Why are projective reps of connected Lie groups with trivial 2nd Lie algebra cohomology group equivalent to reps of the Lie algebra?

Question

I would like to precisely understand how to establish the equivalence between projective representations of a Lie group $G$ and its Lie algebra $\mathfrak{g}$ when $G$ is connected and has trivial second Lie algebra cohomology group.

Answer 1

There are nine steps to establish this equivalence. Essentially, the condition that the Lie group $G$ is connected gives us access to the unique universal cover $\tilde{G}$, which has the nice property of being simply connected. And, the condition that $G$ has trivial second Lie algebra cohomology group lets us apply Bargmann's theorem. I will use the restricted Lorentz group $SO^+(1,3)$ as an example as it is connected and has a trivial second Lie algebra cohomology group.

1. A connected Lie group $G$ has a unique universal cover $\tilde{G}$. By definition $\tilde{G}$ is simply connected and $\mathfrak{g} \cong \tilde{\mathfrak{g}}$.

2. Let $\tilde{G}$ be a simply connected Lie group. The representations of the Lie algebra of a simply connected Lie group $\pi: \tilde{\mathfrak{g}} \rightarrow \mathfrak{gl}(V)$ are in one-to-one correspondence with the representations of the simply connected Lie group $\varphi: \tilde{G} \rightarrow GL(V)$ by the unique mapping $\varphi(e^X) = e^{\pi(X)}$ for all $X \in \tilde{\mathfrak{g}}$.

3. Let $\mathfrak{g} \cong \mathfrak{h}$ be Lie algebra isomorphic by isomorphism $\zeta: \mathfrak{g} \rightarrow \mathfrak{h}$. Representations $\pi: \mathfrak{g} \rightarrow \mathfrak{gl}(V)$ are in one-to-one correspondence with representations $\varphi: \mathfrak{h} \rightarrow \mathfrak{gl}(V)$ by the mapping $\varphi \circ \zeta = \pi$.

4. Bargmann's theorem states that projective representations of a connected Lie group with trivial second Lie algebra cohomology group are equivalent to normal representations of the universal cover of the Lie group.

5. The restricted Lorentz group is connected and has a trivial second Lie algebra cohomology group. Denote the unique universal cover of the restricted Lorentz group by $\text{Spin}(1,3)$.

6. By Bargmann's theorem, projective representations of the restricted Lorentz group $P(\pi): SO^+(1,3) \rightarrow GL(V)$ are equivalent to normal representations $\pi: \text{Spin}(1,3) \rightarrow GL(V)$ of its universal cover.

7. By statement (2) it is equivalent to look for Lie algebra representations $\pi: \mathfrak{spin}(1,3) \rightarrow \mathfrak{gl}(V)$ of the Lie algebra of the universal cover and to look for Lie group representations $\varphi: \text{Spin}(1,3) \rightarrow GL(V)$ of the universal cover.

8. Recall that by definition of a universal cover $\mathfrak{so}(1,3) \cong \mathfrak{spin}(1,3)$. Hence, by statement (3) it is equivalent to look for Lie algebra representations $\pi: \mathfrak{so}(1,3) \rightarrow \mathfrak{gl}(V)$ and to look for Lie algebra representations $\varphi: \mathfrak{spin}(1,3) \rightarrow \mathfrak{gl}(V)$.

9. The statements (6), (7), and (8) establish an equivalence between projective representations of the restricted Lorentz group and representations of the Lorentz algebra.