Reducing first-order boundary data to a split trace

The index of a first-order elliptic boundary problem can be computed after a sequence of explicit deformations. Each deformation has to preserve two structures at once: the interior symbol must remain invertible, and the boundary symbol must remain an isomorphism on the decaying Cauchy data. This lesson constructs the whole sequence. It normalizes the normal coefficient, freezes the collar symbol, collapses all spectral rates while keeping the Calderón projection fixed, rotates the boundary isomorphism after a zero-index stabilization, and reaches the split trace model used by doubling.

The formulas keep every bundle map and every matrix factor in its typed order. In particular, the inverse of the boundary symbol is defined only on the stable bundle, the two off-diagonal blocks of the rotation have different source and target bundles, and a quantized projection is never treated as an exact operator projection.

The named prerequisites are Stable modes and the algebra of boundary data, Cauchy data from jumps and residues, Fredholm boundary problems with first-order Calderón defects, Finite defects under perturbation, Symbols, finite defects, and the index on a closed manifold, The Bott operator, suspension, and reduction of the index to Euclidean space, 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.

1. The starting problem and the reduction theorem

Let $X$ be a compact smooth manifold with boundary $Y$. Let $E,F\to X$ and $G\to Y$ be smooth complex bundles. Consider a first-order elliptic operator and an order-zero boundary operator

$$(P,B)_1:H^1(X;E)\longrightarrow L^2(X;F)\oplus H^{1/2}(Y;G), \qquad B=\mathcal B\gamma_0 . \tag{BR1}$$

Here $\mathcal B\in\Psi_{\mathrm{phg}}^0(Y;E|_Y,G)$. Write $p$ for the principal symbol of $P$. Ellipticity at a nonzero conormal makes

$$c(y,t)=p(y,t,0,dt):E_{(y,t)}\longrightarrow F_{(y,t)} \quad\text{invertible on a sufficiently small collar.} \tag{BR2}$$

After using $c$ to identify the target with the source in that collar, the boundary principal symbol has the form

$$\widetilde p(y,0,\eta,\zeta)=\zeta I_E-a(y,\eta),\qquad q(y,\eta)=\frac{1}{2\pi i}\int_{\Gamma_+} (zI_E-a(y,\eta))^{-1}\,dz . \tag{BR3}$$

The contour surrounds the spectrum of $a(y,\eta)$ in the upper half-plane. The stable Cauchy bundle is $L=\operatorname{ran}q$ over $T^*Y\setminus0$. If $b=\sigma_0(\mathcal B)$, the complementing condition is precisely that $b|_L:L\to\pi^*G$ be a bundle isomorphism.

Stable reduction theorem. There is a bundle $H\to X$, a zero-index auxiliary first-order problem on $H$, and a continuous Fredholm deformation of the stabilized problem to a split bundle-trace problem. If $(P_{\mathrm{sp}},B_{\mathrm{sp}})$ is the endpoint and $\widehat P_{\mathrm{sp}}$ is its geometric double, then

$$\boxed{ \operatorname{ind}(P,B) =\operatorname{ind}(P_{\mathrm{sp}},B_{\mathrm{sp}}) =\operatorname{ind}\widehat P_{\mathrm{sp}} =\operatorname{sind}(\widehat p_{\mathrm{sp}}).} \tag{BR4}$$

The theorem assumes that the original complementing operator $B$ exists. Section 10 records the exact symbol-bundle obstruction exposed when no such $B$ exists.

2. Normal coefficient normalization and collar freezing

Choose a collar $Y\times[0,\delta)_t$. The isomorphism $c^{-1}:F|_{Y\times[0,\delta)}\to E|_{Y\times[0,\delta)}$ is used as a collar frame for the target. This is a change of bundle coordinates, so it does not alter the operator, its domain, its kernel, or its cokernel. In this target frame the collar operator can be written

$$P^b=D_t-A(t)+R_0(t), \qquad \sigma_1(P^b)(y,t,\eta,\zeta)=\zeta I_E-a(y,t,\eta), \tag{BR5}$$

where $A(t)$ is tangential of order one and $R_0(t)$ has order zero. No factor of the original normal coefficient has been discarded: it remains the target identification $c$. Every normalized symbol deformation below is transported back to $F$ by left multiplication with $c$, so the global target bundle stays fixed.

Let $\chi\in C_c^\infty([0,\delta))$, with $\chi=1$ for $0\leq t\leq\delta/3$ and $\chi=0$ for $t\geq2\delta/3$. For $0\leq r\leq1$, define

$$a_r(y,t,\eta) =a(y,t,\eta)+r\chi(t)\bigl(a(y,0,\eta)-a(y,t,\eta)\bigr). \tag{BR6}$$

This is a deformation of the full original principal symbol, not a replacement by a rescaled symbol. To prove ellipticity, restrict to the compact collar cosphere $|\eta|^2+\zeta^2=1$. Since the original symbol is invertible there, there is $M<\infty$ such that

$$\sup\|(\zeta I_E-a(y,t,\eta))^{-1}\|\leq M . \tag{BR7}$$

Uniform continuity gives, after decreasing $\delta$,

$$\|r\chi(t)(a(y,0,\eta)-a(y,t,\eta))\| <\frac{1}{2M} \quad\text{on the collar cosphere.} \tag{BR8}$$

Factor the deformed symbol on the right of the original inverse:

$$\zeta I_E-a_r =(\zeta I_E-a) \left[I_E-(\zeta I_E-a)^{-1} r\chi(t)(a(y,0,\eta)-a(y,t,\eta))\right]. \tag{BR9}$$

The bracket is invertible by its Neumann series. Thus every $a_r$ is elliptic. At $t=0$, the symbol is always $a(y,0,\eta)$, so the stable bundle and $b|_L$ do not change. Transporting this principal deformation back by $c$, and interpolating the lower-order terms, gives an operator-norm continuous Fredholm path on the original source and target bundles. At $r=1$, the tangential operator is independent of $t$ on the smaller collar. All cutoff-transition terms are supported where $t\geq\delta/3$.

3. The smooth stable projection and the fixed-projection collapse

For a fixed $(y_0,\eta_0)$, choose a contour $\Gamma_+$ that separates the upper and lower spectra of $a(y_0,\eta_0)$. Invertibility on this compact contour persists in a parameter neighborhood. Differentiating the resolvent gives

$$\partial_\nu(zI-a)^{-1} =(zI-a)^{-1}(\partial_\nu a)(zI-a)^{-1}. \tag{BR10}$$

Repeated differentiation under the contour integral proves that $q$ is smooth without choosing eigenvalue branches. Two local contours give the same projection on their overlap because both enclose exactly the upper spectrum. Positive homogeneity follows by scaling both the symbol and the contour:

$$q(y,\rho\eta) =\frac{1}{2\pi i}\int_{\rho\Gamma_+} (zI-\rho a(y,\eta))^{-1}\,dz =q(y,\eta),\qquad \rho>0 . \tag{BR11}$$

Choose $Q\in\Psi_{\mathrm{phg}}^0(Y;E|_Y)$ with principal symbol $q$, and choose $\Lambda\in\Psi_{\mathrm{phg}}^1(Y;E|_Y)$ with positive scalar principal symbol $\lambda(y,\eta)I_E$. The operator $Q$ need not satisfy $Q^2=Q$; only its principal symbol enters the following ellipticity calculation.

At the symbol level define, for $0\leq r\leq1$,

$$a^{\langle r\rangle} =(1-r)a+i r\lambda(2q-I_E). \tag{BR12}$$

The contour formula makes $q$ a holomorphic function of $a$, so $aq=qa$. On the primary summand for an eigenvalue $\xi$, with nilpotent part $N_\xi$, the restriction of (BR12) is

$$\left((1-r)\xi+i r\lambda\, \operatorname{sgn}(\operatorname{Im}\xi)\right)I +(1-r)N_\xi . \tag{BR13}$$

Its only eigenvalue stays strictly in the same half-plane as $\xi$. Therefore every $a^{\langle r\rangle}$ has no real eigenvalue, and its upper spectral projection is exactly $q$. This primary-block calculation is the needed proof; convexity of spectra of arbitrary nonnormal matrices would not suffice.

Let $\phi(t)=1$ on a smaller collar and have support inside the region where $A(t)=A(0)$. Quantize the deformation by

$$A_r=(1-r\phi)A +r\phi\bigl(i\Lambda Q-i\Lambda(I_E-Q)\bigr). \tag{BR14}$$

Its principal tangential symbol is (BR12) with $r$ replaced by $r\phi(t)$. Hence $D_t-A_r$ remains elliptic, its stable projection at the boundary remains $q$, and the original $\mathcal B$ remains complementing throughout. The exact order $\Lambda Q$ and $\Lambda(I_E-Q)$ has been retained.

4. Replacing the interior partial operators

In the annulus where a collar cutoff varies, a tangential operator acts in $y$ while $t$ is only a parameter. This is a partial pseudodifferential operator. Its support stays away from $Y$, so it can be approximated by ordinary operators without changing any boundary symbol.

For a tangential order-one symbol $k(y,t,\eta)$, let $\omega(\eta,\varepsilon\zeta)$ be the angular cutoff from the partial-operator approximation theorem, and set

$$k_\varepsilon(y,t,\eta,\zeta) =k(y,t,\eta)\omega(\eta,\varepsilon\zeta). \tag{BR15}$$

After multiplying on both sides by the fixed annular cutoff, $K_\varepsilon$ is an ordinary full-variable operator and, for every real $s$,

$$\|K_\varepsilon-K\|_{H^s\to H^{s-1}} \leq C_s\varepsilon . \tag{BR16}$$

The estimates are uniform in the compact deformation parameter because only finitely many uniform symbol seminorms occur. The image of the parameter interval is a compact subset of the Fredholm locus. A finite operator-norm neighborhood cover therefore supplies one positive tolerance for the entire path. Choose $\varepsilon$ below that tolerance. The ordinary approximating path is Fredholm and has the same index at every parameter. This step changes no collar principal symbol, no boundary map, and no stable bundle.

5. The inverse boundary symbol and its rotation

Let $b:\pi^*E\to\pi^*G$ be the order-zero principal boundary symbol. Its restriction $b|_L:L\to\pi^*G$ is a smooth homogeneous bundle isomorphism. Define

$$s:\pi^*G\longrightarrow\pi^*E,\qquad s=(b|_L)^{-1}, \quad\text{with image in }L. \tag{BR17}$$

Local frames of $L$ turn this into the inverse of a smooth matrix with nonzero determinant, proving smoothness. Since $b$ and $q$ have degree zero, so does $s$. Its complete typed identities are

$$bs=I_G,\qquad qs=s,\qquad sbq=q . \tag{BR18}$$

The last identity follows because $qe\in L$ and $s$ is the inverse of $b$ there. Expressions such as $qb$ or $sq$ are not used because their source and target bundles do not match.

Put $c_\theta=\cos\theta$ and $d_\theta=\sin\theta$. On $\pi^*(E\oplus G)$, define

$$\begin{aligned} b_\theta&=(c_\theta b,\ d_\theta I_G),\\ s_\theta&=\binom{c_\theta s}{d_\theta I_G},\\ q_\theta&= \begin{pmatrix} c_\theta^2q&c_\theta d_\theta s\\ c_\theta d_\theta bq&d_\theta^2I_G \end{pmatrix}, \qquad 0\leq\theta\leq\frac{\pi}{2}. \end{aligned} \tag{BR19}$$

First,

$$b_\theta s_\theta =c_\theta^2bs+d_\theta^2I_G=I_G . \tag{BR20}$$

Direct block multiplication using only (BR18) gives

$$q_\theta^2=q_\theta,\qquad q_\theta s_\theta=s_\theta,\qquad s_\theta b_\theta q_\theta=q_\theta . \tag{BR21}$$

For example, the upper-left block of $q_\theta^2$ is $c_\theta^4q+c_\theta^2d_\theta^2sbq=c_\theta^2q$, and the lower-left block is $c_\theta^3d_\theta bq+c_\theta d_\theta^3bq=c_\theta d_\theta bq$. The other two blocks follow in the same typed order. The last two identities imply

$$\operatorname{ran}q_\theta=\operatorname{ran}s_\theta,\qquad q_0=q\oplus0,\qquad q_{\pi/2}=0\oplus I_G . \tag{BR22}$$

Thus the stable bundle rotates from $L$ to the pulled-back $G$-summand while $b_\theta$ remains its inverse measurement.

6. Extending the boundary bundle and adding zero index

The bundle $G\to Y$ has a smooth complement $G'$ with $G\oplus G'\cong Y\times\mathbb C^N$. This follows from a finite trivializing cover and the explicit embedding-and-orthogonal-complement construction in the Bott lesson. Take

$$H=X\times\mathbb C^N,\qquad H|_Y\xrightarrow{\ \kappa\ }G\oplus G'. \tag{BR23}$$

Let $q_G:H|_Y\to G$ and $\iota_G:G\to H|_Y$ be the projection and inclusion supplied by this splitting, and put

$$\Pi_G=\iota_Gq_G,\qquad \Pi_G^2=\Pi_G,\qquad H|_Y=\operatorname{ran}\Pi_G\oplus\ker\Pi_G . \tag{BR24}$$

Choose an order-one operator $\Lambda_H$, block diagonal near $Y$ for $G\oplus G'$, with the same positive scalar principal symbol $\lambda(y,\eta)I_H$ as $\Lambda$. Form the split auxiliary problem whose collar part is

$$P_H^b=D_t+i\Lambda_H,\qquad B_H=0 . \tag{BR25}$$

Complete it elliptically in the interior as in the split-model theorem. Its stable boundary space is zero, its boundary target is the zero bundle, and that theorem gives

$$\operatorname{ind}(P_H,0)=0,\qquad \operatorname{ind}\bigl((P,B)\oplus(P_H,0)\bigr) =\operatorname{ind}(P,B). \tag{BR26}$$

This is the only stabilization used below. It changes no index.

7. The full two-by-two normal homotopy

Quantize $s$ by $S\in\Psi_{\mathrm{phg}}^0(Y;G,E|_Y)$. Retain $Q$ with symbol $q$. The product $\mathcal BQ:E|_Y\to G$ has principal symbol $bq$. Choose a decreasing collar cutoff $\vartheta(t)$, equal to one near $t=0$, with support where $\phi=1$. For $0\leq\tau\leq\pi/2$, put

$$\alpha=\tau\vartheta(t),\qquad c_\alpha=\cos\alpha,\qquad d_\alpha=\sin\alpha . \tag{BR27}$$

On $\pi^*(E\oplus H)$, extend (BR19) by zero on the $G'$-summand:

$$\mathfrak q_\alpha= \begin{pmatrix} c_\alpha^2q& c_\alpha d_\alpha\,s q_G\\ c_\alpha d_\alpha\,\iota_Gbq& d_\alpha^2\Pi_G \end{pmatrix}. \tag{BR28}$$

The same multiplication as in Section 5, with $q_G\iota_G=I_G$, proves

$$\mathfrak q_\alpha^2=\mathfrak q_\alpha,\qquad \operatorname{ran}\mathfrak q_\alpha =\left\{ \binom{c_\alpha sg}{d_\alpha\iota_Gg}:g\in\pi^*G \right\}. \tag{BR29}$$

The exact block expansion of $I-2\mathfrak q_\alpha$ is

$$I-2\mathfrak q_\alpha= \begin{pmatrix} I_E-2c_\alpha^2q&-\sin(2\alpha)sq_G\\ -\sin(2\alpha)\iota_Gbq&I_H-2d_\alpha^2\Pi_G \end{pmatrix}. \tag{BR30}$$

Every sign and every inclusion in the normal homotopy comes from this identity. With the convention that a $G$-valued term in the $H$-component is inserted by $\iota_G$, define the collar part

$$\begin{aligned} P_\tau^b\binom{u}{v} =\phi\Bigg[ \binom{D_tu}{D_tv} +i\binom{ \Lambda\bigl(u-2c_\alpha^2Qu-\sin(2\alpha)S q_Gv\bigr)} {\Lambda_H\bigl(v-\sin(2\alpha)\iota_G\mathcal BQu -2d_\alpha^2\Pi_Gv\bigr)} \Bigg]. \end{aligned} \tag{BR31}$$

The two off-diagonal blocks are not interchangeable: $Sq_G:H\to E$, whereas $\iota_G\mathcal BQ:E\to H$. Their principal symbols are exactly the off-diagonal blocks in (BR30). Lower-order defects caused by $Q^2-Q$, quantization, and derivatives of the cutoffs do not enter the principal ellipticity test.

The boundary operator is

$$B_\tau(u,v) =\cos\tau\,\mathcal B\gamma_0u +\sin\tau\,q_G\gamma_0v . \tag{BR32}$$

At $\tau=0$, (BR31) is the direct sum of the fixed-projection endpoint (BR14) and $P_H$, while (BR32) is $B$ on the first summand.

8. Ellipticity, boundary bijectivity, and index tracking

The principal tangential generator in (BR31) is

$$a_{\tau,t}=i\lambda(2\mathfrak q_\alpha-I_{E\oplus H}) . \tag{BR33}$$

It has eigenvalue $i\lambda$ on $\operatorname{ran}\mathfrak q_\alpha$ and $-i\lambda$ on its kernel. Thus $\zeta I-a_{\tau,t}$ is invertible for every real $\zeta$ when $(\eta,\zeta)\ne0$, and its upper spectral projection is exactly $\mathfrak q_\alpha$.

At the boundary, $\alpha=\tau$. Define

$$\mathfrak s_\tau g =\binom{\cos\tau\,sg}{\sin\tau\,\iota_Gg}. \tag{BR34}$$

Equations (BR29), (BR32), $bs=I_G$, and $q_G\iota_G=I_G$ give

$$\operatorname{ran}\mathfrak q_\tau =\operatorname{ran}\mathfrak s_\tau,\qquad b_\tau\mathfrak s_\tau =(\cos^2\tau+\sin^2\tau)I_G=I_G . \tag{BR35}$$

Hence $b_\tau$ is bijective on the stable Cauchy bundle for every parameter. The generalized collar Fredholm theorem therefore gives

$$(P_\tau,B_\tau)_1: H^1(X;E\oplus H)\longrightarrow L^2(X;F\oplus H)\oplus H^{1/2}(Y;G) \quad\text{Fredholm}. \tag{BR36}$$

Formula (BR31) is written in the collar target frame from Section 2; multiplication by $c$ returns its first component to $F$. Thus the displayed source and target spaces are fixed along the path. The coefficients depend continuously on $\tau$ in their operator norm. Index local constancy, connectedness of $[0,\pi/2]$, and (BR26) now give

$$\operatorname{ind}(P_\tau,B_\tau) =\operatorname{ind}(P,B) \quad\text{for every }0\leq\tau\leq\frac{\pi}{2}. \tag{BR37}$$

The ordinary interior approximations from Section 4 can be made uniformly along this path. They preserve (BR37), because they lie in one fixed Fredholm neighborhood and are supported away from the boundary symbol.

9. The endpoint is the split trace model

At $\tau=\pi/2$, and where $\vartheta=1$, the off-diagonal terms vanish. Relative to $H|_Y=G\oplus G'$, (BR31)–(BR32) become

$$\begin{aligned} P_{\pi/2}^b &= \bigl(D_t+i\Lambda_E\bigr) \oplus \bigl(D_t-i\Lambda_G\bigr) \oplus \bigl(D_t+i\Lambda_{G'}\bigr),\\ B_{\pi/2}(u,v_G,v_{G'}) &=\gamma_0v_G . \end{aligned} \tag{BR38}$$

The $G$-summand is the complete stable space; the $E$- and $G'$-summands are unstable. To match the split convention with normal coefficient $-I_G$ on the stable summand, apply the target-bundle isomorphism

$$T_{\mathrm{sgn}} =I_E\oplus(-I_G)\oplus I_{G'} . \tag{BR39}$$

On the collar it turns the $G$-block into $-D_t+i\Lambda_G$, while leaving the boundary trace unchanged. It extends to an invertible target-bundle automorphism: keep the splitting on a smaller collar, join the factor $-1$ to $1$ through the nonzero complex phases $e^{i\pi\rho(t)}$, and use the identity beyond the collar. After reordering the summands,

$$E_{\mathrm{sp}}^+=E\oplus G', \qquad E_{\mathrm{sp}}^-=G, \qquad P_{\mathrm{sp}}^b= \begin{pmatrix} D_t+i\Lambda^+&0\\ 0&-D_t+i\Lambda^- \end{pmatrix}, \qquad B_{\mathrm{sp}}=\gamma_0|_{E_{\mathrm{sp}}^-}. \tag{BR40}$$

This accounts for the sign change rather than hiding it in a relabeling. The split doubling theorem applies and yields

$$\operatorname{ind}(P,B) =\operatorname{ind}(P_{\mathrm{sp}},B_{\mathrm{sp}}) =\operatorname{ind}\widehat P_{\mathrm{sp}} =\operatorname{sind}(\widehat p_{\mathrm{sp}}), \tag{BR41}$$

which proves (BR4).

The left panel shows the two analytic deformations that leave $q$ fixed. The middle panel shows the exact stable vector and boundary inverse during the rotation. The right panel records the endpoint summands, the target sign map, and the index chain.

10. The obstruction space and nonzero boundary order

Let $\pi:S^*Y\to Y$ and $L(P)=\operatorname{ran}q$. The exact space of order-zero complementing symbol data is the groupoid

$$\mathfrak C_0(P) =\left\{(G,\beta): G\to Y\text{ a complex bundle},\ \beta:L(P)\xrightarrow{\sim}\pi^*G\right\}. \tag{BR42}$$

Editorial completion of the groupoid definition. An arrow $h:(G,\beta)\to(G_1,\beta_1)$ is a smooth bundle isomorphism $h:G\to G_1$ over $Y$ satisfying $$(\pi^*h)\beta=\beta_1. \tag{BR42a}$$ The identity of $G$ satisfies this equation. If $h$ and $h_1$ are composable arrows, then $(\pi^*(h_1h))\beta=(\pi^*h_1)\beta_1$, the required equation for their composite. Multiplying the displayed equation by $\pi^*h^{-1}$ gives the equation for the inverse. Associativity and the identity laws follow from composition of the underlying bundle maps. This supplies every arrow and proves that the stated collection is a groupoid. All pullbacks and symbol maps remain over the same cosphere bundle.

Given $(G,\beta)$, the formula

$$b=\beta q:\pi^*E\longrightarrow\pi^*G \tag{BR43}$$

extends $\beta$ to a boundary symbol and is complementing. Conversely, every complementing $b$ restricts to such a $\beta$. Thus failure of $L(P)$ to be pulled back is exactly the failure of this groupoid to have an object.

Editorial correction: the complete fiber of the restriction map. The canonical extension $\beta q$ need not equal the original $b$. Put $K(P)=\ker q$; the exact splitting is $\pi^*E=L(P)\oplus K(P)$, with projections $q$ and $I-q$, without an orthogonality assumption. For fixed $(G,\beta)$, every complementing symbol whose restriction is $\beta$ is uniquely $$b=\beta q+c(I-q),\qquad c:K(P)\longrightarrow\pi^*G. \tag{BR43a}$$ Here $c$ is any smooth bundle map on $S^*Y$, extended homogeneously with degree zero to $T^*Y\setminus0$. To prove the assertion, write every vector as $e=qe+(I-q)e$. Then $b(qe)=\beta(qe)$, while $b((I-q)e)=c((I-q)e)$ for $c=b|_{K(P)}$. This proves existence and the formula. Restriction to $K(P)$ recovers $c$, proving uniqueness. Conversely the displayed formula restricts to $\beta$ on $L(P)$, so it is complementing. The fiber is therefore the affine space with origin $\beta q$ and translation space $\Gamma^\infty(S^*Y;\operatorname{Hom}(K(P),\pi^*G))$. If $S^*Y$ is empty, there is one map on each empty pullback and the same assertion has a one-element fiber; it imposes no condition on the finite-dimensional boundary realization.

The left diagram gives the exact two summands in (BR43a). The right diagram is (BR42a), with the identity on $L(P)$. These are the precise morphisms retained by the restriction map; the example (BR51) realizes the free complementary entry as $\kappa$.

The stable $K$-theory obstruction is the defined class

$$\operatorname{ob}_{\partial}(P) =[L(P)]\bmod \operatorname{im}\!\left( \pi^*:K^0(Y)\longrightarrow K^0(S^*Y)\right). \tag{BR44}$$

If a complementing $B$ exists, then $[L(P)]=\pi^*[G]$, so

$$\mathfrak C_0(P)\ne\varnothing \Longrightarrow \operatorname{ob}_{\partial}(P)=0 . \tag{BR45}$$

Consequently a nonzero class in (BR44) forbids even a stably pulled-back local boundary symbol. The converse classification, including realization by a full operator problem, is not proved here and remains a named topological dependency. The reduction theorem does not assert that every elliptic interior operator admits $B$.

If the boundary operator has total order $\mu$, use the exact order-changing isomorphism from the closed-manifold index lesson:

$$J_G^{-\mu}: H^{s-\mu-1/2}(Y;G) \xrightarrow{\ \sim\ } H^{s-1/2}(Y;G), \qquad J_G^{-\mu}\in\Psi_{\mathrm{cl}}^{-\mu}(Y;G). \tag{BR46}$$

Then $B^{(0)}=J_G^{-\mu}B$ has order zero and

$$(P,B^{(0)}) =\bigl(I\oplus J_G^{-\mu}\bigr)(P,B), \qquad \ker(P,B^{(0)})=\ker(P,B), \qquad \operatorname{coker}(P,B^{(0)})\cong\operatorname{coker}(P,B). \tag{BR47}$$

Hence the index is unchanged. If one instead uses an elliptic Fredholm factor $R$ of order $-\mu$ that is not an isomorphism, the composition law gives the required correction

$$\operatorname{ind}(P,RB) =\operatorname{ind}(P,B)+\operatorname{ind}R . \tag{BR48}$$

The exact isomorphism (BR46) avoids this extra term.

11. Three calculations that test the construction

A nonnormal spectral block. Let

$$a= \begin{pmatrix}i&1&0\\0&i&0\\0&0&-2i\end{pmatrix}, \qquad q=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}, \qquad\lambda=1 . \tag{BR49}$$

The fixed-projection path is

$$a^{\langle r\rangle} = \begin{pmatrix} i&1-r&0\\ 0&i&0\\ 0&0&-i(2-r) \end{pmatrix}. \tag{BR50}$$

The upper eigenvalue remains $i$, its nilpotent part decreases to zero, and the lower eigenvalue remains strictly negative imaginary. This calculation shows why the full primary summand, rather than a moving eigenbasis, is the correct object.

A boundary map that differs from its stable restriction. Take $E=\mathbb C^2$, $G=\mathbb C$, $q=\operatorname{diag}(1,0)$, and $b=(1,\kappa)$. Then $s=(1,0)^t$ and $bq=(1,0)$, while $b\ne bq$ when $\kappa\ne0$. Formula (BR19) gives

$$q_\theta= \begin{pmatrix} c_\theta^2&0&c_\theta d_\theta\\ 0&0&0\\ c_\theta d_\theta&0&d_\theta^2 \end{pmatrix}, \qquad s_\theta z=(c_\theta z,0,d_\theta z). \tag{BR51}$$

It is a rank-one projection and $b_\theta s_\theta z=z$. The unused entry $\kappa$ disappears only after the correctly typed factor $bq$; replacing it by $b$ in the lower-left block would be wrong.

Boundary order two. If $B:H^s(X;E)\to H^{s-5/2}(Y;G)$ has total order two, then $J_G^{-2}B$ maps to $H^{s-1/2}(Y;G)$. Both operators have the same kernel and canonically isomorphic cokernels by (BR47). No boundary contribution is lost in the order change.

12. Exercises with complete solutions

1. Prove the homogeneous contour formula. Starting from $a(y,\rho\eta)=\rho a(y,\eta)$, derive (BR11) with every contour factor retained.

Solution. Use $z=\rho w$, $dz=\rho\,dw$, and

$$(zI-\rho a)^{-1} =\rho^{-1}(wI-a)^{-1}. \tag{BR52}$$

The factors $\rho$ and $\rho^{-1}$ cancel, and the contour $\rho\Gamma_+$ becomes $\Gamma_+$. Hence $q(y,\rho\eta)=q(y,\eta)$.

2. Verify the remaining blocks of $q_\theta^2$. Use only the identities in (BR18).

Solution. Besides the upper-left block computed after (BR21), the three blocks are

$$\begin{aligned} (q_\theta^2)_{12} &=c_\theta^3d_\theta qs+c_\theta d_\theta^3s =c_\theta d_\theta s,\\ (q_\theta^2)_{21} &=c_\theta^3d_\theta bq^2+c_\theta d_\theta^3bq =c_\theta d_\theta bq,\\ (q_\theta^2)_{22} &=c_\theta^2d_\theta^2bqs+d_\theta^4I_G =d_\theta^2I_G . \end{aligned} \tag{BR53}$$

Here $qs=s$, $q^2=q$, and $bqs=bs=I_G$. Every composition is type-correct.

3. Recover the off-diagonal coefficient. Why is the coefficient in (BR30) $-\sin(2\alpha)$?

Solution. The off-diagonal entry of $2\mathfrak q_\alpha$ is $2c_\alpha d_\alpha sq_G$. Since $2c_\alpha d_\alpha=\sin(2\alpha)$, subtracting from the identity gives

$$(I-2\mathfrak q_\alpha)_{12} =-\sin(2\alpha)sq_G , \quad (I-2\mathfrak q_\alpha)_{21} =-\sin(2\alpha)\iota_Gbq . \tag{BR54}$$

The minus sign is forced in both blocks.

4. Check the endpoint stable modes. Solve the three scalar normal equations in (BR38) when every $\Lambda$ has frozen eigenvalue $\lambda>0$.

Solution. The equations and their solutions are

$$\begin{array}{c|c|c} \text{block}&\text{solution}&\text{bounded for }t\geq0\\ \hline D_t+i\lambda&e^{\lambda t}a&a=0\\ D_t-i\lambda&e^{-\lambda t}g&\text{every }g\\ D_t+i\lambda&e^{\lambda t}a'&a'=0 . \end{array} \tag{BR55}$$

Thus the stable space is exactly the $G$-summand and its trace is measured by $q_G$.

5. Identify the symbol-level data space and its fiber. Given an isomorphism $\beta:L(P)\to\pi^*G$, prove that (BR43) is complementing, describe all symbols with that restriction, and explain why the canonical extension need not recover their values on $K(P)$.

Solution. For $\ell\in L(P)$, $q\ell=\ell$, so $(\beta q)\ell=\beta\ell$; this restriction is bijective. Conversely, if $b$ is complementing, set $\beta=b|_{L(P)}$. Then $\beta$ is an isomorphism. Writing $e=qe+(I-q)e$ gives $b(e)=\beta qe+(b|_{K(P)})(I-q)e$. Thus the unique free map is $c=b|_{K(P)}$, and every choice of $c$ gives a complementing symbol by the same formula. The canonical extension chooses $c=0$, so it need not recover an original nonzero complementary entry. Therefore (BR42) records exactly the symbol data relevant to the boundary condition.

6. Track a noninvertible order reducer. Suppose $R$ in (BR48) has index $-2$ and $\operatorname{ind}(P,B)=5$. Compute the new index and state why the composition belongs to the declared order-zero class.

Solution. The block target operator $I\oplus R$ has index $-2$. Fredholm composition gives

$$\operatorname{ind}(P,RB)=5-2=3. \tag{BR56}$$

The composition rule gives $RB\in\Psi^0$ from $R\in\Psi^{-\mu}$ and $B\in\Psi^\mu$. This is the stated upper order; a vanishing principal symbol can lower the actual order. The index correction is analytic data of $R$, independent of that order count.

13. Reading notes and references

Stable modes and the algebra of boundary data, Sections 4–5, proves smooth parameter dependence of the Riesz projection and the primary-block deformation that keeps it fixed. Cauchy data from jumps and residues, Sections 8–9, identifies the stable projection and the complementing restriction without choosing roots. Symbols, finite defects, and the index on a closed manifold, Sections 3 and 10, supplies exact order-changing isomorphisms and the operator-norm approximation of partial operators. The Bott operator, suspension, and reduction of the index to Euclidean space, Section 12, supplies the explicit stable complement for $G$. Doubling a boundary problem and computing its index proves the zero index of the auxiliary split problem and the final doubled symbol-index formula.

The construction, block verification, examples, figure, and exercises here are independently written.

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

14. Finite-product and receiving-map proof

This supplement proves the actual finite collar product used by the analytic receiving maps. It keeps the original normal coefficient, complete tangential operators, source distributions, and every boundary row order. The original reduction statements and formulas above remain identifiable. Completing this finite-product proof alone does not close every realization or deformation step of the stable reduction theorem.

The fixed-collar construction and exact symbol estimates are in GI1–GI32 and AL1–AL18. The complete composition proof and half-space receiving maps provide the analytic prerequisites explicitly used below. The following calculation specializes those proofs to the literal finite product, giving every complete coefficient and retaining both error bundles.

NP1. Original operators and actual seed

Let $Y$ be the unchanged compact boundary and let $E_Y,F_Y$ be the separate original collar bundles, with their input densities. Let $m\ge 1$, $D_r=-i\partial _r$, $\delta =-i\partial _r$ on coefficient families, and $P_c=\sum _{j=0}^m A_j(h(r))D_r^j:E_Y\to F_Y$, with $A_j\in \Psi _{\rm tan}^{m-j}$, $A_m=M:E_Y\to F_Y$ invertible. In subsequent formulas $A_j(r)$ denotes exactly this composed family $A_j(h(r))$, so $\delta$ differentiates the composed original family. It never means its coefficient freeze. Boundary rows remain $B_j=\sum _{k=0}^{m-1}B_{jk}\gamma _k$, $B_{jk}\in \Psi _{\rm tan}^{m_j-k}(E_Y,G_j)$, with every original $m_j$ unrestricted. Retain $R=1+|(\eta ,\kappa )|$ and $T=1+|\eta |$.

Use the actual GI24 seed $Q=\sum _i\vartheta _i T_i\vartheta _i:F_Y\to E_Y$, where $\sum _i\vartheta _i^2=1$ and the tangential cutoffs and bundle transfers are the specified original ones. For each local exact symbol polynomial $p_i=\sum _{j=0}^m p_{ij}\kappa ^j$, set $b_{i\ell }=M_i^{-1}p_{i,m-\ell }:E_i\to E_i$, $u_{i0}=I_{E_i}$, and

$$u_{in}=-\sum_{\ell=1}^{\min(m,n)}b_{i\ell}u_{i,n-\ell}, \qquad c_{in}=u_{in}M_i^{-1}:F_i\to E_i, \qquad \mathcal Q_n=\sum_i\vartheta_i\operatorname{Op}_y(c_{in})\vartheta_i. \tag{NP1}$$

Each product in the definition of $u$ is a matrix-symbol product in its stated order. Each product in the definition of $\mathcal Q _n$ is the complete tangential operator product, including its smoothing contribution, its frame transfers, and its actual input density. They are distinct operations. The local exact identity $p_i=M_i\kappa ^m(I+\sum _{\ell =1}^m b_{i\ell }\kappa ^{-\ell })$ proves the recurrence: multiply the matrix inverse expansion by the factor in parentheses and compare a fixed integer power. The high-normal cone constant can be chosen so the sum has norm at most $1/2$. The finite geometric inverse remainder then proves, on both signed tails, for every $L\ge1$ and every $\alpha ,\beta ,v$,

$$\left|\partial_{y,r}^{\beta}\partial_\eta^\alpha\partial_\kappa^v \left(q-\sum_{n=0}^{L-1}q_n\kappa^{-m-n}\right)\right| \le C_{L\alpha\beta v}|\kappa|^{-m-L-v}T^{L-|\alpha|}, \qquad \mathcal Q_n\in\Psi_{\rm tan}^n(F_Y,E_Y). \tag{NP2}$$

Here $q_n$ is the complete left symbol of $\mathcal Q _n$. The statement for the actual cutoffs follows by the exact tangential composition integral NC1: compact cutoff Fourier transforms have every moment, and the complementary shift $|\zeta |\ge \varepsilon |\kappa |$ has arbitrarily rapid decay after integration by parts in its compact base variable. The differentiated local inverse estimate is AL5. For separated compact tangential output/input supports $\alpha (y),\beta (z)$, the original normal amplitude has the same coefficient kernels:

$$\left|\partial_{y,z,r,s}^{\gamma}\partial_\kappa^v \left(k_Q-\sum_{n=0}^{L-1} \chi(r-s)\alpha(y)K_{\mathcal Q_n(r)}(y,z)\beta(z)\kappa^{-m-n} \right)\right| \le C_{\gamma Lv}|\kappa|^{-m-L-v}. \tag{NP3}$$

The actual density and source/target frame factors are part of $K$ and are differentiated in $\gamma$. To prove this estimate split $T\le \varepsilon |\kappa |$ and its complement. On the first region insert NP2 and integrate by parts $N$ times in $\eta$, using the nonzero tangential difference: $N>L+D+d$ makes $\int T^{L+D-N}d\eta$ finite, where $D$ is the requested external derivative cost. On the complement use the original $S^{-m,0}$ bound and choose $N>d+L+D$. Each subtracted $c_{in}$ obeys the identical tail estimate after restoring its full oscillatory integral. This is the AL8–AL10 calculation at the actual complete-cutoff scope. Properness changes away from $r=s$ are compact smooth full kernels and hence rapid in $\kappa$; the separated kernels on $r=s$ are retained. Finally $\mathcal Q _0=\sum _i\vartheta _i M^{-1}\vartheta _i=M^{-1}:F_Y\to E_Y$ exactly.

NP4. Both complete error coefficient sequences

Define the two actual errors on their own bundles:

$$\mathcal R_F=P_cQ-I_{F_Y}:F_Y\to F_Y, \qquad \mathcal R_E=QP_c-I_{E_Y}:E_Y\to E_Y. \tag{NP4}$$

For $n\ge1$ their complete normal coefficients, at $\kappa ^{-n}$, are

$$\begin{aligned} \mathsf F_n&= \sum_{\substack{0\le j\le m,\ h\ge0,\ 0\le\ell\le j\\m-j+h+\ell=n}} \binom j\ell A_j\delta^\ell\mathcal Q_h:F_Y\to F_Y,\\ \mathsf E_n&= \sum_{\substack{0\le j\le m,\ h\ge0,\ \ell\ge0\\m-j+h+\ell=n}} \binom{-m-h}\ell\mathcal Q_h\delta^\ell A_j:E_Y\to E_Y. \end{aligned} \tag{NP5}$$

Both sums are finite because $h+\ell \le n$. For the first identity use the exact finite normal Leibniz formula for $D_r^j$; for the second use the restored normal moment NC5 and $1/\ell !\partial _\kappa ^\ell \kappa ^{-m-h}=\operatorname{binom}(-m-h,\ell )\kappa ^{-m-h-\ell }$. All tangential compositions are complete. At $n=0$ the two products have the distinct coefficients $M\mathcal Q _0=I_F$ and $\mathcal Q _0M=I_E$; subtracting these gives zero, not an omitted boundary map. Thus the first error coefficients are

$$\mathsf F_1=M\mathcal Q_1+mM\delta(M^{-1})+A_{m-1}M^{-1}, \qquad \mathsf E_1=\mathcal Q_1M-mM^{-1}\delta M+M^{-1}A_{m-1}. \tag{NP6}$$

Each coefficient has the sharper order $\mathsf F _n\in \Psi _{\rm tan}^{n-1}(F_Y)$ and $\mathsf E _n\in \Psi _{\rm tan}^{n-1}(E_Y)$. For the top tangential order $n$, every term with $\ell \ge 1$ already has order $n-\ell \le n-1$. The $\ell =0$ terms have principal symbols $\sum _{m-j+h=n}\sigma _{m-j}(A_j)\sigma _h(\mathcal Q _h)$, or the factors in the opposite order. NP1 implies that $\sigma _h(\mathcal Q _h)$ is the coefficient of the inverse of the complete leading tangential homogeneous polynomial: the top homogeneous component of each $c_{ih}$ has that recurrence, scalar-cutoff derivatives have lower tangential order, and $\sum \vartheta _i^2=1$ preserves it. Multiplication of that polynomial with its matrix inverse makes these degree $n$ sums zero in both orders for $n\ge1$. The principal tangential symbol characterization then gives the stated drop. This proof retains the complete coefficient after cancellation; it does not substitute its principal part for the coefficient.

The all-length sharpened remainders are the actual NC8/AC7 ones:

$$\begin{aligned} \mathcal R_F&\sim\sum_{n\ge0}\mathsf F_{n+1}\kappa^{-1-n}, &\mathcal R_E&\sim\sum_{n\ge0}\mathsf E_{n+1}\kappa^{-1-n},\\ \left|\partial_x^\beta\partial_\eta^\alpha\partial_\kappa^v \left(r_\bullet-\sum_{n<L}e_{\bullet,n+1}\kappa^{-1-n}\right)\right| &\le C|\kappa|^{-1-L-v}T^{L-|\alpha|}, &\mathcal R_\bullet&\in\mathcal L^{-1}. \end{aligned} \tag{NP7}$$

Here $e_{F,n}=\mathsf F _n$ and $e_{E,n}=\mathsf E _n$; no equality of their sequences is asserted. The strengthening from a degree-zero cancelled expansion to NP7 uses an actual frequency derivative in the composition remainder. For a local $p_i$ or $\tau _i$, the leading coefficient is the multiplication symbol $M_i$ or $M_i^{-1}$; $\partial _\eta$ kills it and lowers every subsequent tangential order by one. The original enhanced GI5 estimates give the same free derivative improvement globally. A $\partial _\kappa$ derivative lowers every normal power. MC20 expresses the exact local composition error by the integral of $C_t(\partial _{\xi}p_i,D_x\tau _i)$, or $C_t(\partial _{\xi}\tau _i,D_xp_i)$, plus the compact full-frequency cutoff error. NC1–NC7 is uniform in $0\le t\le1$, because scaled base derivatives give factors $t^a\le1$. Hence both exact local errors have NP7. The GI27/GI28 global sums contain these errors and the complete scalar-cutoff commutators; each commutator has the same free derivative improvement. Their $\Delta _i$ terms are zero on the specified actual supports (GI26). Finite summation gives NP7 and its separated kernel estimate with bound $|\kappa |^{-1-L-v}$. This identifies the extra analytic step; cancellation of one leading coefficient alone would not prove it.

NP8. The literal finite product $Q\mathcal R _F$, at every normal coefficient

This is one actual finite operator product on the original bundles,

$$D:=Q\mathcal R_F=\mathcal R_EQ:F_Y\longrightarrow E_Y, \qquad D\in\mathcal L^{-m-1}. \tag{NP8}$$

Equality follows by expanding $(QP_c-I_E)Q=Q(P_cQ-I_F)$ and using associativity of the properly supported actual operators; it neither commutes $Q$ through an error nor identifies $E_Y$ with $F_Y$. Its complete coefficient at $\kappa ^{-m-1-n}$, for $n\ge0$, is

$$\boxed{\displaystyle \mathcal D_n(r)= \sum_{\substack{h,k,\ell\ge0\\h+k+\ell=n}} \binom{-m-h}{\ell}\mathcal Q_h(r)\delta^\ell\mathsf F_{k+1}(r) :F_Y\longrightarrow E_Y,\qquad \mathcal D_n\in\Psi_{\rm tan}^{n}(F_Y,E_Y).} \tag{NP9}$$

The independently typed expression from the other error side is

$$\mathcal D_n= \sum_{p+h+\ell=n} \binom{-1-p}{\ell}\mathsf E_{p+1}\delta^\ell\mathcal Q_h. \tag{NP10}$$

The two expressions are equal because NP8 is an exact operator identity and the common signed-tail coefficients are unique. For uniqueness, hold $\eta$ and the base parameters fixed in the cone and multiply the first nonzero difference coefficient by its $\kappa$ power as $|\kappa |\to \infty$. NP2/NP7 make the remainder tend to zero. The complete tangential left symbols coincide at every $\eta$, and hence their actual operators coincide. The complete source and target remain $F_Y\to E_Y$ in both expressions. Since each summand of NP9 has tangential order at most $h+k=n-\ell \le n$, the stated order follows without replacing a complete tangential factor by a Taylor polynomial.