FINDING A NON-BROKEN CIRCUIT BASIS FOR THE ORLIK-SOLOMON ALGEBRA OF AN ARRANGEMENT OF HYPERPLANES. Final Year Project

Transcription

1 FINDING A NON-BROKEN CIRCUIT BASIS FOR THE ORLIK-SOLOMON ALGEBRA OF AN ARRANGEMENT OF HYPERPLANES Final Year Project January 2012

2 Contents 1 Abstract 2 2 The Theory of Hyperplane Arrangements 4 3 The Non-Broken Circuit Basis 12 4 The Algorithm 21 5 Conclusions 27 1

3 Chapter 1 Abstract A hyperplane is a subspace of dimension (m 1) in a vector space of dimension m. A hyperplane arrangement is a finite collection of hyperplanes in a finite dimensional vector space. Although so easily defined, the study of arrangements leads to profound and beautiful results, often linking diverse areas of mathematics. Their research dates from 1889, where S. Roberts (London Mathematical Society), gave a formula for the number of regions formed by an arbitrary arrangement of n lines in the plane. 1 However, it is only in the last few decades that arrangements have been of increased interest. Peter Orlik and Hiroaki Terao compiled an outstanding resource on the subject Arrangements of Hyperplanes, exploring in particular the links with graph theory, group theory and combinatorics. 2 Of central importance to this subject is the Orlik-Solomon algebra O(A) of a hyperplane arrangement A, the quotient of the exterior algebra E(A) over A by a homogeneous ideal I(A). This is perhaps largely due to the following well-known result introduced by Peter Orlik and Louis Solomon in 1980: Let A, X be an arrangement of hyperplanes and the associated complement. Let M be the associated matroid of the arrangement. Then, the Orlik-Solomon Algebra O(A) is isomorphic to the cohomology ring H (X) = H (X, k). To facilitate the study of the structure of O(A), it was deemed useful to have a standard way to choose a basis. This basis, called the non-broken circuit basis and constructed independently by a number of mathematicians, is discussed in depth in the volume mentioned above. 3 It was noted that in their expert analysis, the authors did not answer the question of how to find such a basis given an arbitrary hyperplane arrangement in a vector space of dimension 3. This 1 Roberts, S.: On the figures formed by the intercepts of a system of straight lines in a plane, and on analogous relations in space of three dimensions. Proc. London Math. Soc (1889) Orlik, P., Hiroaki, T.: Arrangements of Hyperplanes. Springer-Verlag (1991) 3 Ibid, p

4 project seeks to answer that question by way of providing an original algorithm for the procedure. The research is divided as follows. Chapter 2 deals with the general theory of hyperplane arrangments, introduces the idea of an algebra over a field and explains the construction of the Orlik-Solomon algebra. Chapter 3 introduces the idea of a non-broken circuit basis, beginning with a list of definitions. We consider a non-trivial example and deduct many of the fundamental properties of the basis from that example, finishing with a theorem and an original corollary which is vital for the writing of a suitable algorithm. Chapter 4 introduces the algorithm. Its efficiency is demonstrated by taking a 5-dimensional arrangement and finding the non-broken circuit basis both with and without the algorithm. Chapter 5 contains a short conclusion and presents a number of open questions regarding the algorithm. Every example in this report is original, as are the majority of explanations and results. 3

5 Chapter 2 The Theory of Hyperplane Arrangements In this chapter we assume a basic knowledge of linear algebra, and in particular the definition of a vector space and a subspace. We begin with the following definitions: Definition 2.1 Let S = {v 1, v 2,..., v n } be a subset of a vector space V K over a field K. S is said to be linearly dependent iff there exists a set A K, A = {a 1, a 2,..., a n } and with the a i not all zero, s.t. a 1 v 1 + a 2 v a n v n = 0, where 0 V K is the zero vector. If such a set A does not exist, we say S is linearly independent. Fig

6 Definition 2.2 Let M = {v 1, v 2,..., v n } be a subset of a vector space V K over a field K. M is said to span V K if for every x V K a 1, a 2,..., a n K s.t. x = a 1 v 1 + a 2 v a n v n. Definition 2.3 Let B = {v 1, v 2,..., v n } be a subset of a vector space V K over a field K. We say that B is a basis for V K if it is linearly independent and spans V K. Definition Let K be a field and let V K be a vector space of dimension m. A hyperplane H in V K is a subspace of dimension (m 1). Definition 2.5 A hyperplane arrangement A K = (A K, V K ) is a finite set of hyperplanes in V K. For convenience we will in general drop the subscript K. An arrangement will sometimes be denoted an m arrangement to remind the reader of the dimension of the vector space V. We let φ m denote the empty m arrangement. Fig. 2.2 As we can see from the definition, arrangements are easily defined and in the case of 2 space or 3 space arrangements, can often also be easily visualised. Fig. 2.1 shows a typical hyperplane in R 3. The hyperplane here has dimension 2, and would generally be referred to as a plane. We continue with a number of definitions specific to the study of hyperplanes. 1 Orlik, P., Hiroaki, T.: Arrangements of Hyperplanes. Springer-Verlag (1991), p Definitions 2.4, 2.5, 2.6, 2.7, 2.13, 2.14, 2.15 are taken from here. 5

7 Definition 2.6 Each hyperplane H i A is the kernel of a linear function α i : V K. The product Q(A) = H i A is called a defining polynomial of A. By convention Q(φ m ) = 1 is the defining polynomial of the empty arrangement. Definition 2.7 An arrangement A is centreless if H A H =. If C = H A H, A is said to be centred with centre C. In such a centred arrangement, coordinates may be chosen so that each hyperplane contains the origin; we then call A central. In this study we will be restricting our research to central arrangements. From now on, when reference is made to an arrangement A, it will be assumed that A is central. Example 2.8 Define a 3 arrangement A by Q(A) = (x 1 x 3 )(x 2 x 3 )(x 1 x 2 x 3 )(x 1 + x 2 x 3 ). A consists then of 4 planes in R 3 ; see Fig Example 2.9 Define a 2 arrangement A by Q(A) = (x 1 x 2 )(x 1 + x 2 )(2x 1 x 2 )(2x 1 + x 2 )(0.5x 1 x 2 )(0.5x 1 + x 2 ). A consists then of 6 lines in R 2 ; see Fig Note that in this case A is centered with centre {(0, 0)}, so A is in fact central. α i Fig

8 Theorem 2.10 Let A be a 2 arrangement. Suppose B A, B 3. Then B is dependent. Proof: Let H 1, H 2, H 3 B, where α 1 = ax + by, α 2 = cx + dy, α 3 = ex + fy. Setting ex + fy = C 1 (ax + by) + C 2 (cx + dy) and solving yields C 1 = fc de bc ad and C 2 = eb af bc ad where C 1, C 2, K. We can always find bc ad 0 since α i Cα i C K\{0}, and so the result follows. QED In general, given an arrangement A = (A, V ) with V a vector space of dimension m, and where A = p, it follows that all subsets of A of size α 2 are independent, all subsets of size β, 3 β m may or may not be independent, and all subsets of size γ, (m + 1) γ p are dependent. It is more likely in general that subsets of size β will be independent, especially if A consists of randomly chosen hyperplanes in V. Definition 2.11 Given a vector space V, the dual space V of V is defined as the set of all linear functions f : V K, where K denotes the underlying field. V is itself a vector space, where the vectors are linear functions, equipped with the following addition and scalar multiplication: (τ + σ)(x) = τ(x) + σ(x) and (aτ)(x) = a(τ(x)) for all τ, σ, V, x V, and a K. Choose a basis B = {e 1, e 2,...e m } in V ; note now that given any function τ V, τ is completely determined by how it acts on a basis, and in particular, our basis B. In more mathematical terms, suppose that we know that τ(e 1 ) = x 1, τ(e 2 ) = x 2,..., τ(e m ) = x m. Taking an arbitrary vector v = (a 1, a 2,..., a m ) V, we know that v = α 1 e 1 + α 2 e α m e m, where this combination is unique. Using the dual space vector addition and scalar multiplication, we get that τ(v i ) = α 1 τ(e 1 ) + α 2 τ(e 2 ) α m τ(e m ), hence τ is completely determined by how it acts on B. There is a very natural way of constructing a basis for the dual space. Consider the function F i (x i ) where F i (e i ) = 1, F i (e k ) = 0 if k i, e k B. F can also be written as F i (e k ) = δ ik, where δ ik is the Kronecker delta. Then taking again the arbitrary vector v V, we see that F i (v) = F i (a 1, a 2,..., a m ) = F i (α 1 e 1 +α 2 e α m e m ) = (α 1 F i (e 1 ), α 2 F i (e 2 ),..., α m F i (e m )) = α i 1 = a i. Theorem The functions {F 1, F 2,..., F m } constitute a basis for V. Proof: Firstly, they are linearly independent, since λ 1 F 1 + +λ m F m = 0 λ 1 =... = λ m = 0. Secondly, they span the space V. This means that every function F can be written as λ 1 F 1 + +λ m F m = F. We seek λ i such that this equation is true. Let λ i = F (e i ). It then follows that for any v V, we have F = λ 1 F 1 (x) + +λ m F m (x) = F (x). QED Example 2.13 Let V = R 3 and f : R 3 R, f = x 1 + 2x 2 + x 3. Taking the standard basis for R 3, we get f(e 1 ) = 1, f(e 2 ) = 2 and f(e 3 ) = 1. We can then rewrite the function f = 1 F 1 (x)+2 F 2 (x)+1 F 3 (x) = F 1 (x)+2f 2 (x)+f 3 (x). 2 This is a standard linear algebra result. We include it due to its relevance and importance. 7

9 Constructing the Orlik-Solomon algebra of a hyperplane arrangement relies heavily on the notion of linear dependence of hyperplanes. Clearly hyperplanes do not possess a notion of linear dependence from their parent vector space since they are subspaces and not elements of that vector space. It is by viewing hyperplanes as elements of the dual space that we can impose a notion of dependence. From Definition 2.6 we see that given an arbitrary H i in a m arrangement A, H i is the kernel of a polynomial α i of degree 1 defined up to a constant. Note that α i is an element of the dual space, which as explained, is itself a vector space. It follows that the H i in an arrangement inherit a notion of dependence based on the dependence of the α i in the dual space. It is interesting to note that the α i of an arrangement form a vector configuration in the dual space, which in turn determines an oriented matroid. 3 Definition 2.13 L(A) is defined to be to be the set of all nonempty intersections of elements of A. By convention, L(A) contains V as the intersection of the empty collection of hyperplanes. Definition 2.14 The variety of A is defined to be N(A) = H = {v V Q(A)(v) = 0} H A Take for example, the arrangement A as defined in Example 2.9. Then N(A) = {v V Q(A)(v) = 0} = {v R 2 (x 1 x 2 )(x 1 + x 2 )(2x 1 x 2 )(2x 1 + x 2 )(0.5x 1 x 2 )(0.5x 1 + x 2 ) = 0} So we see that in this case if we take any vector v R 2 that is on any of the 6 lines in Fig. 3, then v N(A). This gives us a visual idea of what the variety of an arrangement A actually is. Definition 2.15 The complement of A is defined to be M(A) = V \N(A) So, taking again the arrangement in Example 2.9, it is clear to see that given any v R 2 that is not on any of the 6 lines in Fig. 3, then v M(A). We have now covered the basic definitions pertinent to the study of hyperplane arrangements. Next we will briefly discuss the idea of an algebra on an arrangement, as a prelude to our goal of defining the Solomon-Orlik algebra of an arrangement. Vector Space Algebras General vector spaces do not possess a multiplication operation. When we introduce a vector multiplication operation to a vector space, we get what is 3 Björner, Vergnas, Sturmfels, White, Ziegler: Oriented Matroids. Cambridge University Press (1999) p 10 8

10 termed an algebra over a field. Two formal methods of adding products to any vector space V are as follows: For any vectors v 1 and v 2 V, let v 1 v 2 = v 2 v 1. The resulting algebra is called the symmetric algebra of V, written S(V ). Alternatively, for any vectors v 1 and v 2 V, let v 1 v 2 = v 2 v 1 The resulting algebra in this case is termed the exterior algebra of V, written E(V ). The Orlik-Solomon algebra is formed by taking a quotient of E(A), where A is an arrangement, so for this reason the exterior algebra is important to our study. It is worth noting here that the formal name for this multiplication operation is anti-commutativity. From this it follows that given v i V, v i 2 = 0, since v i v i = v i v i. Furthermore, the exterior product is associative, and E(V ) is a graded algebra, meaning that when elements of different degrees are multiplied, the degrees add (similar to multiplication of polynomials). One final but vital consequence of the exterior product to note here is that given an arbitrary element x = x 1 x 2 x n of E(A), with A an m arrangement and n m, it is possible to rearrange the x i into any desired order, give or take a minus sign. It follows that a basis for E(A) will be all elements of length l m with all the x i in some order which we impose. Since the O(A) is an ideal of the E(A), this property is inherited by the O(A), meaning that once we impose some linear order on the hyperplanes H i, a basis for O(A) will contain only those elements which have the corresponding α i in that order. Refer to Example 2.16 to see an example of this. We will discuss more about linear order in Chapter 3. Let us now consider the following construction. Let A = {α 1, α 2,..., α t } be a finite hyperplane arrangement, with α i being the polynomial of degree 1 whose kernel is the hyperplane H i. Define the sets A i as follows: A 0 = {ɛ} A 1 = {α 1, α 2,..., α t } A 2 = {a 1 a 2 a 1, a 2 A 1 }. A n = {a 1 a 2 a n a 1, a 2,..., a n A} Now consider A = t i=0 Ai. Following from the properties of the exterior algebra discussed above, this represents a basis for E(A 1 ), the exterior algebra of the arrangement A 1. The size of this basis, A 1 = 2 t. A has a multiplication. Let w A n and let v A m. Then wv A n+m, and is obtained by just writing the word v after the word w. 9

11 The Orlik Solomon Algebra The Orlik-Solomon algebra O(A) is the quotient of the exterior algebra by a homogeneous ideal I(A), O(A) = E(A)/I(A). For the homogeneous ideal I(A), consider the summation p ( 1) k 1 α i1 α ik α ip. k=1 The hat symbol α ik means omit the kth element in that particular product. This summation is performed for every dependent subset {α i1,..., α ip } of A, and all such summations are included in the ideal I(A). This completes the definition of the Orlik-Solomon algebra on an arrangement of hyperplanes. 4 Fig. 2.4 Having covered all the necessary theory, it will be very helpful to now consider the following elementary example. Example 2.16 Consider the 2 arrangement A z in Fig We let A z = {H 1, H 2, H 3, H 4 }, with corresponding α i being the polynomials whose kernels are the 4 hyperplanes. Suppose A z is subject to the Orlik-Solomon algebra. We now seek to study the elements of that O(A z ). From Theorem 2.10 any set containing more than two of the hyperplanes will be dependent. So the dependent subsets of A z are {H 1, H 2, H 3 }, {H 1, H 2, H 4 }, {H 1, H 3, H 4 }, {H 2, H 3, H 4 }. These then are the candidates for the summation of the ideal. Knowing that all elements in the ideal get sent to zero, we get the following relations: p ( 1) k 1 α i1 α ik α ip k=1 4 Ibid, 1, p (to see the construction of the Orlik-Solomon algebra in greater detail). 10

12 (1)... α 2 α 3 α 1 α 3 + α 1 α 2 = 0 (2)... α 2 α 4 α 1 α 4 + α 1 α 2 = 0 (3)... α 3 α 4 α 1 α 4 + α 1 α 3 = 0 (4)... α 3 α 4 α 2 α 4 + α 2 α 3 = 0 We also know that from the exterior algebra, E(A z ), we get that α 1 α 2 = α 2 α 1 α 1 α 3 = α 3 α 1 α 1 α 4 = α 4 α 1 α 2 α 3 = α 3 α 2 α 2 α 4 = α 4 α 2 α 3 α 4 = α 4 α 3 Furthermore, any α 2 i = 0, and any word of the form α iα j α k = 0. Multiply, for example, relation (1) above on the left by α 1 : = α 1 α 2 α 3 α 1 α 1 α 3 + α 1 α 1 α 2 = 0 α 1 α 2 α 3 0 α α 2 = 0 α 1 α 2 α 3 = 0 and similarly for the other words of the form α i α j α k. This is a very important observation, and one which we will now generalise. In general, given an m arrangement A where A = p > m, O(A) will contain no elements of length l > m, and hence the summation of the ideal will only be performed on dependent subsets S of A where S m + 1. This follows from Theorem 2.10 and its corollaries. Suppose now that we seek a basis for O(A z ). We then have narrowed down the candidates for the basis elements to the set: {ɛ, α 1, α 2, α 3, α 1 α 2, α 1 α 3, α 1 α 4, α 2 α 3, α 2 α 4, α 3 α 4 } However, from the summation of I(A z ), we know we have four relations, hence three of the α i α j will not be in a basis, since a basis must be linearly independent. The question is, which should we leave out? It is here that the non-broken circuit part of the basis comes into play, which we will discuss in the next chapter. 11

13 Chapter 3 The Non-Broken Circuit Basis In this chapter we introduce and define the non-broken circuit basis for the Orlik-Solomon algebra of an arrangement of hyperplanes. We will also discuss some examples, with a view to understanding the selection process of the basis elements, and also to appreciate the need for an algorithm to compute the basis. Let A be a hyperplane arrangement. We introduce an arbitrary linear order in A. We consider a number of necessary definitions. In this report we will also introduce a number of new notation conventions 1, which will aid in the construction of an algorithm, as follows: Definition 3.1 An m set is a subset of A of cardinality m. Definition 3.2 An m elt is an element of O(A) of length m. Definition 3.3 An m set {H 1, H 2,..., H m } is a circuit if it is minimally dependent. Therefore {H 1, H 2,..., H m } is dependent, but for 1 k m the (m 1) set {H 1, H 2,..., Ĥk,..., H m } is independent. If this is the case, then we also say that its corresponding m elt α 1 α 2 α m is a circuit. Definition 3.4 We say that H A is the maximal element H A, H H, H H. Depending on the context, we will also refer to α, the linear map associated with H, as the maximal element. Definition 3.5 Given an m set S (or its corresponding m elt), let maxs be the maximal element of S in the linear order in A. Definition 3.6 We say that an m elt S is extendable if maxs is not the maximal element (of A). Definition 3.7 An m set (or its corresponding m elt) S is a broken circuit if it is formed by removing maxp from some circuit P. 1 Definitions 3.1, 3.2, 3.6, 3.8, 3.10, 3.11 are original. 12

14 Definition 3.8 An m BC is an m elt which is a broken circuit. Definition 3.9 We say that an element x = x 1 x 2 x n contains the broken circuit y = y 1 y 2 y m, m < n, if the set {y 1, y 2,..., y m } {x 1, x 2,..., x n }. Definition 3.10 We say that an m set or an m BC is necessary if it is not eliminated by an n BC, where n < m. Definition 3.11 Given a set M 1 of m sets of A the set M 2 of sufficient (m + 1) sets of M 1 is defined to be the set of all (m + 1) sets of A which are extensions of sets in M 1. Definition The Non-Broken Circuit Basis for O(A) is defined to consist of all elements of O(A) which are not broken circuits and do not contain a broken circuit. NBCB is an acronym for non-broken circuit basis. Example 3.13 Let A z be the hyperplane arrangement of Example 2.16 from Chapter 2. Suppose that we seek a NBCB for O(A z ). There are four circuits, i.e. four minimally dependent subsets of A z (we use only the indices for clarity 3, following the same linear order used in chapter 2 ): 123, 124, 134, 234 It is worth noting here that in any m arrangement A, every dependent 3 set is a circuit, since any 3 set is minimally dependent. In particular, note that in a 2 arrangement, all 3 sets are circuits, since they are dependent by dimension of V K. This follows from Theorem We therefore get the following list of broken circuits from these circuits, all of which will not be in the NBCB: 12, 13, 23 Therefore the NBCB, which we will denote B z is as follows: B z = {ɛ, α 1, α 2, α 3, α 1 α 4, α 2 α 4, α 3 α 4 } The Cube Example Already from the previous example we can see some patterns in the choice of elements for the NBCB. However, Example 3.12 is a 2 arrangement, and Orlik and Terao have produced a general result about such arrangements, which we reproduce below: 4 2 Orlik, P., Hiroaki, T.: Arrangements of Hyperplanes. Springer-Verlag (1991), p For the complete proof that the NBCB is actually a basis check this reference. 3 From now on, we will generally use the indices to refer to hyperplanes; also depending on the context e.g. 123 could refer to the 3 set or the 3 elt. 4 Ibid, 2, p

15 Theorem 3.14 Suppose A = {H 1,..., H n } is a 2 arrangement. Then the O(A) is spanned by the empty hyperplane, the α i and all α k α n for 1 k n. In other words, these are precisely the elements of the NBCB for O(A). This fact is reflected in the NBCB obtained in Example Our report, however, is concerned with arrangements of higher dimension. To see the difficulties posed by a somewhat non-trivial example of a hyperplane arrangement, we will consider the case of the nine hyperplane symmetries of the cube in R 3 (usual basis) with vertices (±1, ±1, ±1). The symmetry group of the cube is the Coxeter group of type B 3. 5 See Fig Fig. 3.2 The defining polynomial for this 3 arrangement, which we will denote by A c, is: Q(A c ) = xyz(x + y)(x y)(x + z)(x z)(y + z)(y z). We impose a linear order as follows: x y z x + y x y x + z x z y + z y z We begin by noting that the basis E(A c ) will contain 2 9 = 512 elements. This calculation is precisely similar to that performed when calculating the size of 5 Ibid, p 12 14

16 the power set P (A) of a set A. Moving on from this, we now seek to consider the subsets of A c : which are independent, and which are not, for it is this information that is fundamental to listing the elements of the O(A c ) and for finding the circuits. It is clear that any subset of size 2 will be independent; of which there are 46. Also clear is that any subset of size 4 will be dependent; of which there are 382. We now must separate the set of 3 sets in two categories: independent and dependent. There are 84 3 sets of A c, and we now seek to classify them in an ordered fashion. Firstly, consider B = {{v 1, v 2, v 3 } v 1 {1, 2, 3}, v 2, v 3 {4, 5, 6, 7, 8, 9}}. B = 45 and upon inspection we find that the dependent elements of B are: D 1 = {1, 4, 5} D 2 = {2, 4, 5} D 3 = {1, 6, 7} D 4 = {3, 6, 7} D 5 = {2, 8, 9} D 6 = {3, 8, 9} We define D to be the set of dependent 3 sets of A c, and say D i D. Secondly, consider C = {{v 1, v 2, v 3 } v 1, v 2 {1, 2, 3}, v 3 {4, 5, 6, 7, 8, 9}}. C = 18 and upon inspection we find that the dependent elements of C are: D 7 = {1, 2, 4} D 8 = {1, 2, 5} D 9 = {1, 3, 6} D 10 = {1, 3, 7} D 11 = {2, 3, 8} D 12 = {2, 3, 9} Again, we add these elements to the set D. Thirdly, consider F = {{v 1, v 2, v 3 } v 1, v 2, v 3 {4, 5, 6, 7, 8, 9}}. F = 20, and with this set it is not as easy to see which elements, if any, are dependent. Upon inspection we see that H 5 + H 8 = H 6, H 7 + H 8 = H 4, H 6 + H 9 = H 4 and H 5 + H 9 = H 7 yielding that the elements: D 13 = {5, 6, 8} D 14 = {4, 7, 8} D 15 = {4, 6, 9} D 16 = {5, 7, 9} are dependent. We add these elements to the set D. At this stage we have exhausted 83 of the 84 3 sets. The final remaining 3 set is the somewhat trivial {1, 2, 3} which is clearly independent. Thus we conclude that 16 of the 84 3 sets happen to be dependent - notice the relatively small number - even with such a symmetric m arrangement as A c, we find that a great majority of m sets are independent. We are now ready to proceed to the next step of the 15

17 process. Recall the definition of a circuit: a minimally dependent set, and that a broken circuit (BC) is obtained by removing maxs from a circuit S, with respect to the predefined numbering process. Since all of the elements in D are in fact circuits, we remove maxd i from each D i to get a list of the 2 BCs: 12, 13, 14, 16, 23, 24, 28, 36, 38, 46, 47, 56, 57 Since the non-broken circuit basis cannot contain any broken circuit, or any element which within its numbering contains a broken circuit (see Definition 3.9), we conclude that none of these elements are in the basis. Furthermore, we need not extend any of these elements to the 3 elt case, since any 3 elt which contains a 2 BC within its numbering will not be in the NBCB 6. Thus the 2 elts that are in the NBCB: 15, 17, 18, 19, 25, 26, 27, 29, 34, 35, 37, 39, 45, 48, 49, 58, 59, 67, 68, 69, 78, 79, 89 The next step is to consider in some systematic fashion all possible 3 elt extensions of these successful 2 elts and to determine whether each either fails or passes the test to be in the basis. Notice firstly the interesting fact that all of the following 2 elts, which are not extendable (see Definition 3.6), are in the basis: 19, 29, 39, 49, 59, 69, 79, 89 Notice also that each of the following 2 elts has precisely one extension: 18, 48, 58, 68, 78 i.e. postmultiplying with the maximal element 7 i.e. hyperplane no.9 in the linear order. Furthermore, it follows that none of these extensions can be broken circuits (because it is impossible to remove the maximal element from a 4 elt to arrive at any of them, since they already contain the maximal element). Hence are all in the basis. 189, 489, 589, 689, 789 What remains now is to consider the all possible 3 elt extensions of the remaining 2 elts in the NBCB: The extensions are as follows: 15, 17, 25, 26, 27, 34, 35, 37, 45, 67 6 This observation is fundamental to writing the algorithm. 7 See Definition

18 Table 3.1 Note that none of these 3 elts are dependent. The pertinent question now is: What could hinder these elements from being in the NBCB of O(A c )? Here it is important to recall Definition 3.9. It has so happened that while extending the remaining 2 elts in the NBCB, we have generated a number of 2 BCs within the resulting 3 elts. The underlined 3 elts in Table 3.1 contain a 2 BC, and hence will not be in the NBCB. For example, the element 278 contains the 2 BC 28. The search is now confined to the remaining 16 3 elts which are not underlined. However, it is possible that one or more of these 3 elts could be a broken circuit, in which case such 3 elts could not be in the NBCB. To fully explore this possibility, we must consider the 4 sets of O(A c ): which are dependent, and then of those, which are minimally dependent. There are a total of ( 9 4) = sets. However, it should be clear that we need only consider those 4 sets which are obtained by extending the 3 sets listed here. No other 4 elts could give rise to one of the 3 elts in Table 3.1 becoming a 3 BC. Of the remaining 16 3 elts, 10 are not extendable: 159, 179, 259, 269, 279, 349, 359, 379, 459, 679 It follows that these are in the NBCB. We now check if the remaining 158, 178, 267, 345, 458, 678 are circuits. The extensions are as follows: Table 3.2 Since A c is a 3 arrangement, it follows that all of these are dependent. We now check which of these are circuits. Recall that a circuit is a minimally dependent subset of A c. This means that if we remove any one of the indices, what remains will be an independent set. We check as follows: I 159 I 189 I 589 I 158: IS A 3 BC 17

19 On the top is the dependent 4 set under inspection; beneath are the sets obtained by removing any one single element from the set; to their left is indicated D if dependent, I if independent (this information can be readily retrieved from the set D of dependent 3 sets calculated earlier). In this particular example we find that 1589 is minimally dependent, and hence 158 is a 3 BC; this result is written at the bottom of the column. For those 3 elts which have more than one extension e.g. 345, if for any arbitrary extension 345x we get that 345x is minimally dependent, it follows that 345 is a 3 BC. 8. We will not bore the reader, but if this procedure is carried out on all 10 4 elts in Table 3.2, it turns out that all are circuits, implying that 158, 178, 267, 345, 458, 678 are all 3 BCs and hence none of them are in the basis. So B c, the NBCB of O(A c ), subject to our imposed linear order, is B c = {ɛ, 1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 17, 18, 19, 25, 26, 27, 29, 34, 35, 37, 39, 45, 48, 49, 58, 59, 67, 68, 69, 78, 79, 89, 189, 489, 589, 689, 789, 159, 179, 259, 269, 279, 349, 359, 379, 459, 679} B c = 48. There are a number of important things to note from this major example. Firstly, the 3 elts B c are precisely those elements obtained by postmultiplying all extendable 2 elts B c with the maximal element. This is not a coincidence, but rather gives rise to the following general theorem: 9 Theorem 3.15 Suppose A = {H 1,..., H n } is an m arrangement, and let B be the NBCB for O(A). Let B k B be the set of all k elts in B, and let B k m be the set containing all extendable elements in B k but with each postmultiplied by the maximal elt. Then for all k, 2 k m 1, we have that B k m B k+1. Corollary 3.16 When k = m 1, we get that B m k = B k+1. In other words, the m elts B are precisely those elements obtained by postmultiplying all extendable (m 1) elts B with the maximal element. Proof of Corollary Let α n be the maximal element. Suppose there exists an m elt D B s.t. maxd α n. Then Dα n is dependent, since it is of length m + 1. We claim that it is minimally dependent. Proof of claim: Consider the m elts obtained by removing an arbitrary α i 8 See Definition Ibid, 5, p 68, Lemma The author postulated Theorem 3.15 during independent research, but subsequently found a related result as referenced. Corollary 3.16 is original and fundamental to the writing of the algorithm. 18

20 from Dα n. Firstly, if we remove α n, we are left with D, which is independent from our initial assumption. Secondly, if we remove any other α i we are left with an m elt of the form M = α k1 α k2 α k(m 1) α n. M cannot be a broken circuit, because no broken circuit can contain α n, since it is the maximal element. Furthermore, M cannot contain a broken circuit. Suppose M did contain a broken circuit C. Then α n / C C D thus contradicting our initial assumption that D B. But if M neither is a broken circuit, nor contains a broken circuit, it follows from Definition 3.12 that M B. M then must be independent, for B contains no dependent elements. We conclude that Dα n is minimally dependent i.e. it is a circuit. Therefore D is an m BC. But this is a contradiction, since D B. This means that our initial assumption was wrong. Hence m elts D k B, maxd k = α n. It remains to show that α k1 α k2 α k(m 1) B (m 1), where α k(m 1) α n we get that α k1 α k2 α k(m 1) α n B m. Suppose there exists α k1 α k2 α k(m 1) B (m 1) such that α k1 α k2 α k(m 1) α n / B m. But this contradicts Theorem Hence the result follows. QED We will not provide the proof for Theorem 3.15 at this time, but consider the usefulness of this result in the light of searching for an algorithm. To have fore-known this result would have saved us at least half of the effort we expended in finding B c. Listing the 2 BCs is all that would have been necessary; finding the NBCB from these would then be very trivial indeed. Before we close this chapter, a note on definitions 3.10 and 3.11 is due, since these are the only definitions that are not directly referenced so far. These definitions will be used in the writing of the algorithm in chapter 4. It is correct to say that there are two restrictions upon eligibility for the NBCB in general. The first is that a basis element cannot be a broken circuit, the second, that an element cannot contain a broken circuit. Given an m arrangement A, m 3, and searching for the NBCB for O(A), our first step will be to determine all dependent 3 sets, which will then give us a list of all 2 BCs, and hence a list of all 2 elts in the NBCB. Our next goal is then to list all 3 elts in the NBCB. We begin by (STEP I) listing all necessary 3 sets, i.e. those 3 sets which do not contain a 2 BC. All 3 sets in this list will be independent. It should be clear that we need only list the necessary 3 sets at this step, for if a 3 set is not necessary, then it contains a broken circuit, and will not be in the NBCB. Next (STEP II) we determine all dependent sufficient 4 sets, i.e. we consider all those 4 sets which are extensions of 3 sets in STEP I (sufficient) and check for dependency. We then (STEP III) determine which, if any, of those 3 sets in STEP I are 3 BCs by checking any dependent 4 sets obtained in STEP II for minimal sufficiency. Example 2.17 Consider M O(A c ), where A c is the cube arrangement, M = {15, 17, 18, 19, 25, 26, 27, 29, 34, 35, 37, 39, 45, 48, 49, 58, 59, 67, 68, 69, 78, 79, 89} M = 23, M is the set of 2 elts in the NBCB. Table 3.3 then contains all sufficient 3 sets (of M). 19

21 Table 3.3 Example 2.18 Let M be as defined above. Recall that we know all 2 BCs of O(A c ). Table 3.4 then contains all necessary 3 sets (of M) Table 3.4 These two examples illustrate well the difference between the two conditions sufficient and necessary. Having developed all the theory, the algorithm will be presented in the next chapter. 20

22 Chapter 4 The Algorithm In this chapter we introduce the algorithm for finding a non-broken circuit basis for the Orlik-Solomon algebra of an arrangement of hyperplanes. Initially we present the algorithm as specific to a 5 arrangement. We take a random 5 arrangement and find the non-broken circuit basis firstly without using the algorithm, or indeed any shortcuts, and then find the basis with the algorithm, as a means of illustrating its efficiency. Finally we present the algorithm formally, and for the general m arrangement case. ALGORITHM FOR COMPUTING THE NBCB FOR O(A) A is a 5 arrangement. Let B be the NBCB. STEP 1. Add {ɛ, α 1, α 2, α 3, α 4, α 5 } to B STEP 2. (i) List necessary 2 sets (ii) Determine dependent sufficient 3 sets (iii) Determine necessary 2 BCs (iv) Determine 2 elts that are in the NBCB; add to B STEP 3. (i) List necessary 3 sets (ii) Determine dependent sufficient 4 sets (iii) Determine necessary 3 BCs (iv) Determine 3 elts that are in the NBCB; add to B STEP 4. (i) List necessary 4 sets (ii) Determine dependent sufficient 5 sets 21

23 (iii) Determine necessary 4 BCs (iv) Determine 4 elts that are in the NBCB; add to B STEP 5. Determine 5 elts that are in the NBCB; add to B. 1 END OF ALGORITHM - B is the NBCB for O(A) Example of the Algorithm at work We now compute a moderately difficult example both with and without the algorithm, to demonstrate the efficiency of the algorithm. Let A K = (A K, V K ) with V K = R 5 and A K = {x, y, z, w, v, x + y, x + y w, w v} where {x, y, z, w, v} denote the coordinate hyperplanes in R 5. As usual we will apply a linear order as follows: x y z w v x + y x + y w w v I. WITHOUT THE ALGORITHM Firstly we determine all dependent 3 sets. This involves first listing the entire set of possible 3 sets and then checking for dependency by inspection: 1 See Corollary D D D Table

24 In Table 4.1 dependent sets are marked with a D on the right, while those with no mark on the right are independent. Therefore out of a total of ( 8 3) = 56 3 sets, we have that (only) 3 are dependent, namely 126, 458 and 467. These are the only circuits of length 3 in the Orlik-Solomon Algebra of this arrangement. It follows directly that 12, 45 and 46 are the only broken circuits of length 2. We therefore deduce that out of a total of ( 8 2) = 28 2 sets, 25 end up in the NBCB. We now determine all dependent 4 sets, ending up with a table as before: D D D D D D 1458 D D D 4567 D 1267 D D 4568 D 1268 D D D D Table 4.2 Again, dependent sets are marked with a D on the right. Therefore out of a total of ( 8 4) = 70 4 sets, we have that 17 are dependent. Notice the increasing number of dependent subsets, as the size of the subset increases. To determine whether these are circuits, we must check if they are minimally dependent. We will not bore the reader, but if this procedure is carried out on all 17, the only one which yields a 3 BC is 5678: I 568 I 578 I 678 I 567: BC Having performed this exercise, we can now determine which 3 elts will be in the basis. The NBCB cannot contain a broken circuit, or, importantly, any 23

25 element which itself contains a broken circuit. From this it follows that of the 56 3 elts in Table 4.1, 38 make it into the NBCB. In Table 4.1, those 3 elts which are eliminated by the 2 BCs are annotated by a symbol on the left, and the single 3 elt which is eliminated by the 3 BC 567 is annotated by a on the left. We now determine all dependent 5 sets, ending up with a table as before: D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D D Table 4.3 Again, dependent sets are marked with a D on the right. Therefore out of a total of ( 8 5) = 56 5 sets, we have that 34 are dependent. Notice again the increasing number of dependent subsets, as the size of the subset increases. To determine whether these are circuits, we must check if they are minimally dependent. Again we will not bore the reader with all the necessary calculations, but if this procedure is carried out on all 34, the only one which yields a BC is 12578: I 1258 I 1278 I 1578 I 2578 I 1257: BC Having performed this exercise, we can now determine which 4 elts will be in the basis. It follows that of the 70 4 elts in Table 4.2, 28 make it into the NBCB. In Table 4.2, those 4 elts which are eliminated by the 2 BCs are annotated by a symbol on the left, and those 4 elts which are eliminated by the 3 BC 567 are annotated by a on the left. Notice that the 4 BC 1257 does not eliminate anything which is not already eliminated. We now write a list of all 6 sets in ordered fashion: 24

26 D D D D D D D D D D D D D D D D D D D D D D D D D D D D Table 4.4 Again, dependent sets are marked with a D on the right. Naturally all of these will be dependent because of the dimension. We perform this final step to determine whether there exist any 5 BCs. Upon inspection, we find that none of these are minimally dependent i.e. there are no 5 BCs. Having performed this exercise, we can now determine which 5 elts will be in the basis. It follows that of the 56 5 elts in Table 4.3, 8 make it into the NBCB. In Table 4.3, those 5 elts which are eliminated by the 2 BCs are annotated by a symbol on the left, and those 5 elts which are eliminated by the 3 BC 567 are annotated by a on the left. In short, the elements of the NBCB for this arrangement are: 1. the empty element, 2. the 8 singletons, 3. all possible 2 elts except 12, 45 and 46, 4. and all elements in the tables 4.1, 4.2 and 4.3 with no symbol on the left. This ends our rigorous unalgorithmical way of calculating the NBCB. It has cardinality 108. II. WITH THE ALGORITHM STEP 1. Add {ɛ, 1, 2, 3, 4, 5, 6, 7, 8} to B STEP 2. (i) All possible 2 elts; Result: 28 (ii) 56 sufficient 3 sets. Result: 3 dependent (iii) Result: 3: 12, 45, 46 (iv) Result: 25 2 elts B STEP 3. (i) All elts in Table 4.1 without on LHS; Result: 39 (ii) 32 sufficient 4 sets. Result: 1 dependent (iii) Result: 1: 567 (iv) Result: 38 3 elts B 25

27 STEP 4. (i) All elts in Table 4.2 without OR on LHS; Result: 28 (ii) 10 sufficient 5 sets. Result: 0 dependent (iii) Result: 0 (iv) Result: 28 4 elts B STEP 5. Result: 8 5 elts B END OF ALGORITHM - B is the NBCB for O(A) B = =108 ALGORITHM FOR COMPUTING THE NBCB FOR O(A) A is a m arrangement, A = p. Let B be the NBCB. STEP 1. Add {ɛ, α 1, α 2,..., α p } to B STEP 2. Beginning at k = 2, and continuing consecutively; for k, 2 k m 1 do (i) List necessary k sets (ii) Determine dependent sufficient (k + 1) sets (iii) Determine necessary k BCs (iv) Determine k elts that are in the NBCB; add to B STEP 3. Determine m elts that are in the NBCB; add to B. 2 END OF ALGORITHM - B is the NBCB for O(A) 2 See Corollary

28 Chapter 5 Conclusions The goal of this project was to write an algorithm for finding a non-broken circuit basis for the Orlik-Solomon algebra of an arrangement of hyperplanes. That goal has been achieved. During the course of my research, I came across no such algorithm of any kind already written, or indeed any previous work on this subject, save that which I have referenced. It is therefore reasonable to assume that this algorithm is the first of its kind. Also achieved in this report is a brief but informative summary of the theory of hyperplanes, arrangements and associated algebras. The two major, fully-worked examples of a non-broken circuit basis in Chapters 3 and 4 are also notable achievements and illustrate strongly the challenges, difficulties and depth of the theory behind this concept. Exactly how efficient the algorithm is has not been thoroughly considered. We can see in Chapter 4 that enormous computational savings are made by implementing the algorithm, but a general formula for calculating the percentage decrease in the number of calculations necessary would be a bonus, if such a formula exists. Proving Theorem 3.15 is another goal that was not reached due to time constraints. There are two open questions which I will now pose, and which are definitely areas for further work. Question 5.1: Given an m arrangement A and seeking a NBCB for O(A). If, at step k(ii) of the algorithm, 2 k m 1, a dependent (k + 1) set is attained, is it always the case with this algorithm that it is a circuit i.e. will yield a k BC? With regard to all the examples which we have seen, the answer is yes, and therefore I am almost certain that the answer is yes in general. However, this remains to be proven. Question 5.2: At each step k(ii), 2 k m 1, can we replace the condition sufficient in the algorithm with the condition necessary and still get accurate results? Once again the answer is yes with regard to all the examples that we have seen, but it this true in general? A brief comparative review of tables 3.3 and 3.4 will remind the reader of the further computational savings that would be made with the algorithm, if this were the case. 27

29 Bibliography 1. Roberts, S.: On the figures formed by the intercepts of a system of straight lines in a plane, and on analogous relations in space of three dimensions. Proc. London Math. Soc (1889) Referenced in: Chapter 1, [1]; page(s) referenced: Orlik, P., Hiroaki, T.: Arrangements of Hyperplanes. Springer-Verlag (1991) Referenced in: Chapter 1, [2] Referenced in: Chapter 1, [3]; page(s) referenced: Referenced in: Chapter 2, [1]; page(s) referenced: Referenced in: Chapter 2, [4]; page(s) referenced: Referenced in: Chapter 3, [2]; page(s) referenced: Referenced in: Chapter 3, [4]; page(s) referenced: Referenced in: Chapter 3, [5]; page(s) referenced: 12 Referenced in: Chapter 3, [9]; page(s) referenced: Björner, Vergnas, Sturmfels, White, Ziegler: Oriented Matroids. Cambridge University Press (1999) Referenced in: Chapter 2, [3]; page(s) referenced: 10 28