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$mathbb R$-points of semisimple real algebraic groups, connectivity, and Cartan involutions: some questions


A question about Lie algebras corresponding to Lie groups and algebraic groupsHow to classify all $theta$-stable Cartan sub algebras?Question on unitary representation of non-compact simple Lie groupsMaximal tori in Lie vs algebraic groupsScalar product on Lie algebra of compact Lie groupProve identity map of a Lie algebra is unique Cartan involution when Killing form is negative definiteMost general definition of Borel and parabolic Lie algebras?Is every connected complex semisimple Lie group the complexification of a compact Lie group?Semisimple linear algebraic groups in positive characteristicSymmetric space associated to a connected semisimple linear algebraic $mathbbQ$-group













2












$begingroup$


I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.



Let $G = mathbf G(mathbb R)$, which is a real Lie group.



  • $G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?


  • A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?


  • Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?


  • Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?


  • Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.



    Let $G = mathbf G(mathbb R)$, which is a real Lie group.



    • $G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?


    • A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?


    • Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?


    • Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?


    • Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?










    share|cite|improve this question









    $endgroup$














      2












      2








      2


      0



      $begingroup$


      I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.



      Let $G = mathbf G(mathbb R)$, which is a real Lie group.



      • $G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?


      • A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?


      • Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?


      • Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?


      • Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?










      share|cite|improve this question









      $endgroup$




      I am reading about Cartan involutions on semisimple real Lie groups and have a point of confusion I am trying to reconcile with linear algebraic groups. Let $mathbf G$ be a linear algebraic group over $mathbb R$. Assume that $mathbf G$ is semisimple, i.e. $mathbf G times_mathbb R operatornameSpec mathbb C$ is semisimple as an algebraic group (is Zariski-connected and has trivial radical). This implies that $mathbf G$ is also connected in the Zariski topology.



      Let $G = mathbf G(mathbb R)$, which is a real Lie group.



      • $G$ is in general not connected, right? For example, $mathbf G = operatornameSO(p,q)$. Does $G$ have finitely many components?


      • A real Lie group is defined to be semisimple if it is connected and if its Lie algebra is semisimple (has nondegenerate Killing form). The connected component $G^0$ of $G$ is a semisimple real Lie group, right?


      • Let $mathfrak g$ be the Lie algebra of the algebraic group $mathbf G$. Then do we have $mathfrak g(mathbb R) = operatornameLie(G)$?


      • Let $theta$ be a Cartan involution on $operatornameLie(G)$. There is a corresponding involution Lie group automorphism $Theta$ of $G^0$ whose differential is $theta$, as in the Wikipedia article on Cartan decomposition. Does $Theta$ extend to an automorphism of $G$?


      • Let $K$ be the maximal compact subgroup of $G^0$ obtained as the fixed points of $Theta$. Is there a way to realize $K$ as the intersection with $G^0$ of a canonical maximal compact subgroup of $G$?







      algebraic-geometry representation-theory lie-groups lie-algebras algebraic-groups






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      share|cite|improve this question










      asked Mar 21 at 17:45









      D_SD_S

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      14.2k61653




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          In a slightly different order:



          • For future visitors to this problem, here an answer about connectivity on MO.


          • Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.


          • In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)


          • One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            In a slightly different order:



            • For future visitors to this problem, here an answer about connectivity on MO.


            • Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.


            • In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)


            • One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              In a slightly different order:



              • For future visitors to this problem, here an answer about connectivity on MO.


              • Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.


              • In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)


              • One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                In a slightly different order:



                • For future visitors to this problem, here an answer about connectivity on MO.


                • Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.


                • In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)


                • One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.






                share|cite|improve this answer









                $endgroup$



                In a slightly different order:



                • For future visitors to this problem, here an answer about connectivity on MO.


                • Yes, the functor from real linear algebraic groups to real Lie groups respects Lie algebras. See these notes by Milne, section III.2.


                • In characteristic zero, a linear algebraic group is semisimple iff its Lie algebra is semisimple. By the above point, $G^0$ is a semisimple real Lie group. (See the same notes, section II.4)


                • One can include a fixed choice of Cartan involution in the defintion of reductive real Lie group, and hence in the definition of semisimple real Lie group (as a reductive group with finite centre). A good choice is inverse-conjugate transpose, and then $K$ is essentially just the intersection of $G^0$ with the relevant orthogonal or unitary group, all as subgroups of some ambient $mathrmGL(n,mathbbR)$. This is the approach taken by Knapp in his book Representation Theory of Semisimple Groups. This simplification might help answer the last part of your question.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 29 at 4:48









                Stefan DawydiakStefan Dawydiak

                41429




                41429



























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