Elliptic complexes, diagonal traces, and fixed points

Written and dedicated to the public domain by Codex, September 2026 (CC0).

An elliptic complex has a finite-dimensional space of harmonic representatives even though its individual differentials may be rectangular. That finite space makes an Euler index possible. A commuting operator with a singular kernel can still have a trace along the diagonal when its singular directions avoid the diagonal conormal. The same chain homotopy then turns that distributional supertrace into a cohomological one. For pullback by a smooth self-map, the trace becomes a signed count of nondegenerate fixed points, including maps whose derivative is singular.

The needed course results are Detecting regularity without choosing coordinates for bundle kernels, adjoints, wave fronts, and controlled convergence; Symbols, finite defects, and the index on a closed manifold for the full elliptic alternative on every Sobolev scale; and Traces that survive passage to cohomology for finite-rank and graded trace algebra. We prove the extra microlocal diagonal argument here. The chosen density, all half-density factors, both Laplacian summands, source-to-target bundle maps, and every determinant sign stay visible.

1. The original complex and exactness of its symbol

Let $X$ be a compact smooth manifold without boundary and let $E_0,\ldots,E_N$ be smooth finite-rank complex bundles. Give every $E_j\otimes\Omega_X^{1/2}$ a Hermitian metric and put $E_{-1}=E_{N+1}=0$. For one real order $m$, retain each actual bundle map $$D_j\in\Psi^m_{\mathrm{cl}} (X;E_j\otimes\Omega_X^{1/2},E_{j+1}\otimes\Omega_X^{1/2}), \quad -1\le j\le N,\qquad D_{-1}=D_N=0,\qquad D_jD_{j-1}=0. \tag{EC1}$$ Write $d_j(x,\xi):E_{j,x}\to E_{j+1,x}$ for its homogeneous principal symbol, with the original cotangent variable $\xi\ne0$. The symbol complex is elliptic exactly when $$\ker d_j(x,\xi)=\operatorname{ran}d_{j-1}(x,\xi) \quad(0\le j\le N,\ (x,\xi)\in T^*X\setminus0). \tag{EC2}$$ At the endpoints this includes injectivity of $d_0$ and surjectivity of $d_{N-1}$; the zero bundles in (EC1) supply the exact types. No rank equality between adjacent bundles is assumed. For the de Rham example, fix a positive smooth density $\mu$, take $E_j=\Lambda^jT^*X$, $m=1$, and use the correctly typed half-density differential $D_j=M_{\mu^{1/2}}dM_{\mu^{-1/2}}$. Its original principal symbol is $$d_j(x,\xi)\omega=i\,\xi\wedge\omega \quad\text{under }D=-i\partial. \tag{EC3}$$ The common nonzero factor $i$ does not alter the kernel. To prove exactness directly, choose $v\in T_xX$ with $\xi(v)=1$. Interior multiplication obeys the exact identity $\iota_v(\xi\wedge\omega)+\xi\wedge\iota_v\omega=\omega$. If $\xi\wedge\omega=0$, then $\omega=\xi\wedge\iota_v\omega$; at degree zero this says the kernel is zero, and at top degree it proves surjectivity. Thus (EC2) holds in every degree.

2. The full Laplacian symbol and Hodge representative

Choose the distributional Hermitian adjoints $D_j^*$, preserving the half-density bundle types, and define $$\Delta_j=D_{j-1}D_{j-1}^*+D_j^*D_j,\qquad \ell_j=d_{j-1}d_{j-1}^*+d_j^*d_j :E_j\longrightarrow E_j. \tag{EC4}$$ Both terms have order $2m$. For every $v\in E_{j,x}$, $$\langle\ell_jv,v\rangle =\|d_{j-1}^*v\|^2+\|d_jv\|^2. \tag{EC5}$$ If (EC2) holds and the right side vanishes, then $v\in\ker d_j =\operatorname{ran}d_{j-1}$ and $v\perp\operatorname{ran}d_{j-1}$, hence $v=0$. Conversely, if $\ell_j$ is injective for every $j$, decompose any $v\in\ker d_j$ orthogonally in the finite-dimensional fiber as $v=v_{\mathrm{ran}}+v_\perp$ with $v_{\mathrm{ran}}\in\operatorname{ran}d_{j-1}$ and $v_\perp\perp\operatorname{ran}d_{j-1}$. The complex identity gives $d_jv_{\mathrm{ran}}=0$, so $d_jv_\perp=0$, and $d_{j-1}^*v_\perp=0$; (EC5) makes $v_\perp=0$. Thus (EC2) is equivalent to ellipticity of every complete $\Delta_j$, without deleting either summand.

Elliptic Fredholm theory in the current owned global-index lesson applies to $\Delta_j:H^{s+2m}(E_j)\to H^s(E_j)$ for every real $s$. Its kernel is smooth and finite dimensional. For a smooth kernel element, the integrated version of (EC5) shows that this kernel is exactly $$K_j=\ker D_j\cap\ker D_{j-1}^* =\ker\Delta_j\subset C^\infty(X;E_j\otimes\Omega_X^{1/2}). \tag{EC6}$$ Fix an $L^2$-orthonormal basis $k_{j,1},\ldots,k_{j,r_j}$ of $K_j$ and keep the full finite-rank projection $$H_j f=\sum_{\alpha=1}^{r_j} \langle f,k_{j,\alpha}\rangle\,k_{j,\alpha}. \tag{EC7}$$ For a distribution $f$, the bracket is its distribution-smooth pairing with $k_{j,\alpha}$; $H_j$ has a smooth kernel and acts coherently on every Sobolev order. If $r_j=0$, the sum is zero.

The adjoint kernel of $\Delta_j:H^{s+2m}\to H^s$ is the same smooth $K_j$, with its correct anti-dual Sobolev realization. Its range is the annihilator of $K_j$, by the exact elliptic alternative. Therefore $$\Delta_j+H_j:H^{s+2m}(E_j)\xrightarrow{\;\cong\;}H^s(E_j) \quad\text{for every }s. \tag{EC8}$$ For surjectivity, split $f=(f-H_jf)+H_jf$; the first summand annihilates $K_j$ and is $\Delta_jg_0$. Replace $g_0$ by $g_0-H_jg_0$ and add $H_jf$. For injectivity, pair $(\Delta_j+H_j)g=0$ with each $k_{j,\alpha}$ to get $H_jg=0$, and then $\Delta_jg=0$ gives $g\in K_j$, hence $g=0$. The bounded inverse theorem applies to the displayed Banach isomorphism. This argument uses $\Delta_j$ with its original order $2m$; it does not assume $m>0$ or an $L^2$ inverse for a negative-order operator.

Let $$F_j=(\Delta_j+H_j)^{-1}(I-H_j) =(\Delta_j+H_j)^{-1}-H_j. \tag{EC9}$$ The second equality follows because $(\Delta_j+H_j)H_j=H_j$. The full elliptic parametrix of $\Delta_j+H_j$ has order $-2m$, and the difference from its actual inverse is smoothing: if $B_j$ is one parametrix, then $(\Delta_j+H_j)^{-1} =B_j+(\Delta_j+H_j)^{-1}(I-(\Delta_j+H_j)B_j)$; the second term maps every Sobolev order to every higher order, and applying the corresponding adjoint identity gives the same bound in the input variable, hence a smooth kernel. Thus $F_j\in\Psi^{-2m}$, with exact identities $$\Delta_jF_j=F_j\Delta_j=I-H_j,\qquad F_jH_j=H_jF_j=0. \tag{EC10}$$ The range of $F_j$ is in the annihilator of $K_j$. These operators agree on all Sobolev realizations by uniqueness in (EC8).

The complex identity proves the exact intertwining, retaining both Laplacian summands: $$D_j\Delta_j =D_jD_j^*D_j =\Delta_{j+1}D_j,\qquad D_jH_j=H_{j+1}D_j=0,\qquad D_jF_j=F_{j+1}D_j. \tag{EC11}$$ For the last equality, intertwine $\Delta+H$ and its actual inverse, then multiply by $I-H$. Adjoints give $D_j^*F_{j+1}=F_jD_j^*$.

3. The exact pseudodifferential chain homotopy, its smoothing remainder, and cohomology

With the endpoint convention $G_{-1}=G_N=0$, define the degree-lowering maps $$G_j=D_j^*F_{j+1}=F_jD_j^*: \mathcal D'(E_{j+1})\longrightarrow\mathcal D'(E_j), \qquad G_j\in\Psi^{-m}. \tag{EC12}$$ Equation (EC11) retains the factor order needed to compute $$D_{j-1}G_{j-1}+G_jD_j =(D_{j-1}D_{j-1}^*+D_j^*D_j)F_j =I-H_j. \tag{EC13}$$ Thus $H_j=I-D_{j-1}G_{j-1}-G_jD_j$ is the actual smooth-kernel chain homotopy remainder; it commutes with the differential because both $D_jH_j$ and $H_{j+1}D_j$ vanish. It induces the identity on cohomology.

For every $s\in\mathbb R$, if $f\in H^s(E_j)$ and $D_jf=0$ distributionally, (EC13) gives $$f=H_jf+D_{j-1}(G_{j-1}f),\qquad G_{j-1}f\in H^{s+m}(E_{j-1}). \tag{EC14}$$ The exact Sobolev shift is $s+m$, since $G_{j-1}$ has order $-m$. The harmonic part is the unique representative of the cohomology class: if a harmonic $k$ equals $D_{j-1}v$ for a distribution $v$, then $\langle k,k\rangle=\langle v,D_{j-1}^*k\rangle=0$, where elliptic regularity first makes $k$ smooth, so $k=0$. Conversely the difference between a closed $f$ and $H_jf$ has the explicit primitive in (EC14). Smooth input has a smooth primitive because $G_{j-1}$ preserves smooth sections. Every distribution belongs to some $H^s$ on compact $X$, so the same statement holds for the full distributional complex. Therefore smooth, Sobolev and distributional cohomology in degree $j$ are all canonically $K_j$, of finite dimension $r_j$.

The Sobolev complex in that statement has its exact degree shifts. For each fixed real $t$, put $$\begin{aligned} \mathcal H^j_t&=H^{t-jm}(X;E_j\otimes\Omega_X^{1/2}),& D_j&:\mathcal H^j_t\longrightarrow\mathcal H^{j+1}_t,\\ \frac{\ker(D_j:\mathcal H^j_t\to\mathcal H^{j+1}_t)} {D_{j-1}\mathcal H^{j-1}_t} &\ \xrightarrow{\ [f]\mapsto H_jf\ }\ K_j.& \end{aligned} \tag{EC14a}$$ The denominator is contained in the numerator by $D_jD_{j-1}=0$. It is exactly the subspace $\{f\in\mathcal H^j_t:D_jf=0,\ H_jf=0\}$: one inclusion follows from $H_jD_{j-1}=0$, and the other follows from (EC14), whose primitive lies in $H^{t-jm+m}=\mathcal H^{j-1}_t$. Both equations defining this subspace are continuous on $\mathcal H^j_t$, so the boundary space is closed. Consequently the quotient is a finite-dimensional Hausdorff space, and the displayed map and its inverse $k\mapsto[k]$ are continuous. The inclusions from smooth sections and into distributions commute with these maps because the actual $H_j$ and $G_j$ agree on all realizations. This gives the precise canonical comparison, including every negative, zero and positive differential order $m$.

Let $D=\sum_j D_j$ on the graded sum $\bigoplus_j E_j$, and $A=D+D^*$, with its actual even-to-odd component $A_{\mathrm{ev}}$. The full square $A^2=\bigoplus_j\Delta_j$, since $D^2=(D^*)^2=0$. Its principal symbol square is $\bigoplus_j\ell_j$, so $A_{\mathrm{ev}}$ is elliptic. The even kernel is $\bigoplus_{j\ \mathrm{even}}K_j$, and the adjoint odd kernel is $\bigoplus_{j\ \mathrm{odd}}K_j$, by the integrated energy identity. Hence $$\chi(D_\bullet)=\sum_{j=0}^N(-1)^j\dim K_j =\operatorname{ind}A_{\mathrm{ev}}. \tag{EC15}$$ No Fredholm index is assigned to an individual rectangular $D_j$ merely because the whole symbol sequence is exact.

4. Diagonal traces for singular kernels and the required cyclicity

Write $\Delta_X=\{(x,x):x\in X\}\subset X\times X$. For a continuous map $R:C^\infty(X;E)\to\mathcal D'(X;E)$, its distributional Schwartz kernel $K_R$ is a section of the output-input bundle with the input density factor. The diagonal trace is defined when $$\operatorname{WF}(K_R)\cap N^*\Delta_X=\varnothing,\qquad N^*\Delta_X =\{(x,x;\xi,-\xi):\xi\in T_x^*X\setminus0\}. \tag{EC16}$$ Indeed, the pullback $\Delta_X^*K_R$ then exists as a distributional density on $X$. Its fiber trace is a density, and compactness permits the exact scalar $$\operatorname{Tr}_{\Delta_X}R =\int_X\operatorname{tr}_E(\Delta_X^*K_R). \tag{EC17}$$ The finite chart proof of this pullback is Fourier restriction: in coordinates near the diagonal, a test of the pullback requires integrating the full Fourier transform along the normal-frequency directions $(\xi,-\xi)$. Condition (EC16) gives rapid decay in a cone containing those directions, while polynomial distribution bounds handle the complementary compact-frequency part. Cutoffs and changes of coordinates preserve the condition and give the same distribution. This construction retains the actual density and bundle trace, not merely a formal value $K_R(x,x)$.

We need two continuity facts and prove the estimates that matter. Let a compactly localized kernel $K$ have wave front disjoint from $N^*\Delta_X$. In a chart-product cutoff, convolve $K$ with a smooth approximate identity in both variables, or multiply its full Fourier transform by a compactly supported multiplier $\rho(\varepsilon(\xi,\eta))$ equal to one near zero. The resulting smooth $K_\varepsilon$ tends to $K$ weakly. On any closed cone disjoint from $\operatorname{WF}(K)$, all localized Fourier transforms of $K$ decrease faster than every power. The multiplier is uniformly bounded; for a fixed weighted rapid-decay seminorm split at radius $L$, use pointwise convergence on the bounded part and a stronger rapid-decay bound $C_ML^{N-M}$ on the tail. Thus $$K_\varepsilon\longrightarrow K \quad\text{in every wave-front-topology seminorm on cones disjoint from }\operatorname{WF}(K). \tag{EC18}$$ Finite chart partitions and smoothing off-diagonal overlaps give a global smoothing family with the same property.

Let $A\in\Psi^r(X)$ be properly supported. The statement needed here concerns a cone around $N^*\Delta_X$, where both output and input covectors are nonzero. A pseudodifferential kernel can be split into a part supported in an arbitrarily small neighborhood of its diagonal and a smooth remainder. For the first part, output composition has a localized full Fourier transform given by an integral in an intermediate output frequency $\xi'$; the Fourier transform in $x$ of its smooth-symbol coefficient decreases faster than any power of $|\xi-\xi'|$, with at most a fixed polynomial factor in $|\xi'|$. Split into $|\xi-\xi'|\le c|(\xi,\eta)|$ and its complement. In the first region, a smaller conic neighborhood of the diagonal conormal stays inside a larger one, so use the arbitrary rapid-decay bound supplied by (EC16). In the complement, the rapidly decreasing coefficient absorbs the polynomial distribution bound and every requested output power. The input-composition calculation is the same in $\eta$.

The smooth remainder of $A$ needs a separate check: after output composition it can retain a singular input covector, but its output covector is zero; after input composition the reverse holds. Neither one-sided direction meets $N^*\Delta_X$. In particular the stronger assertion that either composition preserves the entire wave front of $K$ would be false. The exact two-sided inclusion that follows from the near-diagonal calculation is $$\begin{aligned} \operatorname{WF}(AK)\cap T^*_{\ne0,\ne0}(X\times X) &\subseteq \operatorname{WF}(K),\\ \operatorname{WF}(KA)\cap T^*_{\ne0,\ne0}(X\times X) &\subseteq \operatorname{WF}(K), \end{aligned} \quad T^*_{\ne0,\ne0} =\{(x,y;\xi,\eta):\xi\ne0,\ \eta\ne0\}. \tag{EC19}$$ The same split, with the uniform bounds in (EC18), proves convergence of the composed smooth approximants in the conic seminorms around $N^*\Delta_X$. Consequently (EC16) holds for $AK$ and $KA$, and their diagonal traces converge. We claim no preservation of the wave-front seminorms at one-sided directions. To see why the restriction is necessary, take a circle, a point $y_0$, and smooth functions $f,g,h$ with $\operatorname{supp}f\cap\operatorname{supp}g=\varnothing$ and $\int hf\ne0$. Put $K(z,y)=f(z)\delta(y-y_0)$ and let $A$ have the smooth rank-one kernel $g(x)h(z)$. Then $AK(x,y)=(\int hf)g(x)\delta(y-y_0)$, which has input-only wave-front directions over $\operatorname{supp}g\times\{y_0\}$ where $K$ has none. Both kernels remain transverse to $N^*\Delta_X$, exactly as (EC19) requires.

For any $A\in\Psi^r(E,F)$ and kernel map $B:F\to E$ satisfying (EC16) in the corresponding typed compositions, these facts give the exact microlocal cyclicity $$\operatorname{Tr}_{\Delta_X}(AB) =\operatorname{Tr}_{\Delta_X}(BA). \tag{EC20}$$ To verify it without treating $A$ as bounded on $L^2$, use the smooth approximants $B_\varepsilon$. Both $AB_\varepsilon$ and $B_\varepsilon A$ have smooth kernels. Expand $B_\varepsilon$ in a finite Fourier-basis approximation, whose smooth-kernel remainder tends to zero in every smooth seminorm; for each finite-rank term the distributional adjoint identity gives equality of the two traces. The smooth-kernel remainder is trace-norm small after applying $A$, because its double Fourier coefficients decrease faster than the polynomial matrix growth of $A$. Thus the trace equality holds for $B_\varepsilon$. The continuity just proved sends both sides to (EC20). Fiber types and matrix trace are retained by the finite-rank identity.

5. Supertrace of a commuting distributional chain map

Let $R_j:C^\infty(E_j)\to\mathcal D'(E_j)$ be continuous for every $j$, assume $$D_jR_j=R_{j+1}D_j \quad\text{on smooth sections,}\qquad \operatorname{WF}(K_{R_j})\cap N^*\Delta_X=\varnothing. \tag{EC21}$$ The chain identity extends by continuity where compositions of distributions are defined. A closed smooth $f$ maps to a closed distribution, so (EC14) defines its action on cohomology. The induced finite-dimensional map on $K_j$ is $H_jR_j|_{K_j}$. Since $H_j$ has a smooth finite-rank kernel, $H_jR_jH_j$ has a smooth finite-rank kernel and $$\operatorname{Tr}(H_jR_jH_j) =\operatorname{Tr}_{K_j}(H_jR_j|_{K_j}). \tag{EC22}$$

To compare it with the singular-kernel diagonal trace, retain the degree-lowering operator in both terms: $$L_j=G_{j-1}R_j+H_{j-1}R_{j-1}G_{j-1}:E_j\to E_{j-1}. \tag{EC23}$$ Using (EC13), the chain identity (EC21), $D_{j-1}H_{j-1}=0$, and $H_jR_jD_{j-1}=H_jD_{j-1}R_{j-1}=0$, calculate in its actual order $$\begin{aligned} D_{j-1}L_j+L_{j+1}D_j &= (D_{j-1}G_{j-1}+G_jD_j)R_j +H_jR_jG_jD_j\\ &= (I-H_j)R_j+H_jR_j(I-H_j)\\ &=R_j-H_jR_jH_j. \end{aligned} \tag{EC24}$$ Each kernel appearing on the right and in the first line has a diagonal trace. For $G R$ and $R G$, this follows from (EC19) and the original (EC21); for a term with $H$, one side is smooth and cannot have a nonzero diagonal conormal in both variables. Apply (EC20) to $D_j$ and $L_{j+1}$, sum (EC24) with the original alternating signs, and reindex only after all source and target types are fixed: $$\begin{aligned} \sum_j(-1)^j\operatorname{Tr}_{\Delta_X} (R_j-H_jR_jH_j) &=\sum_j(-1)^j \bigl(\operatorname{Tr}_{\Delta_X}(D_{j-1}L_j) +\operatorname{Tr}_{\Delta_X}(L_{j+1}D_j)\bigr)\\ &=\sum_j\bigl((-1)^{j+1}+(-1)^j\bigr) \operatorname{Tr}_{\Delta_X}(L_{j+1}D_j)=0. \end{aligned} \tag{EC25}$$ The endpoint maps in (EC1) make the dropped end terms exactly zero. Equations (EC22)–(EC25) prove the full supertrace identity $$\sum_{j=0}^N(-1)^j\operatorname{Tr}_{\Delta_X}R_j =\sum_{j=0}^N(-1)^j \operatorname{Tr}_{K_j}(H_jR_j|_{K_j}). \tag{EC26}$$ This is an equality of a distributional diagonal trace and the actual induced cohomology trace, not an assertion that each nonsmoothing $R_j$ is trace class.

6. The vanishing high-frequency tail used in diagonal limits

The controlled-kernel argument also gives the exact vanishing-parametrix statement. Let $a_\varepsilon(x,\xi)$ be bounded in every symbol seminorm of one fixed $S^r$ class and tend to zero with all derivatives on each compact $(x,\xi)$ set. Its quantized kernels have wave front contained in $N^*\Delta_X$, converge to zero weakly, and converge to zero in every wave-front-topology seminorm on cones outside $N^*\Delta_X$: $$K_{\operatorname{Op}(a_\varepsilon)} \longrightarrow0 \quad\text{in }\mathcal D'(X\times X) \text{ and in the controlled topology off }N^*\Delta_X. \tag{EC27}$$ For weak convergence, pair the oscillatory kernel with a compact smooth test, take its Fourier transform in $x-y$, and use the test’s arbitrary rapid decay against the uniform polynomial symbol bound; pointwise convergence then gives dominated convergence. Away from $N^*\Delta_X$, the finite-chart kernel estimate from the owned geometric microlocal lesson gives $C_M\langle(\xi,\eta)\rangle^{-M}$ uniformly for every $M$. On bounded frequencies use the already-proved weak convergence of localized kernels; on the tail choose $M$ larger than the requested weighted seminorm. This proves (EC27), including all chart and cutoff derivatives.

For a concrete high-frequency tail, take the full symbol $f_j(x,\xi)\in S^{-2m}$ of $F_j$ in each fixed chart and a smooth radial scalar $\chi$ which is zero near the origin and one outside a larger ball. Use the fixed cotangent metric and the same chart partition in every member of the family. Define the localized symbols and their patched operator by $$f_{j,\varepsilon}(x,\xi) =\chi(\varepsilon\xi)f_j(x,\xi),\qquad F_{j,\varepsilon}=\operatorname{Op}(f_{j,\varepsilon}). \tag{EC28}$$ For fixed $\xi$, every derivative tends to zero as $\varepsilon\downarrow0$. Each cutoff derivative $\varepsilon^{|\beta|}\partial^\beta\chi(\varepsilon\xi)$ is supported where $|\xi|\asymp\varepsilon^{-1}$ and has the exact $\langle\xi\rangle^{-|\beta|}$ symbol cost, so the family is uniformly bounded in $S^{-2m}$. Formula (EC27) applies for every real $m$.

Here is the needed trace-limit argument with both conic factors retained. After a finite partition of unity on $X\times X$, the trace of $R F_{j,\varepsilon}$ is the distributional pairing of $K_R(x,z)$ with the transposed kernel $K_{F_{j,\varepsilon}}(z,x)$, including the bundle composition and both density factors. The pairing is defined because the second wave front is contained in the diagonal conormal and (EC16) excludes its negative from the first. In Fourier coordinates, divide the frequency integral into a closed cone around that conormal and its complement. On the first cone, the localized transform of $K_R$ has arbitrary rapid decay, while the transform of $K_{F_{j,\varepsilon}}$ has a uniform polynomial bound from its $S^{-2m}$ seminorms. On the complementary closed cone, the latter has uniform arbitrary rapid decay by (EC27), while the former has a fixed polynomial distribution bound. Choose the rapid-decay powers greater than the respective polynomial powers plus $2\dim X+1$; both products then have an integrable majorant. The localized transforms of $K_{F_{j,\varepsilon}}$ tend pointwise to zero by weak convergence in (EC27), so dominated convergence makes every localized pairing tend to zero. The same proof with the kernels in reverse order treats $F_{j,\varepsilon}R$. Consequently $$\operatorname{Tr}_{\Delta_X}(R F_{j,\varepsilon}) \longrightarrow0,\qquad \operatorname{Tr}_{\Delta_X}(F_{j,\varepsilon} R) \longrightarrow0. \tag{EC29}$$ Proper cutoffs obey the same cone split. If a chosen representation of $F_j$ also has a smoothing remainder, apply the same high-frequency cutoff to a rapidly decreasing complete symbol for each compactly localized piece of that remainder; those tails tend to zero in every smooth-kernel seminorm. An unchanged smoothing remainder is not part of $F_{j,\varepsilon}$. This establishes the high-frequency tail removal used in the diagonal-trace route while (EC24) supplies a direct exact homotopy identity.

7. The full noninvertible-map fixed-point calculation

Choose a Riemannian metric and keep the positive smooth density $\mu$ fixed in (EC3). In the de Rham complex, let $\phi:X\to X$ be any smooth map; no inverse or local diffeomorphism is assumed. The ordinary pullback on $j$-forms is a chain map because $d\phi^*=\phi^*d$, by the coordinate chain rule. Conjugate it by the same multiplication with $\mu^{1/2}$ used for $D_j$, so $D_jR_j=R_{j+1}D_j$ has the exact half-density types: $$R_j=M_{\mu^{1/2}}\circ\phi^* \circ M_{\mu^{-1/2}}. \tag{EC30}$$ In one coordinate chart, its kernel coefficient relative to the input coordinate density is exactly $$K_{R_j}(x,y) =\left(\frac{\mu(x)}{\mu(\phi(x))}\right)^{1/2} \Lambda^j(d\phi_x)^*\,\delta(y-\phi(x)). \tag{EC31}$$ The quotient of densities is evaluated in the chosen coordinate trivialization, and on the diagonal at a fixed point it equals one. The graph map $x\mapsto(x,\phi(x))$ is an embedding even when $d\phi_x$ is singular, because its first component is the identity. Thus the delta kernel has wave front contained in its graph conormal: $$\operatorname{WF}(K_{R_j})\subset \{(x,\phi(x);-d\phi_x^{\,T}\eta,\eta): \eta\in T^*_{\phi(x)}X\setminus0\}. \tag{EC32}$$ The smooth matrix and density factor can only remove, not add, those directions. At a fixed point $x=\phi(x)$, intersection with $N^*\Delta_X$ requires $(-d\phi_x^T\eta,\eta)=(\xi,-\xi)$. Hence $d\phi_x^T\eta=\eta$. The exact transversality condition is therefore $$\det(I-d\phi_x)\ne0 \quad\text{at every fixed point }x. \tag{EC33}$$ This condition makes (EC16) hold for all $R_j$. It also makes each zero of the local map $F(x)=x-\phi(x)$ isolated by the inverse function theorem. The fixed set is closed in compact $X$, hence is finite.

At each fixed point $x_\alpha$, the standard delta change-of-variables formula, applied to the actual map $F$ with its full Jacobian, gives $$\delta(x-\phi(x)) =\sum_\alpha \frac{\delta(x-x_\alpha)} {|\det(I-d\phi_{x_\alpha})|}. \tag{EC34}$$ To verify it, choose disjoint small neighborhoods on which $F$ is a diffeomorphism, test against a compact smooth function, change variables $z=F(x)$, and evaluate the Jacobian at $z=0$. Outside those neighborhoods the delta has no support. The density ratio in (EC31) becomes one at each $x_\alpha$; no Jacobian from a nonexistent inverse of $\phi$ is introduced.

Because $\phi(x_\alpha)=x_\alpha$, the fiber map $(d\phi_{x_\alpha})^*$ acts on the same cotangent space and has a trace. Summing the complete exterior algebra, including degree zero and top degree, gives $$\sum_{j=0}^{\dim X}(-1)^j \operatorname{tr}\Lambda^j((d\phi_{x_\alpha})^*) =\det(I-(d\phi_{x_\alpha})^*) =\det(I-d\phi_{x_\alpha}). \tag{EC35}$$ For a diagonalizable matrix this is the product $\prod_i(1-\lambda_i)$ by expanding it over all subsets of eigenvalues. Both sides are polynomial in the matrix entries, and diagonalizable complex matrices are dense, so the identity holds for every matrix. The second determinant equality follows from equality of a matrix and its transpose determinants, without assuming $d\phi$ is invertible.

Now apply the proved supertrace identity (EC26) to the chain map (EC30), and evaluate its diagonal traces with (EC31)–(EC35). If $L(\phi)$ is the original alternating trace of $\phi^*$ on de Rham cohomology, then $$\begin{aligned} L(\phi) &=\sum_j(-1)^j \operatorname{Tr}\bigl(\phi^*:H^j_{\mathrm{dR}}(X)\to H^j_{\mathrm{dR}}(X)\bigr)\\ &=\sum_\alpha \frac{\det(I-d\phi_{x_\alpha})} {|\det(I-d\phi_{x_\alpha})|}\\ &=\sum_{\phi(x_\alpha)=x_\alpha} \operatorname{sgn}\det(I-d\phi_{x_\alpha}). \end{aligned} \tag{EC36}$$ The equality includes a smooth constant map, whose derivative is zero and whose single fixed point contributes $+1$. It also includes maps with singular derivative at a fixed point, provided $I-d\phi$ is invertible there. No diffeomorphism hypothesis entered the kernel, transversality or determinant calculation.

8. Worked examples with the full operators retained

The circle complex. Give $S^1=\mathbb R/(2\pi\mathbb Z)$ its density $dx$, and identify the half-density factors using $dx^{1/2}$. The de Rham complex is $$0\longrightarrow C^\infty(S^1) \mathop{\longrightarrow}^{D_0=\partial_x} C^\infty(S^1)\,dx\longrightarrow0. \tag{EC37}$$ The adjoint of $\partial_x$ for this density is $-\partial_x$. Thus both complete Laplacians are $-\partial_x^2$. In each degree, the harmonic projection takes the zeroth Fourier coefficient, and the Green operator keeps every nonzero Fourier coefficient with multiplier $n^{-2}$: $$H_j\!\left(\sum_{n\in\mathbb Z}u_ne^{inx}\right)=u_0,\qquad F_j\!\left(\sum_{n\in\mathbb Z}u_ne^{inx}\right) =\sum_{n\ne0}\frac{u_n}{n^2}e^{inx}, \quad j=0,1. \tag{EC38}$$ The factors $dx$ and $dx^{1/2}$ are understood in the degree-one and both half-density slots, respectively. Here $G_0=D_0^*F_1$ has multiplier $1/(in)$ on $n\ne0$, so $G_0D_0=I-H_0$ and $D_0G_0=I-H_1$ on the corresponding spaces. Both harmonic spaces have dimension one, and (EC15) gives $1-1=0$.

A smooth map that is not locally invertible. Fix $p\in X$ and put $\phi(x)=p$ for every $x$ in a nonempty compact $X$. Its only fixed point is $p$, where $d\phi_p=0$ and $\det(I-d\phi_p)=1$. Formula (EC31) retains the delta distribution at the constant input point; it does not require the inverse of $\phi$. The cohomology pullback factors through $H^*(\{p\})$: it is zero in positive degree, and on locally constant degree-zero functions it sends every component value to the value at the component containing $p$. This degree-zero linear map has trace one. Thus both sides of (EC36) equal one, even if $X$ is disconnected.

A covering map with several fixed points. For $k\ge2$, let $\phi_k:S^1\to S^1$ be $e^{ix}\mapsto e^{ikx}$. Its fixed points satisfy $(k-1)x\in2\pi\mathbb Z$, so there are exactly $k-1$ of them. The actual tangent derivative at each is multiplication by $k$, and each local sign is $\operatorname{sgn}(1-k)=-1$. The right side of (EC36) is $1-k$. On cohomology, the degree-zero trace is one and the degree-one trace is $k$, so the left side is also $1-k$. The calculation retains the absolute Jacobian in (EC34) and the signed exterior determinant in (EC35); neither can be removed before taking their quotient.

9. Exercises and full solutions

1. A rectangular symbol complex. Let $V_0=\mathbb C^2$, $V_1=\mathbb C^4$, and $V_2=\mathbb C$. Define $d_0(a,b)=(a,2b,0,0)$ and $d_1(w_1,w_2,w_3,w_4)=3w_3$. Calculate the complete Laplacians, harmonic spaces, and Euler characteristic. Does exactness hold at $V_1$?

Solution. Direct multiplication gives $d_1d_0=0$ and $$\ell_0=\operatorname{diag}(1,4),\qquad \ell_1=\operatorname{diag}(1,4,9,0),\qquad \ell_2=(9). \tag{EC39}$$ Therefore $K_0=K_2=0$ and $K_1=\mathbb C(0,0,0,1)$; the Euler characteristic of this finite complex is $-1$. Its symbol sequence is not exact at $V_1$, since $\ker d_1=\mathbb C e_1\oplus\mathbb C e_2\oplus\mathbb C e_4$, while $\operatorname{ran}d_0=\mathbb C e_1\oplus\mathbb C e_2$. The zero eigenvalue in $\ell_1$ detects precisely the missing exactness. This finite-fiber example is an algebraic test of (EC2)–(EC5), not an elliptic operator on a manifold.

2. The graded trace with one surviving class. On $0\to\mathbb C\mathop{\to}^{d_0}\mathbb C^2\to0$, put $d_0(t)=(t,0)$. Let $R_0=(a)$ and $R_1=\begin{pmatrix}a&b\\0&c\end{pmatrix}$. Compute the harmonic projection, the degree-lowering $G_0$, and both sides of the chain-homotopy and supertrace identities.

Solution. The Laplacians are $\ell_0=(1)$ and $\ell_1=\operatorname{diag}(1,0)$. Thus $H_0=0$, $H_1=\operatorname{diag}(0,1)$, $F_0=(1)$, $F_1=\operatorname{diag}(1,0)$, and $G_0=(1,0)$. The chain condition is $d_0R_0=R_1d_0$. Formula (EC23) gives $L_1=G_0R_1=(a,b)$. In degree zero, $L_1d_0=(a)=R_0-H_0R_0H_0$. In degree one, $$d_0L_1=\begin{pmatrix}a&b\\0&0\end{pmatrix} =R_1-H_1R_1H_1. \tag{EC40}$$ The alternating ordinary traces are $a-(a+c)=-c$, while the only cohomology class is $e_2$ in degree one and $R_1$ acts on it by $c$. Its alternating trace is $-c$. This checks the off-diagonal term $b$ rather than discarding it before the homotopy calculation.

3. Why one-sided wave fronts matter. Construct a kernel $K$ transverse to the diagonal conormal and a smoothing operator $A$ for which $\operatorname{WF}(AK)$ is not contained in $\operatorname{WF}(K)$. Explain why the diagonal trace is still defined.

Solution. On a circle choose nonzero smooth $f,g,h$ with $\operatorname{supp}f\cap\operatorname{supp}g=\varnothing$ and $\int hf\ne0$, then set $K(z,y)=f(z)\delta(y-y_0)$ and $A(x,z)=g(x)h(z)$. The product is $AK(x,y)=(\int hf)g(x)\delta(y-y_0)$. Its wave front has input covector nonzero and output covector zero over $\operatorname{supp}g\times\{y_0\}$, whereas $K$ vanishes there. Such one-sided covectors can never equal $(\xi,-\xi)$ with $\xi\ne0$, so (EC16) still holds. This is why (EC19) asserts only the two-sided inclusion needed for diagonal restriction.

4. The identity map is outside the isolated-fixed-point formula. Let $\phi=I_X$ on a positive-dimensional compact $X$. Which exact hypothesis of (EC36) fails? Is its cohomological Lefschetz number undefined?

Solution. Every point is fixed and $I-d\phi_x=0$, so (EC33) fails everywhere; the graph conormal meets $N^*\Delta_X$, and (EC34) cannot be applied. The cohomological Lefschetz number is still defined and equals $\sum_j(-1)^j\dim H_{\mathrm{dR}}^j(X)$. The failure concerns the isolated-point distributional evaluation in this theorem, not the cohomology trace.

10. Sources and exact scope

The three linked course prerequisites at the start supply the cited bundle calculus, Fredholm alternative, and trace-algebra results. Sections 1–7 prove the extra elliptic-complex, diagonal-kernel, and fixed-point statements used here; Section 8 evaluates them with the original operators retained. The fixed-point formula assumes nondegenerate fixed points and does not claim an isolated-point formula when that condition fails. The diagonal trace requires (EC16); ordinary distributional convergence by itself does not justify restricting a singular kernel to the diagonal.

11. Complete operative proofs without a heat or spectral prerequisite

This supplement retains the original complex, all real orders, both summands of every Laplacian, the actual half-density bundles, every original kernel and the original fixed-point condition. The finite errors and the all-real-order elliptic alternative (I1)–(I7) are proved in the earlier global elliptic lesson. Bundle assembly, adjoints, symbol inversion and Sobolev bounds are proved in (G15)–(G20), (G30)–(G32), (G38)–(G43) of the earlier geometric lesson. Its (WF1)–(WF15) prove Fourier cutoffs and the actual covector maps. We use these specific earlier proofs. We supply the Hilbert and kernel-trace interfaces below, rather than importing an unbounded spectral assertion through another reference. No heat kernel, heat expansion or unbounded eigenfunction expansion enters this lesson.

11.1. The actual Hilbert geometry and trace completion

Here are the Hilbert facts required by (EC7)–(EC10), their anti-dual realization and the finite-rank trace argument. For a closed subspace M of a Hilbert space and a vector x, let δ be its distance from M and choose u_l in M with $\|x-u_l\|^2\to\delta^2$. The full parallelogram identity gives $$\|u_l-u_k\|^2 =2\|x-u_l\|^2+2\|x-u_k\|^2 -4\left\|x-\frac{u_l+u_k}{2}\right\|^2 \leq 2\|x-u_l\|^2+2\|x-u_k\|^2-4\delta^2. \tag{LC1}$$ Thus the sequence is Cauchy, and completeness and closedness give a minimizer u in M. Vary it by tv and itv for every v in M and real t. Expanding the full squared norm shows that both the real and imaginary parts of $\langle x-u,v\rangle$ are zero. The decomposition $x=u+(x-u)$ is therefore orthogonal and unique. Uniqueness makes the projection linear, and its norm is at most one by the same squared-norm identity.

For a nonzero bounded linear functional f, its kernel is closed. Its orthogonal complement is one dimensional: if f(w)≠0, every x is the sum of $x-f(x)w/f(w)$ in the kernel and $f(x)w/f(w)$ in the line generated by w. If v is a nonzero vector in the orthogonal complement, the actual representing vector is $\overline{f(v)}v/\|v\|^2$, for the original convention linear in the first variable. This gives f(x) as its inner product, with equality of the functional and vector norms. The zero functional has representing vector zero. Applying this construction to $x\mapsto\langle Ax,y\rangle$ gives the bounded adjoint, its exact pairing identity and $\|A^*\|=\|A\|$. This proof gives the anti-dual realization in (I7) directly; it uses no spectral calculus or positive square root.

The Fourier spaces here are separable. A finite family of charts and frames reduces their dense smooth subspace to compact smooth vector functions. Fourier truncation and smooth cutoffs approximate these in every fixed Sobolev norm, by the original Fourier inversion and weighted L² dominated convergence. Rational finite Fourier sums in fixed cubes give a countable dense set. Successive orthogonalization of that set, retaining the full differences and their norm factors, supplies a complete orthonormal sequence. Its finite projections converge to each vector because their spans are dense; Pythagoras then proves Parseval. The same reasoning applies to a zero space with the empty sequence. Only this bounded Hilbert geometry is needed.

For a rank-one operator $u\langle\,\cdot\,,v\rangle$, finite projections and Parseval give its trace $\langle u,v\rangle$. If $$T=\sum_l u_l\langle\,\cdot\,,v_l\rangle, \qquad \sum_l\|u_l\|\|v_l\|<\infty, \quad \operatorname{Tr}T=\sum_l\langle u_l,v_l\rangle, \tag{LC2}$$ the operator series converges in norm. For every orthonormal basis $(e_i)$, the double absolute sum is bounded by $\sum_l\sum_i|\langle u_l,e_i\rangle\langle e_i,v_l\rangle| \leq\sum_l\|u_l\|\|v_l\|$. Interchanging the absolutely convergent sums proves (LC2) and its independence of the basis and representation.

The trace-norm estimate used below also has a finite proof. A finite-rank T acts between finite-dimensional spans of its input and output vectors. A positive Hermitian matrix has an orthonormal eigenbasis by successive maximization of its real quadratic form on the compact unit sphere: its first variation in each real and imaginary tangent direction gives the eigenvector equation, the perpendicular space is invariant by the Hermitian identity, and induction applies there. Apply this to the original finite matrix T* T. For each positive eigenvalue σ_l² keep the unit eigenvector e_l and the actual vector $f_l=Te_l/\sigma_l$; these f_l are orthonormal by the full T* T identity. The remaining eigenvectors lie in the kernel. Thus $T=\sum_l\sigma_l f_l\langle\,\cdot\,,e_l\rangle$. The sum of these σ_l equals $\sup_{\|C\|\leq1}|\operatorname{Tr}(CT)|$: the displayed sum gives the upper bound, and the finite-rank contraction sending each f_l to e_l and vanishing on their perpendicular space gives equality. It follows that this finite trace norm is subadditive and that a rank-one term has norm $\|u\|\|v\|$. Finite partial sums in (LC2) are therefore Cauchy in trace norm, with tail bounded by the full omitted sum. The completion injects into bounded operators: if a trace-norm Cauchy sequence T_l tends to zero in operator norm, fix l and a finite-rank norm-one C_l attaining its finite trace norm. For large k, $|\operatorname{Tr}(C_lT_k)|\leq\operatorname{rank}(C_l)\|T_k\|\to0$; the Cauchy bound on $\operatorname{Tr}[C_l(T_l-T_k)]$ then makes $\|T_l\|_1$ arbitrarily small for all sufficiently large l. Thus its completion has no nonzero element with zero operator image. This is the trace-class completion, and $|\operatorname{Tr}T|\leq\|T\|_1$ survives there by continuity. It constructs exactly the trace completion needed for the smooth kernels below without any unbounded spectral prerequisite.

11.2. All-real-order inverse, smoothing kernel and homotopy domains

The finite-fiber proof (EC5) retains both squared norms. On each positive-dimensional component choose any fixed cotangent norm only to test the actual homogeneous principal symbol on its compact unit cosphere. Positivity and continuity give a positive minimum for its smallest quadratic-form value there. The original homogeneity is $\ell_j(x,t\xi)=t^{2m}\ell_j(x,\xi)$ for t>0. The complete inverse therefore has its original order −2m, including m<0. This is ellipticity on the typed spaces $H^{s+2m}\to H^s$, not an assertion of a bounded inverse on L² when m is negative. On a zero-dimensional component all spaces have finitely many finite-dimensional fibers; the same range and kernel argument is finite linear algebra.

The distributional adjoint of the realization $\Delta_j:H^{s+2m}\to H^s$ has domain H^(−s) and target H^(−s−2m) in the anti-dual bundles. The chosen smooth Hermitian metric is the actual order-zero identification with the original E_j. Formal selfadjointness and the earlier elliptic regularity identify its nullspace with the same smooth K_j. The closed-range alternative (I7), using the Hilbert representation just proved, gives precisely its annihilator. Each functional in (EC7) is bounded on every H^s because its smooth test vector belongs to the matching H^(−s) anti-dual. The original surjectivity and injectivity argument in (EC8) now gives a bounded inverse on every displayed pair of Sobolev spaces; uniqueness makes all these inverses agree on common distributions.

Write $Q_j=\Delta_j+H_j$. For its actual inverse and a full smoothing-error parametrix B_j of order −2m, retain $Q_j^{-1}=B_j+Q_j^{-1}(I-Q_jB_j)$. For every M>0 the second term maps H^(−M) boundedly into H^M: the smoothing factor maps H^(−M) into H^(M−2m), and the actual inverse supplies the remaining 2m shift. This proves smoothness of its kernel directly. In compact input and output charts take cutoff Fourier vectors at integer indices p and q. Its kernel coefficient b_pq satisfies $$|b_{pq}|\leq C_M\langle p\rangle^{-M}\langle q\rangle^{-M} \quad\text{for every }M, \tag{LC3}$$ by testing the H^(−M)→HM bound on the input vector and taking the output Fourier coefficient. Fixed cutoffs and frames retain their bounded Sobolev factors. Choosing M larger than each desired pair of derivative orders plus the chart dimensions makes the full double Fourier series and all those derivatives uniformly convergent. Its sum is the kernel, by Fourier testing, hence is smooth. This also proves the smooth-kernel assertion in (EC9) without assuming a spectral representation of Q_j.

All identities (EC10)–(EC13) now hold on distributions by the actual pseudodifferential extensions. For example $H_{j+1}D_j=0$ follows by testing against the smooth K_(j+1), whose vectors are killed by D_j*. The complex identity kills the omitted cross terms in D_jΔ_j=Δ_(j+1)D_j. Since D_j also kills H_j, intertwining the actual inverses gives D_jF_j=F_(j+1)D_j in their original order. Its adjoint follows by testing smooth pairs. The two terms in (EC13) therefore have exactly the common order zero and original source-to-target types.

For $f\in H^{t-jm}$ closed under D_j, its primitive G_(j−1)f has order H^(t−jm+m), exactly the preceding degree’s space. The kernel of the cohomology projection H_j is the range of that primitive, as (EC14) proves. It is closed since D_j and H_j are continuous on the stated spaces. The finite harmonic representatives and both comparisons with smooth sections and distributions consequently have the complete Hausdorff and domain assertions of (EC14a). No ordinary unshifted Sobolev complex was substituted. The full graded operator D+D* has principal symbol square ⊕ℓ_j; therefore each of its even-to-odd and odd-to-even symbols is the actual inverse of the other up to that invertible square. Their finite-fiber ranks agree on positive-dimensional components. Elliptic regularity makes its kernel smooth, and its full energy is the sum of the original $\|D_jf_j\|^2$ and $\|D_{j-1}^*f_j\|^2$. The two kernel dimensions in (EC15) follow, in the actual anti-dual realizations. Rectangular individual D_j receive no extra Fredholm assertion.

11.3. The complete diagonal Fourier restriction

Put d=dim X in this subsection. In one product chart use the actual invertible coordinate map $$u=x,\qquad v=y-x,\qquad \xi\,dx+\eta\,dy=(\xi+\eta)\,du+\eta\,dv. \tag{LC4}$$ Its determinant is one. The diagonal conormal becomes the directions (α,β)=(0,β), β≠0. After compact cutoffs and local frames, let k(u,v) be one coefficient of the original kernel with its two half-density factors transformed by this map. The pullback at v=0 has the following exact testing formula for a smooth compact φ: $$\left\langle k(u,0),\varphi(u)\right\rangle =(2\pi)^{-2d}\int_{\mathbb R^d}\int_{\mathbb R^d} \widehat k(\alpha,\beta)\widehat\varphi(-\alpha) \,d\alpha\,d\beta. \tag{LC5}$$ The entire inverse Fourier constant belongs to the full 2d-dimensional transform. Transversality gives a closed cone around α=0, say |α|≤c|β| above a fixed radius, on which the localized transform of k has arbitrary inverse-power decay. Such a cone and cutoff exist on each sufficiently small diagonal neighborhood by a finite angular cover and the earlier cutoff proof WF2. On this cone choose its decay order greater than 2d+1 to make (LC5) integrable. On its complement |β|≤c^(−1)|α|, a compact distribution has $|\widehat k|\leq C\langle(\alpha,\beta)\rangle^M$. The volume of the permitted β region is bounded by C′⟨α⟩^d. Taking the test’s full Schwartz bound with exponent greater than M+2d+1 gives absolute convergence there as well. Bounded frequencies are harmless.

The same bounds prove a finite-order estimate in φ and continuity when k has a common polynomial bound, weak convergence, and convergence of its rapid conic seminorms near α=0. Indeed a common finite-order estimate gives equicontinuity of the localized Fourier transforms on each bounded frequency ball, by differentiating their original exponential test functions. A finite net and pointwise weak convergence imply uniform convergence on that ball. The two tail bounds just given then make the full integrals converge. Strong distribution seminorms on bounded test families have the same estimate, using a higher common test-derivative bound.

For smooth k, Fourier inversion makes (LC5) the actual restriction. Cutoff independence and coordinate compatibility follow from the already proved qualified pullback WF11–WF14, or directly by inserting a common smooth regularization in the same integrable formula. Thus local constructions agree on overlaps. The two half-density factors restrict to the density |du| on the diagonal, and the original input and output frames now lie over the same point. Their finite matrix trace is invariant under the actual similarity transition. This proves (EC16)–(EC17) as a globally defined distributional density and its compact integral, without assigning a value to an arbitrary singular coefficient at (x,x).

11.4. Full controlled smoothing, with its spatial cutoffs

For a compact coefficient k in q=2d coordinates choose $\rho\in C_c^\infty(\mathbb R^q)$ equal to one near zero, and form $k_\varepsilon=\mathcal F^{-1}[\rho(\varepsilon\zeta)\widehat k(\zeta)]$. It is smooth. The original finite-order estimate bounds its localized transforms by a common polynomial. For a compact smooth cutoff b, the full localized formula is $$\widehat{b k_\varepsilon}(\zeta) =(2\pi)^{-q}\int\widehat b(\zeta-\theta) \rho(\varepsilon\theta)\widehat k(\theta)\,d\theta. \tag{LC6}$$ Take a smaller regular cone W inside a larger one V for k, with the spatial cutoff supported in a neighborhood on which that larger regularity holds. On θ in V, the transform of k decreases to every order. On θ outside V and ζ in W, $|\zeta-\theta|\geq c(|\zeta|+|\theta|)$; the Schwartz decrease of b̂ beats the full polynomial bound of k̂. Since ρ is uniformly bounded, these two integrals bound every weighted rapid seminorm uniformly in ε. The same estimates apply to their differences from k after replacing ρ by ρ−1. On bounded θ the multiplier tends uniformly to zero; on the tail first use a larger decay exponent. Consequently each requested weighted seminorm of b(k_ε−k) tends to zero. If the regularity uses an additional spatial cutoff b_1 equal to one near supp b, split k=b_1k+(1−b_1)k. The second part has a fixed positive spatial separation from b. Its regularizing convolution kernel and every derivative are bounded there by $C_L\varepsilon^{-q-r}(1+\operatorname{distance}/\varepsilon)^{-L}$. Applying the compact distribution’s full derivative bound and taking L larger than q+r and that bound’s order makes this part tend to zero in every C^r seminorm. This supplies the spatial localization step in (EC18), rather than asserting it from a frequency multiplier alone.

Weak convergence follows by Fourier testing against a Schwartz transform and dominated convergence with the common polynomial bound. It is also strong on bounded compact test families: all their Fourier transforms have common Schwartz bounds, so the same bounded-frequency and tail split is uniform for such a family. Use finite chart products, a smooth partition and cutoffs equal to one on each original piece’s support to patch this regularization. Smooth frame multipliers obey (LC6). The earlier exact diffeomorphism covector proof WF5–WF7, with its finite nonstationary estimates, transports the rapid cone bounds under chart changes; all constants on the fixed chart supports are uniform in this family. This proves the global controlled convergence in (EC18).

Here the controlled topology uses the compact distribution seminorms and the localized rapid Fourier seminorms on closed cones disjoint from the original wave front. The displayed estimates prove precisely these seminorm assertions, including the stronger bounded-test assertion just given. No equality of arbitrary topologies merely sharing a sequential convergence rule is assumed.

11.5. Both pseudodifferential compositions and one-sided directions

Localize the actual properly supported A in a product chart. Split its kernel into a piece in an arbitrarily small neighborhood of x=z and a smooth remainder. In left-symbol coordinates the first piece is a finite amplitude reduction of $(2\pi)^{-d}\int e^{i(x-z)\theta}a(x,\theta)\,d\theta$, with the original order r and all proper cutoffs. Its Fourier transfer coefficient â(ξ−ξ′,ξ′), including those cutoffs, obeys, for every L, $$|\widehat a(\xi-\xi',\xi')| \leq C_L\langle\xi'\rangle^{r_+} \langle\xi-\xi'\rangle^{-L}, \qquad r_+=\max(r,0). \tag{LC7}$$ For an amplitude depending also on z, the extra compact smooth cutoff transform is another rapidly decreasing convolution factor; the same bound follows by convolution and choosing its decay order larger. Integration by parts in the compact x coefficient proves (LC7), retaining every differentiated symbol and cutoff. The use of r_+ is an upper bound for the full original order, not a replacement for that symbol.

Output composition has the exact local form $(2\pi)^{-d}\int\widehat a(\xi-\xi',\xi')\widehat k(\xi',\eta)\,d\xi'$, up to the already described cutoff convolutions. Fix a smaller cone about ξ=−η inside a larger regular cone for the original k. In the close region $|\xi-\xi'|\leq c_0|(\xi,\eta)|$, choose c_0 small enough that (ξ′,η) stays in the larger cone. Then |(ξ′,η)| is comparable to |(ξ,η)| and arbitrary rapid decrease of k̂ beats the polynomial factor of â, with L>d+1 controlling the remaining integral. In the far region, $|\xi-\xi'|>c_0|(\xi,\eta)|$. The inequalities $|\xi'|\leq|\xi|+|\xi-\xi'|$ and $|\eta|\leq C|\xi-\xi'|$ show that the full polynomial bounds of both factors are bounded by a fixed power of ⟨ξ−ξ′⟩. Take L larger than that power, the requested output decay, and d+1. The integral then decreases to the requested order in the output frequency. The same proof in η treats input composition. It works with arbitrary derivatives and with the uniform bounds and bounded-frequency convergence of Section 11.4.

For any candidate with two nonzero covectors outside WF(k), choose sufficiently small spatial and angular neighborhoods where its rapid Fourier estimate holds. The near-diagonal support of A can be made small enough to keep the original base variables in those neighborhoods. The preceding argument then proves the two-sided inclusion (EC19). Its smooth remainder has a distinct behavior: differentiating in its output variable leaves smooth parameter-dependent tests of the original distribution, while any remaining singular covector can only be in the input variable. The reversed composition has only output-variable covectors. These one-sided directions do not meet the diagonal conormal, where both covectors are nonzero. This proves preservation of (EC16) and convergence of both composed diagonal traces by (LC5). It does not assert the stronger whole-wavefront inclusion, and the rank-one counterexample in Section 4 and exercise 3 proves exactly its failure.

11.6. Cyclicity with the full smooth-kernel trace estimate

We justify the Fourier expansion used in (EC20) without an unbounded eigenbasis. In one circle coordinate the full Fejér kernel is $$\mathcal F_N(t)=\frac1{N+1}\left|\sum_{l=0}^Ne^{ilt}\right|^2, \qquad \frac1{2\pi}\int_0^{2\pi}\mathcal F_N(t)\,dt=1. \tag{LC8}$$ It is nonnegative. Away from t=0 modulo 2π, the actual geometric sum gives $\mathcal F_N(t)\leq4/[(N+1)|1-e^{it}|^2]$, so its mass there tends to zero. Splitting a continuous periodic function’s convolution into a small neighborhood and its complement proves uniform convergence, by uniform continuity. Applying the same convolution to each derivative proves C^r convergence for smooth functions. Products of these kernels handle every finite cube dimension. Repeated integration by parts gives rapid decrease of smooth Fourier coefficients; their Fourier series and every derivative converge absolutely, and the Fejér convergence identifies this sum with the original function. Testing distributions against that complete test expansion also identifies every distribution from its polynomially bounded Fourier coefficients.

For a smooth bundle kernel B in compact chart products, extend its original coefficient by zero to larger cubes of period 2L in each coordinate. Keep the actual Fourier vectors $e_p(x)=(2L)^{-d/2}e^{i\pi p\cdot x/L}$. Each coefficient is $b_{pq}=\int\int k_B(x,y)\overline{e_p(x)}e_q(y)\,dx\,dy$, so the full reconstruction is $\sum b_{pq}e_p(x)\overline{e_q(y)}$. The two vector factors retain their (2L)^(−d/2) factors. Insert fixed input and output cutoffs equal to one on the old supports. In local frames each summand is the original output section $\chi_x e_p e_a\otimes|dx|^{1/2}$ times the input functional obtained by integrating $\chi_y\overline{e_q}\,e^b(f(y))\otimes|dy|^{1/2}$. Its pairing with the input half-density is a density. The Hermitian duality of Section 11.1 identifies this smooth functional with its actual input vector; no density or frame-dual factor is deleted.

The coefficients satisfy (LC3) by direct integration by parts. The smooth cutoff Fourier vectors have at most polynomial Sobolev norms in their indices. Since A and its actual adjoint are bounded on the original Sobolev pairs, $\|Au_p\|_2$ and $\|A^*v_q\|_2$ also have fixed polynomial bounds. Therefore both sums $\sum|b_{pq}|\|Au_p\|_2\|v_q\|_2$ and $\sum|b_{pq}|\|u_p\|_2\|A^*v_q\|_2$ converge; their complete omitted tails tend to zero. Section 11.1 proves that these are actual trace-norm bounds. Each finite term has $\operatorname{Tr}(Au\langle\cdot,v\rangle)=\langle Au,v\rangle =\langle u,A^*v\rangle =\operatorname{Tr}(u\langle A\cdot,v\rangle)$. Thus the full smooth-kernel traces of AB and BA agree with their actual diagonal integrals, preserving the output and input bundles even when their ranks differ. Finite chart sums give the global statement.