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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?

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4 Answers 4

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Let $C$ be a 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 $\dim X$ 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) 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 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.

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Rigid symmetric monoidal abelian categories over a field $k$ with a fiber functor (i.e., a strong monoidal exact functor) to $K$-vector spaces for some extension $K$ of $k$ (the functor is assumed to exist, but not fixed) are are often called tannakian categories; a tannakian category with a fixed fiber functor to $k$-vector spaces is referred to as a neutral tannakian category. I will adhere to this terminology.

It was proved by Grothendieck, Saavedra-Rivano and Deligne that the 2-category of tannakian categories over a fixed field $k$ is equivalent to the 2-category of affine gerbes over $k$. With each affine gerbe $\Gamma$ you associate the category of vector bundles on $\Gamma$. Conversely, given a tannakian category $\mathcal A$, the objects of the corresponding gerbe over a $k$-algebra $A$ are fiber functors from $\mathcal A$ to finitely generated projective $A$-modules.

So, if $\Gamma$ is an affine gerbe, fiber functors from the category of vector spaces on $\Gamma$ to vector spaces on $k$ correspond to objects of $\Gamma(k)$; hence, to give an example of a tannakian category without fiber functors it is enough to give an example of an affine gerbe without a rational point. There are many such examples; for example, you can take a surjective homomorphism of algebraic groups $E \to G$, and a $G$-torsor $P$ over $k$ that does not lift to an $E$-torsor. The gerbe of liftings of $P$ to an $E$ torsor is an affine gerbe, and does not have any rational points.

As to examples of symmetric rigid monoidal abelian categories without fiber functors to any extension of $k$, the standard one is the category of $\mathbb Z/2\mathbb Z$-graded vector spaces.

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Why can't you just forget the grading? –  Will Sawin Feb 22 '13 at 5:45
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Forgetting the grading doesn't preserve the symmetric monoidal structure (for the "skew" or "super" symmetric monoidal structure on graded vector spaces). If you use the non-skew structure, forgetting the grading does work, and indeed that category is equivalent to representations of the algebraic group $\mu_2$. –  Eric Wofsey Feb 22 '13 at 5:52
    
One should take finitely generated super vector spaces in order to get rigidity. –  Martin Brandenburg Feb 27 '13 at 10:33

There's a naturally occurring one, namely, the category of motives over a field is a rigid monoidal category without a fiber functor because the rank of the motive of an algebraic variety is its Euler characteristic, which may be negative. To make the category Tannakian, you have to assume that the Kunneth components of the diagonal are algebraic, and change the commutativity constraint.

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Representations of quantum groups at roots of unity provide a particularly helpful example (the $x^2=1+x$ example above corresponds to quantized sl(2) at some low index root of unity). These are rigid monoidal abelian categories (not symmetric) whose objects in general have noninteger dimensions, so they clearly cannot have a functor to Vect. However, the objects are representations of a Hopf algebra, so there is a functor to Vect of most of the structure. The tensor product is the ordinary tensor product of representations (again a representation for a Hopf algebra), but then you quotient by a subrepresentation (which can be thought of very loosely as the nonsemisimple part).

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