Defining the cup product in Ext using a Kunneth formula - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T07:24:13Z http://mathoverflow.net/feeds/question/104048 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104048/defining-the-cup-product-in-ext-using-a-kunneth-formula Defining the cup product in Ext using a Kunneth formula Joseph Victor 2012-08-05T20:25:31Z 2012-08-05T20:25:31Z <p>I want to make a Kunneth product of sorts on Ext. In particular, letting $C_*$ be a $R$-free resolution for $k$ over a $k$-hopf algebra $R$, elements in $Ext_R(k,k)$ are represented by maps in $Hom_R(C_*,k)$. Of course, we can consider $Hom_R(C_* \otimes C_*,k)$, the cohomology of this being the same since $k\otimes k \cong k$. </p> <p>What I want then is a sort of Kunneth product:</p> <p>$\times :Hom_R(C_*,k)\otimes Hom_R(C_*,k)\to Hom_R(C_*\otimes C_*,k)$</p> <p>I want it to be the case that this defines the cup product (aka Yoneda product). If we get a magical map </p> <p>$\Delta : C_*\to C_*\otimes C_*$,</p> <p>then </p> <p>$\cup : Ext_R(k,k)\otimes Ext_R(k,k)\to Ext_R(k,k)$</p> <p>can be given, on the co-chain level,</p> <p>$(a\cup b)(\sigma) = (a\times b)(\Delta \sigma)$</p> <p>Now, $\Delta$ is possibly defined up to homotopy by the requirement that it commutes with the augmentation in $R$. </p> <p>My question is this: is there a way to describe this Kunneth product using any $C_*$, or do I need to use a Bar resolution of sorts like in group cohomology. Does this method even make sense? Can you help flesh this out? </p> <p>Thanks!</p>