Cohomology mod 3 of the classifying space of the exceptional Lie group $E_6$, II : The Weyl group invariants

0.22 MB·English

by  Mamoru Mimura

COHOMOLOGY MOD 3 OF THE CLASSIFYING SPACE OF THE
EXCEPTIONAL LIE GROUP E , II : THE WEYL GROUP
6
INVARIANTS
MAMORU MIMURA, YURIKO SAMBE, AND MICHISHIGE TEZUKA
Abstract. We calculate the Weyl group invariants with respect to a maximal torus of
the exceptional Lie group E6.
2
2010 Mathematics Subject Classification. 55R35,55R40, 55T20
1
0 Key words and phrases, exceptional Lie group, Weyl group
2

n
a 1. Introduction.
J

7
The present paper is a sequel to [MST3].
1
Let E6 be thecompact, simply connected, exceptional Lie groupof rank 6, T its maximal
]
T torus, BX the classifying space of X for X = E or T, and W(E ) the Weyl group
6 6
A
which acts on H∗(BT;Z), and hence on H∗(BT;Z ), as usual. As is well known, the
. 3
h invariant subalgebra H∗(BT;Z )W(E6) contains the image of the homomorphism Bi∗ :
t 3
a H∗(BE ;Z ) → H∗(BT;Z ).
m 6 3 3
[ The Rothenberg-Steenrod spectral sequence {Er(X),dr} for X a compact associative H

1 -space has
v E (X) = Cotor (Z ,Z ) with A = H∗(X;Z ),
4 2 A p p p
1 E (X) = GrH∗(BX;Z )
∞ p
4
3 where p is a prime number.
.
1 The E -term Cotor (Z ,Z ) for X = E and p = 3 is calculated in [MS] (see also
0 2 A p p 6
2 [MST3]). In this series of papers, it will be shown that the spectral sequence associated
1
v: with E6 collapses and the ring structure of H∗(BE6;Z3) will be determined (cf. [KM]
i where some of the ring structure of H∗(BE ;Z ) was determined).
X 6 3
The present paper is organized as follows. In Section 2, the action of the Weyl group
r
a W(E ) on H∗(BT;Z ) will be described in terms of some generators of H∗(BT;Z ). From
6 3 3
this, 6 elements in H∗(BT;Z )W(E6) are immediately obtained, that is, x , x , y , x , y
3 4 8 10 20 22
and y . In Section 3, another set S of elements in H∗(BT;Z ) which is invariant as a set
26 3
under W(E ) is considered. The elementary symmetric functions on the elements of S are
6
W(E )-invariant and we find another element x . In Section 4, it is shown firstly that
6 36
H∗(BT;Z )W(E6) ⊂ Z (x )[x ,y ,x ,y ,x ].
3 3 4 8 10 20 22 36
Then 6 more generators of H∗(BT;Z )W(E6) are totally determined explicitly. We prove
3
Partially supported by the Grant-in-Aid for Scientific Research (C) 21540104, Japan Society for the
Promotion of Science.
1
2 M.MIMURA,Y.SAMBE, ANDM.TEZUKA
Theorem 4.13 H∗(BT;Z )W(E6) is generated by the following thirteen elements:
3
x , x , y , x , y , y , x , x , x , y , y , y , y .
4 8 10 20 22 26 36 48 54 58 60 64 76
We thank T. Torii for suggesting us appropriate references on Galois theory and M. Nak-
agawa for pointing a fact stated in [S]. We also thank A. Kono for reading the manuscript
and giving us valuable suggestions. Last but not least, we thank Prof. H. Toda for tutoring
us the relation between the Dynkin diagram and the generators of the Weyl group.
This paper has taken a long time to write. In fact, the results of the paper were an-
nounced in [T].
2. Some invariant elements in low degrees.
Let T ⊂ E be a fixed maximal torus of E , V the universal covering of T, V∗ the dual
6 6 T T
of V . According to Bourbaki [B], the Dynkin diagram is given as follows:
T
α α α α α
1 3 4 5 6
◦ ◦ ◦ ◦ ◦
◦ α
2
where α ∈ V∗ for i = 1,2,...,6 are the simple roots of E . Let h , i be the invariant
i T 6
metric on the Lie algebra of E , V and V∗, normalized in such a way that
6 T T
hα ,α i = 2,
i i
hα ,α i = −1 if i 6= j and α and α are connected,
i j i j
hα ,α i = 0 otherwise.
i j
Let β be the corresponding fundamental weights: hβ ,α i = δ . Let us denote by R
j j i ij i
the reflection with respect to the hyperplane α = 0. Then
i
R (β ) = β − hα ,α iβ and R (β ) = β for i 6= j.
i i i i j j i j j
j
X
Denote by U the centralizer of the torus T1 defined by hα ,ti = 0 for i = 2,3,...,6 and
i
t ∈ T. Then U is a closed connected subgroup of maximal rank and of local type D ×T1
5
such that D ∩T1 = Z (see [V]). The Weyl groups W(E ) and W(U) are generated by
5 4 6
R , R , ..., R and R , ..., R , respectively.
1 2 6 2 6
Put
τ = β ,
6 6
τ = R (τ ) = β −β ,
5 6 6 5 6
τ = R (τ ) = β −β ,
4 5 5 4 5
(2.1) τ = R (τ ) = β +β −β ,
3 4 4 2 3 4
τ = R (τ ) = β +β −β ,
2 3 3 1 2 3
τ = R (τ ) = −β +β ,
1 1 2 1 2
x = β = 1 (τ +τ +···+τ ).
2 3 1 2 6
COHOMOLOGY MOD 3 OF THE CLASSIFYING SPACE OF E6, II 3
Then β are linear combinations of τ ’s and x as follows:
i j
β = x−τ , β = x, β = −x+τ +τ +τ +τ ,
1 1 2 3 3 4 5 6
β = τ +τ +τ , β = τ +τ , β = τ .
4 4 5 6 5 5 6 6 6
Further we put
t = x−τ and t = τ − 1 t for i = 1,...,5.
1 i i+1 2
Then we have
t = β ,
1
t = 1 β +β −β ,
1 2 1 2 3
t = −1 β +β +β −β ,
(2.2) 2 2 1 2 3 4
t = −1 β +β −β ,
3 2 1 4 5
t = −1 β +β −β ,
4 2 1 5 6
t = −1 β +β ,
5 2 1 6
and
β = t,
1
β = 3 t+ 1 t + 1 t + 1 t + 1 t + 1 t ,
2 4 2 1 2 2 2 3 2 4 2 5
β = 5 t− 1 t + 1 t + 1 t + 1 t + 1 t ,
(2.2)′ 3 4 2 1 2 2 2 3 2 4 2 5
β = 3 t+t +t +t ,
4 2 3 4 5
β = t+t +t ,
5 4 5
β = 1 t+t .
6 2 5
Denote t +t +t +t +t by c . Then we have
1 2 3 4 5 1
(2.3) c = t +t +···+t = τ +τ +···+τ − 5t = 2x− 3t.
1 1 2 5 2 3 6 2 2
The R -operations are given by
i
R R R R R R
1 2 3 4 5 6
t 1 t−t + 1 c
4 1 2 1
t −3 t+ 1 t + 1 c −t t
1 8 2 1 4 1 2 2
t 3 t+ 1 t +t − 1 c −t t t
2 8 2 1 2 4 1 1 1 3
t 3 t+ 1 t +t − 1 c t t
3 8 2 1 3 4 1 2 4
t 3 t+ 1 t +t − 1 c t t
4 8 2 1 4 4 1 3 5
t 3 t+ 1 t +t − 1 c t
5 8 2 1 5 4 1 4
Z
where the blanks indicate the trivial action. Taking coefficients in , we have
3
R (t) = t−(t +c ),
1 1 1
(2.4) R (t ) = −t +c ,
1 1 1 1
R (t ) = t −(t +c ) for i = 2,3,4,5.
1 i i 1 1
Roots or weights will be considered in the usual way (cf. [BH]) as elements of H∗(T),
H∗(T;Z ), H∗(BT) or H∗(BT;Z ). Then the weights β ,...,β generate H∗(BT) as well
3 3 1 6
4 M.MIMURA,Y.SAMBE, ANDM.TEZUKA
as H∗(BT;Z ) and so do the elements t,t ,...,t :
3 1 5
H∗(BT;Z ) = Z [t,t ,...,t ] with t,t ,t ,...,t ∈ H2(BT;Z ).
3 3 1 5 1 2 5 3
Z
From now on until the end of Section 2 all coefficients are in .
3
The Weyl group W(E ) acts on BT and hence on H∗(BT;Z ), and the invariant sub-
6 3
algebra H∗(BT;Z )W(E6) contains the image of the homomorphism Bi∗ : H∗(BE ;Z ) →
3 6 3
H∗(BT;Z ) induced by the natural map Bi : BT → BE . The action of W(U) is the same
3 6
as the usual action of W(SO(10)). Thus we obtain
Im Bi∗(H∗(BE ;Z )) ⊂ H∗(BT;Z )W(E6)
6 3 3
(2.5)
⊂ H∗(BT;Z )W(U) = Z [t,p ,p ,c ,p ,p ]
3 3 1 2 5 3 4
where c = σ (t ,t ,...,t ) and p = σ (t2,t2,...,t2) are the elementary symmetric func-
i i 1 2 5 i i 1 2 5
tions on t and t2, respectively.
i j
WeshalldetermineH∗(BT;Z )W(E6),thatis, R -invariantelementsinZ [t,p ,p ,c ,p ,p ].
3 1 3 1 2 5 3 4
We obtain, from the simple relation between c ’s and p ’s, that
i j
c = p −c2,
2 1 1
c = −p +c4 +c2p +c c +p2,
(2.6) 4 2 1 1 1 1 3 1
c c = p −c2,
3 5 4 4
c2 = p +c6 +c3c −c2p −c c p +c c −p3 +p p .
3 3 1 1 3 1 2 1 3 1 1 5 1 1 2
5
The equality 0 = (t −t ) gives rise to
1 i
i=1
Q
t5 = c −t c +t2c −t3c +t4c .
1 5 1 4 1 3 1 2 1 1
Put b = t +c , then we have
1 1
b5 = (−c5 −c3p −c2c +c p2 −c p +c )
(2.7) 1 1 1 1 3 1 1 1 2 5
+ b(−c4 −c2p −p2 +p )+b2(c3 +c )−b3(c2 +p ).
1 1 1 1 2 1 3 1 1
From (2.4), and replacing c , c , c c , c2 and b5 by (2.6) and (2.7), we have
2 4 3 5 3
R (t) = t−b,
1
R (b) = −b,
1
R (c ) = c ,
1 1 1
R (p ) = p ,
1 1 1
R (p ) = p ,
1 2 2
(2.8) R (c ) = c −b(c2 +p )+b3,
1 3 3 1 1
R (c ) = c +b(p2 −p )−b3p ,
1 5 5 1 2 1
R (p ) = p +b(−c3p −c p2 +c p +c )+b2(c2p +p )−b3c p ,
1 3 3 1 1 1 1 3 1 5 1 1 2 1 1
R (p ) = p +b(−c5p +c3p2 +c3p −c2c p +c p2 −c p +c p )
1 4 4 1 1 1 1 1 2 1 3 1 3 1 3 2 5 1
+ b2(−c4p −c2p )+b3(c3p −c p2 +c p +c p +c )
1 1 1 2 1 1 1 1 1 2 3 1 5
+ b4(−c2p −p ).
1 1 2
Thus the following three elements are clearly R -invariant and hence W(E )-invariant:
1 6
(2.9) x = p , x = p −p2, y = c −tx −t3x .
4 1 8 2 1 10 5 8 4
COHOMOLOGY MOD 3 OF THE CLASSIFYING SPACE OF E6, II 5
Put
(2.10) h = p +ty −t2x −t4x and h = p +t3y −t4x −t6x .
12 3 10 8 4 16 4 10 8 4
Then we see that
(2.11) R (h ) = h +d x and R (h ) = h −d x ,
1 12 12 8 4 1 16 16 8 8
where
(2.12) d = b(−t3 −c3 −c x +c )+b2(c2 +x )+b3(t−c )−b4
8 1 1 4 3 1 4 1
and
(2.13) R (d ) = −d .
1 8 8
It follows that
Z [t,p ,p ,c ,p ,p ] = Z [t,x ,x ,y ,h ,h ].
3 1 2 5 3 4 3 4 8 10 12 16
Further, we put
(2.14) h = t(h −x2)+t3(−h +x x )+t5x −t7x +t9
18 16 8 12 4 8 8 4
which is taken so as to satisfy
(2.15) R (h ) = h +d y .
1 18 18 8 10
It follows from (2.11) and (2.15) that the following three elements are R -invariant and
1
hence W(E )-invariant:
6
x = h x +h x ,
20 12 8 16 4
(2.16) y = h y −h x ,
22 12 10 18 4
y = h y +h x ,
26 16 10 18 8
among which holds the following relation:
(2.17) −x y +y x +y x = 0.
20 10 22 8 26 4
Summing up we have found the first six W(G)-invariant elements:
x , x , y , x , y , y .
4 8 10 20 22 26
3. The element x .
36
In order to obtain another R -invariant element, we consider the following set S (cf.
1
[TW]). Put
(3.1) w = 2τ −x for i = 1,2,...,6
i i
and denote by S the set
{ w +w for i < j, x−w , −x−w }.
i j i i
We see that the set S is invariant as a set under the action of W(E ). Therefore the
6
elementary symmetric functions σS on the 27 elements of S are invariant under the action
i
of W(E ).
6
6 M.MIMURA,Y.SAMBE, ANDM.TEZUKA
We shall calculate
27
P = 1+ σS = (1+y)
j
(3.2) y∈S
j=1
X Q
= (1+w +w )· (1+x−w )· (1−x−w )
i j j j
1≤i<j≤6 1≤j≤6 1≤j≤6
Q Q Q
and decompose the result by degree to obtain σS.
j
Since we have
x = −c , w = t−c and w = t+c −t for i > 1
1 1 1 i 1 i−1
by (3.1), (2.1), (2.2)′ and (2.3), the polynomial P is expressed in terms of c ’s and t. (The
i
calculation is carried out by Mathematica and the result has 2600 terms. )
We have from (2.6), (2.9) and (2.10) :
c = −c2 +x ,
2 1 4
c = c4 +c c +c2x −x ,
4 1 1 3 1 4 8
c = y +x t+x t3,
5 10 8 4
(3.3) c2 = h +t4x +t2x −ty +x x +t3x c +tx c −x2c2 −x c c −x c2
3 12 4 8 10 4 8 4 1 8 1 4 1 4 1 3 8 1
+y c +c6 +c3c ,
10 1 1 1 3
c y = h +t6x +t4x −t3y −x2 −t4x c2 −t3x c3 −t3x c −t2x c2 −tx c3
3 10 16 4 8 10 8 4 1 4 1 4 3 8 1 8 1
−tx c +ty c2 −h c2 +x x c2 +x c6 −x c3c −x c c −y c3 +c8.
8 3 10 1 12 1 4 8 1 4 1 4 1 3 8 1 3 10 1 1
Rewrite P making use of (3.3) and we find no c (as well as c , c and c ) in the result P
3 2 4 5 1
and P ∈ Z [c ,t,x ,x ,y ,h ,h ].
1 3 1 4 8 10 12 16
We use two more relations to get rid of c ’s.
1
Replacing c2 in the equality c2y −c ·c y = 0, we obtain a relation:
3 3 10 3 3 10
(3.4) h c = t9x +t7x2 +t7x −t6y −t5x x +t3h x +t3h +t3x2x
16 3 4 4 8 10 4 8 12 4 16 4 8
+ th x +tx x2 −ty2 +h y +x x y +t7x c2 +t6x c3 +t6x c
12 8 4 8 10 12 10 4 8 10 4 1 4 1 4 3
+ t5x c4 +t5x c2 −t4x2c3 −t4x x c −t4x c5 −t4x c2c +t4x c3
4 1 8 1 4 1 4 8 1 4 1 4 1 3 8 1
+ t4x c −t4y c2 −t3h c2 −t3x3c2 −t3x2c4 +t3x x c2 −t3x c3c
8 3 10 1 12 1 4 1 4 1 4 8 1 4 1 3
+ t3x c4 −t3x c c +t3c8 −t2x x c3 −t2x2c −t2x c5 −t2x c2c
8 1 8 1 3 1 4 8 1 8 1 8 1 8 1 3
− t2y c4 +th c4 −th c2 −tx2x c2 +tx x c4 +tx y c3 −tx c8
10 1 12 1 16 1 4 8 1 4 8 1 4 10 1 4 1
+ tx c5c +tx2c2 +tx y c −tx c6 +tx c3c −tc10 −h x c3
4 1 3 8 1 8 10 1 8 1 8 1 3 1 12 4 1
+ h x c +h c5 +h c2c −h x c −h c3 −x3c5 −x2x c3
12 8 1 12 1 12 1 3 16 4 1 16 1 4 1 4 8 1
− x2y c2 −x2c7 −x x2c +x x c5 −x x c2c −x y c4 −x c9
4 10 1 4 1 4 8 1 4 8 1 4 8 1 3 4 10 1 4 1
+ x c6c +x2c +x c7 −x c4c +y2 c −y c6 −c11 −c8c ,
4 1 3 8 3 8 1 8 1 3 10 1 10 1 1 1 3
COHOMOLOGY MOD 3 OF THE CLASSIFYING SPACE OF E6, II 7
Rewriting c2, c y andc h intheequality c2y2 −(c y )2 = 0gives riseto thefollowing
3 3 10 3 16 3 10 3 10
relation.
(3.5) c16 = − t12x2 +t10x3 +t10x x −t9x y −t8x2 +t7x2y −t7x y
1 4 4 4 8 4 10 8 4 10 8 10
+ t6h x2 +t6h x +t6x3x −t6x x2 −t6y2 −t5x x y
12 4 16 4 4 8 4 8 10 4 8 10
− t4h x x +t4h x −t4x2x2 −t4x y2 −t3h x y −t3h y
12 4 8 16 8 4 8 4 10 12 4 10 16 10
− t3x2x y −t3x2y +t2h x2 +t2x x3 −t2x y2 −th x y
4 8 10 8 10 12 8 4 8 8 10 12 8 10
− tx x2y −ty3 +h y2 −h2 −h x2 +x x y2 −x4 −t10x2c2
4 8 10 10 12 10 16 16 8 4 8 10 8 4 1
− t8x2c4 +t8x x c2 −t7x y c2 −t6h x c2 −t6x4c2 −t6x3c4
4 1 4 8 1 4 10 1 12 4 1 4 1 4 1
+ t6x2x c2 +t6x2c6 −t6x x c4 +t6x c8 −t6x2c2 −t5x y c4
4 8 1 4 1 4 8 1 4 1 8 1 4 10 1
− t5x y c2 +t4h x c4 −t4h x c2 −t4h x c2 +t4x3x c2 −t4x3c6
8 10 1 12 4 1 12 8 1 16 4 1 4 8 1 4 1
+ t4x2x c4 +t4x2c8 +t4x x2c2 −t4x x c6 −t4x c10 +t4x c8 −t4y2 c2
4 8 1 4 1 4 8 1 4 8 1 4 1 8 1 10 1
− t3h x x c +t3h y c2 −t3h x2c +t3x4c5 −t3x3x c3 +t3x3y c2
12 4 8 1 12 10 1 16 4 1 4 1 4 8 1 4 10 1
+ t3x3c +t3x2x c5 +t3x2y c4 −t3x x2c3 −t3x x y c2 +t3x x c7
4 1 4 8 1 4 10 1 4 8 1 4 8 10 1 4 8 1
+ t3x y2 c −t3x y c6 +t3x y c4 −t3y c8 +t2h x c4 −t2h x c2
4 10 1 4 10 1 8 10 1 10 1 12 8 1 16 8 1
− t2x2x2c2 −t2x2x c6 −t2x x2c4 +t2x x c8 +t2x2c6 −t2x c10
4 8 1 4 8 1 4 8 1 4 8 1 8 1 8 1
− t2y2 c4 −th x2c −th y c4 −th x x c +th y c2 +tx3x c5
10 1 12 8 1 12 10 1 16 4 8 1 16 10 1 4 8 1
− tx2x2c3 +tx2x y c2 +tx2x c7 +tx2y c6 +tx x2c5 +tx x y c4
4 8 1 4 8 10 1 4 8 1 4 10 1 4 8 1 4 8 10 1
− tx y c8 −tx3c +tx2c7 +tx y2 c −tx y c6 +ty c10 −h2 c4
4 10 1 8 1 8 1 8 10 1 8 10 1 10 1 12 1
− h h c2 −h x2c6 −h x x c4 +h x c8 −h x2c2 −h x y c
12 16 1 12 4 1 12 4 8 1 12 4 1 12 8 1 12 8 10 1
− h x c6 −h c10 −h x2c4 −h x y c −h x c6 +h x c4 +h c8
12 8 1 12 1 16 4 1 16 4 10 1 16 4 1 16 8 1 16 1
− x4c8 +x3x c6 +x3y c5 −x3c10 −x2x y c3 −x2y2 c2 +x2y c7
4 1 4 8 1 4 10 1 4 1 4 8 10 1 4 10 1 4 10 1
+ x x3c2 +x x y c5 −x c14 +x3c4 −x2y c3 +x y2 c2
4 8 1 4 8 10 1 4 1 8 1 8 10 1 8 10 1
+ x y c7 +x c12 +y3 c −y2 c6.
8 10 1 8 1 10 1 10 1
Rewrite c16 in P and the result P has no c and P ∈ Z [t,x ,x ,y ,h ,h ].
1 1 2 1 2 3 4 8 10 12 16
The highest degree of t is 27. We use the relations obtained from (2.14), (2.16) and
(2.17) to lower the degree of t to 8 :
(3.6) t9 = h +t7x −t5x +t3h −t3x x −th +tx2,
18 4 8 12 4 8 16 8
(3.7) h x = −h y +y ,
18 8 16 10 26
h x = h y −y ,
18 4 12 10 22
h x = −h x +x ,
16 4 12 8 20
y x = x y −y x ,
26 4 20 10 22 8
(3.8) h x = h y −h y .
18 20 12 26 16 22
8 M.MIMURA,Y.SAMBE, ANDM.TEZUKA
The result P is as follows.
3
(3.9)
P = 1+h x y −h x2x2 +h x2y t3 +h x y +h x x −h x y +h2 x x
3 12 4 26 12 4 8 12 4 10 12 8 22 12 20 4 12 20 10 12 4 8
−h3 −h h y2 +h x y3 −h x2y t3 +h x4 +h2 x2 +h3 −h y y
12 16 18 10 16 8 10 16 8 10 16 8 16 8 16 18 10 26
+h y3 t3 −h3 −x x +x x y3 +x x y3 t6 +x x y4 −x x y4 t+x x2y
18 10 18 4 8 4 8 10 4 8 10 4 8 10 4 8 10 4 8 10
−x x2y2 +x x2y t+x x3y2 +x x4 +x x4t6 −x x4y t−x y y +x y2
4 8 10 4 8 26 4 8 10 4 8 4 8 4 8 10 4 10 22 4 10
+x y3 −x y −x2x +x2x y2 −x2x y3 +x2x y3 t4 +x2x y −x2x2y
4 10 4 26 4 8 4 8 10 4 8 10 4 8 10 4 8 26 4 8 10
+x2x3y t3 +x2x4 +x2x4t4 −x2y −x2y2 −x2y4 +x2y4 t3 +x2y −x2y t3
4 8 10 4 8 4 8 4 10 4 10 4 10 4 10 4 22 4 22
+x2y −x3 −x3x t8 −x3x y t3 −x3x2 −x3x2t4 −x3x3 −x3x3t6 +x3y t7
4 26 4 4 8 4 8 10 4 8 4 8 4 8 4 8 4 10
+x3y2 −x3y2 t2 +x3y3 −x3y3 t6 +x3y −x3y t−x4x +x4x t6 +x4x y
4 10 4 10 4 10 4 10 4 22 4 22 4 8 4 8 4 8 10
−x4x y t−x5x +x5x t4 −x5y +x5y t3 +x6 −x6t6 −x y +x y2
4 8 10 4 8 4 8 4 10 4 10 4 4 8 10 8 10
−x y2 y −x y3 t8 +x y4 −x y4 t3 −x y −x2 −x2y y −x2y2 −x2y3
8 10 26 8 10 8 10 8 10 8 26 8 8 10 26 8 10 8 10
−x2y3 t4 +x2y +x2y t3 −x3 +x3y +x3y t7 +x3y2 −x3y2 t2 −x4t8
8 10 8 26 8 26 8 8 10 8 10 8 10 8 10 8
−x4y t3 −x4y2 +x5 −x5t4 −x x +x x x2 +x x2 −x x2x −x x3
8 10 8 10 8 8 20 4 20 4 8 20 4 20 4 8 20 4
+x x3t2 −x x −x x y2 −x x2 −x x2y t−x x3 +x x3t2 −x y
20 4 20 8 20 8 10 20 8 20 8 10 20 8 20 8 20 10
−x y2 −x y3 +x y3 t2 −x y +x y −x2 −y y +y2 y −y2 y
20 10 20 10 20 10 20 22 20 26 20 10 26 10 22 10 26
−y3 +y3 y −y3 y t−y4 +y4 t7 +y5 −y5 t2 +y y −y2
10 10 22 10 22 10 10 10 10 22 26 26
Now we decompose P by degree to obtain σS.
3 j
The σS for j ≤ 17 are as follows:
j
σS = 0 for 1 ≤ i ≤ 5 and i = 7,10,11,13,
i
σS = −x3 −x x ,
6 4 4 8
σS = −x2x −x2,
8 4 8 8
σS = −x2y −x y ,
9 4 10 8 10
σS = −x x +x6 −x4x +x y2 −x3,
12 20 4 4 4 8 4 10 8
(3.10)
σS = x (x2 −x )−x5x −x3x2 −x2y2 +x y2 ,
14 20 4 8 4 8 4 8 4 10 8 10
σS = x y +y (x2 +x )−x5y +x x2y −y3 ,
15 20 10 22 4 8 4 10 4 8 10 10
σS = −x x3 +x3y2 ,
16 20 4 4 10
σS = x x y +y (x3 −x x )−y x
17 20 4 10 22 4 4 8 26 8
+ x4x y −x2x2y +x y3 +x3y .
4 8 10 4 8 10 4 10 8 10
The element σS is expressed as
18
σS = − x x2x −x x2 −y x y −y y −x3x3 +x2x y2 +x x4
18 20 4 8 20 8 22 4 10 26 10 4 8 4 8 10 4 8
− x2y2 −h3 +h2 x x +h x x −h x2x2 −t8x3x +t7x3y
8 10 12 12 4 8 12 20 4 12 4 8 4 8 4 10
− t6x6 +t6x4x +t4x5x −t4x3x2 +t3h x2y −t3y x2
4 4 8 4 8 4 8 12 4 10 22 4
+ t3x5y −t3x3x y +t2x x3 −t2x3y2 −ty x3 −tx4x y
4 10 4 8 10 20 4 4 10 22 4 4 8 10
COHOMOLOGY MOD 3 OF THE CLASSIFYING SPACE OF E6, II 9
which cannot be expressed in terms of the R -invariant elements already obtained. We put
1
(3.11) g = − t8x +t7y −t6x3 +t6x x +t4x2x −t4x2 +t3h
24 8 10 4 4 8 4 8 8 18
+ t3x2y −t3x y +t2x −t2y2 −ty −tx x y
4 10 8 10 20 10 22 4 8 10
and
(3.12) x = − g x3 +h3 −h2 x x −h x x +h x2x2.
36 24 4 12 12 4 8 12 20 4 12 4 8
Then
R (h3 −h2 x x −h x x +h x2x2)
1 12 12 4 8 12 20 4 12 4 8
= (h3 −h2 x x −h x x +h x2x2)
12 12 4 8 12 20 4 12 4 8
+ x3(d3 −d2x −d h +d x2)
4 8 8 8 8 16 8 8
and
(3.13) R (g ) = g +d3 −d2x −d h +d x2
1 24 24 8 8 8 8 16 8 8
hold. The element x is shown to be R -invariant and we have
36 1
σS = − x +x (−x2x −x2)−y x y −y y
18 36 20 4 8 8 22 4 10 26 10
(3.14)
− x3x3 +x2x y2 +x x4 −x2y2 .
4 8 4 8 10 4 8 8 10
For the sake of completeness, we list the result of σS for j > 18, where x and x are the
j 48 54
elements found in [MS] and will be studied in §4.
σS = −x2 +x x x2 +x2x4 +x5 −x y2 −x x2y2 −y4 ,
20 20 20 4 8 4 8 8 20 10 4 8 10 10
σS = x x x y +x3y3 +x x y3 −x y −x x2y +y2 y +x2y ,
21 20 4 8 10 4 10 4 8 10 20 22 4 8 22 10 22 8 26
σS = −x x3 +x3y2 ,
22 20 8 8 10
σS = −x2x y3 −x2y3 +x y −y2 y ,
23 4 8 10 8 10 20 26 10 26
σS = x −x x y2 +x x3y2 −x2y4 +x y4 +y y ,
24 48 20 8 10 4 8 10 4 10 8 10 22 26
σS = −x y3 +y5 ,
25 20 10 10
σS = −x4y2 +x x y4 +y3 y −x2y y −y2 ,
26 8 10 4 8 10 10 22 8 10 26 26
σS = −x .
27 54
4. The invariant subalgebra H∗(BT;Z )W(E6)
3
Notation Let Ω be a field extension over a field k. For elements x , ..., x of Ω,
1 n
we denote an algebra generated by x , ..., x over k by k[x ,...,x ] and its quotient
1 n 1 n
field by k(x ,...,x ). In this section, we note that we use the notations k[x ,...,x ] and
1 n 1 n
k(x ,...,x ) by the above extended meaning.
1 n
For a field k and Ω, we take Z and Z (t, t , ...,t ) respectively. We put K =
3 3 1 6
Z (x ,x ,y ,x ,y ,x ) and L = Z (t,x ,x ,y ,h ,h ). We see K ⊂ L.
3 4 8 10 20 22 36 3 4 8 10 12 16
Lemma 4.1. We have L = K(t). The extension degree [L : K] is at most 27.
10 M.MIMURA,Y.SAMBE, ANDM.TEZUKA
Proof. From the definition, we see K ⊂ L and t ∈ L. It means K(t) ⊂ L, where K(t) is a
field generated by K and t in L. To show L ⊂ K(t), it is sufficient to show that h ∈ K
12
Because it implies that h = (x −h x )/x ∈ K(t).
16 20 12 8 4
Replacing h by its defining expression (2.14) in y = h y −h x , we obtain
18 22 12 10 18 4
(4.1) y = (y +tx +t3x )h −t(x −x x2)−t3x2x −t5x x +t7x2 −t9x .
22 10 8 4 12 20 4 8 4 8 4 8 4 4
Hence
(4.2) (y +tx +t3x )h = y +t(x −x x2)+t3x2x +t5x x −t7x2 +t9x .
10 8 4 12 22 20 4 8 4 8 4 8 4 4
Thus h ∈ K(t) and so does h . We have shown that L = K(t).
12 16
Now we shall find a equation of t over K.
(3.16)′ 0 = x +g x3 −h3 +h2 x x +h x x −h x2x2.
36 24 4 12 12 4 8 12 20 4 12 4 8
Replacing g by its defining expression (3.11), h x by x −h x and h x by −y +
24 16 4 20 12 8 18 4 22
h y , we obtain a polynomial in t, x , x , y , x , y , x and h . Since its degree with
12 10 4 8 10 20 22 36 12
respect to h is 3, we multiply (3.12)′ by (y +tx +t3x )3. Then by making use of (4.2)
12 10 8 4
we obtain a polynomial in Z [t,x ,x ,y ,x ,y ,x ] of degree 27 with respect to t:
3 4 8 10 20 22 36
0 = x y3 +x y x y2 +y2 x x y −y3 −y x2x2y2 +tx3x4y2
36 10 20 22 4 10 22 4 8 10 22 22 4 8 10 4 8 10
−tx4x y4 +tx2 x y2 +tx x2x2y2 +tx y x x y +ty2 x x2
4 8 10 20 4 10 20 4 8 10 20 22 4 8 10 22 4 8
−ty x2x3y −ty x3y3 −t2x3y5 +t2x x3y3 −t3x3x y4
22 4 8 10 22 4 10 4 10 20 4 10 4 8 10
−t3x4x3y2 +t3x5y4 −t3x2 x x2 −t3x3 −t3x x2x4 +t3x x3x y2
4 8 10 4 10 20 4 8 20 20 4 8 20 4 8 10
−t3x y x2y +t3x x3 +t3y2 x2x +t4x3x2y3 +t4x5x y3
20 22 4 10 36 8 22 4 8 4 8 10 4 8 10
−t4x2 x2y +t4x x2y3 +t4x y x2x −t4y x2x y2 −t5x3x3y2
20 4 10 20 4 10 20 22 4 8 22 4 8 10 4 8 10
+t5x x3x3 −t6x4x y3 −t6x6y3 −t6x x2x2y −t6x x4x y
20 4 8 4 8 10 4 10 20 4 8 10 20 4 8 10
+t6x y x3 +t6y x2x3 −t6y x3y2 +t6y x4x2 −t7x3x5 +t7x3y4
20 22 4 22 4 8 22 4 10 22 4 8 4 8 4 10
(4.3) +t7x4x2y2 −t7x5x4 +t7x2 x3 +t7x x3y2 −t7x x4x2 +t9x3x2y2
4 8 10 4 8 20 4 20 4 10 20 4 8 4 8 10
−t9x4x4 −t9x5x y2 +t9x6x3 +t9x x2y2 +t9x x3x2
4 8 4 8 10 4 8 20 4 10 20 4 8
+t9x x5x +t9x x3 −t9y x2x y +t9y x4y −t10x3x3y
20 4 8 36 4 22 4 8 10 22 4 10 4 8 10
−t10x4y3 +t10x5x2y −t10x7x y +t10x x2x y −t10x x4y
4 10 4 8 10 4 8 10 20 4 8 10 20 4 10
−t10y x2x2 +t10y x4x −t10y x6 −t11x6y2 +t11x x6 +t12x3y3
22 4 8 22 4 8 22 4 4 10 20 4 4 10
−t12x4x2y +t12x8y −t12x x3y −t12y x3x −t12y x5
4 8 10 4 10 20 4 10 22 4 8 22 4
−t13x3x y2 +t13x5y2 +t13x6x2 +t13x8x +t13x x3x −t13x x5
4 8 10 4 10 4 8 4 8 20 4 8 20 4
+t15x3x3 −t15x4y2 +t15x7x −t15x9 +t15x x4 +t18x3x y
4 8 4 10 4 8 4 20 4 4 8 10
+t18x5y +t19x3x2 +t19x5x +t21x4x +t21x6 −t27x3.
4 10 4 8 4 8 4 8 4 4
Hence we conclude [L : K] = [K(t) : K] ≤ 27. (cid:3)
Corollary 4.2. The elements x , x , y , x , y , x are algebraically independent over
4 8 10 20 22 36
Z .
3