Global conformal group in 2D Euclidean space

Question

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}

Answer 1

This is e.g. explained in Ref. 1:

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.

2. 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.

3. The corresponding connected components connected to the identity are $$\begin{align}{\rm Conf}_0(1,1)~\cong~&SO^+(2,2;\mathbb{R})/\{\pm {\bf 1}_{4\times 4}\}\cr ~\cong~& PSL(2,\mathbb{R})\times PSL(2,\mathbb{R}) \end{align}\tag{3M}$$ and $$\begin{align} {\rm Conf}_0(2,0)~\cong~&SO^+(3,1;\mathbb{R})\cr ~\cong~& PSL(2,\mathbb{C}), \end{align}\tag{3E}$$ 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:

1. 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 $$ \begin{align}&\overline{\mathbb{R}^{1,1}}\cr &\cong(\mathbb{S}^1\times \mathbb{S}^1)/\mathbb{Z}_2 \cr &\cong\left\{(x^0,x^1)\in\mathbb{R}^2 \mid (x^0,x^1)\sim(x^0\!+\!2,x^1)\sim(x^0,x^1\!+\!2)\sim(x^0\!+\!1,x^1\!+\!1)\right\}\cr &\stackrel{x^{\pm}=\frac{1}{2}(x^0\pm x^1)}{\cong}\left\{(x^+,x^-)\in\mathbb{R}^2 \mid (x^+,x^-)\sim(x^+\!+\!1,x^-)\sim(x^+,x^-\!+\!1)\right\}\cr &\cong\mathbb{S}^1\times \mathbb{S}^1 ,\end{align}\tag{4M}$$ with Minkowski metric $$\begin{align}\mathbb{g}~~~~~=~~~~~&\mathrm{d}x^0\odot\mathrm{d}x^0-\mathrm{d}x^1\odot\mathrm{d}x^1 \cr~\stackrel{x^{\pm}=\frac{1}{2}(x^0\pm x^1)}{=}&~4\mathrm{d}x^+\odot\mathrm{d}x^- \end{align}.\tag{5M}$$

Answer 2

The complexified global conformal algebra is indeed generated (over $\mathbb{C}$) by $L_0,L_{\pm 1},\bar L_0, \bar L_{\pm 1}$. But the real global conformal algebra is $sl(2,\mathbb{C})$, with the generators (over $\mathbb{R}$) $$ L_n+\bar L_n \quad ,\quad i(L_n-\bar L_n) $$ For example, $i(L_0-\bar L_0)$ generates rotations $z\to e^{i\theta} z$, which also act on $\bar z$ as $z\to e^{-i\theta}\bar z$. More generally, the global conformal group is $SL(2,\mathbb{C})/\mathbb{Z}_2$, which acts as $z\to \frac{az+b}{cz+d}$ with $a,b,c,d\in\mathbb{C}$.