The Steinberg module and the Hecke algebra [expository notes]

15 Pages·2012·0.353 MB·English

by  Neil P. Strickland

THE STEINBERG MODULE AND THE HECKE ALGEBRA
N.P.STRICKLAND
1. Introduction
This note aims to give a self-contained exposition of the Steinberg module and the Hecke algebra for
GL (F ), aiming towards the applications in algebraic topology. We have tried to use methods that are
n p
elementary, or failing that, familiar to topologists.
I have not tried to investigate the history of these ideas but it seems likely that it goes something like
this.
(a) People worked out how to solve problems by matrix calculations, including row-reduction.
(b) Thinking more abstractly, other people worked out the ideas presented here.
(c) Generalisingfurther, peopledevelopedastillmoreabstracttheoryofCoxetergroups, BN pairsand
so on.
There is plenty of current literature written at levels (a) and (c), but not so much at level (b). Everything
that we say is well-known to those who have digested the more general theory, so our purpose is purely
expository.
To avoid trivialities, we assume n>1 unless otherwise stated.
2. Length of permutations
For any permutation σ ∈Σ , we put
n
L(σ)={(i,j)|0<i<j ≤n and σ(i)>σ(j)}
L+(σ)=L(σ)(cid:113){(i,i)|0<i≤n}
={(i,j)|0<i≤j ≤n and σ(i)≥σ(j)}
l(σ)=|L(σ)|.
The integer l(σ) is called the length of σ.
Example 2.1. For σ =(135)(24) we have
L(σ)={(1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)},
so l(σ)=7.
It is sometimes convenient to reformulate this slightly. We put
L(σ)={{i,j}|(i,j)∈L(σ)}.
Equivalently, L(σ) is the set of subsets T ⊆{1,...,n} such that |T|=2 and σ: T →− σT is order-reversing.
As i<j for all (i,j)∈L(σ) we see that L(σ) bijects with L(σ), so |L(σ)|=l(σ).
Remark 2.2. Put ∆ = (cid:81) (x − x ) ∈ Z[x ,...,x ]. One can then see that for σ ∈ Σ we have
i<j j i 1 n n
σ.∆=(−1)l(σ)∆. Usingthis, wefindthatthemapσ (cid:55)→(−1)l(σ) isanontrivialhomomorphismΣ →− {±1},
n
which must therefore be the same as the signature.
Lemma 2.3. l(σ−1)=l(σ).
Proof. σ induces a bijection L(σ)→− L(σ−1). (cid:3)
Definition 2.4. For i=1,...,n−1 we let s denote the transposition that exchanges i and i+1.
i
Proposition 2.5. l(σ) is the least r such that σ can be written in the form s s ...s .
i1 i2 ir
1
Lemma 2.6. For permutations σ,τ ∈Σ we have
L(στ)=L(τ)∆τ−1L(σ)
∗
(where A∆B is the symmetric difference, (A∪B)\(A∩B).)
Proof. Consider a pair of permutations σ,τ. The composite
τ σ
{i,j}−→τ {i,j}={τ(i),τ(j)}−→σ τ {i,j}
∗ ∗ ∗
is order-reversing iff precisely one of the two composed maps is order-reversing. (cid:3)
Lemma 2.7. If σ(k)<σ(k+1) then l(σs )=l(σ)+1, otherwise l(σs )=l(σ)−1.
k k
Proof. Now take τ = s in the previous lemma, so L(τ) = {{k,k+1}}. It follows that l(σs ) is l(σ)−1 if
k k
{k,k+1}∈s−1L(σ), and l(σ)+1 otherwise. As s {k,k+1}={k,k+1} we have {k,k+1}∈s−1L(σ) iff
k∗ k∗ k∗
{k,k+1}∈L(σ) iff σ(k)>σ(k+1), as required. (cid:3)
Proof of Proposition 2.5. Ifσ =s s ...s ,thenitisimmediatefromthelemmathatl(σ)≤r. Conversely,
i1 i2 ir
suppose we have l(σ) = r. If r = 0 then σ is order-preserving and so must be the identity, which is
compatible with the claim in the proposition. If r > 0 then σ is not order-preserving, so there must exist
k with σ(k+1) < σ(k). The lemma tells us that l(σs ) = r−1. We may thus assume inductively that
k
σs = s s ...s for some i ,...,i . It follows that σ = s ...s s , which is an expression of the
k i1 i2 ir−1 1 r−1 i1 ir−1 k
required form. (cid:3)
Definition 2.8. Consider a word w =s s ...s in the variables s ,...,s . The corresponding permu-
i1 i2 ir 1 n−1
tation then has length at most r. If the length is precisely r, we say that w is reduced.
Definition2.9. Wewriteρforthepermutationρ(i)=n+1−i,whichsatisfiesρ2 =1. Notethatρ(i)>ρ(j)
iff i<j, so l(ρ)=n(n−1)/2.
Definition 2.10. For m≤k we put tk =s s ...s (to be interpreted as the identity when m=k).
m m m+1 k−1
In cycle notation, this is
tk =(m, m+1, ...,k−1, k).
m
Proposition 2.11. For any σ ∈Σ there is a unique sequence m ,...,m with 1≤m ≤k and
n 1 n k
σ =tn tn−1 ...t2 t1 .
mn mn−1 m2 m1
Moreover, we have l(σ)=(cid:80)n (k−m ).
k=1 k
Proof. Put m = σ(n) and τ = (tn )−1σ. Then τ(n) = n, so τ can be regarded as an element of Σ , so
n mn n−1
by induction we have
τ =tn−1 tn−2 ...t2 t1
mn−1 mn−2 m2 m1
for some m ,...,m , and l(τ)=(cid:80)n−1(k−m ). It follows that
n−1 1 k=1 k
σ =tn τ =tn tn−1 ...t2 t1 .
mn mn mn−1 m2 m1
We also have
L(tn )={{i,n}|m ≤i<n},
mn n
so
τ−1L(tn )={{τ−1(i),n}|m ≤i<n},
∗ mn n
whereas L(τ) contains no pairs of the form {i,n}. Thus L(τ) and τ−1L(tn ) are disjoint, and
∗ mn
n
(cid:88)
l(σ)=|L(τ)|+|τ−1L(tn )|=l(τ)+(n−m )= (k−m ).
∗ mn n k
k=1
(cid:3)
Definition 2.12. Let Σ(cid:101)n be the group freely generated by symbols s1,...,sn−1 subject to the relations
s2 = 1 and s s s = s s s and s s = s s whenever |i−j| > 1. It is straightforward to check that
i i i+1 i i+1 i i+1 i j j i
the corresponding relations hold in Σn, so there is a canonical map (cid:15): Σ(cid:101)n → Σn sending si to si. We will
use the notation tkm for the element smsm+1...sk−1 in Σ(cid:101)n as well as the corresponding element in Σn.
2
Proposition 2.13. The map (cid:15): Σ(cid:101)n →Σn is an isomorphism.
Proof. Let Xn ⊆Σ(cid:101)n be the set of elements of the form tnmntnm−n1−1...t1m1 with 1≤mk ≤k, so |Xn|≤n!. We
seefromProposition2.11that(cid:15): X →Σ issurjective, andhencebijectivebyacountingargument. Itwill
n n
therefore suffice to show that Xn = Σ(cid:101)n. Note that Xn = (cid:83)nm=1tnmXn−1, and we may assume by induction
that Xn−1 = Σ(cid:101)n−1, so Xn is closed under right multiplication by Σ(cid:101)n−1. As Σ(cid:101)n is generated by Σ(cid:101)n−1 and
s , it will suffice to show that X is closed under right multiplication by s . This can be deduced from
n−1 n n−1
Lemma 2.14 below, after noting that s commutes with tk for all k <n−1. (cid:3)
n−1 m
Lemma 2.14. If k ≤n and l<n then in Σ(cid:101)n we have
(cid:40)
tn tn−1 for k ≤l
tntn−1s =tntn = l+1 k
k l n−1 k l tntn−1 for k >l.
l k−1
Proof. (a) It is immediate from the definitions that tntn−1s =tntn.
k l n−1 k l
(b) We claim that for l < n we have (tn )−1tn = tn−1(tn)−1. We first give a proof for the case l = 5
l+1 l l l
and n=10, writing k instead of s for brevity, and e for the identity permutation.
k
(t10)−1t10 =987656789
6 5
=987565789 (656=565)
=598767895 ([5,7]=[5,8]=[5,9]=e)
=598676895 (767=676)
=569878965 ([6,8]=[6,9]=e)
=569787965 (878=787)
=567989765 ([7,9]=e)
=567898765 (989=898)
=t9(t10)−1
5 5
This pattern can be converted to a formal proof as follows. The claim is clear when l=n−1, so we
can work by downwards induction on l. Consider the relation s s s =s s s . We multiply on
l+1 l l+1 l l+1 l
therightbytn andontheleftby(tn )−1,notingthat(tn )−1s =(tn )−1 ands s tn =tn
l+2 l+2 l+2 l+1 l+1 l l+1 l+2 l
and that s commutes with tn . This gives
l l+2
(tn )−1tn =s (tn )−1s tn s .
l+1 l l l+2 l+1 l+2 l
We can use the relation s tn =tn and the induction hypothesis to convert the right hand side
l+1 l+2 l+1
to s tn−1(tn )−1s , and from the definitions, this is the same as tn−1(tn)−1.
l l+1 l+1 l l l
(c) Now suppose that k ≤ l. Note that tm = tltm and that tl commutes with tm . We can therefore
k k l k l+1
multiply the identity in (b) on the left by tl to get (tn )−1tn =tn−1(tn)−1. This can be rearranged
k l+1 k k l
to give the case k ≤l of the lemma.
(d) Now suppose we have n ≥ i > j. Take k = j and l = i−1 in (c) to get (tn)−1tn = tn−1(tn )−1.
i j j i−1
From the definitions, the right hand side can be rewritten as tn(tn−1)−1, and the equation can then
j i−1
be rearranged to give tntn =tntn−1. Up to a change of notation, this is the same as the second case
i j j i−1
of the lemma.
(cid:3)
(cid:96)
Definition 2.15. LetW bethesetofwordsoflengthr inletterss ,...,s , andputW = W , whichis
r 1 n r r
thefreemonoidgeneratedbys ,...,s . Let∼ betheequivalencerelationonW generatedbythefollowing
1 n r r
rules:
• us s v ∼us s v if |i−j|>1
i j j i
• us s s v =us s s v if |i−j|=1.
i j i j i j
3
(cid:96)
Put M = W / ∼ and M = M , so M is the quotient monoid of W by the relations s s = s s (for
r r r r r i j j i
|i−j|>1) and s s s =s s s (for |i−j|=1). We then have
i j i j i j
Σ=M/(cid:104)s2 =1|i=1,...,n−1(cid:105)
i
Let
π(cid:48) (cid:47)(cid:47)
W(cid:65) M
(cid:65)π(cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:65)(cid:32)(cid:32) (cid:126)(cid:126)(cid:125)(cid:125)(cid:125)(cid:125)(cid:125)π(cid:125)(cid:125)(cid:48)(cid:48)(cid:125)
Σ
be the obvious projection maps. Recall that a word w =s ···s ∈W is reduced iff l(π(w))=r. We write
(cid:96)p1 pr
R forthesetofreducedwordsoflengthr,andputR= R . Notethatifu∈R andv ∈W andu∼ v
r r r r r r
then v ∈R .
r
Theorem 2.16. Let u,v ∈R be such that π(u)=π(v); then u∼ v (or equivalently π(cid:48)(u)=π(cid:48)(v)).
r r
The proof will be given after some preliminaries.
Definition 2.17. Given σ,τ ∈Σ, we say that σ could end with τ if the following equivalent conditions are
satisfied:
(a) There exists ρ∈Σ such that σ =ρτ and l(σ)=l(ρ)+l(τ).
(b) l(στ−1)=l(σ)−l(τ).
(c) There are reduced words u,v such that π(v)=τ and π(uv)=σ.
(d) L(τ)⊆L(σ).
(It is straightforward to check that (a) to (c) are equivalent, and it follows from Lemma 2.6 that (a) is
equivalent to (d).)
Lemma 2.18. Suppose that σ ∈Σ and that σ could end with s , and also σ could end with s , where j (cid:54)=i.
i j
(a) If |i−j|>1 then σ could end with s s =s s .
i j j i
(b) If |i−j|=1 then σ could end with s s s =s s s .
i j i j i j
Proof. Asσcouldendwiths wehaveL(s )={{i,i+1}}⊆L(σ),soσ(i)>σ(i+1). Similarlyσ(j)>σ(j+1).
i i
If |i−j| > 1 one checks that L(s s ) = {{i,i+1},{j,j +1}}, so σ could end with s s . Suppose instead
i j i j
that |i−j|=1; we may assume without loss that j =i+1. We have σ(i)>σ(i+1)=σ(j)>σ(j+1), so
{{i,i+1},{i+1,i+2},{i,i+2}}⊆L(σ).
The permutation τ = s s s = s s s is then the 3-cycle (i,i+1,i+2), so L(τ) = {{i,i+1},{i+1,i+
i j i j i j
2},{i,i+2}}, so σ could end with τ, as claimed. (cid:3)
Proof. Proof of Theorem 2.16 We work by induction on r, noting that the cases r = 0 and r = 1 are easy.
We may thus suppose that r > 1 and that u = xs , v = ys for some x,y ∈ R and i,j ∈ {1,...,n−1}.
i j r−1
Put
σ =π(u)=π(v)=π(x)s =π(y)s .
i j
If i = j we see that π(x) = π(y), so x ∼ y by the induction hypothesis, so u = xs ∼ ys = ys = v as
i i j
required. If|i−j|=1weseefromLemma2.18thatthereexistsz ∈R withσ =π(zs s s )=π(zs s s ).
r−3 i j i j i j
The i=j case now tells us that u=xs ∼zs s s and zs s s ∼ys =v, and from the definitions we have
i i j i j i j j
zs s s ∼zs s s so u=v. A very similar argument works when |i−j|>1. (cid:3)
i j i j i j
3. Standard notation
In some parts of our exposition, it will be convenient to work with an arbitrary finite-dimensional vector
space W over F . In others, it will be convenient to take W =F n. In that context, we let e ,...,e be the
p p 1 n
4
standard basis, and put E =F {e ,...,e }. We let G=GL (F ) be the automorphism group of F n, and
i p 1 i n p p
put
T ={g ∈G|ge ∈F e for all i}
i p i
={invertible diagonal matrices }
B ={g ∈G|gE =E for all i}
i i
={invertible upper triangular matrices}
U ={g ∈B |ge =e (mod E )}
i i i−1
={upper unitriangular matrices}.
Given g ∈ G we let g denote the element in the i’th row and j’th column of the corresponding matrix. If
ij
n=3, a typical element of G then has the form
(cid:104)g11 g12 g13(cid:105)
g21 g22 g23
g31 g32 g33
(cid:80) (cid:80) (cid:80)
With this convention, we have (gh) = g h and g.e = g e and (g.x) = g x . Moreover, we
ik j ij jk j i ij i i j ij j
have
T ={g ∈G|g =0 whenever i(cid:54)=j}
ij
B ={g ∈G|g =0 whenever i>j}
ij
U ={g ∈B |g =1 for all i}.
ii
WeregardΣ asasubgroupofGintheusualway,soσ.e =e and(σ.x) =x . Thecorresponding
n i σ(i) i σ−1(i)
matrix elements are σ =δ . Note that B fits in a split extension U →− B →− T, and we have
ij i,σ(j)
|T|=(p−1)n
|U|=pn(n−1)/2
|B|=(p−1)npn(n−1)/2.
4. Jordan permutations
Let W be a vector space of dimension n over F . We write Flag(W) for the set of complete flags
p
U = (U < U < ... < U = W) (where necessarily dim(U ) = i for all i). Now suppose we have two flags
0 1 n i
U,V ∈Flag(W). For 0<i,j ≤n we put
U ∩V
Q = i j .
ij (U ∩V )+(U ∩V )
i−1 j i j−1
One checks that the natural map from this to
U +(U ∩V )
i−1 i j
U +(U ∩V )
i−1 i j−1
is an isomorphism. Thus, the groups Q ,...,Q are the quotients in the filtration of the one-dimensional
i,1 i,n
space U /U by the groups (U +(U ∩V ))/U . It follows that for each i there is a unique index
i i−1 i−1 i j i−1
j = σ(i) such that Q (cid:54)= 0. Similarly, for each j, there is a unique i = τ(j) such that Q (cid:54)= 0. It
i,σ(i) τ(j),j
follows that σ and τ are inverse to each other, so both lie in Σ . We write δ(U,V) for σ, and note that
n
δ(V,U)=δ(U,V)−1.
Example 4.1. We claim that δ(σE,E) = σ. This is more or less clear except perhaps for the fact that it
is σ and not σ−1. To check this we put
U =σE =span{e ,...,e }=span{e |σ−1(k)≤i}
i i σ(1) σ(i) k
and V =E =span{e ,...,e }, and we consider the quotients
j j 1 j
U ∩V
Q = i j
ij (U ∩V )+(U ∩V )
i−1 j i j−1
5
Put A ={e |σ−1(k)≤i} and B ={e |k ≤j}. Then A ∩B is a basis for U ∩V , so the set
i k j k i j i j
C =(A ∩B )\((A ∩B )∪(A ∩B )
ij i j i−1 j i j−1
is a basis for Q . However, we have
ij
C =(A \A )∩(B \B )={e }∩{e },
ij i i−1 j j−1 σ(i) j
so Q =0 unless j =σ(i).
ij
Example 4.2. Take W =F 5. Given elements a,...,g ∈F , put
p p
(cid:34)a(cid:35) (cid:34)e(cid:35) (cid:34)1(cid:35) (cid:34)0(cid:35) (cid:34)0(cid:35)
b f 0 1 0
u1 = c u2 = g u3 = 0 u4 = 0 u5 = 1 .
d 1 0 0 0
1 0 0 0 0
Then Q is spanned by u , Q is spanned by u , Q is spanned by u , Q is spanned by u , Q is
15 1 24 2 31 3 42 4 53
spanned by u , and all other Q ’s are zero. It follows that δ(U,E)=(153)(24).
5 ij
Remark 4.3. As the definition of δ(U,V) is completely natural, we have δ(gU,gV) = δ(U,V) for all
U,V ∈Flag(W) and g ∈Aut(W).
Remark 4.4. It is not true in general that
δ(T,V)=δ(T,U)δ(U,V).
Indeed, if n = 2 then Flag(V) is natually identified with the set PV of one-dimensional subspaces of V. If
we let ρ be the nontrivial element of Σ , then δ(L,L) = 1, and δ(L,M) = ρ whenever L (cid:54)= M. It follows
2
that if L, M and N are all distinct, then
δ(L,N)(cid:54)=δ(L,M)δ(M,N).
We now consider the cases where δ(U,V) is the identity or one of the adjacent transpositions s . These
i
could also be extracted from the more general theory in the next section but it is instructive to treat them
directly.
Lemma 4.5. Suppose we have flags U,V with δ(U,V)=σ. If U =V then σ(i)=i iff U =V .
i−1 i−1 i i
Proof. Put A = U = V , so dim(A) = i−1. We have U ∩V = V ∩V = V = A and similarly
i−1 i−1 i−1 i i−1 i i−1
U ∩V =A so Q =(U ∩V )/A. Thus σ(i)=i iff Q (cid:54)=0 iff U ∩V >A. As U and V have dimension
i i−1 ii i i ii i i i i
i and A has dimension i−1, we have U ∩V >A iff U =V . (cid:3)
i i i i
Corollary 4.6. If δ(U,V)=1 then U =V. (cid:3)
Proposition 4.7. We have δ(U,V)=s iff U =V for all j (cid:54)=i but U (cid:54)=V .
i j j i i
Proof. Suppose that δ(U,V) = s . Lemma 4.5 tells us that U = V for j < i, but U (cid:54)= V . Put A =
i j j i i
U =V . As U (cid:54)=V we see that U ∩V must be strictly smaller than U , but it contains A, which has
i−1 i−1 i i i i i
codimension one in U , so U ∩V =A. Using this we find that Q =(U ∩V )/A. As δ(U,V)=s we
i i i i,i+1 i i+1 i
have Q (cid:54)=0, so dim(U ∩V )≥dim(A)+1=i, but dim(U )=i so U ∩V =U , so U <V . Of
i,i+1 i i+1 i i i+1 i i i+1
course we also V <V and
i i+1
dim(U +V )=dim(U )+dim(V )−dim(U ∩V )=i+i−(i−1)=i+1=dim(V ),
i i i i i i i+1
so V = U +V . Symmetrically, we have U = U +V . For j > i+1 we have s (j) = j, so we can use
i+1 i i i+1 i i i
Lemma 4.5 to show that U =V for all such j.
j j
We leave the converse to the reader. (cid:3)
5. Schubert cells
Given a flag V ∈Flag(W) and a permutation σ ∈Σ we put
n
Y =Y(σ,V)={U |δ(U,V)=σ}⊂Flag(W).
We write Y(σ) for Y(σ,E)⊂Flag(F n). We also put
p
X =X(σ)=U ∩U(σρ)−1,
where ρ(i)=n+1−i as in Definition 2.9.
6
Lemma 5.1. A matrix g =(g )n ∈M (F ) lies in X(σ) iff we have
ij i,j=1 n p

1 if i=j

g = arbitrary if (i,j)∈L(σ−1)
ij
0 otherwise.
In particular, we have |X(σ)|=pl(σ).
Proof. Wehaveg ∈U iffg =1,andg =0wheneveri>j. Next,wehaveg ∈Uτ iffτgτ−1 ∈U iffa =0
ii ij ij
whenever τ(i) > τ(j). In particular, we have g ∈ U(σρ)−1 = Uρσ−1 iff g = 0 whenever ρσ−1(i) > ρσ−1(j),
ij
or equivalently σ−1(i)<σ−1(j). The claim follows. (cid:3)
Proposition 5.2. The map φ: g (cid:55)→ gσE gives a bijection X(σ) →− Y(σ). In particular, we have |Y(σ)| =
pl(σ), and thus |Y(σ,V)|=pl(σ) for any V.
Proof. First note that X(σ)⊆U ⊆B, so for g ∈X(σ) we have g−1E =E, so
δ(gσE,E)=δ(σE,g−1E)=δ(σE,E)=σ,
so the flag φ(g)=gσE lies in Y(σ) as claimed.
Next, observe that the stabiliser of σE in G is Bσ−1, so the stabiliser in X(σ)=U ∩Uρσ−1 is contained
in the group
H =Bσ−1 ∩Uρσ−1 =(B∩Uρ)σ−1.
NowB consistsofuppertriangularmatrices,andUρ consistsoflowerunitriangularmatrices,soB∩Uρ =1,
so H =1. It follows that X(σ) acts freely on σE, so φ: X(σ)→− Y(σ) is injective.
Now put
T =F {e |m≤σ(i), σ−1(m)≥i}≤E .
i p m σ(i)
Let (cid:15) : F n →− F be the i’th coordinate projection.
i p p
Consider a flag V ∈ Y(σ). We claim that there is a unique element v ∈ V ∩T such that (cid:15) (v ) = 1,
i i i σ(i) i
and moreover that v ,...,v is a basis for V over F . We will prove this by induction on i, so we assume
1 i i p
that the corresponding fact holds for all j <i. Put
S =F {v |j <i and σ(j)<σ(i)}≤V ∩E
i p j i σ(i)
The leading terms of the vectors v in S are precisely the vectors e with m<σ(i) but i>σ−1(m). Using
j i m
this, we see that E = S ⊕T , and thus that V ∩E = S ⊕L for some (unique) subspace L ≤ T .
σ(i) i i i σ(i) i i i i
We next claim that S = V ∩E . This is straightforward, given that {v ,...,v } is a basis for V
i i−1 σ(i) 1 i−1 i−1
and the leading term in v is e . It follows that the space
j σ(j)
L (cid:39)(V ∩E )/(V ∩E )
i i σ(i) i−1 σ(i)
has dimension at most one. On the other hand, because δ(V,E) = σ we see that (cid:15) : L →− F must be
σ(i) i p
surjective, so there is a unique element v as described.
i
Nowdefineg: F n →− F n byg(e )=v ,sogσ(e )=v ,sogσ(E)=V. Astheleadingtermofv
p p i σ−1(i) i i σ−1(i)
is e , we have g ∈U. We also have
i
gσ(e )=v ∈T ≤F {e |σ−1(m)≥k}=F {e ,e ,...,e },
k k k p m p σ(k) σ(k+1) σ(m)
soσ−1gσ isalower-triangularmatrix, soρσ−1gσρisupper-triangular, sog ∈Bρσ−1. Wealsoknowthatthe
diagonal entries in g are all equal to 1, so the same is true of the diagonal entries in ρσ−1gσρ, so g ∈Uρσ−1.
This means that g ∈X(σ), and φ(g)=V. We conclude that φ is surjective as well as injective. (cid:3)
Example 5.3. Consider the permutation σ =(153)(24)∈Σ , corresponding to the matrix
5
 0 0 1 0 0 
 0 0 0 1 0 
σ= 0 0 0 0 1 
 0 1 0 0 0 
1 0 0 0 0
Then
L(σ−1)={(1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)},
7
so X(σ) is the set of matrices of the form
 1 0 0 b a 
 0 1 0 d c 
g= 0 0 1 f e 
 0 0 0 1 g 
0 0 0 0 1
For such g, we have
 a e 1 0 0 
 b f 0 1 0 
gσ= c g 0 0 1 
 d 1 0 0 0 
1 0 0 0 0
The columns of this are the vectors u as in Example 4.2, so φ(g) is the flag U considered in Example 4.2.
i
Corollary 5.4. Consider triples (V,U,W) where V is an n-dimensional vector space over F and U and W
p
are complete flags in V. Then such triples are classified up to isomorphism by the invariant δ(U,W)∈Σ. In
particular, if U,W,W(cid:48) ∈ Flag(V) then there is an automorphism g ∈ Aut(V) with gU = U and gW = W(cid:48)
if and only iff δ(U,W)=δ(U,W(cid:48)).
Proof. Let (V,U,W) be a pair as above, and put σ = δ(U,W). It will suffice to show that (V,U,W) (cid:39)
(F n,σE,E). Choose elements w ∈ W \W and define f: F n → V by f(a) = (cid:80) a w . This gives an
p i i i−1 p i i i
isomorphism F n → V sending E to W. This must send some other flag F to U. By naturality we have
p
δ(F,E) = δ(U,W) = σ, so F ∈ Y(σ), so F = xσE for some x ∈ X(σ) ≤ B. The map x−1 now gives an
isomorphism (F n,F,E)→(F n,σE,E). (cid:3)
p p
Corollary 5.5. For a triple (V,U,W) as above, we have δ(U,W)=σ iff there exists a basis v ,...,v such
1 n
that for all i we have
U =span{v ,...,v }
i σ(1) σ(i)
W =span{v ,...,v }.
i 1 i
If so, then the number of such bases is (p−1)npl(σ−1ρ).
Proof. AbasisisthesamethingasanisomorphismF n →V;abasiswithpropertiesasaboveisthesameas
p
an isomorphism (F n,σE,E)→(V,U,W). Thus, such bases exist iff δ(U,W)=σ, and if so, the number of
p
such bases is |Aut(F n,σE,E)|. This automorphism group is just B∩Bσ−1, which is conjugate to Bσ∩B.
p
This fits into a short exact sequence
X(σ−1ρ)=Uσ∩U →Bσ∩B →T,
so |Bσ∩B|=|T||X(σ−1ρ)|=(p−1)npl(σ−1ρ) as claimed. (cid:3)
6. The Bruhat decomposition
(cid:96)
Proposition 6.1. We have G= BσB. Moreover, for each σ ∈Σ we have a bijection
σ∈Σn n
X(σ)×B =(U ∩Uρσ−1)×B →− BσB
given by (g,b)(cid:55)→gσb. We thus have
|BσB|=pl(σ)|B|=pl(σ)+n(n−1)/2(p−1)n.
Proof. Given g ∈G, put π(g)=δ(gE,E)∈Σ . If b,b(cid:48) ∈B then bE =E =b(cid:48)E, so
n
π(b−1gb(cid:48))=δ(b−1gb(cid:48)E,E)=δ(gb(cid:48)E,bE)=δ(gE,E)=π(g).
Moreover, if σ ∈ Σ ≤ G then σE = F {e | j ≤ i}, from which it follows directly that π(σ) = σ. This
n i p σ(j)
means that π(BσB)={σ}.
Conversely, suppose that π(h) = σ. Propsition 5.2 tells us that there is a unique element g ∈ X(σ)
such that gσE = hE. If we put b = (gσ)−1h we find that b ∈ B and h = gσb. In particular, this shows
that h ∈ BσB, so π−1{σ} = BσB, so G = (cid:96) BσB. We also see that our map X(σ)×B →− BσB is
σ∈Σn
surjective. To see that it is also injective, suppose we have h = g(cid:48)σb(cid:48) for some other g(cid:48) ∈ X(σ) and b(cid:48) ∈ B.
As bE =E =b(cid:48)E we see that gσE =g(cid:48)σE. Proposition 5.2 now tells us that g =g(cid:48), and as gσb=g(cid:48)σb(cid:48) we
also have b=b(cid:48). (cid:3)
8
Corollary 6.2. |B∩Bρσ−1|=|BσB|/|U|.
Proof. We have B =T ×U and Tτ =T for all τ ∈Σ , so
n
|B∩Bρσ−1|=|T|.|U ∩Uρσ−1|=|T|.pl(σ).
On the other hand,
|BσB|=pl(σ)|B|=|U||T|pl(σ),
and the claim follows easily. (cid:3)
Corollary 6.3. For each σ ∈Σ we have a bijection
n
B×X(σ−1)=B×(U ∩Uρσ)→− BσB
given by (b,g)(cid:55)→bσg.
Proof. We have a bijection φ: X(σ−1) × B → Bσ−1B given by φ(g,b) = gσb. Define χ: G → G by
χ(g) = g−1. As B and X(σ−1) are groups, they are preserved by χ. We also have χ(Bσ−1B) = BσB. We
therefore have a bijection B×X(σ−1)→BσB given by
(b,g)(cid:55)→χ(φ(χ(g),χ(b)))=(g−1σ−1b−1)−1 =bσg.
(cid:3)
Proposition 6.4. If l(στ) = l(σ) + l(τ) then there is a bijection φ: X(σ) × X(τ) → X(στ) given by
φ(g,h)=ghσ−1 =gσhσ−1.
Proof. Recallfirstthattheconditionl(στ)=l(σ)+l(τ)isequivalenttothefollowing: foranyT ⊆{1,...,n}
with |T|=2, at most one of the two maps
τ σ
T −→τ(T)−→στ(T)
is order-reversing.
We now show that φ(g,h) ∈ X(στ), or equivalently that ghσ−1 ∈ U ∩Uρτ−1σ−1. We are given that
g ∈ U and h ∈ Uρτ−1, so it will suffice to show that hσ−1 ∈ U and g ∈ Uρτ−1σ−1. For the first of these,
recall that the (i,j)’th matrix element in hσ−1 is h . Thus, if hσ−1 (cid:54)∈U, we have some i>j with
σ−1(i),σ−1(j)
h (cid:54)= 0. As h ∈ X(τ) we must have σ−1(i) < σ−1(j) and τ−1σ−1(i) > τ−1σ−1(j). This means
σ−1(i),σ−1(j)
that both the maps
{τ−1σ−1(i),τ−1σ−1(j)}−→τ {σ−1(i),σ−1(j)}−→σ {i,j}
are order-reversing, contrary to our hypothesis; so hσ−1 ∈ U after all. For the second claim, suppose that
g (cid:54)∈Uρτ−1σ−1, so gστ (cid:54)∈Uρ. Then there must exist i<j with g (cid:54)=0. As g ∈X(σ), this means that
στ(i),στ(j)
στ(i)<στ(j) and σ−1στ(i)>σ−1στ(j), or equivalently τ(i)>τ(j). This means that both the maps
τ σ
{i,j}−→{τ(i),τ(j)}−→{στ(i),στ(j)}
are order-reversing, which again gives a contradiction. This completes the proof that φ(g,h)∈X(στ).
We now show that φ is surjective. Consider an arbitrary element k ∈X(στ). Then k ∈B, so kσ ∈BσB.
Corollary 6.1 tells us that there is a unique pair (g,b) ∈ X(σ)×B with gσb = kσ. Similarly, we have
bτ ∈BτB, so there is a unique (h,c)∈X(τ)×B with bτ =hτc. It follows that
kστ =gσbτ =gσhτc=φ(g,h)στc
We can now apply Corollary 6.1 to the permutation στ to deduce that k = φ(g,h) and c = 1. This shows
that φ is surjective, but
|X(στ)|=pl(στ) =pl(σ)+l(τ) =|X(σ)×X(τ)|,
so φ is actually a bijection. (cid:3)
Proposition 6.5. If l(στ)=l(σ)+l(τ) then BσBτB =BστB.
9
Proof. It is clear that BστB ⊆BσBτB, so it will suffice to show that
|BσBτB|≤|BστB|=|B|pl(στ) =|B|pl(σ)pl(τ).
For this we note that BσB =X(σ)σB and BτB =BτX(τ−1) so
BσBτB =X(σ)σBτX(τ−1).
As |X(σ)|=pl(σ) and |X(τ−1)|=pl(τ−1) =pl(τ), this gives the required inequality. (cid:3)
7. Parabolic subgroups
Definition 7.1. ForanysetI ⊆{1,...,n−1},weputΣ =(cid:104)s |i∈I(cid:105)≤Σ . Supposethat{0,...,n}\I =
I i n
{i ,i ,...,i } with 0=i <i <···<i =n. Then Σ preserves the sets,
0 1 r 0 1 r I
(0,i ],(i ,i ],...,(i ,i ].
1 1 2 r−1 r
In fact, it is the largest subgroup of Σ that preserves these sets, so
n
Σ (cid:39)Σ ×Σ ×···×Σ .
I i1 i2−i1 ir−ir−1
Definition 7.2. Next, for any n-dimensional vector space V over F we let Flag (V) be the set of flags
p I
W =(0=W <W <W <···<W =V)
i0 i1 i2 ir
with dim(W )=j for all j ∈Ic. This gives a functor from n-dimensional vector spaces to sets.
j
InthecaseV =F n wehaveanobviousflagE ∈Flag (F n)givenbyE =span{e ,...,e }forallj ∈Ic.
p I p j 1 j
We let P denote the stabiliser of this flag. We also write P for P i} and P for P i,j}, and so on. The
I i { ij {
subgroups P are called standard parabolic subgroups of G. More generally a subgroup P ≤G is parabolic if
I
it contains a conjugate of B.
For example, we have P = B and P = G. In the case n = 7 and I = {1,3,4,6}, the group P
∅ {1,...,n−1} I
consists of invertible matrices of the following shape:
 ∗ ∗ ∗ ∗ ∗ ∗ ∗ 
 ∗ ∗ ∗ ∗ ∗ ∗ ∗ 
 0 0 ∗ ∗ ∗ ∗ ∗ 
 0 0 ∗ ∗ ∗ ∗ ∗ 
 0 0 ∗ ∗ ∗ ∗ ∗ 
 
 0 0 0 0 0 ∗ ∗ 
0 0 0 0 0 ∗ ∗
Here Ic ={0,2,5,7}, and the sizes of the diagonal blocks are 2−0, 5−2 and 7−5.
Proposition 7.3. P =BΣ B.
I I
Proof. Firstly, itisclearfromthedefinitionsthatΣ ≤P andB ≤P soBΣ B ⊆P . Conversely, suppose
I I I I I
thatg ∈P andputU =gE ,sofora∈Ic wehaveU =E . Letσ bethepermutationsuchthatg ∈BσB.
I j j a a
Recall from Section 6 that this is characterised by characterised by Q (cid:54)=0, where
i,σ(i)
Q =(U ∩E )/((U ∩E )+(U ∩E )).
ij i j i−1 j i j−1
Suppose we have a < i ≤ b, with a,b ∈ Ic. We claim that a < σ(i) ≤ b, or in other words that Q = 0
ij
for j ≤ a or j > b. Indeed, if j ≤ a then E ≤ E = U ≤ U , so (U ∩E ) ≤ (U ∩E ), so Q = 0.
j a a i−1 i j i−1 j ij
Similarly, if j > b then E ≥ E = U ≥ U , so (U ∩E ) ≤ (U ∩E ), so Q = 0. This shows that σ
j−1 b b i i j i j−1 ij
preserves the interval (a,b] as claimed, for all a,b∈Ic. It follows that σ ∈Σ . (cid:3)
I
Lemma 7.4. If σ(k)>σ(k+1) then σ ∈BσBs B.
k
Proof. Define g ∈ G by g(e ) = e +e and g(e ) = e for i (cid:54)= k. We claim that g ∈ Bs B. In the case
k k k+1 i i k
n=2 and k =1 this follows from the equation
[11][01](cid:2)1 1 (cid:3)=[10].
01 10 0−1 11
The general case follows by taking block sums with suitable identity matrices. Next, we claim that the
element g =σgσ−1 lies in B. Indeed, we have b(e )=e and b(e )=e for all other i, so the claim
σ(k) σ(k+1) i i
follows from the fact that σ(k+1)<σ(k). We now have σ =b−1σg ∈Bσg and g ∈Bs B so σ ∈BσBs B
k k
as claimed. (cid:3)
10