As far as I know there are different ways to categorify the notion of vector space/module. These appear (for example) when trying to find extended TQFTs. There are at least two ways (presented at $2$-vector-space at nLab and more generally in the article on $(\infty,n)$-modules):

  • The 2-category $\mathsf{Vect}_2$ whose objects are $k$-linear categories, $1$-morphisms are functors and $2$-morphims are natural transformations;
  • The 2-category $1\text{-}\mathsf{Alg}$ whose objects are associative $k$-algebras, morphisms are bimodules and $2$-morphisms are bimodules morphisms.

The two are linked by some kind of Tannaka duality: there's a functor $1\text{-}\mathsf{Alg} \to \mathsf{Vect}_2$ (described here) that maps an algebra $A$ to its category of modules $\mathsf{RMod}_A$, an $(A,B)$ bimodule $N$ to the functor $- \otimes_A N$, and a bimodule morphism to the associated natural transformation.

I think this is fully faithful but not essentially surjective. Am I right? How to describe the essential image of this functor? A category in the image has to at least be monoidal and rigid, for example. Is the image possibly the fully dualizable objects?

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    $\begingroup$ Not all categories in the image are monoidal, think of a noncommutative $A$. $\endgroup$ Oct 16, 2014 at 10:13
  • $\begingroup$ One needs commutative or an $(A,B^{op})$-bimodule. Otherwise the tensor functor will go to left modules instead of right modules. The image should be described by using a $k$-linear version of Morita's theorem: There should exist a compact progenitor. What do you mean by fully faithful for a $2$-functor? $\endgroup$ Oct 16, 2014 at 13:02
  • $\begingroup$ @Fernando You're right, I'm not sure what I was thinking. Conditions include being abelian, though. $\endgroup$ Oct 16, 2014 at 13:03
  • $\begingroup$ @Julian Are you sure? If $M$ is a right $A$-mod and $N$ is an $(A,B)$-bimodule then $M \otimes_A N$ is a right $B$-mod no? For fully faithful I meant that $\mathrm{Map}(A,B) \to \mathrm{Map}(F(A), F(B))$ should have been an equivalence of categories, but I've been told since then it's not true (indeed all functors of the form $- \otimes_A N$ preserve colimits, for example). $\endgroup$ Oct 16, 2014 at 13:05
  • $\begingroup$ Yes, sure. I was mixing up the tensor products, sorry. $\endgroup$ Oct 16, 2014 at 21:16

1 Answer 1


As I already noted in the comments: A $k$-linear category is equivalent to a module category iff it is abelian, cocomplete, and there exists a compact projective generator. This is a version of Morita's theorem, which can be found e.g. in this article by Bernhard Keller. For the question about which functors are equivalent to tensor functors the answer is given by Eilenberg-Watts theorem stating that these are exactly the functors which are right exact and preserve small coproducts (see e.g. Athens following article in nlab: http://ncatlab.org/nlab/show/Eilenberg-Watts+theorem )

  • $\begingroup$ Thanks. That first statement is true for abelian categories, right? (ie. if $C$ is abelian, then $C$ is a category of representations iff...). $\endgroup$ Oct 17, 2014 at 6:25
  • $\begingroup$ @Najib Idrissi Yes, I changed accordingly. $\endgroup$ Oct 18, 2014 at 5:19
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    $\begingroup$ The category of finite dimensional $k$ vector spaces is Abelian and has a compact projective generator. How do you distinguish this from, say, the category of all vector spaces? $\endgroup$ Oct 20, 2014 at 12:07
  • $\begingroup$ @ChrisSchommer-Pries Thanks for pointing out this mistake. Of course the category should be cocomplete, otherwise there is the question what compact should mean if you don't have coproducts. I don't know of a theorem describing what properties describe finite dimensional vector spaces (or more general finitely generated, finitely presented, or coherent modules over a ring). $\endgroup$ Oct 21, 2014 at 15:27
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    $\begingroup$ There is the following characterization which might be useful (k is a field): an abelian k-linear category is equivalent to the category of finite dimensional representations of a finite dimensional algebra iff (1) All homs are finite dimensional (2) Every object has finite length (3) there are enough projectives (e.g. if there is a projective generator) and (4) up to isomorphism there are finitely many simple objects. This is well-known in the right communities, and a proof can be found here arXiv:1406.4204. $\endgroup$ Oct 22, 2014 at 7:04

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