Prove that $[frak{g}$ $,cen(I)] neq cen(I)$












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$begingroup$


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



Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.



Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.




Prove that $[frak{g}$$,cen(I)] neq cen(I)$.




Where I'm at:



First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.



Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.



Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.



Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.



Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.



Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$



Any hints?






abstract-algebra lie-algebras







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


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



    Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.



    Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.




    Prove that $[frak{g}$$,cen(I)] neq cen(I)$.




    Where I'm at:



    First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.



    Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.



    Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.



    Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.



    Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.



    Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$



    Any hints?






    abstract-algebra lie-algebras







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


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



      Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.



      Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.




      Prove that $[frak{g}$$,cen(I)] neq cen(I)$.




      Where I'm at:



      First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.



      Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.



      Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.



      Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.



      Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.



      Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$



      Any hints?






      abstract-algebra lie-algebras







      share|cite|improve this question











      $endgroup$




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



      Suppose that $I triangleleft frak{g}$ is an ideal of co-dimension $1$, and that ${0} neq I cap Z(frak{g})$.



      Moreover, define $cen(I) = {g in frak{g} : forall$ $x in I, [x,g] =0}$.




      Prove that $[frak{g}$$,cen(I)] neq cen(I)$.




      Where I'm at:



      First, it can be easily shown that $cen(I)$ is an ideal itself, and hence we get that $[frak{g}$$,cen(I)] subset cen(I)$. So we must show that the inclusion is strict.



      Since $I$ is of co-dimension $1$ we have that $[frak{g},g] subset$ $I$. Therefore $[frak{g}$$,cen(I)] subset I$.



      Clearly, if $Z(frak{g})$ isn't contained in $I$ we can write that $frak{g} =$ $kx_0 oplus I$ for $x_0 in Z(frak{g})$. This gives $[frak{g}$$,cen(I)] = [kx_0 oplus I, cen(I)] = 0$ and we'd be done, since $I cap Z(frak{g})$ is non-zero and is contained in $cen(I)$.



      Thus we can assume: ${0} neq Z(frak{g}) subset$ $I$.



      Here I got stuck; I was trying to use Lie's lemma, and find a functional that is non zero on $cen(I)$, and has a common eigenvector. This would give the required inequality. However it is not clear to me how I may do so.



      Moreover, I'm having trouble interpreting the assumption that ${0} neq I cap Z(frak{g})$



      Any hints?






      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 2 at 16:14







      Mariah

















      asked Jan 2 at 15:49









      MariahMariah

      1,3831518




      1,3831518






















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          $begingroup$

          Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.






          share|cite|improve this answer









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            $begingroup$

            Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.






                share|cite|improve this answer









                $endgroup$



                Let $x$ an element which is not in $I$, $[g,cen(I)]=[kxoplus I,cen(I)]=[kx,cen(I)]$, this implies that $[g,cen(I)]$ is the image of the restriction of $ad_x$ defined by $ad_x(y)=[x,y]$ to $cen(I)$, but the kernel of the restriction of $ad_x$ to $cen(I)$ is not trivial since it contains $Z(I)$, thus $ad_x$ is not surjective and $[g,cen(I)]neq cen(I)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 16:16









                Tsemo AristideTsemo Aristide

                56.8k11444




                56.8k11444






























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