Exact quadratic normal form for a perturbed linear rotor ($H\circ \Phi =E+\omega \cdot y^{\prime }+\frac{1}{2}Q+{ϵ}^2\tilde{V}$)

For $H=\omega \cdot y+\frac{1}{2}|y{|}^2+ϵV(x)$ with a Diophantine frequency vector, solve ${\mathcal{D}}_{\omega }u=⟨V⟩-V$ and apply translated conjugacy of a Diophantine vector field to $w=ϵ\nabla u$. Set $\eta =ϵ\nabla u+a$ and use the symplectic cotangent lift with a closed momentum shift. The transformed Hamiltonian is exactly $E+\omega \cdot y^{\prime }+\frac{1}{2}{y}^{\prime T}D{\chi }^{-1}D{\chi }^{-T}{y}^{\prime }+\frac{1}{2}|\eta \circ \chi {|}^2$, where $E=ϵ⟨V⟩+\omega \cdot a$. Since $a=O({ϵ}^2)$, the last term is an angular potential of order ${ϵ}^2$ and the quadratic coefficient is identity plus $O(ϵ)$. The construction is on a smaller angular strip and momentum ball.