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?

  • 2
    $\begingroup$ A non-vanishing temperature already breaks Lorentz symmetry as explained here. $\endgroup$
    – Dilaton
    Commented Oct 20, 2014 at 19:46
  • $\begingroup$ Related $\endgroup$
    – MannyC
    Commented Sep 6, 2019 at 13:54

1 Answer 1


AdS/CFT correspondence tells you (among other things) the dual geometry for each state of the boundary theory. As @Arnold says (check the link in the comment), a finite temperature state ("spontaneously") breaks Lorentz, and hence, conformal invariance. That's okay because excited states could break many symmetries of the theory (eg: p-shell of electron cloud around a Hydrogen atom breaks SO(3) symmetry). So it's not like you're losing the conformal symmetry describing the dynamics.

(In fact, I wonder whether one might be able to use something like the ideas behind spurion analysis -- on the finite temperature state, since the temperature is the only thing "breaking" the conformal symmetry, spontaneously :-?)


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