# Definition of reducible representation

A reducible representation of a group $$g \rightarrow D(g)$$ is one which leaves a subspace $$U$$ invariant, i.e. $$D(g)|u\rangle \in U, \space \forall |u\rangle \in U$$.A completely reducible representation is one that can be broken down into a direct sum of irreducible representations.

In Howard Georgi's book "Lie Algebras in Particle Physics", he defines irreducible representations in terms of projection operators (page 5 Equation 1.11) in terms of projection operators P that project onto the invariant subspace:

$$PD(g)P = D(g)P$$ where, presumably

$$P = \sum_{\alpha} |\alpha \rangle \langle\alpha|$$.

Furthermore, Georgi defines completely reducible representations to be those in which both $$P$$ and $$1-P$$ project on to an invariant(under the action of $$D(g)$$) subspace.

I'm struggling to see how Georgi's definitions are equivalent to the first.

• Would Mathematics be a better home for this question? – Qmechanic Feb 8 '20 at 8:55

1. Leaving a subspace invariant means that $$D(g)u\in U$$ for all $$g\in G, u\in U$$. Since $$P_U v \in U$$ for the projector $$P_U$$ onto $$U$$ and any $$v\in V$$, you have that $$D(g)P_U v \in U$$ for all $$v\in V$$. So applying $$P_U$$ again to $$D(g)P_U v$$ does nothing, since the latter is already in $$U$$, there is nothing to project away.
2. $$1 - P$$ is the projector onto a subspace $$U^\ast$$ disjoint from $$U$$, and $$V = U \oplus U^\ast$$ (this is a standard fact about projectors).