This is a rather naive question, but I was just wondering.
I know that the local conformal algebra of 2d Euclidean space is the direct sum L0⊕¯L0,
The global conformal algebra is generated by the generators {L±1,L0}∪{¯L±1,¯L0} and is, thus, the direct sum sl(2,R)⊕¯sl(2,R).
Answer
This is e.g. explained in Ref. 1:
The conformal compactifications of the 1+1D Minkowski (M) plane and the 2+0D Euclidean (E) plane are1 ¯R1,1 ≅ S1×S1
and ¯R2,0 ≅ S2,respectively.The (global) conformal groups are Conf(1,1) ≅ O(2,2;R)/{±14×4}
and Conf(2,0) ≅ O(3,1;R)/{±14×4},with 4 and 2 connected components, respectively.The corresponding connected components connected to the identity are Conf0(1,1) ≅ SO+(2,2;R)/{±14×4} ≅ PSL(2,R)×PSL(2,R)
and Conf0(2,0) ≅ SO+(3,1;R) ≅ PSL(2,C),respectively. Here PSL(2,F)≡SL(2,F)/{±12×2}. See also this related Phys.SE post.
References:
- M. Schottenloher, Math Intro to CFT, Lecture Notes in Physics 759, 2008; Subsections 1.4.2-3, Sections 2.3-5, 5.1-2.
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1 In more detail the conformal compactification of the 1+1D Minkowski plane is ¯R1,1≅(S1×S1)/Z2≅{(x0,x1)∈R2∣(x0,x1) ∼ (x0+2,x1) ∼ (x0,x1+2) ∼ (x0+1,x1+1)} x±=12(x0±x1)≅ {(x+,x−)∈R2∣(x+,x−) ∼ (x++1,x−) ∼ (x+,x−+1)}≅S1×S1,
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