Let $\mathsf{T}$ be a rigid abelian tensor category and suppose that we're given fiber functors $\omega\_M:\langle M\rangle \to \mathsf{vect}\_k$ for every object $M$ of $\mathsf{T}$. Is there a canonical way to obtain a fiber functor on $\mathsf{T}$?



Given fiber functors on all the subcategories of the form $\langle M\rangle$, can we obtain a fiber functor on the whole category?
I think the answer is no in general.


Consider, for example, a tannakian category $M$ over a field $k$, and
suppose for simplicity that it is a
union $M=\bigcup M\_{n}$, $M\_{n}\subset M\_{n+1}$, of neutral tannakian
categories. Is $M$ neutral? In other words, if we assume that each $M\_{n}$ has
a $k$-valued fibre functor, does this imply that $M$ does?


Suppose first that $k$ is algebraically closed, and choose a $k$-valued fibre
functor $\omega\_{n}$ on $M\_{n}$. Because $k$ is algebraically closed,
$\omega\_{n+1}|M\_{n}$ is isomorphic to $\omega\_{n}$. In fact, given $\omega
\_{n}$, we can modify $\omega\_{n+1}$ so that $\omega\_{n+1}|M\_{n}=\omega\_{n}$.
Thus there exists a fibre functor $\omega$ on $M$ such that $\omega
|M\_{n}=\omega\_{n}$ (axiom of dependent choice!).


When we try to do this with $k$ not algebraically closed, we obtain a sequence
of torsors $Hom(\omega\_{n},\omega\_{n+1}|M\_{n})$. Of course, by making a
different choice of fibre functors, we get a different sequence of torsors,
but if, for example, the fundamental groups of $P\_{n}$ of the $M\_{n}$ are
commutative, then we get in this way a well-defined element of
$\varprojlim^{1}H^{1}(k,P\_{n})$, which is an obstruction to $M$ being neutral.


Consider the category $M$ of motives of weight 0 over the field with $p$
elements ($p$ prime) and assume the Tate conjecture. We can write
$M=\bigcup M\_{n}$ as above. It is not known whether the $M\_{n}$ are neutral,
even though the obstructions coming from the Brauer group of $k$ vanish.
Assuming they are neutral, Kontsevich has given a heuristic argument that, $M$ will not be neutral.


This answer has been abstracted from arXiv:math/0607569, which see for more details.


