Unique faithful $7$-dimensional representation of semisimple Lie Algebra with $G_2$ root system












2












$begingroup$


I am asked to show that if $mathfrak{g}$ is a semisimple Lie Algebra with root system of type $G_2$, then it has a unique, $7$-dimensional faithful representation.



To start, let $omega_1, omega_2$ be the fundamental weights of the root system $G_2$ and suppose that $lambda = m_1omega_1 + m_2omega_2$ is a dominant weight. Then by using the Weyl Dimension Formula, we see that:



$dim(V(lambda))$ = $frac{1}{120}(m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1 + 3m_2+5)$



Using this formula we see that $V(omega_1)$ is the only irreducible representation of $mathfrak{g}$ and there are no irreducible representations of dimensions $2,3,4,5$ or $6$. This in turn means there are no representations of dimensions $2,3,4,5$ or $6$ not of the form $V(0) ^{otimes n}$



Finally, we note $V(0)^{otimes 7}$ is not faithful, and so we must be required to show that the representation $V(omega_1)$ is faithful.



It is here that I am stuck since I do not know how to do this.



My immediate thought is that if we assume we do not have injectivity then we can find a subrepresentation, though I am unsure of how to do this, or how to use the fact that the representation is specifically $V(omega_1)$ and not just any $7$-dimensional vector space. Below is my attempt



Let $phi$ be the representation $phi : mathfrak{g} rightarrow mathfrak{gl}(V(omega_1))$ and suppose that for some $x neq y, x,y in mathfrak{g}$ we have that $phi(x) = phi(y)$



If $x = ky$ for some $kin mathbb C$ then by linearity of $phi$, we have $phi(x) = 0$



Let $v_1, dots, v_n$ be a basis for $mathfrak{g}$ with $v_1 = x$. Then as $phi$ is a homomorphism, we must have that $dim(phi(mathfrak{g})) < dim(mathfrak{g})$.



[I am unable to pursue this approach any further].



In the case that $x,y$ are linearly independent, then I have even fewer ideas.



I think ultimately, treating $mathfrak{gl}(V(omega_1))$ as a vector space of matrices, then I would like to show that in $ phi(mathfrak{g})$ there is an entire row or column that is not in use. Therefore, we may ignore the corresponding basis vector(s) of $V(omega_1)$ to get a subrepresentation and thus a contradiction.



However, I am unsure of how to achieve this and would appreciate any help.






representation-theory lie-algebras root-systems







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial
    $endgroup$
    – user8268
    Jan 9 at 20:14












  • $begingroup$
    @user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $Phi$ is irreducible which implies $mathfrak{g}$ is simple.
    $endgroup$
    – user366818
    Jan 9 at 20:48
















2












$begingroup$


I am asked to show that if $mathfrak{g}$ is a semisimple Lie Algebra with root system of type $G_2$, then it has a unique, $7$-dimensional faithful representation.



To start, let $omega_1, omega_2$ be the fundamental weights of the root system $G_2$ and suppose that $lambda = m_1omega_1 + m_2omega_2$ is a dominant weight. Then by using the Weyl Dimension Formula, we see that:



$dim(V(lambda))$ = $frac{1}{120}(m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1 + 3m_2+5)$



Using this formula we see that $V(omega_1)$ is the only irreducible representation of $mathfrak{g}$ and there are no irreducible representations of dimensions $2,3,4,5$ or $6$. This in turn means there are no representations of dimensions $2,3,4,5$ or $6$ not of the form $V(0) ^{otimes n}$



Finally, we note $V(0)^{otimes 7}$ is not faithful, and so we must be required to show that the representation $V(omega_1)$ is faithful.



It is here that I am stuck since I do not know how to do this.



My immediate thought is that if we assume we do not have injectivity then we can find a subrepresentation, though I am unsure of how to do this, or how to use the fact that the representation is specifically $V(omega_1)$ and not just any $7$-dimensional vector space. Below is my attempt



Let $phi$ be the representation $phi : mathfrak{g} rightarrow mathfrak{gl}(V(omega_1))$ and suppose that for some $x neq y, x,y in mathfrak{g}$ we have that $phi(x) = phi(y)$



If $x = ky$ for some $kin mathbb C$ then by linearity of $phi$, we have $phi(x) = 0$



Let $v_1, dots, v_n$ be a basis for $mathfrak{g}$ with $v_1 = x$. Then as $phi$ is a homomorphism, we must have that $dim(phi(mathfrak{g})) < dim(mathfrak{g})$.



[I am unable to pursue this approach any further].



In the case that $x,y$ are linearly independent, then I have even fewer ideas.



I think ultimately, treating $mathfrak{gl}(V(omega_1))$ as a vector space of matrices, then I would like to show that in $ phi(mathfrak{g})$ there is an entire row or column that is not in use. Therefore, we may ignore the corresponding basis vector(s) of $V(omega_1)$ to get a subrepresentation and thus a contradiction.



However, I am unsure of how to achieve this and would appreciate any help.






representation-theory lie-algebras root-systems







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial
    $endgroup$
    – user8268
    Jan 9 at 20:14












  • $begingroup$
    @user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $Phi$ is irreducible which implies $mathfrak{g}$ is simple.
    $endgroup$
    – user366818
    Jan 9 at 20:48














2












2








2





$begingroup$


I am asked to show that if $mathfrak{g}$ is a semisimple Lie Algebra with root system of type $G_2$, then it has a unique, $7$-dimensional faithful representation.



To start, let $omega_1, omega_2$ be the fundamental weights of the root system $G_2$ and suppose that $lambda = m_1omega_1 + m_2omega_2$ is a dominant weight. Then by using the Weyl Dimension Formula, we see that:



$dim(V(lambda))$ = $frac{1}{120}(m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1 + 3m_2+5)$



Using this formula we see that $V(omega_1)$ is the only irreducible representation of $mathfrak{g}$ and there are no irreducible representations of dimensions $2,3,4,5$ or $6$. This in turn means there are no representations of dimensions $2,3,4,5$ or $6$ not of the form $V(0) ^{otimes n}$



Finally, we note $V(0)^{otimes 7}$ is not faithful, and so we must be required to show that the representation $V(omega_1)$ is faithful.



It is here that I am stuck since I do not know how to do this.



My immediate thought is that if we assume we do not have injectivity then we can find a subrepresentation, though I am unsure of how to do this, or how to use the fact that the representation is specifically $V(omega_1)$ and not just any $7$-dimensional vector space. Below is my attempt



Let $phi$ be the representation $phi : mathfrak{g} rightarrow mathfrak{gl}(V(omega_1))$ and suppose that for some $x neq y, x,y in mathfrak{g}$ we have that $phi(x) = phi(y)$



If $x = ky$ for some $kin mathbb C$ then by linearity of $phi$, we have $phi(x) = 0$



Let $v_1, dots, v_n$ be a basis for $mathfrak{g}$ with $v_1 = x$. Then as $phi$ is a homomorphism, we must have that $dim(phi(mathfrak{g})) < dim(mathfrak{g})$.



[I am unable to pursue this approach any further].



In the case that $x,y$ are linearly independent, then I have even fewer ideas.



I think ultimately, treating $mathfrak{gl}(V(omega_1))$ as a vector space of matrices, then I would like to show that in $ phi(mathfrak{g})$ there is an entire row or column that is not in use. Therefore, we may ignore the corresponding basis vector(s) of $V(omega_1)$ to get a subrepresentation and thus a contradiction.



However, I am unsure of how to achieve this and would appreciate any help.






representation-theory lie-algebras root-systems







share|cite|improve this question











$endgroup$




I am asked to show that if $mathfrak{g}$ is a semisimple Lie Algebra with root system of type $G_2$, then it has a unique, $7$-dimensional faithful representation.



To start, let $omega_1, omega_2$ be the fundamental weights of the root system $G_2$ and suppose that $lambda = m_1omega_1 + m_2omega_2$ is a dominant weight. Then by using the Weyl Dimension Formula, we see that:



$dim(V(lambda))$ = $frac{1}{120}(m_1+1)(m_2+1)(m_1+m_2+2)(m_1+2m_2+3)(m_1+3m_2+4)(2m_1 + 3m_2+5)$



Using this formula we see that $V(omega_1)$ is the only irreducible representation of $mathfrak{g}$ and there are no irreducible representations of dimensions $2,3,4,5$ or $6$. This in turn means there are no representations of dimensions $2,3,4,5$ or $6$ not of the form $V(0) ^{otimes n}$



Finally, we note $V(0)^{otimes 7}$ is not faithful, and so we must be required to show that the representation $V(omega_1)$ is faithful.



It is here that I am stuck since I do not know how to do this.



My immediate thought is that if we assume we do not have injectivity then we can find a subrepresentation, though I am unsure of how to do this, or how to use the fact that the representation is specifically $V(omega_1)$ and not just any $7$-dimensional vector space. Below is my attempt



Let $phi$ be the representation $phi : mathfrak{g} rightarrow mathfrak{gl}(V(omega_1))$ and suppose that for some $x neq y, x,y in mathfrak{g}$ we have that $phi(x) = phi(y)$



If $x = ky$ for some $kin mathbb C$ then by linearity of $phi$, we have $phi(x) = 0$



Let $v_1, dots, v_n$ be a basis for $mathfrak{g}$ with $v_1 = x$. Then as $phi$ is a homomorphism, we must have that $dim(phi(mathfrak{g})) < dim(mathfrak{g})$.



[I am unable to pursue this approach any further].



In the case that $x,y$ are linearly independent, then I have even fewer ideas.



I think ultimately, treating $mathfrak{gl}(V(omega_1))$ as a vector space of matrices, then I would like to show that in $ phi(mathfrak{g})$ there is an entire row or column that is not in use. Therefore, we may ignore the corresponding basis vector(s) of $V(omega_1)$ to get a subrepresentation and thus a contradiction.



However, I am unsure of how to achieve this and would appreciate any help.






representation-theory lie-algebras root-systems




representation-theory lie-algebras root-systems






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 19:20







user366818

















asked Jan 9 at 19:05









user366818user366818

1,000410




1,000410








  • 2




    $begingroup$
    Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial
    $endgroup$
    – user8268
    Jan 9 at 20:14












  • $begingroup$
    @user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $Phi$ is irreducible which implies $mathfrak{g}$ is simple.
    $endgroup$
    – user366818
    Jan 9 at 20:48














  • 2




    $begingroup$
    Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial
    $endgroup$
    – user8268
    Jan 9 at 20:14












  • $begingroup$
    @user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $Phi$ is irreducible which implies $mathfrak{g}$ is simple.
    $endgroup$
    – user366818
    Jan 9 at 20:48








2




2




$begingroup$
Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial
$endgroup$
– user8268
Jan 9 at 20:14






$begingroup$
Just use the fact that your Lie algebra is simple, so the kernel of your representation is either zero and thus the representation is faithful, or the kernel is everything and the representation is thus trivial
$endgroup$
– user8268
Jan 9 at 20:14














$begingroup$
@user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $Phi$ is irreducible which implies $mathfrak{g}$ is simple.
$endgroup$
– user366818
Jan 9 at 20:48




$begingroup$
@user8268 Thank you! I seem to have over complicated the issue and forgotten then $G_2$ connected implies $Phi$ is irreducible which implies $mathfrak{g}$ is simple.
$endgroup$
– user366818
Jan 9 at 20:48










1 Answer
1






active

oldest

votes


















2












$begingroup$

It is clear that the simple module of smallest dimension, different from the trivial one, is $L(omega_1)$, which has dimension $7$. Since ${mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.






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%2f3067834%2funique-faithful-7-dimensional-representation-of-semisimple-lie-algebra-with-g%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









    2












    $begingroup$

    It is clear that the simple module of smallest dimension, different from the trivial one, is $L(omega_1)$, which has dimension $7$. Since ${mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      It is clear that the simple module of smallest dimension, different from the trivial one, is $L(omega_1)$, which has dimension $7$. Since ${mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It is clear that the simple module of smallest dimension, different from the trivial one, is $L(omega_1)$, which has dimension $7$. Since ${mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.






        share|cite|improve this answer









        $endgroup$



        It is clear that the simple module of smallest dimension, different from the trivial one, is $L(omega_1)$, which has dimension $7$. Since ${mathfrak g}_2$ is a simple Lie algebra, the module is faithful. Every ${mathfrak g}_2$-module is semisimple, i.e., direct sum of simple modules.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 9 at 20:48









        Dietrich BurdeDietrich Burde

        79.9k647103




        79.9k647103






























            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%2f3067834%2funique-faithful-7-dimensional-representation-of-semisimple-lie-algebra-with-g%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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
            Read more