Steenrod squares in terms of chain maps

Question

$\DeclareMathOperator\Sq{Sq}$The Steenrod squares $\Sq^i: H^n({-};\mathbb{F}_2) \to H^{n+i}({-};\mathbb{F}_2)$ are fundamental cohomological operations. By the Yoneda lemma, they induce a map between the Eilenberg–MacLane spaces $K(\mathbb{F}_2; n) \to K(\mathbb{F}_2; n +i)$. By the Dold–Kan correspondence, this map should be expressible as a chain map (if I'm not mistaken): $$\widehat{\Sq^i}: \mathbb{F}_2[-n] \to \mathbb{F}_2[-(n+i)].$$

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Question: How do we explicitly describe $\widehat{\Sq^i}$? Is it much harder to do this for other finite fields $\mathbb{F}_q$?

Answer 1

You do have a map of spaces $Sq^i : K(\mathbb F_2 , n) \to K(\mathbb F_2, n+i)$, but it is not a map of topological abelian groups unless $i=0$ or $i=1$. The Dold-Kan correspondence says that maps of chain complexes correspond to maps of topological abelian groups. So there is no map of chain complexes realizing $Sq^i$ for $i\neq 0, 1$.

For $i=0$, there is -- it's the identity map. For $i=1$, $Sq^1$ is the Bockstein, and is again represented by a map of chain complexes, but it's not $\mathbb F_2$-linear, only $\mathbb Z$-linear.

In fact, if $A$ is a free resolution of $\mathbb F_2$ as a chain complex, you can see that there are no $\mathbb Z$-linear maps $A \to A$ of degree $>1$ because $\mathbb Z$ has homological dimension 1. You can see that there are no $\mathbb F_2$-linear maps of degree $>0$ because $\mathbb F_2$ has homological dimension $0$.

On the spectrum side, these observations correspond to the fact that $Sq^i : H \mathbb F_2 \to H\mathbb F_2 [i]$ is not an $H\mathbb F_2$-linear map for $i > 0$, and not $H\mathbb Z$-linear for $i > 1$.