# $D(x)(I-P)\neq(I-P) \Rightarrow$ Not completely reducible?

In section 1.8 of Georgi's Lie Algebras In Particle Physics the author considers $$D(x) = \begin{pmatrix} 1 & x\\ 0 & 1 \end{pmatrix}$$ as a representation of the integers under addition. It is reducible because the column space of $$P = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}$$ is an invariant subspace of the representation: $$D(x)P = P.$$ He then states

However, $$D(x)(I-P)\neq(I-P)$$ so it is not completely reducible.

and this is where I don't follow anymore, because he does not prove or otherwise motivate the conclusion.

It seems like it might relate to the proof, a few pages later, that every representation of a finite group is completely reducible. For such group representations it is shown that $$I-P$$ projects onto an invariant subspace, and this is used in the proof. But of course this is not enough to be able to claim that $$D(x)(I-P)=(I-P)$$ is a necessary condition for complete reducability.

• There is a similar question here, but the accepted answer does not seem sufficient (from what I can see).
– ummg
Apr 11, 2021 at 4:51
• Would Mathematics be a better home for this question? Apr 11, 2021 at 7:58

Georgi's calculation gives a hint about why the representation may be indecomposable, without rigorous proof. After all, there are other subspaces that one might consider. Consider, for example, the four dimensional representation with block diagonal form \begin{align*} D_4(x):=\begin{bmatrix}\mathbb 1_2 & \mathbb 1_2x\\ \mathbb 0_2 & \mathbb 1_2\end{bmatrix}. \end{align*} This includes $$P_4(a,b,c,d)=(a,b,0,0)$$ as a subrepresentation, and $$D_4(x)(\mathbb 1_4-P_4) \neq (\mathbb 1_4 - P_4)$$, but the representation decomposes into two copies of the original $$D(x)$$ representation. With that in mind, what additional ingredient could be needed in order to prove indecomposability of $$D(x)$$?