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It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements

It All Depends on How You Slice It: An

Introduction to Hyperplane Arrangements

Paul Renteln

California State University San Bernardino and Caltech

April, 2008


Outline

Hyperplane Arrangements

Counting Regions

The Intersection Poset

Finite Field Method




I Begin with Rn = {(x1, x2, . . . , xn) : xi ∈ R}.I An affine hyperplane is the set of points in Rn satisfying an

equation of the form a1x1 + · · ·+ anxn = b.

I A hyperplane arrangement is simply a collection of affine

hyperplanes.



A Hyperplane Arrangement

An arrangement of four lines in the plane



Another Hyperplane Arrangement

An arrangement of three planes in three space


Counting Regions

Regions

I A region of the arrangement A is a connected component of

the complement

Rn −⋃

H∈AH

I Regions can be bounded or unbounded.

I The total number of regions is r(A), and the number of

bounded regions is b(A).


Counting Regions

Region Counting

r(A) = 10 b(A) = 2 (shaded)


Counting Regions

Deletion and Restriction

I Is there a better way to count regions?

I Yes!

Definition

I Let A be an arrangement and H ∈ A a hyperplane.

I A′ = A\H is called the deleted arrangement.

I A′′ = {K ∩H : K ∈ A′} is called the restricted arrangement.

I (A,A′,A′′) is called a triple of arrangements.


Counting Regions

A Triple (A,A′,A′′) of Arrangements

A and H A′ A′′


Counting Regions

Region Counting Recurrence

Theorem (Zaslavsky, 1975)

r(A) = r(A′) + r(A′′)


Counting Regions

Proof by Example (Don’t try this at home!)

A and H

r(A) = 11 =

A′

r(A′) = 7 +

A′′

r(A′′) = 4


Counting Regions

Arrangements in General Position

I Let’s apply this result.

I An arrangement A is in general position if you can move the

hyperplanes slightly and not change the number of regions.


Counting Regions

Two Arrangements

in general position not in general position


Counting Regions

Counting Regions of General Position Line Arrangements

I Start with an arrangement A of k lines in general position in

the plane, and choose a particular line H.

I By hypothesis, H meets A′ in k − 1 points, which divide H

into k regions. So r(A′′) = k.

I Hence r(A) = r(A′) + k, where A′ contains k − 1 lines.

I By continuing to delete lines in this way, we get

r(A) = r(∅) + 1 + 2 + · · ·+ (k − 1) + k.

I When no lines are present there is one region, so r(∅) = 1.

I Hence, for a general position line arrangement we have

r(A) = 1 +k∑

i=1

i = 1 +k(k + 1)

2= 1 + k +

(k

2

).


Counting Regions

Counting Regions of a General Position Line Arrangement

r(A) = 1 + k +(

k

2

)= 1 + 4 +

(42

)= 11X



The Number of Regions in an Arbitrary Arrangement

I We can use the recurrence (and induction) to show that

r(A) = 1 + k +(k

2

)+(k

3

)+ · · ·+

(k

n

)for k hyperplanes in general position in n dimensional space.

(Ludwig Schlafli, 1901)

I But what if the hyperplanes are not in general position?

I Zaslavsky developed a powerful tool to compute r(A) in

general. To describe this, we must take a long detour...



Partially Ordered Sets

A partially ordered set (poset) is a set P and a relation ≤satisfying the following axioms (for all x, y, and z in P ):

1. (reflexivity) x ≤ x.

2. (antisymmetry) x ≤ y and y ≤ x implies x = y.

3. (transitivity) x ≤ y and y ≤ z implies x ≤ z.

4. Posets are represented by their (Hasse) diagrams.



Hasse Diagrams

Some posets



The Intersection Poset of an Arrangement

I The intersection poset L(A) of the arrangement A has as its

elements all the intersections of all the hyperplanes.

I It is ordered (for good reason) by reverse inclusion, so

A ⊆ B ⇔ A ≥ B.

I The minimum element is the ambient space Rn.



An Arrangement and Its Intersection Poset

4

21 3R2

1 2 3 4

1 ∩ 2 ∩ 3 1 ∩ 4 2 ∩ 4 3 ∩ 4

A labeled arrangement A Its intersection poset L(A)



The Mobius Function for Posets

I The (closed) interval [x, y] of a poset is the set of all points

between x and y, including endpoints:

[x, y] = {z : x ≤ z ≤ y}.

I The Mobius function µ(x, y) is defined (recursively) on the

interval [x, y] by the two properties:

1. µ(x, x) = 1.

2. x < y ⇒∑z∈[x,y] µ(x, z) = 0



Some Mobius Function Values

1

-1 -1 -1 -1

2 1 1 1

The values of the Mobius functions µ(0, x)



The Characteristic Polynomial

I We define the characteristic polynomial associated to the

arrangement A by

χ(A, q) =∑

x∈L(A)

µ(0, x)qdim(x)

.



The Characteristic Polynomial of an Arrangement

1

-1 -1 -1 -1

2 1 1 1

χ(A, q) =∑

x∈L(A) µ(0, x)qdim(x) = q2 − 4q + 5.



Zaslavsky’s Theorem

Theorem (Zaslavsky, 1975)

With the definitions above

r(A) =∑

x∈L(A)

|µ(0, x)| = |χ(A,−1)|

b(A) =

∣∣∣∣∣∣∑

x∈L(A)

µ(0, x)

∣∣∣∣∣∣ = |χ(A, 1)|



Zaslavsky’s Theorem–continued

Idea of Proof.

One can show that, for any triple (A,A′,A′′) of arrangements,

χ(A, q) = χ(A′, q)− χ(A′′, q),

from which it follows that |χ(A,−1)| and r(A) satisfy the same

recurrence. As they agree on the empty set, the claim follows. A

similar argument works for b(A).



Counting Regions via Zaslavsky’s Theorem

1

-1 -1 -1 -1

2 1 1 1

χ(A, q) = q2 − 4q + 5

r(A) = |χ(A,−1)| = 10 b(A) = |χ(A, 1)| = 2


Finite Field Method

Finite Field Method

I Recall that the defining equation of a hyperplane can be

written a1x1 + · · · anxn = b for some real numbers

{a1, a2, . . . , an, b}.I In many cases of interest the numbers a1, a2, . . . , an, b are

integers.

I When this holds there is a particularly nice way to compute

the characteristic polynomial.

I For any positive integer q let Aq denote the hyperplane

arrangement A with defining equations reduced mod q.

I Then we have the following result.


Finite Field Method

The Characteristic Polynomial for Integral Arrangements

Theorem (Crapo and Rota (1971), Orlik and Terao (1992),

Athanasiadis (1996), Bjorner and Ekedahl (1996))

For q a sufficiently large prime

χ(A, q) = #

Fnq −

⋃H∈Aq

H

where Fn

q denotes the vector space of dimension n over the finite

field with q elements.


Finite Field Method

Remarks

I Identifying Fnq with {0, 1, . . . , q − 1}n = [0, q − 1]n, χ(A, q) is

the number of points in [0, q − 1]n that do not satisfy modulo

q the defining equations of any of the hyperplanes in A.

I We need large q to avoid lowering the rank of the defining

matrix, but as both sides are polynomials in q, the two sides

will agree for all q.


Finite Field Method

Reflection Arrangements

I Consider the following families of arrangements:

An = {xi − xj = 0|1 ≤ i ≤ j ≤ n}Dn = An ∪ {xi + xj = 0|1 ≤ i ≤ j ≤ n}Bn = Dn ∪ {xi = 0|1 ≤ i ≤ n}

I These are examples of reflection arrangements associated to

finite Coxeter groups of types An−1, Dn, and Bn, respectively.


Finite Field Method

Computing the Characteristic Polynomial I

I What is χ(An)?

I According to the finite field method, we want the number of

points in [0, q − 1]n satisfying xi 6= xj for all 1 ≤ i ≤ j ≤ n.

I This is the same thing as asking for vectors (x1, x2, . . . , xn)all of whose entries are distinct mod q.

I Well, we can pick x1 in q ways, then x2 in q − 1 ways, and so

on. Thus

χ(An, q) = q(q − 1)(q − 2) · · · (q − n+ 1).

I It follows that r(An) = n! (and b(An) = 0).


Finite Field Method

Computing the Characteristic Polynomial II

I What is χ(Bn)?

I Now we want to count the points satisfying xi 6= xj ,

xi 6= −xj , and xi 6= 0.

I Since we do not allow 0, there are only q− 1 (nonzero) choices

for the first entry, q − 3 nonzero choices for the second entry

(because we must avoid the first entry and its negative), etc..

I Thus

χ(Bn, q) = (q − 1)(q − 3) · · · (q − 2n+ 1).

I It follows that r(Bn) = 2nn! (and b(Bn) = 0).


Finite Field Method

The Catalan Arrangement

Cn = {xi − xj = −1, 0,+1|1 ≤ i ≤ j ≤ n}

C3 (projected)


Finite Field Method

The Catalan Arrangement

I Using the finite field method you can show that

χ(Cn, q) = q(q − n− 1)(q − n− 2) · · · (q − 2n+ 1)

I Hence

r(Cn) = n!Cn and b(Cn) = n!Cn−1

where

Cn =1

n+ 1

(2nn

)is the nth Catalan number.

I As of April 1, 2008, Richard Stanley has listed 164

combinatorial interpretations of Cn on his website.


Finite Field Method

Fundamental Open Question

When does the characteristic polynomial factor completely over the

integers?

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