One way to define the Steenrod Operations is to use the cup-i product, as in Mosher and Tangora's book. It basically says, given the chain complex from mod-2 homology $C\_\ast$, define


$D\_0 : C\_\ast\to C\_\ast\otimes C\_\ast$, 


so that the cup product is given (on cocycles)


$(u\cup v)(\sigma) = (u\otimes v)(D\_0\sigma)$


and then for higher $i$, define $D\_i$ so that 


$D\_{i-1}+\rho D\_{i-1} = D\_i\partial + \partial D\_i$


where $\rho$ is the flipping map. Then the $\cup\_i$ product is just


$(u\cup\_i u)(\sigma) = (u\otimes u)(D\_i\sigma)$


And then define for $[u]\in H^n$


$Sq^{2n-i}([u]) = [u\cup\_{i}u]$


This definition seems perfectly well-defined as a binary operation, and yet wherever I've seen it done it has only even been used as a unary operation. 


Is there a reason why this is the case, why either the product is undefined as a binary product or not useful as a binary product or just too hard to use?  

Is this a dumb question?


Thanks,
-Joseph



Why are cup-i products and Steenrod Squares often (always?) unary?
I use $\cup\_i$ products as binary products in my work on the algebraic theory of surgery


<http://www.maths.ed.ac.uk/~aar/papers/ats2.pdf>


They give the higher symmetry properties {$\phi\_s|s \geq 0$} of the Poincare duality chain equivalence 
$$\phi\_0=[M] \cap - : C(M)^{m-\*} \to C(M)$$
of an $m$-dimensional manifold $M$, with
$$d\phi\_s+\phi\_sd^\*+\phi\_{s-1}+\phi\_{s-1}^\*=~0~(\phi\_{-1}=0)$$
up to sign. 


