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.

  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 8, 2020 at 8:55

1 Answer 1

  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).


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