Subject: lambda algebra question (for the list)
Date: Fri, 12 Mar 2004 15:46:45 -0800
From: John H Palmieri <palmieri@math.washington.edu>

Here's a simple question: take the Lambda algebra at the prime 2, and
mod out by the ideal J generated by (lambda_{n} + lambda_{2n+1} : n >= 0).
It seems that in Lambda/J, every product lambda_i lambda_j is
trivial.  Can anyone prove this?

Notes:
  - Lambda is bigraded (by Adams filtration and stem), but J is only
    homogeneous in the first grading, so Lambda/J is graded just by
    filtration.
  - For each i, lambda_i is congruent mod J to lambda_j where j is
    even.  So it suffices to show that lambda_{2i} lambda_{2j} is zero
    in Lambda/J for all i and j.
  - The derivation on Lambda, sending lambda_n to lambda_{n+1}, does
    not send J to J, and so does not induce a derivation on
    Lambda/J.  So you can use it in the obvious way to approach this
    problem.
  - I've done a bunch of calculations -- writing out tables of Adem
    relations and imposing the relations in J -- and I can show this
    way that lots of products in Lambda/J are zero.  I don't know how
    to prove it in general, though.

--
John H. Palmieri
Dept of Mathematics, Box 354350    mailto:palmieri@math.washington.edu
University of Washington           http://www.math.washington.edu/~palmieri/
Seattle, WA 98195-4350