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Adem reduction spans by admissible square composites

Statement

Assume AC, inherited from the cited bundle, cohomology, or operation suppliers. In degree $N \geq 0$ , every element of $A_{A d e m}$ is a finite linear combination of the admissible words $s_{i_{1}} \dots s_{i_{k}}$ with $∣ I ∣ = N$ . Consequently the same is true in $A_{S q}$ .

Facts & Assumptions

Given: AC; a total degree $N \geq 0$ and the free associative graded algebra $T$ on the symbols $s_{1}, s_{2}, \dots$ with its quotient $A_{A d e m}$ by the two-sided Adem ideal, words being normalized sequences of positive entries padded on the right with zeros to length $N$ .

[F1]

The Adem relations hold in the square algebra: for $0 < a < 2 b$ , the element $s_{a} s_{b} + \sum_{j} (\binom{b - j - 1}{a - 2 j}) s_{a + b - j} s_{j}$ acts as zero, and the quotient map $T \to A_{A d e m} \to A_{S q}$ is well defined (Adem relations for Steenrod squares, The mod-two square algebra, admissible sequences, and excess).

[F2]

Admissibility, total degree and normalization of sequences are defined by the finite word calculus of the square algebra, and every word in degree $N$ has at most $N$ positive entries (The mod-two square algebra, admissible sequences, and excess).

Proof

technique · direct

1.1givenF2

The unit is the only degree-zero word. For $N > 0$ , every normalized word has at most $N$ entries. Pad its sequence on the right with zeros to length $N$ and order these finite sequences lexicographically, reading from the left. There are finitely many such sequences of total sum $N$ .

2.1step 1.1F1algebra

If a word is not admissible, choose a positive adjacent pair $(a, b)$ with $a < 2 b$ . Each nonzero summand in the Adem replacement has the pair $(a + b - j, j)$ , where $0 \leq j \leq ⌊ a / 2 ⌋ < b$ . Its first changed entry is therefore $a + b - j > a$ . If $j = 0$ , remove $s_{0} = 1$ and normalize; this removal occurs after the strictly increased entry, so the normalized padded word remains lexicographically larger. Total degree is preserved, and normalized length still is at most $N$ .

3.1step 1.1step 2.1F1algebra∎

Descending induction on this finite ordered set proves that each word is a sum of admissible words: terminal words cannot have a replaceable pair, while each replacement uses only words already covered by the induction. Sums over words are finite. Applying the well-defined quotient-to-operation map proves the second assertion. This proves spanning only; independence follows below.

Depends on

Used by

Low-degree admissible Steenrod monomials Example
Admissible composites present the mod-two square algebra Theorem

Dependency tree · two levels

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Sources

Allen Hatcher, Algebraic Topology (standard reference, not scraped)