# Connected components of conformal group ${\rm Conf}(p,q)$ containing $P$, $T$ and conformal inversion are same or different?

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?

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.