# Representation of homogeneous Lorentz transformation

In Page 63, Section 2.5 of Weinberg's QFT Volume 1, on "One-particle states", he considers the representation of homogeneous Lorentz transformation, $$U(\Lambda, 0) \equiv U(\Lambda)$$ $$U(\Lambda) \Psi_{p, \sigma}=\sum_{\sigma^{\prime}} C_{\sigma^{\prime} \sigma}(\Lambda, p) \Psi_{\Lambda p, \sigma^{\prime}}$$ then he claims that,

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 $$C_{\sigma^{\prime} \sigma}(\Lambda, p)$$ 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.

Now my question: the effect of $$U(\Lambda)$$ is to bring the state $$\Psi_{p, \sigma}$$ to $$\Psi_{\Lambda p, \sigma^{\prime}}$$, so the space that the $$C_{\sigma'\sigma}$$ acts is different for different $$\Lambda$$. But this "conclusion" is weird to me since I think the representation space of a group should be the same for the group elements.

This is an example of induced representation. Consider two groups, $$K < G$$. Let a representation $$D(K)$$ act in a vector (usually, Hilbert) space $${\mathbb{V}}$$. Based on this, we now wish to construct a representation of $$G$$. In mathematics, this (so-called "induced") representation is denoted with $$\operatorname{Ind}_K^GD$$ or simply $$D(K)\uparrow G$$.
Such a representation will be a fiber bundle whose base is the quotient space $$G/K$$, with copies of $${\mathbb{V}}$$ playing the role of fibers. Simply speaking, we take multiple copies of $${\mathbb{V}}$$ and agree that the action of $$G$$ is two-fold. It permutes the unit vectors within each copy of $${\mathbb{V}}$$, and it also permutes the copies of $${\mathbb{V}}$$.
• So in this case, $G$ is the full Lorentz group, and $K$ is the Little group of it. The representation space of $K$ is $\mathbb{V}$. The $G$ operation can take vectors from one $\mathbb{V}$ to another, is that correct understanding? Sep 6, 2020 at 10:08