I am reading the paper from X. Wen about quantum orders and symmetric spin liquids. It can be found here: https://arxiv.org/abs/cond-mat/0107071

The Hamiltonian he is writing about looks like this: \begin{align} H_{MF} = \sum_{<ij>} \Psi_i^\dagger \, U_{ij} \Psi_j + \text{h.c.} + \text{const.} + \text{Lagrange-multiplier terms} \end{align} where \begin{align} \Psi_i^\dagger := (f_{i\uparrow}^\dagger,\, f_{i\downarrow} ) \end{align} and $U_{ij}$ is some complex $2\times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $\Psi_i \to G(i)^\dagger\Psi(i)$, where each $G(i) \in SU(2)$. A symmetry-transformation $T$ is of the form $\Psi_i \to \Psi_{T^{-1}(i)}$, where (I guess) $T$ has to be a bijective function from the lattice to the lattice.

He introduces something called projective symmetry group (PSG). Its elements are pairs $(G_T,T)$ of a Gauge-transformation together with a symmetry such that the Hamilton operator is invariant under the transformation $G_T T$. This means $(G_T,T) \in PSG$ has to satisfy \begin{align} \sum_{<ij>} G_TT(\Psi_i)^\dagger\, U_{ij}\, G_TT(\Psi_j)&= \sum_{<T(i)T(j)>} \Psi_i^\dagger\, G_T(T(i))U_{T(i)T(j)} G_T(T(j))^\dagger\, \Psi_j^\dagger \\&=^! \sum_{<ij>} \Psi_i^\dagger \, U_{ij} \Psi_j \end{align} Thus, \begin{align} G(T(i))U_{T(i)T(j)}G(T(j))^\dagger = U_{ij} \end{align} I do understand the definition. And I hope I got it right, how Gauge transformation and symmetries act on the basis. But I don't understand why it is calle a group.

So basically I am looking for the group operation \begin{align} \cdot: PSG \times PSG \to PSG, \qquad (G_T,T) \cdot (G_S, S) = ?! \end{align} It is nowhere mentioned, and I searched a lot. It is probably very easy, but I just can not figure it out. I tried $(G_TG_S,TS)$ and $(G_TG_S, ST)$. Both are of the right form, Gauge and symmetry, but they don't seem to leave the Hamilton invariant. At least I can't see it. I know, that e.g. $G_TTG_SS$ will not change $H_{MF}$, but it is not of the form $(\text{Gauge}, \text{Symmetry})$. So what to do?


1 Answer 1


Don't bother. I think I found the answer. I will write it down here, cause it is hardly found anywhere. Maybe it helps someone in future.

\begin{align} \cdot : PSG \times PSG \to PSG, \, \qquad (G_T,T)\cdot(G_S,S) := (G_TTG_ST^{-1},TS) \end{align}

This has the required form, since \begin{align} G_TTG_ST^{-1} (\Psi_i) = G_T(i)^\dagger G_S(T(i))^\dagger \, \Psi_i \end{align} is a Gauge transformation. It is two lines to check associativity. And inverse elements are given by \begin{align} (G_T,T)^{-1} = (T^{-1}G_TT,T^{-1}) \end{align} where the neutral element is $(id,id)$. This also makes $IGG$ a subgroup and $SG = PSG / IGG$ as claimed in the paper. So I am quite happy now. :)

  • $\begingroup$ Do you think the inverse should be $(T^{-1} G^{-1}_T T, T^{-1})$? $\endgroup$ May 25, 2023 at 5:30

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