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It is well-known that 2d global conformal invariance constrains the 2, 3-point functions to some very simple form, and 4-point function must be $$ f(\eta, \bar \eta) \prod_{i < j}z_{ij}^{...} \bar z_{ij}^{...} $$ where $\eta$ is the harmonic cross ratio.

The derivation of the above results relies on the conformal Ward identity, for example the infinitesimal one $$ \delta_\epsilon \langle X \rangle = - \sum_{i} \epsilon(w)\partial_{w_i} + \partial \epsilon(w) h_i \ , $$ and the fact that $\delta_\epsilon \langle X \rangle = 0$ when $\epsilon$ is a generator of the global conformal transformations.

But what about local conformal transformations? Can we place some constraints on the correlation functions by looking at the local ones? I can think of some possibilities.

  • In Di Francesco's book, the global conformal transformations are referred to as "real symmetry" (above eq (5.49)), and perhaps the local ones aren't real symmetry, and shouldn't be considered in this context.

  • Even if one consider the local ones, $\delta_\epsilon \langle X\rangle$ depends on the chosen $X$, and therefore no universal statement can be made.

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    $\begingroup$ Local conformal transformations do not preserve the vacuum state so it cannot be used to constrain correlation functions. They can (and are) be used to constrain operator product expansions of the theory. $\endgroup$
    – Prahar
    Commented Oct 20, 2022 at 8:47
  • $\begingroup$ @Prahar Ok, do you mean $L_{-n}|0\rangle \ne 0$ (when $n > 1$)? If so (correct me if I'm wrong), does $L_{n \ge -1}|0\rangle$ implies any invariance of the vacuum $|0\rangle$? $\endgroup$
    – user31415926
    Commented Oct 20, 2022 at 15:34
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    $\begingroup$ It does not. $|0\rangle$ is preserved by $L_n$ for $n>-1$ and $\langle 0|$ is preserved by $L_n$ for $n<1$. Both states are preserved only by $L_{-1}$, $L_0$ and $L_1$. Consequently, only these generates can be used to constrain correlators $\endgroup$
    – Prahar
    Commented Oct 20, 2022 at 17:17
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    $\begingroup$ Another way to see what Prahar is saying is to derive Ward identities from Ward identities of stress-tensor. In global case the only singularities are OPE, in local case you have singularities of the conformal killing vector fields. The result is that local transformations allow you to express correlators with more T insertions in terms of correlators with fewer. $\endgroup$
    – Peter Kravchuk
    Commented Oct 20, 2022 at 20:05

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Local conformal symmetry does not impose any constraint on a given correlation function. But it determines correlation functions of descendant fields in terms of correlation functions of primary fields. See Wikipedia: https://en.wikipedia.org/wiki/Two-dimensional_conformal_field_theory#Conformal_Ward_identities

In practice however, correlation functions of two-dimensional CFTs tend to have decompositions into Virasoro conformal blocks that are particularly "simple": for example, in minimal models, there are finitely many blocks. See my blog post on this subject: http://researchpracticesandtools.blogspot.com/2020/09/does-this-covariant-function-belong-to.html

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  • $\begingroup$ I was reading your note Conformal field theory on the plane, and was confused by the eq (2.2.13). Does that statement come from the fact that combined function $\epsilon(z) \langle T(z) \prod V(z_i) \rangle$ has vanishing residue (in $z$) at the $\infty$, because $T(z \to \infty) \sim 1/z^4$, and $\epsilon$ is at most quadratic? $\endgroup$
    – user31415926
    Commented Nov 4, 2022 at 10:03
  • $\begingroup$ Yes, indeed. Integrating something that has zero residue at infinity over a contour that goes around infinity, we find zero. $\endgroup$
    – Sylvain Ribault
    Commented Nov 5, 2022 at 12:22

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