Consider the Schwinger-fermion approach $\mathbf{S}_i=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$ to spin-$\frac{1}{2}$ system on 2D lattices. Just as Prof.Wen said in his seminal paper on PSG, the enlarged Hilbert space and gauge redundancy complicate our symmetry analyses.
Now let's take the translation-symmetry as an example. The unitary translation-symmetry operator $D$ is defined as $D\psi_iD^{-1}=\psi_{i+a}$, where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$ and $a$ is the lattice vector. As we know, the transformation $\psi_i\rightarrow \widetilde{\psi_i}=G_i\psi_i(G_i\in SU(2))$ doesn't change the spin operators and the projective opearator $P=\prod_{i}(2\hat{n}_i-\hat{n}_i^2)$(Note here $P\neq \prod _i(1-\hat{n}_{i\uparrow}\hat{n}_{i\downarrow})$). Similarly, in the new basis $\widetilde{\psi_i}$, we can define another translation-symmetry operator $\widetilde{D}$ as: $\widetilde{D}\widetilde\psi_i\widetilde{D}^{-1}=\widetilde\psi_{i+a}$. But $D\widetilde\psi_iD^{-1}=G_i\psi_{i+a}\neq \widetilde\psi_{i+a}$, which means that $\widetilde{D}\neq D$, the translation operators depend on the 'fermion basis' we choose. Does this imply the translation operators unphysical?
But the translation operators should be physical, so are they equivalent in the physical subspace, say for any physical spin-state $\phi=P\phi$, does $\widetilde{D}\phi=D\phi$? If this is true, then how to prove it?
Thanks in advance.