What's so special about the forgetful functor from G-rep to Vect? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:05:33Z http://mathoverflow.net/feeds/question/38089 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38089/whats-so-special-about-the-forgetful-functor-from-g-rep-to-vect What's so special about the forgetful functor from G-rep to Vect? Theo Johnson-Freyd 2010-09-08T19:03:15Z 2010-09-08T20:12:20Z <p>The following is some version of Tannaka-Krein theory, and is reasonably well-known:</p> <blockquote> <p>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 <code>$\operatorname{End}_\otimes(\operatorname{Forget})$</code> denote the monoid of monoidal natural transformations of $\operatorname{Forget}$. Then it is a group, and there is a canonical isomorphism <code>$\operatorname{End}_\otimes(\operatorname{Forget}) \cong G$</code>.</p> </blockquote> <p>The following is probably also reasonably well-known, but I don't know it myself:</p> <blockquote> <p>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 $G\text{-Rep} \to \text{Vect}$.</p> </blockquote> <p>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?</p> http://mathoverflow.net/questions/38089/whats-so-special-about-the-forgetful-functor-from-g-rep-to-vect/38092#38092 Answer by Jacob Lurie for What's so special about the forgetful functor from G-rep to Vect? Jacob Lurie 2010-09-08T19:22:54Z 2010-09-08T20:12:20Z <p>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.</p> http://mathoverflow.net/questions/38089/whats-so-special-about-the-forgetful-functor-from-g-rep-to-vect/38094#38094 Answer by David Jordan for What's so special about the forgetful functor from G-rep to Vect? David Jordan 2010-09-08T19:30:08Z 2010-09-08T19:30:08Z <p>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:</p> <p><a href="http://mathoverflow.net/questions/500/finite-groups-with-the-same-character-table" rel="nofollow">http://mathoverflow.net/questions/500/finite-groups-with-the-same-character-table</a></p> <p>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.</p>