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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?

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Maybe use the Alexander-Whitney map... Is it not just the usual tensor product? –  Dylan Wilson Apr 20 '12 at 7:07
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Nice to see you again, Harry! –  David Roberts Apr 20 '12 at 7:27
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@Dylan: No, the Alexander-Whitney map is a natural quasi-isomorphism $$X\otimes_\Delta Y\to X\otimes Y.$$ The question I'm asking is "opposite" to the description of the tensor product of chain complexes transported to the category of simplicial abelian groups, which is given by the formula $$(A\otimes_{\mathrm{Ch}} B)_i = \mathrm{Tot}(A \otimes_{\Delta \times \Delta} B)_i,$$ where the RHS the total simplicial abelian group of the bisimplicial abelian group $$(A \otimes_{\Delta \times \Delta} B)_{jk}=A_j\otimes B_k.$$ –  Harry Gindi Apr 20 '12 at 7:45
    
Likewise, David! –  Harry Gindi Apr 20 '12 at 9:52
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The shuffle map is a natural quasi-inverse of the Alexander-Whitney map. Since it is not surjective, there is no hope of it characterizing the simplicia tensor product. –  Harry Gindi Apr 20 '12 at 11:31
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1 Answer

up vote 9 down vote accepted

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 $N$.)

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Since the sequence in $\{N,NE,E\}$ can be directly translated into sequences of degeneracy maps on either side of the tensor, couldn't we just explicitly decorate the elements $a\otimes b$ with appropriate sequences of degeneracies, so applying the differential as an alternating sum of face maps works every time (as opposed to looking back at the sequence)? I was considering some simple examples, and this works pretty handily. Moreover, it gives us a way to determine which factor a given tensor lives in. –  Harry Gindi Apr 23 '12 at 11:02
    
@Harry, if you're comfortable with expressing and validating things in terms of explicit degeneracy operators, then that's certainly fine. I am not, and so I tend to try to express things differently. –  Tyler Lawson Apr 23 '12 at 12:39
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@Harry: Under this boundary map, the element you've written down isn't a cycle. The problem likely stems from the following. In order to convert a chain complex to a simplicial abelian group, I've used the "intersection of kernels" definition of the normalized complex; in order to convert a simplicial object to a chain complex, I've used a "mod by degeneracies" definition. Both of these I did because it's easier. The problem is that the definition involving the intersection of kernels isn't compatible with simplicial sets... –  Tyler Lawson Apr 23 '12 at 15:49
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Specifically, if you take a simplicial set $K$ and take the first kind of normalized complex for $\mathbb Z[K]$, most of the actual simplices of $K$ don't immediately arise as basis elements; they have to be "converted" by subtracting a degenerate element to arrive at something that has most boundary operators equal to zero. –  Tyler Lawson Apr 23 '12 at 15:53
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For example, in the case of $f \otimes g$ in degree 1, you're likely thinking of this as representing the product of the two 1-simplices represented by $f$ and $g$, with $d_1 = x_1 \otimes y_1$ and $d_0 = x_0 \otimes y_0$. But it doesn't; the lift of $f$ and $g$ from the "quotient" complex to the "intersection" complex are $f - s_0(x_1)$` and $g - s_0(y_1)$`. That's why their tensor product is behaving strangely. –  Tyler Lawson Apr 23 '12 at 15:59
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