THE SPECIAL ORTHOGONAL GROUPS SO (2n ) AS FRAMED BOUNDARIES

Remarks on Framed Bordism Classes of Classical Lie Groups

“…As is seen above it holds that J(ρ − ψ 2 ρ) (3) = 1 so that J(2λ 2 ρ) (3) = J((12 + 9)ρ) (3) . Since…”

“…But it is regrettable that the case n = 5 does so yet. Since [SO(2n + 1), R] (3) and [Sp(n), R] (3) have the same order by Lemma 2.6 of [6] we consider only the case G = SO(2n + 1) below.…”

“…n = 3 + 2 ( ≥ 2). Putting µ =J(ρ) (3) where ρ is as above we have from the first and second formulas of Proposition 2.1, a) From now on we work modulo 3 due to (3.3). Again from (3.2) by induction on k we have (3.5) κ(µ 3k + µ 3k+1 ) = 0 and κ(µ 3k+1 + µ 3k+2 ) = 0 for k ≥ 0.…”

“…For this it is known [7] that 72[G, R] = 0 and more previously in [2] it is conjectured that [G, R] = 0 if rank G ≥ 10 or so. We have [SO(2n), R] = 0 (n ≥ 2) already in [3] and so we are interested in such a conjecture for the cases G = SO(n), SU(n) or Sp(n).…”

“…Put α =J(Ad G ) (3) . Then it can be deduced from (1.7) that (3.6) κ((1 + α)µ k ) = (−1) κ(R k ) and κ((1…”