3

As we known (see this post), the global conformal group for $\mathbb{R}^{p,q}$ is $$ {\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \}$$

The global conformal group ${\rm Conf}(p,q)$ has 4 connected components if both $p$ and $q$ are odd, and 2 connected components if $p$ or $q$ are even.

The connected component that contains the identity element is $$ {\rm Conf}_0(p,q)~\cong~\left\{\begin{array}{ll} SO^+(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \} &\text{if both $p$ and $q$ are odd},\cr SO^+(p\!+\!1,q\!+\!1) &\text{if $p$ or $q$ are even}.\end{array}\right.$$

We also know that spatial reflection ($P : x^1 \rightarrow -x^1$), conformal inversion ($I : x^\mu \rightarrow x^\mu/x^2$) and time reversal ($T : x^0 \rightarrow -x^0$)(exist if both $p$ and $q$ are nonzero) are discrete symmetry.

My questions:

  1. For $\mathbb{R}^D$, isometry group is $O(D)\cong SO(D) \cup P\ SO(D)$ which has $2$ connected components. $ {\rm Conf}(\mathbb{R}^D)~\cong~O(1,D)/\{\pm {\bf 1} \}$ also has two connected components. Does it means that $P$ and $I$ belong to the same connected component? i.e $P\ {\rm Conf}_0(\mathbb{R}^D) = I\ {\rm Conf}_0(\mathbb{R}^D)$ and $P\ {\rm Conf}_0(\mathbb{R}^D) \cap {\rm Conf}_0(\mathbb{R}^D)=\varnothing$?

  2. For $\mathbb{R}^{1,D-1}$ with $D$ odd, isometry group is $$O(1,D-1)\cong SO^+(1,D-1) \cup P\ SO^+(1,D-1) \cup T\ SO^+(1,D-1) \cup PT\ SO^+(1,D-1)$$ which has $4$ connected components. However $ {\rm Conf}(\mathbb{R}^{1,D-1})~\cong~O(2,D)/\{\pm {\bf 1} \}$ only has two connected components. What's the relationship between $ {\rm Conf}_0(\mathbb{R}^{1,D-1})$, $T\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$, $P\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$, $PT\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$ and $I\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$? It seems that some should be different while some should same.

  3. For $\mathbb{R}^{1,D-1}$ with $D$ even, isometry group is $$O(1,D-1)\cong SO^+(1,D-1) \cup P\ SO^+(1,D-1) \cup T\ SO^+(1,D-1) \cup PT\ SO^+(1,D-1)$$ which has $4$ connected components. In this case, $ {\rm Conf}(\mathbb{R}^{1,D-1})~\cong~O(2,D)/\{\pm {\bf 1} \}$ also has $4$ connected components. What's the relationship now between $ {\rm Conf}_0(\mathbb{R}^{1,D-1})$, $T\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$, $P\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$, $PT\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$ and $I\ {\rm Conf}_0(\mathbb{R}^{1,D-1})$?

  4. From question $2$ and $3$, it seems that some different connected components in case $3$ will be same in case $2$, why?

Qmechanic
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maplemaple
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1 Answers1

1

  1. It is convenient to block-decompose elements $$ \Lambda~=~\begin{pmatrix} a & b \cr c& d\end{pmatrix}\tag{1}$$ of the indefinite real orthogonal group $O(p\!+\!1,q\!+\!1)$. The group $O(p\!+\!1,q\!+\!1)$ has $2\times 2=4$ connected components $$G^{++}, \quad G^{+-},\quad G^{-+}, \quad G^{--},\tag{2}$$ characterized by the signs of $\det(a)$ and $\det(d)$. We define $$\begin{align}O^+(p\!+\!1,q\!+\!1) ~:=~&\{\Lambda\in O(p\!+\!1,q\!+\!1)\mid \det(a)>0\}\cr ~=~&G^{++}\cup G^{+-} .\end{align} \tag{3}$$ The full determinant satisfies $$\det(\Lambda)~=~{\rm sgn}\det(a)~{\rm sgn}\det(d).\tag{4}$$ The connected component that contains the identity element is $$SO^+(p\!+\!1,q\!+\!1)=G^{++}.\tag{5}$$

  2. Let us now consider the (global) conformal group $$ {\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \}, \tag{6}$$ cf. e.g. this Phys.SE post. The connected component that contains the identity element is $$ {\rm Conf}_0(p,q)~\cong~\left\{\begin{array}{ll} G^{++}/\{\pm {\bf 1}\} &\text{if both $p$ and $q$ are odd},\cr G^{++} &\text{if $p$ or $q$ are even}.\end{array}\right.\tag{7}$$ The two cases in eq. (7) correspond to whether $-{\bf 1}\equiv PT\in G^{++}$ or not, respectively.

  3. Case $p$ and $q$ are odd: ${\rm Conf}(p,q)$ has 4 connected components: $$\begin{align} G^{++}/\{\pm {\bf 1}\},& \quad G^{+-}/\{\pm {\bf 1}\},\cr G^{-+}/\{\pm {\bf 1}\},& \quad G^{--}/\{\pm {\bf 1}\}. \end{align}\tag{8} $$ Case $p$ and $q$ are even: ${\rm Conf}(p,q)$ has 2 connected components: $$\begin{align} G^{--}/\{\pm {\bf 1}\}~=~& G^{++}/\{\pm {\bf 1}\}~\cong~G^{++}, \cr G^{+-}/\{\pm {\bf 1}\}~=~&G^{-+}/\{\pm {\bf 1}\} .\end{align} \tag{9} $$ Case $p$ odd and $q$ even: ${\rm Conf}(p,q)$ has 2 connected components: $$\begin{align} G^{+-}/\{\pm {\bf 1}\}~=~& G^{++}/\{\pm {\bf 1}\}~\cong~G^{++}, \cr G^{--}/\{\pm {\bf 1}\}~=~&G^{-+}/\{\pm {\bf 1}\} .\end{align} \tag{10} $$ Case $p$ even and $q$ odd: ${\rm Conf}(p,q)$ has 2 connected components: $$\begin{align} G^{-+}/\{\pm {\bf 1}\}~=~& G^{++}/\{\pm {\bf 1}\}~\cong~G^{++}, \cr G^{--}/\{\pm {\bf 1}\}~=~&G^{+-}/\{\pm {\bf 1}\} .\end{align} \tag{11} $$ This should essentially answer OP's questions.

Qmechanic
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