Mixing
decompositions of a representation
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I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
asked Jan 22 at 15:28
A.EA.E
2249
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add a comment |
$begingroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
asked Jan 22 at 15:28
A.EA.E
2249
$endgroup$
add a comment |
$begingroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
asked Jan 22 at 15:28
A.EA.E
2249
$endgroup$
I am reading J.P Serre's book on Linear representations of finite groups. In chapter 2.6 it states:
Let $rho: G rightarrow GL(V)$ be a linear representation of $G$. We are going to define a direct sum decomposition of $V$ which is "coarser" than the decomposition into irreducible representations, but which has the advantage of being $unique$.
Now what does it mean by "coarser" in this context? I believe it means the former is less powerful than the latter, correct?
representation-theory
representation-theory
asked Jan 22 at 15:28
A.EA.E
2249
asked Jan 22 at 15:28
A.EA.E
2249
asked Jan 22 at 15:28
A.EA.E
2249
asked Jan 22 at 15:28
A.EA.E
2249
asked Jan 22 at 15:28
A.EA.E
2249
2249
add a comment |
add a comment |
1 Answer
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It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
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$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
$begingroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
It means that each piece of the former decomposition (that is, the decomposition into irreducible representations) will be part of some piece of the new decomposition.
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:32
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
but this decomposition will not be unique, correct?
$endgroup$
– A.E
Jan 22 at 15:34
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
I can't answer that question, since I don't know which is the specific decomposition defined by Serre.
$endgroup$
– José Carlos Santos
Jan 22 at 15:36
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
ok. is it okay to say "less powerful" instead of "coarser"
$endgroup$
– A.E
Jan 22 at 15:38
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
I'd think "less powerful" is a terrible term.
$endgroup$
– kimchi lover
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
$begingroup$
@kimchilover I agree.
$endgroup$
– José Carlos Santos
Jan 22 at 15:56
|
show 3 more comments
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