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The starting point and notations used here are presented in Two puzzles on the Projective Symmetry Group(PSG)?. As we know, Invariant Gauge Group(IGG) is a normal subgroup of Projective Symmetry Group(PSG), but it may not be a normal subgroup of $SU(2)$, like $IGG=U(1)$. But this may results in a trouble:

By definition, we can calculate the $IGG$ and $IGG'$ of the $SU(2)$ gauge equivalent mean-field Hamiltonians $H(\psi_i)$ and $H(\widetilde{\psi_i})$, respectively. And it's easy to see that for each site $i$, we have $U_i'=G_iU_iG_i^\dagger$, where $U_i'\in IGG'$ and $U_i\in IGG$, which means that $IGG'=G_i\text{ }IGG \text{ }G_i^\dagger$. Now the trouble is explicit, if $IGG$(like $U(1)$) is not a normal subgroup of $SU(2)$, then $IGG'$ may not equal to $IGG$, so does this mean that two $SU(2)$ gauge equivalent mean-field Hamiltonians $H(\psi_i)$ and $H(\widetilde{\psi_i})$ may have different IGGs ? Or in other words, does the low-energy gauge structure depend on the choice of $SU(2)$ gauge freedom?

Thank you very much.

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Again, I just realized that I asked a very naive question and the answer to it is 'no'. Since although $IGG'\neq IGG$, but $IGG'\cong IGG$, therefore, the low-energy gauge structure does not depend on the choice of SU(2) gauge freedom (as we wished).

Furthermore, $PSG'\cong PSG$, where $PSG'$ and $PSG$ are the Projective Symmetry Groups of the two $SU(2)$ gauge equivalent mean-field Hamiltonians.

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