\in {\mathbb R}^3$ to ${\mathbb P}^2$, we get a family $P_z$ of twisted polygons as in Definition \[def:im\]. Namely, we have that $P_1 = P$, and also $[P_z] = R_z[P]$, where $R_z$ is the scaling symmetry .

First, we need to show that if a sequence $V_i$ is a solution of with given initial condition, then $V_i(z) \neq 0$ for any $i$ and every $z \neq 0$, so we can indeed project those vectors to get a sequence of points in ${\mathbb P}^2$. Assume that $V_j(z) = 0$ for some $j$ and $z$. Then, using , we can express all vectors $V_i(z)$ in terms of $V_{j+1}(z), V_{j+2}(z)$. This means that the span of all vectors $V_i(z)$ is at most two-dimensional. But this is impossible, since $V_0, V_1, V_2$ are linearly independent by construction. Further, observe that since the coefficients of equation  are periodic, its solution is quasi-periodic: $V_{i+n}(z) = M_z V_i(z)$ for a certain invertible matrix $M_z$ depending on $z$. Therefore, the projections $v_i(z) \in {\mathbb P}^2$ of the vectors $V_i(z) \in {\mathbb R}^3$ form a twisted polygon whose monodromy is the projective transformation defined by $M_z$. Furthermore since equations  and agree for $z = 1$, and the initial conditions 