# $K$-matrix for $ppq$ Halperin states

In the literature, the $$K$$-matrix which describes a $$(ppq)$$ Halperin state is given as $$\begin{pmatrix} p&q\\ q&p \end{pmatrix}$$ with charge vector $$\vec{q}=(1,1)$$. Are there any equivalent ways to write a different K-matrix for the same $$(ppq)$$ state? One thing I have seen people use is this K-matrix $$\begin{pmatrix} q+1&q+l+1\\ q+l+1&q+1+2l+l^2 \end{pmatrix}$$ with charge vector $$\vec{q}=(1,1)$$ in (http://arxiv.org/abs/1010.4270). They claim that when $$l=1$$ it is a $$(ppq)$$ state. However, apparently, the diagonal entries are different, which should lead to different filling fraction in two layers. What is their rational for identifying this form of the K-matrix?

So after some calculation, I find they might have a typo in their K-matrix, it should be $$\begin{pmatrix} q+1&q+l+1\\ q+l+1&q+1+2l+2l^2 \end{pmatrix}$$ Since K-matrix entries can be obtained from the commutation relation of chiral modes. But still puzzled how that corresponds to a $$(ppq)$$ state.

Update: maybe in their paper, they mean $$l=-1$$, and maybe $$l=-2$$, then everything makes sense.

Different $$K$$ matrices can represent the same topological state if they are related by a similarity transformation (Wen and Zee, PRB 1992). In your case, as long as they give the same filling fraction (or the same $$\sigma_{xy}$$), the same charge and statistics for the excitations, then they represent the same incompressible QH state. To be more precise:
$$\sigma_{xy}=q_i(K^{-1})_{ij}q_j,$$ $$Q_i=-e(K^{-1})_{ij}q_j,$$ $$\theta_{ij}=(K^{-1})_{ij}$$
With the typo you mentioned corrected, we have $$K'^{-1}=\frac{1}{l^2(2q+1)}\begin{pmatrix}q+2l^2+2l+1 & -(q+l+1)\\-(q+l+1)&q+1\end{pmatrix},$$ whereas for the $$(ppq)$$ state with $$p=q+1$$ we have $$K^{-1}=\frac{1}{2q+1}\begin{pmatrix}q+1&-q\\-q&q+1\end{pmatrix}.$$ Comparing the two matrices we can fix $$l$$ to be $$-1$$.