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Say $\mathcal{C}$ is the Abelian category of finitely-generated modules over some $k$-algebra $A$. Then an object $M\in \mathcal{C}$ is finite-dimensional over $k$ if and only if $\text{Hom}(P, M)$ is finite-dimensional for any projective $P\in \mathcal{C}.$

I want a word for this "finiteness" property of $M$ relative to the category $\mathcal{C}$. I also would like to use a similar notion in the dg sense. Here I want to assume we are given the category $\mathcal{P} = \text{Perf}(A)$ of perfect modules and we say that an object $M$ of the ind-completion of $\mathcal{P}$ is "finite" if $\hom(P, M)$ is finite-dimensional in finitely-many homological degrees for any $P\in \mathcal{P}$. (This once again corresponds to the complex of $A$-modules $M$ having finite-dimensional cohomology).

I'd love to call both of these properties "perfectly finite", but don't want to introduce new words if standard nomenclature exists.

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If I understand the question correctly, the answer should be "perfect $k$-modules".

1) Take $P = A$ to see that $M$ should be a perfect $k$-module.

2) Perfect $A$-modules are the same as retracts of finite cell $A$-modules. So to prove that your property is satisfied by perfect $k$-modules it is sufficient to prove it for finite cell $A$-modules $P$. Contravariant $\text{Hom}$ sends colimits to limits, so we see that it is sufficient to require it for $P = A$.

Here we call $V$ a finite cell $A$-module if either $V$ is zero or there is a map $V'\to V$ of $A$-modules, such that the cone of the map is quasiisomorphic to a shift of $A$ and $V'$ is a finite cell module. (So, in fact, it is an inductive definition)

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  • $\begingroup$ No, I don't think that's it: in order to be able to talk about a $\mathfrak{C}$-module as a $k$-module, you need some sort of fiber functor $\mathfrak{C}\to \text{vect}$, which I'm assuming isn't given. $\endgroup$ Commented Sep 19, 2016 at 15:16
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    $\begingroup$ What do you mean by "$\text{Hom}(P, M)$ is finite-dimensional" in the first situation? I thought that it is the dimension of this module over $k$. Indeed $M \in \mathcal{C}$ is finite-dimensional over $k$ if and only if $\text{Hom}(P, M)$ is finite-dimensional as $k$-module. $\endgroup$
    – A K
    Commented Sep 19, 2016 at 15:36
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    $\begingroup$ In the second situation, I thought that $A$ is $k$-dg-algebra, and you are looking for something analogous and by "$\text{hom}(P, M)$ is finite-dimensional in finitely-many homological degrees" again mean that it has cohomology which is finite-dimensional over $k$. $\endgroup$
    – A K
    Commented Sep 19, 2016 at 15:51

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