---
tag: GS8SCGYPX7
kind: statement
role: theorem
name: Adem relations
msc: [55S10]
uses: [VR7RHP41M0, EZYHSE1E8P, H7PCM776GR, 092GEF38YD]
about: [VR7RHP41M0, EZYHSE1E8P]
sources:
  - url: https://en.wikipedia.org/wiki/Steenrod_algebra
---

## Statement

Let $p$ be a prime and consider operations on $H^*(X; \F_p)$ for spaces $X$.
All [[binomial coefficients|092GEF38YD]] below are read modulo $p$, and $\binom{m}{k} = 0$ when $k < 0$ or $k > m$.

1. For $p = 2$, the [[Steenrod squares|VR7RHP41M0]] satisfy, for all integers $a, b > 0$ with $a < 2b$,
$$Sq^a Sq^b = \sum_{j=0}^{\floor{a/2}} \binom{b - 1 - j}{a - 2j} Sq^{a + b - j} Sq^j.$$
2. For $p$ odd, the [[Steenrod reduced powers|EZYHSE1E8P]] satisfy, for all integers $a, b > 0$ with $a < pb$,
$$P^a P^b = \sum_{j=0}^{\floor{a/p}} (-1)^{a + j} \binom{(p-1)(b - j) - 1}{a - pj} P^{a + b - j} P^j.$$
3. For $p$ odd, with $\beta$ the [[Bockstein homomorphism|H7PCM776GR]] of $0 \to \Z/p \to \Z/p^2 \to \Z/p \to 0$, for all integers $a, b > 0$ with $a \le pb$,
$$P^a \beta P^b = \sum_{j=0}^{\floor{a/p}} (-1)^{a + j} \binom{(p-1)(b - j)}{a - pj} \beta P^{a + b - j} P^j + \sum_{j=0}^{\floor{(a-1)/p}} (-1)^{a + j + 1} \binom{(p-1)(b - j) - 1}{a - pj - 1} P^{a + b - j} \beta P^j.$$

## Remarks

One proof computes the operations on the cohomology of a product of Eilenberg–MacLane spaces, or of symmetric products, where all compositions are visible.
Another proof packages the operations into a generating function and uses the Cartan formula.

Example for $p = 2$: taking $a = b = 1$ gives $Sq^1 Sq^1 = \binom{0}{1} Sq^2 = 0$.
Taking $a = 1, b = 2$ gives $Sq^1 Sq^2 = Sq^3$.
Taking $a = b = 2$ gives $Sq^2 Sq^2 = Sq^3 Sq^1$.

The condition $a < 2b$ is needed: when $a \ge 2b$ the composite $Sq^a Sq^b$ is already admissible and is not rewritten.
The relations, together with $Sq^0 = 1$ (and $P^0 = 1$, $\beta^2 = 0$ for odd $p$), give all relations among composites of these operations.
So the mod $p$ Steenrod algebra is the free associative algebra on the operations modulo these relations.

## Explanations

### L1 human

Steenrod operations can be composed, and some composites are equal to sums of other composites.
The Adem relations list exactly which such equalities hold.
They let one rewrite any composite as a sum of composites of a standard form, called admissible.

### L2 human

The Adem relations make the Steenrod algebra computable: the admissible monomials form a basis, which is the Cartan–Serre basis.
They show that $Sq^n$ is decomposable unless $n$ is a power of $2$, so the algebra is generated by the $Sq^{2^k}$.
This fact is the starting point of the Hopf invariant one problem and of the Adams spectral sequence.