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Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $\mathcal{C}$ equipped with a fiber functor $F: \mathcal{C} \to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G \cong Aut(F)$ such that $\mathcal{C} \cong Rep(G)$. This means that any such category is associated with a root datum.

Is there a version of this reconstruction theorem that will tell us when a category $\mathcal{C}$ is the category of finite dimensional representations of a semisimple group? I would like to be able to associate with a Tannakian category a root system, and not just a root datum.

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  • $\begingroup$ The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $D\to Z(G)$ is the same as to give an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne on Tannakian Categories. $\endgroup$
    – anon
    Commented Apr 20, 2019 at 13:36
  • $\begingroup$ To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. See arXiv:0705.1348 $\endgroup$
    – anon
    Commented Apr 20, 2019 at 13:47
  • $\begingroup$ Have you switched "root system" and "root datum"? Usually the latter carries more information (root system + lattice containing it). $\endgroup$
    – LSpice
    Commented Apr 20, 2019 at 14:19
  • $\begingroup$ L.Spice. No I haven't. As stated, the construction gives $Rep(G)$ where $G$ is the simply connected semisimple group with the given root system. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any quotient of $G$. $\endgroup$
    – anon
    Commented Apr 20, 2019 at 16:18
  • $\begingroup$ @anon, sorry, I missed your response because there was no @ on it. I was asking the original poster, not you, about possibly having switched 'root system' and 'root datum' (since the poster says "a root system, and not just a root datum", whereas a root datum is more information than a root system). Indeed, your choice to work with simply connected $G$ seems to be effectively a choice of root datum. $\endgroup$
    – LSpice
    Commented Dec 15, 2019 at 18:39

3 Answers 3

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Another criterion is that there should be only finitely many objects of bounded dimension. This condition might be easy to check in practice from abstract finiteness theorems. The proof is that, if the group is not semi simple, you can take any 1-dimensional character of the identity component and induce up to the main group. Because there are infinitely many characters, infinitely many representations.

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  • $\begingroup$ Could you say something about what bounded dimension means? $\endgroup$
    – Exit path
    Commented Apr 20, 2019 at 0:59
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    $\begingroup$ @leibnewtz I mean for each dimension d, finitely many objects with dimension at most d. Dimension of a representation can be determined by looking at which wedge powers vanish but often is available more directly due to a fiber functor. $\endgroup$
    – Will Sawin
    Commented Apr 20, 2019 at 11:46
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In order for ${\mathcal C}$ to come from an algebraic group rather than a pro-algebraic one, you want ${\mathcal C}$ to be finitely generated. And for semisimplicity, you want the group to have finite center. The center can be read off from the category. Cf. my paper “On the center of a compact group”, Intern. Math. Res. Notes. 2004:51, 2751-2756 (2004) or math.CT/0312257.

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  • $\begingroup$ Ah this is excellent! So that claim is that a semisimple, finitely generated, rigid, symmetric monoidal abelian category with a fiber functor is the category of representations of a semisimple algebraic group if and only if the chain group of the category is finite. Is this correct? $\endgroup$
    – Exit path
    Commented Apr 19, 2019 at 23:54
  • $\begingroup$ I think so. But I’m more into topological groups... $\endgroup$
    – M Mueger
    Commented Apr 19, 2019 at 23:58
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    $\begingroup$ Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$). $\endgroup$
    – Will Sawin
    Commented Apr 20, 2019 at 0:15
  • $\begingroup$ @Will: You’re right. But also connectedness of the group can be seen from the category: It is equivalent to the absence of full tensor subcategories with finitely many simple objects (up to isom.). $\endgroup$
    – M Mueger
    Commented Apr 20, 2019 at 9:04
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    $\begingroup$ @M Mueger but if the group is connected, a simpler criterion is that there do not exist nontrivial $ A , B$ with $ A \otimes B = I$. $\endgroup$
    – Will Sawin
    Commented Apr 20, 2019 at 13:26
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I've decided to turn my comments into an answer.

(a) The conditions characterizing the Tannakian categories attached to connected reductive groups can be found in Chapter 2 (2.20, 2.22, 2.23) of the notes by Deligne and Milne on Tannakian categories.

(b) The center of $G$ is reflected in the gradations on the Tannakian category $Rep(G)$. For example, let $D$ be a diagonalizable algebraic group with character group $M$. To give a homomorphism $D\to Z(G)$ is the same as giving an $M$-gradation on $Rep(G)$. See 5.1 of the notes by Deligne and Milne.

(c) To attach a Tannakian category to a root system, choose a semisimple Lie algebra $L$ with the given root system. Then $Rep(L)$ is a Tannakian category with corresponding group $G$ the simply connected semisimple algebraic group with Lie algebra $L$. The category has a natural gradation by $P/Q$ from which it is possible to read off the category corresponding to any lattice $X$ in $P$ containing $Q$. This gives a complete description of the Tannakian categories corresponding to root systems (better, diagrams) without using algebraic groups. See arXiv:0705.1348 –

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