Is the tensor product of chain complexes a Day convolution?

Question

Recently, Jade Master asked whether the tensor product of chain complexes could be viewed as a special case of Day convolution. Noting that chain complexes may be viewed as $\mathsf{Ab}$-functors from a certain $\mathsf{Ab}$-category $\mathsf{C}$, Yuri Sulyma suggested¹ that maybe we could obtain the tensor product of two chain complexes as a Day convolution by endowing $\mathsf{C}$ with the monoidal structure given by $[n]\otimes_\mathsf{C}[m]\overset{\mathrm{def}}{=}[n+m-1]$.

Questions: Is this affirmation true? More precisely:

1. Given two chain complexes $X_\bullet$ and $Y_\bullet$ on an abelian category $\mathcal{A}$, is their Day convolution as $\mathsf{Ab}$-functors from $(\mathsf{C},\otimes_\mathsf{C})$ the usual tensor product of chain complexes $\otimes_{\mathsf{Ch}(\mathcal{A})}$?
2. If not, is there some other monoidal structure on $\mathsf{C}$ for which Day convolution gives $\otimes_{\mathsf{Ch}(\mathcal{A})}$?
3. If this fails too, is there perhaps another way to view $\otimes_{\mathsf{Ch}(\mathcal{A})}$ as a special case of some general construction in enriched category theory?

---

¹Note that his account is protected and hence his reply is not public.

Answer 1

The answer to the question posed in the title of your post is yes, the tensor product of chain complexes is a Day convolution product. The important thing to note is that, to define a Day convolution monoidal structure on the $\mathcal{V}$-enriched functor category $[\mathcal{C},\mathcal{V}]$ (where $\mathcal{V}$ is a complete and cocomplete symmetric monoidal closed category, e.g. $\mathbf{Ab}$), we needn't demand $\mathcal{C}$ to be a monoidal $\mathcal{V}$-category: it suffices for $\mathcal{C}$ to be a promonoidal $\mathcal{V}$-category. This is the generality at which Day convolution was originally defined in Day's thesis, which may be found here (see also his earlier paper in the Reports of the Midwest Category Seminar IV, where the word "premonoidal" was used).

A promonoidal structure on a small $\mathcal{V}$-category $\mathcal{C}$ consists of tensor product and unit "profunctors", i.e. $\mathcal{V}$-functors $P \colon \mathcal{C}^\mathrm{op}\times\mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{V}$ and $J \colon \mathcal{C} \to \mathcal{V}$, together with associativity and unit constraints subject to the usual two "pentagon" and "triangle" axioms. Given a promonoidal structure on $\mathcal{C}$, we may construct the Day convolution monoidal structure on $[\mathcal{C},\mathcal{V}]$, whose tensor product is given at a pair of $\mathcal{V}$-functors $F,G \in [\mathcal{C},\mathcal{V}]$ by the coend $$F\ast G = \int^{A,B \in \mathcal{C}} P(A,B;-) \otimes FA \otimes GB$$ in $\mathcal{V}$, and whose unit object is the $\mathcal{V}$-functor $J \in [\mathcal{C},\mathcal{V}]$, and so on. This monoidal structure on $[\mathcal{C},\mathcal{V}]$ is biclosed (i.e. the tensor product $\mathcal{V}$-functor has a right $\mathcal{V}$-adjoint -- equivalently, preserves (weighted) colimits -- in each variable). In fact, every biclosed monoidal structure on $[\mathcal{C},\mathcal{V}]$ arises in this way from some promonoidal structure on $\mathcal{C}$. (For instance, one recovers the $\mathcal{V}$-functor $P$ from the tensor product $\ast$ by $P(A,B;C) = (\mathcal{C}(A,-) \ast \mathcal{C}(B,-))C$.)

So, since the $\mathbf{Ab}$-category $\mathbf{Ch}$ of chain complexes is (equivalent to) an $\mathbf{Ab}$-enriched functor category $[\mathcal{C},\mathbf{Ab}]$ (for the $\mathbf{Ab}$-category $\mathcal{C}$ described in the question to which you linked), and since the standard monoidal structure on $\mathbf{Ch}$ is $\mathbf{Ab}$-enriched and biclosed, this monoidal structure must be the Day convolution monoidal structure for some promonoidal structure on $\mathcal{C}$. And it isn't too hard to describe that promonoidal structure. For instance, (presuming I haven't bungled the calculation) the functor $P$ is defined on objects by $$P(i,j;k) = \begin{cases} \mathbb{Z} & \mathrm{if\,\,} i+j=k, \\ \mathbb{Z} \oplus \mathbb{Z} & \mathrm{if\,\,} i+j=k+1, \\ \mathbb{Z} & \mathrm{if\,\,} i+j=k+2, \\ 0 & \mathrm{else}. \end{cases}$$

Answer 2

If you use the category $C$ to represent chain complexes and you mean day convolution using a functor $C \otimes C\to C$ it is not possible. This boils down to whether you can obtain the totalization functor from bi-complexes to chain complexes, as a left adjoint to restriction for some functor $m: C \otimes C \to C$.

You cannot do this because the left adjoint $m_!$ will always take representable projectives to representable projectives. I.e. we will have that $$ m_! (C \otimes C((i,j), -))(r) = C(m(i,j),r).$$

But the totalization of a representable projective of $C \otimes C$ is a direct sum of two different principal projectives of $C$, so no choice of $m$ will work.

What is going wrong is that the totalization functor is given by a unique $(C \otimes C, C)$ bi-module, and this bi-module cannot come from a homomorphism $C \otimes C \to C$, because in some sense it is "multi-valued." To fix this, one could change $C$ to a morita equivalent category, $C'$ for which the bi-module is in fact given by a homomorphism. To construct such a $C'$, we need to choose a collection of generating projectives of ${\rm Ab}^{C}$ which is closed under tensor product. I don't see a particularly nice choice. But skd's comment is that if we use derived Morita equivalence instead of ordinary Morita equivalence, there is a very nice choice of (non-projective) generators, where $C'$ becomes the category $\mathbb N, \leq$.

Answer 3

Note: I'm not sure the construction I give is correct. There is an unresolved problem in proving it is a comonoid. See point 4 in the comonoid section. I don't have the time to resolve this right now.

I want to expand on Alexander Campbell's answer by giving the concrete promonoid as well as the necessary calculations. However my promonoid does not agree with the one he outlined. It's pretty late by now, so I hope I didn't miss anything.

Conventions

Just to get everything straight, let $\mathcal V$ be a closed abelian category with tensor product $\otimes$ and unit $I$. $\mathcal C$ is the free $\mathcal V$-category with objects $\mathbb Z$, morphisms $\partial_n : n \to n - 1$ and relations $\partial_{n-1}\partial_n = 0$. A chain complex is a functor $C \to \mathcal V$.

A $\mathcal V$-profunctor $D \nrightarrow E$ is a $\mathcal V$-functor $E^{op} \otimes D \to \mathcal V$ (the swapped order of $D$ and $E$ is nLab convention). We pretend $\mathcal V\text -\mathbf{Prof}$ is a strict monoidal 2-category with monoid $\otimes$ and unit $\mathcal I$.

We identify objects with their identity arrows.

Definition

We give all higher comultiplications of the comonoid.

For $n \in \{0,1,2,\ldots\}$ we define $(-|-)_n : \mathcal C \nrightarrow \mathcal C^{\otimes n}$, on objects as $$ (k_1, \ldots, k_n| x)_n := \begin{cases} I^{\oplus n} & \text{if } k_1 + \cdots + k_n = x + 1 \\ I & \text{if } k_1 + \cdots + k_n = x \\ 0 & \text{otherwise.} \end{cases}$$

For morphisms in each variable there is only one $\partial$ which has both domain and codomain non zero, in which case the domain is always $I$ and the codomain $I^{\oplus n}$. For the $j$th-contravariant variables the image of $\partial$ is then taken to be the $j$th-inclusion into the direct sum: $$ (k_1,\ldots,\partial_{k_j + 1}, \ldots,k_n|k_1 + \cdots + k_n)_n := \iota_j. $$

In the covariant variable the image of $\partial$ is the diagonal $I \to I^{\oplus n}$ but with signs: $$ (k_1,\ldots, k_n| \partial_{k_1 + \cdots + k_n})_n := \iota_1 + (-1)^{k_1}\iota_2 + (-1)^{k_1 + k_2}\iota_3 + \cdots. $$

We claim that $(-|-)_n$ is a comonoid in $\mathcal V\text-\mathbf{Prof}$ and the day convolution it defines is the tensor product of chain complexes.

Tensor product of chain complexes

Consider chain complexes $C, D : \mathcal C \to \mathcal V$. We have $$ C\otimes D := x \mapsto \int^{m,n} (m,n|x)\otimes Cm \otimes Cn. $$

First we show that $$ (C \otimes D)(x) = \bigoplus_{m + n = x} C(m)\otimes C(n) $$ by showing the cones over the bases of these two colimits are the same.

Fix $x$ and consider $m,n$ with $m+n = x$. The following diagram is exemplary for the diagram defining the coend $C\otimes D$.

and here is the same diagram again:

Consider any cone over this. The component from the center term is already completely determined by the inclusions of its two summands on the left whose components are again determined by those of the top and bottom terms. Thus all components cones are given by those from terms of the form $(m,n|m+n)\otimes C(m) \otimes D(n)$, while nothing constrains these components. Thus the cones are in bijection with those over the base of $\bigoplus_{m+n=x} C(m) \otimes D(n)$ as claimed.

The image of $C\otimes D(\partial)$ is given at $C(m)\otimes D(n)$ by $$ (m,n|\partial_{m+n})\otimes C(m) \otimes D(n) = \begin{cases} C(m) \otimes C(n) &\to (C(m)\otimes D(n))^{\oplus 2}\\ \iota_1 + (-1)^m\iota_2. \end{cases} $$ That is $\partial$ takes a term $C(m) \otimes D(n)$ at the top or bottom of the heart shaped diagram into one of the two term sums at the center of the heart shaped diagram. The two inclusions can then be pulled back and pushed along to the top and bottom term to yield the desired $$ ((C\otimes D)(\partial))_{m,n} = C(\partial_m)\otimes D(n) + (-1)^m C(m)\otimes D(\partial_n). $$

Comonoid in $\mathcal V\text-\mathbf{Prof}$

First note that $(-|-)_1$ is in fact $\mathcal C(-,-)$.

Let $j, k, n \in \{0,1,2,\ldots\}$ and sadly $k-j \leq 1$. We have $$ (m_1,\ldots,m_n|x) \cong \int^y (m_1,\ldots,m_{j-1},y,m_{k+1},\ldots,m_n|x)\otimes(m_j,\ldots,m_k|y). $$

Let $m_1,\ldots,m_n,x \in \mathbb Z$. We distinguish the following cases:

1. $m_1 + \cdots + m_n < x$: If we choose $y$ bigger than $m_j + \cdots + m_k$ the right term of the tensor product vanishes, otherwise the left term does.

2. $m_1 + \cdots + m_n = x$: The only non-zero term is $I\otimes I$.

3. $m_1 + \cdots + m_n = x + 1$: The coend is the colimit of the following diagram:

   Which is the desired $I^{\oplus n}$

4. $m_1 + \cdots + m_n = x + 2$: Here the coend is the colimit of the following diagram:

   If $k-j \leq 1$ this vanishes, but for $k-j >1$ I don't think it does. However since one could express any higher multiplication as one of arity 2 associated such that always $k-j < 1$ the term should also vanish for higher multiplications. So somewhere I must miss something, sadly I don't have the time to resolve this right now.

   I think it might be possible to get rid of these terms by modifying the construction by taking $(m_1,\ldots,m_n,x)' := \mathcal C(m_1+\cdots+m_n,x)\otimes(m_1,\ldots,m_n,x)$. I think if we removed the condition that $\partial\partial = 0$ and also the signs we would get a tensor product of derivations where $\partial(x\otimes y) = \partial x \otimes y + x \otimes \partial y$, where these ghost terms would serve a purpose, as terms of degree $n$ could be constructed from terms of degrees higher than $n+1$. Tensoring with $\mathcal C(-,-)$ would then also ensure that $(-|-)_1 = \mathcal C(-,-)$ even if we do add relations to $\mathcal C$.

5. $m_1 + \cdots + m_n > x + 2$: As in 1. at least one side of the tensor product vanishes.

It remains to check naturality, which is straight forward keeping in mind the presentation of $(m_1,\ldots,m_n|m_1+\cdots+m_n-1)$ as the colimit of 3. above.