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

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?

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Quick summary of the answers below: there are no such "ADJECTIVES" possible, and the claim is false. –  S. Carnahan Sep 9 '10 at 3:17
However, if you're working over the complex numbers and include the adjective symmetric then there is such a result and it's proved in Deligne-Milne. –  Noah Snyder Sep 9 '10 at 5:53

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.

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Interesting! But this is not something I'm used to. In non-algebraic-geometry land, all G-torsors for a given group G are isomorphic, almost by definition. How is the definition different in affine-algebraic land that allows for non-isomorphic torsors-over-k? –  Theo Johnson-Freyd Sep 8 '10 at 20:43
So, I'm going to think out loud while I do the simplest example. Let $k=\mathbb R$, $k'=\mathbb C$, $G=\{\pm 1\}$, then $G\text{-rep}=\text{Vect}[\epsilon]/(\epsilon^2=1)$, so that $\epsilon$ is the nontrivial one-dimensional G-rep, and as a G-rep $\mathbb C=1+\epsilon$. Then your functor is $(a+b\epsilon)\mapsto((a+b\epsilon)(1+\epsilon))^{\epsilon=0}=(a+b+\epsilon(a+b)‌​)^{\epsilon=0}=(a+b)$, which is isomorphic object-by-object to the forgetful functor $(a+b\epsilon)\mapsto(a+b)$. Moreover, there's not enough room for morphisms to be interesting. I think these functors are isomorphic. –  Theo Johnson-Freyd Sep 8 '10 at 21:08
In this context, a $G$-torsor means a $k$-scheme with an action of $G$, which is ($G$-equivariantly) isomorphic to $G$ over some extension field of $k$. So if $k'$ is a Galois extension of k with Galois group $G$, then the spectrum of $k'$ is a $G$-torsor (since $k' \otimes_k k' \simeq k^G$ by virtue of the Galois assumption). –  Jacob Lurie Sep 8 '10 at 21:12
The isomorphism you describe is not an isomorphism of symmetric monoidal functors. To see this, let $k^G$ denote the regular representation of $G$, regarded as commutative algebra via pointwise multiplication. If F: Rep(G) -> Vect(k) is any symmetric monoidal functor, then $F(k^G)$ will inherit the structure of a commutative algebra. In the above example, $F(k^G)$ is the space of $G$-invariants in $k'^{G}$, where $G$ is acting both by permuting the factors and by Galois symmetries.Evaluation at the identity element of $G$ determines an isomorphism of this algebra with $k'$,which is not $k^G$. –  Jacob Lurie Sep 8 '10 at 21:20
But maybe this is a "Law of Small Numbers" problem. –  Theo Johnson-Freyd Sep 8 '10 at 21:29

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:

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.

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The last sentence probably follows from Deligne's theorem which says that a symmetric tensor category satisfying certain growth properties in the length of its objects under iterated tensor product is necessarily RepG for a super group G, and which group it is is determined by the symmetric structure. A very nice exposition, in English, is here: arxiv.org/abs/math/0401347 –  David Jordan Sep 8 '10 at 19:40
Oh, OK! So I should demand that my functor to Vect be symmetric monoidal. The Ostrik paper is a nice one. –  Theo Johnson-Freyd Sep 8 '10 at 20:33
Presumably supergroups appear here for the same kind of reason that supergroupoids appear in '2-Hilbert Spaces' arxiv.org/abs/q-alg/9609018, e.g., where Baez and Dolan prove "a generalized Doplicher-Roberts theorem stating that every symmetric 2-$H^{\ast}$-algebra is equivalent to the category $Rep(G)$ of continuous unitary finite-dimensional representations of some compact supergroupoid $G$." –  David Corfield Jul 19 '11 at 9:23