I have a vague intuition – most likely incorrect – that all (internal) flavor symmetries in a CFT are always spontaneously broken. (Perhaps under some mild assumptions, such as unitarity and that the conformal symmetry itself is unbroken).

Is this actually provable? Or are there any known counterexamples (i.e., a CFT with a faithful global symmetry that is unbroken)? Are there counterexamples for both continuous and discrete symmetries?

I mostly care about $d>2$ since Coleman–Mermin–Wagner won't allow continuous symmetries to break in 2d. But of course if there is anything known about 2d specifically I would also like to hear it.

  • $\begingroup$ I'm a bit ignorant on CFTs, but wouldn't spontaneous breaking require some (relevant?) deformations? If conformal symmetry is exact how can you generate the scale at which the breaking happens? $\endgroup$
    – FrodCube
    Aug 16, 2022 at 12:30
  • $\begingroup$ How about a TQFT? Any TQFT is naturally also a CFT. In a TQFT you can have many vacua, so the global symmetry permuting the vacua would be unbroken. (I think any symmetry in a TQFT would also be unbroken, but I can't immediately cast it in terms of correlation function) $\endgroup$ Aug 16, 2022 at 17:01
  • $\begingroup$ @FrodCube My intuition is that the symmetries are broken at all scales (although, again, my intuition could be totally wrong here). For example, if $\phi$ is any operator that transforms under a symmetry, then under the state-operator map, we get infinitely many states created by $\phi$, all transforming non-trivially under the symmetry, and all having different energies. Thus, all these states break the symmetry, and they exist at all scales. Not sure to what extent this idea makes sense... $\endgroup$ Aug 16, 2022 at 20:41
  • $\begingroup$ @ɪdɪətstrəʊlə If a symmetry permutes vacua, then it doesn't leave them invariant, and therefore I would say that the symmetry is broken. A vacuum preserves a symmetry only if it is neutral under it; but if multiple vacua are permuted, then none of them is netrual, they all transform non-trivially. $\endgroup$ Aug 16, 2022 at 20:43
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    $\begingroup$ All CFTs in $d > 2$ with flavour symmetry I can think of have a unique vacuum. What's wrong with there being excited states at all scales that transform under a symmetry? $\endgroup$ Aug 19, 2022 at 0:15

1 Answer 1


If you have a symmetry which rotates between vacua, it is always possible to take operators which transform trivially under it and build up an algebra from them on top of one arbitrarily chosen vacuum. So you are looking for a construction which yields a CFT of the small type instead of the big type.

How about the $O(2)$ model? To study it with the bootstrap, we assume that getting a non-zero two-point function for the order parameter only requires one vacuum. I.e. the normalization condition looks like $\left < 0 | \phi^i(0) \phi^j(\infty) | 0 \right > = \delta^{ij}$ instead of $\left < 0 | \phi^i(0) \phi^j(\infty) | 0^\prime \right > = \delta^{ij}$. This results in striking agreement with experimental critical exponents.

Could it be that single vacuum $O(2)$ symmetric CFTs with $\Delta_\phi = 0.509$ actually don't exist? Because the observed second order phase transition is governed by a CFT whose structure is rich enough that you lose the order parameter when you try to restrict the number of vacua to one? Yes because numerics do not give a proof. But a wrong assumption sending such a clear signal that we are on the right track would be shocking.


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