# Is a group scheme determined by its category of representations?

More precisely, let $G$ be an affine group scheme over a field $k$, $Rep_k(G)$ be the category of finite dimensional representations of G, and $\omega_0$ be the forgetful functor from $Rep_k(G)$ to the category of vector spaces over $k$. With the usual notation in the theory of Tannakian categories, is the natural map $G\longrightarrow \underline{\text{Aut}}^{\otimes} (\omega_0)$ an isomorphism of group schemes over $k$?

Yes, see Deligne–Milne, "Tannakian Categories", Proposition 2.8, which says precisely this. (It doesn't matter what kind of field $k$ is, but many of the more specific recognition theorems, such as that for reductivity, do.)

I feel like I must be misunderstanding, since I understood that to be the theory of Tannakian categories at its most basic.

N̶o̶.̶ (YES, the answer above is correct.) The following example shows we cannot drop that condition: Work out the example of $G = E$ an elliptic curve/$k$. (The Rep category is severely degenerate in this case.)

[Edited. Some how at that late hour I completely missed the word 'affine' in the OP. Indeed, if the group is affine then his question is answered by the celebrated Tannaka theorems. I'll leave the info here though as it's nice to have examples of what fails.]

Not all is lost though. A more general type of phenomena you may be interested in is that if you start with $T =$Coh($Y$) (coherent sheaves) than you can actually construct a stack $X$ which behaves as $X(S) =$ {fiber functors $T \to S$-mod}, where $S$ is a $k$-algebra.

If your scheme $Y$ was nice enough, then $X$ will actually be $Y$. (If I recall correctly, separatedness is critical.)

In fact this is what you are doing in the well known affine group reconstruction case as well: Under the hood you are using $BG$. This has only one point and the uniqueness of the fiber functor reflects this.

• $E$ is not affine :P – მამუკა ჯიბლაძე Jan 27 '17 at 6:33
• In fact there is something serious and unknown to me here. Let us look at it this way: when your "standard cogenerator" does not work on some particular object you use instead of maps to this cogenerator sections of bundles with fibre the cogenerator. E. g. there are not enough functions on projective varieties, so you use sections of line bundles on them instead. Now switch on the group structure; you have a single object groupoid in schemes with not enough functors to $\text{Vect}$. Following the above principle one should use some kind of $\text{Vect}$-bundles over the groupoid? – მამუკა ჯიბლაძე Jan 27 '17 at 6:40
• @მამუკაჯიბლაძე Here is a counterexample, suggesting that it is hopeless to reconstruct $BE$ from the category of sheaves: $\mathbb P^1$ and $E$ are complete, so that GAGA applies to their categories of coherent sheaves. If you could reconstruct $BE$ from the category of sheaves on $E$, that would suggest that you could describe maps $\mathbb P^1\to BE$ in terms of sheaves. But there are more analytic maps (Hopf surfaces) than algebraic. The total spaces aren't projective, but $\mathbb P^1$ and $E$ are. Lurie isolates this as uniquely bad. – Ben Wieland Jan 27 '17 at 15:13
• @BenWieland Sorry this is way above my level, so what follows may be completely senseless, but I rather had in mind some kind of "meromorphic representations" of $E$ (as a group). Something like a system of line bundles $(L_i)_{1\leqslant i\leqslant n}$ on $E$ and sections $s_{ij}$ of $L_i^*\otimes L_j$, with $s_{ij}(xy)=\sum_ks_{ik}(x)s_{kj}(y)$, where the latter sum is understood as a convolution of sections of $(L_i^*\otimes L_k)\boxtimes(L_k^*\otimes L_j)$... – მამუკა ჯიბლაძე Jan 27 '17 at 16:24
• @მამუკაჯიბლაძე Conceptually, you should consider the 2-category of representations of $E$ on (sufficiently good) abelian categories over $R$ (here $R$ is the base ring which I will assume $=\mathbb C$ for fun). For example, a representation of $E$ on $Vect_{\mathbb C}$ is the same as a representation of cogroup $QCoh(E)$ on $Vect_{\mathbb C}$ ($QCoh(E)$ has obvious multiplication and inherits comultiplication from $E$), which is the same as a line bundle on $E$. Thus linear 2-representations of $E$ correspond to points on the dual elliptic curve. – Anton Fetisov Jan 28 '17 at 0:27