What's so special about the forgetful functor from G-rep to Vect? - MathOverflow

What's so special about the forgetful functor from G-rep to Vect?

2010-09-08T19:03:15Z

The following is some version of Tannaka-Krein theory, and is reasonably well-known:

Let $G$ be a group (in Set is all I care about for now), and $G\text{-Rep}$ the category of all $G$-modules (over some field $\mathbb K$, say). It is a fairly structured category (complete, cocomplete, abelian, $\mathbb K$-enriched, ...) and in particular carries a symmetric tensor product $\otimes$. The forgetful functor $\operatorname{Forget}: G\text{-Rep} \to \text{Vect}$ respects all of this structure, and in particular is (symmetric) monoidal. Let $\operatorname{End}_\otimes(\operatorname{Forget})$ denote the monoid of monoidal natural transformations of $\operatorname{Forget}$. Then it is a group, and there is a canonical isomorphism $\operatorname{End}_\otimes(\operatorname{Forget}) \cong G$ .

The following is probably also reasonably well-known, but I don't know it myself:

Let $G$, etc., be as above, but suppose that we have forgotten what $G$ the category $G\text{-Rep}$ came from, and in particular forgot, at least momentarily, the data of the forgetful functor. We can nevertheless recover it, because in fact $\operatorname{Forget}$ is the unique-up-to-isomorphism ADJECTIVES functor $G\text{-Rep} \to \text{Vect}$.

My question is: what are the words that should go in place of "ADJECTIVES" above? Certainly "linear, continuous, cocontinuous, monoidal" are all reasonable words, although my intuition has been that I can drop "cocontinuous" from the list. But even with all these words, I don't see how to prove the uniqueness. If I had to guess, I would guess that the latter claim is a result of Deligne's, although I don't read French well enough to skim a bunch of his papers and find it. Any pointers to the literature?

Answer by Jacob Lurie for What's so special about the forgetful functor from G-rep to Vect?

2010-09-08T19:22:54Z

If G is an affine algebraic group (for example a finite group), then the category of k-linear cocontinuous symmetric monoidal functors from Rep(G) to Vect_k is equivalent to the category of G-torsors over k. In particular, not every such functor needs to be isomorphic to the identity. For example, if k' is finite Galois extension of k with Galois group G, then the functor $F(V) = (V \otimes_{k} k')^{G}$ will satisfy all the axioms you will think to write down, but is not isomorphic to the identity functor.

Answer by David Jordan for What's so special about the forgetful functor from G-rep to Vect?

2010-09-08T19:30:08Z

One needs to be careful. One cannot recover the group $G$ from the tensor category alone, but only with the data of category, fiber functor. There are examples of non-isomorphic (finite, even) groups with equivalent categories of representations. For instance, see Pasquale Zito's answer to this question:

http://mathoverflow.net/questions/500/finite-groups-with-the-same-character-table

However, as is discussed in the paper Zito links to, remembering the symmetry on the categories recovers the group, up to isomorphism. I'm not sure who it's due to.