Characterising categories of vector spaces - MathOverflow

Characterising categories of vector spaces

2013-01-07T06:08:26Z

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.

Answer by Martin Brandenburg for Characterising categories of vector spaces

2013-01-07T11:14:42Z

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.

Answer by Mariano Suárez-Alvarez for Characterising categories of vector spaces

2013-01-07T17:19:05Z

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.

Answer by John Baez for Characterising categories of vector spaces

2013-01-07T19:39:32Z

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.