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












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












    $begingroup$


    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

      1,5561718




      1,5561718






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059635%2fprove-that-frakg-ceni-neq-ceni%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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

                59k11445




                59k11445






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059635%2fprove-that-frakg-ceni-neq-ceni%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Bressuire

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                    Read more

                    Cabo Verde

                    Image
                      Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                    Read more

                    Gyllenstierna

                    Image
                    .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                    Read more