Lecture Notes in Mathematics Editors: A. Dold, Heidelberg F. Takens, Groningen
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Paul-Hermann Zieschang
An Algebraic Approach to Association Schemes
~ Springer
Author Paul-Hermann Zieschang
Christian-Albrechts-Universifftt zu Kiel Mathematisches Seminar Ludewig-Meyn-StraBe 4 24098 Kiel, Germany
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Ziesehang, Paul-Hermann: An algebraic approach to association schemes / Paul-Hermann Zieschang. - Berlin ; Heidelberg ; New York ; Barcelona ; Budapest ; Hong Kong ; London ; Milan ; Paris ; Tokyo : S p r i n g e r , 1996 (Lecture notes in mathematics ; 1628) ISBN 3-540-61400-1 NE: GT Mathematics Subject Classification (1991): 05-02, 05E30, 51E12, 51E24, 05B25, 16S50 ISSN 0075-8434 ISBN 3-540-61400-1 Springer-Verlag Berlin Heidelberg New Y o r k
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Introduction
Let X be a set. We define
l x :-- {(x, x)
I e x).
Let r C_ X x X be given. We set
r" := {(y, z) I(z, Y) 9 r}, and, for each x 9 X, we define
xr := {y 9 x I (x, y) 9 r}. Let G be a partition of X • X such that 0 ~ G 9 1x, and assume that, for each g 9 G, g* 9 G. Then the pair (X, G) will be called an association scheme if, for all d, e, f 9 G, there exists a cardinal number age! such that, for all y, z 9 X,
(y, z) 9 f ~
lyd N ze* I = ade].l
In these notes, we shall always say scheme instead of association scheme. The pair (X, G) will always denote a scheme. We shall always write 1 instead of l x . The elements of {adeS I d, e, f 9 G} will be called the structure constants of (X, G).2 Occasionally we shall use the expression regularity condition in order to denote the condition which guarantees the existence of the structure constants. The present text provides an algebraic approach to schemes. Similar to the theory of groups, the theory of schemes will be viewed as an elementary 1 This definition of association schemes is slightly more general than the usual one. Usually one requires additionally at least that IX ] be finite. In this case, the term "homogeneous coherent configuration" is common, too; see [14]. (Also in the present text the finiteness of IX I plays an important role.) It is also often required that, for all d, e, f E G, aa~! = a~al; see, e.g., [1]. The term "association scheme" was introduced in [2]. There it is even required that, for each g E G, g* = g, and this additional condition (which implies that, for all d, e, f E G, age! = a~df ) is often included in the definition of association schemes. In order to emphasize the algebraic treatment, association schemes (in the present sense) were called "generalized groups" in [34]. 2 In the literature, the structure constants are also called "intersection numbers".
VI algebraic theory which is naturally connected to certain geometric structures. In fact, the theory of schemes generalizes naturally the theory of groups. Let us first see to what extent the class of groups m a y be viewed as a distinguished class of schemes. For each g E G, we abbreviate ng : : agg, 1.
A n o n - e m p t y subset F of G will be called thin if {1} = {n! I f E F } . (Note that we always have nl = 1.) T h e pair (X, G) will be called thin if G is thin. Let E, F C_ G be given. We define
EF:=lgeVl
:aojg r eEE f E F
and call it the complex product of E and F. It follows readily from the definition of the complex product that, if (X, G) is thin, G ( X , G ) := {{g} I g E G} is a group with respect to the complex multiplication, with {1} as identity element. Conversely, let O be a group. For each 0 E O, we define /} := {((, r/) E O x O 1(0 = 77}, and we set 0 : = { 5 10EO}.Then
7-(o) := (o, 0) is a thin scheme, a Now it is readily verified that, if (X, G) is thin,
T(G(X, G)) ~- (X, O). ~ Moreover, for each group O,
~ T h e elementary proofs of these two facts will be given (iv) in the a p p e n d i x of these notes. W h a t is i m p o r t a n t these two facts allow us to identify each group with its scheme. Therefore, we m a y view the class of groups as a of schemes, namely as the class of thin schemes.
as T h e o r e m A(iii), here for us is that corresponding thin distinguished class
3 This is e..._~y to see. First of all, it is clear that i = l s and that, for each 0 E O, t~* = 0 -1. But also the regularity condition is easily verified for the pair (O, 6). Let fl, 7, e, (, 17 E 19 be given, and assume that (fl,7) E 0- Then, fl~ n 7(" 76 0 if and only if fie = 7( -1 if and only if fie( = 7 if and only if r = r/. Thus, [fl~n 7('1 = ~,r where ~ is the Kronecker delta. 4 In Section 1.7, we shall say what it means for two schemes to be isomorphic.
VII There is still another class of important mathematical objects which can be viewed as a distinguished class of schemes, namely the buildings. Buildings were introduced by J. TITS in [27; (3.1)] as a particular class of chamber complexes. ~ Later, in [28; Theorem 2], Tits characterized the buildings as a particular class of "chamber systems". This characterization indicated already a strong relationship between buildings and schemes. In fact, due to this characterization, it is only a small step to see that, like groups, buildings can be viewed naturally as a distinguished class of schemes. Moreover, the embedding of the buildings into the class of schemes is similar to the one of groups. In other words, there exists a class of schemes which for the buildings plays exactly that role which is played by the class of thin schemes for the groups. Let us give here a rough idea of the definition of this class of schemes. In Section 1.4, we shall define what it means for an element of G to be a "(generalized) involution". The set of all involutions of G will be denoted by Inv(G). For each L C_ Inv(G), we shall define in Section 5.1 what it means for (X, G) to be a "Coxeter scheme with respect to L". Now the pair (X, G) will be called a Coxetev scheme if there exists L C_ Inv(G) such that (X, G) is a Coxeter scheme with respect to L. The Coxeter schemes form the class of those schemes which represent the buildings within the class of all schemes. More precisely, the following three statements hold. First, if (X, G) is a Coxeter scheme, there exists a natural way to construct a building
B(X,G) from (X, G). Secondly and conversely, to each building $, say, there is associated naturally a Coxeter scheme which we shall denote by
Finally, if (X, G) is a Coxeter scheme,
A(z(x, G)) = (x, a). Moreover, for each building 0, =
g.
5 In the hterature, the definition of buildings changes occasionally. In these notes, buildings are always understood to be regular. Viewing buildings as chamber systems (and we shall do that always in this text) we say that a building in the sense of [28] is regularif any two members of one of the defining partitions have the same cardinality. It is an easy exercise in the theory of buildings to prove that buildings in the sense of [27] are regular. It is also obvious that thin buildings in the sense of [28] are regular. In particular, the class of buildings in the sense of the present text contains strictly the class of buildings in the sense of [27].
VllI Occasionally we shall speak of the (B,.A)-correspondence in order to denote the above-mentioned correspondence between buildings and Coxeter schemes. Similarly, the above-mentioned correspondence between groups and thin schemes will be called the (~, T)-eorrespondence. The (B,.A)-correspondence is not as easy to describe as the (9, T)correspondence. Therefore we shall not give the details here. They will be given as Theorem E in the appendix of these notes. There is a simple and natural way to identify Coxeter groups and thin buildings; see [28; (2.3.1)]. Modulo this identification, the (B,A)-correspondence and the (9, T)-correspondence coincide on the class of the thin Coxeter schemes. More precisely, the following two statements are true. If (X, G) is a thin Coxeter scheme, G(X, G) = B(X, G). For each Coxeter group O, T(O) = A(O). As a consequence, thin Coxeter schemes correspond to Coxeter groups. 6 Viewing groups and buildings as the cornerstones of the class of schemes it seems to be promising to develop a structure theory of schemes between these cornerstones. The object of the present lecture notes is in the first place to develop a treatment of schemes analogous to that which has been so successful in the theory of finite groups. As indicated in the first footnote, the condition IX[ E N plays an important role in these notes. Let us say that (X, G) is finite if IX[ E N. The starting point of our approach to schemes is the definition of the complex product. The complex product allows us to treat schemes as algebraic objects. The first chapter is devoted to elementary consequences of the definition of the complex product. Substructures of schemes as well as quotient structures of finite schemes are defined naturally, and we end this introductory chapter with the generalization of the isomorphism theorems [22; w for finite groups due to E. NOETHER. The second chapter begins with a generalization of the fundamental grouptheoretical theorem [16; w due to C. JORDAN and O. H6LDER. As in the theory of finite groups, this theorem allows us to speak of "composition factors". In Sections 2.2, 2.3, and 2.4, we focus our attention on schemes which have thin composition factors. After that, other decompositions of schemes, such as "direct", "quasi-direct", and "semidirect products", are introduced. We include the theorem [9; Theorem 3.17] of P. A. FERGUSON and A. TUR U L L o n "indecomposable" schemes. In the third chapter, we collect various algebraic results which are needed for the representation theory (Chapter 4) and for the theory of generators of schemes (Chapter 5). The fourth chapter gives a general introduction into the representation theory of finite schemes. We start with a generalization of the fundamental group-theoretical theorem [21] of H. MASCHKV.on the semisimplicity of group algebras. After that, our approach is similar to that one given by D. G. 6 ...as is to be expected...
IX HIGMAN in [14]. In this chapter, we restrict ourselves to general structural results ignoring the huge amount of literature which exists in this area. For each L C Inv(G), we shall define in Section 5.1 what it means for (X, G) to be "L-constrained". The pair (X, G) will be called constrained if there exists L C Inv(G) such that (X, G) is L-constrained. In the fifth chapter, we first investigate constrained schemes. The constrained schemes form a class of schemes which is slightly larger than the above-mentioned class of Coxeter schemes. Their definition as well as their treatment seems to be particularly natural. The constrained schemes provide an appropriate framework for showing how smoothly Coxeter schemes are embedded into the class of schemes. From a general algebraic point of view, we consider Section 5.1 as the heart of these notes. Let us call the pair (X,G) thick if {1} = {g E G I ng =- 1}. Then it is particularly easy to exhibit the significance of Theorem 5.1.8(ii). This theorem says in particular that thick constrained schemes and thick Coxeter schemes are the same thing. Therefore, if one is willing to view buildings generally as thick, in other words, if one assumes the definition of [27; (3.1)] for buildings, Theorem 5.1.8(ii) implies that, via the (G, 7-)-correspondence, buildings and thick constrained schemes are the same thing. Since the definition of constrained schemes is particularly simple and natural, Theorem 5.1.8(ii) provides us with probably the most succinct definition of buildings. In the fifth chapter, we include a complete proof of the famous Theorem [8; Theorem 1] of W. FEIT and G. HIGMAN on finite generalized polygons. (We shall present the proof which was given by R. KILMOYER and L. SOLOMON in [20].) Moreover, we give a conceptually alternate and simultaneous approach to the theorems [24; Theorem 2] of S. PAYNE and [23; Satz 1] of U. OTT on polarities of finite generalized quadrangles and finite generalized hexagons via semidirect products. The main focus of the remainder of the fifth chapter is then on an appropriate generalization of Coxeter schemes of "rank" 2. The chapter ends with a generalization of the (algebraic) characterization [34; Theorem B] of finite generalized polygons and Moore geometries; see Theorem 5.8.4. The appendix of these notes is devoted to the embedding of the class of groups as well as the class of buildings into the class of schemes. In other words, we establish explicitly the (G, T)-correspondence and the (/3, .A)correspondence. Let us mention here that, apart from the class of groups and the class of buildings, other classes of mathematical objects as well can be viewed as specific classes of schemes. For the class of distance-regular graphs the embedding was given implicitly by P. DELSARTE in [6; Theorem 5.6]. As a consequence of this, one obtains the (elementary) fact that the class of strongly regular graphs forms a distinguished class of schemes. The fact that Moore geometries can be viewed as a specific class of schemes was partially shown in [34; (2.4)]. In particular, the class of the 2-designs with ~ = 1 forms a
X distinguished class of schemes. As a consequence, a lot of problems and results in graph theory or design theory can be viewed naturally or formulated easily as problems or results on schemes; see, e.g., [32], [33], or [35]. It seems to be a thought-provoking question to ask for other algebraic or geometric objects which can be viewed as classes of schemes. Apart from technical advantages there is at least one other reason for us to view mathematical objects such as groups, buildings, distance-regular graphs, or Moore geometries as schemes. Namely, the language of schemes provides a natural conceptual framework in which the above-mentioned objects may be characterized naturally. Theorem 5.8.4 and Corollary 5.8.5 are characterizations of that type for finite generalized polygons and Moore geometries. A similar characterization of the class of all finite buildings, i.e., a characterization without restriction of the rank, would be a challenging goal. Let us conclude this introduction with three general remarks. First of all, it might be helpful to mention here that the present text is thought to be an introductory monograph. No attempt has been made to give a complete account of the results available on schemes. In particular, the well-worked-out connection between schemes and graphs has been omitted completely. (This connection is discussed extensively in [3] and in [30].) On the other hand, apart from a few elementary facts about vector spaces and groups, the present notes are self-contained. They can be considered as an introduction to the structure theory of schemes. Secondly, the meaning of the symbols Z, IN, and P is fixed for the whole text. We shall denote by Z the set of rational integers. We set
N:= { z ~ z I 0_
Kiel, September 30, 1995
Paul-Hermann Zieschang
Table of C o n t e n t s
1.
Basic Results ............................................. 1.1 S t r u c t u r e C o n s t a n t s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 T h e C o m p l e x P r o d u c t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 G e n e r a t i n g Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Subschemes and Factor Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 C o m p u t i n g in Factor Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 M o r p h i s m s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1 3 10 15 19 22 26
2.
Decomposition Theory .................................... 2.1 N o r m a l Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Strongly N o r m a l Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 T h i n Residues and T h i n Radicals . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 R e s i d u a l l y T h i n Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Direct P r o d u c t s of Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Q u a s i - d i r e c t Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Semidirect P r o d u c t s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33 33 36 39 44 48 54 60
3.
Algebraic Prerequisites ................................... 3.1 T i t s ' T h e o r e m on Free Monoids . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Integers over Subrings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Modules over Associative Algebras . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Associative Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Characters of Associative Algebras . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Tensor P r o d u c t s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65 65 72 77 80 85 87
4.
Representation Theory ................................... 4.1 A d j a c e n c y Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Algebraically Closed Base Fields . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 T h e C o m p l e x Adjacency Algebra . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 R e p r e s e n t a t i o n s a n d Closed Subsets . . . . . . . . . . . . . . . . . . . . . . . 4.5 T h e Case IGI _~ 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97 97 106 109 113 118
XII
5.
Table of Contents Theory of Generators ..................................... 5.1 C o n s t r a i n e d Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Pairs of I n v o l u t i o n s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 F i n i t e C o x e t e r Schemes of R a n k 2 . . . . . . . . . . . . . . . . . . . . . . . . 5.4 T h e T h e o r e m s of O t t and P a y n e . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Length and N o r m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 P a r t i t i o n s o f G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Cosets on W h i c h the Length is C o n s t a n t . . . . . . . . . . . . . . . . . . . 5.8 A C h a r a c t e r i z a t i o n T h e o r e m . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
123 123 134 138 150 155 158 163 167
Appendix
177
.................................................
Index .....................................................
185
References
187
................................................
1. Basic
Results
In this introductory chapter, we develop some basic terminology which will be used throughout the remainder of the text.
1.1 Structure
Constants
Our first lemma is an immediate consequence of the definition of the structure constants of (X, G). (By ~ we mean the Kronecker delta.)
L e m m a 1.1.1 Let e, f E G be given. Then (i) al~! = ~1 = all,. (ii) a~.!l = ~f~fn~..
L e m m a 1.1.2 Let g E G be given. Then (i) For each x E X , Ixgl = ng. (ii) Igl = nglXl. (iii) If IXl ~ r~, ng. = ng.
Proos (i) The definition of g* yields g** = g. Therefore, we have
Ixgl
=
agg, 1 = rig.
(ii) follows from (i). (iii) Assume that IX[ E N. Then, as g C_ X x X, [g[ E N. But from (ii) we also obtain that ng. IxI = Ig*l = Ig[ = nglXI. Therefore, ng. : ng. []
L e m m a 1.1.3 Let d, e, f E G be given. Then (i) For each g E G, E b E G adebab]g = ~cEG adegaeye. (ii) ade]= a~.d.f.. (iii) adyene = aef.dnd.
2
1. Basic Results
Proof. (i) Let y, z E X be such that (y, z) E g. We count in two different ways the elements of e M (yd • z f*). This proves (i). (ii) Let y, z E X be such that (y, z) E f. Then, by definition, ade f "-- [ydnze*[ -- [ze* n yd**l-- ae*d*]*.
(iii) Apply (i) to (f, e*, 1) in the role of (e, f, g), and use (ii).
[]
L e m m a 1.1.4 Let e, f E G be given. Then (i)
a ,j =
(ii) ~ q e c aYge = n]. (iii) ~ g e a ae'fgng : ne'nl"
Proof. (i) Let y, z E X be such that (y, z) e f. We count in two different ways the elements of {(x,g) E ze* • G I (y,x) E g}. This proves (i). (ii) For each g e G, we apply Lemma 1.1.3(ii) to ( f , g , e ) in the role of (d, e, f). Then
Zo gEG
..o. -- n,;
gEG
use (i). (iii) For each g E G, we apply Lemma 1.1.3(iii) to (e*,g) in the role of (d, e). Then y ~ ae..fang : Z 9EG
agy.e*ne, = ne*n];
9~:G
use (i).
[]
Let g E G, let n E I~ \ {0, 1,2}, and let f l , ..., fn E G be given. Then a1~...l. ~ is defined recursively by afx...f,,g : : ~ afl...fr,-xeaef,,9" eEG
The following lemma generalizes Lemma 1.1.3(i), (ii) and Lemma 1.1.4(iii).
L e m m a 1.1.5 Let n E l ~ \ {0,1}, and let fl, ..., fn E G be given. (i) Assume that 3 <_ n, and let g E G be given. Then we have al~...I,g = ~"~eEG afxegaf~...f~e"
(ii) For each 9 E G, afl...f,g --- af~....f~g.. (iii) ~ g e c aY~...Y.ang = n . h ' . . n l . .
1.2 The Complex Product
3
Proof. (i) If n = 3, the claim is just a restatement of L e m m a 1.1.3(i) (with (e, f l , f2, f3) in the role of (c, d, e, f ) ) . Therefore, we assume that 4 g n. Assuming that the claim holds for n - 1, we obtain that
as*"'S,g= Z as,...S,-*cacs.g= Z ( Z cEG Z
as2"S,-:d Z
dEG
aS'dcaS'S"-'d)acs"g=
c6G d6G aSldCacs,g = Z
c6G
as2S"-ld Z
d6G
af, egadS,~e-=
e6G
Zas, eg(Zal,...S._,aaal.e): Zal,egaf,...S.e; e6G
d6G
e6G
use L e m m a 1.1.3(i). (ii) If n = 2, the claim is just a restatement of L e m m a 1.1.3(ii) (with (Ix, f2, g) in the role of (d, e, f ) ) . Therefore, we assume that 3 _< n. Assuming that the claim holds for n - 1, we obtain that asx...Sng :- y ~ afl...f,~_xeae],9 = ~ eEG
af,]e*g, as:_a...f;e* = afg,...f;g.;
e6G
use L e m m a 1.1.300 and (i). (iii) If, = 2, the claim is just a restateme,t of L e m m a 1.1.4(iii)(with ( f l , f2) in the role of (e*, f ) ) . Therefore, we assume that 3 _< n. Assuming that the claim holds for n - 1, we obtain that
Z
aSl...l, gng = Z ( Z
gEG
asI...l.-leael.g)n9 : Z a s ,..-s.-,e Z
gEG eEG
eEG
.f,,.f._,,(.,.f.) =
eEG use L e m m a 1.1.4(iii).
(~ af, ...S,,_,ene)nf,~ :---nfl "" .nf,,; eEG
[]
1.2 The Complex Product Recall that, for all E, F _C G,
Ep
ae.t,gng :
gEG
:= ( .
~ a l ~ ~ aosg ~ 0} eEE SEF
L e m m a 1.2.1 Let D, E, F C G be given. Then (i) If E C F, DEC_OF and ED C_FD. (ii) (DE)F = D(EF).
4
1. Basic Results
Proof. (i) follows immediately from the definition of the complex product. (ii) Let g E (DE)F be given. Then, by definition, there exist b E DE and f E F such that ablg 7s O. Since b E DE, we find d E D and e E E such that ad~b ~ O. It follows that ad~bablg ~ O. Since aaebabfg ~ O, there exists c E G such that adcgae$c ~ 0; see Lemma 1.1.30). From this we conclude that adcg 7s 0 and that a~]c 7s 0. From a~f~ 7s 0 we obtain that c E EF. Thus, as a~g ~ 0, g E D(EF). Since g E (DE)F has been chosen arbitrarily, we have shown that (DE)F C D(EF). Similarly, one obtains that D(EF) C (DE)F. [] From the definition of the complex product we obtain immediately that, for all E, F C_ G,
EF=~
r
OE(E,F).
Moreover, from Lemma 1.1.1(i) we deduce easily that, for each F C G, F{1} = F = {1}F. Therefore, Lemma 1.2.1(ii) says that the set of all non-empty subsets of G is a monoid with respect to the complex multiplication and with {1} as identity element. This monoid will play an important role in the fifth chapter of these notes. For each F C G, we define F* := {f*
fEF}.
It is obvious that F** = F.
L e m m a 1.2.2 Let E, F C G be gwen. Then (EF)* = F'E*.
Proof. For each g E G, we have
g ~ (EF)" r
g* E EF eEE f E F
Z
Z
ay,e.gTs
~=~ Z
eEE f E F
Z aleg ~ 0 ~ 9' E F'E*;
]EF* eEE*
[]
use Lemma 1.1.3(ii).
Let n E N \ {0, 1, 2}, and let F1, ..., F,~ C_ G be given. Then F1 - " Fn is defined recursively by FI...F
n = (F1...Fn_l)Fn.
1.2 The Complex Product T h e following l e m m a generalizes L e m m a 1.2.1(ii) and L e m m a 1.2.2.
L e m m a 1.2.3 Let n E N \ {0, 1}, and let F1, ..., Fn C G be given. Then we have (i) / f 3 < n, F I . . . F , = F I ( F 2 . . . F , ~ ) . (ii) (F1 . . . F,,)* = F,~ . . . F~*.
Proof. (i) If n = 3, the claim is just a restatement of L e m m a 1.2.1(ii). Therefore, we assume that 4 _< n. Assuming that the claim holds for n - 1, we obtain that
FI'"F.
= (FI'"F,~_I)Fn
= (FI(F2""Fn-1))Fn
=
F I ( ( F 2 . . . F , _ I ) F , ) = F1 ( F 2 . . . F , ) ; use L e m m a 1.2.1(ii). (ii) If n = 2, the claim is just a restatement of L e m m a 1.2.2. Therefore, we assume that 3 < n. Assuming that the claim holds for n - 1, we obtain that (FI...F,)* = (F~(F2...F,))* = (F2...F,~)*F( = (Y• . . . F ; ) F ;
= , ,r~" . . . F~;
use (i) and L e m m a 1.2.2.
[]
Let F C_ G be given. For each z E X, we define x F := U z f . ] fi F
For each g E G, we set g F := { g } F
and F g := F { g } .
L e m m a 1.2.4 Let n E N \ { 0 , 1}, and let F1, ..., Fn C G be given. Then, f o r each g C G, the following conditions are equivalent. (a) g E F , . - . F , . (b) Let y, z E X be such that (y, z) E g. Then there exist xo, ..., xn E X such that Xo = y, xn = z, and, for each i E { 1 , . . . , n}, xi E x i - l F i .
(c) There exist ~o,
...,
x . e X such that (~o, x.) C g . . d , for each
i E { 1 , . . . , n } , xi E z i - l F i . (d) There exist go, ..., gn E G such that go = 1, g,~ = g, and, for each i E { 1 , . . . , n } , gi E g i - l F i .
(e) ~(f ...... f~)eF,•215
a:~...:.9# O.
6
1. Basic Results
Proof (a) ::~ (b) The claim is obvious for n = 2. Therefore, we assume that 3 < n. Since g E ( F 1 . . . F , - 1 ) F , , there exist x E X and f E F 1 - ' . F n - 1 such that (y, x) E f and z E x F , . Now, by induction, there exist x0, ..., x , - 1 E X such that x0 = y, x , _ l = x, and, for each i E { 1 , . . . , n - 1}, xi E Xi-lFi. (b) =v (c) This follows from the fact that g # 0. (c) =~ (d) For each i E { 0 , . . . , n } , let gi E G be such that (xo, xi) E gi. Then go = 1, g, = g, and, for each i E { 1 , . . . , n } , agi_,/g , # 0. It follows that, for each i E { 1 , . . . , n}, gi E gi-lFi. (d) =v (e) The claim is obvious for n = 2. Therefore, we assume that 3
af,...f~,_Ig,_ I # O. (fl,...,f,-1)EF1
X...xF,~_I
Since g E gn-lFn, there exists f,~ E Fn such that ag._,f~g ~ 0. It follows that
Z
af1]'~-'g"-'ilg'~-l'f~'g # O.
( f l , . - . , / ~ . ) E F 1 x ... x F ,
Thus, we have
af:t...f=,g = (11 , . . . , f ~ ) f i F l
X ... X F,~
~
~ afl...f,,,_,eaef~.g r O.
( f l ..... .f,~)E F I X ... x F , ~ e f i G
(e) ~ (a) Again, the claim is obvious for n = 2. Therefore, we assume that 3 < n. Let ( f l , . . . , f,~) E F1 • ... • Fn be such that af,...f.g # O. Then there exists e E G such that afl...f._l~ # 0 and a~l.g # 0. From al,...l._,~ # 0 we obtain by induction that e E F 1 - . . F n - 1 . On the other hand, aef,g # 0 implies that g E eFn. Thus, by L e m m a 1.2.1(i), g E F1 ." . F , . [-3
For all e, f E G, we abbreviate
e f :=
e{f}.
Note that, for all E, F C_ G,
= U
eEE
= U
IEF
= U U 4. eEE / E F
1.2 The Complex Product
7
L e m m a 1.2.5
Let e, f E G be given. Then (i) For each g E G, e E fg if and only if f E eg*. (ii) For each g E G, e* E g f if and only if f* E eg. (iii) 1 E e* f if and only if e = f . (iv) Let c, d E G be given. Then c* eMdf* 7s 0 if and only if c d N e f 7s 0.
Proof. (i) Let g E G be given. Then, by Lemma 1.1.3(iii), e E fg r
ayge 7s
r
aeg.y ~ 0
r
f Eeg*.
(ii) Let g E G be given. Then, by Lemma 1.2.2 and (i), e* E g f r
eEfig*
f* Eeg.
r
(iii) By Lemma 1.1.1 (ii), we have 1Ee*f
r
ae.s1 r
r
e=f.
(iv) Assume that c*e fq dr* ~ 0. Let b E c*e fq dr* be given. Then, by definition, ae.~b 7s 0 and adi.b 7s O. By Lemma 1.1.3(iii), aay.b 7s 0 is equivalent to abyd r O. Therefore, ac*ebabId 7s O. Now, by Lemma 1.1.3(i), there exists g E G such that a~'gdaeIg 7s O. It follows that a~Ig 7s 0 and ar ~ O. Thus, by definition, g E e f and d E e'g. The latter is equivalent to g E cd; see (i) and (ii). []
L e m m a 1.2.6 For each g E G, n 9 = 1 if and only i f { l } = g* g.
Proof. From Lemma 1.1.4(ii) we know that E ageg ~ rig. eEG
From Lemma 1.1.1(i) we know that aglg = 1. Thus, ng = 1 if and only if, for each e E G \ { 1}, ageg = O. But, for each e E G, we have ageg=O r use Lemma 1.2.5(i), (ii).
g~ge
r
g~ge*
r
e~g*g;
[]
T h e o r e m 1.2.7 Let d, e, f E G be such that f E de. Assume that nd E N. Then the following conditions are equivalent. (a) ne = 1. (b) {f} = de and adein! = ha.
8
1. Basic Results Proof. (a) =~ (b) If n~ = 1,
~-~ adegng ~. rid; gGG
see Lemma 1.1.4(iii). Therefore, a d e l n f <_ nd. On the other hand, by assumption f E de. Therefore, ade] • O. Thus, by Lemma 1.1.3(iii), na < afe.dnd = aa~lnf . It follows that a a e f n f : rid. Thus, the above equation yields that, for each g E G \ {f}, ad~g : 0. (Recall that we are assuming that nd E 1~.) It follows that {f} : de. (b) ~ (a) Assume that {f} = de. Then, for each g E G \ {f}, aa~g = O. Thus, by Lemma 1.1.4(iii), ad~sn] = ndn~. On the other hand, we are assuming that adeyn] = nd E l~. It follows that n~=l. []
C o r o l l a r y 1.2.8 Let n E N \ {0}, let f l , ..., fn E G, and let g E { f l } ' " { f n } be given. (i) A s s u m e that, f o r each i E { 1 , . . . , n } , ny, = 1. Then we have {g} = { f l } ' " { f n } and ng = 1. (ii) A s s u m e that {g} = { f l } " ' { f ~ } and that ng = 1. A s s u m e f u r t h e r that, f o r each i E { 1 , . . . , n } , n1: = ns,. Then, for each i E { 1 , . . . , n } , n l , = 1.
Proof. If n = 1, the claims are obvious. Therefore, we assume that 2 _< n.
(i) Let us assume that the claim has already been established for n - 1. Then there exists e E G such that {e} = { f l } ' " { f , ~ - a } and ne = 1. From {e} = { f l } ' " { f n - 1 } and g E { f l } ' " { f n } we obtain that g E ef,~. On the other hand, we are assuming that n l , = 1. Thus, by Theorem 1.2.7, {g} = efn = { f l } ' " { f n } and ng < ne. But, as ne = 1, ng < ne yields ng~- 1.
(ii) By hypothesis, ng = 1. Thus, by Lemma 1.2.6, {1} = g*g. It follows that f~.fn C f : . . " f ~ f l " " . f n
: {1};
see Lemma 1.2.3(ii), and use the assumption that {g} = f l " " f~. Therefore, by Lemma 1.2.6, n f , = 1. Thus, by hypothesis, n f: = 1. Since g E f l " " "f~, there exists e E {fx}'" " { A - l } such that g E e l , . From n f : = 1 and Lemma 1.2.6 we obtain that {1} = f n f ~ . Thus, as g E efn,
g/'~ c_ eA/'~ = {e}. It follows that {e} = gf~. From e E g f* we obtain that n~ = 1; see (i). From gf~ C_ {e} we obtain that
1.2 The Complex Product e E { f l } "" " { f n - 1 ) ---- { f l ) " " ' { f n - 1 ) f n f ~ : g f *
9
C {e},
whence {e} : { f l } " " { f ~ - l } . Thus, we are done by induction.
[]
C o r o l l a r y 1.2.9 Assume that ]X[ E l~, and let d, e, f E G be given. Assume that ne = 1. (i) / f f E d e , nd = n 1. (ii) Let g E G be such that ng = 1 and f E 9de. Then {f} = gde, {d} = g*fe*, and n d = n!.
Proof (i) F r o m f E de and L e m m a 1.2.5(i) we obtain that d E fe*. From IXI E N and n~ = 1 we obtain that n~. = 1; see L e m m a 1.1.2(iii). Thus, by T h e o r e m 1.2.7, nd I n!. But f r o m T h e o r e m 1.2.7 we also know that nl I rid.
Therefore,
n d :
n$.
(ii) Since f E 9de, there exists c E de such that f E gc. From c E de we obtain that {c} = de and n d = n~; see T h e o r e m 1.2.7 and use (i). From f E gc we obtain that c* E fig; see L e m m a 1.2.5(ii). Thus, by T h e o r e m 1.2.7 and (i), {c*} = f*g and n S. = no.. It follows that {c} = g*f and n] = n~; see L e m m a 1.2.2 and L e m m a 1.1.2(iii). Together, we have de = g* f and n d - : n!. From de = 9* f we obtain that {f} = gg* f = gde and {d} = dee* = g* f e* ; use L e m m a 1.2.6. []
For each F C G and, for each Y C_ X, we set
Y F := U yF. yEY
L e m m a 1.2.10 Let Y C_ X , and let F C_ G be given. (i) For each Z C Y, Z F C Y F . (ii) For each E C F, Y E C Y F. (iii) For each E C_ G, ( Y E ) F = Y ( E F ) .
Proof. (i) follows immediately from the definitions of Z F and Y F . (ii) follows i m m e d i a t e l y from the definitions of Y E and Y F . (iii) Let x E ( Y E ) F be given. Then, by definition, there exist z E Y E and f E F such that x E z f . Since z E Y E , there exist y E Y and e E E such that zEye.
10
1. Basic Results
Let g E G be such that (y,x) E g. Then, as z E ye M x f * , ae.tg ~ O. Therefore, g E el. Now (ii), Lemma 1.2.1(i), and (i) yield
x E yg C y(ef) C_ y ( E F ) C Y ( E F ) . Conversely, assume that x E Y ( E F ) . Then, by definition, there exist y E Y and g E E F such that (y,x) E g. Since g E E F , there exist e E E and f E F such that aelg 7s O. Therefore, ye M x f* 7s 0. Let z E y e M x f* be given. Then
x E z f C_ z F C_ (ye)F C_ ( Y E ) F ; use (ii) and (i).
1.3 Closed
[]
Subsets
A subset F of G is said to be closed if F * F C F # 0. We shall denote by C(G) the set of closed subsets of G. For the remainder of these notes, we abbreviate c := c ( c ) .
It follows from the definition of C that {1} E C and G E C. Let us derive some elementary facts about closed subsets which in the future we will occasionally quote without reference. Let H E C be given. Then, by definition, H # 0. Let h E H be given. Then, by Lemma 1.2.5(iii) and Lemma 1.2.1(i), 1 E h*h C H*H C_ H. Therefore, we have 1 E H . Since 1 E H, Lemma 1.2.1(i) yields H* = H * I C H*H C H. From this we deduce that H = H** _C H*, whence H* = H. Now we also have that H H = H*H C_ H. Therefore, H H C_ H. For each F C G, we define
X/F
:=
I 9
X}
and
G / F := {gF I g E G}.
T h e o r e m 1.3.1 Let F C_ G be such that 1 E F. Then the following conditions are equivalent. (a) F E C. (b) X / F is a partition of X . (c) G / E is a partition of G.
1.3 Closed Subsets
11
Proof. (a) :=t, (b) Let y, z E X be such that y E zF. We shall be done if we succeed in showing that y F = zF. By hypothesis, F E C. Thus, as y E zF, y F C ( z F ) F = z ( F F ) C zF; use Lemma 1.2.10. But from y E z F and Lemma 1.2.5(i) we also obtain that z E yF* = yF. Thus, we conclude, as before, that z F C_ yF. (b) =r (c) Let d, e E G be such that d E oF. Again, we shall be done if we prove that d F = oF. Since d E eF, there exists f E F such that d E el. This means that ae]d • O. Therefore, there exist y, z E X such that (y, z) E d and yeMzf* ~ I~. Let x E ye fq z f* be given. Since z E z f , (b) yields z F = zF. Let g E d F be given. Then, by Lemma 1.2.5(i), d E gF*. Thus, as (y, z) E d, yg f3 z F ~ 0. Let w E yg M z F be given. From w E z F we obtain that w F = z F = x F . Thus, as 1 E F, w E x F . It follows that g E oF; recall that xEye. Now, as g E d F has been chosen arbitrarily, we have shown that d F C eF. The inclusion eF C_ d F is obtained similarly. (c) :=r (a) Let f E F be given. Then 1 E f * f C_ f * F . On the other hand, by hypothesis, 1 E F. Thus, by (c), f * F = F. Note that f E F has been chosen arbitrarily. Therefore, we have shown that F * F C F. []
Let H E g be given. The elements of X / H (respectively, G / H ) will be called the left cosets of H in X (respectively, G). From Theorem 1.3.1 and Lemma 1.2.2 one obtains easily that {Hg I g E G} is also a partition of G. Occasionally we shall use this fact without any reference. For each F C_ G, we set
F//H := { H i l l I S F}. The elements of G / / H will be called double cosets of H in G.
C o r o l l a r y 1.3.2 Let H E g be given. Then G / / H is a partition of G.
Proos Let e, f E G be such that e E H f H . We shall be done if we succeed in showing that H e l l C_ H f H and that f E H e l l . Since e E H f H , H e l l = ( H e ) H C ( H ( H f H ) ) H = H f H ; see Lemma 1.2.1(i) and Lemma 1.2.3(i). Since e E H f H , there exists g E f H such that e E Hg. Thus, by Lemma 1.2.2, e* E g ' H , whence, by Theorem 1.3.1, g* E e*H. Now Lemma 1.2.2 yields g E He, whence f E gH C H e l l ; use Theorem 1.3.1 once again. []
12
1. Basic Results T h e o r e m 1.3.3 (i) For each 7t C_ C, A H e n H E C. (ii) Let H, K E C be given. Then H K E C if and only if H K = K H .
Proof. (i) follows immediately from the definition of C. (ii) Assume that H K E C. Then H K = (HK)* = K ' H * = K H ; use Lemma 1.2.2. Conversely, if H K = K H , ( H K ) * H K = K * H * H K C K * H K = K * K H C_ K H = H K ; use Lemma 1.2.2 and Lemma 1.2.1. Therefore, H K E C.
[]
L e m m a 1.3.4 Let H, K E C be given. Then the following conditions are equivalent. (a) H fq K -- {1}. (b) For each g E H K , there exist uniquely determined elements h E H and k E K such that g E hk.
Proof. (a) ::~ (b) Let g E H K , let c, d E H, and let e, f E K be such that g E c e n d f . Then, by Lemma 1.2.5(iv), c ' d O e r * # 0. On the other hand, as H E C, c*d C H and, similarly, e f* C_ K. It follows that e*dN el* C_ H N K = {1}. Therefore, 1 E c*d and 1 E e f t . Thus, by Lemma 1.2.5(iii), e = d and e = f . (b) ::~ (a) Let g E H N K be given. Then g E H K and lg = g = gl. Thus, b y ( b ) , g = 1. []
The first part of our next lemma is a special case of [5; Theorem VIII].
L e m m a 1.3.5 [R. DEDEKIND]Let E, F C_ G, and let H E C be given. Then (i) If E C_ H, H N E F = E ( H N F ) and H ( 3 F E = (H O F ) E . (ii) / f E H E C and F H E C, (E (1 F H ) ( F A E H ) = E F N F H A E H .
Proof. (i) It follows from E C_ H E C and Lemma 1.2.1(i) that E ( H ~ F ) C_ HNEF. Conversely, let h E H N E F be given. Since h G E F , there exist e E E and f E F such that h E el. Thus, by Lemma 1.2.5(ii), f* E h*e C_ H. It follows that f E H, whence h E E ( H N F).
1.3 Closed Subsets
13
Since h E H 13 E F has been chosen arbitrarily, we have shown that H 13 E F C_ E ( H 13 F). The second equation is proved similarly. (ii) Since H 6 C, 1 6 H. Therefore, by Lemma 1.2.1(i), E C_ E H and F C_ F H . Therefore, the two equations of (i) yield (E 13 F H ) ( F 13 E H ) = (E 13 F H ) ( E H 13 F) = E H N (E N F H ) F = E H M ( F H 13 E ) F = E H 13 F H 13 EF. This proves (ii).
[]
For each F C G, we set n F := 2-.r hi. ]6F
T h e o r e m 1.3.6 (i) Let z 6 X , and let F C_ G be given. Then we have IzFI = nF. (ii) Let H 6 C, and let h 6 H be given. Then ~ e H ~]]eH a~fh = nil. (iii) Let x 6 X be given. Let H, K E C be such that K C_ H. Then
n H = nKI{YK I Y C xH}l. (iv) For each H 6 C, IXl = nHIX/HI.
Proof. (i) For each f e F, Ixft = n l ; see L e m m a 1.1.2(i). But, for each y 6 x F , there exists a uniquely determined element f 6 F such that y 6 z f . Thus, the claim follows from the definitions of x F and nF. (ii) Let e 6 H, and let f 6 G \ H be given. Then f ~ e*h. Thus, by Lemma 1.2.5(i), (ii), h ~ el, which means that a~fh = 0. Now Lemma 1.1.4(ii)yields
E EaeSh=EEae/h= Ene=nH" e E H ]EH
eEH lEG
eEH
(iii) Let y 6 x H be given. Since K C_ H, y K C_ yH; see Lemma 1.2.10(ii). On the other hand, Theorem 1.3.1 yields yH = xH. Thus, y K C_ xH. Since y 6 x H has been chosen arbitrarily, it follows that U
yK C xg.
yExH
Conversely, as K 6 C, 1 E K. Therefore, we also have that
14
1. Basic Results xH C_ U
yK.
yExH
It follows that xH=
U
yK.
yExH
Thus, the claim follows from (i) and Theorem 1.3.1. (iv) follows from (iii). (Apply (iii) to (G, H) in the role of (H, K).)
[]
T h e o r e m 1.3.7 Let H, K E C be given. Then we have (i) For each x E X, x(H n K) = x t t n xK. (ii) Let y, z E X be such that z E y H K . Then lyH n z K I = UrinE. (iii) n H K n H n K = s l i n K .
Proof. (i) It is clear that x H n x K c_ x ( H A K ) . Conversely, it follows from Lemma 1.2.10(ii) that x(H n K) C xH n xK. Thus, we have x(H n K) = xH n x K . (ii) Since, by hypothesis, z E yHK, yH n zK ~ ~. Let x E yH n z K be given. Then, by Theorem 1.3.1, xH = yH and x K = zK. On the other hand, we know from (i) that x ( H A K ) = x H n x K . Therefore, x(H n K) = yU n zK, so that the claim follows from Theorem 1.3.6(i). (iii) Let x E X be given. We count in two different ways the elements of {(y,z) E X •
lYExHAzK}.
Then the claim follows from (ii).
[]
L e m m a 1.3.8 Let H, K E C be given. Then we have
(i) IX/U n KI <_ IX/UlIX/nl. (ii) Assume that IX/HI, ]X/K I E 1~ and that (IX/HI, ]X/K]) = 1. Then I x / u n KI = IX/HlIX/KI.
Proof. (i) For each x E X, we define x(H N K)O := (xH, xK). We shall be done if we succeed in showing that r is an injective map. Let y, z E X be given.
X/HAK
~
X/H•
1.4 Generating Subsets
15
Assume first that y(H N K ) = z ( H n K ) . Then z 9 yH and z 9 yK. Thus, by Theorem 1.3.1, yH = z H and y K = z K . Assume, conversely, that yH - z H and y K = z K . Then, by Lemma 1.3.7(i), y(H M K ) = yH n y K = z H n z K = z ( H n K). Since y, z 9 X have been chosen arbitrarily, we have shown that r is an injective map. (ii) By Theorem 1.3.3(i) and Lemma 1.3.6(iii), nHng divides nil. By Theorem 1.3.3(i) and Theorem 1.3.6(iv),
nH~KIXIH n 1;1 = nHIXlXl. Therefore, IX~HI divides IX/H n KI. Similarly, we deduce that IX~If] divides I X / H N K I , so that the hypothesis (IX~HI, I X / K I ) = 1 leads to
IXlHIIXlKI <_ IXlH n I~1. []
Thus, the claim follows from (i).
C o r o l l a r y 1.3.9 Let 7t be a finite subset of C. Then (i) IXINHe~ HI <_1-l.~ IX~HI. (ii) Assume that, for each H e 7t, IX~HI 9 N. Then I X l
nue~
Proof. (i) follows from Lemma 1.3.8(i) by induction on I~l(ii) follows from (i).
1.4 Generating
HI 9 N.
[]
Subsets
Let F _C G be given. We define
FenCe
F will be called a generating subset of (F). We set F ~ := {1}. Let n E N \ {0}, and let F1, ..., F,~ q G be such that, for each i E { 1 , . . . , n}, Fi = F. Then we set F '~ :=Fa...F,~. The following theorem will play an important role in the fifth chapter of these notes.
16
1. Basic Results T h e o r e m 1.4.1 Let F C_ G be given. Then (i) (F) = UneN(F* u F) ~. (ii) If IXI E N, (F) = UneN F " .
Proof. (i) Set E := F* U F, and define D:=
U E". hEN
From Lemma 1.2.3(ii) we know that, for each n E H, (E~) * = (E*)" = E n. Therefore, we conclude that D E C. Thus, as F C_ D, (F) C_ D. Conversely, for each n E N, E n _C (E) n _C (E) = (F); use Lemma 1.2.1(i). Therefore, D C_ (F). It follows that (F) = D, and this proves (i). (ii) Set D := U Fn" nEN
Then F C_ D. On the other hand, by (i), D C_ (F). Therefore, it suffices to prove that D E C. To prove this, it suffices to show that F* C_ D. Let f E F, and let y, z E X be such that (y, z) E f . Then, by Lemma 1.2.10, zD C_ ( y f ) D C_ y ( f D ) C_ yD. On the other hand, by Theorem 1.3.6(i), IzOl = n D = lyDI. Thus, as Ixl is assumed to be finite, zD = yD. But, as 1 E D, y E yD. Therefore, y E zD. But y E z f*. Therefore, f* E D. Note that f E F has been chosen arbitrarily. Therefore, we have shown that F* C D. V1
T h e o r e m 1.4.2 Assume that IXI E N. Let F C_ G, and let p E IP be such that, for each f E F, nl <_ p - 1. Then, for each g E (F), p{ ng.
Proof. We set E:={ge(P)
I plng}.
Then, by Lemma 1.1.2(iii),
E*=E. Moreover, our claim is that E = 0. In order to prove this, we first shall show that E F C E.
1.4 Generating Subsets
17
Let g E E F be given. Then, by definition, there exist e E E and f E F such that aelg # O. Since e E E, p ] ne. But, by Lemma 1.1.3(iii), n~ ] a~lgng. Thus, P l aefgng 9 On the other hand, by Lemma 1.1.4(i) and Lemma 1.1.2(iii), aefg < nl" = nl < p - 1. Therefore, as aefg # 0, p ~ aefg.
It follows that p I ng, which, by definition, means that g E E. Since g E E F has been chosen arbitrarily, we have shown that E F C E. Now Theorem 1.4.1(ii) yields E ( F ) C_ E. Assume now that E ~ ~. Then, by Lemma 1.2.5(iii), 1 E E*E. Thus, as E* = E and E C_ (F), 1 E E ( F ) C E, contrary to nl = 1. r"l C o r o l l a r y 1.4.3 Assume that IXI ~ •. Let F c_ G be such that (F) = G, Assume that, for each f E F, n S = 2. Then, for each g E G, ng is a power of 2.
Assume that IXI E N. Let H E C \ {G} be given. Then H is called a maximal closed subset of G if, for each K E C, H _C K implies that K E
{H,a). By P(G) we shall denote the intersection of all maximal closed subsets of G. P(G) is called the Frattini subset of G. (By Theorem 1.3.3(i), P(G) E C.) Let H, K E C be such that H K = G and K ~ G. Then K is called a supplement of H in G. (By definition, G is not a supplement.) In Theorem 2.3.9 we shall find a result which is similar to the second part of the following result. T h e o r e m 1.4.4 (i) Let F C_ G be such that (P(G) U F) = G. Then (F) = a. (ii) P(G) does not have a supplement in G.
Proof. (i) Suppose that (F) • G. Then there exists a maximal closed subset H, say, of G such that (F) C_ H. However, P(G) C_ H by definition of the Frattini subset. Therefore, G = (P(G) U F) C H, a contradiction. (ii) follows immediately from (i). []
For each g E G, we abbreviate
:= <{g)).
18
1. Basic Results We set
Inv(G) := {g E G I I(g)l = 2}. The elements of Inv(G) will be called (generalized) involutions. They will play an important role in the fifth chapter of this text. L e m m a 1.4.5 Let l E Inv(G) be given. Then (i) l* = I. (ii) ll C_ {1,/}. (iii) For each g E G \ { 1, l}, att~ = O. (iv) attl = nt - I. (v) I f nl r 1, t E l l . Proof. (i) Since l E Inv(G), I(/)1 = 2, whence {1,/} E C. Thus, by definition, l* E l*l C {1,l}, which implies that l* = I. (ii) From (i) and {1,/} E C we obtain that ll = l*l C_ {1,/}. (iii) By definition, II = {g E G I aug r 0}. Thus, if aug :/: 0, g E {1,l}; use (ii). (iv) From (iii) and Lemma 1.1.3(iii) we deduce that, for each g E G \ { 1 , l}, agt.t = 0. Thus, by Lemma 1.1.4(i), a l t . t + a u , t = nt. Now (i) yields a l u + a m = nl. But, by Lemma 1.1.1(i), altt = 1. Therefore, alu = nl - 1. (v) is an immediate consequence of (iv). []
Corollary 1.4.6 Let e, f E G be such that e r f , and let l E I n v ( G ) be such that {f} = el. Then, f o r each g E G, a~tg = ~Sg. 1 Proof. Since el C_ {f}, we have that, for each b E G \ {f}, aelb "~ O.
Since e :~ f, we have ael] = 0. Thus, by Lemma 1.1.3(i) and Lemma 1.4.5(iii), aelfaYl] : Z aelbabl] : Z aecfallc : aelfalll. bEG cEG
But, as f E el, Therefore, alt ! = art; = nt - 1; see Lemma 1.4.5(iv). Now Lemma 1.1.4(i) and Lemma 1.4.5(i) yield ael] + nt -- 1 = ael.f + all I ~ ~ adlf : nl, dEG
whence gel] ~ 1. Thus, as ael] ~ 0, gel] = 1.
a Here ~ is the Kronecker delta.
[]
1.5 Subschenms and Factor Schemes
19
1.5 S u b s c h e m e s a n d F a c t o r S c h e m e s Let Y C_ X be given. For each g E G, we define gy : = g n ( Y
x
Y);
for each F _C G, we set
Fy := {/y l / e F } . For each x E X and, for each H E C, we set (X,
G)xH :---- (xH, HxH).
Then the following theorem is obvious.
T h e o r e m 1.5.1 Let x E X , and let H E C be given. Then we have (i) For each h E H, (hxn)* = (h*)xH. (ii) (X,G)xH is a scheme. (iii) Let d, e, f E H be given. Then ad~H~,y~, = ariel.
Let x E X, and let H E C be given. Then (X, G)~H is called the subscheme of (X, G) with respect to x and H.
L e m m a 1.5.2 Let x E X , let H E C, and let F C H be given. Then (i) For each E C_ H, E~HF~H = (EF)~H. (ii) (FxH) = (F>~H. (iii)
FxH E C(H~H) if and only if F E C.
Proof. (i) Let e E E, let f E F, and let h E H be given. Then, by Theorem 1.5.1(iii), hxn E exnfxH if and only if h E el. This proves the claim. (ii) Let n E 1~ be given. Then, by Theorem 1.5.1(i) and (i),
((F~H)* U F~/4)"
=
((F*)~H U FxH)"
=
((F" U F)~H)" = ((F* U F)")~:H.
Therefore, by Theorem 1.4.1(i),
: U
U ((F* nElv, (U hEN
(F" u
u Fx.)" =
UF)")~H =
=
20
1. Basic Results (iii) By Theorem 1.5.1(i) and (i),
(F~:H)*F,:H = ( r * ):~. F =. = (F*F):~H. Therefore,
F~H
9C(HxH) r162
(F.,,)*F,:. c_ p~. (F*F)~H C F~H r F*F C F r F 9 This proves (iii).
[]
For each g 9 G and, for each H 9 C, we set
gg := {(yH, zH) I z 9 yHgH}.
P r o p o s i t i o n 1.5.3 Let e, f 9 G, and let H 9 C be given. Then the following conditions are equivalent. (a) eII El IH # O. (b) H e l l = n f H . (c) en = fn.
Proof. (a) =:~ (b) Let y, z 9 X be such that (yH, zH) 9 e H rh IH. Then, by definition, z 9 yHeH fq y H f H . Let g E G be such that (y, z) 9 g. Then g E H e l l M H f H . But, by Corollary 1.3.2, G / / H is a partition of G. Thus, Hell = H fH. The implications (b) ::~ (c) and (c) :=~ (a) are obvious.
[]
Let H E C, and let F C_ G be given. Referring to Proposition 1.5.3 we shall write
FIIH := { f . I f ~ F}, although, formally, we defined F // H := { H f H I f 9 F} earlier. Let H, K E C be such that K C_ H and H / / K = G / / K . It follows readily from Proposition 1.5.3 that then H --- G. Occasionally we shall use this elementary observation without reference. For each H 9 C, we set
(X, G) H := ( X / H , G / / H ) .
1.5 Subschemes and Factor Schemes
21
For the remainder of this section, we assume that IX] E N.
T h e o r e m 1.5.4 Let H E C be given. Then we have (i) 1X/H = 1H. (ii) For each g E G, (gH). = (g.)H. (iii) (X, G) g is a scheme. (iv) Let d, e, f E G be given. Then 1
adHeHfH ~ - -nH
E Z abc$. bC=HdHcEHeH
(v) For each g E G, ngHn H ~ nHg H.
Proof. (i) follows from Theorem 1.3.1. (ii) follows from the definition of gH and from Lemma 1.2.5(i). (iii), (iv) By Proposition 1.5.3, G / / H is a partition of X / H • X / H . From (i) we know already that 1X[H E G//H. Moreover, from (ii) we know that, for each g E G, (gH). E G//H. Thus, in order to prove (iii), is suffices to verify the regularity condition for (X, G) H. We compute the structure constants explicitly, so that also (iv) will be proved. Let d, e, f E G, and let y, z E X be given. Assume that (yH, zH) E fH. Then, by Theorem 1.3.1, we may even assume that (y, z) E f. Define W := yHdH f3 zHe*H. Then we have
Iwl=
abcl bEHdH cEHeH
Also
x CW r
xH E (yH)(d H) M (zH)((e")*).
Thus, the number
I(yg)(d H) f3 (zH)((eH)*)I -- IWI nH does not depend on the choice of the pair (yH, zH) E fH. (v) follows from (iv) and from Lemma 1.2.5(iii).
[]
We do not know whether or not the hypothesis IX] E N can be deleted in Theorem 1.5.4. For each H E C, (X, G) H will be called the factor scheme of (X, G) over H.
22
1. Basic Results
Let p E II~, and let F C G be given. We shall say that F is p-valenced if, for each f E F, n/ is a power o f p .
L e m m a 1.5.5 Let p E I?, and let H E C be given. Assume that IX[ i s a power of p and that H is thin. Then G is p-valenced if and only if G / / H is p-valenced.
Proof. Let g E G be given. Then, by Theorem 1.5.4(v), n g H n H ~ TIHg H .
For each f E HgH, we conclude from Corollary 1.2.9(ii) that n/ = ng. (Recall that H is assumed to be thin.) Therefore, nHgH : ]HgHing, so that we have n g H n H --~
]HgHing.
By hypothesis, IX] is a power of p. Moreover, by Theorem 1.3.6(iv), IX[ =
nHIX/H]. Therefore, nH is a power of p. We wish to prove that [HgH] is a power of p, too. By hypothesis, H is thin. Therefore, IHI = nil, so that H is a power o f p . From Corollary 1.2.9(ii) we deduce that, for each x E X, the p-group
g((x,
• G((x,
acts transitively on HgH. 2 Thus, IHgHI is a power of p. Now the last equation says that ng is a power of p if and only nan is a power of p. Since g E G has been chosen arbitrarily, we have shown that G is pvalenced if and only if G / / H is p-valenced. []
1.6 Computing in Factor Schemes Throughout this section, we assume that IX] E N.
T h e o r e m 1.6.1 Let n E N \ {0}, and let F1, ..., Fn C_ G be given. Then, for each g E a and, for each H E C, gH E ( F 1 / / H ) ' " (Fn//H) if and only if g E ( H F 1 H ) . . . (HF, H). 2 Recall that 9((X, G)xH) denotes the group which is associated to the thin scheme (X, G)~H via the (~, T)-correspondence.
1.6 Computing in Factor Schemes
23
Proof. If n = 1, the claim follows immediately from Proposition 1.5.3. Therefore, we may assume that 2 < n. Assume first that gH E (F1//H)...(Fn//H). Then there exist e, f E G such that eH E (F1//H)... (F,_I//H), f H E Fn//H, and a~nyng, r O. Since aeHlHgH ~ O, there exist b E Hell and c E H f H such that abca r 0; see Theorem 1.5.4(iv). Thus, by definition,
g E (HeH)(HfH). On the other hand, by induction, eu E (F1//H)...(Fn-~//H) yields e E (HF1H)... (HFn_IH). Moreover, f g E F , / / H implies that f E HF, H; see Proposition 1.5.3. Therefore,
(HeH)(HfH) C (HF1H)... (HF,~H). Together, we obtain that g E (HF1H)... (HF,,H). Conversely, assume that g E (HF~H)... (HF,~H). Then there exist e E (HF1H)...(HF,~_IH) and f E HFnH such that g E ef. Since g E e f, a~fg r O. Thus, by Theorem 1.5.4(iv), a~,lHgn r O. This means that
gH E eH f H. On the other hand, by induction, e G (HF1H)...(HFn-IH) yields en E ( F 1 / / H ) ' " (Fn-1//H). Moreover, f E HF,~H implies that fHE F.//H; see Proposition 1.5.3. Thus, by Lemma 1.2.1(i),
eH f H C_ ( F 1 / / H ) . " (F,,//H). It follows that g n E ( F i f t H ) . . . (F~//H).
[]
Let H E C, and let F C_ G be such that H F H C F. Then F//U E C(G//H) if and only if F E C. C o r o l l a r y 1.6.2
Proof. Assume first that F//H E C(G//H). Let g E F ' F be given. Then g E (HF*H)(HFH). Thus, by Theorem 1.6.1 and Theorem 1.5.4(ii), gH E ( F ' / / H ) ( F / / H ) : (F//H)*(F//H) C_ F//H. Now Theorem 1.6.1 yields g E H F H C_ F. Since g E F*F has been chosen arbitrarily, we have shown that F E C. Conversely, let us now assume that F E C. Let g E G be such that gH E (F//H)*(F//H). By Theorem 1.5.4(ii), we have (F//H)* = F*//H. Therefore, gH E (F*//H)(F//H). Thus, by Theorem 1.6.1,
g E (HF*H)(HFH) = (H*F*H*)(HFH) C_ F*F C F. It follows that gH E F//H.
24
1. Basic Results
Since gH E (F//H)* (F//H) has been chosen arbitrarily, we have shown [] that F N H E C(G//H).
C o r o l l a r y 1.6.3 Let g E G, and let H E C be given. Then UgH = 1 if and only if g*Hg C H.
Proof. First of all, by Theorem 1.5.4(iii) and Lemma 1.2.6, nan
--
1 r
{l."}
--
(gH).gH.
Assume now that rig, -= 1, and let f E g*Hg be given. Then, by Lemma 1.2.1, f E (Hg*H)(HgH). Thus, by Theorem 1.6.1 and Theorem 1.5.4(ii),
Y" E ( g ' ) ' ( g ' ) = ( g " ) ' g " = {1"}. It follows that fH = 1H. Thus, by Proposition 1.5.3, f E H. Since f E g*Hg has been chosen arbitrarily, we have shown that g*Hg C H. Conversely, assume that g*Hg C_ H, and let f E G be such that fH E (gtt).gH We shall be done if we succeed in showing that fH = 1H. From Theorem 1.5.4(ii) we know that (gH). = (g,)H. Therefore, we have fH E (g,)HgH. Thus, by Theorem 1.6.1 and Lemma 1.2.1(ii), f E (Hg*H)(HgH) C_H. Therefore, by Proposition 1.5.3, fH = 1H. []
T h e o r e m 1.6.4 Let H E C, and let F C_ G be such that H C_ (F). Then (FIIH) = (F)IIH.
Proof. Let g E G be such that gH E (F//H). Then, by Theorem 1.4.1(ii), there exists n E 1~ such that gH E (F//H) n. Thus, by Theorem 1.6.1, g E (HFH)". Now, as we assume that H C_ (F), g E (F); see Theorem 1.4.1(ii). It follows that gH E (F)//H. Conversely, let g E G be such that gH E (F)//H. Then there exists f E (F) such that gH = fH. Since f E (F), there exists n E 1~ such that f E Fn; see Theorem 1.4.1(ii). Since gH = fH, g E H f H ; see Proposition 1.5.3. It follows that g E HFnH C (HFH) ~. Therefore, by Theorem 1.6.1, gH E (F//H) '~ C_ (F//H); see Theorem 1.4.1(ii) once again. D T h e o r e m 1.6.5 Let H E C, and let F C_G be such that (F//H) = G//H.
Then (F U H) = G.
1.6 Computing in Factor Schemes
25
Proof. From ( F f f H ) = G f l H we obtain that ( F ) / / H = GffH; see Theorem 1.6.4. Let g E G be given. Then there exists f E (F) such that gH = fH. [] Therefore, by Proposition 1.5.3, g E H f H C_ U ( F) U C_ ( F U H).
C o r o l l a r y 1.6.6 Set H := P(G). Let g E G be such that (gH) = G / / H . Then (g) = G.
Proof. This follows from Theorem 1.6.5 and Theorem 1.4.4(i).
[]
The following theorem describes a relationship between factor schemes of subschemes of (X, G) and subschemes of factor schemes of (X, G).
T h e o r e m 1.6.7 Let x E X be given. Let H, K E C be such that K C H. Then we have (i) K,H E C(H~:H), and ((X, G)~H) K*" is a scheme. (ii) H / / K E 6 ( G / / K ) , and ((X, G)K)(~K)(H//g) is a scheme. (iii) ( ( X , U)xH) "' ~K,, : ((X, G)K)(xK)(H//K).
Proof. (i) Since K E C, K~:H E C(HxH); see Lemma 1.5.2(iii). Therefore, by Theorem 1.5.1(ii) and Theorem 1.5.4(iii),
is a scheme. (ii) Since H E C, S I l K E C(G//K); see Corollary 1.6.2. Therefore, we deduce from Theorem 1.5.4(iii) and Theorem 1.5.1(ii) that
((x, is a scheme. (iii) First of all, we have
xH/I'G:H = {yI'( l Y E xH} = ( x K ) ( H / / K ) . But, for each h E H we also have,
(hxH) g~n --_ (hK)(xK)(H//K), whence
HxH//KxH = I h E H) =
26
1. Basic Results
{(h K )(~m(-//~) I h ~ H}
=
(H//K)(~:K)(H//K). It follows that
((x, a)~.) K=. = ((x, a)K)(.m(.//u). []
Let x E X be given. Let H, K E C be such that K _C H. For the remainder of these notes, we abbreviate
(x, c)~.I< := ( ( x , c ) ~ . ) ~='.
1.7 Morphisms Let (W, F) be a scheme, and let r
XUG
-4 W U F
be a map with X r C W and Gr _C F. Then r is called a morphism from (X, G) to (W, F) if, for all y, z E X and, for each g E G,
(y, z) E g ~
(yr zr E gr
Let us first collect several elementary facts about morphisms.
L e m m a 1.7.1 Let r be a morphism from (X, G). Then (i) For each g E G, g*r = (gr (ii) For all e, f E G, (ef)r C_ eCfr (iii) For each F C_ G, (F)r C_ (Fr
Proof. (i) Let g E G be given. Let y, z E X be such that (y,z) E g. On the one hand, (y, z) E g implies that (z, y) E g*, whence (zr yr E g*r On the other hand, (y, z) E g implies also that (yr162 E gr Therefore, (zr162 E (gr It follows that g*r f-)(gr ~ ~, so that g*r = (gr (ii) Let e, f E G, and let g E e f be given. Then, by definition, a~sg ~ O. Thus, there exist y, z E X such that (y,z) E g and ye n z f * ~ 0. Let x Eye (1 z f* be given.
1.7 Morphisms
27
Since (y, x) 9 e, (y4', x4') 9 e4'. Similarly, as (x, z) 9 f, (x4', z4') 9 re. But (y4,, z4,) 9 g4,, since (y, z) 9 g. Thus, a~r r ~- O. It follows that gr 9 e4,f4'. Since g 9 e f has been chosen arbitrarily, we have proved that (el)4' C_ e4'f4'. (iii) follows from (i) (ii), and from Theorem 1.4.1(i). [] L e m m a 1.7.2 Let (W, F) be a scheme, and let 4' be a morphism from (X, G) to (W, F). Then we have (i) 14' = 1w. (ii) For each E 9 C(F), E4' -1 9 C. (iii) Assume that 4' is surjective. Then, for all d, e 9 F, (de)4' -~ = d4'- l e r -1. Proof. (i) Let z E X be given. Then (x, x) E 1. Therefore, (xr x4,) 9 1r On the other hand, as x4, 9 W, (z4,, xr 9 l w . It follows that 14, = l w . (ii) Let E 9 C(F) be given. Then 1w 9 E. Thus, by (i), 14' 9 E, which means that 1 9 E r In particular, E4,-1 ~= 0. Let c, d 9 E4, -1, and let g 9 c*d be given. Then
g4, 9 (c'd)4' C__c'4'd4' = (c4')*d4' C_ E*E C_ E; use Lemma 1.7.1(i), (ii). It follows that g 9 E4' -1. (iii) Let d, e 9 F, and let g 9 (de)4' -1 be given. Let y, z 9 X be such that (y, z) 9 9. Then (y4', z4') 9 94' 9 de. Thus, by Lemma 1.2.4, there exists w 9 y4'd such that z4' 9 we. Since 4' is assumed to be surjective, there exists x 9 X such that x4' = w. Let b 9 G be such that (y,z) 9 b, and let c 9 G be such that (x, z) 9 c. Then (y4', xr 9 54, and (z4,, z4,) 9 c4,. It follows that 54' = d and c4' = e. Thus, g 9 bc C d4'-ler -1. Conversely, let g 9 dr -1 be given. Then there exist b 9 d4' -1 and c 9 er -1 such that 9 9 bc. Thus, by Lemma 1.7.1(ii), g4, 9 (bc)4, C b4'e4' = de, whence g 9 (de)4' -1.
[]
Let (W, F) be a scheme, and let 8 be a morphism from (X, G) to (W, F). Then we set r := r x W) and r
F).
Note that r (respectively, Ca) is not just the restriction of 4, to X (respectively, G). Also the codomain has changed.
28
1. Basic Results
L e m m a 1.7.3 Let r be a morphism from (X, G). Then (i) If Cx is surjective, Ca is surjective. (ii) / f Ca is injective, Cx is injective. (iii) Assume that Ca is injeetive. Then, for all y, z E X and, for each g E G, (yr zr E gr implies that (y, z) E 9.
Proof. (i) Let (W, F) be the scheme such that r is a morphism to (W, F). Let d E F be given. Then there exist t, u E W such that (t, u) E d. Since Cx is assumed to be surjective, there exist y, z E X such that yr = t and zr = u. Let 9 E G be such that (y, z) E g. Then, as r is assumed to be a morphism, (yr zr E 9r Therefore, (t, u) E 9r It follows that gr = d. (ii) Let y, z E X be such that yr = zr Then, by Lemma 1.7.2(i), (yr162 E 1r Let g E G be such that (y,z) E g. Then, as r is assumed to be a morphism, (yr zr E 9r It follows that 1r = 9r Thus, as Ca is assumed to be injective, we must have 1 = 9. In particular, y = z. (iii) Let y, z E X, and let 9 E G such that (yr zr E 9r be given. Let f E G be such that (y, z) E f. Then (yr zr E fr Therefore, f r = gr But, by hypothesis, Ca is injective. Therefore, f = 9. It follows that (y, z) E g. []
A morphism r from (X, G) is called a fusion if Cx is bijective. We shall not investigate fusions here. They will be used later to define quasi-direct products; see Section 2.6. A morphism r from (X, G) is called a homomorphism if, for all y, z E X and, for each g E G, (yr162 E 9 r
~
3v, w E X : v r 1 6 2
wr162
(v,w) Eg.
Note that, by Lemma 1.7.3(iii), an injective morphism from (X,G) is automatically a homomorphism. Let r be a homomorphism from (X, G). Then we set kerr := {9 E G 19r = 1r The set ker r will be called the kernel of r
T h e o r e m 1.7.4 Let r be a homomorphism from (X, G). Then the following conditions are equivalent. (a) Cx is injective. (b) r is injective. (c) Let y, z E X, and let g E G be such that (yr zr E gr Then (y, z) E g. (d) kerr = {1}.
1.7 Morphisms
29
Proof. (a) ==~ (b) Let e, f E G be such that er = f r Let y, z E X be such that (y, z) E e. Then, as er = r e , (yr zr E fr Thus, as r is assumed to be a homomorphism, there exists v, w E X with vr = yr we = zr and (v,w) E f. Since vr = yr v = y; see (a). Similarly, as wr = zr w = z. Therefore, (y, z) E f. It follows that e = f. (b) =~ (c) follows immediately from Lemma 1.7.3(iii). (c) =~ (d) Let g E ker r be given. Then, by definition, gr = 1r Let y, z E X be such that (y,z) E g. Then, as r is a morphism, (yr162 E gr It follows that (yr zr e 1r Thus, by (c), (y, z) E 1, whence g = 1. Since g E ker r has been chosen arbitrarily, we have proved (d). (d) :=~ (a) Let y, z E X be such that yr = zr Then, by Lemma 1.7.2(i), (yr zr E 1r Let g E G be such that (y, z) E g. Then, as r is a morphism, (yr162 E gr It follows that gr = 1r Thus, by definition, g E kerr Now (d) forces g = 1. In particular, y = z. []
Let (W, F) be a scheme. (X, G) and (W, F) will be called isomorphic if there exists a bijective homomorphism from (X, G) to (W, F). In this case, we shall write (X, G) ~ (W, r ) . For the next two theorems, we assume that IXI E H. T h e o r e m 1.7.5 [HOMOMORPHISMTHEOREM](i) Let H E C be given. For each x E X, we set xr := xH; for each g E G, we set gr := gH. Then r is a surjective homomorphism from (X, G) to (X, G) H which satisfies H = ker r (ii) Let r be a homomorphism from (X, G). Then ker r E C. Moreover, (Xr (Gr162 is a scheme with (Xr (Gr162 ~ (X, G) kerr
Proof. (i) Let y, z E X, and let 9 E G be such that (y, z) G g. Then, as 1 E H, z E yg C yHgH. Thus, by definition, (yH, zH) E gH, which means that (yr zr E gr Since y, z E X and g C G have been chosen arbitrarily, we have shown that r is a morphism. Now let y, z E X, and let g E G be such that (yr zr C gr Then, by definition, (yH, zH) E gH which means that z E yHgH. Thus, by Lemma 1.2.4, there exist v, w E X such that v E yH, w E zH, and (v,w) E g. By Theorem 1.3.1, v E yH is equivalent to vH = yH, and this means that vr = yr Similarly, we deduce that we = zr Thus, r is even a homomorphism from (X, G) to (X, G) u. It follows immediately from the definition of r that r is surjective. By definition, g E kerr if and only if g g = 1F. By Proposition 1.5.3, gH = 1F if and only if g C H. Thus, H = ker r
30
1. Basic Results
(ii) Set H := ker r Then, by Lemma 1.7.2(i), (ii), H E C. Let y, z E X, and let g E G be such that (y,z) E g. Then (yr162 E gr Thus, by Lemma 1.7.20), yr162
r
gr162
gEH
r
r
yH=zH.
In particular,
tx : X I H
-+ Xr
xH ~+ xr
is a bijection. From this, we obtain that, for all e, f E G, er : f r
3v, w,u, z E X : v r 1 6 2 3v, w,y, z E X : v H = y H ,
wr162
(v,~)ee, (~,z) ES r
wH=zH,
Hell = HfH
r
eH
(v,w) Ee, (y,z) E f =
r
fH;
see Lemma 1.2.4 for the third and Proposition 1.5.3 for the fourth equivalence. Thus, tG : G / / H -+ (Gr162 gH ~ (gr162 is a bijection. Let us now prove that tx U tc is a morphism. (From Lemma 1.7.3(iii) we know that then r is even a homomorphism.) Let y, z E X, and let g E G be such that (yH, zH) E gH. Then, by definition, z E yHgH. Thus, by Lemma 1.2.4, there exist v, w E X such that v E yH, w E zH, and (v,w) E g. Since r is a morphism, (v, w) E g implies that (re, we) E gr On the other hand, as v E yH, vH = yH, whence vr = yr Similarly, w E zH implies that we = zr Thus, we conclude that (yr zr E gr []
The homomorphism r in Theorem 1.7.5(i) will be called the natural ho-
momorphism from (X, G) to (X, G) H. T h e o r e m 1.7.6 [FIRST ISOMORPHISMTHEOREM]Let H, K E C be such that K C_ H. Then ((X, G)K) HIIK ~- (X, G) H.
Proof. Let us denote by r (respectively, ~b) the natural homomorphism from (X, G) to (X, G) K (respectively, from (X, G) K to ((X, G)K)H//K). Then r 1 6 2is a surjective homomorphism. On the other hand, we know from Theorem 1.7.5(i) that H / / K = ker r Therefore, we have ker(r162 = {g ~ G l a0 E H / / K } = {g E C I g ~ E H / / K } = H;
1.7 Morphisms
31
use Proposition 1.5.3 for the third equation. Now the claim follows from Theorem 1.7.5(ii). []
Let E, F C G be given, and assume that F • 0. We set NE(F) := {e E E ] Fe = eF}. NE(F) will be called the normalizer of F in E. Let H E C be given. Clearly, from the definition of N c ( H ) we deduce immediately that H C_ NG(H). But, in general, we do not have that N a ( H ) E C. In particular, the first claim of the following Lemma needs a proof.
L e m m a 1.7.7 Let H E C be given. Then (i) N a ( H ) H C_ No(H). (ii) Let F C_ G be such that U A F r 0. Then NH(F) C_ N H ( H N F).
Proof. (i) Let g E N a ( H ) H be given. Then there exists e E NG(H) such that g E ell. Since e E No(H), He = ell. Therefore, g E He. Thus, by Theorem 1.3.1, Hg = He = e H = gH. It follows that g E No(H). (ii) Let h E NH(F) be given. Then, by Lemma 1.3.5(i), h ( H A F) = H M h F = H M F h = (H A F ) h . []
Let H, K E C be such that K C_ H. K is said to be normal in H if H _C NG(K). In this case, we shall write K<~H.
Let H E C be such that {1} # H ~ G. H is called a minimal normal closed subset of G if, for each K E C such that K ~ G, K C_ H implies that K E {{1},H}. The pair (X, G) will be called simple if G is a minimal normal closed subset of G. In other words, (X, G) is simple if G has exactly two normal closed subsets, namely {1} and G. In Theorem 2.1.4, we shall exhibit the importance of simple schemes.
T h e o r e m 1.7.8
[SECONDISOMORPHISMTHEOREM] Let
H, K E C be
such that H C_ N c ( K ) . Then (i) K ~_HK.
(ii) H n K ~ H. (iii) A s s u m e that IX[ E l~. Then, for each x E X , (X,G)KHK ~( X , G ) ~H~K u 9
32
1. B~sic Results
Proof. (i) and (ii) follow immediately from Lemma 1.7.7. (iii) For each y E xH, we define yr := yK. For each h E H, we define (h~H)r := (hK)(xK)(HK//K). Then r
x g U HxH --+ (xlg)(Hl(//Ii) U (HIg//t~()(xK)(HK//K)
is a well-defined map with ( x H ) r C (xK)(HIgffIg) and (H~H)r C_ (HK//Ii)(~:K)(HK//K). We first shall prove that r is a morphism. Let y, z E xH, and let h E H such that (y,z) E h be given. Then z E yh C_ yKhK. Thus, by definition, (ylg, zK) E h K. But, as y E xH, yK E (xlg)(HK//Ig). Similarly, zK E
(xK)(HK//K). Therefore, (yr zr
= (yIg, zig) E (hK)(~:K)(HK//S:) = (h~g)r
Assume now that (yr zr E (h~u)r Then (yK, zK) E h E, whence, by definition, z E y K h K = yhK. In particular, yh N zK r 0. Let w E yh r zK be given. Then w E xH, we = zr and (y, w) E h::g. Thus, we have proved that r is a homomorphism. Let us now show that r is surjective. Let z E x H K be given. Then there exists y E xH such that yK = zK. Therefore, zK = yr Thus, r is surjective. But, by Lemma 1.7.3(i), this yields that even r is surjective. By Proposition 1.5.3, we know that, for each h E H, (hxH)r = ( l x g ) r
r
h g = 1g r
Therefore, ker q) = (H f3 K)xH. Now the claim follows from Theorem 1.7.5(ii).
h E K. []
Let y, z E X, and let H E C be given. In general, it is not true that
(X, G)yH ~-- (X, G)zH. But from Theorem 1.7.8(iii) we obtain immediately the following result. C o r o l l a r y 1.7.9 Assume that IX ] E N. Let H, K E C be such that H C_ N o ( K ) , H K = G, andHClI'( = {1}. Then, foreach x E X, (X,G)~H
(x,a) K Note that, by Lemma 1.7.7(i), the hypothesis H C_ N G ( K ) in Corollary 1.7.9 says exactly that K ~ G.
2. Decomposition Theory
In this chapter, we investigate to what extent it is possible to decompose (X, G) into smaller schemes.
2.1 Normal
Closed
Subsets
Let E, F C G be given, and assume that F r 0. We set
C (F) : : N (e 9 E I ef = r e } .fEF C E ( F ) will be called the ccntralizerof F in E. Note that E C_ C G ( F ) if and only if F C_ C o ( E ) .
L e m m a 2.1.1 Let H, K E C be such that H N K = {1}. Assume that g C N G ( K ) and that K C N c ( H ) . Then H C C a ( K ) .
Proof. Let h E H, let k E K, and let g E hk be given. We shall be done if we succeed in showing that g E kh. Since g E hk, g E hK. Since H C_ N G ( K ) , Kh = hK. Thus, g E Kh. Therefore, there exists f E K such that gEfh. It follows that g E f H . But, as K _C N c ( H ) , H f = f H . Thus, g E H f . Therefore, there exists e E H such that g E ef. Thus, as g E hk, Lemma 1.3.4 [] yields f = k. It follows that g E kh.
For the remainder of this section, we assume that IX] E N.
34
2. Decomposition Theory
T h e o r e m 2.1.2 Let H, K, L E C be such that L C K C_ H. Then we have (i) K I l L ~_ H I l L if and only if, for each h E H, K h L = L h K . (ii) A s s u m e that L <~ H. Then K I l L <~ H I l L if and only if K <1 H.
Proof. (i) From Theorem 1.6.1 we deduce that, for each h E H, (KffL)h L = hL(K//L)
r
Khi = LhK.
This proves the claim. (ii) Let h E H be given. Assume that K h L = L h K . Then K h = K L h = K h L = L h K = h L K = hK.
Conversely, if K h = h K , K h L = h K L = h L K = L h K . (Recall that we assume that Lh = hL.) Now (ii) follows from (i). []
Now we shall generalize two fundamental theorems on finite groups. We start with the generalization of the famous theorem [31; II. w of H. ZASSENHAUS.
T h e o r e m 2.1.3 Let H, H', K , K ' E C be given. Assume that H ~_ H ' and K <~ K ' . Then we have (i) H' n H K ,q H' n H K ' . (ii) For each x E X , (X, ( 2 " ~ H ' n H K "~ I V [u'tH'oK'NHK ~-~]x(H'nHK')
:
V',
u/x(H,nK,)
Proof. By Lemma 1.3.5(i), we have H'nHK=H(H'nK),
H'nHK'=H(H'nK').
Therefore, by Theorem 1.3.3, H InHK,
H'nHK
IEC.
From H ' M H K ' E C we obtain that
(H' n K ' ) ( H ' n H K ) c_ H' n H K ' . Conversely, the equation H ~n H I ( ~ = (H' M I(~)H yields H ' n H K ' C_ (H' n K ' ) ( H ' n H I ( ) ;
see Lemma 1.2.1(i). Therefore, we have
9
2.1 Normal Closed Subsets
35
(H' M K')(H' N H K ) = H' r HK'. Finally, by Lemma 1.7.7(ii),
H' M K' C NH, (HK) C NH, (H' f3 HK). Thus, the theorem follows from Theorem 1.7.8(i), (iii).
[]
Let _<1Sdenote the transitive extension of ~ on C. (Obviously, C is a partially ordered set with respect to ~ , i.e., ~ is a reflexive, anti-symmetric, and transitive relation on C.) A subset 7/ of C is called a subnormal series of G if, for all K, L E 7/, K ~<1 L or L <~<3 K. A maximal subnormal series of G which contains G will be called a composition series of G. Let 7/ be a composition series of G. For each H E 7 / \ {G}, we set H ~ :=
N K. HCKET,I\{H)
H ~ will be called the upper neighbour of H in 7/. Two composition series 7/and/(; of G are called isomorphic, if there exists a bijection ~/: 7 / - + K; such that for each z G X and, for each H E 7 / \ {G}, we have H " r G and (X,
n ~= (x, G)~H.~:. G)~n~ n.
Now we are ready to generalize the fundamental group-theoretical theorem [16; w due to C. JORDAN and O. HOLDER.
T h e o r e m 2.1.4 Any two composition series of G are isomorphic.
Proos Let 7/ and/~ be composition series of G, and let H E 7 / \ {G} be given. Then there exists a uniquely determined element H n in/(: \ {G} such that H ~ C H H npc, H ~ ~=HH ~. Since H n C_ H H '~tc,
H ~ M H H € = H ~. Thus, by Theorem 2.1.3(i),
H <~H~ t O H H ~ <~H n. Since H ~ ~= H H ~, H ~t M H H ~ ~ H ~. Thus, as 7/ is assumed to be a composition series of G, we must have
36
2. Decomposition Theory H = H ~ fq H H " .
Now, by Theorem 2.1.3(ii), we have that, for each x 6 X, H ~ (X, (2XH~nHuPCnHH" (X, G)xHn = Ujx(HnnHn~C ) 9
Define x : AS\ {G) --+ 7t \ {G} analogously to rI. Then H "x: C_ H'TH '7,~u, H ~ ~: H ' H ~ . From the above given isomorphism we deduce that H og (~ H H '7. Therefore, H '#c (t H'TH. Thus, by definition of a, H C H 0~. On the other hand, as above, the fact that AS is a composition series of G yields
H" = H ~ fq HO H 0~. Therefore, H ~ C H "~ would lead to H ~ N H "[ C_ H ~, which contradicts the
above given isomorphism. It follows that H = H 0~ . Now we have shown that rI and g are inverse to each other. In particular, y : 74 \ {G} -+ AS\ {G} is bijective, and by the above given isomorphism, we have that, for each x 6 X , ( X , G)HH n ~= (x, G)~/~.,~. Hn []
Theorem 2.1.4 was first proved as Theorem 4.2 A scheme (IV, F ) will be called a composition exists a composition series 74 of G, an element H 6 that (W, F ) =~ (X, G)~H~. H It follows immediately from Theorem 2.1.2(ii) are simple.
2.2 Strongly
Normal
Closed
in [25]. factor of ( X , G ) if there 74 \ {G}, and x E X such that composition factors
Subsets
Let E, F C_ G be given, and assume that F # 0. Then we set S E ( F ) := {e 6 E I e ' F e C_ F}. Note that, for each H 6 C, H C SG(H).
L e m m a 2.2.1 N G ( H ) C SG(H).
Let H, K 6 C be such that K C_ H. Then SG(K) f3
Proof. Let g @ SG(K) M N G ( H ) be given. Then g*Hg = g*gH C [] g * K g H C K H C H; see Lemma 1.2.1(i).
2.2 Strongly Normal Closed Subsets
37
For the remainder of this section, we assume that IXI E N.
L e m m a 2 . 2 . 2 Let H E C, and let F C_ G be such that F Tk O. Then (i) SH(F) E 6. (ii) SH(F) C_ N H ( F ) .
Proof. Let g E SH(F) be given. Since IX[ is assumed to be finite, there exists n E N such that g* E {9}n; see Theorem 1.4.1(ii). Thus, by Lemma 1.2.3(ii), g E {g*}". It follows that gFg* C_ {g*}"F{g}" C F; see I,emma 1.2.1. Therefore, we have g* E Stt(F). Since g E SH(F) has been chosen arbitrarily, we have shown that S.(F)" = (i) Let g E S H ( F ) * S H ( F ) be given. Then there exist, d, e E SH(F) such that g E d*e. By the above paragraph, d* E SH(F). Therefore, using Lemma 1.2.1 once again, we obtain that g*Fg C_ e'dFd*e C_ F. This means that g S.(F). (ii) Let g E SH(F) be given. By Lemma 1.2.5(iii), 1 Egg*. Thus, by Lemma 1.2.1, Fg C_ gg*Fg C_ gF. But, as g E SH(F), the introductory observation (or (i)) yields that also g* E SH(F). Therefore, we have gF C_ [] gFg* g C_ Fg, too.
Let H, K E C be such that K C_ H. K is said to be strongly normal in H if H C_ So(K). In this case, we shall write
K <~ H. Recall that a non-empty subset F, say, of G is called thin if { 1} = {n! I f E F}. Moreover, we say that (X, G) is thin if G is thin.
T h e o r e m 2.2.3 Let H, K E C be such that K C_ H. Then (i) / f K <~ H, K <J H. (ii) K _<~-H if and only if, for each x E X, (X, G)xHK is thin.
Proof. (i) follows from Lemma 2.2.2(ii). (ii) Assume that K _~1 H. Then, by definition, H C_ SG(K). This means that, for each h E H, h*Kh C K. Therefore, for each h E H, nh~," = 1; see Corollary 1.6.3. On the other hand, by Theorem 1.5.1(iii), we have that, for each h E H and, for each x E X, n(hl<)(~K)(H//K) ~ nhK.
38
2. Decomposition Theory
Now Theorem 1.6.7(iii) yields that, for each x E X, (X, G)~KHis thin. Conversely, assume that for each x E X, (X, G)xgH is thin. Then, by Theorem 1.6.7(iii) and Theorem 1.5.1(iii), {1} = {nh~," I h E H}. Therefore, by Corollary 1.6.3, we have that, for each h E H, h*Kh C_ K. But, by definition, this means that K _
T h e o r e m 2.2.4 Let H, K E C be such that K C_ H. Then SC//K ( H / / K ) --- S c ( H ) / / K .
Proof. Let g E G be given. Then, by Theorem 1.6.1 and Theorem 1.5.4(ii), g g E SG//K(H//K) r (gK)'(H//K)gK C H//K r Kg*HgK C H r g*Hg C_ H r g E SG(H) gK E S c ( H ) / / K . Now assume that gK E S G ( H ) / / K . Then, by definition, there exists f E SG(H) such that gK = f K . Thus, by Proposition 1.5.3 and Lemma 2.2.2(i), g E K f K C_ H f H C_ Sa(H). []
Let H E C be given. We set (S,)~
:--- {1},
and, for each n E l~ \ {0}, we define (SH)n({1}) : : SH((SH)n-I({1})).
T h e o r e m 2.2.5 For all m, n E N, we have (SG//(SG),,({1}))n({1}) : (SG)m+n({1})//(SG)m({1}).
2.3 Thin Residues and Thin Radicals
39
Proof. T h e claim is obvious for n = 0. Therefore, we assume that 1 < n. By induction, we also assume that the claim holds for n - 1. Then (Sv//(s~)~({1}))" ({ 1 }) = Sc//(sa).~({ x}) ((Sa//(so).~({ 1}))"- 1({ 1 })) = SG//(S~),-({1})((SG) m + " - l ( { 1 } ) / / ( s a ) m ( { x } ) )
=
Sa((Sa)'~+"-I({1}))//(Sc)m({1}) = (sa) "~+=({U)//(sa)"~({1}); []
use T h e o r e m 2.2.4 for the third equation.
2.3 T h i n R e s i d u e s and T h i n R a d i c a l s Let H E C be given. We set 0 ~ ( H ) :=
i
K.
K
0 ~ ( H ) is called the thin residue of H. T h e third p a r t of the following theorem gives a useful characterization of the thin residue of H .
Theorem
2.3.1
(i) 0 ~ (ii) O~
Ec. <3t H.
(iii)
O~
Let H E C be given. Then we have
= (UheH h'h).
Proof. (i) follows from L e m m a 1.3.3(i). (ii) For each h E H , we have
h*O~
N
K)hC_ N
K ,~t H
h*KhC
K
N
K=O~
K
Thus, the claim follows f r o m (i). (iii) Set
J:=(U
h'h),
hEH
and let e, f E H be given. First of all, we shall prove that
f*e*ef C_J.
40
2. Decomposition Theory
Let h E e f be given. Then, by Lemma 1.2.5(i), e E h f*, whence h*ef C h * h f * f C_ J. Since h E e f has been chosen arbitrarily, this yields f*e*ef C_ J; see Lemma 1.2.2. Since e, f E H have been chosen arbitrarily, we conclude that J <~ H. Thus, by definition, O ~ (H) C J. C_ O~ Thus, Conversely, by (ii), for each h E H, h*h C_ h*O~ by (i), J C 0 ~ []
The pair (X, G) is called primitive, if { 1} is a maximal closed subset of G. Assume that (X, G) is primitive and not thin. It seems to be an interesting question whether or not, in this case, we even have
U g*g=G. gEG
We do not know the answer.
C o r o l l a r y 2.3.2
Let H, K E C be such that K C_ H. Then O~
C
O~
Proof. This follows immediately from Theorem 2.3.1 (iii).
[]
P r o p o s i t i o n 2.3.3 Let H, K E C be such that H K = G. (i) I f O ~ C_ H, O~ = Noo(K)CL,~, H L. (ii) 0 ~ (K) C H if and only if 0 ~ (G) C_ H.
Proos (i) Let us define L::={LC_G IO~
CL_~H}.
Let g E G, and let L E s be given. Since g E G and G = H K = K H , there exist h E H and k E K such that g E kh. Thus, as L E ~:, Theorem 2.3. l(iii) yields
g*g C h*k*kh C_ h*O~
C_ h*Lh C_ L.
Now, as g ~ G has been chosen arbitrarily, we conclude that O~ C L; see Theorem 2.3.1 (iii). But also L E /~ has been chosen arbitrarily. Therefore, we have at least that
2.3 Thin Residues and Thin Radicals O~
41
C_ A L. LEt:
In particular, O ~ (G) C H. Therefore, we have O ~ (G) <1~ H. On the other hand, by Corollary 2.3.2, O ~ C O~ Therefore, O~ E/2. It follows that N n C= 0 '~(G). LEs
(ii) follows from (i) and Corollary 2.3.2.
[]
The term "thin residue" is justified by the following theorem. Under the hypothesis that IXI E N this theorem says that the thin residue of G is the uniquely determined smallest closed subset of G the factor scheme of which is thin.
T h e o r e m 2.3.4 Assume that IX[ E N. Then we have (i) (X, G) ~ is thin. (ii) Let H E C be such that (X,G) H is thin. Then O~
C_H.
Proof. (i) follows readily from Theorem 2.2.3(ii) and Theorem 2.3.1(ii). (ii) follows from Theorem 2.2.3(ii). []
Let H E C be given. We set (O~)~
:= H,
and, for each n E 1~ \ {0}, we define (O~
:= O~
T h e o r e m 2.3.5 Assume that IXI E N. Let x E X be given, and let H, K C C be such that K C H. Then we have (i) O~ = (O~ (ii) For each n E N, _-
42
2. Decomposition Theory
Proof. (i) For each F C K, we have F~H ~_n KxH if and only if F _<3~K; see Lemma 1.5.2(i), (iii). Therefore,
~
=
A
L=. :
LxH4_tK.H
l]
L=. : ( A
L~K
L)=. =
L~_tK
(o~
(ii) The claim is obvious for n = 0. Therefore, we assume that 1 _< n. By induction, we also assume that the claim holds already for n - 1. Then
(o~
=
O0((o0)n-l(KxH)) :
O0(((O0)n-l(K)).H) (O• ( ( o 0 ) n - l ( f ~ ) ) ) x H
=
=
((o~ use (i) for the third equation.
[]
T h e o r e m 2.3.6 Assume that IX] E N, and let H E C be given. (i) Let K E C be such that K C_ O~ Then O~(HIIK) = O~(H)ffK. (ii) Let n E I~ be such that H C_ ( 0 e)"(G). Then
(O~
= (O~
Proos (i) Let L E C be such that K C_ L C_ H. Then, n ~ H r and
H _C Sa(L) t:~ H / / K C_S a ( L ) / / K
L//K ~ #//K ~ H//K c Sc//~(L//K).
On the other hand, by Theorem 2.2.4, we have that
SG//K(LffK) = SG(L)ffK. Thus,
L ~_~H r L//K ~_~H//g. Now recall that, by Theorem 2.3.1(ii), O ~
O~
_<3~H. Therefore,
~_~H/iS<.
Thus, by definition,
o~(n//K) c o~(~)//K.
2.3 Thin Residues and Thin Radicals
43
Let L C H be such that O~ = L//K. Then, by Theorem 2.3.1(ii), L//K <~ H//K. Therefore, L <11 H, so that, by definition, O ~ (H) C__L. It follows that O~ C_L//K = O~ (ii) The claim is obvious for n = O. Therefore, we assume that 1 _< n. By induction, we also assume that the claim holds already for n - 1. Then (o~
=
OO((od)n-l(G//g)) :
O~176
=
0'~ ((o'~)"-1(C,))11 g =
(O~ use (i) for the third equation.
[]
Let H E 17 be given. We set Oo(H) := {h E H ]nh = 1}. Oo (H) is called the thin radical of H. Clearly, G is thin if and only if O0 (G) = G. But, in general, we do not even have that O0(G) E C. On the other hand, if IXI E N, we do have that, for each H E 17, O0(H) E 17; see the second part of the following theorem. For the remainder of this section, we assume that IX[ E 1~.
T h e o r e m 2.3.7 Let H E C be given. Then (i) o 0 ( g ) = SH({1}). (ii) Oo(H) E C.
Proof. (i) Let h E Oo(H) be given. Then, by Lemma 1.2.6, h*{1}h = h*h = {1}. Thus, h E SH({1}). Conversely, for each h E SH({1}), h*h = h*{1}h C_ {1}, whence, by Lemma 1.2.6, h E O0 (H). (ii) follows from (i) and Lemma 2.2.2(i). []
We conclude this section with two consequences of Theorem 2.3.7.
T h e o r e m 2.3.8 For each H E C, Oo(G//H) =- Sc(H)//H.
44
2. Decomposition Theory
Proos This follows readily from Theorem 2.2.4 and Theorem 2.3.7(i). [] Let H E C be such that HOo(G) = G. Then we have (In particular, irOn(G) has a supplement in G, O~ #
T h e o r e m 2.3.9
O~
= O~
G.) Proos Apply Proposition 2.3.3(i) to Oe(G) in the role of K, and use Theorem 2.3.7(ii). []
In the following section, we investigate the case where each composition factor of (X, G) is thin.
2.4 Residually Thin Schemes Throughout this section, we assume that IX[ E N. The pair (X, G) will be called residually thin if each composition factor of (X, G) is thin. For each H E C, we set
O~
:= ['-] (OO)'~(H). nEN
T h e o r e m 2.4.1
(X, G) is residually thin if and only irOn(G) = {1}.
Proos Assume first that O~(G) r {1}, and let 7-/be a composition series of G with O~(G) E 7/. Then there exists H E 7/ such that H ~t = O e ( G ) . (Recall that H ~t denotes the upper neighbour of H in 7/.) It follows that O~ ~) = H n. Therefore, by Theorem 2.2.3(ii), there exists z E X such that (X, G)xHn U is not thin. Thus, by definition, (X, G) is not residually thin. Assume now that O e ( G ) = {1}, and set K; := {(OO)'~(G) ] n E N}. Then {1} E /(: and, by Theorem 2.3.1(ii) and Theorem 2.2.3(i),/C is a subnormal series of G. Let 7 / b e a composition series of G such that/(: C_ 7/, and let H E 7 / \ {G} be given. Then, as { 1, G} E/C, there exists n E N such that
(o~)"+~(c) c_ H ~ H ~ _c (o~)"(c).
2.4 Residually Thin Schemes
45
By Lemma 2.2.1, this yields H __~ H u. Thus, by Theorem 2.2.3(ii), for each x E X , (X, G)xHn H is thin. Since H E 7i \ (G} has been chosen arbitrarily, we have proved that (X, G) is residually thin; see Theorem 2.1.4. []
L e m m a 2.4.2 Let x E X, and let H E C be given. Then O0(HrH) = (O0(H))~:H.
Proos We have OO(Hx,,) =
N (O~ hEN
= N ((O0)n(U))~H = I N (O~)'*(U)]~g = nEN
hEN
( O ~ (H))~:H; see Theorem 2.3.5(ii) for the second equation.
O~
[]
T h e o r e m 2.4.3 Let H, K E C be such that H C_ N c ( K ) . Assume that = {1}. = {1} = O e ( K ) . Then O ~
Proof. Let x ~ X be given. Then, as O ~ = {1}, OO(HrH) = {1}; see Lemma 2.4.2. Thus, by Theorem 2.4.1, (X, G),H is residually thin. In particular, (X, G),H~Ig is residually thin, so that, by Theorem 1.7.8(iii), (X, G)KHK must be residually thin. From O ~ = {1} we deduce similarly that (X, G),K is residually thin. It follows that (X, G)xHK is residually thin, so that, again with the help of Theorem 2.4.1 and Lemma 2.4.2, we get O ~ = {1}. []
From Theorem 2.4.3 we deduce that there exists a uniquely determined biggest element H E C subject to H _ G and O ~ ( H ) = (1}. We shall denote this closed subset by H(G). Clearly, (X, G) is residually thin if and only if H(G) = G. For each H E C, we set O e ( H ) :=
U (SH)'({1}). hEN
Note that, by Lemma 2.2.2(i), O o ( H ) E C.
T h e o r e m 2.4.4 If Oo(G) = G, O~
= {1).
46
2. Decomposition Theory
Proof. Assume that Oo(G) = G. Then, as IGI E 1~, there exists n E N\{0} such that (So)n({1}) = G. Therefore, by definition, (O~)~ = (SG)n({1}). We now shall prove that, for each i E { 1 , . . . , n } , (O~)I(G) _C (Sa)'*-i({1}). Assume, by induction, that
(OO)i-l(G) C_ (SG)"-~+I({1}) for some i E { 1 , . . . , n } . By definition, (Sc)'~-i({1}) <~ (SG)n-i+I({1}). Therefore, (O'~)i(G) = O ~ 1 7 6
use Corollary 2.3.2. It follows that (O~
C_ O~ = {1).
C (Sc)'-i({1}); []
The converse of Theorem 2.4.4 does not hold. The Coxeter group of order 48 provides a counterexample.
Theorem 2.4.50o(G//O,~(G)) = 0o((7)//0o((7).
Proof. Set m = 1 in Theorem 2.2.5, and use Theorem 2.3.7(i). Then, for each n E N, = (SG)"+I//o0(G).
This proves the theorem.
[]
For the remainder of this section, let p E P be given. If IXI is a power of p, there is a variety of ways to express that (X, G) is residually thin. In particular, in this case, the converse of Theorem 2.4.4 holds, as the following theorem shows.
T h e o r e m 2.4.6 Let p E I? be given, and assume that IX I is a power of p. Then the following conditions are equivalent.
(a) O~(G) = G. (b) O ~ = {1}. (c) G is p-valenced) 1 Recall that a subset F of G is defined to be p-valenced if, for each f E F, n! is a power of p.
2.4 Residually Thin Schemes
47
Proof. (a) ::~ (b) holds generally; see Theorem 2.4.4. (b) ::V (c) Let (X, G) be a counterexample in which IXI is minimal. Then there exists n E N \ {0} such that (OO)'~+I(G) = {1} -r (OO)n(G). Set H := (O~ Then {1} # H C_ Oo(G). From Theorem 2.3.6(ii) we obtain that (O'~)'~(G//H) = {1}. Thus, by definition, we also have that
O~
= {1}.
On the other hand, from Theorem 1.3.6(iv) we conclude that IX~HI is a power of p and strictly less than IX]. Therefore, the choice of (X, G) yields that G//H is p-valenced. Now Lemma 1.5.5 says that G is p-valenced, too. This contradicts the choice of (X, G). (c) ~ (a) Let (X, G) be a counterexample in which ]X] is minimal. By hypothesis, ]X I is a power of p and, by Theorem 1.3.6(iv), ~geG ng= ]X I. Thus, as G is assumed to be p-valenced, O0(G) r {1). (Recall that nl = 1.) On the other hand, the hypothesis that G is p-valenced implies also that G//Oo(G) is p-valenced; see Lemma 1.5.5. Thus, the choice of (X, G) leads to
Oo(G//O~(G)) = G//Oo(G). (By Theorem 1.3.6(iv), ]X/Oo(G)] is a power ofp.) But, by Theorem 2.4.5, we also have that
o o ( a / / o o (a)) = o o ( a ) / / o o ( c ) . Therefore, Oo(G) = G, contrary to the choice of (X, G).
[]
Let H E C be given. Then H is called a closed p-subset if nH is a power of p and if H is p-valenced. We shall denote by
the set of all closed p-subsets of G. The pair (X, G) will be called a p-scheme if it is residually thin and if IX] is a power of p. Note that, if (X, G) is a p-scheme, each Composition factor (W, F), say, of (X, G) satisfies ]W] = p = IF[. We conclude this section with two corollaries of Theorem 2.4.6.
C o r o l l a r y 2.4.7 Let p E P be given. Then G E gp if and only if (X, G)
is a p-scheme.
48
2. Decomposition Theory
Proof. This follows immediately from Theorem 2.4.6 and from Theorem 2.4.1. [] C o r o l l a r y 2.4.8 Letp E F, and let H, K E Cp be such that H C_ N o ( K ) . Then H K E Cp.
Proof. By hypothesis, nH and nK are powers of p. Thus, by Theorem 1.3.7(iii), n H K is a power of p. Let x E X be given. Since H is assumed to be p-valenced, HxH is /9_ valenced, too. Thus, by Theorem 2.4.6, O ~ = {1}. Therefore, by Lemma 2.4.2, O ~ = {1}. Similarly, we deduce that O ~ ( K ) = {1}. Thus, by Theorem 2.4.3, O~ = {1}. Using Lemma 2.4.2 and Theorem 2.4.6 once again, we conclude that H K is p-valenced. Thus, we have proved that H K E Cp. [] Let p E IF be given. From Corollary 2.4.8 we deduce that there exists a uniquely determined biggest normal closed p-subset of G. Following the theory of finite groups, it might be reasonable to denote this closed subset by Op(G). Clearly, by definition, we have Op(G) C_ H(G). It seems to be promising to investigate the way in which the structure of (X, G) depends on {Op(G) I P e I'}.
2.5 D i r e c t P r o d u c t s of C l o s e d Subsets Let H E C, let n E l~l \ {0}, and let H1, ..., Hn E (J be given. We shall say that H is the direct product of ill, ..., Hn if H 1 . . . H , , = H, if, for each i E { 1 , . . . , n } , Hi
n(
U
Hi)
= {1},
is{1 .....~}\{i} and if, for all i, j E { 1 , . . . , n } , Hi C_ N c ( H j ) . To denote that H is the direct product of H1, ..., H,~, we shall write H1 x . . . x H , = H . Moreover, for each i E { 1 , . . . , n}, we shall abbreviate
2.5 Direct Products of Closed Subsets
49
je{1 .....,~}\{i} If H is the direct product of H1, ..., Hn, we have that, for each i E { 1 , . . . , n}, Hil:Ii = H, Hi r3/:/i = {1}, and Hi C_ Na([ti). Moreover, for all i, j E { 1 , . . . , n } with i # j, we have Hi C_ C a ( H i ) ; see Lemma 2.1.1.
T h e o r e m 2.5.1 Let H E C, let n E 1~\{0}, and let H1, ..., Hn E C be such that H I " " H n = H. Assume that, for all i, j E { 1 , . . . , n } , Hi C N e ( H j ) . Then the following conditions are equivalent.
(a) H1 x . . . x H n = H . (b) Let h E H be given. Then there exists, for each i E { 1 , . . . , n } , a uniquely determined element hi E Hi such that h E { h i } . . . {h,}.
Proof. The claim is obvious for n = 1. Therefore, we assume that 2 _< n. (a) ~ (b) Since h E H,~/;/,~, there exist uniquely determined elements e E Hn and f E/~n such that h E e f; see Lemma 1.3.4. By induction, there exists, for each i E { 1 , . . . , n - 1}, a uniquely determined element hi E Hi such that f E { h l } . . . { h n - 1 } . (b) ~ (a) The only fact we have to prove is that, for each i E { 1 , . . . , n}, Hi (3 f/i = {1). From the hypothesis we conclude that, for all i, j E { 1 , . . . , n}, HiHj = Hj Hi. Therefore it is enough to prove that H,~ O/2/, = { 1}. Let g E H , n/7/, be given. Since /~/n = H 1 . - . Hn-1 and g E /2/,~, g E H1 .-. H n - 1. Thus, there exists, for each i E { 1 , . . . , n - 1}, an element hi E Hi such that g E { h l } - . - { h n _ l } . On the other hand, g E H , . Thus, the uniqueness assumed in (b) yields g=l. []
L e m m a 2.5.2 L e t H E C, let n E I ~ \ { 0 } , and let H1, ..., H,~ E C be such that Hx x ... x Hn = H. Then we have (i) For each h E H, h E (H1 r) h [ I 1 ) . . . (Hn n h f I , ) . ^(ii) Let K E C such that K C_ H. Assume that (H~ n K f t a ) . . . ( H , , n K g , ) = H. Then, for each i e { 1 , . . . , n}, IiI:Ii = H.
Proos (i) Let h E H be given. Then, as H1...H,~ = H, there exists, for each i E { 1 , . . . , n } , an element hi E Hi such that
h see Lemma 1.2.4.
{h,}...{h.};
50
2. Decomposition Theory
Let i E { 1 , . . . , n } be given. Then, for each j E { 1 , . . . , n } \ { i } , Hi13Hj = {1}, Hi C_ N a ( H j ) , and H A C_ N c ( I t i ) . Thus, for each j E { 1 , . . . , n } \ {i}, hihj = hjhi; see Lemma 2.1.1. Therefore, h E hi[Ii. Thus, by Theorem 1.3.1, hi E hlSli. It follows that hi E Hi O hlr2Ii. Since i E { 1 , . . . , n } has been chosen arbitrarily, we have proved that h E (H1 13 h / ~ l ) ' ' ' (Hn f3 hi?In). (ii) Let i E { 1 , . . . , n } , and let h E Hi be given. Then h ~_ H, so that, by hypothesis, h E (H1 13 K/2/1) .-' (H, 13 K/2/,). Therefore, there exists, for each j E { 1 , . . . , n}, an element hj E Hj 13 KI?tj such that h E { h i } . . . {h,}. Thus, by Theorem 2.5.1, h = hl. It follows that h E K/:h. Since h E Hi has been chosen arbitrarily, we have proved that Hi C_ K.//i. It follows that H = HiI?Ii C_ KI2Ii. []
Let H E C be such that {1} ~ H. Then H will be called decomposable if there exist K, L E C such that If ~ {1} ~ L and K • L = H. H will be called indecomposable if it is not decomposable. By definition, {1} is neither decomposable nor indecomposable.
T h e o r e m 2.5.3 Assume that [GI E N,and let H E C be such that {1} H. M•
(i) Let K , L E C be such that K • L = H. Let M , N E C be such that ThenK•215 (ii) H is the direct product of indecomposable closed subsets.
Proof. (i) Since K L = H and M N = L, K M N = H. We also have K 13 M N = K 13 L = {1}. Let n E N f'l K M be given. Then there exist k E K and m E M such that n E km. Thus, by Lemma 1.2.5(ii), k* E ran* C_ M N . It follows that kEKfqL= {1}.Therefore, m = n . Thus, a s M M N = { 1 } , n = 1. Since n E N 13 K M has been chosen arbitrarily, we have shown that N 13 K M = {1}. Similarly, we obtain that M 13 K N = {1}. Since M x N = L, M C N a ( N ) and N C_ NG(M). Since K x L = H, M N = L C N a ( K ) . With the help of Lemma 2.1.1 we obtain that
K C_ CG(L) C_ C a ( M ) C N a ( M ) . Similarly, we deduce that K C_ N a ( N ) . (ii) follows from (i) by induction.
I-1
2.5 Direct Products of Closed Subsets
51
The pair (X, G) is called commutative if, for all e, f E G, ef = fe.
P r o p o s i t i o n 2.5.4 Assume that iXI E N, and assume that (X,G) is commutative. Let H E C be given. Let m, n E N \ {0}, let K1, ..., Km E C, and let L1, ..., Ln E C be given. Assume that, for each i E { 1 , . . . , m ) , Ki is indecomposable and that, for each j E { i , . . . , n}, Lj is indecomposable. Assume that K1 x . . . x
Lj
:
Km : H = L1 x . . . x
Then, for each i E { 1 , . . . , m } , H = Lj X Ki.
Ln.
there exists j E { 1 , . . . , n } such that Ki
•
Proof. Let i E {1,..., m) be given, and set
/~', := (L1 n
K~tl)--. (L,~ n K i L , ) .
For each j E {1,..., n}, we set Lj := (K1 N L j f ; 1 ) . . . ( K m nLj[(m). Let us first assume that/-~'i_# H. By Lemma 2.5.2(i), K~ C_ K~. Therefore, by Lemma 1.3.5(i),
L
= K,
(k~ n/~', )
On the other hand, /q
n (fq n/~'~) c_/q n fi~ = {1).
Thus, as (X, G) is assumed to be commutative, we have K~ • (k~ n f~,) : ~'~. By definition, we also have (L1 n Ii~L1) • . . . • (Ln n K~Ln) =/~'~.
By Theorem 2.5.3(i), we may refine both of the above representations of f~'i to direct products of indecomposable closed subsets. Thus, as ]/~'iI < ]H I - 1, induction yields j E { 1 , . . . , n } and C C Lj n K~Lj such that C is indecomposable and From C C_/~'i and C n (f~'i n/~'i) = {1} we conclude that C n/~'i : {1}. On the other hand, we have that
52
2. Decomposition Theory
Therefore, C • I['I = H. From Cf3 [4i = {1} we conclude that C N (Lj n/~'i) = {1}. From C['(i = H and C C_ Lj we obtain that C ( L j A h'i) = Lj ~ C[4~ = Lj; see Lemma 1.3.5(i). Thus, C x (Lj N/~'i) = Lj. But, by hypothesis, Lj is indecomposable. Therefore, Lj -- C. Thus, as C • /~'i = H, Lj • Ki = H. From Lj = C and C C_ K i L j we obtain that Lj C_ Ki]-j. Thus, H = Lj [~j C_ K i L j . It follows that Ki Lj = H. On the other hand, as Lj • J~1i = H = Ki • Ki, Theorem 1.3.7(iii) yields hE, = nLj. (Recall that we are assuming that ]XI E N.) Therefore, using the same theorem once again, H = K i Z j yields that Ki
x L j -~.
H.
Let us now assume that /(i = H. Our first claim is that there exists j E { 1 , . . . , n } such that Lj[ii = H. Assume false. Then, for each t E { 1 , . . . , n}, H # Lt ;see Lemma 2.5.2(ii). Thus, similar to the first case, we obtain s E { 1 , . . . , m} such that Lt x [~, = H = K , • Lt. It follows that s # i. Note that K1//Ks x ... x Ks_l//K,
• Ks+l//K, • ... x Km//Ks
= H//K, =
L1//K, • 2 1 5
L , _ I / / K , • L,+I//K, • 2 1 5
L,//K,.
Thus, by induction, there exists j E { 1 , . . . , n} such that Ki//K,
x Lj//K~ = H//K, = Lj//K, x Ki//K,.
(Recall that s r i.) From the last equation we deduce that H = LjK,(I~'i n l~'s) = Ljf~i; use Lemma 1.3.5(i) for the second equation. This contradicts our hypothesis. Thus, we have shown that, for some j E { 1 , . . . , n } , Lj[~(i = H. On the other hand, fQ = H implies that K i L j = H; see Lemma 2.5.2(ii).
2.5 Direct Products of Closed Subsets
53
Now recall that Lj x L j -~ H = I(.i x f4[i.
Thus, by Theorem 1.3.7(iii), nK, = nL,. Therefore, Ki x Lj = H = Lj X [?(i.
This proves the proposition.
[]
The following theorem was proved as Theorem 3.11 in [9]. T h e o r e m 2.5.5 [P. A. FERGUSON-A. TURULL] A s s u m e that IXI E N, and assume that (X, G) is commutative. Let m, n E N, let H1, ..., Hm E C, and let K1, ..., Kn E C be given. A s s u m e that, for each i E {1,..., m}, Hi is indecomposable and that, for each j E { 1 , . . . , n}, K j is indecomposable. A s s u m e that H1 x . . . x
H,~ = G =
K1 x . . . x
K,.
Then we have ( i ) m -= n .
(ii) There exists a permutation ~r of {1,..., n} such that, for each x E X and, for each i E {1,..., n}, (X, G)~H, "~ (X, C ) , K , . . Proof. From Proposition 2.5.4 we obtain i E {1,..., m} and j E { 1 , . . . , n}
such that H i x I ~ = G = Is j x ~I i.
It follows that Hi x K j = Hi x [-Ii
and Hi x I:Ii = K j x I:Ii.
From the last of these equations we obtain that, for each x E X,
(x, a).m
(x, a ) ~
- (x,
see Corollary 1.7.9. From the equation before the last one we obtain, by the same corollary, that (X, a),,~, --- (X, (7,)'~' ~ (X, a ) , e . (Here it is needed that (X, G) is commutative.) Set iTr := j. Then, by induction, the claim follows from (X,G)~q,
(x,
[]
54
2. Decomposition Theory
2.6 Quasi-direct
Products
Let n E N \ {0} be given. For each i E { 1 , . . . , n}, let (Xi, G,) be a scheme. By P((X 1 X...
X Xn)
• (X 1 x ... x Xn))
we shall denote the power set of (X~ x ... x X n ) x (X1 x . . . x X n ) , i.e., the set of all subsets of (X1 x . . . x X , ) x (XI x ... x X~). We define a map r : G~ x . . . x a .
--+ P ( ( X 1 x . . . x
X.) x (Xl x ...x
X.))
by
((y~,...,y,,),(z~,...,z,d) e ( g ~ , . . . , g , ) ~ r
Vie {1,...,n} :
(v~,zd eg~.
We set (Xl,a!.)
x ... x (Xn,(Tn)
:'- (Xl
x ... x Xn,(al
x ... x Gn)g).
T h e o r e m 2.6.1 Let n e N \ {0} be given. For each i E { 1 , . . . , n } , let ( X i , Gi) be a scheme. T h e n we have (i) l x , • 2 1 5 = ( I x , , . . . , lx,)q. (ii) F o r each (gx,.-.,gn) E G1 • . . . • G , , ((g~,...,g,)~)" = M,...,g*,)~.
(iii) (X~, G~) x . . . • ( X n , G,~) is a scheme. (iv) F o r all ( d l , . . . , d , ) , (el . . . . , e , ) , (fl . . . . , f,,) e G1 • a(dt,...,d,)r
..... f,~)r -----a d l e l f l
x G,~,
9. . ad,~e,~j~.
(v) For all (el, . . . , e,~), ( f l , . . . , f , ) e G1 • . . . • G n , (el . . . .
, e n ) q ( f l , . . . , fn)q =
N
{(91,'' .,gn)ff [gi E e i f i } .
ie{1,...,n}
Proof.. (i) Let (Yl,..., Y,~), (zl . . . . , zn) E X1 • x X,~ be given. Then we have ( ( y l , . . . , yn), ( z l , . . . , z,)) E ( I x , , . . . , lx~)r if and only if, for each i E { 1 , . . . , n}, yi = zi. The latter condition says that ( Y l , . . . , Y,~) = ( z l , . . . , z,,). Thus, lx,•215 = ( I x . . . . . , lx.)r
(ii) Let (Yl, . . . . Yn), ( z l , . . . , Zn) E X1 • , . . x X n be given. Then we have ((Yl . . . .
,Yrt), ( Z 1 , . . . , Z n ) )
~ (gl,.'.,gn)~
2.6 Quasi-direct Products if and only if, for each i E { 1 , . . . , n}, equivalent to the fact that, for each i definition, means that ( ( z l , . . . , zn ), (Yl, Since ( y l , . . . , y,~), ( z l , . . . , z,~) E X1 ily, it follows that
55
(y~, z~) E g~. The latter condition is E { 1 , . . . , n}, (z~, y~) E g*. This, by . . . , Yn ) ) E (g~, . . . , g~ )g. x . . . • Xn have been chosen arbitrar-
= (iii), (iv) From (i) we know already that lx~•215 E (G1 • ... • Gn)g. Moreover, from (ii) we know that, for each (gl,---,g,,) E G~ x ... • G,~, ((gl,...,g,~)g)* E (G1 x ... x Gn)g. Thus, in order to prove (iii), it suffices to verify the regularity condition for (X1, G1) ><-... x (X,~, G,~). We compute the structure constants explicitly, so that also (iv) will be proved. Let (dl,...,d,~), ( e l , . . . , e n ) , ( f ~ , . . . , f n ) E G1 • ... x G,~, and let ( x , , . . . , x n ) , (yl,..-,y,~), ( Z l , . . . , z n ) E X1 • . . . • X,~ be given. Assume that
((Yl,...,Yn),(Zl,..-,Zn))
e
(fl,.-.,fn)r
Then, by definition, (xl,...,xn)
E(yl,
..,Y,*)(dl,...,d,~)r162
if and only if, for each i E { 1 , . . . , n}, xi E yidi O zie~. Therefore, (Yl,...,Y,)(dl,...,d,)q yldl
N Zle 1 x
A (zl,...,z,~)((el,...,e,)r . . . x
=
y n d n A z n e n.
Thus, we conclude that a(dl ..... d.~)r
..... e,,)r
..... fr162 =
adlel.fl
" " " ad,~e.f..
[]
(v) follows immediately from (iv).
Let n E N \ {0} be given. For each i E { 1 , . . . , n } , let ( X i , G i ) denote a scheme. The scheme (X1, G1) • ... • (Xn, G,~) will be called the direct product of the schemes (X1, G1), ..., (X,~, G,,). Set (W,F) := (X1,G1) x ... x (X,,G,~). For each i E {1,... ,n}, we define F, : =
N
I g: = lx }.
je{1 .....n}\{i} The pair (X, G) will be called a quasi-direct product of the schemes (X1, G1), ..., (X,,, G , ) if there exists a fusion r from (X, G) to (W, F) such that, for each i E { 1 , . . . , n } and, for each f E Fi, Ire -1] = 1.
56
2. Decomposition Theory Note that (W, F) itself is also a quasi-direct product of the schemes
(Xl,Cl),
...,
(Xn,an).
The relationship between direct products of closed subsets and direct products of schemes is described in the following theorem which generalizes [9; Proposition 3.13]. T h e o r e m 2.6.2 Let n E N \ {0} be given. For each i E { 1 , . . . , n } , let (Xi, Gi) be a scheme. Then the following conditions are equivalent. (a) (X, G) is a quasi-direct product of (X1, a l ) , ..., (Xn, an). (b) There exist H1, ..., H,~ E C with Hx x ... x H , = G such that, for each i E { 1 , . . . , n } and, for each x E X , (Xi,Gi) ~- (X,G)xH,. Proof (a) =~ (b) We define ( W , F ) :~-~ ( X 1 , G 1 ) • . . . x ( X n , a n ) .
For each i E { 1 , . . . , n}, we define Fi :--
r") { ( g l , . . . ,g,)r I gj = 1xj}. /eD .....n}\{i)
Since (X, G) is assumed to be a quasi-direct product of the schemes (X1, G1), ..., (X,~, G,), there exists a fusion r from (X, G) to (W, E) such that, for each i E { 1 , . . . , n } and, for each f E Fi, If5 -1] = 1. Let i E { 1 , . . . , n} be given. We set Hi := Fir -~. Then, as Fi E C(F), Hi E d; see Lemma 1.7.2(ii). By definition, r is bijective. Thus, by Lemma 1.7.3(i), r is surjective. Therefore, by Lemma 1.7.2(iii), Hi ..- Hn = G. Let i E { 1 , . . . , n } be given. Then (Hi
n(
U
Hj))r c_
jE{1 ....... }\1i}
Hie
n(
U
Hj)r C
jE{1,...,n}\{i} Hie n (
U
Hie) _C
je{x ...... }\(i} Fin
(
U
je{1 ..... ,}\{i}
Fj) = {1r
2.6 Quasi-direct Products
57
use Lemma 1.7.1(iii). Therefore, as {1} = (1r162-1,
Hi M (
U
Hj) = {1}.
j E { 1 ..... n } \ { i }
Let i, j E { 1 , . . . , n } be given. We wish to show that Hi C_ N a ( H j ) . Let g E Hi be given, and set f := gr Then f E Fi. From Theorem 2.6.1(v) we deduce easily that F j f = fFj. On the other hand, we have {g} = f r Thus, by Lemma 1.7.2(iii),
Hjg
= Fjr162
-1 = (Fjf)r
-1 = (fFj)r
-1 = re-
1F.r .1
=
gHj.
Since g E Hi has been chosen arbitrarily, we have shown that Hi C_
Na(Uj). So far, we have proved that H1 x ... x H,~ -- G. Let finally i E { 1 , . . . , n } , and let x E X be given. We set Wi := xCFi. For each y E xHi, we set yr := yr For each g E Hi, we set (9xH.)~ := gr Clearly, as r is a morphism, r is a morphism from (X, G)~:H, to (Wi, Fi). Let e, f E Hi be such that (e,H,)r = ( . f x g i ) r Then er = r e . But, as f r Fi, I(fr162 = 1. Therefore, e = f . This proves that r is injective. Thus, by Lemma 1.7.3(ii), (iii), r is an injective homomorphism. Let w E Wi be given. Then, by definition, w E zCFi. Therefore, there exists f E Fi such that (xr w) E f. Since r is surjective, there exist y E X such that yr = w. Let g E G be such that (x,y) E g. Then (xr = ( x r 1 6 2 E gr It follows that gr = f E Fi, so that g E Hi. In particular,
y E xHi. Since w E I/Vi has been chosen arbitrarily, we have shown that CxH, is surjective. Thus, by Lemma 1.7.3(i), ~ is surjective, too. It follows that r is an isomorphism from (X, G)xH, to (Wi, Fi). On the other hand, it is clear that (Wi, Fi) ~ (Xi, Gi). (b) ~ (a) Let x E X be given. We shall be done if we succeed in showing that (X, G) is a quasi-direct product of the schemes (X, G)xH1, ..., (X, G)~H.. We define ( W , F ) := (X,G),H, x ... x (X,G),H,. Let y E X, and let i E { 1 , . . . , n } be given. Then, as Hi x /2/i = G,
y E xHi[fi and Hi M/2/i = {1}. Thus, by Theorem 1.3.7(ii), IxHi M yHi = 1. Let Yi E X be such that {Yi} = xHi VI yI2Ii. We set yr := ( y l , . . . , y,). Let g E G be given. Then we find, for each i E {1 . . . . ,n}, a uniquely determined element hi E Hi such that g E { h i } . . . {h,~}; see Theorem 2.5.1. We set gr :=
58
2. Decomposition Theory Then r
XUG
-+ W U F
is a well-defined map with Xr = W and Gr C F. Moreover, the restriction of r to X is injective. For each / E { 1 , . . . , n}, we define F, :=
N
{((gl).H,,..., (g,~).H.)r
= 1.Hi}.
je{1 ..... n}\{i}
It follows from the definition of r that, for each i E { 1 , . . . , n } and, for each f E Fi, I f r = 1. Therefore, we shall be done if we succeed in showing that r is a morphism. To this end, let y, z E X , and let g E G such that (y, z) E g be given. Let ( Y l , . . . , Y n ) E xH1 x . . . x xH,~ be such that yr = ( y l , . . . ,
and let ( z l , . . . , z ~ ) E xH1 • ... • xH,~ be such that zr = (zl,...,
Then, by definition, we have that, for each i E { 1 , . . . , n}, Yi E xHi ~ zi. Thus, by Theorem 1.3.1, zi E yiHi. Let hi E Hi be such that (yi,zi) E hi. Then, by definition, (yr
---- ( ( Y l , . . . , Y n ) , ( Z l , . . . , Z n ) )
e ((hl)xH,,...,(hn)xH.)g.
Now we shall be done if we succeed in showing that gr =
Let i E {1, . . . , n} be given. From the definition of Yi we have that y E Yi fli C_ zi hi I:li = zi Hi ~ h i* 9
9
Therefore, zi~Ii N yhi ~ O. Let w E zi~Ii A yhi be given. From the definition of zi we have that z E zi-f-Ii. Thus, as w E zi/:/i, Theorem 1.3.1 yields z E w[Ii C yhi I2Ii. It follows that g E hi/:/i. Since i E { 1 , . . . , n } has been chosen arbitrarily, we obtain that g E [] h i . . . h ~ . Thus, by definition, gr = ( ( h l ) : ~ g l , . . . , (hn)xH,,)~.
The pair (X, G) is said to be indeeomposable if G is indecomposable. The first part of the following theorem generalizes the first part of [9; Theorem 3.17]. Its second part is the second part of [9; Theorem 3.17].
2.6 Quasi-direct Products
59
Theorem 2.6.3 Assume that IX I E 1~. Then we have (i) I f { l } # G, (X, G) is a quasi-direct product of indecomposable schemes. (ii) [P. A. FERGUSON-A. TURULL] Assume that (X, G) is commutative. Let m, n E 1~ \ {0} be given. For each i E { 1 , . . . , m}, let (Wi, Fi) be an indeeomposable scheme; for each j e { 1 , . . . , n}, let (Xj, Gj) be an indeeomposable scheme. Assume that (X,G) is a quasi-direct product of the schemes (W1, F1), ..., (Wr,,, Fro) as well as a quasi-direct product of the schemes (X1, G1), ...,
(X.,G.).
Then m = n and there exists a permutation ~r o f { l , . . . , n} such that, for each x e X and, for each i E { 1 , . . . , n}, (Wi, Fi) ~- (Xi,, Gi,). Proof. (i) By Theorem 2.5.3(ii) there exists n E 1~ \ {0} and there exist H1, ..., Ha E C such that Hi•215 and such that, for each i E { 1 , . . . , n}, Hi is indecomposable. Let x E X be given. Then, by Theorem 2.6.2, (X,G) is a quasi-direct product of the schemes (X, G)xH,, ..., (X, G)xH.. Let i E { 1 , . . . , n} be given. Since Hi is indecomposable, (Hi)zH, is also indecomposable; see Lemma 1.5.2(i). Thus, by definition, (X, G)~H. is indecomposable. (ii) Since, by hypothesis, (X, G) is a quasi-direct product of the schemes (W1, F O, ..., (Win, Fro), there exist H1 . . . . , Hm E C with
H1 x . . . • such that, for each i E {1,..., m} and, for each x E X,
(x, see Theorem 2.6.2. Similarly, there exist K1, ..., K , E C with K1 x ... • K , = G such that, for each j E {1,... ,n} and, for each x E X,
Let x E X, and let i E { 1 , . . . , m } be given. By hypothesis, (Wi,Fi) is indeeomposable. Therefore (X, G)~H, is indecomposable. Thus, by definition, (H~)~H, is indeeomposable, so that, by Lemma 1.5.2(i), Hi is indecomposable. Similarly, we see that, for each j E { 1 , . . . , n}, Kj is indecomposable. Now the claim follows from Theorem 2.5.5. [] We do not know whether or not Theorem 2.6.3(ii) is true also for noncommutative schemes.
60
2. Decomposition Theory
2.7 Semidirect
Products
An isomorphism from (X, G) to (X, G) is called an a u t o m o r p h i s m of (X, G). It follows easily from Lemma 1.7.3(iii) that the set of all automorphisms of (X, G) forms a group with respect to composition of mappings. We shall denote this group by Aut(X, G). Let 8 be a group. We say that 8 acts on (X, G) if there exists a group homomorphism from 8 to Aut(X, G). Assume now that 8 acts on (X, G). Then, by definition, there exists a group homomorphism r from O to Aut(X, G). Let 0 E 8 be given. For each x E X, we abbreviate xO := z(Or
Similarly, for each g E G, we set
go := 9(0r By p ( ( x • 8) • ( x • 8)) we shall denote the power set of (X • O) • (X • 8), i.e., the set of all subsets of (X • 8 ) • (X • 8). We define a map
~o: c • 2 1 5 2 1 5 2 1 5 by ((u, r (z, 7)) e (g, e)~o r
(ye, z) e g, r = 7.
Just in order to familiarize ourselves with the definition of co let us note the following. Let us set q := r Let y, z E X, and let (, r/E 8 be given. Then, for each 0 E 8,
((u, (), (z, 7)) ~ (1, e)~ r
~e = z, r =
and, for each g E G, ((y, (), (z, r])) E (g, 1)r r
(y, z) E g, ( = r].
We set (X, G ) 8 := (X x 8, (G • 8)~;o).
T h e o r e m 2.7.1 Then we have
(i) 1xxo = (1, 1)~.
Let (9 be a group acting on (X,G), and set q := qo-
2.7 Semidirect Products
61
(ii) For each g E G and, for each 0 E O, ((g,O)r = (g*0-1,0-1)~. (iii) (X, G)O is a scheme. (iv) Let d, e, f E G, and let ~, ~, 71 E ~ be given. Then a( d,e)~(e,r)r ],O)~ -~ adCeff~er ,O.2 (V) Let e, f E G, and let ~, 71 E 6) be given. Then (e, ~)~(f, r/)g ---- { (g, Cr/)~ I g e erlf}.
Proof. (i) follows immediately from the definition of ~. (ii) Let y, z E X, and let ~', r/E O be given. Then, by Lemma 1.7.1(i), (yO, z) e g ~
(y, zO-1) Eg0 -1 ~
(zO-l,y) E g'O -1.
On the other hand,
C0=rl ~
r]0-1 = ~ .
Therefore, the definition of ~ yields
((u, r (z,
E
((z,
(U, r
E
This proves (ii). (iii), (iv) It follows immediately from the definition of ~ that (G • O)~ is a partition of (X • O) • (X • O). From (i) we know already that l x • E (G • O)~. Moreover, from (ii) we know that, for each 9 E G and, for each 0 E O, ((9,0)Q* E (G • O)~. Thus, in order to prove (iii), is suffices to verify the regularity condition for (X, G)O. We compute the structure constants explicitly, so that also (iv) will be proved. Let y, z E X, let d, e, f E G, and let/3, 7, c, ~, r/E O be given. Assume
that ((y,/3), (z, V)) E (f, U)~. Then, by definition, (yr/, z) E f and/37 = % Let W _C X, and let A C O be such that
W x A = (y,/3)(d, e)~ n
(z,7)((e,r
We shall be done if we succeed in showing that IW x A I = adr162 Note first that this equation follows from the two implications W x A r ~ =~ IW • Al < aaceS , e~ = rl and r = rl ~
adcef <_-IW •
Therefore we shall prove these two implications. 2 Here ~ is the Kronecker delta.
AI.
62
2. Decomposition Theory
In order to prove the first one, we assume that W • A # ~. Let z E W, and let 0 E A be given. Then, by definition,
(ye, x) C d, (x~,z) E e, fie = O, O~ = 7. From 1% = O, Or = 7, and fir/= 7 we obtain that ci = f - 1 7 = rl. From (ye, x) E d we obtain that (ye~,x~) E d~. Therefore, as eC = r/, (yr/, x~) E d~. But we also have (xC, z) E e. Therefore,
x~ E yr]d~ M ze*. Since x E W has been chosen arbitrarily, we have shown that W~ C_ y~ld~ N ze*. Thus, as ~ is bijective, ]W] < ly~?d~ N ze" I = adCeJ. (Recall that (yr/, z) E f.) On the other hand, as 0 E ,4 has been chosen arbitrarily, the equation f c = 0 yields I,4]-< 1. Therefore, IW • ,4] < adCefLet us now prove that
cC = ~ => hacks <_ IW x ,41. Let x E yrl~-ldf3 z~-le*~ -1 be given. Then we have (y~l~-l,x) E d and (z~ -1, x) E e*( -1. From (yr/~ -1 , x) E d and e~ = r/we obtain that (yc, x) e d. From (zr -1, x) E e*C-1 we obtain that (z, x~) E e*, which means that
(xr z) 9 e. Finally, as f ~ / = 7 and c~ = 7/, fe~ = 7. Thus, by definition, (x, fie) 9 W • ,4. Since x 9 yq~-ldM z~-le*~ -1 has been chosen arbitrarily, we have shown that
lyuC- d n zi- e*i- l < IW • z l. But, as (yr/, z) 9 f , we also have
adce! : lYTId~ N ze*l = l y ~ - l d n zC-%*~ -1 I. (Recall that ~" is bijective.) Thus, we have aden! <_ ]W • ,41. (v) follows immediately from (iv).
[]
Let O be a group acting on (X, G). Then the scheme (X, G)O will be called the semidirect product of (X, G) with 0.
T h e o r e m 2.7.2
(x, G)o.
Let @ be a group acting on (X, G), and set (W, F) :=
2.7 Semidirect Products
63
Then there exist H, K 9 C(F) with H K = F, H n I( = {1}, H
(x, a) ~- (w, F)~H and
7-(o) ~ (w, F)~K?
Proof. We define H := {(g, 1); I g E G} and
I; := {(1, o)~ I e 9 o } . From Theorem 2.7.1(ii), (v) we obtain immediately that H, K 9 C(F). Theorem 2.7.1(v) tells us also that, for each g 9 G and, for each 0 9 O,
(g,O)~ 9 (9 0-1 , 1)r
C HK.
Therefore, H K = G. It follows immediately from the definitions of H and K that H MK = {1}. Let g 9 G, and let 0 9 O be given. Then, by Theorem 2.7.1(ii), (v), ((1, O)~)'(g, 1)~(1, 0)~ = (1, 0-1)~(g0, 0)~ = {(gO, 1)~}. (Recall that, by Lemma 1.7.2(i), 10 = 1.) Thus, as g 9 G and 0 9 O have been chosen arbitrarily, we have proved that K C_ S F ( H ) . Now, as H C_ S F ( H ) 9 C(F), we obtain that F C_ S F ( H ) . This means that U ~ F. From Theorem 2.7.1(ii), (v) we also obtain that, for each 0 9 O,
((1,O)~)*(1,O)q = (1,O-1)q(1,O)q = {(1, 1)~}. Thus Lemma 1.2.6 tells us that, for each 0 9 O, n0,0)~ = 1. By definition, this means that K C_ O0 (F). Let w 9 W be given. Since W = X x O, we find x 9 X and 0 9 O such that w = (x, 0). We still have to prove that (X, G) ~ (W, F)~H and that T(O) ~- (W, F)wK. Let us first prove that (X, G) -~ (W, F)wH. For each y 9 X, we define
(v, 0)r := v. For each g 9 G, we define 3 Recall that T(O) denotes the thin scheme which is associated to the group 0 via the (~, T)-correspondence.
64
2. Decomposition Theory ((g, 1)~)wHr := g.
Then, as w H = {(y, 0) I Y E X}, r
w H U H w H -+ X U G
is a well-defined bijective map with (wH)r = X and (HwH)r = G. But from the definitions o f r and ~, we obtain that, for all y, z E X and, for each g E G,
((v, 0)r (z, 0)r E ((a, 1)~)~.~
r
(y,z) Eg r ((u, 0), (z, 0)) E ((g, 1)~)~.. Therefore, r is an isomorphism. Let us finally prove that T(O) ~ (W, F),~K. For each r/E O, we define
(x~, 07)r := and ((1, ~)~)wKr := ~. Then, as wig = {(xT/, 071) I ~ E O}, r
wKUK~K
-+ OU(~
is a well-defined bijective map with ( w K ) r = O and (K,~K)r = O. But from the definitions of r and g, we obtain that, for all c, 77, 4 E O,
((x~, 0~)r (x4, 04)r E ((1,
~)~)~r
r
o r / = 4 CV ((xe, 0c), (x4, 04)) E ((1, r/)~)~K. Therefore, r is an isomorphism.
[]
3. Algebraic Prerequisites
In this chapter, we prove basic algebraic results which will be needed in the fourth and in the fifth chapter of these notes.
3.1 Tits'
Theorem
on Free Monoids
In this section we deal with free monoids. Our main goal is Theorem 3.1.5(ii), a result which usually is called "Tits' solution of the word problem for Coxeter groups". Here this solution is presented as a result about certain equivalence relations defined on the elements of a free monoid. Throughout this section, L will be a set. We shall denote by F(L) the free monoid constructed on L. The multiplication of F(L) will be denoted b y . , the identity element of F(L) will be denoted by 1. A map from {N C L I INI = 2} to ( N \ {0, 1}) U {R0} is called a Coxeter map of L. For the remainder of this section, m denotes a Coxeter map of L. Let h, k E L be such that h ~ k, and let j E N be given. If j = 0, we define fj(h,k) := 1. Now assume that 1 < j, and let / 1 , . . . , /j E {h,k} be such that, for each i E { 1 , . . . , j ) , l~ = h if and only i f 2 { i . Then we define
fj(h,k) :~---/1 * " ' ' * b " For all h, k E L with h r k and m({h, k}) E N, we abbreviate fro(h, k) := fm({h,k})(h, k). We shall now define two equivalence relations on F(L) which we wish to compare. We set ,~m :-- {(a* fro(h, k)* b , a * fm(k,h) , b ) I h,k E L, h # k, m((h,k}) E N, a , b E F(L)}
66
3. Algebraic Prerequisites
and .7":={(a,l,l,b,a,b)
llEL,
a, b E F ( L ) } .
We shall denote by Qm the uniquely determined smallest equivalence relation on F(L) which contains ~:,~. By 7~m we shall denote the uniquely determined smallest equivalence relation on F(L) which contains Cm U Y'. Clearly, we have Qm _C Rm. Let d, e E F(L) be given. Occasionally we shall write d .--m e if (d, e) E Q,~ and d~m e if (d,e) E T~m. Note that d ~m e implies that d ~,~ e. (This is just a restatement of our observation that Q,~ c_ T~m.) The main result of this section, Theorem 3.1.5(ii), says that, if d and e are "reduced" elements of F(L), the converse is true too. Let us denote by ~L the uniquely determined monoid homomorphism from F(L) to (the additive monoid) 1~ with LAL C {1}. For the remainder of this section, we abbreviate A := AL.
Let n E 1~ \ {0}, and let 11, ..., In E L be given. We set (11 * .--*ln)* := 1, * . . - * 1 1 , and, for each i E { 1 , . . . , n}, we define si(/1 * " . * ln) = 11 * . . . * li. We also set 1" := 1 and s0(ll * ' . . * l n ) := 1. Let e E F(L) \ {1} be given. For each i E {1,...,eA}, we set t i ( e ) : : si(e) * si-l(e)*.
For each f E F(L) \ {1}, we define a ( f , e ) := I{i E { 1 , . . . , e A } l ( f , t , ( e ) ) E R~,~}I and e(f,e) := (-1) ~(f'e).
L e m m a 3.1.1 e(f, d) - e(f, e).
Let d, e, f E F(L) \ { 1 } be given. Then, i f d
~'m e,
3.1 Tits' Theorem on Free Monoids
67
Proof. Clearly, we may assume that (d, e) E gm U .T. We assume first that in fact (d, e) E Sin. Then, by definition, there exist h, k E i with h ~ k and m({h,k}) E N, and there exist a, b E F(L) such that d = a * f m ( h , k ) * b and e = a * f m ( k , h ) * b . Suppose that a r 1. Then, for each i E {1,... ,aA}, ti(d)
= ti(a)
= ti(e).
For each i E { 1 , . . . , m ( { h , k } ) } , ta~+~(d) = a , ti(fm(h, k ) ) , a* =
a * f2i-l(h, k) ,
a* ~m
a *
f2m({h,k})_2i+l(k, h) * a *
=
a * tm({h,k})-i+l(fm(k, h)) * a* = ta),+m({h,k})-i+l (e).
Suppose that b r 1. Then, for each i E { 1 , . . . , b A } , taX+m({h,k})+i(d) = a * fro(h, k) * ti(b) * fro(h, k)* * a* "~m a * f,~ (k, h) * ti(b) * fro(k, h)* * a* = t,x+,~({k,h})+i(e). Thus, if (d, e) E gin, we have a ( f , d) = a ( f , e). Let us now assume that (d, e) E .T. Then, by definition, there exist l E L a n d a , b E F ( L ) such that d = a * l * l * b a n d e = a * b . Suppose that a r 1. Then, for each i E {1,...,aA}, ti(d) = ti(a) = ti(e).
Note also that ta~+a (d) = a * l * a*
'~m a * l * l * I * a* --
t~+2(d).
Suppose that b r 1. Then, for each i E {1,...,bA}, t~+2+i(d)
= a * l * 1 * ti(b)
* 1 * 1 * a* "~m a * t i ( b )
Now we are done also in the case where (d, e) E .T.
For each f E F(L), we set
7~m(f) :=
{eE
F(L)[ (f,e)ET~m}
and fA' : - min{eA I e E zero(f)}. We define Fro(L) :-- {f E F(L) [ fA' = fA}. The elements of Fro(L) are called m-reduced.
* a* = t a ~ + i ( e ) .
[]
68
3. Algebraic Prerequisites
L e m m a 3.1.2 Let f E F(L) \ {1} be given. Then f E F,~(L) if and only ifl{Ttm(ti(f)) I i E {1,...,fX}}{ = fX.
Proof. Assume first that I{T~m(t,(f)) I i ~ {1, ...,fA}}l < f A - 1. Then there exist i, j E { 1 , . . . , f A } such that i _< j - 1 and t~(f) ~,~ tj(f). Thus, sj(f) ~,~ ti(f) * Sj_l(f ) = Si-l(f) * si(f)* * s j - l ( f ) . But (si(f)* * Sj_l(f)))~ t < j - i - 1. Therefore, sj(f)X t < j - 2. It follows that fA t < fA - 2. Thus, by definition, f ~ F,~ (L). Assume now that {{R,~(ti(f)) l i E {1,...,fA}}{ = fA. Let i E {1,... ,fA} be given. Then, by definition, c r ( t i ( f ) , f ) = 1, whence c(ti(f), f) = - 1 . Let e E Fro(L) be such that e ~,~ f. Then, by Lemma 3.1.1, e(ti(f),e) = r Therefore, e(ti(f), e ) = - 1 , so that cr(ti(f), e) # 0. Since i E { 1 , . . . , f $ } has been chosen arbitrarily, the assumption that {{7~m(ti(f)) { i E { 1 , . . . , f $ } } [ = fA forces that fA _< e$. The latter means that f E Fro(L). [3
L e m m a 3 . 1 . 3 Let l E L, and let f E Fm(L) be such that l * f ~ Fm(L). Then there exists i E { 1 , . . . , f $ } such that si(l * f) ~,,~ si(f).
Proos By hypothesis, l . f ~ Fro(L). Thus, by Lemma 3.1.2, there exist i, j E { 1 , . . . , f A + 1} such that i < j - 1 and t i (l *
f) ~m tj (l * f) = l * t j _ l (f) * I.
Suppose that 1 ~ i. Then l * t i - l ( f ) * l = t i ( l , f ) . It follows that l * t i _ l ( f ) * l ~m l * t j - l ( f ) * l, which implies that t i - l ( f ) ~,~ t j - l ( f ) . Thus, by Lemma 3.1.2, f ~ Fro(L), contrary to the hypothesis. Therefore, 1 = i. Then l = ti(l * f). It follows that l ~,~ l* t j - l ( f ) *l = l , s j - l ( f ) * sj_2(f)* * l, which implies that sj_ 1(l * f) ~,~ sj_ 1 (f).
[]
L e m m a 3.1.4 Let h, k E L be such that h ~s k, and let f E Fro(L) be given. Assume that h * f, k * f @ Fro(L).
3.1 Tits' Theorem on Free Monoids
69
Then there exists c E Fro(L) such that m( { h, k})+cA = fA and f,~ ( h, k ) * C ~rn f.
Proof. We shall prove that, for each j E { 1 , . . . , m({h, k})}, there exists cj E F,r,(L) such that j + cjA = fA, fj(h, k) , cj ~-,m f, and fs(k, h) * cj ~m f. The claim is obvious for j = 1. (Recall that we assume that h * f, k 9f Fro(L), and use Lemma 3.1.3.) Let j E { 2 , . . . , m({h, k})} be given, and assume that there exists cj-1 E Fro(L) with j - 1 + cj-lA = fA and f j - l ( h , k) , cj-1 ~,~ f. Since f E Fro(L), f j - l ( h , k ) * cj_l E Fro(L). On the other hand, as k . f q~ Fro(L), fj(k, h) * cj-1 ~ F,~(L). Thus, by Lemma 3.1.3, there exists i E { 1 , . . . , f A } , such that si(fj (k, h) * cj-1) ~,n s i ( f j - l ( h , k) * cj-1). Suppose that i _< j - 1. Then i <_ m({h, k}) - 1 and fi(k, h) ~rn fi(h, k). Since this is impossible, we must have j _< i. Let e E Fro(L) be such that si(fj_l(h, k) * cj_,) 9e = f j - l ( h , k) * cj-1. Then si(fj (k, h) * cj-1) * e ~m s i ( f j - l ( h , k) * ej-1) * e ~m f. Thus, as j < i, there exists cj E Fm (L) such that fj (k, h) * cj m,, f. Now we set c := cm({h,k}), and the lemma is proved.
[]
We now shall denote by 3rn the uniquely determined smallest reflexive and transitive relation on F(L) which contains s U 9r. For each f E F(L), we set &n(f) := {e E F(L) ] (f,e) E Sin}. Clearly, we have S.~ C_~.~. The following theorem was proved as TMor~me 3 in [26].
T h e o r e m 3.1.5 [J. TITS] (i) Let d, e E F(L) be such that d ~m e. Then S,,, (d) N Sm (e) • O. (ii) Let d, e E Fm (L) be such that d m,~ e. Then d ~,~ e.
70
3. Algebraic Prerequisites
Proof. (i) We define 79 := {(d~e) 9 F(L) • F(L) I d ~,~ e, Sin(d)O Sin(e) = 0}, and we assume that 79 ~ 0. Then there exists (d,e) max{dA, eA} is minimal. Set
9 79 such that
n := min{max{dA, cA} I (d, e) 9 79}, and choose (d, e) 9 79 such that max{dA, cA} = n. Our first claim is that dA = eA. In order prove this, we suppose that dA r eA. Without loss of generality, we may assume that eA < dA. Then, as d ~m e, we must have that eA < dA - 2. It is obvious that Sm(d)A C { 0 , . . . , dA}. We shall now show that even S,n(d)A = {dA}. Suppose that there exists c E S,,~(d) such that cA < d A - 1. Then max{eA, eA} < d A - 1 = n - 1. (Recall that eA < d A - 2.) Therefore, (c, e) ~ P. But, since c 9 S,,(d), (d, c) 9 Sr~ _C 7~,,~. Therefore, c ~,~ d. Thus, as d ~,rn e, c ~'m e. Therefore, Sin(c) n Sin(e) r 0. But, as c 9 Sin(d), Sin(c) C_ Sin(d). Therefore, S,~ (d) n Sm (e) r 0, contrary to the choice of (d, e) as an element of 7~. This contradiction confirms that Sm(d)A = {dA}. Since eA < d A - 2 , d r 1. Thus, there exist l E L and b E F(L) such that d=l*b. Set f:=l*e. Then fA=eA+l
Suppose first that Sm (b) O Sm (f) ~ 0. Let a E Sm (b) A Sm (f) be given. Since Sm(d)A = {dA} and d = / * b , S,~(b)A = {bA}. In particular, aA = bA. Thus, as a E Sin(b), b E Sin(a). Thus, as a E Sin(f), b E Sin(f). But fA < bA. Therefore we also have f E Sm (b). Thus, as e E Sr,, (l * l * e), we obtain the contradiction that e E Srn (d). Suppose now that Sm (b) (1 S,n (f) = ~. Then, as
b~*m l *l * b = l * d ~ m l * e = f , (b, f) E P. But fA <_ bA = d A - 1 = n - 1, contrary to the choice of (d, e) E P.
3.1 Tits' Theorem on Free Monoids
71
Thus, we have shown that dA = cA. Our second claim is that {d, e} C_ F,~(L). In order to prove this we suppose that {d,e} ~ F,,~(L). Then, as d ~,,~ e and dA = eA, {d,e} n Fro(L) = 0. Thus, there exists a E Fm (L) such that d ~m a ~,n e and aA < dA - 2. Since aA < d A - 2 = n - 2, (d,a) ~ P. Thus, as d ~,~ a, we must have Sm (d) n S,,~(a) • 0. Let d' E S,,~(d) M Sm (a) be given. Similarly, we find e ' E S , ~ ( a ) n S,~(e). Recall that a E Fro(L). Thus, d' "~r~ e'. In particular, e' E 8m(d'). But d' E Sm (d), whence e' E S,, (d) M Sm (e). Since this contradicts the choice of (d, e) E P, we have shown that {d, e} C_ f,~ (L). Since d r 1, there exist h E L and d' E F , , (L) such that d=h,d
I.
Similarly, there exist k E L and e ~ E Fro(L) such that e= k*e'. The choice of (d,e) E 7~ forces h r k. Now, by Lemma 3.1.4, there exist c E Fro(L) such that dA = m({h, k}) + cA = eA and frn(h,k)* c ~m d, f , ~ ( k , h ) * c ~ , ~ e. It follows that f,n({h,k})-l(k, h) * c ~,~ d', fm({k,h})-l(h, k) * c ~rn e'.
Thus, the choice of (d, e) E 7~ forces Srn(f,~({n,k})-l(k, h) * c) = Sm(d'), S m ( f m ( { k , h ) ) - l ( h , k) * c) = 8m(e'), whence Sin(d) = S,~(f,~(h, k) 9c) = S m ( f m ( k , h ) * c) = S,,~(e). This contradiction proves that P = 0. (ii) follows immediately from (i).
[]
C o r o l l a r y 3.1.6 Let l E L~ and let f E F,,(L) be such that l * f ~ Fro(L). Then there exists e E Fro(L) such that l 9 e ..~,~ f .
Proof. This follows from Lemma 3.1.3 and Theorem 3.1.5(ii).
[]
72
3. Algebraic Prerequisites
C o r o l l a r y 3.1.7 Let f E F(L) be given. Then the following conditions are equivalent. (a) f ~ F ~ ( L ) . (b) There exists I E L and there exist a, b E F(L) such that f "~,n a*l*l*b.
Proof. (a) =:~ (b) Let e E Fro(L) be such that e ~rn f. Then, by Theorem 3.1.5(i), S,~(e) N Sin(f) # 0. Thus, as e E F,~(L), e E Sin(f). F r o m e E Fro(L), f ~ Fro(L), and e ~,~ f we also obtain that eA < f A - 1 . In particular, e 7~m f. Thus, as e E Sin(f), there exist l E L and a, b r F(L) such that f "~m a * l , l * b. (b) ::~ (a) Assume that there exist 1 E L and a, b E Fro(L) such that f'm a*l*l*b. Then fA = ( a , l * l * b ) A and f ~,~ a * b . Therefore, f ~ Fro(L). []
3.2 Integers
over Subrings
In this section, R will denote a field. For all S, T C_ R, we define ST:={
E st I P C - S x T ' (s,t)EP
[PIEN\{0}}"
Let S be a subring of R. 1 An element r E R is called integral over S if there exists n E l~ \ {0} such that r n E S{r' [ i E { O , . . . , n - 1}}. By I(S) we shall denote the set of all those elements of R which are integral over S.
P r o p o s i t i o n 3.2.1 Let S be a subring of R. (i) Let A C_ I(S) be such that IA]E 1~ \ {0}. Then there exists B C_ R with IBI E 1~ and A C S B such that S B is a subring of R. (ii) Let B C_ R be such that ]B I E 1~. Assume that S B is a subring of R. Then S B C I(S).
Proos (i) Since, by hypothesis, A C_ I(S), there exists, for each a E A, an element n(a) E 1~ \ {0} such that 1 Subrings contain the unit by definition.
3.2 Integers over Subrings
73
a n(a) E S{a i I i E { 0 , . . . , n ( a ) - 1}}. Set
ai(a) li(a) E { 0 , . . . , n ( a ) -
B := { H
1}}.
aEA
Then ]B] E N, A C_ SB, and S B is closed under addition. Let us now prove that S B is closed under multiplication too. Let c E A, and let b E B be given. Since R is distributive, we shall be done if we succeed in showing that cb E SB. Since b E B, there exist, for each a E A, an element i(a) E { 0 , . . . , n ( a ) - l } such that b = I T hi(a)" aEA
Set d :=
1-I hi(a)" aeA\{c}
Then cb = ci(c)+ld. Ifi(c) < n(c) - 2, cb E B. If i(c) = n(c) - 1,
cb = cn( )d
S{c' I i e { 0 , . . . , n(c) - 1}}{d} C SB.
(ii) Let r E S B be given. Since, by hypothesis, S B is a subring of R, there exists {s~a I c, d E B} C_ S such that, for each c E B, cr = E Scdd. dEB
The characteristic polynomial of the matrix (Scd)c d m a p s r onto 0. Setting n := IBI we conclude that r" E S{r' ] i E { 0 , . . . , n - 1}}. Therefore, we have r E I(S). []
T h e o r e m 3.2.2 For each subring S of R, I(S) is a subring of R.
Proof. Let s, t E I(S) be given. Set A = {s,t} in Proposition 3.2.10). Then we obtain from Proposition 3.2.1(ii) that s - t, st E I(S). []
It is not hard to see that Theorem 3.2.2 is also true under the weaker hypothesis that R is a commutative ring with unit.
L e m m a 3.2.3 Assume that char(R) = 0. Let Q denote the prime field of R, and let Z denote the subring of R generated by 1. Set I := I(Z). Then l M Q = Z.
74
3. Algebraic Prerequisites
Proof. It is clear that Z C_ I (3 Q, so let us prove that 1 ('1Q c Z. Let r E I O Q be given. Since r E Q, there exist s, t E Z such that t r 0, (s, t) = 1, and r = st -1. Since r E I, there exists n E 1~ \ {0} such that r" E Z { r i [ i E { 0 , . . . , n -
1}}.
Thus, s" E t Z { s i t " - 1 - i [ i E { 0 , . . . , n 1}} C tZ. It follows that t I s". But we have chosen s and t E Z such that (s, t) = 1. Therefore, t = 1. It follows that r E Z. []
Our next result is elementary, but crucial in the proof of T h e o r e m 5.3.5.
L e m m a 3.2.4 Assume that char(R) = 0 and that R is algebraically closed. Let Z denote the subring of R generated by 1. Let r E R be such that r + r -1 E Z. Assume that {n E N \ {0} ] r n = 1 } # 0 . Then r 4 = 1 or r 6 = 1 .
Proof. For each i E 1~ \ {0}, we have r i'{'l -~ r -(i'{-1) = ( r -~- r - 1 ) ( r i -~- r - i ) -- ( r i - 1 -~ t - ( i - I ) ) .
Thus, as we assume that r + r -1 E Z, induction yields that, for each i E N, r i + r - i E Z. (Note that r ~ + r - ~ E Z.) By hypothesis, there exists n E N \ {0} with r n = 1. Therefore, the set { r i + r - i I i E N} is finite. Let j E N \ {0} be such that (r j + r - J ) 2 is m a x i m a l (with respect to the natural ordering of Z). Then
(r + r-l)2(, z + r-J) = + t-(j+2)) +
+
+ 2(r + r-J) _<
+ r-J). Thus, as 2 < r / + r - j , we obtain that (r + r - l ) 2 _< 4. Assume that (r + r - l ) 2 = 4. Then r 2 - 2 + r -2 = 0. It follows that (r 2 - 1 ) 2 = r 4 - 2 r 2 + 1 = 0 , w h e n c e r 2=1. Assume that ( r + r - 1 ) 2 = 1. Then r 2 + l + r -2 = 0. It follows that r 6-1=(r 4+r 2+l)(r 2-1)=0,whencer 6=1. Assume that r + r -1 = 0. Then r 4 - 1 = (r 2 + 1)(r 2 - 1) = 0, whence r 4 : 1. []
L e m m a 3.2.5 Assume that R is algebraically closed. Let r E R be such that r 2 ~ l, and let s E N be given. Then (i) / f r 2'+1 = 1,
3.2 Integers over Subrings $
}--~(r i + r - i ) ~ = 2s + 3 i--0 and
$
Z(r
i+1 -- r - ( i + 1 ) -1- r i -- r - i ) 2 ---- --(28 q- 1 ) ( r + r - 1 + 2).
i=o
(ii) If r ~" = 1, s--1 Z[(r/+1
-- r - ( i + l ) ) 2 -t- ( r i - - r - / ) 21
=
-4s
i=0 and
s-1
~(~,+1
_ ~-(,+1))( r, _ r-,) = - 8 ( r + ~-1).
i=0
Proof. (i) A n elementary computation shows that 28
ri(r-
1) = r 2 , §
_ 1.
i--0
Thus, our assumption that r 2s+1 = 1 leads to 2s
r
- 1) = 0.
i=0
But, by hypothesis, r ~- 1. Therefore, 2s r i =0. Z i----0
From this we obtain immediately that 8
$
Z ( r + ~-,)2 = ~ ( ~ 2 , + 2 + ~ - 2 ' ) = 28 + 3, i=0
i=0
which is our first claim. But we also obtain that $
Z(r
2(i+1) + r - 2 ( i + 1 ) ) --~ r -~- r - 1 -- 1,
i=0 $
i=0
and
75
76
3. A l g e b r a i c
Prerequisites $
2i+1 + r -(2i+1)) = 1.
E(r i=o
These last three equations yield 8
}-~(~i+1
_ ~ - ( i + 1 ) + ~i _ ~ - i ) 2 =
i=0 $
a
E [ r 2 ( i + l ) + r -2(i+1) + r 2i + r - 2 / ] + 2 E [ - - 1 + r = i + * - r - r -1 + r -(2i+1)i=o
1] =
i=o
r + r -1 - l + l + 2 [ - ( s + l) + l - (s + l ) ( r + r - 1 ) - (s + l)] = - ( 2 s + 1)(r + r -1 + 2). (ii) An elementary computation shows that s--1
Er2i(r2 i=O
-
1) = r 2' -
1.
Thus, our assumption that r 2. = 1 leads to s--1
~
~ i ( ~ _ 1) = 0.
i=0
But, by hypothesis, r 2 r 1. Therefore, 5--1
E r i = O. i=0
Multiplying this equation by r 2 we obtain E r (i+11 : 0. i=0
The last two equations yield S--1
~[(~,+1
_ ~-(,+1/)~ + (~, _ r-')q
=
i=0 8-1
E[r
2(i§
-- 2 -t- r -2(i+1) + r 2i -- 2 -t- r -2i] = --4s,
/=0
which is our first claim. Multiplying the first of the above equations by r we obtain that 3--1
E i=0
r i+1 : O.
3.3 Modules over Associative Algebras
77
Therefore, we also obtain that S--1
E(r/+I
--
r-(i+l))(r
i _ r - i ) --_
i=0 s-1
E[r
2i+1 -- r -- r - 1 Jr- r - ( 2 i + 1 ) ]
---- --8(r
-Jr r - l ) ,
i=0
[]
which is our second claim.
3.3 M o d u l e s over A s s o c i a t i v e A l g e b r a s In this section, R is a field. Let A be a vector space over R. For all b, e E A, we define bc E A such that the following conditions hold.
- Vb, c, d E A : b(cd) = (bc)d, b(c + d) = bc + bd, (b + c)d = bd + cd.
-VrERVb, eEA: (rb)c=r(be)=b(rc). - 31EAVaEA:
al=a=la.
Then A is called an (associative) algebra over R. For the remainder of this section, A is an algebra over R such that dimR(A) E N \ {0}. Let V be a vector space over R. For each v E V and, for each a E A, we define va E V such that the following conditions hold.
-Vt, uEVVaEA: (t+u)a=ta+ua. - VvE VVb, eE A: v(b+e)=vb+ve, - VrERVvEVVaEA:
-VvEV:
v(be)=(vb)e.
(rv)a=r(va)
=v(ra).
vl=v.
Then V is called a (right) A-module. For the remainder of this section, V denotes an A-module such that dimR(V) E N \ {0}. For each U C_ V and, for each B C_ A, we set UB:={
ublPC
E
U x B, I P I E N \
{0}}.
(u,b)eP
An R-subspace U of V is called a submodule of V, if UA C_ U. We shall write U _~ V to indicate this state of affairs. Assume that U < V. For each v E V and, for each a E A, we set
(U + v)a := U + va.
78
3. Algebraic Prerequisites
Then the factor space V / U is an A-module. It will be called the factor module of V with respect to U. Let W be an A-module, and let r E Homn(V, W) be given. Then r is called an A-homomorphism if, for all v E V and a E A, (vr
= (va)r
The set of all A-homomorphisms from V to W will be denoted by HOmA (V, W). The A-modules V and W are called isomorphic, if there exists a bijective A-homomorphism from V to W. In this case, we shall write V ~ W.
T h e o r e m 3.3.1 (i) Let r be an A-homomorphism from V. Then ker r < V, and V r is an A-module with V r ~- V~ ker r (ii) Let T, U < V be given. Then we have T fq U < V, T + U < V, and
(T + U)/T ~- U/(T n U). Proof. (i) It is obvious that ker r _< V. For each v E Y and, for each a E A, (vr is an A-module. Set U := ker r Then, for all t, u E V, s r = tr r
(s-t)r
r
= (va)r E Vr Therefore, Yr
s-tEU
r
U+s=U+t.
Therefore, ~: V / U -~ Vr
U+v
~
vr
is a bijection. The fact that L is an A-homomorphism follows easily from the fact that r is an A-homomorphism. (ii) It is obvious that T M U < V and that T + U < V. It is also straightforward that L: U --~ ( T + U ) / T ,
u ~-~ T + u
is a surjective vector-space homomorphism with kert = T fq U. Finally, for each u E U and, for each a E A, (ut)a = (T -t- u)a = T + ua = (ua)t. Therefore, t is an A-homomorphism. Now (ii) follows from (i).
f-]
Let U < V be such that U ~- V. U is called a maximal submodule of V if, for each T < V, U C T implies that T E {U, V}.
3.3 Modules over Associative Algebras
79
V is called irreducible if 0 is a maximal submodule of V. V is called completely reducible if V (as a vector space) is the direct sum of irreducible submodules of V. (0 is completely reducible.)
P r o p o s i t i o n 3.3.2 The following statements are equivalent. (a) V is the sum of irreducible submodules of V. (b) For each U < V there exists T <_ V such that T ~ U = V. (c) 0 is the intersection of the maximal submodules of V. (d) V is completely reducible.
Proof. (a) =~ (b) Let S be a set of irreducible submodules of V such that V=~~S. SES Let U _~ V be given, and let T_~ V be maximal subject to T A U -- 0. Assume that T + U ~ V. Then there exists S E S such that S ~ T + U. As S is irreducible, this yields S n (T § U) = 0. Thus, by Theorem 3.3.1(ii),
S + T / ( ( S + T) n U) ~ (S + T + U)/U = (S + U)/U § (T + U)/U ~- S + T. It follows that (S + T) n U -- 0, contrary to the choice of T. (b) =~ (c) Let us denote by U the intersection of the maximal submodules of V. Then U _< V. Thus, by (b), there exists T < V such that T (9 U -- V. Assume that U ~ 0. Then T ~ V. Thus, there exists a maximal submodule M, say, of V such that T C_ M. This leads to the contradiction that V .C_
T+UCM. (c) ::~ (d) Let U _< V be minimal such that V/U is completely reducible. We shall be done if we succeed in showing that U is contained in each maximal submodule of V. Suppose that V possesses a maximal submodule S, say, such that U q~ S. Then U + S -- V. From this we conclude that U/(U n S) r S/(U n S) = V/(U n S). But with the help of Theorem 3.3.1(ii) we also obtain that
U/(U n S) ~- V / S and that
s/(u n s)
v/u.
Now the choice of U yields U n S = U. This means that U C_ S, contrary to the choice of S. (d) ::v (a) is obvious. []
80
3. Algebraic Prerequisites
3.3.3 Let S be a set of irreducible submodules of V such that S. Then, for each irreducible submodule U of V, there exists S E S such that U~-S. Lemma
V
=
Proof. Let U be an irreducible submodule of V. Then by Proposition 3.3.2, there exists T < V such that T G U = V. Since U ~ 0, T :~ V. Thus, there exists S E S such that S q~ T. Since S is irreducible, we conclude that S N T = 0. But, as U is irreducible, T is a maximal submodule of V. Therefore, S T = V. It follows that S @ T = V. Now we conclude from Theorem 3.3.1(ii) that U "~ V / T ~- S. []
V will be called homogeneous if V (as a vector space) is the direct sum of irreducible submodules of V which are isomorphic in pairs. From L e m m a 3.3.3 we obtain immediately
Proposition 3.3.4 Assume that V is completely reducible. Let W denote the set of the maximal homogeneous submodules of V. Then (i) V = ~ w e w W. (ii) For each irreducible submodule U of V, there exists exactly one W E kV such that U C W. (iii) Let S and T be two irreducible submodules of V. Then S ~ T if and only if there exists W E kV such that S, T C W.
3.4 Associative
Algebras
In this section, R is a field, and A is an algebra over R such that dimn(A) E
N \ {0}. Obviously, A is an A-module with respect to the multiplication of A. We shall denote this A-module by A b. A is called semisimple if A b is completely reducible. A submodule V of A b is called an ideal of A if A V C_ V. In this case, we shall write V <1 A. Let B ~ A be such that 0 :~ B. B is called a minimal ideal of A if, for each C <1 A, C C_ B implies that C E {0, B}. A is called simple if A is a minimal ideal of A. The following result is a consequence of [29; Theorem 10] and [29; Theorem
17].
3.4 Associative Algebras
81
T h e o r e m 3.4.1 [J. H. M. WEDDERBURN]Assume that A is semisimple. Let 13 denote the set of the minimal ideals of A. Then we have (i) B is the set of the maximal homogeneous submodules of A ~.
(ii) A = t~BeB B. (iii) Each element of 13 is a simple algebra over R. Proof. (i) Let W denote the set of the maximal homogeneous submodules of A ~. We have to prove that }d2 = B. Let us first prove that W C_ B. Let W E 14', and let a E A be given. Let V be an irreducible submodule of W. For each v E V, we set V,~a : ~ a v .
Then, as A is associative, )~a E HomA(V, {a}V). (Note that { a } V < A ~.) Therefore, as V is irreducible, we have either { a } V V or { a } V = 0; see Theorem 3.3.1(i). Thus, by Proposition 3.3.4(ii), (iii),
{a)V c_ w . Since a E A has been chosen arbitrarily, we conclude that A V C W . Since V has been chosen as an arbitrary irreducible submodule of W, we conclude that A W C_ W . (Recall that W is homogeneous. In particular, W is the sum of its irreducible submodules.) This means that W <3 A. Suppose, by way of contradiction, that W r B. Then, as W <3 A, there exists B E B such that B C W # B. Thus, there exists an irreducible submodule V, say, of W such that V V/B = 0. It follows that V B -= O. Let T be an irreducible submodule of B. Then, as W is homogeneous, there exists t E HomA(V, T) with Vt = T. Now, as V B = O, T B = ( V t ) B = (vB)
= 0.
Since T has been chosen arbitrarily, we have shown that B 2 = 0. By Proposition 3.3.2, there exists C _< A ~ such that C G B = A. Then C B = O. Thus, as B 2 = 0, B = {1}B C_ A B = ( C + B ) B = O, contrary to B E B. This contradiction shows that W E B. Since W E }/Y has been chosen arbitrarily, we have shown that I'Y C_ B. Let B E B be given. Then, by Proposition 3.3.4(i),
B=AB=
~_, W B . WEW
For each W E W, W B C_ W and, as W C_ 13, W B W B E {0, W}.
<3 A. Therefore,
82
3. Algebraic Prerequisites
It follows that B E W. Since B E/3 has been chosen arbitrarily, we have shown that/3 C_ W. (ii) follows immediately from (i) and from Proposition 3.3.4(i). (iii) From (ii) we obtain that, for each B E B, there exists 1/3 E B such that 1 = ~-"] lB. BE/3
Let C E/3 be given. Then, for each c E C, c l c = c ( y ] ~ 1B) = e l = c . BEI3
Similarly, one obtains that, for each c E C, 1cc = c. From (ii) we deduce that ideals of C are also ideals of A. Therefore, as C is minimal, C is simple. []
T h e o r e m 3.4.2 Assume that A is semisimple. Then (i) Each irreducible A-module is isomorphic to a submodule of A ~. (ii) Let V be an A-module such that dimR(V) E IN \ {0}. Then V is completely reducible.
Proof. (i) Let V be an irreducible A-module. Let v E V \ 0 be given. For each a E A, we set aAv := va. Then A~ E HomA (A b, V). Since 1A~ = v, 0 # A~Av < V. Thus, as V is assumed to be irreducible, we must have AbA~ = V. Therefore, by Theorem 3.3.1(i), ker)% < A ~ and V ~- A ~/ker A~. Since kerA, < A ~, there exists T _< A ~ such that T G ker A~ = A~; see Proposition 3.3.2. In particular, by Theorem 3.3.1(ii), T ~ A b/ker A~. It follows that V ~ T. (ii) Let T be a basis of V. By hypothesis, A ~ possesses a set S of irreducible submodules such that A ~ = ~ s e s S. Thus, V= VA-
~{t}A=
~-~ y ~ { t } S .
tET
t E T SEs
Let t E T, and let S E 8 be given. For each s E S, we set
3.4 Associative Algebras
83
s)~t,a := ts. Then )~t,s 9 Homa(S, {t}S). (Note that { t } S < V.) Therefore, as S is irreducible, we have either { t } S ~- S or { t ) S = 0; see Theorem 3.3.1(i). Therefore, the claim follows from Proposition 3.3.2. []
For each A-module V, we set EndA(V) := HomA(V, V). The following lemma is often referred to as "Schur's lemma".
L e m m a 3 . 4 . 3 Let V be an irreducible A-module. Then, if R is algebraically closed, Enda(V) ~- R (as an algebra over R).
Proof. Let r 9 EndA(V) be given. By hypothesis, dimn(V) 9 1~ \ {0}, and R is algebraically closed. Therefore, r has an eigenvalue r, say, in R. Set (i := r r l . Then, (f 9 Enda(V) and ker(f r 0. As V is assumed to be irreducible, it follows that kerr = V, i.e., (f = 0, i.e., r = r l . []
T h e o r e m 3.4.4 Let V be an irreducible A-module. Assume that R is algebraically closed and that A is simple. Then A ~- Endn(V) (as an algebra over R).
Proof. By Theorem 3.4.2(i), we may assume that V < A ~. Set B:={aEAIV{a}=0}. Then, obviously, 1 ~ B ~ A. Therefore, as A is assumed to be simple, B = 0. Thus, A is isomorphic to a subalgebra of EndR(V). We shall now prove that each element of EndR(V) is multiplication from the right hand side by an appropriate element of A. For all t, u E V, we set u,~t := tu. Then, as A is associative, {A. I v E V} C_ Enda(V). Thus, for each t E V, there exists rt E R such that, for all v E V, tv = rtv; see Lemma 3.4.3.
3. Algebraic Prerequisites
84
Let r E EndR(V) be given. Then, for all t, u E V, =
=
=
Since V # O, 0 # A V <3A. Thus, as A is assumed to be simple, A V = A. Therefore, there exists P C_ A • V such that tPI E N \ {0} and 1 ~
~ aY. (a,v)eP
Now we have that, for each u E V, ur = ( u l ) r
(ua)(v l = u( (a,v)eP
(a,v)eP
(a,v)eP
a(v@ (a,v)fiP
This means that r is multiplication by the element ~(a,~)eP a(vr
[]
We shall denote by J(A) the intersection of the maximal submodules of A ~. (3(A) is usually called the Jacobson radical of A.)
T h e o r e m 3.4.5 (i) For each irreducible A-module V, VJ(A) = 0. (ii) J(A) = 0 if and only irA is semisimple.
Proos (i) Let v E V \ 0 be given. For each a E A, we set a)t v : ~ ~)a.
Then )~ E HomA(A ~, V). Since 1)~ = v, 0 # A~)~ _< V. Thus, as V is assumed to be irreducible, we must have A~)~ = V. Therefore, by Theorem 3.3.1(i), ker $~ < A ~ and V ~- A ~/ker ~ . It follows that ker )~ is a maximal submodule of A ~. Therefore, by definition, J(A) C ker)~. This means that vJ(A) = 0. Recall that v E V \ 0 has been chosen arbitrarily. Therefore, we have shown that VJ(A) = 0. (ii) follows immediately from Proposition 3.3.2. (We apply that proposition to the A-module A ~.) []
3.5 Characters of Associative Algebras 3.5 Characters
of Associative
85
Algebras
In this section, R is a field, and A is an algebra over R such that dimR(A) E
\ {0}. Let V be an A-module such that 0 ~ V. Then, for each a E A, 8v(a) : V -q V, v ~-~ va is a linear map. The (linear) map Xv : A ~
R, a ~
tr6v(a)
is called the character of A afforded by V. (Here trSv (a) denotes the trace of the linear map 8v (a).)
T h e o r e m 3.5.1 Let V and W be A-modules such that V ~s 0 ~ W and V ~ W . Then, X v = X w .
Proof. Let a E A be given, and let S be a basis of V. Then, there exists {rtt, I t, u E S} C R such that, for each t E S, ta : ~ rtuU. uES
In particular, X v ( a ) = ~-~ses rss. On the other hand, since V ~ W, there exists a bijective A-homomorphism t from V to W. In particular, St is a basis of W, and~ for each t E S,
(t,)a -= (ta), = (~-~ rt~u), = ~ uES
rtu(u,).
uES
It follows that x w ( a ) = ~ s e s r,s. Recall that a E A has been chosen arbitrarily. Therefore, we have shown that Xv = X w . []
A character of A which is afforded by an irreducible A-module will be called irreducible. The set of all irreducible characters of A will be denoted by Irr(A). An element X E Irr(A) is called linear if X(1) = 1. Let us now assume that A is semisimple, and let us denote by B the set of the minimal ideals of A.
86
3. Algebraic Prerequisites
Let B E 13 be given. Then, by Theorem 3.4.1(i), B is a homogeneous A-module. Thus, by Theorem 3.5.1, there exists exactly one CB E Irr(A) such that, for each irreducible submodule V of B, Xv = V s . For each B E B, there exists 1B E B such that 1---- ~-'~ 1B; BEt3
see Theorem 3.4.1(ii).
T h e o r e m 3.5.2 Assume that A is semisimple. Let B denote the set of the minimal ideals of A. Then we have (i) {r [ B E 13} = Irr(A). (ii) Let C, n E 13 be such that C ~ D. Then r (iii) Let B E 13, and let a E A be given. Then r
= {0}. = CB(lna).
Proof. (i) Let X E Irr(A) be given. Then, by definition, there exists an irreducible A-module V such that )iv = X. By Theorem 3.4.2(i), there exists an irreducible submodule W, say, of A ~ such that V ~ W. Since V ~- W , Xw = Xv; see Theorem 3.5.1. By Proposition 3.3.4(ii) and Theorem 3.4.1(i), there exists exactly one B E B such that W C_ B. Since W <_ B, Xw = CB. It follows that X = CB. (ii) Let d E D be given. Then, as C{d} C_ C D = 0, Co(d) = 0. (iii) From (ii) we obtain that
CB(a) = CB(la) = C B ( Z l e a ) = Z CB(1ca) = CB(1Ba). CED CEB []
Theorem 3.5.2(i) says that r
B -~ Irr(A), B ~
CB
is surjective. In the first part of the following theorem, we shall see that ~ is even a bijection if char(R){ lcm{x(1) I )f E Irr(A)}.
T h e o r e m 3.5.3 Assume that A is semisimple. Assume further that char(R) ~ lcm{x(1 ) [ X E Irr(A)}. Then we have (i) Let 13 denote the set of the minimal ideals of A. Then 1131= I l r r ( A ) l .
3.6 Tensor Products
87
(ii) Irr(A) is a linearly independent subset of Homn(A, R). (iii) If R is algebraically closed,
XA =
x(1)x. xEIrr(A)
In particular, dimn(A) = ~xr
2"
Proof. (i), (ii) From Theorem 3.5.2(ii), (iii) we deduce that, for all C, DEB, r = 3CDr 2 Thus, the claims are consequences of Theorem 3.5.20). (iii) follows from Theorem 3.4.1, from Theorem 3.4.4, and from (i).
3.6 Tensor
[]
Products
In this section, R will be a field. V and W will be vector spaces over R such that d i m n ( V ) :~ 0 # d i m a ( W ) . Let T be a basis for V, and let U be a basis for W. For each t E T and, for each u E U, let
t|
TxU
~
R
denote the uniquely determined m a p which, for each (t ~, u') E T • U, satisfies
(t', u')(t | u) = ~,,t~,u. 3 We define
~/"~R W "~- {
~
rtu(t ~ U) I rtu E R}. 4
(t,u)r V QR W is called the tensor product of I / a n d W. Note that
{t|
I (t,u) E T x U}
is a basis for V | W. Let v E V, and let w E W be given. Since v E V, there exists {rvt [ t E T} C R such that 2 Here ~ is the Kronecker delta. 3 Here ~ is the Kronecker delta. 4V| W is the vector space of all R-valued functions on T • U. It is the "free R-module based on T • U ' .
88
3. Algebraic Prerequisites
=~
V
rvtt.
tET
Since w E W, there exists {rwu I u E U} C_ R such that W
= ~
rwuU.
uEU
We define
v | ~ :=
~
,'~,,'~.(t | u).
(t,u)ET•
The following lemma will be needed for the proof of Theorem 3.6.6.
L e m m a 3.6.1 Assume that direR(V), direR(W) E N. Let T' be a basis for V, and let U' be a basis for W. Then {t' | u' [ (t',u') E T' • U'} is a basis for V | W .
Proos Let t E T, and let u E U be given. Since t E T, there exists {rtt, I t' E T'} C R such that t =
Z rtt'tt" t~ET ~
Since u E U, there exists {ruu, ] u' E U'} C R such that
= ~
ruu tILl.
ulEU ~
Thus, by definition, t| u =
~
,',,,,',,,,(t'
| u')
(t',u')ET'xU'
In particular, V | W is generated by {t' | u' I (t', u') E T' x U'}. On the other hand, we have that
t{t' | u' I (t', This proves the lemma.
u')
~ T' x U'}I = dimn(V |
W). []
Let ~r E EndR(V), and let r E EndR(W) be given. By cr| we denote the uniquely determined element of EndR(VQR W) which, for each (t, u) E T • U, satisfies (t | u ) ( ~ | T) = tr * ur.
3.6 Tensor Products
89
L e m m a 3.6.2 Let ~r E EndR(V), and let v E EndR(W) be given. Then, for each v E V and, for each w E W , (v | w)(c~ | r) = v~ | w r .
Proof. Let v E V, and let w E W be given. Since v E V, there exists {rvt I t E T } C R such that v :- E Vvtt. tET
Since w E W, there exists { r ~
I u E U} C_ R such that W ~ E rwuU. uEU
For each t E T there exists {rtt,(cr) I t' ~ T} C_ R such that tO" --- E rtt'(~ t~ET
Similarly, for each u E U there exists {r~.,(r) I u~ E U} _C R such that
~
ur=
r~,(r)u'.
utEU
It follows that
(t | u)(~ | r) = ta @ uv =
,..,(,.)r | ~ t~ET
,-~,u-)u' =
u~6U
,.,. (,.)r,,,,,(~-)(t' |
,/).
(t',uOET•
Therefore,
(v | w)(~ | r) =
(~ ~.,t | tET
[
~
I 2 ,..~,,)(o- | ~-) = uEU
,.,,,,.,~,,(t o ~)1(o- |
=
(t,u)~.TxU
r,,m~uI(t e ~)(o | T)]
=
(t,u)ITxU
E
(t,u)ETxU
,-~,.,,,,
~
,-.,(,.),-,,,,,(T)(t'| u').
(t',u')ETxU
On the other hand, vo" @ wT" -~
90
3. Algebraic Prerequisites Z rvttO" | Z rwuUT : tET uEU tET
Z r,,,U/t' | Z
Z
t'fiT
u'fiU
ufiU
Z
| ZEZ
t'ET tilT
u~fiU uEU
(tJ,ul)fiTxU t i l t
: =
uEU
[]
T h e o r e m 3.6.3 Let A and B be algebras over R. Assume that V is an A-module and that W is a B-module. (i) For all a, a' E A and, for all b, b' E B, define (a | b)(a' @ b') := aa' | bb'. Then A | B is an R-algebra. (ii) For all v C V, w E W and, for all a E A and b E B, define
(v | w)(a | b) Then V |
W is an A |
:=
va | wb.
B-module.
Proof. (i) is obvious. (ii) Let v E V, let w E W, let a, a' E A, and let b, b' E B be given. Then, by Lemma 3.6.2,
(v | w)[(a | b)(a' @ b')] = (v | w)(aa'| bb') = v(aa') | w(bb') = (va)a' @ (wb)b' = (va | wb)(a' | b') = [(v | w)(a | b)](a' | b').
The other equations which have to be confirmed follow similarly.
[]
T h e o r e m 3.6.4 Let A be an algebra over R, and let S be an extension field of R. Let t, u E S be given. For all b, c E A, define (t @ b)(u | e) := tu @ be. For each a E A, define l(u | a) := tu | a. Then A s := S | A is an algebra over S with d i m s ( A s ) = dimn(A).
Proof. By Theorem 3.6.3, As is an algebra over R. Defining
3.6 Tensor Products
91
t(u | a) = (t | l)(u @ a) = t u @ a for all t, u E S and, for each a E A, As becomes an algebra over S.
I--1
For the remainder of this section, we assume that R is the field of complex numbers. For each r E R, we shall (as usual) denote by ~ the complex conjugate of r. We also assume that dimn(V), dimR(W) E N. Let us denote by ~T the uniquely determined hermitian form on V such that, for all t, t' E T,
~T (t, t')
=
~tt,,
where 6 is the Kronecker delta. We define ~u analogously. The tensor product ~T| Of~T and (u is defined as follows. It is the uniquely determined hermitian form on V| W such that, for all t, t' E T and u, u' E U, ~T|
| u , t ' | U') : ~tt'~uu',
where, again, ~ denotes the Kronecker delta.
Lemma3.6.5
Let v, v' E V, and let w, w' E W be given. Then
~T|
| w, v' | w') = ~r(~, ~')@(~, ~ ' )
Proos Since T is a basis for V, there exists {rvt ] t E T} C_ R such that y = ~ rvtt tET
and {r,,t I t E T} C_ R such that VI :
~
rv't 'tl.
t'ET
Similarly, as U is a basis for W, we find {r~. ] u E U} C_ R such that W ~ ~ rwu~ uEU
and {u~,u I u E U} C R such that rwlu~'lt .
WI =
u'EU
3. Algebraic Prerequisites
92
Then
(T| tET
| w, V' | W') =
uEU
t'ET
E
(t,u)ETxU
}2
u~EU
(t',u')ETxU
F.
(t,u)ETxU
(t',u')ETxU Z rvtrwurv,trw,u (t,u)ETxU
and
~r(v, v')~v(w, w') = tET tET
t~.T UET
uEU uEU
u~EU ulEU
E rvtrvlt E rwurw'u 9 tET uEu
[]
Let us briefly recall some elementary facts from Linear Algebra. It follows immediately from the definition of ~T that, for each v E V,
0 _< ~r(v, v) and that
O=~r(v,v) ~
v=0.
Thus, ~T is positive-definite. In other words, V is a (finite dimensional) Hilbert space with respect to (TLet cr E EndR(V) be given. It is easy to see (and well-known) that there exists a uniquely determined element c~* E EndR(V) such that, for all t,
uEV, ~r(t~, ~) = ~r(t, u~'); see [18; Hauptsatz II.5.2 a)]. The endomorphism a* is called the adjoint of a
relative to (T. Assume that cr is hermitian (which, by definition, means that a* = cr). Then V possesses an orthonormal basis S, say, and there exists a set {rs I s E S} C_ R of real elements such that, for each s E S, s~r = rss; see [18; Hauptsatz II.6.7]. In particular, all eigenvalues of c~ are real. It is not too hard to see
3.6 Tensor Products
93
that all eigenvalues of a are non-negative if and only if, for each v E V, 0 < (T(Vo., v); see [18; Satz II.7.3]. In this case, c, is called p o s i t i v e - s e m i d e f i n i t e .
We shall use both of these facts in the proof of the second part of the following theorem. Let r E EndR(W) be given. We shall denote by r" the adjoint of r relative to ~u. Moreover, as, by definition, a N 7- E Endn(V | W), it is reasonable to speak of the "adjoint" of ~r | 7- "relative to" (T| We shall denote that endomorphism by (o. | r)*.
T h e o r e m 3.6.6 L e t o. E Endn(V), and let 7- E Endn(W) be given. T h e n we have
(i) (o" | 7.)" = o.* | 7-*. (ii) A s s u m e that o.* = ~ and7"* = 7-. T h e n (~| 7-)* = a @ r and, i f cr and 7- are p o s i t i v e - s e m i d e f i n i t e ,
so is cr | 7".
Proof. (i) Let t, t' E T, and let u, u' E U be given. Then, by Lemma 3.6.5,
~7,|
|
@|
~)(o. | 7.), t' | u')
=
(to" | uT-, t' | u') =
@ (t,~, t')@ (~7-, ~') and
~r|
o ,,, (~' o ,/)(~" o 7-')) = @|
| u,t'o." | u%') =
~- (t, e ~ ' ) @ (u, u'7-'). It follows that (~r | 7-)* = (r" | r ' . (ii) The first claim follows immediately from (i). Assume that ~r is positive-semidefinite. Then, by the first of the abovementioned arguments from Linear Algebra, V possesses an orthonormal basis S, say, and there exists a set {re I s E S} C R of real elements such that, for each 8 E S, 80" =
7"~8
and 0 _< rs. By Lemma 3.6.1, {s | u I (s,u) E S • U} is a basis for V | W. Let { r ~ I (s, u) E S x U} C R be given. Then, by Lemma 3.6.2 and Lemma 3.6.5,
(s,u)ESxU
~T|
~ (s,u)6_SxU
(s,u)eSxU
,,,~[(8|
u)(~ | 7-)1,
~ (s,u)ESxU
~,~(8 | u)) =
94
3. Algebraic Prerequisites
Z
rsu
(s,u)ESxU
E (s',u')6SxU
~-Tv,(r|
|
~)(~ | ~), s' | .')
=
r-'~7-~u,~T| (SO" @llT", st @ U') -~ E rsu E (s,u)6SxU (s',u')eSxU rsu
~.-;v@ (s~, s')@ (uT, =') =
E
(,,u)~SxV
(,',u')Esxu
s6Su6Uu'6U
sES
u E U u'6U
u6U
s6S
u6U
Thus, if r is positive-semidefinite, ~ | r is positive-semidefinite. (Recall that, for each s 9 []
In order to prove the last theorem of this section, we assume that V = W and that T = U. We set S : = T and
V ~ := R { s |
I s 9 S}.
By t we denote the uniquely determined element of H o m n ( V , V ~ such that, for each s 6 S, st = s | s. It is clear that t is bijective. Let rr denote the projection of V @u V on V ~ with respect to (the complement) R{t @ u I t, u G S, t # u}.
L e m m a 3 . 6 . 7 (i) Let t, u E V be given. Then ~s(t, u) : (s|162 ut). (ii) Let (rtu I (t,u) E S x S} C_ R, and let {r= ] s 6 S} C_ R be given.
Then
~=|
~
,.,,,(t|174
(t,u)ESxS
(t,u)ESxS
=6S
=ES
Proof. (i) It is enough to assume that t, u E S. But then, by definition, ~s(t, u) : ~t,, : ~,, : ~ s | (ii) We have
| t, ~, |
,,).
3.6 Tensor Products
~=|
~
,.,u(t|174
(t,u)ESxS
E
95
",~
(t,u)ESxS
s6S
E~=|
|
sES
E
~,=|
=) =
rssr-7
sES
and
~=|
~
r.,(teu).,~,',(=|
(t,u)6SxS
(t,u)6SxS
seS
s6S
E
r=,~~s,
sES
[]
Let ~, r E Endn(V) be given. Since a E Endn(V), we find {rtu(~r) I t, u E such that, for each t 6 S,
S} C R
u6.S
Since r 6 Endn(V), there exists {rt~(r) I t, u 6 S} _C R such that, for each tES, tr = E
rt~(v)u.
uES
By a o r we denote the uniquely determined element of Endn(V) which, for each t E S, satisfies
~(~ o 7-) = ~ r,~(~)r,=(7-)u uES
Note that a o r depends on S.
T h e o r e m 3.6.8 Let G, r E EndR(V) be given. Then we have (i) For each v E V, vt(c~ | r)~ = v(a o r)t. (ii) If a and 7" are positive-semidefinite hermitian, so is a o 7".
Proof. Since cr 6 EndR(V), there exists {rtu(~) for each t 6 T,
t, u E S} _C R such that,
96
3. Algebraic Prerequisites to" = E
rtu(o-)u.
uES
Since r 6 Endn(V), there exists
{rtu(v) [ t, u
E S} C R such that, for each
tES, ~r = E vt~,(v)u. uES (i) For each s E S, we have
ts
u6S
E ~,,(o-)E ~,~(T)(t| u)~ t6S
=
u6S
r . ~,(o-)~,, (T)(t | t) tES and
( E rst(o-)r,t(T)t)t = tEs 8((7 o 7")t. This proves (i). (ii) Let v E V be given. Then, by (i), Lemma 3.6.7 and Theorem 3.6.6(ii),
~S(~)(O- 0 7"), V) = ~S(VL(O- ~ 7")Trt -1, v) =
~| This proves (ii).
e r)~. v,) : ~s|
| T), ~,) ___o. []
4. R e p r e s e n t a t i o n
Theory
In this chapter, we assume that IX[ E 1~. In the first four sections of this chapter R will denote a field.
4.1 Adjacency
Algebras
Let us denote by R X the vector space of all R-valued functions on X. 1 We write
Rx := { Z r:vx
R)
:vEX
Then X is a basis for R X . X is called the standard basis of R X . Let g E G be given. We shall denote by (rgR the uniquely determined element of EndR(RX) which, for each x E X, satisfies xc~=
~-~y. yE:vg
In the first four sections of this chapter, we always abbreviate O'g :---- O'gR
.
The following lemma is probably the most important consequence of the definition of a scheme.
L e m m a 4.1.1 (i) For all e, f E G, (re~rl = ~--]~gecaetg~rg. (ii) Let n E l~ \ {0, 1}, and let f l , ..., fn E G be given. Then we have O'fl
9 9 9O'f,, ~- Z g f G
afx--.f~g~
Proos (i) For each x E X, we have x R X is the "free R-module based on X".
98
4. Representation Theory
xcrecrl= Z
y~I = Z
yExe
Z ael9 Z gEG
Z
Z=Z
yExe zEy.f
Z
aelgz=
gEG zExg
z-= Z aelgXtrg = X ( Z aelg(Tg).
zEzg
gEG
gEG
(ii) Assume that 3 _< n. Then, by induction,
~ i , . . . ~I. = ( ~
ai,..,I._,~o)~I.
=
eEG
Z
Y: a<, ..<._,, y:
eEG
gEG
eEG
Z ( Z aIl""I~-leael"g)O'g = Z aIx f"g~ gEG eEG gEG use (i).
[]
For each F C_ G, we set
RF :=
{ Z rfo'!
I r! 9 R}.
IEF (Note that R0 = 0.) Obviously, RF is a vector space over R with basis {crl ] f 9 F}. From Lemma 4.1.1(i) it follows that RG is not only a vector space over R but even an algebra (over R). It is called the adjacency algebra of (X, G) over R. It is obvious that RX is an RG-module. The vector space RX is called the standard module of RG. The character Xnx of RG afforded by the standard module (of RG) will be called the standard character of RG. With the help of thc standard basis of RX we may compute explicitly the values of the standard character. (By we mean the Kroneeker delta.)
L e m I n a 4.1.2 (i) For each g E G, XnX(~g) = ~:glXI. (ii) For all e, f E G, XRX(ae.c~f ) = 8eyle*l. (iii) Let {rg [ g E G} C R be given, and set cr : : ~ g e c rg~rg. Then, for each g E G, Xnx (#g. a) = rglg* I.
Proos (i) follows from the definition of crg. (ii) Let e, f E G be given. Then we obtain from Lemma 4.1.1(i), from (i), from Lemma 1.1.1(ii), and from Lemma 1.1.2(ii) that
4.1 Adjacency Algebras
99
x~x(o'~.~s)= xR~(~ a~.sg~g)=~ a~.s~x,,~(og)= gEG
gEG
E ae*fg~lglXl = ae*fxlXI = r~efrte*IXl
=
~ef [e* I"
gEG
(iii) follows from (ii) because RG is distributive and Xnx is linear.
[]
The third part of the following theorem was proved in [21].
T h e o r e m 4.1.3
(i) Set J := {g E G [ char(R) I Ig*l}. T h e . ,
3(RG)
RJ. (ii) Assume that, for each g E G, char(R) { Igl- Then RG is semisimpte. (iii) [H. MASCHKE] Assume that ( X , G ) i s thin. Then, /fchar(R) { IXI, RG is semisimple.
Proof. (i) Let ~ E J ( R G ) be given. Since ~ E RG, there exists {rg I g E
G} C_R such that (3" = E FgCrg" gC:G
Let g E G be such that rg ~ O. We shall be done if we succeed in showing that char(R) divides Ig*lTo that end, let O = Vo <_ VI <_ . . . <_ V. = R X
be a series of submodules of RX, such that, for each i E { 1 , . . . , n}, ~ / V i - 1 is irreducible. Then, by Theorem 3.4.5(i), V~rg. c~ _C Y~-x for each i E { 1 , . . . , n}. Therefore, Xt~x (~g" ~) = O. On the other hand, by Lemma 4.1.2(iii), Xnx(Crg. a) = rg[g*[. It follows that rglg*l = 0. Thus, as rg # 0, b*l = 0. This means that char(R) divides
b'l.
(ii) follows from (i) and Theorem 3.4.5(ii). (iii) follows from (ii) and Lemma 1.1.2(ii).
[]
For the remainder of this section, we assume that
char(R) ~ 1-I Igl. gEG
Thus, by Theorem 4.1.3(ii), we have that RG is semisimple (so that we may apply the results of Section 3.4).
100
4. Representation Theory
Since R G is semisimple, there exists, for each X E Irr(RG), precisely one minimal ideal B• of R G such that, for each irreducible submodule V of (RG) ~ such that V C Bx, Xy = X; see Theorem 3.5.3(i). Set ~X : = 1Bx"
Then, by Theorem 3.5.30) , 1 xEIrr(RG)
For each X E Irr(RG), let m x E N be such that XRX =
E
rnxx"
xEIrr(RG)
L e m m a 4.1.4 Let X E Irr(RG) be given. (i) For each r E B x,
mx x(%" o')
0.= gEG F_,
Ig'l
gea
lg*]
(ii) We have O"9 .
(iii) We have char(R) Jf mx.
Proos (i) Since 0. E RG, there exists {rg ] g E G} C_ R such that 0. =
E 790.9" gEG
We shall be done if we succeed in showing that, for each g E F,
mxx(0.g. 0.) ~" -
Ig'l
Let 9 E G be given. Then, as 0. E B x <] RG, 0.g.0. E B• Therefore, by Theorem 3.5.2(ii), Xnx (0.g"0.) = mxx(ag. 0.). On the other hand, by Lemma 4.1.2(iii), Xnx(0.g*0.) = rgIg*l. Therefore, we have rnxX(0.g.0. ) = rgl9*l, so that we have shown (i). (ii) follows from (i) and Theorem 3.5.2(iii). (iii) follows from (ii), because e• # 0. []
4.1 Adjacency Algebras
101
T h e o r e m 4.1.5 [ORTHOGONALITYRELATIONS]Let r r E Irr(RG) be given. Then we have the following. (i) For each g E G,
r
)
eEG JEG
(ii) We have
1 eer
r
gEG
Proof. (i) From Lemma 4.1.4(ii) and Lemma 4.1.1(i) we conclude that
~ r = (~ -,~(~.)~)(~ mr162 le'l
eEG
eEGfEG
fEG
=
If'l
le*llf*l
le'llf*l
gEC eEGfEG
On the other hand, as er E Be and er E B e, % % = &r162162162 Using once again that
~ =~
:c
mr
19"1
o-~,
we obtain that, for each g E G, r162
mr162
eEG f E G
Recall that, by Lemma 4.1.4(iii), me # 0 # me. Therefore we conclude that, for each g E G, a~f.glg*I
r
eEG f E G
Finally, recall that, by Lemma 1.1.3(ii), (iii) and by Lemma 1.1.2(ii),
a~f.glg*l = af~.g. Ig*l= ag.,:Ifl. Thus, we have proved (i). (ii) follows from (i) by setting g* = 1; use Lemma 1.1.1(i).
[]
102
4. Representation Theory
We do not need Theorem 4.1.5(i) in order to prove Theorem 4.1.5(ii). From Lemma 4.1.4(ii), (iii) we have r
=~--'--
1 r
(
Theorem 4.1.5(ii) is now obtained by applying r to this equation. As in the proof of Lemma 4.1.4(i), (ii), one has to use Theorem 3.5.2(ii), (iii). In the remainder of this section, we do not always really need that, for each g E G, char(R) ~ Igl, but for the sake of simplicity we retain this hypothesis. Besides the standard character, the adjacency algebra R G possesses still another natural character which we want to introduce now. Let us define temporarily V'.-~. E X . xEX Then, for each g E G, we have VO'g ~- Z E y ~ rig, Z x -- ng, V. xEX yExg xEX It follows that R v is a submodule of the RG-module R X . We shall denote by l n a the character afforded by the RG-module Rv, in other words, we set 1RG := XRv. The character l n a is called the principal character of RG. 2
L e m m a 4.1.6 (i) For each g E G, lna(6~g) = ng,. (ii) For each X E Irr(RG) \ {lna}, EgEG~(O'g) ~- O. (iii) We have m i n d = 1.
Proof. (i) follows immediately from the definition of l n a . (ii) Set r = l n a in Theorem 4.1.5(ii), and use (i) and Lemma 1.1.2(ii). (iii) Set r = l n a = r in Theorem 4.1.5(ii), and use (i), Lemma 1.1.2(ii),
and Theorem 1.3.6(iv).
[]
Let us denote by 7 the uniquely determined element of E n d n ( R G ) which, for each g E G, satisfies
In the literature, the principal character is also called "index character".
4.1 Adjacency Algebras
103
We have
L e m m a 4.1.7
= 1 E EndR(RG).
gEC.,
Proof. For each f E G, we have 1
0"] E gEG
-~gO'fO'g*'~O'g=
gEG
5 gEG
1
~gO'g*"fO'g = E
= E eEG
o..1
gEG ng
g
see Lemma 4.1.1(i) and Lemma 1.1.1(ii).
[]
Let V be an RG-module. It is easy to prove that in general, for each r E nndR(V), E
l ~ r g ' C a g E EndRa(V).
gEG The argument is exactly the same as that of the proof of Theorem 4.1.9 below. It is natural to ask which elements of EndRG(V) are obtained by "averaging" elements of EndR(V). The following theorem says that, if V is irreducible, the identity on V is always obtained this way. 3 The proof of this theorem uses a simplified version of the main idea of a theorem due to M. IKEDA on so-called "Frobenius algebras"; see [19]. Originally this idea is due to W. GASCHiJTz; see [12]. T h e o r e m 4.1.8 [M. IKEDA]For each irreducible RG-module V there exists r E EndR(V) such that E
1
g.r
= 1 E EndRa(V).
gEG ng
Proof. We still assume that, for each g E G, char(R) ~ ]g]. Thus, by Theorem 4.1.3(ii), RG is semisimple. Let V be an irreducible RG-module. Then, by Theorem 3.4.2(i), we may assume that V ~_ (RG) ~. Since RG is semisimple, there exist W ~_ (RG) ~ such that 3 Actually the irreducibility of V can be dispensed with.
104
4. Representation Theory V 9 W = (RG)b;
see Proposition 3.3.2. Let rr denote the projection of (RG) ~ onto V with respect to W, i.e., for each v E V and, for each w E W, we define (v + w)Tr := v. Then, for each g E
G, rO'g
= o'g~.
Set r := rTrr. Then r E Endn(V) and, for each v E V, gEG
gEG
' g.7P~TrO'g =w ngCr
Z
gEG
see L e m m a 4.1.7.
[]
For each ~r E RG, we set r
:: ~
1 --crg, cr~rg.
gEG ng
Let us also define
gEG
T h e o r e m 4.1.9 ~ is a linear map from R G to Z(RG).
Proof. It is easy to see that ( is linear. Thus, we only have to prove that
nGr c z(na). For each g E G and, for each ~r E RG, we have 1
1
eEG~ ne
eEG ne
~ 8 .lEG 8 o~o.~.~.~
eEG
]EG eEG
.T
= ~ ~2 ~1a ~ , . ~ , . ~
=
JEGeEG
lEG nf
eEG
1
leg see L e m m a 4.1.1(i) and L e m m a 1.1.3(ii), (iii).
[]
4.1 Adjacency Algebras T h e o r e m 4.1.10
105
(i) For each e E G,
O'er: E [ E ag*eg!' lEG gEG ng jcr]. (ii) We have 1~ E RO e(a). (iii) The matrix
( E ag'eg! )e! gEG ng is stochastic ( i,, the se,,se that each of its col,,m,-s,,ms is
IXl).
Proos (i) Let e E G be given. Then, by Lemma 4.1.1 (ii),
CreW: E - ~ 91 0"9"0"e0"9: E 1 E ag'eg!O'! = E [ E ag~ gfiG gEG ng lEG !fig gfiG ng (ii) From Lemma 1.1.1(i) we conclude that, for all f, y E G, ag~ f : ag,g]. Thus, by (i), 1(
:
ag. g!
!EG gEG For each f E G, define rf
:: E ag.g!. g6G rig
Then
I~ = E
rf(Tf.
.lEG
Let f E G be such that r! :/= 0. Then there exists g E G such that ag~ • O. This means that f E g*g. Thus, by Theorem 2.3.1(iii), we obtain that f E O e(G). Since f E G has been chosen arbitrarily such that r! :/: 0, we conclude that 1r E RO e(G). Off) For each f E G, we have
eEG gEG d6.G
E EIEo
g6G d6G e6G
.g,.l
ng
= E E a,g! = Ixl; g6G d6G
here we needed Lemma 1.1.3(ii), Lemma 1.1.40), Lemma 1.1.2(iii), and Theorem 1.3.6(ii), (iv). []
106
4. Representation Theory
4.2 Algebraically
Closed
Base
Fields
In this section, we again assume that char(R) ~ H
[gl.
gEG
Thus, by Theorem 4.1.3(ii), we have that RG is semisimple. In addition, we shall assume that R is algebraically closed. Recall that, for each O- E RG, O-( : = Z
- -1
gE G n g O'g,
O-O-g,
The following result was proved as (7.1) in [14].
T h e o r e m 4.2.1
[D. G.
HIGMAN]RG(
= Z(RG).
Proof. From Theorem 4.1.9 we know already that (RG)( C_Z(RG). Thus, we only have to prove that Z(RG) C (RG)(. Let X E Irr(RG) be given. By Theorem 3.4.1(ii), Theorem 3.4.4, and Lemma 3.4.3 we shall be done if we succeed in showing that ex E (RG)(. To that end, let V be an irreducible submodule of (RG) ~such that V C B x . Then, by Theorem 3.4.4, B x ~ EndR(V) (as an algebra). Thus, by Theorem 4.1.8, there exists O- E B x such that, for each v E V, v(o-() = v. But, as B• ~_ RG, a( E B• Thus, O-( = ex. [] C o r o l l a r y 4.2.2
The matrix
( Z ag*egf)ef gEG ng has rank [Irr(nG)]. Proof. From Theorem 3.4.1(ii), (iii) and Theorem 3.4.4 we conclude that d i m n ( Z ( R G ) ) = IIrr(nG)l . Thus, the claim follows from Theorem 4.1.10(i) and Theorem 4.2.1. [] Let X E Irr(RG) be given, and let V• be a fixed irreducible RG-module which affords X. Fix a basis S• say, of V• Then, for each O- E RG, there exists
4.2 Algebraically Closed Base Fields
107
such that, for each t E S x,
uES x
(Note that {rtXu [ t, u E the irreducible module V• find a basis with respect to Let v, w 6 S• be given. there exists
S• does not depend on the particular choice of For each RG-module isomorphic to V• we may which a yields the same coefficients.) Then, by Theorem 3.4.1(iii) and Theorem 3.4.4,
exw C B• such that, for all t, u E S x, X 'X = gt,.&,v. rt,,(%,.) (In other words, we~,~ = v and, for each s E S x \ {w}, s e x = 0.)
L e m m a 4.2.3 Let X E I r r ( R G ) , and let t, u E S x be given. (i) For each ~ E R C , X ( ~ e s = r s (ii) We have ~
Proof. (i) For each s E
Ig'l
Sx ,
yes•
yes x
Therefore,
x(~q0 = r~(~) (ii) By definition, r and Lemma 4.1.4(i).
E B x. Thus, the desired equality follows from (i) []
The following theorem generalizes Theorem 4.1.5 in the case where R is algebraically closed. Its proof is similar to that of Theorem 4.1.5.
T h e o r e m 4.2.4 [ScHuR and let v, w E S~ be given. (i) For each g E G,
RELATIONS]Let
r r E Irr(RG), let t, u E Sr
108
4. Representation Theory
eEG ] E G
me
(ii) We have 1
r .
1 me
9
= 5r gEG
Proof. (i) From Lemma 4.2.3(ii) and Lemma 4.1.1(i) we obtain
'~
~
eEG
le'l r
E
If*l
L_,
lEG
rr
~f) =
r
I.'llf*l
E
eEG : E G
r
EIE E g6G e6G f 6 G
r~
cr
I:llf*l
a.:g]..
But, by definition, er 9 Be and e~r 9 Be. Thus ~t~ "r e~pvw= 5r162 Lemma 4.2.3(ii) once again, we also have
gec
Using
Ig'l
Therefore, for each g 9 G,
.ec:ea
I:ll:*l
Ig*l
Recall that, by Lemma 4.1.4(iii), me :~ 0 ~ me. Therefore, for each g 9 G, we must have
Finally, recall that, by Lemma 1.1.3(ii), (iii) and by Lemma 1.1.2(ii), a.s'~lg*l = aS:g" Ig'l = ~:~:l/I (ii) follows from (i) by setting g* = 1; use Lemma 1.1.1(i).
E]
L e m m a 4.2.5 Let Z denote the subring of R generated by 1. Let g E G, and let X 9 Irr(RG) be give.. Then X(ag) 9 I ( Z ) .
4.3 The Complex Adjacency Algebra
109
Proof. Clearly, X(~r9) is a sum of characteristic roots of eg. Each characteristic root of ~rg is an element of I(Z). Thus, the claim follows from Theorem 3.2.2. []
4.3 T h e C o m p l e x A d j a c e n c y A l g e b r a In this section, we assume that R is the field of complex numbers.
Let A be a linear character of RG.
L e m m a 4.3.1
(i) For each g ~ C, IA(,'g)l _< 1RC(~g). (ii) Let F C_ G be given. Assume that, for each f E F, A(e]) = lnG(~l)Then, for each g E (F), A(~g) = lnG(eg). (In particular, if (F) = G, ,~ = lnc.)
Proof. (i) Let V be an RG-module which affords )~. By Lemma 4.1.4(iii), we may assume that V < R X . Let {rx I z E X} C R be such that 0
Exex r~x ~ V. Let g E G be given, and set := ~(~g). Then
Z srxx = s / Z r~x/ = / ~ xEX
xEX
xEX
xEX
rxx~ =
xEX
yExg
~:C:X yExg ~
It follows that, for each x E X, 8r x =
~ ry. yExg*
Pick a z E X such that, for each x E X, [rxI _< IrzI . Then, by Lemma 1.1.2(i),
Isllr~l = rsr~r= I Z yEzg*
,'~r _< Z
Ir~l _< n~.lr~l
yEzg ~
But r~ ~ 0. Therefore, [,~(crg)I = Is I _< ng.. Now the claim follows from Lemma 4.1.6(i). (ii) If F = ~, (F) = {1}. In this case, the claim holds obviously.
110
4. Representation Theory
Let us assume that F ~ ~, and let g E G \ F be given. Then, by Theorem 1.4.1(ii), there exists n E N \ {0, 1} such that g E F ". Therefore, there exist f l , . . . , f,~ E F such that g E f l "'" f,~. Now Lemma 1.2.4 implies that a fl...f,,g • O.
By assumption, we have that, for each i E { 1 , . . . , n}, A(a],) = l n v ( a L ) . Therefore, by Lemma 4.1.1(ii), Lemma 1.1.5(iii), (ii), and Lemma 4.1.6(i), =
eEG
eEG
A(al, . . . a s . ) = A ( a f , ) . . .
A(al.) = us; ... ns- =
Z al,~...lze.ne.= 2 al'l"ene" = Z a J ' s " e l R a ( a e ) ' eF_G
eEG
eEG
But from (i) we know that, for each e E G, IA(cre)l _< lna(ae). On the other hand, for each e E G, as,...i,,e E N. Thus, for each e E G with as1 ..She r 0, we must have A(c%) = lna(a~). In particular, A(ag) = 1RG(Crg). []
Let us denote by ~ the uniquely determined hermitian form on RX such that, for all y, z E X, ~(u, z) = ~y~, where (f is the Kronecker delta. It follows immediately from the definition of ~ that, for each v E RX,
o ___~(v, v) and
0=~(v,v)
*~ v = 0 .
Thus, ( is positive-definite. In other words, RX is a (finite dimensional) Hilbert space with respect to ~. For each a E EndR(RX), we shall denote by a* the adjoint of a relative to ~. For each r E R, we shall denote by ~ the complex conjugate of r. Let ~r E EndR(RX) be given. It is straightforward to verify that, for each r E Endn(RX), (a + r)* = a ' + r * and, for each r E R, ( r a ) ' = ~a*;
see [18; Hauptsatz II.5.2 f)]. We shall use these equations in the proof of the second part of the following lemma.
4.3 The Complex Adjacency Algebra
111
L e m m a 4.3.2 (i) For each g E G, (~rg)* = ~rg.. (ii) Let {rg [ g E G} C_ R be given. Then
gEG
gEG
(iii) Let {s 9 [ g E G} C_ R be given, and set r := ~ SgO'g. gEG
Let {tg [ 9 E G} C R be given, and set 7"
:: ~
tgO'g
gEG
and
:= ~
p
8g~gCrg,
gEG
Then, if 6t* = ~r and v* = r, p* = p.
Proof. (i) For all y, z E X, we have
~(Y~,'z)=~(~ ~'z)= ~ ~(~'~)= o ~:Eyg
xEyg
if (y, z) E g if (y, z) ~ g
and 1 if (z,y) E9* xEzg*
~Ezg ~
(ii) Using the above-mentioned equalities (i) yields =
gEG
:
gEG
:
gEG
gEG
gEG
(iii) Let g E G be given. If ~r* = ~r, Sg = sg'-7; see (ii). Similarly, if r* = r, tg = tg.. Thus, using (ii) once again, we obtain that
9ea
9ea
Thus, we have proved (iii).
9Ea
gea []
Lemma 4.3.2(i) says in particular that, for each g E G, <Xg is hermitian if and only if g* = 9.
112
4. Representation Theory
Let X E Irr(RG) be given. By Lemma 4.1.4(iii), we may assume that Vx ~ R X . In particular, ~ (restricted to V• • Vx) is a positive-definite (hermitian) form on V• Therefore, Vx possesses an orthonormal basis with respect to (restricted to Vx • Vx) ; see [18; Hauptsatz II.4.14 a)]. (Occasionally this fact is referred to E. SCHMIDT.) Let us now assume that S x is orthonormal with respect to ~. Let ~ E EndR(RX) be given. Let {rX,((r) I t, u E S x} C R be such that, for each t E S x,
E
uESx
Assume that cr is a positive-semidefinite hermitian map. Then ~r restricted to Vx is a positive-semidefinite hermitian map. Therefore, as S x is assumed to be orthonormal with respect to ~, ( r txu ) t u is a positive-semidefinite hermitian matrix; see [18; Bemerkung 11.7.2 a)]. We shall use this argument in the proof of the second part of the following theorem. We are now ready to establish the so-called "Krein condition".
T h e o r e m 4.3.3 Let r r E Irr(RG) be given, and assume that Sr and Sr ave ovthonovmal with respect to ~. Let s E Sr and let s' E Sr be given. (i) The map
E gec
rr
(r
Ig*12
%
is positive-semidefinite hermitian. (ii) Let X E Irr(RG) be given, and assume that S• respect to ~. Then
)r,,,, Ig'l=
is ortho9onal with
[rL
is a positive-semidefinite hermitian matrix.
Proof. (i) For all t, u E Sr we have
Therefore, r is hermitian. Moreover, e r has only 0 and 1 as eigenvalues. Therefore, er 8 is positive-semidefinite. Similarly, we obtain that cr is a positive-semidefinite hermitian map. Thus, by Theorem 3.6.8(ii), cr 8 o r162 is positive-semidefinite hermitian. Let us now prove that
4.4 Representations and Closed Subsets rr
r
-~.
113
0 ~8131.
Let y, z E X, and let g E G such that (y, z) E g be given. Then, by Lemma 4.2.3(ii), the coefficient of z in yer is
~L(~.) Similarly, the coefficient of z in yet, s, is r~r (~g 9)
"~r
Ig*l
r Therefore, the coefficient of z in y(e~, o G,,') is
m~mr
r r o"
]g, 12
(ii) follows from (i) considering the above-mentioned argument from Linear Algebra. []
C o r o l l a r y 4.3.4 given. Then we have
Let r
r E HomR(RG, R), and let X E Irr(RG) be
0 < g~E G ig.-~(~g.)r
)x(~)
Proos This follows immediately from Theorem 4.3.3(ii).
4.4
[]
Representations and Closed S u b s e t s
For each F C G, we shall denote by R[F] the smallest subalgebra of EndR(RX) which contains {~r! [ f E F U {1}}.
T h e o r e m 4.4.1 Let H ~. C be given. Then we have (i) R H = R[H]. (ii) Let F C_ H be such that R[F] = R[H]. Then (F) = H.
114
4. Representation Theory
Proof. (i) Let e, f E H be given. We shall be done if we succeed in showing that 0.e0.I E R H . Since H E C, we have that, for each g E G \ H , a~,g = 0. Thus, by Lemma 4.1.1(i),
0.~0.. = E ~
= E oe.h0.h~ R .
gEG
hEH
(ii) Let h E H be given. Since, by hypothesis, R[F] = R[H], there exist f l , . . . , f,~ E F such that 0 . I , " "0.S. ~ R ( H \ {h}); see (i). Thus, by Lemma 4.1.1(ii), al,...1,h # O. Therefore, by Lemma 1.2.4, h E fl "" 'fn C_ F n. Since h E H has been chosen arbitrarily, the claim follows from Theorem 1.4.1(ii). []
Let F C_ G be such that F # O. We define 0.F :---- E
0.-f"
]EF
Recall also that we set
r/F :=
E f E F r/].
L e m m a 4.4.2 Let H E C be given. Then we have (i) 0.~/= nn0.n. (ii) For each R G - m o d u l e V, V = ker 0.H @ {V E V ] V0.H
= nHV}.
Proof. (i) We have
0.~/= ( X 0.e)(E 0..)= X X X oe-0.~ = eEH
E(E
]EH
eEH ] 6 H gEH
E ao,,)0.,-- E""0., ="'0.';
gEH eEH f E H
gEH
see Lemma 4.1.1(0 and Theorem 1.3.6(ii). (ii) Let v E V be given. Set t := n H V -- V0.H and u := V0.H. Then n H v = t + u. On the other hand, with the help of (i), one obtains t0.H = ( n H v -- V0.H )0.H "- nHV0.H -- v0.2ii = v ( n H 0 . n -- 0"21f) = 0
and r
= V0.2H -~ nH?)0.H ~ n H U .
Now the claim follows from the fact that nn • O.
C o r o l l a r y 4.4.3 Let l E Inv(G) be given. Then (i) 0.~ = n,1 + (n, - 1)0.,.
[]
4.4 Representations and Closed Subsets
115
(ii) For each RG-module V, v = {v
v I
= - v } 9 {v
v I
= n,v}.
Proof. Since I E Inv(G), {1, l} E C. Thus, (i) follows from Lemma 4.4.2(i), and (ii) follows from Lemma 4.4.2(ii). []
T h e o r e m 4.4.4 Let L C Inv(G) be such that ILl = 2. Assume that R is algebraically closed and that R[L] = R[G]. Then, for each RG-module V, dimn(V) _< 2.
Proof. Let V be an irreducible RG-module, and let h, k E L be such that
h#k. Since R is assumed to be algebraically closed, there exists v E V \ 0 such that RV~hCrk = Rv. Let us define U := Rv + Rv~rh. Then U # 0. From Corollary 4.4.3(i) we obtain that (RVO'h)Crh = Rva2h C Rv + RvO'h = U,
whence U~h C U. Similarly, Corollary 4.4.30) yields Uc~k _C U. Recall that we are assuming that R[L] = R[G]. Therefore, V is a submodule of V. But U r 0 and V is assumed to be irreducible. Therefore, U = V. It follows that dimn(V) _< 2. []
T h e o r e m 4.4.5 Let x E X , and let H E C be given. Then we have R H "~ R ( H , H ) (as algebras over R).
Proof. This is an immediate consequence of Lemma 4.1.1(i) and Theorem 1.5.1(iii). []
P r o p o s i t i o n 4.4.6 Assume that char(R) = 0. Let H E C, and let g E G be given. (i) n ( g g ) is an nil-module. (ii) For each h E H, XR(gH)(~h) = ~ ] r alhl" (iii) 1RH has multiplicity 1 in XR(gH).
Proof. Let f E gH, and let h E H be given. Then, by Lemma 4.1.1(i),
116
4. Representation Theory 0"]0"h ~ ~
alheO'e"
e6G
(i) Let e E G be such that alh~ ~ O. Then, by definition, e E fh C_ fH. But, as f E gH, f H = gH; see Lemma 1.3.1. Therefore, e E gH. Since e E G has been chosen arbitrarily with alhe 7~ O, the above equality yields that ~rl~h E R(gH). (ii) follows from the above equality together with the fact that {or/ ] f E gH} is a basis of R(gH). (iii) By (ii),
y~XR(gH)(~h)= y ~ y ~ alh ! -: ~ hEH
hEH .fEgH
~
a!h ! : n i l .
]EgH hEH
But, for each X E I r r ( R H ) \ {lnH}, Theorem 4.4.5 and Lemma 4.1.6(ii) yield x ( ~ h ) = 0. hEH
Thus, the claim follows from
hEH
[]
and from the hypothesis that char(R) = 0.
For each H E C, we define
n . := 0
e nal
h = nh }.
hEH
Theorem
4.4.7
Assume that char(R) = O. Then, for each H E C,
IG/H] = dimn(RH). Proof. Let T C_ G be such that TH = G and ITI = IG/H]. Then, by Theorem 1.3.1, {tH ] t E T} is a partition of G. In particular, RG=OR(tH). tET
Let ~ E RH be given. Since ~ E RG and RG = ~ t E T R(tH), there exists, for each t E T, an element c~(t) E R(tH) such that ~ = ~-]~teTcat). Let h E H be given. Then, as ~r E RH,
4.4 Representations and Closed Subsets E
o'(t)o'h = o'0"h = nhtr = nh E
tET
O'(t) = E
tEH
117
nhtr(t)"
tET
But, for each t E T, nhtr (t) E R ( t H ) and, by Proposition 4.4.6(i), tx(t)Crh E R ( t H ) . Thus, as R G = ~)teT R ( t H ) , for each t E T, cr(t)crh = nhcr (t). Since h E H has been chosen arbitrarily, we obtain that, for each t E T, ~r(t) E RH. It follows that
R,, = G R(tH) n R,,. tET
Thus, the claim follows from Proposition 4.4.6(iii).
[]
Let {9 be a group acting on (X, G). We shall now see that the standard module R X of R G is a module also for the adjacency algebra of the semidirect product (X, G)@ of (X, G) with O. Occasionally the existence of this module leads to severe restrictions on the structure of (X, G); see Theorem 5.4.4 for instance. For each 0 E 0, we set Fixx(O) := {z E X I xo = x}.
T h e o r e m 4.4.8 Let O be a group acting on (X, G), set ~ := co, and set
(W, F) := (X, G)@. For each x E X , for each g E G, and, for each 0 E 0 , define xCr(g,O)~ := E
y"
yExOg
Then we have (i) For each x • X and, for each g E G, xCr(g,1)r = xag. (ii) For each x E X and, for each 0 E 0 , xcr(1,e)r = zO. (iii) For each {rr [ x E X } C_ R and, for each {r(g,0) [ (g, 0) E G • O} C_ R, we define
rEX
(g,O)EGxO
rEX (g,O)EGxe
Then R X is an RF-module. (iv) Let r denote the character of R F afforded by R X . Let g E G, and let 0 E O be given. Then
r
-- I{x E Z I (zO, z) ~ g}l.
(In particuZar, for each 0 ~ e, r162
= IFixx(e)l.)
118
4. Representation Theory
Proos (i) and (ii) follow immediately from the definition of xc~(9,0)r where xEX, gEG, and0EO. (iii) Let e, f E G, and let (, 7] E O be given. We first wish to prove that, for each x E X, First of all, with the help of Lemma 4.1.1(i) and Theorem 2.7.1(iv), we obtain
O'(e,(),;O'(.f,r/)r ----
~
a(e,()~(J,tl)r162
: ~ ae,fgO'(g,(,)r gEG
(g,O)EGxO
Therefore, for each x E X,
x(er(e,r162162
=xZ
ae,fgcr(g,r162= E aev/gXa(a,r162=
gEG
gEG
Zoo., Z y= E Z ao.,y gEG
yEx(~g
gEGyEx(~g
On the other hand, we have, for each x E X,
yEx(e
~_,
Z
y~Ex(~e~ zEyrj]
yEx(e zEyv]f
z= Z
Z
ae,sg z"
gEG zEx(~g
Thus, we have proved the desired equation. Now looking at the definition of an RF-module, we see that the above equation suffices in order to show that R X is an RF-module. (iv) follows immediately from the definition of x~(g,e)r where x E X, g E G, and O E O. []
4.5 The
Case
[ G I < 5.
In this section, we shall see how the representation-theoretical results of the previous sections can be applied successfully if ]G] is small. We start with a result due to D. G. Higman; see [14; (4.2)].
T h e o r e m 4.5.1 [D. G. HIGMAN] (i) Let R be an algebraically closed field with char(R) = 0. Then, if IIrr(RG)] = 2, Ial = 2. (ii) If lal < 5, then (X, G) is commutative.
4.5 The Case IGI _ 5.
119
Proof. (i) Let X E Irr(RG) \ {1RG} be such that { l n c , X} = Irr(RG). R For each g E G, we abbreviate % := crg. Let g E G \ {1} be given. Then, by Lemma 4.1.6(i), (iii),
ng. + mxX(Crg ) = m l , G I R e ( a a ) + mxX(ag ) = XRx(o'g) = O. It follows that
x(~g) = - %. mx"
Let Q denote the prime field of R, and let Z denote the subring of R generated by 1. Then X(o'g) E Q. On the other hand, by Lemma4.2.5, X(~rg) E I ( Z ) , whence, by Lemma 3.2.3, X(ag) E Z. It follows that
X(era) < - 1 . On the other hand, by Theorem 3.5.3(iii), IGI = dimR(RG) = 1 + X(1) 2, whence [ G [ - 1 = X(1) 2. Therefore, by Lemma 4.1.6(ii), 0= Z
gEV
X(erg ) = )C(1) +
~
X(ag) _< X(1) - (1(:;I- 1) = x(1) - x(1) 2.
gEG\{1}
It follows that X(1) = 1, and from this we obtain that IGI = 2. (ii) is an immediate consequence of (i) and Theorem 3.5.3(iii).
[]
The second part of the following theorem was proved as Theorem 6 in [7]. Its third part was first proved in [17].
T h e o r e m 4.5.2 Assume that there exist e, f E G \ {1} such that e* = e # f and { 1 , e , f } = G. (i) We have (a,ee - a,,z)(IXl - 1) + 2ne E Z . x/(ae,, - a~,l)2 + 4(n~ - a,el) (ii) [P. ERD6s-A. RISNYI-V. $6s] If a ~ = 1, then ar # 1. (iii) [A. J. HOFFMAN-R. R. SINGLETON]I f aeee ---- 0 and ae~l = 1, then
(n~, IXI) z {(2, 5), (3,10), (7, 50), (57,3250)}. Proof. (i) Let R be an algebraically closed field with char(R) = 0.
For each g E G, we abbreviate ~r9 := cry. Let X e I r r ( R G ) \ { l n c } be given. Then, by Theorem 3.5.3(iii), X is linear. Thus, by Lemma 4.1.1(i), X(~r~)2 = x((r~) = x(n~l + ae~Cre + a~I(rl) =
120
4. Representation Theory ne + aeeeX(O',) + aeeyX(Crl).
But, by Lemma 4.1.6(ii), X(~/) = - X ( ~ , ) - 1. (Recall that, by hypothesis, {1, e, f} = G.) Therefore, x(~s
= n~ + a . ~ x ( ~ )
+ a.s(-x(~.)
- i),
whence
a.~
x(~o) ~ - (~.~ -
- (n~ - a e 4 ) : 0.
Prom Theorem 3.5.3(iii) we know that there exist r r E I r r ( R G ) \ {1ha} such that r # r and {laG, r r = Irr(RG). Without loss of generality, we may assume that 1
r
= ~[a,o,- '~eeS +
r
=
and
Prom L e m m a
~ / ( a . e - a . j ) 2 + 4(he - a . s ) ]
1 ~[~.~aee:- ~(aeee-- aee:)'+4(ne- aees)]. 4.1.6(iii) we know that mIRG = i. Therefore,
1 + m ~ -I- m r =
mxX(1) = Xnx(1) = IXI-
E
xEIrr(RG) From Lemma 4.1.60) we know that lna(cr~) = h e . . Therefore, as e" = e, n~ + mr162
+ mr162
=
~ m• xEIrr(RG)
= XRX(O'~) = O.
Now we have r .~=
89176 - a . 1 + g ( a . , x/(a~
[IXl
1+
- 1) + ne
=
r162176
- a.1) ~ + 4(n. - a.S)](IXl-
1) + ne
- aeel) ~ + 4(n~ - a ~ l )
(a~
- a.s)(IXl-
1) + 2-~
v ~ ( a . ~ - aoos) ~ + 4(n~ - , , . S ) ]"
(ii) Assume, by way of contradiction, that aeee = 1 = a e e l . Then, by (i), divides ne. Therefore, n~ = 2. On the other hand, a ~ ! # 0 yields a l e e # 0; see Lemma 1.1.3(iii), and use the hypothesis that e* = e. Moreover, by Lemma 1.1.1(i), we also have that alee # O. Thus, by Lemma 1.1.4(i), 3 _< n~. = n~, contradiction. (iii) Let us assume that a ~ = 0 and that a ~ ! = 1. By Lemma 1.1.1(i), a l ~ = 1. Thus, as a ~ -- 0, Lemma 1.1.4(i) yields a l e e = h e . -- 1. Since e* = e, this means that a l e . e = ne -- 1.
4.5 The Case
IGI < 5.
121
Now, as a ~ ! = 1, Lemma 1.1.3(iii) yields n ! = (ne - 1)he. Thus, as {1,e,f}
=a,
IXl = , ~ + 1.
From (i) we obtain that ~ - 3 4nvf4-h--f:~ - 3 divides n~ (ne - 2). Assume that ne :~ 2, and set
divides I X [ -
1-
2n~. Therefore,
s:= 4 4 - ~ - 3 . Then
1
ne = 4 (s 2 + 3)
and s divides
~
(s 2 + 3)(s 2 - 5).
In particular, s divides 15, whence s E {3, 5, 15}. (Recall that, by hypothesis, aee! ---- 1, whence f E ee; therefore, by Lemma 1.2.9(i), ne # 1.) Now the 1 2 claim follows from ne = ~(s + 3) and IX[ = ne2 + 1. []
It is a well-known open problem whether IX[ = 3250 can actually occur in Theorem 4.5.2(iii).
5. T h e o r y of Generators
Throughout this chapter, L will be a set of involutions of G. After the first section, we shall always assume that ILl = 2.
5.1 Constrained
Schemes
Let F(L) denote the free monoid constructed on L. We shall denote by , the multiplication of F(L). The identity element of F(L) will be denoted by 1. By "~L we shall denote the uniquely determined monoid homomorphism from F(L) to (the additive monoid) N such that, for each l E L, l)~ L : 1. Then, by definition, 1)~L ----0. For the remainder of this section, we abbreviate )~ :----)~L. Let R(G) denote the monoid of all non-empty subsets of G with respect to the complex multiplication. Recall that { 1} is the identity element of R(G). (This follows easily from Lemma 1.1.1(i).) Let us denote by PL the uniquely determined monoid homomorphism from F(L) to R(G) such that, for each 1 E L, lpL = {l}. Then, by definition, lpL = {1}. For the remainder of this chapter, we abbreviate P :--=PL. For each j E N, we set j)~-I := {f E F(L) I f~ = J}L e m m a 5.1.1 For each j E N, we have LJ = Of~j,~_l fp. Proof. If j E {0, 1}, the claim is obvious. If 2 _< j, the claim follows from Lemma 1.2.4. [] Let h, k E L be such that h ~ k, and let j E l~ be given. If j -- 0, we define f j ( h , k ) := 1. Now assume that 1 <: j, and let l l , . . . , /j E {h, k} be such that, for each i E { 1 , . . . , j } , l~ = h if and only i f 2 ~ i . Then we define
124
5. Theory of Generators
fj(h,k) :=/1 * ...*lj. For all h, k 9 L with h # k, we set
ML({h,k}) := {j 9 N \ {0} I fj(h,k)pnfj(k,h)p # O} and
mL({h, k}) :=
minML({h,k}) R0
if ML({h,k}) # 0 if ML({h, k}) = 0.
It follows immediately from the definition of m L that of L. For the remainder of this chapter, we abbreviate m
:~
mL
is a Coxeter map
m L.
For all h, k 9 L with h # k and m({h, k}) 9 N, we abbreviate
fro(h, k) := fm({h,k))(h, k). We set •m := {(a*fm(h,k) *b,a*fm(k,h)*b) I
h,k E L, h # k, m({h,k}) E N, a , b E F(L)}. By Qm we shall denote the uniquely determined smallest equivalence relation on F(L) which contains Cm. For all d, e E F(L), we shall write d "~m e if (d, e) 9 Q,~. Let us denote by Fm (L) the set of t:he m-reduced elements of F(L). We shall say that the pair (X, G) is L-constrained if (L) = G and if, for each f 9 Fm (L), Ifpl = 1.
L e m m a 5.1.2 Assume that (X, G) is L-constrained, and let d, e E F(L) be such that d ~.,~ e. Then dp= ep.
Proos Clearly, we may assume that (d, e) E ,~m. This means that there exist h, k E L with h # k and m({h, k}) E N, and that there exist a, b E F(L) such that d=a*fm(h,k)*b and e = a , fm(k,h) 9b. The definition of m implies that
fm(h,k)pNfm(k,h)p # ~.
5.1 C o n s t r a i n e d Schemes
125
On the other hand, we have fro(h, k), f,~(k, h) E Fro(L). Therefore, as (X, G) is assumed to be L-constrained, we also have Ifm(h, k)p I = 1 = If,~(k, h)p I. It follows that fro(h, k)p = fro(k, h)p, and this yields dp = apfm(h, k)pbp = apf,~ (k, h)pbp = ep. []
Let g E (L) be given. Then we know from Theorem 1.4.1(i) that {jEN
l g E L j} #O.
(Recall that, by Lemma 1.4.5(i), L* = L.) We define #L(g) := min{j E l~ ]g E LJ}. For the remainder of this chapter, we abbreviate P : : PL.
For each N C L, we denote by f ( g ) the submonoid of F(L) generated by N. Note that F(N) is the free monoid constructed on N. Note further that, for each f E F(N), fPN = fP and fAN = fA. The first part of the following proposition requires Corollary 3.1.7, which in turn depends on Theorem 3.1.5(i).
P r o p o s i t i o n 5.1.3 Assume that (X, G) is L-constrained, and let N C_ L be given. Then we have (i) Let f q F(N) be such that fA E PN(fP). Then f E F,~(L). (ii) For each g E (N), there exists f E F(N)nF,~(L) such that #N(g) = fA and {g} = fp.
Proof. (i) Let f E F(N) \Fro(L) be given. First of all, as f ~ Fro(L), there exist l E L and a, b E Fm (L) such that f "~m a * l *l * b; see Corollary 3.1.7. Thus, by Lemma 5.1.2, f p = apllbp. On the other hand, by Lemma 1.4.5(ii), apllbp C_ apbp U aplbp, so that we have f p C_ apbp U aplbp.
126
5. Theory of Generators
Now recall that f 9 F(N). Since f "~m a * l 9 l * b, this implies that l 9 N and that a, b 9 F(N). Therefore, we obtain from Lemma 5.1.1 that apbp C_ N f)'-2 and aplbp C_ N f'x-1. It follows that fp C N f~- 2 O N f)'- 1. Thus, for each g 9 fp, ttN(g) _< fA -- 1. In particular, fA ~ PN(fP). (ii) Let g 9 (N) be given, and set j := ttN (g). Then, by definition, g 9 NJ. Thus, by Lemma 5.1.1, there exists f 9 F(N) such that j = fA and g 9 fp. It follows that fA 9 ttN(fp), so that, by (i), f 9 Fro(L). Now recall that, by hypothesis, (X,G) is L-constrained. Therefore, we obtain from f 9 Fro(L) also that {g} = fp. O
Our next result is a general structure theorem on L-constrained schemes. All parts of this theorem depend on Corollary 3.1.7.
T h e o r e m 5.1.4 Assume that (X, G) is L-constrained. (i) Let F C G be such that Oo(G) C_ F. Assume that F 2 C F and that, for each f E F, f * f C_ F. Then F C_ (L A F). (ii) Let H E C be such that Oe(G) C_ H. Then (L Cl H) = H. (iii) Assume that O,~(G) = {1}. Then, for each H E d, (L f3 H) = H. In particular, C = {(N) I N C_ L}. (iv)
(v) oo (a) 9 c. (vi) Oo (G) = G if and only if L C Oo (a). (vii) Oo(G) = {1} if and only if L N Oo(G) = 0.
Proof. (i) Let us define N := L N F. We want to show that F C_ (N). Suppose, by way of contradiction, that F q: (N). Then F \ (N) r 0. Let f E F \ (N) be such that p(f) is minimal. First of all, by Proposition 5.1.3(ii), there exists f E F,~(L) such that p(f) = fA and {f} = fp. Since f ~t (N), f # 1. Thus, as f E fp, 1 # f. Therefore, there exist l E L and e E Fro(L) such that f=e*l. Since e E Fro(L), there exists g E G such that {g} = ep. (Recall that (X, G) is assumed to be L-constrained.) Thus, {f} = f p = ( e * l ) p = e p l = g l . Our claim is now that l E F. If nt = 1, this follows from the hypothesis that Oo(G) C_ F. Assume that nl # 1. Then we obtain from Lemma 1.4.5(i),
5.1 Constrained Schemes
127
(v), Lemma 1.2.5(iii), and Lemma 1.2.2 that l EII C_ l*g*gt = f * f . But, by hypothesis, f * f C_ F. Thus, we have proved that IEF. Since f E gl, g E fl; see Lemma 1.2.5(i) and Lemma 1.4.5(i). Thus, as we are assuming that F 2 C_ F, we obtain that g E F. On the other hand, by Lemma 5.1.1, g E ep C_ L e~, so that we obtain p(g) _< eA = f A - 1 = p ( f ) - 1. Therefore, the choice of f forces g ~t F \ (N). Thus, as g E F, g E (N). It follows that f E gl C_ (N). (Recall that l E L M F = N.) This contradiction proves (i). (ii) From (i) we obtain that H C_ (L M H). On the other hand, as H E C, we have (L M H) _C H, too. Therefore, (L M H) = H. (iii) follows immediately from (ii). (iv) We set N := L M Oo(G). Note first that, by Corollary 1.2.8(i), Oo(G) ~ C_ O,~(G). Moreover, for each g E Oo(G), g*g = {1} C_ O0(G); see Lemma 1.2.6. Thus, by (i), O0 (G) C_ (N). Conversely, as N C_ O0 (G), Corollary 1.2.8(i) yields that, for each n E 1~, N'* C_ Oo(G). But, by Theorem 1.4.1(i),
(N) = U Nn. hEN
(Recall that, by Lemma 1.4.5(i), N* = N.) Therefore, we have also ( S ) C_
oo(a) (v), (vi), and (vii) are immediate consequences of (iv).
[]
The pair (X, G) will be called a Coxeter scheme with respect to L if (X, G) is L-constrained and if, for all d, e E Fro(L), d p = ep implies that d "~m e. If (X, G) is a Coxeter scheme with respect to L, ILl is called the rank of (X, a ) . In the following theorem, we give two characterizing conditions for (X, G) to be a Coxeter scheme with respect to L. The proof of the implication (c) =~ (a) requires Corollary 3.1.6. Proposition 5.1.3(ii) is needed for the proof of the implication (a) ::~ (b). Note that condition (b) of the following theorem is the converse of Proposition 5.1.3(i) in the case where N = L.
T h e o r e m 5.1.5 Assume that (X, G) is L-constrained. Then the following conditions are equivalent. (a) (X, G) is a Coxeter scheme with respect to L. (b) For each f E Fro(L), fA E p(fp).
128
5. Theory of Generators (c) For each f E Fm (L), l p = fp implies that 1 = f.
Proof (a) =:~ (b) Let f E F,~(L), and let g E f p be given. Then, as ( X , G ) is assumed to be L-constrained, {g} = fp. By Proposition 5.1.3(ii), there exists e E Fro(L) such that/~(g) = eA and {g} = ep. From {g} = ep we obtain that f p = ep. Thus, by (a), f "~m e. In particular, fA = eA. Thus, as #(g) = eA and g E fp, fA E # ( f p ) . (b) ~ (c) Let f E F,~(L) be such that l p = fp. Then, by (b), fA E /~(fp) = p ( l p ) = {0}. Therefore, 1 = f. (c) =V (a) We define 7) := { ( d , e ) E Fro(L) • Fro(L) I d p = ep, d 7~rn e}, and we assume, by way of contradiction, that 7) :~ 0. We choose (d, e) E 7) such that dA is minimal. Suppose that d = 1. Then l p = ep and 1 7L,n e, contrary to (c). Thus, d :~ 1. Therefore, there exist l E L and b E Fro(L) such that d=l*b. Set f:=l*e. F r o m d = l * b we obtain that d p = lbp. Thus, by L e m m a 1.4.5(i) and L e m m a 1.2.5(iii),
bp C llbp = Idp = lep = ( l . e)p = fp. Suppose constrained, choice of (d, Suppose F,~(L) such
first that f E F,~(L). Then, as (X, G) ]fPl = 1. It follows that b p = fp. Thus, e) forces b ,-~0~ f, c o n t r a r y to b)~ = dX now that f ~ Fro(L). Then, by Corollary that l * c--~m e.
is assumed to be Las b)~ = d ) ~ - 1, the 1 < e~ - 1 = f,~ - 2. 3.1.6, there exists c E
Thus, by L e m m a 5.1.2,
Icp = (l* c)p = ep = d p = lbp. It follows that,
bp C_ llbp = llcp C_ cp U ep; see L e m m a 1.4.5(i), (ii) and L e m m a 1.2.5(iii). Thus, b p E {cp, ep}. Now, as b)~ = d ) ~ - 1, the choice of ( d , e ) yields b ~r~ c or b ~,~ e. But, as b 1 = d l - 1 < e $ - 1, we cannot have b "~m e. On the other hand, b ~,~ c leads to the contradiction d = l * b "~m l * c ,~,,~ e. I-1
5.1 Constrained Schemes
129
Recall that lp = {1}. Thus, condition (c) of Theorem 5.1.5 just says that, for each f E Fr~(L) \ {1}, 1 ~ fp.
Lemmah.l.6 Let x E X , and let N C L be given. Set L' := N~(N), and let us denote by 7r the uniquely determined monoid homomorphisrn from F ( N ) to F(L') such that, for each l E N, l~r = Ix(N).1 For each f E F ( N ) , we abbreviate fl := fTr. (i) Set p' := PL'. Then, for each f E F ( N ) , f~p' = (fP)x(N). (ii) Set m' := m L , . Then, for all h, k E N with h # k, m'({h', k'}) =
m({h,k}).
(iii) Set m' := mL,, and let d, e E F ( N ) be given. Then d' "m, e' if and only if d ~,~ e. (iv) Set m' := mL,. Then, for each f E F(N), f' E F,~, (L') if and only if f E F,~(L).
Proof. (i) Let f E F ( N ) be a counterexample in which fA is minimal. Then f r 1. Therefore, there exist l E N and e E F ( N ) such that f=l*e. From this we obtain immediately that f'p' = ((l * e)')p' : ( l ' , e')p' = l'e'p' and that fp = (1 * e)p = lep. On the other hand, as eA = fA - 1, the choice of f E F ( N ) yields e'p' : (ep)x(N). Finally, by Lemma 1.5.2(i), l~(N)(ep)x(N) = (lep)~(g). Thus, we obtain f, pl = l'e' p' = l~(N) (ep)~(~) : (lep)~(N) = (fP)~(N), contrary to the choice of f E F ( N ) . (ii) Let h, k E N be such that h # k, and let j E 1~ be given. Then fj(h', k ' ) = (fj(h, k))'. Thus, by (i), fj(h', k')p' = (f/(h, k))'p' = (fj (h, k)p)+(N). Similarly, we deduce fj (k', h')p ~ = (fj (k, h)p)~(N). Thus, it follows from the definitions of rn' and m that m'({h', k'}) -- m({h, k}). (iii) is an elementary consequence of (ii). (iv) follows from (iii) and Corollary 3.1.7. [] 1 Clearly, 7r is bijective.
130
5. Theory of Generators
We emphasize that the proof of Lemma 5.1.6(iv) does not really depend on Corollary 3.1.7. It is also possible to deduce the statement directly from the definitions of Fro, (L r) and Fro(L), but this needs notation which is of no use for the remainder of the present text.
T h e o r e m 5.1.7 Let x E X, and let N C L be given. (i) If (X, G) is L-constrained, (X, G),(N) is N~(g)-constrained. (ii) If (X, G) is a Coxeter scheme with respect to L, (X, G),(N) is a Cox-
eter scheme with respect to Nx(N). Proof. Set L r := Nx(N), and let us denote by 7r the uniquely determined monoid homomorphism from F ( N ) to F(L r) such that, for each l E N, 17r = Ix(N). F o r each f E F ( N ) , we abbreviate fl := f~r. Let us also set pr := PL, a n d rr/r :---- r n L i ,
(i) First of all, we know from Lemma 1.5.2(ii) that (n') = (N),(N). Let f E F ( N ) be such that f ' E F,~,(L'). Then, by Lemma 5.1.6(iv), f E F,~(L). Therefore, as (X, G ) i s assumed to be L-constrained, ]fp] = 1. Thus, by Lemma 5.1.6(i), ]f'p'] = 1. (ii) Let d, e E F ( N ) be such that d', e r E Fml(L') and d'p' = e'p'. We shall be done if we succeed in showing that d r "~m, e t. From d', e' E F,~,(L') we obtain that d, e E Fro(L); see Lemma 5.1.6(iv). From drp r = e'p' we obtain that dp = ep; see Lemma 5.1.6(i). Thus, as (X, G) is assumed to be a Coxeter scheme, we conclude that d "~,n e. Now we obtain from Lemma 5.1.6(iii) that d r "~m, e'. []
Assume that (X, G) is L-constrained. Then, by Theorem 5.1.4(vii),
Oo(G)={1}
r
LMO0(G)=O
r
lq~{nt [ I E L } .
Thus, the hypothesis of the second part of the following theorem can be expressed in various different ways.
T h e o r e m 5.1.8 Assume that (X, G) is L-constrained. (i) (X, G) is a Coxeter scheme with respect to L if and only if, for each x E X, (X, G)xoo(a) is a thin Coxeter scheme with respect to ( L N O 0 ( G ) ) x o o ( c ) . (ii) If Oo(G) = {1}, ( X , G ) is a Coxeter scheme with respect to L.
Proof. (i) Let us assume first that (X, G) is a Coxeter scheme with respect to L. Then we obtain from Theorem 5.1.4(iv) and from Theorem 5.1.7(ii) that, for each x E X , (X,G),oo(G) is a Coxeter scheme with respect to (L n O0 (a))~o~(a).
5.1 Constrained Schemes
131
Suppose now that (X, G) is not a Coxeter scheme with respect to L. Then, by Theorem 5.1.5, there exists f E F,~(L) \ {1} such that lp = fp. From l p = fp and lp = {1} we obtain that {1} = fp. Set j := f)~. Then, as 1 :~ f, j :~ 0. Thus, there exist 11, . . . , / j E L such that f =ll *...*lj. It follows that fp = {11 } . . . {lj }. Therefore, {1}={ll}...{b}. Thus, we conclude from Corollary 1.2.8(ii) that { l l , . . . ,lj} C Oo(G). Let x E X be given. Then the last equation yields l~o,(a) E {la}~o0(a)''' {/j}~o~(a) = {(ll)~o,(a)} " " {(/j)~oo(c)}; use Lemma 1.5.2(i) and the fact that {ll . . . . ,lj} C_ Oo(G). Now set L ~ := (L N Oo(G))~o~(a) and m' := mL,. Then, as f E Fro(L), we obtain from Theorem 5.1.4(iv) and from Lemma 5.1.6(iv) that (ll)~o0(a) * ' " * (//)zoo(c) E F,~,(L'). Therefore, by Theorem 5.1.5, (X,G)xoo(a) is not a Coxeter scheme with respect to (L M Oo(G))~oo(a). (ii) follows from (i). []
In view of Theorem 5.1.8(i) it might be interesting to know which groups are, via the (~,T)-correspondence, associated to the thin radical of an Lconstrained scheme which is not thin. 2 In Theorem 5.1.8(ii), the condition 1 ~ {hi I l E L} cannot be deleted. Assume that IX] = 4 = IGI and that L = G \ {1}. Then m assumes only the value 2. It is obvious that, for each f E F(L), ]fp] = 1. On the other hand, for all h, k, l E L with {h,k,l} = L, we have (h* k)p = hk = {l} = lp and h*kT~ml. T h e o r e m 5.1.9 Assume that (X, G) is a Coxeter scheme with respect to L, and let N C L be given. Then the following conditions are equivalent. (a) For each h E g and, for each k E L \ N, hk = kh. (b) For each e (g) and, for each f E (n \ g ) , eI = Ie. (c) ( g ) _~ G.
Proos (a) ::~ (b) Let e E (N), and let f E (L \ N) be given. Then, by Proposition 5.1.3(ii), there exist d E F(N) with {e} = d p and e E F(L \ N) with {f} = Pp. Now (a) yields 2 Note that (X, G) is L-constrained if (L) = G and if (X, G) is thin.
132
5. Theory of Generators e f = dpep = epdp = f e.
(b) =r (c) Let g E G be given. Then, by Proposition 5.1.3(ii), there exists f E F(L) such that (g} -- fp. Thus, by (b), (N)g = (N)fp = fp(N) = g(N).
(c) =~ (a) Let h E N, and let k E L \ N be given. First of all, as h r k, h * k E Fro(L). Thus, as (X, G) is assumed to be L-constrained, there exists g E G such that { g } = hk.
Since h E N, we now obtain from (c) that g E g k C_ ( g ) k = k ( N ) . Thus, by definition, there exists f E (N) such that g E k f . Since f E (N), Proposition 5.1.3(ii) yields f E F(N) M Fro(L) such that {f} = fp. It follows that g E k f p. Assume that k * f ~ Fro(L). Then, by Corollary 3.1.6, there exists c E Fro(L) such that k * e " " m f . But, as k E L \ N and f E F(N), this is impossible. Therefore, we have k * f E Fro(L). Thus, as (X,G) is assumed to be L-constrained, {g} = kfp = (k, f)p. On the other hand, we have {g} = hk = (h * k)p. Thus, (h * k)p = (k * f)p. Now recall that (X, G) is assumed to be a Coxeter scheme with respect to L. Therefore, we may conclude that h * k ,~,~ k * f. It follows that h = f, whence hk = { g } = k f p = kh.
[]
T h e o r e m 5.1.10 Assume that (X, G) is a Coxeter scheme with respect to L. Then we have (i) For all M , N C L, ( M M N ) = (M) M (N). (ii) For each N C_ L, N = i N (N). (iii) Let M , N C_ L be such that ( M ) = (N). Then i = N . (iv) Assume that Oo(G) = {1}. Then, as a lattice, C is isomorphic to the lattice of all subsets of L.
Proof. (i) It is obvious that (M I"1N) C_ (M) M (N). So let us prove that (M) 71 (N) C (M 71N). Let g E ( M ) N ( N ) be given. Since 9 E (M), there exists d E F ( M ) A F m ( L ) such that {g} = dp; see Proposition &l.3(ii). Similarly, since g E (N), there exists e E F ( N ) N Fro(L) such that {g} = ep.
5.1 Constrained Schemes
133
It follows that d p = ep. Thus, as (X, G) is assumed to be a Coxeter scheme with respect to L, we obtain that d --~,~ e. In particular, d, e E F ( M N N). Therefore, g E (M M N). (ii) Let l E L M (N) be given. Then, by Proposition 5.1.3(ii), there exists f E F ( N ) A F , , ( L ) such that {l} = fp. Since, by definition, lp = {l}, we obtain that lp = fp. Now, as (X, G) is assumed to be a Coxeter scheme with respect to L, we conclude that l "~m f. Since f E F(N), this implies that I E N. Since I E L N ( N ) has been chosen arbitrarily, we have shown that LM(N) C_ N. On the other hand, it is obvious that N C L M (N). (iii) follows from (ii). (iv) follows from (iii) and Theorem 5.1.4(iii). [] Assume that (X, G) is L-constrained, and let R be a field. The essential consequence of the second part of the following theorem is that then RILl = R[G]. This result will be used in Sections 5.3 and 5.4. T h e o r e m 5.1.11 Assume that (X,G) is L-constrained. Let g E G be given, and set j := It(g). (i) If2 <_ j, there exist 11, ..., lj E L such that, for each e E G, aq...tje = (~ge .3 R = (ii) If 1 < j, there exist 11, ..., lj E L such that, for each field R, c~g 0 "R 90"~. ll ""
Proof. (i) By Proposition 5.1.3(ii), there exists f 6 F,~ (L) such that j = fA and {g} = fp. Since 2 _< j, g # 1. Thus, as g 6 fp, f # 1. Therefore, there exist l E L and e E Frn(L) such that f=e*l. Since e 6 F,~(L), there exists f 6 G such that {f} = ep. (Recall that (X, G) is assumed to be L-constrained.) Thus, {g} = f p = ( e * l ) p = e p l = f l . Now Corollary 1.4.6 yields that, for each e E G, a fie
:
~ge.
(Note that It(f) = #(g) - l, so that f r g.) Since g C fl, j - 1 <_ It(f). On the other hand, as eA = j - 1 and f E ep, f E LJ-1; see Lemma 5.1.1. Thus, by definition, It(f) < j - 1. It follows that It(f) -- j - 1. a Here 5 is the Kronecker delta.
134
5. Theory of Generators
Assume first that j = 2. Then /t(f) = 1, which means that f E L. But, for each e E G, we have alte = gge. Therefore, we are done in this case. Assume now that 3 < j. Then 2 < # ( f ) . Thus, by induction, there exist 11, ..., lj-1 E L such that, for each d E G, all...lj_ld = (~fd.
Set lj := l, and let e E G be given. Then alx...lje = Z alx...lj_adadlje = a]le = ~9e" dfi G
(ii) There is nothing to prove if j = 1. Thus, the claim follows from (i) and L e m m a 4.1.1(ii). []
5.2 Pairs of Involutions In this section, we assume that [L] = 2, and we set m := re(L). Let h, k E L be such that h # k.
L e m m a 5.2.1 Let i, j E 1~ \ {0}, and let g E fi(h, k)p*fj(h, k)p be given. (i) Assume that j < i and 2 ~ i. Then there exists s E { i - j , . . . , i + j - 1} such that g E f~(h, k)p. (ii) Assume that j <_ i and 2 I i. Then there exists s E { i - j, . . . , i + j - 1} such that g E fs(k, h)p. (iii) Assume that i < j and 2 ~ i. Then either g E fj_,(k, h)p or there exists s E { j - i + 1 , . . . , i + j - 1} such that g E fs(h,k)p. (iv) Assume that i < j and 2 I i. Then either g E fj-i(h, k)p or there exists s E { j - i + 1,.. .,i + j - 1} such that g E f,(k,h)p.
Proof. By L e m m a 1.4.5(i), (ii), fi(h, k)p*fj(h, k)p = fi_, (k, h)p*hhfj_i(k, h)p C_ fi- 1(k, h)p*fj_ l(k, h)p U fl- 1(k, h)p* hfj _ ,(k, h)p. If2~i, f i - i ( k , h)p * hfj_y(k, h)p = fi+j-l (h, k)p; i f 2 ] i, f i - l ( k , h)p'hfj_l(k, h)p = fi+j_i(k, h)p.
5.2 Pairs of Involutions
135
Thus, we m a y assume that g C ri_,(k, h)p*f~_~(k, h)p. (i), (ii) Assume first that j = 1. Then f j - l ( k , h)p = {1}. Therefore, if2 { i,
g c fi_l(~, h)p" = f~_l(h, k)p = f~_j(h, k)p. Similarly, i f 2 I i, g E f~_j(k, h)p, so that we are done in the case where j = 1. Assume now that 2 _< j. If 2 { i, 2 I i - 1. Thus, by induction, there exists s E { i - j , . . . , i + j - 3} such that g E f , ( h , k ) p . If 2 I i, 2 { i - 1. Thus, by induction, there exists s E {i - j , . . . , i + j - 3} such that g E f~(k, h)p. (iii), (iv) Assume first that i = 1. Then f i - l ( k , h)p = {1}. It follows that
g E fj-l(k,h)p = fj_i(k,h)p, so that we are done in this case. Assume now that 2 < i. If 2 Jf i, 2 I i - 1. Thus, by induction, either g E fj-i(k, h)p or there exists s E { j - i + l , . . . , i + j - 3 } such that g E f,(h, k)p. If 2 I i, 2 { i - 1. Thus, by induction, either g E fj_~(h, k)p or there exists [] s E {j- i+ 1,...,i+j3} such that g E fs(k,h)p.
Lemma 5.2.2 (i) For each j C l~ \ {0}, fj(h,k)pL C_ fj-l(h,k)p U rj(h, k)p u f~+,(h, k)p. (ii) For each n E N, L n C_ U~.=o[fj(h,k)pUfj(k,h)p]. Proof. (i) follows immediately from L e m m a 1.4.5(ii). (ii) The claim holds obviously if n = 0. Assume that 1 _< n, and let g E L n be given. T h e n there exists f E L '~-' such that g E fL. By induction hypothesis, there exist j E { 0 , . . . , n - 1} such that f E fj(h, k)p U fj(k, h)p. I f j = O, f = 1. T h e n g E L = h(h,k)pUf~(k,h)p. If 1 < j, (i) yields
g e G_l(h,k)pUG(h,k)pu~+~(h,k)pU
~_l(k,h)pu~(k,h)pU~+,(k,h)p. []
L e m m a 5.2.3 Let i, j E N be given. (i) Iffi(h, k)p rl fj(k, h)p # 0, 1 E fi+j(h, k)p U fi+j (k, h)p. (ii) Assume that fi(h, k)p Clfj(h, k)p r 0. Then, if 1 < i < j, there exists s E {j-i,...,i+j1} such that 1 E fs(h,k)pUfs(k,h)p.
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5. Theory of Generators
Proof. (i) By assumption, fi(h, k)p O fj(k, h)p 7~ 0. Let 9 E fi(h, k)p A fj(k, h)p be given. Then, by Lemma 1.2.5(iii), 1 6 g'9 C fi(h,k)p*fj(k,h)p C_fi+j(h,k)pOfi+j(k,h)p. (ii) By assumption, fi(h,k)pAfj(h,k)p r 0. Let 9 E fi(h,k)pAfj(h,k)p be given. Then, by Lemma 1.2.5(iii),
1 E g*9 C fi(h, k)p*fj(h, k)p. Thus, the claim follows from Lemma 5.2.1(iii), (iv).
L e m m a 5.2.4 (i) We have m E N if and only if {j E l ~ \ { 0 } fj(h, k)p) # o. (ii) /fro E N, 2m -= min{j E 1~\ {0} I 1 E fj(h,k)p}.
[]
I1 E
Proof. Assume that m E N. Then, by definition, fr~(h,k)pAf,,,(k,h)p # 0. Thus, by Lemma 5.2.3(i), 1 E f2m(h, k)p. Let us now assume that {j E 1~ \ {0} ] 1 E fj(h, k)p} # 0. We set j := min{j E l~\ {0} ] 1 E fj(h,k)p}. Assume first that 2 ~ j. Then, as 1 ~ f1(h, k)p, 3 <_j. Moreover, fj(h, k)p = fj-l(h,k)ph. Thus, as 1 E fj(h,k)p, there exists f E fj_1(h,k)p such that 1 E fh. Now, by Lemma 1.2.5(iii) and Lemma 1.4.5(i), h = f, whence h E fj-l(h, k)p. Therefore, there exists e E fj_~(k, h)p such that h E he. Thus, by Lemma 1.2.5(ii) and Lemma 1.4.5(i), (ii), e* E {1, h}. If e* = 1, 1 E fj-2(k,h)p. If e* = h, h E fj_~(k,h)p, whence 1 E hh C fj-l(k, h)p. In both of these cases, we obtain the contradiction that j < 2. Thus, we must have that 2 ] j. Therefore, there exists i E 1~ such that 2i = j. It follows that 1 E f/(h, k)pfi(k, h)p'. Thus, there exists e E f/(h, k)p and f E fi(k, h)p such that 1 E ef*. Now Lemma 1.2.5(iii) yields e = f. It follows that
fi(h,k)pAfi(k,h)p # ~. Thus, by definition, we must have that m < i. From this we obtain that 2m < j, so that we have proved the lemma in the case where 2 I J[] Lemma
5.2.5
If [(L)[ E 1~, m E N.
Proof. If I(L)I E N, there exist i, j E l~ \ {0} such that i # j and fi(h, k)pn fj(h, k)p ~ ~. Thus, by Lemma 5.2.3(ii), there exists s E {1, . . . , i + j - 1} such that 1 E f,(h, k)p U f,(k, h)p. Now the claim follows from Lemma 5.2.4(i). Yl
5.2 Pairs of Involutions
137
T h e o r e m 5.2.6 Assume that m E N. (i) Let i, j E { 0 , . . . , m } be such that fi(h,k)p N f j ( k , h ) p # 0. Then we have i = j e {0, m}. (ii) Let i, j E { 0 , . . . , m } be such that f i ( h , k ) p M f j ( h , k ) p # 0. Then we have i = j. (iii) Let j E { 1 , . . . , m ) be such that ]fj(h,k)p I = 1. Then we have If -l(h, k)pl = 1. (iv) 2m < I(n)l. (v) Assume that (L) = G. Then 2m = Ial if and only if (X, G) is Lconstrained if and only if ( X, G) is a Coxeter scheme with respect to L. (vi) Assume that 2m = I(L)I, that 2 { m, and that IXl e 1~. Then we have n h
~
n k .
Proof. (i) If i = 0, fi(h,k)p = {1}. Therefore, 1 E fj(k,h)p. Thus, as j E { 0 , . . . , m}, L e m m a 5.2.4(ii) yields j = 0. Similarly, j = 0 implies that i = 0. Thus, we may assume that 0 ~ {i, j}. By L e m m a 5.2.3(i), 1 E fi+j (h, k)pUfi+j(k, h)p. Thus, as i + j # 0, L e m m a 5.2.4(ii) yields 2m < i + j. From the hypothesis that i, j E { 0 , . . . , m} we therefore deduce that i = m = j. (ii) For the same reason as in (i) we m a y assume that 0 ~ {i, j}. Without loss of generality, we assume that i < j. Then, by L e m m a 5.2.3(ii), there exist s E {j - i , . . . , i + j - 1} such that
1 E f,(h,k)pUf,(k,h)p. But, as i, j e { 0 , . . . , m ) , i + j - 1 < 2 m - 1. Thus, L e m m a 5.2.4(ii) yields s = 0, whence i = j. (iii) Since we are assuming that [fj (h, k)p[ = 1, we find g E fj (h, k)p such that {g} = fj(h, k)p. Let e, f E f j - l ( h , k)p be given. Assume that 2 { j . Then eh, f h C f j _ l ( h , k ) p h = f j ( h , k ) p = {g}. Thus, eh = {g} = fh. It follows that f E f h h = ehh = {e} U eh = {e,g}; see L e m m a 1.4.5(i), (ii). If f = g, f j - l ( h , k)p M fj(h, k)p # O, contrary to (ii). Therefore, e = f . Similarly, the assumption that 2 [ j leads to e = f. (iv) follows from (i) and (ii). (v) Let us first prove that, (X, G) is L-constrained if 2m = ]G]. We have ]Fm(i)[ = 2m + 1 and fro(h, k)p M fro(k, h)p # ~. Therefore, if 2m = [G[, we obtain from (i) and (ii) that, for each f E F,~(L), Ifp[ = 1. This means that (X, G) is L-constrained. Assume now that ( X , G ) is L-constrained. Then, for each g E G, there exists f E Fro(L) such that {g} = fp; see Proposition 5.1.3(ii). On the other
138
5. T h e o r y of G e n e r a t o r s
hand, we have IFr,,(L)I = 2m + 1 and frn(h, k)p N fro(k, h)p # 0. Therefore, we conclude that IGI < 2m. Thus, by (iv), we must have 2m = IGI. From Lemma 5.2.4(ii) we obtain that, for each f E Fro(L) \ {1}, 1 ~ fp. Thus, if (X, G) is assumed to be L-constrained, Theorem 5.1.5 yields that (X, G) is even a Coxeter scheme with respect to L. (vi) Since, by assumption, 2 { m, there exists j E 1~ \ {0} such that m=2j+l. Since, by assumption, 2m = I
fm(h,~)p = {g) =f~(k,h)p; see (i) and (ii). Thus, by Lemma 1.2.4, Lemma 1.1.5(iii), and Theorem 5.1.11(i), Now the claim follows from the hypothesis that nh, nk E l~.
5.3 Finite
Coxeter
Schemes
of Rank
[]
2
In this section, we assume that [LI = 2, that (X, G) is a Coxeter scheme with respect to L, and that IX{ E N. We set m := re(L). Since IX[ E l~, ]G I E N. Thus, by Lemma 5.2.5, m E N, so that, by Theorem 5.2.6(v), 2m = IGI . The main result of this section is Theorem 5.3.5 9 This theorem says that, if (X, G) is not thin, m E {2, 3, 4, 6, 8, 12}. The main idea of the proof of Theorem 5.3.5 is to apply Theorem 4.1.5(ii).
L e m m a 5.3.1 Let R be an algebraically closed field with char(R) = 0. For each i E {1,2}, set mi := l x E Irr(nG) X ( 1 ) = i } . Then (i) A1 U A2 = Irr(RG). (ii) / f 2 ~ m , [All = 2 and IA~[ = m-1 2 " (iii) If2 I m, Iall = 4 and [A2I = ~ - 1
Proof. (i) Since (X, G) is assumed to be a Coxeter scheme with respect to L, (X, G) is L-constrained. Thus, by Theorem 5.1.11(ii), R[L] = R[G], so that we are done by Theorem 4.4.4. (ii), (iii) Since 2m = IG], (i) and Theorem 3.5.3(iii) yield 2m = IA~I + 41A21.
5.3 Finite Coxeter Schemes of Rank 2
139
By Theorem 5.1.11(ii), each element of Ax is completely determined by its values on {cq I l E L}. Thus, by Corollary 4.4.3(ii), IAll < 4. On the other hand, as l n a E A1, 1 < IAI]. Therefore, the above equation yields IAll = 2 if 2 ~ m and [mx[ = 4 if2 [m. The value for [A2[ is now obtained from the above equation. []
For the remainder of this section, we set n := I ' I v/~. IEL
By assumption, IL] = 2. Therefore, {fj(h,k) I h , k E L, h # k, j E { 0 , . . . , m } } = Fro(L) and, for all h, k E L with h # k, f,~(h, k) ~,~ f,~(k, h). By assumption, (X, G) is a Coxeter scheme with respect to L. In particular, (X, G) is L-constrained. Thus, for all h, k E L with h # k and, for each j E { 0 , . . . , m}, there exists gj(h) E G such that {gj(h)} = fj(h, k)p. Now, as 2m = IG], Theorem 5.2.6(i), (ii) yields {gj(l) I l E L, j E { 0 , . . . , m } } = G
and, for all h, k E L, gin(h) = gin(k). In the following proposition, we completely compute the non-linear irreducible characters of RG, provided that R is an algebraically closed field with char(R) = 0.
P r o p o s i t i o n 5.3.2 Let R be an algebraically closed field with char(R) = O, and let X E Irr(RG) be such that X(1) = 2. Then there exists r E R with r 2 # 1 = r m such that, for each i E I~ with i < y "~ the following holds. (i) For all h, k E L with h # k, ni
X(~Rg~,+,(h)) : r - r - 1 [(nh -- 1)(r/+l -- r -(/+U) + (nk -- 1 ) ~ ( r ' -- r-')]. (ii) For each l E L, ~.R
' " = n'(r' + r-').
Proos For each g E G, we abbreviate ~r9 := a 9R .
140
5. Theory of Generators
Let V be an irreducible R G - m o d u l e such that X v = X, and let h, k E L be such that h -fi k. Let us denote by s and t the eigenvalues of (iv ( ~ h ~ k ) . First of all, d e t ( ~ V ( q h q k ) ) = St. On the other hand, for each l E L, - 1 and nt are the eigenvalues of (fv(crt); see Corollary 4.4.3(ii) and T h e o r e m 5.1.11(ii). Thus, det(Jv(cth)) = --nh and det((~y(~rk)) = - n k . It follows that s~ = n 2.
~ Set r : = g . 8 Then, as st = n 2 , r 2 -- T' Suppose that r 2 = 1. Then s = t. Thus, Chr induces a scalar multiplication on V, c o n t r a r y to the irreducibility of V. Therefore, we conclude that r~r Set c : = m if 2 ~ m and e := T if 2 I m . Define a := (~rhak) c. From gin(h) = gin(k) we deduce easily that c~ah = CrhC~ and cr(rk = (rkcr. Thus, by T h e o r e m 5.1.11(ii), (iv(a) E n n d n a ( V ) . Therefore, by L e m m a 3.4.3, c~ induces a scalar multiplication on V. It follows that s c = U. Therefore, as r 2 = ~, we obtain rm=l.
From st = n ~ we obtain that rt = n. Therefore, we have s § t -- n ( r § r - 1 ) .
It follows that
,~v (,~h,~k) = = (~ + t),~v (~'h"k) -- , t l = n ( r + , ' - ~ ) 6 v (,'hC'k) -- n21. Thus, we obtain that, for each j E { 4 , . . . , m},
5V(O'gj(h) ) ---- n(r § r-1)~v(Ogj_2(h))
-- n26V(O'gj_,(h));
see T h e o r e m 5.1.11(ii). For i E {0, 1} the equations in question hold obviously. Let i E N \ {0, 1} with i _< ~ be given. We assume that the equations in question hold for i - 1 and i - 2. T h e n the last equation yields
X(%2,+,(h)) = hi-1
n(r + r-1)~[(nh
hi-2
n ~ [ ( n h r--r
-1
and
- 1)(r i-1 - r -(i-1))
- r-(i-1))]-
§ ( n k -- 1) nh (r i - 2 - r - ( i - 2 ) ) ] = n
ni r
- 1)(r i - r - i ) + (nk - 1) nh (ri-ln
r-1
[(nh - 1)(r i+1 - r -(i+1)) + (nk - 1) n_____(r h i -- r - i ) ] n
5.3 Finite Coxeter Schemes of Rank 2 x(~g~,(~)) = .~(r + r-1)(H -1 + r-('-~)) - ~ ( r ~-2 + r-(~-~)) = ~'(~' +
141
~-~). []
Let R be an algebraically closed field with char(R) = 0, and let X E Irr(RG) be such that X(1) = 2. Then we shall denote by R• the (non-empty) set of all those elements of {r E R I r~ 7~ 1 = r m} which satisfy, for each i E N with i _< ~ , conditions (i) and (ii) of Proposition 5.3.2.
L e m m a 5.3.3 Let R be an algebraically closed field with char(R) = 0. Then we have (i) Let X E Irr(RG) be such that X(1) = 2, and let r E R x be given. Then r -1 E R x. (ii) For each r E R with r 2 • 1 = r '~, there exists X E Irr(RG) such that X(1) = 2 and r E R•
Proof. (i) follows immediately from Proposition 5.3.2.
(ii) From L e m m a 5.3.1(ii), (iii) we deduce easily that 1 I{x ~ I r r ( R a ) I X(1) = 2}1 = 71{r E R I r2 # 1 - rm}l . []
Therefore, the claim follows from (i).
P r o p o s i t i o n 5.3.4 Let R be an algebraically closed field with char(R) = O, let X E Irr(RG) be such that X(1) = 2, and let r E R x be given. Then (i) I f 2 { m ,
1 R R ~Z ~-r x(~r)~(~ )=
( n - 1) 2 2rail - n(r + r - S - 2) .].
(ii) If 21m, 1 R R geG ~ ' X ( ~ g ' ) X ( ~ g ) = 2m[1
1 7" - - r - l )
2
[(nh -- 1) 2 + (nk -- 1) 2 ] 7/,h nk
(nh -- 1)(nk -- 1 ) ( r + r - 1 ) ] . n(r - r - ' ) 2
R Proof, For each g E G, we abbreviate o'g := c~g.
142
5. Theory of Generators
(i) Assume that 2 ~ m. Then, by T h e o r e m 5.2.6(vi), nh = n = nk. Moreover, there exists s E 1~ such that 2 s + 1 = m. Now we obtain f r o m Proposition 5.3.2 and L e m m a 3.2.50) that, for each l E L,
5x(~g.)x(r gs
,
=
g*
2
x(~g~,/t)) 2
2 X(ag~,+l(L))2
4 =
n rn
i=0
F_, 2(,.' + i=0 8
2(,,- 1) ~ y-]~(,.~+~
,.-~+~/+,.~
,.-;)2_
i=o
(n-112 (r'+ln ( ~ - - r_--i-)2 + 2)
r -O+U
2(n-
'
1) 2
2m[1 -
r'
+
(n-
r-') 2
-
' r -1
+
1) 2
+
2)
-
-4= 4
=
];
n(r + r -1 --2)
note that r "+1 = r - ' (and r - ( ' + 1 ) = 7"). (ii) Assume that 2 I m. T h e n there exists s E N such that 2s = m. Now we obtain f r o m Proposition 5.3.2 and L e m m a 3.2.5(ii) that, for all h, k E L with h # k, gEG
~--]~[2 (.h,~)i i=O
+
g"
(nhnk)inh . +
(nknh)ink ]+
(nhnk)"
4=
8--1
2(ri + r-i)2+ i=0 $--1
Z
1
nh(r -- r - l ) 2 [(nh -- 1)(r i+1 -- r -(i+U) + (nk -- 1 ) - ~ ( r i -- r - i ) 1 2 +
i=0 s--1
1 n k ( r - - r - l ) 2[(nk - 1)(ri+l - r - ( i + l ) ) + (nh - l ) - - ~ ( r i -
i=O
( r ' + r - S ) 2 --
2m+
4
=
r-i)]2+
5.3 Finite Coxeter Schemes of Rank 2
,-1
1
[(nh -- 1) 2
i=0
(~
+
- 1 ) 2 ][(,,+,
,-1 4(nh -- 1)(na -- 1)
2rail
(~ _
;-1)2
~-')=]+
nk
nh Z i=0 1
_ ~-(,+*))~ + (r~ _
143
g~__--~-l)-2
(g'Tl--r-(i+l))(gl
[(nh_-l)2+(na-1) ~h ~-2
-- r-i) ----"
(nh--1)(nk--1)(r+r-1)]. ~(~ - ,'-1) '
2]
[]
Recall that, by Theorem 5.1.4(vi), Oo(G)=G
r
LC_Oo(G) r
{1}={nl I lEi}.
Thus, the hypothesis of the first part of the following theorem can be expressed in several different ways. Also, by Theorem 5.1.4(vii), Oe(G)={1}
r
LMOo(G)=0
r
lqL {nl I I E i } ,
which gives us different ways of expressing the hypothesis of the second, the third, and the fourth part of the following theorem. The following theorem was proved as Theorem 1 in [8].
T h e o r e m 5.3.5 [W. FEIT-G. HIGMAN] (i) If Oe(G) r {2, 3,4, 6,8, 12}. (ii) / f Oe(C) = {I} and m = 6, n E Z. (iii) I f O o ( G ) = {1} and m = 8, v ~ n E Z. (iv) I f O e ( G ) = {1}, m # 12.
G, m ~_
Proof. Let R be an algebraically closed field with char(R) = 0, let Q denote the prime field of R, and let Z denote the subring of R generated by 1. Set I := I(Z). Suppose first that 2 ~ m. Then 3 _< m. Thus, by Lemma 5.3.1(ii), there exists )C E Irr(RG) such that ;~(1) = 2. Let r E R• be given. Then we obtain from Proposition 5.3.40) and Theorem 4.1.5(ii) that 2m[1
i
( n - 1) 2 E Q. n(r + r-1 _ 2)]
But our hypothesis Oe(G) r G means that n r 1. Therefore, we conclude thatr+r-lEQ. On the other hand, by Theorem 3.2.2, r + r - 1 E I. Thus, by Lemma 3.2.3, r + r -1 E Z. Now Lemma 3.2.4 yields m = 3.
144
5. Theory of Generators
Suppose now that 2 [ m and that 4 < m. Then, by L e m m a 5.3.1(iii), {X E I r r ( R G ) ] X(1) = 2} 7~ 0. Let r E I r r ( R G ) be such that r = 2, and let r E Re be given. Then, by L e m m a 5.3.3(ii), there exists r E I r r ( R G ) such that r = 2 and - r G R e . Now, by Proposition 5.3.4(ii) and T h e o r e m 4.1.5(ii), 2m[2
2
tr(nh-
(r - r - 1 ) 2
1)2 + (nk -- 1)2]]
nh
E Q.
Thus, as we are assuming that O0 (G) :fi G, we must have that -1)2 E Q
It follows that r2+r-~EQ. On the other hand, by T h e o r e m 3.2.2, r 2 + r -2 E 1. Thus, by L e m m a 3.2.3, r 2 + r -2 E Z. Now L e m m a 3.2.4 yields m E {4, 6, 8, 12}. Thus, we have proved (i). From (r - r - l ) 2 E Q, Proposition 5.3.4(ii), and T h e o r e m 4.1.5(ii) we also obtain (nh - 1)( k - 1)(r + .(r _
EQ.
In order to prove (ii), (iii), and (iv), let us now assume that Oo(G) = {1}. T h e n nh ~ 1 # nk, so that we obtain r + r -1 - - E Q . n
Assume that m = 6. Then ( r + r - 1 ) 2 = 1. Thus, we must have that n E Q. Therefore, by L e m m a 3.2.3, n E Z. Assume that m = 8. By L e m m a 5.3.3(ii), we m a y assume that r is a primitive 8-th root of unity. Then (r + r - l ) 2 = 2. Thus, we must have that x/~n E Q. Therefore, by L e m m a 3.2.3, v ~ n E Z. Assume finally that m = 12. Suppose first that r is a primitive 12-th root of unity. T h e n (r + r - l ) 2 = 3. Thus, we must have that v ~ n E Q. Therefore, by L e m m a 3.2.3, x/-3n E Z. Suppose now that r is a primitive 6-th root of unity. Then (r + r - l ) 2 = 1. Thus, we must have that n E Q. Therefore, by L e m m a 3.2.3, n E Z, c o n t r a r y to x/~n E Z. []
Let us now have a closer look onto the cases m = 3, m = 4, m = 6, m = 8, and m = 12. For the remainder of this section, R denotes an algebraically closed field with char(R) = 0. For each i E {1, 2}, we set Ai := {X E I r r ( R G ) [ )/(1) = i}. (Recall that, by L e m m a 5.3.1(i), A1 U A2 = I r r ( R G ) . )
5.3 Finite Coxeter Schemes of Rank 2
145
From Lemma 5.3.1(ii), (iii) and Corollary 4.4.3(ii) we know that there exists st E A1 such that, for each l E L, s t ( a n ) = - 1 . Let r be a primitive m-th root of unity. Then, by Lemma 5.3.3(ii), there exists, for each i E N such that i < m@, an element Xi E Irr(RG) such that Xi(1) = 2 and r i E R• Moreover, for all i, j ~ N such that i, j < - ~ and i # j , Xi 7k X j .
T h e o r e m 5.3.6 A s s u m e that m = 3. Then (i) IXl = (n + 1)(n 2 + n + 1). (ii) rnst = n a. (iii) {Xl} = A2 and rex, = n ( n + 1).
Proof. (i) follows from Theorem 1.3.6(i).
(ii) By Theorem 4.1.5(ii), IXI = ~
2
1-L-st(cr~. )st(aRg ) = 1 + - +
2 + - 1- = [X[ "
use (i). (iii) From Lemma 5.3.1 we know that {iRe, st, x1} = Irr(G). Thus, we conclude that Xnx(1) = m l n G l n a ( 1 ) + m , t s t ( 1 ) + mx~Xl(1 ) = 1 + n 3 + mxlxl(1); use (ii) and Lemma 4.1.6(iii). On the other hand, by (i),
xRx(1) = IXl = n 3 + 2 n 2 + 2 n + Therefore, as
Xi(1) =
1.
2, mx, = n ( n + 1).
[]
It is a well-known open question, whether or not m = 3 implies that n is a prime power. For the remainder of this section, we assume that 4 < m, and we choose h, k E L such that h :/: k. Then, by Lemma 5.3.1(ii) and Corollary 4.4.3(ii), there exists Ah E A1 such that ah(a
) = nh,
= -1
and there exists Ak E A1 such that Ak(aht r
Ak(akR ) = n k .
Without loss of generality, we assume that nh < nk. The last part of the following theorem is part of [15; Theorem 3.2].
146
5. Theory of Generators T h e o r e m 5.3.7 Assume that m = 4. Then (i) IXl = (ha + 1)(nk + 1)(nhnk + 1). 2 2 (ii) We have mst = nhn k, n~(nhnk + 1) m~
h
--
n h 4- n k
and mA.k ~--
n2h(nhnk + 1)
nh + nk
(iii) {Xl} = A2 and nank(nh + 1)(nk + 1) reX1 ----
nh + nk
(In particular, nh + nk divides nhnk(nh + 1)(nk + 1).) (iv) [D. G. HICMAN] If nh # 1, nk < n~.
Proof. (i) follows from Theorem 1.3.6(i).
(ii) By Theorem 4.1.5(ii),
IXl
-
-
1
-
rrtst
n
:
gEG rig.
1 (1 + nh)(1 +
)(1+
1
nhnk
IXl
) = nhn 2 k2;
use (i). The values of rrtx h and mx k are obtained similarly. (iii) From Lemma 5.3.1(iii) we obtain that {Xl} = A2. The value of m• is easily computed with the help of (i), (it), and Lemma 4.1.6(iii). (iv) Assume that R is the field of complex numbers. Then, by Corollary 4.3.4, 0
--ng. 1
-(1-
)(l+nk)(1--~hk~)=
2
~ ( n h -- 1)(1 + nk)(n~ -- na). Thus, if na # 1, na < n~.
[]
Assume that m = 4 and that O0(G) = {1}. Then, if (X, G) is isomorphic to a known example, we have (nh,nk) E {(q-- 1 , q + 1), (q,q), (q, q2), (q2,q3)},
5.3 Finite Coxeter Schemes of Rank 2
147
where q is a prime power. The last part of the following theorem is the main result of [13].
T h e o r e m 5.3.8 A s s u m e that m = 6. Then (i) IXl = (ha + 1)(nk + 1)(n~hn 2 + nhnk + 1). (ii) We have rest = n h3n 3k, n ka( n h.-n k~ + nhnk + 1)
m)~h = and
n2h + n h n k "l- n~
n h3 ( n h2n2 k + nhnk + 1) rnxk = n~ + n h n k + n~
(iii) We have {X1, X2} = A~, n h n k ( n h + 1)(nk + 1)(nhn 2 k2 + nhn~ + 1) mx1 = 2 [ ( n l n ~ + n~ - n ~ n ~ + ~k + n h n ~ ) + (n~ - 1)(n~ - 1)~], and
~ k~ + n h n k + 1) n h n k ( n h + 1)(na + 1) (nhn m x : = 2[(n~na 4- na - n a n k 4- nk 4- nan~r - (nh - 1)(na - 1)n]" (iv) [W. H. HAEMERS] I f n h • 1, nk ~ n 3.
Proof. (i) follows from Theorem 1.3.6(i).
(ii) By Theorem 4.1.5(ii),
~.lx--!- ~
1---st(~
)~t(o-~) =
gE
1 Ixl 1 ( 1 + ~ 1 +_-r:-s_~)= 3 3;
(1+~)(1+
nh~k
nhn k
nhn k
use (i). The values of mxh and mxk are obtained similarly. (iii) From Lemma 5.3.1(iii) we obtain that {X1, X2} = A2. The values of m• and rex2 are easily computed with the help of Theorem 4.1.5(ii) and Proposition 5.3.4(ii). (iv) Assume that R is the field of complex numbers. The values of X2 are easily computed with the help of Proposition 5.3.2. Thus, by Corollary 4.3.4,
0 < V" ~kI~.)2x~(~) _ -
-
gEG~
n2*
--
148
5. T h e o r y of G e n e r a t o r s
nh -- 1 2n 2 + ~n------~--+ nk - 1 - - F + n h
.~(.~
-
1) - ( n ~ -
1).
(nk - 1)n nSh
(,~
-
1).~
- (.k -
2nk
1)n
(nh - 1)nkn + 2nkn = n~ n~
nl•-a
(nh + 1)(nk + 1)(n~ -- l)(n~ -- n). []
Thus, if nh # 1, n < n~, i.e., nk _< n].
Assume that m = 6 and that O0(G) = {1}. Then, if (X, G) is isomorphic to a known example, we have (nh,nk) E {(q, q), (q, q3)},
where q is a prime power. The last part of the following theorem is part of [15; Theorem 3.2].
T h e o r e m 5.3.9 Assume that m = 8. Then (i) IXl = (-h + 1)(nk -4- 1)(nhnk -4- 1)(n~n~ + 1). 4 4k, (ii) We have rest = nhn n4(nhnk + 1)(n~n~ + 1)
and
2 k2 -t- 1) n4(nhnk + 1)(nhn
m~, =
(.h + .~)("~ + "~)
(iii) We have {X1, X2, X3} = As, 2 2 + 1) + 1)(-k + 1)(--h~k + 1)("k"k rex1 = 4[(n~nk + nh - 2nhnk + nk + nhn~) + (nh - 1)(nk - 1)v~n] '
nhnk(-h
m•
nnnk(nh + 1)(nk + 1)(nt, n k2 + = 2(nh + nk)
1) '
and nhnh(nh + 1)(nk + 1)(nhnk + 1)(n~n~ + 1) m,~3 =
4[(n~-k
(In particular,
+ '~h - 2 , , h , ~ k + , ~ + , , ~ , , ~ )
n h • nk.) (iv) [D. G. HIGMAN] If nh # 1, nk < n2h.
-- ( n ~ -- 1 ) ( - ~
-- 1 ) V % ]
"
5.3 Finite Coxeter Schemes of Rank 2
149
Proof. (i), (ii), and (iii) are obtained in the same way as the corresponding parts of T h e o r e m 5.3.8. (iv) Assume that R is the field of complex numbers. Then, by Corollary 4.3.4,
O_<E-----y----Xk(crgR)3 (1 -gEG
rig,
)(1 + n k ) ( 1 - -
)(1 + -nk -
=
h
1
n~(,,i - 1)(1 + nk)(n~ - ,~)(.~ + n~). Thus, if nh # 1, nk <_ n2h.
[]
Assume that m = 8 and that O o ( G ) = {1}. Then we know from T h e o r e m 5.3.5(iii) that nh # nk. Moreover, if (X, G) is isomorphic to a known example, we have (nh, nk) = (q, q2), where q is a power of 2.
T h e o r e m 5 . 3 . 1 0 A s s u m e that m = 12. Then nh = 1 and (i) IX[ = 2(nk + 1)2(n~ + n 2 + 1). (ii) We have m , t = n~, mxh = n~, and m~ k = 1. (iii) We have { X l , X2, X3, )(.4, )(.5} : As, m•
nk
= -~-(nk + 1)2(nk2 + nk + 1), nk
rex2 = v ( n ~
+ 1)2(n~ - "~ + 1),
2nk , 4
~ (nk+n~+l)'
m• nk
m•
= T ( n k + l)2(n~ - nk + 1),
m•
= --~-(nk + 1)2(nl + nk + 1).
and
nk
Proof. From T h e o r e m 5.3.5(iv) we obtain that nh = 1. All other claims are obtained in exactly the same way as the corresponding parts of T h e o r e m 5.3.9. []
Assume that m = 12. Then, if (X, G) is isomorphic to a known example, nk is a prime power.
150
5. Theory of Generators
5.4 The
Theorems
of Ott and Payne
As in the previous section, we assume in the present section that ILl = 2, that (X, G) is a Coxeter scheme with respect to L, and that IXl E r~. As before, we set m := m(L), and we obtain that m E N, that 2m = IGI, that
{fj(h,k) I h, k E L, h~t k, j E { O , . . . , m ) } = Fm(L), and that, for all h, k E L with h ~t k, fro(h, k) "-~mfro(k, h). By assumption, (X, G) is a Coxeter scheme with respect to L. In particular, for all h, k E L with h ~ k and, for each j E { 0 , . . . , m } , there exists gj(h) E G such that {gj(h)} = fj(h, k)p. Now, as 2m = IGI, Theorem 5.2.6(i), (ii) yields {gj(l) I l E L, j E { 0 , . . . , m } ) = G
and, for all h, k e L, g.~(h) = gr.(k). An element rr E Aut(X, G) is called a polarityof (X, G) if, for all h, k E L with h r k, hrr = k and kzr = h.
L e m m a 5.4.1 Let I E L be given, and let rr be a polarity o f ( X , G). Then we have the following. (i) For each j E { 0 , . . . , m}, gj(l)rr = gj(lr). (ii) Let z E X, and let j E { 1 , . . . , m } be such that (x, xr) E gj(l). Then 2[j orj--m.
Proof. (i) follows from Lemma 1.7.1(ii) by induction.
(ii) If (z, x~r) e gj(l), g~(l)" ~ (~., ~) = ( ~ , J )
e g~(l)~ = gj(l~);
see (i). Therefore, gj(l)* = gj(lzr). But, by definition, lrr ~t I. Therefore 2 I j or j = m; see Theorem 5.2.6(i). []
As in the previous section, we define n :--- H v/-~. IEL
Note that, for each l E L, nl~ = nl and 17r ~t l. Therefore, for each l E L, n~
hi,
Recall that Fixx(rr) := {z E X I xrr = z).
5.4 The Theorems of Ott and Payne
151
L e m m a 5.4.2 Assume that 2 I m. Then, for each polarity 7r of (X, G), IFixx(rr)l = n ml~ + 1.
Proos Let rr be a polarity of (X, G), and let us abbreviate Y := Fixx (r). By hypothesis, 2 I m. Therefore, there exists s E Z such that 2s = m. Our first claim is
x: 0 U i=OIEL
Let z E X be given. Then there exist 1 E L and j E { 0 , . . . , m} such that
(x, xrr) E gj(l). Thus, by L e m m a 5.4.1(ii), 2 i J. Let i E Z be such that 2i = j. Since j < m, {gj(l)} = gi(l)gi(17r)*. Thus, by L e m m a 1.2.4, there exists y e xgi(l) N xTrgi(lrr). It follows that
u~ e x~g~(l~) n xg~(t). Let g E G be such that (y, yTr) E g. Then, by L e m m a 1.2.4, g E g, (l)'gi (l) fq gi (lr)'gi (lrr). Now L e m m a 5.2.1(i), (ii) and Theorem 5.2.60) yield g = 1, which means that y E Y. Therefore X
E ug,(I)" c Y{g~(1),g,(l.)}.
Since x E X has been chosen arbitrarily, we have proved our first claim. Let y, z E Y be such that y r z. Then there exist I E L and j E { 1 , . . . , m } such that (y, z) E gj(l). From Theorem 5.2.6(i) we conclude that j = m. Therefore, for all h, k E L with h # k and, for all i, j E { 1 , . . . , s}, we must have either i = s = j or Y{gi(h)} M Y{gj(k)} = 0. Conversely, let y E Y, and let x E yg,(l) be given. Then there exists a uniquely determined element z E Y such that x E zg,(br). From Theorem 5.1.11(i), L e m m a 1.2.4, and L e m m a 1.1.5(iii) we deduce that, for each l E L and, for each i E { 1 , . . . , s } , ng,(i) = n i.
Therefore, s-1
Ixl : IYI(1 + 2n + . . . + 2n '-~ + n ~ = IYl(n + 1)
~
n'
i=0
and
rn--1
IXl = (n + 11 ~
$-1
n' = (~ + 1)(~ s + 1) y ~ ~'.
i=0
Thus, we conclude that ]Y] = n ~ + 1.
i=O
[]
152
5. Theory of Generators Let us assume that (X, G) possesses a polarity ~r, and let us define O :=
{1, 7r}, (W, F) := (X, G)O, and ~ := co.4 For each l 9 L and, for each j 9 { 1 , . . . , rn}, we shall abbreviate
]j+(l) := (gj(l), 1)~ and
17(0 := (g~(l), ~)~. Note that then
{ f + ( l ) , f f (l) It 9 L, j 9 { 0 , . . . , m } } = F. In the following proposition, we completely compute the character of R F afforded by the RF-modul RX, provided that 2 [ m.
P r o p o s i t i o n 5.4.3 Assume that (X, G) possesses a polarity 7r. Set 0 :=
{1,~} and (W,F) := (X,G)O. Let R be a field, and let r denote the character of R F afforded by R X . Let I 9 L, and let j 9 { 1 , . . . , m} be given. Then we have (i) I f j # O, r R = O. (ii) I f 2 l m
R
and 2 { j , r
) = O.
R
(iii) If2 I m and 2 I j, r
) = (n"~/2 + 1)nJ/2.
R Proof. For each g G G, we abbreviate ag := ag. (i) For each z G X, xal+(O = x~gj(l);
see Theorem 4.4.8(i). Therefore, if j -~ 0,
r
= xRx(%(,))
= o.
(ii), (iii) By Theorem 4.4.8(iv),
r
= I{~ 9 x I ( ~ , x) 9 g~(l)}l.
Thus, by Lemma 5.4.1(ii),
2{j
~ r
4 Recall that (X, G)O denotes the semidirect product of (X, G) with O.
5.4 The Theorems of Ott and Payne
153
Assume now that 2 I j, and let i E Z be such that 2i = j. Note that xgi(l) contains a fixed point of a" if and only if (x,xTr) E gj(1). Thus, by Lemma 5.4.2, 2 I J =:V r = (n "~/2 + 1)n j/2.
[]
The first part of the following theorem was proved as Theorem 2 in [24]. Its second part is [23; Satz 1].
T h e o r e m 5.4.4 Assume that (X, G) possesses a polarity and that n ~ 1. Then (i) [S. E. PAYNE] / f m = 4, V / ~ E Z. (ii) [U. Oww] / f m = 6, v f ~ E Z. Proof. Let 7r be a polarity of (X, G), and let us define O := {1, 7r), (w, F) := (x,
a)o,
and r := co. From the definition of (W, F) we obtain that IF] = 4m. From Theorem 2.7.1(v) we deduce easily that (h, a)q (1, 7r)~ E Inv(F). The same reference then shows that (W, F) is a Coxeter scheme with respect to {(h, 1)~, (1, 7r)r and that {f~- (h)} = ((1, 7r)g(h, 1)~)2(1, lr)g, {f~- (k)} = (h, 1)~(1, ~')q(h, 1)r {f4 (h)} --- ((1, 7r)q(h, 1)g))4(1, ~')g, {f4 (k)} = ((h, 1)~(1,/r)g))3(h, 1)~, and, if 6 _< m, {f6 (h)} = ((1, ~r)r
1)r
7r)r
Let R be an algebraically closed field with char(R) = 0, let Q denote the prime field of R, and let Z denote the subring of R generated by 1. Set I := I(Z). For each 9 E G, we abbreviate r := ~rgn. Let r E R be a primitive 2m-th root of unity. Then, by Lemma 5.3.3(ii), there exists X E Irr(RF) such that X(1) = 2 and r E R x(i) Assume that m = 4. From Proposition 5.3.2(i) (applied to (W, F ) i n the role of (X, G)) we compute easily that = (- - 1)v
= x(:s:(k/)
154
5. Theory of Generators
and that X(o'f~-(h)) = O. Moreover, by Proposition 5.4.3(iii), = n(n 2 + 1) = r
r
Therefore, using Proposition 5.4.3(i), (ii), we obtain
~
+x(~.)r
=
2lxl + ~[x('~1;(h)) + X('~J;(k))]r 2
2IX] + ~z(n - 1)v/~r
= --
2iX] + 2_(. _ a)(. ~ + 1)v'~. n
Since we assume that n r 1, we now conclude from Theorem 4.1.5(ii) that x / ~ E Q. Therefore, by Lemma 3.2.3, v ' ~ E Z. (ii) Let us now assume that m = 6. From Proposition 5.3.2(i) (applied to (W, F) in the role of (X, G)) we compute easily that
x(~1;(h)) = (" - 1 ) v ~ = ~(~s:(k)), that
X(~s:(h)) = .(n - :)v~. = x(~s:(k)), and that X(o'lz(n)) = 0. Moreover, by Proposition 5.4.3(iii),
r
= ('~ + a). = r
r
= (n 3 + 1)n 2 = r
that Therefore, using Proposition 5.4.3(i), (ii), we obtain 1
=
gEG
1
21Xl + U[x(~s:(h)) +
x(o.s;(k))]r
1
21Xl
~[x(~s:(h)) + X(~I:(k))]r = 2 (n - 1)Vf~r + 2--(-n~- 1)v/~r +
=
4
2IX I
+ - ( n - al(n ~ + 1 ) v ' ~ , n
As in (i) we now obtain that x / ~ E Z.
[]
5.5 Length and Norm
155
5.5 Length and N o r m In this section, we assume that (L) = G, that [LI = 2, and that re(L) E 1% (We do not assume any longer that (X, G) is a Coxeter scheme with respect to L.) As in the previous three sections, we set m := re(L). For all h, k E L with h r k and, for each j E l~, we define
Rj(h) := fj(h,k)p.
L e m m a 5.5.1 For each g E G and, for each l E L, {j E N I g E n~(l)} # 0.
Proof. Let g E G be given. Then, as we are assuming that (L) = G, 9 E (L). Thus, by Theorem 1.4.1(i), there exists n E I~ such that g E L". (Recall that, by Lemma 1.4.5(i), L* = L.) Now, by Lemma 5.2.2(ii), there exist h E L and j E N such that g E Rj(h). Choose j E 1~ minimal subject to g E Rj(h). Clearly, we may assume that jr Let k E L be such that {h,k} = L, and set i := 2m. Then, by Lemma 5.2.4(ii), 1 E Ri(k). Therefore, g E lg C_ Ri(h)*Rj(h). If j _< i, the claim follows from Lemma 5.2.1(ii). If i <_ j, the claim follows from Lemma 5.2.1(iv). (Recall that i r 0. Therefore, the choice of j forces g ~ Rj_i(h).) [] Let g E G be given. For each l E L, we set Itt(g) := min{j E l~ ]g E Rj(/)}, and we shall call Itg(g) the l-length of g. Note that, by Lemma 5.2.2(ii),
It(g) = min{Itt (g) I I E L}. Note also that, for each 1 E L, #l(1) = 0 and Itt(l) = 1. The following four results are devoted to the /-length of the elements of G, where l E L.
156
5. Theory of Generators
L e m m a 5.5.2 (i) Let h, k E L be such that h 7k k. Then, for each g E G \ {1}, 2m _< #h(g) +Pk(g). (ii) For each l E i and, for each j E {0,..., m}, p(Rj(l)) = {j}.
Proof (i) Set i : = tth(g) and j := Pk(g). Then g E Ri(h) M Rj(k). Thus, by Lemma 5.2.3(i), 1 E Ri+i(h) U Ri+j(k). Therefore, by Lemma 5.2.4(ii), 2m < i + j. (Recall that, by hypothesis, g # 1. Therefore, i # 0 # j.) (ii) The claim is obvious for j = 0. Therefore, we assume that 1 < j. Let g E Rj(l) be given. We shall be done if we succeed in showing that p(g) = j. Since g E Rj (/), /~t (g) _< j. Set i := Pt (g). Then i _< j and, by definition, g E Ri(l). It follows that ni(l) M Rj(l) # 0 and that
i + j <_2m. (Recall that, by hypothesis, j _< m.) Suppose that i = 0. Then Ri(l) = {1}, so that 1 E Rj(l). Thus, as j <_ m, Lemma 5.2.4(ii) yields j = 0, contrary to our hypothesis that 1 _< j. It follows that 1 _< i. Now we conclude from Lemma 5.2.3(ii) and Lemma 5.2.4(ii) that i = j. This means that #t(g) = j. From/tt (g) = j we obtain in particular that/~t (g) < m. Thus, we conclude from (i) that/~(g) = Pl(g). It follows that/~(g) = j. []
L e m m a 5.5.3 Let g E G be given. (i) Let l E n be given. Then, if2 { #t(g), Pt(g*) <_Pt(g). (ii) Let h, k E n be such that h # k. Then, if2 I Ph(g), #k(g*) <_ Ph(g).
(iii) p(g*) = I~(g). Proof. (i) Set j := ttt(g). Then g E Rj(l) and, by assumption, 2 { j . Thus, g* E Rj(l)* = Rj(l), whence Pt(g*) _< j. (ii) Set j := Ph(g). Then g E Rj(h) and, by assumption, 2 I J. Thus, g* E Rj(h)* = Rj(k), whence pa(g*) < j. (iii) follows from (i) and (ii). [] L e m m a 5.5.4 Let e, f E G be given. (i) Let l E L be given. Then, if f E eL and e # 1, Pt(f) <_ #t(e) + 1. (ii) Let n E N be such that f E eL m. Then p(f) < #(e) + n.
5.5 Length and Norm
157
Proof. (i) Set j := ~at(e). Then e C Rj(l) and, as e # 1, j # 0. Thus, by Lemma 5.2.2(i), f E eL C_ Rj(l)L C_ Rj_l(l) U Rj(l) U Rj+I(I), whence ~at(f) <_ j + 1. (ii) Set j :=/a(e). Then e E UteL Rj(l) C_ LJ. Thus, f E LJL n = L j+n, whence/a(f) < j + n. []
We define E(L) := {g E (L) ] ]{/at(g) ]l E L}I = 1}.
L e m m a 5.5.5 (i) E(L)* = E(L). (ii) 1 E E(L). (iii) {1} :fi E(L). (iv) m = min/a(E(/) \ {1}).
Proof. (i) Let e E E(L) be given, and set j :=/a(e). Then e E NteL Rj(l) = NleL Rj(l)*, whence e* E NtcL Rj(l). Now Lemmab.5.3(iii) forces e* E E(L). (ii) For each l E L, /at(i) = 0. Therefore, 1 E E(L). (iii), (iv) Let h, k E L be such that h :fi k. Then the definition of m yields that Rm(h) n Rm(k) ~ 0. Let e E R,,(h) • Rm(k) be given. From e E R,~(h) we obtain that/ah(e) < m; from e E Rm(k) we obtain that /ak(e) < m. Moreover, since e E Rm(h), Lemma 5.2.4(ii) yields e # 1. Thus, by Lemma 5.5.2(i),/ah(e) = m = #k(e). It follows that e E E(L). So far, we have shown that {1} :~ E(L) and that m E/a(E(L) \ {1}). But, by Lemma 5.5.2(i), we also must have m _< min/a(E(L) \ {1}). []
Let g E G be given. For all h, k E L such that h r k, we set
(g) :=/ah (g) -/ak (g). Let l E L be given. For each g E G, we say that t~t(g) is the l-norm of g. Note that E(L) contains exactly those elements of G on which the/-norm vanishes. The last two results of this section are devoted to the/-norm of the elements of G.
Lemmab.5.6
Let g E G, and let l E L be given.
158
5. Theory of Generators (i) If2 { p(g), 0 <_ ut(g*)ut(g). (ii) If2 l #(g), uj(g*)ut(g) < O.
Proof. (i) Assume first that tq(g) < 0. Then p(g) = #~(g). Thus, by Lemma 5.5.3(i), (iii), ~tt(g*)
Proof. Assume, by way of contradiction, that ut (e)ut (f) < - 1 . Then there exist h, k E L such that ph(e) < pk(e) - 1 and #k(f) < Ph(f) - 1. If 1 E {e, f}, the claim holds obviously. Therefore, we assume that e # 1 -~ f. Then, by Lemma 5.5.4(i), ph(e) < pk(e) -- 1 <_Pk(f) < Ph(f) -- 1 <_ph(e). It follows that/~h(e) = pk(f). We set j := ph(e). Since, by hypothesis, f E eL, we have f E eh or f E ek. Let us assume first that f e eh. If 2 { j, f E eh C_ Rj(h)*h. Thus, by Lemma 5.2.1(i), f E Rj-l(h) U Rj(h), contrary to j + 1 < t~h(f). If 2 I J, e E fh C Rj(h)*h. Thus, by Lemma 5.2.1(ii), e E Rj-l(k) U Rj(k), contrary toj+l
5.6 Partitions of G As in the previous section, we assume in the present section that (L) = G, that IL] = 2, and that m(L) E N. As before, we set m := re(L). Moreover, as before, we abbreviate, for all h, k E L with h # k and, for each jell,
Rj(h) := fj(h, k)p.
5.6 Partitions of G
159
For each l 9 L and, for each i 9 {-1, 0, 1}, we define
a~(O := {g 9 a I min#(gt) = #(g) + it.
L e m m a 5.6.1 Let l 9 L be given. (i) G_l(l) U G 0 ( / ) U GI(/) = G. (ii) 1 9 Gl(l). (iii) l 9 G_~(l). (iv) G - l ( / ) = {g 9 G I ~,(g*) _< 0} \ {1}. (v) a0(t) = {g 9 G I -~(g') -- 1}.
Proof (i) follows from Lemma 5.5.4(ii). (ii) follows immediately from the definition of G1 (l). (iii) follows immediately from Lemma 1.4.5(i) and Lemma 1.2.5(iii). (iv) Let us first prove that G - i ( / ) C_ {g E G I ul(g*) <_ 0} \ {1}. Let g E G-l(l) be given. Then, by (ii), g ~ 1. Thus, wc shall be done if we succeed in showing that vt (g*) _< 0. If g E E(L), g* E E(L); see Lemma 5.5.5(i). Then v~(g*) = 0. Therefore, we assume that g ~ E(L). Set j := #(g). Then, as g # 1, 1 <_ j. If j = 1, 1 E gl, whence, by Lemma 1.2.5(iii), g* = I. In this case, we obviously have vt(g*) <_ O. Therefore, we assume that 2 _< j. By assumption, there exists f E gl such that #(f) = j - 1. Assume first that vt(g) <_ O. Then, as g ~ E(L), vl(g) _< - 1 . Thus, as f E gl, vl(f) <_ 0; see Lemma 5.5.7. It follows that j - 1 = #l(f), whence
f 9 Rj-I(0.
Suppose that 2 ] j . Then, as 2 <_ j, g 9 II C_ Rj_l(l)*l. Thus, by Lemma 5.2.1(i), g 9 Rj_2(l) U Ri_l(l ). It follows that #,(g) _< j - 1, contrary to j = #(g). Therefore, we have 2 r j = #(g). Thus, as vt(g) _~ - 1 , Lemma 5.5.6(i) yields vl (g*) < 0. Assume now that 1 ~_ vl(g). Then, as f E gl, 0 ~_ vt(f); see Lemma 5.5.7. Let h, k 9 L be such that I = h r k. Then j - 1 = #k(f), whence f 9 Rj-l(k). Suppose that 2 ~ j . Then, as 2 _~ j, g E fh C_ Rj_~(h)*h. Thus, by Lemma 5.2.1(ii), g 9 Rj-2(k) U Rj-l(k). It follows that Pk(g) _~ J - 1, contrary to
j = ~(g). Therefore, we have 2 ] j = #(g). Thus, as 1 < vl (g), Lemma 5.5.6(ii) yields
vl(g*) ~_ O. Thus, we have shown that
G_~(l) c_ {g ~ G I ~,(g*) <- O} \ {1}.
160
5. Theory of Generators
Let us now prove that {g E O [ v~(g*) _< 0} \ {1} _C G_I(/). Let g 9 G \ {1} be such that t,l(g*) < 0. Then #(g*) = p,(g*). Set j := it(g). Then j # 0 and, by Lemma 5.5.3(iii), #(g*) = j. It follows that #~(g*) = j. Thus, g" 9 Rj(1). Assume first that 2 ~ j. Then g 9 Rj(l). Thus, as j # 0, there exists f 9 Rj-s(l) such that g 9 fl. From f 9 Rj-l(1) we deduce that p(f) < /~l(f) < j - 1. Moreover, as g 9 fl, f E gl; see Lemma 1.2.5(i) and Lemma 1.4.5(i). Thus, g 9 G-l(l). Similarly, the assumption that 2 l J leads to g 9 G_l(l). Thus, we have shown that
{g 9 o I
< o} \ {i} c_
(v) Let us first prove that Go(l) C_ {g E G [ t,~(g*) = 1}. Let g E Go(l) be given. Then, by (ii), g # 1. Thus, by (iv), 1 < t,t(g*). Set j := p(g). Then, by definition, there exists f E gl such that p(f) = j. From f E gl we obtain that g* E l f*; see Lemma 1.2.5(ii). Moreover, p(f) = j yields p(f*) = j; see Lemma 5.5.3(iii). Let h, k E L b e s u c h t h a t l = h # k . Suppose that #h(f*) = j. Then f* E Rj(h), whence g* E hRj(h). Thus, by Lemma 5.2.1(iii), g* E Rj-l(k) U Rj(h). (Note that, as g # 1, j # 0.) If g* G Rj_l(k), p(g*) < j - 1, contrary to Lemma 5.5.3(iii). If g* E Rj(h), Ph(g') <_ j. But, by Lemma 5.5.3(iii), #(g') = j. Therefore, t,h(g*) <_ 0, contrary to 1 < t~l(g*). Thus, we must have pk(f*) = j. It follows that f* E Rj(k), whence g* 9 hRj(k) = Rj+l(h). Thus, #h(g*) < j + 1, which means that g,(g*) < j + 1. It follows that ~1(9") _< 1. Thus, we have shown that
Oo(1) c_ {g E O I -,(g*) = I}. Let us now prove that {g E G I vt(g*) = 1} _C Go(l). Let g E G be such that t,~(g*) = 1. Set j := p(g*), and let h, k E L be such that l = h # k. Then, Ph(g*) = j + l, whence g* E Rj+l(h) = hRj(k). Therefore, there exists f E Rj(k) such that g* E hr. By Lemma 1.2.5(ii), the latter says that f* E gl. On the other hand, from f E Rj(k) we deduce that # ( f ) _< j. Thus, by Lemma 5.5.3(iii), #(f*) _< #(g). It follows that g q~ Gl(l). Since vl(g*) = 1, (iv) tells us also that g ~ G-l(l). Therefore, we must have g E G0(l); see (i). Thus, we have shown that
{g E G I vl(g*) = 1} C_ Go(l). []
C o r o l l a r y 5.6.2 (i) G \ {1} = UIcL G-l(l).
5.6 Partitions of G
161
(ii) E(L) \ {1} = NteL a - l ( / ) . (iii) For each l E L, E(L) M Go(l) = O.
Proof. (i) follows from Lemma 5.6.1(iv). (ii) Let g E E(L) \ {1} be given. Since g E E(L), g* E E(L); see Lemma 5.5.5(i). Thus, for each l E L, vt(g*) = 0. It follows that g E N~r G-i(/); see Lemma 5.6.1(iv). Conversely, let g E Ntr G - l ( / ) be given. Then g # 1 and, for each I E L, vl(g*) = 0; see Lemma 5.6.1(iv). It follows that g* E E(L). Thus, by Lemma 5.5.5(i), g n(L). (iii) is an immediate consequence of (ii) and Lemma 5.6.1(ii). []
L e m m a 5.6.3 For each g E UIeL Go(l), there exists e E E(L) \ {1} such that #(e) < p(g) and g* E eUIEL R,(g)-,(~)+l(/).
Proof. Let l E L, and let g E Go(l) be given. Set j := #,(g*). Then g* E Rj(l) and, by Lemma 5.6.1(ii), (iii), j ~ {0, 1}. Thus, by definition, there exist 11, ..., lj C L such that g* E ll " .lj and, for each i E { 1 , . . . , j}, l~ = l if and only if 2 { i. Since g* E 11 . . . l j , Lemma 1.2.4 yields go, ..., gj E G such that go = 1, gj = g*, and, for each i E { 1 , . . . , j } , gi E gi-lli. Since gl -- l, v/(gl) _< - 1 . On the other hand, as gj -- g*, gj E Go(l). Thus, by Lemma 5.6.1(v), vl(gj) -- 1. Set I := {i E { 1 , . . . , j } [ 0 _< vt(g,)}. Then I # 0. Set n := m i n i . Then 2 < n and ut(g,,-1) < - 1 . But g,~ E g n - l L . Thus, by Lemma 5.5.7, ut(gn) = 0. This means that g,~ E E(L). Set e := gn. Since gj -- g*, pt(gj) = j. Thus, by induction, Lemma 5.5.4(i) yields that, for each i E { 1 , . . . , j } , i < Pi(gi). On the other hand, as gl = l, Pt(gl) = 1. Therefore, by induction, Lemma 5.5.4(i) yields that, for each i E { 1 , . . . , j } , Pt(gi) < i. It follows that, for each i E { 1 , . . . , j } , Pl(gi) = i. In particular, pl(e) = n. Thus, as e E E(L), ,(e) = .. Since u,(gn) = 0 and ut(gj) = 1, n < j - 1 . Since ul(g*) = 1 and pl(g*) = j, #(g*) - - j - 1. Thus, by Lemma 5.5.3(iii), /~(g) -- j - 1. It tbllows that p(e) < p(g), the first part of our claim.
162
5. Theory of Generators
Now we apply Lemma 1.2.4 to (go,g,~,...,gj) in order to obtain that g* E eln+l . . . l j . Recall that, for each i E {n + 1,... ,j}, Ii = l if and only if 2 { i. Therefore,
g" e e U nj_.(O. IEL
Now the lemma is proved because we have j = It(y) + 1 and n = It(e).
Lemmah.6.4
Let g E G be such that It(g) < m - 1 .
[]
Then we have
(i) g ~ U, eL Go(l). (ii) g E UtCL Gl(l).
Proof. (i) is an immediate consequence of Lemma 5.6.3 and Lemma 5.5.5(iv). (ii) Since It(g) _< m - 1, g ~ E(L) \ {1}; see Lemma 5.5.5(iv). Thus, by Corollary 5.6.2(ii), there exists I E L, such that g q~ G-l(l). But, by (i), we also have g q~ Go(l). Thus, by Lemma 5.6.1(i), g E GI(/). []
L e m m a 5.6.5 Let c, d, e, f E G \ E(L) be such that c E L d n e L f E LeV)dL. (i) /fit(c) = It(e), It(d) = It(f). (ii) /fc*e M L Md*f # 0 and It(e) + 1 = #(e), It(d) + 1 = It(f).
and
Proof. Since c, d, e, f ~ E(L), Lemma 5.5.7 yields that, for each l E L, 1 <
1< 1< 1 < vl(d)vl(f). (i) Let l E L be given, and assume that It(e) = It(e). Then, by Lemma 5.5.6, 1 < v,(e*)vt(c)vl(e*)vl(e). Now we deduce from the above four equations that
1 <_ vl(d*)vl(d)vt(f*)vt(f). Thus, by Lemma 5.5.6, 2 I It(d) - It(f). On the other hand, by hypothesis, f E dL. Thus, by Lemma 5.5.4(ii), It(d) - It(f) E {-1, 0, 1}. It follows that #(d) - #(f).
5.7 Cosets on Which the Length is Constant
163
(ii) Applying (i) to (d, f, c, e) in the role of (c, e, d, f) we obtain that p(d) #(f). But, by hypothesis, f E dL. Thus, by Lemma 5.5.4(ii), # ( f ) E {p(d) - 1,#(d) § 1). Suppose that p ( f ) = #(d) - 1, and let l E e*e n L M d*f be given. From I E d*f we deduce that f* E I'd*; see Lemma 1.2.5(ii). Thus, by Lemma 1.2.2, f E dl. Therefore, as p(f) = p(d) - 1, d E G-l(l). Thus, by Lemma 5.6.1(iv), vl(d*) <_ - 1 . On the other hand, as l E c'e, c E el; see Lemma 1.2.5(ii) and Lemma 1.4.5(i). Moreover, by hypothesis, p(c) = p ( e ) - 1. Therefore, e E G - l ( / ) , which, by Lemma 5.6.1(iv), implies that vt(e*) <_ - 1 . It follows that 1 <_ vl(d*)u,(e*). Thus, by the first of the above equations, 1 _< v~(c*)v,,(e*). Now the third one yields 1 <_ vt(c*)vt(c)vl(e*),l(e). But, as p(c) § 1 = p(e), this contradicts Lemma 5.5.6. Therefore, we have #(f) = #(d) + 1.
[]
5.7 C o s e t s on W h i c h the L e n g t h is C o n s t a n t As in the previous section, we assume in the present section that (L) = G, that ILl = 2, and that re(L) E l~. As before, we set m := m(L). Moreover, as before, we abbreviate, for all h, k E L with h # k and, for each jell, Rj(h) := fj(h, k)p. For each l E L, we define
a ~ ( l ) := {g E a I I~(gq))l -- 1}.
L e m m a 5.7.1 (i) For each I E L and, for each 9 E Goo(l), g(l> C Goo(l). (ii) For each l E L, Goo(l) c Go(l). (iii) Assume that IG I E N. Let g E G be such that #(g) = maxp(G). Then g E UIeL Goo(l) U E(L).
164
5. Theory of Generators
Proof. (i) Let f E g(l) be given. Then, by Theorem 1.3.1, f(1) = g(l), whence I ~ ( f ( / ) ) l = I ~ ( g ( / ) ) l -- 1. I t f o l l o w s t h a t f e G~(l). (ii) follows immediately from the definitions of Goo(l) and Go(l). (iii) For each l E L, g q~ G~(l) implies that a ~ G_x(/); see Lemma 5.5.4(ii). Thus, if a ~ UteL Goo (l), g E E(L); see Corollary 5.6.2(ii). [] For each g E G, we define D(g) := {f E gL ] It(f) = #(g) - 1}. From Corollary 5.6.20) we know that, for each g E G \ {1}, D(g) :/: 0.
L e m m a 5.7.2 (i) For each l E L, D(/) = {1}. (ii) For each g E G \ L, D(g) N E(L) = 0.
Proof. (i) follows immediately from the definition of D(/). (ii) Let g E G be such that D(g)AE(L) :~ 0. Let e E D(g)ME(L) be given. Since e E D(g), Lemma 5.5.4(ii) yields e ~ (']leL G_~(l). Thus, as e E E(L), e = 1; see Corollary 5.6.2(ii). It follows that g E L. []
L e m m a 5.7.3 Let l E L, let g E Goo(l), and let f E D(g*)* be given. Then we have (i) f ~ G_l(l). (ii) It(f(l) \ E(L)) = {p(g) - 1}. (iii) f E E(L)I U G~(l). (iv) /f#(g) ~/~(E(L)), f E Coo(1).
Proof. Since f E D(g*)*, f* E D(g*). Thus, by definition, f* Eg*L and/~(f*) =/z(g*) - 1. From p(f*) =/~(g*) - 1 we obtain that ~(f)
= ~ ( g ) - 1;
see Lemma 5.5.3(iii). (i) Since g E Goo(l), g E G0(l); see Lemma 5.7.1(ii). Therefore, by Lemma 5.6.1(v), ul(g*) = 1. But f" E g*L. Thus, by Lemma 5.5.7, 0 < ut(g*)ul(f*). It follows that 0 < ul(f*). Assume, by way of contradiction, that f E G-l(l). Then, by Lemma 5.6.1(iv), ul(f*) <_ O. Thus, ul(f*) = 0, which means that f* E E(L). But
5.7 Cosets on Which the Length is Constant
165
we also have that f* E D(g*). Thus, by Lemma 5.7.2, f* = 1, contrary to f E G - l ( / ) ; see Lemma 5.6.1(ii). (ii) Let d E fl \ E(L) be given. We shall be done if we succeed in showing that #(d) = #(g) - 1. Since f* E g ' L , f E Lg. Thus, d E Lgl. Therefore, there exists c E gl such that d E Lc. It follows that c E LdngL. On the other hand, as d E fl, f E dl. Thus,
f E Lg O dL. Finally, as c E gl and g E G ~ (l),
, ( c ) = #(g). By Lemma 5.7.1(ii) and Corollary 5.6.2(iii), g ~ E(L). Since e E gl and g E Goo(l), e E G~(I); see Lemma 5.7.1(i). Thus, using the above two references once again, we obtain also that c ~ E(L). Assume that f E E(L). Then, by (i) and Corollary 5.6.2(ii), f = 1. Thus, g E L, contrary to g E G~(l). Therefore, we must have f ~ E(L). Now, as d ~ E(L), Lemma 5.6.5(i) yields /~(d) = # ( f ) . It follows that #(d) = #(g) - 1. (iii), (iv) Assume that f ~ G~(l). Then, by definition, Ip(f(l))l ~ 1. Thus, there exists e E f(l) such that p(e) :fl p ( f ) = p(g) - 1. Now, by (ii), e E E ( / ) , whence f E el C_ E(L)I. This proves (iii). On the other hand, as f E el, #(e) ~ #(f) yields #(e) E {p(f) - 1,#(f) + 1}; see Lemma 5.5.4(ii). Assume that #(e) = #(f) - 1. Then e E D ( f ) N E(L), whence, by Lemma 5.7.2(ii), f E L. It follows that p(g*) = 2, which, by Lemma 5.5.3(iii), implies that #(g) = 2. On the other hand, as g E G~(l), Lemma 5.7.1(ii) yields g E Go(l). Thus, by Lemma 5.6.4(i), m _
For each l E L, we define
G~o(l) := {g E G~(l) I#(g) = m}.
P r o p o s i t i o n 5.7.4 Let h, k E L be such that h ~ k, and assume that G ~ ( h ) r O. Then, if ]X] E N, we have the following. (i) nh + 1 < nk.
(ii) 2 2fm. (iii) G ~ ( k ) = 0.
166
5. Theory of Generators
(iv) There exists e E E(L) such that p(e) = m and It(e(h) \ E(L)) = { m - 1}.
Proof. (i), (ii) Let g E G ~ (h), and let y, z E X be such that (y, z) E g. Let l E L be such that It(g) = Itt(g) For each x E z(h), set Yx := yl fq xRm-l(k). We shall first prove that, for each x E z(h), Yx r 0. Let x E z(h), and let f E G be such that (y,x) E f . Then f E g(h). Thus, by Lemma 5.7.1(i), f E G~(h). Now, by Lemma 5.7.1(ii) and Lemma 5.6.1(v), vh(f*) = 1. In particular, #k(f*) <_ #h(f*), and this implies that
It(f*) = I t k ( f ' ) . Since f E G~o(h), It(f) = m. Therefore, by Lemma 5.5.3(iii), It(f*) = m. Thus Itk(f*) = rn. It follows that f* E Rm(k). Therefore, as m r 0, there exists 11 E L such that f* E Rm-l(k)l I. Thus, by Lemma 1.2.2 and Lemma 1.4.5(i), f e llRm-l(k) *. It follows that
Itt'(f) <_ m -= It(f). On the other hand, we have It(g) = Itt(g). Thus, by Lemma 5.5.7, we must have It(f) = Itt (f),
too. (Recall that f e g(h).) Note that, by Lemma 5.7.1(ii) and Corollary 5.6.2(iii), f ~ E(L). Therefore, we conclude that 11 = I. Therefore, f E IRm-l(k)*, which implies that
Y~r We now shall prove that, for all x, x I E z(h), Yx M Y~:, ~ 0 implies that X ~ _ X I.
Let x, x' E z(h) be such that Y~ M Yx' # O. Let w E Y~ n Y~, be given. Let e, e', f , f ' E G be such that (y,x) E f , (y,x') E f ' , (w,x) E e, and
(w, z') ~ e'. Since w E Yx C x R m - l ( k ) , e* E Um-l(k). Thus, It(e*) < m 1. On the other hand, as f E g(h) and g E G ~ ( h ) , It(f) = m. Thus, by Lemma 5.5.3(iii), It(f*) = m. But f* E e*l. Thus, by Lemma 5.5.4(ii), m < It(e*) + 1. It follows that It(e*) = m - 1. Thus, by Lemma 5.5.3(iii), It(e) = m - 1. Similarly, we deduce that It(e I) = m - 1. Assume, by way of contradiction, that x ~ x ~. Then e ~ E eh. Thus, as It(e l) = It(e), e ~ Gl(h). But as It(e) = m - 1, we also have that e ~ G0(h); see Lemma 5.6.4(i). Therefore, by Lemma 5.6.1(i), e E G-l(h). This means that there exists d E eh such that -
It(d)
=
u(~)
-
i =
,-,, -
2.
5.8 A Characterization Theorem
167
On the other hand, as d 9 eh, e 9 dh. Thus, as f 9 le, f 9 ldh. Therefore, there exists c 9 ld such that f 9 ch. It follows that #(c) _< m - 1 and that c 9 f(h), contrary to f 9 G~o(h). Thus, we have shown that x = z'. Now we conclude from
U Y~ _c yz xqz(h)
that nh 4- 1 -- Iz(h)l _< lyll = n,. It follows that h ~ I. (Recall that we are assuming that IX] E I~l. Therefore, nh, nt 9 1~.) This means that l -- k, whence #(g) = /~k(g). This means that 0 < ~h(g). On the other hand, by Lemma 5.7.1(ii) and Lemma 5.6.1(v), vh(g*) = 1. Thus, we have 0 < vh(g*)~h(g). Now we obtain from Lemma 5.5.6(ii) that 2 { m. (Clearly, by Lemma 5.7.1(ii) and Corollary 5.6.2(iii), g ~t e(L).) (iii) follows from (i). (iv) Let g 9 e ~ ( h ) , and let f 9 D(g*)* be given. Then, by Lemma 5.7.3(iii), f 9 E ( L ) h U G~(h). But from #(g) = m and f 9 D(g*)* we also obtain that it(f) -- m - 1; see [,emma 5.5.3(iii). Thus, by Lemma 5.6.4(i), f ~ Go(h), whence, by Lemma 5.7.1(ii), f ~ G~(h). Thus, there exists e 9 E(L) such that f 9 eh. Thus, by Theorem 1.3.1, e(h) = f(h). Therefore, by Lemma 5.7.3(ii), p(e(h) \ E(L)) = {m - 1}. Moreover, by Lemma 5.5.5(iv) and Lemma 5.5.4(ii), m _< p(e) <_ p ( f ) 4- 1 = m. (Note that f ~ h because m ~ 2.) []
5.8 A Characterization
Theorem
As in the previous three sections, we assume in the present section that (L) = G and that ILl = 2. As before, we set m := re(L). In addition, we assume now that
Ixl 9 N, so that, by Lemma 5.2.5, m 9 Under this hypothesis, we shall completely describe the case where ]E(L) I = 2. In this case, the structure of (X, G) is severely restricted by the following proposition.
168
5. Theory of Generators P r o p o s i t i o n 5.8.1 If ]E(L)] = 2, m = m a x p ( G ) .
Proof. Since, by hypothesis, IE(L)I = 2, there exists e E E(L) \ {1} such that {e} = E(L) \ {1}. Since {e} = E(L) \ {1}, Lemma 5.5.5(iv) says that = m.
Set n := m a x # ( G ) . Then m < n. We now shall assume, by way of contradiction, that m+l
(z0, z.-m+2) E g*, (z0, z l ) E e,
and, for each j E { 2 , . . . , n -
m + 2}, zj E z j - l l j - 1 .
For each j E { 1 , . . . , n -
m + 2 } , let g u E G be such that (z0, zj) E glj.
(Note that gll = e and g1,,~-,~+2 = g*.) Then, for each j E { 2 , . . . , n - m + 2 } , (z0,zj-1) E gl,j-1 and zj E zoglj N Z j - l l j - 1 . It follows that, for each j E { 2 , . . . , n - m + 2}, aglflj_lgl,j_ 1 ~ O. For each i E { 2 , . . . , n -
m + 2}, we set
5.8 A Characterization Theorem
169
gil :~ gli" In order to simplify the notation, let us also set Y0 := Zl and Yl := z0. Our first claim is now that there exist y2, ..., y,~-m+2 E X such that, for each i E { 2 , . . . , n m + 2}, * Yi E y i - l l i - 1 N Yogi1.
From {e} = E ( L ) \ {1} we conclude that e' = see L e m m a Therefore, (Y0, yl) E e = g11. Thus, as ag12hgll ~ 0, there exists an element Y2 E yxll M Yog~l. This proves our claim for i = 2. Let i E { 3 , . . . , n - m + 2} be given, and assume, by induction hypothesis, that we have already found Yi-1 E X such that Yi-1 E yi-21i-2 M Yoga-l,1. Then, in particular, (Yo,Yi-1) E gl,i-1. Thus, as ag,,,l,_~,g~,,_~ ~ O, there exists an element Yi E y i - l l i - 1 f'l Yoga'l, so that we have shown our first claim. O u r second claim is now that, for each j E { 2 , . . . , n - m + 2}, P ( . q l j ) ---- 7n -- 2 -t- j.
From g l , n - m + 2 = g* and p(g) = n we obtain that
]2(gl,n_rn+2) : n; see L e m m a 5.5.3(iii). This proves our claim for j = n - rn + 2. Let j E { 2 , . . . , n m + 1} be given. Then gl,n-m+2
E
gljlj "" "In-m+1 C gljL n-m+2-j.
Thus, as P(gl,n-m+2) = n, L e m m a 5.5.4(ii) yields m - 2 + j < P(glj). Recall that g~2 E eL and that e E E(L) \ {1}. Thus, by L e m m a 5.5.4(ii) and L e m m a 5.7.2, #(g12) _< m, so that our claim is proved for j = 2. Let us now assume that 3 < j. Then
glj E g~212.. "lj-1 C g12L j-2. Thus, by L e m m a 5.5.4(ii),
P(gu) <- m - 2 + j. This proves our second claim. From L e m m a 5.5.3(iii) we now obtain that, for each i E { 2 , . . . , n - m + 2 } , we have P(gil) = m - 2 + i, too. (Recall that, for each i E { 2 , . . . , n - m + 2}, gil := g~i') Recall that g ...... +2,1 = g E G ~ ( h ) . Thus, by induction, we obtain that, for each i E { 2 , . . . , n - rn + 2}, gil E G ~ (h); see L e m m a 5.7.3(iv). In particular, g2~ E G ~ ( h ) . Thus, by Proposition 5.7.4(ii),
170
5. Theory of Generators 2~rn,
and Proposition 5.7.4(iv) yields
\ {e}) = {m - 1}. Since g21 e G~o(h), g21 r e; see L e m m a 5.7.1(ii) and Corollary 5.6.2(iii). Thus, as gl~ 9 gl~k C_ e(k} and P(gl~) = m, the last equation implies that
h•k. For all i, j 9 { 2 , . . . , n - rn 4- 2}, let gij 9 G be such that (y~, zj) 9 gij. Suppose that g ~ = e. Then, as h r k, g~3 9 e(h). Thus, by the above equation, g23 = e or tt(g~3) = m - 1. If g23 -- e, g32 9 D(g31) A E ( L ) , c o n t r a r y to L e m m a 5.7.2. (Recall that #(g31) = m + 1 and that p(e) = m.) If P(g23) = m - 1 , L e m m a 5.5.3(iii) and L e m m a 5.hA(ii)lead to the contradiction P(gl3) _< m. Therefore, we must have g22 ~ e. Assume that tt(g22) = m - 1. Since 2 f m , 2 [ m 1. Thus, g~2 9 R m - l ( k ) or g ~ 9 R m - l ( k ) . In the first case, g12 9 kg~2 C_ kR,~_l(k) C_ R,~-2(h) U R m - l ( k ) , which yields #(g~2) ~ m - 1, contradiction. In the second case, we obtain similarly the contradiction tt(g~l) _< m - 1. Thus, we have P(g~2) :P m - 1. From it(g21) = m and g2~ 9 g21L we now conclude that
P(g~2) 9 { m , m + 1}; see L e m m a 5.5.4(ii). Assume that P(g~2) = m + 1. Then, as 11 = k, {g~2,g~2} _C G - l ( k ) . Therefore, by L e m m a 5.6.1(iv), vk(g~2), gk(g22) <_ O. It follows that 0 < gk(g~2)Pk(g~2). On the other hand, as 2 ] m + 1, L e m m a 5.5.6(ii) yields vk(g~)v~(g~2) <_ O. Thus, uk(g~)vk(g22) = 0, which means that g22 9 E ( L ) . It follows that g22 = e. This contradiction yields #(g22) = m. Our final claim is now that, for all i, j 9 { 2 , . . . , n -
m + 2},
lt(gij) = m -- 4 § i § j. (This i m m e d i a t e l y will lead us to a contradiction.) We proceed by induction on i + j. Let i, j 9 { 2 , . . . , n m + 2} be given. Assume first that j = 2. If, in addition, i = 2, we are done. (We just showed that/-t(g22) = m.) If 3 _< i,
P(gi-l,j-1) = m -- 3 + i
5.8 A Characterization Theorem
171
and #(gi,j-1) = m -- 2 + i. On the other hand, by induction hypothesis,
~(gi-l,j) = m-4+i-
1 +j
= m - 3 +i.
(Recall that we are assuming that j = 2.) Thus, by L e m m a 5.6.5(i),
P(gij) = m -
2+ i = m-
4+i
+j.
(Apply this l e m m a to ( g i - l , j - l , g i , j - l , g i - l , j , g i j ) in the role of (c, d, e, f).) We obtain the desired conclusion similarly if we assume that i = 2. Therefore, we now assume that 3 _ i, j. T h i s time, we have, by induction hypothesis, that
~.(g~-la-1) + 1 = #(g~_~,.~) and that p ( g i , j - 1 ) = m - 5 + i + j. Thus, by L e m m a 5.6.5(ii), , ( g i j ) = p(gi,j-1) + 1 = m - 4 + i + j. (Apply L e m m a 5.6.5(ii) t o (ffi-l,j-l,gi,j-l,gi-l,j,gij) instead of (e, d, e, f).) This proves our final claim. From our a s s u m p t i o n 1 _< n - m we now obtain that
~(g) + 1 = n + 1 < 2n - m = ~(gn-m+2,n-m-l-2), c o n t r a r y to the choice of g.
Theorem
5.8.2
[]
A s s u m e that IE(L)I = 2. Then we have
(i) U~L U~{o ..... m} Rj(t) = a. (ii) Let ~ e E(L) be such that {1, ~} = E(L). The,, there e~ist h, k ~ L such that h • k, R m ( h ) = {e}, and Rm(k) = Go~(h) U {e} C e(k>. (iii) Let I E L, and let j E { 0 , . . . , m 1} be given. Then IRj(l)] = 1.
Proof. (i) Let g E G be given. Set j := #(g), and let l E L be such that p(g) = / ~ ( g ) . T h e n g E Rj(l) and, by Proposition 5.8.1, j < m. (ii) We define M:={geGl#(g) =m}. Then, by definition, M C_ (.JteL R,~(l). On the other hand, for each l E L, # ( R m ( l ) ) = {m}; see L e m m a 5.5.2(ii). Thus, we have
172
5. Theory of Generators
M=URm(l). IEL
Note also that, by Proposition 5.8.1, M = {g E G I P(g) = max#(G)}. Therefore, by Lemma 5.7.1(iii), U C_ U Go~(l) U {e}. IEL
Assume first that, for each l E L, G~r = 0. Then M C_ {e}, so that we are done in this case. (Recall that, by Lemma 5.5.5(iv), #(e) = m, i.e.,
eEM.) Assume now that there exists h E L such that Goo(h) # 0. Let I E L be given. From Lemma 5.7.1(ii) we know that Goo(1) .C Go(l). Therefore, by Proposition 5.8.1 and Lemma 5.6.4(i),
G~ (l) = Goo(1). It follows that Go~ (h) # 0. Now we are in the position to apply Lemma 5.7.4. From Lemma 5.7.4(ii) we obtain that
2~m, Let k E L be such that {h, k} = L. Then, by Lemma 5.7.4(iii), Go~ (k) = 9. (Recall that G~(k) = Goo(k).) Thus,
M = ao~(h) u {e}, so that, in particular, Rm(k) C_ Go~(h) U {e}. We still have to prove that Rm(h) C_ {e} and that Goo(h) c_ e(k). Let us first prove that Go~(h) r e(k). Let g E Go~(h) be given. Then we conclude from M = Go~(h) U {e}, M* = M, and e* = e that g* E Go~(h). Thus, by Lemma 5.7.1(ii) and Lemma 5.6.3,
gEeL. (Recall that G~(h) = G~(h). Therefore, #(g*) = m.) Assume that g E eh. Then, by Lemma 1.2.5(i) and Lemma 1.4.5(i), e E gh. But, by Corollary 5.6.2(ii), e E G-l(h). Thus, there exists f E eh such that # ( f ) = m - 1. Now we have f , g E eh. Thus, by Theorem 1.3.1, f E g(h), contrary to g E G~(h). Therefore, we must have that g E e(k). Since g E G ~ (h) has been chosen arbitrarily, we have shown that
a (h) C
5.8 A Characterization Theorem
173
Now let g E R,~(h) be given. Then, as 2 { m, there exists f E R,~_l(h) such that g E fh. It follows that f E gh and/~(f) < ph (f) < m - 1, whence g E G-l(h). Thus, as Rm(h) C M C_ Gw(h)U{e},g = e; use Lemma 5.7.1(ii) once again. Since g E R m (h) has been chosen arbitrarily, we have shown that R,~ (h) C_ {e}. Thus, as/t(e) = m, we have n
(h) =
(iii) By Theorem 5.2.6(iii), we shall be done if we succeed in showing that I R m - a ( l ) l = 1.
Let f E R,~_I(I) be given. Then, by Lemma 5.5.2(ii), p ( f ) = m - 1. Thus, by Lemma 5.6.4(ii), there exists g E f L such that p(g) = m. Let e E E(L) be such that {1, e} = E(L). We wish to prove that f E eL. If g = e, this follows from g E f L . Therefore, we assume that g # e. Then, by (ii), there exist h, k E L such that h # k, g E Go(h), and g E ek. (Recall that #(g) = m.) From g E G~(h), f E gL, and p ( f ) + 1 = p(g) we conclude that f E gk. Thus, as g E ek, f E ek C_ eL. Since f E R,~_ 1(l) has been chosen arbitrarily, we have shown that
Rm-l(l) C eL. Now Lemma 5.6.4(i) yields IRm_i(l)l = 1.
[]
The following result gives us an algebraic condition which implies that IE(L)I = 2.
T h e o r e m 5.8.3 Let R be an algebraically closed field with char(R) = 0. Assume that R[L] = R[G]. Then ]E(L)] = 2.
Proof. Let h, k E L be such that h 7~ k. We first shall prove that IGI < [G/(h)l + IG/(k)l and
I l G / ( h ) l - IG/(k)ll< 1. For each i E { l, 2}, let Mi denote the sum of the/-dimensional submodules of (RG) ~. Then, by Theorem 4.4.4,
RG = M1 • M2. Let l E L be given. Set crt := cry, and define
R(t) := {a E R G I ~r~t = nl~}.
174
0. •
5. Theory of Generators Let 0. E R(t) be given. Then there exist 0.1 E M1 and ~r2 E Ms such that 0.1 -t'- 0"2. Since 0. E Rq), 0.0.t = hi0. = n~(0.1 + 0.2) = n w l
+ hi0.2.
On the other hand, 0.0.i = (0.1 + 0.2)0.~ = 0.10.1 + 0.20.t.
Thus, as R G = M1 (])M2, we obtain that o'10.t = nt0.1 and ~2~rt = nw2. Since 0. E R G has been chosen arbitrarily, we have shown that Rq) = (M~ ,q Rq)) | (Ms r3 Rq)). Let i E {1, 2} be given. We set mi := dimn(Mi), and, for each l E L, we define d~ := dimn(Mi VI Rq)). Since R G = M1 | M2, this notation yields I a l = m i + m2.
Moreover, for each l E L,
la/(t)l = d{ +
d~;
recall that R(t) = (M1 n R(l)) @ (Ms A Rq)), and use Theorem 4.4.7. From Corollary 4.4.3(ii) we obtain that, for each l E L, 1
~m2 = d'~. From Corollary 4.4.3(ii) we also deduce that ml <_ E
dll.
IEL
(Recall that R G possesses a 1-dimensional right ideal which affords the character lnv.) Thus, we have
lal _< y~(d~ + d~) = ~ la/q)l. IEL
IEL
From Corollary 4.4.3(ii) we finally deduce that, for each l E L, d~ E {1, 2}. Thus, I I G l ( h ) l - I G l ( k ) l l = Idhl - d~l < 1. Let us now define
5.8 A Characterization Theorem
175
c := U a/(O, IEL
1" := {(g(h),g(k)) I g 9 a } , and F := (C, r). Then, by Theorem 4.4.1(ii), the graph F is connected. Since m 9 N, F cannot be a tree. Thus,
Icl _< i~i. Moreover, the definition of r yields
ki _< lal. Finally, as Ie I <_ JG/(h)l + IG/(k)l, we have
ICl <_ ICl. It follows that
Icl = Id = far. In particular, for each g 9 G, g(h) N g(k} = {g}. Therefore, we conclude that IE(G)I
= 2.
[]
In the present section, we assume that IXI E N. Thus, by Theorem 5.2.6(v), (X, G) is a Coxeter scheme with respect to L if and only i f 2 m = lal. The pair (X, G) will be called a Moore scheme with respect to L i f 2 m + l =
laB. The last two results of this section will be given without proof. The reason for stating them here is to show that, to some extent, the case IE(L)I = 2 can be discussed further. With the help of Theorem 5.8.3 we obtain the following characterization of finite Coxeter schemes with respect to L and of Moore schemes with respect to L.
T h e o r e m 5.8.4 (i) Assume that HE(L)[ = 2 and that ( X , G ) is not a Coxeter scheme with respect to L. Then there exists h E L such that G ~ (h) 0 and, setting t := OgeO~(h) g, (X, [ a \ a ~ ( h ) ] U {t})
is a Moore scheme with respect to L. (ii) Assume that ( X , G ) is a Coxeter scheme or a Moore scheme with respect to L. Then [E(L)[ = 2.
Without proof, we mention here that, if (X, G) is a Moore scheme with respect to L, then m E {3, 5}. The long and difficult proof for this fact follows
176
5. Theory of Generators
from [10; Theorem 3], [4; Theorem 2], [11; Theorem 1], and from the natural embedding of the class of Moore geometries into the class of schemes. As a corollary of Theorem 5.8.3 and Theorem 5.8.4(i), we obtain the following result, which is [34; Theorem B].
C o r o l l a r y 5.8.5 Let R be an algebraically closed field with char(R) = 0, A s s u m e that R[L] = R[G]. Then (X, G) is a Coxeter or a Moore scheme with respect to L.
We now shall leave the structure theory of schemes.
Appendix
In the introduction of these notes, we claimed that groups as well as buildings can be viewed as distinguished classes of schemes. The purpose of this appendix is to prove these claims. Let us start with groups. Let O be a group. For each 0 E O, we define ::{(r162 We set O := {0 I 0 9 e } and
7(0) := (o, 8). We set 6 ( X , a ) := {{g} I g 9 a } .
T h e o r e m A (i) Assume that (X,G) is thin. Then ~ ( X , G ) is a group with respect to the complex multiplication and with {1} as identity element. (ii) Let 0 be a group. Then 7-(0) is a thin scheme. (iii) If (X, G) is thin, T(G(X, G)) -~ (X, G). (iv) For each group O, G(T(O)) -~ O.
Proof. (i) From Theorem 1.2.7 we conclude that the complex product of two elements of 6(X, G) is again an element of ~(X, G). Thus, by Lemma 1.2.1(ii), 6(X, G) is a semigroup. From Lemma 1.1.1(i) we obtain that {1} is the identity element ofF(X, G). Finally, as (X, G) is assumed to be thin, we deduce from Lemma 1.2.6 that, for each g E G, {g*}{g} = {1} = {g}{9"}. (ii) 1 First of all, it is clear that i = l o and that, for each 0 E O,
~, __ 0_'-'-i, But also the regularity condition is easily verified for the pair (O, 8). Let/3, 7, c, ~', r/ E O be given, and assume that (/3, 7) E 7). Then, /3g M 7~* r O if 1 We repeat the argument given in the introduction of these notes.
178
Appendix
and only if fie
I gn 7r
=
7( -1
if and only if 3e(" = 7 if and only if e~" = r/. Thus,
= a<,. 2
(iii) Set O := ~ ( X , G ) , and fix v E S . For each z E X, we denote by z r the uniquely determined element of O such that z E v(zr For each g E G, we define
gr := {g}. Then
XUG
r
~
OU~9
is a well-defined bijective map with X r C O and Gr _C 8. Let y, z E X, and let g E G such that (y,z) E 9 be given. By Lemma 1.7.3(iii), we shall be done if we succeed in showing that (yr zr E gr Let e E G be such that (v,y) E e, and let f E G be such that (v, z) E f . Then, by Lemma 1.2.4, f E {e}{g}. From this we conclude that {f} = {e}{g}; see Theorem 1.2.7, and recall that (X, G) is assumed to be thin. Thus, by definition,
({e}, {S)) E {g). On the other hand, as (v,y) E e, {e} = yr Similarly, as (v,z) E f , {f} = zr Therefore, we conclude that (yr zr E gr (iv) For each 0 E O, we define
0r := {g}. Then r is a well-defined bijective map. Let ~', r/E O be given, and set 0 := fir/. We shall be done if we succeed in showing that 0r = ffCr/r First of all, since O / = 0, (if, 0) E_~. Moreover, we also have (1, r E ~ and (1, 0) E 0. Thus, by Lemma 1.2.4, 0 E ~f/. But, by (ii), n~ = 1 = n~. Thus, by Theorem 1.2.7, {g) = It follows that 0r = ~r162
[]
Let us now see how buildings can be viewed as a distinguished class of schemes. Let C be a set, and let r C C • C be given. We shall say that r is regular if, for all a, b E C, la,'l = Ib,'l. Assume that r is regular, and let c E C be given. Then Icrl is called the valency of r. For each s C_ C x C, we define 2 Here 5 is the Kronecker delta.
Appendix
179
r o s : = {(a,b) E c x c I arnbs* # 0}. T h e element 7" is called an incidence relation of C if lcnr=O#r=r* and
rorC
1cUr.
We shall denote by Inc(C) the set of the incidence relations of C. 3
L e m m a B Let C be a set, and let R be a partition of C x C such that 0 ~ R. Let I E Inc(C) be given, assume that I is regular, and let n denote the valency of I. Let s, t E R be such that s o l = t # s. Let a, b E C, and let r E R be such that (a, b) 5: r. Then l a s n bll = s and
]at n bll =
n n -1 0
if r = s, ifr =t, if r e {s,t}.
Proof. Clearly, if r = t, as n bl # O. Conversely, assume that as 0 bl # 0. Then (a,b) ~ r n (sol) = r n t , whence r - t. Assume that 2 _< [as n bl[. Let d, e E as n bl be such that d r e. Then, since l = l* and l o I _C 1c U l, (d, e) E l. Thus,
(a,e) E ~ n ( s o l ) = s n t . It follows that s -" t, c o n t r a r y to our hypothesis. T h u s we have shown that las n bl I = 5rt. Let us now c o m p u t e l a t n bl I. If r = s, bl C a(s ol) = at, whence l a t n bl I = n. Now assume that at n bl # 0 and that r r s. Let d E at n bl be given. Since t C sol, asndlr Let e E a s n d l b e g i v e n . Then e l C a ( s o l ) = a t . On the other hand, as (e,b) E l o l _C 1c U1 and s ~ r, b E el. Thus, r = t and bl \ {e} C_ el C at. It follows that [at n bll = n - 1. [] 3 There is an obvious bijection between Inc(C) and the set of the non-identity equivalence relations of C. For each r E Inc(C), l c # l c O r and l c U r is an equivalence relation of C. Conversely, for each equivalence relation r of C such that l c # r, r \ l c E Inc(C).
180
Appendix
Let C be a set, and let L C Inc(C) be given. Then the pair (C, L) will be called a chamber system. 4 It is called regular if each element of L is regular. Let (C, L) be a c h a m b e r system. Recall that we denote by F ( L ) the free monoid constructed on L. By * we denote the multiplication of F ( L ) . T h e identity element of F ( L ) is denoted by 1. Let n E 1~ \ {0}, let co, ..., c,~ E C, and let ll, ..., l,~ E L be given. Set f := la * . " * l , . T h e n ( c o , . . . , cn) is called a gallery of type f (from co to en) if, for each i E { 1 , . . . , n } , (ci-l,ci) E li.5 For each c E C, (c) will be called a gallery of type 1. Recall that we denote by R ( G ) the monoid of all n o n - e m p t y subsets of G with respect to the c o m p l e x multiplication. Moreover, for each L C_ I n v ( G ) , PL denotes the uniquely determined monoid h o m o m o r p h i s m from F ( L ) to R ( G ) such that, for each l E L, lpL = {l}. L e m m a C Let L C I n v ( G ) be given. Then we have (i) (X, L) is a regular chamber system. (ii) Let y, z E X , and let g E G be such that (y, z) E g. Let f ~ r ( L ) be given. Set p := PL. Then 9 E fp if and only if (X, L) possesses a gallery of
type f from y to z. Proof. (i) From L e m m a 1.4.5(i), (ii) we obtain that Inv(G) C_ I n c ( X ) . This proves (i). (ii) Clearly, we have g E l p if and only if g = 1 if and only if y = z. Therefore, the claim is obvious if 1 = f. A s u m e now that 1 ~ f. T h e n there exist n E 1~ \ {0} and 11, ..., In E L such that f = 11 * - - - * in. I f g E fp, g E 11 . . . I n . Thus, by L e m m a 1.2.4, there exist z0, ..., xn E X such that z0 = y, xn = z, and, for each i E { 1 , . . . , n } , ( x i - l , x i ) E l i . T h i s m e a n s that there exists a gallery of type f f r o m y to z. Conversely, assume that there exists a gallery of type f f r o m y to z. T h e n there exist z0, ..., Xn E X such that x0 = y, xn = z, and, for each i E { 1 , . . . , n } , ( x i - l , x i ) E Ii. Thus, by L e m m a 1.2.4, g E l t . . . l ~ . (Recall that (y, z) E g.) It follows that g E fp. []
Let (C, L) be a c h a m b e r system, and let m be a Coxeter m a p of L. 4 Formally, our definition of chamber systems differs slightly from the usual one; see, e.g., [28]. Usually, the pair (C, L) is called a chamber system if L is a family of partitions on C. But, as mentioned earlier, non-identity partitions on C and incidence relations of C correspond to each other uniquely. Thus, our chamber systems correspond to those chamber systems in the sense of [28] which neither contain the identity partition nor have repeated partitions. 5 What we call a "gallery" here is the same as a "simple gallery" in [28].
Appendix
181
Recall that Fm (L) denotes the set of the m-reduced elements of F(L). We shall denote by Fro(L) the Coxeter group defined by m. By em we shall denote the natural monoid homomorphism from F(L) to Fro(L). The pair (C, L) is called a building of type m if (C, L) is regular and if there exists a map 5 from C x C to Fro(L) such that, for all a, b E C and, for each f E F~n(L), 6(a, b) = fCm if and only if (C, L) possesses a gallery of type f from a to b. The pair (C, L) is called a building if there exists a Coxeter map m of L such that (C, L) is a building of type m. 6 Assume now that (C, L) is a building. Then, by definition, there exist a Coxeter map m of L and a map 6 from C x C to Fro(L) as above. For each 3, E Fro(L), we define r-~ := 5 - 1 ( 7 ) .
For each 3, E Fro(L), we shall denote by 3,~ the length of 3, as an element of the Coxeter group Fro(L).
L e m m a D Let m be a Coxeter map, and let ( C , L ) be a building of type m. Let T, A E Fro(L) be such that Ae = 1 and Tt~ + 1 = (TA)L Then r. t o r), = r~),.
Proos Let us abbreviate Let f E Fm (L) be such that fr and let l E L be such that 1r = A.
Then the hypothesis 3't + 1 = (3,,k)e implies that f * l E Fro(L). Now let a, b E C be given, and let us assume first that (a, b) E r-yx. Then, by definition, 5(a, b) = 3,A = f r 1 6 2= ( f , / ) r Thus, as f , l E Fro(L), (C, L) possesses a gallery ( c o , . . . , en), say, of type f * l from a to b. Set c := cn-1. Then, ( c 0 , . . . , c , - 1 ) is a gallery of type f from a to c, and (c~-1, c,) is a gallery of type l from e to b. Therefore, by definition, 5(a, c) = f r = 3, and 5(c, b) = Ir = A. It follows that (a, c) E r~ and (e, b) E r~. Thus, (a, b) E r~ o r~. Let us now assume that (a, b) E r~ o r~. Then there exists c E C such that (a, e) E r~ and (c, b) E r~. It follows that 5(a, e) = 3' = f r and ~(c, b) = A = 1r 6 As indicated in the introduction of these notes, our definition of buildings is midway between the two definitions given in [27] and [28].
182
Appendix
Thus, by definition, (C, L) possesses a gallery ( c o , . . . , c,~), say, of type f from a to c and a gallery of type l from c to b. Now ( c o , . . . , e,,, b) is a gallery of type f , l from a to b. Thus, as f , l E F,~(L),
8(a, b) = (f * l)r = f r 1 6 2= 3~A. It follows that (a, b) E r-yx.
[]
For each building (C, L), we set R(C, L) := {r~ I ~ E s
and
.A(C, L) := (C, R(C, L)). For each L C Inv(G), we define
BL(X, G) := (X, L). The second part of the following theorem uses Theorem 3.1.5(ii).
T h e o r e m E (i) Let L C I n v ( a ) be given, and assume that ( X , G ) is a Coxeter scheme with respect to L. Then BL(X, G) is a building of type m L .7 (ii) Let m be a Coxeter map, and let (C, L) be a building of type m. Then .A(C, L) is a Coxeter scheme with respect to L. Moreover, m = mL. (iii) Let L C_ Inv(G) be given, and assume that (X, G) is a Coxeter scheme with respect to L. Then A ( B L ( X , G)) = (X, G). (iv) For each building (C, L), BL(A(C, L)) = (C, L).
Proof. (i) From Lemma C(i) we know already that ISL(X, G ) i s a regular chamber system. Let us abbreviate m := mL and p := PL. Let y, z E X, and let g E G be such that (y, z) E g. Then, by Proposition 5.1.3(ii), there exists f E F,,,(L) such that g E fp. We define 8(y, z) := fern. Let us first convince ourselves that 8 is a well-defined map from X • X to Fro(L). Let y, z E X , let g E G such that (y,z) E g, and let d, e E Fro(L) such that g E dp n ep be given. Assuming that (X, G) is a Coxeter scheme with respect to L, we have, in particular, that (X, G) is L-constrained. Therefore, as dp (] ep ~ 0, we obtain that dp = ep. This gives d .~,~ e. (We use once again that (X, G) is a Coxeter scheme with respect to L.) It follows that dem = ecru. 7 Recall that mL is a Coxeter map which was introduced in the beginning of Section 5.1.
Appendix
183
Let y, z E X, and let f E F,~(L) be given. We have to prove that ~f(y, z) = fern if and only if I3L (X, G) possesses a gallery of type f from y to z. Let g E G be such that (y,z) E g. Assume first that ~(y, z) = fern. Then the definition of ~ yields g E fp. Thus, by Lemma C(ii), BL(X, G) possesses a gallery of type f from y to z. Conversely, assume that BL (X, G) possesses a gallery of type f from y to z. Then, by Lemma C(ii), g E fp. Now, as (y, z) E g, we have $(y, z) = fern. (ii) Since (C, L) is assumed to be a building of type m, there exists a map ~i from C x C to Fro(L) such that, for all a, b E C and, for each f E Fro(L), 6(a, b) = fern if and only if (C, L) possesses a gallery of type f from a to b. Let us abbreviate R := R(C, L). Clearly, ~i is surjective. Therefore, ~ ~ R. Moreover, 1c = r1 E R and, for each 3' E Fro(L), (r~)* = r.y-, E R. Thus, in order to prove that A(C, L) is a scheme, we only have to verify the regularity condition for .A(C, L). Let r, .q, I E R be given. Let a, b E C be such that (a,b) E r. In order to prove that ,4(C, L) satisfies the regularity condition, we have to show that the cardinality of ag f3 bl* does not depend on the choice of (a, b) E r. Let 3' E Fm (L) be such that
and let A E Fro(L) be such that l=rx. There is nothing to prove if l c = I. Thus, we may assume that 1c ~ I. Moreover, by induction on At, we may assume that At = 1. Assume first that 3'g + 1 = (TA)g. Set t := rT~. Then, by Lemma D,
g o l = r~or~ = r ~ = t. Thus, by Lemma B, lag M bl I = 6~t. (Set s := g.) Assume now that (TA)g + 1 = 7g. Set s := rTx. Then, by Lemma D,
s o l = r.~xor~ = r.y = g . Thus, by Lemma B, lag M bl] does not depend on the choice of (a, b) E r. (Set t := g, and use the hypothesis that l is regular.)
We now have proved that .A(C, L) is a scheme, and we still have to prove that .A(C, L) is a Coxeter scheme with respect to L. (Note that L C_ Inv(R).) Let R(R) denote the monoid of all non-empty subsets of R with respect to the complex multiplication (in R). Let us denote by p' the uniquely determined monoid homomorphism from F(L) to R(R) such that, for each l E L, lp' =
{1}.
We first shall prove that .A(C, L) is L-constrained, in other words that, for each f E Fro(L), Ifp'l = 1. (It is clear that (L) = R.)
184
Appendix
Let f E Fro(L) be given. Then fern E F,n(L). Set r := ( f - l ( f e m ) . Then rER. Let a, b E C be such that (a, b) E r. Then, by definition, $(a, b) = fern. Thus, by definition, (C, L) possesses a gallery of type f from a to b. Thus, by L e m m a C(ii), r E fp~. Let s E fpl be given. Then s E R. Therefore, there exists e E Fro(L) such that s = 5-1(eCru). Let a, b E C be such that (a,b) E s. Then, by definition, ~(a, b) = er Thus, (C, L) possesses a gallery of type e from a to b. Thus, by L e m m a C(ii), s E ep'. But, as s E fp', (C, L) possesses a gallery of type f from a to b; use L e m m a C(ii) once again. Thus, ecru = fr whence r = s. Since s E fp~ has been chosen arbitrarily, we have shown that {r} = fp~. Since f E Frn(L) has been chosen arbitrarily, we have shown that .A(C, L) is L-constrained. Let d, e E Fro(L) be such that dp' = ep'. Let r E dp ~, and let a, b E C be such that (a, b) E r. Then, by L e m m a C(ii), (C, L) possesses a gallery of type d from a to b as well as a gallery of type e from a to b. It follows that d e m = er Therefore, d ~,~ e. Thus, by Theorem 3.1.5(ii), d "~m e. Since d, e E Fro(L) have been chosen arbitrarily, we now have proved that .A(C, L) is a Coxeter scheme with respect to rn. (iii) Let us abbreviate R := R ( X , L). Then we just have to prove that G = R. Let R ( R ) denote the monoid of all non-empty subsets of R with respect to the complex multiplication. Let us denote by p' the uniquely determined monoid h o m o m o r p h i s m from F(L) to R ( R ) such that, for each l E L, lp' =
{t}.
Let g E G, and let r E R be such that g N r ~ 0. We shall prove that g = r. Let y, z E X be such that (y, z) E g N r. By Proposition 5.1.3(ii), there exists f E F ~ (L) such that {a} = fp. Thus, (X, L) possesses a gallery of type f from y to z; see L e m m a C(ii). From L e m m a C(ii) we now deduce that r E fp'. Thus, as (X, R) is L-constrained, we obtain that = fp'. Let v, w E X be such that (v, w) E g, and let s E R be such that (v, w) E s. Since (v, w) E g, (X, L) possesses a gallery of type f from v to w; see L e m m a C(ii). Thus, using L e m m a C(ii) once again, we conclude that s E f p ' = {r}. It follows that r = s. In particular, (v, w) E r. Since v, w E X have been chosen arbitrarily such that (v,w) E g, we conclude that g C_ r. Similarly, we obtain that r C_ g. It follows that g = r. Since g E G and r E R have been chosen arbitrarily, we have shown that
G=R. (iv) Since (C, L) is assumed to be a building, ,4(C, L) is a Coxeter scheme with respect to L; see (ii). Thus, by definition, BL(.A(C, L)) = (C, n). []
Index
Acting (group) on (a scheme) 60 Adjacency algebra (of a finite scheme) over (a field) 98 Adjoint (vector-space endomorphism) relative to (a hermitian form) 92 A-homomorphism (of an A-module) 78 Algebra over (a field) 77 A-module (for an algebra A) 77 Association scheme V Automorphism (of a scheme) 60 (B, .A)-correspondence VIII Building of type m 181 Building 181 Centralizer (of a non-empty subset of G) in (a subset of G) 33 Chamber system 180 Character (of an algebra A) afforded by (an A-module) 85 Closed p-subset (of G) 47 Closed (subset of G) 10 Commutative (scheme) 51 Completely reducible (A-module) 79 Complex multiplication (of two subsets of G) Vl Complex product (of two subsets of G) VI Composition factor (of a scheme) 36 Composition series (of G) 35 Constrained (scheme) IX Coxeter map 65 Coxeter scheme with respect to (a set of involutions) 127 Coxeter scheme VII Decomposable (closed subset of G) 50 Direct product (of closed subsets of G) 48 Direct product (of schemes) 55
Double coset (of a closed subset of G) in (G) 11 Factor module (of an A-module) with respect to (a submodule) 78 Factor scheme (of (X, G)) over (a closed subset of G) 21 Finite (scheme) VIII Frattini subset (of G) 17 Fusion (of a scheme) 28 Gallery of type f 180 Generating subset 15 (G, T)-correspondence VIII Homogeneous (A-module) 80 Homomorphism (of a scheme) 28 Ideal (of an algebra) 80 Incidence relation 179 Indecomposable (closed subset of G) 5O Indecomposable (scheme) 58 Integral (ring element) over (a subring) 72 Involution 18 Irreducible (character) 85 Irreducible (A-module) 79 Isomorphic (A-modules) 78 Isomorphic (composition series) 35 Isomorphic (schemes) 29 Jacobson radical (of an algebra) Kernel (of a homomorphism)
84
28
L-constrained (scheme) 124 Left coset (of a closed subset of G) in (X (respectively, G)) 11 Linear (character) 85 /-length (of an element of G) 155
186
Index
/-norm (of an element of G)
Supplement (of a closed subset of G) in (a) 17
157
Maximal (closed subset of G) 17 Maximal (submodule) 78 Minimal (ideal of all algebra) 80 Minimal (normal closed subset of G) 31 Moore scheme with respect to (a set of involutions) 175 Morphism from (a scheme) to (a scheme) 26 m-reduced (element of a free monoid) 67 Natural (homomorphism) 30 Normal (closed subset of G) in (a closed subset of G) 31 Normafizer (of a non-empty subset of G) in (a subset of G) 31
Upper neighbour (of a closed subset of G) in (a composition series of G) 35 Valency (of a regular binary relation) 178
Polarity (of a scheme) 150 Positive-semidefinite (vector-space endoxnorphism) 93 Primitive (scheme) 40 Principal character 102 p-scheme 47 p-valenced (subset of G) 22 Quasi-direct product (of schemes)
Tensor product (of two hermitian forms) 91 Tensor product (of two vector spaces) 87 Thick (scheme) IX Thin (non-empty subset of G) VI Thin (scheme) VI Thin radical (of a closed subset of G) 43 Thin residue (of a closed subset of G) 39
55
Rank (of a Coxeter scheme) 127 Regular (binary relation) 178 Regular (chamber system) 180 Regularity condition V Residually thin (scheme) 44 Scheme V Semidirect product (of a scheme) with (a group) 62 Semisimple (algebra) 80 Simple (algebra) 80 Simple (scheme) 31 Standard basis 97 Standard character 98 Standard module 98 Strongly normal (closed subset of G) in (a closed subset of G) 37 Structure constant (of a scheme) V Submodule (of an A-module) 77 Subnormal series (of G) 35 Subscheme (of (X, G)) with respect to (an element of X and a closed subset of G) 19
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