A note on Samelson products in the exceptional Lie groups

A note on Samelson products in the exceptional Lie groups Hiroaki Hamanaka and Akira Kono October 23, 2008 1 Introduction Samelson products have been studied extensively for the classical groups ([5], [9], [10]), but few results are known for exceptional Lie groups. In [13], Ōshima determines the Samelson product π n (G 2 ) π 11 (G 2 ) π n+11 (G 2 ) for n = 3, 11. Let G(l) be the compact, simply connected, exceptional simple Lie group of rank l, where l = 2, 4, 6, 7, 8. Define the set of integers N(l) and the prime r(l) as in the following table. l G(l) N(l) r(l) 2 G 2 {2, 6} 7 4 F 4 {2, 6, 8, 12} 13 6 E 6 {2, 5, 6, 8, 9, 12} 13 7 E 7 {2, 6, 8, 10, 12, 14, 18} 19 8 E 8 {2, 8, 12, 14, 18, 20, 24, 30} 31 If p is a prime and p r(l), then G(l) is p-regular (see [14]), that is, there is a homotopy equivalence j N(l) S 2j 1 (p) G(l) (p), where (p) stands for the localization at the prime p in the sense of Bousfield and Kan [3]. For k N(l) define ɛ 2k 1 π 2k 1 (G(l) (p) ) = Z (p) by the composition S 2k 1 i j N(l) S 2j 1 (p) G(l) (p), where i is the canonical inclusion. The purpose of this paper is to show: 1

Theorem 1.1. If k 1, k 2 N(l) satisfy k 1 +k 2 = r(l)+1, then the Samelson product ɛ 2k1 1, ɛ 2k2 1 0 in π 2r(l) (G(l) (r(l)) ). Theorem 1.2. If k 1 and k 2 N(l) satisfy k 1 + k 2 = r(l) + 1, then the Samelson product ɛ 2k1 1, ɛ 2k2 1, ɛ 2r(l) 3 0 in π 4r(l) 3 (G(l) (r(l)) ). Corollary 1.1. The nilpotency class of the localized self homotopy group [G(l), G(l)] r(l) greater than or equal to 3. is In order to prove Theorem 1.1, we use the following lemma which will be proved in 3 and 4. Let p be a prime greater than 5. Then we have H (BG(l); F p ) = F p [y 2j ; j N(l)], y 2j = 2j. (1.1) Lemma 1.1. Modulo ( H (BG(l); Z/p)) 3, we have ξ 1 y 4 y 60 + ξ 2 y 16 y 48 + ξ 3 y 24 y 40 + ξ 4 y 28 y 36 (l, p) = (8, 31) P 1 ξ 1 y 4 y 36 + ξ 2 y 12 y 28 + ξ 3 y 16 y 24 + ξ 4 y20 2 (l, p) = (7, 19) y 4 ξ 1 y 4 y 24 + ξ 2 y 12 y 16 + ξ 3 y 10 y 18 (l, p) = (6, 13) ξ 1 y 4 y 24 + ξ 2 y 12 y 16 (l, p) = (4, 13) for ξ j (Z/p). Proof of Theorem 1.1. The proof for l = 2 is done in [13]. Put l = 4, 6, 7. Then we follow the proof of [8, Theorem 1.1]. Consider the map ɛ 2k : S2k BG(l) (p) which is the adjoint of ɛ 2k 1. Suppose that the Whitehead product [ɛ 2k 1, ɛ 2k 2 ] = 0 for k 1, k 2 N(l) and k 1 + k 2 = p + 1 (p = r(l)). Then we have a homotopy commutative diagram: S 2k 1 S 2k 2 ɛ 2k 1 ɛ 2k 2 BG(l)(p) S 2k 1 S 2k 2 θ BG(l) (p), where the left vertical arrow is the inclusion. It is clear that P 1 θ (y 4 ) = 0. On the other hand, we have θ (P 1 y 4 ) 0. Then we obtain [ɛ 2k 1, ɛ 2k 2 ] 0 and thus, by adjointness of Whitehead products and Samelson products, we have established ɛ 2k1 1, ɛ 2k2 1 0. Proof of Theorem 1.2. If p = r(l), G(l) is p-regular and then π 2p (G(l) (p) ) = j N(l) π 2p (S 2j 1 (p) ) = π 2p (S(p)) 3 2

for a dimensional reason (see [15]). If k 1, k 2 N(l) and k 1 + k 2 = p + 1, there is an integer ξ k 1,k 2 Z (p) satisfying a homotopy commutative diagram S 2p α 1 S 3 ξ ɛ 2k1 1,ɛ 2k2 1 G(l) (p), k 1,k 2 ɛ 3 where α 1 is a generator of the p-primary component of π 2p (S 3 ) which is isomorphic to Z/p. In particular, we have a homotopy commutative diagram: S 2p α 1 S 3 ξ ɛ 3,ɛ 2p 1 Then there is a homotopy commutative diagram: G(l) (p) 2,p 1 ɛ 3 S 2p S 2p 3 α 1 1 S 2p 3 S 3 S 2p 3 S 2p α 1 S 3 ɛ 2k1 1,ɛ 2k2 1 ɛ 2p 3 ξ k 1,k ɛ 3 ɛ 2p 3 ξ 2 k 1,k ɛ 3,ɛ 2p 3 2 ξ k 1,k ξ 2 2,p 1 ɛ 3 γ G(l) (p) G(l) (p) G(l) (p) G(l) (p) G(l) (p), where γ is the commutator map of G(l) (p). Since α 1 (α 1 1 S 2p 3) 0 in π 4p+1 (S(p) 3 ) (see [14] and [15]), we have established Theorem 1.2. Proof of Corollary 1.1. Define θ i [G(l) (p), G(l) (p) ] for i N(l) by the composition q G(l) (p) S 2k 1 1 ɛ 2k 1 1 G(l) (p), where q is the projection. If l 4, there are k 1, k 2 N(l) satisfying k 1 < k 2 < p = r(l) and k 1 + k 2 = p + 1. Then it follows from Theorem 1.2 that the commutator [[θ k1, θ k2 ], θ p 1 ] in [G(l) (p), G(l) (p) ] = [G(l), G(l)] (p) is nontrivial ([3]) and therefore the proof is completed. 2 W (E 8 )-invariant polynomials Let p be a prime > 5. Then we have H (BG(l); Z (p) ) = Z (p) [ỹ 2j ; j N(l)], ρ(ỹ i ) = y i, where ρ is the modulo p reduction and y i are as in (1.1). Let T be a maximal torus of G(l) and let W (G(l)) be the Weyl group of G(l). Consider the fibre sequence: G(l)/T BT λ l BG(l) 3

Then λ l (ỹ 2j) is W (G(l))-invariant and the sequence {λ l (ρ(ỹ 2j)); j N(l)} is a regular sequence. Let α i (i = 1,..., 8) be the simple roots E 8. Then, as is well known, the dominant root α of E 8 is given by α = 2α 1 + 3α 2 + 4α 3 + 6α 4 + 5α 5 + 4α 6 + 3α 7 + 2α 8. Recall that the completed Dynkin diagram of E 8 is: α 1 α 3 α 4 α 5 α 6 α 7 α 8 α α 2 Let e i be the dual of (0,..., 0, 1, i 0,..., 0) R 8 for i = 1,..., 8. Then we can put α 1 = 1 2 (e 1 + e 8 ) 1 2 (e 2 + e 3 + e 4 + e 5 + e 6 + e 7 ) α 2 =e 1 + e 2 α i =e i 2 e i 3 (3 i 8). Hence we have α = e 7 + e 8 (see [7]). Put t 1 = e 1, t 8 = e 8 and t i = e i for i = 2,..., 7. are generator of H (BT ). Let ϕ i and ϕ be the reflections on the hyperplanes α i = 0 and α = 0 respectively. Then it is well known that W (E 8 ) is generated by ϕ 1,..., ϕ 8. Let W be the subgroup of W (E 8 ) generated by ϕ 2,..., ϕ 8, ϕ. Namely, W is the Weyl group of Ss(16) in E 8 which is a compact connected simple Lie group of type D 8. Put ϕ = ϕ 1. Then it is straight forward to check ϕ(t i ) = t i 1 (t 4 1 + + t 8 ) (2.1) for i = 1,..., 8. We can regard t 1,..., t 8 are a basis of H 2 (BT ; Z (p) ). Define polynomials c i and p i by 8 (1 + t i ) = i=1 8 i=0 c i and 8 8 (1 t 2 i ) = ( 1) i p i i=1 i=0 respectively. Then since W is the Weyl group of Ss(16) in E 8 as is noted above, we have H (BT ; Z (p) ) W = Z (p) [p 1,..., p 7, c 8 ]. 4

It follows from (2.1) that 8 ϕ(c i ) = i=0 8 (1 + ϕ(t i )) = i=1 8 (1 1c 4 1 + t i ) = i=1 8 (1 1c 4 1) 8 k c k k=0 and then we obtain ϕ(c 1 ) = c 1, ϕ(c 2 ) = c 2 and ϕ(c i ) c i 1(8 k + 1)c 4 i 1c 1 mod (c 2 1) (2.2) for i = 3,..., 8. In particular, we have ϕ(p 1 ) = p 1 and the ideals (c 1 ), (c 2 1, p 1 ) = (c 2 1, c 2 ), (c 2 1, p 2 1) = (c 2 1, c 2 2) are W (E 8 )-invariant. Then it follows from (2.2) that ϕ(p j ) p j + h j c 1 mod (c 2 1) for i = 2,..., 8, where h 2 = 3c 2 3, h 3 = 1(5c 2 5 + c 3 c 2 ), h 4 = 1(7c 2 7 + 3c 5 c 2 c 4 c 3 ), h 5 = 1(5c 2 7c 2 3c 6 c 3 + c 5 c 4 ), h 6 = 1(5c 2 8c 3 3c 7 c 4 + c 6 c 5 ), h 7 = 1(3c 2 8c 5 c 7 c 6 ). Summarizing, we have established: Lemma 2.1. Modulo (c 2 1), we have ϕ(p 4 ) p 4 + 1 2 (c 7c 1 + 3c 5 c 2 c 1 c 4 c 3 c 1 ), ϕ(p 2 2) p 2 2 + 6c 4 c 3 c 1 + 3c 3 c 2 2c 1, ϕ(c 8 ) c 8 1 4 c 7c 1, ϕ(p 3 p 1 ) p 3 + 5c 5 c 2 c 1 + c 3 c 2 2c 1, ϕ(p 2 p 2 1) p 2 + 6c 3 c 2 2c 1. Corollary 2.1. If f H 16 (BT ; Z (p) ) satisfies ϕ(f) f such that f = α f 16 + α p 4 1, mod (c 2 1), then there exist α, α Z (p) where f 16 = 120p 4 + 10p 2 2 + 1680c 8 36p 3 p 1 + p 2 p 2 1. We also have established: Lemma 2.2. Modulo (c 2 1, c 2 2) = (c 2 1, p 2 1), we have ϕ(p 0 ) p 6 5c 2 8c 3 c 1 + 3c 2 7c 4 c 1 1c 2 6c 5 c 1, ϕ(p 5 p 1 ) p 5 p 1 3c 6 c 3 c 2 c 1 + c 5 c 4 c 2 c 1, ϕ(p 4 p 2 ) p 4 p 2 + 3c 8 c 3 c 1 + 7c 7 c 4 c 1 + 3c 6 c 3 c 2 c 1 + 3c 5 c 4 c 2 c 1, 3c 5 c 2 3c 1 + 1 2 c2 4c 3 c 1, ϕ(c 8 p 2 ) c 8 p 2 + 3c 2 8c 3 c 1 1c 2 7c 4 c 1, ϕ(p 2 3) p 2 3 + 10c 6 c 5 c 1 + 2c 6 c 3 c 2 c 1 + 10c 5 c 4 c 2 c 1 5c 5 c 2 3c 1 c 3 3c 2 c 1, ϕ(p 3 p 2 p 1 ) p 3 p 2 p 1 + 6c 6 c 3 c 2 c 1 + 10c 5 c 4 c 2 c 1 3c 3 3c 2 c 1, ϕ(p 3 2) p 3 3 + 18c 2 4c 3 c 1. 5

Corollary 2.2. If f H 24 (BT ; Z (p) ) satisfies ϕ(f) f such that f β f 24 mod (p 2 1), where mod (c 2 1, p 2 1), there exists β Z (p) f 24 = 60p 6 5p 5 p 1 5p 4 p 2 + 110c 8 p 2 + 3p 2 3 p 3 p 2 p 1 + 5 36 p3 2. Now we make a choice of generators of H (BE 8 ; Z (p) ) as follows: Theorem 2.1. We can choose ỹ 16 and ỹ 24 such that λ 8(ỹ 16 ) = f 16 and λ 8(ỹ 24 ) = f 24 mod (p 2 1). Proof. First of all, we can choose ỹ 4 such that λ 8(ỹ 4 ) = p 1. Then since ϕ(λ 8(ỹ 16 )) = λ 8(ỹ 16 ), it follows from Corollary 2.1 that we can choose ỹ 16 such that λ 8(ỹ 16 ) = α f 16 for some α Z (p). Suppose that (α, p) = p. Then {λ 8(ρ(ỹ 4 )), λ 8(ρ(ỹ 16 ))} is not a regular sequence and this is a contradiction. Thus we obtain (α, p) = 1. The case of ỹ 24 is quite similar. 3 Proof of Lemma 1.1 for l = 8 We abbreviate the modulo 31 reduction of t i, c i, p i by the same t i, c i, p i respectively. We write the modulo 31 reduction of f i by f i for i = 16, 24. Put T n = t 2n 1 + + t 2n 8. Then, by Girard s formula ([11]), we have: ( 1) k T k = k ( 1) (i i1+ +i8 1 + i 8 1)! p i 1 i i 1 +2i 2 + +8i 1! i 8! 1 p i 8 8. (3.1) 8 On the other hand, we have λ 8(y 4 ) = p 1 = T 1, P 1 T 1 = 2T 16 and then λ 8(P 1 y 4 ) = 2T 16 (3.2) We denote the subalgebra F 31 [p 1,..., p 7, c 8 ] of H (BT ; F 31 ) by R. Note that Imλ 8 R. Define an algebra homomorphism π 1 : R F 31 [x 1, x 5 ]/(x 2 1) by π 1 (p i ) = 0 (i = 2, 3, 4, 7), π 1 (p 1 ) = x 1, π 1 (p 5 ) = x 5, π 1 (p 6 ) = 1 12 x 1x 5, π 1 (c 8 ) = 0. Put φ 1 = π 1 λ 8. Then we have φ 1 (y 2 4) = and, by Theorem 2.1, φ 1 (y 16 ) = φ 1 (y 24 ) = 0. Hence, for a degree reason, we can put P 1 y 4 = ξ 1 y 4 y 60 + γ 1 for ξ 1 F 31 and γ 1 Kerφ 1. It follows from (3.1) that 2T 16 p 1 p 3 5 p 2 5p 6 mod Kerπ 1 and hence we obtain π 1 (2T 16 ) = 11 12 x 1x 3 5 0. Thus, by (3.2), we have established φ 1 (P 1 y 4 ) = π 1 (2T 16 ) 0 which implies ξ 1 0. Define an algebra homomorphism π 2 : R F 31 [x 4, x 4] by π 2 (p i ) = 0 (i = 1, 2, 3, 5, 6, 7), π 2 (p 4 ) = x 4, π 2 (c 8 ) = x 4. 6

Put φ 2 = π 2 λ 8. Then we have φ 2 (y 4 ) = 0 and, for a degree reason, φ 2 (y 24 ) = φ 2 (y 28 ) = 0. Thus, for a dimensional reason, we can put P 1 y 4 = ξ 2 y 16 y 48 + ξ 2y 4 16 + γ 2 for ξ 2, ξ 2 F 31 and γ 2 Kerφ 2. Now, by (3.1), we have 2T 16 = 1 4 p4 4 p 2 4p 8 + 1 2 p2 8 mod Kerπ 2 and then π 2 (2T 16 ) = 1 4 x4 4 x 42 x 2 4 + 1 2 x 44, where p 8 = c 2 8. This implies that if ξ 2 = 0, then ξ 2 0. On the other hand, it follows from Theorem 2.1 that φ 2 (y 16 ) = 120(x 4 + 14x 4). Suppose that ξ 2 = 0. Then ξ 2 0 as above and thus, by (3.2), φ(p 1 y 4 ) = ξ 2(120(x 4 + 14x 4)) 4 = 1 4 x4 4 x 42 x 2 4 + 1 2 x 44. This is a contradiction and therefore ξ 2 0. Define an algebra homomorphism π 3 : R R 3 = K[x 5, x 6 ] by π 3 (p i ) = 0 (i = 1, 2, 3, 4, 7), π 3 (p 5 ) = x 5, π 3 (p 6 ) = x 6, π 3 (c 8 ) = 0. Put φ 3 = π 3 λ 8. Then it follows that φ 3 (y 4 ) = φ 3 (y 16 ) = φ 3 (y 28 ) = 0 and hence we can put P 1 y 4 = ξ 3 y 24 y 40 + γ 3 for ξ 3 F 31 and γ 3 Kerφ 3. By (3.1), we have 2T 16 = p 2 5p 6 mod Kerπ 3 and then π 3 (2T 16 ) = x 2 5x 6 0 which implies ξ 3 0 by (3.2). Define π 4 : R F 31 [x 3, x 7 ] by π 4 (p i ) = 0 (i = 1, 2, 4, 5), π 4 (p 3 ) = x 3, π 4 (p 6 ) = 1 20 x2 3, π 4 (p 7 ) = x 7, π 4 (c 8 ) = 0. Put φ 4 = π 4 λ 8. Then we have φ 4 (y 4 ) = φ 4 (y 16 ) = 0 and, by Theorem 2.1, φ 4 (y 28 ) = 0. Thus, for a dimensional reason, we can put P 1 y 4 = ξ 4 y 28 y 36 + γ 4 for ξ 4 F 31 and γ 4 Kerφ 4. It follows from (3.1) that 2T 16 = 2p 3 p 6 p 7 + p 3 3p 7 mod Kerπ 4 and then π 4 (2T 16 ) = 2 20 x3 3x 7 + x 3 3x 7 = 11 10 x3 3x 7 0. Thus, for (3.2), we obtain ξ 4 0. Now we have established Lemma 1.1 for l = 8. 4 Proof of Lemma 1.1 for l = 4, 6, 7 Let us first recall the construction of G(l) for l = 4, 6, 7. Consider the following commutative diagram of the natural inclusions: SU(2) SU(3) G 2 Spin(4) Spin(6) Spin(7) Spin(16) 7

Note that we have the canonical map i : Spin(16) Ss(16) E 8 as in the previous section. Then i(g 2 ), i(su(3)), i(su(2)) are the closed subgroups of E 8 and we know that F 4, E 6, E 7 are the identity component of the centralizers of the image of the above i(g 2 ), i(su(3)), i(su(2)) in E 8 respectively (see [1]). Consider the natural inclusion Spin(k) Spin(16 k) Spin(16) for k = 4, 5, 6, 7. Then we obtain a commutative diagram of inclusions: Spin(9) j 1 Spin(10) j 2 Spin(11) Spin(12) i 1 i 2 i 3 i 4 F 4 k 1 E 6 k 2 E 7 E 7 Next we recall classical results due to [2], [6] and [16] on the cohomology of homogeneous spaces given by the above inclusions: Lemma 4.1. The integral cohomology of E 6 /F 4, E 6 /Spin(10) and E 7 /E 6 are given as follows. 1. H (E 6 /F 4 ; Z) = Λ(x 9, x 17 ), x j = j. 2. H (E 6 /Spin(10); Z) = Z[x 8 ]/(x 3 8) Λ(x 17 ), x j = j. 3. H (E 7 /E 6 ; Z) = Z{1, z 10, z 18, z 37, z 45, z 55 } Z/2{z 28 }, z j = j, where R{a 1, a 2,...} stands for a free module with a basis a 1, a 2,... over a ring R. Hereafter, we let p be a prime greater than 5. Then the mod p cohomology of BG(l) is given by (1.1). Then, by the standard spectral sequence argument together with 4.1, we obtain: Lemma 4.2. 1. We can choose generators y 2i H (BE 6 ; F p ) such that { k1(y 0 i = 5, 9 2i ) = i N(4) y 2i 2. We can choose generators y 2i H (BE 7 ; F p ) such that y10 2 i = 10 k2(y 2i ) = y 10 y 18 i = 14 i N(4). Recall that we have H (BSpin(2l + 1); F p ) = F p [p 1,..., p l ], H (BSpin(2l); F p ) = F p [p 1,..., p l 1, c l ], y 2i 8

where p i is the i-th universal Pontrjagin class and c l is the Euler class. Let T l be the standard maximal torus of Spin(2l + ɛ) for ɛ = 0, 1 and let t 1,..., t l be the standard generators of H 2 (BT l ; F p ). Then the canonical map λ : BT l BSpin(2l + 1) satisfies l ( 1) j λ (p j ) = j=0 l (1 t 2 i ), (4.1) i=1 where p 0 = 1 and c 2 l = p l (see [4]). Specializing to our case, we have j 1(p k ) = p k (k = 1, 2, 3, 4), j 2(p k ) = p k (k = 1, 2, 3, 4), j 1(c 5 ) = 0, j 2(p 5 ) = c 2 5. It follows from [4, (3) in 19] that we can choose y 2i H (BF 4 ; F p ) for i = 2, 6, 8 such as 4.1 The case of l = 7 i 1(y 4 ) = p 1, i 1(y 12 ) = 6p 3 + p 2 p 1, i 1(y 16 ) = 12p 4 + p 2 2 1 2 p2 1p 2. (4.2) We put l = 7 and p = 19 in the above observation. Put T n = t 2n 1 + + t 2n 5. Then we have Girard s formula (3.1). By (4.1), we have i 3y 4 = p 1 = T 1 and then i 3(P 1 y 4 ) = P 1 i 3(y 4 ) = P 1 T 1 = 2T 10. (4.3) Define an algebra homomorphism π 1 : F 19 [p 1,..., p 5 ] F 19 [x 1, x 2, x 5 ]/(x 2 1, x 3 2, x 2 5) by π 1 (p i ) = x i (i = 1, 2, 5), π 1 (p 3 ) = 1 6 x 2x 1, π 1 (p 4 ) = 1 12 x2 2 Then we have π 1 (p 4 1 2 ) = π 1 (p 1 p 3 ) = 0. Put φ 1 = π 1 i 3. Then it follows from Lemma 4.2, (4.2) and a degree reason that φ 1 (y 2 4) = φ 1 (y 12 ) = φ(y 16 ) = φ 1 (y 20 ) = 0. Hence, for a degree reason, we can put P 1 y 4 = ξ 1 y 4 y 36 + γ 1 for ξ 1 F 19 and γ 1 Kerφ 1. On the other hand, by (4.3), we have 2T 10 3p 1 p 2 2p 5 p 1 p 4 p 5 p 2 p 3 p 5 mod Kerπ 1 and then π 1 (2T 10 ) = 35x 12 1x 2 2x 5 0. Thus, by (4.3), we have obtained ξ 1 0. Define an algebra homomorphism π 2 : F 19 [p 1,..., p 5 ] F 19 [x 2, x 3, x 5 ]/(x 2 2, x 2 3, x 2 5) by π 2 (p i ) = x i (i = 2, 3, 5), π 2 (p i ) = 0 (i = 1, 4). Put φ 1 = π 2 i 3. Then it follows from Lemma 4.2, (4.2) and a degree reason that φ 2 (y 4 ) = φ 2 (y 16 ) = φ 2 (y 2 20) = 0 and hence we can put P 1 y 4 = ξ 2 y 12 y 28 + γ 2 9

for ξ 2 F 19 and γ 2 Kerφ 2. Now, by (3.1), we have π 2 (2T 10 ) = x 2 x 3 x 5 and then, by (4.3), ξ 2 0. Define algebra homomorphisms π 3 : F 19 [p 1,..., p 5 ] F 19 [x 2 ] and π 4 : F 19 [p 1,..., p 5 ] F 19 [x 5 ] by π 3 (p 2 ) = x 2, π 3 (p i ) = 0 (i 2), π 4 (p i ) = 0, π 4 (p 5 ) = x 5 (i 5). Put φ i = π i i 3 for i = 3, 4. Then it follows from Lemma 4.2, (4.2) and a degree reason that φ 3 (y 4 ) = φ 3 (y 12 ) = φ 3 (y 20 ) = 0 and φ 4 (y 4 ) = φ 4 (y 12 ) = φ 4 (y 16 ) = 0. Thus we can put P 1 y 4 = ξ 3 y 16 y 24 + γ 3 = ξ 4 y 2 20 + γ 4 for ξ 3, ξ 4 F 19, γ 3 Kerφ 3 and γ 4 Kerφ 4. On the other hand, by (3.1), we have 2T 10 = 1 5 p5 2 mod Kerπ 3 and 2T 10 = 1 2 p2 5 mod Kerπ 4. Then π 3 (2T 10 ) = 1 5 x5 2 0 and π 4 (2T 10 ) = 1 2 x2 5 0 which imply ξ 3 0 and ξ 4 0. Therefore the proof of Lemma 1.1 for l = 7 is completed. 4.2 The case of l = 4, 6 We first consider the case l = 4 and p = 13. As in the above sections, we put T n = t 2n 1 + t 2n 4. Then we have Girard s formula (3.1) and i 1(P 1 y 4 ) = 2T 7. (4.4) Define an algebra homomorphism π 1 : F 13 [p 1,..., p 4 ] F 13 [x 1, x 2 ]/(x 2 1) by π 1 (p 1 ) = x 1, π 1 (p 2 ) = x 2, π 1 (p 3 ) = 1 6 x 1x 2, π 1 (p 4 ) = 1 12 x2 2. Put φ 1 = π 1 i 1. Then it follows from (4.2) that φ 1 (y 2 4) = φ 1 (y 12 ) = φ 1 (y 16 ) = 0 and thus we can put P 1 y 4 = ξ 1 y 4 y 24 + γ 1 for ξ 1 F 13 and γ 1 Kerφ 1. By (3.1), we have 2T 7 = (3p 1 p 3 2 2p 1 p 2 p 4 2p 2 2p 3 + p 3 p 4 ) mod Kerπ 1 and then π 1 (T 7 ) 0. Thus, for (4.4), we have obtained ξ 1 0. We define an algebra homomorphism π 1 : F 13 [p 1,..., p 4 ] F 13 [x 3, x 4 ] by π 2 (p i ) = 0 (i = 1, 2), π 2 (p i ) = x i (i = 3, 4). Put φ 2 = π 2 i 1. Then, by (4.2), we have φ 2 (y 4 ) = 0 and then we can put P 1 y 4 = ξ 2 y 12 y 16 + γ 2 for ξ 2 F 13 and γ 2 Kerφ 2. It follows from (3.1) that 2T 10 = p 3 p 4 mod Kerπ 2 and then π 2 (2T 7 ) 0 which implies ξ 2 0 by (4.4). Thus the proof of Lemma (1.1) for l = 4 is completed. 10

Next we consider the case l = 6 and p = 13. By Lemma 4.2 and the above result for l = 4, we only have to show ξ 3 0 in Lemma 1.1. As is seen in [12, Theorem 5.18], we have P 1 σ(y 4 ) = ξ 3 σ(y 28 ) in H (E 7 ; F 13 ) for some ξ 3 0 F 13, where σ denotes the cohomology suspension. By Lemma 4.2, we have k2(y 28 ) = y 10 y 18 and then we obtain P 1 y 4 ξ 3 y 10 y 18 mod (y 4, y 12, y 16 ) in H (BE 6 ; F 13 ). Therefore the proof of Lemma 1.1 is accomplished. References [1] J.F. Adams, Lectures on Exceptional Lie Groups, Chicago Lectures in Mathematics. University of Chicago Press, Chicago, IL, 1996. [2] S. Araki, On the non-commutativity of Pontrjagin rings mod 3 of some compact exceptional groups, Nagoya Math. J. 17 (1960), 225-260. [3] A.K. Bousfield and D.M. Kan, Homotopy Limits, Completions and Localizations, Lecture Notes in Mathematics, 304, Springer-Verlag, Berlin-New York, 1972. [4] A. Borel and F. Hirzebruch, Characteristic classes and homogeneous spaces I, Amer. J. Math. 80 (1958), 458-538. [5] R. Bott, A note on the Samelson products in the classical groups, Comment. Math. Helv. 34 (1960), 249-256. [6] L. Conlon, On the topology of EIII and EIV, Proc. Amer. Math. Soc. 16 ()1965), 575-581. [7] A. Kono, Hopf algebra structure of simple Lie groups, J. Math. Kyoto Univ. 17 (1977), 259-298. [8] A. Kono and H. Ōshima, Commutativity of the group of self homotopy classes of Lie groups, Bull. London Math. Soc. 36 (2004), 37-52. [9] A.T. Lundell, The embeddings O(n) U(n) and U(n) Sp(n), and a Samelson product, Michigan Math. J., 13 (1966), 133-145. [10] M. Mahowald, A Samelson product in SO(2n), Bol. Soc. Math. Mexicana, 10 (1965), 80-83. [11] J.W. Milnor and J.D. Stasheff, Characteristic classes, Ann. of Math. Studies 76, Princeton Univ. Press, Princeton N.J., 1974. [12] M. Mimura and H. Toda, Topology of Lie groups I, II, Translations of Mathematical Monographs 91, American Mathematical Society, Providence, RI, 1991. 11

[13] H. Ōshima, Samelson products in the exceptional Lie group of rank 2, J. Math. Kyoto Univ. 45 (2005), 411-420. [14] J.P. Serre, Groupes d homotopie et classes de groupes abéliens, Ann. of Math. 12 (1953), 258-294. [15] H. Toda, Composition Methods in Homotopy Groups of Spheres, Ann. of Math. Studies 49 Princeton University Press, Princeton, N.J., 1962. [16] T. Watanabe, The integral cohomology ring of the symmetric space EVII, J. Math. Kyoto Univ. 15 (1975), 363-385. H. Hamanaka, Department of Natural Science, Hyogo University of Teacher Education, Yashiro, Hyogo 673-1494, Japan E-mail: hammer@sci.hyogo-u.ac.jp A. Kono, Department of Mathematics, Kyoto University, Kyoto 606-8502, Japan E-mail: kono@math.kyoto-u.ac.jp 12