Real and complex symplectic normal forms of functions

A canonical coordinate change preserves Poisson brackets. Multiplication by a nonzero function gives additional freedom, and for a complex symbol that freedom can remove an entire sequence of transverse errors. We develop the local models, including the exact marked covector, the homogeneity of every multiplier, and the distinction between a simple bracket and fixed higher contact.

We retain the conventions and geometry of Phase space and generating families, the nonzero restricted primitive theorem in Homogeneous submanifold normal forms, and the full coordinate completion in Prescribed canonical coordinates and isotropic fibers. The exact prerequisites are finite-coordinate flows, NF1–NF7, tangent and commuting flows, F0–F1, and inverse and implicit functions, P2–P3. The measure and norm proofs, M3–M8 give Fatou, dominated convergence, product integration, completeness and smooth density; Fourier proofs L1–L3 give the actual inverse maps, distributional compatibility and measurable multipliers. B1–B2 prove the weighted Fourier norms and their local frequency estimate. The proof map supplies exact locators and all earlier inputs. Section 4 proves smooth division by a complex function of one real variable, using an almost-analytic extension and a smooth complex root. The geometric source is Hörmander III, Section 21.3, Theorems 21.3.1–21.3.6, including the proof ending on; the division theorem is credited to Hörmander I, Theorem 7.5.6. The present proof supplies the needed one-variable division and parameter solver.

1. Allowed changes and the real normal form

Write

Thus $H_{\xi_j}=\partial_{x_j}$, $H_{x_j}=-\partial_{\xi_j}$, and $\{\xi_j,x_k\}=\delta_{jk}$. A function of degree $m$ satisfies $Rf=mf$. Its Hamilton field satisfies

The first identity follows by differentiating the coordinate expression for $H_f$; the second follows either from that expression or by applying $R$ to the bracket.

Fix $c\in T^*\mathbb R^n\setminus0$. Our changes are multiplication $p\mapsto ap$ by a nowhere zero smooth homogeneous complex function and a local homogeneous canonical map $\chi$, with the convention that $\chi$ carries model coordinates to a neighborhood of $c$. All conclusions are local conic germs. The multiplier can change degree: if $p$ has degree $m$, multiplying by $|\xi|^{1-m}$ makes it degree one.

Theorem 1.1 (real principal-type model). Suppose $p$ is real, smooth and homogeneous of degree one, $p(c)=0$, and $H_p(c)$ and $R(c)$ are independent. There are homogeneous canonical coordinates, with $c$ represented by $(0,e_n)$, such that

Proof. Prescribe the degree-one momentum $p_1=p$, no position coordinates, and the marked values $p_j(c)=\delta_{jn}$, $q_j(c)=0$. The homogeneous coordinate-completion theorem applies precisely because the given Hamilton field and $R$ are independent. Its unused position index is $n$, whose prescribed momentum is nonzero. The completed coordinate map has the stated mark and retains $p$ as an actual function. Its inverse is $\chi$. $\square$

The hypothesis includes the marked compatibility. In dimension one it cannot hold at a zero: $dp(R)=p=0$, so $H_p$ and $R$ lie in the same one-dimensional symplectic orthogonal. For example $p=x\xi$ near $(0,1)$ has $H_p=-R$ at its zero. A homogeneous transformation could not turn this zero into the model momentum at $(0,e_1)$, where that momentum equals one.

If a real symbol initially has degree $m$, the positive preliminary multiplier scales its Hamilton field at a zero by a positive number, so it preserves nonradiality and gives (1.3) for the multiplied symbol.

2. A positive weighted bracket normalizer

The complex models will require a pair of functions with bracket one. A scalar multiplier can accomplish this even when the functions have fractional degrees.

Theorem 2.1 (weighted normalization). Let $p,q$ be real smooth functions on a symplectic manifold, $p(c)=q(c)=0$, and $C=\{p,q\}>0$ near $c$. If $\alpha,\beta>0$ and $\alpha+\beta=1$, there is a unique positive smooth local germ $u$ such that

In a conic symplectic manifold, if $p,q$ have degrees $m,m'$, this germ extends homogeneously with

Proof. The full product rule, using $\alpha+\beta=1$, gives

The two normal coordinates are independent because $C\ne0$. Put $X=V/C$. Crucially,

Choose submersion coordinates $(p,q,z)$ near $c$. The field $X$ vanishes when $p=q=0$. Taylor division therefore writes its tangential component as $pA+qB$, with $A,B$ smooth vector functions. For its backwards flow $\Phi_{-s}$,

Take nested relatively compact $z$ neighborhoods, with the smaller closure inside the larger. Bounds for $A,B$ on the larger box give

Make the initial normal box small enough that this bound is less than the distance between the two $z$ boundaries. The backwards orbit remains in the larger box for every $s$, so the local flow extends to all $s\ge0$. The integrable right side of (2.5) also makes $z_s$ converge.

We need estimates for every derivative in the initial data, not only orbit convergence. In the nonautonomous $z$ equation, each derivative of its right side with respect to $(p,q,z)$ is bounded by $K_l e^{-\delta s}$, where $\delta=\min(\alpha,\beta)>0$, on the fixed box. First derivatives of $z_s$ satisfy a linear equation whose coefficient has finite integral and whose forcing is integrable. The integral inequality gives a uniform bound in $s$. Explicitly, if $y(s)\le B+\int_0^s a(t)y(t)\,dt$, where $B$ bounds the initial value and the total forcing integral and $a\ge0$ is integrable, put $Y(s)=B+\int_0^s a(t)y(t)\,dt$. Then $Y'\le aY$, so multiplication by $\exp(-\int_0^s a)$ and the fundamental theorem give $y(s)\le B\exp(\int_0^\infty a)$. Replace $B$ by $B+\epsilon$ and let $\epsilon\downarrow0$ if $B=0$. Apply this to the norms of the differentiated finite-time integral equations from NF2–NF5. The bound does not depend on the final time. Inductively, a derivative of order $l$ satisfies the same linear equation, plus a finite sum of products of bounded lower derivatives with derivatives of the right side. That forcing is again integrable. Thus every finite-order derivative of $z_s$ is uniformly bounded on a smaller initial box. The explicit derivatives of $p_s,q_s$ have the same property.

Define

The positive function $C$ has upper and lower positive bounds on the larger box. The derivative estimates just proved allow differentiation of (2.7) to every order, with an integrable bound $K_l e^{-s}$. Consequently $u$ is smooth and positive, and $u|_{\{p=q=0\}}=1/C$.

To verify the equation, put $F=1/C$. Differentiation along $X$, followed by integration by parts, gives

Thus $Cu+Vu=1$, as required. If $v+Xv=0$, then

Shrink the initial box so the entire backwards orbit lies in the domain of the proposed smooth germ $v$. Its closure is compact there, so $v$ is bounded on the orbit. Letting $s\to\infty$ in (2.9) gives $v(w)=0$. This proves uniqueness of the scalar equation among all smooth germs and hence uniqueness in (2.1).

For the conic assertion let $\delta_t$ multiply covectors by $t>0$. Pullback of a bracket under $\delta_t$ obeys

Set $d=1-m-m'$. Applying (2.10) to (2.1) shows that $t^{-d}\delta_t^*u$ satisfies that same equation, since the powers of $t$ in the bracket cancel. Uniqueness gives $\delta_t^*u=t^du$ for $t$ close to one on a common germ. Extend along the positive covector ray; these identities make the local extensions agree on overlaps. Equation (2.2) follows by adding degrees. $\square$

The formula remains local in the tangential variables. The backwards time is unbounded, but estimate (2.6) keeps the entire integral inside a fixed local neighborhood. No data are prescribed on the singular fixed set beyond the forced value $1/C$.

3. A nonzero complex bracket and its marked sign

Write $p=p_1+ip_2$, with real $p_j$. At a zero of $p$, every derivative of a scalar multiplier $a$ is accompanied by a factor $p$ or $\bar p$, so

The sign of a nonzero bracket is therefore invariant under our changes.

We first supply the homogeneous auxiliary momentum used in this and the higher-contact construction.

Lemma 3.1 (a commuting positive momentum). Suppose $\phi,\psi$ are real homogeneous functions of degrees $a,b$, $a+b=1$, $\phi(c)=\psi(c)=0$, and $\{\phi,\psi\}=1$. There is a positive degree-one function $h$, with $h(c)=1$, such that

Moreover

can be retained as the first momentum, first position and last momentum of a homogeneous canonical coordinate system with mark $(0,e_n)$.

Proof. The fields $H_\phi,H_\psi$ commute because their bracket is the Hamilton field of the constant one. Their span is symplectic: $\omega(H_\phi,H_\psi)=\{\phi,\psi\}=1$. The zero set $Z=\{\phi=\psi=0\}$ is a transverse section for their two commuting flows. The radial field lies in $TZ$, and is nonzero there. Choose a positive degree-one function $h_0$ on $Z$, equal to one at $c$, by using a transverse section to its radial flow and extending its positive datum.

The product of the two Hamilton flows from $Z$ is a local diffeomorphism. Extend $h_0$ constantly in both flow times. This proves (3.2), positivity and the marked value. Dilations carry each flow to the same flow with a rescaled time, because of (1.2), and carry $Z$ to itself. The extension of $t^{-1}\delta_t^*h$ has the same datum and the same two zero transport equations as $h$. Flow uniqueness proves its equality to $h$. Thus $h$ has degree one.

The product rule in (3.3) gives

At $c$, $H_P=H_\phi$, $H_Q=H_\psi$. The field $H_h$ is orthogonal to their symplectic span and nonzero, since $dh(R)=h=1$. If $R$ were in the span of these three fields, pairing with $H_P,H_Q$ would force its first two coefficients to vanish. Then $R$ would be a multiple of $H_h$, contradicting $dh(R)=1$ and $dh(H_h)=0$. Hence the three prescribed Hamilton fields and $R$ are independent. The homogeneous coordinate theorem applies with $Q=q_1$, $P=p_1$, $h=p_n$, and the unused position index $n$, whose momentum value is one. It completes the coordinates and preserves these functions exactly. $\square$

Theorem 3.2 (simple complex normal form). Let $p$ be smooth and homogeneous near $c$, $p(c)=0$, and $\{p_1,p_2\}(c)\ne0$. There is a nonzero homogeneous multiplier $a$ and a homogeneous canonical map such that

The model mark is $(0,e_n)$ for positive bracket and $(0,-e_n)$ for negative bracket.

Proof. The nonzero bracket makes $dp_1,dp_2$ independent. Homogeneity gives $dp_j(R)=0$ at the zero. The nondegenerate Hamilton plane therefore cannot contain $R$; in particular the fields and the radial direction are independent.

First make the degree one by a positive multiplier, which preserves the bracket sign at the zero. In the positive case apply Theorem 2.1 with weights $1/2,1/2$. Then

Lemma 3.1 with $b=1/2$ provides coordinates

Their exact identity is

All factors are positive smooth on the chosen conic germ; the multiplier has the required degree.

For negative bracket perform the positive construction on $\bar p$, then conjugate the resulting identity. This gives $\eta_1-i y_1\eta_n$ at $(0,e_n)$. The simultaneous canonical flip $(y_n,\eta_n)\mapsto(-y_n,-\eta_n)$ produces (3.5) at $(0,-e_n)$. The model's bracket at either mark is $\xi_n=\pm1$, so (3.1) also proves that the two marked signs cannot be exchanged by an admissible equivalence at the same mark. $\square$

4. Remove one momentum from the imaginary part

Smooth complex division. If a complex $C^\infty$ function $F(t,z)$, with real $t,z$, has $F(0,0)=0$ and $\partial_tF(0,0)\ne0$, then every smooth $G(t,z)$ has a representation

on a neighborhood chosen from $F$ and the original domain, with $Q,S$ smooth and complex valued. The remainder is independent of $t$. This is the $k=1$ case of Hörmander I, Theorem 7.5.6; its hypotheses are those of Theorem 7.5.5. Neither the quotient nor the remainder is asserted to be unique. The following proof treats complex functions in real variables; it does not assume that their real zeros form a parameter graph.

Proof. First extend a smooth function $a(t,z)$ almost analytically in $t$. Multiply it by a fixed real cutoff equal to one on the patch in use. For $v=t+iw$, choose a fixed smooth cutoff $\chi$, equal to one near zero, and put

All coefficients and their derivatives are bounded on a fixed larger compact patch. A derivative of total order $l<k$ of the $k$-th term is bounded by $C_{kl}\varepsilon_k^{k-l}$. Choose positive radii decreasing to zero so that these bounds are at most $2^{-k}$ for every $l\le k/2$. Each fixed derivative series then converges uniformly after finitely many terms. The sum is smooth on a fixed complex domain; only the radii and its seminorm bounds depend on $a$. Its normal jets at $w=0$ are the formal analytic jets. Thus, with $\partial_{\bar v}=(\partial_t+i\partial_w)/2$, Taylor's integral remainder gives

The derivatives here include all real parameters $z$. No holomorphic continuation of an arbitrary smooth function is claimed.

Apply this construction to $F$, writing its extension as $\widetilde F(v,z)$. At the mark, its real derivative in $(t,w)$ is complex multiplication by $\lambda=\partial_tF(0,0)\ne0$. In particular it is invertible. A direct parameter argument produces a smooth complex root $T(z)$, with $T(0)=0$. Indeed the real map

has derivative zero at the mark. On a sufficiently small closed complex disc its derivative norm is at most $q<1$, uniformly on a smaller parameter patch. Shrink that patch so that its value at $v=0$ has modulus at most $(1-q)/2$ times the disc radius. It maps the disc into itself and its fixed point lies in the interior. Iteration from zero has geometrically summable successive differences, so completeness of $\mathbb C$ gives its unique fixed point $T(z)$. The same contraction estimate first bounds parameter differences of $T$ by a constant times the parameter differences. Taylor's formula in the fixed-point identity then proves differentiability, with real derivative

The inverse exists throughout the disc by the contraction bound. This formula has continuous coefficients. Induction differentiating it proves smoothness of every order. Thus $\widetilde F(T(z),z)=0$, with no real-zero or holomorphic-root assumption.

We next divide by the known graph $t-T(z)$. For any almost-analytic extension $A$ as above, the segment $v_s=T+s(t-T)$ stays inside its fixed domain after one shrink depending on $T$. Set $R(z)=A(T(z),z)$. The full real chain rule gives

The sign and the error term follow from $t-\overline T=(t-T)+2i\operatorname{Im}T$. Every derivative of $E$ is bounded by every positive power of $|\operatorname{Im}T|$, because the imaginary part of $v_s$ is $(1-s)\operatorname{Im}T$ and the antiholomorphic derivative has all the preceding flat estimates. Put $s_0=|t-T(z)|^2\ge|\operatorname{Im}T|^2$. For $s_0>0$, define

Each derivative of this expression costs only finitely many negative powers of $s_0$. The arbitrary positive powers in the bounds for $E$ absorb them, so every derivative tends to zero faster than every power of $s_0$. Its zero extension across $s_0=0$ is smooth even if that zero set is singular: multiply the expression by a smooth cutoff vanishing for $s_0\le\delta$ and equal to one for $s_0\ge2\delta$. On the transition region all differentiated cutoff factors cost finitely many powers of $\delta^{-1}$, while the flat estimates give arbitrarily many positive powers. These smooth functions and all their derivatives converge uniformly to the zero extension. The fundamental theorem of calculus identifies those derivative limits with the derivatives of the limit. Consequently

holds smoothly and exactly, including real zeros of the graph.

For $a=F$, use the particular extension $\widetilde F$ which constructed $T$. Its remainder is zero, so $F=(t-T)U$. Differentiating at the mark gives $U(0,0)=\partial_tF(0,0)\ne0$; shrink once so $U$ never vanishes. Apply the same graph division to $a=G$, obtaining $G=(t-T)V+S(z)$. Then $Q=V/U$ proves (4.1). The real and complex neighborhoods were fixed before choosing $G$; its extension radii and coefficient bounds may depend on $G$. This proves exactly the needed common-neighborhood division. $\square$

Example (a nonreal root). For $F(t,z)=t-iz$ and $G(t,z)=t^2$, the root is $T(z)=iz$, and the exact division is

The remainder depends only on $z$. For $z\ne0$ there is no real root in $t$; the proof uses the smooth complex root instead.

Theorem 4.1 (one-momentum reduction). Suppose $p$ is smooth and homogeneous, $p(c)=0$, and $H_{\operatorname{Re}p}(c)$ is independent of $R(c)$. There are a nonzero homogeneous multiplier $a$ and homogeneous canonical coordinates at $(0,e_n)$ such that

The smooth real $f$ has degree one and is independent of $\eta_1$.

Proof. Make $p$ degree one by a positive multiplier. Theorem 1.1 then makes $\operatorname{Re}p=\xi_1$ at the mark $(0,e_n)$. In particular $\partial_{\xi_1}p\ne0$.

Apply (4.1) to divide the real function $\xi_1$ by $p$:

For homogeneity, perform this division on the positive $\xi_n=1$ slice, in the real variable $\xi_1/\xi_n$, with all remaining slice coordinates as parameters. Extend $q$ to degree zero and $r$ to degree one. This preserves (4.3). At the mark $r=0$, and differentiation in $\xi_1$ gives $q(c)=1/\partial_{\xi_1}p(c)\ne0$. Shrink so $q$ is nonzero.

Prescribe

These functions have degrees zero and one and bracket $\{\eta_1,y_1\}=1$. At the mark $R$ has its nonzero last-momentum component, $H_{y_1}=-\partial_{\xi_1}$, and $H_{\eta_1}$ has $x_1$ component one. Thus their fields and $R$ are independent. The homogeneous coordinate theorem completes them with mark $(0,e_n)$.

Equation (4.3) becomes $qp=\eta_1-i r_2$. The identity $H_{y_1}r_2=-\partial_{\xi_1}r_2=0$ is invariant under the coordinate change. In the new coordinates $H_{y_1}=-\partial_{\eta_1}$, so $f=-r_2$ is independent of $\eta_1$, as asserted. The product of $q$ and the initial positive multiplier is $a$. $\square$

The function $f$ may depend on $y_1$. That dependence carries the contact information in the next section.

5. Fixed higher contact, a filtration and the multiplier law

Let $p=p_1+ip_2$, $p(c)=0$, and $H_{p_1}(c)\ne0$. The hypersurface $V_1=\{p_1=0\}$ is foliated by its nonzero Hamilton field. We assume a fixed integer $k>1$: in a small flow box, $p_2$ has a zero of exactly order $k$ on every nearby Hamilton curve. On the resulting nearby zero set,

The assumption includes existence of the fixed-order zero on each nearby curve. A single nonzero kth derivative at $c$, or a statement concerning only those curves that happen to have zeros, is insufficient.

Lemma 5.1 (the smooth zero set and its brackets). Under this assumption the zero set $V=\{p_1=p_2=0\}$ is smooth of codimension two. Near $V$, choose a smooth function $g$ with $\{p_1,g\}\ne0$ such that

On $V$, $H_{p_2}=bH_{p_1}$, and every iterated Poisson bracket of at most $k$ factors chosen from $p_1,p_2$ vanishes.

Proof. Use flow-box coordinates $(t,z)$ on $V_1$, with $H_{p_1}=\partial_t$. The function $\partial_t^{k-1}p_2$ has a simple zero at $c$. The implicit function theorem gives its unique nearby zero graph $t=T(z)$. The fixed-order zero supplied on each nearby curve has all lower derivatives zero and must therefore be this graph. Shrinking the flow box, Taylor's formula below also shows that there are no other zeros in it.

All derivatives of order $<k$ vanish on this graph. Taylor's integral formula gives

on $V_1$. The last factor is smooth and nonzero. Extend $g=t-T(z)$ and that factor to a neighborhood of $V_1$. The difference between $p_2$ and $c_0g^k$ vanishes on $p_1=0$, so real Taylor division gives (5.2). Its zero set is $\{p_1=g=0\}$, and the two differentials are independent because $\{p_1,g\}\ne0$. Since $k>1$, differentiating (5.2) on $V$ gives $dp_2=b\,dp_1$, proving the Hamilton-field identity.

For the bracket assertion put

These are ideals of smooth local functions. Direct product rules give

For the second inclusion use

The only potentially low-order contribution of $g^{k-1}H_g$ acting on $p_1A+g^jB$ is $g^{k-1}\{g,p_1\}A$, and $k-1\ge j-1$. The other terms retain $p_1$ or at least $g^{j-1}$.

Both $p_1,p_2$ lie in $\mathcal D_k$. A right-nested bracket of $\ell$ factors lies in $\mathcal D_{k-\ell+1}$; for $\ell\le k$ this vanishes on $V$. Any bracket tree is a linear combination of right-nested brackets with the same factors: apply the Jacobi identity to move a bracket in the first argument successively to the right, and induct on its number of internal brackets. This proves the assertion for every tree. $\square$

Theorem 5.2 (finite-contact multiplier law). For every smooth complex multiplier $a=A+iB$, on $V$,

If $a\ne0$ and $A-bB\ne0$, the transformed symbol has the same fixed order $k$. For odd $k$ its kth derivative sign is unchanged.

Proof. Put $F=A-bB$, $G=B+bA$, and write (5.2) as

The product rule extends (5.5) to $H_f\mathcal D_j\subset\mathcal D_{j-1}$. Indeed $H_f=A H_{p_1}-B H_{p_2}+p_1H_A-p_2H_B$, and the last two terms acting on $\mathcal D_j$ still lie in $\mathcal D_{j-1}$. Thus $H_f^{k-1}(p_1U+p_2W)$ vanishes on $V$ for arbitrary smooth $U,W$.

The full first bracket has the form

This follows by expanding all factors; every term differentiating $A$ or $B$ contains $p_1$ or $p_2$. Also

After $k-1$ applications of $H_f$, the first two terms and all extra terms in (5.9) vanish on $V$. In the last term, every surviving derivative must hit a different one of its $k-1$ factors $g$; a derivative hitting its coefficient would leave a zero factor. Since $H_fg=F H_{p_1}g$ on $V$, the result is

The same Taylor product rule gives $H_{p_1}^kp_2=k!c_0(H_{p_1}g)^k$. This proves (5.7) with all differentiated multiplier terms accounted for.

If $F\ne0$, the equation $f=0$ solves $p_1=(Bc_0/F)g^k$. On that hypersurface,

Its factor is nonzero, and $H_fg=F H_{p_1}g\ne0$ on the zero set. Here $df=F\,dp_1\ne0$ on $V$, so the implicit function theorem makes $f=0$ a smooth hypersurface. Inside it, $g=0$ is transverse to $H_f$. The transverse-flow chart F0 identifies a neighborhood with a time interval times that zero section. Thus every nearby Hamilton curve meets it, and (5.12) gives exactly a kth-order zero there. If $k$ is odd, $k-1$ is even, so the factor in (5.7) is positive. If $F=0$, $H_f=0$ on $V$; that case fails the nonzero real-field hypothesis. $\square$

6. The homogeneous finite-contact normal form

Theorem 6.1 (fixed-contact model). Let $p=p_1+ip_2$ be homogeneous and satisfy the fixed-order hypothesis of Section 5 with $k>1$. There is a nonzero homogeneous multiplier and a homogeneous canonical map such that

The mark is $(0,e_n)$, except when $k$ is odd and $H_{p_1}^kp_2(c)<0$; that case has mark $(0,-e_n)$. The exceptional negative odd case cannot have the displayed model at the positive mark.

Proof. First $H_{p_1}(c)$ cannot be radial. If it were a nonzero multiple of $R(c)$, homogeneity would make its field radial along the covector ray through $c$. Flow uniqueness would keep its orbit on that ray. Since $p_2$ vanishes along the ray, it would have infinite, rather than finite, zero order on this orbit.

A positive preliminary degree change gives degree one and preserves the nearby Hamilton curves in $p_1=0$, their zero order and the leading derivative sign. Use Theorem 1.1 to make $p_1=\xi_1$, then Theorem 4.1 to reduce the symbol to

We check that the division multiplier is admissible for finite contact. On $V$, $H_{p_2}=bH_{\xi_1}$, so $\partial_{\xi_1}p=1+ib$. The multiplier $q$ in (4.3) satisfies $q(c)=(1+ib(c))^{-1}$. Therefore

It stays nonzero after shrinking. Theorem 5.2 preserves fixed contact, and its positive factor at $c$ preserves the kth derivative sign in this particular reduction. Canonical changes preserve all the Hamilton identities.

Rename these coordinates $(x,\xi)$. The equation $\partial_{x_1}^{k-1}f=0$ has a unique nearby solution

by the nonzero kth derivative. Each nearby $H_{\xi_1}=\partial_{x_1}$ curve has its fixed-order zero there. Because $f$ is independent of $\xi_1$, this describes it for all $\xi_1$ in the patch. Its lower derivatives all vanish on the graph. Taylor's formula gives

Uniqueness of the graph and homogeneity give $\deg X=0$, $\deg F=1$. Initially assume the kth derivative is positive. Then $F>0$, and

This is a signed smooth coordinate; taking a nonnegative root of $f$ would lose that property.

Apply Theorem 2.1 to $\xi_1,g$ with

The normalizer has degree $-1/k$. The functions $\phi=u^\alpha\xi_1$, $\psi=u^\beta g$ have degrees $\alpha,\beta$ and bracket one. Lemma 3.1 supplies $h>0$, degree one and commuting with them, and retains

as canonical coordinates at the positive mark. The exact identity, including its multiplier, is

Indeed $k\beta=\alpha$ and $1-k\beta=\beta$. The product $u^\alpha h^\beta$ has degree zero.

For negative kth derivative, apply the positive construction to $\bar p$ and conjugate the resulting identity. This gives $\eta_1-i y_1^k\eta_n$ for a nonzero multiplier times $p$, at $(0,e_n)$. If $k$ is even, multiply by $-1$ and flip the canonical pair $(y_1,\eta_1)$; the result is the positive displayed model at that same mark. If $k$ is odd, flip the last pair $(y_n,\eta_n)$; the result is (6.1) at $(0,-e_n)$. For odd $k$, (5.7) has positive factor for every admissible multiplier and the positive model has kth derivative $k!>0$. This proves the obstruction at the positive mark. $\square$

7. A smooth elliptic solver with parameters

The remaining normal form has a complex Hamilton field tangent to a two-dimensional characteristic leaf. Solving it needs a common open set for all transverse parameters and all derivative orders. We prove that input.

Lemma 7.1 (two-variable parameter solvability). Let

have smooth complex coefficients, where $x\in\mathbb R^2$ and $\lambda\in\mathbb R^d$. Suppose the real map $(\tau_1,\tau_2)\mapsto a(0,0)\tau_1+b(0,0)\tau_2$ is invertible. Every smooth $f(x,\lambda)$ has a joint smooth local solution $L_\lambda u=f$ on a fixed smaller product neighborhood. The solution can be chosen linearly and continuously in $f$ in the smooth seminorms on fixed compact subsets.