In CFT, one constructs the generators of various confromal transformations, and re-expresses them in terms of $J_{ab}$'s that manifestly satisfy commutation relations of $so(d+1,1)$ (taking Euclidean spacetime). Then it is claimed, everywhere I looked and in particular in Di Francesco et al page 98, that this forms an isomorphism between the conformal group in $d$ dimensions and $SO(d+1,1)$.

My question is: why $SO(d+1,1)$ and not $O(d+1,1)$?

  • 2
    $\begingroup$ Dilatations are part of the $SO(d+1, 1)$ algebra. They have det $=1$. $\endgroup$
    – M.Jo
    Nov 18 '19 at 17:24
  • $\begingroup$ Yes, you are right, was sloppy. Edited. Still, how can one see that it is SO and not O? $\endgroup$
    – Yoni
    Nov 18 '19 at 17:56
  • 2
    $\begingroup$ The exponential mapping links the generators of a Lie algebra to the conncted component of identity of a Lie group. $\endgroup$
    – DanielC
    Nov 18 '19 at 18:25
  • $\begingroup$ ((cosθ, sinθ),(sinθ,-cosθ)) is orthogonal and so in O(2); but not in SO(2). You really don't want to take its logarithm. $\endgroup$ Nov 18 '19 at 22:51
  • $\begingroup$ Ok, is there anything else you are willing to share? No, I did not get it. What's the point in counting the generators if SO(p,q) is disconnected? Of course the group dimension is larger than the generators, but how is that useful information in this case? Also, and maybe I misunderstand here, but O and SO share the same algebra $\endgroup$
    – Yoni
    Nov 20 '19 at 22:20

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