It is well known that in $\mathbb{Z}_2$-valued simplicial cohomology (and other cohomologies)
$$ Sq^1 = \beta\;,$$
where $Sq^1$ is the first Steenrod square and $\beta$ is the Bockstein homomorphism for the short exact sequence
$$ \mathbb{Z}_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}_2\;.$$
I was wondering whether there is a way to also interpret $Sq^2$ or even higher Steenrod squares in a similar fashion. One could think that instead of
$$\beta = g^{-1} d h^{-1}$$
for a short exact sequence of abelian groups
$$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C$$
one could try something like
$$g^{-1}dh^{-1}di^{-1}$$ for an exact sequence
$$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C\overset{i}{\rightarrow} D\;.$$
However, if I'm not mistaken, the long exact sequence splits into two short exact sequences, and the above is just the product of the two corresponding Bockstein homomorphisms. E.g., when applied to the long exact sequence
$$ \mathbb{Z}_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\cdot 2}{\rightarrow} \mathbb{Z}_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}_2$$
we get $\beta^2$ for the short exact sequence given in the beginning, which is trivial.
Crossed module extensions are a more non-trivial analogue to ordinary group extensions which also come with an action of $C$ on $B$. It is known that crossed module extensions give rise to group 3-cocycles in $H^3(BD, A)$ just as ordinary extensions give rise to 2-cocycles. If $D$ acts trivially on $A$, the group 3-cocycle can be used to map a (simplicial) 1-cocycle to a 3-cocycle, and the "higher Bockstein" should generalize this to a map from $i$-cocycles to $i+2$-cocycles. However, I'm not sure how to use the action of $C$ on $B$ to define sich a higher Bockstein.
So I basically have two questions: 1) Is there a non-trivial analogue of a Bockstein operation for some sort of longer exact sequences/crossed module extensions? 2) If yes, is any of these operations equal to $Sq^2$ (in general, or maybe at least when applied to a specific degree)?
Best Answer
$\DeclareMathOperator{\Sq}{Sq}\newcommand{\Z}{\mathbb{Z}}$The short version is that every cohomology operation can be interpreted as a Bockstein operator for an "exact sequence" (read: fiber sequence) of grouplike $E_\infty$ spaces.
Any cohomology operation $\delta:H^*(-;A)\Rightarrow H^{*+k}(-;B)$ such as $\Sq^i$ gives a morphism of Eilenberg-MacLane spectra $f_\delta:HA\to \Sigma^k HB$. You can take the fiber of $f_\delta$ to obtain a spectrum $F$ which defines a generalized cohomology theory $E_F^*:X\mapsto \pi_{-*} F^X$, and there is a resulting long exact sequence $$ \dots\to E_F^i(X)\to H^i(X;A)\xrightarrow{\delta}H^{i+k}(X;B)\to E_F^{i+1}(X)\to\dots $$ If $\delta = \Sq^1$, the fiber is again an Eilenberg-MacLane spectrum (namely $F\simeq H\Z/4$). If the degree of $\delta$ is bigger than $1$, it will have two non-zero homotopy groups, namely $A$ in degree $0$ and $B$ in degree $k-1$.
The connection to the crossed modules you mention is that applying the functor $\Omega^{\infty-1}$ to the fiber of $\Sq^2$ gives rise to a $2$-group, i.e. a homotopy type $X$ whose homotopy groups vanish outside degrees $1$ and $2$. As you mention, these are classified by the two groups $A = \pi_1(X),B = \pi_2(X)$, the action of the former on the latter, and a $k$-invariant in $H^3(A;B)$, which together can be packaged into the datum of a crossed module. However, these deloop once if and only if the action is trivial and the $k$-invariant vanishes. It is still possible to find reasonably easy algebraic models for spaces whose homotopy groups vanish outside degrees $k$ and $k+1$ (given by braided ($k=2$) and symmetric ($k\ge 3$) monoidal Picard (every object has a tensor inverse) groupoids), although I do not know a definition of $\Sq^2$ in this language.
Questions in the comments
$$ H^*(-;A)\to H^*(-;A\otimes Z/2)\xrightarrow{\Sq^2} H^{*+2}(-;A\otimes\Z/2)\to H^{*+2}(-;B) $$
The relation to Picard groupoids is a consequence of the Homotopy hypothesis, and given a braided Picard groupoid $C$, you can associate to it its abelian group $\pi_0 C$ of isomorphism classes and the (abelian!) group $\pi_1 C$ of automorphisms of the unit $1$, together with the map $q: \pi_0 C\to \pi_1 C$ which sends $x$ to the composition
$$ 1\cong x\otimes x^{-1}\xrightarrow{\beta_{x,x^{-1}}} x^{-1}\otimes x\cong 1 $$
It's a fun exercise to show that $q$ is quadratic in the above sense. It is not straightforward to give an inverse to this construction, i.e. construct the braided Picard groupoid from the quadratic map; for a reference in the symmetric setting, see Cegarra, A. M.; Khmaladze, E., Homotopy classification of graded Picard categories, Adv. Math. 213 (2007).
A chain level representative of $\Sq^2$ (and higher Steenrod squares) can be found in Ralph M. Kaufmann, Anibal M. Medina-Mardones. Cochain level May-Steenrod operations.