# Spontaneously breaking $SO(3)$ symmetry without $\mathbb Z_2 \times \mathbb Z_2$?

Are there any examples of Hamiltonians (probably spin models) with an $SO(3)$ symmetry which is spontaneously broken down in the ground state without breaking the subgroup of $\pi$-rotations?

• A state with cubic symmetry would be $\mathbb{Z}_2\times \mathbb{Z}_2$ symmetric but not $SO(3)$ symmetric wouldn't it? Or have I misunderstood what you are asking? – By Symmetry May 17 '17 at 12:30
• Regarding v1 of the question, that's clearly not the case. As a simple counterexample, try $$\psi(x,y,z) = x^2y^2z^2 e^{-r^2}.$$ – Emilio Pisanty May 17 '17 at 12:44
• @EmilioPisanty I agree the way I presented the first version was very doubtful. My real point of interest is finding an example where a state like yours would arise by spontaneous symmetry breaking . I no longer think it is impossible, but at the same time I cannot recall ever encountering an example. – Ruben Verresen May 17 '17 at 12:52
• For v2 of the question, atoms with isotropic interactions can form a solid with cubic symmetry, although this may not be simplest case to analyse. – By Symmetry May 17 '17 at 12:53
• @BySymmetry Haha that does indeed seem a painfully straight-forward example of what I was asking. Thanks! It would be interesting to see if there is some example for spins on a lattice. – Ruben Verresen May 17 '17 at 12:56