Crossing direction and the necessary finite-type bound

A finite-order zero can either force regularity or hide a concentrated approximate solution. The direction of an odd crossing distinguishes these behaviors. A crossing from positive to negative produces a rapidly decreasing null profile. Correcting that profile to any finite accuracy rules out every positive fractional gain. At crossings with the permitted direction, weighted concentration gives a different obstruction: the gain cannot exceed the reciprocal of the bracket type plus one.

We first turn a nearby crossing of minimum odd order into an exact smooth factorization. We then construct approximate null solutions, including arbitrary order-zero terms. Finally we prove the missing weighted argument when a transverse Hamilton field points along a dilation ray, and assemble the necessary finite-type and orientation conditions.

The estimate, openness and multiplier results are in Detecting a fractional gain in a cone. We use its order reduction, Canonical transport of fractional regularity, and the exact reduction in Smooth complex preparation and conic flow coordinates. The packet estimates are in Weighted packets and the limit of the gain. The two weighted Taylor alternatives and the rotated-bracket equivalence are in Sign-constrained weighted Taylor geometry and Rotated brackets and finite-jet flow coordinates. Lemma 2.1 of Homogeneous submanifold normal forms supplies the exact real homogeneous momentum coordinate used below. Its real geometric construction is reused.

Basic references are Lerner [L], Evans and Zworski [E], and Hörmander [H]. We use \(D=-i\partial\), \(H_fg=\{f,g\}\), and \(\omega=d\xi\wedge dx\). Operators are scalar and properly supported. Their symbols are ordinary \(S_{1,0}\) symbols with a smooth homogeneous principal representative modulo one lower order. No expansion of the lower symbol into homogeneous terms is required.

1. Minimum odd order makes a root persist

An adverse zero of a real function \(Q(t,y)\) is a zero of finite odd order \(r\) in \(t\) with

The function is positive just before this zero and negative just after it.

Lemma 1.1 (a persistent minimum-order root). Suppose an adverse zero of order \(r\) is present, and no adverse zero of smaller odd order occurs in the working neighborhood. On a smaller product neighborhood there are smooth real functions \(a(y)\) and \(b(t,y)<0\) such that

The marked root is \(t=a(y_*)\). If \(Q\) has degree one under a dilation of the parameters that fixes \(t\), then \(a\) has degree zero and \(b\) has degree one, on the corresponding smaller conic germ.

Proof. Translate the marked \(t,y\) to zero. Shrink an interval so that \(\partial_t^rQ<0\) throughout its product with a parameter neighborhood. At its two endpoints the marked function has opposite signs, positive on the left. Those endpoint signs persist for every nearby parameter.

For each such parameter there is at least one positive-to-negative zero of finite odd order at most \(r\). Here is why finiteness and the order bound hold. A function with nonzero \(r\)-th derivative throughout an interval has at most \(r\) distinct zeros, by repeated Rolle's theorem. At any zero at least one derivative through order \(r\) is nonzero. The opposite endpoint signs force at least one of the finitely many zeros to be a positive-to-negative crossing. Its first nonzero derivative has odd order and is negative. Minimum order now makes that order exactly \(r\).

The equation \(\partial_t^{r-1}Q=0\) has a unique nearby solution \(t=a(y)\): its \(t\)-derivative is \(\partial_t^rQ<0\), and the implicit function theorem gives existence and smoothness. Every adverse zero of order \(r\) just found must equal this solution. Consequently

for every nearby parameter. Taylor's integral formula gives (1.2), with

For a homogeneous \(Q\), positive dilation multiplies \(\partial_t^{r-1}Q\) by the same positive factor. Uniqueness of its root preserves \(a\); (1.2) or (1.4) then gives the degree of \(b\). ∎

Minimum order is a neighborhood property. The order of one isolated crossing need not persist when parameters change.

2. Straighten the root while retaining the real momentum

Suppose the real principal symbol has become \(\tau\). Write the full complex principal representative as

The real function \(q\) can depend on \(\tau\). Its restriction to \(\tau=0\) controls the imaginary part on the real bicharacteristics through characteristic points.

Assume that restriction has the factorization

Real division in \(\tau\) gives

where \(c=\int_0^1q_\tau(t,s\tau,w)\,ds\) is smooth and real. Thus the nonzero degree-zero multiplier \(1/(1+ic)\) gives

All functions are defined on a smaller cone where the segment in (2.3) stays in the domain. Homogeneity of \(q\) makes \(c\) degree zero and \(C\) degree one.

Let \(\varphi_a^\sigma\) be the transverse Hamilton flow of the degree-zero real function \(a\). The exact map

is symplectic and homogeneous. It fixes the marked point after the time translation, and retains \(\tau=\sigma\). Since \(a\) is constant along its own flow,

Verification. The form on the transverse parameter \(v\) is preserved by the flow. Its extra \(\sigma,v\) pairing is \(\iota_{H_a}\omega_w=-da\). This cancels the \(d\sigma\wedge da\) from \(d\tau\wedge dt\), leaving

Because \(a\) has degree zero, its Hamilton field has degree minus one: \( [R_w,H_a]=-H_a\). Equivalently

Equations (2.5) and (2.8) prove homogeneity. The map is locally invertible, also directly from its displayed formula. Only a local flow is needed.

After this change, (2.4) is exactly

The coefficient can depend on the new momentum. We retain that dependence in the construction that follows.

3. Correct a decreasing null profile to any finite accuracy

Lemma 3.1 (concentrated approximate null solutions). Let \(n\geq2\), and let \(L\in\Psi^1\) have, near the characteristic point \(c=(0,e_n)\), principal representative

In particular \(\xi_1(c)=0\), while the transverse covector is nonzero. Here \(C\) is smooth homogeneous of degree one and the full symbol differs from \(\ell_1\) by \(S^0\). For every \(N>0\), there are compactly supported smooth functions \(u_h\), \(0<h<h_N\), concentrating at the ray of \(c\), such that

For any proper order-zero \(A\) elliptic at \(c\), and any \(0<\varepsilon<1\),

for small \(h\). The support can lie in any prescribed base neighborhood of zero. Separated microlocal cutoffs applied to these functions are \(O(h^M)\) in every fixed Sobolev norm, for every \(M\).

Proof. Put \(a=1/(r+1)\), and use the unitary dilation and modulation

Normalized frequency displacements are \(h^{1-a}\eta=h^{ra}\eta\). On these scales the leading operator, after multiplication by \(h^a\), is

Its null profile is

Since \(r+1\) is even and \(\beta>0\), this is a nonzero Schwartz function. For any polynomial \(F\),

Consequently \(L_0\) maps polynomial multiples of \(v_0\) onto all polynomial multiples of \(v_0\): integrate the polynomial on the right in \(y_1\), choosing integration constant zero. This surjectivity is what permits all the corrections.

We give the finite symbol expansion and its remainder, rather than just solving a formal equation. For every fixed integer \(J\), on any fixed finite-dimensional space of polynomial multiples of \(v_0\),

The \(L_j(h)\) are polynomial differential operators, with coefficients uniformly bounded for \(0<h\leq h_J\). They preserve the space of polynomial multiples of \(v_0\), allowing its degree to increase by a fixed amount.

Here are the bounds establishing (3.8). The scaled principal part is exactly

in the left Fourier formula. Taylor terms with base multi-index \(\gamma\) and frequency multi-index \(\nu\) have power \(h^{a(|\gamma|+r|\nu|)}\). Retain those of weighted degree at most \(J\). For the \(S^0\) remainder \(b_0(x,\xi)\), the scaled symbol is

Its central Taylor coefficients include

which are uniformly bounded by the \(S^0\) estimates. The displayed power includes the prefactor \(h^a\) in (3.10). Thus its coefficients can depend on \(h\), without any homogeneous lower expansion.

On a fixed small neighborhood of the normalized phase-space center, Taylor's integral remainder and the same estimates after any fixed number of \(\eta\)-derivatives bound the discarded symbols by

for a fixed finite \(K\). Choose a sufficiently long Taylor expansion to control the derivatives needed below. Equivalently apply the weighted Taylor remainder argument of the packet lesson, with base weight one and frequency weight \(r\). Differentiation in \(\eta\) contributes \(h^{ra}\) per derivative, compensating the lowered Taylor order. The extra factor \(y_1^r\) in (3.9) adds only a fixed polynomial power.

Integrating by parts in \(\eta\), as in Lemma 2.2 of that lesson, gives the \(L^2\) bound in (3.8): Schwartz decay of \(\widehat v\) absorbs the \(\eta\)-powers, and arbitrarily many integrations give an integrable \(y\)-bound. Outside the small base neighborhood, \( |y|\geq c h^{-a}\), so that same decay absorbs every fixed power of \(h^{-1}\). Outside the normalized frequency neighborhood, \( |\eta|\geq c h^{-ra}\); the full symbol estimates give at most polynomial growth in \(h^{-1},\eta\), which the Schwartz tail absorbs to any order. This includes the bounded-frequency region of the original symbol. Cutoff derivatives have the corresponding nonnegative scale factors and obey the same bounds. Local properness corrections have smooth kernels; integration against the nonzero central frequency makes them \(O(h^M)\) for every \(M\). These arguments are uniform over bounded polynomial coefficients in the chosen finite-dimensional profile space. They prove (3.8) for the full properly localized operator.

Define polynomials recursively, with \(v_j=F_j(h,y)v_0\), by

Equation (3.7) solves every step. Polynomial degrees are bounded for fixed \(J\), and the coefficients remain bounded uniformly in \(h\). With

all powers through \(h^{Ja}\) cancel in (3.8). Hence

Take \(Ja\geq N\). Multiplying \(U_hv_h\) by a fixed compact base cutoff equal to one near zero changes both its norm and its equation by \(O(h^M)\) for every \(M\). Indeed the profiles and all their derivatives are uniformly Schwartz; outside that cutoff their scaled base arguments are at least \(c h^{-a}\). Mapping by an order-one operator loses only a fixed power of \(h\). This gives the required \(u_h\), and \(v_h\to v_0\) in every fixed Schwartz seminorm.

Finally the scaled left-symbol formula for \(h^\varepsilon\langle D\rangle^\varepsilon A\), on bounded \(y\)-sets, converges to \(a_0(c)v_0\). The domination and tail proof is precisely Lemma 3.1 of the packet lesson, now with a uniformly Schwartz profile converging to \(v_0\). Its nonzero limit proves (3.3). The same Fourier tail and separated-support argument gives the asserted microlocal concentration. ∎

The corrections include arbitrary order-zero terms. Keeping only \(v_0\) would in general leave a bounded error, and would not establish (3.2) to arbitrary accuracy.

4. An adverse finite odd crossing excludes a fractional gain

Theorem 4.1. Let \(p\) be the homogeneous principal symbol of a scalar operator \(P\) of any real order. Suppose \(p(c)=0\), and for some positive odd integer \(k\),

Then \(P\) cannot have microlocal loss \(0<\delta<1\) at \(c\).

Proof. Reduce the order to one by a positive homogeneous elliptic factor. On the hypersurface \(\operatorname{Re}p=0\), the new real Hamilton field is that positive factor times the old one. At the first nonzero derivative along that field, differentiating the factor gives only lower vanished derivatives. Multiplication of the imaginary part also contributes a positive factor. Therefore the order \(k\) and the negative sign in (4.1) survive this reduction.

The real Hamilton field in (4.1) is nonradial at the marked point. If it were a multiple of the radial field there, homogeneity would make it tangent to the entire marked ray, with that ray invariant under its local flow. Both parts of \(p\) vanish along this ray. Every derivative of the imaginary part along that flow would then be zero, contradicting (4.1).

The exact real homogeneous canonical-pair construction gives \(\operatorname{Re}p=\tau\). Thus (4.1) becomes an adverse zero of \(Q(t,w)=\operatorname{Im}p(t,0,w)\). Assume for contradiction that \(P\) has the stated loss. Openness gives the same loss at all points of a sufficiently small conic neighborhood.

Among adverse zeros in that neighborhood choose the smallest odd order \(r\) that occurs, and a point where it occurs. Such an order exists and is at most \(k\). Shrink about this point inside the good neighborhood. Lemma 1.1 gives the exact factorization (2.2). Equations (2.3)–(2.9) give a nonzero elliptic multiplier and an exact homogeneous canonical map with principal symbol of the form (3.1).

Proper graph quantizations and their two-sided microlocal inverses transfer the assumed loss to an order-one operator \(L\) with this principal symbol. The new characteristic point has zero time momentum and a nonzero transverse covector, so \(n\geq2\) and the central-point hypotheses of Lemma 3.1 hold. All the terms left by quantization have order zero; that lemma includes them. On sufficiently small input and output cones the conic estimate must therefore hold:

with \(A\) elliptic at the new point and \(u\) supported in a fixed small compact neighborhood. Microlocally smoothing errors from shrinking the graph cones have arbitrarily small norms on the concentrating tests and do not change the argument. Lemma 3.1 makes the right side bounded and the left side grow at least as \(h^{-(1-\delta)}\). This is a contradiction. The minimum-order point was allowed to move within the common good neighborhood; thus it excludes the assumed loss at the original point. ∎

In particular, a fractional gain forces

The theorem applies also to any nonzero constant rotation (zp). Multiplying the operator by \(z\) preserves its loss. Higher odd crossings can violate the gain even when (4.3) holds everywhere.

The loss and transport proofs cited above use ordinary symbol bounds, proper conic inverses, Sobolev mapping, and a principal representative modulo one lower order. Their Baire, commutator and lower-term steps require no polyhomogeneous expansion of \(P\). The graph prerequisite's continuity, parametrix and Egorov results are also for ordinary \(S_{1,0}\) symbols. Consequently the argument covers the full symbol class stated in the introduction, including nonpolyhomogeneous \(S^{m-1}\) terms.

5. The remaining transverse direction can be radial

Use the exact homogeneous reduction

at \(c=(0;0,e_{n-1})\). Bracket length and the first characteristic sign survive the nonzero factor and canonical map. Suppose every bracket with at most \(k\) leaves vanishes at \(c\), and (4.3) holds nearby. We prove that the principal germ admits balanced positive weights with vanishing weight \(k\).

For \(k=1\), give every position and frequency displacement weight one. The characteristic value alone gives weight-one vanishing. Assume henceforth \(k\geq2\). Since \(H_\tau=\partial_t\),

If \(d(\partial_t^j q)(c)=0\) whenever \(2j<k\), the one-direction weighted theorem gives

Otherwise let \(s\) be the first index with nonzero differential and \(2s<k\). The differential has no (dt) component, by (5.2), and no \(d\tau\) component. It is therefore a nonzero transverse differential. Put

It is real homogeneous of degree one, and \(f(c)=0\).

The nonradial alternative. If \(H_f(c)\) is not radial, the real homogeneous momentum-coordinate construction makes \(f=\eta_2\) exactly by a transverse canonical map independent of \(t,\tau\). Extend that map by the unchanged first pair. The mixed weighted theorem now applies, with

Its precise hypotheses hold: all lower differentials vanish by the choice of \(s\); the slice derivative is exactly \(\eta_2\); bracket vanishing is canonical invariant; and \(q=0\) implies \(q_t\geq0\). A nonzero zero-valued homogeneous momentum requires a radial pair among the other transverse coordinates. This alternative consequently needs at least two transverse pairs.

Lemma 5.1 (the radial alternative). If \(H_f(c)\) is a nonzero radial multiple, the original homogeneous coordinates already give the required weights after reordering the transverse pairs:

Here the marked radial momentum is \(\rho=\eta_2=1\); its Taylor variable is the displacement \(\rho-1\). The leading position \(z=x_2\) has weight \(k-s\), rather than \(s+1\).

Proof. At the marked point the transverse canonical one-form is \(dz\). Radiality of \(H_f\) means

No nonlinear straightening of this radial field is asserted. Work on the real positive slice \(\rho=1\), and write

The function \(Q\) has \(Q=0\Rightarrow Q_t\geq0\), pure time derivatives zero below \(k\), all differentials of \(\partial_t^jQ\) zero for \(j<s\), and

Temporarily give \(t\) weight one and every slice variable other than \(t\) weight \(k-s\). Every Taylor monomial of \(Q\) has weight at least \(k\). Pure time terms are controlled by (5.2). A term linear in a slice variable needs time power at least \(s\), by the vanished differentials. Two slice variables already cost \(2(k-s)>k\).

Keep the weight \(k-s\) for \(z\), and lower a common weight \(\theta\geq k/2\) for the remaining variables \(Z=(x'',\eta''/\rho)\) as far as weight-\(k\) vanishing permits. Only finitely many jets can impose a restriction, since all weights stay positive. If \(\theta>k/2\), a nonzero weight-\(k\) monomial involving \(Z\) must occur. It is linear in \(Z\), contains no \(z\), and has time power different from \(s\), by (5.9). The leading weighted polynomial is therefore

with at least one \(A_\nu\ne0\) and no \(t^s\) term in any \(A_\nu\). Each \(A_\nu\) is a polynomial. If there are no remaining transverse variables, the lowering step is absent.

Finite Taylor remainder estimates make the weighted rescalings converge in \(C^1\) on every compact set. Their zero-set slope is nonnegative, and the closedness lemma in the weighted geometry lesson gives that same property to \(F\). Every finite \(z,Z\) is accessible in this polynomial limit: the original slice neighborhood expands under positive dilation of its centered variables. At \(t\ne0\), solve \(F=0\) for the free variable \(z\). The accessible-affine-zero lemma gives

Thus \(A_\nu\) is a multiple of \(t^s\) on each half-line. Its polynomial character and the absence of that power make it zero everywhere, a contradiction. Hence the minimum transverse weight is \(k/2\). Raising it to \((k+1)/2\) preserves vanishing and removes those variables from the leading part. We have proved weight-\(k\) vanishing of \(Q\) for slice weights \(1,k-s,(k+1)/2\).

Return to (5.8). Give \(\rho-1\) weight \(s+1>0\). Multiplication by \(\rho=1+(\rho-1)\) and substitution \(\eta''/\rho\) preserve weighted vanishing. This can be checked on each finite Taylor polynomial: expanding \(1/\rho\) adds only nonnegative powers of a variable of positive weight; the substituted transverse monomial retains at least its original weight. The smooth remainder has the same weighted estimate. Thus \(q\), and then \(\tau+iq\), vanish to weight \(k\) with (5.6). Each canonical pair has sum \(k+1\). The displacement notation is used only for Taylor derivatives at a nonzero covector; the actual coordinates and operator transformation remain homogeneous. ∎

The two alternatives exchange the weights of the special canonical pair. Treating a radial field as an ordinary homogeneous momentum would miss this exchange.

6. Assemble the necessary type and orientation conditions

For a characteristic symbol \(p=p_1+ip_2\), let \(\kappa(c)\) be the largest integer through which every iterated bracket word vanishes, with each leaf chosen from \(p_1,p_2\). Set \(\kappa=0\) at noncharacteristic points and allow infinity. Define

Theorem 6.1 (necessary finite-type conditions). If \(P\) has microlocal loss \(0<\delta<1\) at \(c_0\), there is a conic neighborhood \(V\) on which the same loss holds and:

1. at every \(c\in V\), some \(T_j(z;p,c)\ne0\) with \(j\leq\delta/(1-\delta)\);
2. if the least integer for which \(T_j(z;p,c)\) is not identically zero in \(z\) is odd, then \(T_j(z;p,c)\geq0\) for every \(z\).

Equivalently the finite bracket depth satisfies

The nonvanishing in condition 1 can be checked at \(c_0\) and then persists after shrinking \(V\). Condition 2 is a neighborhood requirement.

Proof. Openness supplies \(V\). Theorem 4.1 gives the first characteristic sign (4.3) throughout \(V\). At a characteristic point with the stated loss the complex Hamilton field is nonradial, by Corollary 5.1 of the packet lesson. Exact homogeneous complex reduction therefore gives (5.1), with unchanged bracket depth, sign, and loss.

Suppose all brackets through \(k\geq1\) leaves vanish at this point. Section 5 covers every possible differential: either no early transverse differential is present, or its first nonzero transverse Hamilton vector is nonradial, or it is radial. In each case balanced positive weights give weight-\(k\) vanishing. The full packet theorem, including its frequency tails, yields

Infinite bracket depth is impossible, since (6.3) for all \(k\) contradicts \(\delta<1\). Taking \(k=\kappa(c)\) at a characteristic point proves (6.2). The rotated-bracket equivalence under (4.3) identifies this depth with the least \(j\) for which (6.1) is nonzero for some \(z\). At a noncharacteristic point that least order is zero. This proves condition 1.

If the least order is odd and an endpoint is negative for some fixed \(z\ne0\), all lower endpoints for that rotation are zero. Theorem 4.1 applied to (zp) contradicts the same loss at that point. Every least odd endpoint is therefore nonnegative, proving condition 2.

Finally each fixed finite endpoint is smooth in \(c\). A nonzero endpoint at \(c_0\) stays nonzero nearby; this proves the last nonvanishing assertion. It does not remove the neighborhood quantifier from condition 2. ∎

This theorem proves necessity. The converse requires quantitative estimates for variable symbols satisfying the full orientation condition; it does not follow from weighted Taylor vanishing alone.

7. Exercises with complete solutions

Exercise 1 — the odd order can drop nearby, 8 points. For \(Q(t,y)=-t^5+y^2t\), find the adverse order at the origin and at its nonzero real roots when \(y\ne0\). Explain why a minimum-order point must be chosen before using Lemma 1.1.

Solution. At the origin the first nonzero time derivative is \(Q_{ttttt}=-5!<0\), so the adverse order is five. For \(y\ne0\), the nonzero roots are \(t=\pm\sqrt{|y|}\). At either root \(t^4=y^2\), and

They are adverse simple zeros. Near the origin the minimum adverse order is one. The root at zero has positive slope \(y^2\) for \(y\ne0\), so the order-five adverse root does not persist as such. Near any chosen nonzero simple root the implicit function theorem instead gives the persistent order-one branch required by the lemma.

Exercise 2 — an exact time-section map, 10 points. In two canonical pairs take \(a(z,\eta)=z^2\), and

Find (2.5), verify its one-form, and write the exact transformed principal symbol.

Solution. The transverse field is \(H_a=-2z\partial_\eta\). Hence

Its one-form is

It is therefore symplectic, and its base variables have degree zero while both momenta scale with the same positive factor. The transformed symbol is exactly

The coefficient depends on \(\sigma\), as allowed in Lemma 3.1. It has negative imaginary sign in a sufficiently small cone about \(\eta=1,\sigma=z=0\).

Exercise 3 — lower terms are corrected, 10 points. Use \(r=3\), \(C(c)=2-3i\), and a real constant lower term \(\lambda\). Find the null profile, the first two polynomial corrections arising from \(\lambda\), and the scale of the first transverse-frequency correction for

Solution. Here \(a=1/4\), and the leading profile is \(v_0=\exp(-(3+2i)y_1^4/4-y_2^2/2)\). The lower constant enters the scaled operator at \(h^{1/4}\lambda\). Equations (3.7) and (3.13) give

Indeed their \(L_0\)-images are \(-\lambda v_0\) and \(-\lambda v_1\), respectively. The central frequency of \(D_z\) is \(h^{-1}\), while its displacement is \(h^{-1/4}D_{y_2}\). After multiplying by \(h^{1/4}\), the latter term is \(h^{3/4}(2-3i)y_1^3D_{y_2}\). It first appears at correction index three, and is included in that recursive step.

Exercise 4 — radial weights, 12 points. Near \(\rho=1\), take

Check the characteristic first sign, find the first nonzero transverse differential, and give the weights and resulting necessary loss.

Solution. If \(t=0\), then \(q=0\) and \(q_t=0\). At a zero with \(t\ne0\), \(z=-2t^5\), and \(q_t=5\rho t^6\geq0\). The first transverse differential is that of \(\partial_t^2q(0,z,\rho)=\rho z\), namely \(dz\) at the marked point. Its field is \(-\partial_\rho=-R\), so the radial alternative has \(s=2,k=7\). The weights are

Both leading imaginary monomials have weight seven. Their products with \(\rho-1\) have higher weight. All brackets through seven leaves vanish by the leading-bracket theorem, while \(H_\tau^7q(c)=7!\ne0\). Thus \(\kappa=7\), and every fractional-loss estimate must have \(\delta\geq7/8\). The early nonzero transverse differential does not reduce that bracket depth.

Exercise 5 — compare the nonradial pair, 10 points. In three canonical pairs take

Compare its special weights with Exercise 4 and compute the complete seventh constant-rotation endpoint.

Solution. The first transverse function is \(f=\eta_2\), whose field is \(\partial_{x_2}\), independent of the radial field. The nonradial weights are

The special pair's weights are exchanged relative to the radial case. The characteristic sign follows from \(\eta_2=-2t^5\rho\) at nonzero-time zeros, giving \(q_t=5t^6\rho\geq0\); at \(t=0\) the slope is zero. In the mixed leading model \(b=t^7\), so \(B=b_t-2b/t=5t^6\). Formula (3.6) in the weighted geometry lesson, with rotation \(z_0=A+iB_0\), gives

It is nonnegative and nonzero at \(z_0=1\). All lower endpoints vanish and \(\kappa=7\), again forcing loss at least (7/8).

Exercise 6 — orientation must hold nearby, 12 points. Let \(e(z)=e^{-1/z^2}\) for \(z\ne0\), extended by zero, and

Determine the bracket depth at \(t=z=0,\rho=1\), check the characteristic first sign nearby, and decide whether loss (9/10) is possible at that point.

Solution. Flatness of \(e\) makes every bracket jet at the marked point agree with that of \(\tau+i\rho t^9\). All words through nine leaves vanish and \(H_\tau^9q(c)=9!\ne0\), so \(\kappa=9\). Its least odd endpoint is \(9!(A^2+B_0^2)A^8\geq0\). The numerical necessary bound permits \(\delta=9/10\).

The nearby characteristic first sign also holds. At \(t=0\) the slope is zero. At a nonzero root \(t^4=e(z)\), it is \(\rho(9t^8-5e(z)t^4)=4\rho e(z)^2\geq0\). Nevertheless, whenever \(z\ne0\), the root \(t=0\) has adverse order five, with \(\partial_t^5q=-5!\rho e(z)<0\). Such points approach the marked point. Theorem 4.1 excludes every loss below one at them; openness excludes loss (9/10), or any other loss below one, at the marked point. Pointwise type and pointwise orientation do not replace the neighborhood condition.

References

• [L] Nicolas Lerner, Metrics on the Phase Space and Non-Selfadjoint Pseudodifferential Operators, Birkhäuser, 2010. Open author's manuscript, 2009. Ordinary symbol estimates, characteristic signs and finite-type geometry.
• [E] Lawrence C. Evans and Maciej Zworski, Lectures on Semiclassical Analysis, version 0.2. Open university-hosted lecture notes. Concentrated states and finite semiclassical symbol expansions.
• [H] Lars Hörmander, The Analysis of Linear Partial Differential Operators IV: Fourier Integral Operators, Springer, 1985. Scalar subellipticity and characteristic geometry.

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