Temperature in CFT

Question

Non-vanishing Temperature can break conformal symmetry(Can anyone show this point explicitly), my question is that in AdS/CFT the temperature of boundary field theory is non-zero, why the boundary field theory whose conformal symmetry is breaking is still a conformal field theory?

This post imported from StackExchange Physics at 2014-10-20 10:36 (UTC), posted by SE-user Yuan Sun

Comments

Non-vanishing temperature already breaks Lorentz symmetry as the concept of temperature requires a distinguished time direction. However, only the state has the broken symmetry, not the theory itself. This should answer your second question, too.

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 I am not being very clear about " only the state has the broken symmetry, not the theory itself." Could you give more comment about this? 

I read the followings just now in Freedman's book "Supergravity" (Page 528) which is helpful.

"If the gravity solution is pure AdS$_{d+1}$, these correlators are covariant under conformal transformations. If the bulk solution looks like AdS$_{d+1}$ near the boundary but differs in the interior, then conformal symmetry is broken. The boundary data can then act as sources for operators, which perturb a conformal theory and induce renormalization group flows. An alternative is that conformal symmetry is broken by non-vanishing onepoint functions <O(x)>, which indicate that the boundary CFT$_d$ is not in its ground state.If the bulk spacetime includes a black hole with non-zero Hawking temperature, then the gravity dual is a CFT in a thermal ensemble."

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"only the state has the broken symmetry, not the theory itself". This means that even for a conformal theory whose ground state has the full conformal symmetry, nevertheless all thermal states have (necessarily) broken Lorentz (and hence conformal) symmetry. The expectations are taken with respect to the thermal (KMS) state, and because of the broken symmetry, a vector operator can have a non-vanishing one-point function.