Why is this representation completely reducible?

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So I'm reading Georgi's Lie Algebras in Particle Physics. With $S$ be a similarity transformation: $D(g) rightarrow D'(g) = S^-1D(g)S$, on page 6, he said:
We will show later that any representation of a finite group is completely reducible.
For example, for $(1.5)$, take
beginequation
tag1.15
S
= frac13
beginpmatrix
1
& 1
& 1
\
1
& omega^2
& omega
\
1
& omega
& omega^2
endpmatrix
endequation
where
beginequation
tag1.16
omega = e^2 pi i /3
endequation
then
beginequation
tag1.17
D'(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D'(a)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega
& 0
\
0
& 0
& omega^2
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega2
& 0
\
0
& 0
& omega
endsmallmatrix
right).
endequation
(Original scanned image here.)
Here's $(1.5)$:
HereâÂÂs another representation of $Z_3$:
beginequation
tag1.5
D(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D(a)
= left(
beginsmallmatrix
0
& 0
& 1
\
1
& 0
& 0
\
0
& 1
& 0
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
0
& 1
& 0
\
0
& 0
& 1
\
1
& 0
& 0
endsmallmatrix
right).
endequation
(Original scanned image here.)
I don't see how this $(1.17)$ can be completely reducible. Isn't it equivalent to the $Z_3$ representation in $(1.5)$ above, thus irreducible instead?
Thank you!
Edit: Answer: So since it's not clear $D(g)$ is irreducible or not, we use the similarity transformation $S$ to create an equivalent representation $D'(g)$. We see that $D'(g)$ is in block-diagonal form, and thus is obviously a completely reducible representation. Thus both $D(g)$ and $D'(g)$ are completely reducible.
group-theory representation-theory lie-groups lie-algebras
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up vote
1
down vote
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So I'm reading Georgi's Lie Algebras in Particle Physics. With $S$ be a similarity transformation: $D(g) rightarrow D'(g) = S^-1D(g)S$, on page 6, he said:
We will show later that any representation of a finite group is completely reducible.
For example, for $(1.5)$, take
beginequation
tag1.15
S
= frac13
beginpmatrix
1
& 1
& 1
\
1
& omega^2
& omega
\
1
& omega
& omega^2
endpmatrix
endequation
where
beginequation
tag1.16
omega = e^2 pi i /3
endequation
then
beginequation
tag1.17
D'(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D'(a)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega
& 0
\
0
& 0
& omega^2
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega2
& 0
\
0
& 0
& omega
endsmallmatrix
right).
endequation
(Original scanned image here.)
Here's $(1.5)$:
HereâÂÂs another representation of $Z_3$:
beginequation
tag1.5
D(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D(a)
= left(
beginsmallmatrix
0
& 0
& 1
\
1
& 0
& 0
\
0
& 1
& 0
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
0
& 1
& 0
\
0
& 0
& 1
\
1
& 0
& 0
endsmallmatrix
right).
endequation
(Original scanned image here.)
I don't see how this $(1.17)$ can be completely reducible. Isn't it equivalent to the $Z_3$ representation in $(1.5)$ above, thus irreducible instead?
Thank you!
Edit: Answer: So since it's not clear $D(g)$ is irreducible or not, we use the similarity transformation $S$ to create an equivalent representation $D'(g)$. We see that $D'(g)$ is in block-diagonal form, and thus is obviously a completely reducible representation. Thus both $D(g)$ and $D'(g)$ are completely reducible.
group-theory representation-theory lie-groups lie-algebras
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
So I'm reading Georgi's Lie Algebras in Particle Physics. With $S$ be a similarity transformation: $D(g) rightarrow D'(g) = S^-1D(g)S$, on page 6, he said:
We will show later that any representation of a finite group is completely reducible.
For example, for $(1.5)$, take
beginequation
tag1.15
S
= frac13
beginpmatrix
1
& 1
& 1
\
1
& omega^2
& omega
\
1
& omega
& omega^2
endpmatrix
endequation
where
beginequation
tag1.16
omega = e^2 pi i /3
endequation
then
beginequation
tag1.17
D'(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D'(a)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega
& 0
\
0
& 0
& omega^2
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega2
& 0
\
0
& 0
& omega
endsmallmatrix
right).
endequation
(Original scanned image here.)
Here's $(1.5)$:
HereâÂÂs another representation of $Z_3$:
beginequation
tag1.5
D(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D(a)
= left(
beginsmallmatrix
0
& 0
& 1
\
1
& 0
& 0
\
0
& 1
& 0
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
0
& 1
& 0
\
0
& 0
& 1
\
1
& 0
& 0
endsmallmatrix
right).
endequation
(Original scanned image here.)
I don't see how this $(1.17)$ can be completely reducible. Isn't it equivalent to the $Z_3$ representation in $(1.5)$ above, thus irreducible instead?
Thank you!
Edit: Answer: So since it's not clear $D(g)$ is irreducible or not, we use the similarity transformation $S$ to create an equivalent representation $D'(g)$. We see that $D'(g)$ is in block-diagonal form, and thus is obviously a completely reducible representation. Thus both $D(g)$ and $D'(g)$ are completely reducible.
group-theory representation-theory lie-groups lie-algebras
So I'm reading Georgi's Lie Algebras in Particle Physics. With $S$ be a similarity transformation: $D(g) rightarrow D'(g) = S^-1D(g)S$, on page 6, he said:
We will show later that any representation of a finite group is completely reducible.
For example, for $(1.5)$, take
beginequation
tag1.15
S
= frac13
beginpmatrix
1
& 1
& 1
\
1
& omega^2
& omega
\
1
& omega
& omega^2
endpmatrix
endequation
where
beginequation
tag1.16
omega = e^2 pi i /3
endequation
then
beginequation
tag1.17
D'(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D'(a)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega
& 0
\
0
& 0
& omega^2
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& omega2
& 0
\
0
& 0
& omega
endsmallmatrix
right).
endequation
(Original scanned image here.)
Here's $(1.5)$:
HereâÂÂs another representation of $Z_3$:
beginequation
tag1.5
D(e)
= left(
beginsmallmatrix
1
& 0
& 0
\
0
& 1
& 0
\
0
& 0
& 1
endsmallmatrix
right)
quad
D(a)
= left(
beginsmallmatrix
0
& 0
& 1
\
1
& 0
& 0
\
0
& 1
& 0
endsmallmatrix
right)
quad
D'(b)
= left(
beginsmallmatrix
0
& 1
& 0
\
0
& 0
& 1
\
1
& 0
& 0
endsmallmatrix
right).
endequation
(Original scanned image here.)
I don't see how this $(1.17)$ can be completely reducible. Isn't it equivalent to the $Z_3$ representation in $(1.5)$ above, thus irreducible instead?
Thank you!
Edit: Answer: So since it's not clear $D(g)$ is irreducible or not, we use the similarity transformation $S$ to create an equivalent representation $D'(g)$. We see that $D'(g)$ is in block-diagonal form, and thus is obviously a completely reducible representation. Thus both $D(g)$ and $D'(g)$ are completely reducible.
group-theory representation-theory lie-groups lie-algebras
edited Aug 25 at 9:15
Jendrik Stelzner
7,57221037
7,57221037
asked Aug 25 at 2:20
D. Tran
33519
33519
add a comment |Â
add a comment |Â
1 Answer
1
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1
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The representation (1.5) is not irreducible. The vector $e_1 + e_2 + e_3$ is an invariant vector with respect to the representation (1.5), so (1.5) is reducible.
The representation (1.17) is completely reducible, because all of the representing matrices for all the group elements are diagonal. Thus $V = langle e_1 rangle oplus langle e_2 rangle oplus langle e_3 rangle$, and each element of this decomposition is invariant with respect to all group elements.
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The representation (1.5) is not irreducible. The vector $e_1 + e_2 + e_3$ is an invariant vector with respect to the representation (1.5), so (1.5) is reducible.
The representation (1.17) is completely reducible, because all of the representing matrices for all the group elements are diagonal. Thus $V = langle e_1 rangle oplus langle e_2 rangle oplus langle e_3 rangle$, and each element of this decomposition is invariant with respect to all group elements.
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
add a comment |Â
up vote
1
down vote
accepted
The representation (1.5) is not irreducible. The vector $e_1 + e_2 + e_3$ is an invariant vector with respect to the representation (1.5), so (1.5) is reducible.
The representation (1.17) is completely reducible, because all of the representing matrices for all the group elements are diagonal. Thus $V = langle e_1 rangle oplus langle e_2 rangle oplus langle e_3 rangle$, and each element of this decomposition is invariant with respect to all group elements.
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The representation (1.5) is not irreducible. The vector $e_1 + e_2 + e_3$ is an invariant vector with respect to the representation (1.5), so (1.5) is reducible.
The representation (1.17) is completely reducible, because all of the representing matrices for all the group elements are diagonal. Thus $V = langle e_1 rangle oplus langle e_2 rangle oplus langle e_3 rangle$, and each element of this decomposition is invariant with respect to all group elements.
The representation (1.5) is not irreducible. The vector $e_1 + e_2 + e_3$ is an invariant vector with respect to the representation (1.5), so (1.5) is reducible.
The representation (1.17) is completely reducible, because all of the representing matrices for all the group elements are diagonal. Thus $V = langle e_1 rangle oplus langle e_2 rangle oplus langle e_3 rangle$, and each element of this decomposition is invariant with respect to all group elements.
answered Aug 25 at 4:15
Ted
20.9k13158
20.9k13158
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
add a comment |Â
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
I'm dumb so I'm not sure what you mean. Are you talking about the trivial representation? I fail to see how I can decompose $(1.5)$ into block diagonal form to prove that it's completely reducible...
â D. Tran
Aug 25 at 4:19
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
(1.17) is the decomposition of (1.5) into block diagonal form. When you apply a similarity transformation $S$, you change the basis (and hence the matrices) with respect to which the representation is expressed: $D'(g) = S^-1 D(g) S$. This transforms (1.5) into (1.17), and shows that (1.5) is completely reducible.
â Ted
Aug 25 at 4:24
add a comment |Â
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