Tannaka duality for semisimple groups












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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.






ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category







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    2












    $begingroup$


    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.






    ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      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.






      ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category







      share|cite|improve this question









      $endgroup$




      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.






      ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category




      ag.algebraic-geometry rt.representation-theory ct.category-theory tannakian-category






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      share|cite|improve this question











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      asked 2 hours ago









      leibnewtzleibnewtz

      55428




      55428






















          2 Answers
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          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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?
            $endgroup$
            – leibnewtz
            56 mins ago










          • $begingroup$
            I think so. But I’m more into topological groups...
            $endgroup$
            – M Mueger
            52 mins ago










          • $begingroup$
            Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
            $endgroup$
            – Will Sawin
            34 mins ago



















          1












          $begingroup$

          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.






          share|cite|improve this answer









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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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?
              $endgroup$
              – leibnewtz
              56 mins ago










            • $begingroup$
              I think so. But I’m more into topological groups...
              $endgroup$
              – M Mueger
              52 mins ago










            • $begingroup$
              Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
              $endgroup$
              – Will Sawin
              34 mins ago
















            2












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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?
              $endgroup$
              – leibnewtz
              56 mins ago










            • $begingroup$
              I think so. But I’m more into topological groups...
              $endgroup$
              – M Mueger
              52 mins ago










            • $begingroup$
              Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
              $endgroup$
              – Will Sawin
              34 mins ago














            2












            2








            2





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            M MuegerM Mueger

            1635




            1635












            • $begingroup$
              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?
              $endgroup$
              – leibnewtz
              56 mins ago










            • $begingroup$
              I think so. But I’m more into topological groups...
              $endgroup$
              – M Mueger
              52 mins ago










            • $begingroup$
              Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
              $endgroup$
              – Will Sawin
              34 mins ago


















            • $begingroup$
              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?
              $endgroup$
              – leibnewtz
              56 mins ago










            • $begingroup$
              I think so. But I’m more into topological groups...
              $endgroup$
              – M Mueger
              52 mins ago










            • $begingroup$
              Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
              $endgroup$
              – Will Sawin
              34 mins ago
















            $begingroup$
            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?
            $endgroup$
            – leibnewtz
            56 mins ago




            $begingroup$
            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?
            $endgroup$
            – leibnewtz
            56 mins ago












            $begingroup$
            I think so. But I’m more into topological groups...
            $endgroup$
            – M Mueger
            52 mins ago




            $begingroup$
            I think so. But I’m more into topological groups...
            $endgroup$
            – M Mueger
            52 mins ago












            $begingroup$
            Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
            $endgroup$
            – Will Sawin
            34 mins ago




            $begingroup$
            Nothing here forces the group to be connected, and this finite center criterion holds only for connected groups (try $O(2)$).
            $endgroup$
            – Will Sawin
            34 mins ago











            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 35 mins ago









                Will SawinWill Sawin

                68.7k7140285




                68.7k7140285






























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