Tannaka duality for semisimple groups Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?Tannaka formalism and the étale fundamental groupIs there a ``path'' between any two fiber functors over the same field in Tannakian formalism?Counter example in Tannaka reconstruction?Recovering classical Tannaka duality from Lurie's version for geometric stacksTannaka DualityCan one explain Tannaka-Krein duality for a finite-group to … a computer ? (How to make input for reconstruction to be finite datum?)Tannakian Formalism for the Quaternions and Dihedral GroupTannakian theory for Lie algebrasIs it possible to reconstruct a finitely generated group from its category of representations?

Tannaka duality for semisimple groups



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?Tannaka formalism and the étale fundamental groupIs there a ``path'' between any two fiber functors over the same field in Tannakian formalism?Counter example in Tannaka reconstruction?Recovering classical Tannaka duality from Lurie's version for geometric stacksTannaka DualityCan one explain Tannaka-Krein duality for a finite-group to … a computer ? (How to make input for reconstruction to be finite datum?)Tannakian Formalism for the Quaternions and Dihedral GroupTannakian theory for Lie algebrasIs it possible to reconstruct a finitely generated group from its category of representations?










2












$begingroup$


Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. 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 $mathcalC$ 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.










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. 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 $mathcalC$ 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.










    share|cite|improve this question









    $endgroup$














      2












      2








      2





      $begingroup$


      Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. 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 $mathcalC$ 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.










      share|cite|improve this question









      $endgroup$




      Tannakian formalism tells us that for any rigid, symmetric monoidal, semisimple category $mathcalC$ equipped with a fiber functor $F: mathcalC to Vect_k$ for a field $k$ (of characteristic $0$) there exists a reductive algebraic group $G cong Aut(F)$ such that $mathcalC cong Rep(G)$. 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 $mathcalC$ 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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      leibnewtzleibnewtz

      55428




      55428




















          2 Answers
          2






          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
            2 hours ago










          • $begingroup$
            I think so. But I’m more into topological groups...
            $endgroup$
            – M Mueger
            2 hours 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
            2 hours 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









          $endgroup$












          • $begingroup$
            Could you say something about what bounded dimension means?
            $endgroup$
            – leibnewtz
            1 hour ago











          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "504"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328495%2ftannaka-duality-for-semisimple-groups%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          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
            2 hours ago










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










          • $begingroup$
            I think so. But I’m more into topological groups...
            $endgroup$
            – M Mueger
            2 hours 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
            2 hours 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 2 hours 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
            2 hours ago










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










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












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




          $begingroup$
          I think so. But I’m more into topological groups...
          $endgroup$
          – M Mueger
          2 hours 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
          2 hours 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
          2 hours 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









          $endgroup$












          • $begingroup$
            Could you say something about what bounded dimension means?
            $endgroup$
            – leibnewtz
            1 hour 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









          $endgroup$












          • $begingroup$
            Could you say something about what bounded dimension means?
            $endgroup$
            – leibnewtz
            1 hour ago













          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 2 hours ago









          Will SawinWill Sawin

          68.7k7140285




          68.7k7140285











          • $begingroup$
            Could you say something about what bounded dimension means?
            $endgroup$
            – leibnewtz
            1 hour ago
















          • $begingroup$
            Could you say something about what bounded dimension means?
            $endgroup$
            – leibnewtz
            1 hour ago















          $begingroup$
          Could you say something about what bounded dimension means?
          $endgroup$
          – leibnewtz
          1 hour ago




          $begingroup$
          Could you say something about what bounded dimension means?
          $endgroup$
          – leibnewtz
          1 hour ago

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to MathOverflow!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f328495%2ftannaka-duality-for-semisimple-groups%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          How to create a command for the “strange m” symbol in latex? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How do you make your own symbol when Detexify fails?Writing bold small caps with mathpazo packageplus-minus symbol with parenthesis around the minus signGreek character in Beamer document titleHow to create dashed right arrow over symbol?Currency symbol: Turkish LiraDouble prec as a single symbol?Plus Sign Too Big; How to Call adfbullet?Is there a TeX macro for three-legged pi?How do I get my integral-like symbol to align like the integral?How to selectively substitute a letter with another symbol representing the same letterHow do I generate a less than symbol and vertical bar that are the same height?

          Българска екзархия Съдържание История | Български екзарси | Вижте също | Външни препратки | Литература | Бележки | НавигацияУстав за управлението на българската екзархия. Цариград, 1870Слово на Ловешкия митрополит Иларион при откриването на Българския народен събор в Цариград на 23. II. 1870 г.Българската правда и гръцката кривда. От С. М. (= Софийски Мелетий). Цариград, 1872Предстоятели на Българската екзархияПодмененият ВеликденИнформационна агенция „Фокус“Димитър Ризов. Българите в техните исторически, етнографически и политически граници (Атлас съдържащ 40 карти). Berlin, Königliche Hoflithographie, Hof-Buch- und -Steindruckerei Wilhelm Greve, 1917Report of the International Commission to Inquire into the Causes and Conduct of the Balkan Wars

          Чепеларе Съдържание География | История | Население | Спортни и природни забележителности | Културни и исторически обекти | Религии | Обществени институции | Известни личности | Редовни събития | Галерия | Източници | Литература | Външни препратки | Навигация41°43′23.99″ с. ш. 24°41′09.99″ и. д. / 41.723333° с. ш. 24.686111° и. д.*ЧепелареЧепеларски Linux fest 2002Начало на Зимен сезон 2005/06Национални хайдушки празници „Капитан Петко Войвода“Град ЧепелареЧепеларе – народният ски курортbgrod.orgwww.terranatura.hit.bgСправка за населението на гр. Исперих, общ. Исперих, обл. РазградМузей на родопския карстМузей на спорта и скитеЧепеларебългарскибългарскианглийскитукИстория на градаСки писти в ЧепелареВремето в ЧепелареРадио и телевизия в ЧепелареЧепеларе мами с родопски чар и добри пистиЕвтин туризъм и снежни атракции в ЧепелареМестоположениеИнформация и снимки от музея на родопския карст3D панорами от ЧепелареЧепелареррр