So I'm reading Georgi's Lie Algebras in Particle Physics. With $S$ be a similarity transformation: $D(g) \rightarrow D'(g) = S^{-1}D(g)S$, 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
\begin{equation}
\tag{1.15}
S
= \frac{1}{3}
\begin{pmatrix}
1
& 1
& 1
\\
1
& \omega^2
& \omega
\\
1
& \omega
& \omega^2
\end{pmatrix}
\end{equation}
where
\begin{equation}
\tag{1.16}
\omega = e^{2 \pi i /3}
\end{equation}
then
\begin{equation}
\tag{1.17}
D'(e)
= \left(
\begin{smallmatrix}
1
& 0
& 0
\\
0
& 1
& 0
\\
0
& 0
& 1
\end{smallmatrix}
\right)
\quad
D'(a)
= \left(
\begin{smallmatrix}
1
& 0
& 0
\\
0
& \omega
& 0
\\
0
& 0
& \omega^2
\end{smallmatrix}
\right)
\quad
D'(b)
= \left(
\begin{smallmatrix}
1
& 0
& 0
\\
0
& \omega2
& 0
\\
0
& 0
& \omega
\end{smallmatrix}
\right).
\end{equation}(Original scanned image here.)
Here's $(1.5)$:
Here’s another representation of $Z_3$:
\begin{equation}
\tag{1.5}
D(e)
= \left(
\begin{smallmatrix}
1
& 0
& 0
\\
0
& 1
& 0
\\
0
& 0
& 1
\end{smallmatrix}
\right)
\quad
D(a)
= \left(
\begin{smallmatrix}
0
& 0
& 1
\\
1
& 0
& 0
\\
0
& 1
& 0
\end{smallmatrix}
\right)
\quad
D'(b)
= \left(
\begin{smallmatrix}
0
& 1
& 0
\\
0
& 0
& 1
\\
1
& 0
& 0
\end{smallmatrix}
\right).
\end{equation}(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.
Best 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.