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I think we all agree on what is the good notion of being finite dimensional for a vector space. But what is the correct analogue for chain complexes?

To be clear on the definitions, I am working with $\mathbb{Z}$-graded chain complexes, i.e. collections $$C = (C_n)_{n\in\mathbb{Z}}$$ of vector spaces endowed with differentials $$d:C_n\longrightarrow C_{n-1}$$ such that $d^2=0$. The dual of such a chain complex is $$C^\vee:=((C_{-n})^\vee)_{n\in\mathbb{Z}}$$ and the tensor product of two chain complexes is given by $$C\otimes D := \left(\bigoplus_{p+q=n}C_p\otimes D_q\right)_{n\in\mathbb{Z}}.$$ I have two possibilities for a definition of "finite dimensional" chain complex:

  1. The sum of the dimensions of all $C_n$ is finite.
  2. All the $C_n$ are finite dimensional.

The first definition looks a bit too strict to me, as it forces the chain complex to be concentrated in only finitely many degrees. However, a property one might like from a finite dimensional object is that $(C\otimes D)^\vee\cong C^\vee\otimes D^\vee$ canonically. if we take the second definition, then this is not true. Is there a third possible way to go? If not, opinions (with motivation!) on what is the correct definition between the two I gave above are more than welcome.

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    You can produce a free resolution of any finite dimensional vector space, by means of a koszul resolution, which will have finitely many terms. This produces a complex which will be concentrated in finitely many degrees. You might use this as a starting point for what you seek. You might also consider finite cohomological (or horological) dimension to be sufficient. – 2017-02-04
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    @JohnMartin Do you mean "finite dimensional algebra" (or "module", or something)? Because in itself a vector space is already free, no? – 2017-02-04
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    Yes. Sorry I intended to write module. – 2017-02-08
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    @JohnMartin Yeah, I know about projective resolutions and Koszul resolutions, but the real goal of my question was to understand what are the good assumptions on the underlying chain complex so that the dual of an algebra is a coalgebra (I'm talking algebra over an operad, here). The usual assumption is number (1), but somehow it looks too strong to me. – 2017-02-08

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I think the good analogue of being finite dimensional I was looking for one an a half year ago, when I asked this question, is for a chain complex to be finite dimensional in every degree and with degrees bounded below, i.e. we ask that there is some $N\in\mathbb{Z}$ such that $C_n = 0$ for all $nessentially finite dimensional (efd) chain complexes, even though this is probably not a good nomenclature.

They satisfy the following properties I can think of off the top of my head:

  1. Tensor products and direct sums of efd chain complexes are efd.
  2. A choice of basis for an edf chain complex $C$ gives an isomorphism $C\cong C^\vee$.
  3. If $C,D$ are edf chain complexes, there is a canonical isomorphism $(C\otimes D)^\vee\cong C^\vee\otimes D^\vee$.