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 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).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.