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 $S^1$?

  • $\begingroup$ Often see where? Which page? $\endgroup$
    – Qmechanic
    Jun 17, 2019 at 23:02
  • $\begingroup$ One example: "A Mathematical Introduction to Conformal Field Theory" by Schottenloher $\endgroup$ Jun 17, 2019 at 23:08
  • $\begingroup$ Another example is this wikipedia page: en.wikipedia.org/wiki/Conformal_geometry except there's also a semidirect product with $\mathbb{Z}$. $\endgroup$ Jun 17, 2019 at 23:25
  • 2
    $\begingroup$ On the wiki page it explicitly says that it's working with the compactified space. $\endgroup$
    – Aaron
    Jun 17, 2019 at 23:39
  • $\begingroup$ 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. $\endgroup$ Jun 17, 2019 at 23:45

1 Answer 1


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.


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


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

  • $\begingroup$ 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. $\endgroup$ Jun 19, 2019 at 22:32
  • $\begingroup$ There are also inversions in 1+1D. $\endgroup$
    – Qmechanic
    Jun 20, 2019 at 12:09

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