An introduction to hyperplane arrangements

114 Pages·2006·0.554 MB·English

by  Stanley R.P

An Introduction to Hyperplane
Arrangements
Richard P. Stanley

Contents
An Introduction to Hyperplane Arrangements 1
Lecture 1. Basic de(cid:12)nitions, the intersection poset and the characteristic
polynomial 2
Exercises 12
Lecture 2. Properties of the intersection poset and graphical arrangements 13
Exercises 30
Lecture 3. Matroids and geometric lattices 31
Exercises 39
Lecture 4. Broken circuits, modular elements, and supersolvability 41
Exercises 58
Lecture 5. Finite (cid:12)elds 61
Exercises 81
Lecture 6. Separating Hyperplanes 89
Exercises 106
Bibliography 109
3

IAS/Park CityMathematics Series
Volume 00, 0000
An Introduction to Hyperplane
Arrangements
Richard P. Stanley1;2
1versionofFebruary26,2006
2
TheauthorwassupportedinpartbyNSFgrantDMS-9988459. HeisgratefultoLaurenWilliams
for her careful reading of the original manuscript and many helpful suggestions, and to H(cid:19)el(cid:18)ene
BarceloandGuangfengJiangforanumberofofadditionalsuggestions. Studentsin18.315,taught
atMITduringfall2004,alsomadesomehelpfulcontributions.
1
2 R. STANLEY,HYPERPLANE ARRANGEMENTS
LECTURE 1
Basic de(cid:12)nitions, the intersection poset and the
characteristic polynomial
1.1. Basic de(cid:12)nitions
The following notation is used throughout for certain sets of numbers:
N nonnegative integers
P positive integers
Z integers
Q rational numbers
R real numbers
R positive real numbers
+
C complex numbers
[m] the set 1;2;:::;m when m N
f g 2
Wealsowrite[tk](cid:31)(t)forthecoe(cid:14)cientoftk inthepolynomialorpowerseries(cid:31)(t).
For instance, [t2](1+t)4 =6.
A (cid:12)nite hyperplane arrangement A is a (cid:12)nite set of a(cid:14)ne hyperplanes in some
vector space V =Kn, where K is a (cid:12)eld. We will not consider in(cid:12)nite hyperplane
(cid:24)
arrangements or arrangements of general subspaces or other objects (though they
have many interesting properties), so we will simply use the term arrangement for
a(cid:12)nitehyperplanearrangement. MostoftenwewilltakeK =R,butaswewillsee
even if we’re only interested in this caseit is useful to consider other (cid:12)elds as well.
To make sure that the de(cid:12)nition of a hyperplane arrangement is clear, we de(cid:12)ne a
linear hyperplane to be an (n 1)-dimensional subspace H of V, i.e.,
(cid:0)
H = v V : (cid:11) v =0 ;
f 2 (cid:1) g
where (cid:11) is a (cid:12)xed nonzero vector in V and (cid:11) v is the usual dot product:
(cid:1)
((cid:11) ;:::;(cid:11) ) (v ;:::;v )= (cid:11) v :
1 n 1 n i i
(cid:1)
An a(cid:14)ne hyperplane is a translate J of a linear hypXerplane, i.e.,
J = v V : (cid:11) v =a ;
f 2 (cid:1) g
where (cid:11) is a (cid:12)xed nonzero vector in V and a K.
2
If the equationsof the hyperplanesof A are givenby L (x)=a ;:::;L (x)=
1 1 m
a , where x=(x ;:::;x ) and each L (x) is a homogeneous linear form, then we
m 1 n i
call the polynomial
QA(x)=(L1(x) a1) (Lm(x) am)
(cid:0) (cid:1)(cid:1)(cid:1) (cid:0)
the de(cid:12)ning polynomial of A. It is often convenient to specify an arrangement
by its de(cid:12)ning polynomial. For instance, the arrangement A consisting of the n
coordinate hyperplanes has QA(x)=x1x2 xn.
(cid:1)(cid:1)(cid:1)
Let A be an arrangement in the vector space V. The dimension dim(A) of
A is de(cid:12)ned to be dim(V) (= n), while the rank rank(A) of A is the dimension
of the space spanned by the normals to the hyperplanes in A. We say that A is
essential if rank(A)=dim(A). Suppose that rank(A)=r, and take V =Kn. Let
LECTURE1. BASIC DEFINITIONS 3
Y be a complementary space in Kn to the subspace X spanned by the normals to
hyperplanes in A. De(cid:12)ne
W = v V : v y =0 y Y :
f 2 (cid:1) 8 2 g
If char(K)=0 then we can simply take W =X. By elementary linear algebra we
have
(1) codim (H W)=1
W
\
for all H A. In other words, H W is a hyperplane of W, so the set A :=
W
2 \
H W : H A isanessentialarrangementinW. Moreover,thearrangementsA
f \ 2 g
andA are\essentiallythe same,"meaningin particularthat they havethe same
W
intersectionposet(asde(cid:12)nedinDe(cid:12)nition1.1). LetuscallA theessentialization
W
of A, denoted ess(A). When K =R and we take W =X, then the arrangementA
isobtainedfromA by\stretching"the hyperplaneH W A orthogonallyto
W W
\ 2
W. Thus if W? denotes the orthogonal complement to W in V, then H0 A if
W
2
and only if H0 W? A. Note that in characteristicp this type of reasoningfails
(cid:8) 2
since the orthogonal complement of a subspace W can intersect W in a subspace
of dimension greaterthan 0.
Example1.1. LetAconsistofthelinesx=a ;:::;x=a inK2(withcoordinates
1 k
x and y). Then we can take W to be the x-axis, and ess(A) consists of the points
x=a ;:::;x=a in K.
1 k
Now let K = R. A region of an arrangement A is a connected component of
the complement X of the hyperplanes:
X =Rn H:
(cid:0)
H2A
[
Let R(A) denote the set of regions of A, and let
r(A)=#R(A);
thenumberofregions. Forinstance,thearrangementAshownbelowhasr(A)=14.
It is a simple exercise to show that every region R R(A) is open and convex
2
(continuing to assume K = R), and hence homeomorphic to the interior of an n-
dimensional ball Bn (Exercise 1). Note that if W is the subspace of V spanned by
thenormalstothehyperplanesinA,thenR R(A)ifandonlyifR W R(A ).
W
2 \ 2
We say that a region R R(A) is relatively bounded if R W is bounded. If A
2 \
is essential, then relatively bounded is the same as bounded. We write b(A) for
4 R. STANLEY,HYPERPLANE ARRANGEMENTS
the number of relatively bounded regions of A. For instance, in Example 1.1 take
K = R and a < a < < a . Then the relatively bounded regions are the
1 2 k
(cid:1)(cid:1)(cid:1)
regions a < x < a , 1 i k 1. In ess(A) they become the (bounded) open
i i+1
(cid:20) (cid:20) (cid:0)
intervals(a ;a ). TherearealsotworegionsofAthatarenotrelativelybounded,
i i+1
viz., x<a and x>a .
1 k
A (closed) half-space is a set x Rn : x (cid:11) c for some (cid:11) Rn, c R. If
f 2 (cid:1) (cid:21) g 2 2
H isahyperplaneinRn,then the complementRn H hastwo(open)components
whose closures are half-spaces. It follows that the(cid:0)closure R(cid:22) of a region R of A is
a (cid:12)nite intersection of half-spaces, i.e., a (convex) polyhedron (of dimension n). A
bounded polyhedron is called a (convex) polytope. Thus if R (or R(cid:22)) is bounded,
then R(cid:22) is a polytope (of dimension n).
An arrangement A is in general position if
H ;:::;H A; p n dim(H H )=n p
1 p 1 p
f g(cid:18) (cid:20) ) \(cid:1)(cid:1)(cid:1)\ (cid:0)
H ;:::;H A; p>n H H = :
1 p 1 p
f g(cid:18) ) \(cid:1)(cid:1)(cid:1)\ ;
Forinstance, if n=2 then a set of lines is in generalposition if no twoare parallel
and no three meet at a point.
Let us considersome interestingexamples of arrangementsthat will anticipate
some later material.
Example1.2. LetA consistofmlinesingeneralpositioninR2. Wecancompute
m
r(A ) using the sweep hyperplane method. Add a L line to A (with A L in
m k K
[f g
general position). When we travel along L from one end (at in(cid:12)nity) to the other,
everytime weintersect aline in A wecreateanewregion,and wecreateonenew
k
region at the end. Before we add any lines we have one region (all of R2). Hence
r(A ) = #intersections+#lines+1
m
m
= +m+1:
2
(cid:18) (cid:19)
Example 1.3. The braid arrangement B in Kn consists of the hyperplanes
n
B : x x =0; 1 i<j n:
n i j
(cid:0) (cid:20) (cid:20)
Thus B has n hyperplanes. To count the number of regions when K =R, note
n 2
that specifying which side of the hyperplane x x = 0 a point (a ;:::;a ) lies
i j 1 n
(cid:0) (cid:1) (cid:0)
on is equivalent to specifying whether a < a or a > a . Hence the number of
i j i j
regions is the number of ways that we can specify whether a < a or a > a for
i j i j
1 i < j n. Such a speci(cid:12)cation is given by imposing a linear order on the
(cid:20) (cid:20)
a ’s. In other words, for each permutation w S (the symmetric group of all
i n
2
permutations of 1;2;:::;n), there corresponds a region R of B given by
w n
R = (a ;:::;a ) Rn : a >a > >a :
w 1 n w(1) w(2) w(n)
f 2 (cid:1)(cid:1)(cid:1) g
Hence r(B )=n!. Rarely is it so easy to compute the number of regions!
n
NotethatthebraidarrangementB isnotessential;indeed,rank(B )=n 1.
n n
(cid:0)
When char(K) does not divide n the space W Kn of equation (1) can be taken
(cid:18)
to be
W = (a ;:::;a ) Kn : a + +a =0 :
1 n 1 n
f 2 (cid:1)(cid:1)(cid:1) g
Thebraidarrangementhasanumberof\deformations"ofconsiderableinterest.
We will just de(cid:12)ne some of them now and discuss them further later. All these
LECTURE1. BASIC DEFINITIONS 5
arrangementslie in Kn, and in all of them we take 1 i<j n. The readerwho
(cid:20) (cid:20)
likes a challenge can try to compute their number of regions when K = R. (Some
are much easier than others.)
generic braid arrangement: x x = a , where the a ’s are \generic"
i j ij ij
(cid:15) (cid:0)
(e.g.,linearlyindependentovertheprime(cid:12)eld,soK hastobe\su(cid:14)ciently
large"). Theprecisede(cid:12)nitionof\generic"willbegivenlater. (Theprime
(cid:12)eld ofK isitssmallestsub(cid:12)eld,isomorphictoeitherQorZ=pZforsome
prime p.)
semigeneric braid arrangement: x x =a ,wherethea ’sare\generic."
i j i i
(cid:15) (cid:0)
Shi arrangement: x x =0;1 (so n(n 1) hyperplanes in all).
i j
(cid:15) (cid:0) (cid:0)
Linial arrangement: x x =1.
i j
(cid:15) (cid:0)
Catalan arrangement: x x = 1;0;1.
i j
(cid:15) (cid:0) (cid:0)
semiorder arrangement: x x = 1;1.
i j
(cid:15) (cid:0) (cid:0)
threshold arrangement: x +x =0(notreallyadeformationofthe braid
i j
(cid:15)
arrangement, but closely related).
An arrangement A is central if H = . Equivalently, A is a translate
H2A 6 ;
of a linear arrangement (an arrangement of linear hyperplanes, i.e., hyperplanes
T
passingthroughtheorigin). Manyotherwriterscallanarrangementcentral,rather
than linear, if 0 H. If A is central with X = H, then rank(A) =
2 H2A H2A
codim(X). If A is central, then note also that b(A)=0 [why?].
T T
There are two useful arrangements closely related to a given arrangement A.
If A is a linear arrangement in Kn, then projectivize A by choosing some H A
to be the hyperplane at in(cid:12)nity in projective space Pn(cid:0)1. Thus if we regard 2
K
Pn(cid:0)1 = (x ;:::;x ) : x K; not all x =0 = ;
K f 1 n i 2 i g (cid:24)
where u v if u=(cid:11)v for some 0=(cid:11) K, then
(cid:24) 6 2
H =( (x ;:::;x ;0) : x K; not all x =0 = )=Pn(cid:0)2:
f 1 n(cid:0)1 i 2 i g (cid:24) (cid:24) K
The remaining hyperplanes in A then correspond to \(cid:12)nite" (i.e., not at in(cid:12)nity)
projectivehyperplanesinPn(cid:0)1. Thisgivesanarrangementproj(A)ofhyperplanes
K
in Pn(cid:0)1. When K = R, the two regions R and R of A become identi(cid:12)ed in
K (cid:0)
proj(A). Hence r(proj(A))= 1r(A). When n=3, we can draw P2 as a disk with
2 R
antipodal boundarypoints identi(cid:12)ed. The circumferenceof the disk representsthe
hyperplaneatin(cid:12)nity. Thisprovidesagoodwaytovisualizethree-dimensionalreal
lineararrangements. Forinstance,ifAconsistsofthethreecoordinatehyperplanes
x =0, x =0, and x =0, then a projective drawing is given by
1 2 3
1
2
3
6 R. STANLEY,HYPERPLANE ARRANGEMENTS
Thelinelabellediistheprojectivizationofthehyperplanex =0. Thehyperplane
i
at in(cid:12)nity is x = 0. There are four regions, so r(A) = 8. To draw the incidences
3
amongalleightregionsofA,simply\re(cid:13)ect"theinteriorofthedisktotheexterior:
2
1
2
3
1
Regarding this diagram as a planar graph, the dual graph is the 3-cube (i.e., the
vertices and edges of a three-dimensional cube) [why?].
For a more complicated example of projectivization, Figure 1 shows proj(B )
4
(where we regard B as a three-dimensional arrangement contained in the hyper-
4
plane x +x +x +x = 0 of R4), with the hyperplane x = x labelled ij, and
1 2 3 4 i j
with x =x as the hyperplane at in(cid:12)nity.
1 4
We now de(cid:12)ne an operation which is \inverse" to projectivization. Let A be
an (a(cid:14)ne) arrangementin Kn, given by the equations
L (x)=a ; :::; L (x)=a :
1 1 m m
Introduce a new coordinate y, and de(cid:12)ne a central arrangementcA (the cone over
A) in Kn K =Kn+1 by the equations
(cid:2)
L (x)=a y; :::; L (x)=a y; y =0:
1 1 m m
For instance, let A be the arrangement in R1 given by x = 1, x =2, and x = 3.
(cid:0)
The following (cid:12)gure should explain why cA is called a cone.