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In a pseudo-Euclidean space $\mathbf{E}^{p,q}$, the conformal group is $\mathrm{Conf}(p, q) ≃ \mathrm{O}(p + 1, q + 1) / \mathbb{Z_2}$. See this. Or some notes say that $\mathrm{Conf}(p, q) ≃ \mathrm{SO}(p+1,q+1)$.

But we know $\mathrm{O}(p,q)$ has $4$ connected components and $\mathrm{SO}(p,q)$ has $2$ connected components for $p,q\ge1$. See this.

So if this notation $\mathrm{Conf}(p, q) ≃ \mathrm{O}(p + 1, q + 1) / \mathbb{Z_2}≃\mathrm{SO}(p+1,q+1)$ is correct, then $\mathrm{Conf}(p,q)$ should have $2$ connected components.

However from every textbook of CFT, they firstly derived the Lie algebra (commutation relation) of conformal symmetries. It's isomorphic to $\mathfrak{so}(p+1,1+1)$ or $\mathfrak{o}(p+1,q+1)$ (It doesn't matter because at Lie algebra level they are same.) Then they will say the conformal group is $\mathrm{SO}(p+1,q+1)$ or $\mathrm{O}(p+1,q+1)/\mathbb{Z}_2$. (Francesco's CFT p98)

It puzzles me. Since if they only want to talk about the connected component of conformal group, then neither $\mathrm{SO}(p+1,q+1)$ nor $\mathrm{O}(p+1,q+1)/\mathbb{Z}_2$ is connected. If they want to talk about the full conformal symmetry including the discrete conformal symmetry $P,T$ and inversion $x^\mu \rightarrow x^\mu/x^2$, then I think the full conformal group should be $\mathrm{O}(p+1,q+1)$. Is it right?

Emilio Pisanty
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maplemaple
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  • This question (v5) is answered in my Phys.SE post here. – Qmechanic Jan 16 '18 at 22:51
  • What do you mean by conformal group? You probably know that conformal transformations do not actually honestly act on $\mathbb{R}^{p,q}$, but only on an appropriate completion. The discrete questions of the type you're asking depend on the completion one uses. – Peter Kravchuk Jan 16 '18 at 22:55
  • I.e. note that Qmechanic's referenced answer uses conformal compactification, which in case $pq\neq 0$ is not very physically interesting due to existence of closed timelike curves. – Peter Kravchuk Jan 16 '18 at 23:02
  • @PeterKravchuk I don't know. Maybe common textbooks treat these problems too hand-waving. They directly give the conformal group by introducing the conformal algebra. I'm reading Qmechanic's answer and reference. – maplemaple Jan 16 '18 at 23:45
  • Related post by OP: https://physics.stackexchange.com/q/465947/2451 – Qmechanic Sep 13 '21 at 01:17

1 Answers1

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Firstly, the answer is that $\mathrm{Conf}(p,q)$ is connected. By definition, the conformal group $\mathrm{Conf}(p,q)$ is the connected component that contains the identity in the group of conformal diffeomorphisms of the conformal compactification of $\mathbb{R}^{p,q}$.

To my feeling, the confusion comes from the definition of these groups. Let's recall that $\mathrm{SO}(p+1,q+1)$ is defined as the connected component of the identity in group $\mathrm{O}(p+1,q+1)$. Notice that: -1 may lie in this group. And $\mathrm{SO}(p+1,q+1)$ is a subgroup of $$\{\Lambda\in \mathrm{Conf}(p+1,q+1) : \det \Lambda =1\}.$$ These two facts may against your instincts obtained from usual linear algebra with the metric $(+,\cdots,+)$, but they are true.

What you have written $\mathrm{O}(p+1,q+1)/\{\pm 1\}$ is actually the group of all conformal transformations of the conformal compactification of $\mathbb{R}^{p,q}$. The conformal group is the connected subgroup containing $1$ in this group: (i) if $-1$ is not in the connected component containing 1 of $\mathrm{O}(p+1,q+1)$, we have $\mathrm{Conf}(p,q)\simeq \mathrm{SO}(p+1,q+1)$, (ii) if $-1$ is contained in the connected component containing 1 of $\mathrm{O}(p+1,q+1)$, we have $$\mathrm{Conf}(p,q)\simeq \mathrm{SO}(p+1,q+1)/\{\pm 1\}.$$

Emilio Pisanty
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Zhian Jia
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