Why are cup-i products and Steenrod Squares often (always?) unary? - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T16:47:11Zhttp://mathoverflow.net/feeds/question/104380http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/104380/why-are-cup-i-products-and-steenrod-squares-often-always-unaryWhy are cup-i products and Steenrod Squares often (always?) unary?Joseph Victor2012-08-09T23:39:32Z2012-08-11T14:42:34Z
<p>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 $C_*$, define</p>
<p>$D_0 : C_*\to C_*\otimes C_*$, </p>
<p>so that the cup product is given (on cocycles)</p>
<p>$(u\cup v)(\sigma) = (u\otimes v)(D_0\sigma)$</p>
<p>and then for higher $i$, define $D_i$ so that </p>
<p>$D_{i-1}+\rho D_{i-1} = D_i\partial + \partial D_i$</p>
<p>where $\rho$ is the flipping map. Then the $\cup_i$ product is just</p>
<p>$(u\cup_i u)(\sigma) = (u\otimes u)(D_i\sigma)$</p>
<p>And then define for $[u]\in H^n$
$Sq^{2n-i}([u]) = [u\cup_{i}u]$</p>
<p>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. </p>
<p>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?<br>
Is this a dumb question?</p>
<p>Thanks,
-Joseph</p>
http://mathoverflow.net/questions/104380/why-are-cup-i-products-and-steenrod-squares-often-always-unary/104391#104391Answer by Andrew Ranicki for Why are cup-i products and Steenrod Squares often (always?) unary?Andrew Ranicki2012-08-10T07:32:50Z2012-08-10T07:32:50Z<p>I use $\cup_i$ products as binary products in my work on the algebraic theory of surgery</p>
<p><a href="http://www.maths.ed.ac.uk/~aar/papers/ats2.pdf" rel="nofollow">http://www.maths.ed.ac.uk/~aar/papers/ats2.pdf</a></p>
<p>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. </p>
http://mathoverflow.net/questions/104380/why-are-cup-i-products-and-steenrod-squares-often-always-unary/104490#104490Answer by Peter May for Why are cup-i products and Steenrod Squares often (always?) unary?Peter May2012-08-11T14:42:34Z2012-08-11T14:42:34Z<p>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 <code>$H^*(G/H)$</code> as the torsion product over <code>$H^*(BG)$</code> of $R$ and <code>$H^*(BH)$</code>, where coefficients are taken in a suitable commutative ring $R$. This even works for suitable $H$-spaces. See
<a href="http://www.math.uchicago.edu/~may/BOOKS/GugMay.pdf" rel="nofollow">http://www.math.uchicago.edu/~may/BOOKS/GugMay.pdf</a> and
<a href="http://www.math.uchicago.edu/~may/PAPERS/MNApril20.pdf" rel="nofollow">http://www.math.uchicago.edu/~may/PAPERS/MNApril20.pdf</a></p>