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A ring prespectrum gives a graded product on stable homotopy groups

Statement

Let $E$ be a sequential prespectrum with a unital multiplication. If $x \in π_{p} (E)$ and $y \in π_{q} (E)$ are represented at stages $m, n$ by

$f : S^{m + p} \to E_{m}, g : S^{n + q} \to E_{n},$

then the smash $f \land g$ , followed by $μ_{m, n}$ and the canonical sphere coordinate rearrangement, defines a product

$π_{p} (E) \times π_{q} (E) ⟶ π_{p + q} (E) .$

It is well defined, associative, and unital. If the multiplication has the commutativity datum of the preceding definition, then

$x y = (- 1)^{p q} y x .$

Facts & Assumptions

[F1]

The pairing has both displayed structure-compatibility homotopies; the second includes the suspension-coordinate twist (Pairings and unital multiplication of sequential prespectra).

[F2]

Smash associators, twists, and units are coherent canonical homeomorphisms (Canonical associativity, symmetry, and unit maps for smash products).

[F3]

The commutativity datum includes the level-block permutation, whose degree is $(- 1)^{a b}$ for blocks of dimensions $a, b$ (Pairings and unital multiplication of sequential prespectra).

Proof

Given: A unital ring prespectrum $E$ and stable classes $x, y$ represented as in the statement.

1.1

Define the product on representatives. Use [F2] to identify $S^{m + n + p + q}$ with $S^{m + p} \land S^{n + q}$ in the fixed order, then set [F1] $[f] \cdot [g] = [μ_{m, n} (f \land g)] \in π_{m + n + p + q} (E_{m + n}) .$

Changing $f$ or $g$ through a based homotopy changes this composite through a based homotopy.

1.2

Check the colimit relation. Advance $f$ once. The first comparison in [F1] homotopes the resulting composite to the one-fold suspension of the product. Advancing $g$ once gives the same conclusion from the second comparison; its explicit twist is exactly the coordinate rearrangement needed to put the new $S^{1}$ at the front. Hence replacing either representative by a bonding-map representative does not change the colimit class. Repetition and common-stage comparison prove full well-definedness.

1.3

Associativity and the unit. For three representatives, the two products are the two composites $μ (μ \land 1)$ and $μ (1 \land μ)$ after the same canonical reassociation of sphere coordinates. The specified associativity homotopy and [F2] identify them. The class of $η : S^{0} \to E_{0}$ is a two-sided identity by the two unit homotopies.

1.4

Compute the commutativity sign. Compare the two representative maps after putting both in one fixed order: degree coordinates $p, q$ followed by level coordinates $m, n$ . The parity contributions are [F1] $(m + p) (n + q) (swap the two full source blocks),$ $m n (the target level-block permutation), m q + n p (restore the fixed stable ordering) .$

By [F3] their sum modulo two is $(m + p) (n + q) + m n + m q + n p \equiv p q$ . The commutativity homotopy therefore gives $x y = (- 1)^{p q} y x$ , independent of the chosen stages. $□$

Depends on

Used by

Scope boundary for stable homotopy theory Remark

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Sources

J. P. May, A Concise Course in Algebraic Topology (standard reference, not scraped)