A question about special linear group












2












$begingroup$


Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    3 hours ago
















2












$begingroup$


Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    3 hours ago














2












2








2





$begingroup$


Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$




Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices




linear-algebra matrices orthogonal-matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









TotoroTotoro

1417




1417








  • 1




    $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    3 hours ago














  • 1




    $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    3 hours ago








1




1




$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
3 hours ago




$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
3 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

These are exactly the finite-order elements:




  1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


  2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







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: "504"
    };
    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%2fmathoverflow.net%2fquestions%2f322542%2fa-question-about-special-linear-group%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









    3












    $begingroup$

    These are exactly the finite-order elements:




    1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


    2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      These are exactly the finite-order elements:




      1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


      2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        These are exactly the finite-order elements:




        1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


        2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







        share|cite|improve this answer









        $endgroup$



        These are exactly the finite-order elements:




        1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


        2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Andy PutmanAndy Putman

        31.6k7134214




        31.6k7134214






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to MathOverflow!


            • 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%2fmathoverflow.net%2fquestions%2f322542%2fa-question-about-special-linear-group%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

            Fluorita

            Image
            Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
            Read more

            Hulsita

            Image
            Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
            Read more

            Península de Txukotka

            Image
            Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
            Read more