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In $D=2$, we can have locally analytic transformations that cannot be globally well-defined.

However, for CFTs in $D>2$, we have only the global group. Why is that?

Also, is it a statement that depends on topology? Any references on that?

Qmechanic
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1 Answers1

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  1. Let $\overline{\mathbb{R}^{p,q}}$ denote the conformal compactification of $\mathbb{R}^{p,q}$. Let $n:=p+q$ denote the dimension.

  2. (The pseudo-Riemannian generalization of) Liouville's rigidity theorem states that if $n\geq 3$, then all local conformal transformations of $\overline{\mathbb{R}^{p,q}}$ can be extended to global conformal transformations. For a proof, see e.g. this Phys.SE post.

  3. The upshot of Liouville's rigidity theorem is that for $n\geq 3$ there are only relatively few local conformal transformations of $\overline{\mathbb{R}^{p,q}}$ [which consist of (composition of) translations, similarities, orthogonal transformations and inversions], and these are easily extendable to global ones.

  4. On the other hand, for $n=2$, there are many more local conformal transformations. See also e.g. this Phys.SE post.

  5. Similar rigidity results (in the smooth case) hold on any conformal manifold $M$, cf. Wikipedia.

Qmechanic
  • 201,751
  • Thanks for this great answer. I knew how the distinction between d=2,3 arises in the derivation of conformal maps, but wanted to know if that was explained in detail somewhere. This helps. – Srivatsan Balakrishnan Feb 03 '15 at 21:41
  • There is still the question about topology, and the business of finding conformal transformations in manifolds that are not conformally related to flat space. – Srivatsan Balakrishnan Feb 03 '15 at 21:45