Defining the Lie bracket on a tensor product Lie algebra












0












$begingroup$


So, my question is the following:



Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.






vector-spaces lie-algebras







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
    $endgroup$
    – md2perpe
    Aug 12 '18 at 17:13










  • $begingroup$
    As far as I know there is no way to do this.
    $endgroup$
    – Qiaochu Yuan
    Aug 12 '18 at 17:13










  • $begingroup$
    @QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
    $endgroup$
    – G K
    Aug 12 '18 at 17:20






  • 1




    $begingroup$
    Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
    $endgroup$
    – Tobias Kildetoft
    Aug 12 '18 at 17:28










  • $begingroup$
    @TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
    $endgroup$
    – G K
    Aug 12 '18 at 17:30
















0












$begingroup$


So, my question is the following:



Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.






vector-spaces lie-algebras







share|cite|improve this question









$endgroup$












  • $begingroup$
    Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
    $endgroup$
    – md2perpe
    Aug 12 '18 at 17:13










  • $begingroup$
    As far as I know there is no way to do this.
    $endgroup$
    – Qiaochu Yuan
    Aug 12 '18 at 17:13










  • $begingroup$
    @QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
    $endgroup$
    – G K
    Aug 12 '18 at 17:20






  • 1




    $begingroup$
    Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
    $endgroup$
    – Tobias Kildetoft
    Aug 12 '18 at 17:28










  • $begingroup$
    @TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
    $endgroup$
    – G K
    Aug 12 '18 at 17:30














0












0








0





$begingroup$


So, my question is the following:



Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.






vector-spaces lie-algebras







share|cite|improve this question









$endgroup$




So, my question is the following:



Suppose that we have two Lie algebras $(mathfrak{g}_1,[bullet,bullet]_1)$ and $(mathfrak{g}_2,[bullet,bullet]_2)$. Then we can define the tensor product of these algebras, namely the Lie algebra $$(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2}).$$
The underlying vector space $mathfrak{g}_1otimesmathfrak{g}_2$ is constructed using the map $otimes:mathfrak{g}_1timesmathfrak{g}_2tomathfrak{g}_1otimesmathfrak{g}_2$ and consists of the vectors ${X_1otimes X_2|X_1inmathfrak{g}_1,X_2inmathfrak{g}_2}$. My question is on how to define the Lie bracket $[bullet,bullet]_{1otimes 2}$ correctly, so that the vector space $mathfrak{g}_1otimesmathfrak{g}_2$ becomes a Lie algebra.






vector-spaces lie-algebras




vector-spaces lie-algebras






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 12 '18 at 16:28









G KG K

1198




1198












  • $begingroup$
    Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
    $endgroup$
    – md2perpe
    Aug 12 '18 at 17:13










  • $begingroup$
    As far as I know there is no way to do this.
    $endgroup$
    – Qiaochu Yuan
    Aug 12 '18 at 17:13










  • $begingroup$
    @QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
    $endgroup$
    – G K
    Aug 12 '18 at 17:20






  • 1




    $begingroup$
    Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
    $endgroup$
    – Tobias Kildetoft
    Aug 12 '18 at 17:28










  • $begingroup$
    @TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
    $endgroup$
    – G K
    Aug 12 '18 at 17:30


















  • $begingroup$
    Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
    $endgroup$
    – md2perpe
    Aug 12 '18 at 17:13










  • $begingroup$
    As far as I know there is no way to do this.
    $endgroup$
    – Qiaochu Yuan
    Aug 12 '18 at 17:13










  • $begingroup$
    @QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
    $endgroup$
    – G K
    Aug 12 '18 at 17:20






  • 1




    $begingroup$
    Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
    $endgroup$
    – Tobias Kildetoft
    Aug 12 '18 at 17:28










  • $begingroup$
    @TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
    $endgroup$
    – G K
    Aug 12 '18 at 17:30
















$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13




$begingroup$
Have you looked at what happens when $mathfrak{g}_1$ and $mathfrak{g}_2$ consist of matrices? Perhaps you could take the result and generalize it.
$endgroup$
– md2perpe
Aug 12 '18 at 17:13












$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13




$begingroup$
As far as I know there is no way to do this.
$endgroup$
– Qiaochu Yuan
Aug 12 '18 at 17:13












$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20




$begingroup$
@QiaochuYuan So, we just suppose that $(mathfrak{g}_1otimesmathfrak{g}_2,[bullet,bullet]_{1otimes 2})$ is a Lie algebra but we cannot define the Lie bracket?
$endgroup$
– G K
Aug 12 '18 at 17:20




1




1




$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28




$begingroup$
Why should we suppose that it is a Lie algebra? Sure, if the Lie bracket really is a commutator, we can do this by just taking the usual tensor product of algebras. But why would we expect there to be some universal way to do this in general?
$endgroup$
– Tobias Kildetoft
Aug 12 '18 at 17:28












$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30




$begingroup$
@TobiasKildetoft I see, so the underlying vector space is well defined but there is no notion of a Lie algebra of that kind since we cannot in general define a Lie product. Is this correct?
$endgroup$
– G K
Aug 12 '18 at 17:30










1 Answer
1






active

oldest

votes


















0












$begingroup$

There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
$$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.






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%2f2880510%2fdefining-the-lie-bracket-on-a-tensor-product-lie-algebra%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









    0












    $begingroup$

    There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
    $$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
    For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
      $$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
      For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
        $$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
        For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.






        share|cite|improve this answer









        $endgroup$



        There is a way to define the Lie bracket on the tensor product as follows. Suppose that $mathfrak{g}_1$ and $mathfrak{g}_2$ are Lie algebras with two bilinear maps $B_1:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_1$ and $B_2:mathfrak{g}_1times mathfrak{g}_2longrightarrow mathfrak{g}_2$. Then with some compatibility conditions one can define the Lie bracket on the tensor product by
        $$ [g_1otimes g_2, g_1'otimes g_2']:= B_1(g_1,g_2)otimes B_2(g_1',g_2')quad text{for } g_1,g_1' in mathfrak{g}_1 text{ and } g_2,g_2' inmathfrak{g}_2.$$
        For example if $mathfrak{g}_1$ and $mathfrak{g}_2$ are both ideals of some Lie algebra with bilinear maps given by Lie multiplication then above definition works without any extra constraint. For details see the paper https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089500008107.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 15:55









        moutheticsmouthetics

        50137




        50137






























            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%2f2880510%2fdefining-the-lie-bracket-on-a-tensor-product-lie-algebra%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