I'm wondering if there is more than the empty theory (no local fields, identically vanishing stress energy tensor) that can have central charge $c$ equals to $0$?

My intuition tells me no, the stress energy tensor would transform as a primary field and so would only contain primary fields but there are none except for the identity operator. So there could only be (topological) degrees of freedom on the boundary. Is this reasoning wrong?

ps: I'm only thinking at unitary theories, otherwise, sure it could be possible to combine in a balanced way fields carrying positive and negative central charge.


Your argument is correct. Indeed (in $d$ dimensions) when $c = 0$, unitarity implies that $T_{\mu \nu} \equiv 0$, but by assumption any local CFT must have a stress tensor $T_{\mu \nu}$ satisfying $$ T_{\mu \nu}(x) \times \mathcal{O}(y) \sim \Delta \mathcal{O}(y)$$ (a Ward identity) where $\Delta$ is the dimension $\mathcal{O}.$ So unless $\Delta = 0$ (the unit operator), you fail to satisfy this condition.

  • $\begingroup$ Do you mean to restrict yourself to even dimensions? In odd dimensions there is no conformal anomaly. $\endgroup$ – Matthew Feb 17 '14 at 20:35
  • $\begingroup$ No, here I define $c$ as follows: $\langle T(x) T(0) \rangle = c / (x^2)^d$ (omitting indices). Maybe you need an extra factor of $1/2$ to agree with other people's conventions. $\endgroup$ – Vibert Feb 17 '14 at 22:03

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