# Why are cup-i products and Steenrod Squares often (always?) unary?

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

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I think it does get used as a binary product. For instance, $u \cup_i v$ measures how far $u \cup_{i-1} v$ is from being commutative. So all the $\cup_i$ together are telling you information about the classical $\cup$ product (which is $\cup_0$) in the same way that all the levels of $A_\infty$ together give you a homotopy associative product. I think the main reason to move from $\cup_i$ to $Sq^i$ is that for the application Mosher and Tangora want (division algebras) they care about operations on cohomology, i.e. from $H^*$ to $H^*$. – David White Aug 10 '12 at 0:16
The cup_i-product of two closed chains may be nonclosed, if I remember correctly. – Pelle Salomonsson Aug 10 '12 at 9:46
While the $\cup_i$'s are useful, they may see less attention due to the fact that they do not satisfy a nice list of properties, but the steenrod operations do. – Sean Tilson Aug 12 '12 at 19:42

I use $\cup_i$ products as binary products in my work on the algebraic theory of surgery
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.
Under suitable hypotheses, Gugenheim and I use the binary $\cup_1$ and especially the fact that it is a graded derivation (Hirsch formula) as the key to giving a calculation of $H^*(G/H)$ as the torsion product over $H^*(BG)$ of $R$ and $H^*(BH)$, where coefficients are taken in a suitable commutative ring $R$. This even works for suitable $H$-spaces. See http://www.math.uchicago.edu/~may/BOOKS/GugMay.pdf and http://www.math.uchicago.edu/~may/PAPERS/MNApril20.pdf