Cfrtjryk

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_{iuparrow}^dagger,, f_{idownarrow} )
end{align}
and $U_{ij}$ is some complex $2times 2$-matrix. A Gauge-transformation $G$ is a transformation of the form $Psi_i to G(i)^daggerPsi(i)$, 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?

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. :)