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

Question

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 $\mathfrak{so}(1,3)$?

Answer 1

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.

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