# Why is $\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(S^1) \times \rm{Diff}(S^1)$ and not $\rm{Diff}(\mathbb{R}) \times \rm{Diff}(\mathbb{R})$?

The Minkowski metric for $$\mathbb{R}^{1,1}$$ is $$ds^2 = dt^2 - dx^2 = du dv$$ for coordinates $$u = t + x \hspace{1cm} v = t - x$$

If you do any coordinate transformation that acts independently one $$u$$ and $$v$$,

$$u = f(u') \hspace{1 cm} v = g(v')$$ then the metric transforms as $$ds^2 = du dv = d(f(u')) d(f(v')) = f'(u') g'(v') du' dv'.$$

Notice that the metric in $$(u', v')$$ coordinates is just a local rescaling of the original metric $$dudv$$. Therefore this is a conformal transformation. In fact, all conformal transformations of $$\mathbb{R}^{1,1}$$ are of this form.

Why then do I often see written $$\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(S^1) \times \rm{Diff}(S^1)$$? Shouldn't it just be $$\rm{Conf}(\mathbb{R}^{1,1}) = \rm{Diff}(\mathbb{R}) \times \rm{Diff}(\mathbb{R})$$? Why is $$\mathbb{R}$$ "compactified" into $$S^1$$?

• Often see where? Which page? – Qmechanic Jun 17 at 23:02
• One example: "A Mathematical Introduction to Conformal Field Theory" by Schottenloher – user1379857 Jun 17 at 23:08
• Another example is this wikipedia page: en.wikipedia.org/wiki/Conformal_geometry except there's also a semidirect product with $\mathbb{Z}$. – user1379857 Jun 17 at 23:25
• On the wiki page it explicitly says that it's working with the compactified space. – Aaron Jun 17 at 23:39
• But what's the point of working in compactified space for $\mathbb{R}^{1,1}$? It doesn't seem to me like anything is gained, but I feel like there must be some point. – user1379857 Jun 17 at 23:45

## 1 Answer

TL;DR: The compactified definition follows a long & useful tradition in e.g. projective & conformal geometry, where one wants to treat infinity on the same footing as a point.

The upshot is that the local conformal groupoid $${\rm LocConf}(1,1)\tag{A}$$ for the 1+1D Minkowski plane is defined in mathematics as the set of locally defined conformal transformations on its conformal compactification$$^1$$ $$\overline{\mathbb{R}^{1,1}}~\cong~\mathbb{S}^1\times \mathbb{S}^1 .\tag{B}$$ $${\rm LocConf}(1,1)$$ contains 4 connected components. The connected component that contains the identity element is $${\rm LocConf}_0(1,1)~\cong~ {\rm LocDiff}^+(\mathbb{S}^1)\times {\rm LocDiff}^+(\mathbb{S}^1).\tag{C}$$ For more details, see e.g. Ref. 1 and this & this Phys.SE posts.

References:

1. M. Schottenloher, Math Intro to CFT, Lecture Notes in Physics 759, 2008; p. 37.

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$$^1$$ It should be mentioned that the conformal compactification contains closed timelike loops, cf. above comment by Cinaed Simson.

• In dimensions other than 1+1 D I can see the point for compactification because it allows you to include inversions. However, in 1+1 D I see no benefit to this compactification. The reason it disturbs me is because the Witt Algebra for example can be thought of as vector fields on $S^1$, which are different from vector fields on $\mathbb{R}$. Wouldn't the compactification change important facts about the algebra of diffeomorphisms? Again, I can see the usefulness in higher dimensions, but not in 1+1. – user1379857 Jun 19 at 22:32
• There are also inversions in 1+1D. – Qmechanic Jun 20 at 12:09