# Notes Regarding the Complex Structure

I know I said I wasn’t going to do this, but it isn’t that hard, and for completeness I should actually explain how the posts on real Lie groups/algebras relate to the complex case.

Let $V$ be a finite-dim real vector space. Then we call $J$ a complex structure of $V$ if $J\in End_{\mathbb{R}}(V)$ and $J^2=-id_V$. Given such a pair $(V, J)$, we can turn $V$ into a complex vector space $\widehat{V}$ by defining scalar multiplication by $(a+bi)v=av+J(bv)$. Likewise, we get a complex Lie algebra out of a real one if the complex structure also satisfies $[u, J(v)]=J([u,v])$ for all $u,v\in\frak{g}$. This is the associated complex Lie algebra to $\frak{g}$ denoted $\widehat{\frak{g}}$.

We can of course go the other direction much more easily. Given a complex Lie algebra $\frak{g}$, then restricting the $\mathbb{C}$-action to $\mathbb{R}$ gives us a real Lie algebra $\frak{g}_\mathbb{R}$. Note that $\widehat{\frak{g}_\mathbb{R}}=\frak{g}$ under the complex structure $J: \frak{g}_\mathbb{R}\to\frak{g}_\mathbb{R}$ by $u\mapsto iu$.

Suppose that $G$ and $H$ are complex Lie groups. Then if we have a morphism $\phi: G\to H$ in the category of complex Lie groups, i.e. a holomorphic map that is also a group homomorphism, then we can regard $d\phi: \mathcal{L}(G)\to\mathcal{L}(H)$ as a morphism of complex Lie algebras. Then $\phi$ can be regarded as a map of real Lie groups whose differential commutes with the complex structure on Lie algebras. The other way works as well. Given a real Lie group map whose differential is $\mathbb{C}$-linear, we have that $\phi$ is actually a map of complex Lie groups.

So far this is a fast overview. I don’t want to spend more than one post on this, but if you want to see more about any of these things, just comment.

The main purpose of bringing this up is that if we have a complex lie group $G$, then we’ll denote the underlying real Lie group as $G_\mathbb{R}$. By the posts I’ve already done, we can explicitly construct $exp_{G_\mathbb{R}}: \frak{g}\to G$. Note this is only a real holomorphic map. By the above statements, if $d(exp_{G_\mathbb{R}})$ is $\mathbb{C}$-linear, then it is actually complex holomorphic.

Just as in the exponential map post, we define the $\mathbb{C}$-linear map $\alpha: \mathbb{C}\to\frak{g}$ by $\alpha(z)=zv$ where $v\in T_eG$. Since $\mathbb{C}_\mathbb{R}$ is simply connected, there is a unique lift of this map to all of $\mathbb{C}$ which we call $\phi_v: \mathbb{C}\to G$ such that $d\phi_v=\alpha$. Which means that $\phi_v=exp_{G_\mathbb{R}}\circ \alpha$. i.e. $\phi_v(z)=exp_{G_\mathbb{R}}(zv)$.

Thus $d\phi_v=\alpha$ is complex holomorphic giving the 1-parameter subgroup in $G$ satisfying $d\phi_v(1)=v$ and exponential map $exp_G(v)=\phi_v(1)$. We have essentially proved the theorem that $exp_G=exp_{G_\mathbb{R}}$ which means the previous posts on the subject still apply.

As motivation for later, we’ll now do the example. Let $\mathbb{C}^n$ be the (only) simply connected complex Lie group of dimension $n$. We have global coordinates $z_1, \ldots, z_n$ since this is a vector space. Thus $\frac{\partial}{\partial z_1}\Big|_0, \ldots, \frac{\partial}{\partial z_n}\Big|_0$ forms a basis for $\mathcal{L}(\mathbb{C}^n)$.

We have that $[\frac{\partial}{\partial z_i}, \frac{\partial}{\partial z_j}]=0$, so the Lie algebra is abelian. i.e. we can identify $\mathcal{\mathbb{C}^n}$ with $\mathbb{C}^n$. The exponential map is just the identity.

The other example is to take a real basis $e_1, \ldots, e_{2n}$ of $\mathbb{C}^{n}$ and let $\displaystyle D=\{\sum_{i=1}^{2n}m_ie_i : m_i\in \mathbb{Z}\}$ and quotient $\mathbb{C}^n/D$. This is a real torus and will be incredibly important in when we return to compact complex Lie groups.