Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products - MathOverflow

Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products

2012-02-10T22:44:21Z

See the definition of the Alexander-Whitney transformation:

http://ncatlab.org/nlab/show/Alexander-Whitney+map

and the Eilenberg-Zilber transformation:

http://ncatlab.org/nlab/show/Eilenberg-Zilber+map

Is there a natural or obvious way to extend them to higher tensor powers i.e to, lets say

$$ \Delta_{A_1,\ldots,A_n} : C(\otimes_{j=1}^n A_j) \to \otimes_{j=1}^n C(A_j) $$ and $$ \nabla_{A_1,\ldots,A_n} : \otimes_{j=1}^n C(A_j) \to C(\otimes_{j=1}^nA_j) $$

or to the infinite tensor power series, such that the adjointness is still there?

(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)

Answer by Ralph for Extend Alexander-Whitney and Eilenberg-Zilber map to n-fold tensor products

2012-02-11T00:48:20Z

Yes, there is a generalization to a finite number of simplicial complexes. A reference is Corollary 2.2 in the paper

Eilenberg, MacLane: On the groups $H(\Pi,n)$, II: Methods of Computation, Ann. of. Math. 60(1954), No. 1, 49 - 139.

Using the definitions from nLab, the maps are given as follows:

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)

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.