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 $\mathsf{T}$?
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1$\begingroup$ I don't quite know if there is a canonical way, but the obvious thing would be that if, for every pair of objects $M,N$,$\omega_M(X)\cong\omega_N(X)$ for every object in both $\langle M\rangle$ and $\langle N\rangle$. Which seems something like a "global section of the sheaf of functors to vector spaces", in a non rigourous sense. Then in order to be canonical, you would want the isomorphism $\omega_M(X)\cong\omega_N(X)$ to be canonical in some sense, which would probably be classified by some sort of cohomology-like object $\endgroup$– AidanCommented Feb 14, 2022 at 18:02
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$\begingroup$ One example I have in mind is differential Galois theory. If $\mathsf{T}$ is the category of differential modules, then fiber functors $\omega_M$ are determined by Picard-Vessiot extensions. Then we can take the direct limit of all such extensions to obtain a fiber functor of the whole category. But I'm not sure if this works in other cases. $\endgroup$– GabrielCommented Feb 14, 2022 at 19:53
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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.