Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:23:40Z http://mathoverflow.net/feeds/question/88156 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88156/extend-alexander-whitney-and-eilenberg-zilber-map-to-n-fold-tensor-products Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products Nevermind 2012-02-10T22:44:21Z 2012-02-11T01:02:44Z <p>See the definition of the Alexander-Whitney transformation:</p> <p><a href="http://ncatlab.org/nlab/show/Alexander-Whitney+map" rel="nofollow">http://ncatlab.org/nlab/show/Alexander-Whitney+map</a></p> <p>and the Eilenberg-Zilber transformation:</p> <p><a href="http://ncatlab.org/nlab/show/Eilenberg-Zilber+map" rel="nofollow">http://ncatlab.org/nlab/show/Eilenberg-Zilber+map</a></p> <p>Is there a natural or obvious way to extend them to higher tensor powers i.e to, lets say</p> <p>$$\Delta_{A_1,\ldots,A_n} : C(\otimes_{j=1}^n A_j) \to \otimes_{j=1}^n C(A_j)$$ and $$\nabla_{A_1,\ldots,A_n} : \otimes_{j=1}^n C(A_j) \to C(\otimes_{j=1}^nA_j)$$</p> <p>or to the infinite tensor power series, such that the adjointness is still there?</p> <p>(My first obvious guess is to simply iterate them using associativity of the usual tensor product but I'm not sure if it is that simple due to concerns about braiding and singns)</p> http://mathoverflow.net/questions/88156/extend-alexander-whitney-and-eilenberg-zilber-map-to-n-fold-tensor-products/88167#88167 Answer by Ralph for Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products Ralph 2012-02-11T00:48:20Z 2012-02-11T01:02:44Z <p>Yes, there is a generalization to a finite number of simplicial complexes. A reference is <em>Corollary 2.2</em> in the paper</p> <blockquote> <p>Eilenberg, MacLane: On the groups $H(\Pi,n)$, II: Methods of Computation, Ann. of. Math. 60(1954), No. 1, 49 - 139.</p> </blockquote> <p>Using the definitions from nLab, the maps are given as follows: </p> <p>1) Let $a_i \in A_i$ be homogen. $$\nabla(a_1 \otimes \cdots \otimes a_n) = a_1 \nabla a_2 \nabla ... \nabla a_n$$ (well-definied since $\nabla$ is associative)</p> <p>2) Let $a_i \in (A_i)_m$. $$\Delta(a_1 \otimes \cdots \otimes a_n) = \sum \displaystyle \otimes_{i=1}^n\displaystyle\tilde{d}^{m-j_i}d_0^{j_{i-1}}a_i$$ where the sum is taken over $0 \le j_1 \le \cdots \le j_{n-1} \le m$ and $\tilde{d}^{m-j_n},\;d_0^{j_0}$ has to be interpreted as identity. </p>