$frak{g}^{bot}$ $=Z(frak{g})$ if and only if $frak{g}$ is reductive












1












$begingroup$


Let $frak{g}$ be a finite dimensional Lie algebra over $k$, a field of characteristic $0$.



Recall that $frak{g}$ is called reductive if the center $Z(frak{g})$ is equal to the radical, rad$(frak{g})$, the largest solvable ideal of $frak{g}$.



Let $kappa$ denote the killing form, and $frak{g}^{bot} = {$ $x in frak{g}:$ $forall y in frak{g}$, $kappa(x,y) = 0 }$ its radical.




Show $frak{g}^{bot}$ $=Z(frak{g})$ if and only if $frak{g}$ is
reductive.





Assume first that $frak{g}$ is reductive.



By definition $Z(frak{g}) subset$ $frak{g}^{bot}$, so we need the converse inclusion. By Cartan's criterion for solvability $frak{g}^{bot}$ is solvable, hence, as it is an ideal, $frak{g}^{bot} subset$ rad$(frak{g})$ as required.



The other direction is what I'm having issues with:



I know that if $I$ is a solvable ideal of $frak{g}$, the last term in its derived series is abelian. Also it is true that for any abelian ideal $J$, $kappa(J, frak{g}) = 0$ as for $j in J, x in frak{g}$ $ad(j)ad(x)$ is nilpotent.



So if we assume $frak{g}^{bot}$ $=Z(frak{g})$, we get that $J subset Z(frak{g})$, and moreover $I cap Z(frak{g}) neq {0}$.



Questions:




  1. Can I continue this attempt for the unsolved direction?


  2. If not, can you provide a hint to continue?







abstract-algebra lie-algebras







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $frak{g}$ be a finite dimensional Lie algebra over $k$, a field of characteristic $0$.



    Recall that $frak{g}$ is called reductive if the center $Z(frak{g})$ is equal to the radical, rad$(frak{g})$, the largest solvable ideal of $frak{g}$.



    Let $kappa$ denote the killing form, and $frak{g}^{bot} = {$ $x in frak{g}:$ $forall y in frak{g}$, $kappa(x,y) = 0 }$ its radical.




    Show $frak{g}^{bot}$ $=Z(frak{g})$ if and only if $frak{g}$ is
    reductive.





    Assume first that $frak{g}$ is reductive.



    By definition $Z(frak{g}) subset$ $frak{g}^{bot}$, so we need the converse inclusion. By Cartan's criterion for solvability $frak{g}^{bot}$ is solvable, hence, as it is an ideal, $frak{g}^{bot} subset$ rad$(frak{g})$ as required.



    The other direction is what I'm having issues with:



    I know that if $I$ is a solvable ideal of $frak{g}$, the last term in its derived series is abelian. Also it is true that for any abelian ideal $J$, $kappa(J, frak{g}) = 0$ as for $j in J, x in frak{g}$ $ad(j)ad(x)$ is nilpotent.



    So if we assume $frak{g}^{bot}$ $=Z(frak{g})$, we get that $J subset Z(frak{g})$, and moreover $I cap Z(frak{g}) neq {0}$.



    Questions:




    1. Can I continue this attempt for the unsolved direction?


    2. If not, can you provide a hint to continue?







    abstract-algebra lie-algebras







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $frak{g}$ be a finite dimensional Lie algebra over $k$, a field of characteristic $0$.



      Recall that $frak{g}$ is called reductive if the center $Z(frak{g})$ is equal to the radical, rad$(frak{g})$, the largest solvable ideal of $frak{g}$.



      Let $kappa$ denote the killing form, and $frak{g}^{bot} = {$ $x in frak{g}:$ $forall y in frak{g}$, $kappa(x,y) = 0 }$ its radical.




      Show $frak{g}^{bot}$ $=Z(frak{g})$ if and only if $frak{g}$ is
      reductive.





      Assume first that $frak{g}$ is reductive.



      By definition $Z(frak{g}) subset$ $frak{g}^{bot}$, so we need the converse inclusion. By Cartan's criterion for solvability $frak{g}^{bot}$ is solvable, hence, as it is an ideal, $frak{g}^{bot} subset$ rad$(frak{g})$ as required.



      The other direction is what I'm having issues with:



      I know that if $I$ is a solvable ideal of $frak{g}$, the last term in its derived series is abelian. Also it is true that for any abelian ideal $J$, $kappa(J, frak{g}) = 0$ as for $j in J, x in frak{g}$ $ad(j)ad(x)$ is nilpotent.



      So if we assume $frak{g}^{bot}$ $=Z(frak{g})$, we get that $J subset Z(frak{g})$, and moreover $I cap Z(frak{g}) neq {0}$.



      Questions:




      1. Can I continue this attempt for the unsolved direction?


      2. If not, can you provide a hint to continue?







      abstract-algebra lie-algebras







      share|cite|improve this question











      $endgroup$




      Let $frak{g}$ be a finite dimensional Lie algebra over $k$, a field of characteristic $0$.



      Recall that $frak{g}$ is called reductive if the center $Z(frak{g})$ is equal to the radical, rad$(frak{g})$, the largest solvable ideal of $frak{g}$.



      Let $kappa$ denote the killing form, and $frak{g}^{bot} = {$ $x in frak{g}:$ $forall y in frak{g}$, $kappa(x,y) = 0 }$ its radical.




      Show $frak{g}^{bot}$ $=Z(frak{g})$ if and only if $frak{g}$ is
      reductive.





      Assume first that $frak{g}$ is reductive.



      By definition $Z(frak{g}) subset$ $frak{g}^{bot}$, so we need the converse inclusion. By Cartan's criterion for solvability $frak{g}^{bot}$ is solvable, hence, as it is an ideal, $frak{g}^{bot} subset$ rad$(frak{g})$ as required.



      The other direction is what I'm having issues with:



      I know that if $I$ is a solvable ideal of $frak{g}$, the last term in its derived series is abelian. Also it is true that for any abelian ideal $J$, $kappa(J, frak{g}) = 0$ as for $j in J, x in frak{g}$ $ad(j)ad(x)$ is nilpotent.



      So if we assume $frak{g}^{bot}$ $=Z(frak{g})$, we get that $J subset Z(frak{g})$, and moreover $I cap Z(frak{g}) neq {0}$.



      Questions:




      1. Can I continue this attempt for the unsolved direction?


      2. If not, can you provide a hint to continue?







      abstract-algebra lie-algebras




      abstract-algebra lie-algebras






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 17:14







      Mariah

















      asked Jan 3 at 16:12









      MariahMariah

      1,5561618




      1,5561618






















          1 Answer
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          $begingroup$

          Let $I$ be a solvable ideal, and $I,[I,I]=I^2,...,[I^{n-1},I^{n-1}]=I^n$ its derived series. Suppose $I^nneq 0$, you have already shown that $I^n$ is in the center.



          Remark that for every $x in I^{n-1}, y,zin g$, $(ad_xcirc ad_y)(z)=[x,[y,z]]$ since $x$ is an element of the ideal $I^{n-1}$ we deduce that $a=[x,[y,z]]in I^{n-1}$.



          We have $(ad_xcirc ad_y)^2(z)=(ad_x(ad_y(a))=[x,[y,a]]$ since $x$ and $[y,a]$ are in $I^{n-1}$, we deduce that $b=[x,[y,a]]in I^n$. This implies that $(ad_xcirc ad_y)^3(z)=(ad_x(ad_y(b))=0$, we deduce that $ad_xcirc ad_y$ is nilpotent and $kappa(x,y)=0$. This implies by hypothesis that $xin Z(g)$. We deduce that $I^{n-1}subset Z(g)$. We can apply a recursive argument to show that $I^{n-2},...,I^2,I subset Z(g)$, this implies that $R(g)subset Z(g)$.






          share|cite|improve this answer











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            $begingroup$

            Let $I$ be a solvable ideal, and $I,[I,I]=I^2,...,[I^{n-1},I^{n-1}]=I^n$ its derived series. Suppose $I^nneq 0$, you have already shown that $I^n$ is in the center.



            Remark that for every $x in I^{n-1}, y,zin g$, $(ad_xcirc ad_y)(z)=[x,[y,z]]$ since $x$ is an element of the ideal $I^{n-1}$ we deduce that $a=[x,[y,z]]in I^{n-1}$.



            We have $(ad_xcirc ad_y)^2(z)=(ad_x(ad_y(a))=[x,[y,a]]$ since $x$ and $[y,a]$ are in $I^{n-1}$, we deduce that $b=[x,[y,a]]in I^n$. This implies that $(ad_xcirc ad_y)^3(z)=(ad_x(ad_y(b))=0$, we deduce that $ad_xcirc ad_y$ is nilpotent and $kappa(x,y)=0$. This implies by hypothesis that $xin Z(g)$. We deduce that $I^{n-1}subset Z(g)$. We can apply a recursive argument to show that $I^{n-2},...,I^2,I subset Z(g)$, this implies that $R(g)subset Z(g)$.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Let $I$ be a solvable ideal, and $I,[I,I]=I^2,...,[I^{n-1},I^{n-1}]=I^n$ its derived series. Suppose $I^nneq 0$, you have already shown that $I^n$ is in the center.



              Remark that for every $x in I^{n-1}, y,zin g$, $(ad_xcirc ad_y)(z)=[x,[y,z]]$ since $x$ is an element of the ideal $I^{n-1}$ we deduce that $a=[x,[y,z]]in I^{n-1}$.



              We have $(ad_xcirc ad_y)^2(z)=(ad_x(ad_y(a))=[x,[y,a]]$ since $x$ and $[y,a]$ are in $I^{n-1}$, we deduce that $b=[x,[y,a]]in I^n$. This implies that $(ad_xcirc ad_y)^3(z)=(ad_x(ad_y(b))=0$, we deduce that $ad_xcirc ad_y$ is nilpotent and $kappa(x,y)=0$. This implies by hypothesis that $xin Z(g)$. We deduce that $I^{n-1}subset Z(g)$. We can apply a recursive argument to show that $I^{n-2},...,I^2,I subset Z(g)$, this implies that $R(g)subset Z(g)$.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Let $I$ be a solvable ideal, and $I,[I,I]=I^2,...,[I^{n-1},I^{n-1}]=I^n$ its derived series. Suppose $I^nneq 0$, you have already shown that $I^n$ is in the center.



                Remark that for every $x in I^{n-1}, y,zin g$, $(ad_xcirc ad_y)(z)=[x,[y,z]]$ since $x$ is an element of the ideal $I^{n-1}$ we deduce that $a=[x,[y,z]]in I^{n-1}$.



                We have $(ad_xcirc ad_y)^2(z)=(ad_x(ad_y(a))=[x,[y,a]]$ since $x$ and $[y,a]$ are in $I^{n-1}$, we deduce that $b=[x,[y,a]]in I^n$. This implies that $(ad_xcirc ad_y)^3(z)=(ad_x(ad_y(b))=0$, we deduce that $ad_xcirc ad_y$ is nilpotent and $kappa(x,y)=0$. This implies by hypothesis that $xin Z(g)$. We deduce that $I^{n-1}subset Z(g)$. We can apply a recursive argument to show that $I^{n-2},...,I^2,I subset Z(g)$, this implies that $R(g)subset Z(g)$.






                share|cite|improve this answer











                $endgroup$



                Let $I$ be a solvable ideal, and $I,[I,I]=I^2,...,[I^{n-1},I^{n-1}]=I^n$ its derived series. Suppose $I^nneq 0$, you have already shown that $I^n$ is in the center.



                Remark that for every $x in I^{n-1}, y,zin g$, $(ad_xcirc ad_y)(z)=[x,[y,z]]$ since $x$ is an element of the ideal $I^{n-1}$ we deduce that $a=[x,[y,z]]in I^{n-1}$.



                We have $(ad_xcirc ad_y)^2(z)=(ad_x(ad_y(a))=[x,[y,a]]$ since $x$ and $[y,a]$ are in $I^{n-1}$, we deduce that $b=[x,[y,a]]in I^n$. This implies that $(ad_xcirc ad_y)^3(z)=(ad_x(ad_y(b))=0$, we deduce that $ad_xcirc ad_y$ is nilpotent and $kappa(x,y)=0$. This implies by hypothesis that $xin Z(g)$. We deduce that $I^{n-1}subset Z(g)$. We can apply a recursive argument to show that $I^{n-2},...,I^2,I subset Z(g)$, this implies that $R(g)subset Z(g)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 5:45









                Pratyush Sarkar

                2,7901227




                2,7901227










                answered Jan 3 at 18:02









                Tsemo AristideTsemo Aristide

                58.9k11445




                58.9k11445






























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