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On conjugation in the mod p-steenrod algebra

In this paper we prove a formula involving the canonical anti-automorphism χ of the mod-p Steenrod algebra. We would like to sincerely thank referee for many helpful comments on this paper | Turk J Math 24 (2000) , 359 – 365. ¨ ITAK ˙ c TUB On Conjugation in the Mod-p Steenrod algebra Ismet Karaca and Ilkay Yaslan Karaca Abstract In this paper we prove a formula involving the canonical anti-automorphism χ of the mod-p Steenrod algebra. Key Words: Steenrod algebra, anti-automorphism, Milnor basis 1. Introduction and Main Result Let A be a mod-p Steenrod algebra. Let R = (r1 , r2 , . . . ) be a sequence of nonnegative integers with finitely many nonzero terms. Let P(R) denote the corresponding Milnor basis element in A so that the elements P(R) form an additive basis for the subalgebra ∞ P pi − 1 r i Ap of A generated by the Steenrod powers P i , i ≥ 0. We define |R| = i=1 and e(R) = ∞ P ri . Thus, considered as a mod-p cohomology operation, P(R) raises the i=1 dimension of a cohomology class by 2 |R| and has excess 2e(R). The anti-automorphism χ of Ap plays a fundamental part in our argument, and we find it convenient to write θb = (−1)dimθ χ(θ) for every element θ ∈ Ap . We are interested in an explicit conjugation formula for the Steenrod operations of Ap in the form X(k, n) = P(pk n)P(pk−1 n) · · · P(pn)P(n), AMS Mathematics Subject Classification: Primary 55S10, 55S05 359 KARACA, KARACA where k and n are nonnegative integers. So the following formula is the mod-p analogue of Theorem 3.1 in [6]. Theorem 1.1 For all positive integers j and i, we have b pi+1 − 1) = X(i, pj+1 − 1). X(j, We will introduce the following useful notation: each natural number a has a unique p-adic expansion a= ∞ X αi (a)pi i=0 with 0 ≤ αi (a) l, suppose that (1) m + n = pk − pl (2) m 0, we have p−1 X p−1 l l−1 b b b ))P(pm+(p−1)pl )− ). P(m+wpl−1 )·P(n−wp P(m)·P(n) = P(n−(p−1)(m+p w w=1 Proof. (i) Let l = 0. Using Proposition 1.3, we have X |R| + e(R) P(R), pm b P(m) · P(n) = R and b − (p − 1)m − 1) · P(pm + 1) = P(n X e(R) P(R), pm + 1 R where |R| = (p − 1)(pk − 1) and 1 ≤ e(R) ≤ pk − 1. In order to prove these sums are equivalent in .