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In a Conformal Field Theory in arbitrary dimensions, four-point functions are constrained to take the form

$$\langle O_1(x_1)O_2(x_2)O_3(x_3)O_4(x_4)\rangle = K(x_i)F(z,\bar z)$$

where $K(x_i)$ is a conformally-covariant prefactor and $(z,\bar z)$ are the cross ratios defined by $$u = \dfrac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2},\quad v=\dfrac{x_{14}^2x_{23}^2}{x_{13}^2x_{24}^2},\quad u=z\bar z,\quad v=(1-z)(1-\bar z).$$

The function $F(z,\bar z)$ is constrained by crossing symmetry in any number of dimensions, but apart from that it is arbitrary. Now in a two-dimensional CFT apart from the global conformal symmetry we have the local Virasoro symmetry as well. So it would seem to me that there had to be some signature of this Virasoro symmetry in the corresponding $F(z,\bar z)$.

In other words: in any number of dimensions a four-point function is given by prescribing some function $F(z,\bar z)$. In that case, if I were given some $F(z,\bar z)$ how can I check whether this particular choice is compatible with Virasoro symmetry - and so a possible two-dimensional four-point function - or not, and hence not a two-dimensional four-point function? What is the signature of Virasoro symmetry left in $F(z,\bar z)$?

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Here is my blog post about this very question:

http://researchpracticesandtools.blogspot.com/2020/09/does-this-covariant-function-belong-to.html

Short answer: without more assumptions, any $F(z,\bar z)$ may belong to a CFT (or several CFTs) with Virasoro symmetry. However, for the CFT to be interesting and tractable, you need to be able to decompose $F(z, \bar z)$ into a reasonable family of conformal blocks. Here reasonable might mean finite, or discrete, or diagonal, for example.

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  • $\begingroup$ Thanks @SylvainRibault! Your blog post addresses exactly my question. So it seems like in the end, even some $F(z,\bar z)$ coming from a higher-dimensional CFT would a priori make sense as a two-dimensional four-point function right? By the way, I got interested in your comment about higher symmetry algebras. Is there an analogue of Virasoro blocks for $w_{1+\infty}$? $\endgroup$
    – Gold
    Commented Jun 14, 2023 at 12:37
  • $\begingroup$ If by 'making sense' you mean not being manifestly inconsistent, sure, pretty much anything goes. If you mean a function from which you can extract useful information, by decomposing it into Virasoro blocks, then probably no, unless some nontrivial relation predicts that higher-dimensional correlators mean something in 2d. $\endgroup$ Commented Jun 14, 2023 at 15:33
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    $\begingroup$ You can define conformal blocks for any W-algebra and in principle compute them, but in general this is awfully complicated. $\endgroup$ Commented Jun 14, 2023 at 15:34

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