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When a conformal block has dimensions and spin that violates its unitary bounds, does that make the block equal to zero. I'm asking because I'm trying to calculate 3D conformal blocks via a recursion relation and get blocks in the relation that violate unitary bounds.

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  • $\begingroup$ No, it doesn't. $\endgroup$ Jul 5 '18 at 6:32
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Conformal blocks are analytic functions of conformal dimensions, analogous to characters of representations. Their definition has nothing to do with unitarity. So no, violating unitarity bounds does not make blocks vanish.

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  • $\begingroup$ Got it, thank you very much. However now that we are on the subject, I do have one more quick question, does a conformal block of negative spin vanish? $\endgroup$ Jul 5 '18 at 16:48
  • $\begingroup$ If you mean conformal spin, the answer is no: conformal spin is the difference between left and right dimensions, and there is actually a symmetry between negative and positive conformal spins. $\endgroup$ Jul 5 '18 at 17:44
  • $\begingroup$ Oops, my comment was valid only in 2d. In general dimension I believe that the spin is defined as a positive integer (or half-integer) and negative spins need not a priori make sense. $\endgroup$ Jul 5 '18 at 18:34

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