1,2,3,4)in a representation of the matroid B1,1,1,1. The answer seems
to be no. By varying p2 continuously in the construction of the previous exer-cise, we can keep the last three of those four fixed while varying the first contin-uously; so there can’t be any algebraic constraint relating the four cross ratios. In that example, of course, the last three are not only fixed, but also equal. Generi-cally, however, we know that no two of the four cross ratios are equal, since that is what happened in the example in Proposition 10.1-2. Indeed, in that example, the six possible cross ratios that we can get by taking the four A-points in some order together with the six similar cross ratios from the other three lines constitute 24 distinct scalars. Thus, generically, even if we allow ourselves to reorder the points, no cross ratio along any column line coincides with any cross ratio along any other column line. On the other hand, it can also happen that all four cross ratios are equal, with no reordering allowed, as we see in the next section.
Exercise 10.3-3 Are there representations of the budget matroid B1,1,1,1in which all sixteen points lie on a common quadric? In such a representation, the four col-umn lines must be generating lines of that quadric, all drawn from the same family.
[Answer: Yes — at least, the authors’ Newton-Raphson search converged on a convincing approximation to such a representation.]
Exercise 10.3-4 Continuing the theme of the preceding exercise, suppose that the sixteen points{(Ai,Bi,Ci,Di)}i∈[1..4]form a representation of the matroid B1,1,1,1 in which all sixteen points lie on a common quadric. Show that the four cross ra-tios A(1,2,3,4), B(1,2,3,4), C(1,2,3,4), and D(1,2,3,4)along the four column lines must be distinct, and hence this representation is not very symmetric.
[Hint: Given three skew lines, all of their common transversals form a one-parameter family that is called — at least, was once called [29] — a regulus. The lines of a regulus sweep out a ruled quadric surface, forming one of that surface’s two families of generating lines. The other family of generating lines of the same ruled quadric is called the complementary regulus.
Getting back to the exercise, suppose, by way of contradiction, that the cross ratios A(1,2,3,4) and B(1,2,3,4) are equal. The four lines{AiBi}i∈[1..4] then belong to a common regulus whose complement — call it R — contains both a and b. The lines c and d are constrained in two ways: They are polars of each other in the sin-gle null system N = N((Ai,Bi), (Aj,Bj), (Ak,Bk))that results from any subset
140 CHAPTER 10. THE BUDGET MATROID B1,1,1,1
{i, j,k} ⊂ {1,2,3,4}, and they also belong to some regulus R0 that contains both a and b. Show that those two constraints imply that the lines(c,d)are harmonic conjugates of the lines(a,b)in the single regulus R0 = R. But, if R0 = R, the lines(AiBi)meet the lines c and d, which cannot happen because, for example, the set{A1,B1,C1,C2}is forbidden to be coplanar.]