(also nonabelian homological algebra)
category with duals (list of them)
dualizable object (what they have)
ribbon category, a.k.a. tortile category
monoidal dagger-category?
A natural tensor product of chain complexes that makes the category of chain complexes into a closed monoidal category.
Let $R$ be a commutative ring and $\mathcal{A} = R$Mod the category of modules over $R$. Write $Ch_\bullet(\mathcal{A})$ for the category of chain complexes of $R$-modules.
For $X, Y \in Ch_\bullet(\mathcal{A})$ write $(X \otimes Y)_\bullet \in Ch_\bullet(\mathcal{A})$ for the chain complex whose component in degree $n$ is given by the direct sum
over all tensor products of components whose degrees sum to $n$, and whose differential is given on elements $(x,y)$ of homogeneous degree by
The tensor product of chain complexes is equivalently the total complex of the double complex which is the objectwise tensor product:
For $X, Y \in Ch_\bullet(\mathcal{A})$ write $X_\bullet \otimes Y_\bullet \in Ch_\bullet(Ch_\bullet(\mathcal{A}))$ for the double complex whose component in degree $(n_1, n_2)$ is given by the tensor product
whose horizontal differential is $\partial^{hor} \coloneqq \partial^X \otimes id_Y$ and whose vertical differential is $\partial^{vert} \coloneqq id_{X} \otimes \partial^Y$.
The tensor product of chain complexes, def. 1 is isomorphic to the total complex of the double complex of def. 2:
By direct unwinding of the definitions.
For $R$ some ring, let $I_\bullet \in Ch_\bullet(R Mod)$ be the chain complex given by
where $\partial^I_0 = (-id, id)$.
This is the normalized chain complex of the simplicial chain complex of the standard simplicial interval, the 1-simplex $\Delta_1$, as follows: we may think of
as the $R$-linear span of two basis elements labelled “$(0)$” and “$(1)$”, to be thought of as the two 0-chains on the endpoints of the interval. Similarly we may think of
as the free $R$-module on the single basis element which is the unique non-degenerate 1-simplex $(0 \to 1)$ in $\Delta^1$.
Accordingly, the differential $\partial^I_0$ is the oriented boundary map of the interval, taking this basis element to
and hence a general element $r\cdot(0 \to 1)$ for some $r \in R$ to
We now write out in full details the tensor product of chain complexes of $I_\bullet$ with itself, according to def. 1:
By definition and using the above choice of basis element, this is in low degree given as follows:
where in the last line we express a general element as a linear combination of the canonical basis elements which are obtained as tensor products $(a,b) \in R\otimes R$ of the previous basis elements. Notice that by the definition of tensor product of modules we have relations like
etc.
Similarly then, in degree-1 the tensor product chain complex is
And finally in degree 2 it is
All other contributions that are potentially present in $(I \otimes I)_\bullet$ vanish (are the 0-module) because all higher terms in $I_\bullet$ are.
The tensor product basis elements appearing in the above expressions have a clear geometric interpretation: we can label a square with them as follows
This diagram indicates a cellular square and identifies its canonical singular chains with the elements of $(I \otimes I)_\bullet$. The arrows indicate the orientation. For instance the fact that
says that the oriented boundary of the bottom morphism is the bottom right element (its target) minus the bottom left element (its source), as indicated. Here we used that the differential of a degree-0 element in $I_\bullet$ is 0, and hence so is any tensor product with it.
Similarly the oriented boundary of the square itself is computed to
which can be read as saying that the boundary is the evident boundary thought of as oriented by drawing it counterclockwise into the plane, so that the right arrow (which points up) contributes with a +1 prefactor, while the left arrow (which also points up) contributes with a -1 prefactor.
For $X,Y \in$ Top two topological spaces, the Eilenberg-Zilber theorem asserts a quasi-isomorphism
between the singular chain complex of the product topological space and the tensor product of chain complexes of the separate singular chain complexes.
As for any total complex of a double complex, the tensor product of chain complexes is naturally a filtered chain complex, either by the degree of the first of by that of the second chain complex factor.
Let $R$ be a commutative ring. For $A, B \in R$Mod, the two ways of computing the Tor left derived functor coincide
and hence we can consistently write $Tor_n(A,B)$ for either.
Let $Q^A_\bullet \stackrel{\simeq_{qi}}{\to} A$ and $Q^B_\bullet \stackrel{\simeq_{qi}}{\to} B$ be projective resolutions of $A$ and $B$, respectively. The corresponding tensor product of chain complexes $Tot (Q^A_\bullet\otimes Q^B_\bullet)$, hence by prop. 1 the total complex of the degreewise tensor product of modules double complex carries the filtration by horizontal degree as well as that by vertical degree.
Accordingly there are the corresponding two spectral sequences of a double complex, to be denoted here $\{{}^{A}E^r_{p,q}\}_{r,p,q}$ (for the filtering by $A$-degree) and $\{{}^{B}E^r_{p,q}\}_{r,p,q}$ (for the filtering by $B$-degree). By the discussion there, both converge to the chain homology of the total complex.
We find the value of both spectral sequences on low degree pages according to the general discussion at spectral sequence of a double complex - low degree pages.
The 0th page for both is
For the first page we have
and
Now using the universal coefficient theorem in homology and the fact that $Q^A_\bullet$ and $Q^B_\bullet$ is a resolution by projective objects, by construction, hence of tensor acyclic objects for which all Tor-modules vanish, this simplifies to
and similarly
It follows for the second pages that
and
Now both of these second pages are concentrated in a single row and hence have converged on that page already. Therefore, since they both converge to the same value:
For instance section 2.7 of
Last revised on January 19, 2015 at 15:13:27. See the history of this page for a list of all contributions to it.