Proving the trace of a representation is equal to zero
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
edited Dec 9 '18 at 21:07
André 3000
12.6k22243
12.6k22243
asked Dec 9 '18 at 20:08
empmothempmoth
133
133
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
add a comment |
$begingroup$
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 $0$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
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.
$endgroup$
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Dec 9 '18 at 20:27
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
add a comment |
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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?
$endgroup$
– empmoth
Dec 9 '18 at 20:54
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
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.
$endgroup$
– Ted
Dec 9 '18 at 22:55
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
$begingroup$
You're right I can't believe I didn't see that. Thanks for the help!
$endgroup$
– empmoth
Dec 10 '18 at 0:21
add a comment |
$begingroup$
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 $0$.
$endgroup$
add a comment |
$begingroup$
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 $0$.
$endgroup$
add a comment |
$begingroup$
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 $0$.
$endgroup$
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 $0$.
answered Dec 9 '18 at 21:15
Derek HoltDerek Holt
53.4k53571
53.4k53571
add a comment |
add a comment |
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