---
tag: HEV7170SEK
kind: definition
name: Steenrod algebra
msc: [55S10]
defines: JWTTXSGD0E
uses: [VR7RHP41M0, 8KB4SQGC0C, 092GEF38YD]
sources:
  - url: https://en.wikipedia.org/wiki/Steenrod_algebra
---

## Definition

Consider symbols $\operatorname{Sq}^i$ of degree $i$, for $i \ge 0$.
The mod $2$ *Steenrod algebra* $\mathcal{A}_2$ is the [[graded|8KB4SQGC0C]] associative algebra over $\Z/2\Z$ generated by them, subject to these relations.

1. $\operatorname{Sq}^0 = 1$.
2. The Adem relations hold:
$$\operatorname{Sq}^a \operatorname{Sq}^b = \sum_{j=0}^{\floor{a/2}} \binom{b - 1 - j}{a - 2j} \operatorname{Sq}^{a + b - j} \operatorname{Sq}^j \quad \text{for all } 0 < a < 2b,$$
where the [[binomial coefficients|092GEF38YD]] are taken modulo $2$.

## Remarks

The algebra acts on the mod $2$ cohomology of every space, with $\operatorname{Sq}^i$ acting as the [[Steenrod square|VR7RHP41M0]] of the same name.
It is isomorphic to the algebra of all stable cohomology operations on mod $2$ cohomology.

Example: for $a = b = 1$, the sum has the single term $j = 0$ with coefficient $\binom{0}{1} = 0$, so $\operatorname{Sq}^1 \operatorname{Sq}^1 = 0$.
For $a = 1$ and $b = 2$, the coefficient is $\binom{1}{1} = 1$, so $\operatorname{Sq}^1 \operatorname{Sq}^2 = \operatorname{Sq}^3$.
Thus $\operatorname{Sq}^3$ is not needed as a generator; the algebra is generated by the $\operatorname{Sq}^{2^k}$, $k \ge 0$.

Non-example: the free associative algebra over $\Z/2\Z$ on the symbols $\operatorname{Sq}^i$ is not the Steenrod algebra, since in it $\operatorname{Sq}^1 \operatorname{Sq}^1 \ne 0$.

A basis of $\mathcal{A}_2$ as a vector space is given by the admissible monomials $\operatorname{Sq}^{i_1} \cdots \operatorname{Sq}^{i_r}$ with $i_s \ge 2 i_{s+1}$ for all $s$ and $i_r \ge 1$, together with $1$.