Monday, 15 October 2018

mathematical physics - Global conformal group in 2D Euclidean space


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,

where L0 and ¯L0 are two independent Witt algebras. The respective conformal group is ZˉZ, where Z consists of all the holomorphic and ˉZ of all the anti-holomorphic coordinate transformations.


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).

I have read that the global conformal group is the group SL(2,C)/Z2, however shouldn't it be the group SL(2,R)/Z2ׯSL(2,R)/Z2?



Answer



This is e.g. explained in Ref. 1:





  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.




  2. 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.




  3. 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:




  1. M. Schottenloher, Math Intro to CFT, Lecture Notes in Physics 759, 2008; Subsections 1.4.2-3, Sections 2.3-5, 5.1-2.


--


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,

with Minkowski metric g = dx0dx0dx1dx1 x±=12(x0±x1)= 4dx+dx.


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