Proving the trace of a representation is equal to zero












2














I'm having some trouble in beginner's representation theory and am pretty lost about this problem:



Let ($rho$, $V$) be a representation of $G$, so $rho$: $G$ $to$ $GL(V)$ is a group homomorphism. Let $H$$subset$$G$ be a normal subgroup of index 2. Suppose $V$ is $G$-irreducible, but not $H$-irreducible. Prove tr($rho$($g$))=0 for $g$$notin$$H$.



I thought about starting by saying $G/H$ is isomorphic to $C_2$, and so you could say that some character



$lambda$=
begin{cases}
1, & text{if $g in H$} \
-1, & text{if $g notin H$}
end{cases}



But I don't think this is very helpful, and I feel like this is not the way to go about the proof. Any help would be appreciated!






abstract-algebra group-theory representation-theory characters







share|cite|improve this question





























    2














    I'm having some trouble in beginner's representation theory and am pretty lost about this problem:



    Let ($rho$, $V$) be a representation of $G$, so $rho$: $G$ $to$ $GL(V)$ is a group homomorphism. Let $H$$subset$$G$ be a normal subgroup of index 2. Suppose $V$ is $G$-irreducible, but not $H$-irreducible. Prove tr($rho$($g$))=0 for $g$$notin$$H$.



    I thought about starting by saying $G/H$ is isomorphic to $C_2$, and so you could say that some character



    $lambda$=
    begin{cases}
    1, & text{if $g in H$} \
    -1, & text{if $g notin H$}
    end{cases}



    But I don't think this is very helpful, and I feel like this is not the way to go about the proof. Any help would be appreciated!






    abstract-algebra group-theory representation-theory characters







    share|cite|improve this question



























      2












      2








      2


      1





      I'm having some trouble in beginner's representation theory and am pretty lost about this problem:



      Let ($rho$, $V$) be a representation of $G$, so $rho$: $G$ $to$ $GL(V)$ is a group homomorphism. Let $H$$subset$$G$ be a normal subgroup of index 2. Suppose $V$ is $G$-irreducible, but not $H$-irreducible. Prove tr($rho$($g$))=0 for $g$$notin$$H$.



      I thought about starting by saying $G/H$ is isomorphic to $C_2$, and so you could say that some character



      $lambda$=
      begin{cases}
      1, & text{if $g in H$} \
      -1, & text{if $g notin H$}
      end{cases}



      But I don't think this is very helpful, and I feel like this is not the way to go about the proof. Any help would be appreciated!






      abstract-algebra group-theory representation-theory characters







      share|cite|improve this question















      I'm having some trouble in beginner's representation theory and am pretty lost about this problem:



      Let ($rho$, $V$) be a representation of $G$, so $rho$: $G$ $to$ $GL(V)$ is a group homomorphism. Let $H$$subset$$G$ be a normal subgroup of index 2. Suppose $V$ is $G$-irreducible, but not $H$-irreducible. Prove tr($rho$($g$))=0 for $g$$notin$$H$.



      I thought about starting by saying $G/H$ is isomorphic to $C_2$, and so you could say that some character



      $lambda$=
      begin{cases}
      1, & text{if $g in H$} \
      -1, & text{if $g notin H$}
      end{cases}



      But I don't think this is very helpful, and I feel like this is not the way to go about the proof. Any help would be appreciated!






      abstract-algebra group-theory representation-theory characters




      abstract-algebra group-theory representation-theory characters






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 21:07









      André 3000

      12.4k22042




      12.4k22042










      asked Dec 9 '18 at 20:08









      empmoth

      133




      133






















          2 Answers
          2






          active

          oldest

          votes


















          2














          Let $chi$ be the character of $rho$.
          From orthogonality of characters,
          $$sum_{gin G}|chi(g)|^2=|G|$$
          since $rho$ is irreducible on $G$, and
          $$sum_{gin H}|chi(g)|^2=m|H|=frac m2|G|$$
          where $mge2$ is an integer,
          since $rho$ is reducible on $H$.
          Then $|G|gefrac m2 |G|$, so $m=2$ and we have
          $$sum_{gnotin H}|chi(g)|^2=0$$
          etc.






          share|cite|improve this answer





















          • Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
            – empmoth
            Dec 9 '18 at 20:54










          • Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
            – Ted
            Dec 9 '18 at 22:55












          • You're right I can't believe I didn't see that. Thanks for the help!
            – empmoth
            Dec 10 '18 at 0:21



















          2














          Here is an alternative solution that does not directly use character theory. We are told that $H$ does not act irreducibly on $V$, so let $W$ be a nonzero subspace of $V$ of smallest dimension that is invariant under the action of $H$.



          Now choose $g in G setminus H$. The normality of $H$ in $G$ implies that $g(W)$ is also invariant under that action of $H$. Also, since $g^2 in H$, $g^2(W) = W$.



          Now $W + g(W)$ is invariant under $G$ and so, since $G$ acts irreducibly on $V$, we have $V = W + g(W)$. Also, by minimality of $W$, we must have $W cap g(W) = {0}$, so $V = W oplus g(W)$.



          Now, all elements of $G setminus H = gH$ interchange the $H$-invariant subspaces $W$ and $g(W)$ so, by choosing a basis of $V$ that consists of a union of bases of $W$ and of $g(W)$, we see that the matrices of the elements of $G setminus H$ with respect to this basis have trace $0$.






          share|cite|improve this answer





















            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%2f3032914%2fproving-the-trace-of-a-representation-is-equal-to-zero%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            Let $chi$ be the character of $rho$.
            From orthogonality of characters,
            $$sum_{gin G}|chi(g)|^2=|G|$$
            since $rho$ is irreducible on $G$, and
            $$sum_{gin H}|chi(g)|^2=m|H|=frac m2|G|$$
            where $mge2$ is an integer,
            since $rho$ is reducible on $H$.
            Then $|G|gefrac m2 |G|$, so $m=2$ and we have
            $$sum_{gnotin H}|chi(g)|^2=0$$
            etc.






            share|cite|improve this answer





















            • Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
              – empmoth
              Dec 9 '18 at 20:54










            • Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
              – Ted
              Dec 9 '18 at 22:55












            • You're right I can't believe I didn't see that. Thanks for the help!
              – empmoth
              Dec 10 '18 at 0:21
















            2














            Let $chi$ be the character of $rho$.
            From orthogonality of characters,
            $$sum_{gin G}|chi(g)|^2=|G|$$
            since $rho$ is irreducible on $G$, and
            $$sum_{gin H}|chi(g)|^2=m|H|=frac m2|G|$$
            where $mge2$ is an integer,
            since $rho$ is reducible on $H$.
            Then $|G|gefrac m2 |G|$, so $m=2$ and we have
            $$sum_{gnotin H}|chi(g)|^2=0$$
            etc.






            share|cite|improve this answer





















            • Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
              – empmoth
              Dec 9 '18 at 20:54










            • Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
              – Ted
              Dec 9 '18 at 22:55












            • You're right I can't believe I didn't see that. Thanks for the help!
              – empmoth
              Dec 10 '18 at 0:21














            2












            2








            2






            Let $chi$ be the character of $rho$.
            From orthogonality of characters,
            $$sum_{gin G}|chi(g)|^2=|G|$$
            since $rho$ is irreducible on $G$, and
            $$sum_{gin H}|chi(g)|^2=m|H|=frac m2|G|$$
            where $mge2$ is an integer,
            since $rho$ is reducible on $H$.
            Then $|G|gefrac m2 |G|$, so $m=2$ and we have
            $$sum_{gnotin H}|chi(g)|^2=0$$
            etc.






            share|cite|improve this answer












            Let $chi$ be the character of $rho$.
            From orthogonality of characters,
            $$sum_{gin G}|chi(g)|^2=|G|$$
            since $rho$ is irreducible on $G$, and
            $$sum_{gin H}|chi(g)|^2=m|H|=frac m2|G|$$
            where $mge2$ is an integer,
            since $rho$ is reducible on $H$.
            Then $|G|gefrac m2 |G|$, so $m=2$ and we have
            $$sum_{gnotin H}|chi(g)|^2=0$$
            etc.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 20:27









            Lord Shark the Unknown

            101k958132




            101k958132












            • Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
              – empmoth
              Dec 9 '18 at 20:54










            • Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
              – Ted
              Dec 9 '18 at 22:55












            • You're right I can't believe I didn't see that. Thanks for the help!
              – empmoth
              Dec 10 '18 at 0:21


















            • Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
              – empmoth
              Dec 9 '18 at 20:54










            • Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
              – Ted
              Dec 9 '18 at 22:55












            • You're right I can't believe I didn't see that. Thanks for the help!
              – empmoth
              Dec 10 '18 at 0:21
















            Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
            – empmoth
            Dec 9 '18 at 20:54




            Thanks for the answer! Sorry if this is a dumb question, but can you explain a tiny bit the conclusion? Why does m=2 line mean that $sum_{gnotin H}$$lvert$$chi$($g$)$rvert$^2 = 0?
            – empmoth
            Dec 9 '18 at 20:54












            Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
            – Ted
            Dec 9 '18 at 22:55






            Because $sum |chi(g)|^2$ over all $g in G$ is $|G|$ by the orthogonality relations, so if the sum over $g in H$ is $|G|$ then the sum over $g not in H$ must be 0.
            – Ted
            Dec 9 '18 at 22:55














            You're right I can't believe I didn't see that. Thanks for the help!
            – empmoth
            Dec 10 '18 at 0:21




            You're right I can't believe I didn't see that. Thanks for the help!
            – empmoth
            Dec 10 '18 at 0:21











            2














            Here is an alternative solution that does not directly use character theory. We are told that $H$ does not act irreducibly on $V$, so let $W$ be a nonzero subspace of $V$ of smallest dimension that is invariant under the action of $H$.



            Now choose $g in G setminus H$. The normality of $H$ in $G$ implies that $g(W)$ is also invariant under that action of $H$. Also, since $g^2 in H$, $g^2(W) = W$.



            Now $W + g(W)$ is invariant under $G$ and so, since $G$ acts irreducibly on $V$, we have $V = W + g(W)$. Also, by minimality of $W$, we must have $W cap g(W) = {0}$, so $V = W oplus g(W)$.



            Now, all elements of $G setminus H = gH$ interchange the $H$-invariant subspaces $W$ and $g(W)$ so, by choosing a basis of $V$ that consists of a union of bases of $W$ and of $g(W)$, we see that the matrices of the elements of $G setminus H$ with respect to this basis have trace $0$.






            share|cite|improve this answer


























              2














              Here is an alternative solution that does not directly use character theory. We are told that $H$ does not act irreducibly on $V$, so let $W$ be a nonzero subspace of $V$ of smallest dimension that is invariant under the action of $H$.



              Now choose $g in G setminus H$. The normality of $H$ in $G$ implies that $g(W)$ is also invariant under that action of $H$. Also, since $g^2 in H$, $g^2(W) = W$.



              Now $W + g(W)$ is invariant under $G$ and so, since $G$ acts irreducibly on $V$, we have $V = W + g(W)$. Also, by minimality of $W$, we must have $W cap g(W) = {0}$, so $V = W oplus g(W)$.



              Now, all elements of $G setminus H = gH$ interchange the $H$-invariant subspaces $W$ and $g(W)$ so, by choosing a basis of $V$ that consists of a union of bases of $W$ and of $g(W)$, we see that the matrices of the elements of $G setminus H$ with respect to this basis have trace $0$.






              share|cite|improve this answer
























                2












                2








                2






                Here is an alternative solution that does not directly use character theory. We are told that $H$ does not act irreducibly on $V$, so let $W$ be a nonzero subspace of $V$ of smallest dimension that is invariant under the action of $H$.



                Now choose $g in G setminus H$. The normality of $H$ in $G$ implies that $g(W)$ is also invariant under that action of $H$. Also, since $g^2 in H$, $g^2(W) = W$.



                Now $W + g(W)$ is invariant under $G$ and so, since $G$ acts irreducibly on $V$, we have $V = W + g(W)$. Also, by minimality of $W$, we must have $W cap g(W) = {0}$, so $V = W oplus g(W)$.



                Now, all elements of $G setminus H = gH$ interchange the $H$-invariant subspaces $W$ and $g(W)$ so, by choosing a basis of $V$ that consists of a union of bases of $W$ and of $g(W)$, we see that the matrices of the elements of $G setminus H$ with respect to this basis have trace $0$.






                share|cite|improve this answer












                Here is an alternative solution that does not directly use character theory. We are told that $H$ does not act irreducibly on $V$, so let $W$ be a nonzero subspace of $V$ of smallest dimension that is invariant under the action of $H$.



                Now choose $g in G setminus H$. The normality of $H$ in $G$ implies that $g(W)$ is also invariant under that action of $H$. Also, since $g^2 in H$, $g^2(W) = W$.



                Now $W + g(W)$ is invariant under $G$ and so, since $G$ acts irreducibly on $V$, we have $V = W + g(W)$. Also, by minimality of $W$, we must have $W cap g(W) = {0}$, so $V = W oplus g(W)$.



                Now, all elements of $G setminus H = gH$ interchange the $H$-invariant subspaces $W$ and $g(W)$ so, by choosing a basis of $V$ that consists of a union of bases of $W$ and of $g(W)$, we see that the matrices of the elements of $G setminus H$ with respect to this basis have trace $0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 '18 at 21:15









                Derek Holt

                52.6k53570




                52.6k53570






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f3032914%2fproving-the-trace-of-a-representation-is-equal-to-zero%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