It is a well known fact that given a real simply connected Lie group $G$ there is a correspondence between smooth complex representations of $G$ and complex representations its Lie algebra $\mathfrak{g}$ — which in turn correspond to complex representations of the complexification $\mathfrak{g}_{\mathbb{C}} = \mathfrak{g} \otimes \mathbb{C}$. Explicitely, if $\rho : G \to \operatorname{GL}_n(\mathbb{C}$ is a complex representation of $G$, its derivative at the identity $\rho_* = d \rho_1 : \mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$ is a Lie algebra homomorphism — so it is a complex representation of $\mathfrak{g}$ — and the map $\operatorname{Hom}_\mathbb{R}(G, \operatorname{GL}_n(\mathbb{C})) \to \operatorname{Hom}_\mathbb{R}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$ that takes a representation $\rho$ to $rho_*$ is a bijection. This is a particular case of the fact that the map $\operatorname{Hom}_\mathbb{R}(G, H) \to \operatorname{Hom}_\mathbb{R}(\mathfrak{g}, \mathfrak{h})$ that takes a smooth group homomorphism $\varphi: G \to H$ to $\varphi_* = d \varphi_1 : \mathfrak{g} \to \mathfrak{h}$ is well defined and is a bijection for every simply connected $G$.
The usual proof of this fact relies heavely in the existance of the exponential map $\operatorname{exp} : \mathfrak{g} \to G$ and the Campbell-Hausdorff formula.
I wonder if the same holds for complex Lie groups, i.e. complex manifolds endowed with a group structure whose operations are holomorphic maps.
Given a complex Lie group $G$, there is a complex Lie algebra $\mathfrak{g}$ associated with $G$ — the algebra of holomorphic left invariant vector fields. An analytic representation of $G$ is a finite-dimensional complex vector space $V$ endowed with a holomorphic group homomorphism $\rho : G \to \operatorname{GL}_n(\mathbb{C})$. If $G$ is simply connected, is there a one-to-one correspondence between (isomorphism classes of) analytic representations of $G$ and representations of $\mathfrak{g}$?
If this is the case, does the usual proof for the real case also works for the complex case? Is there a homolomorphic map $\operatorname{exp} : \mathfrak{g} \to G$ satisfying the same conditions satisfied by the real exponential map of Lie groups, and does the Campbell-Housdorff formula also hold in this case?
Best Answer
The correspondence holds for complex Lie groups. I'm not sure if the proof of the real case can be reproduced, but we can instead use the real case in our favor. Given a complex simply connected Lie group $G$ of complex dimension $n$, we can regard $G$ as a real Lie group of real dimension $2 n$. The main idea is to use the fact that we can obtain the real Lie algebra of $G$ -- the algebra of smooth left invariant fields -- by restricting the scalars of the complex Lie algebra of $G$ -- the algebra of holomorphic left invariant vector. This is because every invariant vector field is holomorphic.
Since $G$ is simply connected, the map $\operatorname{Hom}_{\mathbb{R}}(G, \operatorname{GL}_n(\mathbb{C})) \to \operatorname{Hom}_{\mathbb{R}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$ that takes a smooth group homomorphism $\varphi : G \to \operatorname{GL}_n(\mathbb{C})$ to the real Lie algebra homomorphism $\varphi_* : \mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$ is a bijection. Now consider the subset $\operatorname{Hom}_\mathbb{C}(G, \operatorname{GL}_n(\mathbb{C})) \subset \operatorname{Hom}_{\mathbb{R}}(G, \operatorname{GL}_n(\mathbb{C}))$ of holomorphic group homomorphisms. We will show that our map takes $\operatorname{Hom}_\mathbb{C}(G, \operatorname{GL}_n(\mathbb{C}))$ to the subset $\operatorname{Hom}_\mathbb{C}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C})) \subset \operatorname{Hom}_{\mathbb{R}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$ of complex Lie algebras homomorphisms. Given a smooth homomorphism $\varphi : G \to \operatorname{GL}_n(\mathbb{C})$, if $\varphi$ is holomorphic then its derivative $\varphi_*$ is $\mathbb{C}$-linear, so it lies in $\operatorname{Hom}_{\mathbb{C}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$. On the other hand, if $\varphi_*$ is $\mathbb{C}$-linear then $\varphi$ is holomorphic, so that $\varphi$ lies in $\operatorname{Hom}_\mathbb{C}(G, \operatorname{GL}_n(\mathbb{C}))$.
We should note that this is a particular case of the fact of a more general theorem which states that the map $\operatorname{Hom}_{\mathbb{C}}(G, H)) \to \operatorname{Hom}_{\mathbb{C}}(\mathfrak{g}, \mathfrak{h})$ is well defined and is a bijection for all complex simply connected $G$ and all complex $H$. Indeed, our proof still works if we replace $\operatorname{GL}_n(\mathbb{C})$ with any $H$. I hope I'm not missing any detail in here, but I was convinced by this proof.