For those spin liquids with SU(2) spin-rotation symmetry or time-reversal(TR) symmetry , the Spin Density Wave (SDW) order parameters are always zero, say $\left \langle \mathbf{S}_i \right \rangle=\mathbf{0}$, due to the spin-rotation symmetry or TR symmetry.
But if a spin-liquid state is neither spin-rotation symmetric nor TR symmetric, e.g. the exact chiral spin liquid ground state of the generalized Kitaev model on the decorated honeycomb lattice, is there any possibility that the SDW order parameters $\left \langle \mathbf{S}_i \right \rangle\neq \mathbf{0}$ ? Therefore, if a spin-liquid state has nonzero order parameters $\left \langle \mathbf{S}_i \right \rangle$, why we still call it a "spin liquid" rather than a "SDW phase"?
Remarks: The spin-rotation mentioned here should be understood as the continuous $SU(2)$ or $SO(3)$ one, say all the spin-rotation transformations. Although the Kitaev model(on the decorated honeycomb lattice) and its ground state break this continuous spin-rotation symmetry, they still possess the $\pi$ spin-rotation symmetry about $S_x,S_y$ and $S_z$ spin-axes, and this is enough to ensure $\left \langle \mathbf{S}_i \right \rangle=\mathbf{0}$.
Thanks in advance.