irreducible contragredient representation
$begingroup$
I am trying to understand the following statement:
Suppose $rho: G longrightarrow text{GL}_mathbb{C}(V)$ an irreducible representation of a finite group $G$. Then its contragredient representation $rho ^*: G longrightarrow text{GL}_mathbb{C}(V^*)$ is also an irreducible representation.
I have seen a site where they prove it using this:
Suppose $U$ is a $G$-subspace of $V^*$, i.e. $rho^*(g)(U) subseteq U$ for all $g in G$. Define $$W := {v in V | f(v)=0, forall f in U},.$$ Then, $W$ is a $G$-subspace of $V$.
PROOF:
We have $f rho(g)=rho^*(g^{-1})(f) =0$ over $W$ for all $f in U$, because $rho^*(g^{-1})(U) subseteq U$. So $rho(g)(W) subseteq W$.
Can anyone help me understanding this proof?
Thank you very much. Let me know any improvement of the post that can be done.
group-theory finite-groups representation-theory proof-explanation dual-spaces
edited Dec 11 '18 at 16:08
Batominovski
1
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
$endgroup$
|
show 2 more comments
$begingroup$
I am trying to understand the following statement:
Suppose $rho: G longrightarrow text{GL}_mathbb{C}(V)$ an irreducible representation of a finite group $G$. Then its contragredient representation $rho ^*: G longrightarrow text{GL}_mathbb{C}(V^*)$ is also an irreducible representation.
I have seen a site where they prove it using this:
Suppose $U$ is a $G$-subspace of $V^*$, i.e. $rho^*(g)(U) subseteq U$ for all $g in G$. Define $$W := {v in V | f(v)=0, forall f in U},.$$ Then, $W$ is a $G$-subspace of $V$.
PROOF:
We have $f rho(g)=rho^*(g^{-1})(f) =0$ over $W$ for all $f in U$, because $rho^*(g^{-1})(U) subseteq U$. So $rho(g)(W) subseteq W$.
Can anyone help me understanding this proof?
Thank you very much. Let me know any improvement of the post that can be done.
group-theory finite-groups representation-theory proof-explanation dual-spaces
edited Dec 11 '18 at 16:08
Batominovski
1
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
$endgroup$
1
$begingroup$
Which part is the problem? The proof is very short and direct, so it is hard to see which part could be an issue.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:52
$begingroup$
That is the problem for me. I do not see anything in the proof. If anyone can see anything... Idk.
$endgroup$
– idriskameni
Dec 10 '18 at 13:54
1
$begingroup$
What do you mean? The proof precisely includes those things needed for the statement, so I don't see how you don't see anything in the proof.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:56
$begingroup$
Could you rewrite it and explain it step by step?
$endgroup$
– idriskameni
Dec 10 '18 at 13:57
1
$begingroup$
Not unless you can pinpoint which part is an issue. The proof is already very much step by step.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:58
|
show 2 more comments
$begingroup$
I am trying to understand the following statement:
Suppose $rho: G longrightarrow text{GL}_mathbb{C}(V)$ an irreducible representation of a finite group $G$. Then its contragredient representation $rho ^*: G longrightarrow text{GL}_mathbb{C}(V^*)$ is also an irreducible representation.
I have seen a site where they prove it using this:
Suppose $U$ is a $G$-subspace of $V^*$, i.e. $rho^*(g)(U) subseteq U$ for all $g in G$. Define $$W := {v in V | f(v)=0, forall f in U},.$$ Then, $W$ is a $G$-subspace of $V$.
PROOF:
We have $f rho(g)=rho^*(g^{-1})(f) =0$ over $W$ for all $f in U$, because $rho^*(g^{-1})(U) subseteq U$. So $rho(g)(W) subseteq W$.
Can anyone help me understanding this proof?
Thank you very much. Let me know any improvement of the post that can be done.
group-theory finite-groups representation-theory proof-explanation dual-spaces
edited Dec 11 '18 at 16:08
Batominovski
1
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
$endgroup$
I am trying to understand the following statement:
Suppose $rho: G longrightarrow text{GL}_mathbb{C}(V)$ an irreducible representation of a finite group $G$. Then its contragredient representation $rho ^*: G longrightarrow text{GL}_mathbb{C}(V^*)$ is also an irreducible representation.
I have seen a site where they prove it using this:
Suppose $U$ is a $G$-subspace of $V^*$, i.e. $rho^*(g)(U) subseteq U$ for all $g in G$. Define $$W := {v in V | f(v)=0, forall f in U},.$$ Then, $W$ is a $G$-subspace of $V$.
PROOF:
We have $f rho(g)=rho^*(g^{-1})(f) =0$ over $W$ for all $f in U$, because $rho^*(g^{-1})(U) subseteq U$. So $rho(g)(W) subseteq W$.
Can anyone help me understanding this proof?
Thank you very much. Let me know any improvement of the post that can be done.
group-theory finite-groups representation-theory proof-explanation dual-spaces
group-theory finite-groups representation-theory proof-explanation dual-spaces
edited Dec 11 '18 at 16:08
Batominovski
1
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
edited Dec 11 '18 at 16:08
Batominovski
1
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
edited Dec 11 '18 at 16:08
Batominovski
1
edited Dec 11 '18 at 16:08
Batominovski
1
edited Dec 11 '18 at 16:08
Batominovski
1
1
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
asked Dec 10 '18 at 13:11
idriskameniidriskameni
568318
568318
1
$begingroup$
Which part is the problem? The proof is very short and direct, so it is hard to see which part could be an issue.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:52
$begingroup$
That is the problem for me. I do not see anything in the proof. If anyone can see anything... Idk.
$endgroup$
– idriskameni
Dec 10 '18 at 13:54
1
$begingroup$
What do you mean? The proof precisely includes those things needed for the statement, so I don't see how you don't see anything in the proof.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:56
$begingroup$
Could you rewrite it and explain it step by step?
$endgroup$
– idriskameni
Dec 10 '18 at 13:57
1
$begingroup$
Not unless you can pinpoint which part is an issue. The proof is already very much step by step.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:58
|
show 2 more comments
1
$begingroup$
Which part is the problem? The proof is very short and direct, so it is hard to see which part could be an issue.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:52
$begingroup$
That is the problem for me. I do not see anything in the proof. If anyone can see anything... Idk.
$endgroup$
– idriskameni
Dec 10 '18 at 13:54
1
$begingroup$
What do you mean? The proof precisely includes those things needed for the statement, so I don't see how you don't see anything in the proof.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:56
$begingroup$
Could you rewrite it and explain it step by step?
$endgroup$
– idriskameni
Dec 10 '18 at 13:57
1
$begingroup$
Not unless you can pinpoint which part is an issue. The proof is already very much step by step.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:58
1
1
$begingroup$
Which part is the problem? The proof is very short and direct, so it is hard to see which part could be an issue.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:52
$begingroup$
Which part is the problem? The proof is very short and direct, so it is hard to see which part could be an issue.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:52
$begingroup$
That is the problem for me. I do not see anything in the proof. If anyone can see anything... Idk.
$endgroup$
– idriskameni
Dec 10 '18 at 13:54
$begingroup$
That is the problem for me. I do not see anything in the proof. If anyone can see anything... Idk.
$endgroup$
– idriskameni
Dec 10 '18 at 13:54
1
1
$begingroup$
What do you mean? The proof precisely includes those things needed for the statement, so I don't see how you don't see anything in the proof.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:56
$begingroup$
What do you mean? The proof precisely includes those things needed for the statement, so I don't see how you don't see anything in the proof.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:56
$begingroup$
Could you rewrite it and explain it step by step?
$endgroup$
– idriskameni
Dec 10 '18 at 13:57
$begingroup$
Could you rewrite it and explain it step by step?
$endgroup$
– idriskameni
Dec 10 '18 at 13:57
1
1
$begingroup$
Not unless you can pinpoint which part is an issue. The proof is already very much step by step.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:58
$begingroup$
Not unless you can pinpoint which part is an issue. The proof is already very much step by step.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:58
|
show 2 more comments
0
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1
$begingroup$
Which part is the problem? The proof is very short and direct, so it is hard to see which part could be an issue.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:52
$begingroup$
That is the problem for me. I do not see anything in the proof. If anyone can see anything... Idk.
$endgroup$
– idriskameni
Dec 10 '18 at 13:54
1
$begingroup$
What do you mean? The proof precisely includes those things needed for the statement, so I don't see how you don't see anything in the proof.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:56
$begingroup$
Could you rewrite it and explain it step by step?
$endgroup$
– idriskameni
Dec 10 '18 at 13:57
1
$begingroup$
Not unless you can pinpoint which part is an issue. The proof is already very much step by step.
$endgroup$
– Tobias Kildetoft
Dec 10 '18 at 13:58