A cancellation plane or a uniform bracket lower bound

The canonical cell has a distinguished frequency direction. Far enough in that direction, the linear coefficient already forces a large bracket. Near it, the answer depends on a polynomial Wronskian: a large Wronskian jet keeps the bracket scale uniformly large, while a small jet produces one fixed frequency plane on which that scale falls below half its value at the center.

We prove this alternative on a spatial cylinder whose radius is larger than the radius used to distinguish the two gradient cases. The larger radius will allow nearby admissible centers and a covering argument. Both the finite bracket family and the extension of the cylinder need explicit estimates.

Use the hypotheses and notation of An explicit canonical cell for a large transverse gradient, the estimates in The linear frequency coefficient and the bracket scale, and the full polynomial comparisons in Polynomial brackets and the residual jet. All additional arguments below use Taylor's formula, the product rule and finite polynomial jets.

1. The larger cylinder and the exact cancellation frequency

Write \(t=x_1,z=x_2,\xi=\xi_2,u=(x'',\xi'')\), and let \(v=(\xi,u)\). As before,

The parameter \(\rho\geq1\) is fixed first. Constants in the conclusions are independent of \(\rho,\lambda\); the required upper bound for \(\lambda\) may depend on \(\rho\).

Choose a fixed exponent \(\kappa_2\) and define

The initial bracket bound gives

for small \(\lambda\). Indeed \(M\geq c(\lambda^{-2}/\rho)^{1/(k+1)}\), so \(R_2/L\leq C\rho^{-k/(2(k+1))}\lambda^{1/(k+1)-\kappa_2}\). The last inequality follows by comparing the two maxima that define the reciprocal scales.

Let \(c(t,z)=\partial_\xi Q(t,z,0)\). The selected orbit is an orbit of \(\partial_t^s q(0,\cdot)\); that function is constant along its own Hamiltonian flow. Consequently,

where \(c_1(0)\) has uniform positive upper and lower bounds. Define the exact frequency

The denominator is nonzero. The center time-jet bound and (1.1) give

The bracket scale is

Every \(Q_I\) here is evaluated at \((t,z,\xi,u)\) on \(\tau=0\). Brackets are formed before evaluating on that sheet. All statements about \(Q\) retain its canonical-cell domain \(|Mt|<1,|bz|<1,|v|<L\).

Theorem 1.1 (geometric alternative). There is a uniform constant \(\gamma>0\) such that one of the following holds for sufficiently small \(\lambda\):

at every frequency in the canonical-cell domain; or

The exact plane in (1.9) lies in that domain by (1.3) and (1.6). In particular the alternative covers all frequencies \(|\xi|<R_2\), as needed for localization. The alternatives are not asserted to be mutually exclusive.

2. A polynomial model with an explicit finite family

Make the canonical change in the first two pairs

and leave the remaining canonical pairs unchanged. Products of the two scaling factors in each pair are one, so the change is symplectic. Put

Thus the original leaves are \(M\sigma,Mh\). Exact symplectic invariance gives

in these coordinates, where the leaves defining \(h_I\) are \(\sigma,h\).

Let \(F_0\) be the Taylor polynomial of \(F\) at zero through ordinary degree \(k\). Let \(G_0\) be the time Taylor polynomial of \(G(T,0)\) through degree \(\ell\). For every fixed derivative order \(d\),

Here \(C^d\) denotes the maximum over partial derivatives of order at most \(d\).

Proof of (2.4). The transformed high-derivative bound in the canonical-cell proof gives

If \(i+j\geq k+1\), its right side is at most \(C_{ij}\rho/M\). Taylor's integral remainder proves the first estimate for derivatives of order at most \(k\); derivatives of larger order are already small, and the polynomial derivatives vanish. Center coefficients of \(F_0\) are bounded by \(C\rho\), using the weighted center jets.

For the second estimate, the normalized coefficient in the preceding lesson is exactly \(G\). Its spatial variable there is \(bz=(b/B_2)Z\). Since \(b/B_2\leq1\), composing that estimate with this linear change does not enlarge any spatial derivative. The normalized time-polynomial remainder and every mixed derivative are bounded by \(C_dR/L\). The coefficients of \(G_0\) are uniformly bounded, and

This proves all the claims. ∎

Set

With weights \(1,m_2\) on \(T,Z\), every monomial of the ordinary-degree-\(k\) polynomial \(F_0\) has weight at most \(km_2<m_1\). Also \(\deg G_0<m_2\) and \(m_2\leq m_1\). These explicit choices satisfy every degree hypothesis of the polynomial lemma. Its constants depend only on \(k\) and the fixed coefficient normalization of \(G_0\), hence are uniform in \(\rho\).

The selected time derivative of \(F_0\) at \(T=0\) is constant in \(Z\). Indeed (1.4) makes every coefficient with time order \(s\) and positive spatial order vanish. Therefore the polynomial lemma's shift is the constant

It is the exact frequency from (1.5), rather than an approximate root of the model.

3. Compare every required bracket and discard longer words

Fix a sufficiently large uniform number \(N\). We first work on

Notice that \(|A_2\xi|=|\eta|/\rho\). Physical frequencies in (3.1) satisfy \(|\xi|\leq NR\), because \(\rho B_2=A_2^{-1}<R\).

For any fixed \(d\), with derivatives in the unscaled remaining canonical coordinates \(u\),

The norm includes all mixed derivatives in \(T,Z,\eta,u\). Each term in \(\delta\) tends to zero for fixed \(\rho\).

Proof. Write \(Q=\mathcal R+\xi c(t,z)\). The canonical residual bounds, after (2.1), give

For \(r=|\alpha|=0\), subtract the trace at \(\eta=u=0\) and integrate its first derivatives. The difference, including any \(T,Z\) derivatives, is at most \(C\rho(|B_2\eta|+|u|)/L\leq C_N\rho R_2/L\). If \(r+|\alpha|\geq1\), the corresponding derivative of that trace is zero; (3.3), \(b/B_2\leq1\), \(B_2/L<R/(\rho L)\) and \(L\to\infty\) give the bound in (3.2).

The linear term of \(h\) is \(\eta G(T,Z)\). Its difference from \(\eta G_0(T)\) and every mixed derivative are bounded by \(C_{d,N}\rho R/L\), using (2.4) and \(|\eta|\leq\rho N\). Since \(R<R_2\), this is included in (3.2). The remaining difference \(F-F_0\) is bounded by \(C_d\rho/M\). ∎

Both leaves and their fixed derivatives are bounded by \(C_{d,N}\rho\) on (3.1) and \(|\sigma|<1\). Expand a bracket word by the product rule. A word with \(r\) leaves is a finite sum of products of leaf derivatives, with total derivative order \(2(r-1)\). In a difference of two such products, at least one factor is a derivative of \(h-\mathsf L_2\). Thus (3.2), with \(d=2(m_1-1)\), implies

The leaf \(\sigma\) is identical in both families. No bound independent of \(\rho\) is claimed for \(\varepsilon_1\); the convergence is taken after fixing \(\rho\). All comparison constants below remain uniform.

Words longer than the family defining \(M\) need a separate estimate. The original anisotropic bracket estimate gives \(|q_I|\leq C_I\lambda^{-2}\) for every fixed word. The canonical map sends the domain here into the original domain. By (2.3), symplectic invariance and \(\lambda^{-2}\leq C\rho M^{k+1}\),

If \(m_1=k+1\), this range is empty. Combining (3.4)–(3.5), define

Both families are evaluated at the indicated coordinates on \(\sigma=0\). Then, throughout (3.1),

To see both directions, the shorter model words differ from the actual ones by at most \(\rho\varepsilon_1\); every longer model word is at most \(\rho(\varepsilon_1+C/M)\). Taking maxima gives (3.7). The polynomial lemma also makes every word longer than \(m_1\) zero, so no unbounded family has been introduced.

If \(0<h_*\leq1\), the definition of the root scale implies

Indeed each normalized root uses an exponent \(1/|I|\) with \(1\leq|I|\leq k+1\). This keeps the family maximum distinct from its leaf-dependent roots.

4. The two Wronskian regimes on the initial square

Set

The sum is finite. Exact Taylor translation of the full polynomial jet in both variables gives, for a uniform \(C_J\),

For example, write each derivative at \((T,Z)\) as its finite two-variable Taylor sum at zero, and sum absolute values. The reverse estimate uses the translation by \((-T,-Z)\). The coefficient sums are bounded because the degree of \(E_0\) has a bound depending only on \(k\). Using only \(|E_0|\), instead of the full jet, would not give (4.2).

The polynomial lemma and \(|T|<1\) give uniform two-sided bounds

Choose a uniform \(\theta>0\) so small that

Increase the fixed \(N\) in (3.1), if necessary, so that \(|H_0|\leq\rho N\) and the large-frequency coefficient estimate gives \(\mu\geq M\) whenever \(|A_2\xi|\geq N\). Both requirements use uniform constants.

Small Wronskian regime. If \(e_0<\theta\rho\), take the exact plane \(\eta=-H_0\), equivalently \(\xi=-\Xi\). Equations (4.2)–(4.4) give \(\mathcal B/\rho\leq\tfrac12 3^{-(k+1)}\). For sufficiently small \(\lambda\), (3.7) makes \(\mathcal H\leq3^{-(k+1)}\). Thus

Non-small Wronskian regime. If \(e_0\geq\theta\rho\), equations (4.2)–(4.3) yield \(\mathcal B/\rho\geq c_*:=\theta/(C_BC_J)>0\) everywhere on (3.1). For small \(\lambda\), (3.7) gives \(\mathcal H\geq c_*/2\). By (3.8), after decreasing a uniform constant if needed,

Frequencies with \(|A_2\xi|\geq N\) satisfy the stronger lower bound \(\mu\geq M\) in the full canonical cell, by the coefficient theorem. This covers every frequency in that domain on the initial square.

At the center, \(\mathcal H=1\) by the definition of \(M\). Equation (3.7) and (4.3) also give \(|H_0|+e_0\geq c\rho\) for small \(\lambda\). Therefore a small Wronskian regime can occur because the shifted frequency at the center is large. The falling scale on the cancellation plane is compatible with its original center value.

5. Extend both regimes to the larger spatial radius

If \(B_2'=B_2\), no extension is needed. Otherwise

For \(|A_2\xi|\leq N\), the full symbol estimates take the form

throughout \(|Mt|<1,|z|<R_2,|u|<R_2\). Frequencies here have \(|\xi|\leq NR\), so this cylinder is inside the canonical-cell domain for small \(\lambda\).

Apply the differentiated anisotropic bracket estimate with amplitude \(\rho\), bracket scale \(M\), time scales \(M,(\rho M)^{-1}\), distinguished transverse scales \(L^{-1},A_2\), and the remaining scales \(L^{-1}\). Its contraction factors are

They are bounded by one for small \(\lambda\). For every word in the original defining family,

The constants are uniform. Integrating from \(z=0\) to any \(|z|<R_2\), while keeping \(t,\xi,u\) fixed, gives

In the small Wronskian regime, every defining normalized word at \(z=0,\xi=-\Xi\) has magnitude at most \(3^{-(k+1)}\), by (4.5). Make the error in (5.5) strictly less than \(2^{-(k+1)}-3^{-(k+1)}\). Every such word then has magnitude strictly less than \(2^{-(k+1)}\) on the full larger cylinder. Taking its leaf-dependent root proves (1.9), with strict inequality.

In the non-small regime, at \(z=0\) at least one defining normalized word has magnitude at least \(c_*/2\). Make every error in (5.5) less than \(c_*/4\). Its magnitude remains at least \(c_*/4\); (3.8) gives a uniform lower root bound on the larger cylinder for \(|A_2\xi|\leq N\). The large-frequency theorem covers every remaining frequency in the canonical-cell domain, because \(|z|<R_2< L=b^{-1}\). Taking the minimum of these uniform lower constants proves (1.8).

This completes the proof of the geometric alternative, including the longer initial \(x_2\) radius, the possible extension to \(R_2\), every defining bracket word, and the exact cancellation plane. ∎

6. Admissible neighborhoods of the large-gradient type

Definition 6.1. If the lower-bound alternative (1.8) holds for a large-gradient center, its admissible neighborhood of type II is the image under the exact canonical map of

In the original coordinates this is

The center is \((0,0)\) in the chosen coordinates. A translated center uses the corresponding time origin and its own \(M,B_2'\) and canonical map. The canonical invariance of the bracket scale transfers the lower bound to every point of \(\Omega_0\). All four restrictions are part of the definition, including the distinguished frequency restriction separate from \(|u|<R_2\). In the longer spatial branch the neighborhood can follow a long segment of the selected Hamiltonian orbit; its image need not be an ordinary Euclidean ball.

The definition selects centers with a uniform lower bound over their whole indicated neighborhood. Proving that every point of the large-gradient case lies within a bounded multiple of \(R\) of such a center, and constructing cutoffs with the required motion bounds, are subsequent steps.

7. Graded exercises with solutions

Exercise 1 — basic. Take \(k=3,s=1\). Find the explicit weights and word cutoff in (2.7). Why would simply identifying the polynomial lemma's cutoff with \(k+1\) be unjustified?

Solution. Here \(\ell=1,m_2=2,m_1=7\). A Taylor monomial \(Z^3\) has ordinary degree three but weight six. It fits the strict bound seven and need not fit a strict bound four. The polynomial comparison may therefore use words with five, six or seven leaves. Estimates (3.4)–(3.5), rather than an assumed equality of cutoffs, make those longer words negligible in the actual scale comparison.

Exercise 2 — intermediate. For a polynomial \(E(T,Z)=T^2+Z\), compare its value and its full jet at the origin. Explain why using only the value would misclassify the two regimes.

Solution. The value is zero. The nonzero derivatives there are \(E_Z=1\) and \(E_{TT}=2\), so \(J(E)(0,0)=3\). A zero value does not mean a small full jet. Polynomial translation preserves quantitative control of the full jet, while the value can vanish at isolated points or along a curve.

Exercise 3 — intermediate. Suppose \(k=2\), and each defining normalized word has magnitude at most \(1/27\) on the initial cancellation plane. How much additive error can be allowed in extending it while still concluding \(\mu<M/2\)?

Solution. The largest leaf count is three. It suffices that each extended normalized word be strictly smaller than \(2^{-3}=1/8\). Thus any uniform error strictly below \(1/8-1/27=19/216\) works. The cubic-root margin also controls one- and two-leaf words, since their roots of a number below \(1/8\) are smaller than one half.

Exercise 4 — intermediate. Let \(F_0(T,Z)=T^3+T+Z\), \(G_0=1\), and choose \(s=0\). Determine whether the constant-plane argument in this lesson applies directly, and identify the exact additional hypothesis.

Solution. The polynomial lemma itself applies with sufficiently large weights, but its shift is \(H_0(Z)=F_0(0,Z)=Z\). This is not a constant plane. The geometric construction additionally supplies \(\partial_T^sF_0(0,Z)\) constant along the selected Hamiltonian orbit. That identity, established before the polynomial reduction, is what makes \(H_0=\Xi/B_2\) independent of \(Z\). Without it the polynomial comparison alone cannot supply the fixed frequency plane in (1.9).

Exercise 5 — advanced. Take any fixed \(\rho\geq1\), \(k=1\), \(a=\lambda^{-1}\), and \(q(t,z,\xi)=a^2t+a\xi+\rho\). Use the identity canonical map. Compute the model in (2.2), its shift and Wronskian, and the actual scale on its cancellation plane.

Solution. At the center \(M=a/\sqrt\rho\), \(s=0\), \(A_2=1/\sqrt\rho\), and \(B_2=1/\sqrt\rho\). Thus

The Wronskian is in the non-small regime. On \(\xi=-\Xi\), the value of \(q\) is \(a^2t\), while its time bracket is \(a^2\). Hence \(\mu\geq(a^2/\rho)^{1/2}=M\) everywhere. At \(t=0\) the value vanishes but the bracket scale stays exactly \(M\). This example also shows why the cancellation of one selected value does not suffice for the small-scale alternative.

Exercise 6 — advanced. If \(B_2'<B_2\), compute the distinguished transverse contraction factor. Explain why no derivative of the root scale \(\mu\) is needed to extend either alternative.

Solution. The contraction factor is \(\rho A_2b=\rho A_2/L=1/(B_2L)<R_2/L\to0\). We differentiate the smooth bracket functions \(Q_I\), integrate their normalized changes in (5.5), and only then take the finitely many roots. Their absolute values and roots need not be smooth at zeros. Working with the bracket functions avoids assuming such smoothness and keeps the strict margin in the small regime.

References

The geometric alternative and the admissible neighborhood follow the local analysis in Hörmander, The Analysis of Linear Partial Differential Operators IV, Chapter 27, Section 27.4, especially Proposition 27.4.8 and Definition 27.4.9. The proof above spells out the finite family in the polynomial reduction, all mixed Taylor errors, the exact constant shift, and both spatial-radius branches. The preceding lessons provide the stated scalar bracket and canonical-cell estimates.

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