Consider the category $FdVect_k$ of finite dimensional $k$-vector spaces, for some given field. It is abelian, semisimple, in that each object is a finite sum of simple objects (of which there is only one up to isomorphism), and also compact closed with simple tensor unit which is a progenerator.

Can we characterise $FdVect_k$ as a category of vector spaces purely by properties of the category such as above? I don't mean to demand for instance that $End(I)$ is a field, for instance, and I suspect that this may stop any such characterisation. But this I don't mind, seeing as if rings which are 'nice enough' cannot be distinguished from fields in this way, then so be it.

I should point out that if someone says 'but what about Morita equivalence?', then I'm not sure that's right answer, since I'm looking for equivalence as a compact closed semisimple abelian ... category, not just as a bare category - but I may be wrong on this point.

  • $\begingroup$ I should point out this tangentially relevant paper: arxiv.org/abs/0807.2927 $\endgroup$
    – David Roberts
    Jan 7, 2013 at 6:56
  • $\begingroup$ Let f be an endomorphism of I. The object I is simple, so f has kernel 0 or all of I. If the kernel is 0, then f has a left inverse since every injection is split in a semisimple category. Using images instead of kernels gives right inverses as well. This shows that End(I) is a division ring. $\endgroup$ Jan 7, 2013 at 7:48
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    $\begingroup$ I think Morita equivalent rings should have equivalent module categories that preserve all the extra structure in sight as well. $\endgroup$ Jan 7, 2013 at 7:59
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    $\begingroup$ @Mike: What is this extra structure? If $R$ is a non-commutative ring, then $\mathsf{Mod}(R)$ is not monoidal. And for commutative rings Morita-equivalence is just isomorphism. $\endgroup$ Jan 7, 2013 at 11:52
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    $\begingroup$ @Mariano: Indeed, or R could be Morita-equivalent to a commutative ring. A more precise (and correct) statement is that the monad corresponding to the forgetful functor $\mathsf{Mon}(R) \to \mathsf{Set}$ is monoidal if and only if $R$ is commutative. $\endgroup$ Jan 7, 2013 at 17:45

3 Answers 3


Let $C$ be an abelian monoidal category such that $1 \in C$ is simple, and each object is a finite sum of copies of $1$, i.e. isomorphic to $1^{\oplus n}$ for some $n \in \mathbb{N}$. It is well-known that $k:=\mathrm{End}(1)$ is a commutative ring and that $C$ is $k$-linear. By Schur's Lemma $k$ is even a field. Now, consider the functor $\hom(1,-) : C \to \mathsf{Mod}(k)$. It maps $1^{\oplus n}$ to $k^n$, thus factors as an essentially surjective functor $C \to \mathsf{Mod}_f(k)$. It is also fully faithful, because $\hom(1^{\oplus n},1^{\oplus m}) \cong \prod_n \prod_m \hom(1,1) = k^{n \times m}$. It has a canonical lax monoidal structure given by $\hom(1,x) \otimes \hom(1,y) \xrightarrow{\otimes} \hom(1 \otimes 1,x \otimes y) \cong \hom(1,x \otimes y)$. This is an isomorphism: Since both sides commute with finite direct sums in $x$ and $y$, it is enough to verify this for $x=y=1$, where it is clear. Thus, $C \cong \mathsf{Mod}_f(k)$ as monoidal categories.

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    $\begingroup$ nice! related paper: Finite, Connected, Semisimple Rigid Tensor Categories are Linear - Greg Kuperberg $\endgroup$ Jan 7, 2013 at 12:03
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    $\begingroup$ If you add a contravariant identity-on-objects involution to the requirements (complete, compact, semisimple, monoidal with simple unit), the scalars will be an involutive field $k$, and the category will be (equivalent to) that of finite-dimensional $k$-Hilbert spaces. See tac.mta.ca/tac/volumes/22/13/22-13abs.html. $\endgroup$ Jan 7, 2013 at 13:18
  • $\begingroup$ Thanks for that reference, Chris. This is exactly the sort of thing I'm thinking about, just in the plain vector space setting. $\endgroup$
    – David Roberts
    Jan 7, 2013 at 22:47
  • $\begingroup$ So if "C [is] an abelian monoidal category such that 1∈C is simple, and each object is a finite sum of copies of 1", then C is Mod_f(k) for some field k? Excellent. Thanks, Eduardo! Link: arxiv.org/abs/math/0209256 $\endgroup$
    – David Roberts
    Jan 7, 2013 at 23:16
  • $\begingroup$ The classification I have proven above is quite trivial, but the one by Greg Kuperberg is more general and profound. $\endgroup$ Jan 7, 2013 at 23:53

In Schur Functors I, Todd Trimble and I proved that $FdVect_k$ is the free symmetric monoidal $k$-linear Cauchy complete category on no objects. Here a $k$-linear category is Cauchy complete if it has direct sums (that is, biproducts) and all idempotents split. So, we don't need to mention compact closedness, abelianness, semisimplicity....

More precisely:

Proposition. For any field $k$, if $C$ is a symmetric monoidal $k$-linear Cauchy complete category, there exists exactly one symmetric monoidal $k$-linear functor $i:FdVect_k \to C$, up to symmetric monoidal $k$-linear isomorphism.

  • $\begingroup$ Do we really need that $C$ is Cauchy complete? I think we only need that finite direct sums exist. $\endgroup$ Jan 7, 2013 at 23:59
  • $\begingroup$ The Cauchy completeness assumption is important for our development of the abstract theory of Schur functors; it's not particularly critical for this particular proposition. $\endgroup$
    – Todd Trimble
    Jan 8, 2013 at 7:46
  • $\begingroup$ Todd, yes I agree with that. $\endgroup$ Jan 9, 2013 at 10:16

The category of f.d. vector spaces is the unique fusion category of Perron-Frobenius dimension $1$, if I recall correctly. Pavel Etingof, Dmitri Nikshych, Viktor Ostrik classified categories with prime $\operatorname{PFdim}$ $p$ as $\mathsf{Vec}_{C_p}^\omega$, twists by cocycles of the cat. of reps of the cyclic group $C_p$) and $1$ should be prime :-)

This does start assuming the category is $k$-linear, and you dit not want that, though.

  • $\begingroup$ (The condition of rigidity that comes with being a fusion category s automatic when the only simple object is the identity object, so this characterization is not that far away from Martin's) $\endgroup$ Jan 7, 2013 at 17:21

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