# Why do we look for the representations of $\mathfrak{so}(1,3)$ when looking for projective representations of $SO^+(1,3)$?

In a relativistic quantum field theory, one classifies quantum fields by looking for finite dimensional projective representations of the restricted Lorentz group $$SO^+(1,3)$$ over the target space $$V$$ of a quantum field $$\Phi: \mathbb{R}^{1,3} \rightarrow V$$. Why do we pursue this object by looking for normal representations of the Lie algebra of $$SO^+(1,3)$$, denoted $$\mathfrak{so}(1,3)$$?

It can be proven that there is a sort of equivalence between finite dimensional projective representations of a connected Lie group $$G$$ and finite dimensional normal representations of the Lie algebra of (the universal cover of $$G$$) $$G$$. The following is an outline of the logic.

1. The restricted Lorentz group $$SO^+(1,3)$$ is connected. Hence, it has a universal cover. And, the universal cover is unique (Hall, Prop 5.13).
2. By Bargmann's theorem, it is equivalent to find projective representations of $$SO^+(1,3)$$ and normal representations of its universal cover $$\text{Spin}(1,3)$$.
3. By definition of a universal cover, their Lie algebras are isomorphic $$\mathfrak{\text{Spin}}(1,3) \cong \mathfrak{so}(1,3)$$. Hereafter, we denote both by $$\mathfrak{so}(1,3)$$.
4. By definition, the universal cover of a Lie group is simply-connected. Hence, every Lie algebra representation $$\pi: \mathfrak{so}(1,3) \rightarrow \mathfrak{gl}(V)$$ induces a Lie group representation of the universal cover $$\Phi: \text{Spin}(1,3) \rightarrow GL(V)$$ (Hall, Thm 5.6).$$^\dagger$$

Altogether, because $$SO^+(1,3)$$ is connected, representations of its Lie algebra induce representations of its universal cover, which are precisely its projective representations.

$$^\dagger$$ I emphasize that this claim is not true for groups that are not simply-connected. However, its converse is always true.

[1] Brian C. Hall, Lie Groups, Lie Algebras, and Representations.