Coordinate transport and directional localization

This is a bounded modified selection from AN03-U012, Sections 2, 7–8 and 18.1–18.3, with connecting arguments for the present intrinsic class.

Original principal author and publisher: AN-03 course-writing task / AN-03 local course project, 2026. Earlier modification: AN-03 course-writing task and OpenAI Codex. This selection and its connecting arguments: GPT-6 Astra (OpenAI), Ultra, 4 October 2026; publisher: AN-04 local course project. Original text: CC0.

The free human sources are Gerd Grubb's author-hosted Chapter 8, Section 8.1, and Lars Hörmander's freely readable Fourier integral operators. I, Section 2.5. Only the coordinate-amplitude and Fourier directional arguments specified below are used. Every proof needed here is supplied in the programme.

T0. Inputs, test functions and the distributional coordinate map

The earlier analytic inputs are P1, B0–B6, P2, O0–O6, and P3, M0–M8 and L0–L3. They supply the exact ordinary amplitude expansion, its differentiated remainders, proper support, smoothing kernels, Fourier inversion and the local $B^s_{2,\infty}$ estimates. Smooth finite-dimensional calculus, finite cutoffs and compactness have the same exact U001 proofs as C0. The compact parameter integral and integration-by-parts arguments use the earlier U001 contract FTC-TAYLOR-COMPACT-PARAMETERS and the exact P17.2–P17.5 proofs. Compact chart substitution is proved in U001 P21.3–P21.4. No new theorem about general distribution kernels or arbitrary smooth-map pullbacks is assumed.

Write $\langle u,\varphi\rangle$ for the bilinear pairing of a scalar distribution with a compact smooth test density, represented as $\varphi(y)\,dy$ in coordinates. Continuity on tests with support in a fixed compact $K$ implies a finite-order estimate

Indeed a zero neighborhood making the pairing less than one contains an intersection of finitely many test-seminorm balls. Their maximum is bounded by a constant times the displayed seminorm for the largest of their derivative orders. Rescaling a nonzero test proves (T1); the zero-seminorm case follows by arbitrary rescaling.

For a compactly supported distribution $v$, choose $\tau\in C_c^\infty$ equal to one near its support and define $\widehat v(\xi)=\langle v,\tau e^{-iy\cdot\xi}\rangle$. The value is independent of $\tau$, since a test vanishing near the support pairs to zero. To justify that last assertion from the definition of support, cover its compact test support outside $\operatorname{supp}v$ by finitely many open sets on each of which the distribution vanishes, and use the U001 finite partition. The product rule and (T1) give, for every fixed $\alpha$,

Here the derivative corresponds to the test multiplier $(-iy)^\alpha$, on the same fixed compact support; the order $M$ in (T1) works for all fixed $\alpha$. Taylor's formula in the parameter $\xi$, in each of the finitely many required test seminorms, justifies these derivatives. This also makes $v$ a tempered distribution.

Let $\kappa:U_y\to V_x$ be a smooth diffeomorphism and $g=\kappa^{-1}$. The distribution representing $u\circ g$ is defined by

The test on the right has compact support in $U$. On every fixed output compact set, the product and chain rules bound all its test seminorms by finitely many seminorms of $\psi$. Thus the map on tests is continuous, and (T3) defines a distribution. The determinant is smooth and nonzero; its absolute value is smooth because its sign is locally constant. The determinant chain rule proves $T_\lambda T_\kappa=T_{\lambda\circ\kappa}$ and $T_gT_\kappa=I$, by direct substitution in (T3). The completed change-of-variables proof U001 P21.3–P21.4 identifies (T3) with ordinary composition for smooth functions, after compact localization. P3 M8 identifies these compact integrals with Lebesgue integrals. This is the function identity needed for the kernel argument. The same test bounds give continuity on distributions. For the strong dual topology, the image of a bounded test set is bounded: the inverse image of each zero neighborhood under a continuous linear map is a zero neighborhood. This proves the required bound on each strong-dual seminorm.

T1. Transport an ordinary symbol with both Jacobians

Let $A$ be a properly supported ordinary operator of order $d$ on $U$. Its transported operator is

Then $A^\kappa$ is a proper ordinary operator of the same order. For a full local symbol $a$, its principal symbol is

The assertion includes matrix symbols in their fixed matrix order.

Proof. P2 O5 reduces each compact localization to finitely many compact amplitude kernels plus a compact smooth kernel. A smooth kernel transports to

which is again smooth, with compact support for a compact kernel. So consider one symbol kernel near the diagonal in a convex base neighborhood. Put

The fundamental theorem of calculus proves $\kappa(y)-\kappa(w)=L(y,w)(y-w)$, and $L(y,y)=D\kappa(y)$. On a sufficiently small neighborhood of a compact diagonal piece, $\det L$ stays away from zero by continuity and compactness. The cofactor inverse formula and the compact parameter derivative proof give bounded derivatives of $L$ and $L^{-1}$ there.

Use $x=\kappa(y)$, $z=\kappa(w)$ and $\xi=L(y,w)^T\eta$ in the kernel integral. With a compact cutoff $\chi(y,w)=1$ near the relevant diagonal piece, the new amplitude is

The numerator is the frequency Jacobian. The denominator is the input base Jacobian. Neither can be omitted.

This identity also holds for the distribution kernels: first insert a smooth frequency cutoff tending to one, where the changes of variables and test pairings are ordinary integrals. Such transformed cutoffs have the form $r(\varepsilon L^T\eta)$, with $r=1$ near zero. Every base derivative is uniformly bounded: differentiating the argument produces $\varepsilon\eta$, which is bounded on the support of a derivative of $r$, since $L$ is uniformly invertible. Higher product and chain rules give the same bound. For fixed frequency these cutoffs and all base derivatives tend to those of one. Against a compact kernel test, integrate by parts in $z$ using $(1-\Delta_z)^k/\langle\eta\rangle^{2k}$, exactly as in P2 O4. Base derivatives preserve amplitude order, so the resulting integrand has a uniform integrable bound $C\langle\eta\rangle^{d-2k}$ when $2k>d+n$. Dominated convergence from P3 proves the same kernel limit as the untransformed regularization. P2 O0 product-test detection identifies the two kernels.

Here are all derivative estimates needed in (T8). On these compact sets, invertibility gives $\langle L^T\eta\rangle\asymp\langle\eta\rangle$. A frequency derivative differentiates $a$ in frequency and loses one order. A base derivative differentiating $L^T\eta$ introduces one linear factor in $\eta$ and one frequency derivative of $a$, for net order zero. Direct base derivatives of $a$, and derivatives of the cutoffs, determinant ratio and inverse chart, cost zero. Repeated chain and product rules therefore give

Every constant uses finitely many derivatives on the fixed compacts. The compact-amplitude theorem P2 O4 now gives a left symbol $b$ with, for every integer $N\ge1$,

This includes every differentiated remainder estimate. At the diagonal the two Jacobians in (T8) cancel. The $N=1$ formula therefore gives (T5). A finite cover of the compact diagonal part, with the finite partition already supplied in U001, finishes the local assertion; the omitted off-diagonal kernels are smooth by P2 O4.

Finally the support relation transforms by the homeomorphism $\kappa\times\kappa$. For a compact set in one output factor, its inverse image is compact, the original proper projection bounds the other factor by a compact set, and $\kappa$ maps that compact set to a compact set. This proves both proper projections. Transposing the continuous test maps identifies (T4) on all distributions, using T0. $\square$

T2. Conic smoothing and composition

A full symbol is smoothing at $(y_0,\xi_0)$, $\xi_0\ne0$, if it is in $S^{-N}$ for every $N$ on one open base and frequency cone about that point, with all differentiated estimates. Define the essential support $\operatorname{ess}(A)$ to be the complementary closed conic set of a local full symbol.

This definition is independent of the full-symbol representative: after compact kernel localization, a symbol is recovered by the Fourier transform in the difference variable of its kernel. Fourier inversion and P2 O0 show uniqueness. Altering a kernel localization away from the diagonal changes that compact kernel by a smooth compact kernel, whose difference-variable transform is $S^{-\infty}$ by integration by parts. These are exactly the changes of local representatives used here.

It is also independent of coordinates. In (T10), each coefficient is a finite sum of derivatives of $a$ at $(y,D\kappa(y)^T\eta)$, multiplied by smooth coefficients and polynomials in $\eta$. If $a$ is smoothing on a cone there, every coefficient is smoothing on a fixed smaller transformed cone. The remainder is in $S^{d-N}$ for arbitrary $N$ on that same cone. Hence $b$ is smoothing there. Applying the assertion to $g$ proves the converse. The resulting cotangent transformation is

The full product formula P2 O3 also gives

Indeed, on a cone where either factor is smoothing, every term in each finite differential product expansion is smoothing. The remainder is in $S^{d+d'-N}$ for every $N$, with all derivatives. This proves smoothing on that same cone. Local compact cutoffs and P2 O6 account for the proper composition and smooth off-diagonal remainders. A compact localized symbol with empty essential support is globally $S^{-\infty}$: a finite cover of its compact base and unit-frequency directions supplies each fixed seminorm bound. Its kernel is smooth by P2 O4. These facts also justify all uses of “equal modulo smoothing” below.

T3. Coordinate and frame invariance of iterated regularity

For a closed conic Lagrangian $\Lambda$, use the original intrinsic definition: every word in proper order-one operators whose symbols restrict to order zero on $\Lambda$ must take $u$ into $B^s_{2,\infty,\mathrm{loc}}$, with $s=-m-n/4$, including the empty word.

This definition is invariant under a base diffeomorphism and a smooth invertible change of finite-rank frame. To prove it, T1 and (T5) carry precisely the admissible operators to the admissible operators for (T11). Ordinary conic symbol orders are preserved: every derivative of the base-dependent linear frequency substitution gains at most the frequency power cancelled by the corresponding frequency derivative, as in (T9). The order-zero error in an order-one principal symbol stays order zero. Conjugation preserves products exactly,

P1 B5 preserves the local Besov space for this exact real exponent $s$, including the supremum endpoint. Thus (T13) proves one implication, and the inverse chart proves the other.

For a frame matrix $F(x)$, multiplication by $F$ and $F^{-1}$ preserves the local Besov space by P1 B4. The transported operator $FLF^{-1}$ is proper, and P2 gives principal symbol $F l F^{-1}$ modulo $S^0$. Its restriction to the Lagrangian is therefore of order zero exactly when that of $l$ is. The product cancellation in (T13) applies with $F$ in place of $T_\kappa$. This proves the frame assertion without commuting matrix factors. In the half-density frame the additional factor is $|\det Dg(x)|^{1/2}$, smooth and nonzero; it is the same case, with the smooth-root proof in U001.

All assertions are local in the base. Here is why local operator tests suffice. For a fixed compact output and a word of length $N$, choose nested compact neighborhoods within the chart and scalar cutoffs $\chi_0,\ldots,\chi_N$, with $\chi_{j+1}=1$ near $\operatorname{supp}\chi_j$. Replace the localized word by $\chi_0L_1\chi_1L_2\cdots L_N\chi_Nu$. In the telescoping difference, each factor $\chi_jL_{j+1}(1-\chi_{j+1})$ has its variables separated. Properness and P2 O4 make its output smooth; every factor to its left preserves smooth functions by P2 O5. The difference is consequently smooth. Compact kernels inside the chart extend by zero and remain ordinary proper operators, while multiplication of their principal symbols by these scalar cutoffs preserves the vanishing condition. A finite base partition on a compact set proves equivalence with the original local definition. No global elliptic inverse has entered this coordinate argument.

W1. Fourier cutoffs and the definition of wavefront

A nonzero covector $(y_0,\xi_0)$ is regular for $u$ if some compact smooth $\chi=1$ near $y_0$ makes $\widehat{\chi u}$ rapidly decreasing in an open cone about $\xi_0$. Rapid decrease here means every power bound for its values. This defines the complement of $\operatorname{WF}(u)$.

For a compact distribution $v$ and $b\in C_c^\infty$, Fourier inversion on the test, with (T1), gives

The integral converges absolutely by (T2) and Schwartz decrease of $\widehat b$. It converges in each fixed test seminorm before pairing: differentiated Fourier inversion supplies additional polynomial factors, defeated by further Schwartz decay. Thus passing the distribution through the integral is justified by (T1).

If $\widehat v$ is rapidly decreasing in a cone $V$, then $\widehat{bv}$ is rapidly decreasing in every cone $W$ whose angular closure is contained in $V$. For $\xi\in W,\eta\notin V$,

To prove the positive constant, normalize $|\xi|+|\eta|=1$ and minimize on the resulting closed bounded set, including zero endpoints. A zero minimum would give two equal nonzero vectors in disjoint angular sets; both zero is excluded by the normalization. Homogeneity restores (W2). In (W1), on $\eta\in V$ use rapid decrease of both factors and $1+|\xi|\le(1+|\xi-\eta|)(1+|\eta|)$. Taking both decay exponents above $N+n+1$ yields $C_N(1+|\xi|)^{-N}$ after integration. On the complementary region, take the decay exponent of $\widehat b$ above $N+M+n+1$ and apply (T2) and (W2). This proves the assertion with every $N$.

The same convolution argument holds for a tempered distribution whose Fourier transform is a polynomially bounded function: dual Fourier inversion gives (W1) and the same absolutely convergent integrals. This version will be used for a Fourier multiplier output.

A cutoff merely nonzero at $y_0$ gives the same definition: multiply by its smooth reciprocal on a smaller neighborhood and apply the result just proved. The regular set is open and conic, and the definition is unchanged on restricting to a smaller base open set. Smooth multiplication cannot enlarge wavefront; a smooth invertible matrix cannot change the union of the component wavefront sets, by applying this assertion to its entries and then to its inverse.

A distribution is smooth near a point exactly when all directions there are regular. For the nontrivial implication, cover the unit sphere by finitely many of the regular direction cones, shrink the base cutoff to the intersection of their neighborhoods, and apply (W1). The resulting compact distribution has rapidly decreasing Fourier values in every direction. Its inverse Fourier integral and every derivative are absolutely convergent, so it is smooth. The inverse distribution identity in P3 identifies this function with that distribution. The converse follows by integration by parts for compact smooth functions.

W2. The complete nonstationary bound for a chart

For a real phase $\Phi(y,\xi,\eta)$, linear in the two frequency vectors, assume on the fixed compact amplitude support $K$ that

For $I=\int a(y)e^{i\Phi}\,dy$, use the full operators

The superscript $t$ is the bilinear transpose. All derivatives of each coefficient are $O(R^{-1})$. Indeed divide the phase by $R$; its derivatives are bounded, its squared gradient is bounded below by $c^2$, and repeated reciprocal, product and chain rules give the claim, retaining the single outside factor $R^{-1}$. By induction, $(L^t)^Na$ is a finite sum with $N$ such factors and amplitude derivatives of order at most $N$. Compact integration and integration by parts give

For $R<1$, the direct compact integral gives the same bound with $R$ replaced by $1+R$. Any fixed further parameter derivative introduces only finitely many frequency factors; take that many additional integrations in (W4). This proves the differentiated versions used below, not just an estimate of values.

W3. Diffeomorphisms transport exactly the cotangent direction

The coordinate map (T3) satisfies

Proof. Suppose $u$ is regular at $(y_0,\xi_0)$. Put $x_0=\kappa(y_0)$ and $\eta_0=D\kappa(y_0)^{-T}\xi_0$. Choose $v=\chi u$ compactly supported, with $\chi=1$ near $y_0$, whose transform is rapidly decreasing on a cone $V$ about $\xi_0$. Take $b$ supported sufficiently close to $x_0$, with $g(\operatorname{supp}b)$ inside that neighborhood. Equations (T3) and Fourier inversion on its test give

For fixed $\eta$, the inner transform decays faster than any power for large $\xi$, either by ordinary test-function Fourier decay or by W2. Thus the outer integral and its derivation from (T1) are justified.

The phase gradient is $\xi-D\kappa(y)^T\eta$. On the compact support, the matrix and its inverse have bounded norms. Consequently this gradient is bounded below by $c(|\xi|+|\eta|)$ whenever $|\xi|$ is sufficiently small or sufficiently large relative to $|\eta|$. For comparable lengths, shrink the base support and an output cone $W$ about $\eta_0$ so that the directions of $D\kappa(y)^T\eta$, $\eta\in W$, lie in a cone whose angular closure is inside $V$. Compact separation as in (W2) gives the same lower bound if $\xi\notin V$. All higher phase derivatives have the upper bounds (W3).

On these separated regions, (W5) and (T2) give an integral bounded by

The last estimate follows by $\xi=(1+|\eta|)\zeta$, leaving the integrable factor $(1+|\zeta|)^{M-N}$. Choose $N$ for any prescribed output decay. On the remaining comparable region $\xi\in V$, use arbitrary rapid decrease of $\widehat v$ and the constant bound for the compact integral $I$. Its integration volume is at most $C(1+|\eta|)^n$, so arbitrary output decay follows again. This proves regularity at $(x_0,\eta_0)$. Apply the proved implication to the inverse map $g$ and the exact composition law in T0. The two inclusions give (W6). $\square$

W4. Ordinary operators and compact microlocal cutoffs

For a proper ordinary operator $P$,

Here is a direct proof that requires no elliptic parametrix. Localize the output compactly. P2 O5 confines the input to a compact set; input terms separated from the output are smooth by P2 O4. For the remaining symbol piece, P1 B4 gives the Fourier formula

For a compact distribution $v$, this formula follows by Fourier inversion on tests and the finite-order bound, or by the regularizations in P2; the displayed integral converges absolutely for each fixed $\xi$, by (T2) and a sufficiently large $N$.

If the input is regular at the point under consideration, choose the input cutoff inside that regular neighborhood. In (W10) split into the cone where $\widehat v$ is rapidly decreasing and its complement. The first part is controlled by rapid decrease and the Schwartz factor; the second uses (W2). The estimates in the proof of W1, with $M+d$ in place of the polynomial exponent, give rapid output decrease on a smaller cone. If instead $p$ is smoothing on a base and frequency cone, choose the output cutoff inside that base set. In the good frequency cone, integration by parts in the base variable gives, for arbitrary $N,J$,

This defeats the polynomial input. Outside it, angular separation and (W10) give the same rapid output bound. Smooth kernel remainders preserve regularity. This proves both exclusions in (W9), including for matrix entries.

Given $\rho=(y_0,\xi_0)$, choose nested small base and angular neighborhoods. Smooth finite cutoffs from U001, homogenized off zero, give a real smooth frequency cutoff $\gamma(\xi)$, zero near $\xi=0$, homogeneous of degree zero for large $|\xi|$, supported in the larger angular cone and equal to one in the smaller cone for large frequency. Homogenization is explicit: apply a smooth cutoff to $\xi/|\xi|$ and multiply by a radial cutoff; the smooth-root proof makes this map smooth off zero. Take compact $\varphi,\chi$, with $\varphi=1$ near $y_0$ and $\chi=1$ near $\operatorname{supp}\varphi$. Set

Its kernel has compact support in both base variables, so it is proper. P2 O4 gives a full symbol $\varphi(y)\gamma(\xi)+S^{-\infty}$: in its amplitude expansion every positive $z$-derivative of $\chi(z)$ vanishes on the support of $\varphi$, and every remainder has arbitrarily negative order. Thus $\operatorname{ess}(Q)$ lies in the prescribed base/angular neighborhood and $Q$ is elliptic at $\rho$, with principal symbol one nearby.

Moreover, for every distribution $u$,

Near $y_0$, where $\chi=\varphi=1$, the difference equals $v-\gamma(D)v$, $v=\chi u$. Its Fourier transform is $(1-\gamma)\widehat v$, zero at high frequency in the smaller cone. Multiplying by a smaller base cutoff preserves rapid decrease by W1, in its polynomial-Fourier-transform version. This proves (W13). If $u$ was regular throughout the chosen cone, (W9) and the essential-support bound make $Qu$ smooth.

W5. Iterated regularity has wavefront contained in the Lagrangian

If $u$ satisfies the intrinsic word condition of T3, then

We give the proof with one fixed output cutoff for every word length, so that arbitrarily high local regularity is obtained on one neighborhood.

Fix $\rho\notin\Lambda$. Closedness of $\Lambda$ permits a product of base and angular neighborhoods whose closure is disjoint from $\Lambda$. Choose $Q$ as in W4, supported in a smaller such product. Choose a proper scalar operator

with the cutoffs equal to one on a neighborhood of the essential support of $Q$, but still with its principal support disjoint from $\Lambda$. The principal symbol of $L$ therefore vanishes on $\Lambda$; multiply by the identity for a vector bundle. On a fixed cone about $\operatorname{ess}(Q)$, its full symbol equals $\langle\xi\rangle$ modulo $S^{-\infty}$, by the same amplitude argument as W4. P2's full product expansion shows, for every integer $N\ge1$, that the symbol of $L^N$ equals $\langle\xi\rangle^N$ modulo $S^{-\infty}$ on that cone. Every term with a positive base derivative of this frequency-only symbol vanishes.

Let $q$ be a compact-base full symbol of $Q$. Quantize $q(y,\xi)\langle\xi\rangle^{-N}$ with a proper kernel cutoff equal to one near the diagonal; call the result $B_N$, of order $-N$. Its full symbol has that value modulo $S^{-\infty}$. On the cone just chosen, the composition expansion gives the full symbol $q$ for $B_NL^N$, modulo smoothing. Outside $\operatorname{ess}(Q)$, $q$ and every derivative are smoothing, so every product coefficient and remainder is smoothing there as well. These two open sets cover all directions. Compact localization and the finite-cover argument of T2 imply

All cutoffs may be taken in one fixed compact coordinate region; their differences away from the diagonal are smooth by P2 O4.

The word condition gives $L^Nu\in B^s_{\mathrm{loc}}$. P1 B6 applied to $B_N$, and (W16), give $Qu\in B^{s+N}_{\mathrm{loc}}$ for every $N$. After any compact output cutoff, P1 B1 embeds this into $H^{s+N-1}$. For any derivative order $k$, choose $N$ with $s+N-1>k+n/2$; weighted Cauchy–Schwarz then makes the inverse Fourier integral and its derivatives of order at most $k$ absolutely convergent. This is the same Fourier proof as P1 B2 and shows $Qu$ is smooth. Equation (W13) now makes $u$ regular at $\rho$, proving (W14). $\square$

Two normalization checks with complete solutions

Exercise T1. In one dimension let $\kappa(y)=2y$ and $A=D_y=-i\partial_y$. Determine the transported operator and explain both Jacobians in (T8).

Solution. Since $T_gf(y)=f(2y)$, $A^\kappa f(x)=2D_xf(x)$. Here $L=2$, so the symbol is $a(2\eta)=2\eta$. The frequency Jacobian is 2 and the input base Jacobian is $1/2$; their product is one. Omitting either would give the wrong coefficient. For comparison, $T_\kappa\delta_0=2\delta_0$ as a scalar distribution: (T3) evaluates the test $2\psi(2y)$ at $y=0$. This scalar-density factor is distinct from the principal-symbol cancellation. $\square$

Exercise T2. For $\kappa(y_1,y_2)=(y_1+(y_1^2+y_2^2)/2,y_2)$, find the image of the covector $(\lambda,0)$ at $(0,s)$, $\lambda>0$, and check the principal symbol of the transported $D_{y_2}$.

Solution. On this curve,

The new base point is $(s^2/2,s)$. Formula (T5) takes the old symbol $\xi_2$ to $(D\kappa^T\eta)_2=s\eta_1+\eta_2$. Direct differentiation of $f(\kappa(y))$ gives $D_{y_2}(f\circ\kappa) =y_2(D_{x_1}f)\circ\kappa+(D_{x_2}f)\circ\kappa$. The coefficient is on the left, as required by left quantization. On the transformed conormal covector $(\lambda,-s\lambda)$, this principal symbol vanishes. Thus the operator and wavefront cotangent conventions agree with the exact model in C6. $\square$

Supplied scope and remaining localization

This component proves ordinary coordinate transport, invariance of the intrinsic word condition under charts and frames, Fourier wavefront covariance, proper microlocal cutoffs, pseudolocality, and (W14). The converse using an arbitrary elliptic order-zero test still needs the full conic parametrix and finite conic reconstruction. Those proofs, the prescribed nondegenerate and clean phase converse, the refined symbol order theorem and global Maslov data remain in the original course scope. This component does not clear publication.