Polynomial brackets and the residual jet

A weighted polynomial model can retain an entire bracket family in a small amount of scalar data. The data consist of one shifted frequency and the full derivative jet of a Wronskian. The Wronskian measures the part of the spatial polynomial that cannot be absorbed into the frequency coefficient.

We will prove both comparisons needed for this reduction: the bracket maximum is comparable to that scalar data, and the residual polynomial jet is comparable to the Wronskian jet. The bounds are uniform over the normalized frequency coefficients, allow zeros of those coefficients, and hold for all real spatial points with explicit polynomial factors in time.

The bracket convention and leaf count come from An adaptive scale for repeated brackets. The coefficient normalization motivating this model is proved in The linear frequency coefficient and the bracket scale. This lesson uses polynomial algebra, Taylor translation and compactness in finite-dimensional coefficient spaces.

1. The model and both comparisons

Use canonical coordinates $(t,z,\tau,\eta)$, with $\{\tau,t\}=\{\eta,z\}=1$. Fix positive integers $m_2\leq m_1$. Give $t,z$ weights $1,m_2$, respectively. Let $F(t,z)$ and $G(t)$ be real polynomials with

\deg_{\mathrm{wt}}F<m_1,\qquad \deg G<m_2. \tag{1.1}

Assume, for a fixed $C\geq1$ and some $s$,

\max_j|G^{(j)}(0)|\leq C,\qquad |G^{(s)}(0)|\geq C^{-1}. \tag{1.2}

In particular $0\leq s<m_2$ and $G$ is not identically zero. No coefficient bound is imposed on $F$.

Set $L_1=\tau,L_2=F(t,z)+G(t)\eta$, and form words $L_{(a)}=L_a,L_{(a,I)}=\{L_a,L_I\}$. Define

\begin{gathered} E=G F_t-F G',\\ H(z)=F^{(s)}(0,z)/G^{(s)}(0),\\ R(t,z)=F(t,z)-G(t)H(z),\\ \mathcal J(P)(t,z)= \sum_{\alpha\in\mathbb N^2}|D_{t,z}^{\alpha}P(t,z)|,\\ \mathcal B(t,z,\eta)= \max_{1\leq|I|\leq m_1}|L_I(t,z,0,\eta)|. \end{gathered} \tag{1.3}

The jet sums are finite because their arguments are polynomials.

Theorem 1.1 (weighted polynomial reduction). Every word with more than $m_1$ leaves is identically zero. There are constants $C_1,N$, depending only on $m_1,m_2,C$, such that for every real $t,z,\eta$,

\begin{gathered} C_1^{-1}(1+|t|)^{-N}\\ \times\bigl(|\eta+H(z)|+\mathcal J(E)(t,z)\bigr)\\ \leq\mathcal B(t,z,\eta)\\ \leq C_1(1+|t|)^N\\ \times\bigl(|\eta+H(z)|+\mathcal J(E)(t,z)\bigr), \end{gathered} \tag{1.4}

and

\begin{gathered} C_1^{-1}(1+|t|)^{-N}\\ \times\mathcal J(E)(t,z)\\ \leq\mathcal J(R)(t,z)\\ \leq C_1(1+|t|)^N\\ \times\mathcal J(E)(t,z). \end{gathered} \tag{1.5}

The same constants can be used for every allowed choice of $s$, since there are only finitely many such choices.

We prove the nilpotence, the uniform inverse controlling (1.5), and then both bracket bounds in (1.4).

2. Count weights before shifting the frequency

Give $\tau,\eta$ weights $m_1-1,m_1-m_2$. All four weights are nonnegative. Both $L_1,L_2$ have weighted degree at most $m_1-1$. Each Poisson contraction differentiates in a canonical pair whose two weights sum to $m_1$. Thus bracketing a polynomial of degree at most $m_1-1$ with one of degree at most $m_1-r$ gives degree at most $m_1-r-1$.

Induction shows that a word with $r$ leaves has degree at most $m_1-r$. For $r>m_1$, that degree is negative. A polynomial in variables of nonnegative weight cannot have a nonzero monomial of negative weight, so the word is zero. This also covers $m_1=m_2$, when the weight of $\eta$ is zero.

Now make the global canonical change

\eta_*=\eta+H(z),\qquad t_*=t,\quad z_*=z,\quad\tau_*=\tau. \tag{2.1}

The difference of canonical one-forms is $H(z)\,dz=d\int_0^zH(u)\,du$, so the map is symplectic. It preserves the sheet $\tau=0$, and replaces $L_2$ by $R(t,z)+G(t)\eta_*$. Moreover

R^{(s)}(0,z)=0,\qquad G R_t-R G'=E. \tag{2.2}

Nilpotence is preserved by symplectic invariance. We proved it before the shift because the shift need not preserve the original weights. The ordinary degree in $t$ of $R$, however, remains less than $m_1$, since $m_2\leq m_1$. Its total ordinary degree is bounded in terms of $m_1,m_2$. The same is true of $E$.

For the rest of the proof work in these shifted coordinates, write the frequency as $\eta_*$, and write the bracket maximum as $\mathcal B_*$. Thus $\mathcal B_*(t,z,\eta_*)=\mathcal B(t,z,\eta_*-H(z))$.

3. The Wronskian has a uniform inverse after removing its kernel

For a polynomial $p(t)$ of degree at most $D$, Taylor translation gives

p^{(i)}(t)= \sum_{r=0}^{D-i}\frac{t^r}{r!}p^{(i+r)}(0), \qquad p^{(i)}(0)= \sum_{r=0}^{D-i}\frac{(-t)^r}{r!}p^{(i+r)}(t). \tag{3.1}

Hence its full time jet at zero and at $t$ bound one another by $C_D(1+|t|)^D$. Apply this also to every $z$ derivative of a polynomial in $(t,z)$. A common degree bound gives the same comparison for the full spatial jets.

Fix an allowed $s$, and let $V_s$ be the vector space of real polynomials $f(t)$ of degree less than $m_1$ satisfying $f^{(s)}(0)=0$. For a normalized $G$, define

T_Gf=Gf'-fG'. \tag{3.2}

Lemma 3.1 (uniform coefficient inverse). In fixed coefficient norms, $\|f\|\leq C'\|T_Gf\|$ for all $f\in V_s$, uniformly over the $G$ satisfying (1.2).

Proof. First $T_G$ is injective on $V_s$. If $T_Gf=0$, then $(f/G)'=0$ on any nonempty interval where $G\ne0$. Thus $f=cG$ on that interval. Equality of polynomials makes this identity global. The condition $f^{(s)}(0)=0$ and the nonzero $G^{(s)}(0)$ give $c=0$.

The coefficient set of $G$ is compact: its finitely many center derivatives are bounded, and the lower bound on the chosen derivative defines a closed subset. If a uniform inverse bound failed, there would be normalized $G_\nu$ and $f_\nu\in V_s$ with $\|f_\nu\|=1$ but $\|T_{G_\nu}f_\nu\|\to0$. Pass to convergent coefficient subsequences. Their limits satisfy $T_Gf=0$, $f\in V_s$ and $\|f\|=1$, contradicting injectivity. If $V_s=\{0\}$, the estimate is immediate. ∎

For each fixed $z$ and every $j$, the polynomial $f(t)=\partial_z^jR(t,z)$ lies in $V_s$, by (2.2). Since $G$ is independent of $z$, $T_Gf=\partial_z^jE(t,z)$. Use center derivative norms as the fixed coefficient norms, and sum over all $j$ up to the common total-degree bound. This gives

\mathcal J(R)(0,z)\leq C'\mathcal J(E)(0,z). \tag{3.3}

Translation (3.1) from $t$ to zero and back proves the upper bound for $\mathcal J(R)$ in (1.5). For the other bound, differentiate $E=GR_t-RG'$. The product rule expresses every derivative of $E$ as derivatives of $R$ multiplied by derivatives of $G(t)$. Those derivatives are bounded by $C'(1+|t|)^{m_2-1}$, uniformly over the normalized $G$. Summing proves the other inequality. Both hold also when $E$ is identically zero; then injectivity forces $R=0$.

This argument never divides by $G(t)$ at the point being estimated. Zeros of $G$ are allowed.

4. Recover the full Wronskian jet from the brackets

Write

U_r=\partial_t^rR+G^{(r)}\eta_*. \tag{4.1}

For $0\leq r<m_1$, these are the words obtained by applying $r$ brackets with $L_1$ to $L_2$. Their Hamilton fields satisfy $H_{U_r}=(\operatorname{ad}H_{L_1})^rH_{L_2}$. The identity

E=G\,U_1-G'\,U_0 \tag{4.2}

first controls its time derivatives: repeated $H_{L_1}=\partial_t$ makes each $\partial_t^iE$ a finite sum of bracket words with coefficients that are polynomials in derivatives of $G$.

On a function independent of $\tau,\eta_*$, the Hamilton field $H_{U_r}$ acts as $G^{(r)}(t)\partial_z$. Its action on a coefficient depending only on $t$ is zero. Therefore, for every $i,j$,

H_{U_r}^{\,j}\partial_t^iE =\bigl(G^{(r)}(t)\bigr)^j \partial_z^j\partial_t^iE. \tag{4.3}

Expand the left side using the commutator expression for $H_{U_r}$, or equivalently the Jacobi identity. It is a finite sum of bracket words with coefficients polynomial in derivatives of $G(t)$. Words longer than $m_1$ vanish; the others are bounded by $\mathcal B_*$. For every derivative appearing in the finite jet of $E$, this yields

|G^{(r)}(t)|^j\,|\partial_t^i\partial_z^jE| \leq C'(1+|t|)^{N'}\mathcal B_*. \tag{4.4}

The constants involve only the degree bounds and the normalization of $G$, not the coefficients of $F$.

At every $t$, some derivative of $G$ has a quantitative lower bound. Apply Taylor translation to $G^{(s)}(0)$ and use (1.2):

\begin{gathered} \max_{0\leq r<m_2}|G^{(r)}(t)|\\ \geq c(1+|t|)^{-(m_2-1)}. \end{gathered} \tag{4.5}

Choose such an $r$ at the point in (4.4). Divide by its $j$-th power and sum the finitely many jet derivatives. No differentiation of the chosen index is needed. We obtain

\mathcal J(E)(t,z) \leq C'(1+|t|)^{N'}\mathcal B_*. \tag{4.6}

The shifted frequency is controlled independently. Because $R^{(s)}(0,z)=0$, we have $U_s(0,z,\eta_*)=G^{(s)}(0)\eta_*$. Each $U_r$ has time degree less than $m_1$. Translation of this identity to $t$, followed by (1.2), gives

|\eta_*|\leq C'(1+|t|)^{m_1-1} \max_{0\leq r<m_1}|U_r(t,z,\eta_*)| \leq C'(1+|t|)^{m_1-1}\mathcal B_*. \tag{4.7}

Combining (4.6) and (4.7) proves the lower bracket bound in (1.4).

5. Bound every bracket by the residual jet

We now check the structure of every nonzero word, rather than assuming that its coefficients are uniformly bounded.

A word containing one $L_2$ and otherwise $L_1$ leaves is, up to sign, one of the $U_r$. Its bound is $|D_t^rR|+|G^{(r)}(t)\eta_*|$. When another $L_2$ is first bracketed with such a word, the frequency cancels:

\{L_2,U_r\} =G\,\partial_z\partial_t^rR -G^{(r)}\partial_zR. \tag{5.1}

The result is independent of both frequencies and linear in derivatives of $R$. Further brackets with $L_1$ differentiate it in $t$; further brackets with $L_2$ apply $G(t)\partial_z$. Induction therefore expresses every such word as a finite linear combination of derivatives of $R$, with coefficients polynomial in derivatives of $G(t)$. A word containing only $L_1$ vanishes unless it is the single leaf $\tau$, which is zero on the sheet under consideration.

There are finitely many words of length at most $m_1$. All their coefficient polynomials are bounded by $C'(1+|t|)^{N'}$. It follows that

\mathcal B_* \leq C'(1+|t|)^{N'} \bigl(|\eta_*|+\mathcal J(R)(t,z)\bigr). \tag{5.2}

Use the already proved upper bound in (1.5) to replace $\mathcal J(R)$ by a polynomial factor times $\mathcal J(E)$. This gives the upper bound in (1.4). Finally substitute $\eta_*=\eta+H(z)$, increase $C_1,N$ to cover all the finite estimates and all possible $s$, and return to the original coordinates. This completes both comparisons and the proof of Theorem 1.1. ∎

The full derivative jet is essential. A zero value of $E$ at a point does not mean the bracket family is zero there. The polynomial factors in $t$ are also essential when the comparison is required on the entire real line.

6. Exercises with complete solutions

Exercise 1 — basic. Take $m_1=4,m_2=2$, $F=t^3+tz,G=1+t$, and $s=0$. Compute $H,E,U_1$ and $\{L_2,U_1\}$. Check the weighted hypotheses and a fourth-leaf time word.

Solution. The monomials $t^3,tz$ have weight $3<4$, and $G$ has degree $1<2$. Its center derivatives are $1,1$, so (1.2) holds with $C=1$. Here $H=0$, and $\begin{gathered} E=(1+t)(3t^2+z)-(t^3+tz)\\ =2t^3+3t^2+z. \end{gathered}$ We have $U_1=3t^2+z+\eta$. Since $F_z=t$, (5.1) gives $\{L_2,U_1\}=(1+t)-t=1$. The word with three time brackets and one $L_2$ leaf is $\partial_t^3L_2=6$, with four leaves. Every word with more than four leaves is zero, as the theorem asserts.

Exercise 2 — intermediate. Let $m_1=4,m_2=2$, $F=tz+t^2,G=t$, and $s=1$. Compute $H,R,E$, and explain how the inverse argument works at the zero $t=0$ of $G$.

Solution. The weights of $tz,t^2$ are $3,2<4$. Since $G'(0)=1$, $H(z)=z$. Hence $R=t^2$ and $E=t(z+2t)-(tz+t^2)=t^2$. The identity $T_G(t^2)=t(2t)-t^2=t^2$ is polynomial and is valid at $t=0$ as well as elsewhere. Injectivity was proved by dividing on any interval where $G\ne0$, then using a global polynomial identity; it does not require division at the point of evaluation. Here $\mathcal J(R)=\mathcal J(E)=t^2+2|t|+2$.

Exercise 3 — intermediate. Show why both the residual constraint and the lower normalization in (1.2) are needed for a uniform inverse. Use $G=1+t^2$ for the first issue, and $G_\delta=\delta$, $f=t$, $0<\delta<1$, for the second.

Solution. For $m_1=m_2=3,s=0$, the nonzero polynomial $f=G=1+t^2$ has $T_Gf=0$. It is excluded from $V_0$ because $f(0)=1$. Removing that constraint leaves the kernel generated by $G$.

For $m_1=2,m_2=1,s=0$, $f=t$ satisfies $f(0)=0$, but $T_{G_\delta}f=\delta$. Its inverse ratio is proportional to $\delta^{-1}$, which is unbounded. The upper coefficient bound remains valid with $C=1$; the lower bound $|G_\delta(0)|\geq C^{-1}$ fails. Nonzero $G$ alone does not supply a uniform inverse.

Exercise 4 — advanced. Take $m_1=5,m_2=3$, $F=z,G=1+t^2,s=0$. Compute the shifted residual and its weight. Verify that the frequency shift is canonical and explain why nilpotence must be established before applying it.

Solution. We have $H=z$, so $R=z-(1+t^2)z=-t^2z$, whose weight is $2+3=5$. It no longer has weight strictly below $m_1$. The frequency shift $\eta_*=\eta+z$ changes the one-form by $z\,dz=d(z^2/2)$, so it is canonical. Also $E=-2tz=T_GR$, as required. The original symbols have the prescribed weights and therefore words longer than five vanish. That conclusion transfers by symplectic invariance. Reapplying the original strict-weight hypothesis to the shifted residual would be invalid, although its ordinary time degree remains less than five and the inverse proof applies.

Exercise 5 — advanced. Let $m_1=m_2=2$, $F=1,G=1+t,s=0$. Compute $R,E$. On $t>0$, choose the shifted frequency $\eta_*=t/(1+t)$ and compute the bracket maximum. Show why polynomial factors in $t$ cannot generally be replaced by constants.

Solution. Here $H=1,R=-t,E=-1$, so $\mathcal J(R)=|t|+1$ and $\mathcal J(E)=1$. The ratio of those jets is unbounded. The shifted second symbol is $-t+(1+t)\eta_*$, which vanishes at the chosen frequency. Its time derivative is $-1+\eta_*=-1/(1+t)$. All other nonzero words on $\tau=0$ are this derivative up to sign; longer words vanish. Thus

\begin{gathered} \mathcal B_*=\frac1{1+t},\\ |\eta_*|+\mathcal J(E)\geq1. \end{gathered} \tag{6.1}

A constant lower comparison would fail as $t\to\infty$. The polynomial factors in both (1.4) and (1.5) describe this permitted global behavior.

Exercise 6 — advanced. Take $m_1=4,m_2=2,F=tz,G=1,s=0$. At $t=z=\eta=\tau=0$, compare the value of $E$, its full jet and a three-leaf bracket. Decide whether $\mathcal J(E)$ can be replaced by $|E|$ in the theorem.

Solution. We have $H=0,R=tz,E=z$. At the origin $E=0$, but $\partial_zE=1$, so $\mathcal J(E)=1$. The first time bracket is $\{L_1,L_2\}=z$, and $\{L_2,z\}=1$, a three-leaf word. Thus $\mathcal B\geq1$, even though both $|\eta+H|$ and $|E|$ vanish there. Replacing the full jet by the value would make the proposed upper bracket bound zero and would be false. The mixed derivative $\partial_t\partial_zR=1$ is the corresponding residual-jet information.

References

The same results are treated in Hörmander, The Analysis of Linear Partial Differential Operators IV, Chapter 27, Section 27.4, especially Lemma 27.4.7 and its two polynomial comparisons. The proof and exercises above are independently organized. They retain strict original weights, the normalization of $G$, zeros of $G$, the whole derivative jets, both inequalities in each comparison, and uniform constants without bounding the coefficients of $F$. The subsequent geometric alternative and admissible covering remain separate steps.

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