Steenrod reduced power

Builds on Cohomology operation Singular cohomology Cup product

Definition

Let $p$ be an odd prime, and write $H^n(X)$ for the singular cohomology group $H^n(X; \F_p)$ of a space $X$ . The Steenrod reduced powers are the unique family of homomorphisms $P^i \colon H^n(X) \to H^{n + 2i(p-1)}(X), \qquad i \ge 0,\ n \ge 0,$ defined for all spaces $X$ , with the following properties.

Each $P^i$ is natural: $f^* \circ P^i = P^i \circ f^*$ for every continuous map $f \colon Y \to X$ .
Each $P^i$ commutes with the suspension isomorphism $H^n(X) \cong H^{n+1}(SX)$ in reduced cohomology.
$P^0$ is the identity.
If $x \in H^{2i}(X)$ , then $P^i x = x^p$ , the $p$ -th cup product power of $x$ .
If $x \in H^n(X)$ and $2i > n$ , then $P^i x = 0$ .
(Cartan formula) $P^k(x \smile y) = \sum_{i + j = k} P^i x \smile P^j y$ for all $x, y \in H^*(X)$ .

Remarks

Existence of such a family is a theorem of Steenrod; uniqueness follows from the axioms by evaluating on products of Eilenberg–MacLane spaces. The operations also exist on cohomology of pairs and commute with the coboundary maps of pairs.

Example: on $X = \C P^\infty$ , with $H^*(X) = \F_p[x]$ and $\deg x = 2$ , clauses 4 and 6 give $P^i(x^m) = \binom{m}{i} x^{m + i(p-1)}$ . For instance $P^1 x = x^p$ , which is nonzero.

Non-example: let $Q^0$ be the identity and $Q^i = 0$ for $i > 0$ . This family satisfies clauses 1, 2, 3, 5 and 6. It fails clause 4: on $\C P^\infty$ one has $Q^1 x = 0$ , while $x^p \neq 0$ .

The operation $P^i$ is a cohomology operation of type $(\F_p, n; \F_p, n + 2i(p-1))$ . For $p = 2$ the analogous operations are the Steenrod squares $Sq^i$ , with $Sq^{2i}$ playing the role of $P^i$ .

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Tag	`AMDN6PK5RA`
URI	`m2://AMDN6PK5RA`
Kind	definition
Writing	W0 stub
MSC	55S10, 55S05
Defines	Steenrod reduced power
Uses	Cohomology operation, Singular cohomology, Cup product

Steenrod reduced power

Definition

Remarks

Reading path

Further reading

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