A Tannakian category is defined to be a rigid monoidal abelian category over a field with an exact tensor functor to the category of vector spaces. (It is always equal to the category of representations of some pro-algebraic group.)


Is there such a thing as a rigid monoidal abelian category over a field without an exact tensor functor to the category of vector spaces?


What could such a category look like?



Rigid monoidal abelian category without an exact tensor functor to Vect
Let $C$ be a [fusion category](http://ncatlab.org/nlab/show/fusion+category) with simple objects $X\_i$. If $X$ is an object, then $X \otimes X\_i = \bigoplus N\_{ij} X\_j$ for some matrix $N\_{ij}$ of non-negative integers. The largest non-negative real eigenvalue of this matrix is called the [Frobenius-Perron dimension](http://arxiv.org/abs/math/0401246) $\dim X$ of $X$. It is preserved by exact tensor functors (Corollary 4.6 in the linked paper) and coincides with the ordinary dimension for finite-dimensional vector spaces. It follows that a fusion category with an object whose Frobenius-Perron dimension is not an integer does not admit an exact tensor functor to $\text{FinVect}$. If I'm not mistaken, tensor functors preserve dualizable objects, so any tensor functor to $\text{Vect}$ must land in $\text{FinVect}$ anyway, and the conclusion follows. 


In particular there is a fusion category with an object $X$ satisfying $X^{\otimes 2} \cong 1 \oplus X$ whose Frobenius-Perron dimension is $\frac{1 + \sqrt{5}}{2}$. 


**Edit:** Here is maybe a simpler way of putting it. Associated to any fusion category $C$ is its Grothendieck ring, and an exact tensor functor between fusion categories gives a homomorphism of Grothendieck rings. So it suffices to find a fusion category whose Grothendieck ring does not admit a homomorphism to $\mathbb{Z}$, and clearly a fusion category with an object $X$ satisfying $X^{\otimes 2} \cong 1 \oplus X$ has this property.


**Edit #2:** It seems that in the literature a Tannakian category is required to be *symmetric* monoidal. In this case, ignore everything I said above. 


Deligne (see e.g. [this exposition by Ostrik](http://arxiv.org/abs/math/0401347)) showed that over an algebraically closed field of characteristic $0$ together with some other mild hypotheses, which hold in particular for symmetric fusion categories, any such category admits a *super* fiber functor to $\text{sVect}$. From here one can classify symmetric fusion categories in terms of finite groups; see section 2.12 of [On braided fusion categories I](http://arxiv.org/abs/0906.0620) by Drinfeld, Gelaki, Nikshych, and Ostrik. $\text{sVect}$ itself does not admit an ordinary fiber functor, as Angelo pointed out; the obvious functor doesn't preserve symmetries. 


