The matrix Weyl index in the original phase coordinates

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

The preceding lesson constructed the full compact relative projector from both ordered Weyl errors and compared its classical Chern form with the cutoff differential form. Here we identify its formal trace using the exact flat algebra, the original cyclic cocycle and the higher algebraic index theorem. Two orientation signs arise in different places and cancel. We then transfer the resulting matrix formula from the isotropic symbol metric to the original product metric and left quantization.

Read Relative Weyl projectors and the Chern cutoff form for (RP1)–(RP16), (CP1)–(CP14) and (BN1)–(BN5). Read Radial compression and index transport for product-metric symbols for the full RC/PT endpoint argument. The Scaled Weyl parametrices and the surviving differential degree lesson defines the separate finite ordered Weyl coefficient; the present source-theorem route proves its integrated value without deleting any of its terms. The Weyl kernels, operator traces, and a finite trace-class test lesson fixes the original $(2\pi)^{-n}$ operator-trace normalization.

The external theorem used below is the higher algebraic index theorem of M. Pflaum, H. Posthuma and X. Tang. Its degree-zero specialization is stated at the point of use with the actual compact relative pair and full curvature. The local top Riemann–Roch input cited in their proof is due to B. Feigin, G. Felder and B. Shoikhet. Our proof here establishes the exact product map, cocycle value, characteristic side and analytic receiving map for the original course symbols.

1. The full source product and flat connection

1.1 The full product and parameter dictionary

Keep the original phase-space coordinates $z=(x_1,\xi_1,\ldots,x_n,\xi_n)$, original orientation $\Omega_{\mathrm{AN}}=\bigwedge_{j=1}^n(dx_j\wedge d\xi_j)$, full matrix product order and formal parameter $\lambda$. The full compact-symbol product is: $$f\#_\lambda g =\sum_{k\ge0}\frac{\lambda^k}{(2i)^k k!} \left[ \sum_{j=1}^n \bigl(\partial_{\xi_j}\otimes\partial_{x_j} -\partial_{x_j}\otimes\partial_{\xi_j}\bigr) \right]^k(f\otimes g)\big|_{\mathrm{matrix\ multiplication}}. \tag{FM1}$$ The source writes $\omega=\sum_jdp_j\wedge dq_j$ and $$u\star_\hbar v =\sum_{k\ge0}\frac{\hbar^k}{2^k k!} \left[ \sum_{j=1}^n \bigl(\partial_{p_j}\otimes\partial_{q_j} -\partial_{q_j}\otimes\partial_{p_j}\bigr) \right]^k(u\otimes v)\big|_{\mathrm{matrix\ multiplication}}. \tag{FM2}$$ For smooth compact symbols, (FM2) means the coefficientwise extension of the source’s polynomial product: at each fixed $k$ it contains finitely many derivatives, so it is defined without asserting convergence of the infinite $\hbar$-series.

Define the coordinate swap and coefficientwise parameter substitution $$F(x,\xi)=(p=\xi,q=x),\qquad \lambda=i\hbar,\qquad (\Phi f)(p,q;\hbar)=f(q,p;i\hbar). \tag{FM3}$$ This is an isomorphism of Laurent coefficient rings because $i\ne0$. The source bidifferential operator in (FM2), acting on $\Phi f\otimes\Phi g$, is exactly this course operator in (FM1) after $x=q,\xi=p$. Also $(i\hbar)^k/(2i)^k=\hbar^k/2^k$ for every nonnegative $k$. The matrix factors remain in their written order. Thus each finite coefficient agrees, including $k=0$ and every noncommutative derivative word: $$\Phi(f\#_\lambda g) =(\Phi f)\star_\hbar(\Phi g). \tag{FM4}$$ The inverse is $(\Phi^{-1}u)(x,\xi;\lambda)=u(\xi,x;\lambda/i)$. Hence (FM4) is an algebra isomorphism of the actual formal smooth matrix-symbol algebras, not merely a comparison of their first Poisson brackets.

The support statement is equally exact. Every term of either product is a local differential expression, so for coefficientwise compact $f,g$, $$\operatorname{supp}(\Phi f)=F(\operatorname{supp}f),\qquad \operatorname{supp}(f\#_\lambda g) \subseteq\operatorname{supp}f\cap\operatorname{supp}g . \tag{FM5}$$ The first equality uses the inverse linear diffeomorphism and nonzero parameter substitution; the second applies coefficient by coefficient. No completion changes the original support.

1.2 One exact flat Fedosov realization with the negative curvature

The source describes its general Fedosov curvature as the full closed formal form $\Omega_F=-\omega+\hbar\omega_1+\hbar^2\omega_2+\cdots$, CyclicWeyl.tex lines 365–376. To see concretely that the product (FM2) has a flat Fedosov realization whose curvature retains the stipulated negative leading term, take the trivial Weyl bundle over the same $(p,q)$-space. Write $P_j,Q_j$ for fiber coordinates, with the same fiber product (FM2), so $[P_j,Q_k]_\star=\hbar\delta_{jk}$. Use the flat base connection $d_{p,q}$, zero curvature $\widetilde R=0$, and the explicit one-form $$A_0=\sum_j(Q_j\,dp_j-P_j\,dq_j),\qquad B=-2\sum_jp_j\,dq_j,\qquad A=A_0+B . \tag{FM6}$$ Here $B$ is central in the fiber Weyl algebra; it changes the chosen curvature representative but contributes nothing to the commutator derivation. For every fiber section $U$, $$D U=d_{p,q}U+\hbar^{-1}[A,U]_\star =\sum_j\left[ dp_j(\partial_{p_j}-\partial_{P_j})U +dq_j(\partial_{q_j}-\partial_{Q_j})U \right]. \tag{FM7}$$ Indeed $[Q_j,U]_\star=-\hbar\partial_{P_j}U$ and $[P_j,U]_\star=\hbar\partial_{Q_j}U$ exactly: a linear fiber coordinate has no second or higher derivatives. Thus (FM7) has no hidden higher-order term. Its unique horizontal lift with value $u(p,q;\hbar)$ at $P=Q=0$ is the coefficientwise Taylor section $\widehat u(p,q;P,Q;\hbar)=u(p+P,q+Q;\hbar)$; direct substitution into (FM7) gives $D\widehat u=0$.

The source curvature formula for the flat base connection is $\Omega_F=dA+(2\hbar)^{-1}[A,A]_\star$, with the graded bracket of one-forms. Since $P,Q$ are fiber variables independent of the base, $dA_0=0$. The central $B$ satisfies $dB=-2\omega$ and has no graded commutator contribution. The remaining product is $$\hbar^{-1}A_0\wedge_\star A_0 =\hbar^{-1}\sum_j \bigl((-Q_j)\star P_j+P_j\star Q_j\bigr)\,dp_j\wedge dq_j =\hbar^{-1}\sum_j[P_j,Q_j]_\star\,dp_j\wedge dq_j =\omega . \tag{FM8}$$ In the first displayed coefficient, $(-Q_j)\star P_j+P_j\star Q_j$ is exactly $[P_j,Q_j]_\star$; all cross-index contributions cancel by the separate coordinate commutators. Consequently $$\Omega_F=-2\omega+\omega=-\omega,\qquad D^2=\hbar^{-1}[\Omega_F,-]_\star=0 . \tag{FM9}$$ This explicit representative has all higher $\omega_j=0$. The full general source curvature remains in (FC4)–(FC7); the central shift stays in the cocycle computation below.

The central term is a choice of lift of the derivation-valued connection, not a degree-zero perturbation of its noncentral Fedosov correction. The source’s exact derivation sequence in CyclicWeyl.tex lines 37–43 has the constants as kernel. In the present flat model the actual derivation is $d-\delta$ in (FM7), with zero higher noncentral correction. Both $A_0$ and $A_0+B$ give this same derivation and hence the same horizontal algebra at every coefficient. The antecedent FFSrevised.tex lines 2189–2194 expressly permits a central one-form change of this lift before choosing its Chern–Weil representative. We keep the complete $A_0+B$, including $dB=-2\omega$, rather than identify the two curvature forms pointwise.

For the particular degree-zero pairing used here, the equality of the trace densities under this central change is also direct. In the full determinant contraction (WR9), every one of the $2n$ connection slots is differentiated once in a fiber coordinate. Any term with $B$ in one such slot vanishes because $B$ is fiber constant. Thus the complete cochain on $A_0+B$ equals that on $A_0$; no matrix word or noncentral term is removed. On the characteristic side the two closed curvature representatives are $\Omega_0=+\omega$ and $\Omega_1=-\omega=\Omega_0+dB$. For the full compact closed relative Chern form $\gamma$ of (FC3), put $\Omega_s=\Omega_0+s\,dB$. The exact transgression, retaining the full exponential and its inverse parameter, is

$$\frac{d}{ds}\bigl[\gamma\wedge e^{-\Omega_s/(2\pi i\hbar)}\bigr]_{2n} =d\left[-\frac{1}{2\pi i\hbar}\,\gamma\wedge B\wedge e^{-\Omega_s/(2\pi i\hbar)}\right]_{2n-1}. \tag{FC8}$$

Indeed $d\gamma=0=d\Omega_s$, every degree of $\gamma$ is even, and $dB$ commutes with the other two-forms; differentiating the finite top-degree exponential gives exactly the displayed factor. The rank-difference term of $\gamma$ is identically zero, and every remaining coefficient has the original compact support $K$. The primitive in (FC8) is therefore compactly supported despite the noncompact $B$. Stokes followed by integration over $s\in[0,1]$ proves equality of the two characteristic integrals, coefficient by coefficient. Consequently the source pairing on the actual flat connection and the source negative-curvature representative used in (MB3) agree by an explicit cochain calculation and an exact compact transgression. No assumption that the central term belongs to the source’s degree-at-least-three noncentral correction is needed.

For horizontal lifts, fiber differentiation of $\widehat u$ is base differentiation of $u$. Therefore $$\widehat u\star_{\mathrm{fiber}}\widehat v =\widehat{u\star_\hbar v},\qquad (\widehat u\star_{\mathrm{fiber}}\widehat v)|_{P=Q=0} =u\star_\hbar v . \tag{FM10}$$ This establishes the source-form Fedosov product for the explicit connection at every formal order and shows why (FM4), not an assumed gauge equivalence, is the exact algebra map used below.

1.3 Transport of the actual compact relative projector

Let $e_\infty,e_0\in M_{2\nu}$ be the specific formal idempotents of RP11, built from the original ordered $a,b=\psi a^{-1}$ and both error sides. RP11 proves $$e_\infty\#_\lambda e_\infty=e_\infty,\qquad e_0\#_\lambda e_0=e_0,\qquad \operatorname{supp}(e_\infty-e_0)_k \subseteq K:=\operatorname{supp}(1-\psi) \quad\text{for every coefficient }k . \tag{FM11}$$ Apply (FM4) to the whole matrix idempotents, not merely to their zeroth terms. With $E_\infty=\Phi e_\infty$ and $E_0=\Phi e_0$, one obtains $$E_\infty\star_\hbar E_\infty=E_\infty,\qquad E_0\star_\hbar E_0=E_0,\qquad \operatorname{supp}(E_\infty-E_0)_k\subseteq F(K). \tag{FM12}$$ The reference rank, all original matrix off-diagonal terms, powers of $i$, and both original error orders survive. The horizontal-lift map of (FM7)–(FM10) realizes $E_\infty,E_0$ as flat Fedosov idempotents for this explicit connection. Their difference is coefficientwise compact in the same fixed set $F(K)$, so every local source density evaluated on the difference has a legitimate compact support. This supplies the actual compact relative pair to the source theorem.

The orientation remains separate from the algebra map: $$F^*(\omega^n/n!)=(-1)^n\Omega_{\mathrm{AN}},\qquad \deg_{\mathrm{or}}F=(-1)^n . \tag{FM13}$$

2. Degree-zero pairing and the exact cyclic sign

Let $\Psi=\Psi_{2n}^{2n}$ be the source trace density of CyclicWeyl.tex definition dfn:psi. For a compactly supported flat formal symbol $f$, use the source’s own integral functional, as written in HAnaInThm.tex lines 541–548: $$T_\Psi(f):=\int_M\Psi(f). \tag{NS1}$$ The source’s definition dfn:chi, at $i=r=0$ and $\alpha_0=1$, gives directly $$\chi_{0,M}^{0}(1)(f) =\int_M1\wedge\Psi_{2n}^{2n}(f) =T_\Psi(f). \tag{NS2}$$ Its subsequent construction, CyclicWeyl.tex equation consmor2, includes the explicit factor $(2\pi\sqrt{-1})^{-n}$. At degree zero there is one summand, so no cyclic-periodicity convention or higher cochain can change the following exact identity: $$\mathsf Q_M^0(1)(f) =\frac{1}{(2\pi i)^n}\chi_{0,M}^{0}(1)(f) =\frac{1}{(2\pi i)^n}T_\Psi(f). \tag{NS3}$$ This statement uses only the source definitions and is valid for the compact-support algebra on the actual flat $M=\mathbb R^{2n}$. The source extends the Weyl cocycles to finite matrices by the ordered ordinary trace, IndThms.tex lines 163–171. Thus (NS3) holds with $f$ replaced by a compactly supported matrix symbol and $T_\Psi$ replaced by its matrix-trace extension.

For the relative pair $(e_\infty,e_0)$ of the preceding lesson, the difference is coefficientwise supported in the fixed compact $K=\operatorname{supp}(1-\psi)$. Locality of $\Psi$ keeps its evaluation on the difference supported there. At cyclic degree zero, the source Chern chain is exactly $c_0(e)=e$, IndThms.tex lines 21–44; the higher $c_j$ do not enter. By linearity and the source pairing definition at lines 123–134, $$\big\langle\mathsf Q^0(1),e_\infty-e_0\big\rangle =\frac{1}{(2\pi i)^n} \int_{\mathbb R^{2n}} \Psi^{(2\nu)}(e_\infty-e_0). \tag{NS4}$$ No finite-parametrix error, matrix order, rank reference or support term has been discarded. Equation (NS4) is the exact source cyclic pairing-to-source trace-density morphism needed before a comparison with RP16.

2.1 The operator that the source’s proof specifies

Write $z_{2s-1}=P_s,z_{2s}=Q_s$ for $1\leq s\leq n$. On $2k$ ordered tensor slots $r=0,\ldots,2k-1$, let $D_{ra}=\partial_{z_a}^{(r)}$. All the $D_{ra}$ commute, even when different partial derivatives act on the same slot. The source’s Poisson operator is $$\alpha_{rs}=\sum_{\ell=1}^n (D_{r,2\ell-1}D_{s,2\ell} -D_{r,2\ell}D_{s,2\ell-1}),\qquad \alpha_{sr}=-\alpha_{rs}. \tag{WR1}$$ The plus sign before the displayed minus is ordinary addition; no $1/2$ is included in $\alpha$. For a $k$-element subset $I\subset\{1,\ldots,n\}$, let $D_I$ be the $2k$-by-$2k$ matrix with rows the tensor slots and columns $P_\ell,Q_\ell$ for $\ell\in I$, in increasing plane order. Define the full alternating operator $$\Pi_{2k}^{[n]}=\sum_{\substack{I\subset\{1,\ldots,n\}\\|I|=k}} \det D_I,\qquad \Pi_0^{[n]}=1,\qquad \Pi_{2k}^{[n]}=0\quad(k>n). \tag{WR2}$$ For $k=n$, this is precisely the full determinant with its slots relabelled $0,\ldots,2n-1$, including every one of the $(2n)!$ signed permutations and no factorial multiplier. The superscript $[n]$ records the ambient number of symplectic planes; it is not a matrix rank.

Here is an exact operator identity for $0\leq k<n$. Let $(\Pi_{2k}^{[n]})^i$ act on the $2k$ ordered slots of a $2k+2$-slot tensor other than slots $0$ and $i$. Then $$\boxed{\quad \Pi_{2k+2}^{[n]} =\sum_{i=1}^{2k+1}(-1)^i \alpha_{i0}(\Pi_{2k}^{[n]})^i . \quad} \tag{WR3}$$ To prove it without a normalization convention, introduce exterior basis symbols $e_0,\ldots,e_{2k+1}$ over the commutative ring generated by the $D_{ra}$. Put $v_a=\sum_r D_{ra}e_r$. Direct expansion of (WR1) gives $$\beta:=\sum_{r<s}\alpha_{rs}e_r\wedge e_s =\sum_{\ell=1}^{n}v_{2\ell-1}\wedge v_{2\ell}. \tag{WR4}$$ In $\beta^{k+1}/(k+1)!$, the square of each decomposable plane term vanishes. A set $I$ of $k+1$ distinct planes occurs $(k+1)!$ times before division. The coefficient of $e_0\wedge\cdots\wedge e_{2k+1}$ is therefore $\sum_{|I|=k+1}\det D_I=\Pi_{2k+2}^{[n]}$. Equivalently, this coefficient is the Pfaffian of the skew matrix $(\alpha_{rs})$, with its fully written normalization $$\operatorname{Pf}_{2k+2}(\alpha) =\frac{1}{2^{k+1}(k+1)!} \sum_{\sigma\in S_{2k+2}}\operatorname{sgn}(\sigma) \prod_{j=1}^{k+1} \alpha_{\sigma(2j-2),\sigma(2j-1)} . \tag{WR5}$$ This follows directly by expanding $\beta^{k+1}$: each pairing appears in $2^{k+1}(k+1)!$ ordered and internally oriented forms. Sorting the pair containing slot $0$ gives the Pfaffian Laplace expansion $$\operatorname{Pf}_{2k+2}(\alpha) =\sum_{i=1}^{2k+1}(-1)^{i+1} \alpha_{0i}\operatorname{Pf}_{2k}(\alpha_{\widehat0\widehat i}). \tag{WR6}$$ The removed-slot Pfaffian equals $(\Pi_{2k}^{[n]})^i$ by the same (WR4) calculation on those ordered remaining slots. Since $\alpha_{0i}=-\alpha_{i0}$, (WR6) is exactly (WR3). This proof also shows why repeated choices of one symplectic plane cancel rather than add a hidden $k!$: their exterior square is zero.

Pflaum–Posthuma–Tang explicitly take $(\hbar\alpha)^{\wedge1}=\hbar\alpha$ in their one-plane display, CyclicWeyl.tex lines 124–134. In the proof of $b\tau_{2k}=\tau_{2k+1}$, their lines 232–257 assert, on the actual ordered slots, that $$\sum_{i=1}^{2k+1}(-1)^i\hbar\alpha_{i0} \big((\hbar\alpha)^{\wedge k}\big)^i =(\hbar\alpha)^{\wedge(k+1)} . \tag{WR7}$$ They first write the larger $\sum_s$ expression, then explicitly identify its $s=0$ part with the right side; (WR7) is that identification, not an inferred equality from the theorem alone. Starting from their displayed degree-one operator, (WR7) uniquely determines every subsequent power by induction. Equation (WR3), including the $k=0$ base $\alpha_{01}=\Pi_2^{[n]}$, proves the source-specified power is $$\boxed{\quad (\hbar\alpha)_{\mathrm{PPT}}^{\wedge k} =\hbar^k\Pi_{2k}^{[n]} \quad\text{on every ordered set of \(2k\) slots}, \qquad 0\leq k\leq n . \quad} \tag{WR8}$$ The source’s separate remark at lines 113–115 that its top cocycle is Feigin–Felder–Shoikhet’s up to $(-1)^n$ agrees with (WR8). A convention using the undivided exterior power would instead give $k!\Pi_{2k}^{[n]}$, fail (WR7) already at $k=2$, and disagree with that top-cocycle remark. The factor of two from an undivided exterior square at $k=2$ is thus resolved by the source’s own recursion.

2.2 Exact value of the source cochain for the written connection

The source definition at CyclicWeyl.tex lines 95–112 and (WR8), with the full ordered exponential product $\mathcal E$, give an equality of cochains in every dimension: $$\tau_{2n}^{\mathrm{PPT}} =(-1)^n\hbar^n\mu_{2n} \int_{\Delta^{2n}}\mathcal E(u)\Pi_{2n}^{[n]}\,du_1\cdots du_{2n} =\tau_{2n}^{\det}. \tag{WR9}$$ No $n!$ is omitted: (WR5) and (WR8) specify it. For the full FM6 connection $$A=\sum_{j=1}^n(Q_j\,dp_j-P_j\,dq_j)-2\sum_{j=1}^np_j\,dq_j, \quad D=d-\sum_j(dp_j\partial_{P_j}+dq_j\partial_{Q_j}), \quad \Omega_F=-\omega, \tag{WR10}$$ and each compact formal matrix symbol $u$, its horizontal section is $\widehat u(p,q;P,Q)=u(p+P,q+Q)$. The central shift in $A$ supplies the written curvature and has zero fiber derivative; it is retained, not removed from the connection.

At degree $2n$, the source shuffle in CyclicWeyl.tex lines 393–411 has a length-zero first chain and only the identity $(0,2n)$-shuffle. Its definition of $\Psi$ at lines 550–570 therefore gives $$\Psi_{2n}^{2n}(\widehat u) =\hbar^{-2n}\tau_{2n}^{\mathrm{PPT}} (\widehat u,A,\ldots,A). \tag{WR11}$$ All $2n$ inverse powers, the ordinary ordered matrix trace and the full $A$ are present. By (WR9), this is the source cochain on the written connection. The determinant contraction can be calculated without a normalization shortcut:

Write $$\beta_{2j-1}:=\partial_{P_j}A=-dq_j,\qquad \beta_{2j}:=\partial_{Q_j}A=dp_j . \tag{FD5}$$ Each factor $A$ is differentiated exactly once in every summand of (the full determinant). Exterior multiplication of the resulting differential-form slots gives $$\begin{aligned} \bigwedge \Pi_{2n} (\widehat u\otimes A^{\otimes2n}) &=\widehat u\sum_{\sigma\in S_{2n}} \operatorname{sgn}(\sigma)\, \beta_{\sigma(1)}\wedge\cdots\wedge\beta_{\sigma(2n)}\\ &=(2n)!\,\widehat u\, \beta_1\wedge\cdots\wedge\beta_{2n}\\ &=(2n)!\,\widehat u\, \bigwedge_{j=1}^n(dp_j\wedge dq_j) =(2n)!\,\widehat u\,\frac{\omega^n}{n!}. \end{aligned} \tag{FD6}$$ The second equality uses the sign of the permutation once from the determinant and once from exterior permutation, so every one of the $(2n)!$ summands has the same sign. The third equality retains each original ordered coordinate pair: $(-dq_j)\wedge dp_j=dp_j\wedge dq_j$.

After (the full determinant) acts, every $A$-slot has fiber-constant coefficients $\beta_j$. Every positive-degree term in any exponential of (the ordered exponential) differentiates one such constant slot and vanishes, including the terms involving slot zero. Thus $\mathcal E$ acts as the identity on this exact tensor. The simplex volume is $1/(2n)!$, and $\mu_{2n}$ evaluates $\widehat u$ at $P=Q=0$, yielding $u$.

These equations retain all $(2n)!$ permutations, the ordered simplex volume $1/(2n)!$, and the vanishing of every positive exponential term after the $A$-slots become fiber-constant. Hence, coefficientwise for every finite matrix rank $m$, $$\boxed{\quad \Psi_{2n}^{2n,(m)}(\widehat u) =(-1)^n\hbar^{-n}\operatorname{tr}_m(u)\frac{\omega^n}{n!}. \quad} \tag{WR12}$$ There is no assumption that $u$ is scalar, idempotent or only a zeroth-order symbol. At each formal order the source expression is local in finitely many derivatives, so a fixed coefficientwise compact support remains compact.

Retain the source normalization $(2\pi i)^{-n}$, the exact algebra map $p=\xi,q=x,\lambda=i\hbar$ of FM3–FM4, and both separate orientation signs of FM13. Their product is $+1$ in the phase-space integral. Thus for every compact formal finite-matrix symbol $f$ in the original ordered Moyal algebra, $$\boxed{\quad \mathsf Q^0(1)(\widehat{\Phi f}) =(2\pi i)^{-n}\int_{\mathbb R^{2n}_{p,q}} \Psi_{2n}^{2n,(m)}(\widehat{\Phi f}) =(-1)^n(2\pi\lambda)^{-n} \int_{\mathbb R^{2n}_{x,\xi}} \operatorname{tr}_m(f)\,\Omega_{\mathrm{AN}} =(-1)^n\mathcal T_\lambda(f). \quad} \tag{WR13}$$ The expression $\mathsf Q^0(1)(\widehat{\Phi f})$ denotes evaluation on the horizontal Fedosov section; it does not pretend that a base function is already in the source’s section domain. In particular, for the actual relative idempotent pair, $$\mathsf Q^0(1)(\widehat{\Phi e_\infty} -\widehat{\Phi e_0}) =(-1)^n\mathcal T_\lambda(e_\infty-e_0). \tag{WR14}$$ The difference has the fixed compact coefficientwise support proved in the preceding lesson and FM12, so both sides are defined. For $n=1$, (WR13) is exactly the one-plane negative value; for $n=2$ it is positive, and for $n=3$ negative. This parity is proved by (WR3)–(WR12), not extrapolated from three examples.

3. The complete curvature and Chern orientation

Let $P\in C^\infty(\mathbb R^{2n};M_{2\nu}(\mathbb C))$ be CP1. The constant reference projector is $e_0=\operatorname{diag}(0,I_\nu)$. CP1 gives $$P^2=P,\qquad e_0^2=e_0,\qquad P=e_0\ \text{outside }K=\operatorname{supp}(1-\psi), \qquad \operatorname{rank}P=\operatorname{rank}e_0=\nu. \tag{FC1}$$ The last rank equality follows from the exact global conjugation $P=U_0 pU_0^{-1}$ in CP1, including where $a$ is noninvertible. Thus the image bundles $V_P,V_0$ agree by the identity outside $K$, not merely by an unspecified exterior isomorphism.

Differentiate $P^2=P$. Multiplying by $P$ on both sides gives $P(dP)P=0$; replacing $P$ by $I-P$ gives $(I-P)(dP)(I-P)=0$. Therefore $dP$ is off-diagonal for the splitting $\operatorname{im}P\oplus\ker P$, and every odd power $(dP)^{2k+1}$ has zero full matrix trace. Since $d(dP)=0$, direct exterior differentiation yields $$d\,\operatorname{tr}_{2\nu}[P(dP)^{2k}] =\operatorname{tr}_{2\nu}[(dP)^{2k+1}]=0 \quad(k\ge0). \tag{FC2}$$ For $k\ge1$, these forms vanish outside $K$ because $dP=0$ there. At $k=0$, $\operatorname{tr}(P-e_0)=\nu-\nu=0$ identically. Hence the relative Chern-character representative in whatever fixed source convention is a finite sum of compactly supported closed even forms $$\gamma=\operatorname{Ch}(V_P-V_0) =\sum_{k=1}^{n}\gamma_{2k},\qquad \gamma_{2k}=c_k\operatorname{tr}_{2\nu}[P(dP)^{2k}], \quad d\gamma_{2k}=0,\quad \operatorname{supp}\gamma_{2k}\subseteq K . \tag{FC3}$$ Here $c_k$ records the source Chern normalization, calculated exactly in (CN1)–(CN2). Equation (FC2) proves the needed closure independently of that normalization. The source’s theorem uses precisely the difference of bundles determined by the zeroth idempotents, which are $V_P,V_0$ here.

Keep the source’s entire closed formal Fedosov two-form. Its leading term is negative the source symplectic form (CyclicWeyl.tex lines 365–376): $$\Omega_F=-\omega+\hbar\Omega_1+\hbar^2\Omega_2+\cdots, \qquad d\Omega_F=0 \tag{FC4}$$ on $\mathbb R^{2n}$. No term has been set to zero, even though the base is flat. Each coefficient has an explicit global primitive by the radial Poincaré homotopy: for a closed two-form $\beta$, let $$(H\beta)_z(v) =\int_0^1 s\,\beta_{sz}(z,v)\,ds,\qquad \eta_F=\sum_{j\ge0}\hbar^j H\Omega_j \quad(\Omega_0=-\omega). \tag{FC5}$$ Differentiate $s^2\beta_{sz}(v,w)$ in $s$, use the coordinate closedness equation for $\beta$, and integrate from zero to one. The zero endpoint is zero, while the other is $\beta_z(v,w)$; this proves $d(H\beta)=\beta$ and hence $d\eta_F=\Omega_F$ coefficient by coefficient. The primitive need not decay, but its product with a compactly supported $\gamma_{2k}$ is compactly supported.

For every $j\ge1$, closedness of $\gamma_{2k}$ and $\Omega_F$ gives the exact compact-support identity $$\gamma_{2k}\wedge\Omega_F^j =d\bigl(\gamma_{2k}\wedge\eta_F \wedge\Omega_F^{j-1}\bigr). \tag{FC6}$$ The sign is $+1$ because $\deg\gamma_{2k}=2k$ is even. The primitive is compactly supported by (FC3), so its integral over the original $\mathbb R^{2n}$ orientation is zero by Stokes. This handles every formal $\hbar$-coefficient and every negative $\hbar$-power from the source exponential without absorbing or dropping the original $\omega$, the corrections $\Omega_j$, or their constants.

Choose the exact flat Euclidean connection on $T\mathbb R^{2n}$. Its curvature is zero, so its $\hat A$-form is $1$; the source characteristic class is unchanged by this connection choice. In the source theorem’s $k=0,\alpha_0=1$ characteristic expression, expand the complete exponential and select top degree. Every term containing a positive power of $\Omega_F$ integrates to zero by (FC6), and the rank-difference term is zero by (FC1). Thus the exact reduction is $$\int_{\mathbb R^{2n}} \bigl[\hat A(T\mathbb R^{2n})\, \operatorname{Ch}(V_P-V_0)\, \exp(-\Omega_F/(2\pi i\hbar))\bigr]_{2n} =\int_{\mathbb R^{2n}}\gamma_{2n}. \tag{FC7}$$

IndThms.tex lines 385–387 defines the invariant Chern polynomial on matrices as $\operatorname{tr}(\exp X)$. Its source pairing formula at lines 431–475 has the global factor $(2\pi i)^{-n}$ and applies this polynomial to the vector-bundle curvature and $-\Omega_F/\hbar$. In the flat tangent case, where the chosen exact Euclidean connection has $\hat A=1$, distributing that factor among the homogeneous terms gives the entire top-degree source characteristic integrand $$\frac{1}{(2\pi i)^n} \sum_{k=0}^{n} \frac{1}{k!(n-k)!} \operatorname{tr}_{V_P-V_0}(F^{\,k}) \wedge\left(-\frac{\Omega_F}{\hbar}\right)^{n-k}. \tag{CN1}$$ Here the $k=0$ trace means the rank difference, $F_P,F_0$ are the two actual bundle curvatures, and the displayed polynomial is finite by degree. In particular, the source’s degree-$2k$ Chern normalization is $$\gamma_{2k} =\frac{1}{k!(2\pi i)^k} \bigl(\operatorname{tr}_{V_P}F_P^k -\operatorname{tr}_{V_0}F_0^k\bigr). \tag{CN2}$$ Indeed the $k$-th factor $(2\pi i)^{-k}$ in (CN2) times the remaining $(2\pi i)^{-(n-k)}$ in $\exp[-\Omega_F/(2\pi i\hbar)]$ reconstructs exactly the global $(2\pi i)^{-n}$ in (CN1). No $\hbar$, $2\pi$, factorial or source curvature sign has been absorbed.

Choose the Grassmann connections allowed by IndThms.tex lines 150–154: $$\nabla_P s=P\,ds\quad(Ps=s),\qquad \nabla_0s=e_0\,ds\quad(e_0s=s). \tag{CN3}$$ They agree under the identity outside $K=\operatorname{supp}(1-\psi)$, because $P=e_0$ there. Differentiate $P^2=P$ and multiply on both sides to get $P(dP)P=0$; likewise $(I-P)(dP)(I-P)=0$. Thus $dP$ is off-diagonal and $(dP)^2$ commutes with $P$. For a section $s=Ps$, $$\begin{aligned} \nabla_P^2s &=P\,d(P\,ds) =P(dP)\wedge ds\\ &=P(dP)\wedge[(dP)s+P\,ds] =P(dP)^2s . \end{aligned} \tag{CN4}$$ Consequently $F_P=P(dP)^2P=P(dP)^2$ on $V_P$, $F_0=0$, and for every $k\ge1$ $$\operatorname{tr}_{V_P}(F_P^k) =\operatorname{tr}_{2\nu}\!\left(P(dP)^{2k}\right). \tag{CN5}$$ The identity follows by multiplying the commuting factors $P$ and $(dP)^2$; it keeps the full matrix order and all $2k$ one-form factors. The traces vanish outside $K$ because $dP=0$ there. The rank difference at $k=0$ is zero by CP1.

Equation (FC6) proves that for $k<n$, every term of (CN1) with a positive $\Omega_F$-power has a compactly supported primitive and integrates to zero, coefficient by coefficient in $\hbar$. Substituting (CN5) in the surviving $k=n$ term yields, on the source phase space $M_{\mathrm{src}}$ oriented by $\omega^n/n!$, the exact value $$\int_{M_{\mathrm{src}}}^{\omega} \bigl[\hat A\operatorname{Ch}(V_P-V_0) \exp(-\Omega_F/(2\pi i\hbar))\bigr]_{2n} =\frac{1}{n!(2\pi i)^n} \int_{M_{\mathrm{src}}}^{\omega} \operatorname{tr}_{2\nu}\!\left(P_{\mathrm{src}}(dP_{\mathrm{src}})^{2n}\right). \tag{CN6}$$ Now apply the same coordinate map $F(x,\xi)=(p=\xi,q=x)$ as (FM3). Its degree from the original orientation $\Omega=dx_1\wedge d\xi_1\wedge\cdots\wedge dx_n\wedge d\xi_n$ to the source symplectic orientation is $(-1)^n$. The source projector is $P_{\mathrm{src}}=P_{\mathrm{AN03}}\circ F^{-1}$, so naturality of $d$ and matrix trace gives $F^*\operatorname{tr}[P_{\mathrm{src}}(dP_{\mathrm{src}})^{2n}]=\operatorname{tr}[P_{\mathrm{AN03}}(dP_{\mathrm{AN03}})^{2n}]$. Unlike the source trace-density density $f\,\omega^n/n!$, this Chern form supplies no second standalone volume-form sign. The oriented change of variables is therefore $$\begin{aligned} \int_{M_{\mathrm{src}}}^{\omega} \bigl[\hat A\operatorname{Ch}(V_P-V_0) e^{-\Omega_F/(2\pi i\hbar)}\bigr]_{2n} &=\frac{(-1)^n}{n!(2\pi i)^n} \int_{M_{\mathrm{AN03}}}^{\Omega} \operatorname{tr}_{2\nu}\!\left(P(dP)^{2n}\right)\\ &=(-1)^n(2\pi)^{-n}\frac{2}{i^n n!} \int_{M_{\mathrm{AN03}}}^{\Omega} (1-\psi)^n\operatorname{tr}_\nu[(db\wedge da)^n]. \end{aligned} \tag{CN7}$$

4. Applying the theorem to the actual relative projector

4.1 The actual algebra, projectors and source theorem hypotheses

Keep the original isotropic metric $g_z=h(z)(|dx|^2+|d\xi|^2)$, $h(z)=(1+|z|^2)^{-1}$, and the original full $a\in S(1,g;\operatorname{End}\mathbb C^\nu)$, uniformly invertible outside a compact set. Let $\psi$ be the written exterior cutoff, $b=\psi a^{-1}$ on the invertibility region extended by zero, $t=1-\psi$, and $\Omega_{\mathrm{AN}}=dx_1\wedge d\xi_1\wedge\cdots\wedge dx_n\wedge d\xi_n$. These are the objects of IP1–IP3, not a normalized replacement. The preceding lesson constructs from both ordered errors the full formal idempotents $e_\infty,e_0\in M_{2\nu}(C^\infty(\mathbb R^{2n}_{x,\xi})[[\lambda]])$ with $$e_\infty\#_\lambda e_\infty=e_\infty,\qquad e_0\#_\lambda e_0=e_0,\qquad \operatorname{supp}[\lambda^j](e_\infty-e_0) \subseteq K:=\operatorname{supp}(1-\psi)\quad(j\ge0). \tag{MB1}$$ The individual projectors need not be compactly supported. The coefficientwise support statement is the precise hypothesis needed for the source’s relative pairing.

Equations (FM1)–(FM12) prove an algebra identity, not a formal equivalence claim: with $F(x,\xi)=(p=\xi,q=x)$ and $\lambda=i\hbar$, the complete ordered $\#_\lambda$ product becomes the source’s written Weyl product $\star_\hbar$ at every coefficient. The explicit FM6 Fedosov one-form, its derivation, curvature and horizontal lift are $$A=\sum_{j=1}^{n}(Q_j\,dp_j-P_j\,dq_j)-2\sum_{j=1}^{n}p_j\,dq_j,\quad D=d-\sum_{j=1}^n(dp_j\partial_{P_j}+dq_j\partial_{Q_j}),\quad \Omega_F=-\omega,\quad \widehat u(p,q;P,Q)=u(p+P,q+Q). \tag{MB2}$$ The central term in $A$ is essential to the displayed curvature. The two horizontal source projectors $\widehat E_\infty=\widehat{\Phi e_\infty}$ and $\widehat E_0=\widehat{\Phi e_0}$ remain idempotent; their difference is coefficientwise supported in $F(K)$ as a section because every Taylor coefficient differentiates a compactly supported base coefficient. Both lie in the source formal Laurent algebra $\mathcal A^{((\hbar))}$; in fact they have no negative $\hbar$-powers before the trace normalization.

At cyclic degree zero take the source theorem’s closed total-form sequence $\alpha=(\alpha_0)=(1)$. Pflaum–Posthuma–Tang’s theorem thm:higher-algind, IndThms.tex lines 412–424, applies to these two actual projectors with compact difference and states $$\left\langle\mathsf Q^0(1), \widehat E_\infty-\widehat E_0\right\rangle =\int_{\mathbb R^{2n}_{p,q}}^{\omega} \left[ \hat A(T\mathbb R^{2n})\, \operatorname{Ch}(V_P-V_0)\, \exp\!\left(-\frac{\Omega_F}{2\pi i\hbar}\right) \right]_{2n}. \tag{MB3}$$ Here $V_P,V_0$ are determined by the zeroth source projectors, not by an exterior-only rank surrogate. Equation (CP1) proves $P=[\lambda^0]e_\infty=U_0pU_0^{-1}$ globally, $e_0=\operatorname{diag}(0,I_\nu)$, $\operatorname{rank}P=\operatorname{rank}e_0=\nu$, and $P=e_0$ outside $K$. Thus the theorem’s bundle and support hypotheses hold. The base is an ordinary symplectic manifold, so no orbifold specialization or unstated properness argument is needed; the source theorem itself expressly assumes compact projector difference on $M$.

4.2 Both sides in the original coordinates, with every sign

Equations (NS3)–(NS4) identify the source degree-zero pairing with $(2\pi i)^{-n}\int\Psi_{2n}^{2n}$ on a compact relative difference. Equations (WR1)–(WR14) prove the original source wedge normalization from its displayed recurrence and evaluate that same $\Psi$-shuffle for the full $A$ of (MB2). Its all-dimensional identity on the actual projectors is $$\left\langle\mathsf Q^0(1), \widehat E_\infty-\widehat E_0\right\rangle =(-1)^n\mathcal T_\lambda(e_\infty-e_0),\qquad \mathcal T_\lambda(f) =(2\pi\lambda)^{-n} \int_{\mathbb R^{2n}_{x,\xi}} \operatorname{tr}_{2\nu}(f)\,\Omega_{\mathrm{AN}} . \tag{MB4}$$ The two signs in the coordinate-swapped trace-density integral cancel separately: $F^*(\omega^n/n!)=(-1)^n\Omega_{\mathrm{AN}}$ and $\deg_{\mathrm{or}}F=(-1)^n$. The remaining $(-1)^n$ is the source cocycle sign. Every coefficient of the difference has fixed compact support, so (MB4) is an equality in $\mathbb C((\lambda))$, not an integral of the individual noncompact projectors.

On the characteristic side, choose the exact flat Euclidean connection for $T\mathbb R^{2n}$, giving $\hat A=1$. Keep the full closed source curvature $\Omega_F=-\omega+\hbar\Omega_1+\cdots$; for the particular connection (MB2), all later coefficients happen to vanish, but Equations (FC5)–(FC6) prove the stronger exact statement needed here: every positive power of $\Omega_F$ multiplied by a compactly supported closed positive-degree relative Chern form has a compactly supported primitive and integrates to zero, coefficient by coefficient, including its negative $\hbar$-powers. The rank-difference $0$-form is exactly zero. Therefore FC7 reduces the entire exponential, without suppressing it from (MB3), to the top source Chern form.

Equations (CN1)–(CN6) obtain that form from the source’s complete $(2\pi i)^{-n}$ factor and the actual Grassmann curvature $F_P=P(dP)^2$: $$\int_{\mathbb R^{2n}_{p,q}}^\omega [\hat A\,\operatorname{Ch}(V_P-V_0) e^{-\Omega_F/(2\pi i\hbar)}]_{2n} =\frac{1}{n!(2\pi i)^n} \int_{\mathbb R^{2n}_{p,q}}^\omega \operatorname{tr}_{2\nu} [P_{\rm src}(dP_{\rm src})^{2n}] . \tag{MB5}$$ In contrast to the trace density, this Chern form has no separate symplectic-volume factor. Naturality gives $F^*\operatorname{tr}[P_{\rm src}(dP_{\rm src})^{2n}] =\operatorname{tr}[P(dP)^{2n}]$, while the oriented-domain change contributes one $(-1)^n$. Equations (CN7) and (CP14) then give, with the original cutoff and all matrix words retained, $$\begin{aligned} \int_{\mathbb R^{2n}_{p,q}}^\omega [\hat A\,\operatorname{Ch}(V_P-V_0) e^{-\Omega_F/(2\pi i\hbar)}]_{2n} &=(-1)^n\,\frac{1}{n!(2\pi i)^n} \int_{\mathbb R^{2n}_{x,\xi}}^{\Omega_{\rm AN}} \operatorname{tr}_{2\nu}[P(dP)^{2n}]\\ &=(-1)^n(2\pi)^{-n}\frac{2}{i^n n!} \int_{\mathbb R^{2n}_{x,\xi}}^{\Omega_{\rm AN}} (1-\psi)^n\operatorname{tr}_{\nu}[(db\wedge da)^n]. \end{aligned} \tag{MB6}$$ The one Chern orientation sign is the same parity as the one source-cocycle sign in (MB4); neither sign has been set to $+1$ individually.

Combining the two evaluations of the same source pairing (MB3)–(MB6) cancels those two factors $(-1)^n$, which are invertible over $\mathbb C((\lambda))$. The exact resulting formal identity is $$\boxed{\quad \mathcal T_\lambda(e_\infty-e_0) =\frac{1}{n!(2\pi i)^n} \int_{\mathbb R^{2n}}^{\Omega_{\rm AN}} \operatorname{tr}_{2\nu}[P(dP)^{2n}] =(2\pi)^{-n}\frac{2}{i^n n!} \int_{\mathbb R^{2n}}^{\Omega_{\rm AN}} (1-\psi)^n\operatorname{tr}_{\nu}[(db\wedge da)^n]. \quad} \tag{MB7}$$ The right side has no $\lambda$: the rank term is zero and FC6 kills every curvature power after integration. This is an equality of the original full formal trace with the exact classical relative form, not only an equality of their leading symbols.

4.3 The analytic index and the further coefficient consequence

Equation (RP16) proves from the original two error orders and the actual analytic Fredholm index that $$\operatorname{ind}a^w =(2\pi)^{-n}\int_{\mathbb R^{2n}}^{\Omega_{\rm AN}} \operatorname{tr}_{2\nu}[\lambda^n] (e_\infty-e_0). \tag{MB8}$$ Taking $[\lambda^0]$ of the full left side of (MB7) reproduces precisely (MB8), since $(2\pi\lambda)^{-n}$ shifts the coefficient by $n$. The right side of (MB7) is constant. Thus the original matrix analytic index formula is proved for every $n\ge1$ under the stated $S(1,g)$ and exterior invertibility hypotheses: $$\boxed{\quad \operatorname{ind}a^w =(2\pi)^{-n}\frac{2}{i^n n!} \int_{\mathbb R^{2n}}^{\Omega_{\rm AN}} (1-\psi)^n \operatorname{tr}_{\nu}[(db\wedge da)^n]. \quad} \tag{MB9}$$ The preceding lesson proves the complete scalar cutoff primitive, its exact beta moment $(n!)^2/(2n)!$, and Stokes with the original outward boundary orientation. Applying that already proved identity to the written right side of (MB9), without a convention change, gives the original boundary coefficient $$\boxed{\quad \operatorname{ind}a^w =-(-2\pi i)^{-n}\frac{(n-1)!}{(2n-1)!} \int_{\partial B}^{\Omega_{\rm AN}} \operatorname{tr}_{\nu}[(a^{-1}da)^{2n-1}]. \quad} \tag{MB10}$$ The boundary lies in the exterior invertibility region and carries the orientation induced by $\Omega_{\rm AN}$. No rank, cutoff term, matrix order, factorial, power of $i$, or coordinate sign has been discarded.

There is a further exact consequence of (MB7). Write $e_\infty-e_0=\sum_{j\ge0}\lambda^j d_j$, each $d_j$ supported in $K$. Since the full Laurent series $\mathcal T_\lambda(e_\infty-e_0)$ is constant, coefficient comparison in $\mathbb C((\lambda))$ gives $$\int_{\mathbb R^{2n}}^{\Omega_{\rm AN}} \operatorname{tr}_{2\nu}d_j=0\quad(j\ge0,\ j\ne n), \qquad (2\pi)^{-n}\int_{\mathbb R^{2n}}^{\Omega_{\rm AN}} \operatorname{tr}_{2\nu}d_n=\operatorname{ind}a^w . \tag{MB11}$$ RP16 had independently established the vanishing for $j<n$; (MB11) supplies every $j>n$ through the source theorem. The conclusion concerns integrated matrix traces, not pointwise vanishing of those coefficients.