complexification of $SO(2)$

Question

While computing the complexification of Lie group $SO(2)$, I get the result is all the matrix of the following form $$\left(\begin{array}{cc} \frac{e^{t-\sqrt{-1}\theta}+e^{-t+\sqrt{-1}\theta}}{2} & \frac{e^{-t+\sqrt{-1}\theta}-e^{t-\sqrt{-1}\theta}}{2\sqrt{-1}} \\ -\frac{e^{-t+\sqrt{-1}\theta}-e^{t-\sqrt{-1}\theta}}{2\sqrt{-1}}& \frac{e^{t-\sqrt{-1}\theta}+e^{-t+\sqrt{-1}\theta}}{2} \\ \end{array} \right),$$ for $t$ and $\theta$ are both real numbers. I want to know what are this matrices like, or there is another description. Specially $SO(2)$ can act on $\mathbb{C}$ as a rotation? How could its complexification acts on $\mathbb{C}$ naturally?

Answer 1

Consider the map $\mathbb{C}^{\ast}\rightarrow SO_2(\mathbb{C})$ given by $$ t\mapsto \begin{pmatrix} \frac{t+t^{-1}}{2} & \frac{i(t-t^{-1})}{2} \cr -\frac{i(t-t^{-1})}{2} & \frac{t+t^{-1}}{2} \end{pmatrix} $$ This is a group isomorphism. So we have a simple description.

Answer 2

I struggled with this for a while so I just wanted to complement Dietrich's answer with a way of coming up with this isomorphism.

The way that I found natural is to note that the vector representation of $SO(2,\mathbb{C})$ on $\mathbb{C}^2$ is not irreducible. In fact, following the technique suggested in Decompose the representation $V$ of $SO_2$ into irreducible representations, one sees that it splits into a direct sum of 1-dimensional irreducible representations $\mathbb{C}^2=V_+\oplus V_-$. On $V_{\pm}=\text{span}_\mathbb{C}\{(1,\mp i)\}$ an element of the form $e^{aJ}\in SO(2,\mathbb{C})$, with $$J=\begin{pmatrix}0 & -1\\1 & 0\end{pmatrix},$$ acts like $t^{\pm1}=e^{\pm ia}=\cos(a)+i\sin(a)\in\mathbb{C}^\times$. Moreover, $$e^{aJ}=\begin{pmatrix}\cos(a) & -\sin(a)\\ \sin(a) & \cos(a)\end{pmatrix}=\begin{pmatrix}\frac{t+t^{-1}}{2} & -\frac{t-t^{-1}}{2i}\\\frac{t+t^{-1}}{2i} & \frac{t+t^{-1}}{2} \end{pmatrix}$$