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Let's restrict to the case of spin-1/2 system. As we know, a spin-liquid (SL) state is the ground state of a lattice spin Hamiltonian with no spontaneous broken symmetries (sometime it may spontaneously break time-reversal symmetry and is called a chiral SL), where two essential symmetries of a SL state are lattice translation and spin-rotation symmetries.

Since, traditionally, we usually describe a SL state by using a spin Hamiltonian with the full $SU(2)$ spin-rotation symmetry (e.g., Heisenberg model), and the corresponding SL state is hence also $SU(2)$ symmetric, i.e., a RVB type SL. While, the honeycomb Kitaev model provides us an exact SL ground state with $Q_8$ spin-rotation symmetry, where $Q_8$ is a finite subgroup of $SU(2)$, indicating that the Kitaev SL does NOT belong to the RVB type.

Thus, my question is: Generally speaking, what is the minimal spin-rotation symmetry required for a spin Hamiltonian to describe a SL ground state? Is $Q_8$ group the minimal one? Thank you very much.

[My motivation for this question is that for a spin Hamiltonian without any spin-rotation symmetry, whether or not can it possess a SL ground state? And does the existence of a SL state with some spin-rotation symmetry imply the occurrence of emergent symmetries?]

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