Do 2d CFTs define healthy 4d QFTs?

Question

When doing 2d CFTs we typically complexify coordinates and formally consider $mathbb C^2$, with the understanding that, in the end, we are to restrict to the real slice $bar z=z^*$. If we do not impose this, but regard $z,bar z$ as truly independent, does the resulting object define a healthy four-dimensional QFT? If so, is such theory Poincaré invariant? (i.e., does $mathfrak{vir}otimesoverline{mathfrak{vir}}$ contain a subgroup isomorphic to $mathfrak{so}(4)subset!!!!!!small+ mathbb R^4$?) Is it even conformal? (i.e., same as before but with $mathfrak{so}(1,5)$.)

Sylvain Ribault · Accepted Answer

Correlation functions in 2d CFT are single-valued on $mathbb{C}$ but not in $mathbb{C}^2$.
For example, in minimal models, four-point functions are of the type
$$ Z(z) = sum_{i=1}^n c_i F_i(z)F_i(bar z) $$
where $F_i$ is a hypergeometric function or generalization thereof, and $z$ is the cross-ratio of the four positions. In a free boson CFT, they look like
$$ Z(z) = z^a bar z^a $$
where the exponent $a$ is generally not an integer.
If $z$ and $bar z$ are no longer complex conjugates, these expressions become ambiguous. An interpretation in terms of QFT on $mathbb{C}^2$ seems difficult.

Do 2d CFTs define healthy 4d QFTs?

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