Recall that a chain complex is a (finite) diagram of the form $$ V = \{ \dots \to V_3 \overset{d_3}\to V_2 \overset{d_2}\to V_1 \overset{d_1}\to V_0 \to 0 \} $$ where the $V_n$ are (finite-dimensional) vector spaces and for each $n$, $d_n \circ d_{n+1} = 0$. If $V$ and $W$ are chain complexes, a chain map $f: V \to W$ is a map $f_n : V_n \to W_n$ for each $n$ such that all the obvious squares commute — "$[d,f]=0$" — and the pair (chain complexes, chain maps) defines a category. In fact, it is a 2-category: the 2-morphisms between $f,g : V \rightrightarrows W$ are the chain homotopies, i.e. a system of maps $h_n: V_n \to W_{n+1}$ such that "$[d,h] = f-g$". The category of chain complexes has a biproduct (both a product and a coproduct) $\oplus$ given by the pointwise direct sum.

I thought I knew what the tensor product of chain complexes was. Namely, if $V$ and $W$ are chains, then the usual thing is to define $$ (V\otimes W)\_n = \bigoplus_{k=0}^n V_k \otimes W_{n-k} $$ and the chain maps are the sums of the obvious tensor products of differentials, decorated with signs.

But now I'm not sure why this is the tensor product picked. Namely, if I have a linear category, I think that a tensor product $V \otimes W$ should satisfy the following universal property: for any $X$, $\hom(V \otimes W,X)$ should be naturally isomorphic to the space of bilinear maps $V \times W \to X$. Now, I've never really known how to write down the word "bilinear" in a general category, without refering to individual points. But I think I do know what the "set" $V \times W$ is when $V$ and $W$ are chains — it's the set underlying $V \oplus W$ — and then I think I do know what bilinear maps should be.

In any case, then it's clear that the usual tensor product is not this. For example, if $V,W$ have no non-zero terms above degree $n$, then the bilinear maps $V \times W \to X$ I think cannot be interesting above degree $n$, whereas the above $\otimes$ has terms in degree $2n$.

In any case, in HDA6, Baez and Crans consider two-term chain complexes $V_1 \to V_0$ (they argue that these are the same as "2-vector-spaces"), and then construct a different tensor product, given by: $$ V\otimes W = \{ (V_1 \otimes W_1) \oplus (V_1 \otimes W_0) \oplus (V_0\otimes W_1) \to (V_0 \otimes W_0) \} $$ where the differential is the sum of the obvious tensor products of differentials and identity maps. They then assert that this tensor product satisfies the correct universal property, although they leave the details to the reader.

This leads naturally to:

Question: What is the precise universal property that $\otimes$ ought to have, and what "product" of chain maps satisfies this universal property?

  • $\begingroup$ Of course, there are many good reasons to consider variations on what I've called a "chain complex", but those issues are orthogonal to my question. $\endgroup$ Mar 12 '10 at 6:03
  • $\begingroup$ I think that this should be by the universal property of the tensor product, that is, $Hom(M\otimes N, -) \cong Hom(M,(-)^N)$, where $(-)^N$ is the internal hom. To be honest, I'm not actually sure that this works for complexes, but it seems like it should. $\endgroup$ Mar 12 '10 at 6:12
  • 1
    $\begingroup$ @fpqc: That works for complexes (with the correct sign convention on the Hom-complex, something people are often not very careful about). $\endgroup$ Mar 12 '10 at 6:24
  • $\begingroup$ @Theo: the unique occurrence of "n^2" in the question---should it be "2n"? $\endgroup$ Mar 12 '10 at 7:10
  • 1
    $\begingroup$ I suspected Baez and Crans' proof left to readers was actually wrong (Proposition 17 page 16-17 of arxiv.org/pdf/math/0307263v6). The secret is that the "natural functor" $i:V\times V'\to V\otimes V'$ is not a functor. $\endgroup$
    – Ma Ming
    Nov 18 '11 at 13:48

I think you need to revise your treatment of degree when working with bilinear maps, since the notion of "bilinear" requires more information from $V \times W$ than just the fact that it is an object in the category. For a simple case, try forgetting the differentials, and just work with graded vector spaces, or comodules over $k[\mathbb{Z}]$, where $k$ is your base field. The correct notion of tensor product forces a bilinear map from $V \times W$ to respect total degree.

  • $\begingroup$ In fact, the notion of bilinearity can be defined when the product is not an object of the category. In Tom Leinster's book Higher Categories, Higher Operads he discusses this in his discussion of multicategories. $\endgroup$ Mar 12 '10 at 6:45
  • $\begingroup$ Yes, this machinery is also treated (in the symmetric case) in Beilinson and Drinfeld's Chiral Algebras under the name pseudo-tensor category. $\endgroup$
    – S. Carnahan
    Mar 12 '10 at 7:05

Several people have mentioned that $\otimes N$ should left adjoint to a suitable $Hom(N,-)$. There are numerous instances of this sort of construction (including one using joins of augmented simplicial sets mentioned in another question w.r.t Lurie's HTT). There is a good discussion of the ideas in MacLane's Homology and in Hilton and Stammbach. The sign conventions are, however, often not fully explained. (They work, but why? is the feeling one gets.)

The point in Baez and Crans is more that there are variants that have good more or less geometric interpretations and they give sensible answers. (There is also work by Andy Tonks on tensor products of crossed complexes that makes this clear.)

The discussion in older sources such as Spanier is also good for explaining why the signs are important.

You ask about bilinearity. This is really a strange idea. It can be handled in more generality than the simple form you mention but it can often be best interpreted as coming from the evaluation map $ Hom(A,B)\times A \to B$ in which $(f,x)$ gets sent to $f(x)$. If you look at the adjunction that was mentioned between $\otimes$ and $Hom$, then from $\phi : C\to Hom(A,B)$ you do a naive thing and get $C\times A \to Hom(A,B)\times A\to B$ and that will be bilinear, ugh! Thus the representability of bilinear maps is the same universal property as the one everyone has been mentioning.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.