Any quotient representation of completely reducible is completely reducible. [closed]
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Prove that every quotient representation of a completely reducible representation is completely reducible.
Could anyone give me a hint for this?
representation-theory invariant-theory
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
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closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill Jan 23 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
Prove that every quotient representation of a completely reducible representation is completely reducible.
Could anyone give me a hint for this?
representation-theory invariant-theory
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
$endgroup$
closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill Jan 23 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that every quotient representation of a completely reducible representation is completely reducible.
Could anyone give me a hint for this?
representation-theory invariant-theory
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
$endgroup$
Prove that every quotient representation of a completely reducible representation is completely reducible.
Could anyone give me a hint for this?
representation-theory invariant-theory
representation-theory invariant-theory
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
asked Jan 23 at 11:44
IntuitionIntuition
1,089825
1,089825
closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill Jan 23 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill Jan 23 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Adrian Keister, Lee David Chung Lin, Davide Giraudo, David Hill
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
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votes
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Hint : if $M$ is completely reducible, $N$ is a subrepresentation, and $K$ a subrepresentation of $M/N$, we wish to find a subrepresentation $S$ of $M/N$ such that $Koplus S = M/N$.
If $pi: Mto M/N$ is the projection map, what can you say about $pi^{-1}(K)$ ? What does the hypothesis on $M$ tell us ?
answered Jan 23 at 12:12
MaxMax
15k11143
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How you will show that every invariant subspace has an invariant complement ?
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– Intuition
Jan 23 at 21:38
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For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
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– Intuition
Jan 23 at 21:41
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$pi^{-1}(K)$ is an invariant subspace, can you see why ?
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– Max
Jan 23 at 21:43
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No I can not Max
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– Intuition
Jan 23 at 22:12
1
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You were talking about a representation... ?
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– Max
Jan 24 at 10:36
|
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint : if $M$ is completely reducible, $N$ is a subrepresentation, and $K$ a subrepresentation of $M/N$, we wish to find a subrepresentation $S$ of $M/N$ such that $Koplus S = M/N$.
If $pi: Mto M/N$ is the projection map, what can you say about $pi^{-1}(K)$ ? What does the hypothesis on $M$ tell us ?
answered Jan 23 at 12:12
MaxMax
15k11143
$endgroup$
$begingroup$
How you will show that every invariant subspace has an invariant complement ?
$endgroup$
– Intuition
Jan 23 at 21:38
$begingroup$
For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
$endgroup$
– Intuition
Jan 23 at 21:41
$begingroup$
$pi^{-1}(K)$ is an invariant subspace, can you see why ?
$endgroup$
– Max
Jan 23 at 21:43
$begingroup$
No I can not Max
$endgroup$
– Intuition
Jan 23 at 22:12
1
$begingroup$
You were talking about a representation... ?
$endgroup$
– Max
Jan 24 at 10:36
|
show 5 more comments
$begingroup$
Hint : if $M$ is completely reducible, $N$ is a subrepresentation, and $K$ a subrepresentation of $M/N$, we wish to find a subrepresentation $S$ of $M/N$ such that $Koplus S = M/N$.
If $pi: Mto M/N$ is the projection map, what can you say about $pi^{-1}(K)$ ? What does the hypothesis on $M$ tell us ?
answered Jan 23 at 12:12
MaxMax
15k11143
$endgroup$
$begingroup$
How you will show that every invariant subspace has an invariant complement ?
$endgroup$
– Intuition
Jan 23 at 21:38
$begingroup$
For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
$endgroup$
– Intuition
Jan 23 at 21:41
$begingroup$
$pi^{-1}(K)$ is an invariant subspace, can you see why ?
$endgroup$
– Max
Jan 23 at 21:43
$begingroup$
No I can not Max
$endgroup$
– Intuition
Jan 23 at 22:12
1
$begingroup$
You were talking about a representation... ?
$endgroup$
– Max
Jan 24 at 10:36
|
show 5 more comments
$begingroup$
Hint : if $M$ is completely reducible, $N$ is a subrepresentation, and $K$ a subrepresentation of $M/N$, we wish to find a subrepresentation $S$ of $M/N$ such that $Koplus S = M/N$.
If $pi: Mto M/N$ is the projection map, what can you say about $pi^{-1}(K)$ ? What does the hypothesis on $M$ tell us ?
answered Jan 23 at 12:12
MaxMax
15k11143
$endgroup$
Hint : if $M$ is completely reducible, $N$ is a subrepresentation, and $K$ a subrepresentation of $M/N$, we wish to find a subrepresentation $S$ of $M/N$ such that $Koplus S = M/N$.
If $pi: Mto M/N$ is the projection map, what can you say about $pi^{-1}(K)$ ? What does the hypothesis on $M$ tell us ?
answered Jan 23 at 12:12
MaxMax
15k11143
answered Jan 23 at 12:12
MaxMax
15k11143
answered Jan 23 at 12:12
MaxMax
15k11143
answered Jan 23 at 12:12
MaxMax
15k11143
15k11143
$begingroup$
How you will show that every invariant subspace has an invariant complement ?
$endgroup$
– Intuition
Jan 23 at 21:38
$begingroup$
For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
$endgroup$
– Intuition
Jan 23 at 21:41
$begingroup$
$pi^{-1}(K)$ is an invariant subspace, can you see why ?
$endgroup$
– Max
Jan 23 at 21:43
$begingroup$
No I can not Max
$endgroup$
– Intuition
Jan 23 at 22:12
1
$begingroup$
You were talking about a representation... ?
$endgroup$
– Max
Jan 24 at 10:36
|
show 5 more comments
$begingroup$
How you will show that every invariant subspace has an invariant complement ?
$endgroup$
– Intuition
Jan 23 at 21:38
$begingroup$
For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
$endgroup$
– Intuition
Jan 23 at 21:41
$begingroup$
$pi^{-1}(K)$ is an invariant subspace, can you see why ?
$endgroup$
– Max
Jan 23 at 21:43
$begingroup$
No I can not Max
$endgroup$
– Intuition
Jan 23 at 22:12
1
$begingroup$
You were talking about a representation... ?
$endgroup$
– Max
Jan 24 at 10:36
$begingroup$
How you will show that every invariant subspace has an invariant complement ?
$endgroup$
– Intuition
Jan 23 at 21:38
$begingroup$
How you will show that every invariant subspace has an invariant complement ?
$endgroup$
– Intuition
Jan 23 at 21:38
$begingroup$
For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
$endgroup$
– Intuition
Jan 23 at 21:41
$begingroup$
For the first question I do not know the answer .... For the second question the hypothesis tell us that every invariant subspace has an invariant complement .... but then what Max ?
$endgroup$
– Intuition
Jan 23 at 21:41
$begingroup$
$pi^{-1}(K)$ is an invariant subspace, can you see why ?
$endgroup$
– Max
Jan 23 at 21:43
$begingroup$
$pi^{-1}(K)$ is an invariant subspace, can you see why ?
$endgroup$
– Max
Jan 23 at 21:43
$begingroup$
No I can not Max
$endgroup$
– Intuition
Jan 23 at 22:12
$begingroup$
No I can not Max
$endgroup$
– Intuition
Jan 23 at 22:12
1
1
$begingroup$
You were talking about a representation... ?
$endgroup$
– Max
Jan 24 at 10:36
$begingroup$
You were talking about a representation... ?
$endgroup$
– Max
Jan 24 at 10:36
|
show 5 more comments
