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
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          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





















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            2 Answers
            2






            active

            oldest

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            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






























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