Steenrod algebra

Remarks

The algebra acts on the mod $2$ cohomology of every space, with $\operatorname{Sq}^i$ acting as the Steenrod square 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 $1$.