Let $x\_n,y\_n,\cdots$ be cocycles in $Z^n(X,\mathbb{Z}\_2)$ (not cohomology classes in $H^n(X,\mathbb{Z}\_2)$).
Let $Sq^k(x)\equiv x\_n \cup\_{n-k} x\_n$ be the Steenrod square (This definition is valid for cocycles).
Is the following expression valid for cocycles
$$
Sq^k(x\_n\cup y\_m) =\sum\_{i=0}^k Sq^{i}(x\_n)\cup Sq^{k-i}(y\_m)
$$
The above Cartan Formula is true for cohomology classes. I like to ask if it is even true for cocycles?



Cartan Formula for Steenrod square on cocycles
This is not necessarily true. The first warning one should give is that there are multiple definitions of "$\cup\_{i}$" on cocycles, and different ones satisfy different identities. However, this is unlikely to hold in general.


One of the few examples that can be calculated model-independently is the following.
If $n=m=1$ and $k=2$ the left-hand side is $x\cup y \cup x \cup y$, while the right-hand side is $x \cup x \cup y \cup y$, which are almost always distinct cochains using the Alexander-Whitney product (this assumes you define $Sq^k x = 0 $ for $k > |x|$).


One of the simpler example that I can find for the explicit versions of the cup-$i$ operations defined by McClure and Smith in "Multivariable cochains and little n-cubes" is when $k=0$. I'm finding that the left-hand side usually has an extremely large number of cross-terms which don't ultimately contribute. In their notation, for example, when $n=m=1$, the right-hand side (in their notation) is
$$
\langle 131242 \rangle(x, y, x, y)
$$
while the left-hand side is
$$
\langle 131234 \rangle(x, y, x, y) +
\langle 123234 \rangle(x, y, x, y) +
\langle 134124 \rangle(x, y, x, y) +
\langle 123424 \rangle(x, y, x, y)
$$
and the fact that all of these are equivalent to $x \cup y$ is non-obvious.


