Explicit description of the "simplicial tensor product" of chain complexes Recall that there is an equivalence of categories (Dold-Kan) $$N:\mathrm{s}\mathbf{Ab}\simeq \operatorname{Ch}_{\geq 0}(\mathbf{Ab}):\Gamma$$ between simplicial abelian groups and (connective) chain complexes, where $N$ sends a simplicial abelian group to its associated normalized chain complex.  
Using this equivalence of categories, we can, by transport of structure, give an unorthodox tensor product on the category of chain complexes.  We may define this by the formula $X\otimes_\Delta Y=N(\Gamma(X)\otimes \Gamma(Y)),$ where the tensor product on the righthand side is the tensor product (taken pointwise) of simplicial abelian groups.
Then my question: Is there an explicit description of this tensor product in terms of the chain complexes themselves?  
 A: The bad news is that in degree $n$, this tensor product has $3^n$ terms.
The functor $\Gamma$ can be roughly described as follows.  If we write $[n]$ for the ordered set $0 < 1 < \cdots < n$, then
$$
\Gamma(C)_n = \bigoplus_{k} \bigoplus_{\phi\colon [n] \twoheadrightarrow [k]} C_k.
$$
The face maps have two characters.  The map $d_i$ for $i > 0$ simply deletes the element $i$ from the ordered set $[n]$, and reindexes; if the resulting map $[n-1] \to [k]$ is no longer surjective, the corresponding factor maps to zero.  By contrast, the map $d_0$ deletes $0$ and reindexes, but if the corresponding map $\phi$ is no longer surjective its image is isomorphic to $[k-1]$, and we apply the boundary map.
When you take the tensor product of $\Gamma(C)$ and $\Gamma(D)$ levelwise, you get a direct sum indexed by pairs of surjections $[n] \twoheadrightarrow [p]$ and $[n] \twoheadrightarrow [q]$.
The functor $N$ then will take the quotient of this by the subcomplex of degenerate ones; those where the maps $[n] \twoheadrightarrow [p]$ and $[n] \twoheadrightarrow [q]$ factors through a surjection $[n] \twoheadrightarrow [m]$.  In practice, the pairs which are not degenerate are those for which the map $[n] \to [p] \times [q]$ is injective.
As a result, we have that
$$
N(\Gamma(C) \otimes \Gamma(D))_n = \bigoplus_\phi C_p \otimes D_q
$$
where the sum is indexed by injections $[n] \to [p] \times [q]$ where composing with either projection is surjective.
In practice, you can index this direct sum by $n$-tuples of strings of elements from$\{N, NE, E\}$, representing a path of length $n$; $p$ and $q$ are determined by the height and width of the path.
Unfortunately, a chain complex isn't very useful without its differential, and that's more complicated to describe.  The boundary map is the alternating sum of face maps; each face map deletes $i$ from the ordered set $0 < \cdots < n$ and reindexes.  If $i > 0$ and one of the resulting projections to $[p]$ or $[q]$ is no longer surjective, the corresponding factor maps to zero; if $i = 0$ then one or both of the maps to $[p]$ or $[q]$ misses zero, the appropriate image(s) are isomorphic to $[p-1]$ or $[q-1]$, and we apply the boundary map on those factors.
UPDATE: One way to write this is in a group homology style.  Then we can view elements of $\Gamma(C)_n$ as decorated with these $n$-tuple paths, the width and height determine which group we land in, and the boundary map takes an alternating sum of deleting commas with the understanding that $NN = EE = 0$.  So, for example, taking the boundary of this element in $C_3 \otimes D_2$:
$$
d(a \otimes b)_{(N,E,N,NE)} = (da) \otimes b_{(E,N,NE)} - a \otimes b_{(NE,N,NE)} + a \otimes b_{(N,NE,NE)} - a \otimes b_{(N,E,NNE)}.
$$
(The last term involving $NNE$ is dropped because it is zero.  The first term had a boundary map applied to the first factor because it was $N$.)
