Let $X$ be some space, $C^\*(X,R)$ its cochain complex. Then there is a multiplication 


$$
\mu : C^\*(X,R) \otimes C^\*(X,R) \rightarrow C^\*(X,R)
$$


inducing the cup product, and a homotopy 


$$H : C^\*(X,R) \otimes C^\*(X,R) \rightarrow C^{\*-1}(X,R)$$


"witnessing" the graded commutativity of the cup product, i.e. such that $dH(x,y) - Hd(x,y) = \mu(x,y) \pm \mu(y,x)$. 


I'm interested in the cohomology operation $x \mapsto H(x,x)$. Note that this goes $H^p(X,R) \rightarrow H^{2p-1}(X,R)$. For $R = \mathbb{Z}/2$, I think this is the Steenrod square $Sq\_1 = Sq^{p-1}$ (notation from chapter 2 [this book](http://www.maths.ed.ac.uk/~aar/papers/moshtang.pdf)), where $p = |x|$. As a first step, I'd like to understand it when $R = \mathbb{Z}/4$. This is some lift of $Sq\_1$, but what more is known about it? Slightly more concretely, what methods are there for computing this kind of cohomology operation?


(I have reason to believe that this operation is NOT a ``Steenrod square'' for $\mathbb{Z}/4$, i.e. does not appear in the cohomology of the spectrum $H \mathbb{Z}/4$. But I also have reason to wish that what I just said were not true.) 



What is known about this cohomology operation?
I'm going to refer to $H$ as the cup-$1$ product.


So when $p = |x|$ is odd and $x$ is a cycle, the boundary formula says
$$d(x \cup\_1 x) = 2x^2$$
which is not necessarily zero. In this case, we don't get a cohomology operation unless (e.g.) when $2 = 0$ in $R$, and then the cohomology operation is "mostly like $Sq^{p-1}$".


When $p$ is even, then $d(x \cup\_1 x) = 0$ and so we do get a cohomology operation (once we verify that it's well-defined, of course). However, in this case the next product (the cup-2 product) has an identity 
$$
d(x \cup\_2 x) = 2 x\cup\_1 x.
$$
Here are some consequences.


* If 2 is invertible in the ring $R$, then $x \cup\_1 x$ is automatically zero in cohomology.
* If $R = \Bbb Z$, then $x \cup\_2 x$ is a cochain which reduces, mod 2, to $Sq^{p-2} r(x)$ (the Steenrod square on the mod-2 reduction $r(x)$). However, the integral Bockstein $\beta$ of $Sq^{p-2} r(x)$ is calculated by taking a cocycle representative over $\Bbb Z/2$, lifting it to a cochain representative over $\Bbb Z$, taking the coboundary, and dividing by $2$. This means that $x \cup\_1 x$ is a representative for $\beta Sq^{p-2} r(x)$. (Note that if we then mod-2 reduce, we see it become $Sq^1 Sq^{p-2} r(x) = Sq^{p-1} r(x)$ because $p$ is even.)
* If $R = \Bbb Z/4$, ~~I believe that~~ a similar calculation shows that $x \cup\_1 x$ is a representative for $\beta' Sq^{p-2} r(x)$ where $r$ is reduction mod $2$ and $\beta'$ is the Bockstein associated to the exact sequence
$$0 \to \Bbb Z/4 \to \Bbb Z/8 \to \Bbb Z/2 \to 0.$$
~~However, in my time available I can only make this work if $x$ lifts to mod-$8$ cohomology, and so there's possibly a correction factor.~~ EDIT: This computation works, but you need to use a cup-3 product to show that there are no extra correction factors.


More information on the cup-$i$ products is available in the McClure-Smith's paper (<http://arxiv.org/abs/math/0106024>) referenced in Henrik Rüping's answer, and in Mosher and Tangora's book.


