Reducing arbitrary-order boundary problems to first order

An elliptic boundary problem of order $m>1$ can be reduced to a first-order problem, but the reduction is not obtained by deleting $m-1$ factors. Those factors carry a genuine operator on the global Sobolev space. The correct construction first adds zero-index equations, deforms the resulting block operator through a Cayley system whose stable space is unchanged, rewrites every boundary measurement on that stable space, and only then cancels a common zero-index right factor.

This lesson proves that construction with the original bundles, Sobolev exponents, coefficient order, and factor order visible throughout. The normal order of a boundary operator is always distinguished from its total order. Lower-order discrepancies are recorded as compact maps, and the first-order interior operator keeps its target bundle $F$ even though $F$ is identified with $E$ in a collar.

The named prerequisites are Stable modes and the algebra of boundary data, Composition in the mixed symbol calculus, Fredholm boundary problems with first-order Calderón defects, Finite defects under perturbation, Reducing first-order boundary data to a split trace, and Doubling a boundary problem and computing its index. We use $D_t=-i\partial_t$, an inward collar coordinate $t\geq0$, and complex bundles. Every displayed product acts from right to left.

1. The arbitrary-order problem and the reduction theorem

Let $X$ be a compact smooth manifold with boundary $Y$. Let $E,F\to X$ and $G_j\to Y$ be smooth complex bundles. Fix $m>1$, and let $P$ be an elliptic mixed operator of order $m$. For $s\geq m$, consider

$$(P,\mathbf B)_s:\bar H^s(X;E)\longrightarrow \bar H^{s-m}(X;F)\oplus \bigoplus_{j=1}^{J}H^{s-m_j-1/2}(Y;G_j). \tag{AR1}$$

Here $B_j$ has total order $m_j$, while its normal order $r_j$ satisfies $r_j<m$. In a collar it has the exact form

$$B_j^b u=\sum_{\ell=0}^{r_j}B_{j\ell}\gamma_\ell u, \qquad B_{j\ell}\in\Psi_{\mathrm{phg}}^{m_j-\ell}(Y;E|_Y,G_j), \qquad \gamma_\ell u=(D_t^\ell u)|_{t=0}. \tag{AR2}$$

There is no hypothesis $m_j<m$. Only the number of normal derivatives is restricted. Assume that the principal boundary map is bijective on the stable Cauchy bundle, so that $(P,\mathbf B)_s$ is Fredholm.

Arbitrary-order reduction theorem. After adding $m-1$ zero-index equations, there is a Fredholm homotopy to a block problem with a common right factor of order $m-1$. Cancelling that factor gives a first-order elliptic boundary problem $(D_*,\boldsymbol\beta)$. Applying the first-order stable reduction and the geometric double gives

$$\boxed{ \operatorname{ind}(P,\mathbf B)_s =\operatorname{ind}(D_*,\boldsymbol\beta)_{s-m+1} =\operatorname{ind}\widehat D_{\mathrm{sp}} =\operatorname{sind}(\widehat d_{\mathrm{sp}}).} \tag{AR3}$$

Every equality will be proved below. In particular, the first equality includes the indices of all auxiliary factors; none is silently omitted. For a tangential row $\beta_j$ on $Y$, the notation $(D_*,\boldsymbol\beta)$ denotes the actual boundary realization $u\mapsto(D_*u,(\beta_j\gamma_0u)_j)$. The same convention applies to tangential rows $\widetilde B_j$ in a first-order realization; every boundary composition below displays the trace.

2. A common right factor and its stable inverse

Choose a positive scalar elliptic tangential operator $\Lambda^+\in\Psi^1_{\mathrm{phg}}(Y;E|_Y)$, constant in the collar variable, with principal symbol $\lambda(y,\eta)I_E$, where $\lambda(y,\eta)>0$ for $\eta\ne0$. Write

$$L=D_t+i\Lambda^+,\qquad R=D_t-i\Lambda^+, \qquad L_\eta=D_t+i\lambda(y,\eta),\quad R_\eta=D_t-i\lambda(y,\eta). \tag{AR4}$$

First consider a problem that already has the collar factorization

$$P^b=(D_t-A)L^{m-1},\qquad B_j^b=\widetilde B_j\gamma_0L^{m-1},\qquad \widetilde B_j\in \Psi_{\mathrm{phg}}^{m_j+1-m}(Y;E|_Y,G_j). \tag{AR5}$$

The leading normal coefficient identifies $F$ with $E$ only in this collar. Globally, the first operator in the product still maps $E$ to $F$.

Freeze the principal coefficients at $(y,\eta)\in T^*Y\setminus0$. If $u$ is a decaying solution of the product equation, set

$$w=L_\eta^{m-1}u. \qquad (D_t-a(y,\eta))w=0,\qquad b_j(y,\eta,D_t)u(0)=\widetilde b_j(y,\eta)w(0). \tag{AR6}$$

The inverse of one $L_\eta$ factor on exponentially decreasing functions is explicit:

$$(T_\lambda g)(t) =-i\int_0^\infty e^{-\lambda s}g(t+s)\,ds, \qquad L_\eta T_\lambda g=g. \tag{AR7}$$

If $\|D_t^kg(t)\|\leq M_ke^{-\delta t}$, then differentiation under the integral gives

$$\|D_t^kT_\lambda g(t)\| \leq\frac{M_k}{\lambda+\delta}e^{-\delta t}. \tag{AR8}$$

The homogeneous equation $L_\eta^qf=0$ has exactly the functions $e^{\lambda t}p(t)$, where $p$ is a vector polynomial of degree less than $q$. None is bounded unless it is zero. It follows that

$$\mathcal M^+\big((D_t-a)L_\eta^{m-1}\big) \xrightarrow[\cong]{\ L_\eta^{m-1}\ } \mathcal M^+(D_t-a),\qquad w\longmapsto T_\lambda^{m-1}w \text{ is the inverse.} \tag{AR9}$$

This proves both injectivity and surjectivity; a dimension count is not being used. The complementing map for the product problem is therefore the composite of this isomorphism with the reduced boundary map:

$$\mathcal M^+\big((D_t-a)L_\eta^{m-1}\big) \xrightarrow{L_\eta^{m-1}} \mathcal M^+(D_t-a) \xrightarrow{(\widetilde b_j)_j} \bigoplus_j(G_j)_y. \tag{AR10}$$

Consequently the higher-order boundary symbol is complementing exactly when the reduced first-order boundary symbol is complementing.

3. The global zero-index factor and the cancellation theorem

Let $\phi,\psi\in C_c^\infty([0,\delta))$ satisfy $0\leq\phi,\psi\leq1$, with $\phi=1$ near $0$, the support of $\phi$ contained in the factorization collar, and $\psi=1$ on the support of $\phi$. Choose a positive scalar elliptic interior operator $\Lambda$ of order one on $E$. Define

$$Q=\psi(D_t+i\Lambda^+)+i(1-\psi)\Lambda(1-\psi): \bar H^q(X;E)\longrightarrow\bar H^{q-1}(X;E). \tag{AR11}$$

Its principal symbol in the transition region is

$$q(x,\xi)=\psi(t)\bigl(\xi_t+i\lambda^+(y,\eta)\bigr) +i(1-\psi(t))^2\lambda(x,\xi). \tag{AR12}$$

For real $\xi_t$ and a nonzero covector, its imaginary scalar part satisfies

$$\operatorname{Im}q =\psi\lambda^++(1-\psi)^2\lambda>0 \quad\text{unless }\psi=1,\ \eta=0, \quad\text{where }q=\xi_t\ne0. \tag{AR13}$$

Thus $Q$ is elliptic. Its collar equation is $Lu=0$, whose stable space is zero, so it needs no boundary operator. It is the split first-order model with the entire bundle in the growing summand. The shift estimate and its adjoint estimate from the split model give

$$Q_q:\bar H^q(X;E)\longrightarrow\bar H^{q-1}(X;E) \text{ Fredholm},\qquad \operatorname{ind}Q_q=0 \quad(q\geq1). \tag{AR14}$$

Now choose a first-order mixed operator $\widetilde P:E\to F$ whose collar and interior pieces have the types

$$\widetilde P^b=\phi(D_t-A):E\longrightarrow F,\qquad \widetilde P^i\in\Psi_{\mathrm{phg}}^1(X^\circ;E,F),\qquad \widetilde P=\widetilde P^b+\widetilde P^i. \tag{AR15}$$

For $s\geq m$, the composed boundary problem is the bounded map

$$(\widetilde P,\widetilde{\mathbf B})_{s-m+1}Q_s^{m-1} =\bigl(\widetilde P Q^{m-1}, (\widetilde B_j\gamma_0Q^{m-1})_j\bigr): \bar H^s(X;E)\longrightarrow \bar H^{s-m}(X;F)\oplus \bigoplus_jH^{s-m_j-1/2}(Y;G_j). \tag{AR16}$$

The index of a composition of Fredholm maps is additive. Since every $Q_q$ has index zero,

$$\operatorname{ind}\bigl(\widetilde P Q^{m-1}, \widetilde{\mathbf B}Q^{m-1}\bigr)_s =\operatorname{ind}(\widetilde P,\widetilde{\mathbf B})_{s-m+1} +(m-1)\operatorname{ind}Q =\operatorname{ind}(\widetilde P,\widetilde{\mathbf B})_{s-m+1}. \tag{AR17}$$

Suppose the product principal symbol is sufficiently close to that of $P$, while its boundary principal symbol is the one in (AR5). The straight segment between the two principal symbols then remains in the open elliptic set, and its quantization gives a Fredholm path. At the product endpoint, two quantizations with that same principal symbol differ in the interior by order at most $m-1$; a boundary remainder one total order lower has the compact mappings

$$\begin{aligned} K&:\bar H^s(X;E)\longrightarrow \bar H^{s-m+1}(X;F)\hookrightarrow\bar H^{s-m}(X;F),\\ R_j&:\bar H^s(X;E)\longrightarrow H^{s-m_j+1/2}(Y;G_j) \hookrightarrow H^{s-m_j-1/2}(Y;G_j). \end{aligned} \tag{AR18}$$

The embeddings are compact because $X$ and $Y$ are compact. Operator-norm openness of the Fredholm set, followed by the compact straight-line path for the retained lower-order remainders, proves the cancellation theorem

$$\boxed{ \operatorname{ind}(P,\mathbf B)_s =\operatorname{ind}(\widetilde P,\widetilde{\mathbf B})_{s-m+1}.} \tag{AR19}$$

The theorem applies only after the global factor $Q^{m-1}$ and the compact comparison have been constructed. The stable-space bijection (AR9) alone does not prove an operator index equality.

4. Stabilizing an arbitrary problem

Return to the original problem (AR1). Put $\mathcal E=E^m$ and $\mathcal F=F\oplus E^{m-1}$. On $U=(U_0,\ldots,U_{m-1})$, first form

$$\mathcal P_0U=\bigl(PU_0,Q^mU_1,\ldots,Q^mU_{m-1}\bigr), \qquad \mathcal B_j^0U=B_jU_0. \tag{AR20}$$

It acts as

$$(\mathcal P_0,\boldsymbol{\mathcal B}^0)_s: \bar H^s(X;\mathcal E)\longrightarrow \bar H^{s-m}(X;\mathcal F)\oplus \bigoplus_jH^{s-m_j-1/2}(Y;G_j). \tag{AR21}$$

This is a direct sum of the original problem and $m-1$ copies of $Q^m$. Therefore

$$\operatorname{ind}(\mathcal P_0,\boldsymbol{\mathcal B}^0)_s =\operatorname{ind}(P,\mathbf B)_s +(m-1)m\operatorname{ind}Q =\operatorname{ind}(P,\mathbf B)_s. \tag{AR22}$$

The literal composition $Q^m$ need not be presented in the permitted mixed operator form. The mixed composition theorem supplies an allowed operator $Q_{[m]}$ with the same collar expression and an arbitrarily small global norm error:

$$Q_{[m]}^b=L^m,\qquad \|Q_{[m]}-Q^m\|_{\bar H^s\to\bar H^{s-m}}<\varepsilon. \tag{AR23}$$

Choose $\varepsilon$ below the Fredholm stability radius for all auxiliary copies. The path

$$Q^m_\rho=(1-\rho)Q^m+\rho Q_{[m]},\qquad0\leq\rho\leq1, \tag{AR24}$$

is Fredholm and has index zero. Hence we may use the allowed $Q_{[m]}$ blocks from now on while retaining (AR22). The replacement has been recorded as a norm-controlled path, rather than identifying two different operator classes.

5. Cayley coefficients with the full collar polynomial retained

Shrink the collar once. On its compact cosphere the original principal symbol has a uniform inverse bound. A cutoff interpolation from its coefficient family at $t$ to the family at $t=0$ is therefore elliptic when the collar is sufficiently thin, by the same inverse-factor estimate used for first-order collar freezing. Extend the coefficients at $t=0$ constantly across the smaller collar and retain the original operator outside the cutoff support. This is an explicit elliptic homotopy of the original principal symbol, not an identification of the two operators.

Use the invertible leading normal coefficient as a collar identification $F\simeq E$. In that frame write the complete collar operator

$$P^b=\sum_{k=0}^{m}P_k^bD_t^k,\qquad P_k^b\in\Psi_{\mathrm{phg}}^{m-k}(Y;E,E),\qquad P_m^b=I_E. \tag{AR25}$$

Let $F_+\in\Psi_{\mathrm{phg}}^{-1}(Y;E)$ be a two-sided parametrix of $\Lambda^+$. For $0\leq j,k\leq m$, set

$$c_{jk}=\frac{i^{k-m}}{2^m} \sum_{r=0}^{j}(-1)^{j-r} \binom{k}{r}\binom{m-k}{j-r}, \qquad \sum_{j=0}^{m}c_{jk}=\delta_{mk}. \tag{AR26}$$

Define the order-zero tangential operators, with the factor order fixed as written,

$$A_j=\sum_{k=0}^{m}c_{jk}P_k^bF_+^{m-k} \in\Psi_{\mathrm{phg}}^0(Y;E,E). \tag{AR27}$$

The numerical identity in (AR26) and $F_+^0=I_E$ give the exact operator identity

$$\sum_{j=0}^{m}A_j =\sum_{k=0}^{m}\left(\sum_{j=0}^{m}c_{jk}\right) P_k^bF_+^{m-k} =P_m^b=I_E. \tag{AR28}$$

No parametrix remainder is used in this calculation. At principal-symbol level, the Cayley formula gives

$$p^b(y,\eta,z) =\sum_{j=0}^{m}a_j(y,\eta) (z-i\lambda)^j(z+i\lambda)^{m-j}, \qquad a_j=\sigma_0(A_j). \tag{AR29}$$

Quantize the right side without changing its factor order:

$$P_{\mathrm{Cay}}^b =\sum_{j=0}^{m}A_jR^jL^{m-j}, \qquad K_{m-1}=P^b-P_{\mathrm{Cay}}^b\in\Psi_{\mathrm{mix}}^{m-1}. \tag{AR30}$$

The remainder contains every parametrix defect, commutator, and lower total-order coefficient. It is not erased. The collar path

$$P_\rho^b=P^b-\rho K_{m-1} =(1-\rho)P^b+\rho P_{\mathrm{Cay}}^b,\qquad0\leq\rho\leq1, \tag{AR31}$$

has fixed principal symbol. With a collar cutoff it extends by the original operator outside the collar. Its contribution $K_{m-1}:\bar H^s\to\bar H^{s-m}$ is compact by (AR18), so the Fredholm index remains fixed. We now work at the exact endpoint $P_{\mathrm{Cay}}^b$, while (AR31) retains the relation to the original operator.

6. The quantized block homotopy

Choose $\chi\in C_c^\infty([0,\delta))$, equal to one on a smaller boundary collar, with support inside the region where (AR30) is valid. For $0\leq\tau\leq1$, write

$$\sigma(t)=\tau\chi(t),\qquad0\leq\sigma(t)\leq1. \tag{AR32}$$

On the collar define an order-$m$ operator matrix $\mathfrak P_\tau$ on $U=(U_0,\ldots,U_{m-1})$. Its first row is

$$\begin{aligned} (\mathfrak P_\tau U)_0={}& \bigl((1-\sigma^m)P_{\mathrm{Cay}}^b +\sigma^mA_0L^m\bigr)U_0\\ &+\sum_{j=1}^{m-2}\sigma^{m-j}A_jL^mU_j +\sigma\bigl(A_{m-1}L^m+A_mRL^{m-1}\bigr)U_{m-1}, \end{aligned} \tag{AR33}$$

and its remaining rows are

$$(\mathfrak P_\tau U)_j =L^mU_j-\sigma RL^{m-1}U_{j-1}, \qquad1\leq j<m. \tag{AR34}$$

For $m=2$, the sum in (AR33) is empty. The endpoint formulas are

$$\mathfrak P_0 =\operatorname{diag}(P_{\mathrm{Cay}}^b,L^m,\ldots,L^m), \qquad \mathfrak P_1^b=\mathfrak F_*L^{m-1} \quad\text{where }\chi=1, \tag{AR35}$$

with the common factor on the right. In the transition region derivatives of $\chi$ and all quantization commutators have lower total order and remain in the displayed operator; none changes the principal symbol.

Let $S(U_0,\ldots,U_{m-1})=(0,U_0,\ldots,U_{m-2})$. If $C_\sigma$ is the leading normal coefficient of $\mathfrak P_\tau$, define $H_\sigma=I+N_\sigma$, where the only nonzero off-diagonal blocks of $N_\sigma$ are

$$(N_\sigma)_{0j}=h_j =\sigma^{m-j}\sum_{k=j}^{m}A_k,\qquad1\leq j<m. \qquad C_\sigma=H_\sigma(I-\sigma S). \tag{AR36}$$

Since $N_\sigma^2=0$, $S^m=0$, and (AR28) is exact, the inverse is the finite operator matrix

$$C_\sigma^{-1} =\left(\sum_{r=0}^{m-1}\sigma^rS^r\right)(I-N_\sigma), \qquad (C_\sigma^{-1})_{ij}= \begin{cases} \sigma^{i-j}I-\sigma^ih_j,&i\geq j\geq1,\\ -\sigma^ih_j,&1\leq j>i,\\ \sigma^iI,&j=0. \end{cases} \tag{AR37}$$

Each entry is linear in the $A_k$ after replacing $I$ by $\sum A_k$. Thus this calculation remains valid in the noncommutative order-zero operator algebra. Editorial correction of the global target map. The matrix $C_\sigma$ and its inverse above are operators in the collar target frame. They do not identify $F$ with $E$ over all of $X$. Let $c:E|_C\to F|_C$ be the retained normal-coefficient identification on the collar $C$, and put $$J_c=\operatorname{diag}(c,I,\ldots,I):\mathcal E|_C\longrightarrow\mathcal F|_C, \qquad M_\sigma=J_cC_\sigma J_c^{-1}, \qquad M_\sigma^{-1}=J_cC_\sigma^{-1}J_c^{-1}. \tag{AR37a}$$ Every product is a composition of tangential operators on collar sections. No factor is moved through a coefficient. Direct multiplication gives both inverse identities because the intervening $J_c^{-1}J_c$ and $C_\sigma^{-1}C_\sigma$ cancel in their displayed order. Where $\chi=0$, one has $\sigma=0$, $N_0=0$, and $C_0=I$, hence $M_0=I_{\mathcal F}$. The difference $M_\sigma-I$ is supported in the collar where $c$ is defined, so extend $M_\sigma$ and its displayed inverse by the identity outside that collar.

These are bounded inverse operators on every restriction Sobolev space $\bar H^q(X;\mathcal F)$. To check the assertion locally, tangential order-zero families are bounded in tangential Sobolev norms; each normal derivative gives a finite sum of their normal coefficient derivatives, again of tangential order zero, applied to normal derivatives of the input. The integer whole-space Sobolev estimate follows by summing these terms. The transposed families have the same property, giving the negative integer estimates by duality; interpolation gives every real exponent. Extend the smooth family across the collar boundary before applying this argument, and then restrict. Since the operator acts at each fixed normal coordinate, it preserves extension differences vanishing in the interior, proving the quotient-space bound. A finite chart partition completes the global estimate. The entries depend continuously on $\tau$ in these bounds, and the exact inverses remain as written.

Thus the global comparison with the original target retained is

$$\widehat{\mathfrak P}_\tau=M_\sigma^{-1}\mathfrak P_\tau, \qquad \mathfrak P_\tau=M_\sigma\widehat{\mathfrak P}_\tau. \tag{AR38}$$

In the collar frame this is exactly $C_\sigma^{-1}J_c^{-1}\mathfrak P_\tau$; multiplication by $J_c$ returns it to $\mathcal F$. Postcomposition by the global invertible $M_\sigma^{-1}$ identifies the original range quotient with the new one and leaves the kernel unchanged. Its index is zero, so this step adds no index contribution and changes no boundary measurement.

The square is an exact identity of operators on collar sections, with vertical maps $J_c$. Outside the collar the target automorphism is the identity. The full formulas and inverse proof are (AR37a)–(AR38).

At a frozen nonzero tangential covector, write $p(z)$ for the original monic normal polynomial and $L(z)=z+i\lambda$. Block elimination gives the exact identity

$$\det\mathfrak p_\sigma(z) =\det p(z)\,L(z)^{m(m-1)\operatorname{rank}E}. \tag{AR39}$$

Indeed, the lower-right block has determinant $L^{m(m-1)\operatorname{rank}E}$; its Schur complement is $(1-\sigma^m)p+\sigma^m\sum a_jR^jL^{m-j}=p$. Polynomial continuation covers $L=0$. For real $z$, ellipticity of $p$ and $\lambda>0$ imply

$$\det\mathfrak p_\sigma(z)\ne0 \qquad(0\leq\sigma\leq1). \tag{AR40}$$

The same identity applies pointwise with $\sigma=\tau\chi(t)$. It proves ellipticity throughout the collar transition, while the operator remains the original stabilized problem outside the support of $\chi$.

7. The stable space along the block homotopy

Freeze again at $(y,\eta)$, now in the region $\chi=1$. The lower rows of the equation say $L_\eta^{m-1}(L_\eta U_j-\tau R_\eta U_{j-1})=0$. The expression in parentheses is bounded for a stable solution, and the bounded kernel of $L_\eta^{m-1}$ is zero. Therefore

$$L_\eta U_j=\tau R_\eta U_{j-1},\qquad L_\eta^jU_j=\tau^jR_\eta^jU_0 \quad(1\leq j<m). \tag{AR41}$$

Substitution into the first row, with all $A_j$ left of the scalar factors, gives $p(D_t)U_0=0$. Hence

$$\Pi_\tau:\mathcal M^+(\mathfrak p_\tau)\longrightarrow \mathcal M^+(p),\qquad U\longmapsto U_0 \tag{AR42}$$

is injective. Its inverse is constructive. For $u\in\mathcal M^+(p)$, define

$$U_0=u,\qquad U_j=\tau T_\lambda R_\eta U_{j-1} =\tau^j(T_\lambda R_\eta)^ju,\qquad1\leq j<m. \tag{AR43}$$

The estimate (AR8), applied successively, proves that every component and all its derivatives decrease exponentially. The recurrences hold, and substitution in the first row gives

$$(1-\tau^m)p(D_t)u +\tau^m\sum_{j=0}^{m}a_jR_\eta^jL_\eta^{m-j}u =p(D_t)u=0. \tag{AR44}$$

Thus (AR42) is surjective. On compact subsets of $T^*Y\setminus0$, the integral formula and its parameter derivatives are uniformly bounded because $\lambda$ has a positive lower bound. The stable spaces therefore form smooth bundles, and

$$\begin{array}{ccc} \mathcal M^+(\mathfrak p_\tau)&\xrightarrow{\Pi_\tau}&\mathcal M^+(p)\\ \downarrow&&\downarrow\\ T^*Y\setminus0&=&T^*Y\setminus0 \end{array} \quad\text{is a smooth bundle isomorphism for every }0\leq\tau\leq1. \tag{AR45}$$

The auxiliary factor adds only the lower-half-plane root $-i\lambda$, with its full multiplicity recorded in (AR39). It adds no stable Cauchy data.

8. Transporting every boundary measurement

Retain the complete boundary operator (AR2). Its frozen normal polynomial is

$$b_j(y,\eta,z)=\sum_{\ell=0}^{r_j}b_{j\ell}(y,\eta)z^\ell, \qquad r_j<m,\qquad b_{j\ell}\text{ homogeneous of degree }m_j-\ell. \tag{AR46}$$

The degree-$m-1$ Cayley basis gives unique symbols $\beta_{jk}$ such that

$$b_j(z)=\sum_{k=0}^{m-1}\beta_{jk}R(z)^kL(z)^{m-1-k}, \quad \beta_{jk}=\sum_{\ell=0}^{r_j}c'_{k\ell} b_{j\ell}\lambda^{\ell+1-m}, \tag{AR47}$$

where

$$c'_{k\ell}=\frac{i^{\ell+1-m}}{2^{m-1}} \sum_{a=0}^{k}(-1)^{k-a} \binom{\ell}{a}\binom{m-1-\ell}{k-a}. \tag{AR48}$$

Because $b_{j\ell}$ has degree $m_j-\ell$, every term in $\beta_{jk}$ has degree $(m_j-\ell)+(\ell+1-m)=m_j+1-m$. Quantization and descending-order correction therefore give

$$\beta_j=(\beta_{j0},\ldots,\beta_{j,m-1}) \in\Psi_{\mathrm{phg}}^{m_j+1-m} (Y;E^m|_Y,G_j). \tag{AR49}$$

Editorial correction of the frozen-symbol identity. Let $\beta_j^{\mathrm{pr}}=(\beta_{j0},\ldots,\beta_{j,m-1})$ denote the row of symbols in (AR47), before quantization. At $\tau=1$, (AR41) gives the following exact equality of functions for a frozen stable solution $U$:

$$b_j(y,\eta,D_t)U_0 =\sum_{k=0}^{m-1}\beta_{jk}(y,\eta)R_\eta^kL_\eta^{m-1-k}U_0 =\sum_{k=0}^{m-1}\beta_{jk}(y,\eta)L_\eta^{m-1}U_k =\beta_j^{\mathrm{pr}}(y,\eta)L_\eta^{m-1}U. \tag{AR50}$$

The formula containing $\tau^{-k}$ is needed only to derive this endpoint identity; it is never used at $\tau=0$. Define the final boundary homotopy solely at the elliptic interior endpoint:

$$\mathcal B_j^\kappa U =(1-\kappa)B_jU_0+\kappa\beta_j\gamma_0L^{m-1}U, \qquad0\leq\kappa\leq1. \tag{AR51}$$

If $\mathcal C^+$ is the stable trace bundle of the frozen principal polynomial $\mathfrak p_1$, let $\mathfrak b_j^\kappa$ denote the principal boundary map of (AR51). For the stable solution determined by a vector in $\mathcal C^+$, evaluation of (AR50) at zero gives

$$\left.\mathfrak b_j^\kappa(y,\eta)\right|_{\mathcal C^+} =\bigl[U\mapsto b_j(y,\eta,D_t)U_0(0)\bigr] =\bigl[U\mapsto\beta_j^{\mathrm{pr}}(y,\eta)(L_\eta^{m-1}U)(0)\bigr]. \tag{AR52}$$

This identity holds for every $\kappa$; it is a principal-symbol identity, while the full quantized operator remains (AR51) with its lower-order terms. Thus the boundary principal map is literally constant along $\kappa$, rather than merely a convex path between invertible maps. The component orders are exact:

$$\bar H^s(X;E^m) \xrightarrow{L^{m-1}} \bar H^{s-m+1}(X;E^m) \xrightarrow{\beta_j\gamma_0} H^{s-m_j-1/2}(Y;G_j). \tag{AR53}$$

The complementing condition and Fredholm index are therefore fixed throughout (AR51). A quantized lower-order boundary remainder maps first into $H^{s-m_j+1/2}$ and is compact into the target in (AR53), so its retained straight-line correction also leaves the index fixed.

9. The endpoint factor and its global cancellation

On the smaller collar, combine (AR35), (AR38), and (AR51). In the collar target frame write $\mathfrak P_1^{\mathrm{fr}}=J_c^{-1}\mathfrak P_1=\mathfrak F_*^{\mathrm{fr}}L^{m-1}$. The explicit first-order factor with the original target is $D_*^b=J_cC_1^{-1}\mathfrak F_*^{\mathrm{fr}}$. Thus there is a first-order mixed operator matrix $D_*^b:E^m\to F\oplus E^{m-1}$ such that

$$\widehat{\mathfrak P}_1^b=D_*^bL^{m-1}, \qquad \mathcal B_j^1=\beta_j\gamma_0L^{m-1}. \tag{AR54}$$

At frozen principal-symbol level the scalar factor can be written on either side. At operator level, (AR54) is the required right factor; it has not been commuted through the tangential coefficients.

Let $Q_{\mathcal E}$ be the operator (AR11) on $\mathcal E=E^m$. Its principal symbol $q_{\mathcal E}$ is invertible off the zero section. If $\widehat{\mathfrak p}_1$ is the complete endpoint principal symbol, define the first-order symbol

$$d_*(x,\xi)=\widehat{\mathfrak p}_1(x,\xi) q_{\mathcal E}(x,\xi)^{1-m}: \pi^*\mathcal E\longrightarrow\pi^*\mathcal F. \tag{AR55}$$

It agrees with the symbol of $D_*^b$ in the collar. Quantize it by a global first-order mixed operator $D_*:\mathcal E\to\mathcal F$ with that collar form. The exact endpoint and the composition retain a lower-order difference:

$$\widehat{\mathfrak P}_1-D_*Q_{\mathcal E}^{m-1}=K_{m-1},\qquad \mathcal B_j^1-\beta_j\gamma_0Q_{\mathcal E}^{m-1}=R_j, \tag{AR56}$$

where $K_{m-1}$ and $R_j$ have the compact mappings in (AR18). The full symbols are joined first inside the open elliptic set; then the retained compact remainders are joined by their straight-line paths. The cancellation theorem gives

$$\operatorname{ind}(\widehat{\mathfrak P}_1,\boldsymbol{\mathcal B}^1)_s =\operatorname{ind}(D_*,\boldsymbol\beta)_{s-m+1} +(m-1)\operatorname{ind}Q_{\mathcal E} =\operatorname{ind}(D_*,\boldsymbol\beta)_{s-m+1}. \tag{AR57}$$

The reduced map has the unchanged boundary targets

$$(D_*,\boldsymbol\beta)_{s-m+1}: \bar H^{s-m+1}(X;\mathcal E)\longrightarrow \bar H^{s-m}(X;\mathcal F)\oplus \bigoplus_jH^{s-m_j-1/2}(Y;G_j). \tag{AR58}$$

This verifies the source, interior target, and every boundary target after cancellation.

10. The complete index chain

Combining stabilization, the lower-order collar comparison, the block homotopy, the boundary homotopy, and cancellation gives

$$\begin{aligned} \operatorname{ind}(P,\mathbf B)_s &=\operatorname{ind}(\mathcal P_0,\boldsymbol{\mathcal B}^0)_s\\ &=\operatorname{ind}(\widehat{\mathfrak P}_1,\boldsymbol{\mathcal B}^1)_s\\ &=\operatorname{ind}(D_*,\boldsymbol\beta)_{s-m+1}. \end{aligned} \tag{AR59}$$

If a component $\beta_j$ has nonzero order $\mu_j=m_j+1-m$, choose an exact invertible order reducer $J_{G_j}^{-\mu_j}$ on its target. Then

$$\beta_j^{(0)}=J_{G_j}^{-\mu_j}\beta_j,\qquad \operatorname{ind}(D_*,\boldsymbol\beta^{(0)}) =\operatorname{ind}(D_*,\boldsymbol\beta) +\sum_j\operatorname{ind}J_{G_j}^{-\mu_j} =\operatorname{ind}(D_*,\boldsymbol\beta). \tag{AR60}$$

The last equality uses invertible reducers, whose indices are zero. The first-order stable reduction deforms this problem to a split trace problem without changing its index. Its zero-index reflected complement glues to a closed doubled operator $\widehat D_{\mathrm{sp}}$, and ordinary operator-norm approximation gives its symbol index:

$$\operatorname{ind}(D_*,\boldsymbol\beta) =\operatorname{ind}(D_{\mathrm{sp}},B_{\mathrm{sp}}) =\operatorname{ind}\widehat D_{\mathrm{sp}} =\operatorname{sind}(\widehat d_{\mathrm{sp}}). \tag{AR61}$$

Equations (AR59)–(AR61) prove (AR3) for the original problem. The stabilization is specialized back by (AR22), and the cancelled factors are specialized back by (AR57); the conclusion is not confined to the terminal block model.

The left panel depicts (AR20)–(AR24), the middle panels depict (AR33)–(AR54), and the last panel depicts (AR57)–(AR61). The blue strand is the stable Cauchy bundle carried by the first-coordinate isomorphism (AR42). The amber strands are the growing auxiliary modes of $L^m$; they never enter the stable boundary map.

11. Two exact calculations

Take $m=2$, $\lambda=1$, and

$$p(z)=(z-2i)(z+3i)=z^2+iz+6. \tag{AR62}$$

For $L=z+i$ and $R=z-i$, direct coefficient comparison gives

$$p(z)=-L(z)^2+\frac72R(z)L(z)-\frac32R(z)^2, \qquad -1+\frac72-\frac32=1. \tag{AR63}$$

The block path is

$$\begin{aligned} (\mathfrak p_\tau U)_0 &=\bigl((1-\tau^2)p-\tau^2L^2\bigr)U_0 +\tau\left(\frac72L^2-\frac32RL\right)U_1,\\ (\mathfrak p_\tau U)_1&=L^2U_1-\tau RLU_0. \end{aligned} \tag{AR64}$$

At $\tau=1$, every term has the common right factor $L$. If the boundary measurement is $B(z)=z$, then

$$z=\frac12L(z)+\frac12R(z),\qquad B(D)U_0=\frac12LU_0+\frac12LU_1 \quad\text{on }\mathcal M^+(\mathfrak p_1), \tag{AR65}$$

because $LU_1=RU_0$. This is the endpoint identity (AR50) in a case where every coefficient can be checked by multiplication.

For a separate order calculation, let $m=3$ and let a boundary row have total order $m_j=5$ but normal order two:

$$B_j=B_{j0}\gamma_0+B_{j1}\gamma_1+B_{j2}\gamma_2,\qquad \operatorname{ord}B_{j0}=5,\quad \operatorname{ord}B_{j1}=4,\quad \operatorname{ord}B_{j2}=3. \tag{AR66}$$

Every Cayley coefficient $\beta_{jk}$ belongs to $\Psi^{5+1-3}=\Psi^3$, and $\beta_j\gamma_0L^2$ belongs to the declared total-order-five boundary class. Cancellation of a leading component can lower its actual order. This verifies directly that a large total boundary order is compatible with the reduction; the essential restriction is the normal degree $2<3$.

12. A typed correction and the exact limits of the argument

The global interior part of the reduced first-order operator in (AR15) must have type $E\to F$. Writing it as $E\to E$ is valid only after choosing a global bundle isomorphism $F\simeq E$, while ellipticity supplies such an identification only in the boundary collar. This is a bundle-target correction, not a change to the collar calculation.

The argument also proves more generality than an order comparison might suggest: the integers $m_j$ are unrestricted. Formula (AR49) remains valid for positive, zero, or negative $m_j+1-m$. What must remain below $m$ is the normal degree $r_j$, because the endpoint Cayley basis in (AR47) has degree $m-1$.

Three limits remain explicit. The construction assumes an existing complementing boundary system; it does not remove the stable-bundle obstruction. It proves stable cancellation, not equality of all solutions of a product equation. Finally, it uses a zero-index global completion $Q$; a local factor $L^{m-1}$ without that completion does not justify the global index equality.

13. Exercises with complete solutions

Exercise 1. Verify the sign and uniqueness in (AR7).

Solution. Put $I(t)=\int_0^\infty e^{-\lambda s}g(t+s)\,ds$. Integration by parts gives $I'=-g+\lambda I$. For $f=-iI$,

$$D_tf+i\lambda f=-if'+i\lambda f=g. \tag{AR67}$$

The difference of two bounded solutions solves $L_\eta h=0$, hence equals $ce^{\lambda t}$, so $c=0$.

Exercise 2. Prove directly that the transition symbol (AR12) has no real zero.

Solution. If $0\leq\psi<1$, then $\psi\lambda^++(1-\psi)^2\lambda>0$, so the imaginary part is nonzero. If $\psi=1$ and $\eta\ne0$, it equals $\lambda^+>0$. If $\psi=1$ and $\eta=0$, a nonzero covector has $\xi_t\ne0$, and

$$q=\xi_t\ne0. \tag{AR68}$$

Exercise 3. Derive the sum identity in (AR28) without assuming that $F_+$ is an exact inverse.

Solution. Sum (AR27) over $j$, retain the displayed order, and use only the numerical identity in (AR26):

$$\sum_jA_j =\sum_k\delta_{mk}P_k^bF_+^{m-k} =P_m^bF_+^0=I_E. \tag{AR69}$$

No product $F_+\Lambda^+$ occurs.

Exercise 4. Check the inverse in (AR37).

Solution. Since $N_\sigma^2=0$, $H_\sigma^{-1}=I-N_\sigma$. Since $S^m=0$,

$$(I-\sigma S)^{-1}=\sum_{r=0}^{m-1}\sigma^rS^r,\qquad C_\sigma^{-1}C_\sigma =(I-\sigma S)^{-1}H_\sigma^{-1}H_\sigma(I-\sigma S)=I. \tag{AR70}$$

Multiplying the two finite block matrices gives the component formula in (AR37) without interchanging any $A_k$.

Exercise 5. Explain why the boundary path (AR51) cannot lose the complementing condition.

Solution. On the frozen stable trace bundle, evaluation of (AR50) at zero makes the two principal boundary maps equal. Put $T_0U=b_j(y,\eta,D_t)U_0(0)$ and $T_1U=\beta_j^{\mathrm{pr}}(y,\eta)(L_\eta^{m-1}U)(0)$. Then $T_0=T_1$, so for every $\kappa$,

$$(1-\kappa)T_0+\kappa T_1=T_0. \tag{AR71}$$

The complete boundary map on the stable bundle is therefore the original bijection for the entire path.

Exercise 6. Locate every zero contribution in the index chain.

Solution. The $m-1$ stabilization blocks each contain $Q^m$, so their total contribution is $(m-1)m\operatorname{ind}Q=0$. Cancellation removes $Q^{m-1}$, contributing $(m-1)\operatorname{ind}Q=0$. Exact target reducers and the reflected complementary half are invertible or zero-index. Thus

$$\operatorname{ind}(P,\mathbf B) =\operatorname{ind}(D_*,\boldsymbol\beta) =\operatorname{ind}\widehat D_{\mathrm{sp}}, \tag{AR72}$$

with the homotopy equalities supplied by (AR31), (AR33)–(AR40), and (AR51)–(AR53).

14. Reading notes and references

The finite-dimensional Cayley identities, their noncommutative leading-matrix factorization, and the explicit stable lift are proved in Stable modes and the algebra of boundary data. The global approximation and compact composition steps use Composition in the mixed symbol calculus and Finite defects under perturbation. The first-order endpoint is treated in Reducing first-order boundary data to a split trace, and its closed-manifold realization is proved in Doubling a boundary problem and computing its index.

This is an independent teaching derivation. Its historical statement route is recorded separately from the proof, and exact correspondence to an original-author source remains open.

Written and dedicated to the public domain by Codex under CC0 1.0.

15. Editorial receiving proofs for the global reduction

The following arguments supply the analytic maps used in (AR14), (AR23), (AR38), (AR56), and (AR60). They retain the original operators and their factor order. The normal polynomial degree $m$ is an integer: the $m$ components in (AR20), the coefficient $P_m^b$, and the powers in the stated construction all use that same integer $m>1$.

15.1. The actual zero-index factor and every intervening Sobolev space

The operator $Q$ in (AR11) is exactly the split operator (DI2) of Doubling a boundary problem and computing its index, with $$E^+=E,\qquad E^-=0,\qquad \phi=\psi,\qquad P^b=\psi(D_t+i\Lambda^+),\qquad P^i=i(1-\psi)\Lambda(1-\psi),\qquad B=0. \tag{ARX1}$$ Both occurrences of $1-\psi$ remain. The transition symbol is consequently (AR12), and the frozen stable space is the zero bundle by (DI4). The empty boundary map is the unique isomorphism from that zero stable bundle to the zero measurement bundle. Thus the Fredholm and dual regularity theorem in Fredholm boundary problems with first-order Calderón defects applies with precisely this boundary datum.

For clarity, the index-zero proof includes the cokernel. Use a product density in the collar to compute the pairing; equivalent smooth densities do not change the realization or its index. The original inward coordinate and linear-first pairing give $$2\operatorname{Im}(\psi D_tu,u) =\|\gamma_0u\|_Y^2+\int\psi'(t)|u|^2\,dV. \tag{ARX2}$$ For each positive scalar principal operator $A$, the retained identity (DA1) is $(A+A^*)/2=R_A^*R_A+H_A$, with $H_A$ of order zero. Applying it to $\Lambda^+$ and to $\Lambda$ on the actual input $(1-\psi)u$, and retaining the $\psi'$ and all patching terms, yields $$2\operatorname{Im}(Qu,u)\ge \|\gamma_0u\|_Y^2-C\|u\|_X^2. \tag{ARX3}$$ Choose $T>C/2$, enlarging $C$ once for the adjoint. Then $Q+iTI$ has zero kernel. A vector $v$ in its cokernel at the realization $\bar H^1\to L^2$ is smooth by dual regularity. Green’s identity gives $$(Q^*-iT)v=0,\qquad \gamma_0v=0. \tag{ARX4}$$ Indeed the boundary term is $i(\gamma_0u,\gamma_0v)$, and arbitrary smooth traces force the displayed condition. The operator $-Q^*$ has inward normal coefficient $-1$, positive scalar imaginary principal terms, and the complete differentiated-cutoff terms of (DA3)–(DA5). Apply its version of (ARX3) under $\gamma_0v=0$. The shift $+iT$ then gives $v=0$. The shifted operator is therefore bijective. The path $Q+irTI$, $0\le r\le1$, keeps both principal symbols fixed and remains Fredholm. Its index is zero throughout. The same smooth kernel and dual obstruction spaces represent every realization $q\ge1$, proving (AR14) without deriving an index from ellipticity alone.

The actual powers have the separately typed factorization $$Q_s^m=Q_{s-m+1}\circ Q_{s-m+2}\circ\cdots\circ Q_s: \bar H^s(E)\longrightarrow\bar H^{s-m}(E), \tag{ARX5}$$ $$Q_s^{m-1}=Q_{s-m+2}\circ\cdots\circ Q_s: \bar H^s(E)\longrightarrow\bar H^{s-m+1}(E). \tag{ARX6}$$ For $s\ge m$, every source exponent of a factor is at least one. Fredholm product additivity gives zero index for both powers. The same proof on $\mathcal E=E^m$ gives $Q_{\mathcal E}^{m-1}$ and its zero contribution in (AR57). Stabilization still contributes exactly $(m-1)m\operatorname{ind}Q$; cancellation contributes exactly $(m-1)\operatorname{ind}Q_{\mathcal E}$.

15.2. The collar freeze and the pure normal covectors after target normalization

Use the original collar pullbacks of $E|_Y$ and $F|_Y$, and write $p_t(y,\eta,z):E_y\to F_y$ for the complete homogeneous principal polynomial. Its inverse on the compact real unit cosphere is uniformly bounded. Smoothness in $t$ gives a collar thickness for which $$\sup_{y,\,|\eta|^2+z^2=1,\,0\le t\le\delta_0} \|p_t^{-1}(p_0-p_t)\|=\theta<1. \tag{ARX7}$$ With a cutoff $\chi_0$ supported in that collar, retain the exact path and ordered inverse $$p_{r,t}=p_t+r\chi_0(t)(p_0-p_t),\qquad p_{r,t}^{-1} =[I+r\chi_0p_t^{-1}(p_0-p_t)]^{-1}p_t^{-1}, \tag{ARX8}$$ $$[I+r\chi_0p_t^{-1}(p_0-p_t)]^{-1} =\sum_{\nu=0}^\infty[-r\chi_0p_t^{-1}(p_0-p_t)]^\nu. \tag{ARX9}$$ The ratio is at most $\theta$. Homogeneity extends invertibility from the unit cosphere to every nonzero real covector. At $t=0$, the path is exactly $p_0$, so the stable polynomial and every original boundary principal measurement are unchanged. Quantize the coefficient differences in their original orders, retain the original operator outside the cutoff support, and keep all lower terms. The resulting norm-continuous Fredholm path is the freezing operation used in Section 5. It does not require an unproved freezing assertion from another unit.