Finite temperature spontanous symmetry breaking and Goldstone bosons

Question

I recently asked (and then attemped to answer) a question about spontaneous symmetry breaking in the Heisenberg model: Spontanous symmetry breaking in the Heisenberg model?

The question and then the conclusion I came to in the answer can be summarized as follows*:

Spontaneous symmetry breaking is when the a ground state $|GS\rangle$ of the Hamiltonian does not posses the same symmetry as the Hamiltonian itself and the reason we see spontaneously broken systems is due to imperfections (e.g. symmetry breaking fields).

Looking at the 1D Ising model the grounds states have either all spins up or all spins down. Thus we have a spontnous symmetry breaking of the $Z_2$ symmetry of the Hamiltonian.

That said at any finite temperature: $$\lim_{h\rightarrow 0}\lim_{N\rightarrow \infty}\frac{1}{N} \sum_i\mathrm{Tr}(\rho_e \sigma_i)=0$$ where $\rho_e=e^{-\beta H}/T$. I.e. the thermal average of the magnetization is zero in the limit of the symmetry breaking field $h$ going to zero and the volume $V$ going to infinity. This means the symmetry breaking does not show at finite temperature.

I.e. we appear to have the following:

1. The 1D Ising model does have Spontaneous symmetry breaking.
2. At any finite temperature the symmetry breaking is not manifest.

I have seen several sources (e.g. here; pg1) state that (exact quote from linked source):

Ising model cannot have spontaneous symmetry breaking at finite temperature,...

I assume that this is an abuse of terminology and what is meant is that the spontaneous symmetry breaking does not manifest itself.

I am not sure the way I am using the terminology in the above is correct and as such want to ask the following clarifying question:

If, for a system, there is spontaneous symmetry breaking of a continuous symmetry which is not manifest at that temperature $T$. Will the system have Goldstone modes at temperature $T$?

_{*If this is wrong please feel free to answer that question correctly.}

_{**Sorry for the long rambling before the actual question - I am trying to prevent it being an XY problem.}

Answer 1

There actually is no spontaneous symmetry breaking at all in the 1D Ising model, even at zero temperature. The reason why is somewhat technical - look into "Mermin-Wagner theorem". I don't understand your distinction between "not having SSB" and "SSB not manifesting itself". Those mean the same thing. So I don't understand you main question.

(Also, the Ising model doesn't have a continuous symmetry so there are no Goldstone bosons anyway.)