A finite formal logarithm has Hamiltonian coefficients on a star-shaped domain

Fix $a\in U$ and $\epsilon >0$ with $\bar{B}(a,4\epsilon )\subset U$. On this ball, [F8] bounds each required derivative of $F=X_H$. Choose $B\ge 1$ bounding the norms of $F$ and the kick and drift fields, and $L\ge 1$ bounding $\mathrm{\parallel }DF\mathrm{\parallel }$. For centers in $\bar{B}(a,3\epsilon )$ use midpoint stage radius $\rho =\epsilon$. Choose $h_0>0$ with $h_{0}B\mathrm{/}2<\epsilon$ and $h_{0}L\mathrm{/}2\le 1\mathrm{/}2$; all these balls then satisfy the uniform branch conditions of [F2]. Any other admitted midpoint ball gives the same root: its Picard iterates start at the same $z$ and obey the same recurrence, so their limit is also the limit for the ball of radius $\epsilon$. Choose $T>0$ with $2BT<\epsilon$ and decrease $h_0$ to at most $T$. Starting in $\bar{B}(a,2\epsilon )$, exact trajectories stay in $\bar{B}(a,3\epsilon )$ for $\mathrm{\mid }t\mathrm{\mid }\le T$ by the speed bound and [F7]. Numerical endpoints and kick/drift stages also stay there for $n\mathrm{\mid }h\mathrm{\mid }\le T$: each explicit step has total substep displacement at most $2B\mathrm{\mid }h\mathrm{\mid }$, and a midpoint step has displacement at most $B\mathrm{\mid }h\mathrm{\mid }$. The branch balls and every comparison segment therefore lie in the compact convex ball $\bar{B}(a,4\epsilon )$. All hypotheses needed for the local symplecticity and order conclusions of [F1] and [F2] hold on these smaller neighborhoods. No single convex tube covering all of $U$ is needed.

Fix $N$. Smooth kick and drift formulas give smooth step germs at $h=0$ throughout $U$. For midpoint apply [F4] to $G(z,Z,h)=Z-z-(h\mathrm{/}2)F(Z)$ at $(z,z,0)$, where $D_{Z}G=I$. Its smooth solution germ is the exact root in step 1.1, and uniqueness identifies these germs on overlaps. Thus the Taylor coefficients $d_k$ are smooth on all of $U$, and on any fixed compact subset a finite cover by the neighborhoods of step 1.1 gives ${\mathrm{\Phi }}_h(z)=z+{\sum }_{k=1}^{N+1}{h}^k{d}_k(z)+O(h^{N+2})$, also with a uniform first-state-derivative remainder by smoothness. Consistency gives $d_1=X_H$; the verified local order estimates of [F1] and [F2] identify the coefficients through degree $r$ with those of the exact flow.

Seek $Y_h=X_H+{\sum }_{m=1}^N{h}^m{X}_m$. The time-$h$ flow expansion is $z+{\sum }_{k=1}^{N+1}\frac{h^k}{k!}(Y_h\cdot \mathrm{\nabla }{)}^{k-1}{Y}_h(z)+O(h^{N+2})$. At coefficient $h^{m+1}$, the first term $h{Y}_h$ contributes $X_m$; every term with $k\ge 2$ can involve only $X_1,\dots ,X_{m-1}$ at that degree. Thus set $X_m=d_{m+1}-P_m$, where $P_m$ is the already determined coefficient from those higher flow terms. This finite recursion gives smooth fields on $U$ and matches coefficients through degree $N+1$. Since step 2.1 identifies the exact-flow coefficients through degree $r$, it gives $X_1=\cdots =X_{r-1}=0$.

For fixed parameter $h$, the pullback $Q_{h,t}$ of the time-$t$ flow of $Y_h$ obeys ${\mathrm{\partial }}_t(Q_{h,t}f)=Q_{h,t}({\mathcal{L}}_{Y_h}f)$ and $Q_{h,0}=I$. Coefficient recursion in $t$ therefore gives $Q_{h,t}={\sum }_{k=0}^{N+1}{t}^k{\mathcal{L}}_{Y_h}^k\mathrm{/}k!$ modulo terms of total order $N+2$; at $t=h$, this is $\exp (h{\mathcal{L}}_{Y_h})$. Step 3.1 gives ${\mathrm{\Phi }}_h$ and its comparison flow equal Taylor coefficients through degree $N+1$, so [F6] makes their pullbacks equal modulo $h^{N+2}$. Since $P_h-I=O(h)$, the finite polynomial logarithm in the Statement is valid modulo $h^{N+2}$. The formal identity $\log (\exp A)=A$ in the truncated algebra then gives ${\log }_N(P_h)=h{\mathcal{L}}_{Y_h}$ modulo $h^{N+2}$.

Fix $m\in \{r,\dots ,N\}$ and assume $X_r,\dots ,X_{m-1}$ are Hamiltonian. The truncated field $Y_h^{[m-1]}=X_H+{\sum }_{j=r}^{m-1}{h}^j{X}_j$ is Hamiltonian, so by [F3] its time-$h$ flow ${\psi }_h$ is symplectic. Step 3.1 gives ${\mathrm{\Phi }}_h-{\psi }_h=h^{m+1}{X}_m+O(h^{m+2})$ and, after differentiating the smooth expansions, $D{\mathrm{\Phi }}_h-D{\psi }_h=h^{m+1}D{X}_m+O(h^{m+2})$.

The symplectic matrix identity for both maps and $D{\psi }_h=I+O(h)$ imply $0=h^{m+1}(D{X}_m^{T}J+JD{X}_m)+O(h^{m+2})$. Divide by $h^{m+1}$ and take $h\to 0$ to obtain $D{X}_m^{T}J+JD{X}_m=0$.

The one-form ${\alpha }_m={\iota }_{X_m}\omega$ has coefficient column $-J{X}_m$, so its derivative matrix is $-JD{X}_m$. Step 5.1 says this matrix is symmetric; hence ${\alpha }_m$ is closed. Fix a star center $z_{\ast }$ and define $H_m(z)={\int }_0^1{\alpha }_m(z_{\ast }+t(z-z_{\ast }))[z-z_{\ast }]dt$. For any direction $v$, differentiation under the integral gives $t(D{\alpha }_m(z_{\ast }+t(z-z_{\ast }))[v])[z-z_{\ast }]+{\alpha }_m(z_{\ast }+t(z-z_{\ast }))[v]$. Closedness swaps the two arguments in the first term, making the integrand $\frac{d}{dt}(t{\alpha }_m(z_{\ast }+t(z-z_{\ast }))[v])$. Integration yields $D{H}_m(z)[v]={\alpha }_m(z)[v]$. Thus $d{H}_m={\alpha }_m$ and [F5] makes $X_m$ Hamiltonian on all of $U$.

Applying the coefficient induction in step 4.2, the matrix consequence in step 5.1, and the primitive construction in step 6.1 successively for $m=r,\dots ,N$ supplies all finite Hamiltonians $H_j$. Step 4.1 identifies the resulting vector-field coefficients with the truncated formal logarithm. The construction makes no assertion about convergence as $N\to \mathrm{\infty }$.