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Why is this universal cover isomorphic to $textSL_2(mathbbC)$?



The 2019 Stack Overflow Developer Survey Results Are InIs the universal cover of an algebraic group an algebraic group?Betti Numbers and the Néron-Severi group.Why aren't those Cartier Divisors equivalent?Gluing sheaves togetherWhy is this a torusStatement of Chow's theoremCovering Maps Abelian Lie groupsDoes the singularities of black holes not being simply connected imply that the universal coverings of quantum field theory fail?Universal homeomorphisms between non-isomorphic schemesAbout the proof of Borel-Weil-Bott theorem: Two derived pushforward sheaves are isomorphic by descent theory










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In Lurie's proof of the Borel-Weil theorem http://www.math.harvard.edu/~lurie/papers/bwb.pdf, he states that the universal cover of a Levi factor of $S$, where $Ccup U'B=SB$, is isomorphic to $textSL_2(mathbbC)$. Why is this the case? Could someone please give a reference for this?
Here $C$ is the Bruhat cell corresponding to a simple root.










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$endgroup$
















    1












    $begingroup$


    In Lurie's proof of the Borel-Weil theorem http://www.math.harvard.edu/~lurie/papers/bwb.pdf, he states that the universal cover of a Levi factor of $S$, where $Ccup U'B=SB$, is isomorphic to $textSL_2(mathbbC)$. Why is this the case? Could someone please give a reference for this?
    Here $C$ is the Bruhat cell corresponding to a simple root.










    share|cite|improve this question











    $endgroup$














      1












      1








      1





      $begingroup$


      In Lurie's proof of the Borel-Weil theorem http://www.math.harvard.edu/~lurie/papers/bwb.pdf, he states that the universal cover of a Levi factor of $S$, where $Ccup U'B=SB$, is isomorphic to $textSL_2(mathbbC)$. Why is this the case? Could someone please give a reference for this?
      Here $C$ is the Bruhat cell corresponding to a simple root.










      share|cite|improve this question











      $endgroup$




      In Lurie's proof of the Borel-Weil theorem http://www.math.harvard.edu/~lurie/papers/bwb.pdf, he states that the universal cover of a Levi factor of $S$, where $Ccup U'B=SB$, is isomorphic to $textSL_2(mathbbC)$. Why is this the case? Could someone please give a reference for this?
      Here $C$ is the Bruhat cell corresponding to a simple root.







      algebraic-geometry algebraic-groups






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      edited Mar 23 at 17:33







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      asked Mar 23 at 16:55









      theathea

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          $begingroup$

          There is a lot going on in that sentence. A general reference for the structure theory of reductive algebraic groups would be the book by Springer, or Borel, or any of the online notes circulating. The ones by Herzig and Murnaghan are both useful for this stuff. Specifically, see theorem 180 and the rest of Ch. 6 in these notes. The example
          to keep in mind that makes reasoning about these things easy is the standard block-upper-triangular parabolic subgroups of $mathrmGL_n$.



          Anyway, by construction the Lie algebra of $S$ is isomorphic to $mathfraksl_2,mathbbC)$, so by the theory of Lie groups, $S$ is covered by $mathrmSL(2,mathbbC)$. In more detail, if $alpha=epsilon_1-epsilon_2$ is the positive root in question for $mathrmGL_n$, then I think $S$ is can be taken to be generated by the upper-triangular matrices with 1s on the diagonal and and the matrices with 1s on the diagonal, and a single other entry at $(2,1)$, corresponding to $-alpha$.






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            $begingroup$

            There is a lot going on in that sentence. A general reference for the structure theory of reductive algebraic groups would be the book by Springer, or Borel, or any of the online notes circulating. The ones by Herzig and Murnaghan are both useful for this stuff. Specifically, see theorem 180 and the rest of Ch. 6 in these notes. The example
            to keep in mind that makes reasoning about these things easy is the standard block-upper-triangular parabolic subgroups of $mathrmGL_n$.



            Anyway, by construction the Lie algebra of $S$ is isomorphic to $mathfraksl_2,mathbbC)$, so by the theory of Lie groups, $S$ is covered by $mathrmSL(2,mathbbC)$. In more detail, if $alpha=epsilon_1-epsilon_2$ is the positive root in question for $mathrmGL_n$, then I think $S$ is can be taken to be generated by the upper-triangular matrices with 1s on the diagonal and and the matrices with 1s on the diagonal, and a single other entry at $(2,1)$, corresponding to $-alpha$.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              There is a lot going on in that sentence. A general reference for the structure theory of reductive algebraic groups would be the book by Springer, or Borel, or any of the online notes circulating. The ones by Herzig and Murnaghan are both useful for this stuff. Specifically, see theorem 180 and the rest of Ch. 6 in these notes. The example
              to keep in mind that makes reasoning about these things easy is the standard block-upper-triangular parabolic subgroups of $mathrmGL_n$.



              Anyway, by construction the Lie algebra of $S$ is isomorphic to $mathfraksl_2,mathbbC)$, so by the theory of Lie groups, $S$ is covered by $mathrmSL(2,mathbbC)$. In more detail, if $alpha=epsilon_1-epsilon_2$ is the positive root in question for $mathrmGL_n$, then I think $S$ is can be taken to be generated by the upper-triangular matrices with 1s on the diagonal and and the matrices with 1s on the diagonal, and a single other entry at $(2,1)$, corresponding to $-alpha$.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                There is a lot going on in that sentence. A general reference for the structure theory of reductive algebraic groups would be the book by Springer, or Borel, or any of the online notes circulating. The ones by Herzig and Murnaghan are both useful for this stuff. Specifically, see theorem 180 and the rest of Ch. 6 in these notes. The example
                to keep in mind that makes reasoning about these things easy is the standard block-upper-triangular parabolic subgroups of $mathrmGL_n$.



                Anyway, by construction the Lie algebra of $S$ is isomorphic to $mathfraksl_2,mathbbC)$, so by the theory of Lie groups, $S$ is covered by $mathrmSL(2,mathbbC)$. In more detail, if $alpha=epsilon_1-epsilon_2$ is the positive root in question for $mathrmGL_n$, then I think $S$ is can be taken to be generated by the upper-triangular matrices with 1s on the diagonal and and the matrices with 1s on the diagonal, and a single other entry at $(2,1)$, corresponding to $-alpha$.






                share|cite|improve this answer









                $endgroup$



                There is a lot going on in that sentence. A general reference for the structure theory of reductive algebraic groups would be the book by Springer, or Borel, or any of the online notes circulating. The ones by Herzig and Murnaghan are both useful for this stuff. Specifically, see theorem 180 and the rest of Ch. 6 in these notes. The example
                to keep in mind that makes reasoning about these things easy is the standard block-upper-triangular parabolic subgroups of $mathrmGL_n$.



                Anyway, by construction the Lie algebra of $S$ is isomorphic to $mathfraksl_2,mathbbC)$, so by the theory of Lie groups, $S$ is covered by $mathrmSL(2,mathbbC)$. In more detail, if $alpha=epsilon_1-epsilon_2$ is the positive root in question for $mathrmGL_n$, then I think $S$ is can be taken to be generated by the upper-triangular matrices with 1s on the diagonal and and the matrices with 1s on the diagonal, and a single other entry at $(2,1)$, corresponding to $-alpha$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 26 at 5:18









                Stefan DawydiakStefan Dawydiak

                37929




                37929



























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