An identity for the coefficients of characteristic polynomials of hyperplane arrangements
aa r X i v : . [ m a t h . M G ] S e p AN IDENTITY FOR THE COEFFICIENTS OF CHARACTERISTICPOLYNOMIALS OF HYPERPLANE ARRANGEMENTS
ZAKHAR KABLUCHKO
Abstract.
Consider a finite collection of affine hyperplanes in R d . The hyperplanes dissect R d into finitely many polyhedral chambers. For a point x ∈ R d and a chamber P the metricprojection of x onto P is the unique point y ∈ P minimizing the Euclidean distance to x . Themetric projection is contained in the relative interior of a uniquely defined face of P whosedimension is denoted by dim( x, P ). We prove that for every given k ∈ { , . . . , d } , the numberof chambers P for which dim( x, P ) = k does not depend on the choice of x , with an exceptionof some Lebesgue null set. Moreover, this number is equal to the absolute value of the k -th coefficient of the characteristic polynomial of the hyperplane arrangement. In a specialcase of reflection arrangements, this proves a conjecture of Drton and Klivans [A geometricinterpretation of the characteristic polynomial of reflection arrangements, Proc. Amer. Math.Soc. , 138(8): 2873–2887, 2010]. Introduction and statement of results
Introduction.
The starting point of the present paper was the following conjecture ofDrton and Klivans [7, Conjecture 6]. Consider a finite reflection group W acting on R d . Themirror hyperplanes of the reflecting elements of W dissect R d into isometric cones or chambers.Let C be one of these cones. Take some k ∈ { , . . . , d } . A point x ∈ R d is said to havea k -dimensional projection onto C if the unique element y ∈ C minimizing the Euclideandistance to x is contained in a k -dimensional face of C but not in a face of smaller dimension.For example, points in the interior of C have a d -dimensional projection onto C . Drton andKlivans [7, Conjecture 6] conjectured that for a “generic” point x ∈ R d the number of pointsin the orbit { gx : g ∈ W} having a k -dimensional projection onto C is constant, that isindependent of x . Moreover, they conjectured that this number is equal to the absolute value a k of the coefficient of t k in the characteristic polynomial of the reflection arrangement. Drtonand Klivans [7] observed that in the case of reflection groups of type A their conjecture followsfrom the work of Miles [17], proved it for reflection groups of types B and D , and gave furtherpartial results on the conjecture including numerical evidence for its validity in the case ofexceptional reflection groups. Somewhat later, Klivans and Swartz [14] proved that if x ischosen at random according to a rotationally invariant distribution on R d , then the conjectureof Drton and Klivans is true on average , that is the expected number of points in the orbit { gx : g ∈ W} having a k -dimensional projection onto C equals a k .The aim of the present paper is to prove the conjecture of Drton and Klivans in a much moregeneral setting of arbitrary affine hyperplane arrangements. After collecting the necessarydefinitions in Section 1.2 we shall state our main result in Section 1.3. Mathematics Subject Classification.
Primary: 52C35, 51M20. Secondary: 52A55, 51M04, 52A22,60D05, 52B11, 51F15.
Key words and phrases.
Hyperplane arrangement, metric projection, chambers, reflection arrangement, char-acteristic polynomial, normal cone, conic intrinsic volume.
Definitions. A polyhedral set in R d is an intersection of finitely many closed half-spaces.A bounded polyhedral set is called a polytope . If the hyperplanes bounding the half-spaces passthrough the origin, the intersection of these half-spaces is called a polyhedral cone , or just acone. We denote by F k ( P ) the set of all k -dimensional faces of a polyhedral set P ⊂ R d , forall k ∈ { , . . . , d } . For example, F ( P ) is the set of vertices of P , while F d ( P ) = { P } provided P has non-empty interior. The set of all faces of P of whatever dimension is then denoted by F ( P ) = ∪ dk =0 F k ( P ). The relative interior of a face F , denoted by relint F , consists of all pointsbelonging to F but not to a face of strictly smaller dimension. It is known that any polyhedralset is a disjoint union of the relative interiors of its faces: P = [ · F ∈F ( P ) relint F. (1.1)For more information on polyhedral sets and their faces we refer to [18, Sec. 7.2, 7.3], [19]and [24, Chap. 1 and 2]. Polyhedral sets form a subclass of the family of closed convex sets;for the face structure in this more general setting we refer to [20, § § P and a point x ∈ R d , there is a uniquely defined point minimizingthe Euclidean distance k x − y k among all y ∈ P . This point, denoted by π P ( x ), is called the metric projection of x onto P . For example, if x ∈ P , then π P ( x ) = x . By (1.1), the metricprojection π P ( x ) is contained in a relative interior of a uniquely defined face F of P . If thedimension of F is k , we say that the point x has a k -dimensional metric projection onto P andwrite dim( x, P ) = k .Next we need to recall some basic facts about hyperplane arrangements referring to [23]and [3, Sec. 1.7] for more information. Let A = { H , . . . , H m } be an affine hyperplane ar-rangement in R d , that is a collection of pairwise distinct affine hyperplanes H , . . . , H m in R d .In general, the hyperplanes are not required to pass through the origin, but if they all do,the arrangement is called linear . The connected components of the complement R d \ ∪ mi =1 H i are called open chambers , while their closures are called closed chambers of A . The closedchambers are polyhedral sets which cover R d and have disjoint interiors. The collection of all closed chambers will be denoted by R ( A ). If not otherwise stated, the word “chamber” alwaysrefers to a closed chamber in the sequel. The characteristic polynomial of the affine hyperplanearrangement A may be defined by the following Whitney formula [23, Thm. 2.4]: χ A ( t ) = X B⊂A : ∩ H ∈B H = ∅ ( − B t dim (cid:18) ∩ H ∈B H (cid:19) . (1.2)Here, B denotes the number of elements in B . The empty set B = ∅ , for which the corres-ponding intersection of hyperplanes is defined to be R d , contributes the term t d to the abovesum. The classical Zaslavsky formulae [23, Thm. 2.5] state that the total number of cham-bers is given by R ( A ) = ( − d χ A ( − − rank A χ A (1), where rank A is the dimension of the linear space spanned by the normals tothe hyperplanes of A . For the coefficients of the characteristic polynomial it will be convenientto use the notation χ A ( t ) = d X k =0 ( − d − k a k t k . (1.3) This convention deviates from the standard notation [23], where R ( A ) is the collection of open chambers,but will be convenient for the purposes of the present paper. OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 3
Main result.
We are now ready to state a simplified version of our main result . Theorem 1.1.
Let A be an affine hyperplane arrangement in R d whose characteristic polyno-mial χ A ( t ) is written in the form (1.3) . Take some k ∈ { , . . . , d } . Then, { P ∈ R ( A ) : dim( x, P ) = k } = a k for every x ∈ R d outside certain exceptional set which is a finite union of affine hyperplanes. Example 1.2 (First Zaslavsky’s formula) . Let us show that the first Zaslavsky formula R ( A ) =( − d χ A ( −
1) is a consequence of Theorem 1.1. Take some point x not belonging to the excep-tional set. On the one hand, for every chamber P ∈ R ( A ) there is a unique face whose relativeinterior contains the metric projection π P ( x ), hence interchanging the order of summation weget d X k =0 { P ∈ R ( A ) : dim( x, P ) = k } = X P ∈R ( A ) d X k =0 { dim( x,P )= k } = X P ∈R ( A ) R ( A ) . On the other hand, the sum on the left-hand side equals P dk =0 a k by Theorem 1.1. Altogether,we arrive at R ( A ) = P dk =0 a k , which proves the first Zaslavsky formula. The second Zaslavskyformula will be discussed in Example 1.7. Example 1.3 (Reflection arrangements) . Consider a finite reflection group W acting on R d .Let χ ( t ) be the characteristic polynomial of the associated reflection arrangement and C oneof its chambers. Drton and Klivans [7, Conjecture 6] conjectured that for a “generic” point x ∈ R d the number of group elements g ∈ W with dim( gx, C ) = k is equal to the absolute valueof the coefficient of t k in χ ( t ), for all k ∈ { , . . . , d } . This conjecture is an easy consequence ofTheorem 1.1. Indeed, since every g ∈ W is an isometry, dim( gx, C ) equals dim( x, g − C ). If g runs through all elements of W , then g − C runs through all chambers of the reflection arrange-ment, and the conjecture follows from Theorem 1.1. Note that the characteristic polynomialsof the reflection arrangements are known explicitly; see, e.g., [3, p. 124].Let us now restate Theorem 1.1 in a more explicit form involving a concrete description ofthe exceptional set. First we need to define the notions of tangent and normal cones. Letpos A := ( m X i =1 λ i a i : m ∈ N , a , . . . , a m ∈ A, λ , . . . , λ m ≥ ) . denote the positive hull of a set A ⊂ R d . The tangent cone of a polyhedral set P at its face F ∈ F ( P ) is defined by T F ( P ) = pos { p − f : p ∈ P } = { u ∈ R d : ∃ δ > f + δu ∈ P } , (1.4)where f is an arbitrary point in relint F . It is known that this definition does not depend on thechoice of f and that T F ( P ) is a polyhedral cone. Moreover, T F ( P ) contains the linear subspaceaff F − f , where aff F is the affine hull of F , i.e. the minimal affine subspace containing F .For a polyhedral cone C ⊂ R d its dual or polar cone is defined by C ◦ = { z ∈ R d : h z, y i ≤ y ∈ C } , After the first version of this paper has been uploaded to the arXiv, we have been informed by GiovanniPaolini and Davide Lofano that the same theorem has been obtained in their paper [15, Corollary 5.13]. Itseems that our proof is quite different (and more elementary).
ZAKHAR KABLUCHKO where h· , ·i denotes the standard Euclidean scalar product on R d . The normal cone of apolyhedral set P at its face F ∈ F ( P ) is defined as the dual of the tangent cone: N F ( P ) = ( T F ( P )) ◦ . By definition, N F ( P ) is a polyhedral cone contained in (aff F ) ⊥ , the orthogonal complement ofaff F . Here, the orthogonal complement of an affine subspace A ⊂ R d is the linear subspace A ⊥ = { z ∈ R d : h z, y i = 0 for all y ∈ A } . Now, the metric projection of a point x ∈ R d onto a polyhedral set P satisfies π P ( x ) ∈ F fora face F ∈ F ( P ) if and only if x ∈ F + N F ( P ). Here, A + B = { a + b : a ∈ A, b ∈ B } is theMinkowski sum of the sets A, B ⊂ R d , which in our special case is even orthogonal meaningthat every vector from N F ( P ) is orthogonal to every vector from F . Similarly, we have π P ( x ) ∈ relint F ⇐⇒ x ∈ (relint F ) + N F ( P ) . Let int A denote the interior of a set A , and let ∂A = A \ int A be the boundary of A . Weare now ready to restate our main result in a more explicit form. Theorem 1.4.
Let A be an affine hyperplane arrangement in R d whose characteristic polyno-mial χ A ( t ) is written in the form (1.3) . Then, for every k ∈ { , . . . , d } we have ϕ k ( x ) := X P ∈R ( A ) X F ∈F k ( P ) F + N F ( P ) ( x ) = a k , for all x ∈ R d \ E k , (1.5) where the exceptional set E k is given by E k = [ P ∈R ( A ) [ F ∈F k ( P ) ∂ ( F + N F ( P )) . (1.6) Also, we have ϕ k ( x ) ≥ a k for every x ∈ R d . An equivalent representation of the set E k , implying that it is a finite union of affine hyper-planes, will be given below; see (2.26) and (2.27). Example 1.5.
Let us consider a simple example showing that the exceptional set cannot beremoved from the statement of Theorem 1.4. Consider an arrangement A consisting of thecoordinate axes { x = 0 } and { x = 0 } in R . There are four chambers and the characteristicpolynomial is given by χ A ( t ) = ( t − . It is easy to check that the functions ϕ and ϕ definedin (1.5) are given by ϕ ( x , x ) = 1 + { x =0 } + { x =0 } + { x = x =0 } , ϕ ( x , x ) = 2 ϕ ( x , x ) . These functions are strictly larger than a = 1 and a = 2 on the exceptional set E = E = { x = 0 } ∪ { x = 0 } . Remark 1.6 (Similar identities) . It is interesting to compare Theorem 1.4 to the followingidentity: For every polyhedral set P ⊂ R d we have d X k =0 X F ∈F k ( P ) ( − k F − N F ( P ) ( x ) = ( , if P is bounded , , if P is unbounded and line-free, (1.7)for all x ∈ R d , without an exceptional set. Various versions of this formula valid outside certainexceptional sets of Lebesgue measure 0 have been obtained starting with the work of McMul-len [16, p. 249]; see [22, Proof of Theorem 6.5.5], [8, Hilfssatz 4.3.2], [9], [10, Corollary 2.25on p. 89]. The exceptional set has been removed independently in [21] (for polyhedral cones) OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 5 and in [11] (for general polyhedral sets). Cowan [4] proved another identity for alternatingsums of indicator functions of convex hulls. The exceptional set in Cowan’s identity has beensubsequently removed in [12].1.4.
Conic intrinsic volumes and characteristic polynomials.
As another consequenceof our result we can re-derive a formula due to Klivans and Swartz [14, Theorem 5] whichexpresses the coefficients of the characteristic polynomial of a linear hyperplane arrangementthrough the conic intrinsic volumes of its chambers. Let us first define conic intrinsic volumes;see [22, Section 6.5] and [2, 1] for more details. Let ξ be a random vector having an arbitraryrotationally invariant distribution on R d . As examples, one can think of the uniform distributionon the unit sphere in R d or the standard normal distribution. The k -th conic intrinsic volume ν k ( C ) of a polyhedral cone C ⊂ R d is defined as the probability that the metric projection of ξ onto C belongs to a relative interior of a k -dimensional face of C , that is ν k ( C ) = P [dim( ξ, C ) = k ] , k ∈ { , . . . , d } . To state the formula of Klivans and Swartz [14, Theorem 5], consider a linear hyperplanearrangement, i.e. a finite collection A = { H , . . . , H m } of hyperplanes in R d passing throughthe origin. The hyperplanes dissect R d into finitely many polyhedral cones (the chambers ofthe arrangement). The formula of Klivans and Swartz [14, Theorem 5] states that for every k ∈ { , . . . , d } the sum of ν k ( C ) over all chambers is equal to the absolute value of the k -thcoefficient of the characteristic polynomial χ A ( t ), namely X C ∈R ( A ) ν k ( C ) = a k , for all k ∈ { , . . . , d } . (1.8)For proofs and extensions of the Klivans-Swartz formula see [13, Theorem 4.1], [1, Section 6]and [21, Eq. (15) and Theorem 1.2]. To see that (1.8) is a consequence of our results, note thatby Theorem 1.1 applied with x replaced by ξ , X C ∈R ( A ) { dim( ξ,C )= k } = a k with probability 1 . Taking the expectation and interchanging it with the sum yields (1.8). Thus, in the setting oflinear arrangements, our result can be seen as an a.s. version of the Klivans-Swartz formula.
Example 1.7 (Second Zaslavsky’s formula) . Let us use Theorem 1.4 in combination withidentity (1.7) to sketch a proof of the second Zaslavsky formula for the number b ( A ) of boundedchambers, namely b ( A ) = P dk =0 ( − k a k , where we assumed for simplicity that the arrangement A has full rank d (the general case can be easily reduced to this one). Let ξ be a randomvector uniformly distributed on the unit sphere in R d . Applying identity (1.7) to each chamber P ∈ R ( A ) and the point x = Rξ , where R >
0, and taking the expectation, we get X P ∈R ( A ) d X k =0 ( − k X F ∈F k ( P ) P [ Rξ ∈ F − N F ( P )] = b ( A ) . It is an exercise to check that for every F ∈ F ( P ),lim R →∞ ( P [ Rξ ∈ F − N F ( P )] − P [ Rξ ∈ F + N F ( P )]) = 0 . Note that the characteristic polynomial in the notation of Klivans and Swartz [14, p. 420] corresponds toour χ A ( t ) /t d − rank( A ) , where rank( A ) = d − dim ∩ mi =1 H i is the rank of the arrangement. ZAKHAR KABLUCHKO
It follows that b ( A ) = lim R → + ∞ X P ∈R ( A ) d X k =0 ( − k X F ∈F k ( P ) P [ Rξ ∈ F + N F ( P )]= lim R → + ∞ d X k =0 ( − k E X P ∈R ( A ) X F ∈F k ( P ) F + N F ( P ) ( Rξ )= d X k =0 ( − k a k by Theorem 1.4.1.5. Extension to j -th level characteristic polynomials. Let us finally mention one simpleextension of the above results. Let L ( A ) be the set of all non-empty intersections of hyperplanesfrom A . By convention, the whole space R d is also included in L ( A ) as an intersection of theempty collection. Take some j ∈ { , . . . , d } and let L j ( A ) denote the set of all j -dimensionalaffine subspaces in L ( A ). The restriction of the arrangement A to the subspace L ∈ L ( A ) isdefined as A L = { H ∩ L : H ∈ A , H ∩ L = L, H ∩ L = ∅ } , which is an affine hyperplane arrangement in the ambient space L . Note that it may happenthat H ∩ L = H ∩ L for some different H , H ∈ A , in which case the corresponding hyperplaneis listed just once in the arrangement A L .Now, the j -th level characteristic polynomial of A may be defined as χ A ,j ( t ) = X L ∈L j ( A ) χ A L ( t ) . (1.9)We refer to [1, Section 2.4.1] for this and other equivalent definitions. Note that in the case j = d we recover the usual characteristic polynomial χ A ( t ). For the coefficients of the j -th levelcharacteristic polynomial we use the notation χ A ,j ( t ) = j X k =0 ( − j − k a kj t k . (1.10)Recall that R ( A ) denotes the set of all closed chambers generated by the arrangement A . For j ∈ { , . . . , d } , let R j ( A ) be the set of all j -dimensional faces of all chambers, that is R j ( A ) = [ P ∈R ( A ) F j ( P ) . The j -th level extension of Theorem 1.4 reads as follows. Theorem 1.8.
Let A be an affine hyperplane arrangement in R d whose j -th level characteristicpolynomial χ A ,j ( t ) is written in the form (1.10) . Then, for every j ∈ { , . . . , d } and k ∈{ , . . . , j } we have X P ∈R j ( A ) X F ∈F k ( P ) F + N F ( P ) ( x ) = a kj , for all x ∈ R d \ E kj , where the exceptional set E kj is given by E kj = [ P ∈R j ( A ) [ F ∈F k ( P ) ∂ ( F + N F ( P )) . (1.11) OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 7
Proof of Theorem 1.8 assuming Theorem 1.4.
Consider any L ∈ L j ( A ) and apply Theorem 1.4to the hyperplane arrangement A L in the ambient space L . This yields X P ∈R j ( A ): P ⊂ L X F ∈F k ( P ) F +( N F ( P ) ∩ L ) ( z ) = a k,L , for all z ∈ L \ E k,L , (1.12)where L := L − π L (0) is a shift of the affine subspace L that contains the origin, the a k,L ’s aredefined by the formulae χ A L ( t ) = j X k =0 ( − j − k a k,L t k (1.13)and the exceptional sets E k,L ⊂ L are given by E k,L = [ P ∈R j ( A ): P ⊂ L [ F ∈F k ( P ) ∂ L ( F + ( N F ( P ) ∩ L )) . (1.14)Here, ∂ L denotes the boundary operator in the ambient space L . Note that in (1.12) the normalcone of F ∈ F k ( P ) in the ambient space L is represented as N F ( P ) ∩ L , where N F ( P ) denotesthe normal cone in the ambient space R d . Also, we have the orthogonal sum decomposition N F ( P ) = ( N F ( P ) ∩ L ) + L ⊥ . Hence, we can rewrite (1.12) as X P ∈R j ( A ): P ⊂ L X F ∈F k ( P ) F + N F ( P ) ( x ) = a k,L , for all x ∈ R d \ ( E k,L + L ⊥ ) . (1.15)Since each j -dimensional face P ∈ R j ( A ) is contained in a unique affine subspace L ∈ L j ( A ),we can take the sum over all such L arriving at X P ∈R j ( A ) X F ∈F k ( P ) F + N F ( P ) ( x ) = X L ∈L j ( A ) X P ∈R j ( A ): P ⊂ L X F ∈F k ( P ) F + N F ( P ) ( x ) = X L ∈L j ( A ) a k,L for all x ∈ R d outside the following exceptional set: [ L ∈L j ( A ) ( E k,L + L ⊥ ) = [ L ∈L j ( A ) [ P ∈R j ( A ): P ⊂ L [ F ∈F k ( P ) (cid:0) ∂ L ( F + ( N F ( P ) ∩ L )) + L ⊥ (cid:1) = [ P ∈R j ( A ) [ F ∈F k ( P ) ∂ ( F + N F ( P )) = E kj . Here, we used that ∂ L ( A ) + L ⊥ = ∂ ( A + L ⊥ ) for every set A ⊂ L . It follows from (1.9), (1.10),(1.13) that X L ∈L j ( A ) a k,L = a kj , which completes the proof. (cid:3) Remark 1.9.
Using almost the same argument as in Section 1.4, Theorem 1.8 yields the fol-lowing j -th level extension of the Klivans-Swartz formula obtained in [1, Theorem 6.1] and [21,Eq. (15)]: X P ∈R j ( A ) ν k ( P ) = a kj , for all j ∈ { , . . . , d } , k ∈ { , . . . , j } . (1.16)The remaining part of this paper is devoted to the proof of Theorem 1.4. ZAKHAR KABLUCHKO Proof of Theorem 1.4
Step 1.
We start with a proposition which, as we shall see in Remark 2.3 below, impliesTheorem 1.4 for linear hyperplane arrangements in the special case k = 0. Recall that the dualcone of a polyhedral cone C ⊂ R d is defined by C ◦ = { z ∈ R d : h z, y i ≤ y ∈ C } . It is known that C ◦◦ = C ; see [1, Prop. 2.3]. The lineality space of a cone C is the largest linearspace contained in C and is explicitly given by C ∩ ( − C ). It is known that the linear spacespanned by the dual cone C ◦ coincides with the orthogonal complement of the lineality spaceof C ; see, e.g., [1, Prop. 2.5] for a more general statement. In particular, the lineality space of C is trivial (i.e. equal to { } ) if and only if C ◦ has non-empty interior. Proposition 2.1.
Let A be a linear hyperplane arrangement in R d . Then, X C ∈R ( A ) C ◦ ( x ) = a , for all x ∈ R d \ E ∗ , (2.1) where a is defined by (1.2) and (1.3) and the exceptional set E ∗ is given by E ∗ := [ L ∈L ( A ) \{ } L ⊥ = [ C ∈R ( A ) ∂ ( C ◦ ) . (2.2) Proof.
Since for linear arrangements C
7→ − C defines a bijective self-map of R ( A ) and since( − C ) ◦ = − ( C ◦ ), we have X C ∈R ( A ) C ◦ ( x ) = X C ∈R ( A ) − C ◦ ( x ) , for all x ∈ R d . Therefore, it suffices to prove that X C ∈R ( A ) ( C ◦ ( x ) + − C ◦ ( x )) = 2 a , for all x ∈ R d \ E ∗ . Since every cone C ∈ R ( A ) is full-dimensional, implying that the dual cone has trivial linealityspace C ◦ ∩ ( − C ◦ ) = { } , it suffices to prove that X C ∈R ( A ) C ◦ ∪− C ◦ ( x ) = 2 a , for all x ∈ R d \ E ∗ . Let L ( x ) := { λx : λ ∈ R } be the 1-dimensional line generated by x ∈ R d \{ } . Then, x ∈ C ◦ ∪ − C ◦ if and only if L ∩ C ◦ = { } . It therefore suffices to prove that X C ∈R ( A ) { L ( x ) ∩ C ◦ = { }} = 2 a , for all x ∈ R d \ E ∗ . In a slightly different form, this result is contained in [21, Thm. 1.2, Eq. (16)]. For completeness,we provide a proof. By the Farkas lemma [1, Lemma 2.4], L ( x ) ∩ C ◦ = { } is equivalent to L ( x ) ⊥ ∩ int C = ∅ . Thus, we need to show that X C ∈R ( A ) { L ( x ) ⊥ ∩ int C = ∅ } = 2 a , for all x ∈ R d \ E ∗ . (2.3)By the first Zaslavsky formula, the total number of chambers of A is given by R ( A ) =( − d χ A ( −
1) = P dk =0 a k . By the second Zaslavsky formula, χ A (1) = 0 (because there are OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 9 no bounded chambers in a linear arrangement). Hence, P dk =0 ( − k a k = 0 and it follows that R ( A ) = 2 P [ d/ k =0 a k . In view of this, it suffices to show that X C ∈R ( A ) { L ( x ) ⊥ ∩ int C = ∅ } = 2 [ d/ X k =1 a k , for all x ∈ R d \ E ∗ . (2.4)This identity is known [13, Thm. 3.3] provided that the hyperplane L ( x ) ⊥ is in general positionwith respect to the arrangement A . By definition [13, Sec. 3.1] , the general position conditionmeans that for every L ∈ L ( A ) with L = { } , we have dim( L ∩ L ( x ) ⊥ ) = dim L −
1. This isthe same as to require that L is not a subset of L ( x ) ⊥ or, equivalently, that x / ∈ L ⊥ . So, theabove identity holds for all x ∈ R d \ ∪ L ∈L ( A ) \{ } ( L ⊥ ), which completes the proof of (2.1). Thesecond representation of the exceptional set E ∗ in (2.2) was mentioned just for completeness.We shall prove it in Lemma 2.8 without using it before. (cid:3) Lemma 2.2.
Let A be a linear hyperplane arrangement in R d . Then, X C ∈R ( A ) C ◦ ( x ) ≥ a , for all x ∈ R d . Proof.
The proof of Proposition 2.1 applies with minimal modifications. Indeed, by [13, Lemma3.5, Eq. (38)] (note that R ( A ) denotes the collection of open chambers there), the equalityin (2.4) has to be replaced by the inequality ≤ , which means that the equality in (2.3) shouldbe replaced by ≥ . The rest of the proof applies. (cid:3) Remark 2.3.
With Proposition 2.1 at hand, we can prove Theorem 1.4 for k = 0 providedthe arrangement A is linear. Assume first that A is essential, i.e. it has full rank meaning that ∩ H ∈A H = { } . Then, F = { } is the only 0-dimensional face of every chamber C ∈ R ( A ). Thenormal cone of C an this face is N { } ( C ) = C ◦ . Hence, the case k = 0 of Theorem 1.4 followsfrom Proposition 2.1 and Lemma 2.2. In the case of a non-essential linear arrangement, thatis if L ∗ := ∩ H ∈A H = { } , Theorem 1.4 becomes trivial for k = 0 as there are no 0-dimensionalfaces and the zeroth coefficient of χ A ( t ) vanishes by its definition (1.2). Proposition 2.1 alsobecomes trivial since the dual cone C ◦ of every chamber C is contained in L ⊥∗ , which coincideswith the exceptional set ∪ L ∈L ( A ) \{ } ( L ⊥ ). Since a = 0, both sides of (2.1) vanish for x / ∈ L ⊥∗ . Remark 2.4.
In the special case of reflection arrangements, Proposition 2.1 can be found inthe paper of Denham [6, Thm.2]; see also [5] for a related work.
Step 2.
We are interested in the function ϕ k ( x ) = X P ∈R ( A ) X F ∈F k ( P ) F + N F ( P ) ( x ) , x ∈ R d . First of all note that in the case k = d we trivially have ϕ d ( x ) = 1 for all x ∈ R d \ ∪ H ∈A H .In the following, fix some k ∈ { , . . . , d − } . Recall that R j ( A ) = ∪ P ∈R ( A ) F j ( P ) is the setof all j -dimensional faces of all chambers of A (without repetitions). Interchanging the orderof summation, we may write ϕ k ( x ) = X F ∈R k ( A ) X P ∈R ( A ): F ∈F k ( P ) F + N F ( P ) ( x ) . Recall also that L ( A ) is the set of all non-empty intersections of hyperplanes from A and that L k ( A ) is the set of all k -dimensional affine subspaces in L ( A ). Since each k -dimensional face F ∈ R k ( A ) is contained in a unique k -dimensional affine subspace L ∈ L k ( A ), we may splitthe sum in the above formula for ϕ k ( x ) as follows: ϕ k ( x ) = X L ∈L k ( A ) ϕ L ( x ) , (2.5)where for each L ∈ L k ( A ) we define ϕ L ( x ) = X F ∈R k ( A ): F ⊂ L X P ∈R ( A ): F ∈F k ( P ) F + N F ( P ) ( x ) . (2.6) Step 3.
In this step we shall prove that for every k ∈ { , . . . , d − } and every L ∈ L k ( A ) thefunction ϕ L ( x ) defined in (2.6) is constant outside the exceptional set E ( L ) := E ′ ( L ) ∪ E ′′ ( L ) , (2.7)where E ′ ( L ) := [ L k − ∈L k − ( A ): L k − ⊂ L ( L k − + L ⊥ ) and E ′′ ( L ) := [ L k +1 ∈L k +1 ( A ): L k +1 ⊃ L ( L ⊥ k +1 + L ) . (2.8)For k = 0 we put E ′ ( L ) := ∅ . Note that E ( L ) is a finite union of affine hyperplanes. Moreover,we shall identify the value of the constant in terms of the characteristic polynomial of somehyperplane arrangement in L ⊥ , the orthogonal complement of L . The final result will be statedin Proposition 2.5 at the end of this step.First we need to introduce some notation. Recall that h· , ·i denotes the standard Euclideanscalar product on R d . Let the affine hyperplanes H , . . . , H m constituting the arrangement A be given by the equations H i = { z ∈ R d : h z, y i i = c i } , i ∈ { , . . . , m } , for some vectors y , . . . , y m ∈ R d \{ } and some scalars c , . . . , c m ∈ R . Every closed chamberof the arrangement A can be represented in the form P = { z ∈ R d : ε ( h z, y i − c ) ≤ , . . . , ε m ( h z, y m i − c m ) ≤ } with a suitable choice of ε , . . . , ε m ∈ {− , +1 } . Conversely, every set of the above form definesa closed chamber provided its interior is non-empty . Note in passing that the interior of thischamber is represented by the corresponding strict inequalities as follows:int P = { z ∈ R d : ε ( h z, y i − c ) < , . . . , ε m ( h z, y m i − c m ) < } . Finally, the chambers determined by two different tuples ( ε , . . . , ε m ) and ( ε ′ , . . . , ε ′ m ) havedisjoint interiors. Indeed, if the tuples differ in the i -th component, then any point z in therelative interior of one chamber satisfies h z, y i i < c i , whereas the points in the relative interiorof the other chamber satisfy the converse inequality.Fix some k -dimensional affine subspace L ∈ L k ( A ), where k ∈ { , . . . , d − } . It can bewritten in the form L = { z ∈ R d : h z, y i i = c i for all i ∈ I } for a suitable subset I ⊂ { , . . . , m } . Without restriction of generality we may assume that L passes through the origin (otherwise we could translate everything). It follows that c i = 0 for i ∈ I . Moreover, after renumbering (if necessary) the hyperplanes and their defining equations,we may assume that the linear subspace L is given by the equations L = { z ∈ R d : h z, y i = 0 , . . . , h z, y ℓ i = 0 } (2.9) OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 11 for some ℓ ∈ { d − k, . . . , m } . Finally, without loss of generality we may assume that H i ∩ L is a strict subset of L for all i ∈ { ℓ + 1 , . . . , m } since otherwise we could include the definingequation of H i into the list on the right-hand side of (2.9).Take any point x ∈ R d \ E ( L ), where we recall that E ( L ) is defined by (2.7) and (2.8). Theorthogonal projection of x onto the linear subspace L , denoted by π L ( x ), is contained in therelative interior of some uniquely defined face G ∈ ∪ kp =0 R p ( A ) with G ⊂ L . In fact, we evenhave G ∈ R k ( A ) because if the dimension of G would be strictly smaller than k , we could findsome L k − ∈ L k − ( A ) with G ⊂ L k − ⊂ L . This would contradict the assumption x / ∈ E ′ ( L ).So, we have π L ( x ) ∈ relint G, G ∈ R k ( A ) , G ⊂ L. Then, the definition of ϕ L ( x ) given in (2.6) simplifies as follows: ϕ L ( x ) = X P ∈R ( A ): G ∈F k ( P ) G + N G ( P ) ( x ) . (2.10)Indeed, for every F ∈ R k ( A ) and P ∈ R ( A ) with F ⊂ L , F ∈ F k ( P ) and F = G we have x / ∈ F + N F ( P ), which follows from the fact that x ∈ relint G + L ⊥ , while relint G ∩ F = ∅ and N F ( P ) ⊂ L ⊥ . This means that all terms with F = G do not contribute to the right-hand sideof (2.6).By changing, if necessary, the signs of some y i ’s and the corresponding c i ’s, we may assumethat the face G is given as follows: G = { z ∈ L : h z, y ℓ +1 i ≤ c ℓ +1 , . . . , h z, y m i ≤ c m } = { z ∈ R d : h z, y i = 0 , . . . , h z, y ℓ i = 0 , h z, y ℓ +1 i ≤ c ℓ +1 , . . . , h z, y m i ≤ c m } . (2.11)The relative interior of G is given by the following strict inequalities:relint G = { z ∈ L : h z, y ℓ +1 i < c ℓ +1 , . . . , h z, y m i < c m } = { z ∈ R d : h z, y i = 0 , . . . , h z, y ℓ i = 0 , h z, y ℓ +1 i < c ℓ +1 , . . . , h z, y m i < c m } . (2.12)Let now P ∈ R ( A ) be a closed chamber such that G ∈ F k ( P ). Then, there exist some ε , . . . , ε ℓ ∈ {− , +1 } such that P is given by P = P ε ,...,ε ℓ := { z ∈ R d : ε h z, y i ≤ , . . . , ε ℓ h z, y ℓ i ≤ , h z, y ℓ +1 i ≤ c ℓ +1 , . . . , h z, y m i ≤ c m } . (2.13)Conversely, if for some ε , . . . , ε ℓ ∈ {− , +1 } the interior of the set P ε ,...,ε ℓ defined above is non-empty, then P ε ,...,ε ℓ is a chamber in R ( A ) and it contains G as a k -dimensional face. Hence,we can rewrite (2.10) as follows: ϕ L ( x ) = X ε ,...,ε ℓ ∈{− , +1 } :int P ε ,...,εℓ = ∅ G + N G ( P ε ,...,εℓ ) ( x ) . Write x = π L ( x ) + π L ⊥ ( x ) as a sum of its orthogonal projections π L ( x ) and π L ⊥ ( x ) onto L and L ⊥ , respectively. Since π L ( x ) ∈ G ⊂ L and N G ( P ε ,...,ε ℓ ) ⊂ L ⊥ , we arrive at ϕ L ( x ) = X ε ,...,ε ℓ ∈{− , +1 } :int P ε ,...,εℓ = ∅ N G ( P ε ,...,εℓ ) ( π L ⊥ ( x )) . (2.14) Let us now characterize first the tangent and then the normal cone of the face G in thepolyhedral set P ε ,...,ε ℓ . Take some z ∈ relint G . Then, by (2.12), h z , y i = 0 , . . . , h z , y ℓ i = 0 , h z , y ℓ +1 i < c ℓ +1 , . . . , h z , y m i < c m . (2.15)By definition, see (1.4), the tangent cone is given by T G ( P ε ,...,ε ℓ ) = { u ∈ R d : ∃ δ > z + δu ∈ P ε ,...,ε ℓ } . It follows from this definition together with (2.13) and (2.15) that T G ( P ε ,...,ε ℓ ) = { u ∈ R d : h u, ε y i ≤ , . . . , h u, ε ℓ y ℓ i ≤ } . Note that the linear span of y , . . . , y ℓ is L ⊥ by (2.9). The tangent cone T G ( P ε ,...,ε ℓ ) containsthe linear space L . Let us now restrict our attention to the space L ⊥ and define the cone T ε ,...,ε l := { u ∈ L ⊥ : h u, ε y i ≤ , . . . , h u, ε ℓ y ℓ i ≤ } = T G ( P ε ,...,ε ℓ ) ∩ L ⊥ ⊂ L ⊥ . (2.16)Then, the tangent cone T G ( P ε ,...,ε ℓ ) can be represented as the direct orthogonal sum T G ( P ε ,...,ε ℓ ) = L + T ε ,...,ε l , T ε ,...,ε l ⊂ L ⊥ . Taking the polar cone, we obtain the normal cone of the face G in the polyhedral set P ε ,...,ε ℓ : N G ( P ε ,...,ε ℓ ) = L ⊥ ∩ T ◦ ε ,...,ε l . (2.17)That is, N G ( P ε ,...,ε ℓ ) is just the dual cone of T ε ,...,ε l taken with respect to the ambient space L ⊥ . Although we shall not use this fact in the sequel, let us mention that the normal cone canbe represented as the positive hull N G ( P ε ,...,ε ℓ ) = pos( ε y , . . . , ε ℓ y ℓ ) = { λ ε y + . . . + λ ℓ ε ℓ y ℓ : λ , . . . , λ ℓ ≥ } . In the following, we shall argue that those cones of the form T ε ,...,ε l that have non-emptyinterior are the chambers of certain linear hyperplane arrangement A ( L ) in L ⊥ . The dual conesof these chambers are the normal cones N G ( P ε ,...,ε ℓ ). It is crucial that this arrangement iscompletely determined by y , . . . , y ℓ and does not depend on G ⊂ L . Applying Proposition 2.1,we shall prove that ϕ L ( x ) is constant outside some explicit exceptional Lebesgue null set.Let us be more precise. First of all, note that the vectors y , . . . , y ℓ are pairwise different.Indeed, if two of them would be equal, say y = y , then (in view of c = c = 0) thecorresponding hyperplanes H and H would be equal, which is prohibited by the definition ofthe hyperplane arrangement. Therefore, the orthogonal complements of the vectors y , . . . , y ℓ (taken with respect to the ambient space L ⊥ ) are also pairwise different and define a linearhyperplane arrangement in L ⊥ which we denote by A ( L ) := { L ⊥ ∩ y ⊥ , . . . , L ⊥ ∩ y ⊥ ℓ } . (2.18)Since the linear span of y , . . . , y ℓ is L ⊥ by (2.9), this arrangement is essential, that is theintersection of its hyperplanes is { } . The chambers of the arrangement A ( L ) are those of thecones T ε ,...,ε ℓ , ( ε , . . . , ε ℓ ) ∈ {− , +1 } ℓ , defined in (2.16), that have non-empty interior in L ⊥ .Note also that A ( L ) is uniquely determined by the choice of L ∈ L k ( A ) and does not dependon G .Now we claim that for ( ε , . . . , ε ℓ ) ∈ {− , +1 } ℓ the relative interior of the cone T ε ,...,ε ℓ isnon-empty if and only if the interior of the polyhedral set P ε ,...,ε ℓ is non-empty. If int P ε ,...,ε ℓ is non-empty, then it has dimension d , G ∈ F k ( P ), and the tangent cone T G ( P ε ,...,ε ℓ ) is strictlylarger than the linear space L (because the latter has dimension k < d ). It follows from (2.16)that relint T ε ,...,ε ℓ = ∅ . Conversely, if relint T ε ,...,ε ℓ = ∅ , then T ε ,...,ε ℓ has the same dimension OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 13 as L ⊥ , while G has the same dimension as L . It follows that the dimension of P ε ,...,ε ℓ is d , thusits interior is non-empty.From the above it follows that the formula for the function ϕ L stated in (2.14) can be writtenas the following sum over the chambers of the arrangement A ( L ): ϕ L ( x ) = X C ∈R ( A ( L )) C ◦ ( π L ⊥ ( x )) . (2.19)We are now going to apply Proposition 2.1 to the hyperplane arrangement A ( L ) in theambient space L ⊥ . This is possible provided π L ⊥ ( x ) does not belong to the exceptional set E ∗ defined in Proposition 2.1. In our setting of the ambient space L ⊥ , the exceptional set is givenby E ∗ = [ M ∈L ( A ( L )) \{ } ( M ⊥ ∩ L ⊥ ) . (2.20)Each linear subspace M ∈ L ( A ( L )) \{ } has the form M = ( ∩ i ∈ I y ⊥ i ) ∩ L ⊥ for some set I ⊂{ , . . . , ℓ } . Then, the corresponding orthogonal complement M ⊥ ∩ L ⊥ has the form lin { y i : i ∈ I } , where lin A denotes the linear subspace spanned by the set A . Moreover, the condition M = { } is equivalent to the condition lin { y i : i ∈ I } 6 = L ⊥ . Since the linear span of the vectors y , . . . , y ℓ is L ⊥ by (2.9), any linear subspace of the form lin { y i : i ∈ I } 6 = L ⊥ is contained ina linear subspace of the form lin { y i : i ∈ I ′ } , for some I ′ ⊂ { , . . . , ℓ } satisfying the followingcondition: dim lin { y i : i ∈ I ′ } = I ′ = dim L ⊥ − d − k − . (2.21)Therefore, we have E ∗ = [ I ′ ⊂{ ,...,ℓ } :(2.21) holds lin { y i : i ∈ I ′ } . Given I ′ ⊂ { , . . . , ℓ } such that (2.21) holds, define the linear subspace L k +1 := { z ∈ R d : h z, y i i = 0 for all i ∈ I ′ } ⊂ R d . (2.22)Then, L k +1 is non-empty since L ⊂ L k +1 and, moreover, the dimension of L k +1 equals k + 1,that is L k +1 ∈ L k +1 ( A ) (recall that the case k = d has been excluded from the very beginning).Conversely, every L k +1 ∈ L k +1 ( A ) containing L can be represented in the form (2.22) withsome I ′ ⊂ { , . . . , ℓ } satisfying (2.21). Taking into account that lin { y i : i ∈ I ′ } = L ⊥ k +1 , itfollows that E ∗ = [ L k +1 ∈L k +1 ( A ): L k +1 ⊃ L L ⊥ k +1 ⊂ L ⊥ . (2.23)Proposition 2.1 applies to all x ∈ R d such that π L ⊥ ( x ) / ∈ E ∗ . This is equivalent to the conditionthat x is outside the set [ L k +1 ∈L k +1 ( A ): L k +1 ⊃ L ( L ⊥ k +1 + L ) , which coincides with the set E ′′ ( L ) introduced in (2.8).Applying Proposition 2.1 and Lemma 2.2 with the ambient space L ⊥ to the right-hand sideof (2.19), we arrive at the following result. Proposition 2.5.
Let A be an affine hyperplane arrangement in R d . Fix some k ∈ { , . . . , d − } and L ∈ L k ( A ) . Then, the function ϕ L defined in (2.6) satisfies ϕ L ( x ) = a ( L ) , for every x ∈ R d \ E ( L ) , where the exceptional set E ( L ) is given by (2.7) and (2.8) , and a ( L ) is ( − d − k times thezeroth coefficient of the characteristic polynomial of the linear arrangement A ( L ) in L ⊥ definedby (2.18) . Also, for all x ∈ R d we have ϕ L ( x ) ≥ a ( L ) .Step 4. Let us first write down a more explicit expression for a ( L ) appearing in Proposition 2.5.Recalling the definition of the characteristic polynomial, see (1.2), we can write χ A ( L ) ( t ) = X J ⊂{ ,...,ℓ } ( − J t dim L ⊥ − rank { y j : j ∈ J } , where rank { y j : j ∈ J } denotes the dimension of the linear span of a system of vectors { y j : j ∈ J } . Taking the zeroth coefficient of this polynomial and multiplying it with ( − d − k , wecan write Proposition 2.5 as follows: ϕ L ( x ) = a ( L ) = ( − d − k X J ⊂{ ,...,ℓ } :lin { y j : j ∈ J } = L ⊥ ( − J , for all x ∈ R d \ E ( L ) . Recalling the representation of L stated in (2.9), we see that a set of vectors { y j : j ∈ J } with J ⊂ { , . . . , ℓ } contributes to the above sum if and only if L = ∩ j ∈ J H j . Moreover, aset J ⊂ { , . . . , m } which is not completely contained in { , . . . , ℓ } cannot satisfy L = ∩ j ∈ J H j since H j ∩ L is a strict subset of L for all j ∈ { ℓ + 1 , . . . , m } ; see the discussion after (2.9).Therefore, we can rewrite the above sum as follows: ϕ L ( x ) = ( − d − k X J ⊂{ ,...,m } : L = ∩ j ∈ J H j ( − J = ( − d − k X B⊂A : ∩ H ∈B H = L ( − B , for all x ∈ R d \ E ( L ) . Taking the sum over all k -dimensional affine subspaces L ∈ L k ( A ) generated by the arrangement A and recalling (2.5), we arrive at ϕ k ( x ) = X L ∈L k ( A ) ϕ L ( x ) = ( − d − k X B⊂A :dim( ∩ H ∈B H )= k ( − B , (2.24)for all x ∈ R d such that x / ∈ [ L ∈L k ( A ) E ( L ) = [ L ∈L k ( A ) ( E ′ ( L ) ∪ E ′′ ( L )) (2.25)with E ′ ( L ) = [ L k − ∈L k − ( A ): L k − ⊂ L ( L k − + L ⊥ ) , E ′′ ( L ) = [ L k +1 ∈L k +1 ( A ): L k +1 ⊃ L ( L ⊥ k +1 + L ) . (2.26)By the definition of the characteristic polynomial χ A ( t ), see (1.2) and (1.3), the right-hand sideof (2.24) is nothing but a k . So, ϕ k ( x ) = a k for all x ∈ R d satisfying (2.25). If (2.25) is notsatisfied, we can use the inequality ϕ L ( x ) ≥ a ( L ) to prove that ϕ k ( x ) ≥ a k . Step 5.
To complete the proof of Theorem 1.4, it remains to check the following equality of theexceptional sets: [ L ∈L k ( A ) ( E ′ ( L ) ∪ E ′′ ( L )) = [ P ∈R ( A ) [ G ∈F k ( P ) ∂ ( G + N G ( P )) , (2.27)for all k ∈ { , . . . , d − } . We need some preparatory lemmas. OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 15
Lemma 2.6.
Let A = { H , . . . , H m } be a linear hyperplane arrangement in R d . Suppose that A is of full rank meaning that ∩ mi =1 H i = { } . Then, ∪ C ∈R ( A ) ( C ◦ ) = R d .Proof. By Lemma 2.2 it suffices to show that a >
0. By Proposition 2.1, the function ϕ ( x ) = P C ∈R ( A ) C ◦ ( x ) is Lebesgue-a.e. equal to a , hence a ≥
0. Since the arrangement is of fullrank, the lineality space of each chamber is trivial, that is C ∩ ( − C ) = { } . This impliesthat the dual cone C ◦ has non-empty interior, hence the the function ϕ ( x ) cannot be a.e. 0implying that a = 0. (cid:3) Lemma 2.7.
Let C ⊂ R d be a polyhedral cone with a trivial lineality space, that is C ∩ ( − C ) = { } . Let v ∈ R d \{ } be a vector. Then, at least one of the cones pos( C ∪{ + v } ) or pos( C ∪{− v } ) has a trivial lineality space.Proof. It follows from C ∩ ( − C ) = { } that there exists ε ∈ {− , +1 } such that εv / ∈ − C . Weclaim that pos( C ∪ { εv } ) has a trivial lineality space. To prove this, take some w such thatboth + w and − w are contained in pos( C ∪ { εv } ). We then have w = z + λ εv = − z − λ εv for some z , z ∈ C and λ , λ ≥
0. If λ = λ = 0, then z = − z implying that z = z = 0and thus w = 0. So, let λ + λ >
0. Then, we have εv = − ( z + z ) / ( λ + λ ) ∈ − C, acontradiction. (cid:3) Lemma 2.8.
Let A = { H , . . . , H m } be a linear hyperplane arrangement in R d . Then, [ L ∈L ( A ) \{ } L ⊥ = [ C ∈R ( A ) ∂ ( C ◦ ) . (2.28) Proof. If A is not essential meaning that L ∗ := ∩ mi =1 H i = { } , then the left-hand side of (2.28)equals L ⊥∗ . On the other hand, the cones C ◦ are contained in L ⊥∗ , satisfy ∂ ( C ◦ ) = C ◦ , andcover the space L ⊥∗ by Lemma 2.6 applied to the ambient space L ⊥∗ , thus proving that (2.28)holds.In the following let A be of full rank meaning that ∩ mi =1 H i = { } . Let H = y ⊥ , . . . , H m = y ⊥ m for some vectors y , . . . , y m ∈ R d \{ } . The linear span of y , . . . , y m is R d since the arrangementhas full rank. Any subspace L ∈ L ( A ) has the form L = ∩ i ∈ I H i = lin { y i : i ∈ I } ⊥ for some set I ⊂ { , . . . , m } . The corresponding orthogonal complement is L ⊥ = lin { y i : i ∈ I } . It followsthat [ L ∈L ( A ) \{ } L ⊥ = [ I ⊂{ ,...,m } :lin { y i : i ∈ I }6 = R d lin { y i : i ∈ I } . To complete the proof, we need to show that [ I ⊂{ ,...,m } :lin { y i : i ∈ I }6 = R d lin { y i : i ∈ I } = [ C ∈R ( A ) ∂ ( C ◦ ) . (2.29)To prove the inclusion ⊂ , let v ∈ lin { y i : i ∈ I } 6 = R d for some I ⊂ { , . . . , m } . By firstextending I and then excluding the superfluous linearly dependent elements, we may assumethat M := lin { y i : i ∈ I } has dimension d − { y i : i ∈ I } are linearlyindependent. We can find ε i ∈ {− , +1 } , for all i ∈ I , such that v ∈ pos { ε i y i : i ∈ I } . Let M + and M − be the closed half-spaces in which the hyperplane M dissects R d . Let J , respectively J , be the set of all j ∈ { , . . . , m }\ I such that y j ∈ M , respectively y j ∈ R d \ M . The conepos { ε i y i : i ∈ I } ⊂ M has a trivial lineality space because { ε i y i : i ∈ I } is a basis of M .By inductively applying Lemma 2.7 in the ambient space M , we can find ε j ∈ {− , +1 } , forall j ∈ J , such that the cone D := pos { ε i y i : i ∈ I ∪ J } ⊂ M has a trivial lineality space. Furthermore, for every j ∈ J we can find ε j ∈ {− , +1 } such that ε j y j ∈ int M + . With thesigns ε , . . . , ε m ∈ {− , +1 } constructed as above, we consider the cone C := { z ∈ R d : h z, ε y i ≤ , . . . , h z, ε m y m i ≤ } . (2.30)The dual cone is the positive hull C ◦ = pos { ε y , . . . , ε m y m } . (2.31)By construction, C ◦ ⊂ M + and C ◦ ∩ M = D . Also, the cone C ◦ has a trivial lineality spacebecause ± w ∈ C ◦ would imply ± w ∈ C ◦ ∩ M = D , which implies w = 0 because D has atrivial lineality space by construction. By duality, C has non-empty interior. It follows that C is a chamber of the arrangement R ( A ). By construction, C ◦ ⊂ M + and v ∈ C ◦ ∩ M , hence v ∈ ∂ ( C ◦ ), thus completing the proof of the inclusion ⊂ in (2.29).To prove the inclusion ⊃ in (2.29), take any C ∈ R ( A ) and any v ∈ ∂ ( C ◦ ). Then, C and C ◦ must be of the same form as in (2.30) and (2.31). Moreover, since C has non-empty interior,the lineality space of the cone C ◦ is trivial. If v ∈ ∂ ( C ◦ ), then v ∈ F for some face F ∈ F ( C ◦ )of dimension d −
1. Let I be the set of all i ∈ { , . . . , m } with ε i y i ∈ F . Then, we havelin { ε i y i : i ∈ I } = lin F , which contains v and does not coincide with R d . It follows that v belongs to the left-hand side of (2.29), thus completing the proof. (cid:3) Now we are in position to prove (2.27). We have [ P ∈R ( A ) [ G ∈F k ( P ) ∂ ( G + N G ( P )) = [ L ∈L k ( A ) [ G ∈R k ( A ): G ⊂ L [ P ∈R ( A ): G ∈F k ( P ) (( ∂G + N G ( P )) ∪ ( G + ∂N G ( P )))= [ L ∈L k ( A ) ( H ′ ( L ) ∪ H ′′ ( L ))with H ′ ( L ) = [ G ∈R k ( A ): G ⊂ L (cid:16) ∂G + [ P ∈R ( A ): G ∈F k ( P ) N G ( P ) (cid:17) , H ′′ ( L ) = [ G ∈R k ( A ): G ⊂ L (cid:16) G + [ P ∈R ( A ): G ∈F k ( P ) ∂N G ( P ) (cid:17) . We claim that H ′ ( L ) = E ′ ( L ). To prove this it suffices to show that for every G ∈ R k ( A )such that G ⊂ L we have ∪ P ∈R ( A ): G ∈F k ( P ) N G ( P ) = L ⊥ . In (2.17) we characterized the normalcones N G ( P ) as the dual cones of the chambers of some essential (full rank) linear hyperplanearrangement A ( L ) in L ⊥ . These dual cones cover L ⊥ by Lemma 2.6, thus proving the claim.It remains to show that H ′′ ( L ) = E ′′ ( L ). To this end, it suffices to prove that for every G ∈ R k ( A ) such that G ⊂ L we have [ P ∈R ( A ): G ∈F k ( P ) ∂N G ( P ) = [ L k +1 ∈L k +1 ( A ): L k +1 ⊃ L L ⊥ k +1 . (2.32)Again, recall from (2.17) that the normal cones N G ( P ) are the dual cones of the chambers ofthe linear full-rank hyperplane arrangement A ( L ) = { y ⊥ ∩ L ⊥ , . . . , y ⊥ ℓ ∩ L ⊥ } in L ⊥ . ApplyingLemma 2.8 to this arrangement, we obtain [ P ∈R ( A ): G ∈F k ( P ) ∂N G ( P ) = [ M ∈L ( A ( L )) \{ } ( M ⊥ ∩ L ⊥ ) . The right-hand side coincides with the set E ∗ defined in (2.20). Thus, the claim (2.32) followsfrom the identity already established in (2.23). (cid:3) OEFFICIENTS OF CHARACTERISTIC POLYNOMIALS OF HYPERPLANE ARRANGEMENTS 17
Acknowledgements
Supported by the German Research Foundation under Germany’s Excellence Strategy EXC2044 – 390685587, Mathematics M¨unster: Dynamics - Geometry - Structure.
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Zakhar Kabluchko: Institut f¨ur Mathematische Stochastik, Westf¨alische Wilhelms-Universit¨atM¨unster, Orl´eans-Ring 10, 48149 M¨unster, Germany
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