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

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?

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.
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.