Analytic double fibrations and Bolker visibility

In submersion coordinates $(x,v)$, a compactly supported test density on $Y$ is $\psi (x,v)\mathrm{\mid }dxdv\mathrm{\mid }$. Its fiber integral is $\tilde{\psi }(x)=\int \psi (x,v)\mathrm{\mid }dv\mathrm{\mid }$. The integrand has compact support, so differentiation under the integral gives a smooth function; its support lies in the compact projection of $\mathrm{supp}\psi$. Thus $\tilde{\psi }(x)\mathrm{\mid }dx\mathrm{\mid }$ is a test density and transposing this operation defines $p^{\ast }u$.

If $q$ is smooth and proper on $\mathrm{supp}V$ and $K$ is compact in $M$, then $Q_K=q^{-1}(K)\cap \mathrm{supp}V$ is compact. Choose one smooth cutoff ${\beta }_K$ equal to one near $Q_K$. For every test function $\phi$ supported in $K$, smoothness of $q$ makes ${\beta }_K(\phi \circ q)$ smooth with support in the fixed compact set $\mathrm{supp}{\beta }_K$, so define $⟨q_{\ast }V,\phi ⟩=⟨V,{\beta }_K(\phi \circ q)⟩$. If another cutoff is used, the difference of the two test functions vanishes on a neighborhood of $\mathrm{supp}V$: it vanishes near $Q_K$ because both cutoffs equal one there, and at every point of $\mathrm{supp}V\setminus Q_K$, continuity of $q$ gives a neighborhood mapped outside the closed set $K$, where $\phi \circ q=0$. Hence [F4] gives the same pairing; choosing a common cutoff for the union of two compact test supports also proves linearity. The chain and product rules bound every fixed-support seminorm of ${\beta }_K(\phi \circ q)$ by finitely many fixed-support seminorms of $\phi$, so continuity of $V$ proves continuity of $q_{\ast }V$. A finite cover by coordinate balls and [F12] supplies ${\beta }_K$. Multiplication by a smooth density is defined by the stated transpose formula, so the typed pushforward is well-defined.

For tests supported in a fixed compact $Q\subseteq G\times X$, ambient local finiteness gives $\mathrm{\mid }⟨K_Z,\mathrm{\Phi }⟩\mathrm{\mid }\le {\sup }_Q\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }{\int }_{Z\cap Q}\mathrm{\mid }{\omega }_Z\mathrm{\mid }$. Thus the integral is finite and continuous in the fixed-support test topology. If $Z$ is closed, $Z\cap Q$ is compact in $Z$, and finitely many relatively compact charts bound this integral for every smooth density. In coordinates where $Z=\{t=0\}$, the functional is integration of the restricted test function against ${\omega }_Z$. Under a change of coordinates on $Z$, the coefficient transforms by the absolute tangential Jacobian [F5], so the functional is independent of the defining chart.

A covector annihilating $T_{z}Z$ transforms by the inverse transpose rule [F6], so $N^{\ast }Z$ and $C_Z$ are coordinate independent. If a conormal has the form $(\eta ,0)$, it annihilates the image of $d{\pi }_G:T_{(g,x)}Z\to T_{g}G$, which is all of $T_{g}G$ because ${\pi }_G$ is a submersion; therefore $\eta =0$. Applying the same argument to ${\pi }_X$ shows that $(0,\xi )$ can only be zero.

The analytic conditions are relative to the fixed atlases. On overlaps of induced analytic charts on $Z$, the nonzero coordinate Jacobian has locally constant sign, so its absolute value is analytic. The density transition rule therefore preserves analyticity of its coefficient. The defining submersions and analytic chart changes specify the same analytic embedded submanifold, while the underlying smooth distribution operations are unchanged.

On overlapping submersion charts, the full absolute Jacobian factors into the base Jacobian and the fiber-density quotient. The fiber integral therefore transforms as a base density. This proves that the transposed functional in step 1.1 is chart independent.

Let $\phi$ be any compactly supported test density on $M$; the zero test is immediate. Its support $K$ is compact. Cover $K$ by finitely many smaller coordinate balls $B_i$ whose closures lie in coordinate balls $U_i$. Using [F12] in each chart, choose ${\beta }_i$ supported in $U_i$ and equal to one on $B_i$, and put $\sigma ={\sum }_i{\beta }_i$. Then $\sigma >0$ on a neighborhood of $K$. A second finite cover by coordinate balls and [F12] gives a compactly supported smooth $\theta$ in $\{\sigma >0\}$ equal to one near $K$. Define ${\chi }_i=\theta {\beta }_i\mathrm{/}\sigma$ on $\{\sigma >0\}$ and extend by zero outside. Each ${\chi }_i$ is smooth and compactly supported in $U_i$, and ${\sum }_i{\chi }_i=\theta$. Hence $u(\phi )={\sum }_{i}u({\chi }_i\phi )$, since $\theta =1$ near $K$. Each term is a chartwise pairing with a compactly supported Euclidean distribution, so [F9] and [F10] give its tempered and smooth-function realizations. The finite sum therefore recovers the original functional and its packet pairings, without assuming $\mathrm{supp}u$ compact or choosing global coordinates.

If $K=\mathrm{\varnothing }$, set ${\chi }_N=0$ for all $N$; the empty set is a neighborhood of $K$, so every assertion holds. Assume henceforth $K\mathrm{\ne }\mathrm{\varnothing }$. Consider all pairs of concentric balls $B_r(x)⋐B_R(x)⋐U$ with $x\in K$ and $0<r<R$. Their smaller balls cover $K$, so compactness supplies finitely many pairs $B_{r_j}(x_j)⋐B_{R_j}(x_j)$, $1\le j\le m$, whose smaller balls cover $K$. Choose their finitely many bumps $b_j$ by [F1], translated to $x_j$, with $0\le b_j\le 1$, $b_j=1$ on $B_{r_j}(x_j)$ and compact support in $B_{R_j}(x_j)$. The smooth function $g=1-{\prod }_{j=1}^m(1-b_j)$ satisfies $0\le g\le 1$, equals one on the open set $O={\bigcup }_{j=1}^m{B}_{r_j}(x_j)$ containing $K$, and has compact support $S⋐U$. Only finitely many bumps are selected.

Choose a nonnegative smooth bump ${\rho }_0$ supported in the unit ball and equal to one on the ball of radius $1\mathrm{/}2$, using [F1]. Its compactly supported integral is finite by [F3] and positive: a closed nondegenerate rectangle inside $B_{1\mathrm{/}2}(0)$ has positive volume and ${\rho }_0=1$ there. Normalize $\rho ={\rho }_0\mathrm{/}\int {\rho }_0$, so $\int \rho =1$, $\rho \ge 0$, and $\mathrm{supp}\rho \subset {\bar{B}}_1(0)$. Since $K⋐O$ and $S⋐U$, choose $\epsilon >0$ smaller than one third of each distance $\mathrm{dist}(K,O^c)$ and $\mathrm{dist}(S,U^c)$, omitting the second restriction when $U={\mathbb{R}}^n$.

For each $N\ge 1$, put $\delta =\epsilon \mathrm{/}N$, ${\rho }_{\delta }(x)={\delta }^{-n}\rho (x\mathrm{/}\delta )$, and recursively ${\eta }_1={\rho }_{\delta }$, ${\eta }_{j+1}={\eta }_j\ast {\rho }_{\delta }$ for $1\le j<N$. Compact-rectangle Fubini [F4] and change of variables give $\int {\eta }_N=1$ and $\mathrm{supp}{\eta }_N\subset {\bar{B}}_{\epsilon }(0)$; [F4] also permits reassociation of these compact convolution integrals. Define ${\chi }_N=g\ast {\eta }_N$. The integral defining ${\chi }_N$ is over a fixed compact rectangle. Differentiating the smooth kernel under this compact integral is justified by uniform convergence of its difference quotients on that rectangle, so ${\chi }_N$ is smooth. Positivity and unit mass give $0\le {\chi }_N\le 1$. If $\mathrm{dist}(x,K)<\epsilon$, then every $y\in \mathrm{supp}{\eta }_N$ satisfies $\mathrm{dist}(x-y,K)<2\epsilon <\mathrm{dist}(K,O^c)$, hence $g(x-y)=1$ and ${\chi }_N(x)=1$. Also $\mathrm{supp}{\chi }_N\subset S+{\bar{B}}_{\epsilon }(0)⋐U$.

For smooth compactly supported $f,h$, write $\mathrm{\parallel }h{\mathrm{\parallel }}_1={\int }_{{\mathbb{R}}^n}\mathrm{\mid }h\mathrm{\mid }$. Fubini on a rectangle containing their supports and the triangle inequality give $\mathrm{\parallel }f\ast h{\mathrm{\parallel }}_1\le \int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int \mathrm{\mid }f(x-y)\mathrm{\mid }\mathrm{\mid }h(y)\mathrm{\mid }dydx=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }h{\mathrm{\parallel }}_1\mathrm{.}$ Also ${\mathrm{\partial }}_j(f\ast h)=({\mathrm{\partial }}_{j}f)\ast h=f\ast ({\mathrm{\partial }}_{j}h)$ by differentiating under the compact integral and integration by parts; there is no boundary term because the factors have compact support. For $k=\mathrm{\mid }\alpha \mathrm{\mid }\le N$, distribute the $k$ coordinate derivatives to $k$ distinct convolution factors in ${\eta }_N$. Scaling gives $\mathrm{\parallel }{\mathrm{\partial }}_j{\rho }_{\delta }{\mathrm{\parallel }}_1={\delta }^{-1}\mathrm{\parallel }{\mathrm{\partial }}_j\rho {\mathrm{\parallel }}_1,$ so repeated use of the preceding inequality yields $\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }{\eta }_N{\mathrm{\parallel }}_1\le (AN{)}^k,A={\epsilon }^{-1}\max (1,{\max }_j\mathrm{\parallel }{\mathrm{\partial }}_j\rho {\mathrm{\parallel }}_1)\mathrm{.}$ For $k=0$ this is $\mathrm{\parallel }{\eta }_N{\mathrm{\parallel }}_1=1$.

Since $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, $\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{\chi }_N(x)\mathrm{\mid }=\mathrm{\mid }g\ast ({\mathrm{\partial }}^{\alpha }{\eta }_N)(x)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }{\eta }_N{\mathrm{\parallel }}_1\le (AN{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{.}$ Thus the lemma holds with $C=1$ (or any larger fixed constant), uniformly for every $\mathrm{\mid }\alpha \mathrm{\mid }\le N$. The cutoffs are smooth and no analytic bump is used.

Because $H$ is complex symmetric and invertible, some vector $v$ has $v^{T}Hv\mathrm{\ne }0$: otherwise polarization would give $u^{T}Hv=0$ for every $u,v$, contradicting invertibility. Extend $v$ to a basis and use this as a complex linear coordinate change, so $H_{11}\mathrm{\ne }0$. From now on, compose each coordinate change with the preceding ones; $\mathrm{\Psi }$ denotes the final map from the original coordinates. The holomorphic equation ${\mathrm{\partial }}_{z_1}\varphi (z_1,z^{\mathrm{\prime }})=0$ has a unique holomorphic solution $z_1=\psi (z^{\mathrm{\prime }})$ near $0$ by [F3] and [F4], since ${\mathrm{\partial }}_{z_1}\varphi (0)=0$ and ${\mathrm{\partial }}_{z_1}^2\varphi (0)=H_{11}\mathrm{\ne }0$.

Set $t=z_1-\psi (z^{\mathrm{\prime }})$. Taylor's formula in $t$, with the linear term zero, gives $\varphi (\psi (z^{\mathrm{\prime }})+t,z^{\mathrm{\prime }})=\varphi (\psi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})+t^{2}a(t,z^{\mathrm{\prime }}),a(t,z^{\mathrm{\prime }})={\int }_0^1(1-s){\mathrm{\partial }}_{z_1}^2\varphi (\psi (z^{\mathrm{\prime }})+st,z^{\mathrm{\prime }})ds\mathrm{.}$ The integral is holomorphic by locally uniform convergence of the power series [F2]; $a(0,0)=H_{11}\mathrm{/}2\mathrm{\ne }0$. Shrink the polydisc so $a=a(0,0)(1+b)$ with $\mathrm{\mid }b\mathrm{\mid }<1$. The binomial series for $(1+b{)}^{1\mathrm{/}2}$ converges absolutely by [F2], so $w_1=t\sqrt{2a(t,z^{\mathrm{\prime }})}$ is holomorphic and has nonzero $t$-derivative at $0$.

By [F4], $(t,z^{\mathrm{\prime }})\mapsto (w_1,z^{\mathrm{\prime }})$ is biholomorphic near $0$, and the phase becomes $\varphi (\psi (z^{\mathrm{\prime }}),z^{\mathrm{\prime }})+w_1^2\mathrm{/}2$. The Hessian of the reduced phase in $z^{\mathrm{\prime }}$ is the Schur complement $H_{22}-H_{21}{H}_{11}^{-1}{H}_{12}$. It is invertible because the block determinant formula gives $\det H=H_{11}\det (H_{22}-H_{21}{H}_{11}^{-1}{H}_{12})$.

For real symmetric positive-definite $H$, [F8] gives $H=L{L}^{\ast }$ with $L$ lower triangular and positive diagonal. Since $H$ is real, $\bar{L}$ is another such factorization; uniqueness in [F8] gives $L=\bar{L}$, so $L$ is real and $H=L{L}^T$. On each ball $B_R$, the map $x\mapsto L^{T}x$ sends a compact Jordan set to a compact Jordan set; [F11] changes variables by $\mathrm{\mid }\det L{\mathrm{\mid }}^{-1}$. Letting $R\to \mathrm{\infty }$ and using [F12] gives the corresponding nonnegative improper-integral equality. Since $\det L>0$, [F9]–[F10] give $\det H=(\det L{)}^2$. Repeated compact-rectangle Fubini [F15] factors the standard Gaussian integral on $[-R,R{]}^n$ into one-dimensional integrals, and [F12] lets $R\to \mathrm{\infty }$. By [F13] and the substitution $t=x\mathrm{/}\sqrt{2}$ in [F14], each one-dimensional integral of $e^{-x^2\mathrm{/}2}$ is $\sqrt{2\pi }$. Finally, $z=\sqrt{\lambda }w$ scales the integral by ${\lambda }^{-n\mathrm{/}2}$, proving the formula for real positive-definite $H$ with the positive determinant root.

Repeating steps 1.1–2.1 on the reduced phase proves the normal form by induction on $n$, with the constant phase as the $n=0$ case. Because $\mathrm{\Psi }$ is the composite from the original coordinates, differentiating $\varphi (z)=\varphi (0)+\frac{1}{2}\mathrm{\Psi }(z{)}^T\mathrm{\Psi }(z)$ twice at $0$ gives the original Hessian identity $H=(D\mathrm{\Psi }(0){)}^{T}D\mathrm{\Psi }(0)$; taking determinants gives the asserted square relation. For a continuous parameter family, continuity of $\det H$ lets one restrict to a neighborhood whose determinant image lies in a disc avoiding zero, where a square-root branch can be fixed from its base value.

For complex symmetric $H$ with positive-definite real part, the Gaussian integral is holomorphic in its entries: on each compact subset of this matrix domain, the integrand and its entry derivatives are bounded by a polynomial times $e^{-c\mathrm{\mid }x{\mathrm{\mid }}^2}$ for some $c>0$, so [F6] permits differentiation under the integral. This domain is convex, hence connected, and $\det H\mathrm{\ne }0$ there because $Hv=0$ would imply $\mathrm{Re}(v^{\ast }Hv)=v^{\ast }(\mathrm{Re}H)v>0$ for $v\mathrm{\ne }0$. For $H_t=I+t(H-I)$, [F16] and determinant multiplicativity give $\frac{d}{dt}\det H_t=(\det H_t)\mathrm{tr}(H_t^{-1}(H-I))$. Thus $L(H)={\int }_0^1\mathrm{tr}(H_t^{-1}(H-I))dt$ depends holomorphically on $H$, satisfies $e^{L(H)}=\det H$, and is zero at $I$. When $H$ is real positive definite, its segment $H_t$ is real positive definite, so $L(H)$ is real and the branch $e^{L(H)\mathrm{/}2}$ is positive. The real Gaussian formula from step 3.2 and the complex integral formula agree on a real open set of positive-definite matrices. In a polydisc centered at any point of that set, apply the one-variable identity theorem successively in the independent entries of a complex symmetric matrix; then [F5] extends the identity from that polydisc to the connected matrix domain.

The holomorphic $n$-form in the statement is closed: in complex coordinates its exterior derivative has repeated holomorphic differentials and no antiholomorphic coefficients. The difference of two homologous admissible contours is the boundary of the swept chain. By [F7], its integral is the boundary integral on the truncation of that chain; the stated decay makes this term tend to zero as the truncation expands. Therefore the contour integrals agree.

The chart representatives are smooth submersions by [F2]–[F3], and their analytic representatives have local holomorphic extensions by [F5]–[F6]. Hence the fiber and incidence data in the statement are well-defined in the stated smooth and analytic categories.

The pullback ${\pi }_X^{\ast }u$ exists by [F1]. Multiplying it by the nowhere vanishing analytic density ${\omega }_Z$ gives a distributional density; multiplication by the smooth function $\kappa$ preserves that type. If $\mathrm{supp}u\subset K$ with $K$ compact, the support of this product lies in ${\pi }_X^{-1}(K)$, on which ${\pi }_G$ is proper by hypothesis.

The proper pushforward in [F1] therefore defines $Ru$ as a distributional density on $G$. In coordinates, transposing fiber integration in the two submersions shows that it pairs a data test function with the integral over the corresponding incidence fiber, weighted by $\kappa {\omega }_Z$; the density Jacobians cancel by [F4]. The cotangent relation of the kernel is precisely $C_Z$ by the sign convention in [F1].

For fixed $x,\lambda ,\omega$, the function of $y$ in the displayed pairing is smooth and supported in $\mathrm{supp}\chi$, hence it is a test density. Therefore the packet is a well-defined complex number by [F1].

Replacing ${\xi }_0$ by a positive multiple leaves the unit direction ${\xi }_0\mathrm{/}\mathrm{\mid }{\xi }_0\mathrm{\mid }$ unchanged; the definition quantifies over all $\lambda \ge 1$ and an open cone of unit directions. Thus analytic regularity is conic in the cotangent variable, and its complement defines a conic subset of $T^{\ast }X\setminus 0$.

By [F2]–[F3], an analytic coordinate change near $x_0$ extends to a holomorphic map on a smaller complex neighborhood, so its cotangent action and Gaussian packet phase are locally defined there. The definition makes no claim that the numeric packet is unchanged under this change; the next lemma proves equivalence for the packet families actually used on this page.

Choose $r>0$ so that ${\chi }_N=1$ on $B_{2r}(0)$ for every $N$, the phase and amplitude are holomorphic near ${\bar{B}}_{2r}(0)$, and the stated positivity holds. The positive-phase hypothesis and compactness of $\mathrm{supp}a\setminus B_r(0)$ give $\mathrm{Im}(\mathrm{\Phi }-{\mathrm{\Phi }}_0)\ge d>0$ there. Since $a$ is bounded and ${\chi }_N$ is bounded by one, the absolute value of that contribution after multiplication by $e^{-i\lambda {\mathrm{\Phi }}_0}$ is at most $\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }U\mathrm{\mid }{e}^{-\lambda d}$, uniformly in $N$. Thus only the integral over the fixed local ball affects the power series; the complement contributes $O(e^{-d\lambda })$.

Put $\varphi =-i(\mathrm{\Phi }-{\mathrm{\Phi }}_0)$, so $\mathrm{Re}\varphi =\mathrm{Im}(\mathrm{\Phi }-{\mathrm{\Phi }}_0)$ and ${\varphi }^{\mathrm{\prime }\mathrm{\prime }}(0)=H$. Apply [F1] to obtain a holomorphic Morse coordinate $w=\mathrm{\Psi }(x)$ with $\varphi (x)=\frac{1}{2}w\cdot w$. The transformed real ball is a local contour $\mathrm{\Gamma }$ through $0$. The transformed amplitude $b(w)=a({\mathrm{\Psi }}^{-1}(w))\det D{\mathrm{\Psi }}^{-1}(w)$ is holomorphic in a fixed polydisc and has coefficients bounded by $M{R}^{-\mathrm{\mid }\alpha \mathrm{\mid }}$ for some $M,R>0$ by [F3]–[F5].

Differentiating the Morse identity in [F1] twice at $0$ gives $A^{T}A=H$, where $A=D\mathrm{\Psi }(0)$. Taylor's formula gives $\mathrm{Re}(\varphi (tv))=\frac{1}{2}{t}^2{v}^T(\mathrm{Re}H)v+o(t^2)$; comparing with $\mathrm{Re}\varphi (tv)\ge c{t}^2\mathrm{\mid }v{\mathrm{\mid }}^2$ and dividing by $t^2\mathrm{/}2$ as $t\to 0$ yields $v^T(\mathrm{Re}H)v\ge 2c\mathrm{\mid }v{\mathrm{\mid }}^2$. For a real vector $v$, write $Av=u+iw$ with $u,w\in {\mathbb{R}}^n$. Then $\mathrm{\mid }u{\mathrm{\mid }}^2-\mathrm{\mid }w{\mathrm{\mid }}^2=\mathrm{Re}(v^{T}Hv)>0$ when $v\mathrm{\ne }0$. In particular, the real projection on $A{\mathbb{R}}^n$ is injective, hence bijective by dimension, and $A{\mathbb{R}}^n=\{u+iKu:u\in {\mathbb{R}}^n\}$ for a real linear map $K$. For nonzero $u$, applying the strict inequality to $v=A^{-1}(u+iKu)$ gives $\mathrm{\mid }Ku\mathrm{\mid }<\mathrm{\mid }u\mathrm{\mid }$; compactness of the unit sphere gives $\mathrm{\parallel }K\mathrm{\parallel }<1$. The real inverse theorem [F6] makes the nonlinear contour $\mathrm{\Gamma }$ a graph $u+i(Ku+r(u))$, with $r(u)=O(\mathrm{\mid }u{\mathrm{\mid }}^2)$. For sufficiently small $r_0$, the homotopy graphs $u+i(Ku+tr(u))$ remain in the holomorphic chart and satisfy $\mathrm{Re}(w\cdot w)\ge c_{\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{u}\mathrm{r}}\mathrm{\mid }u{\mathrm{\mid }}^2$ for $\mathrm{\mid }u\mathrm{\mid }\le r_0$, because the linear margin $\mathrm{\mid }u{\mathrm{\mid }}^2-\mathrm{\mid }Ku{\mathrm{\mid }}^2\ge (1-\mathrm{\parallel }K{\mathrm{\parallel }}^2)\mathrm{\mid }u{\mathrm{\mid }}^2$ dominates the $O(\mathrm{\mid }u{\mathrm{\mid }}^3)$ error. The homotopy $u+itKu$, $0\le t\le 1$, also has $\mathrm{Re}(w\cdot w)=\mathrm{\mid }u{\mathrm{\mid }}^2-t^2\mathrm{\mid }Ku{\mathrm{\mid }}^2\ge (1-\mathrm{\parallel }K{\mathrm{\parallel }}^2)\mathrm{\mid }u{\mathrm{\mid }}^2$. On the fixed outer boundary these bounds, together with boundedness of the transformed amplitude and Jacobian, make the integrals $O(e^{-c_1\lambda })$. [F7] and the contour statement in [F1] therefore deform $\mathrm{\Gamma }$ to the real Gaussian contour, with only that exponential boundary term.

Truncate the Taylor series after total degrees below $2N$. On a smaller ball $\mathrm{\mid }w\mathrm{\mid }\le r_0<R$, the Cauchy bound [F5] and a geometric-series estimate give $\mid b(w)-{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }<2N}{b}_{\alpha }{w}^{\alpha }\mid \le C_2^{N+1}\mathrm{\mid }w{\mathrm{\mid }}^{2N}\mathrm{.}$ Integrating this remainder against $e^{-\lambda \mathrm{\mid }w{\mathrm{\mid }}^2\mathrm{/}2}$ and using $z=\sqrt{\lambda }w$ bounds it by $C_3^{N+1}{\lambda }^{-N-n\mathrm{/}2}{\int }_{{\mathbb{R}}^n}\mathrm{\mid }z{\mathrm{\mid }}^{2N}{e}^{-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}2}dz\le C_4^{N+1}N!{\lambda }^{-N-n\mathrm{/}2}\mathrm{.}$ For the last inequality, convexity gives $({\sum }_{j=1}^n{z}_j^2{)}^N\le n^{N-1}{\sum }_{j=1}^n\mathrm{\mid }{z}_j{\mathrm{\mid }}^{2N}$. Fubini factors each summand into one-dimensional Gaussian moments; integration by parts gives $M_N=(2N-1)M_{N-1}$ with $M_0=\sqrt{2\pi }$, since the polynomial times Gaussian boundary term tends to zero. Hence $M_N=(2N-1)!!\sqrt{2\pi }\le 2^{N}N!\sqrt{2\pi }$, and the displayed integral is at most $(2n{)}^N(2\pi {)}^{n\mathrm{/}2}N!$. Absorb this fixed-dimensional factor into $C_4^{N+1}$. The extension of the Taylor polynomial $P_{<2N}(w)={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }<2N}{b}_{\alpha }{w}^{\alpha }$ from $\mathrm{\mid }w\mathrm{\mid }\le r_0$ to the full real Gaussian integral needs a separate estimate uniform in $N$. For $\mathrm{\mid }w\mathrm{\mid }\ge r_0$ and $k<2N$, $\mathrm{\mid }w{\mathrm{\mid }}^k\le r_0^{k-2N}\mathrm{\mid }w{\mathrm{\mid }}^{2N}$. Cauchy's coefficient bound and the at most exponential count of degree-$k$ multi-indices give $D_N:={\sum }_{k<2N}{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k}\mathrm{\mid }{b}_{\alpha }\mathrm{\mid }{r}_0^{k-2N}\le C_5^{N+1}$, since $r_0,R$ are fixed positive constants and the sum has $2N$ terms after grouping degrees. Consequently ${\int }_{\mathrm{\mid }w\mathrm{\mid }\ge r_0}\mathrm{\mid }{P}_{<2N}(w)\mathrm{\mid }{e}^{-\lambda \mathrm{\mid }w{\mathrm{\mid }}^2\mathrm{/}2}dw\le C_5^{N+1}{\lambda }^{-N-n\mathrm{/}2}{\int }_{{\mathbb{R}}^n}\mathrm{\mid }z{\mathrm{\mid }}^{2N}{e}^{-\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{/}2}dz\le C_6^{N+1}N!{\lambda }^{-N-n\mathrm{/}2}$. Thus the polynomial extension tail is controlled by the factorial remainder for every $N$, rather than by an exponential constant independent of $N$. The actual amplitude away from the local critical region and the contour boundary still have the exponential estimates from step 1.1 and [F1]; no extension of that amplitude outside its holomorphy neighborhood is used. Combining these estimates proves the stated normalized error for every $N$.

Expand $b(w)={\sum }_{\alpha }{b}_{\alpha }{w}^{\alpha }$ and group by total degree. Odd monomials integrate to zero over ${\mathbb{R}}^n$ because the Gaussian and domain are invariant under $w\mapsto -w$. Each monomial of degree $2j$ contributes a constant times ${\lambda }^{-n\mathrm{/}2-j}$ by the change of variables $z=\sqrt{\lambda }w$. In one dimension, integration by parts gives $I_k=\int t^{2k}{e}^{-t^2\mathrm{/}2}dt=(2k-1)I_{k-1}$ and $I_0=\sqrt{2\pi }$, hence $I_k=(2k-1)!!\sqrt{2\pi }$. Products give the $n$-dimensional moments. If $\alpha =2k$ and $\mathrm{\mid }k\mathrm{\mid }=j$, then ${\prod }_l(2k_l-1)!!\le (2j-1)!!\le 2^{j}j!$; all monomials with an odd coordinate vanish. After scaling, every degree-$2j$ moment is therefore at most $C_0^{j+1}j!{\lambda }^{-j}$. The number of degree-$2j$ multi-indices is at most $C_n^{j+1}$ and $\mathrm{\mid }{b}_{\alpha }\mathrm{\mid }\le M{R}^{-2j}$, so the coefficient of ${\lambda }^{-j}$ is bounded by $C_1^{j+1}j!$. The Jacobian at zero contributes $(\det H{)}^{-1\mathrm{/}2}$ on the chosen branch, and the degree-zero coefficient is $a(0)$.