Abstract

Let p ⩾ 7 be an odd prime. Based on the Toda bracket 〈α1β1p−1, α1β1, p,γs〉, the authors show that the relation α1β1p−1h2,0γs = βp/p−1γs holds. As a result, they can obtain α1β1ph2,0γs = 0 ∈ π*(S0) for 2 ⩽ s ⩽ p − 2, even though α1h2,0γs and β1α1h2,0γs are not trivial. They also prove that β1p−1α1h2,0γ3 is nontrivial in π*(S0) and conjecture that β1p−1α1h2,0γs is nontrivial in π*(S0) for 3 ⩽ s ⩽ p − 2. Moreover, it is known that \(\beta _{p/p - 1} \gamma _3 = 0 \in Ext_{BP_* BP}^{5,*} (BP_* ,BP_* )\), but βp/p−1γ3 is nontrivial in π*(S0) and represents the element β1p−1α1h2,0γ3.

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