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A standard QFT cannot be defined as a set of Poincare-invariant correlation functions because this does not take into account the possibility of non-perturbative effects (e.g. instantons)

Can we define a CFT as a set of conformally invariant correlation functions?

What is the correct definition of a CFT?

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1 Answer 1

The exact correlation functions as defined by a lattice simulation do take into account all nonperturbative effects, they contain all the physics. It is only the expansion of the correlation functions that doesn't take instantons into account. So yes, you can define a CFT by its correlation functions.

This is true for the usual situation of fields which can be added naturally and averaged. This excludes cases where the fields are of the sigma model type--- you can't add points on a manifold. In this case, an ad-hoc solution is to embed the sigma model into a larger R^n where you can define addition of points, and then you can define the correlation functions.

The definition of CFT is when you have conformal invariant correlation functions.

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Hi Ron, i think correlation functions are only required to be conformally covariant. –  user10001 Oct 9 '12 at 2:16
    
@dushya: What's the mistake? I didn't get it. –  Ron Maimon Oct 9 '12 at 3:01
    
Sorry, I meant invariance is stronger condition than covariance; and in CFT correlation functions may not be conformally invariant. right ? –  user10001 Oct 9 '12 at 3:05
    
@dushya: Oh, yes, of course--- I assume the "abstract object" correlation function is invariant, which means it's actual value is covariant--- it's mathematician speak, which in this case, I like better, but you are right. –  Ron Maimon Oct 9 '12 at 14:28
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