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Lie's Third Theorem: For every finite-dimensional real or complex Lie algebra $\mathfrak{g}$, there exists a unique (up to isomorphism) simply connected Lie group $G$ whose Lie algebra is isomorphic to $\mathfrak{g}$.
Proof:- Apply Ado's Theorem:
By Ado's Theorem, there exists a finite-dimensional vector space $V$ and a faithful (injective) Lie algebra homomorphism $\rho: \mathfrak{g} \hookrightarrow \mathfrak{gl}(V)$. This allows us to identify $\mathfrak{g}$ with a Lie subalgebra of $\mathfrak{gl}(V)$, the Lie algebra of the general linear group $GL(V)$. - Identify $\mathfrak{gl}(V)$ with $GL(V)$:
The Lie algebra $\mathfrak{gl}(V)$ corresponds to the Lie group $GL(V)$, which consists of invertible linear transformations on $V$. Since $\rho(\mathfrak{g}) \subset \mathfrak{gl}(V)$ is a Lie subalgebra, we aim to construct a Lie subgroup of $GL(V)$ with Lie algebra $\rho(\mathfrak{g})$. - Construct the connected Lie subgroup $G'$:
By the Closed Subgroup Theorem, every Lie subalgebra of $\mathfrak{gl}(V)$ corresponds to a unique connected Lie subgroup of $GL(V)$. Explicitly, the subgroup $G' \subset GL(V)$ is generated by the image of the exponential map applied to $\rho(\mathfrak{g})$:
$$
G' = \langle e^{X} \mid X \in \rho(\mathfrak{g}) \rangle.
$$
This $G'$ is connected, and its Lie algebra is $\rho(\mathfrak{g})$. - Take the universal cover of $G'$:
The group $G'$ may not be simply connected. However, every connected Lie group $G'$ has a universal covering group $G$, which is simply connected. The universal cover $G$ is constructed by "unwinding" the topology of $G'$, and the covering map $\pi: G \to G'$ is a Lie group homomorphism. - Verify the Lie algebra of $G$:
The Lie algebra of a universal cover is isomorphic to the Lie algebra of the original group. Thus, $\text{Lie}(G) \cong \text{Lie}(G') \cong \rho(\mathfrak{g})$. Since $\rho$ is faithful, $\mathfrak{g} \cong \rho(\mathfrak{g})$, so $\text{Lie}(G) \cong \mathfrak{g}$. - Uniqueness of $G$:
Suppose there exists another simply connected Lie group $\tilde{G}$ with $\text{Lie}(\tilde{G}) \cong \mathfrak{g}$. By the Lie group-Lie algebra correspondence, the isomorphism of Lie algebras lifts to a unique isomorphism of Lie groups $\tilde{G} \to G$. Thus, $G$ is unique up to isomorphism.
Conclusion:
Using Ado's Theorem to embed $\mathfrak{g}$ into $\mathfrak{gl}(V)$, constructing the connected Lie subgroup $G'$, and taking its universal cover $G$, we obtain a simply connected Lie group with Lie algebra $\mathfrak{g}$. The uniqueness follows from the correspondence between simply connected Lie groups and their Lie algebras. This completes the proof of Lie's Third Theorem. |
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