Explicit description of the "simplicial tensor product" of chain complexes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:43:42Z http://mathoverflow.net/feeds/question/94606 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94606/explicit-description-of-the-simplicial-tensor-product-of-chain-complexes Explicit description of the "simplicial tensor product" of chain complexes Harry Gindi 2012-04-20T07:03:22Z 2012-04-23T12:47:54Z <p>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. </p> <p>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.</p> <p>Then my question: Is there an explicit description of this tensor product in terms of the chain complexes themselves? </p> http://mathoverflow.net/questions/94606/explicit-description-of-the-simplicial-tensor-product-of-chain-complexes/94640#94640 Answer by Tyler Lawson for Explicit description of the "simplicial tensor product" of chain complexes Tyler Lawson 2012-04-20T13:16:26Z 2012-04-23T12:47:54Z <p>The bad news is that in degree $n$, this tensor product has $3^n$ terms.</p> <p>The functor $\Gamma$ can be roughly described as follows. If we write $[n]$ for the ordered set $0 &lt; 1 &lt; \cdots &lt; n$, then <code>$$\Gamma(C)_n = \bigoplus_{k} \bigoplus_{\phi\colon [n] \twoheadrightarrow [k]} C_k.$$</code> The face maps have two characters. The map <code>$d_i$</code> 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 <code>$d_0$</code> 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.</p> <p>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 $[n] \twoheadrightarrow [q]$.</p> <p>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.</p> <p>As a result, we have that <code>$$N(\Gamma(C) \otimes \Gamma(D))_n = \bigoplus_\phi C_p \otimes D_q$$</code> where the sum is indexed by injections $[n] \to [p] \times [q]$ where composing with either projection is surjective.</p> <p>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.</p> <p>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 $0 &lt; \cdots &lt; n$ 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.</p> <p>UPDATE: One way to write this is in a group homology style. Then we can view elements of <code>$\Gamma(C)_n$</code> 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 <code>$C_3 \otimes D_2$</code>: <code>$$d(a \otimes b)_{(N,E,N,NE)} = (da) \otimes b_{(E,N,NE)} - a \otimes b_{(NE,N,NE)} + a \otimes b_{(N,NE,NE)} - a \otimes b_{(N,E,NNE)}.$$</code> (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 $N$.)</p>