The stable Picard group of $\mathcal{A}(2)$

Abstract

Using a form of descent in the stable category of A(2)-modules, we show that there are no exotic elements in the stable Picard group of A(2), i.e. that the stable Picard group of A( 2) is free on 2 generators. AcknowledgmentsAuthors would like to thank Bob Bruner for some fruitful conversations.Convention. Through out this paper, F will denote the field with two elements. Every algebraic structure is implicitly over the base field F, and tensor products are taken over F. The Hopf algebras under consideration in this paper are connected, cocommutative finite dimensional graded Hopf algebras, unless explicitly specified otherwise.

Iterated doubles of the Joker and their realisability

“…Notice that as well as the Sq 1 and Sq 2 actions we also have Sq 4 (x 3 y 1 ) = x 6 y 2 so this agrees with Joker ′ 1 [4] as an A-module. We will begin by showing that there is a factorisation…”

“…Now take a minimal CW complex equivalent to the fibre of the map X ′ → kO 7 and let X 4 be its 8-skeleton. By a straightforward calculation with either of the Serre or Eilenberg-Moore spectral sequences and making use of the kO results of Examples B.3, we find that H * (X) realises the A-module Joker 1 [4]. Proof.…”

“…it follows that Joker 0 and Joker 1 are dual up to a degree shift, i.e., (1.2) Joker 1 ∼ = DJoker 0 [4].…”

“…To realise Joker 1 [4], we start with an unstable complex S 3 ∪ η 3 e 5 ∪ 2 e 6 which exists since the suspension of the Hopf map S 3 → S 2 gives an element η 3 ∈ π 4 (S 3 ) of order 2, see [17, chapter V].…”

“…• More details on the Joker and its homological algebra can be found in [5, appendix A.8]. For a recent result which highlights the special significance of the Joker see [4]. Incidentally, the use of the name Joker appears to be due to Frank Adams, although…”