Rotated brackets and finite-jet flow coordinates

Repeated brackets use two Hamilton fields in every possible order. A directional test instead follows just the real part of one complex rotation of the symbol. These tests look weaker: they sample powers of one field, rather than all bracket words. A nonnegative first bracket on the characteristic set makes them sufficient to detect the same finite type.

We use the Hamilton convention and the two weighted Taylor theorems in Sign-constrained weighted Taylor geometry. Phase space and generating families supplies symplectic linear algebra, ordinary Darboux coordinates and Hamiltonian flow invariance. The function-preserving real coordinate construction is in Homogeneous submanifold normal forms, Section 2 and the ordinary construction in Theorem 3.2. We use its real geometry and prove the additional finite-jet control needed for a complex symbol.

Basic references are Lerner [L], Evans–Zworski [E] and Hörmander [H3]. All coordinates in this lesson are ordinary local symplectic coordinates. The final equivalence applies to arbitrary smooth scalar symbols; homogeneous operator coordinates require a further construction.

1. What one rotated field measures

Let \(p=p_1+ip_2\) be a smooth complex function on a symplectic manifold, with real \(p_1,p_2\). At a marked point \(c\), define the directional tests

Our Poisson bracket is \(H_fg=\{f,g\}\). If \(z=a+ib\), then

In particular

For a word \(I=(i_1,\ldots,i_\ell)\) in \(1,2\), write

As in the weighted geometry lesson, \(\kappa(c)\) is the largest depth through which all such values vanish, and \(\kappa=0\) at a noncharacteristic point. Thus

We allow infinite depth. The sign hypothesis for this lesson is

on a neighborhood of \(c\). It concerns the full nearby characteristic set, even when the directional tests are evaluated at a single point.

2. Lower vanishing freezes a multiplier at the endpoint

For an integer \(r\geq1\), let \(\mathcal I_r\) be the ideal of smooth functions generated by the bracket functions \(p_I\) with \(|I|\leq r\). Products with arbitrary smooth coefficients are included. A bracket tree can be reduced to right-nested words using antisymmetry and the Jacobi identity, so it gives the same ideal if every bracket arrangement is allowed.

Lemma 2.1. Replace each leaf of an \(\ell\)-leaf bracket by a smooth linear combination of \(p_1,p_2\). Its result is the corresponding expansion using constant leaf coefficients evaluated at \(c\), modulo terms that vanish at \(c\), whenever \(\mathcal I_{\ell-1}\) vanishes there. Consequently, if every bracket with at most \(j\) leaves vanishes at \(c\), a smooth complex multiplier \(u\) satisfies

Proof. Start with the product rule

Here \(F,G\) may themselves be bracket functions. The first term has the combined full leaf count. Every other term has a bracket function with strictly fewer leaves as a factor; the remaining expression is a smooth coefficient. The same observation gives

Induction on the bracket tree now expresses its full-length terms with coefficients equal to products of its leaf coefficients. All terms differentiating a leaf coefficient belong to \(\mathcal I_{\ell-1}\). Evaluation at \(c\) discards those terms, and the full-length coefficients become their values at \(c\). Taking \(\ell=j+1\), and the real and imaginary combinations in (1.2), proves (2.1). The case \(j=0\) is immediate without a lower-vanishing hypothesis. ∎

The same expansion with an invertible smooth matrix of leaf coefficients, and then its inverse, proves that the ideals of vanishing through any fixed depth are unchanged. In particular a nonzero complex multiplier preserves \(\kappa\). Symplectic coordinate changes preserve every bracket exactly. These facts also follow from Section 2 of the weighted geometry lesson.

The sign hypothesis is preserved by a nonzero multiplier because

Equation (2.1) is a more precise assertion than unqualified multiplier invariance of an individual higher directional test. It is the lower vanishing that removes derivatives of the multiplier.

A bracket with \(\ell\) leaves uses at most \(\ell-1\) derivatives of any leaf. This follows by induction from the first-order product rule for the Poisson bracket. Therefore its value is determined by the \((\ell-1)\)-jets of the leaves. If two symbols have the same \(N\)-jet, all their bracket values with at most \(N+1\) leaves agree. This observation will let us use a coordinate reduction of any prescribed finite accuracy.

3. Prepare a complex Taylor polynomial, then correct its real flow

Write \(\mathfrak m^{N+1}\) for the smooth functions whose derivatives through total order \(N\) vanish at the origin. Multiplying them by a smooth function, or composing them with a smooth local diffeomorphism fixing the origin, preserves that property.

Lemma 3.1 (finite polynomial preparation). Suppose \(P(0)=0\), with coordinates \((\tau,y)\), and

For every \(N\geq1\), there are complex polynomials \(A(\tau,y)\), \(\phi(y)\), with \(A(0)=d\), \(\phi(0)=0\), such that

Proof. Take the ordinary Taylor polynomial through degree \(N\). Work temporarily with polynomials modulo terms of degree greater than \(N\); the variable \(\tau\) is a formal variable in this calculation. Construct \(\phi\) degree by degree. If its terms below degree \(r\) have been chosen, the degree-\(r\) term of \(P(\phi(y),y)\) changes by \(d\phi_r\) when the next homogeneous term \(\phi_r\) is added. Choose \(\phi_r\) to be minus the current degree-\(r\) term divided by \(d\). Starting at degree one, this makes

Polynomial subtraction gives a factorization of

by \(\tau-\phi(y)\). Truncate its quotient to the needed degree and call it \(A\). Its constant coefficient is \(d\). Equation (3.2), and the ordinary smooth Taylor remainder, give (3.1). This is finite polynomial algebra: no evaluation of the original smooth function at a complex argument is being asserted. ∎

The real geometry needed next is the function-preserving flow construction in the stated prerequisite. We record its finite-jet estimate explicitly.

Lemma 3.2 (a flow chart with the identity jet). In canonical coordinates \((t,\tau,w)\), where \(w\) consists of the remaining canonical pairs, let

with \(r,E\) real. There is a symplectic map \(\Phi\), fixing zero, such that

Proof. On the hypersurface \(t=0\), solve \(r(0,v,w)=\sigma\) for \(v=v(\sigma,w)\). The implicit function theorem applies because \(r_\tau(0)=1\), and

Define the flow chart

Hamiltonian flow keeps \(r\) constant, so \(r\circ\Phi=\sigma\). The parameter derivative is \(\partial_t\Phi=H_r\). On the initial hypersurface, the symplectic form restricts to the form of the \(w\) pairs: every term involving \(dt\) vanishes. The Hamiltonian flow preserves this restriction. Moreover

because \(\iota_{H_r}\omega=-dr\). These identities show that the pullback form is \(d\sigma\wedge dt+\omega_w\). Thus the chart is symplectic; its derivative at zero is invertible.

For the extra estimate, \(H_r=\partial_t+V\), where the coefficients of \(V\) vanish to order \(N\). Smooth dependence of the local flow and its integral equation give, uniformly for small parameters,

Indeed the trajectory stays bounded by a fixed multiple of the size of these parameters, and integrating the bound on \(V\) gives (3.6). Combine this with (3.4). Each component of \(\Phi-\operatorname{id}\) is smooth and bounded by a constant times total parameter size to power \(N+1\); its Taylor polynomial therefore has no term of degree at most \(N\). Rename \(\sigma\) as \(\tau\). This proves (3.3). ∎

Proposition 3.3 (finite-jet complex flow reduction). Let \(p(c)=0\), \(dp(c)\ne0\). For every \(N\geq1\), a nonzero smooth complex multiplier and a local symplectic change give

Here \(q\) is real and independent of \(\tau\). The marked point is zero. If (1.6) holds for \(p\), then exactly

near zero. The \(N\)-jet of \(P\) is that of \(\tau+iq\).

Proof. Use Darboux coordinates centered at \(c\). A symplectic linear change makes \(\partial_\tau p(0)\ne0\): choose a real tangent vector on which \(dp\) is nonzero, put it in a symplectic basis, and use it as the momentum direction of the first pair. Lemma 3.1 gives (3.1). The polynomial \(A\) is nonzero on a smaller neighborhood. Multiply by \(A^{-1}\), and write

with \(h,b\) real. We obtain

First straighten the real momentum \(r_0=\tau-h(t,w)\) by the real function-preserving flow chart. Its time coordinate is still \(t\), since \(\{r_0,t\}=1\). More explicitly, let \(w(t)\) solve

The map

is the same canonical pair flow construction, with \(r_0\circ\Phi_0=\sigma\). To check its symplectic character, restrict the form to its initial \(t=0\) hypersurface, where it is \(\omega_v\), and use the Hamiltonian flow identities in Lemma 3.2. This is the real chart supplied by the prerequisite. Crucially \(w(t)\) is independent of \(\sigma\). Hence \(b\circ\Phi_0\) stays independent of the new momentum.

After this chart, (3.9) has the form

Its exact real part is \(\tau+\operatorname{Re}R_0\). Apply Lemma 3.2 to it. Since this second canonical map has the identity \(N\)-jet, the new imaginary part, say \(Q(t,\tau,w)\), still has the same \(N\)-jet as a function independent of \(\tau\). Its real part is now exactly \(\tau\).

Set \(q(t,w)=Q(t,0,w)\). The integral form of Taylor's formula in \(\tau\) gives

The independence of the \(N\)-jet implies \(B\in\mathfrak m^N\), proving (3.7). All changes made before (3.7) were exact symplectic maps or nonzero multipliers, so (1.6) is preserved. The characteristic set of (3.7) is exactly \(\tau=q=0\), and there

This proves (3.8). Finally \(\tau B\) has no jet of degree at most \(N\), as claimed. ∎

The multiplier and coordinates may depend on \(N\). Proposition 3.3 supplies every requested finite accuracy together with an exact characteristic-set sign condition. It does not assert a single smooth chart removing the momentum dependence to infinite order.

4. The weighted endpoints detect every remaining bracket

Suppose a normalized germ \(\tau+iq(t,w)\) has all bracket values with at most \(K\) leaves zero, where \(K\geq2\), and has (3.8). The weighted geometry lesson gives two alternatives.

If

its one-direction theorem has \(c=\partial_t^K q(0)\) and endpoint

The value conditions \(\partial_t^j q(0)=0\), \(j<K\), already follow from the assumed bracket vanishing. If every endpoint (4.2) is zero, take \(a=1,b=0\) to get \(c=0\). That theorem then gives vanishing of every bracket with \(K+1\) leaves.

Otherwise choose the smallest \(s\) for which \(d(\partial_t^s q)(0)\ne0\) and \(2s<K\). Earlier differential jets vanish. The \(t\) component of this nonzero differential is zero: \(\partial_t^{s+1}q(0)=0\), since \(s+1<K\). Its nonzero part therefore lies in the transverse symplectic slice \(t=\tau=0\). The real function-preserving coordinate construction on that slice makes

Extend this slice map independently of \(t,\tau\). It is symplectic, preserves all the earlier value and differential conditions, and leaves (3.8) intact. When there is no transverse pair this alternative cannot occur.

The mixed theorem of the weighted geometry lesson applies with \(k=K\). Its nonnegative slope polynomial \(B_0(t,z)\) gives

Here \(B_0\) denotes that theorem's polynomial, distinct from the flat correction \(B\) in (3.7). If every value (4.4) is zero, take any \(a\ne0\) and let \(b\) range over the real line. The second argument ranges over the entire real line. Thus \(B_0\) vanishes for all \(t\ne0,z\), and polynomial continuity makes it identically zero. The mixed theorem again gives vanishing of all \(K+1\)-leaf brackets.

This proves the following exact consequence of the two preceding weighted theorems:

Lemma 4.1. For a normalized germ with (3.8), if every bracket through \(K\) leaves and every constant-rotation endpoint \(T_K\) vanish, then every bracket through \(K+1\) leaves vanishes.

The proof has checked every required normalization. In the mixed alternative, the slice momentum in (4.3) is zero at the marked point because \(\partial_t^s q(0)=0\). Both coordinate constructions are ordinary local ones; no radial condition is being added to this lemma.

5. All words, constant rotations, and smooth rotations agree

Theorem 5.1 (rotated-bracket equivalence). Let \(p\) be any smooth complex scalar function satisfying (1.6) near \(c\). For every integer \(k\geq0\), the following are equivalent:

• \(\kappa(c)>k\);
• \(\left(H_{\operatorname{Re}(fp)}\right)^j\operatorname{Im}(fp)(c)=0\) for every smooth complex \(f\) and every \(0\leq j\leq k\);
• \(T_j(z;p,c)=0\) for every constant \(z\in\mathbb C\) and every \(0\leq j\leq k\).

Proof. If all bracket words through \(k+1\) leaves vanish, the product-rule expansion in Lemma 2.1 makes every test using a smooth multiplier vanish. This implication allows multipliers that are zero at the point. Smooth tests include constant tests. It remains to recover the bracket words from the constant tests.

For \(k=0\), taking \(z=1,i\) says exactly \(p(c)=0\). For \(k=1\), (1.3) adds \(\{p_1,p_2\}(c)=0\), which is precisely all two-leaf vanishing. These bases do not require the sign hypothesis.

Induct on \(k=K\geq2\). The tests through \(K-1\), by induction, already give vanishing of every bracket through \(K\) leaves. If \(dp(c)=0\), the initial Hamilton fields are zero at \(c\). Every bracket of length at least two is the action of one of those fields on a smooth function, so every such value is zero. We are done in that case.

Suppose \(dp(c)\ne0\). Apply Proposition 3.3 with \(N=K+2\). Let the resulting exact transformed symbol be \(P\), and put

The ideal invariance under the nonzero multiplier and symplectic map gives vanishing of the \(K\)-leaf brackets for \(P\). Lemma 2.1, using that lower vanishing, transfers the assumed \(K\)-th constant endpoints to \(P\): multiplying all constants by the fixed nonzero multiplier value still ranges over all of \(\mathbb C\).

Symbols \(P,P_*\) have the same \(N\)-jet. Their bracket values and directional endpoints through \(K+1\) leaves agree. Also \(q\) has the exact sign condition (3.8), rather than merely a formal sign condition on its jet. Lemma 4.1 therefore proves that every \(K+1\)-leaf value for \(P_*\), and hence for \(P\), is zero. Undoing the multiplier and symplectic change gives that same conclusion for \(p\). This completes the induction and the proof. ∎

At a noncharacteristic point all three conditions fail, already at \(j=0\). At a characteristic point with \(dp=0\), all three hold for every finite \(k\). Thus no principal-type hypothesis was silently imposed on the theorem.

Corollary 5.2. Under (1.6), finite bracket type is equivalent to the existence of a nonzero constant-rotation test. More precisely, if \(\kappa(c)=r<\infty\), then all \(T_j\) with \(j<r\) vanish, and some \(T_r\) is nonzero. Conversely, if \(r\) is the least order at which some constant-rotation test is nonzero, then \(\kappa(c)=r\).

Proof. Apply Theorem 5.1 at \(k=r-1\) and \(k=r\), using the convention at \(r=0\). ∎

This is a geometric finite-type equivalence. It supplies neither a subelliptic estimate nor the neighborhood orientation condition for every lowest odd endpoint. Those analytic and sign requirements enter the full finite-type characterization separately.

6. Exercises with complete solutions

Exercise 1 — prepare an actual complex polynomial, 8 points. Let

Find \(A,\phi\) in Lemma 3.1 through total degree four. State exactly what is discarded.

Solution. Choose \(A=1+i\tau\), \(\phi=-i t^3\). Their product is

It equals \(P-t^5\) exactly. The remainder is \(t^5\), whose derivatives through degree four vanish at zero. The unit \(A\) is nonzero near zero. This is a degree-four preparation, not a claim that the displayed \(\phi\) is an exact root of the full polynomial.

Exercise 2 — the size of a correcting flow, 8 points. On one canonical pair let

Construct the map in Lemma 3.2 for \(N=3\), and verify its symplectic form and identity jet.

Solution. Since \(r(0,v)=v\), the initial value is \(v=\sigma\). Its Hamilton field is \(\partial_t-t^3\partial_\tau\), so

Direct substitution gives \(r\circ\Phi=\sigma\), and

The correction has degree four, so the map has the identity three-jet. The sign of the frequency correction agrees with \(H_r\tau=-t^3\).

Exercise 3 — a smooth multiplier with varying coefficients, 10 points. Set

Find \(\kappa(0)\), and compute \(T_5(a+ib;up,0)\) without differentiating the full variable-coefficient field five times. Identify the lower vanishing that permits your calculation.

Solution. The one-direction theorem with \(K=5\) gives \(c=5!\), so \(\kappa=5\); all words through five leaves vanish, and \(H_\tau^5t^5(0)=5!\ne0\). Lemma 2.1 applies at the six-leaf endpoint. Since \(u(0)=1\),

The smooth unit preserves \(\kappa\) and the characteristic-set sign condition. Freezing it at an earlier endpoint without the required lower vanishing would need a separate justification.

Exercise 4 — the full mixed rotation polynomial, 10 points. Use

Find \(T_9(a+ib;p,0)\) as a polynomial in \(a,b\). Determine the least nonzero rotation order and compare it with the all-word type.

Solution. The mixed theorem has \(K=9,s=2\) and \(B_0=7t^8+t^2z^2\). In (4.4) the second argument is \(-a^2b/6\). Hence

It is nonnegative and is nonzero at \(a=1,b=0\). All earlier rotation orders vanish because all words through nine leaves vanish. Thus the least rotation order is nine, and Theorem 5.1 gives \(\kappa=9\). For example \(T_9(1+i)=9!\cdot253/126\); the factorial and transverse coefficient both matter.

Exercise 5 — a critical point of the symbol, 8 points. Let

at the origin of a canonical plane. Compute its bracket type there and all directional tests. Is the sign hypothesis needed for this calculation at the point?

Solution. Both initial Hamilton fields are zero at the origin, because \(dp(0)=0\). Every right-nested word with at least two leaves has one of these zero fields acting as its outermost factor, so its value is zero. The leaves themselves vanish too. Thus \(\kappa=\infty\). A rotated symbol also has value and differential zero there, so every directional test vanishes, including those with smooth multipliers. This pointwise calculation needs no sign hypothesis. In fact \(\{t^2,\tau^2\}=-4t\tau\), and the characteristic set is only the origin, where it is zero.

Exercise 6 — finite type does not settle the sign geometry, 10 points. For

find the bracket type and one nonzero constant-rotation test at the origin. Does Theorem 5.1 exclude the flat adverse crossing exhibited in the preceding weighted geometry lesson?

Solution. The mixed model has \(K=13,s=4\) and \(B_0=9t^{12}+t^2z^2\). Taking \(a=1,b=0\) in (4.4) gives

Every bracket through thirteen leaves vanishes, so \(\kappa=13\) and this is the least nonzero rotation order. With fixed \(z\ne0,\xi_2=0\), however, the imaginary part has leading term \(-t^3z^2\) near zero, and changes from positive to negative along increasing \(t\). Its zero slope is stationary at that crossing. The equivalence detects bracket type under the first slope condition; it does not replace the stronger neighborhood sign geometry required for an operator estimate.

References

• [L] Nicolas Lerner, Semi-classical estimates for non-selfadjoint operators, Asian Journal of Mathematics 11 (2007), 217–250. Open author's version. Characteristic signs and finite-type geometry.
• [E] Lawrence C. Evans and Maciej Zworski, Lectures on Semiclassical Analysis. Open university-hosted lectures. Real principal-type canonical coordinates and Hamiltonian flow methods.
• [H3] Lars Hörmander, The Analysis of Linear Partial Differential Operators III: Pseudo-Differential Operators, Springer, 2007 reprint. Publisher's record. Section 21.1: symplectic coordinates, canonical transformations and the Darboux theorem.

Written by GPT-6.1 Sol (OpenAI), at Ultra reasoning effort, September 2026. Self-checked by the writing AI. Public domain (CC0).