A MATHEMATICA REPRESENTATION OF SOME ... - Neil Strickland

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12 N. P. STRICKLANDProposition 30. There is a unique element ρ ′′ ∈ 〈σ ′ , 8ι 14 , 2σ 14 〉 1 that satisfies H(ρ ′′ ) = µ 13 .Proof. Put B = 〈σ ′ , 8ι 14 , 2σ 14 〉 1 and let ξ be any element of B. We know from [3, Equations 10.12]thatH(ξ) = µ 13 + xν 13 ◦ ν 16 ◦ ν 19 + yη 13 ◦ ɛ 14for some integers x and y. Next, we know from [3, Lemma 5.14] that H(σ ′ ) = η 13 , so H(σ ′ ◦ɛ 14 ) =η 13 ◦ ɛ 14 and H(σ ′ ◦ ν 14 ) = η 13 ◦ ν 14 , which is the same as ν 13 ◦ ν 16 ◦ ν 19 by [3, Lemma 6.3]. Itfollows that the element ρ ′′ = ξ − xσ ′ ◦ ν 14 − yσ ′ ◦ ɛ 14 has H(ρ ′′ ) = µ 13 . The indeterminacyof B is σ ′ ◦ Σπ21 13 + π15 7 ◦ (2σ 15 ). We see from [3, Theorem 7.1] that π15 7 has exponent 2 andΣπ21 13 = Z 2 ɛ 14 ⊕ Z 2 ν 14 ; it follows that the indeterminacy in B is generated by σ ′ ◦ ν 14 and σ ′ ◦ ɛ 14 .This shows that ρ ′′ ∈ B, and that ρ ′′ is the unique element in B with H(ρ ′′ ) = µ 13 .□Proposition 31. H(ɛ ′ ) = ɛ 5 (mod ρ (4) ).Proof. Using [3, Theorem 12.7 and Lemma 12.3] we see that the map Σ: π 2 19 −→ π 3 20 is as follows:Z 2 (η 2 µ 3 σ 12 ) ⊕ Z 2 (η 2 η 3 ɛ 4 ) −→ Z 2 (η 3 µ 4 σ 13 ) ⊕ Z 4 (ɛ ′ ) ⊕ Z 2 (µ 3 )η 2 µ 3 σ 12 ↦→ η 3 µ 4 σ 13η 2 η 3 ɛ 4 ↦→ 2ɛ ′ .As the EHP sequence is exact, we see that H induces a monomorphism from Z 2 (ɛ ′ ) ⊕ Z 2 (µ 3 ) tothe group π20 5 = Z 2 (ɛ 5 ) ⊕ Z 2 (ρ (4) ). We also learn from [3, Lemma 12.2] that H(µ 3 ) = ρ (4) . Theproposition follows.□Proposition 32. There is a unique element ζ ′ ∈ π 6 22 such that H(ζ ′ ) = ζ 11 and Σζ ′ = σ ′ ◦ η 14 .Proof. We know from [3, Lemma 12.1] that there is an element in π22, 6 which we temporarilycall ξ, such that H(ξ) = ζ 11 (mod 2ζ 11 ) and Σξ = σ ′ ◦ η 14 ◦ ɛ 15 (mod 2π23). 7 We see from [3,Theorem 12.6] that 2π23 7 = 0 and thus that Σ(nξ) = σ ′ ◦ η 14 ◦ ɛ 15 for any odd n. As H(ξ) = ζ 11(mod 2ζ 11 ), we can choose n so that H(nξ) = ζ 11 , and we then put ζ ′ = nξ. Uniqueness is left tothe reader.□Proposition 33. H(κ 7 ) = ɛ 13 .Proof. We first note from [3, Theorem 7.1] that π21 13 = Z 2 ɛ 13 ⊕ Z 2 ν 13 , and the P homomorphismkills both generators by [3, Equations 7.27], so H : π21 7 −→ π21 13 must be surjective. We haveπ21 7 = Z 8 σ ′ ◦ σ 14 ⊕ Z 4 κ 7 by [3, Theorem 10.3], and H(σ ′ ◦ σ 14 ) = H(σ ′ ) ◦ σ 14 = ɛ 13 + ν 13by [3, Lemmas 5.14 and 6.4]. It follows that H(κ 7 ) must be ɛ 13 or ν 13 . I think that Λ-algebrarepresentatives are as follows:ν 13 ∼ λ 53ɛ 13 ∼ λ 233κ 7 ∼ λ 6233 + λ 4721 + λ 3623 + λ 3443H(κ 7 ) ∼ λ 233 .In particular, H(κ 7 ) has Adams filtration 3 so it cannot equal ν 13 , and must therefore equal ɛ 13instead.□Proposition 34. (Σσ ′ ) ◦ ν 15 ◦ ν 18 = ν 8 ◦ σ 11 ◦ ν 18 .Proof. The claimed equation takes place in the groupπ 8 21 = Z 2 (ν 8 ◦ σ 11 ◦ ν 18 ) ⊕ Z 2 (σ 8 ◦ ν 15 ◦ ν 18 ).This has exponent two so we need not worry about signs.We will show that both sides are equal to P (ν 17 ◦ ν 20 ). For the right hand side, this is provedby Toda, just above Equations 7.28. For the left hand side, we have P (ν 17 ◦ ν 20 ) = w 8 ◦ ν 15 ◦ ν 18 ,and w 8 = ±(2σ 8 − Σσ ′ ) by [3, Equations 5.16]. As ν 15 ◦ ν 18 is a suspension we can distribute, andas π21 8 has exponent 2 we are just left with (Σσ ′ ) ◦ ν 15 ◦ ν 18 as claimed.□Proposition 35. σ ′ ◦ ν 14 = ν 7 ◦ σ 10 (mod 2ν 7 ◦ σ 10 )
A MATHEMATICA REPRESENTATION OF SOME UNSTABLE HOMOTOPY GROUPS OF SPHERES 13Proof. This takes place in π 17 S 7 = Z 2 (η 7 ◦ µ 8 ) ⊕ Z 8 (ν 7 ◦ σ 10 ), which maps to π10 S = Z 2 (η ◦ µ). Weknow that νσ = 0 stably, so σν = 0 stably, but Σ 2 σ ′ = 2σ 9 so σ ′ ν also vanishes in the stable group.It follows that σ ′ ◦ ν 14 = mν 7 ◦ σ 10 for some integer m. The part of the relevant unstable AdamsE 2 term that contributes to π 17 S 7 has rank 4, with a single Z 2 in each of filtrations 3, 4, 5 and 6,the first generator corresponding to λ 433 in the lambda algebra. One checks that σ ′ is representedby λ 43 + λ 61 and ν 14 is represented by λ 3 so σ ′ ◦ ν 14 is represented by (λ 43 + λ 61 )λ 3 = λ 433 . Thismeans that σ ′ ◦ ν 14 has minimal Adams filtration and so cannot be divided by 2. It follows thatm is odd.□4. QuestionsWhat are the following elements, in terms of Toda’s generators for the groups in which theylive?• σ ′ ν 14 ∈ π 17 S 7• ν 9 σ 12 ∈ π 19 S 9• ν 7 ν 15 ν 18 ∈ π 21 S 7• σ 10 ɛ 17 ∈ π 25 S 10• σ ′′′ σ 12 ∈ π 19 S 5• H(ɛ ′ ) ∈ π 20 S 5• η 13 η 14 µ 15 ∈ π 24 S 13• P (σ 9 ) ∈ π 14 S 4• η 9 ɛ 10 ∈ π 25 S 9• ν 5 σ 8 η 15 ∈ π 16 S 9(This is only a sample of the open questions in Toda’s range.)References[1] H. J. Baues. Commutator calculus and groups of homotopy classes, volume 50 of London Mathematical SocietyLecture Notes. Cambridge University Press, 1981.[2] M. E. Mahowald. The Metastable Homotopy of S n , volume 72 of Memoirs of the American MathematicalSociety. American Mathematical Society, 1967.[3] H. Toda. Composition Methods in Homotopy Groups of Spheres. Number 49 in Annals of Mathematics Studies.Princeton University Press, Princeton, 1962.
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Page 10 and 11: 10 N. P. STRICKLANDProof. We know f

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