Does (spontaneous) symmetry breaking imply long-range order and vice-versa? Crystalline solids have a long-range order (where symmetry is broken) but liquids have only a short-range order (where no symmetry is broken). Ferromagnets have a long-range magnetic order while a paramagnet lacks it. The converse also seems to be true, for example, in the Kosterlitz-Thouless transition there is no symmetry breaking and there is no long-range order (but quasi long-range). By long-range order, I understand that below some critical temperature $T_c$ the two-point correlation function of the order parameter (density) becomes a constant (independent of position).
Is this a generic feature? In other words, does long-range order necessarily imply the symmetry-breaking? And does the symmetry-breaking necessarily imply the long-range order?
 A: There are some subtleties, but the answer is basically "yes" in local, translationally invariant systems, because of the cluster decomposition property. Empirically, virtually any "realistic" physical system satisfies the property that
$$\lim_{|x - y| \to \infty} \left[ \langle A(x)\, B(y) \rangle - \langle A(x) \rangle \langle B(y) \rangle \right] = 0$$
for any local operators $A(x)$ and $B(y)$. In other words, correlations decay with distance, so expectation values of faraway observables are uncorrelated. (One can derive this result from various technical locality assumptions.) If $m(x)$ is a local symmetry-breaking order parameter, then $\langle m(x) \rangle \equiv \bar{m}$ is constant by translational invariance, so if we let $A(x) = B(x) = m(x)$ in the identity above then we have
$$\lim_{|x - y| \to \infty} \langle m(x)\, m(y) \rangle = \bar{m}^2.$$
The left-hand side being nonzero defines long-range order, and the right-hand side being nonzero defines spontaneous symmetry breaking, so we see that either implies the other.
