I am learning about complexification of real Lie groups. From the Wikipedia article, I see that I have a lot to learn.
I learned from my earlier question that the real Lie group $SU(n)$ is a real form of $SL(n,\mathbb{C})$, or in other words, $SL(n,\mathbb{C})$ is the complexification of $SU(n)$.
For the corresponding Lie algebras I verified that given $C\in \mathfrak{sl}(n,\mathbb{C})$ I can find $A,B\in \mathfrak{su}(n)$ such that $C=A+iB$. And similarly, if $A,B\in \mathfrak{su}(n)$, then $A+iB \in \mathfrak{sl}(n,\mathbb{C})$. Which I understand to mean that $\mathfrak{sl}(n,\mathbb{C})=\mathfrak{su}(n) \bigoplus \mathfrak{su}(n)$, and, in fact, $\mathfrak{sl}(n,\mathbb{C})=\mathfrak{su}(n) \bigotimes \mathbb{C}$.
But as regards the Lie groups, I fail to similarly break down a special linear matrix $C\in SL(2,\mathbb{C})$ into two special unitary matrices $A,B\in SU(2)$. For example, when I solve:
$$\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}a & b\\ -b^* & a^*\end{pmatrix} + i \begin{pmatrix}c & d\\ -d^* & c^*\end{pmatrix}$$ I get: $$\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & \frac{1}{2}\\ -\frac{1}{2} & 1\end{pmatrix} + i \begin{pmatrix}0 & -i\frac{1}{2}\\ -i\frac{1}{2} & 0\end{pmatrix}$$ The last two matrices are in $U(2)$ but not in $SU(2)$ because their determinants are not equal to $1$.
So there is something more subtle going on here with the complexification.
Please, as concretely as possible, how does complexification of $SU(2)$ yield $SL(2,\mathbb{C})$? Additionally, I am curious what this example could help me understand in the Wikipedia article. Thank you!
