# nLab tensor product of chain complexes

### Context

#### Homological algebra

homological algebra

and

nonabelian homological algebra

diagram chasing

### Theorems

#### Monoidal categories

monoidal categories

# Contents

## Idea

A natural tensor product of chain complexes that makes the category of chain complexes into a closed monoidal category.

## Definition

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.

###### Definition

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

$(X \otimes Y)_n := \oplus_{i + j = n} X_i \otimes_R Y_j$

over all tensor products of components whose degrees sum to $n$, and whose differential is given on elements $(x,y)$ of homogeneous degree by

$\partial^{X \otimes Y} (x, y) = (\partial^X x, y) + (-1)^{deg(x)} (x, \partial^Y y) \,.$

## Properties

### As a total complex of a double complex

The tensor product of chain complexes is equivalently the total complex of the double complex which is the objectwise tensor product:

###### Definition

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

$X_{n_1} \otimes Y_{n_2} \in \mathcal{A}$

whose horizontal differential is $\partial^{hor} \coloneqq \partial^X \otimes id_Y$ and whose vertical differential is $\partial^{vert} \coloneqq id_{X} \otimes \partial^Y$.

###### Proposition

The tensor product of chain complexes, def. 1 is isomorphic to the total complex of the double complex of def. 2:

$Tot_\bullet (X_\bullet \otimes Y_\bullet) \simeq (X \otimes Y)_\bullet \,.$
###### Proof

By direct unwinding of the definitions.

## Examples

### Square as tensor product of interval with itself

###### Example

For $R$ some ring, let $I_\bullet \in Ch_\bullet(R Mod)$ be the chain complex given by

$I_\bullet = \left[ \cdots \to 0 \to 0 \to R \stackrel{\partial^{I}_0}{\to} R \oplus R \right] \,,$

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

$I_0 = R \oplus R \simeq R[ \{(0), (1)\} ]$

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

$I_1 = R \simeq R[\{(0 \to 1)\}]$

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

$\partial^I_0 : (0 \to 1) \mapsto (1) - (0)$

and hence a general element $r\cdot(0 \to 1)$ for some $r \in R$ to

$\partial^I_0 : r\cdot(0 \to 1) \mapsto r\cdot (1) - r\cdot(0) \,.$

We now write out in full details the tensor product of chain complexes of $I_\bullet$ with itself, according to def. 1:

$S_\bullet \coloneqq I_\bullet \otimes I_\bullet \,.$

By definition and using the above choice of basis element, this is in low degree given as follows:

\begin{aligned} S_0 &= I_0 \otimes I_0 \\ & = (R \oplus R) \otimes (R \oplus R) \\ & \simeq R \oplus R \oplus R \oplus R \\ & = \left\{ r_{00} \cdot ((0),(0)') + r_{01} \cdot ((0),(1)') + r_{10} \cdot ((1),(0)') + r_{11} \cdot ((1),(1)') | r_{\cdot, \cdot} \in R \right\} \end{aligned} \,,

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

$r ( (0), (1)') = (r(0), (1)') = ((0), r(1)')$

etc.

Similarly then, in degree-1 the tensor product chain complex is

\begin{aligned} (I \otimes I)_1 & = (I_0 \otimes I_1) \oplus (I_1 \otimes I_0) \\ & \simeq R \otimes (R \oplus R) \oplus (R \oplus R) \otimes R \\ & \simeq R \oplus R \oplus R \oplus R \\ & \simeq \left\{ r_{0} \cdot ((0),(0\to 1)') + r_{1} \cdot ((1), (0 \to 1)') + \bar r_0 \cdot ((0\to 1), (0)') + \bar r_1 \cdot ((0 \to 1), (1)') | r_{\cdot}, \bar r_{\cdot} \in R \right\} \end{aligned} \,.

And finally in degree 2 it is

\begin{aligned} (I \otimes I)_2 & \simeq I_1 \otimes I_1 \\ & \simeq R \otimes R \\ & \simeq R \\ & \simeq \left\{ r\cdot ((0 \to 1), (0 \to 1)') | r \in R \right\} \end{aligned} \,.

All other contributions that are potentiall 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

$\array{ ((0),(1)') &&\underset{((0\to 1),(0))}{\to}&& ((1),(1)') \\ \\ {}^{\mathllap{((0),(0\to 1)')}}\uparrow &&\righttoleftarrow^{((0 \to 1), (0\to 1)')}&& \uparrow^{\mathrlap{((1),(0 \to 1)')}} \\ \\ ((0),(0)') &&\underset{((0\to 1),(0)')}{\to}&& ((1),(0)') } \,.$

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

\begin{aligned} \partial^{I \otimes I} ((0 \to 1), (0)') & = (\partial^I (0 \to 1), (0)') + (-1)^1 ((0\to 1), \partial^I (0)) \\ & = ( (1) - (0), \;(0)' ) - 0 \\ & = ((1), (0)') - ((0), (0)') \end{aligned}

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

\begin{aligned} \partial^{I \otimes I} ((0 \to 1), (0 \to 1)') &= (\partial^I (0 \to 1), (0 \to 1)') - ((0 \to 1), \partial^I(0 \to 1)) \\ & = ((1)- (0), (0 \to 1)') - ((0 \to 1), (1)' - (0)') \\ & = ((1), (0 \to 1)') - ((0), (0 \to 1)') - ((0 \to 1), (1)') + ((0 \to 1), (0)') \end{aligned} \,,

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.

### Singular chain complex

For $X,Y \in$ Top two topological spaces, the Eilenberg-Zilber theorem asserts a quasi-isomorphism

$C_\bullet(X \times Y) \to (C(X) \otimes C(Y))_\bullet$

between the singular chain complex of the product topological space and the tensor product of chain complexes of the separate singular chain complexes.

### Filtering and spectral sequences

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.

###### Proposition

Let $R$ be a commutative ring. For $A, B \in R$Mod, the two ways of computing the Tor left derived functor coincide

$(L_n ((-)\otimes_R B))(A) \simeq (L_n (A \otimes_R (-)))(B)$

and hence we can consistently write $Tor_n(A,B)$ for either.

###### Proof

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

${}^A E^0_{p,q} = {}^B E^0_{p,q} \coloneqq Q^A_p \otimes_R Q^B_q \,.$

For the first page we have

\begin{aligned} {}^A E^1_{p,q} & \simeq H_q(C_{p,\bullet}) \\ & \simeq H_q( Q^A_p \otimes Q^B_\bullet ) \end{aligned}

and

\begin{aligned} {}^B E^1_{p,q} & \simeq H_q(C_{\bullet,p}) \\ & \simeq H_q( Q^A_\bullet \otimes Q^B_p ) \end{aligned} \,.

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

\begin{aligned} {}^A E^1_{p,q} & \simeq Q^A_p \otimes H_q(Q^B_\bullet) \\ & \simeq \left\{ \array{ Q^A_p \otimes_R B & if\; q = 0 \\ 0 & otherwise } \right. \end{aligned}

and similarly

\begin{aligned} {}^B E^1_{p,q} & \simeq H_q(Q^A_\bullet) \otimes_R Q^B_p \\ & \simeq \left\{ \array{ A \otimes_R Q^B_p & if\; q = 0 \\ 0 & otherwise } \right. \end{aligned} \,.

It follows for the second pages that

\begin{aligned} {}^A E^2_{p,q} & \simeq H_p(H^{vert}_q(Q^A_\bullet \otimes Q^B_\bullet)) \\ & \simeq \left\{ \array{ (L_p( (-)\otimes_R B ))(A) & if \; q = 0 \\ 0 & otherwise } \right. \end{aligned}

and

\begin{aligned} {}^B E^2_{p,q} & \simeq H_p(H^{hor}_q(Q^A_\bullet \otimes Q^B_\bullet)) \\ & \simeq \left\{ \array{ (L_p ( A \otimes_R (-) ))(B) & if \; q = 0 \\ 0 \; otherwise } \right. \end{aligned} \,.

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:

$L_p((-)\otimes_R B)(A) \simeq {}^A E^2_{p,0} \simeq {}^A E^\infty_{p,0} \simeq {}^B E^2_{p,0} \simeq L_p(A \otimes_R (-))(B) \,.$

## References

For instance section 2.7 of

Revised on July 8, 2014 09:19:40 by Anonymous Hero (141.211.63.38)