# Different invariant gauge groups (IGG) on different lattices with the same form mean-filed Hamiltonian?

Suppose that we use the Schwinger-fermion ($\mathbf{S_i}=\frac{1}{2}f_i^\dagger\mathbf{\sigma}f_i$) mean-field theory to study the Heisenberg model on 2D lattices, and now we arrive at the mean-field Hamiltonian of the form $H_{MF}=\sum_{<ij>}(\psi_i^\dagger u_{ij}\psi_j+H.c.)$ with $u_{ij}=t\sigma_z$($t>0$), where $\psi_i=(f_{i\uparrow},f_{i\downarrow}^\dagger)^T$, and $\sigma_z$ is the third Pauli matrix.

Now let's find the IGG of $H_{MF}$, by definition, the pure gauge transformations in IGG should satisfy $G_iu_{ij}G_j^\dagger=u_{ij}\Rightarrow G_j=\sigma_zG_i\sigma_z$ on link $<ij>$—(1) . Specifically, consider the IGGs on the following different 2D lattices:

(a)Square and honeycomb lattices(unfrustrated): These two lattices can be both viewed as constituted by 2 sublattices denoted as $A$ and $B$. Due to Eq.(1), it's easy to show that for both of these two lattices the gauge transformations $G_i$ in the same sublattice are site-independent while those in different sublattices differ by $G_A=\sigma_zG_B\sigma_z$ and $IGG=SU(2)$.

(b)Triangular and Kagome lattices(frustrated):Due to Eq.(1), it's easy to show that for both of these two lattices the gauge transformations $G_i$ are global (site-independent) and $G_i=\bigl(\begin{smallmatrix} e^{i\theta }& 0\\ 0& e^{-i\theta } \end{smallmatrix}\bigr)$ which means that $IGG=U(1)$.

So my question is: The same form mean-field Hamiltonian $H_{MF}$ may has different $IGGs$ on different lattices?

Yes, it is possible. In my understanding, the most easy way to identify IGG is to study the loop of ansatz with the same basepoint. Starting from $SU(2)$ formulation, if loops are colinear, then $IGG=SU(2)$. If loops are coplanar, then $IGG=U(1)$. Otherwise, $IGG=Z_2$. The reason to consider loop is because only gauge flux (more accuracy, Wilson loop) is observable. You can easily see the difference between loops in honeycomb/square lattice with loops in Kagome/triangular lattice. And this is the true reason for your question. However, there is one thing to be notice: there may exists hidden symmetry beyond the formulation. For example, even if we use $U(1)$ slave boson formulation, we can still have $SU(2)$ spin liquid. However, it is very difficult to extract the hidden symmetry, I think.
• @ Shenghan Jiang, thanks for your good answer. But I'm studying PSG at the very beginning and got a lot of puzzles, like by 'loop of ansatz ', what does it mean? And are there only $SU(2), U(1), Z_2$ three kinds of IGGs? From the math side, for example, can $IGG=Z_4$? – Kai Li Sep 10 '13 at 18:20
• @ Shenghan Jiang, from the math viewpoint, we can construct many subgroups of $SU(2)$ (math.stackexchange.com/questions/488309/…), so physically, I think all these subgroups may become the possible IGGs in proper physical models. What's your opinion? Thanks. – Kai Li Sep 10 '13 at 18:25