Determinant of adjoint representation












1












$begingroup$


Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.



I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.



I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.










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$endgroup$












  • $begingroup$
    What is $AN$? A Borel subgroup? Or a parabolic subgroup?
    $endgroup$
    – D_S
    Jan 13 at 23:00










  • $begingroup$
    Also, are you assuming that $G$ is split?
    $endgroup$
    – D_S
    Jan 13 at 23:00
















1












$begingroup$


Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.



I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.



I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.










share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $AN$? A Borel subgroup? Or a parabolic subgroup?
    $endgroup$
    – D_S
    Jan 13 at 23:00










  • $begingroup$
    Also, are you assuming that $G$ is split?
    $endgroup$
    – D_S
    Jan 13 at 23:00














1












1








1





$begingroup$


Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.



I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.



I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.










share|cite|improve this question











$endgroup$




Let $G$ be a semisimple Lie group with Iwasawa decomposition $G=KAN$ and consider the determinant of the adjoint representation $operatorname{Ad}$ of $AN$.



I want to determine what the derived representation looks like on $mathfrak{a}$ (on $mathfrak{n}$ it is obviously zero). I suspect that one can calculate this values using the root space decomposition w.r.t the root system $(mathfrak{g},mathfrak{a})$.



I know that the derivative of the determinant is the trace. Does this help?
Thanks for any hints.







lie-groups lie-algebras root-systems lie-derivative






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edited Jan 13 at 21:40







Luke Mathwalker

















asked Jan 13 at 17:33









Luke MathwalkerLuke Mathwalker

376213




376213












  • $begingroup$
    What is $AN$? A Borel subgroup? Or a parabolic subgroup?
    $endgroup$
    – D_S
    Jan 13 at 23:00










  • $begingroup$
    Also, are you assuming that $G$ is split?
    $endgroup$
    – D_S
    Jan 13 at 23:00


















  • $begingroup$
    What is $AN$? A Borel subgroup? Or a parabolic subgroup?
    $endgroup$
    – D_S
    Jan 13 at 23:00










  • $begingroup$
    Also, are you assuming that $G$ is split?
    $endgroup$
    – D_S
    Jan 13 at 23:00
















$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00




$begingroup$
What is $AN$? A Borel subgroup? Or a parabolic subgroup?
$endgroup$
– D_S
Jan 13 at 23:00












$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00




$begingroup$
Also, are you assuming that $G$ is split?
$endgroup$
– D_S
Jan 13 at 23:00










1 Answer
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$begingroup$

My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).






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    $begingroup$

    My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).






      share|cite|improve this answer









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        0





        $begingroup$

        My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).






        share|cite|improve this answer









        $endgroup$



        My question is answered in Knapps book Lie groups beyond an introduction on page 472 (Integration, Application to Reductive Lie Groups). The answer is $2rholog(a)$, where $rho$ is the half sum of positive roots (counted with multiplicities).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 23:07









        Luke MathwalkerLuke Mathwalker

        376213




        376213






























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