I have yet found anywhere the condition for a representation to be completely reducible. However. when I reach page 9 in Georgi's book Lie Algebras for Particle Physics, he said:
$$D(x)(I-P) \ne (I-P)$$
so it is not completely reducible.
Is this a condition for completely reducible representation?
In particular, the entire page said:
The integers form an infinite group under addition:
$$xy=x+y$$
Here's a representation:
$$D ( x ) = \left( \begin{array} { l l } { 1 } & { x } \\ { 0 } & { 1 } \end{array} \right)$$
This representation is reducible, but you can show that it is not completely reducible and is not equivalent to a unitary representation. It is reducible because
$$D(x)P=P$$
where
$$P = \left( \begin{array} { l l } { 1 } & { 0 } \\ { 0 } & { 0 } \end{array} \right)$$
However,
$$D(x)(I-P) \ne (I-P)$$
so it is not completely reducible.
But if the group is finite, it turns into a cyclic group (say 9 elements from $0$ to $9$, and every number exceeds $9$ get round-down to its 1st unit, i.e. $17$ to $7$), and is completely reducible, yes? Yet nothing changes, and $D(x)(I-P) \ne (I-P)$ still. What am I misunderstanding here? Thank you!
Best Answer
I don't think that this is correct. Perhaps that what Georgi had in mind was this: with respect to the usual inner product, $P$ and $I-P$ are orthogonal. Therefore, if the representations was completely reducible, then, since the space $\langle P\rangle$ spanned by $P$ is stable (that is, for each integer $x$ and each $M\in\langle P\rangle$, $D(x)M\in\langle P\rangle$), $\langle P\rangle^\perp$ is stable too. But $\langle P\rangle^\perp=\langle I-P\rangle$ and it turns out that it isn't stable.
It would be better to show (and that's not hard) that if $M\notin\langle P\rangle$, then $\langle M\rangle$ is not stable. This proves that the representation is not completely reducible.