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MacDonald characters of Weyl groups of rank ≤4 Andreassian, Agnes 1973-03-22

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MACDONALD CHARACTERS OF WEYL GROUPS OF. RANK <4 by AGNES ANDREASSIAN B. S.,M. S.,M.A. , American University of Beirut, 1968 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in the department of MATHEMATICS V7e accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA August 1973 In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may he granted by the Head of my Department or by his representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission. Department of MoiMruL^AAi- cs The University of British Columbia Vancouver 8, Canada Date R^dt 53 , 19 73 Supervisor: Dr.B.Chang ABSTRACT In this thesis we obtain all the ordinary irreducible characters of Weyl groups of rank <_4 by using MacDonald' s method.This method enables us to find almost all the characters, and the remaining ones may be obtained by combining MacDonald characters and characters of exterior products of the reflection representation. iii TABLE OF CONTENTS INTRODUCTION A2 A3 A4 D . B2 and C2 and C^ B4 and C4 BIBLIOGRAPHY 1 4 5 7 11 18 20 26 41 61 iv LIST OF TABLES Table I: Character Table for (A2) . . .4 Table II .-Character Table for (A3) . . .6 Table III: Character Table for (A4) . . .10 Table IV: Character Table for (D4) . . .16 Table V: Character Table for (B2)=(C2) . . .19 Table VI: Character Table for (B3)=(C3) . .24 Table VII: Character Table for (B4)=(C4) . .38 Table VIII: Character Table for (F.) . .58 V ACKNOWLEDGMENT I would like to express my gratitude to Dr. B.Chang for suggesting the topic of this thesis and for all the help he so readily offered during its preparation.I would also like to thank Dr.R.Ree for reading the thesis. 1 INTRODUCTION In a paper entitled "Some Irreducible Representations of Weyl Groups" [5],I.G.MacDonald describes the following construction which gives many,but in general not all of the irreducible representations of a Weyl group. Let V be a finite-dimensional vector space over the rational field,and suppose V has a positive-definite inner product.Let V* be the dual of V.Let R be a root system of a Weyl group (R) and let S be a subsystem of R.For a given ordering on R,let n(S) denote the product of all the positive roots of S.Then II (S) is. a homogeneous rational-valued polynomial function on V.The space of all rational-valued polynomial functions on V is the symmetric algebra E=Sym(V*).Let the Weyl group (R) act on E,and let P(S) denote the subspace of E spanned by the polynomial functions g(n(S)) for all g in (R).MacDonald proves that P(S) is an absolutely irreducible (R)-module. The purpose of this thesis is to use this construction to compute the irreducible characters of the Weyl groups of rank <4.We wish to find out which characters can be so obtained and whether the missing characters are related to those obtained.The characters that are obtained by this method will be called •MacDonald characters. Using this method we find all the irreducible characters in the case of the groups (A^).We find that neither the subsystems of alone nor the subsystems of C^ alone provide all the irreducible characters of the group (B^)=(C^).However, using the subsystems of both B. and C. we obtain all the.' - -• irreducible characters.In the case of (D.),the method gives 2 eleven of the thirteen irreducible characters.The exterior products of the reflection representation provide the missing characters.In the case of (F^) we find seventeen characters.The remaining eight can be written as products of two MacDonald characters. We describe below the notation used and the method of obtaining all the subsystems of a given root system of a Weyl group. We use Dynkin diagrams to represent root systems and the set of positive roots in terms of orthonormal basis as given by N.Bourbaki [1] .As in [2] ,in listing the subsystems of and ,we distinguish between a diagram with long roots and one with shorter roots by using S to denote a Dynkin diagram with long roots and S one with shorter roots.We do the opposite for subsystems of C^. The conjugacy classes and the orders of the centralizers for (A^) , (B^) or (C^) for i=2,3,4,and (D^) are staightforward to find.For (F^) we use the conjugacy classes obtained by R.W.Carter [2],and in all cases we use Carter's notation for conjugacy classes.For the sake of convenience,we also use permutation notation and sign changes. In the tables,the entries under h^ denote the numbers of elements in the conjugacy classes.denotes a character. For any root a,w denotes the reflection defined by a.We use the notation V(a,,...,a ) to denote the Vandermonde 1 ' n determinant of order n, . n-1 a n Given a root system R,we use Dynkin1 s method [3] to obtain all the subsystems,We start by adjoining to the root system R the minimal root of (R).The diagram so obtained is called the extended Dynkin diagram for R.By deleting roots from the extended Dynkin diagram we obtain subsystems. Repeating the process with the subsystems obtained we eventually find all the subsystems of the given root system. l.A2. Subsystems of ; The extended Dynkin diagram for is -(x1-x3) x1-x2 x2_x3 From this diagram we obtain the following subsystems (1) A0: Xl X2 X2 X3 (2) A, Xl X2 We note that II(A2) =-V (x1 ,x2 ,x^) where V(x-^,x2,x3) is the Vandermonde determinant of order 3. Table I: Character Table for (A2) A, Al Conjugacy Class Representative Characteristic Polynomial h. I \ \ $ (1) x2-2x+l 1 1 1 2 Al (12) x2 -1 3 1 -1 0 A, (123) x2 + .x+1 2 1 1 -1 5 2.A3. Subsystems of A., The extended Dynkin diagram for A^ is r (XT_-X4) From this diagram we obtain the following subsystems: (1) A^ : o o o xl x2 X2 X3 X3~X4 (2) 2AX (3) A2: (4) A1 P(A^): We have Xl X2 . X3_X4 Xl~x2 X2 X3 Xl x2 II (A^ ) =V (x^ ,x2 ,x3 ,x^ ) , the Vandermonde determinant of order 4. P(2A^): In this case,it is convenient to write the polynomials in terms of the roots (where ai~xi-xj_+i) rather than the x^'s.We have II (2A^)-a-^OL^ r and so P(2A^) is spanned by: b1=a1a3, ^2~a2(an+a2+a3^' b3=(a-^+a2) (a2+a3 ) =b-^+b2 . P(A2): The group (A3) acting on H(A2) gives rise to the following polynomials: b-^V (x1 ,x2 ,x3) , b2=V(x2,x3,x4) , b3=V(x1,x3,x4), b4=V(x1,x2,x4). Then 1 1 1 1 1 1 1 1 xl x2 x3 x4 x2 X2 x3 x2  4 shows b1-b2+b3-b4=o and {b1,b2,b3} may be taken as a basis for P(A2). Table II:Character Table for (A-,) A3 2AX A, Al Conjugacy Class Representative Characteristic Polynomial h. l *o Xl x2 x4 $ (1) x3-3x2+3x-l 1 1 1 2 3 3 Al (12) x3- x2- x+1 6 1 -1 0 -1 1 A, (123) x3 -1 8 1 1 -1 0 0 A3 (1234) x3 + x2+ x+1 6 1 -1 0 1 -1 2AX (12)(34) x3+ x2- x-1 3 1 1 2 -1 -1 3 .A, Subsystems of A^: The extended Dynkin diagram for A^ is (xrx5) From this diagram we obtain the following subsystems (i) A. : n _ 0 o (2) A3: xl~x2 x2 x3 X3~X4 X4 X5 (3) A2+Ar Xl X2 X2 X3 x3~x4 (4) 2A±: Xl X2 X2 X3 X4~X5 (5) A, Xl X2 X3 X4 (6) A, Xl x2 X2~X3 Xl X2 P(A^): We have n(A4)=V(X1,X2,X3,X4,x5). P(A3): The (A4)-orbit of II (A3) consists of +b± (i=l,...,5) where b1=V(x1,x2,x3,x4) , b2=V(x2,x3,x4,x5) , b3=V (x, x3 , x4 , x^) , b4=V(xl'X2'X4,X5* ' b^=V (x-^,X2 f x3 f x^) . As in the case of P(A2) of (A^),we have and {b^,b2,b3} forms a basis for P(A3) P(A2+A1): We have n (A2+A1)=- (x4-x5)V(x1,x2,x3) Hence P(A2+A^) is spanned by: bl= (x4- X5 V (xx,x2 rx3) V (x3- X5 V(x1,x2 b3= (x2- X5 V (xlfx3 rX4) b4 = X5 V(x2,x3 (xr X4 )V(x2,x3 .x5) V (x2- X4 V (x1,x3 rX5) (x3- X4 V (x1,x2 rX5) V (xr X3 >V(x2,x4 rX5) b9= (x2- x3 )V(xlfx4 'X5) ho x2 )V(x3,x4 rX5) We have b1-b2+b3-b4=o which can be seen from the identity: X1"X5 x2 X5 X3 x5 x4-x 1 • 1 1 1 Xl x2 x3 x4 x2  Xl x| X3 x4 Similarly,we have 10 2 5 6 1_ 1_ 1_ T_ 1— 1' 2' 3' 5' 6 are easily seen to be linearly independent, and hence form a basis for P(A2+A^). P(2A^): Writing the polynomials in terms of the roots a where a.=x.-x.,,,we have IT (2A1) '=a1a3 • It is easy to see that P(2A^) is spanned by the following linearly independent elements: ala3' ala4' a2a4, a2 ^al+a2+a3^' a3 (a2+a3+a4 ) • P(A2): A basis for P(A2) is given by the polynomials: -IT (A2)=V (x1,x2,x3), V(x1,x2,x4) , • V(x1,x2,x5), V(x1,x3,x4), V (xirx3 ,x5)', V(x1,x4,x5). 10 Table III: Character Table for (A.) <S> A4 A3 Al A+A 2 1 2AX A2 Conjugacy Class Representative Characteristic Polynomial h. l *0 Xl x2 *3 x4 *5 (1) x'*-4x3 + 6x2-4x+l 1 1 1 4 4 5 5 6 Al (12) x"-2x3 +2x-l 10 1 -1 -2 2 -1 1 0 A2 (123) x1*- x3 - x+1 20 1 1 1 1 -1 -1 0 A3 (1234) x* -1 30 1 -1 0 0 1 -1 0 A4 (12345) x"*+ x3 + x2+ x+1 24 1 1 -1 -1 0 0 1 (12)(34) x" -2x2 +1 15 1 1 0 0 1 1 -2 A2+Al (123) (45) xlf+ x3 - x-1 20 1 -1 1 -1 -1 1 0 4.D 4 Subsystems of D The extended Dynkin diagram for D^ is Q *3 *4 xrx2 o- o x0+x. x2-x3 3 4 - (x1+x2) From this diagram we find the following subsystems: (1) D (3) A (7) A c >X3 X4 D4: i *• v */ X1"X2 X2 "X3 x3+x4 4A1: 0 0 0 X1"X2 xl +x2 x3-x4 A3 : o X1~X2 X2 "x3 x3~x4 Al: /-> o Xl"X2 X2 "X3 x3+x4 A": • j '"" '" \J \j x3+x4 x2 -x3 x3~x4 3AX: o o o X1~X2 X3 ~x4 x3+x4 A2: o X1~X2 X2 "x3 2A±: 0 0 X1~X2 X3 +X4 2A£: 0 0 X1~X2 xl +X2 x3+x4 12 (10) 2A": x o o xx-x2 x3-x4 (11) A,: 1 o X1~X2 P(D^): We have n(D4)=V(x?L,x|,x|,x|) . •P(4A1): A basis for P (4AX) consists of n (4A1) = (x21-x|) (x2-x2) , (x'-x2) (x'-x*). P(A3): The (D4)-orbit of ^A^ consists of + b.^ (i=l,,..,4) where b1=V(x1,x2,x3,x4), b2=V(x-^fx2,—x3,—x4), b3= V(x1,-x2,x3,-x4), b4=V(xl'-X2'~X3'X4*' A simple calculation of determinants shows that b1+b2+b3-b4=o and b^,b2,b3 are linearly independent. P (Al) : We have II (A^)=V(x1,x2,x3,-x4) , and so P(-&3) is spanned by: b^—V (x-j^, x2 ,x3 , —x4 ) , 4 b2= V (x-^ , x2 , —x3 , x4 ) , b3~V (x-j^, — x2 , x3 , x4 ) , b4=V(-x1,x2,x3,x4). As above,a simple calculation shows that b1+b2+b3+b4=0 and {b-^,b2,,b3} forms a basis for P(A3). P (A3') : We find n(A3')=-V(x2,x|,x|) . This case is analogous to that of P(A2) of (A3),and we obtain a basis consisting of: V(x2,x2,x2), V(x*,x2,x*), V(xJ,x*,x|). p (3A^) : We have II (3A1) = (x1-x2) (x3-x4) . The element (12) [-1 1 -1 l] of '(D4) acting on n(3A1) gives (xx+x2) (x3-x4). Then (x1~x2) (x3~x4) + (xl+x2) (x3~x4)=2xl(x3~x4} is in P(3A1).Since P(3A1) is irreducible,we may take the (D4)-orbit of x1(x|-x4) instead of that of II(3A1) to span P(3A1).We can now easily write down a basis as follows: x. xn X, X. X. X. x, 2 _ 2) 2 2 . X3~X4;' xl"x3J' 2 2 x3~x4;' X1~X2)' 2 _ 2 > x2 4; ' xl x2;' X4 ^X2 XP " P (A2) : We have H (A2)=-v(x1,x2,x3) We note that N(A2) is in P(3A^) above, for Hence P(A2)CP(3A^).But both spaces are irreducible,hence P (A2)=P(3A1) . P(2A1): We have n(2AX)=(x1-x2)(x3+x4). Writing this in terms of the roots of D^,where a-j_=xi-xi+i for i=l,2,3 and a^x^+x^we can easily find a basis for P(2A1),such as ala4' a3 ^al+2a2+a3+a4) i a2(a1+a2+a4). P(2A|): A basis for P(2A|) is given by 2 2 xl~x2' 2 2 xl x3' 2 2 Xl X4' P(2A£): We have n (2A^) = (x1-x2) (x3-x4) • Again,it is more convenient to use the rather, than the x We may take ala3' a4 (a2.+2a2+a3+a4^ ' a2(a1+a2+a3) as a basis for P(2A^). In calculating the characters,the following fact enables us to simplify our computation. The group (D^) contains a central involution z(the element corresponding to 4A^).Then for any character x and element g of the group,we have X(zg)=ex(g) where Xd) Furthermore, e=l or -1 according as II (S) is of even or odd degree. Denote by [x] an element in the conjugacy class designated by X in the table.Then we have the following conjugacy relations: z=[4Ai] , [2A|1 ~z[2Aj] , [Aj ~Z [3Aj , LD4(al>] ~s[Val>] ' [A2] ~z[D4] , [2A").z[2Ag, [2Aj ~Z [2AJ , [AJ] ~Z [A'] , [A3]~Z[A3]' [A-] -z [A'] . Using the subsystems of we obtain all but two of the irreducible characters of (D.). Table IV: Character Table for (D„). $ D4 4A1 2A1 2A^ 2A^ Conjugacy Class Representatives Characteristic Polynomial h. X Xl x2 x3 x4 X5 $ (1) [1111] x4-4x3+6x2 -4x+l 1 1 1 2 3 3 3 (1) [-1-111] x^ -2x2 +1 6 1 1 2 -1 3 -1 4A1 (1) [-1-1-1-1] x't + 4x3 + 6x2 +4x+l 1 1 1 2 3 3 3 Al (12) [1111] x"-2x3 +2x-l 12 1 -1 0 1 1 1 A3 ' (12) [-11-11] -1 24 1 -1 0 -1 1 -1 3Al) A" J (12) [-1-1-1-1] x"+2x3 -2x-l 12 1 -1 0 1 1 1 3 J A2 (123) [llll] x"- x3 - x+1 32 1 1 -1 0 0 0 D4 (123) [-1-1-1-1] xh+ x3 + x+1 32 1 1 -1 0 0 0 2A1 (12) (34) [llll] xh -2x2 +1 6 1 1 2 -1 -1 3 2A1 (12) (34)[-l-lll] xh -2x2 +1 6 1 1 2 3 -1 -1 D4 (ax) (12) (34)[-ll-ll] <t 2 x +2x +1 12 1 1 2 -1 -1 -1 A3 (1234) [llll] <• x -1 24 1 -1 0 -1 -1 1 A3 (1234)[-l-lll] X -1 24 1 -1 0 1 -1 -1 Table IV: Character Table for (D.) Continued. A3 A3 A3 Al 3A, A2 X6 X7 X8 x9 X10 xll X12 3 3 3 4 4 6 8 -1 -1 3 0 0 -2 0 4A1 3 3 3 -4 -4 6 -8 Al -1 -1 -1 2 -2 0 0 a " ' 3 1 1 -1 0 0 0 0 3Al) -1 -1 -1 -2 2 0 Aj J A2 0 0 0 1 1 0 -1 D4 0 0 0 -1 -1 0 1 2A£ 3 -1 -1 0 0 -2 0 2AX -1 3 -1 0 0 -2 0 -1 -1 -1 0 0 2 0 A3 -1 1 1 0 0 0 0 A3 1 -1 1 0 0 0 0 5. B2 and C2 Subsystems of B2: The extended Dynkin diagram for B2 is -JO xl*~x2 x2 (x±+x2) From this diagram we obtain the following subsystems (1) B2: ,, (2) 2A1 (3) Ax: (4) A1 X1~X2 X2 xrx2 -(x1+x2) Xl"X2 x2 P (B2) : We have H (B2)=x1x2(x£-x|) P (2A1) : P(2A1) is spanned by n(2A1)=-(x2-x2) . Subsystems of C2: The extended Dynkin diagram for C2 is •3 -2x1 xl~x2 2x2 We obtain the following subsystems: (2) 2AX: (3) A1 (4) Ax: X1~"X2 2x2 9 9 -2x^ o Xl~x2 2x2 Combining the P(S) for the subsystems S of B0 and of C0, we obtain all the characters of (B2)=(C2). Table V: Character Table for (B0)=(Co). B2 2A1 Al Al $ C2 2A1 Al Al Conjugacy Class Representative Characteristic Polynomial h. l xo Xl x2 X3 X4 B2 c2 (1) [11] x2-2x+l 1 1 1 1 1 2 Al (1) [-11] x2 -1 2 1 -1 1 -1 0 2A1 (1) [-1-1] x2+2x+l 1 . 1 1 1 1 -2 Al Al (12) [11] x2 -1 2 1 -1 -1 1 0 B2 C2 (12) [-11] x2 +1 2 1 1 -1 -1 0 Note:The first line of the table gives the subsystems of B_ and the second the subsystems of Cr 20 6. and C^. Subsystems of B.,: The extended Dynkin diagram for B^ is ?-(x1+x2) X1~X2 X2~X3 We obtain the following subsystems: (1) B0: (2) A3: (3) 2A1+A1 (4) A. (5) B, (6) A1+A1: (7) 2An (8) An (9) A-o-Xl"X2 x1-x2 Xl x2 x2 X3 Xl x2 Xl X2 xl x2 x2 X3 x2 x3 -(x1+x2) 0 o xl~x2 x2~x3 o.. - -—7*— ® X, X 3 o -(x1+x2) -o -(x1+x2) x. X. P(B3): We have n(B3)=~XlX2X3V(xifX2'x3)* P(A3): It is easy to see that P(^3) is spanned by IT (A3)=V(x2 ,x2 ,x2) P (2A1+A1) : We have n (2A1+A1)=-x3 (x^-x2) . Hence P(2A^+A^) is spanned by: xl (x2~X3)' X2 ^i-5^ ' i 2 _ 2 \ X3 1 x2 ' which are linearly independent. P(A2): We have n (A2)=-V(xlrx2,x3) and V(xlfx2,x3)-wx^ (V(x1,x2,x3) ) =V(x1/x0,x^)-V(x1,x2,-x3) = 2 •l'"2'*3 110 xl x2 x3 x! x2 0 =2x3(x2-x2) =-2n(2A1+A1) This shows that P(A2)=P(2A1+A1) P(B2): We find n (B2)=X2X3(x2~x3) . Thus P(B2) is spanned by the following linearly independent elements: X2X3(x2~x3)' X1X2 {X1~X2} P(A1+A1): The (B3)-orbit of IlfA^A^) consists of x^Xj-x^) and +x.(x.+x, ) where i,j,k are distinct elements of {1,2,3) — 1 j K Therefore X1X2' X1X3' X2X3 form a basis for P(A1+A1). P (2A1) : We have n(2A1)=-(x2-x2) , and a basis for P(2A^) is given by: 2 2 xl_x2' 2 2 xl x3* Subsystems of C3 The extended Dynkin diagram for C3 is 3= 2xl xl x2 X2 x3 2x3 From this diagram we obtain the following subsystems (1) 3 o (2) C2+A-L (3) A2 —o 2~X3 u xl~x2 X2~X3 2x3 •2x1 x2~x3 2x3 Xl"x2 X2 X3 x2 X3 2x3 <5>2V -2x-^ 2x^ <6> Al+Al: (7) Ai: (8) A2: 2x„ Xl X2 We next consider subsystems of the above.From the extended Dynkin diagram of C^+A^, 9 w > Q / -n •2x1 ~^x2 x2~x3 2x3 we obtain the subsystem (9) 3A^: & 0 "~2x^ ^^3 P (C2+A1) : We have II (C2+Ai^ =-8xiX2X3 ^X2~X3^ * Hence a basis for P (C^+A-^) is given by; X-j^X^X^ (x^—:X2) e X1X2X3(xl~X3)' P(3A1): P (3A1) is spanned by IT (3A1) =8x1x2x3. We e note that X3 AN(3 x^fO^tained from subsystems of C^, cannot be obtained using subsystems of B^,whereas x2 an<^ X5 obtained from subsystems of B^,cannot be obtained from C^. The group (B^) contains a non-trivial central element, the element corresponding to 2A^+A^.As in the case of (D^), denoting this central element by z and an element in the conjugacy class X by [x] ,we observe the following conjugacy relations: [A-J] ~Z [2Aj , [Aj ~Z [AI+AJ , [Aj ~Z [BJ . Table VI: Character Table for (B-) = (C.J $ B3 A3 <D C3 3A1 Conjugacy Class Representative Characteristic Polynomial h. X X 0 Xl x2 X3 B3 C3 $ $ (1) [111! x3 -3x2+3x-l 1 1 1 1 1 Al Al (1) [ll-l] x3- x2- x+1 3 1 -1 1 -1 2AX 2A1 (1) [1-1-1] x3+ x2- x-1 3 1 1 1 1 2A1+A] 3A1 (1) [-1-1-1] x3+3x2+3x+l 1 1 -1 1 -1 Al Al (12) [111] x3- x2- x+1 6 1 -1 -1 1 B2 C2 (12) [-111] x3- x2+ x-1 6 1 1 -1 -1 Al+Al Al+Al (12) [ll-l] x3+ x2- x-1 6 1 1 -1 -1 A3 C2+Al (12) [l-l-l] x3+ x2+ x+1 6 1 -1 -1 1 A2 A2 (123) [111] x3 -1 8 1 1 1 1 B3 C3 (123) [-111] +1 8 1 -1 1 -1 25 Table VI: Continued 2A1 Al Al A2 2A1+A1 B2 • VA1 C2+Al Al Al A2 C2 Al+Al x4 X5 X6 X7 X8 x9 (1) [111] 2 2 3 3 3 3 (1) [11-1] -2 2 1 1 -1 -1 (1) [1-1-1] 2 2 -1 -1 -1 -1 (1) [-1-1-1] -2 2 -3 -3 3 3 (12) [111] ' o • 0 1 -1 -1 1 (12) [-111] 0 0 1 -1 1 -1 (12) [11-1] 0 0 -1 1 -1 1 (12) [1-1-1] 0 0 -1 1 1 -1 (123) [111] -1 -1 0 0 0 0 (123) [-111] 1 -1 0 0 0 0 Note:The first line of the table gives the subsystems of B and the second the subsystems of C-.. 7. B4 and C4. Subsystems of B4: The extended Dynkin diagram for B4 is 9-(x1+x2) Xl X2 x2 X3 X3 X4 From this diagram we obtain the following subsystems (1) B„: „ „ X1~X2 (2) A3+A1: ^ x2 X3 X3 X4 Xl x2 X2~X3 (x1+x2) (3) D, 9- (x1+x2) -o xl~x2 X2~X3 X3 X4 (4)B„+2A, : 2. 1 O X1~X2 -(x1+x2) X3~X4 (5) B3: -Cr x2 x3 X3~X4 x (6) B2+A1: az Xl X2 X3~X4 x (7) A2+A1 (8) A. X1~X2- X2 X3 © x, Xl X2 x2 X3 -(x1+x2) (9) A' Xl x2 x2 x3 X3 x4 (10) 2A1+A1: Q X1~X2 - (x1+x2) 27 (11) 3A±: Q X1"X2 "(X1+X2) X3"X4 (12) A2: (13) 2A : (14) 2A| X2"X3 X3 X4 xrx2 ~(x1+x2) xl X2 X3~X4 (15) A1+A1: D X2~X3 X4 (16) B2: Q ^ X3~X4 X4 (17) A± (18) Ax: Xl"X2 X4 From the extended Dynkin diagram for B2+2A-L, O O Cn x1-x2 -(x1+x2) x3~x4 x4 -(x3+x4) we obtain the subsystem: (19) 4A 1 xl~x2 ~(x1+x2) x3-x4 -(x3+x4) P(B4): P(B4) is spanned by H(B4)=x1x2x3x4V(x2fx2,x3,x2) P(A3+A^): We have n (A3+A1)=x4V(x2_,x2,,x2) , and so PCA^+A^ is spanned by: xjV(x2'^3'^4^' x2V(x2,x*,x2), x^V (X-^ , x2 f x^) , x^V (x-^ f x2 , x^) . It is easy to see that these are linearly independent. P(D4): P(D^) is spanned by one element: n(D4)=V(x2,x2,x2,x2). P(B2+2A1): We find IT (B2+2A1)=-x3x4 (x£-x2) (x2-x4). It can be seen that the following elements are linearly independent and span P (B2+2A^); x.^x4 (x^—x2) (x3~x4) , x.^x2 (x^—x2) (x3_x4) f ^2X4 ^xl~~X3 ^ ^x2~~X4 ^ ' x^x4(x^—x4) (x^-x3) , X2X3 ^Xl—x4^ ^x2~x3 ^ * P (B3) : We find ir(B3)=x2x3x4v(x2/X2,x2), hence P(B3) is spanned by: X2X3X4^^X2'X3'X4^' X1X3X4^^xl'X3'X4^' X2.X2X4^ ^Xl' X2 ' X4 ^ ' x-j^x2^3V (xi, x2/• x^) • We observe that these are linearly independent. P(B2+A1): We have n (B2+A1)=x3x4(x|-x4) (x1~x2). Since (I-wx ) (x3x4(x3~x4) (x1-x2)=2x1x3x4(x3-x4) , P(B2+A^) is spanned by the (B4)-orbit of x^x^x^(x^-xp We can now easily write down a basis as follows: X1X2X3 <*i -*2 X1X2X3 "«! X1X2X4 -x| xix2X4 -4 X1X3X4 "*3 X1X3X4 -4 X2X3X4 (x| "*3 X2X3X4 (x» P(A2+A1); We have 11 (A2+A1)=x4V(x1,x2,x3) In this case it is easier to consider the (B4)-orbit of (I-wx )(n(A2+A1))=2x1x4(x2-x2) rather than that of II (A^+A^) .Then a basis for P(A2+A1) is given by: X1X2(x3"x4'' X1X3(x2~X4)' X1X4(x2~X3)' X2X3(xl_x4)' X2X4(X1 X3)' x^x^ (x.^—x2)• P(A3): We find n(A3)=V(x2 ,x2,x2) . The (B4)-orbit in this case is the same as the (A3)-orbit 1T(A2) with each x^ replaced by x? . Therefore,a basis for P(A3) is given by: V(x2,x2,x2)., Vfx* ,x*,x2) , VCx^x2 ,x2) . P(A3): We have n (A3)=V(X1,X2,X3,X4) . Let Q=(I-wv ) (n(A') ) , X3 R=(I-wv ) (Q) , X4 S=(I+v/ ) (R) . X2 Writing these in determinant form we see that S=8II (B2+2A1) . Therefore P(A^)=P(B2+2A1). P(2A1+A1): It can be seen that the following elements are linearly independent and that they span P(2A1+A1): -n(2A1+A1)=x4(x2-x2), •x4 (x-^—x3) f Xl ^2~*l) ' Xl^X2*~X4 ^ ' 31 x2 (xl x3* ' X2 ^Xl~X4 ^ ' (x.^—x2) / x^ (x-^""X^) . P (3A1) : We have n(3AX)=-(x^-x2)(x3-x4), and .2 „2 (I-wx )(n(3A1))=2x4(x£-x^) =7-211 (2A1+A1) . Therefore P (3A1)=P (2A1+A]L) . P(A2): We have and n(A2)=-V(x2,x3,x4), (I-w ) (wXi-X3 (n (A2} ) )=2x4 (xl_x2) =-211 (2A1+A1) Hence P (A2) =P (2A1+A]L) . P(2A-^): A basis for P(2A^) is given by -n (2A1)=xJ-x2, 2 2 xl~x3' 2 2 x1-x4. P(2A|): Let c1=II(2A|)=(x1-x2) (x3~x4), .C2=WX2(C1) = (X1+X2) (X3~X4) ' C3=WX4(C1) = (X1~X2) <x3+x4}' We have C4=Wx2(c3)=(Xl+X2)(X3+X4} C1+C2+C3+C4=4X1X3' and we obtain the following basis for P(2A£): xlx2' X1X3' XlX4f  X2X3 ' X2X4 ' X3X4* P(A1+A1): We have n (A1+A1) =x4 (x2-x3) , showing that P(A-j+A^)=P(2A£) P(B2): We find n(B2)=x3x4(x2-x4), and P(B2) is spanned by: X1X2(xl~X2 i 2_ 2  XlX3lXl X3 X1X4 ^X1~X4 X2X3(X2~X3 X2X4(X2_X4 ( 2 _ 2  x3X4lX3 X4 It is clear that these are linearly independent, P (4A1) : We have n(4A1) = (xJ-x|) (x3-x2) , V and the (B4)-orbit of IT(4A1) consists of ~<±=1,2 , 3) ,where b2=(x2"x3)(xi~X4)' b3=(xj-x5)(x|-xj). As in P(2AX) of A3, b3=b1+b2 and {b^,b2} forms a basis for P (4A^) Subsystems of : The extended Dynkin diagram for is 2xl xl x2 X2~X3 X3"X4 2x4 From this diagram we obtain the following subsystems: (1) C, '4 xl x2 x2 X3 (2) 2C2: ^ '"' ~2xl X1~X2 (3) C3+Ai: e (5) C2+A1: Q X1~X2 (6) C2+A1 (7) A3: ^ 2xl X2 X3 (4)A1+2A1: o © a —=*\ :3~X4 s 2x o -i 4 *s 2x4 s —<j - N "x4 2x4 2 -JC **™2JC-J^ 2X^ ~x4 2x4 O 6S> 2x-^ x3~x4 2x4 xl~x2 x2 X3 x3 x4 34 (8) C x2~x3 (9) A2+A1: (10) 2A. X1~X2 Xl X2 (11) A^A^ 0 X2~X3 (12) 2A1: (13) A, (14) C. (15) A, (16) A, X3~X4 xl x2 2x. X3 X4 X2~X3 X3 X4 2x, 2x, 2x, 2x, -2x^ 2x4 u x2~x3 X3~X4 < o BI Next,we consider subsystems of the above.From the extended Dynkin diagram for 2C2 1 o *C E -2x- xl x2 2x. -2x. x3~x4 2x, we obtain (17) 4A, : -2x, 2x, 0 -2x. '1 ""2 3 From the extended Dynkin diagram for C^+A^ 2x, SJC^ ^ 2 ~~ ^ 3 ^*3^4 ^ ^* 4 we obtain (18) C2+2A3: . Q o < 25C-^ ^^2 ^3 "^4 2x Finally,from the extended Dynkin diagram for C„+A, Jt—A. „, •2x^ ~2x3 X3-x4 2x we obtain (19) 3Ai: -2x^ -2x3 2x4 P (2C2) : We have n (2C2)=-16x1x2x3x4(x|-x|) (x2-x2) As in P(4A1) of (B4), :1X2X3X4 X-.X«X~X. ( X -T ~ X ~ ) ( X - — x. ) / ^1^2^3^4 (x2~x^) (x^—x4) is a basis for P(2C2). P (C3+A1) : We find II (C3+A1) =16x1x2x3x4V (x2 ,x3 ,x4) . As in P(A3) of (B4) ,a basis for P(C3+A-^) is given by: X1X2X3X4V(X1'X2'X3)' X1X2X3X4^^X2'X3'X4^' X1X2X3X4V(X1'X3'X4K P (A1+2A1) : We have 36 and we find Thus JI (A1+2A1)=-4x1x4 (x2-x3) , (I+wx )(n(A1+2A1))=-8x1x2x4 X1X2X3' X1X2X4' X1X3X4' ' X2X3X4 form a basis for P(A^+2A^) P(4A1): P(4A ) is spanned by II (4A1) =16x1x2x3x4 P (C2+2A1) : We find n(C2+2A1)=16x1x2x3x4(x|-x|) A basis for P (C2+2A^) is given by: X1X2X3X4 ^l"*!* ' Xlx2x3x4(x2-x2), X1X2X3X4(xl~xl* ' P(3A1): We have II (3A1)=8x1x3x4 and P(3A1)=P(A1+2A1). We note that X3, X5/ Xg^ X9,and x13 obtained from subsystems of C4,cannot be obtained using subsystems of B4,while x2' X4» Xg, X-jran& X-^Q robtained from subsystems of B4 ,cannot be obtained from C.. Again,in calculating the characters we use the following relations: z= [4Aj , [Aj-z [2A1+A]] , [2AJ ~z[2Aj , [Aj ~Z [3A.J , [BJ ~z [B2+2AJ , [Ax+Aj -Z^+AJ , [A3].z[A3], [A2]~ZfDJ'' [B^-zfA^Aj , [2A-] ~z[2Aj] , [B2+Aj -z[B2+Aj , [D4(ai)] -^VM , [A3].Z[A3j, [B4]~Z[B41 • 3.8 Table VII: Character Table for (B.) = (C/,). $ B4 D4 C4 4A1 Conjugacy Class Representative Characteristic h. i Xl X2 X3 C4 Polynomial * $ (i) [1111] x4-4x3+6x2-4x+l 1 1 1 1 1 Al (1) [-1111] x4-2x3 +2x-l 4 1 -1 1 -1 2A1 (1) [-1-111] x4 -2x2 +1 6 1 1 1 1 2A1+A1 3AX (1) [-1-1-11] x4+2x3 -2x-l 4 1 -1 1 -1 4A1 4A1 (1) [-1-1-1-1] x'f+4x3 + 6x2+4x+l 1 1 1 1 1 Al Al (12) [mil x"-2x3 +2x-l 12 1 -1 -1 1 B2 C2 (12) [-1111] x"-2x3+2x2-2x+l 12 1 1 -1 -1 Al+Al Al+Al (12) [11-11] x4 -2x2 +1 24 1 1 -1 -1 A3 B2+2Ai A3+A1J C2+Al C2+2A1 (12) [-11-11] (12) [-11-1-1] x" -1 x'* + 2x3 + 2x2 + 2x+l 24 12 1 1 -1 1 -1 -1 1 -1 3A1 A1+2A1 (12) [-1-1-1-1] x"+2x3 -2x-l 12 1 -1 -1 1 A2 A2 (123) [mi] x4- x3 - x+1 32 1 1 1 1 B3 C3 ' (123) [-1111] x4- x3 + x-1 32 1 -1 1 -1 A2+Al A2+Al (123) [lll-l] x4+ x3 - x-1 32 1 -1 1 -1 D4 C3+Al (123) [-1-1-1-1] x4+ x3 + x+1 32 1 1 1 1 2A| 2A£ (12) (34) [1111] x4 -2x2 +1 12 1 1 1 1 B2+Al C2+Al d2) (34) [-1111] x4 -1 24 1 -1 1 -1 Val> 2C2 (12) (34) [-11-11] x4 +2x2 +1 12 1 1 1 1 A3 A3 (1234) [llll] x4 -1 48 1 -1 -1 1 B4 C4 (1234) [-1111] x4 +1 48 1 1 -1 -1 Table VII: Continued 4A1 2Al A3 B3 Al Al 2C2 3+AL C3 Al Al 3A1 \+2Al x4 X5 X6 X7 X8 x9 X1Q Xll K12 ,X13 (1) [llll] 2 2 3 3 3 3 4 4 4 4 (1) [-1111] 2 -2 3 3 -3 -3 2 -2 2 -2 (1) [-1-111] 2 2 3 3 3 3 0 0 0 0 (1) [-1-1-11] 2 -2 3 3 -3 -3 -2 2 -2 2 (1) [-1-1-1-1] 2 2 3 3 3 3 -4 -4 -4 -4 (12) [llll] 0 0 1 -1 -1 1 -2 -2 2 2 (12) [-1111] 0 0 1 -1 1 -1 -2 2 2 -2 (12) [11-11] 0 0 1 -1 1 -1 0 0 0 0 (12) [-11-11] 0 0 1 -1 -1 1 0 0 0 0 (12) [-11-1-1] 0 0 1 -1 1 -1 2 -2 -2 2 (12) [-1-1-1-1] 0 0 1 -1 -1 1 2 2 -2 -2 (123) [mi] -1 -1 0 0 0 0 1 1 1 1 (123) [-1111] -1 1 0 0 0 0 1 -1 1 -1 (123) [lll-l] -1 1 0 0 0 0 -1 1 -1 1 (123) [-1-1-1-1] -1 -1 0 0 0 0 -1 -1 -1 -1 (12) (34) [llll] 2 2 -1 -1 -1 -1 0 0 0 '0 (12) (34) [-1111] 2 -2 -1 -1 1 1 0 0 0 0 (12) (34) [-11-11] 2 2 -1 -1 -1 -1 0 0 0 0 (1234) [llll] 0 0 -1 1 1 -1 0 0 0 0 (1234) [-1111] 0 0 -1 1 -1 1 0 0 0 0 40 Table VII: Continued A3 B2+2A1 A2+Al A1+A1 2A. . B2 B2+Al A 3A| 2A1+A1 A3 A2+A1 2A1 AX+A: 2A. . C2 C2+Al C2+Al V X14 X15 X16 X17 X18 X19 (i) [im] 6 6 6 6 8 8 (i) [-1111] 0 0 0 0 -4 4 (i) [-l-iii] -2 -2 -2 -2 0 0 (i) [-1-1-11] 0 0 0 0 4 -4 (i) [-1-1-1-1] 6 6 6 6 -8 -8 (12) [1111] -2 0 2 0 0 0 (12) [-1111] 0 -2 0 2 0 0 (12) [11-11] 0 2 0 -2 0 0 (12) [-11-11] 2 0 -2 0 0 0 (12) [-11-1-1] 0 -2 0 2 0 0 (12) [-1-1-1-1] -2 0 2 0' 0 0 (123) [llll] 0 0 0 0 -1 -1 (123) [-1111] 0 0 0 0 1 -1 (123) [lll-l] 0 0 0 0 -1 1 (123) [-1-1-1-1] 0 0 0 0 1 1 (12) (34) [1111] 2 -2 2 -2 0 0 (12) (34) [-1111] 0 0 0 0 0 0 (12) (34) [-11-11 -2 2 -2 2 0 0 (1234) [llll] 0 0 0 0 0 0 (1234) [-1111] 0 0 0 0 0 0 Note: The first line in the table gives the subsystems of and the second the subsystems of C.. Subsystems of : The extended Dynkin diagram for F. is O- O o 13 —O xl~x2 X2~X3 X3~X4 x4 ~h (X1+X2+X3+X4 From this diagram we obtain the following subsystems: (1) F4: Q o ^ x2-x3 x3-x4 x4 -3S(x1+x2+x3+x4) (2) B4: o o o > -J* xl x2 X2 x3 X3 X4 X4 (3) A3+A1: Q Q : o 0 xrx2 x2-x3 x3-x4 -J5(x1+x2+x3+x4) (4) A2+A2: 0 Q # @ X1_X2 X2_X3 X4 -J2(x1+x2+x3+x4) (5) C3+Al: o o xx-x2 x3-x4 x4 -h(x1+x2+x3+x4) (6) B3: Q o > x2 X3 x3~x4 x4 (7) C3: x3~x4 x4 -%(x1+x2+x3+x4) (8) 2A1+AlS 0 Xl"x2 • X3~X4 ~ %(x1+x2+x3+x4) (9) A3: Xl x2 x2 X3 X3~x4 (10) A2+A1: ^ X2-X3 x3"x4 (x1+x2+x3+x4) (11) A2+Ai: 0 x2~x3 x4 -%(x1+x2+x3+x4) xl x2 X3 X4 X4 (13) A2 x2"x3 X3 X4 (14) A2: x4 -h (x1+x2+x3+x4) (15) 2Ai: 0 (16) A^A-^ Xl X2 ' X3 X4 x3-x4 -%(x1+x2+x3+x4) <17) B2: :-(18) A1 (19) A±. x3 X4 X4 Xl X2 ® x, 4 Repeating the process with the above systems of roots we find four subsystems that are not congruent under (F4) to any of the above,as seen below. The extended Dynkin diagram for B4 is o - (x1+x2) -o xl x2 x2 x3 x3 X4 x4 from which we obtain (20) D.: o -(x1+x2) The extended Dynkin diagram for C^+A^ is xl~x2 x3"x4 x4 (xl+X2+X3+X4^ Xl+X2 From this diagram we obtain the following two new subsystems (21) B2+2Ai: Q Q "— > (22) 3AX xl~x2 Xl+X2 X3 X4 X4 xl~x2 X3~X4 xl+x2 Finally, from the extended Dynkin diagram for B2+2A-^ o o o > xj~ X2 xl"^~x2 X3—x4 X4 — ^X3"^X4^ we obtain (23) 4A1: xl~x2 xl+x2 X3~x4 (x3+X4) In the group (F^), we have the following coset decomposition (F4) = (B4)U(B4)wrU(B4)wx^wr where r=-h(x^+X2+x3+x4).Consequently the (F4)-orbit of a polynomial II (S) is the union of the (B4)-orbits of n (S) ,wr (H (S) ) f and wv w^ (H (s) ) . x 4 In some cases we find that ±w (n(S)) and ±w w (n(S)) r x4 r are polynomials that are already in the (B4)-orbit of H(S). Then the (F4)-orbit of n(s) is the same as its (B4)-orbit. V7e find this to be the case for the subsystems A2+A2,C3+A^, C3 ,A2+A1 ,A2 ,A2 ,D4 ' 3A1 anc^ 4A^.In 911 tne other cases it turns out that either wr (n(S) )=±n(S) or 44 w (n (s) )=±wv w (n (s)), r x4 and so a basis for P(S) can be found in the union of the (B^)-orbits of two of n(s) ,w (ii(s)) ,wv w„(n(s)). x x4 i P (F^) :: We have n(F„)=:. 1 V(y ,y ,y ,y^)V(x ,x ,x ,x^) ¥096" J- ^ -a « x ^ J * where yi=x1+x2, y2=xl~x2' y3=x3+x4, y4=x3-x4. P (B4) : We have II (B4)=x1x2x3x4V(x2 ,x2,x2 ,x|) We find that X3_X2X3X4^ ^xl'X2 'X3 ' X4 ^ ' -16wr (n(B4) )=16w wr (II (B4) ) -• £(x1-x2) - (x3+x4) 2] [(x1+x2f- (x3-x4) 2] V (x2 ,x2 ,x2 ,x4) form a basis for P(B4). P (A3+A1) : We have IT (A3+A1)=-32 (x1+x2+x3+x4)V(x1,x2,x3,x4) We find wr(n(A3+A1))=-n(A3+A1), wx^wr(H(A3+Ax))=x4V(x|,x2,x2). As in (B4),we obtain from x4V(x2,x2,x2) the following four linearly independent polynomials: We have bl=XlV *X2,X3,X4 ) , b2=X2V(xl'x3'X4 ) , b3 ^xl,x2,X4 ) , b4=x4V(x*,x2,x* ) . 1 1 1 1 xl X2 X3 X4 -2(n(A3+A1))= xi 2 X2 X3 Xl xl *2 X3 X4 =-b l+b2 -b3+b4 • We note that any other polynomial in the (B4)-orbit of H(A3+A1) can be obtained from this by an appropriate change in signs.We can,therefore,express each of these polynomials as a linear combination of b^,...,b4 by making the corresponding change in signs in b^ , . . . b4 . Thus (b-^ ,b2 ,b3 ,b4 } is a basis for P(A3+A^)-P(A2+A2): We find n(A2+A2)=4X4 [(xl+x2+x3) 2_x42] V(x1,x2,x3) , and -w (II(A2+A2) )=w wr (n(A2+A2) )=IT(A2+A2) . X4 Therefore the (F4)-orbit of II (A2+A2) is the same as its (B4)-orbit.Let Q= (l-wx ) (n (A2+A2) )., R=(I+wv ) (Q) . X3 Then R=x1x4(x|-x3) (x2-x2-x3+x4). The (B4)-orbit of R gives us the following linearly independent polynomials spanning P(A2+A2): X1X2 (X3 "X4> (x2+x2-x2-x2), Xlx3 (x2 -X4^ (x2-x2+x2-x2), X1X4 (X2 -x2) (x2-x2-x2+x2), X2X3 (X1 V (-x2+x2+x2-x2), X2X4 <*; -x|> (-Xi+x2~x3+x4)' X3X4 (x2 -xl) (-x2-x2+x2+x2). P (C^+A^ : We have n(C3+A1)=- x3x4(x2-x2) (x2-x2) [(x1-x2)2-(x3+x4)2J [(x1-x2)2-(x3-x4)2} We find -wr (n(c3+A1) )=w wr(n(c3+A1)) 4 =w w (n(C-+An)). x""*^3 2 ^4 Therefore we only need to consider the (B 4)-orbit of n(c3+A1) We find that P (C3+A1} is spanned by: bl=x3X4(xl~x2) / 2 2 \ lx3 X4J [(x1-x2)2- Cx3+x4) 2J[( X1"X2)2- (x3-x4) 2] , b2=X3X4 ^l"^ , 1 2 \ tx3 x4; [(x1+x2)2- (x3+x4) 2][ (x1+x2)2 "(x3-x4)2] , b3=xlx2 ^l-^ / 2 2 \ [(x1+x2)2- (x3-x4) 2] [ '(xrx2)2 ~(x3-x4)2], b4=xlX2(xl""X2) / 2 2 \ {x3 X4J [(x1+x2)2- (x3+x4) 2J [ (xVx2)2 -(x3+x4)2] , b5=XlX4(x2~x3} i 2 2\ (xrx4) [(x1+x4)2-'X2~X3) 2][ -(x2-x3)2] , b6=XlX4(X2_X3) / 2 2 \ (xx-x4) [(x1+x4)2- x2+x3) 2][ (xrx4)2 ~(x2+x3)2] , b7=X2X3(x2~X3} / 2 2 \ [(xrx4)2- x2+x3) 2J[ "(x^)2 -(x2-x3)2] , b8=X2X3 (x2~x3* / 2 2 \ (xrx4) [(x1+x4)2-( x2+x3) 2][ ;(x1+x4)2 -(x2-x3)2] , b9=X2X4 ^Xl-X3^ (2 2 \ ix2-x4; [(x1~x3) 2- ( x2+x4) (X1~X3)2 -(x2-x4)2] , b10=X2X4(XI-X3] <x|-x" ) [(x1+x3) 2- (x2+x4 )2] [(x1+x3) 2-(x2-x4)2] , bll=XlX3(xi"x3} )[(x1+x3)2- (x2-x4 ).2l [(xrx3) 2-(x2-x4)2J , b12=XlX3(X1~X3) (x|-x| ) [(x1+x3)2- (x2+x4 )2J [(x1-x3) 2-(x2+x4)2] . If we substitute 47 yi=x2+x2 y2=x1-x2 y3=x3+x4 Y4=X3~X4 in b^,b2,b3,fc>4 we obtain the simpler forms b.=-%yny0y.y/1v(y12,y2 y2) , b4r=-^iy2y3Y4V(Yi'Y2'Y3) From this we can see that Vbl"b2+b3 Similarly,we find WW b12s=b9"b10+bll' It can be seen that {b^ ,b2 ,b^ ,br-,bg ,b^ ,b^ ,b^^ ,b^^} forms a linearly independent set. P (B3) : We find n (B3)=-x2x3x4V(x2,x2,x2) , -wr (IT (B.J )=w w (II (B ) ) 1 = . — (xx-x2+x3+x4) (x1+x2~x3+x4) (x1+x2+x3-x4)V(-x1,x2,x3,x4) Four linearly independent polynomials are found from the (B4) -orbit of JI (B ) ; b^=x2x3x^V(x2,x3iX4), b2=x^x3x V(x^,x3,x^), b3-x1x2x4V(x2,x2,x2), b4~x^x2X3V(x^,x2,x3). Let 48 Q=8wv (w -(n(B-))) = (x,+x„-x_-x.)(x.-x^+x-.-x.)(x.-x,-x.+x.)V(x,,x0,x,,x4) 4 4 4 = .(?xi.T?Xi xjt?£xixjxk)V(xl'x2'x3'x4) i^j i<j<k This can be written in terms of the following determinants B= 1 1 1 1 xl x2 x3 X4 x2 xl x2  X2 *3 X4 x6 xl x6  X2 X3 1 1 1 i xl X2 X3 x4 x3 xl X2 x3  X3 *4 x5 xl X2 x5  X3 < 1 1 1 1 x2 xl x2  X2 x2  X3 *4 X2 x3  X3 x3  X4 *2 x" X3 x* X4 = EXI t£xixjt£ XiXjxk)V(x1,x2,x3,x4), i=l xj = l Jij,k=l J ift i<j<k 4 4 = (&?x.+2r; x.x xk)V(x1,x2,x3,x4) xj = l J 3j,k=l J i^j i<j<k :(Z]xixixk)v(xl'x2'x3'x4) * iik=l i<j<k We have Let Q=A-2B+3C, Then R=(I+w ) (Q) , x2 S=(I+wv ) (R) , X3 T=(I+w )(S). x4 49 T=< 1 1 1 1 1 1 x2 *3 x2  X4 + 3x3 *2 *; x2  X4 x6  X2 *3 x6  X4 *3 x4  X4 \ =-8 [xx (x2+x2+x2)V(x2,x2 ,x2)-3x3V(x2 ,x2,x2)] =-8x1 [(x2+x2+x2+x2)-4x2] V(x2 ,x2 ,x2) . We can now complete the basis for P(B^) by applying to T: b5=xx [(x2+x2+x2+x2 ) -4x2] V(x2,x2,x2) , b6=x2 [(x2+x2+x2+x2)-4x2]v(x2 ,x2 ,x2) , b?=x3 [(x2+x2+x2+x2)-4x2]v(x2,x2,x2) , bg=x4 [(x2+x2+x2+x2)-4x2]v(x2,x2,x2) . w X . -x. i 3 P(C3): In this case it is simpler to write II (C3) in terms of the y^ as defined in P(F4).We have n(c3)=~ gT Y1Y3y4V(y^,y3,y4) . We note that the elements w ,w , c fr, N x. x.+x.,w ,and w w of (F.) x l j r x4 can be expressed in terms of w ,w ,w , , . . , N yL' y^Yj' -%(y1+y2+y3+y4), and ww,, , , , .,and conversely.Therefore,we can find y4 (yi+y2+y3+y4) the (F.)-orbit of II (C0) by applying w ,w . ,w , , , , , w 4 3 y^ yi±Yj (y1+y2+y3+y4) and ww... y4 ~h (y1+y2+y3+y4) to y1y3y4v(y2,y2,y2). We observe that this.is the same as -w (II (B_) ) with y. replacing-the X£ Hence a basis for P(C3) is given by the polynomials in the basis for P(B^) with y^ replacing the x^. P(2A1+A1): We have 50 n (2A1+A1)=-3s (x1-x2) (x3-x4) (x1+x2+x3+x4) We find wr(n(2A1+A1))=-n(2A1+A1), w„ w_ (II (2A1+A1) )=-x4 (x2-x2) . x4 r As in (B4) X4 (x2-x2), X4 (xi"TX3)' Xl (x2-x2), Xl (x2~X4*' X2 (x2-x2 ) lxl x3; ' X2 (x2-x2), X3 (x2-x2), X3 are linearly independent.lt can be seen that H(2A^+A^) and other polynomials in its (B4)-orbit are all linear combinations of these elements. P(A3): We find n(A3)=V(x1,x2,x3,x4), wr(II(A3))=II<A3) , wx^wr(n(A3) )=V(x2,x2,x2) . As shown for P(A3) of (B4),to find the span of the (B4)-orbit of V(x^,x2,x3,x4),we may consider the orbit of x.^x4 (x^—x2) (x^—x4) • We have seen in P(B2+2A^) of (B4) that the following polynomials are linearly independent: b1=x3x4(x2-x2) (x2-x2), b2=xlx2(xl-x2)(x3-x4)' 51 t>3=x1x3 (x2-x2) (x^-x^) , b4=X2X4(xi~x3) ^x2"x4^' b5=x1x4(x2-x2)(x2-x2), H6=X2X3{Xl~Xl)(X2_X3}-The (B^)-orbit of V(x^,x2,x3) gives us three more linearly independent polynomials,completing the basis for P(A3): b7=V^xl'x2'X3^' b8=V(x2fx2,x2), b9=V(xl'x3'x4)' P(A2+AX): We find II (A2+A1)=%(x1+x2+x3+x4)V(x2,x3,x4) , wr (n (A2+A1) ) =-n (A2+A1) , Wx wr (n (A2+Ai^=-x4V(~xl'x2,x3^ ' 4 We have seen in P(A2+A^) of (B^-) that the span of the (B4)-orbit of x4V(x^,x2,x3) contains the following linearly independent polynomials: bl=XlX2(x3-X4)' b2=XlX3 (X2~XV ' b3=x1x4(x2-x2), b4=X2x3(x!~X4)' VX2X4 (xl_x3)' bg~ x^x. (x — x«) • D 5 4 X Z. We can complete this to a basis of P(A2+A1) by considering the (B4)-orbit of IT (A2+A-^) .Let Q=2wx -x (n (A2+Ai) )"(xi+x2+x3+X4)V(xl'x2'X3) ' R=wx (Q)=(x1+x2+x3-x4)V(x1,x2,x3). Then and Q+R=2(x1+x2+x3)V(x1,x2,x3) 1 1 1 = 2 Xl X2 X3 x3  Xl x3  X2 X3 (I+wx ) (Q+R)=-4x1x2(x2-x2). Therefore,a basis for P(A2+A^) is given by b-^, and . b,. above b b7=XlX2(Xl-X2)' b8=x1x3(x2-x^), VX1X4(X1-X4)' b10:=X2X3 ' bll=X2X4(X2~X4)' b12=X3X4(X3"X4)' P(A2+A1): We have n (A2+A1)=jx4 (x2-X3) [(x1+x2+x3) 2~xl~l ' and (I-w ) (n(A2+A1) )=xxx4 (x2-x2), showing that P (A2+A1)=P (A2+A1) P(B2+A1): We find IT (B2+A1)=x3x (x1-x2).(x3-x4) -wr (n(B2+A1) )=wx wr(n(B2+A1) = h [(x1+x2) 2- (x3-x4) 2] (x2-x2) (x3-x4) . Therefore, the (B)-orbits of II (B^A^ and wx wr (II (B^A^ ) 4 span P(B2+A^).We have already seen that the following polynomials form a basis for P(B0+A,) of (B ): Let bl =X1X2X3 (x2-x2) b2 =X1X2X3 (x2-x2) b3 =X1X2X4 (x2-x2) t b4 =X1X2X4 (x2-X4> b5 =X1X3X4 (x2-lxl x2) r b6 =X1X3X4 Xl> i b7 =X2X3X4 (x|-x2) t b8 =X2X3X4 fx2-lX2 X4> • Q= (xj-xj) (x3- X4> -We have 4wx wr(H(B2+A1))=Q+2b1-2b3. Therefore we may consider Q instead of w w (II (B„+A ) ) . x4 r -si We find h (Q+Wx (Q) ) =x3 (xj-x*) - (x2-x2) (x3 + 3x3x2 ) =x3(x2-x2)(x£+x2-x2-3x2). We can now complete the basis for P(B2+A^) by the following linearly independent polynomials: b9=xx(x|-x3> (•x2_x2_ 2_o 2 N KX1 x2 3 4;' bio=xi( X2 -*v (x.^—x^-3x.^-~x^) / bll=x2( xi -x|) (-xj+x2-x|-3x2) r b12=X2( x2  Xl -*4> (-x2+x2-3x2-x2) r b13=x3( X2 xl -x2) X2J (-x2-x2+x2-3x2) i b14=X3( X2 xl -x2) (-x2-3x|+x2-x|) f b15=X4< X2 1 -x2^ x2; (-x2-x2-3x2+x2) t b16=X4( x2 xl -x2l X3J (-x2-3x2-x2+x2) • P(A2): As in the case of (B4),we have and n(A2)=-v(x2,x3,x4) P(A2)=P(2A1+A1) . P(A2): We find n(A2)-|x4f(Xl+x2+x3)2-x2], -wr (n(A2) )=wx wr(n(A2) )=n(A2). Therefore the (F4)-orbit of n(A2) is the same as its (B4)-orbit.Let Hence Q=(I-wx ) (H (A2) )=x1x4 (x2+x3) (I+wx )(Q)=2x1x2x4 Also, (I+w +wx +wx ) (II (A2) ) =x4 (x2+x2+x2-x2) . It can be seen that II (A2) and any polynomial in its (B4)-orbit may be expressed as a linear combination of polynomials of the form x (X2+X2+X2TX2) and x.x.x, m x D k m' x j k Therefore a basis for P(A2) is given by: x1(x2+x2+x2-x2), x2(x2+x2+x2-x2), x3(x2+x2+x2-x2), x4(x2+x2+x2-x2), XlX2X3f  X1X2X4' X1X3X4' X2X3X4* P (2A^) : We have n (2A1) = (x1-x2) (x3-x4) , wr(n(2A1) )=n(2A1) , wx wr(n(2A ) )=-(x2-x2) . 4 We observe that (x^-x2) (x^-x^) is II(2A|) of (B^) and -(x2-x2) is II (2A^) of (B^).We find that the union of the bases for P(2A1) and P(2A|) of (B4) provides a basis for P(2A1) of (F4) xlX2' X1X3' X1X4' X2X3' X2X4' X3X4' 2 2 Xl~X2' 2 2 xl~X3' 2 2 xi"X4* P (Aj+A.^) : We have n(A1+A1)=-J2 (x3-x4) (x1+x2+x3+x4) , wr (n (A1+A1) ) =-n (A1+A1) , Wx Wr^n (Ai+Ai) ) =~x4 (xi+x2^ • Since -x^(x^+x2) is contained in P(2A^), P (A1+A]L)=P (2A1) . P(B ): We find n (B2)=X3X4(x3-x4). This is one of the elements in the basis of P(A2+A^).Therefore P(B2)=P(A2+A1). • P(D^): As in (E^),P{D^) is spanned by one element n(D4)=V(x2,x2,x2,x2). P (B2+2A-L) : We find n(B2+2A1)=x3x4(x2-x2)(x2-x2). But this is in P(A3),hence P(B2+2A1)=P(A3). P(3AX): We have n(3A1) = (x2-x2) (x3-x4) . We observe that this is in P(2A^+A^),hence P(3A1)=P(2A1+A1). P (4A1) : We have n(4A1)=-(x2-x2) (x2-x2) and w (n(4A1))=w w (II (4A.. ) )=n (4A ) . 3T X IT X X Therefore,as in (B4), (x2-x2)(x2-x2), (x2-x2) (xj-x*) form a basis for P(4A^). Using the above (F4)-modules,we obtain seventeen of the irreducible characters of (F4).The characters that cannot be obtained by the above method can be found by using the following relations: x10=x4x5/ X11=X1X8' X12=X1X9' X14=X2X13, X22=X1X20' The following conjugacy relations are used in calculating the characters of (F^): z= [4AJ , [AJ-Z^+AJ , [2AJ .z[2Aj , [A1+Aj - z [AX+AJ , [Aj.zfDj, [Aj-z^+Aj , [BJ - z [A3+AJ , [A3]~z[A3] , [B2+Aj-.z[B2+Aj , [C3]~4VAl]' [B3].z[A2+ij , [A2+A2-J-.z[P4(AI)] , [D4(a1)]~z[D4(a1)] , 58 Table VIII: Character Table for (F.). F4 D4 B4 Conjugacy Class Representative Characteristic . Polynomial h, l X0 Xl x2 X3 x4 (i) [mi] x"*-4x3+6x2-4x+l 1 .1 1 1 1 2 Al (12) [llll] x4-2x3 +2x-l 12 1 -1 -1 1 -2 Al (i) [-1111] x"-2x3 +2x-l 12 1 -1 1 -1 0 2A1 (i) [-1-111] x1* -2x2 +1 18 1 1 1 1 2 Al+Al (12) [11-11] x" -2x2 +1 72 1 1 -1 -1 0 A2 (123) [llll] x1*- x3 - x+1 32 1 1 1 1 2 A2 x1*- .x3 - x+1 32 1 1 1 1 -1 B2 (12) [-1111] x"-2x3+2x2-2x+l 36 1 1 -1 -1 0 3A1 (12) [-1-1-1-1] x"+2x3 -2x-l 12 1 -1 -1 1 -2 2A1+A1 (1) [-1-1-11] x"+2x3 -2x-l 12 1 -1 1 -1 0 A3 (1234) [llll] x" -1 72 1 -1 -1 1 -2 B2+Al (12) (34) [-1111] x4 -1 72 1 -1 1 -1 0 C3 x4- x3 + x-1 96 1 -1 -1 1 1 B3 (123) [-llll] x1*- x3 + x-1 96 1 -1 1 -1 0 A2+Al xk + x3 - x-1 96 1 -1 -1 1 1 A2+Al (123) [lll-l] xk + x3 - x-1 96 1 -1 1 -1 0 4A1 (1) [-l-l-l-l] x'*+4x3 + 6x2+4x+l 1 1 1 1 1 2 A2+A2 x'* + 2x3 + 3x2+2x+l 16 1 1 1 1 -1 A3+Al (12) [-11-1-1] x',+2x3 + 2x2 + 2x+l 36 1 1 -1 -1 0 C3+Al x"*+ x3 + x+1 32 1 1 1 1 -1 D4 (123) [-1-1-1-1] xV x3 +X+1 32 1 1 1 1 2 D4(ax) (12) (34) [-11-11] xk +2x2 +1 12 1 1 1 1 2 B4 (1234) [-1111] xH +1 144 1 1 -1 -1 0 F4(ai) x"-2x3+3x2-2x+l 16 1 1 1 1 -1 F4 • x* - x2 +1 96 1 1 1 1 -1 4A1 Al Al A3+Al A2+A2 B3 C3 X5 X6 X7 X8 x9 X10 Xll X12 X13 X14' X15 X16 2 2 2 4 4 4 4 4 6 6 8 8 Al 0 0 2 2 -2 0 -2 2 0 0 -4 0 Al 2 -2 0 2 2 0 -2 -2 0 0 0 -4 2A1 2 2 2 0 0 4 0 0 -2 -2 0 0 A1+A1 0 0 0 0 0 0 0 0 2 -2 0 0 A2 -1 -1 2 1 1 -2 1 1 0 0 2 -1 A2 2 2 -1 1 1 -2 1 1 0 0 -1 2 B2 0 0 0 2 -2 0 2 -2 -2 2 0 0 0 0 2 -2 2 0 2 -2 0 0 4 0 2A1+A1 2 -2 0 -2 -2 0 2 2 0 0 0 4 A3 0 0 2 0 0 0 0 0 0 0 0 0 B2+Al 2 -2 0 0 0 0 0 0 0 0 0 0 C3 0 0 -1 1 -1 0 -1 1 0 0 1 0 B3 -1 1 0 1 1 0 -1 -1 0 0 0 1 A2+Al 0 0 -1 -1 1 0 1 -1 0 0 -1 0 A2+Al -1 1 0 -1 -1 0 1 1 0 0 0 -1 4A1 2 2 2 -4 -4 4 -4 -4 6 6 -8 -8 A2+A2 -1 -1 -1 -2 -2 1 -2 -2 3 3 2 2 A3+Al 0 0 0 -2 2 0 -2 2 -2 2 0 0 C3+Al 2 2 -1 -1 -1 -2 -1 -1 0 0 1 -2 D4 -1 -1 2 -1 -1 -2 -1 -1 0 0 -2 1 2 2 2 0 0 4 0 0 2 2 0 0 B4 0 0 0 0 0 0 0 0 0 0 0 0 P4(al} -1 -1 -1 2 2 1 - 2 2 3 3 -2 -2 F4 -1 -1 -1 0 0 1 0 0 -1 -1 0 0 ^2 B2 3AX A2 2A1 A3 C3+A1 A2+Al B2+A1 2A1+A1 Al+Al B2+2A1 A2+Al X17 X18 X19 X20 X21 X22 X23 X24 8 8 9 9 9 9 12 16 Al 0 » 4 3 -3 -3 3 0 0 Al 4 0 3 3 -3 -3 0 0 0 . 0 1 1 1 1 -4 0 Al+Al 0 0 1 -1 1 -1 0 0 A2 -1 2 0 0 0 0 0 -2 A2 2 -1 0 0 0 0 0 -2 B2 0 0 . 1 -1 1 -1 0 0 3A1 0 -4 3 -3 -3 3 0 0 2A1+A1 -4 0 3 3 -3 -3 0 0 A3 0 0 -1 1 1 -1 0 0 B2+Al 0 0 -1 -1 1 1 0 0 C3 0 -1 0 0 0 0 0 0 B3 -1 0 0 0 0 0 0 0 A2+Al 0 1 0 0 0 0 0 0 A2+Al 1 0 0 0 0 0 0 0 4A1 -8 -8 9 9 9 9 12 -16 A2+A2 2 2 0 0 0 0 -3 -2 A3+Al 0 0 1 -1 1 -1 0 0 C3+Al -2 1 0 0 0 0 0 2 D4 1 -2 0 0 0 0 0 2 D4(ax) 0 0 -3. -3 -3 -3 4 0 B4 0 0 -1 1 -1 1 0 0 F4(al> -2 -2 0 0 0 0 -3 2 F4 0 0 0 0 0 0 1 0 BIBLIOGRAPHY [l] N.Bourbaki ,Groupes et Algebres de Lie ,Chapitres 4,5 et 6, Hermann,Paris,1968 . [2] R.W.Carter,Conjugacy Classes in the Weyl Group,Seminar on Algebraic Groups and Related Finite Groups,Springer Lecture Notes No.131,1970. [3] E.B.Dynkin,Semisimple subalgebras of semisimple Lie Algebras.Mat. SbornikN.S. 30 (72),349-462 (1952). (Russian).(Amer. Math. Soc. Transl.,(2) 6 (1965), 111-244). [4]T.Kondo, Characters of the Weyl Group of Type F^J.Fac. • Sci. Univ. Tokyo, Sect. I, 11 (1965), 145-163. [5] I.G.MacDonald,Some Irreducible Representations of Weyl Groups, Bull. London Math. Soc, 4 (1972), 148-150. 

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