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On page 3 of this paper (https://hal.science/hal-00627906v3/document), the authors say that in the $c\to\infty$ limit only the global generators will survive when computing the conformal block. In particular, they say that because the norm is $$||L_{-n}|\Delta_s \rangle||^2=2n\Delta_s+\frac{c}{12}n(n-1)(n+1),$$ which then goes to infinity for $c\to\infty$ and $\,n\geq 2$, this means that the $L_{-n}$, generators will not survive and hence not contribute to the conformal block.

Why does the norm being infinite mean that these generators/descendents don't survive/contribute to the conformal block?

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Roughly speaking, the contribution of the state $L_{-n}|\Delta_s\rangle$ to a conformal block is $$ \mathcal{F}_{\Delta_s}^{(s)} = \cdots + \frac{\langle V_1V_2 L_{-n}|\Delta_s\rangle \langle V_3V_4L_{-n}|\Delta_s\rangle}{||L_{-n}|\Delta_s\rangle ||^2} + \cdots $$ In this contribution the 3pt functions that appear in the numerator are $c$-independent. So if the denominator becomes infinite, the contribution vanishes.

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  • $\begingroup$ Ah right that's very clear, thank you for your answer! $\endgroup$ Mar 17 at 21:37

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