A Counterexample for Lightning Flash Modules over E(e1,e2)

We give a counterexample to Theorem 5 in Section 18.2 of Margolis' book,"Spectra and the Steenrod Algebra", and make remarks about the proofs of some later theorems in the book that depend on it. The counterexample is a module which does not split as a sum of lightning flash modules and free modules.


Introduction
Let k be a field and E(e 1 , e 2 ) be a graded exterior algebra on generators e 1 and e 2 with degrees satisfying 0 < |e 1 | < |e 2 |. Theorem 5 in §18.2 of Margolis [2] states that every graded E(e 1 , e 2 )-module is a coproduct of free modules and lightning flashes. In this note, we give a simple counterexample to this statement. Statement (c) following Proposition 7 of the same section is true, but not because of Theorem 5. The proof of Theorem 8 in §18.3 depends on this statement. The proofs of Proposition 9 and Lemma 10 of the same section also depend on Theorem 5, and are used in Chapter 20. Fortunately, the paper of Adams and Margolis [1] provides correct proofs of these statements that do not rely on Theorem 5.

The counterexample
In this section we display a bounded below module M for E(e 1 , e 2 ) which is not isomorphic to a coproduct of free modules and lightning flashes.
First we note that every module for E(e 1 , e 2 ) can be written as a direct sum of a free module and a module on which e 1 e 2 acts as zero. So we may as well work with modules for E(e 1 , e 2 )/(e 1 e 2 ).
We use the notation of §18.2 of Margolis. Let M(n) be the lightning module L(n, 0, 1) of dimension 2n. Here is a picture of M(n): The shorter arrows represent the action of e 1 , and the longer ones e 2 . Thus a presentation of the module is given by e 1 x i+1 = e 2 x i = y i (0 ≤ i ≤ n − 1), e 1 x 0 = 0, e 2 x n = y n . We arrange that the element x 0 in M(n) is in degree zero, so that x i has degree i(|e 2 | − |e 1 |) and y i has This work was partially supported by the Simons Foundation and by the Mathematisches Forschungsinstitut Oberwolfach. degree |x i | + |e 2 |. Similarly, L(∞, 0) is the infinite lightning flash obtained by letting this diagram continue to the right indefinitely.
Our counterexample is the module To see that it is a counterexample, first note that e 1 M(n) is the linear span of y 0 , . . . , y n−1 , so e −1 2 e 1 M(n) is the linear span of all the basis elements except x n . Here, if U is a linear subspace of a module, we write e −1 2 U for the linear subspace consisting of the vectors whose image under e 2 is in U.
Inductively, we see that for j > 0, (e −1 2 e 1 ) j M(n) is the linear span of the basis elements y 0 , . . . , y n , x 0 , . . . , x n−j . Thus Since a finite sum is always a direct summand of the product, it follows that M has exactly one copy of each M(n) as a summand, and no summand isomorphic to L(∞, 0). Since e 1 M 0 = 0, no summand of the form L(∞, 1), L(n, 1, 0) or L(n, 1, 1) can contribute to M 0 ; and finally (2.1) shows that no summand of the form L(n, 0, 0) can contribute to M 0 , since that intersection is non-zero for such a module. The summands we have identified do not exhaust M 0 , and hence M cannot be a direct sum of lightning flash modules. On the other hand, modules of finite type for E(e 1 , e 2 ) can be shown to be direct sums of lightning flashes, by the method of filtrations of the forgetful functor to graded vector spaces. The proof is similar to but easier than the functorial filtration proof given in Ringel [3].