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 L0⊕¯L0, 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}
Answer
This is e.g. explained in Ref. 1:
The conformal compactifications of the 1\!+\!1D Minkowski (M) plane and the 2\!+\!0D Euclidean (E) plane are^1 \overline{\mathbb{R}^{1,1}}~\cong~\mathbb{S}^1\times \mathbb{S}^1 \tag{1M} and \overline{\mathbb{R}^{2,0}}~\cong~\mathbb{S}^2, \tag{1E} respectively.
The (global) conformal groups are {\rm Conf}(1,1)~\cong~O(2,2;\mathbb{R})/\{\pm {\bf 1}_{4\times 4}\}\tag{2M} and {\rm Conf}(2,0)~\cong~O(3,1;\mathbb{R})/\{\pm {\bf 1}_{4\times 4}\}, \tag{2E} with 4 and 2 connected components, respectively.
The corresponding connected components connected to the identity are {\rm Conf}_0(1,1)~\cong~SO^+(2,2;\mathbb{R})/\{\pm {\bf 1}_{4\times 4}\}~\cong~PSL(2,\mathbb{R})\times PSL(2,\mathbb{R}) \tag{3M} and {\rm Conf}_0(2,0)~\cong~SO^+(3,1;\mathbb{R})~\cong~PSL(2,\mathbb{C}), \tag{3E} respectively. Here PSL(2,\mathbb{F})\equiv SL(2,\mathbb{F})/\{\pm {\bf 1}_{2\times 2}\}. See also this related Phys.SE post.
References:
- M. Schottenloher, Math Intro to CFT, Lecture Notes in Physics 759, 2008; Subsections 1.4.2-3, Sections 2.3-5, 5.1-2.
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^1 In more detail the conformal compactification of the 1\!+\!1D Minkowski plane is \begin{align}\overline{\mathbb{R}^{1,1}}&\qquad\cong\qquad(\mathbb{S}^1\times \mathbb{S}^1)/\mathbb{Z}_2 \cr &\qquad\cong\qquad\left\{(x^0,x^1)\in\mathbb{R}^2 \mid (x^0,x^1)~\sim~(x^0\!+\!2,x^1)~\sim~(x^0,x^1\!+\!2)~\sim~(x^0\!+\!1,x^1\!+\!1)\right\}\cr &~~\stackrel{x^{\pm}=\frac{1}{2}(x^0\pm x^1)}{\cong}~\left\{(x^+,x^-)\in\mathbb{R}^2 \mid (x^+,x^-)~\sim~(x^+\!+\!1,x^-)~\sim~(x^+,x^-\!+\!1)\right\}\cr &\qquad\cong\qquad\mathbb{S}^1\times \mathbb{S}^1 ,\end{align}\tag{4M} with Minkowski metric \mathbb{g}~=~\mathrm{d}x^0\odot\mathrm{d}x^0-\mathrm{d}x^1\odot\mathrm{d}x^1~\stackrel{x^{\pm}=\frac{1}{2}(x^0\pm x^1)}{=}~4\mathrm{d}x^+\odot\mathrm{d}x^- .\tag{5M}
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