Why do we need conformal compactification to define the global conformal group?

Question

First I have the definition of a conformal map. Let $(M,g)$ and $(M',g')$ be two pseudo-Riemannian manifolds of same dimension. Let $U\subset M$ and $V\subset M'$, we say that a smooth map of maximal rank $\Phi : U\to V$ is a conformal map if there is some smooth $\Omega : U\to [0,+\infty)$ such that $$\Phi^\ast g'=\Omega^2 g.\tag{1}$$

Now, intuitively I would imagine that by a global conformal transformation one means just a globally defined conformal map $\Phi : M\to M'$.

My intuition would tell that to define the global conformal group of $M$ we would just need to take all globally defined conformal diffeomorphisms $\Phi : M\to M$.

Still one usually introduces a conformal compactification here. Take for instance this Phys.SE post or rather the book "A Mathematical Introduction to Conformal Field Theory" by M. Schottenloher. A canonical reference taking the same approach is the paper "Relativistic Symmetry Groups" by Penrose.

All these authors are talking about $(M,g) = \mathbb{R}^{p,q}$. To define the global conformal group they pick $\overline{\mathbb{R}^{p,q}}$ the conformal compactification and define $\operatorname{Conf}(p,q)$ the group of globally defined conformal transformations on it.

Why is that? Why not just define $\operatorname{Conf}(p,q)$ to be the group of globally defined conformal diffeomorphisms on $\mathbb{R}^{p,q}$ itself, i.e., maps $\Phi : \mathbb{R}^{p,q}\to \mathbb{R}^{p,q}$ satisfying Eq. (1) with the canonical metric tensor? Why pass to the conformal compactification?

Answer 1

For starters, several conformal transformations, e.g. the special conformal transformations, take finite points $p\in M$ to $\infty\notin M$, which technically violates OP's suggested definition.

For Euclidean space $\mathbb{R}^n$, the conformal compactification $\overline{\mathbb{R}^n}\cong \mathbb{S}^n$ is the one-point compactification, i.e. the $n$-sphere, which indicates that $\infty$ should be treated on equal footing with other points.

The notion of infinity becomes more subtle in the case of indefinite metric, and the conformal compactification helps resolve this, cf. e.g. this Phys.SE post.

Answer 2

I think one can define conformal group of $\mathbb{R}^{p,q}$ itself, and that will be just Poincare plus scaling without inversion (special conformal transformation). But for field theory, Poincare plus scaling imply full conformal symmetry under some assumptions, so I guess that's some physical reason? To consider the full conformal symmetry we need conformal compactification.