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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
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Outline
Hyperplane Arrangements
Counting Regions
The Intersection Poset
Finite Field Method
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Hyperplane Arrangements
Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Hyperplane Arrangements
A Hyperplane Arrangement
An arrangement of four lines in the plane
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Hyperplane Arrangements
Another Hyperplane Arrangement
An arrangement of three planes in three space
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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).
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Counting Regions
Region Counting
r(A) = 10 b(A) = 2 (shaded)
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Counting Regions
A Triple (A,A′,A′′) of Arrangements
A and H A′ A′′
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Counting Regions
Region Counting Recurrence
Theorem (Zaslavsky, 1975)
r(A) = r(A′) + r(A′′)
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Counting Regions
Two Arrangements
in general position not in general position
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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
).
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Counting Regions
Counting Regions of a General Position Line Arrangement
r(A) = 1 + k +(
k
2
)= 1 + 4 +
(42
)= 11X
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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...
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
Hasse Diagrams
Some posets
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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)
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
Some Mobius Function Values
1
-1 -1 -1 -1
2 1 1 1
The values of the Mobius functions µ(0, x)
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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)
.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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)|
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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).
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
The Intersection Poset
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
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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).
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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).
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Finite Field Method
The Catalan Arrangement
Cn = {xi − xj = −1, 0,+1|1 ≤ i ≤ j ≤ n}
C3 (projected)
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
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.
It All Depends on How You Slice It: An Introduction to Hyperplane Arrangements
Finite Field Method
Fundamental Open Question
When does the characteristic polynomial factor completely over the
integers?