This is a rather naive question, but I was just wondering.
I know that the local conformal algebra of 2d Euclidean space is the direct sum \begin{equation} \cal{L}_0\oplus\overline{\cal{L}_0}, \end{equation} where $\cal{L}_0$ and $\overline{\cal{L}_0}$ are two independent Witt algebras. The respective conformal group is $Z\otimes\bar Z$, where $Z$ consists of all the holomorphic and $\bar Z$ of all the anti-holomorphic coordinate transformations.
The global conformal algebra is generated by the generators $\{L_{\pm 1}, L_0\}\cup\{\overline{L}_{\pm 1}, \overline{L}_0\}$ and is, thus, the direct sum \begin{equation} \text{sl}(2,\mathbb{R})\oplus\overline{\text{sl}(2,\mathbb{R})}. \end{equation} I have read that the global conformal group is the group $\text{SL}(2,\mathbb{C})/\mathbb{Z_2}$, however shouldn't it be the group \begin{equation} \text{SL}(2,\mathbb{R})/\mathbb{Z_2}\hspace{0.2cm}\times\hspace{0.2cm}\overline{\text{SL}(2,\mathbb{R})/\mathbb{Z_2}}\qquad ? \end{equation}