Product, coproduct, and convolution of chain complexes

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I am currently reading Gelfand and Manin's Methods of Homological Algebra, and at some point (IV. 10. Exercise 2 I guess? The numbering in the book is quite inconvenient to get around, so I'm mentioning it for completeness, but I'll re-explain my question so that there's no need to look into the book) they say the following:

Let $\mathcal{D}$ be a triangulated category, $X^\bullet$ a finite chain complex of objects in $\mathcal{D}$, and $T \in \text{Tot} X^\bullet$ be its convolution.

I do not understand what they mean by the word convolution. In the previous pages of the book, the letter $T$ usually denotes the translation operator of a triangulated category ($T^k$ would map a complex $X^\bullet$ to $X^\bullet[k]$, which is the same complex shifted by an index of $k$).

I was guessing that it meant taking the convolution of the complex $X^\bullet$ with itself in the "standard" sense of convolution (the one we know for power series for example), which would give another chain complex. So the $n$-th term of that convolution would look something like:

\begin{equation} \coprod_{k \in \mathbb{Z}} X^k \otimes X^{n-k} \end{equation}

Where maybe it's a product $\prod$ instead of a coproduct $\coprod$, and maybe we have another operation instead of $\otimes$. That's just a guess based on what discrete convolutions look like; but maybe it's not exactly these operations.


But even the above guess raises a few questions:

  1. What are the natural notions of products/coproducts on the category of chain complex $\text{Ch}(\mathcal{A})$, given $\mathcal{A}$ an arbitrary abelian category? I have found the following thread; which seems to point that the natural notion of coproduct is the usual direct sum (defined componentwise). That is indeed pretty convenient. But:

Is there a natural notion of product on Ch$(\mathcal{A})$?

Though I'm not expecting it to be necessarily as nice as the coproduct, because taking componentwise products doesn't sound as good.

  1. Is my guess for the definition of convolution correct? Or am I not using the right operations/the definition is not as elementary?

In any case, why is the convolution of $X^\bullet$ an element of $\text{Tot} X^\bullet$ ?


That's it for my question. For context, the statement they then prove with these definitions is that there exists a spectral sequence

\begin{equation} E_1^{p,q} = H^q(X^p) \implies H^n(T) \end{equation}