TITLE: How many tensor products of chain complexes are there?
QUESTION [12 upvotes]: Let $Ch$ be the category of nonnegatively-(homologically-)graded chain complexes of abelian groups. Suppose that $(Ch,\boxtimes)$ is a monoidal biclosed structure.

Assume that the forgetful functor $(Ch, \boxtimes) \to (Gr,\otimes)$ is strong monoidal, where $Gr$ is the category of nonnegatively-graded abelian groups and $\otimes$ its usual tensor product.

Thus $(A\boxtimes B)_n = \oplus_{i+j = n} A_i \otimes B_j$, equipped with some differential, in an associative and unital way.
Question 1: Can one classify all such monoidal structures $(Ch,\boxtimes)$, up to monoidal equivalence over $(Gr,\otimes)$?
Notes:

The usual tensor product $\otimes$ of chain complexes is an example, with $d(x\otimes y) = d x \otimes y + (-1)^{|x|} x \otimes d y$ but not the only one -- e.g. one might define the differential as $d'(x \otimes y) = (-1)^{|y|} d x \otimes y + x \otimes d y$ for a different example.

If it's easier to work over a field instead of $\mathbb Z$, that would be an interesting start.

I've singled out nonnegatively-graded chain complexes, but if the situation changes when passing to unbounded chain complexes, that would be interesting.


Question 2: Are all such monoidal structures $\boxtimes$ symmetric monoidal? In how many ways?

REPLY [3 votes]: Okay, here's a solution when we allow unbounded chain complexes.
First, note that the automorphism group of $Ch$ over $Gr$ is $C_2^\omega$ (or rather $C_2^{\mathbb Z}$ since we've switched to the unbounded case), because swapping the sign of the $n$th differential on all objects of $Ch$ affects neither the underlying graded abelian groups, nor the notion of chain map (and similarly for swapping the sign of $d_n$ for each $n \in S$, where $S \subseteq \mathbb Z$ is an arbitrary subset). And indeed, the two examples of monoidal structures which I gave are related by such an automorphism -- the one where you swap the sign of $d_n : A_n \to A_{n-1}$ whenever $n$ is odd.
Claim: All the monoidal biclosed structures on $Ch$ over $Gr$ are related to the standard one by such an automorphism.
An easy exercise shows that the only nonidentity automorphism of the standard tensor product over $Gr$ is the one which flips the signs of all the differentials, so most of these monoidal structures are different from one another. And apparently they are all symmetric monoidal (in exactly as many ways as the standard monoidal structure -- however many that is).
Proof:
Consider the short exact sequence $S(0) \to D(1) \to S(1)$ (here and in the following, $D(n)$ is the degree $n$ disk and $S(n)$ is the degree $n$ sphere; we take advantage in the following of the fact that these objects have canonical bases which we sometimes silently identify). Tensoring the exact sequence against itself, we get an exact $3\times 3$ grid. Unitality of the monoidal structure tells us that the differentials on $S(0) \boxtimes D(1)$ and $D(1) \boxtimes S(0)$ are canonically isomorphic to the differential on $D(1)$. By looking at the inclusion maps from these objects to $D(1)^{\boxtimes 2}$, we see that the $d_1$ differential on $D(1)^{\boxtimes 2}$ is given by the codiagonal $[1,1]: D(1)_1 \otimes D(0)_0 \oplus D(0)_0 \otimes D(0)_1 \to D(0)_0 \otimes D(0)_0$. This implies that the $d_2$ differential is of the form $[a,-a] : D(1)_1 \otimes D(1)_1 \to D(1)_1 \otimes D(0)_0 \oplus D(0)_0 \otimes D(0)_1$ for some $a \in \mathbb Z$. Conjugating by an automorphism of $Ch$ over $Gr$, we may assume that $a \geq 0$. Quotienting along the maps to $D(1) \boxtimes S(1)$ and $S(1) \boxtimes D(1)$, we see that the differential on the former is given by $a$ and on the latter the differential is $-a$.
I claim that $a = 1$. For if $a = 0$, then $S(1) \boxtimes D(1)$ has a null differential. Tensoring with $S(-1)$, it follows that $D(1)$ has a null differential, a contradiction. On the other hand, if $a > 1$, then there is an exact sequence $0 \to D(2) \to S(1) \boxtimes D(1) \to S(1)/a \to 0$. Again tensoring with $S(-1)$, we obtain an exact sequence $0 \to S(-1) \boxtimes D(2) \to D(1) \to S(0)/a \to 0$. We know all the groups involved here, and because the differential on $D(1)$ is a unit, the differential on $S(-1) \boxtimes D(2)$ would have to be $1/a$, which is of course impossible.
Now just as we used knowledge of the differential on $S(0) \boxtimes D(1)$ to determine the differential on $S(1) \boxtimes D(1)$, we can work our way up to work out that the differential on $S(n) \boxtimes D(1)$ is standard for all $n \geq 0$. A dual argument allows us to work out negative $n$'s as well.
At this point we know the differential on $D(1) \boxtimes D(1)$. It’s the standard one, which splits as a sum $D(2) \oplus D(1)$. We also have canonical isomorphisms $S(n-1) \boxtimes D(1) = D(n)$ and dually $D(1) \boxtimes S(m-1) = D(m)$. Together we can now determine the differential on $D(n) \boxtimes D(m)$. By compatibility with colimits, this determines the entire monoidal structure.

If we stick to just nonnegatively graded complexes, then I have difficulty ruling out the cases $a=0$ or $a > 1$ in the above argument. I would be interested to see the argument pushed through to completion in this case, but I think I have run out of energy to do it myself right now.