Why do single particle states furnish a rep. of the inhomogeneous Lorentz group? Following up on this question: Weinberg says 

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 $C_{\sigma'\sigma}(\Lambda, p)$ 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 inhomogenous Lorentz group.

But why inhomogeneous Lorentz group if, in the first place, we performed a homogeneous Lorentz transformation on the states, via $U(\Lambda)$? I also want to be clear what is meant by the states "furnishing" a representation.
Regarding the above confusion, the same scenario again shows up during the discussion on the little group. Here's a little background: $k$ is a "standard" 4-momentum, so that we can express any arbitrary 4-momentum $p$ as $p^{\mu} = L^{\mu}_{\nu}(p) k^{\nu}$, where $L$ is a Lorentz transformation dependent on $p$. We consider the subgroup of Lorentz transformations $W$ that leave $k$ invariant (little group), and find that:
$U(W)\psi_{k \sigma} = \sum_{\sigma'} D_{\sigma' \sigma}(W)\psi_{k \sigma'}$. Then he says:

The coefficients $D(W)$ furnish a representation of the little group; i.e., for any elements $W$ and $W'$ , we get $D_{\sigma' \sigma}(W'W) = \sum_{\sigma''}D_{\sigma' \sigma''}(W)D_{\sigma''\sigma}(W')$.

So is it that even in the first part about the Lorentz group, $C$ matrices furnish the representation and not $\psi$?
Also, for the very simplified case if $C_{\sigma'\sigma}(\Lambda, p)$ is completely diagonal, would I be correct in saying the following in such a case, for any $\sigma$?
$$U(\Lambda)\psi_{p,\sigma} = k_{\sigma}(\Lambda, p)\psi_{\Lambda p, \sigma}$$
Only in this case it is clear to me that $U(\Lambda)$ forms a representation of Lorentz group, since $\psi_{p,\sigma}$ are mapped to $\psi_{\Lambda p, \sigma}$.
 A: In the  inhomogenous Lorentz group $ISO(1,3)$, you have the space-time translation group $\mathbb{R}^{1,3}$, and the Lorentz group $SO(1,3)$.
You begin to find a representation of the space-time translation group, by choosing a momentum $p$. So your representation must have a $p$ index, 
$$\psi_p \, .$$
After this, you will have to get the full representation, by finding a representation of the Lorentz group compatible with the momentum $p$, this will add another index $\sigma$ which corresponds to the polarization, so you will have a representation,
$$\psi_{p, \sigma} \, ,$$
which is the representation of the inhomogenous Lorentz group.
A: Re the meaning of representation, here is a definition from Peter Woit's "Quantum Mechanics for Mathematicians" lecture notes (available on-line), section 1.3:

Definition (Representation).  A (complex) representation ($\pi, V$) of a group $G$ is a homomorphism $$ \pi: g \in G \rightarrow \pi(g) \in GL(V) $$ 
  where $GL(V)$ is the group of invertible linear
  maps $V \rightarrow V$, with $V$ a complex vector space. 
Saying a map is a homomorphism means 
  $$ \pi(g_1) \pi(g_2) = \pi(g_1g_2) $$
  When $V$ is finite dimensional and we have chosen a basis of $V$, then we have an identification of linear maps and matrices
  $$ GL(V) \simeq GL(n,\boldsymbol{C}) $$
  where $GL(n,\boldsymbol{C})$ is the group of invertible $n$ by $n$ complex matrices.

So the representation is the homomorphism (the operation-preserving map) from the group $U(\Lambda)$ to the transformation matrices (Weinberg's C's and D's), but these matrices require a vector space (the $\psi$s), on which to act.

For the rest, here's my answer (caveat emptor, I'm just a slow student):  
This section 2.5 is titled "One Particle States".  If $C$ turns out to be reducible (block-diagonalizable), the different blocks are independent of one another (no mixing between blocks) and are interpreted as different particles species.  So, for a single particle state a single irreducible block is assumed.  
In this argument it's OK to generalize from homogeneous to inhomogeneous transformations, because translations don't mix $\sigma$'s and hence don't affect the block structure of $C$:
$$U(1,a) \Psi_{p,\sigma} = e^{-ip\cdot a} \Psi_{p,\sigma} $$
Finally, in the case you posit of a completely diagonal $C$, I think you are left with a bunch of particle species with no $\sigma$-mixing at all, i.e. scalars, each with a trivial little group ($k=1$).
