Let p ≥ 7 be an odd prime. Based on the Toda bracket 〈α1β1^p-1, α1β1,p, γs〉, the authors show that the relation α1β1^P-1h2,0γs= βp/p-1γ/s holds. As a resulL they can obtain α1β1^ph2,0γs= 0 ∈ π＊（S^0...Let p ≥ 7 be an odd prime. Based on the Toda bracket 〈α1β1^p-1, α1β1,p, γs〉, the authors show that the relation α1β1^P-1h2,0γs= βp/p-1γ/s holds. As a resulL they can obtain α1β1^ph2,0γs= 0 ∈ π＊（S^0） for 2 ≤ s ≤ p - 2, even though α1h2,0γs and β1α1h2,0γs are not trivial. They also prove that β1^p-1 α1h2,0γ3 is nontrivial in π＊（S^0） and conjecture β1^p-1 α1h2,0γs is nontrivial in π＊（S^0） for 3 ≤s ≤ p - 2. Moreover, it is known that βp/p-1γ3 = 0 ∈ EXtBP＊Bp^5,＊（BP＊, BP＊）, but βp/p-1γ3 is nontrivial in π＊（S^0） and represents the element β1^p-1α1h2,0γ3.展开更多
基金supported by the National Natural Science Foundation of China(Nos.11071125,11261062,11471167)the Specialized Research Fund for the Doctoral Program of Higher Education(No.20120031110025)
文摘Let p ≥ 7 be an odd prime. Based on the Toda bracket 〈α1β1^p-1, α1β1,p, γs〉, the authors show that the relation α1β1^P-1h2,0γs= βp/p-1γ/s holds. As a resulL they can obtain α1β1^ph2,0γs= 0 ∈ π＊（S^0） for 2 ≤ s ≤ p - 2, even though α1h2,0γs and β1α1h2,0γs are not trivial. They also prove that β1^p-1 α1h2,0γ3 is nontrivial in π＊（S^0） and conjecture β1^p-1 α1h2,0γs is nontrivial in π＊（S^0） for 3 ≤s ≤ p - 2. Moreover, it is known that βp/p-1γ3 = 0 ∈ EXtBP＊Bp^5,＊（BP＊, BP＊）, but βp/p-1γ3 is nontrivial in π＊（S^0） and represents the element β1^p-1α1h2,0γ3.
