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 $(1.5)$:




Here’s 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.







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    up vote
    1
    down vote

    favorite
    1












    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 $(1.5)$:




    Here’s 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.







    share|cite|improve this question
























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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 $(1.5)$:




      Here’s 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.







      share|cite|improve this question














      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 $(1.5)$:




      Here’s 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.









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 25 at 9:15









      Jendrik Stelzner

      7,57221037




      7,57221037










      asked Aug 25 at 2:20









      D. Tran

      33519




      33519




















          1 Answer
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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.






          share|cite|improve this answer




















          • 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...
            – 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.
            – Ted
            Aug 25 at 4:24










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          up vote
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          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.






          share|cite|improve this answer




















          • 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...
            – 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.
            – Ted
            Aug 25 at 4:24














          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.






          share|cite|improve this answer




















          • 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...
            – 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.
            – Ted
            Aug 25 at 4:24












          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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...
            – 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.
            – 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...
            – 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.
            – 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...
          – 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...
          – 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.
          – 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.
          – Ted
          Aug 25 at 4:24

















           

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