We now turn from Euclidean 3-space to projective 3-space, which gives us a much richer class of transformations to exploit. We withdraw our decision to set p1 := 3 p2— that decision was motivated only by a quest for beauty, after all — but we keep everything else the same as before, including the constraints p1+p4 =0 and p2+ p3 = 0. The key idea in the projective case is to restrict our choices of the parameters(pi)so that the identity
p1p2p3p4+(p1p2+ p1p3+ p1p4+ p2p3+ p2p4+ p3p4)+9=0 holds. Note that this identity is symmetric in the four variables(pi). If this iden-tity holds, it follows from Exercise 10.3-1 that we have qi = −1/pi for all i.
Thus, converting from Euclidean coordinates(X,Y,Z) to homogeneous coordi-nates [w,x,y,z], we can write our configuration in the form:
A B C D
1 [−p1,1,1,1] [1,p1,−1,1] [1,1,p1,−1] [1,−1,1,p1] 2 [−p2,1,1,1] [1,p2,−1,1] [1,1,p2,−1] [1,−1,1,p2] 3 [−p3,1,1,1] [1,p3,−1,1] [1,1,p3,−1] [1,−1,1,p3] 4 [−p4,1,1,1] [1,p4,−1,1] [1,1,p4,−1] [1,−1,1,p4]
Given a configuration of this form, what column permutations can we achieve via projective symmetries? The six column permutations that fix the A column can all be achieved by the same Euclidean transformations as before. In addition, the projective transformation [w,x,y,z] 7→[x,−w,−z,y] swaps the point Ai with Bi and Ci with Di, for all i in [1. .4]. Thus, we can achieve all 24 column permu-tations via projective symmetries. The symmetries that achieve the even column permutations leave the rows fixed; those that achieve the odd permutations swap row 1 with row 4 and swap row 2 with row 3.
It remains to choose the parameters(pi), being careful to avoid any forbidden incidences. So far, we have three constraints on those four parameters. Perhaps the prettiest way to tie down the final degree of freedom is to require that p1 =1/p2, so that the set{p1,p2,p3,p4}is closed under both negation and inversion. A little
10.4. REPRESENTATIONS WITH PROJECTIVE SYMMETRIES 141 algebra reveals that we then have
p1 = +√ 3+√
2 p2 = +√
3−√ 2 p3 = −√
3+√ 2 p4 = −√
3−√ 2;
and some further algebra verifies that these choices do not lead to any forbidden incidences. The resulting representation of B1,1,1,1seems like a good candidate to win the projective beauty contest.
Chapter 11
Open questions
We wrap up by discussing some open questions about the budget matroids and their relationship to n-dependency.
11.1 Representability in general
Since what makes the budget matroids interesting is that so many of them are rep-resentable, the most tempting challenge about the budget matroids is to determine precisely which of them are representable.
Challenge 11.1-1 Determine which of the budget matroids are representable over the complex numbers. For those that are representable, count the degrees of free-dom involved in choosing a representation — that is, compute the dimension of the set of representations. Also, if possible, show how to construct a generic rep-resentation using a flat-side as the only geometric tool — that is, find a rational parameterization of the set of representations.
One could ask the same questions for the budgetary matroids as well. But the evidence in Chapter 5 suggests that most of them are unrepresentable, so the an-swers are less likely to be interesting.
Of course, the set of representations of a budget matroid has to be more than just a set, in order for its dimension to be well defined. The books by Harris [14], by Reid [46], and by Cox, Little, and O’Shea[7] are fine places to learn algebraic geometry at the level that you need to follow the next few paragraphs. It is simplest to assume, at this point, that our field of scalars is algebraically closed.
Let Bb1,...,bk be a budget matroid and let S denote projective space of dimension b−1, where b :=b1+· · ·+bkis the total budget. Each incidence that the matroid
Bb1,...,bk requires is a polynomial constraint on the homogeneous coordinates of our
bk points in S. Those constraints determine a certain variety W in the Cartesian product Sbkof bk copies of S. (If we like, we can use the Segre embedding [16]
to view the Cartesian product Sbkas a subvariety of projective space of dimension 143
144 CHAPTER 11. OPEN QUESTIONS bbk−1, thus realizing W also as a subvariety of that single, high-dimensional pro-jective space — so W is a propro-jective variety.) Each point in the variety W gives a way of mapping the elements of the matroid Bb1,...,bk to points in S so that all de-pendent sets of elements map to sets of points that are mutually incident — but various independent sets of elements may also map to sets of points that are mu-tually incident. Such a map is called a weak representation.
The variety W of weak representations is typically reducible, but we can de-compose it into its irreducible components. Each of the irreducible components V of W is either good or bad:
good The bulk of the points in V give true representations; only those in a proper subvariety of V have forbidden incidences.
bad All of the points in V give weak representations that share some forbidden incidence.
We define the freedom #(Bb1,...,bk) to be the maximum of the dimensions of the good components — unless all of the components are bad, in which case we set
#(Bb1,...,bk):= ⊥.
Note that there typically exist bad components whose dimension is larger than that of any good component. For example, consider the matroid B1,1,1,1. On the good side, we have #(B1,1,1,1)=20. On the other hand, one way to build a weak representation is to let the three column lines a, b, and c coincide, which is enough to ensure that all of the perfect coplanarities hold. There are four degrees of free-dom in the choice of the line a =b=c, four more in the skew line d, and one in each of the sixteen points along its column line, for a total of 24. Thus, speaking loosely, there are many more weak representations than there are true representa-tions — many more degenerate cases than nondegenerate ones.
Exercise 11.1-2 A simpler flavor of weak representation of the budget matroid B1,1,1,1 has all sixteen of its points lying in a common plane. How many degrees of freedom are involved in choosing one of those?
[Answer: 27, assuming that we still take the ambient projective space to be a 3-space, as opposed to a plane.]
Let Bb1,...,bk be a budget matroid that is representable, let W be its variety of weak representations, and consider those irreducible components V of W that are good. It seems reasonable to hope, even if several good components V exist, that only one of them will have the maximal dimension #(Bb1,...,bk). Whenever that is true, we have a particular, irreducible variety V that we can think of, loosely, as being ‘the variety of representations’ of the matroid. We can then go on to ask whether that variety V is either rational, meaning that V is birationally isomorphic to projective space of dimension #(Bb1,...,bk), or — failing that — perhaps unira-tional, meaning that there exists a dominant rational map from projective space
11.2. THE BUDGET MATROIDS WITH FEW COLUMNS 145