Conjugation symmetry of the KAM construction: real perturbations produce real tori

The truncation and mean objects. Assume $P_j^c=P_j$; by [F1] derivatives, products with the real-coefficient coordinates $I_l$ and finite sums give $Q_j^c=Q_j$. In the Fourier integral over real angles, conjugation exchanges the modes $k$ and $-k$: $(Q_j^c{)}_k(I,\nu )=\overline{(Q_j{)}_{-k}(\stackrel{ˉ}{I},\stackrel{ˉ}{\nu })}$. Since $\mathrm{\mid }k{\mathrm{\mid }}_1\le K_j$ is invariant under this exchange, $(T_{K_j}{Q}_j{)}^c=T_{K_j}(Q_j^c)$, and the $k=0$ coefficient is preserved. Hence $R_j^c=R_j$, ${\widehat{R}}_j^c={\widehat{R}}_j$ and $v_j^c=v_j$.

The generator. Put $G_j:=R_j-{\widehat{R}}_j$, so $G_j^c=G_j$ by step 1.1 and $G_j$ has zero mean; by [F1], $F_j^c$ is holomorphic, has zero mean and the same Fourier support, and $\nu \cdot {\mathrm{\partial }}_{\vartheta }{F}_j^c=(\nu \cdot {\mathrm{\partial }}_{\vartheta }{F}_j{)}^c=G_j^c=G_j$; since $F_j$ is the unique zero-mean holomorphic solution of $\nu \cdot {\mathrm{\partial }}_{\vartheta }w=G_j$ (The linear homological equation $\nu \cdot {\mathrm{\partial }}_{\vartheta }u=v$ on a Diophantine label set, with uniform parameter bounds and conservative loss clause 1), $F_j^c=F_j$.

The frequency retraction. By step 1.1, $v_j^c=v_j$; fix ${\nu }_{+}\in O_{h_j\mathrm{/}4}({\mathrm{\Omega }}_{\alpha ,\tau })$ and put $\psi ({\nu }_{+}):=\overline{{\varphi }_j({\stackrel{ˉ}{\nu }}_{+})}$; then $\psi ({\nu }_{+})+v_j(\psi ({\nu }_{+}))=\overline{{\varphi }_j({\stackrel{ˉ}{\nu }}_{+})}+\overline{v_j({\varphi }_j({\stackrel{ˉ}{\nu }}_{+}))}=\overline{{\stackrel{ˉ}{\nu }}_{+}}={\nu }_{+}$, and $\psi ({\nu }_{+})\in O_{h_j\mathrm{/}2}({\mathrm{\Omega }}_{\alpha ,\tau })$ because that set is conjugation-invariant; by the uniqueness clause of Elimination of the frequency shift by a holomorphic contraction retraction clause 1, $\psi ={\varphi }_j$, so ${\varphi }_j^c={\varphi }_j$ and ${\varphi }_j$ is real on real points.

The flow. By step 2.1 and [F1], $X_{F_j}^c=X_{F_j}$, that is $X_{F_j}(\bar{z},\bar{\nu })=\overline{X_{F_j}(z,\nu )}$; if $t\mapsto z(t)$ solves $z^{\mathrm{\prime }}=X_{F_j}(z,\nu )$ then $t\mapsto \overline{z(t)}$ solves $y^{\mathrm{\prime }}=X_{F_j}(y,\bar{\nu })$ (differentiate and conjugate), with $y(0)=\overline{z(0)}$; by [F3] $t\mapsto \overline{z(t)}$ agrees on its interval with the maximal solution through $\overline{z(0)}$ with parameter $\bar{\nu }$, i.e. $({\mathrm{\Phi }}_t{)}^c={\mathrm{\Phi }}_t$ for the flow of $X_{F_j}$, and evaluating at $t=1$ gives ${\tilde{\mathrm{\Phi }}}_j^c={\tilde{\mathrm{\Phi }}}_j$; in particular, for real $\nu$, ${\tilde{\mathrm{\Phi }}}_j$ maps real points to real points.

Induction and the composed maps. By steps 1.1, 2.1, 3.1, 2.2 the whole one-step operation preserves conjugation symmetry: $P_j^{+}$ is built from $P_j$, $R_j$, $F_j$, the flows of $X_{F_j}$ and Poisson brackets by sums, products, compositions and parameter integrals, all of which preserve $(\cdot {)}^c$ by [F1], and the substitution $\nu \mapsto {\varphi }_j(\nu )$ preserves it by step 2.2; hence $P_{j+1}^c=P_{j+1}$. The extended step $T_j(z,\nu )=({\tilde{\mathrm{\Phi }}}_j(z,{\varphi }_j(\nu )),{\varphi }_j(\nu ))$ is therefore conjugation-symmetric, and so is its extended composition $T_0\circ \cdots \circ T_{j-1}=({\mathrm{\Psi }}_j,{\chi }_j)$. Projecting the two components yields ${\mathrm{\Psi }}_j^c={\mathrm{\Psi }}_j$ and ${\chi }_j^c={\chi }_j$. This proves clauses 1 and 2 by induction on $j$.

The limit. For $\nu \in \mathrm{\Lambda }$ and $\vartheta \in S_{s_0\mathrm{/}4}$, both $(0,\vartheta ,\nu )$ and $(0,\bar{\vartheta },\nu )$ belong to $\mathcal{L}$. Step 4.1 gives ${\mathrm{\Psi }}_j(0,\bar{\vartheta },\nu )=\overline{{\mathrm{\Psi }}_j(0,\vartheta ,\nu )}$; passing to the pointwise limits furnished by [F4] gives ${\mathrm{\Psi }}_{\mathrm{\infty }}(0,\bar{\vartheta },\nu )=\overline{{\mathrm{\Psi }}_{\mathrm{\infty }}(0,\vartheta ,\nu )}$. Likewise ${\chi }_j(\nu )=\overline{{\chi }_j(\nu )}$ for $\nu \in \mathrm{\Lambda }$ gives ${\chi }_{\mathrm{\infty }}(\nu )=\overline{{\chi }_{\mathrm{\infty }}(\nu )}$. These are exactly ${\mathrm{\Psi }}_{\mathrm{\infty }}^c={\mathrm{\Psi }}_{\mathrm{\infty }}$ on $\mathcal{L}$ and ${\chi }_{\mathrm{\infty }}^c={\chi }_{\mathrm{\infty }}$ on $\mathrm{\Lambda }$ under the stated map-level definition. At real angles the phase point is therefore real, so $b(\vartheta ,\nu )$ and $w(\vartheta ,\nu )$ are real. This proves clause 3 without an open complex-parameter domain for the limit.