Weinberg's classification of one-particle states and representations of the Poincare group

Question

A representation of a group $G$ is a pair $(\rho, V)$ where $V$ is a vector space and $\rho : G\to GL(V)$ is a homomorphism. If $V$ is actually a Hilbert space and $\rho : G\to \mathcal{U}(V)$ maps into the unitary group of $V$ and is strongly continuous, then $\rho$ is a unitary representation.

In section 2.5 Weinberg sets out to study the classification of one-particle states. What he does is:

1. He first notices that the unitary representation $(U,\mathscr{H})$ of the Poincare group gives rise to the momentum operators $P^\mu$. He then considers the basis of eigenstates of $P^\mu$, namely $\Psi_{p,\sigma}$.

2. He then notices that $U(\Lambda) = U(\Lambda, 0)$ - the restriction of $U$ to the proper orthochronous Lorentz group - when acting on $\Psi_{p,\sigma}$ yields an eigenvector of $P^\mu$ with eigenvalue $\Lambda^\mu_\nu p^\nu$. Hence it lies in the eigenspace associated to $\Lambda p$ being $$U(\Lambda) \Psi_{p,\sigma}=\sum_{\sigma'} C_{\sigma'\sigma}(\Lambda, p) \Psi_{\Lambda p,\sigma'}.$$

3. He states the following (which is what I'm confused about):

   In general it may be possible by using suitable linear combinations of the $\Psi_{p,\sigma}$ to choose the $\sigma$ labels in such a way that the matrix $C_{\sigma'\sigma}(\Lambda,p)$ is block-diagonal; in other words, so that the $\Psi_{p,\sigma}$ with $\sigma$ within any one block by themselves furnish a representation of the inhomogeneous Lorentz group. It is natural to identify the states of a specific particle type with the components of a representation of the inhomogeneous Lorentz group which is irreducible, in the sense that it cannot be further decomposed in this way.

Now, I'm missing Weinberg's point. It seems he is actually talking about complete reducibility, namely, the property that the representations can be decomposed into irreducible ones.

It seems, however, that for groups like the Poincare group, complete reducibility is a non trivial thing to derive - I've read in this discussion that it actually requires direct integrals.

So it seems that to deal with these complications - which seem to come from the translations - Weinberg is switching to a basis of the generators of translations.

But how does that work really? Why to get into complete reducibility for the Poincare group, it is a good idea to do this? By the way, why this gives complete reducibility? Is he looking at "each eigenspace of $P$ at once"?

Also what he means that with $\sigma$ on each block $\Psi_{p\sigma}$ furnish a representation of the Poincare group? I mean, $U$ is a representation, and if we simply restrict its action to a subspace of $\mathscr{H}$ it will continue to be so.

So, how to understand Weinberg's argument (which seems to be about complete reducibility), and how does it connect to the "more usual" representation theory?

Answer 1

As Prahar already hints at in a comment:

The eigenspace associated to $\Lambda p$ is finite-dimensional because it is simply spanned by the $\Psi_{\Lambda p,\sigma}$ for all possible values of $\sigma$ and the spin label $\sigma$ are drawn from a finite set. So all Weinberg is saying is that the finite-dimensional representation of the Lorentz group given by $C_{\sigma,\sigma'}(\Lambda,p)$ on that eigenspace may be reduced, this is just the standard result that all finite-dimensional representations of the Lorentz group decompose as the direct sum of irreducible representations.