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\begin{document}

\section{From ordered Weyl errors to the exterior index
form}\label{from-ordered-weyl-errors-to-the-exterior-index-form}

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

The scaled index construction leaves one finite expression: the part of
two ordered error powers that contains exactly \(2n\) phase derivatives.
This lesson evaluates that expression directly. Independent coordinate
scales force one derivative in every phase coordinate. Alternation then
cancels every term that differentiates one atomic factor twice. What
remains is the first Poisson bracket in each selected error factor, and
its exact alternating count is the exterior form
\(\operatorname{Tr}[(db\wedge da)^n]\).

Read \href{scaled-weyl-index-degree.pdf}{Scaled Weyl parametrices and the
surviving differential degree} for the finite coefficient and its trace
remainder, \href{metric-operator-bounds.pdf}{When a moving symbol scale
controls an operator} for norm-continuous symbol paths, and
\href{relative-projectors-chern-cutoff.pdf}{Relative Weyl projectors and
the Chern cutoff form} for the matrix exterior identities and cutoff
geometry. We prove the coordinate-scaling step here, so the calculation
does not use the later source-theorem evaluation of the index.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The six exact reductions from the ordered Weyl errors to the exterior index form}]{../figures/ordered_weyl_exterior_reduction.png}}
\caption{The six exact reductions from the ordered Weyl errors to the
exterior index form}
\end{figure}

Scaling selects the one-derivative multi-degree. Alternation supplies
the cancellation, and the exterior count fixes the coefficient.
Equations (1)--(18) prove every reduction in the figure with the
original ordered errors and full coefficients. The reproducible source
is
\href{../figures/ordered_weyl_exterior_reduction.py}{ordered\_weyl\_exterior\_reduction.py}.

\subsection{1. The finite coefficient for every error
power}\label{the-finite-coefficient-for-every-error-power}

Let

\[
 g_z=(1+|z|^2)^{-1}|dz|^2,
 \qquad
 a\in S(1,g;\operatorname{End}\mathbb C^\nu),
 \qquad z=(x,\xi)\in\mathbb R^{2n},
\]

and suppose that \(a(z)\) has a uniformly bounded inverse outside a
compact set. Choose a smooth scalar cutoff \(\psi\) supported in the
invertibility region and equal to one outside a larger compact set. Put

\[
 b=\psi a^{-1},\qquad t=1-\psi,
\]

and retain the two original matrix orders

\[
 r_1=I-b\# a,\qquad r_2=I-a\# b.
 \tag{1}
\]

The matrix products stay in their written order. The two error symbols
belong to \(S(h,g)\), where \(h(z)=(1+|z|^2)^{-1}\). For each integer
\(N>n\), their \(N\)-fold Weyl powers have integrable order \(h^N\), and
the trace-power index identity is

\[
 \operatorname{ind}a^w
   =\operatorname{Tr}\bigl((r_1^{\#N})^w-(r_2^{\#N})^w\bigr).
 \tag{2}
\]

To recall why the finite coefficient exists, scale both phase variables
by \(\varepsilon\). Expanding each ordered product only through degree
\(n\) gives matrix coefficients \(F_{j,k}^{(N)}\), where degree \(k\)
means \(2k\) actual phase differentiations. The omitted symbol lies
uniformly in \(S(h_\varepsilon^{n+1},g_\varepsilon)\); its full trace
norm is \(O(\varepsilon^2)\) by IP22--IP28, with both original factors
and their output domains. The original phase-space integral has the same
order:

\[
 \int_{\mathbb R^{2n}}h_\varepsilon^{n+1}(z)\,dz
   =O(\varepsilon^2).
\]

For \(k\leq n\), every retained word contains \(N\) error factors. If
all factors had positive intrinsic degree, their total degree would be
at least \(N>n\geq k\). Thus one factor is the compactly supported
scalar \(tI_\nu\), and every \(F_{j,k}^{(N)}\) is compactly supported.
Define

\[
 A_k^{(N)}=(2\pi)^{-n}\int_{\mathbb R^{2n}}
  \operatorname{tr}\bigl(F_{1,k}^{(N)}-F_{2,k}^{(N)}\bigr)\,dz
 \tag{3}
\]

The exact scaled trace expansion is

\[
 \operatorname{ind}a^w
   =\sum_{k=0}^{n}A_k^{(N)}\varepsilon^{2k-2n}
      +O(\varepsilon^2).
\]

Its left side is independent of \(\varepsilon>0\). Starting with the
smallest nonzero negative power and multiplying by the power that makes
it constant shows successively that every coefficient below \(n\)
vanishes. Letting \(\varepsilon\) tend to zero then gives

\[
 A_0^{(N)}=\cdots=A_{n-1}^{(N)}=0,
 \qquad
 \operatorname{ind}a^w=A_n^{(N)}.
 \tag{4}
\]

The rest of the lesson evaluates this same \(A_n^{(N)}\) for arbitrary
\(N>n\).

\subsection{2. Independent scales select one derivative per
coordinate}\label{independent-scales-select-one-derivative-per-coordinate}

Expand the degree-\(n\) expression in (3) down to the original factors
\(a,b,t\). Every word contains exactly \(2n\) differentiations. Group
the words by the multi-degree

\[
 \alpha=(\alpha_1,\ldots,\alpha_{2n})\in\mathbb N^{2n},
 \qquad |\alpha|=2n,
\]

where \(\alpha_j\) counts the differentiations with respect to \(z_j\).
Let \(I_\alpha\) be the complete integrated group with that
multi-degree, including every matrix order and numerical factor.

For positive numbers \(\rho_1,\ldots,\rho_{2n}\), set

\[
 D_\rho z=(\rho_1z_1,\ldots,\rho_{2n}z_{2n}),
 \qquad
 a_\rho=a\circ D_\rho,\qquad
 b_\rho=b\circ D_\rho,\qquad
 t_\rho=t\circ D_\rho.
\]

The transformed symbol remains in \(S(1,g)\) and remains uniformly
invertible off a compact set. Join \(D_\rho\) to the identity by a
positive diagonal path. The chain rule gives uniform symbol bounds along
the path, so the corresponding Weyl operators vary continuously in
operator norm. Each is Fredholm by the construction in Section 1;
therefore its index is constant along the path.

For a word of multi-degree \(\alpha\), the chain rule contributes
\(\rho^\alpha\), while the change of variables \(w=D_\rho z\)
contributes \(\prod_j\rho_j^{-1}\). Equations (3)--(4) therefore give
the finite Laurent identity

\[
 \operatorname{ind}a^w
  =(2\pi)^{-n}\sum_{|\alpha|=2n}
       I_\alpha\prod_{j=1}^{2n}\rho_j^{\alpha_j-1}
       \qquad(\rho_j>0).
 \tag{5}
\]

Distinct Laurent monomials are linearly independent. For an explicit
proof, choose an integer vector \(v\) that separates the finitely many
exponents \(\alpha-\mathbf1\), and put \(\rho_j=e^{s v_j}\). Equation
(5) becomes a finite exponential polynomial that is constant for every
real \(s\). Its distinct exponential coefficients vanish. Hence

\[
 I_\alpha=0\quad(\alpha\ne\mathbf1),
 \qquad
 \operatorname{ind}a^w=(2\pi)^{-n}I_{\mathbf1},
 \quad \mathbf1=(1,\ldots,1).
 \tag{6}
\]

This is a statement about each complete multi-degree group. Individual
words inside a group can still cancel one another.

The same result gives the reflection sign without an exterior formula.
Replacing one coordinate \(z_j\) by \(-z_j\) changes the sign of every
surviving word, because that word has exactly one \(z_j\)-derivative.
The absolute Jacobian is one. Thus

\[
 \operatorname{ind}(a\circ R_j)^w=-\operatorname{ind}a^w.
 \tag{7}
\]

An even coordinate permutation lies in the positive-determinant
component and preserves the index. An odd permutation is a
positive-determinant map followed by one reflection. Therefore

\[
 \operatorname{ind}(a\circ P_\tau)^w
   =\operatorname{sgn}(\tau)\operatorname{ind}a^w.
 \tag{8}
\]

No boundary index theorem has entered this argument.

\subsection{3. Alternation creates a sign-reversing
involution}\label{alternation-creates-a-sign-reversing-involution}

Let \(S^1\) be the full matrix differential expression that contributes
to \(I_{\mathbf1}\). Every coordinate appears exactly once. If
\(S^1_\tau\) denotes the expression obtained by permuting all
differentiation labels by \(\tau\in S_{2n}\), equations (6) and (8),
followed by the change of variables back to \(z\), give

\[
 \operatorname{ind}a^w
  ={(2\pi)^{-n}\over(2n)!}
   \int_{\mathbb R^{2n}}\operatorname{tr}S^2(z)\,dz,
 \qquad
 S^2=\sum_{\tau\in S_{2n}}
        \operatorname{sgn}(\tau)S^1_\tau.
 \tag{9}
\]

Fully expand a word and regard its \(2n\) differentiations as labeled
slots assigned to occurrences of the atomic factors \(a,b,t\). Suppose
two slots act on one atomic occurrence. Swap the two coordinate labels
assigned to those slots. Mixed partial derivatives commute, so the
matrix word does not change, while the permutation sign reverses. This
fixed-point-free pairing proves

\[
 \text{every word in which one atomic factor is differentiated twice
 vanishes from }S^2.
 \tag{10}
\]

The argument removes more than visibly repeated derivatives in one
two-factor product. The finite differential coefficients of an
\(N\)-fold Weyl product are generated by

\[
 \left.
 \exp\!\left({1\over2i}
  \sum_{1\le r<s\le N}\sum_{u,v=1}^{2n}
    J^{uv}\partial_{z_{r,u}}\partial_{z_{s,v}}
 \right)
 \prod_{r=1}^{N}f_r(z_r)
 \right|_{z_1=\cdots=z_N=z},
 \tag{11}
\]

read only through the finite degree in question. The constant Poisson
tensor is \(J^{uv}\). Derivatives act entrywise, and the matrix product
remains in the displayed factor order. Iterating the two-factor Weyl
formula proves (11); its constant-coefficient contraction operators
commute, so the result is independent of parenthesization.

An internal coefficient \(C_k(b,a)\) or \(C_k(a,b)\) with \(k\geq2\)
differentiates both atomic factors more than once and disappears by
(10). An internal \(C_1\) differentiates its two atomic factors once
each. Any later contraction that reaches either factor differentiates it
again and also disappears.

Only contractions joining scalar \(t\)-occurrences could remain. Under
(10), their contraction graph is a matching, because every vertex has
degree at most one. Each edge contains

\[
 \{t,t\}=\sum_{j=1}^{n}
  (t_{\xi_j}t_{x_j}-t_{x_j}t_{\xi_j})=0.
\]

The scalar \(t\) commutes with all matrix factors, so this zero factor
can be gathered without changing an order. Every outer contraction
vanishes. This proves that alternation keeps only the undifferentiated
product of first internal Weyl brackets and scalar \(t\)-factors.

\subsection{4. The first brackets are the only
survivors}\label{the-first-brackets-are-the-only-survivors}

Use the fixed convention

\[
 \{b,a\}=\sum_{j=1}^{n}
  (b_{\xi_j}a_{x_j}-b_{x_j}a_{\xi_j}),
 \qquad
 C_1(b,a)={1\over2i}\{b,a\}=-{i\over2}\{b,a\}.
\]

Inside the alternating differential-degree-\(2n\) projection, the two
errors in (1) may therefore be replaced by

\[
 r_1=tI_\nu+{i\over2}\{b,a\},
 \qquad
 r_2=tI_\nu+{i\over2}\{a,b\},
 \tag{12}
\]

Each bracket contributes two atomic differentiations. Exactly \(n\) of
the \(N\) error factors must supply a bracket, and the remaining \(N-n\)
factors supply \(tI_\nu\). Since \(t\) is scalar, all choices of the
\(n\) bracket positions give the same ordered matrix power. Their
complete contribution is

\[
 \binom Nn t^{N-n}\left({i\over2}\right)^n
 \bigl(\{b,a\}^n-\{a,b\}^n\bigr).
 \tag{13}
\]

The two matrix brackets have not been replaced by negatives of one
another. That identity is generally false before the matrix trace.

\subsection{5. Counting the exterior word and fixing the
orientation}\label{counting-the-exterior-word-and-fixing-the-orientation}

First order the coordinates as

\[
 \widetilde\Omega
   =d\xi_1\wedge dx_1\wedge\cdots\wedge
      d\xi_n\wedge dx_n.
\]

A term in \({b,a}^n\) that uses each coordinate once must use every
canonical pair \(j\) once. There are \(n!\) ways to order those pairs
among the \(n\) matrix factors and \(2^n\) ways to choose one of the two
bracket terms in every pair. Thus there are exactly

\[
 2^n n!
 \tag{14}
\]

base words. The sign of each base word is its coordinate-permutation
sign relative to \(\widetilde\Omega\): exchanging the two entries of a
pair contributes the bracket minus sign, while permuting two blocks of
length two is even. For any coordinate word in \((db\wedge da)^n\),
every base word has one unique relabeling to it, and the relabeling sign
times the base sign is the target exterior sign. Hence every exterior
word occurs with the common multiplicity \(2^n n!\).

Applying this count to (13) gives the untraced identity

\[
 S^2\widetilde\Omega
 =n!\,i^n\binom Nn t^{N-n}
  \bigl((db\wedge da)^n-(da\wedge db)^n\bigr).
 \tag{15}
\]

For matrix-valued forms of degrees \(p\) and \(q\), cyclicity of the
finite matrix trace gives

\[
 \operatorname{Tr}(\alpha\wedge\beta)
  =(-1)^{pq}\operatorname{Tr}(\beta\wedge\alpha).
\]

Rotate the first one-form in \((db\wedge da)^n\) past the other \(2n-1\)
one-forms. The sign is \(-1\), so

\[
 \operatorname{Tr}[(da\wedge db)^n]
   =-\operatorname{Tr}[(db\wedge da)^n].
\]

The course orientation is

\[
 \Omega=dx_1\wedge d\xi_1\wedge\cdots\wedge
          dx_n\wedge d\xi_n,
 \qquad
 \widetilde\Omega=(-1)^n\Omega.
\]

Insert (15) into (9). Since \((-1)^ni^n=i^{-n}\), the direct coefficient
is

\[
 \boxed{
 A_n^{(N)}
  =(2\pi)^{-n}{2\,n!\over i^n(2n)!}\binom Nn
   \int_{\mathbb R^{2n}}^{\Omega}
      t^{N-n}\operatorname{Tr}[(db\wedge da)^n]
 }
 \qquad(N>n).
 \tag{16}
\]

The numerator \(2\,n!\) means twice \(n!\); the denominator \((2n)!\) is
the factorial of \(2n\). Choosing \(N=2n\) gives

\[
 \binom{2n}{n}{2\,n!\over(2n)!}={2\over n!},
\]

and therefore

\[
 \operatorname{ind}a^w
  =A_n^{(2n)}
  =(2\pi)^{-n}{2\over i^n n!}
    \int_{\mathbb R^{2n}}^{\Omega}
      (1-\psi)^n\operatorname{Tr}[(db\wedge da)^n].
 \tag{17}
\]

This is the exterior index formula derived directly from the two ordered
error powers.

\subsection{6. The beta moment and the boundary
coefficient}\label{the-beta-moment-and-the-boundary-coefficient}

On the region where \(a\) is invertible, put \(\theta=a^{-1}da\).
Differentiating \(a^{-1}a=I\) gives \(d(a^{-1})=-a^{-1}(da)a^{-1}\), so

\[
 d\theta=-\theta^2,
 \qquad
 db\wedge da=d\psi\wedge\theta-\psi\theta^2.
\]

Graded cyclicity gives \(\operatorname{Tr}(\theta^{2n})=0\), because
rotating one degree-one factor through the other \(2n-1\) factors
changes the sign. It also gives

\[
 d\operatorname{Tr}(\theta^{2n-1})=0.
\]

In the expansion of \((d\psi\wedge\theta-\psi\theta^2)^n\), two copies
containing \(d\psi\) multiply to zero. The term without \(d\psi\) has
trace zero, and the \(n\) terms with one copy agree after the trace.
Therefore

\[
 \operatorname{Tr}[(db\wedge da)^n]
   =n(-1)^{n-1}\psi^{n-1}d\psi\wedge
       \operatorname{Tr}(\theta^{2n-1}).
\]

For an integer \(q\geq0\), define the complete scalar primitive

\[
 G_{n,q}(s)=n\int_0^s u^{n-1}(1-u)^q\,du.
\]

It vanishes at zero, so
\(G_{n,q}(\psi)\operatorname{Tr}(\theta^{2n-1})\) extends smoothly by
zero across the region where \(a\) may fail to be invertible. On a large
outer sphere \(\psi=1\). Stokes' theorem, followed by the closedness of
the odd trace form on the annulus down to the original enclosing sphere
\(\partial B\), gives

\[
 \begin{aligned}
 I_q
 &:=\int_{\mathbb R^{2n}}^{\Omega}
     t^q\operatorname{Tr}[(db\wedge da)^n]\\
 &=(-1)^{n-1}G_{n,q}(1)
    \int_{\partial B}^{\Omega}
       \operatorname{Tr}(\theta^{2n-1})\\
 &=(-1)^{n-1}{n!\,q!\over(n+q)!}
    \int_{\partial B}^{\Omega}
       \operatorname{Tr}(\theta^{2n-1}).
 \end{aligned}
 \tag{18}
\]

For completeness, the full beta factor follows without a table. Write \[
 \Gamma(n)\Gamma(q+1)
 =\int_{s,t>0}s^{n-1}t^q e^{-(s+t)}\,ds\,dt
 =\int_0^1u^{n-1}(1-u)^q\,du
       \int_0^\infty v^{n+q}e^{-v}\,dv.                  \tag{25}
\] The actual change is \(s=uv,t=(1-u)v\), with Jacobian \(v\); all
integrands are nonnegative and both endpoints are retained. The last
integral is \(\Gamma(n+q+1)\). Integration by parts in the original
Gamma integral gives \(\Gamma(n)=(n-1)!\), \(\Gamma(q+1)=q!\), and
\(\Gamma(n+q+1)=(n+q)!\), including \(q=0\). Multiplication by the
original factor \(n\) proves exactly \(G_{n,q}(1)=n!q!/(n+q)!\).

The last equality is the full beta integral \(nB(n,q+1)=n!q!/(n+q)!\).
For \(q=N-n\),

\[
 \binom Nn {n!(N-n)!\over N!}=1.
\]

Thus (16) is independent of the chosen error power after integration.
Substituting (18) gives the boundary coefficient

\[
 \operatorname{ind}a^w
 =-(-2\pi i)^{-n}{(n-1)!\over(2n-1)!}
   \int_{\partial B}^{\Omega}
      \operatorname{Tr}[(a^{-1}da)^{2n-1}].
\]

Every sign comes from a displayed operation: the alternating matrix
trace, the pair-order change from \(\widetilde\Omega\) to \(\Omega\),
and the cutoff primitive.

\subsection{7. Worked checks in one and two phase
planes}\label{worked-checks-in-one-and-two-phase-planes}

For \(n=1\), choose \(N=2\). Equation (16) becomes

\[
 A_1^{(2)}={1\over\pi i}
  \int_{\mathbb R^2}(1-\psi)\operatorname{Tr}(db\wedge da).
 \tag{19}
\]

Equation (18) has
\(I_1=\tfrac12\int_{\partial B}\operatorname{Tr}(a^{-1}da)\), so (19)
gives the winding-number coefficient \(1/(2\pi i)\).

For \(n=2\), first choose the smallest admissible power \(N=3\).
Equation (16) gives

\[
 A_2^{(3)}=-{1\over2}(2\pi)^{-2}
  \int_{\mathbb R^4}(1-\psi)
    \operatorname{Tr}[(db\wedge da)^2].
 \tag{20}
\]

Here (18) gives
\(I_1=-\tfrac13\int_{\partial B}\operatorname{Tr}(\theta^3)\), so the
boundary coefficient is \(1/(24\pi^2)\). If instead \(N=4=2n\), equation
(17) reads

\[
 A_2^{(4)}=-(2\pi)^{-2}
  \int_{\mathbb R^4}(1-\psi)^2
    \operatorname{Tr}[(db\wedge da)^2].
 \tag{21}
\]

Now \(I_2=-\tfrac16\int_{\partial B}\operatorname{Tr}(\theta^3)\),
giving the same \(1/(24\pi^2)\). The densities in (20) and (21) have
different cutoff powers and are generally different at a point. Their
integrated values agree by the exact beta moment.

\subsection{8. Exercises with solutions}\label{exercises-with-solutions}

\textbf{Exercise 1.} Starting from (16), prove in one line that the
choice \(N=2n\) gives the coefficient in (17).

\textbf{Solution.} Use the complete central binomial coefficient:

\[
 {2\,n!\over(2n)!}\binom{2n}{n}
 ={2\,n!\over(2n)!}{(2n)!\over(n!)^2}
 ={2\over n!}.
 \tag{22}
\]

The cutoff power is \(t^{2n-n}=t^n\), so every factor in (17) follows.

\textbf{Exercise 2.} Let a fully expanded alternating word have two
derivative slots assigned to the same atomic factor. Prove that its
complete alternating sum is zero even when the other factors are
matrices.

\textbf{Solution.} Pair each coordinate-label permutation with the
permutation obtained by transposing the labels in those two slots. The
two mixed partial derivatives on the common factor commute. All other
derivatives and the order of every matrix factor remain fixed. The
transposition reverses the permutation sign, so each pair sums to zero.
No matrix commutation is used.

\textbf{Exercise 3.} Why can an outer Weyl contraction between two
scalar \(t\)-factors not survive after the repeated-derivative terms
have been removed?

\textbf{Solution.} The surviving contraction graph has degree at most
one at every scalar occurrence, so it is a matching. Every edge is the
Poisson contraction

\[
 \{t,t\}=\sum_j(t_{\xi_j}t_{x_j}-t_{x_j}t_{\xi_j})=0.
 \tag{23}
\]

Because \(t\) is scalar, the zero edge factor can be moved next to the
remaining factors without changing their matrix order.

\textbf{Exercise 4.} For \(n=3\) and \(N=6\), simplify the scalar
coefficient in (16) before the integral.

\textbf{Solution.} Since \(N=2n\), equation (17) applies. As \(i^3=-i\),

\[
 (2\pi)^{-3}{2\over i^3\,3!}
  ={i\over3(2\pi)^3}.
 \tag{24}
\]

The integral retains the full factor
\((1-\psi)^3\operatorname{Tr}[(db\wedge da)^3]\) and the original
six-dimensional orientation.

\subsection{9. References}\label{references}

\begin{itemize}
\tightlist
\item
  The preceding course lessons give the full trace-class remainder,
  metric operator continuity, relative projectors and cutoff exterior
  identities used here.
\end{itemize}

\subsection{10. Complete direct proof of the finite-to-exterior
receiving
map}\label{complete-direct-proof-of-the-finite-to-exterior-receiving-map}

This supplement supplies every finite support, continuity, permutation,
atomic-slot and endpoint step used above. It keeps the original
\(a,b,\psi,t\), both matrix orders, every factor in (1)--(25), and the
original interleaved orientation. The earlier inputs are the complete
finite product and remainder proofs (DE1)--(DE11), (RP1)--(RP5),
(RA1)--(RA10), (OC1)--(OC31) in \href{scaled-weyl-index-degree.pdf}{the
scaled Weyl lesson}, the Hilbert bound (B26) in
\href{metric-operator-bounds.pdf}{metric operator bounds}, the
finite-defect index perturbation proof (F6) in
\href{fredholm-stability.pdf}{Fredholm stability}, and the actual trace
and powers-of-errors identities in \href{weyl-trace-criterion.pdf}{Weyl
traces} and \href{traces-and-complexes.pdf}{traces and complexes}. The
complete graded trace and Euclidean ball integration proofs (ES1)--(ES7)
are in \href{relative-projectors-chern-cutoff.pdf}{the relative-projector
lesson}. Only these earlier programme proofs and the direct derivation
below are inputs. No exterior index theorem or later formal-to-classical
assertion is used.

\subsubsection{10.1. The exact finite coefficient, support and remainder
for every
N}\label{the-exact-finite-coefficient-support-and-remainder-for-every-n}

Write \(d=2n\), \(z_{2j-1}=x_j,z_{2j}=\xi_j\). Set the actual constant
tensor \[
J^{2j,2j-1}=1,\qquad J^{2j-1,2j}=-1,
\qquad J^{uv}=0\text{ for the other entries}.
\tag{OE1}
\] The binary coefficient and its full coordinate form are \[
\begin{split}
C_k(f,g)
&=\frac1{k!(2i)^k}
  \sum_{u_1,v_1,\ldots,u_k,v_k=1}^{d}
  \left(\prod_{\ell=1}^kJ^{u_\ell v_\ell}\right)
  (\partial_{u_1}\cdots\partial_{u_k}f)
  (\partial_{v_1}\cdots\partial_{v_k}g)\\
&=\left(\frac i2\right)^k
  \sum_{|\alpha|+|\beta|=k}
    \frac{(-1)^{|\beta|}}{\alpha!\beta!}
       (\partial_x^\alpha\partial_\xi^\beta f)
       (\partial_x^\beta\partial_\xi^\alpha g).
\end{split}
\tag{OE2}
\] The second equality is the finite multinomial expansion: each
\(x\)-left, \(\xi\)-right choice has coefficient \(-1/(2i)=i/2\), and
each opposite choice has coefficient \(1/(2i)=-i/2\). Dividing the
multinomial coefficient by \(k!\) retains precisely the two factorials.
The first matrix argument always remains on the left.

Put \(c_{1,0}=c_{2,0}=tI_\nu\), \(c_{1,k}=-C_k(b,a)\),
\(c_{2,k}=-C_k(a,b)\) for \(k\geq1\). Use an auxiliary variable
\(\lambda\) only for finite coefficient extraction. In \(N\) independent
phase slots let \[
L_{rs}=\frac1{2i}\sum_{u,v=1}^{d}
                 J^{uv}\partial_{z_{r,u}}\partial_{z_{s,v}},
\qquad
Q_N=\sum_{1\leq r<s\leq N}L_{rs}.
\tag{OE3}
\] Every derivative operator in this finite sum has scalar constant
coefficients and commutes with the others. The degree-\(k\) coefficient
is exactly \[
F_{j,k}^{(N)}
 =\sum_{\substack{i_1,\ldots,i_N\geq0,\ m\geq0\\
                         i_1+\cdots+i_N+m=k}}
     \frac1{m!}
       \left[Q_N^m\prod_{r=1}^{N}c_{j,i_r}(z_r)
                               \right]_{z_1=\cdots=z_N=z}.
\tag{OE4}
\] This is a finite sum. It is the actual (OC17) coefficient: applying
the diagonal chain rule to a product of its first \(N-1\) slots replaces
each outer left derivative by the sum of the derivatives in those slots.
The new contraction is therefore \(\sum_{r<N}L_{rN}\). The finite
binomial rule combines it with \(Q_{N-1}\) to give \(Q_N\), preserving
\(1/m!\) and each matrix slot. Induction starting at the original binary
(OE2) proves (OE4) and (11). This argument is the complete finite one,
and does not assert convergence of an infinite exponential.

Expand every derivative in (OE4) by the finite Leibniz rule down to
\(a,b,t\). A slot with \(i_r>0\) has two ordered atomic occurrences and
already has exactly \(i_r\) derivatives on each. A slot with \(i_r=0\)
has the one scalar atomic occurrence \(t\). Every contraction
contributes two additional atomic derivatives. Consequently each
degree-\(k\) word has exactly \(2k\) labeled derivative slots. If
\(k<N\), at least one intrinsic degree \(i_r\) is zero; every derivative
of that \(t\)-occurrence is supported in the original compact
\(K=\operatorname{supp}(1-\psi)\). Every individual expanded word is
therefore compactly supported in \(K\). In particular all degree-\(n\)
integrals and all their individual multi-degree groups are absolutely
defined whenever \(N>n\).

For every such fixed integer \(N\), (DE5) at truncation \(M=n+1\) gives
\[
r_{j,\varepsilon}^{\#N}
 =\sum_{k=0}^{n}\varepsilon^{2k}
                F_{j,k}^{(N)}(\varepsilon z)
       +E_{j,\varepsilon}^{(N,n+1)},\qquad
E_{j,\varepsilon}^{(N,n+1)}
       \in S(h_\varepsilon^{n+1},g_\varepsilon)
                         \text{ uniformly}.
\tag{OE5}
\] Its proof permits every \(N\geq1\), rather than only \(n+1\). At each
product the full factor is \((1+h_\varepsilon/4)^{4n}\), with
\(1\leq(1+h_\varepsilon/4)^{4n}\leq(1+\varepsilon^2/4)^{4n}\). The
provider's full finite-seminorm chain, including \(4^{-L}\), the
diagonal factors \(2^{k/2}\) and \(2^J\), is retained in this receiver.
A high intrinsic-degree product has its whole weight
\(h_\varepsilon^{i+j}\) and enters the remainder with ratio
\(h_\varepsilon^{i+j-(n+1)}
\leq\varepsilon^{2(i+j-n-1)}\) when \(i+j\geq n+1\); it is not discarded
or assigned a negative truncation order. The compact zeroth factor is
uniformly order one for the scaled symbol construction and is never
declared uniformly order \(h_\varepsilon\).

The actual trace-norm bound (RA1)--(RA10) gives
\(\|(E_{j,\varepsilon}^{(N,n+1)})^w\|_1\leq C_N\varepsilon^2\), with its
proved original multiplier domains and Hilbert factors. The separate
symbol integral is exactly \[
\int h_\varepsilon^{n+1}\,dz
 =\varepsilon^2\pi^n
                    \frac{\Gamma(1)}{\Gamma(n+1)}.
\tag{OE6}
\] Every compact coefficient has the complete Weyl trace factor
\((2\pi)^{-n}\), and the change \(w=\varepsilon z\) has Jacobian
\(\varepsilon^{-2n}\). The complete power-index proof (IP14), valid for
each \(N\geq n+1\), therefore gives (3)--(4) for every integer \(N>n\).
For the vanishing induction, subtract all already vanishing lower terms,
multiply by \(\varepsilon^{2n-2k}\) at a proposed least nonzero \(k<n\),
and pass to zero; its sole possible constant is \(A_k^{(N)}\), while the
fixed index and every later term tend to zero. It must vanish. Once
these terms are zero, passage to zero gives
\(A_n^{(N)}=\operatorname{ind}a^w\). This proves the complete starting
receiver, with no trace-norm inference from a symbol integral alone.

\subsubsection{10.2. Actual coordinate paths, Laurent independence and
orientation}\label{actual-coordinate-paths-laurent-independence-and-orientation}

Here is the operator-norm path estimate used in Section 2. Let \(L_s\)
be any smooth matrix path on a compact real interval, with \(L_s\)
invertible there. Suppose \(\|L_s\|\), \(\|L_s^{-1}\|\), and
\(\|L_s'\|\) have finite common bounds. Such bounds hold for the
specific positive diagonal paths and coordinate plane rotations used
below. For all \(z\), \[
c\langle z\rangle\leq\langle L_sz\rangle
             \leq C\langle z\rangle,\qquad
|\partial_z^\alpha(a\circ L_s)(z)|
             \leq C_\alpha\langle z\rangle^{-|\alpha|}.
\tag{OE7}
\] The bracket comparisons follow from the bounds for \(L_s\) and its
inverse. The derivative is the full finite tensor contraction of
\((d^{|\alpha|}a)(L_sz)\) with the corresponding original columns of
\(L_s\); no column is replaced in the symbol.

Its parameter derivative has two types of terms. Differentiating a
column replaces that one column by its \(L_s'\) column, keeping the
order and the other columns. Differentiating the symbol derivative
contributes \((d^{|\alpha|+1}a)(L_sz)\) applied also to \(L_s'z\). This
second term has one extra inverse bracket, and
\(|L_s'z|\leq C|z|\leq C'\langle L_sz\rangle\). Thus every term has the
same full \(\langle z\rangle^{-|\alpha|}\) bound, uniformly on the
interval. The fundamental theorem of calculus gives a Lipschitz bound in
every original isotropic symbol seminorm. The actual Hilbert bound (B26)
gives operator-norm continuity. Exterior invertibility is preserved: if
the original inverse bound holds outside radius \(R\), it holds for
\(a(L_sz)\) outside the common radius \(R\sup_s\|L_s^{-1}\|\). The
finite error construction proves Fredholmness of every path operator.
The proved (F6) then makes its index constant on the interval.

For \(L_s=\operatorname{diag}(1+s(\rho_j-1))\), \(0\leq s\leq1\), every
diagonal entry is positive and bounded away from zero. The estimate
proves the exact positive-coordinate index constancy in (5), with no
symplectic restriction on this comparison map.

Let \(S=F_{1,n}^{(N)}-F_{2,n}^{(N)}\) and write its complete expansion
as \(S=\sum_\alpha S_\alpha\), where the original derivative label
counts are \(\alpha\). Let
\(I_\alpha=\int\operatorname{tr}S_\alpha\,dz\); all numerical factors
and matrix orders are in these sums. For each expanded word, the
ordinary chain rule and the positive diagonal Jacobian give exactly \[
\int\operatorname{tr}S_\alpha(a_\rho,b_\rho,t_\rho)\,dz
 =\left(\prod_{j=1}^{d}\rho_j^{\alpha_j-1}\right)I_\alpha.
\tag{OE8}
\] This proves (5) from the actual degree-\(n\) receiver.

For completeness, include the exponent \(0\) in the finite set of all
\(\alpha-\mathbf1\). Choose \(M=4n+2\) and \(v_j=M^{j-1}\). Differences
of two such exponents have coordinates of absolute value at most \(2n\).
If a difference is nonzero, its highest nonzero coordinate has absolute
value at least one, so its highest power exceeds the sum of all possible
lower powers: \(M^r>2n\sum_{j<r}M^j\). Thus the integer dot products
with \(v\) are distinct, including the zero exponent. Set
\(\rho_j=e^{sv_j}\) and move the constant left side of (5) into its zero
exponential coefficient. If \(\sum_\ell c_\ell e^{\mu_\ell s}=0\) for
these distinct real \(\mu_\ell\), differentiation through orders
\(0,\ldots,\ell_{\max}\) at zero gives \(\sum_\ell c_\ell\mu_\ell^k=0\).
For each \(\mu_j\), the finite polynomial \[
p_j(u)=\prod_{\ell\ne j}
                 \frac{u-\mu_\ell}{\mu_j-\mu_\ell}
\tag{OE9}
\] has \(p_j(\mu_\ell)=\delta_{j\ell}\). Taking its full finite linear
combination of the preceding derivative equalities gives \(c_j=0\). This
proves (6), including the constant coefficient, without allowing a
nonzero exponent accidentally to have zero dot product.

Apply this result to the reflected triple
\(a\circ R_j,b\circ R_j,t\circ R_j\). Its selected words each have
exactly one derivative in that reflected coordinate. The derivative
contributes \(-1\) and the absolute Jacobian is one. Hence its selected
integral is \(-I_{\mathbf1}\); the result applied to that triple proves
(7).

We also give an actual path for every even coordinate permutation,
rather than assuming a connectedness theorem. Every real orthogonal
matrix of determinant one is a product of coordinate plane rotations. To
prove this by induction on dimension, act on its first column by
rotations in the \((1,j)\) planes, from \(j=d\) down to \(2\). If both
entries at a step are zero, skip the step. Otherwise choose the rotation
with cosine \(v_1/(v_1^2+v_j^2)^{1/2}\) and the appropriate sine
\(v_j/(v_1^2+v_j^2)^{1/2}\) so that their resulting entries are
\(((v_1^2+v_j^2)^{1/2},0)\). The final first column is \(e_1\).
Orthogonality makes its first row \(e_1^{\mathsf T}\), leaving an
orthogonal determinant-one matrix in the remaining original coordinates.
Induction decomposes that matrix too; in dimension one it is the
identity. Reverse these exact rotations to obtain the original matrix.
Each rotation of angle \(\vartheta\) has the smooth path of angles
\(s\vartheta\). Their finite concatenation has the bounds used in (OE7).
An even permutation has determinant one, so its index is unchanged along
this path. An odd permutation is a single reflection composed with an
even orthogonal permutation, and (7) applied to the transformed triple
supplies its negative sign. This proves (8).

If \(P_\tau e_j=e_{\tau(j)}\), the chain rule says that the selected
expression for the triple composed with \(P_\tau\) is
\(S^1_\tau(P_\tau z)\), where \(S^1=S_{\mathbf1}\). The absolute
permutation Jacobian is one. Equations (6) and (8) therefore give \[
\int\operatorname{tr}S^1_\tau\,dz
  =(2\pi)^n\operatorname{sgn}(\tau)\operatorname{ind}a^w.
\tag{OE10}
\] Multiply by the same permutation sign and sum all \(d!\)
permutations. This proves (9) with its precise divisor, while the
individual expanded words and their support are retained.

\subsubsection{10.3. Complete atomic pairing and every outer
contraction}\label{complete-atomic-pairing-and-every-outer-contraction}

Fix one fully expanded monomial of \(S^1\), retaining its scalar
coefficient, its ordered matrix occurrences and its \(d\) labeled
derivative slots. The labels form a bijection with the original
coordinate list. Under full alternation, its contribution is the finite
sum over all bijections \(\tau\), with coefficient
\(\operatorname{sgn}\tau\). If two slots belong to the same atomic
occurrence, compose each \(\tau\) with the transposition of the labels
in exactly those two slots. The two labels are different, so the
permutation changes and the pairing has no fixed point. The scalar
numerical coefficient is unchanged. On that occurrence, the two
derivatives commute, including every other derivative on it. All other
occurrences, their derivatives and their matrix order are unchanged. The
two evaluated words are therefore identical with opposite signs. This
proves (10) for each complete atomic monomial, not only after its
integral or matrix trace.

If any intrinsic degree \(i_r\geq2\), each of its original \(a,b\)
occurrences already has at least two derivative slots and its entire
alternating contribution is zero. A degree-one slot has exactly its two
once-differentiated \(a,b\) occurrences. Every outer contraction hitting
either of them supplies another derivative slot and is zero by the same
pairing. Thus a surviving outer edge can touch only
intrinsic-degree-zero slots, whose atomic functions are the same scalar
\(t\). No such slot can be touched by two edges, because that would
differentiate its atomic occurrence twice. The surviving outer graph is
a matching.

For a selected edge of this matching, its two differentiated factors are
\(\partial_ut\) and \(\partial_vt\), possibly separated by matrix
factors. They are scalar and can be gathered without reordering any
matrix factor. In the complete alternating sum, transposing the two
labels assigned to these edge slots changes the sign and leaves the
product \((\partial_ut)(\partial_vt)\) unchanged. No other slot receives
either of these labels, because the original multi-degree is
\(\mathbf1\). This is again a fixed-point-free cancellation, and it
proves that every such edge disappears. It is also the finite tensor
version of the exact identity \(\{t,t\}=0\), with both terms of each
original pair retained. This proof respects the one-coordinate
projection: it never gathers an incomplete Poisson sum and declares it
zero.

Consequently every surviving term has no outer contractions, exactly
\(n\) intrinsic-degree-one slots, and \(N-n\) undifferentiated
\(tI_\nu\) slots. Each selected positive slot is
\(-C_1(b,a)=(i/2)\{b,a\}\) or \(-C_1(a,b)=(i/2)\{a,b\}\), respectively.
There are exactly \(\binom Nn\) choices of positions. The scalar \(t\)
commutes through them, while the bracket matrices stay in their original
ordered powers. This proves the complete projected word (13), before the
exterior count.

\subsubsection{10.4. The bijective exterior count and all
constants}\label{the-bijective-exterior-count-and-all-constants}

Write \(B=db\), \(A=da\) only to denote their unchanged matrix one-forms
in this paragraph. In a base word of \(\{b,a\}^n\) with one derivative
in each coordinate, every canonical pair must occur exactly once: a
bracket supplies both coordinates of a pair, and repetition of that pair
would repeat its coordinates. There are \(n!\) choices of their ordered
positions and \(2^n\) choices of the two terms in those positions. These
choices give distinct coordinate label strings and preserve the matrix
slot sequence \(b,a,\ldots,b,a\).

Let \(\ell=(\xi_1,x_1,\ldots,\xi_n,x_n)\) be the original label order
for \(\widetilde\Omega\). A base word string \(w\) has sign
\(\epsilon(w)\) given by its bracket choices. Reordering two length-two
blocks has even sign, while reversing the two entries of one block has
sign minus one. Hence \(\epsilon(w)\) is exactly the permutation sign of
its coordinate string relative to \(\ell\).

For each target string \(u\) that lists all \(d\) coordinate labels, and
each base string \(w\), there is exactly one coordinate relabeling
\(\tau\) with \(\tau(w)=u\). Its sign satisfies
\(\operatorname{sgn}\tau\,\epsilon(w)=\epsilon(u)\). The resulting
matrix word is the target coefficient
\(b_{u_1}a_{u_2}\cdots b_{u_{d-1}}a_{u_d}\), with no interchange of
slots. The exterior expansion of \((db\wedge da)^n\) has exactly these
target words with signs \(\epsilon(u)\). Thus each target occurs once
per base string, with the exact total multiplicity \(2^n n!\). This
proves the pointwise finite identity \[
\operatorname{Alt}\Pi_{\mathbf1}(\{b,a\}^n)
                         \,\widetilde\Omega
      =2^n n!(db\wedge da)^n.
\tag{OE11}
\] The operators \(\Pi_{\mathbf1}\) and \(\operatorname{Alt}\) here mean
the original derivative-label projection and finite signed relabeling
sum, not a new quantization. The identical bijection with the matrix
sequence \(a,b,\ldots,a,b\) gives \(2^n n!(da\wedge db)^n\). Multiplying
by the retained \((i/2)^n\binom Nn t^{N-n}\) gives (15) exactly.

For matrix-valued forms, the finite entry sums prove
\(\operatorname{tr}(\alpha\wedge\beta)
=(-1)^{pq}\operatorname{tr}(\beta\wedge\alpha)\) in degrees \(p,q\).
Rotate just the first one-form of \((db\wedge da)^n\) through the
remaining \(2n-1\). Its trace has sign minus one and becomes the trace
of \((da\wedge db)^n\). The two matrix words were never equated before
this trace. Thus (15) traced is \[
\operatorname{tr}S^2\,\widetilde\Omega
   =2\,n!\,i^n\binom Nn t^{N-n}
                    \operatorname{tr}[(db\wedge da)^n].
\tag{OE12}
\] The measure in (9) is the positive coordinate measure in the retained
orientation \(\Omega\). Since \(\widetilde\Omega=(-1)^n\Omega\), solving
(OE12) for the top form \(\operatorname{tr}S^2\,\Omega\) gives the
additional \((-1)^n\), and \((-1)^ni^n=i^{-n}\). Inserting its full
result in (9) proves (16), with all of \((2\pi)^{-n}\), \(2\,n!\),
\(i^{-n}\), \((2n)!^{-1}\), \(\binom Nn\) and \(t^{N-n}\) still
explicit. The choice \(N=2n\) and the finite central-binomial
cancellation in (22) prove (17), without changing any original word or
cutoff factor.

\subsubsection{10.5. The full primitive, zero extension and boundary
coefficient}\label{the-full-primitive-zero-extension-and-boundary-coefficient}

Let \(U_a\) be the actual open invertibility region. The derivative of
\(a^{-1}a=I_\nu\) proves \(d(a^{-1})=-a^{-1}(da)a^{-1}\). Hence the
original \(\theta=a^{-1}da\) obeys \[
d\theta=-\theta^2,\qquad
d(\theta^r)=-\sum_{j=0}^{r-1}(-1)^j\theta^{r+1}.
\tag{OE13}
\] This follows from the full graded Leibniz rule; every term keeps its
original ordered factors. Graded trace cyclicity makes
\(\operatorname{tr}\theta^{2n}=0\). The full finite signed sum with
\(r=2n-1\) therefore gives \(d\,\operatorname{tr}\theta^{2n-1}=0\).

The original inverse and cutoff give
\(db\wedge da=d\psi\wedge\theta-\psi\theta^2\). Expand its \(n\)-th
power as an ordered sum of all choices of these two two-form factors.
Every word containing two copies of \(d\psi\) is zero: move the scalar
second copy past the intervening one-forms with their exterior signs
until it meets the first, whose wedge square is zero. The word with none
has trace \((-\psi)^n\operatorname{tr}\theta^{2n}=0\). There are \(n\)
words with one copy. Their two-form blocks can be cyclically moved
through the finite trace with sign plus one, so each trace is
\((-1)^{n-1}\psi^{n-1}d\psi\wedge\operatorname{tr}\theta^{2n-1}\). This
proves the complete trace density used in Section 6.

For every integer \(q\geq0\), retain the actual finite scalar polynomial
\[
G_{n,q}(s)
 =n\sum_{\ell=0}^{q}(-1)^\ell\binom q\ell
                        \frac{s^{n+\ell}}{n+\ell},
\qquad
G_{n,q}'(s)=ns^{n-1}(1-s)^q.
\tag{OE14}
\] It is exactly the primitive already defined in Section 6, including
both endpoints, and does not require a monotone or radial cutoff. On
\(U_a\), \[
t^q\operatorname{tr}[(db\wedge da)^n]
 =(-1)^{n-1}
         d\!\left(G_{n,q}(\psi)\operatorname{tr}\theta^{2n-1}\right).
\tag{OE15}
\] Its zero extension is smooth. Indeed \(\operatorname{supp}\psi\) is a
closed subset of \(U_a\); every point outside \(U_a\) has a neighborhood
where \(\psi=0\), hence \(b=0\), \(db=0\), and \(G_{n,q}(\psi)=0\). Both
sides are zero on that neighborhood. No inverse at that point is used.
Outside \(K\), \(\psi=1\), \(d\psi=0\), and the traced top form is zero.
The derivative of the global primitive is therefore compactly supported
even though the primitive itself need not be.

Choose an actual ball whose interior contains \(K\). The proved
Euclidean theorem (ES5)--(ES7) integrates this smooth original primitive
in the interleaved orientation and outward boundary orientation. Its
scalar boundary value is \(G_{n,q}(1)\). There is no inner boundary. If
two containing balls are chosen, its derivative has the same support in
the smaller ball; integrating the same primitive over each proves
equality of its two boundary values. Since \(G_{n,q}(1)\ne0\), the
original odd boundary integrals are equal as well.

The full value of the scalar moment has an elementary proof: for
\(p,q'>0\) integers, integration by parts with \(u^p/p\) gives
\(\int_0^1u^{p-1}(1-u)^{q'-1}du
=(q'-1)/p\int_0^1u^p(1-u)^{q'-2}du\) when \(q'>1\); both boundary terms
are zero. At \(q'=1\) the value is \(1/p\). Repeating and retaining
every factor proves \[
G_{n,q}(1)=n\frac{(n-1)!\,q!}{(n+q)!}
                       =\frac{n!\,q!}{(n+q)!}>0 .
\tag{OE16}
\] This is the same complete moment as the actual two-variable Gamma
change in (25); its Jacobian and endpoints were retained there.
Equations (OE15)--(OE16) now prove (18). For \(q=N-n\), its full
multiplier satisfies \(\binom Nn n!(N-n)!/N!=1\). The coefficient in
(16) therefore becomes, without dropping any contribution, \[
(2\pi)^{-n}\frac{2\,n!}{i^n(2n)!}(-1)^{n-1}
 =(2\pi)^{-n}\frac{(-1)^{n-1}(n-1)!}
                         {i^n(2n-1)!}
 =-(-2\pi i)^{-n}\frac{(n-1)!}{(2n-1)!}.
\tag{OE17}
\] This proves the all-dimensional boundary index form directly from the
finite ordered coefficient.

The numerical checks and solutions above now receive the full proof. At
\(n=1,N=2\), the unreduced coefficient in (16) is \((2\pi)^{-1}\,2/i\),
and the moment is \(1/2\), giving \(1/(2\pi i)\). At \(n=2,N=3\), it is
\(-(2\pi)^{-2}/2\), while (18)'s sign and moment are \(-1/3\); the
product is \(1/(24\pi^2)\). At \(n=2,N=4\), the full coefficient is
\(-(2\pi)^{-2}\), the signed moment is \(-1/6\), and the same boundary
value follows. At \(n=3,N=6\), the unreduced central-binomial expression
in (16) gives precisely (24). The original distinct cutoff densities are
not identified pointwise.

\subsection{11. Further exact consequences and their defect
maps}\label{further-exact-consequences-and-their-defect-maps}

\subsubsection{11.1. The algebraic statement does not require
ellipticity}\label{the-algebraic-statement-does-not-require-ellipticity}

The atomic proof in Sections 10.3--10.4 used only smooth square matrix
functions \(a,b\), a smooth scalar \(t\), and the finite coefficients
with \(c_{j,0}=tI_\nu\), \(c_{1,k}=-C_k(b,a)\), \(c_{2,k}=-C_k(a,b)\)
for \(k\geq1\). It did not use \(ab=ba=(1-t)I_\nu\), a symbol metric or
any inverse. For these independently given smooth functions, define
\(F_{j,n}^{(N)}\) by the same complete finite formula (OE4).

For every \(N\geq n\), its full coordinate-one alternation therefore
satisfies exactly the untraced identity (15), with \(t^{N-n}\). Every
step of the proof is the same finite slot pairing and target-word
bijection, so no analytic hypothesis is being assumed in place of a
calculation. For \(N<n\), after the proved cancellations a surviving
word would require \(n\) intrinsic-degree-one slots among only \(N\)
slots, which is impossible. Its whole alternation is zero. This
strengthens the pointwise finite identity; it does not assert the
analytic trace-power formula at \(N=n\) or below, where the original
trace-class hypothesis \(N>n\) remains necessary.

\begin{figure}
\centering
\pandocbounded{\includegraphics[keepaspectratio,alt={The full atomic cancellation and the eight exact base words in two phase planes}]{../figures/ordered-exterior-atomic-count.png}}
\caption{The full atomic cancellation and the eight exact base words in
two phase planes}
\end{figure}

The example panel retains \(n=2,N=3\), every original label and the
matrix occurrence order \(b,a,b,a\). There are eight base strings. For a
fixed target exterior string, each has exactly one relabeling; its sign
times the base sign is the same exterior sign. The complete proof is
(OE11)--(OE12), rather than a numerical sample of matrices. The
cancellation panel depicts the exact finite pairings of Section 10.3.

\subsubsection{11.2. Every error power and cutoff has an explicit
compact
comparison}\label{every-error-power-and-cutoff-has-an-explicit-compact-comparison}

For any admissible cutoff \(\psi\) and integer \(N>n\), retain the full
original top form, including its numerical coefficient, \[
\kappa_{N,\psi}
 =(2\pi)^{-n}\frac{2\,n!}{i^n(2n)!}
       \binom Nn(1-\psi)^{N-n}
                         \operatorname{tr}[(db_\psi\wedge da)^n],
\qquad b_\psi=\psi a^{-1}.
\tag{OE18}
\] For two such powers \(N,M\) and two such cutoffs \(\psi_1,\psi_0\)
for the same original \(a\), define on \(U_a\) \[
\begin{split}
\Xi_{N,1;M,0}
  ={}&(2\pi)^{-n}\frac{2\,n!}{i^n(2n)!}(-1)^{n-1}\\
    &\quad\times
      \left[\binom NnG_{n,N-n}(\psi_1)
            -\binom MnG_{n,M-n}(\psi_0)\right]
                         \operatorname{tr}\theta^{2n-1}.
\end{split}
\tag{OE19}
\] Every scalar difference is zero near the noninvertible region, since
both scalar arguments there are zero. Outside the union of their
original compact supports, both arguments are one and the full beta
factors in (OE16) give
\(\binom NnG_{n,N-n}(1)=\binom MnG_{n,M-n}(1)=1\). Its zero extension is
therefore globally smooth and compact. Differentiating its full original
coefficient and using the closed odd trace gives the exact pointwise
defect map \[
\kappa_{N,\psi_1}-\kappa_{M,\psi_0}
                         =d\Xi_{N,1;M,0}.
\tag{OE20}
\] The space \(\Omega_c^{2n}(\mathbb R^{2n})/
                         d\Omega_c^{2n-1}(\mathbb R^{2n})\) receives
their equal classes. The full-space integral is a well-defined linear
map on this quotient: each compact primitive's derivative integrates to
zero by the coordinate fundamental theorem and Fubini in (ES5). Thus the
different original error powers and cutoffs have equal integrated index
values through a specified exact differential, without identifying their
densities at a point or discarding their cutoff contributions.

\subsubsection{11.3. The actual bridge to the earlier relative
projector}\label{the-actual-bridge-to-the-earlier-relative-projector}

The preceding direct calculation also closes a receiving map that must
not be assumed merely from a later assertion. Keep the exact original
relative projector \(P\) of (CP1) in
\href{relative-projectors-chern-cutoff.pdf}{the earlier
relative-projector lesson}. Its already proved classical identity (CP11)
has the compact primitive with both endpoint values of \(W_n\), and
(CP12) gives the exact equality \[
\int\operatorname{tr}_{2\nu}[P(dP)^{2n}]
      =2\int(1-\psi)^n
                         \operatorname{tr}_{\nu}[(db\wedge da)^n].
\tag{OE21}
\] Composing only this earlier classical equality with the direct
analytic formula (17) gives \[
\operatorname{ind}a^w
 =(2\pi)^{-n}\frac1{i^n n!}
        \int\operatorname{tr}_{2\nu}[P(dP)^{2n}].
\tag{OE22}
\] Every rank, original block and orientation is that of (CP1); the
local defect before integration remains exactly (CP11).

The earlier independently proved formal trace identity (FC10) supplies,
in its actual compact coefficientwise ideal, \(\tau_{2\nu}(e_\infty-e_0)
=(2\pi)^n\lambda^n\operatorname{ind}a^w\). Its proof uses the actual
finite analytic remainder and every higher-coefficient vanishing
induction; it does not assume a classical Chern identification.
Substitute the now proved (OE22) and retain its exact parameter map
\(\lambda=i\hbar\). This gives the complete formal/classical receiving
identity \[
\begin{split}
\tau_{2\nu}(e_\infty-e_0)
 &=\frac{\lambda^n}{i^n n!}
           \int\operatorname{tr}_{2\nu}[P(dP)^{2n}]\\
 &=\frac{\hbar^n}{n!}
           \int\operatorname{tr}_{2\nu}[P(dP)^{2n}],
                     \qquad\lambda=i\hbar .
\end{split}
\tag{OE23}
\] The left side is a formal coefficientwise integral, and the right
side has precisely its degree-\(n\) formal multiplier. The zero
coefficients in every other degree remain those actually proved in
(FC10). No convergence of the formal series or pointwise identity of
formal and classical symbols is asserted. This is a consequence of the
direct finite coefficient calculation, not an input to it, and it uses
no result from the later formal/classical bridge lesson.

These are complete consequences of the original course formulas. The
bounded source-reading route and exact earlier programme versions are
retained privately. No novelty or exhaustive literature claim is made.

\end{document}