An introduction to arrangements of hyperplanes [Lecture notes]

15 Pages·1999·0.213 MB·English

by  David G. Wagner

AN INTRODUCTION TO ARRANGEMENTS OF
HYPERPLANES
DAVID G. WAGNER
These notes are intended as a \users' guide" to the book Arrange-
ments of Hyperplanes by Orlik and Terao [1]. My aim is to provide
a brief introduction to a few of the main ideas involved, highlighting
the interplay of combinatorics, algebra, topology, and geometry. I cer-
tainly make no attempt to be comprehensive in my treatment. For
some details and many further developments one should go directly to
[1]. Theorem 3.43 in [1] will be referredto here as OT(3.43), and so on.
Many thanks to Tony Geramita for inviting me to present this material
at Queen's University, January 11-13, 1999.
1. Combinatorics
Let k be a (cid:12)eld, and let V be an n-dimensional k-vectorspace. A
hyperplane H in V is an a(cid:14)ne translate of a subspace of codimension
one. A (cid:12)nite set A = fH1;::: ;Hmg of hyperplanes in V is an a(cid:14)ne
hyperplane arrangement. (Similarly, arrangements of hyperplanes in
n n
projective space may be considered. Since P rH ' A , there is no
k k
signi(cid:12)cant novelty in the projective case.)
T
Ana(cid:14)nearrangement A is central if 0 2 A; i.e. if eachhyperplane
in A is a subspace of V. Anoncentral arrangement Amaybe converted
0
to a central one A by the operation of coning: identify V with the
(cid:3) 0 (cid:3) 0(cid:3) 0
hyperplane fz = 1g of V := V (cid:8)kz (here z 2 V is dual to z 2 V )
0 0 0
and for H 2 A let H := spankfv (cid:8)z : v 2 Hg; then de(cid:12)ne A (cid:26) V
by
0 0 (cid:3)
A := fH : H 2 Ag[fker(z )g:
Results on central arrangements can thus be translated into similar
statements on noncentral arrangements. For the purposes of these
notes we therefore concentrate on central arrangements. A central
T
arrangement is essential if f0g = A. By considering the natural in-
T
duced arrangement in the quotient space V= A we may assume that
a central arrangement is essential.
Research supported by the Natural Sciences and Engineering Research Council
of Canada under operating grant OGP0105392.
1
2 DAVID G. WAGNER
a u
(cid:8)H
(cid:17) (cid:8) (cid:10)J H
(cid:17) (cid:8) H
(cid:8) (cid:10) J H
(cid:17) (cid:8) H
(cid:17) (cid:8)u (cid:10)u Ju Hu
Q (cid:17) H H
@H (cid:0) H (cid:0)@ (cid:0)
Q (cid:17) H H
Q(cid:17) b @(cid:0)H (cid:0)H @(cid:0)
(cid:17)Q H H
(cid:0)@ (cid:0)H H(cid:0)@
(cid:17) Q H H
(cid:17) Q (cid:0)u @(cid:0)u H(cid:0)u @Hu
Q a HH b J (cid:10) c (cid:8)(cid:8) d
Q H (cid:8)
Q H J (cid:10) (cid:8)
Q c HHJ(cid:10)(cid:8)u (cid:8)
d
Figure 1. The arrangement with Q(A) = xyz(x (cid:0) y)
and its intersection lattice L(A).
A central arrangement may be described by choosing for each hy-
(cid:3)
perplane H 2 A a linear form (cid:11)H in the dual space V such that
H = ker((cid:11)H). Then A is the variety (zero-locus) of the polynomial
Q
(cid:3)
Q(A) := H2A(cid:11)H in the symmetric algebra S := Sym(V ) of polyno-
mial functions on V. This Q(A) is homogeneous of degree #A and is
de(cid:12)ned only up to a nonzero scalar multiple.
2
The picture at the left of Figure 1 is in the projective plane P at
k
3
in(cid:12)nity of A and represents the four hyperplanes in the 3-dimensional
k
(cid:3)
central arrangement de(cid:12)ned by Q(A) = xyz(x(cid:0)y). (Here V = kx(cid:8)
ky(cid:8)kz.)
The intersection lattice of a central arrangement A is the set L(A)
T
of all subspaces X of V of the form X = B for some subset B (cid:18) A,
T
partially ordered by reverse inclusion. By convention, ? = V, so
this is the unique minimal element of L. Since A is central, L has
T
a unique maximal element T := A (and T = f0g if and only if A
is essential). Figure 1 also illustrates L for the given arrangement.
The partially ordered set L has a join operation X _ Y := X \ Y
given by intersection of sets; since L is (cid:12)nite and also has a unique
minimal element it follows that L is a lattice. The atoms of L are
the hyperlanes H 2 A, and every X 2 L is a join of atoms: that is,
L is atomic. The lattice L has a rank function rk : L ! Z given
by rk(X) := codim(X) := dimk(V) (cid:0) dimk(X): The modular law for
dimensions of subspaces in V yields the lower semimodular inequality
rk(X _Y)+rk(X ^Y) (cid:20) rk(X)+rk(Y)
for all X;Y 2 L. A (cid:12)nite lattice which is atomic, ranked, and lower
semimodular is called a geometric lattice.
HYPERPLANE ARRANGEMENTS 3
For any X 2 L, the interval [X;T] in L is the intersection lattice of
the arrangement
X
A := fX \H : X 6(cid:18) H 2 Ag
in X, called the restriction of A to X. Also, the interval [V;X] in L is
the intersection lattice of the subarrangement
AX := fH : X (cid:18) H 2 Ag
in V. Together with restriction this shows that everyinterval [X;Y]for
X (cid:20) Y in L is itself a geometric lattice. (The same is true in general
for any geometric lattice.)
Many properties of hyperplane arrangements are established induc-
tivelyby themethod of deletion and restriction. For H0 2 A the induc-
0 00 0
tive triple of A with respect to H0 is (A;A;A )inwhichA := ArfH0g
00 H0
is obtained by deletion of H0 and A := A is obtained by restriction
to H0. (For example, we will construct several short exact sequences
associated with an inductive triple of arrangements.)
The M(cid:127)obius function of L is (cid:22) : L ! Z de(cid:12)ned for X 2 L by
induction on rk(X) as follows:
(cid:22)(V) := 1;
and for X > V in L,
X
(cid:22)(X) := (cid:0) (cid:22)(Y):
V(cid:20)Y<X
Figure 2 illustrates the values of the M(cid:127)obius function for the lattice
from Figure 1. More generally, for X (cid:20) Y in L, denote by (cid:22)(X;Y)
the value of the M(cid:127)obius function of the lattice [X;Y] calculated at Y.
Notice that
(cid:26)
X
0 if X 6= Y;
(1.1) (cid:22)(X;Z) =
1 if X = Y;
Z2[X;Y]
by construction.
The Poincar(cid:19)e polynomial of A is
X
rk(X)
(cid:25)(A;t) := (cid:22)(X)((cid:0)t) :
X2L(A)
2 3
So for the example in Figure 1, (cid:25)(A;t) = 1 + 4t + 5t + 2t . In this
example all the coe(cid:14)cients are nonnegative; this is true in general,
which we prove next. In fact this is the Poincar(cid:19)e polynomial of a
skew-commutative graded algebra A(A), the \Orlik-Solomon algebra,"
associated with A (and a commutative ring of coe(cid:14)cients). In the next
section we de(cid:12)ne this algebra and develop some of its properties. In
4 DAVID G. WAGNER
(cid:0)2 u
(cid:8)H
(cid:8) (cid:10)J H
(cid:8) H
(cid:8) (cid:10) J H
2 (cid:8)u (cid:8) 1 (cid:10)u Ju1 HHu1
H H
@H (cid:0) H (cid:0)@ (cid:0)
H H
@(cid:0)H (cid:0)H @(cid:0)
H H
(cid:0)@ H(cid:0) H(cid:0)@
H H
u(cid:0) @u(cid:0) Hu(cid:0) @Hu
H (cid:8)
(cid:0)1 H(cid:0)1J (cid:10)(cid:0)1(cid:8) (cid:0)1
H (cid:8)
H J (cid:10) (cid:8)
H (cid:8)
HJ(cid:10)(cid:8)u
1
Figure 2. The M(cid:127)obius function of L
Section 3 we show that for a complex arrangement A in V the integral
S
singular cohomology ring of the manifold V r A is isomorphic to the
Orlik-Solomon algebra of A with coe(cid:14)cients in Z.
Lemma 1.1 (OT(2.40)). Let A be an arrangement, and let V < Y (cid:20)
Z in L(A). Then
X
(cid:22)(X) = 0:
X: X_Y=Z
Proof. We proceed with Y (cid:12)xed, by induction on rk(Z). For the basis,
Z = Y and the condition X _ Y = Y reduces to X (cid:20) Y. Now, since
V 6= Y we have
X
(cid:22)(X) = 0
X2[V;Y]
by equation (1:1). For the induction step
X X X
(cid:22)(X) = (cid:22)(X)(cid:0) (cid:22)(X)
X: X_Y=Z X: X_Y(cid:20)Z X: X_Y<Z
!
X X X
= (cid:22)(X)(cid:0) (cid:22)(X) :
X(cid:20)Z U<Z X: X_Y=U
The (cid:12)rst summation is zero by (1:1) again, and each inner summation
in the second part is zero by the inductive hypothesis.
Theorem 1.2 (OT(2.47)). Let A be an arrangement. Then for any
Z 2 L(A),
rk(Z)
((cid:0)1) (cid:22)(Z) > 0:
HYPERPLANE ARRANGEMENTS 5
Proof. We proceed by induction on rk(Z); the basis Z = V being
trivial. For the induction step, choose any H 2 A with H (cid:20) Z in L,
and apply Lemma 1.1 with Y = H:
X X
(cid:22)(X) = 0 = (cid:22)(Z)+ (cid:22)(Z):
X: X_H=Z X<Z: X_H=Z
Since L is atomic, Z is a join of atoms, so the last summation is not
empty. By lower semimodularity, if X _H = Z then
rk(Z) = rk(X _H) (cid:20) rk(X _H)+rk(X ^H)
(cid:20) rk(X)+rk(H) = rk(X)+1:
So if X < Z as well, then rk(X) = rk(Z)(cid:0)1. By induction we have
rk(Z)
((cid:0)1) (cid:22)(X) < 0 for all such X, and the result follows.
Corollary 1.3. Let A be an arrangement. Then the coe(cid:14)cients of the
Poincar(cid:19)e polynomial (cid:25)(A;t) are nonnegative integers.
We close this section with a deletion-restriction recurrence relation
0 00
for the Poincar(cid:19)e polynomials of an inductive triple (A;A ;A ) of ar-
rangements.
Lemma 1.4 (OT(2.35)). Let A be an arrangement. For X (cid:20) Y in
L, let U(X;Y) be the set of central subarrangements B (cid:18) A such that
AX (cid:18) B and T(B) = Y. Then
X
#(BrAX)
(cid:22)(X;Y) = ((cid:0)1) :
B2U(X;Y)
Proof. Let (cid:23)(X;Y) denote the value on the right-hand side. Since we
have a disjoint union
[
U(X;Z) = fB (cid:18) A : AX (cid:18) B (cid:18) AYg;
Z2[X;Y]
This yields
X X X
#(BrAX) #C
(cid:23)(X;Z) = ((cid:0)1) = ((cid:0)1) :
Z2[X;Y] AX(cid:18)B(cid:18)AY C(cid:18)AYrAX
For X = Y we obtain (cid:23)(X;X) = 1, and for X < Y the sum is zero
since AX is properly contained in AY. Comparison with the recursion
(1.1) (cid:12)nishes the proof.
Lemma 1.5 (OT(2.55)). Let A be an arrangement. Then
X
#B rk(B)
(cid:25)(A;t) = ((cid:0)1) ((cid:0)t) ;
B(cid:18)A
with the sum over all central subarrangements of A.
6 DAVID G. WAGNER
Proof. From Lemma 1.4 we obtain
X X
#B rk(X)
(cid:25)(A;t) = ((cid:0)1) ((cid:0)t) :
X2LB2U(V;X)
Every central subarrangement B of A is in U(V;X) for exactly one
X 2 L, namely for X = T(B), and for this one, rk(X) = rk(B).
0 00
Theorem 1.6 (OT(2.56)). Let (A;A ;A ) be an inductive triple of ar-
rangements, relative to the hyperplane H0 2 A. Then
0 00
(cid:25)(A;t) = (cid:25)(A;t)+t(cid:25)(A ;t):
0
Proof. Let U be the set of central subarrangements of A which do not
00
contain H0, and let U be the set of central subarrangements of A
which do contain H0. From Lemma 1.5,
X X
#B rk(B) #B rk(B)
(cid:25)(A;t) = ((cid:0)1) ((cid:0)t) + ((cid:0)1) ((cid:0)t) :
B2U0 B2U00
0
The (cid:12)rst of these summations is clearly (cid:25)(A;t), again by Lemma 1.5.
00
For the second summation recall that L is isomorphic to the interval
00
[H0;T] in L, and has rank function rk := rk(cid:0)1. Now
X X X
#B rk(B) #B rk(Y)
((cid:0)1) ((cid:0)t) = ((cid:0)1) ((cid:0)t)
B2U00 Y2L00B2U(H0;Y)
X X
= (cid:0) ((cid:0)1)#(BrAH0)((cid:0)t)rk(Y)
Y2L00B2U(H0;Y)
X
00
1+rk (Y)
= (cid:0) (cid:22)(H0;Y)((cid:0)t)
Y2L00
00
= t(cid:25)(A ;t):
(The penultimate equation follows from Lemma 1.4.)
These proofs remain valid for any ((cid:12)nite) geometric lattice L by
interpreting A as theset of atoms of L and for X 2 L, AX := fH 2 A :
H (cid:20) Xg. (Infact,theresultsofSection2canalso begeneralizedtoany
geometric lattice, but the interpretation as in Section 3 is unknown.)
2. Algebra
We denote by K a commutative ring, over which we will construct
severalalgebras. Consider acentralarrangementAinann-dimensional
k-vectorspace V, and (cid:12)x an arbitrary total order (cid:30) on the hyperplanes
in A. Let feH : H 2 Ag be a set of \symbols" in bijection with
L
A and let E1 := H2AKeH be the free K-module of all K-linear
V L V
m j
combinations of these symbols. Let E := E1 = j=0 E1 be
HYPERPLANE ARRANGEMENTS 7
the exterior algebra of E1. For S = fH1 (cid:30) (cid:1)(cid:1)(cid:1) (cid:30) Hjg (cid:18) A, let
eS := eH1^:::^eHj; in particular e? = 1. There is a unique symmetric
K-bilinear form h(cid:1);(cid:1)i on E making feS : S (cid:18) Ag an orthonormal basis
of E. Let @ : E ! E be the endomorphism of E which is adjoint to the
(cid:0)P (cid:1)
endomorphism H2AeH ^ (cid:1) : E ! E with respect to this bilinear
form. More explicitly, @ is given by
j
X
i(cid:0)1
@(eH1 ^:::^eHj) = ((cid:0)1) (eH1 ^:::^ecHi ^:::^eHj);
i=1
where the notation ecHi indicates that the factor eHi is omitted from
2
the product. It is a well-known and easily-checked fact that @ = 0
and that if E1 6= f0g then (E;@) is an acyclic chain complex. Also, if
V
j j
! 2 Ej := E1 and (cid:17) 2 E then @(!^(cid:17)) = (@!)^(cid:17)+((cid:0)1) !^(@(cid:17)):
T
A subset S (cid:18) A is called independent if rk( S) = #S, otherwise S
is dependent. (This corresponds to linear independence/dependence of
linear forms de(cid:12)ning the hyperplanes in S.) Let I(A) be the ideal of E
generated by the set f@eS : S is a dependent subset of Ag. The quo-
tient K-algebra A(A) := E=I(A) is called the Orlik-Solomon algebra
of A over K.
L
For each X 2 L(A), let EX := S(cid:18)A: TS=X KeS. This gives an
L-grading
M
E = EX
X2L
of E which re(cid:12)nes the usual grading of E by degree:
M
Ej = EX:
X: rk(X)=j
Notice that the ideal I(A) is homogeneous with respect to the L-
T T
grading of E, since if S (cid:18) A is dependent then (S rfHg) = S for
every H 2 S. Therefore the quotient A(A) inherits an L-grading from
E:
M
A(A) = AX(A)
X2L
in which AX := EX=(EX \I(A)).
Lemma 2.1 (OT(3.11)). LetA be acentral arrangement. Then @I(A) (cid:18)
I(A).
Proof. By K-linearity, the following veri(cid:12)cation su(cid:14)ces. Let S;T (cid:18) A
be disjoint, with S dependent. Then @(eT ^ @eS) = (@eT) ^ (@eS) +
#T 2
((cid:0)1) eT ^(@ eS) = (@eT)^(@eS) 2 I(A).
8 DAVID G. WAGNER
By this lemma, we may de(cid:12)ne an endomorphism @ : A(A) ! A(A)
by @(! + I) := @(!) + I for all ! 2 E. Then (A(A);@) is a chain
complex.
Lemma 2.2 (OT(3.13)). Let A be a nonempty central arrangement.
Then the chain complex (A(A);@) is acyclic.
Proof. It su(cid:14)ces to show that ker(@) (cid:18) im(@) in A(A). Fix any H 2 A
and let b = eH + I in A(A). For any a 2 A(A) we have @(b^ a) =
a(cid:0)b^(@a), so if @a = 0 then a = @(b^a).
A subset Q (cid:18) A is a circuit if it is a minimally dependent subset
of A. For S (cid:18) A, let maxS denote the last hyperplane in S, relative
to the (cid:12)xed total order (cid:30) on A. A broken circuit is a subset P (cid:18) A
such that there is an H 2 A r P with maxP (cid:30) H and such that
P [ fHg is a circuit. Let C(A) be the set of all subsets S (cid:18) A such
that S does not contain any broken circuits of A, and let C(A) be the
free K-submodule of E with feS : S 2 C(A)g as a basis. We have an
L-grading on C(A) by putting CX := EX \C(A) for all X 2 L.
The (cid:12)rst main result of this section is that the obvious homomor-
phism : C(A) ! A(A) given by (eS) := eS + I for S 2 C(A) is
a K-module isomorphism. Since preserves the L-grading, to show
that it is an isomorphism it su(cid:14)ces to prove that for each X 2 L, the
restriction X : CX(A) ! AX(A) is an isomorphism.
Lemma 2.3 (OT(3.42)). Let A be a nonempty central arrangement,
and consider @ : E ! E. Then @C(A) (cid:18) C(A) and the chain complex
(C(A);@) is acyclic.
Proof. If S contains no broken circuits of A then every subset of S also
shares this property. From (2.1) it follows that @C(A) (cid:18) C(A). To
completethe proof, it su(cid:14)cesto show that ker(@) (cid:18) im(@)in C(A). To
dothiswerepeattheproof ofLemma2.2,choosing H 2 Atobethelast
hyperplane relative to the total order (cid:30) on A. Since a broken circuit is
obtained by deleting the last element of a circuit, there are no broken
circuits which contain H. Thus, if S contains no broken circuits, then
S[fHg contains no broken circuits. Therefore eH^(cid:1) : C(A) ! C(A).
Finally, for any c 2 C(A) we calculate that @(eH ^c) = c(cid:0)eH ^(@c)
and therefore ker(@) (cid:18) im(@) in C(A).
Lemma 2.4 (OT(3.31), OT(3.40)). Let A be a central arrangement,
and let Y (cid:20) X in L. Then AY(AX) ' AY(A) and CY(AX) = CY(A).
Theorem 2.5 (OT(3.43)). Let A be a central arrangement. The K-
linear homomorphism : C(A) ! A(A) given by (eS) := eS +I for
S 2 C(A) is a K-module isomorphism.
HYPERPLANE ARRANGEMENTS 9
L
Proof. For each 0 (cid:20) j (cid:20) rk(T), let Cj := X: rk(X)=jCX and Aj :=
L
X: rk(X)=jAX, and let j : Cj ! Aj be the restriction of to Cj.
T
We proceed by induction on r := rk(T), where T := A is the
unique maximal element of L(A). For the basis, if rk(T) = 0 then
T = V and A = ?, so C(A) = K = A(A) and is the identity.
For the induction step, for every X 2 L the diagram
X
CX(AX) (cid:0)! AX(AX)
# #
X
CX(A) (cid:0)! AX(A)
commutes, and from Lemma 2:4 the vertical maps are isomorphisms.
By induction on rk(T), it follows that for all X < T in L that X :
CX ! AX is an isomorphism. Thus, for all 0 (cid:20) j < r, it follows that
j : Cj ! Aj is an isomorphism. Finally, we have a commutative
diagram
0 (cid:0)! Cr (cid:0)! Cr(cid:0)1 (cid:0)! (cid:1)(cid:1)(cid:1) (cid:0)! C1 (cid:0)! C0 (cid:0)! 0
# r # r(cid:0)1 # 1 # 0
0 (cid:0)! Ar (cid:0)! Ar(cid:0)1 (cid:0)! (cid:1)(cid:1)(cid:1) (cid:0)! A1 (cid:0)! A0 (cid:0)! 0
in which the rows are the acyclic chain complexes of Lemmas 2:2 and
2:3. Since j is an isomorphism for 0 (cid:20) j < r, the Five Lemma implies
that r is an isomorphism as well.
Thebroken-circuitbasis C(A)andtheideal I(A)providea\standard
monomial theory" for the Orlik-Solomon algebra A(A), which makes
calculation in this ring very convenient. For the example from Figure
1 with the order a (cid:30) b (cid:30) c (cid:30) d on the hyperplanes, the dependent sets
are fa;b;cg and fa;b;c;dg, so fa;b;cg is the only circuit and fa;bg
is the only broken circuit. With E1 := Ka (cid:8) Kb (cid:8) Kc (cid:8) Kd and
V
E := E1, the Orlik-Solomon algebra is thus
A(A) = E=(b^c(cid:0)a^c+a^b; b^c^d(cid:0)a^c^d+a^b^d(cid:0)a^b^c):
The broken-circuit basis of C(A) provides a basis of \standard mono-
mials" for A(A); in this case
f1; a;b;c;d; a^c;a^d;b^c;b^d;c^d; a^c^d;b^c^dg:
The nonstandard monomials are reduced to combinations of the stan-
dard ones using \straightening laws" obtained from the ideal I(A) :
a^b 7! (a(cid:0)b)^c;
a^b^c 7! (a(cid:0)b)^c^c = 0;
a^b^d 7! (a(cid:0)b)^c^d;
a^b^c^d 7! 0^d = 0:
10 DAVID G. WAGNER
Multiplication in A(A) is thus described by multiplying two standard
monomials as in E and reducing the resulting (possibly nonstandard)
monomial using the straightening laws.
Next, we use the broken-circuit bases of Orlik-Solomon algebras to
construct a short exact sequence associated with an inductive triple
0 00 0
(A;A;A ) of arrangements. (Recall that H0 2 A is (cid:12)xed, and A :=
00 H0 0 00 0 00 0 00
ArfH0gandA := A .) Thenotations L;L;L ; A;A;A ; C;C ;C
and so on denote the above constructions applied to the corresponding
0 00 0 00
arrangements A;A;A . In order that the broken circuit bases C;C;C
0 00
be \compatible" choose total orders on A;A;A such that
(cid:15) H0 is the (cid:12)rst hyperplane of A,
0
(cid:15) the order on A (cid:26) A is induced by the order on A, and
0 00 0
(cid:15) for H;K 2 A, if H0 \H (cid:30) H0 \K in A , then H (cid:30) K in A.
0 00
Lemma 2.6 (OT(3.61)). Let (A;A;A ) be an inductive triple of ar-
0
rangements. Then, with the notation above, C (cid:18) C.
0
Proof. Consider anyS (cid:18) A. IfeS 62 C then S contains a brokencircuit
P of A. Thus there is a hyperplane H 2 A such that P [fHg contains
a circuit of A and maxP (cid:30) H. Since H0 is the (cid:12)rst hyperplane of A
0
it follows that H0 (cid:30) H, and so H 2 A, and so S contains the broken
0 0 0
circuit P of A. Therefore,if S 2 C then S 2 C, and henceC (cid:18) C.
0 i
We will identify the cokernel of the inclusion map 0 ! C (cid:0)! C
00 00
as C . To do this, de(cid:12)ne a K-linear map (cid:18) : E ! E as follows. For
0
S = fH1 (cid:30) (cid:1)(cid:1)(cid:1) (cid:30) Hjg (cid:18) A, let
(cid:18)(eS) := 0
and
(cid:18)(eH0 ^eS) := eH0\H1 ^(cid:1)(cid:1)(cid:1)^eH0\Hj:
0 0
Since H0 62 A, it is clear that (cid:18)(C ) = 0.
0 00
Lemma 2.7 (OT(3.62)). Let (A;A;A ) be an inductive triple of ar-
00
rangements. Then, with the notation above, (cid:18)(C) = C .
00
Proof. First we show that (cid:18)(C) (cid:18) C , arguing by contradiction. Sup-
00
pose that there is an S 2 C with (cid:18)(eS) 62 C . Then H0 2 S, from
the de(cid:12)nition of (cid:18), and S contains no broken circuits of A, while
00 00
S := fH0 \ H : H0 6= H 2 Sg does contain a broken circuit P
00 00 00 00 00 00 00
of A . Thus there is an H 2 A with maxP (cid:30) H and P [ fH g
00
a circuit of A . Choose a set P = fH1 (cid:30) (cid:1)(cid:1)(cid:1) (cid:30) Hjg (cid:26) S such that
00
P = fH0 \H1 (cid:30) (cid:1)(cid:1)(cid:1) (cid:30) H0 \Hjg, and also chooses H 2 A such that
00
H0 \H = H . Then P [fH0g is a broken circuit of A; since H0 2 S
we have P (cid:18) S, a contradiction.