Questions about $Y$-systems

Question

Let $\Delta$ be a Dynkin diagram with vertex set $I$ and let $C$ be the Cartan matrix of $\Delta$ and $J$ be the identity matrix of the same size as $C$. The set $I$ of vertices is the disjoint union of two sets $I_{+}$ and $I_{-}$ s.t. there is no edge between any two vertices of $I_{+}$ nor between any two vertices of $I_{-}$. Define $\epsilon(i)$ to be $1$ or $-1$ based on whether $i \in I_{+}$ or $i \in I_{-}$. Also, let $A=2J-C$ denote the incidence matrix of $\Delta$. The system of equations \begin{equation} Y_i(t+1)Y_i(t-1)=\displaystyle\prod_{j \in I}(Y_j(t)+1)^{a_{ij}}, \; t \in \mathbb{Z} \end{equation} is called a $Y$-system. I am reading in this paper (p.19) that the variables $Y_i(k)$ on the lhs of the above equation have a fixed "parity" $\epsilon(i)(-1)^k$. I am not sure if I understand this correctly. We see that if $i \in I_{+}$ and if $k$ is odd then $\epsilon(i)(-1)^k = -1$ and if $k$ is even then $\epsilon(i)(-1)^k = 1$. Else if $i \in I_{-}$ we have that $\epsilon(i)(-1)^k = -1$ if $k$ is even and $\epsilon(i)(-1)^k = 1$ if $k$ is odd. They then say that the $Y$-system decomposes into two independent systems, an even one and an odd one. What do they mean? Do they just take one system for $k$ odd and one system for $k$ even in that decomposition? In what way are they independent? They also say: wlog, we may (therefore) assume that \begin{equation} Y_i(k+1)=Y_i(k)^{-1} \; \text{whenever} \; \epsilon(i)=(-1)^k. \end{equation} Why is this a reasonable assumption to make?

Answer 1

Variables of a given parity only depend on variables of the same parity, through the definition relations of the Y-system.

Therefore one can compute all variables of positive parity without ever needing the value of any variable of negative parity. This is what is meant by decomposition into two independent systems of relations.

The final assumption is one possible way to transfer the variables from one parity to the other parity in such a way that the Y-relations still hold.