# Is there a path'' between any two fiber functors over the same field in Tannakian formalism?

I will take the approach of this question: Tannaka formalism and the étale fundamental group

and think of the etale fundamental group as Tannakian formalism for $\mathbb{F}_1$. Then our "Tannakian category" is the category of finite etale covers, and each fiber functor is a functor from this category to $Sets$ (thought of as finite dimensional spaces over $\mathbb{F}_1$).

For the etale fundamental group, it is true that for any two fiber functors (given by two different geometric points) there is a path'' between them. Meaning: there is a natural isomorphism between these two functors.

My question is whether this independence of the basepoint'' result applies to Tannakian formalism as well:

Is it true that for any two fiber functors $H_1, H_2: \mathcal{C}\rightarrow Vec_K$, there is a natural isomorphism $H_1 \cong H_2$?

I don't think this is true in general. The point is that that there are non-isomorphic groups with equivalent categories of representations; since the category of representations together with the fiber functor determines the group, this gives a counterexample.

This happens under the following circumstances; the following construction was first given, I believe, in Giraud's book on non-abelian cohomology.

Suppose that $G$ is an affine algebraic group over $\mathop{\rm Spec} K$ and $P \to \mathop{\rm Spec}K$ is a $G$-torsor. Call $H$ the group scheme of automorphisms of $P$ as a torsor; then $P$ becomes an $(H, G)$-bitorsor, that is, admits commuting actions of $G$ on the right and of $H$ on the left, and is a torsor for both. Conversely, if $P \to \mathop{\rm Spec}K$ is an $(H, G)$-bitorsor, then $H$ is the automorphism group scheme of $P$ as a $G$-torsor.

Then the categories of representations of $G$ and $H$ are isomorphic. This follows, essentially, from descent theory; if $V$ is a representation of $G$, then the quotient $(P \times_{\mathop{\rm Spec}K} V)/G$ is a vector space on $K$ with an action of $H$; the inverse functor is obtained by exchanging $G$ and $H$ (and right and left actions).

My favorite example of this is the following: if $q$ and $q'$ are non-degenerate quadratic forms in $n$ variable, the orthogonal groups $\mathrm O(q)$ and $\mathrm O(q')$ have equivalent categories of representations (although they are not isomorphic, in general). The bitorsor is the functor of isometries of $q$ and $q'$.

On the other hand, if $K$ is algebraically closed the fiber functors are indeed isomorphic; if memory serves me well, this is in Deligne's paper in Grothendieck's Festschrift, but I don't have it here and can't check right now.

• This is indeed in Deligne's Festschrift article. One can also look at Breen's article in the Motives volume. What Deligne shows is that, for any Tannakian category, any two fiber functors are isomorphic over a field extension. This was the `missing' piece in Saavedra's thesis. – Keerthi Madapusi Pera Jul 31 '11 at 17:24
• I guess that makes sense. Does this mean that $\mathbb{F}_1$ behaves like an algebraically closed field?! I must say that I'm growing very fond of this Tannakian formalism business. – James D. Taylor Jul 31 '11 at 17:25

The obstruction to the existence of such an isomorphism is a (bi)torsor, that has been studied in various real-life situations. An example extracted from

On the relation between Nori Motives and Kontsevich Periods Annette Huber, Stefan Müller-Stach http://arxiv.org/abs/1105.0865

"As already explained by Kontsevich, singular cohomology and algebraic de Rham cohomology are both fiber functors on the same Tannaka category of motives. By general Tannaka formalism, there is a pro-algebraic torsor of isomorphisms between them. The period pairing is nothing but a complex point of this torsor."

Basically, by tannaka duality, you can build a counter-example out of any couple of non-isomorphic objects of a gerbe.