Surjectivity of the completion map for rings of $C^\infty$-functions. (Whitney extension theorem for general filtrations)

TL;DR

This paper investigates the conditions under which the completion map for rings of smooth functions is surjective, extending classical results like Borel's lemma and Whitney extension to more general settings.

Contribution

It establishes necessary and sufficient conditions for the surjectivity of the completion map in general $C^$-rings and filtrations, and constructs representatives with specific analytic and positivity properties.

Findings

Characterization of surjectivity conditions for $C^$-ring completions

Construction of smooth representatives with prescribed properties

Extension of classical extension theorems to broader contexts

Abstract

The classical lemma of Borel reads: any power series with real coefficients is the Taylor series of a smooth function. Algebraically this means the surjectivity of the completion map at a point, $C^\infty(\Bbb{R}^n) \twoheadrightarrow \Bbb{R}[[\underline{x}]]$. Similarly, Whitney extension theorem implies the surjectivity of the completion at closed subsets of $\Bbb{R}^n$. For various applications one needs the surjectivity of completion for general $C^\infty$-rings and general filtrations. We establish the necessary and sufficient conditions for this surjectivity. Moreover, we prove: any element of the completion admits a $C^\infty$-representative that is real-analytic outside of the locus of completion, has any prescribed vanishing rate "at infinity", and the prescribed positivity behaviour at the finite part. Alternatively, one can impose on the smooth representative a set of (compatible) linear conditions.

Equations

$$\begin{array}{ccccc}0\to (\underline{x}{)}^{\infty }\to & C^{\infty }({\mathbb{R}}^n)& \to & \mathbb{R}[[\underline{x}]]& \to 0\\ & \cup & & ||& \\ & C^{\infty }({\mathbb{R}}^n)\cap C^{\omega }({\mathbb{R}}^n\setminus \{o\})& \to & \mathbb{R}[[\underline{x}]]& \to 0\end{array}$$

$$\begin{array}{ccccc}0\to I(Z{)}^{⟨\infty ⟩}\to & C^{\infty }(\mathcal{U})& \to & {\widehat{C^{\infty }(\mathcal{U})}}^{(Z)}& \to 0\\ & \cup & & ||& \\ & C^{\infty }({\mathbb{R}}^n)\cap C^{\omega }(\mathcal{U}\setminus Z)& \to & {\widehat{C^{\infty }(\mathcal{U})}}^{(Z)}& \to 0\end{array}$$

$$I(Z{)}^{⟨j⟩}:={\cap }_{x\in Z}{\mathfrak{m}}_x^j\subset R.$$

$${\partial }_{x_1}^{k_1}\dots {\partial }_{x_n}^{k_n}g$$

$$\begin{matrix}0&\to&\big{\{}\sum g_{j}|\ \sum^{N}_{j=0}g_{j}\in I_{N},\ \forall\ N\big{\}}&\to&\prod^{\infty}_{j=0}I_{j}&\stackrel{{\scriptstyle\pi}}{{\to}}&{\widehat{R}}^{(I_{\bullet})}&\to&0.\end{matrix}$$

$$\begin{array}[]{l}\text{Given a closed subset $Z\subset\mathcal{U}$ and a collection of continuous functions, $\{h_{\underline{k}}\}$, on $Z$,}\\ \text{ does there exist $f\in C^{\infty}(\mathcal{U})$ with the prescribed partial derivatives $\{f^{({\underline{k}})}|_{Z}=h_{\underline{k}}\}_{\underline{k}}$? }\end{array}$$

$$Z\stackrel{h_k}{\to }Ho{m}_{\mathbb{R}}(Sy{m}^k({\mathbb{R}}^n),\mathbb{R}).$$

$$\forall d,k\in \mathbb{N},\forall x,y\in Z:h_k{|}_x-\sum _{0\le \tilde{k}\le d}\frac{1}{\tilde{k}!}{h}_{k+\tilde{k}}{|}_y(\underset{\tilde{k}}{\underbrace{x-y,\dots ,x-y}})=o(||x-y|{|}^d).$$

$$({\oplus }_k{h}_k)\cdot ({\oplus }_{\tilde{k}}{\tilde{h}}_{\tilde{k}})=\underset{k,\tilde{k}}{\oplus }\left(\genfrac{}{}{0ex}{}{k+\tilde{k}}{k}\right)\cdot Sym(h_k\otimes h_{\tilde{k}})\in C^{\infty }(Z).$$

$$\begin{array}{ccccc}0\to I(Z{)}^{⟨\infty ⟩}\to & C^{\infty }(\mathcal{U})& \to & C^{\infty }(Z)& \to 0\\ & \cup & & ||& \\ & C^{\infty }(\mathcal{U})\cap C^{\omega }(\mathcal{U}\setminus Z)& \to & C^{\infty }(Z)& \to 0\end{array}$$

$$\sum _{k=0}^{\infty }\frac{h_k(\stackrel{k}{\overbrace{y,\dots ,y}})}{k!}\in C^{\infty }({\mathcal{U}}_{x_1\dots x_j})[[y_{j+1},\dots ,y_n]].$$

$${\mathcal{U}}_j=\{x||\tilde{\tau }(x)|>{ϵ}_j\}\cap Bal{l}_{\frac{1}{{ϵ}_j}}(o)\cap \mathcal{U}.$$

$$f_{\alpha }-\sum _{j=0}^N{g}_j{|}_{{\mathcal{U}}_{\alpha }}\in I_N\cdot C^{\infty }({\mathcal{U}}_{\alpha }),\text{ for each α and N.}$$

$$\{u_{\alpha}\in C^{\infty}(\mathcal{U})\}_{\alpha}:\quad\quad 0<u_{\alpha}\big{|}_{\mathcal{U}_{\alpha}}\leq 1,\quad\quad\quad\quad u_{\alpha}\big{|}_{\mathcal{U}\setminus\mathcal{U}_{\alpha}}=0,\quad\quad\quad\quad\sum u_{\alpha}=1_{\mathcal{U}}.$$

$$\forall N:f-\sum _{j=0}^N{g}_j=\sum _{\alpha }{u}_{\alpha }{f}_{\alpha }-\sum _{j=0}^N{1}_{\mathcal{U}}\cdot g_j=\sum _{\alpha }(u_{\alpha }{f}_{\alpha }-\sum _{j=0}^N{u}_{\alpha }{g}_j)\in I_N.\text{hfill}■$$

$$\begin{array}{c}0\to I_{\infty }\to R\stackrel{{\hat{}}_R}{\to }{\hat{R}}^{(I_{\bullet })}\\ \downarrow \varphi \downarrow \downarrow \hat{\varphi }\\ 0\to {\tilde{I}}_{\infty }\to \tilde{R}\stackrel{{\hat{}}_{\tilde{R}}}{\to }{\hat{\tilde{R}}}^{({\tilde{I}}_{\bullet })}\end{array}$$

$$Z\subset \mathcal{U}(Z),V(q)\subset \mathcal{U}(V(q)),\mathcal{U}(Z)\cap \mathcal{U}(V(q))=\varnothing .$$

$$V_i(I_1)\subseteq V_i(I_2)\subseteq \cdots \subseteq \underset{j}{\cup }{V}_i(I_j)=:Z_i.$$

$$I_j:=I(X_1)\cap I(X_2{)}^2\cap \cdots \cap I(X_j{)}^j.$$

$$\{I_{j}\cap I(Z_{1})\cap I(Z_{2})^{\langle 2\rangle}\cap\dots\cap I(Z_{j})^{\langle j\rangle}\big{|}_{\mathcal{U}_{0}}\}.$$

$$I_{d_{j}}\big{|}_{\mathcal{U}_{0}}\subseteq I_{j}\cap I(Z_{1})\cap I(Z_{2})^{\langle 2\rangle}\cap\dots\cap I(Z_{j})^{\langle j\rangle}\big{|}_{\mathcal{U}_{0}}\subset R|_{\mathcal{U}_{0}}={\footnotesize\left.\raisebox{1.37775pt}{$C^{\infty}(\mathcal{U}_{0})$}\!/\!\raisebox{-1.37775pt}{$J\cdot C^{\infty}(\mathcal{U}_{0})$}\right.}.$$

$$\tilde{g}_{j}(x)=\Bigg{\{}\begin{array}[]{l}0,\ x\not\in Ball_{\epsilon_{j}}(p_{j})\\ >0,\ x\in Ball_{\epsilon_{j}}(p_{j})\\ j!,\ x=p_{j}\end{array}$$

$$Z_1\cap {\mathcal{U}}_0=V_1(I_1)\cap {\mathcal{U}}_0,Z_2\cap {\mathcal{U}}_0=V_2(I_1)\cap {\mathcal{U}}_0,\dots Z_i\cap {\mathcal{U}}_0=V_i(I_1)\cap {\mathcal{U}}_0.$$

$$p_j\in V_{i+1}(I_{j+1})\setminus V_{i+1}(I_j),Bal{l}_{{ϵ}_j}(p_j)\cap V_{i+1}(I_j)=\varnothing .$$

$${\mathfrak{m}}_{p_j}^i+J\subset R.$$

$$\sum _{j=1}^N{\tilde{g}}_j-\tilde{f}\in {\tilde{I}}_N+J,\text{ for any }N.$$

$$I_j=⟨\prod _{i=1}^j(x-p_i{)}^i⟩\subset C^{\infty }({\mathbb{R}}^1)=:R.$$

$$Z_i:=\{p_0,p_i,p_{i+1},p_{i+2},\dots \}\subset {\mathbb{R}}^n,I_j:=I(Z_1)\cap \cdots \cap I(Z_j{)}^{⟨j⟩}\subset R:=C^{\infty }({\mathbb{R}}^n).$$

$${\cap }_j(\pi (I_j)\cdot {\hat{R}}^{(I_{\bullet })})=\underleftarrow{\lim }(0)=(0)\subset \underleftarrow{\lim }R.$$

$${\tilde{I}}_{\bullet }:=I_{\bullet }+{\mathfrak{m}}_p^i\subset R$$

Full text

For various applications one needs the surjectivity of completion for general $C^{\infty}$-rings and general filtrations. We establish the necessary and sufficient conditions for this surjectivity.

Moreover, we prove: any element of the completion admits a $C^{\infty}$-representative that is real-analytic outside of the locus of completion, has any prescribed vanishing rate “at infinity”, and the prescribed positivity behaviour at the finite part. Alternatively, one can impose on the smooth representative a set of (compatible) linear conditions.

D.Kerner was partially supported by Israel Science Foundation (grant No. 1910/18)

1. Introduction

1.1.

The classical lemma of Borel111Though published in [Borel], it was partially known to Peano, see [Besenyei]. reads: any sequence of real numbers is realizable as the sequence of partial derivatives (at the origin) of a smooth function, moreover this function can be assumed analytic off the origin. Algebraically this means the surjectivity of the completion map $C^{\infty}(\mathbb{R}^{n})\twoheadrightarrow\mathbb{R}[[{\underline{x}}]]$, for the filtration $({\underline{x}})^{\bullet}$. We get two exact sequences:

[TABLE]

This surjectivity goes in notable difference to the completion maps of traditional (Noetherian) rings of Commutative Algebra/Algebraic Geometry.

More generally, Whitney extension theorem gives the necessary and sufficient conditions to extend a function with prescribed “derivatives” on a(ny) closed set $Z\subset\mathcal{U}\subseteq\mathbb{R}^{n}$ to a function smooth on $\mathcal{U}$, §2.2. (Moreover, the function can be taken real analytic on $\mathcal{U}\setminus Z$.) Algebraically this implies the two exact sequences of completion:

[TABLE]

(Here $I(Z)^{\langle\infty\rangle}$ denotes the ideal of functions flat on $Z$. The completion is taken with respect to the differential powers of ideals, we denote this filtration by $I(Z)^{\langle\bullet\rangle}$, see §2.1.ii.)

These results are of everyday use in Analysis/Differential Geometry. In Algebraic Geometry over $C^{\infty}$-rings, [Joyce], [Moerdijk-Reyes], and in $C^{\infty}$-Singularity Theory one uses more general $C^{\infty}$-rings and more general filtrations, $I_{\bullet}$, not necessarily by powers of ideals. The surjectivity of completion does not always hold in this case, and the (necessary/sufficient) conditions are not implied by Whitney extension theorem. The data of derivatives of a function is essentially different from the data of an element of $\widehat{C^{\infty}(\mathcal{U})}\ \!^{(I_{\bullet})}$. Though this surjectivity question is most natural and basic, it was not addressed before. (Perhaps because the methods needed involve both analysis and commutative algebra.) In [Bel.Boi.Ker] we have obtained the first sufficient condition for this surjectivity, but it was far from being necessary.

Our paper extends the classical results in several directions:

• •

We give the necessary and sufficient conditions for the surjectivity of the completion map for general filtrations. Moreover, for an element of the completion $\widehat{f}\in{\widehat{R}}^{(I_{\bullet})}$, we establish the preimage, $f\in R$, which is real analytic (in the maximal allowed region), decays with any prescribed rate at the boundary, $\partial\mathcal{U}$, and is positive off a small neighbourhood of $Z$.

Without requiring the analyticity on $\mathcal{U}\setminus Z$ we can ensure that $f$ satisfies any prescribed system of $\mathbb{R}$-linear conditions that are compatible in a precise sense. This extends the classical moment problem.

• •

We establish this surjectivity for several types of $C^{\infty}$-rings, e.g. the quotient rings, {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}, and localizations, $C^{\infty}(\mathcal{U})[S^{-1}]$. In particular, this addresses the functions on manifolds/$C^{\infty}$-schemes, and the $C^{\infty}$-germs (at points or along closed sets).

The sufficient condition (in §4) has a different form than the necessary condition (in §3), but is trivially weaker than the necessary condition. Therefore both conditions are necessary and sufficient (and equivalent). The necessary/sufficient conditions are stated algebraically (via the equivalence of filtrations by ideals). An interesting question is whether/how to formulate (and prove) these conditions in terms of the classical analysis.

We remark that the surjectivity conditions have nothing to do with $I_{\bullet}$ being finitely/sub-analytically/$o$-minimally generated. See example 3.4 for polynomially generated $I_{\bullet}$ with non-surjective completion. It is impressive that this “algebraization” of Whitney extension theorem is manageable in full generality, in view of the notoriously complicated/pathological behavior of (filtrations by) ideals in $C^{\infty}$-rings.

Finally, the additional properties of the preimage, $f\in R$, are valuable in various applications.

Due to the lack of space we postpone other results/applications (e.g. inversion of Borel-Taylor map for $I_{\bullet}$ and the Tougeron-type approximations) to the next paper.

This surjectivity is the necessary “Step.0” for various problems in Singularity Theory and Commutative Algebra. The standard approach is: pass to the completion, $R\to{\widehat{R}}$, resolve the problem over ${\widehat{R}}$, then pullback the results to $R$. The later step is done via the Artin-Tougeron approximation, which in the $C^{\infty}$-case begins with “Take a $C^{\infty}$-representative of $\widehat{f}\in{\widehat{R}}$.” (See [Bel.Ker.a], [Bel.Boi.Ker] for the $C^{\infty}$-approximation results and further references.) We were particularly motivated by the study of determinacy/algebraizability/deformations/local topology of $C^{\infty}$-maps/schemes with non-isolated singularities, [Bel.Ker.b], [Boi.Gre.Ker], [Bel.Ker.c]. As the filtrations there are not by powers of ideals, the classical Whitney does not help, see e.g. §5.1.vi.

1.2. The structure of the paper

1. §2

is preparatory. In §2.1 we fix the notations and conventions.

1. §2.2

presents the classical Whitney extension theorem as the surjectivity of completion for the special filtration, by differential powers of ideals, $I(Z)^{\langle\bullet\rangle}$. 2. §2.3

recalls the analytic approximation of Whitney. 3. §2.4

we prove the filtered (strengthened) version of Whitney theorem on zero sets. For any filtration $I_{\bullet}$ of $C^{\infty}(\mathcal{U})$, the total zero set $Z:=\overline{\cup{V(I_{j})}}\subset\mathcal{U}$ is defined by one element of $I_{\infty}:=\cap I_{j}$. Moreover, this element can be chosen real-analytic on $\mathcal{U}\setminus Z$. This goes in notable difference to the traditional Noetherian rings of Algebraic Geometry, and can be considered as a $C^{\infty}$-Nullstellensatz. It is used repeatedly later. 4. §2.5

we show the persistence of surjectivity under the change of rings, for taking quotients, R\to{\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{J}\right.}, and localizations, $R\to R[S^{-1}]$. This reduces the surjectivity problem to the very particular ring $C^{\infty}(\mathcal{U})$, for an open $\mathcal{U}\subseteq\mathbb{R}^{n}$, that can be chosen as a ball.

In particular, the surjectivity criteria for $C^{\infty}(\mathcal{U})$ apply, e.g. to the rings of $C^{\infty}$-germs (along closed sets). 5. §2.6

the surjectivity for $C^{\infty}$-rings on manifolds/schemes is reduced to the surjectivity for {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}. 2. §3

gives the necessary condition for the surjectivity, theorem 3.1. The proof goes by tracing the zero loci of the ideals $I_{\bullet}$, and the vanishing orders of $I_{\bullet}$ along the zero loci (with respect to the maximal ideals). Then we show the pointwise stabilization of the filtration {\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}\subset{\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}.

A peculiar consequence is: if the $I_{\bullet}$-completion of $C^{\infty}(\mathcal{U})$ is surjective then it is non-injective, $I_{\infty}:=\cap I_{j}\neq(0)$. 3. §4

gives the sufficient condition for the surjectivity, theorem 4.1. Moreover, we show that any element of $\widehat{C^{\infty}(\mathcal{U})}^{\ (I_{\bullet})}$ admits a representative $f\in C^{\infty}(\mathcal{U})$ that is real-analytic on $\mathcal{U}\setminus Z$, and positive outside of a small neighbourhood of $Z$. The positivity of $f$ cannot be strengthened to the positivity on the whole $\mathcal{U}\setminus Z$, see remark 4.5.

The proof is an explicit construction and (unsurprisingly) uses cutoff functions with derivatives of controlled growth. 4. §5

contains the simplest examples and applications. In §5.1 we derive various classical statements, e.g. multi-Borel lemma, Borel lemma in families, Borel lemma for flat functions. In §5.1.vi we consider a simple non-isolated hypersurface singularity. The corresponding filtration is not by powers of ideals, thus already in this case the needed surjectivity cannot be obtained with the classical Whitney extension theorem.

As immediate applications of the surjectivity we address the “Inverse Artin-Tougeron problem” in §5.2, the contraction of ideals in §5.3, the completion of $C^{\infty}$-modules in §5.4, and lifting the ${\widehat{R}}$-modules to $R$-modules in §5.5. 5. §6

strengthens the surjectivity further, by ensuring special properties of $f$.

1. §6.1:

a representative $f\in C^{\infty}(\mathcal{U})\cap C^{\omega}(\mathcal{U}\setminus Z)$ can be chosen with any (prescribed) rate of decay on $\partial\mathcal{U}$. This strengthens/extends the Borel-type results for spaces of Schwartz functions. 2. §6.2:

a representative $f\in C^{\infty}(\mathcal{U})$ can be chosen to satisfy any prescribed set of linear conditions that are compatible in a precise sense. This extends numerous versions of the classical moment problem.

1.3. Acknowledgement

We thank M.Sodin for the highly useful reference to [Hörmander], A. Fernandez-Boix for the participation in the initial stage of this work, and A. Kiro, E. Shustin for important comments.

2. Preparations

2.1. Notations and conventions

Take an open subset $\mathcal{U}\subseteq\mathbb{R}^{n}$ (possibly topologically non-trivial, e.g. non-contractible), an ideal $\{0\}\subseteq J\subset C^{\infty}(\mathcal{U})$, and the quotient ring R={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}.

1. i.

For any ideal $I\subset R$ we take its (reduced) set of zeros, $V(I)\subseteq V(J)\subseteq\mathcal{U}$. For any subset $Z\subset V(J)\subseteq\mathcal{U}$ denote by $I(Z)\subset R$ the ideal of functions vanishing on $Z$. Thus $I(V(I))\supseteq I$.

For any $\epsilon>0$ define the $\epsilon$-neighbourhood, $\mathcal{U}_{\epsilon}(Z):=\{x|\ dist(x,Z)<\epsilon\}\subset\mathcal{U}$. 2. ii.

Denote the maximal ideal of a point $x\in\mathcal{U}$ by ${\mathfrak{m}}_{x}\subset R$.

For any subset $Z\subset V(J)\subseteq\mathcal{U}$ define the $j$’th differential power of $I(Z)$ as the set of functions that vanish on $Z$ up to $j$’th order (see e.g. [D.S.G.H])

[TABLE]

The inclusion $I(Z)^{\langle j\rangle}\supseteq I(Z)^{j}$ is obvious, and is often proper. The classical example is $Z=V(xy,yz,xz)\subset\mathbb{R}^{3}$. Here $xyz\in I(Z)^{\langle 2\rangle}\setminus I(Z)^{2}$.

The ideal of functions flat on $Z$ is $I(Z)^{\langle\infty\rangle}=\cap_{j}I(Z)^{\langle j\rangle}=\cap_{x\in Z}{\mathfrak{m}}^{\infty}_{x}$.

Denote the completion with respect to the filtration $I(Z)^{\langle\infty\rangle}$ by ${\widehat{R}}^{(Z)}$, as in equation (2). 3. iii.

We denote the partial derivatives by $\partial^{k_{1}}_{x_{1}}\dots\partial^{k_{n}}_{x_{n}}g$ or by $g^{({\underline{k}})}$, using multi-indices. The condition

[TABLE]

is abbreviated to “$|g^{({\underline{k}})}|<...$”.

The total $k$’th derivative is a function valued in (dual) $k$-forms, $\mathcal{U}\stackrel{{\scriptstyle g^{(k)}}}{{\to}}Hom_{\mathbb{R}}(Sym^{k}(\mathbb{R}^{n}),\mathbb{R})$. Denote its operator norm by $||g^{(k)}||$. 4. iv.

For a closed subset $Z\subset\mathcal{U}$ denote by $C^{\infty}(\mathcal{U},Z)$ the ring of germs of smooth functions at $Z$, see example 2.11. 5. v.

Fix a filtration by ideals, $R=I_{0}\supseteq I_{1}\supseteq\cdots$. Denote $I_{\infty}:=\cap I_{j}$. Take the corresponding completion, R\to{\widehat{R}}^{(I_{\bullet})}:=\lim\limits_{\leftarrow}{\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{j}}\right.}. Its elements are (equivalence classes of) Cauchy sequences of functions, $\{f_{j}\}\in R$, such that $f_{j+i}-f_{j}\in I_{j}$, for all $i,j>0$. These elements can be presented also as the formal sums $\sum^{\infty}_{j=0}g_{j}\in\prod I_{j}$. These sums are taken up to the equivalence: $\sum g_{j}\equiv\sum\tilde{g}_{j}$ if $\sum^{N}(g_{j}-\tilde{g}_{j})\in I_{N}$ for any $N$. One has the exact sequence:

[TABLE]

An element $\sum^{\infty}_{j=0}g_{j}\in{\widehat{R}}^{(I_{\bullet})}$ is presented by $f\in R$ if $f-\sum^{N}_{j=0}g_{j}\in I_{N}$ holds for any $N\geq 0$.

Two filtrations, $I_{\bullet}$, $\tilde{I}_{\bullet}$ are called equivalent if they satisfy: $I_{k_{j}}\subseteq\tilde{I}_{j}\subseteq I_{d_{j}}$, where $d_{j},k_{j}\to\infty$ as $j\to\infty$. Equivalent filtrations induce isomorphic completions, as on the diagram.

   $\begin{matrix}R\to{\widehat{R}}^{(I_{\bullet})}\\ \parallel\quad\quad\parallel\quad\ \\ R\to{\widehat{R}}^{(\tilde{I}_{\bullet})}\end{matrix}$

2.2. Whitney extension theorem vs the surjectivity of completion

Recall the classical Whitney extension problem:

[TABLE]

If $f$ exists then the Taylor expansion of each $h_{\underline{k}}$ should be expressible via the functions $\{h_{{\underline{k}}+\tilde{\underline{k}}}\}_{\tilde{\underline{k}}}$.

To write this explicitly it is useful to combine, for each $k$, the partials $\{h_{\underline{k}}\}_{|{\underline{k}}|=k}$ into the total derivative:

[TABLE]

Then the conditions $\{f^{(k)}|_{Z}=h_{k}\}_{k}$ imply the compatibility conditions:

[TABLE]

The Whitney extension theorem says that these compatibility conditions are also sufficient, [Whitney]. Moreover, one can choose $f$ to be real-analytic on $\mathcal{U}\setminus Z$.

2.2.1.

A Whitney jet on a closed subset $Z\subset\mathcal{U}\subseteq\mathbb{R}^{n}$ is a collection of functions compatible as in (7), denote it by $Z\stackrel{{\scriptstyle\oplus h_{k}}}{{\to}}\prod_{k}Hom_{\mathbb{R}}(Sym^{k}(\mathbb{R}^{n}),\mathbb{R})$. (Note that we allow infinite sums here.) Denote by $C^{\infty}(Z)$ the vector space of all the Whitney jets. Define the product on $C^{\infty}(Z)$ by

[TABLE]

(Here $Sym(\dots)$ is the total symmetrization of the tensor, this imposes the factor $\binom{k+{\tilde{k}}}{k}$.)

With this product $C^{\infty}(Z)$ becomes a (commutative, associative, unital) ring. (By the direct check.) Whitney extension theorem is then the exactness of the two sequences (of rings and ideals):

[TABLE]

Lemma 2.1**.**

Whitney extension theorem identifies \widehat{C^{\infty}(\mathcal{U})}^{(Z)}\xrightarrow{\,\smash{\raisebox{-2.79857pt}{\scriptstyle\sim}}\,}{}C^{\infty}(Z), the completion for the filtration $I(Z)^{\langle\bullet\rangle}$ in equation (2). In particular this implies the surjectivity of the completion.

*Proof.  *Define the map $\widehat{C^{\infty}(\mathcal{U})}^{(Z)}\stackrel{{\scriptstyle\Psi}}{{\to}}C^{\infty}(Z)$ by $\sum g_{j}\to(\sum g_{j}|_{Z},\sum g^{(1)}_{j}|_{Z},\dots)$. If $\sum g_{j}\equiv\sum\tilde{g}_{j}\in\widehat{C^{\infty}(\mathcal{U})}^{(Z)}$ then $\Psi(\sum g_{j})=\Psi(\sum\tilde{g}_{j})$, see §2.1.v. Here each sum $\sum g^{(k)}_{j}|_{Z}$ is finite as $g^{(k)}_{j}|_{Z}=0$ for $j>k$. By construction $\Psi$ is a homomorphism of rings. It is injective, as $ker(\Psi)=I(Z)^{\langle\infty\rangle}\equiv 0\in\widehat{C^{\infty}(\mathcal{U})}^{(Z)}$.

For any Whitney jet $\oplus h_{k}$ take its Whitney representative, $f\in C^{\infty}(\mathcal{U})$. Then its completion, $\widehat{f}\in\widehat{C^{\infty}(\mathcal{U})}^{(Z)}$, is mapped by $\Psi$ to $\oplus h_{k}$. Thus $\Psi$ is surjective, hence an isomorphism, and equation (9) is just equation (2). $\blacksquare$

Whitney theorem does not imply the surjectivity for more general filtrations, even if one assume $V(I_{1})=V(I_{2})=\cdots$. For example, take a filtration $C^{\infty}(\mathcal{U})\supset I_{\bullet}$ such that $\{V(I_{j})=Z\}_{j}$, $I_{j}\subset I(Z)^{\langle j\rangle}$, but $I_{\bullet}$ is not equivalent to $I(Z)^{\langle\bullet\rangle}$. Thus $I_{\infty}\subsetneq I(Z)^{\langle\infty\rangle}$. For any element $\sum g_{j}\in{\widehat{R}}^{(I_{\bullet})}$ Whitney theorem ensures a $f\in C^{\infty}(\mathcal{U})$ such that $f-\sum^{N}g_{j}\in I(Z)^{\langle N\rangle}$, for all $N$. But this $f$ is not a representative of $\sum g_{j}$ as it does not satisfy $f-\sum^{N}g_{j}\in I_{N}$.

Besides, filtrations satisfying $I_{j}\subseteq I(Z)^{\langle j\rangle}$ are rather special, see examples 3.2 and 5.1. For the general filtrations the surjectivity question does not seem to be resolvable by Whitney theorem.

2.2.2.

We prove that the surjectivity of completion implies Whitney extension theorem for a large class of sets $Z$. Suppose a closed subset $Z\subset\mathcal{U}\subseteq\mathbb{R}^{n}$ admits a locally finite smooth stratification, i.e. $Z=\coprod^{n}_{j=0}Z_{j}$, where each $Z_{j}$ is the disjoint locally finite (possible empty) union of dimension $j$ manifolds in $\mathcal{U}$. This class of sets is rather large, e.g. any semi-algebraic/sub-analytic set is of this type.

Lemma 2.2**.**

For sets with locally finite smooth stratification the surjectivity of completion implies the Whitney extension theorem.

*Proof.  *Take a Whitney jet $\{h_{k}\}_{k}$ on $Z$. By partition of unity arguments, see e.g. §2.5.1, we can pass to small open subsets of $\mathcal{U}$. Thus we assume the finite smooth stratification of $Z$. Moreover, for each connected component of each $Z_{j}$ we can assume that the normal bundle (inside $\mathcal{U}$) is trivial.

1. Step 1.

$Z_{0}$ is a finite set of points, say $\{p_{i}\}_{i}$. For each $p_{i}$ the restrictions $\{h_{k}|_{p_{i}}\}_{k}$ are the prescribed (total) derivatives of $k$’th orders. By the surjectivity of completion (Borel lemma) there exists $f_{0}\in C^{\infty}(\mathcal{U})$ with these derivatives at $\{p_{i}\}$.

Define $\{h_{1,k}:=h_{k}-f^{(k)}_{0}|_{Z}\}_{k}$, thus $\{h_{1,k}|_{Z_{0}}=0\}_{k}$. It remains to resolve the Whitney extension problem for $\{h_{1,k}\}_{k}$ on $Z$. 2. Step 2.

(Induction step.) Assume we have constructed $f_{0},\dots,f_{j-1}$, as before. We have the Whitney jet $\{h_{j,k}\}_{k}$, these functions vanish on $Z_{0}\cup\dots\cup Z_{j-1}$. Now $Z_{j}$ is a finite disjoint union of submanifolds of $\mathcal{U}$, and it is enough to construct $f_{j}$ on each of them separately. (Note that $f_{j}$ will be flat on the inner boundary of $Z_{j}$.)

Accordingly, let $Z_{j}\subset\mathcal{U}$ be a manifold of $dim=j$. We can assume $Z_{j}$ is small, in particular we trivialize its normal bundle and split the $\mathcal{U}$-coordinates near $Z_{j}$: $x_{1}\dots,x_{j}$ along $Z_{j}$, and $y_{j+1},\dots,y_{n}$ transversal to $Z_{j}$. Take the power series, here $y$ is the multi-variable

[TABLE]

Present it by $f_{j}\in C^{\infty}(\mathcal{U})$, by the surjectivity of completion. Then $f_{j}$ extends the Whitney jet $\{h_{k}\}_{k}$ on $Z_{j}$. Note that $f_{j}$ is flat on $\overline{Z_{j}}\cap(\cup^{j-1}_{i=0}Z_{i})$. Define $\{h_{j+1,k}:=h_{j,k}-f^{(k)}_{j}|_{Z}\}_{k}$, thus $\{h_{j+1,k}\}_{k}$ all vanish on $\cup^{j}_{i=0}Z_{i}$. This finishes the inductive step. $\blacksquare$

Remark 2.3**.**

By the same argument one can obtain the Whitney extension theorem for the following larger class of sets: $Z$ contains a locally finite collection of disjoint manifolds, $\{M_{i}\}$, whose complement is a discrete, at most countable, union of manifolds. Namely, $Z\setminus(\coprod M_{i})=\coprod N_{j}$, where $\coprod N_{j}\subset\coprod\mathcal{U}(N_{j})$, for some disjoint neighbourhoods, $\mathcal{U}(N_{j})\cap\mathcal{U}(N_{i})=\varnothing$ for $i\neq j$.

Sets with locally finite smooth stratifications are obviously of this type. An example without a locally finite smooth stratification: $Z$ is a sequence of points with a locally finite set of condensation points.

2.3. Analytic approximation of smooth functions

For a closed subset $Z\subset\mathcal{U}$ take an exhaustion of $\mathcal{U}\setminus Z$ by compactly embedded opens, i.e. $\varnothing\subseteq\mathcal{U}_{1}\subseteq\cdots\subset\mathcal{U}\setminus Z$ open bounded such that $\overline{\mathcal{U}_{j}}\subset\mathcal{U}_{j+1}$ and $\cup\mathcal{U}_{j}=\mathcal{U}\setminus Z$.

Lemma 2.4**.**

[Whitney, Lemma 6]** For any $\tilde{f}\in C^{\infty}(\mathcal{U}\setminus Z)$ and any sequence of reals $\{\epsilon_{j}\to 0\}$ there exists $f\in C^{\omega}(\mathcal{U}\setminus Z)$ satisfying $||f^{(k)}-\tilde{f}^{(k)}||<\epsilon_{j}$ on $\mathcal{U}\setminus(Z\cup\mathcal{U}_{j})$, for $0\leq k\leq j<\infty$.

If $\tilde{f}\in C^{\infty}(\mathcal{U})$ then the lemma guarantees: $f\in C^{\infty}(\mathcal{U})\cap C^{\omega}(\mathcal{U}\setminus Z)$. Indeed, $\lim_{x\to x_{0}}f^{(k)}|_{x}=\tilde{f}^{(k)}|_{x_{0}}$, for any $k$ and any $x_{0}\in Z$. Thus each $f^{(k)}$ extends continuously onto the whole $\mathcal{U}$.

2.4. Filtered Whitney theorem of zeros

Whitney theorem on zeros reads: any closed subset $Z\subset\mathcal{U}$ is presentable as $V(\tau)\subset\mathcal{U}$ for some $\tau\in C^{\infty}(\mathcal{U})$. (In notable difference to the situation Algebraic Geometry/Commutative Algebra.) We give a filtered version, for $I_{\bullet}\subset C^{\infty}(\mathcal{U})$.

Proposition 2.5**.**

Let $Z\!=\!\overline{\cup_{j}V(I_{j})}\!\subset\!\mathcal{U}$. Then exists $\tau\!\in\!I_{\infty}\cap C^{\omega}(\mathcal{U}\setminus Z)$ such that $\tau|_{Z}\!=\!0$ and $\tau|_{\mathcal{U}\setminus Z}\!>\!0$.

Proof.

1. 1.

For any $j$ we construct $\tau_{j}\in I_{j}$ satisfying $V(\tau_{j})=V(I_{j})$. For any point $p\in\mathcal{U}\setminus V(I_{j})$ exists some $\tau_{p}\in I_{j}$ such that $\tau_{p}(p)\neq 0$. By local compactness this collection $\{\tau_{p}\}$ can be chosen locally finite on $\mathcal{U}$. Then define $\tau_{j}:=\sum(\tau_{p})^{2}\cdot a_{p}$, where $\{a_{p}\}$ are the relevant non-negative cutoff functions. 2. 2.

We construct $\tilde{\tau}\in I_{\infty}\subset C^{\infty}(\mathcal{U})$ satisfying: $V(\tilde{\tau})=Z$ and $\tilde{\tau}>0$ on $\mathcal{U}\setminus Z$. Fix $\{\tau_{j}\in I_{j}\}_{j}$ satisfying $\{V(\tau_{j})=V(I_{j})\}_{j}$. We can assume $0\leq\tau_{j}\leq 1$ and $\tau_{j}$ are flat on $V(I_{j})$. (For example, replace $I_{j}$ by $I_{j}\cap I(V(I_{j}))^{\langle\infty\rangle}$.)

Take a decreasing sequence of positive reals $\{\epsilon_{j}\to 0\}$, and the corresponding small neighbourhoods, $\mathcal{U}_{\epsilon_{j}}(Z)$, see §2.1.i. Define $\tilde{\tau}$ on $\{\mathcal{U}_{\frac{\epsilon_{j-1}+\epsilon_{j}}{2}}(Z)\setminus\mathcal{U}_{\epsilon_{j}}(Z)\}_{j}$ by $\tilde{\tau}(x)=\tau_{1}(x)\cdots\tau_{j}(x)$. Extend it in a $C^{\infty}$ way to $\mathcal{U}\setminus Z$ such that $\tau_{1}\cdots\tau_{j+1}\leq\tilde{\tau}\leq\tau_{1}\cdots\tau_{j}$ on $\mathcal{U}_{\epsilon_{j}}(Z)\setminus\mathcal{U}_{\frac{\epsilon_{j}+\epsilon_{j+1}}{2}}(Z)$.

We have defined a positive function $\tilde{\tau}\in C^{\infty}(\mathcal{U}\setminus Z)$. It is flat on each $V(I_{j})$, and therefore flat on $Z=\cup_{j}V(I_{j})$. Extend $\tilde{\tau}$ to $Z$ by [math]. Thus $\tilde{\tau}\in I_{j}$ for any $j$, and $V(\tilde{\tau})=Z$. 3. 3.

Take a sequence $\{\epsilon_{j}\to 0\}$ and the corresponding bounded opens

[TABLE]

Thus $\cup_{j}\mathcal{U}_{j}=\mathcal{U}\setminus Z$. By lemma 2.4 there exists $\tau\in C^{\infty}(\mathcal{U})\cap C^{\omega}(\mathcal{U}\setminus Z)$ satisfying $|\tau-\tilde{\tau}|<\epsilon_{j}$ on $\mathcal{U}_{j}$. Thus $\tau|_{Z}=0$, $\tau|_{\mathcal{U}\setminus Z}>0$ and $\tau\in I_{\infty}$. $\blacksquare$

Example 2.6**.**

1. i.

In the trivial case of constant filtration, $\{I_{j}=I_{1}\}$, we get: any closed $Z\subset\mathcal{U}$ is the zero locus of a smooth function which is real-analytic off $Z$. 2. ii.

Suppose the zero sets $\{V(I_{j})\}$ stabilize. Then $\cup V(I_{j})=V(I_{\infty})=V(\tau)$ for some $\tau\in I_{\infty}$. In particular, $I_{\infty}\neq 0$, regardless of how fast the ideals $I_{j}$ decrease.

2.5. Persistence of the surjectivity under change of rings

2.5.1. The local-to-global transition

Suppose we want to represent an element of completion $\sum g_{j}\in\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$ by some $f\in C^{\infty}(\mathcal{U})$.

Lemma 2.7**.**

It is enough to resolve this locally at each point of $\mathcal{U}$.

*Proof.  *Suppose for a cover $\mathcal{U}=\cup\mathcal{U}_{\alpha}$ the functions $\{f_{\alpha}\in C^{\infty}(\mathcal{U}_{\alpha})\}_{\alpha}$ represent the elements $\{\sum g_{j}|_{\mathcal{U}_{\alpha}}\in\widehat{C^{\infty}(\mathcal{U}_{\alpha})}^{(I_{\bullet})}\}_{\alpha}$. Thus

[TABLE]

We can assume that the cover is locally finite, by shrinking $\{\mathcal{U}_{\alpha}\}_{\alpha}$ if needed. Take the corresponding partition of unity,

[TABLE]

We can assume each $f_{\alpha}$ is bounded on $\mathcal{U}_{\alpha}$, thus $u_{\alpha}f_{\alpha}\in C^{\infty}(\mathcal{U})$. Define $f:=\sum u_{\alpha}f_{\alpha}$, then $f\in C^{\infty}(\mathcal{U})$, as the sum is locally finite. And $f$ represents $\sum g_{j}$:

[TABLE]

2.5.2.

Take a surjective homomorphism of (commutative, associative) rings, $R\stackrel{{\scriptstyle\phi}}{{\twoheadrightarrow}}{\tilde{R}}$.

Lemma 2.8**.**

Suppose the completion of $R$ is surjective, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$. Then the completion of ${\tilde{R}}$ is surjective, ${\tilde{R}}\twoheadrightarrow\hat{{\tilde{R}}}^{(\tilde{I}_{\bullet})}$, here $\tilde{I}_{\bullet}:=\phi(I_{\bullet})\cdot{\tilde{R}}$. 2. 2.

Suppose the completion of ${\tilde{R}}$ is surjective, ${\tilde{R}}\twoheadrightarrow\hat{{\tilde{R}}}^{(\tilde{I}_{\bullet})}$. Then the completion of $R$ is surjective, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$, here $I_{\bullet}:=\phi^{-1}(\tilde{I}_{\bullet})$.

Proof.

The homomorphism $R\stackrel{{\scriptstyle\phi}}{{\to}}{\tilde{R}}$ induces the homomorphism of completions, \lim\limits_{\leftarrow}{\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{\bullet}}\right.}\stackrel{{\scriptstyle\hat{\phi}}}{{\to}}\lim\limits_{\leftarrow}{\footnotesize\left.\raisebox{1.37775pt}{{\tilde{R}}}\!/\!\raisebox{-1.37775pt}{\tilde{I}_{\bullet}}\right.}. Thus we get the commutative diagram with exact rows.

[TABLE]

If $\phi$ is surjective then $\hat{\phi}$ too. Thus the surjectivity of  $\widehat{}_{R}$ implies that of  $\widehat{}_{\tilde{R}}$. 2. 2.

Starting from the filtration $\tilde{I}_{\bullet}\subset{\tilde{R}}$ we get $I_{\bullet}\subset R$ and the diagram (15). Then the surjectivity of  $\widehat{}_{\tilde{R}}$ implies that of  $\widehat{}_{R}$. $\blacksquare$

Example 2.9**.**

1. i.

The surjectivity $C^{\infty}(\mathcal{U})\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$ implies {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}\twoheadrightarrow\widehat{{\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}}^{(I_{\bullet}(mod\ J))}. 2. ii.

The surjectivity {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}\twoheadrightarrow\widehat{{\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}}^{(I_{\bullet}(mod\ J))} implies $C^{\infty}(\mathcal{U})\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$, provided $I_{\infty}\supseteq J$.

2.5.3. Localization of a ring

Let $S\subset R$ be a multiplicatively closed set, and $I_{\bullet}\subset R$ a filtration. Take its image under the localization, $I_{\bullet}[S^{-1}]\subset R[S^{-1}]$.

Lemma 2.10**.**

Suppose $I_{\infty}\cap S=\varnothing$ and for any $q\in S$ exists $\tau_{q}\in I_{\infty}$ such that $1+\tau_{q}\in(q)\subset R$. Then $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$ iff $R[S^{-1}]\twoheadrightarrow\widehat{R[S^{-1}]}^{(I_{\bullet}[S^{-1}])}$.

Note that $R\to R[S^{-1}]$ is non-surjective and $I_{\bullet}\neq R\cap(I_{\bullet}\cdot[S^{-1}])$. Thus this lemma is not implied by the previous lemma.  

Proof.  $\Rrightarrow$ Take any element $\sum\frac{g_{j}}{q_{j}}\in\widehat{R[S^{-1}]}^{(I_{\bullet}[S^{-1}])}$, here $g_{j}\in I_{j}\subset R$ and $q_{j}\in S$. Choose $\{\tau_{j}\}$ as in the assumption, then $\sum\frac{g_{j}}{q_{j}}$ is equivalent to $\sum\frac{g_{j}(1+\tau_{j})}{q_{j}}\in{\widehat{R}}^{(I_{\bullet})}$. Take its representative, $f\in R$. Then the image of $f$ in $R[S^{-1}]$ is the representative of $\sum\frac{g_{j}}{q_{j}}$.

$\Lleftarrow$ Take an element $\sum g_{j}\in{\widehat{R}}$. Its image in $\widehat{R[S^{-1}]}^{(I_{\bullet}[S^{-1}])}$ is presented by some $\frac{f}{q}\in R[S^{-1}]$, here $q\in S$. Take $\tau$ as in the assumption, then $\frac{f(1+\tau)}{q}\in R$ presents $\sum g_{j}$. $\blacksquare$

Example 2.11**.**

Let $R=C^{\infty}(\mathcal{U})$ and $Z=V(I_{\infty})$. Suppose no element of the set $S$ vanishes at any point of $Z$. Then $C^{\infty}(\mathcal{U})\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$ iff $C^{\infty}(\mathcal{U})[S^{-1}]\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U})[S^{-1}]}^{(I_{\bullet}[S^{-1}])}$​​. Indeed, we have $I_{\bullet}\!\cap\!S\!=\!\varnothing$. In addition, for any $q\!\in\!S$ take separating open neighbourhoods of the closed sets, §2.1.i:

[TABLE]

Then take $\tau\in I_{\infty}$ such that $\tau|_{\mathcal{U}(V(q))}=-1$. It exists e.g. by proposition 2.5. Then $1+\tau\in(q)$, now invoke lemma 2.10.

Some particular cases:

• •

For a point $Z=\{x\}\subset\mathcal{U}$, and $S=\{$all the functions that do not vanish at $x\}$, we get the germs at the point, $C^{\infty}(\mathcal{U})[S^{-1}]=C^{\infty}(\mathcal{U},x)$.

• •

For a closed subset $Z\subset\mathcal{U}$ and $S=\{$all the functions that do not vanish at any point of $Z\}$, we get the germs along $Z$, $C^{\infty}(\mathcal{U})[S^{-1}]=C^{\infty}(\mathcal{U},Z)$.

• •

Let $Z\times\mathcal{V}\subset\mathcal{U}\times\mathcal{V}$, for a closed $Z\subset\mathcal{U}$. Take $S=\{$all the functions that do not vanish at any point of $Z\times\mathcal{V}\}$. Then C^{\infty}(\mathcal{U}\times\mathcal{V})[S^{-1}]=C^{\infty}\big{(}(\mathcal{U},Z)\times\mathcal{V}\big{)}.

Thus the surjectivity question for rings of germs, e.g. $C^{\infty}(\mathcal{U},Z)$, is reduced to the ring $C^{\infty}(\mathcal{U})$.

Remark 2.12**.**

For some rings/filtrations the surjectivity of completion can be proved by the following simple argument. Assume $(R,{\mathfrak{m}})$ is local and $I_{j}\subseteq{\mathfrak{m}}^{d_{j}}$, with $d_{j}\to\infty$. Then the completion $R\to{\widehat{R}}^{({\mathfrak{m}})}$ factorizes through $R\to{\widehat{R}}^{(I_{\bullet})}\to{\widehat{R}}^{({\mathfrak{m}})}$. Thus, if the map $R\to{\widehat{R}}^{({\mathfrak{m}})}$ is surjective and the map ${\widehat{R}}^{(I_{\bullet})}\to{\widehat{R}}^{({\mathfrak{m}})}$ is injective, the map $R\to{\widehat{R}}^{(I_{\bullet})}$ is surjective. However, a necessary condition for the injectivity ${\widehat{R}}^{(I_{\bullet})}\to{\widehat{R}}^{({\mathfrak{m}})}$ is $\cap I_{j}\supseteq{\mathfrak{m}}^{\infty}$. And this does not hold for many filtrations. Thus one must prove the surjectivity separately.

2.6. Surjectivity for rings of functions on manifolds and other spaces

Let $\mathcal{U}\subseteq\mathbb{R}^{n}$ and $X\subset\mathcal{U}$ a closed subset, with the defining ideal $I(X)$. Take the ring of Whitney jets, $C^{\infty}(X)$, see §2.2. By Whitney theorem we identify C^{\infty}(X)={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{I(X)^{\langle\infty\rangle}}\right.}. For sets with dense interior, $X=\overline{Int(X)}$, the elements of $C^{\infty}(X)$ are those elements of $C^{\infty}(Int(X))$ whose derivatives extend continuously onto $X$. In this case $I(X)^{\langle\infty\rangle}=I(X)$ and C^{\infty}(X)={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{I(X)}\right.}.

If $X\subset\mathcal{U}$ is a (closed) $C^{\infty}$-submanifold then the ring of smooth functions on $X$ can be presented as {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{I(X)}\right.}. For a $C^{\infty}$-algebraic scheme this is the ring of $C^{\infty}$-regular functions, [Joyce]. For analytic subset $X\subset\mathcal{U}$ the ring of real-analytic functions is C^{\omega}(X):={\footnotesize\left.\raisebox{1.37775pt}{C^{\omega}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{I(X)}\right.}.

In theorems 3.1 and 4.1 we address the surjectivity for the ring {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}. In particular this gives the necessary and sufficient condition for the surjectivity $C^{\infty}(X)\twoheadrightarrow\widehat{C^{\infty}(X)}^{(I_{\bullet})}$ in all these cases.

3. The necessary condition for the surjectivity of completion map

Let R={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}, for some open $\mathcal{U}\subseteq\mathbb{R}^{n}$ and some ideal $(0)\subseteq J\subset C^{\infty}(\mathcal{U})$. Fix a filtration $I_{\bullet}\subset R$, and denote $Z:=V(I_{\infty})$. For an open subset $\mathcal{U}_{0}\subset\mathcal{U}$ we take the restriction I_{\bullet}\big{|}_{\mathcal{U}_{0}}\subset R|_{\mathcal{U}_{0}}:={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U}{0})}\!/\!\raisebox{-1.37775pt}{C^{\infty}(\mathcal{U}{0})\cdot J}\right.}.

In this section we prove:

Theorem 3.1**.**

Suppose the completion map is surjective, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$. Take any compactly embedded open subset, $\mathcal{U}_{0}\subset\overline{\mathcal{U}_{0}}\subset\mathcal{U}$, where $\overline{\mathcal{U}_{0}}$ is compact in $\mathbb{R}^{n}$. The filtration restricted to $\mathcal{U}_{0}$ is equivalent to a particular form: $I_{\bullet}|_{\mathcal{U}_{0}}\sim I_{\infty}+I_{\bullet}\cap I(Z)^{\langle\bullet\rangle}|_{\mathcal{U}_{0}}$.

First we introduce the loci $Z_{i}\subset\mathcal{U}$ where the ideals $I_{\bullet}$ have prescribed orders with respect to the maximal ideals. The surjectivity of $R\to\widehat{R}\ \!^{(I_{\bullet})}$ imposes heavy restrictions on $I_{\bullet}$ along these loci, lemma 3.3. Using these restrictions we prove theorem 3.1.

As a corollary we get the necessary condition for localized rings, §3.4.

3.1. The loci of prescribed orders

Take the zero locus $V(J)\subseteq\mathcal{U}$. Define the loci of $i$’th order: $V_{i}(I_{j}):=\{x|\ I_{j}\subseteq{\mathfrak{m}}^{i}_{x}\subset R\}\subset V(J)\subseteq\mathcal{U}$. Define their limiting loci:

[TABLE]

These loci $\{Z_{i}\}$ satisfy $V(J)\supseteq Z_{1}\supseteq Z_{2}\supseteq\cdots$, they are not necessarily closed. Define $Z_{\infty}=\cap Z_{i}$.

Example 3.2**.**

1. i.

Take the filtration $\{I_{j}:=I^{j}\}$. Then $Z_{1}=\cdots=Z_{\infty}=V(I)\subseteq V(J)$. 2. ii.

Let $\{X_{j}\}$ be an infinite collection of compact manifolds in $\mathbb{R}^{n}$. Take their defining ideals, $\{I(X_{j})\}_{j}$. Take the filtration

[TABLE]

Then  $Z_{1}=\mathop{\cup}\limits_{j\geq 1}X_{j}\supset Z_{2}=\mathop{\cup}\limits_{j\geq 2}X_{j}\supset\cdots\supset Z_{i}=\mathop{\cup}\limits_{j\geq i}X_{j}\supset\cdots$.

With no assumptions on $I_{\bullet}$ (or on the surjectivity $R\twoheadrightarrow{\widehat{R}}$), the chains in (17) do not necessarily stabilize, and the loci $\{Z_{i}\}$ are not necessarily closed, see the last example.

3.2. Stabilization of the loci of prescribed orders

Lemma 3.3**.**

Let R={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.} and suppose the completion map is surjective, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$.

For any compactly embedded open subset $\mathcal{U}_{0}\subset\overline{\mathcal{U}_{0}}\subset\mathcal{U}$ the restricted filtration I_{\bullet}\big{|}_{\mathcal{U}_{0}} is equivalent to the filtration

[TABLE] 2. 2.

In particular, the restrictions of chains in (17) stabilize, $Z_{i}\cap\mathcal{U}_{0}=V_{i}(I_{j})\cap\mathcal{U}_{0}$ for $j\gg i$. Therefore $\{Z_{i}\subset\mathcal{U}\}$ are closed for all $1\leq i\leq\infty$.

*Proof.  *To check the equivalence of the filtrations it is enough to show: for any $j$ exists $d_{j}<\infty$ such that

[TABLE]

For this it is enough to show: for any $i$ the chain \{V_{i}(I_{j})\big{|}_{\mathcal{U}_{0}}\}_{j} in (17) stabilizes. We prove this by induction on $i$.

Case $i=1$, for $Z_{1}$. Suppose the loci V_{1}(I_{j})\big{|}_{\mathcal{U}_{0}}\subset V(J)\cap\mathcal{U}_{0} do not stabilize, i.e. $I_{j}\not\subseteq I(Z_{1})$, for any $j$. Replace $I_{\bullet}$ by its (equivalent) subsequence that satisfies: V_{1}(I_{1})\big{|}_{\mathcal{U}_{0}}\subsetneq V_{1}(I_{2})\big{|}_{\mathcal{U}_{0}}\subsetneq\cdots. Fix a sequence of points, p_{j}\in V_{1}(I_{j+1})\setminus V_{1}(I_{j})\big{|}_{\mathcal{U}_{0}}\subset V(J)\cap\mathcal{U}_{0}, and small balls, $Ball_{\epsilon_{j}}(p_{j})\cap V_{1}(I_{j})=\varnothing$. Accordingly fix a sequence of elements, $\{g_{j}\in I_{j}\}$, and the corresponding functions $\{\tilde{g}_{j}\in C^{\infty}(\mathcal{U})\}$, satisfying:

[TABLE]

(These are constructed from elements of $I_{j}$ by using the standard bump functions.)

Suppose the element $\sum^{\infty}_{j=0}g_{j}\in{\widehat{R}}^{(I_{\bullet})}$ is presented by some $f\in R$, see §2.1.vi. Then $f$ must be presented by $\tilde{f}\in C^{\infty}(\mathcal{U})$, which is bounded on the compact set $\overline{\mathcal{U}_{0}}$. But this contradicts the construction: $\tilde{f}(p_{N-1})=\sum^{N}_{j=0}\tilde{g}_{j}(p_{N-1})\geq N!$.

The case of arbitrary $i$ is similar. Assuming the statement for $i$, i.e. for $Z_{1},\dots,Z_{i}$, we pass to the (equivalent) filtration satisfying:

[TABLE]

Suppose the loci $\{V_{i+1}(I_{j})\cap\mathcal{U}_{0}\}_{j}$ do not stabilize, i.e. $I_{j}\not\subseteq I(Z_{i+1})^{\langle i\rangle}$ for any $j$. As before, we can assume $V_{i+1}(I_{1})\cap\mathcal{U}_{0}\subsetneq V_{i+1}(I_{2})\cap\mathcal{U}_{0}\subsetneq\cdots$. As before, fix a sequence of points and small balls:

[TABLE]

Fix a sequence of elements $\{g_{j}\in I_{j}\}$ satisfying $ord_{p_{j}}g_{j}=i$. Take their representatives $\{\tilde{g}_{j}\in C^{\infty}(\mathcal{U})\}$ satisfying $ord_{p_{j}}\tilde{g}_{j}=i$. By using bump functions we can assume $\tilde{g}_{j}|_{\mathcal{U}\setminus Ball_{\epsilon_{j}}(p_{j})}=0$. Moreover, we can assume the Taylor expansion $\tilde{g}_{j}(x)=Q_{i}(x-p_{j})+O(x-p_{j})^{i+1}$, where $Q_{i}$ is a homogeneous polynomial of degree $i$, with large coefficients. More precisely, as $J\not\supseteq{\mathfrak{m}}^{i}_{p_{j}}$, we have the vector subspace of positive dimension:

[TABLE]

Thus we choose 0\neq[Q_{i}]\in{\footnotesize\left.\raisebox{1.37775pt}{{\mathfrak{m}}^{i}{p{j}}+J}\!/\!\raisebox{-1.37775pt}{{\mathfrak{m}}^{i+1}{p{j}}+J}\right.} with large coefficients, and then $\tilde{g}_{j}(x)+q$ has large $i$’th derivative for any $q\in J$.

Finally, if $\sum g_{j}\in{\widehat{R}}$ is presented by some $f\in R$ then any representative $\tilde{f}\in C^{\infty}(\mathcal{U})$ satisfies:

[TABLE]

Here the derivative $\tilde{f}^{(i)}$ must be bounded on $\mathcal{U}_{0}$. But, by construction, its values on the sequence of points $\{p_{i}\}$ are unbounded. $\blacksquare$

Example 3.4**.**

This lemma forbids numerous filtrations, even with polynomially generated ideals.

1. i.

Fix a sequence $\{p_{i}\}$ of distinct points in $\mathbb{R}^{1}$, converging to [math]. Let

[TABLE]

Here none of $V_{i}(I_{j})$ stabilizes in $j$. Thus the completion map $R\to{\widehat{R}}^{(I_{\bullet})}$ is not surjective. 2. ii.

If $I_{\infty}=(0)\subset R$ then the completion map is not surjective. Indeed, the surjectivity forces $V_{1}(I_{j})$ to stabilize (lemma 3.3), and then $V_{1}(I_{\infty})=V_{1}(I_{j})\subsetneq V(J)$, by §2.4. This goes in notable difference to the Noetherian rings Commutative Algebra, where the non-stabilizing filtrations usually satisfy $I_{\infty}=(0)$.

Remark 3.5**.**

The restriction to the compactly embedded subsets, (\dots)\big{|}_{\mathcal{U}_{0}}, is important. For example, let $\mathcal{U}=(0,1)\subset\mathbb{R}^{1}$ and I_{j}=\big{\{}f|\ f=0\ on\ [\frac{1}{j},1-\frac{1}{j}]\big{\}}\subset C^{\infty}(0,1). The chain $V_{i}(I_{j})\subset V_{i}(I_{j+1})\subset\cdots$ does not stabilize, for any $i$. But the completion is surjective, $C^{\infty}(0,1)\twoheadrightarrow\widehat{C^{\infty}(0,1)}\ \!^{(I_{\bullet})}$. Indeed, for any sequence $\{g_{j}\in I_{j}\}$ we have $\sum g_{j}\in C^{\infty}(0,1)$, as the sum is locally finite on $(0,1)$.

Remark 3.6**.**

The sequence of subsets $Z_{1}\supseteq Z_{2}\supseteq\cdots$ of (17) does not necessarily stabilize. For example, take a converging sequence $\mathbb{R}^{n}\ni p_{i}\to p_{0}$. Take

[TABLE]

The sequence $\{Z_{j}\}$ does not stabilize, but the completion is surjective, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$. Indeed, fix $\sum g_{j}\in{\widehat{R}}^{(I_{\bullet})}$, here $g_{j}\in I_{j}$. This element is presented by $f\in R$ iff $f$ has the fixed jets, $jet_{i,p_{i}}f=jet_{i,p_{i}}\sum^{i}_{j=1}g_{j}$. By Whitney extension, §2.2, we recover $f$ with such derivatives. The only Whitney compatibility condition to be checked is at $p_{0}$, this reads: each $g_{k}$ is flat at [math]. And this holds as each $g_{k}$ vanishes on the whole sequence $\{p_{i}\}$.

3.3. Proof of theorem 3.1

Take the completion, $R\stackrel{{\scriptstyle\pi}}{{\to}}{\widehat{R}}^{(I_{\bullet})}$.

1. Step 1.

We claim: $\cap_{j}(\pi(I_{j})\!\cdot\!{\widehat{R}}^{(I_{\bullet})})\!=\!(0)$. Indeed, for any $N$ the ideal $\pi(I_{N})\!\cdot\!{\widehat{R}}^{(I_{\bullet})}$ is sent to (0)\subset{\footnotesize\left.\raisebox{1.37775pt}{{\widehat{R}}^{(I_{\bullet})}}\!/\!\raisebox{-1.37775pt}{\pi(I_{N})\cdot{\widehat{R}}^{(I_{\bullet})}}\right.}\cong{\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{N}}\right.}. Thus we get the projective limit of the zero ideals,

[TABLE] 2. Step 2.

Consider the quotient ring {\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}, with its filtration {\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}. For any fixed $i\geq 1$ and any point $p\in V(I_{\infty})\subseteq\mathcal{U}$ the quotient filtration

[TABLE]

stabilizes. (As a descending chain of ideals in the Artinian ring.) Thus we can identify {\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{\infty}+{\mathfrak{m}}^{i}_{p}}\right.} with its $\tilde{I}_{\bullet}$-completion, and identify the $\tilde{I}_{\bullet}$-filtration with its completed version:

[TABLE]

By step 1 the bottom-right corner of this diagram converges to zero, for $\bullet\to\infty$. Therefore $\tilde{I}_{\bullet}$ stabilizes to zero, for any $p\in V(I_{\infty})$ and any $i$. Therefore V(I_{\infty})=\cap Z_{i}({\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.})=Z_{\infty}({\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}).

But the surjectivity of the initial completion, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$, implies the surjectivity of the completion of the quotient, {\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}\twoheadrightarrow{\footnotesize\left.\raisebox{1.37775pt}{{\widehat{R}}^{(I_{\bullet})}}\!/\!\raisebox{-1.37775pt}{{\widehat{I}}{\infty}}\right.}, see example 2.9. Thus, by lemma 3.3, we have equivalence of the filtrations, {\footnotesize\left.\raisebox{1.37775pt}{I{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}|_{\mathcal{U}_{0}}\sim{\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}\cap{\footnotesize\left.\raisebox{1.37775pt}{I(Z)^{\langle\bullet\rangle}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}\!|_{\mathcal{U}_{0}}. Thus $I_{\bullet}\!|_{\mathcal{U}_{0}}\sim I_{\infty}+I_{\bullet}\cap I(Z)^{\langle\bullet\rangle}|_{\mathcal{U}_{0}}$, for Z=Z_{\infty}({\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.})=V(I_{\infty}). $\blacksquare$

Remark 3.7**.**

One would like to strengthen theorem 3.1 and to obtain further special properties of $I_{\bullet}$. For example, suppose all $I_{\bullet}$ are analytically generated, how does the surjectivity restrict the primary decomposition of $I_{j}$? As $V(I_{\bullet})$ stabilizes, the minimal primes stabilize, and the corresponding primary ideals form a filtration. But not much can be said about the embedded primes, because of the sufficient condition, theorem 4.1. For example, let $\{{\mathfrak{q}}_{j,k}\}_{k}$ be the primary ideals corresponding to the minimal primes of $I_{j}$. Then $\{{\mathfrak{q}}_{j,k}\}_{j}$ is a filtration of ${\mathfrak{q}}_{1,k}$, for each $k$. Assume ${\mathfrak{q}}_{j,k}\subseteq{\mathfrak{q}}_{1,k}^{j}$. Then, for any sequence $\{emb_{j}\}$, corresponding to embedded primes, the filtration $emb_{\bullet}\cap(\cap_{k}{\mathfrak{q}}_{\bullet,k})$ induces the surjective completion map.

3.4. The necessary criterion for localizations

Let R={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.} and $S\subset R$ be a multiplicatively closed set. Assume no element of $S$ vanishes at any point of $Z$. Take a filtration $I_{\bullet}\subset R[S^{-1}]$. For any open subset define the restriction I_{\bullet}|_{\mathcal{U}_{0}}\subset R[S^{-1}]|_{\mathcal{U}_{0}}={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U}{0})}\!/\!\raisebox{-1.37775pt}{C^{\infty}(\mathcal{U}{0})J}\right.}[S^{-1}].

Corollary 3.8**.**

If $R[S^{-1}]\twoheadrightarrow\widehat{R[S^{-1}]}^{(I_{\bullet})}$ then $I_{\bullet}|_{\mathcal{U}_{0}}\sim I_{\infty}|_{\mathcal{U}_{0}}+I_{\bullet}\cap I(Z)^{\langle\bullet\rangle}|_{\mathcal{U}_{0}}$ for any compactly embedded open subset $\mathcal{U}_{0}\subset\overline{\mathcal{U}_{0}}\subset\mathcal{U}$.

*Proof.  *For $R\stackrel{{\scriptstyle\phi}}{{\to}}R[S^{-1}]$ take $\phi^{-1}(I_{\bullet})\subset R$. We get $R\twoheadrightarrow{\widehat{R}}^{(\phi^{-1}(I_{\bullet}))}$, by example 2.11. Thus (theorem 3.1)

[TABLE]

For any ideal $\mathfrak{a}\subset R[S^{-1}]$ one has \mathfrak{a}=R[S^{-1}]\cdot\phi\big{(}\phi^{-1}(\mathfrak{a})\big{)}. Indeed, take any system of generators $\{a_{i}\}$ of $\mathfrak{a}$, not necessarily finite. Clear the denominators, thus can assume $\{a_{j}\in\phi(\mathfrak{a})\}$. Thus $\phi(\phi^{-1}(\mathfrak{a}))=\phi(R)\{a_{j}\}$. Therefore $I_{j}=R\phi(\phi^{-1}(I_{j}))|_{\mathcal{U}_{0}}\subseteq I_{\infty}+I_{d_{j}}\cap I(Z)^{\langle d_{j}\rangle}|_{\mathcal{U}_{0}}$. $\blacksquare$

4. The sufficient condition for the surjectivity

By §2.5 the question is reduced to the particular ring $C^{\infty}(\mathcal{U})$. Take a filtration $I_{\bullet}\subset R:=C^{\infty}(\mathcal{U})$, let $Z=V(I_{\infty})\subset\mathcal{U}$. For each point $p\in\mathcal{U}$ we take the ring of germs, $C^{\infty}(\mathcal{U},p)$, see example 2.11, and its (localized) filtration.

Theorem 4.1**.**

Suppose for each point $p\in\mathcal{U}$ the localized filtration $I_{\bullet}|_{(\mathcal{U},p)}$ is (locally) equivalent to a filtration of type $\mathfrak{a}+\sum_{k}\mathfrak{b}_{k,\bullet}$, where

• •

The ideal $\mathfrak{a}$ does not depend on $\bullet$; the summation over $k$ is finite.

• •

The ideals $\{\mathfrak{b}_{k,\bullet}\}$ satisfy: $\mathfrak{c}_{k}\cdot\mathfrak{b}_{k,\infty}\subseteq\mathfrak{b}_{k,j}\subseteq\mathfrak{c}_{k}\cap\mathfrak{b}_{k}^{\langle d_{j}\rangle}$, where the ideals $\mathfrak{c}_{k}$, $\mathfrak{b}_{k}$ do not depend on $j$; the zero loci are constant, $V(\mathfrak{b}_{k,\infty})=V(\mathfrak{b}_{k,j})$; and $d_{j}\to\infty$.

Then the two sequences are exact:

[TABLE]

Moreover, for any neighbourhood $Z\subset\mathcal{U}(Z)$ and any element $\widehat{f}\in{\widehat{R}}^{(I_{\bullet})}$ there exists a representative $f\in C^{\infty}(\mathcal{U})\cap C^{\omega}(\mathcal{U}\setminus Z)$ that satisfies $f>0$ on $\mathcal{U}\setminus\mathcal{U}(Z)$.

*Proof.  *Given $\sum g_{j}\in{\widehat{R}}^{(I_{\bullet})}$, with $g_{j}\in I_{j}\subset C^{\infty}(\mathcal{U})$, we should construct its representation $f\in C^{\infty}(\mathcal{U})$, satisfying $f-\sum\limits^{N}_{j=0}g_{j}\in I_{N}$ for any $N$. There are several steps:

1. Step 1.

We reduce the proof to the ring $C^{\infty}(Ball_{1}(o))$ and a very particular filtration, equation (33). 2. Step 2.

We bound the growth of derivatives of $g_{j}$ on the locus $V(\mathfrak{b})\subset Ball_{1}(o)$. 3. Step 3.

We construct $f$ from $\{g_{j}\}$ using the cutoff functions with controlled growth. 4. Step 4.

We verify that $f$ represents $\sum g_{j}$ in ${\widehat{R}}^{(I_{\bullet})}$. This proves exactness of the first row of (31). 5. Step 5.

For this $f$ we use Whitney approximation theorem to establish the second row of (31).

1. Step 1.

(Simplifying the filtration) By the partition of unity argument, §2.5.1, we can pass to the ring $R:=C^{\infty}(Ball_{1}(o))$ and replace $I_{\bullet}$ by the equivalent filtration as in the assumptions:

[TABLE]

An element of ${\widehat{R}}$ is $a_{0}+\sum_{j\geq 1}(a_{j}+\sum_{k}b_{k,j})$, where $a_{j}\in\mathfrak{a}$ and $b_{k,j}\in\mathfrak{b}_{k,j}$. We should construct a representative $f\in R$ that satisfies: $f-a_{0}-\sum^{N}_{j=1}(a_{j}+\sum_{k}b_{k,j})\in I_{N}$ for any $N$. As $\mathfrak{a}\subseteq I_{j}$ for each $j$, one can omit $\{a_{j}\}$.

Moreover, as the summation $\sum_{k}$ is finite, it is enough to find a representative for each $\sum_{j\geq 1}b_{k,j}$, with $k$ fixed. This reduces the statement to the filtration $\{\mathfrak{b}_{j}\}_{j}$, with $\mathfrak{c}\cdot\mathfrak{b}_{\infty}\subseteq\mathfrak{b}_{j}\subseteq\mathfrak{c}\cap\mathfrak{b}^{\langle d_{j}\rangle}$, $V(\mathfrak{b}_{\infty})=V(\mathfrak{b})$. Finally, pass to an equivalent filtration with $d_{j}=j$.

Summarizing, it is enough to establish the surjectivity $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$ for $R=C^{\infty}(Ball_{1}(o))$ and the filtration of the particular type:

[TABLE] 2. Step 2.

We have $\{g_{j}\in I_{j}\}$ for the specific filtration of the ring $C^{\infty}(Ball_{1}(o))$ as in (33). By slightly shrinking the ball we can assume $g_{j}\in C^{\infty}(\overline{Ball_{1}(o)})$, in particular each derivative of each $g_{j}$ is bounded. Take the zero locus $Z_{\mathfrak{b}}:=V(\mathfrak{b})\subset Ball_{1}(o)$.

For any $0\leq k<j<\infty$ and any $x\in Ball_{1}(o)$ we bound the operator norm of $k$’th derivative:

[TABLE]

(Here $\{C_{g_{j}}\}$ are some constants that depend on $g_{j}$.)

Indeed, for any $x\in Ball_{1}(o)\setminus Z_{\mathfrak{b}}$ take some $z\in Z_{\mathfrak{b}}$ for which $dist(x,z)-dist(x,Z_{\mathfrak{b}})=\epsilon\ll dist(x,Z_{\mathfrak{b}})$. By the assumption $g_{j}\in{\mathfrak{m}}_{z}^{j}$, thus $g^{(k)}_{j}|_{z}=0$ for $k<j$. Therefore the Taylor expansion (in $Ball_{dist(x,z)}(z)$) with remainder gives:

[TABLE]

(See §2.1.iii.) Note the bounds

[TABLE]

The derivatives $||g^{(j)}_{j}||$ are bounded on $Ball_{1}(o)$. Thus $|g_{j}(x)|\leq C_{0}\cdot dist(x,Z_{\mathfrak{b}})^{j}$, for a constant $C_{0}$.

The bounds on the derivatives, $||g^{(k)}_{j}|_{x}||\leq\dots$, are obtained in the same way, by Taylor expanding $g^{(k)}_{j}$ at $z$. For each $k$ we get its constant $C_{k}$. Then define $C_{g_{j}}:=\max\limits_{k<j}(C_{k})$. 3. Step 3.

We use a particular cutoff function with controlled growth of total derivatives:

Theorem 1.4.2 of [Hörmander, pg. 25] For any compact set with its neighbourhood, $Z\subset\mathcal{U}\subset\mathbb{R}^{n}$, and a positive decreasing sequence $\{d_{j}\}$, satisfying $\sum d_{j}<dist(Z,\partial\mathcal{U})$, there exists a smaller neighbourhood, $Z\subset\mathcal{U}(Z)\subsetneq\mathcal{U}$, and a function $\tau\in C^{\infty}(\mathbb{R}^{n})$ satisfying

1. (a)

$\tau|_{\mathbb{R}^{n}\setminus\mathcal{U}}=0$, $\tau|_{\mathcal{U}(Z)}=1$; 2. (b)

for any $k$ and for any $x\in\mathcal{U}$ the norm of $k$’th derivative is bounded: $||\tau^{(k)}|_{x}||\leq\frac{C^{k}\cdot}{d_{1}\cdots d_{k}}$.

(Here the constant $C$ depends only on the dimension $n$.)

In our case the subset $Z_{\mathfrak{b}}\subset Ball_{1}(o)$ is closed and we can assume it is compact by shrinking the ball. Take the neighbourhood, $\mathcal{U}_{\epsilon}(Z_{\mathfrak{b}})$, see §2.1.i. Fix a decreasing sequence of positive numbers $\{\epsilon_{j}\}$, $\epsilon_{j}\to 0$. Assume it decreases fast, so that for each $j$ exists a cutoff function satisfying:

[TABLE]

Define $f(x):=\sum_{j}\tau_{j}(x)\cdot g_{j}(x)$. We claim that $f\in C^{\infty}(Ball_{1}(o))$, provided the sequence $\{\epsilon_{j}\}$ decreases quickly.

The statement $f\in C^{\infty}(\mathbb{R}^{n}\setminus Z_{\mathfrak{b}})$ is obvious, as for any $x\in\mathbb{R}^{n}\setminus Z_{\mathfrak{b}}$ the summation is finite. To check the behaviour on/near $Z_{\mathfrak{b}}$ we bound ${\underline{k}}$’th partial derivatives:

[TABLE]

We assume the sequence $\{\epsilon_{j}\}$ decreases fast to ensure, for $j>|{\underline{k}}|+1$:

[TABLE]

Present f^{({\underline{k}})}(x)=\sum^{|{\underline{k}}|+1}_{j=0}\big{(}\tau_{j}(x)\cdot g_{j}(x)\big{)}^{({\underline{k}})}+\sum_{j>|{\underline{k}}+1|}\dots. Our bounds ensure that the infinite tail converges uniformly on the whole $\mathbb{R}^{n}$. Thus each $f^{({\underline{k}})}$ is continuous. 4. Step 4.

We claim: $\tau_{j}\cdot g_{j}-g_{j}\in\mathfrak{c}\cdot\mathfrak{b}_{\infty}$, for any $j$. Indeed, take $q\in\mathfrak{b}_{\infty}$, satisfying $Z_{\mathfrak{b}}=q^{-1}(0)$. (See proposition 2.5.) As $\tau_{j}\cdot g_{j}-g_{j}$ vanishes on $\mathcal{U}_{\epsilon_{j}}(Z_{\mathfrak{b}})$, the ratio $\frac{\tau_{j}\cdot g_{j}-g_{j}}{q}$ extends to a smooth function on $Ball_{1}(o)$. Moreover, as $q$ is invertible on $Ball_{1}(o)\setminus Z_{\mathfrak{b}}$, we get: $\frac{\tau_{j}\cdot g_{j}-g_{j}}{q}\in\mathfrak{c}$.

Therefore $\tau_{j}\cdot g_{j}-g_{j}\in\mathfrak{a}\cdot\mathfrak{b}_{\infty}$. Hence $f-\sum^{N}_{j=0}g_{j}\in I_{N}$, for any $N$.

Thus the completion map sends $f$ to $\sum g_{j}$. Therefore the first row of (31) is exact. 5. Step 5.

(Exactness of the second row of (31)) Let $I_{\bullet}$ be a filtration of $C^{\infty}(\mathcal{U})$, as in the assumption. Take an element of the completion, $\sum g_{j}\in\widehat{C^{\infty}(\mathcal{U})}\ \!^{(I_{\bullet})}$. In the previous steps we have constructed a representative $\tilde{f}\in C^{\infty}(\mathcal{U})$ of $\sum g_{j}$.

We can assume $\tilde{f}>1$ on $\mathcal{U}\setminus\mathcal{U}(Z)$. Indeed, take some $\tau\in C^{\infty}(\mathcal{U})$ such that $\tau$ vanishes on a small neighbourhood of $Z$, but $\tau>|\tilde{f}|+1$ on $\mathcal{U}\setminus\mathcal{U}(Z)$. Then $\tau\in I_{\infty}$ and we replace $\tilde{f}$ by $\tilde{f}+\tau$.

As the completion is surjective, $C^{\infty}(\mathcal{U})\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U})}\ \!^{(I_{\bullet})}$, the zero sets $V(I_{j})$ stabilize, by lemma 3.3. Thus we assume $Z:=V(I_{1})=V(I_{2})=\cdots=V(I_{\infty})$. By proposition 2.5, $V(I_{\infty})=V(\tau)$, for some $\tau\in I_{\infty}$.

Take a sequence of positive reals, $\{\epsilon_{i}\to 0\}$. Define

[TABLE]

By the approximating lemma 2.4 there exists $f\in C^{\infty}(\mathcal{U})\cap C^{\omega}(\mathcal{U}\setminus Z)$ satisfying: $|f-\tilde{f}|<\epsilon_{i}$ on $\mathcal{U}_{i}$. Therefore $|f-\tilde{f}|<\tau^{2}$ on $\mathcal{U}\setminus Z$. Then $f-\tilde{f}\in I_{\infty}$. Therefore $f$ represents $\sum g_{j}$, is analytic on $\mathcal{U}\setminus Z$ and is positive on $\mathcal{U}\setminus\mathcal{U}(Z)$. $\blacksquare$

Remark 4.2**.**

1. i.

It is peculiar that the sufficient condition of theorem 4.1 seems to impose weaker restriction on $I_{\bullet}$ than the necessary condition of theorem 3.1. The two theorems imply that in fact the two sets of conditions are equivalent. 2. ii.

The class of filtrations allowed by the sufficient condition is rather broad:

• •

the condition is only on the germs of $I_{\bullet}$,

• •

the ideals $\mathfrak{a},\{\mathfrak{c}_{k}\},\{\mathfrak{b}_{k,j}\}$ are not necessarily finitely/sub-analytically generated, 3. iii.

Assuming only “$I_{j}$ are polynomially generated” does not ensure the surjectivity, see example 3.4.

Corollary 4.3**.**

Assume $I_{\bullet}\sim I_{\infty}+I_{\bullet}\cap I(Z)^{\langle\bullet\rangle}$, for $Z=V(I_{\infty})$. Then {\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.} is complete for the filtration {\footnotesize\left.\raisebox{1.37775pt}{I_{\bullet}}\!/\!\raisebox{-1.37775pt}{I_{\infty}}\right.}.

Remark 4.4**.**

If the ideals $\{\mathfrak{c}_{k}\},\{\mathfrak{b}_{k,j}\}$ are analytically generated then locally $\mathfrak{c}_{k}\cap\mathfrak{b}_{k}^{d_{j}}\subset\mathfrak{c}_{k}\cdot\mathfrak{b}^{d_{j}-d_{0}}_{k}$, for some $d_{0}$. (By Artin-Rees in the Noetherian ring $\mathbb{R}\{{\underline{x}}\}$.) Thus the filtration of theorem 4.1 is equivalent to just $\{\mathfrak{a}+\sum_{k}\mathfrak{c}_{k}\cap\mathfrak{b}_{k}^{\bullet}\}$.

For the (highly non-Noetherian) ring {\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}, with no analyticity assumption, no such simplification holds.

Remark 4.5**.**

For any neighbourhood $Z\subset\mathcal{U}(Z)\subset\mathcal{U}$ the theorem ensures a representative $f\in C^{\infty}(\mathcal{U})\cap C^{\omega}(\mathcal{U}\setminus Z)$ that is positive on $\mathcal{U}\setminus\mathcal{U}(Z)$. We cannot strengthen this to $f|_{\mathcal{U}\setminus Z}>0$, nor to $f$ is sign-semi-definite on $\mathcal{U}(Z)$, due to the following immediate examples.

1. i.

Let $I_{\bullet}=(x(x-1))^{\bullet}\subset C^{\infty}(\mathbb{R}^{1})$, and $g_{1}=x(x-1)(x-\frac{1}{2})$, $g_{\geq 2}=0$, $\widehat{f}=\sum g_{j}$. Suppose $f\in C^{\infty}(\mathbb{R}^{1})$ represents $\widehat{f}$, then $f^{\prime}|_{0}=\frac{1}{2}$ and $f^{\prime}|_{1}=\frac{1}{2}$. Thus $f|_{(0,\epsilon)}>0$ and $f|_{(1-\epsilon,1)}<0$ for some $\epsilon>0$. Thus $f$ must have a zero on $(\epsilon,1-\epsilon)$. 2. ii.

Let $I_{\bullet}=(y)^{\bullet}\subset C^{\infty}(\mathbb{R}^{2})$ and $g_{1}=yx$, $g_{\geq 2}=0$, $\widehat{f}=\sum g_{i}$. As before, for any $f$ representing $\widehat{f}$, and for any $x_{0}>0$, exists $\epsilon>0$ such that $f>0$ on $\{x_{0}\}\times(0,\epsilon)$ and $f<0$ on $\{-x_{0}\}\times(0,\epsilon)$. Thus $f$ necessarily vanishes at some point of $\mathcal{U}(Z)\setminus Z$, for any $Z\subset\mathcal{U}(Z)$.

5. Examples and applications

5.1. Examples

1. i.

As the simplest case suppose I_{\bullet}={\mathfrak{m}}^{\bullet}\subset{\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathbb{R}^{n},o)}\!/\!\raisebox{-1.37775pt}{J}\right.},{\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}, or, more generally, the filtration $I_{\bullet}$ is equivalent to ${\mathfrak{m}}^{\bullet}$. We get the Borel lemma on the subscheme $V(J)\subseteq(\mathbb{R}^{n},o),\mathcal{U}$. The analyticity of the representative, $f\in C^{w}(\mathcal{U}\setminus\{0\})\cap C^{\infty}(\mathcal{U})$, recovers [Shiota, Theorem 1]. 2. ii.

Suppose $V(I_{j})=V({\mathfrak{m}})$ and $I_{j}\subseteq{\mathfrak{m}}^{d_{j}}$, with $d_{j}\to\infty$, but $I_{j}\not\supseteq{\mathfrak{m}}^{N_{j}}$, for any $N_{j}<\infty$. (This happens, e.g. when $I_{j}$ is generated by flat functions.) We still get the surjectivity of completion, though not implied by Borel lemma. (Now $I_{\bullet}$ is not equivalent to ${\mathfrak{m}}^{\bullet}$.)

For example, take a flat function $\tau\in{\mathfrak{m}}^{\infty}$, assume $V(\tau)=\{o\}\subset\mathbb{R}^{n}$. Then any formal series $\sum a_{j}\tau^{j}$, $a_{j}\in C^{\infty}(\mathcal{U})$, is presentable by some $f\in C^{\infty}(\mathbb{R}^{n})$, in the sense: $f-\sum^{N}a_{j}\tau^{j}\in(\tau)^{N}$, for any $N\in\mathbb{N}$. 3. iii.

More generally, suppose $V(I_{j})=V(I_{1})=:Z$ and $I_{j}\subseteq I(Z)^{\langle d_{j}\rangle}$, for a sequence $d_{j}\to\infty$. Again, theorem 4.1 implies the surjectivity of completion. Note that we do not assume any regularity/subanalyticity conditions on the closed set $Z$.

If $Z\subset\mathcal{U}$ is a discrete subset then we get a “multi-Borel” lemma. 4. iv.

Take the ring $C^{\infty}(\mathbb{R}^{n}_{x}\times\mathbb{R}^{m}_{y},o)$ with coordinates $x,y$, and the filtration $I_{\bullet}=(y)^{\bullet}$. The $(y)^{\bullet}$-completion map is the Taylor map in $y$-coordinates, and theorem 4.1 ensures its surjectivity:

[TABLE]

Moreover, the preimage can be chosen $y$-analytic for $y\neq 0$. This recovers the classical Borel theorem, see [Hörmander, Theorem 1.2.6, pg. 16] and [Moerdijk-Reyes, Theorem 1.3, pg. 18]. 5. v.

For the ring R={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}\big{(}(\mathbb{R}^{n},o)\times[0,1]\big{)}}\!/\!\raisebox{-1.37775pt}{J}\right.}, we can interpret the elements as the families of function germs. Then we get the surjectivity of completion in families, $R\twoheadrightarrow{\widehat{R}}$. For example, for $I_{j}=({\underline{x}})^{j}$ we get a particular version of Borel lemma in families: any power series $\sum a_{{\underline{m}}}(t){\underline{x}}^{\underline{m}}$, with $a_{{\underline{m}}}(t)\in C^{\infty}([0,1])$, is the ${\underline{x}}$-Taylor expansion of some function germ f_{t}(x)\in C^{\infty}\big{(}(\mathbb{R}^{n},o)\times[0,1]\big{)}. 6. vi.

Many important filtrations are not by (differential) powers of ideals, neither are equivalent to this, see example 3.2. As the simplest case, in Singularity Theory, when studying the germ (at the origin) of a non-isolated hypersurface singularity with singular locus $\{x_{1}=0=x_{2}\}$, one often considers the filtration

[TABLE]

More generally, $I_{j}=x^{2}_{1}(x_{1},y_{1})^{d_{1,j}}+x^{2}_{2}(x_{2},y_{2})^{d_{2,j}}+\cdots+x^{2}_{n}(x_{n},y_{n})^{d_{n,j}}\subset C^{\infty}(\mathbb{R}^{n}_{x}\times\mathbb{R}^{n}_{y},o)$ is a typical filtration for complete intersections with non-isolated singularities.

This filtration satisfies the conditions of theorem 4.1. Thus one can use the surjectivity of completion to pull-back various formal results (over the completion) to the $C^{\infty}$-statements.

For further applications see [Bel.Boi.Ker], [Boi.Gre.Ker].

5.2. An inverse Artin-Tougeron problem

Let $R=C^{\infty}(\mathcal{U}_{x}\times\mathcal{U}_{y})$, for some open sets $\mathcal{U}_{x}\subseteq\mathbb{R}^{n}_{x}$, $o\in\mathcal{U}_{y}\subseteq\mathbb{R}^{m}_{y}$. Take the filtrations $I_{\bullet}$ of $C^{\infty}(\mathcal{U}_{x})$ and $(y)^{\bullet}$ of $C^{\infty}(\mathcal{U}_{y})$, and the corresponding filtration $\tilde{I}_{\bullet}:=R\cdot I_{\bullet}+R\cdot(y)^{\bullet}$. Identify {\widehat{R}}^{(\tilde{I}_{\bullet})}\xrightarrow{\,\smash{\raisebox{-2.79857pt}{\scriptstyle\sim}}\,}{}\widehat{C^{\infty}(\mathcal{U}_{x})^{(I_{\bullet})}}[[y]]. An element $\widehat{F}\in({\widehat{R}}^{(\tilde{I}_{\bullet})})^{q}$ defines the system of (formal) power series equations, $\widehat{F}(y)=0$. Suppose this has a formal solution, \widehat{y}(x)\in\big{(}\widehat{C^{\infty}(\mathcal{U}_{x})^{(I_{\bullet})}}\big{)}^{m}. The classical Artin-Tougeron problem asks for an ordinary solution, $\widehat{F}(y(x))=0$, for some y(x)\in\big{(}C^{\infty}(\mathcal{U}_{x})\big{)}^{m} that is sent by the completion to $\widehat{y}(x)$.

The inverse Artin-Tougeron problem (see e.g. [Hübl] and references therein) asks for $F\in R^{q}$ that is sent to $\widehat{F}$, and $y(x)\in(C^{\infty}(\mathcal{U}_{x}))^{m}$ that is sent to $\widehat{y}$, such that $F(y(x))=0$.

In the $C^{\infty}$-case this is resolved now trivially as follows. Assume the completion maps are surjective, $C^{\infty}(\mathcal{U}_{x})\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U}_{x})^{(I_{\bullet})}}$ and $R\twoheadrightarrow{\widehat{R}}$. Take any representatives, $\tilde{F}$ of $\widehat{F}$ and $y(x)$ of $\widehat{y}(x)$. Then $\tilde{F}(y(x))\in\tilde{I}_{\infty}\cdot(C^{\infty}(\mathcal{U}_{x}))^{q}$. Therefore $F(y):=\tilde{F}(y)-\tilde{F}(y(x))$ is the needed representative of $\widehat{F}$, and $F(y(x))=0$.

5.3. $C^{\infty}$-contraction of ideals

Given an ideal $\mathfrak{a}\subset R$ and a ring homomorphism $R\stackrel{{\scriptstyle\phi}}{{\to}}S$, one often takes the contraction, $\phi^{-1}(S\cdot\phi(\mathfrak{a}))\subset R$. Then $\phi^{-1}(S\cdot\phi(\mathfrak{a}))\supseteq\mathfrak{a}+Ker(\phi)$, the inclusion can be proper. The equality holds for the completion of Noetherian local rings [Bourbaki, Chapter III], but fails otherwise. In our case the surjectivity of completion forces the equality.

Corollary 5.1**.**

If $R\stackrel{{\scriptstyle\phi}}{{\twoheadrightarrow}}{\widehat{R}}^{(I_{\bullet})}$ then \phi^{-1}\big{(}{\widehat{R}}^{(I_{\bullet})}\phi(\mathfrak{a})\big{)}=\mathfrak{a}+Ker(\phi), for any ideal $\mathfrak{a}\subset R$.

Equivalently, $\phi(\mathfrak{a})={\widehat{R}}^{(I_{\bullet})}\phi(\mathfrak{a})$. Thus any ideal in ${\widehat{R}}^{(I_{\bullet})}$ is of the form $\phi(\mathfrak{a})$, with $V(\mathfrak{a})\subseteq Z$. This answers a question of N. Zobin, [Fefferman], initially asked for the filtrations $I(Z)^{\langle\bullet\rangle}$.  

*Proof.  *For any elements $\{a_{i}\}$ of $\mathfrak{a}$ and $\{\widehat{f}_{i}\}$ of ${\widehat{R}}^{(I_{\bullet})}$ the preimage of $\sum\widehat{f}_{i}\cdot\phi(a_{i})$ is $\sum f_{i}\cdot a_{i}$ . $\blacksquare$

5.4. Surjectivity for completion of modules

Let $M$ be a module over the ring R={\footnotesize\left.\raisebox{1.37775pt}{C^{\infty}(\mathcal{U})}\!/\!\raisebox{-1.37775pt}{J}\right.}. A filtration $I_{\bullet}$ of $R$ induces the filtration $M_{\bullet}=I_{\bullet}\cdot M$.

Corollary 5.2**.**

If $M$ is finitely generated and $I_{\bullet}$ satisfies the conditions of theorem 4.1 then the completion map $M\to{\widehat{M}}^{(I_{\bullet})}$ is surjective. 2. 2.

Assume $M$ contains an $R$-regular element, i.e. $Ann_{R}(z)=0$ for some $z\in M$. If the map $M\to{\widehat{M}}^{(I_{\bullet})}$ is surjective then the conditions of theorem 3.1 are satisfied.

*Proof.  *1. Fix some generators $\{z_{\alpha}\}$ of $M$, then an element of ${\widehat{M}}^{(I_{\bullet})}$ is presentable as $\sum_{\alpha}(\sum_{j}g_{j,\alpha}z_{\alpha})$. Here $g_{\bullet,\alpha}\in I_{\bullet}$. Thus if $f_{\alpha}\in R$ is a representative of $\sum g_{\bullet,\alpha}$ then $\sum_{\alpha}f_{\alpha}z_{\alpha}$ represents this element.

2. Let $z\in M$ be a regular element then $R\cong R\cdot\{z\}\subset M$. Thus the surjectivity of $M\to{\widehat{M}}^{(I_{\bullet})}$ implies the surjectivity of $R\to{\widehat{R}}^{(I_{\bullet})}$. $\blacksquare$

Remark 5.3**.**

The finitely-generated/regular assumptions of this corollary are necessary.

1. i.

Let $R=C^{\infty}(\mathcal{U})$ with the filtration by powers of a maximal ideal, ${\mathfrak{m}}^{\bullet}$. Take an infinite direct sum, $M=\oplus R\langle t_{i}\rangle$, with generators $\{t_{i}\}$. Take an element $0\neq x\in C^{\infty}(\mathcal{U})$. Then {\widehat{M}}^{({\mathfrak{m}})}\ni\sum^{\infty}_{n=0}x^{n}\Big{(}\sum^{n}_{j=1}t_{j}\Big{)}. This element has no $M$-representative, as $M$ contains only finite sums, $\sum^{k}_{j=1}f_{j}t_{j}$, for $f_{j}\in C^{\infty}(\mathcal{U})$. 2. ii.

Let $R=C^{\infty}(\mathbb{R}^{n})$, M={\footnotesize\left.\raisebox{1.37775pt}{R}\!/\!\raisebox{-1.37775pt}{(f)}\right.}, for some $0\neq f\in R$. Take a filtration $I_{\bullet}$ satisfying: $I_{N}\subseteq(f)$ for $N\gg 1$. Then ${\widehat{M}}^{(I_{\bullet})}\cong M$, as $R$ and ${\widehat{R}}^{(I_{\bullet})}$-modules, but the completion $R\to{\widehat{R}}^{(I_{\bullet})}$ is not necessarily surjective.

5.5. Lifting ${\widehat{R}}^{(I_{\bullet})}$-modules to $R$-modules

Let $R=C^{\infty}(\mathcal{U})$, with a filtration $I_{\bullet}$. Denote $Z:=V(I_{\infty})$ and define the $Z$-support of an ${\widehat{R}}^{(I_{\bullet})}$-module ${\widehat{M}}$ by $Supp_{Z}({\widehat{M}}):=\{x\in Z|\ M|_{(\mathcal{U},x)}\neq 0\}$.

Corollary 5.4**.**

Assume $I_{\bullet}$ satisfies the conditions of theorem 4.1. Suppose an ${\widehat{R}}^{(I_{\bullet})}$-module ${\widehat{M}}$ is finitely-generated (resp. finitely-presented). Then there exists a finitely-generated (resp. finitely-presented) $R$-module $M$ satisfying: $M\otimes{\widehat{R}}^{(I_{\bullet})}\cong{\widehat{M}}$ and $Supp(M)=Supp_{Z}({\widehat{M}})$.

This extends (partially) [Tougeron.76-2].

*Proof.  *Choose a presentation ${\widehat{R}}^{n}\stackrel{{\scriptstyle\widehat{A}}}{{\to}}{\widehat{R}}^{m}\to{\widehat{M}}\to 0$. Let $\tilde{A}\in\operatorname{Mat}_{m\times n}(R)$ be an $R$-representative of $\widehat{A}$, then $Coker(\tilde{A})\otimes{\widehat{R}}\approx Coker(\tilde{A}\otimes{\widehat{R}})={\widehat{M}}$. To ensure the support take some $\tau\in I_{\infty}$ satisfying $Z=V(\tau)$, see proposition 2.5. Define $A=\begin{bmatrix}\tilde{A}|\tau{1\rm I}\end{bmatrix}\in Mat_{m\times(m+n)}(R)$ and $R^{n+m}\stackrel{{\scriptstyle A}}{{\to}}R^{m}\to M\to 0$. Then $M\otimes{\widehat{R}}\cong{\widehat{M}}$, and $Supp(M)\subseteq Z$. Finally, for each point $x\in Z$ the map $A|_{x}$ is surjective iff $\widehat{A}|_{x}$ is surjective. Thus $Supp(M)=Supp_{Z}({\widehat{M}})$. $\blacksquare$

6. $C^{\infty}$-representatives with further properties

6.1. Representatives with prescribed decay on $\partial\mathcal{U}$

For various applications one wants a representative with fast decay “at infinity”. Fix a (multi-)sequence of functions $q_{\underline{k}}\in C^{\infty}(\mathcal{U})$ satisfying: $\{q_{\underline{k}}>0\}_{\underline{k}}$ on $\mathcal{U}$, and $\{q_{\underline{k}}\to 0\}_{\underline{k}}$ for $x\to\partial\mathcal{U}$. Accordingly define the Schwartz-type vector space of functions decaying with their derivatives,

[TABLE]

For any $\widehat{f}\in\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$ we want to ensure a representative with the prescribed decay on $\partial\mathcal{U}$, i.e. to strengthen the surjectivity of completion to $Sch_{\partial\mathcal{U}}\twoheadrightarrow\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$.

This cannot hold without further assumptions, e.g. $Sch_{\partial\mathcal{U}}$ is not a unital ring, it does not contain $1$. More generally, let $Z=V(I_{\infty})$ and take the element $g_{1}+\sum 0\in\widehat{C^{\infty}(\mathcal{U})}^{(I_{\bullet})}$, where $\lim\limits_{x\to x_{0}}g^{({\underline{k}})}_{1}(x)\neq 0$, for some $x_{0}\in\overline{Z}\cap\partial\mathcal{U}$ and some ${\underline{k}}$. (Here $\overline{Z}$ is the closure in $\mathbb{R}^{n}$.) All the $C^{\infty}(\mathcal{U})$-representatives of $g_{1}$ are of the form $\{g_{1}\}+I_{\infty}$. None of these belongs to $Sch_{\partial\mathcal{U}}$.

Therefore we should a priori impose the needed behaviour on the elements of $C^{\infty}(\mathcal{U})$, at least at the points of $\overline{Z}\cap\partial\mathcal{U}$. Accordingly, we define the space of functions whose germs at the points of $\overline{Z}\cap\partial\mathcal{U}$ have the prescribed decay:

[TABLE]

Thus $Sch_{\overline{Z}\cap\partial\mathcal{U}}\supseteq Sch_{\partial\mathcal{U}}$. If $\overline{Z}\cap\partial\mathcal{U}=\varnothing$ then $Sch_{\overline{Z}\cap\partial\mathcal{U}}=C^{\infty}({\mathcal{U}})$. If $Z=\mathcal{U}$ then $Sch_{\overline{Z}\cap\partial\mathcal{U}}=Sch_{\partial\mathcal{U}}$.

Neither $Sch_{\overline{Z}\cap\partial\mathcal{U}}$ nor $Sch_{\partial\mathcal{U}}$ are unital rings. Still, for a filtration $I_{\bullet}\subset Sch_{\overline{Z}\cap\partial\mathcal{U}}$ with $V(I_{\infty})=Z$, we can take the completion of vector spaces,

[TABLE]

We establish the surjectivity with representatives of prescribed decay rate on $\partial\mathcal{U}$.

Proposition 6.1**.**

Take a filtration $I_{\bullet}\subset Sch_{\overline{Z}\cap\partial\mathcal{U}}$, with $V(I_{\infty})=Z$. Suppose $I_{\bullet}$ satisfies the assumptions of theorem 4.1. Then the following rows are exact:

[TABLE]

Moreover, for any neighbourhood $Z\subset\mathcal{U}(Z)$ and any element $\widehat{f}\in\widehat{Sch_{\overline{Z}\cap\partial\mathcal{U}}}^{(I_{\bullet})}$ there exists a representative $f\in Sch_{\partial\mathcal{U}}\cap C^{\omega}(\mathcal{U}\setminus Z)$ that satisfies $f>0$ on $\mathcal{U}\setminus\mathcal{U}(Z)$.

*Proof.  *We can assume that $\mathcal{U}\subset\mathbb{R}^{n}$ is bounded. (Recall, any open subset of $\mathbb{R}^{n}$ is real-analytically equivalent to a bounded open, e.g. by $x\to\frac{x}{1+||x||^{2}}$.)

Take an element $\sum g_{j}\in\widehat{Sch_{\overline{Z}\cap\partial\mathcal{U}}}^{(I_{\bullet})}$. Here $g_{j}$ does not necessarily belong to $Sch_{\partial\mathcal{U}}$. However, at each point $x\in\overline{Z}\cap\partial\mathcal{U}$ the inclusion holds for germs, $[g_{j}]_{x}\in[Sch_{\partial\mathcal{U}}]_{x}$. Therefore for each $j$ we take a small (relatively) open neighbourhood $\overline{Z}\subset\mathcal{U}_{j}(\overline{Z})\subset\overline{\mathcal{U}}$ such that $g_{j}|_{\mathcal{U}_{j}(\overline{Z})}\in I_{j}\cap Sch_{\partial\mathcal{U}}|_{\mathcal{U}_{j}(\overline{Z})}$.

Take an element $\tau_{j}\in I_{\infty}$ that satisfies $\tau_{j}|_{\overline{\mathcal{U}}\setminus\mathcal{U}_{j}(\overline{Z})}=1$. It exists e.g. by proposition 2.5.

Then the element $\sum g_{j}\in\widehat{Sch_{\overline{Z}\cap\partial\mathcal{U}}}^{(I_{\bullet})}$ is equivalent to $\sum g_{j}(1-\tau_{j})$. By construction: $g_{j}(1-\tau_{j})\in I_{j}\cap Sch_{\partial\mathcal{U}}$, in particular this is flat on $\partial\mathcal{U}$.

Extend $g_{j}(1-\tau_{j})$ to $\tilde{g}_{j}\in C^{\infty}(\mathbb{R}^{n})$ by zero outside of $\mathcal{U}$. Take the filtration $I_{\bullet}\cap{\mathfrak{q}}\cap I(\mathbb{R}^{n}\setminus\mathcal{U})$ of the ring $C^{\infty}(\mathbb{R}^{n})$, here ${\mathfrak{q}}$ is the ideal defined by the $\{q_{\underline{k}}\}$-vanishing conditions. This filtration satisfies the assumptions of theorem 4.1. Thus we have a representative $\tilde{f}\in C^{\infty}(\mathbb{R}^{n})\cap C^{w}(\mathcal{U}\setminus Z)$ of $\sum\tilde{g}_{j}$. Moreover, for any given neighbourhood $Z\subset\mathcal{U}(Z)\subset\mathcal{U}$, we can assume $\tilde{f}>0$ on $\mathcal{U}\setminus\mathcal{U}(Z)$. Finally, the restriction $f:=\tilde{f}|_{\mathcal{U}}\in Sch_{\partial\mathcal{U}}\cap C^{w}(\mathcal{U}\setminus Z)$ is the needed representative of $\sum g_{j}\in\widehat{Sch_{\overline{Z}\cap\partial\mathcal{U}}}^{(I_{\bullet})}$. $\blacksquare$

Example 6.2**.**

Assume $Z\subset\mathcal{U}$ is compactly embedded, then $Sch_{\overline{Z}\cap\partial\mathcal{U}}\!=\!C^{\infty}(\mathcal{U})\!=:\!R$. We get: any $\widehat{f}\!\in\!{\widehat{R}}^{(I_{\bullet})}$​ admits a representative $f\!\in\!Sch_{\partial\mathcal{U}}\!\cap\!C^{w}(\mathcal{U}\!\setminus\!Z)$, decaying (at the prescribed rate) on $\partial\mathcal{U}$.

6.2. Representatives with imposed $\mathbb{R}$-linear conditions

For various applications one needs $C^{\infty}$-representatives of power series with prescribed integrals, e.g. $\{\int^{\infty}_{-\infty}f\cdot x^{k}dx=c_{k}\}$ for the Hamburger moment problem. More generally, take a (possibly infinite) set $\{l_{k}\}$ of $\mathbb{R}$-linear functionals defined on $R=C^{\infty}(\mathcal{U})$, or on a vector subspace of $R$. (We do not assume that $\{l_{k}\}$ are continuous in whichever sense.)

Definition 6.3**.**

1. The conditions $\{(l_{k},c_{k})\}$ are resolvable if exists $f\in R$ satisfying $\{l_{k}(f)=c_{k}\}$.

2. The conditions $\{(l_{k},c_{k})\}$ are $\widehat{f}$-resolvable if exists a representative $f\in R$ of $\widehat{f}$ that satisfies $\{l_{k}(f)=c_{k}\}$.

Resolvability of $\{(l_{k},c_{k})\}$ can be obstructed in various ways.

Example 6.4**.**

1. i.

Define $g_{\infty}(x):=\sum g_{k}(x)\in C^{0}(\mathcal{U})$, suppose the series converges uniformly on $\mathcal{U}$. Define $l_{k}(f):=\int_{\mathcal{U}}g_{k}fd\mathcal{U}$, for $k\leq\infty$, these functionals act on the subspace of functions for which these integrals converge. Then, $\sum l_{k}=l_{\infty}$, imposing restrictions on $\{c_{k}\}$. 2. ii.

Take a convergent sequence of points $x_{k}\to x_{\infty}\in\mathcal{U}$, assume $\{x_{k}\}_{k}$ are pairwise distinct for $k\leq\infty$. Define $l_{k}(f):=f(x_{k})$ for $k\leq\infty$. Thus $l_{k}\to l_{\infty}$, pointwise on $R$. The functionals $\{l_{k}\}$ are linearly independent. But the conditions $\{(l_{k},c_{k})\}$ are resolvable iff $\lim c_{k}=c_{\infty}$. 3. iii.

Take a sequence of points $\{x_{k}\}$ in $\mathcal{U}$. Define $l_{k}(f):=\int_{\mathcal{U}}f\cdot g_{k}dx$, where $\{g_{k}\}$ are bump functions on $\{Ball_{\frac{1}{k!}}(x_{k})\}$. Then $\{l_{k}\}$ are linearly independent. Assume the sequence $\{x_{k}\}$ has a condensation point $x_{0}\in\mathcal{U}$, then the conditions $\{l_{k}(f)=1\}$ force $f$ to explode at $x_{0}$. 4. iv.

Assume $\mathcal{U}$ is bounded and take an exhaustion $\mathcal{U}_{1}\subset\mathcal{U}_{2}\subset\cdots\subset\cup\mathcal{U}_{k}=\mathcal{U}$, by compactly embedded sets. Define $l_{k}(f):=\int_{\mathcal{U}}f\cdot(1-g_{k})dx$, where $\{g_{k}\}$ are bump functions on $\{\mathcal{U}_{k}\}$. Assume $\mathcal{U}\setminus\mathcal{U}_{k}$ decreases fast. Then the conditions $\{l_{k}(f)=1\}$ force $f$ to explode at some point(s) of $\partial\mathcal{U}$. While $f$ remains $C^{\infty}$, the conditions $l_{k}(f)=c_{k}$ become ill-defined, as the integrals diverge.

Therefore any statement on $\widehat{f}$-resolvability of $\{(l_{k},c_{k})\}$ must assume the resolvability of $\{(l_{k},c_{k})\}$.

Lemma 6.5**.**

The conditions $\{(l_{k},c_{k})\}$ are $\widehat{f}$-resolvable iff $\{(l_{k},c_{k})\}$ are resolvable and moreover $\cap Ker(l_{k})\twoheadrightarrow{\widehat{R}}$.

Proof.  $\Rrightarrow$ Take $\{c_{k}=0\}$ and any $\widehat{f}\in{\widehat{R}}$.

$\Lleftarrow$ Take some $g\in R$ satisfying $\{l_{k}(g)=c_{k}\}$. Take some $\tilde{f}\in\cap Ker(l_{k})$ whose image in ${\widehat{R}}$ is $\hat{\tilde{f}}=\widehat{f}-{\widehat{g}}$. Then $g+\tilde{f}$ is the needed solution. $\blacksquare$

Example 6.6**.**

1. i.

Assume the completion is surjective, $R\twoheadrightarrow{\widehat{R}}^{(I_{\bullet})}$, denote $Z=V(I_{\infty})$. Call a functional “punctual” if it is presentable as $l(f):=\sum_{j}c_{{\underline{m}}_{j}}\cdot f^{({\underline{m}}_{j})}|_{x_{j}}$, the values of prescribed partials at prescribed points. If this sum is infinite then $l$ acts on a subspace of $R$ for which the sum converges. Call the points $\{x_{j}\}$ “the base points of $l$”.

Assume the collection $\{l_{k}\}$ consists of an arbitrary number of punctual functionals, all of whose base-points lie in $\mathcal{U}\setminus\mathcal{U}(Z)$, for a neighbourhood $\mathcal{U}(Z)$. Then $\cap Ker(l_{k})\twoheadrightarrow{\widehat{R}}$. Hence $\{(l_{k},c_{k})\}$ are $\widehat{f}$-resolvable iff $\{(l_{k},c_{k})\}$ are resolvable. 2. ii.

Assume the collection $\{l_{k}\}$ consists of an arbitrary number of punctual functionals, with base-points in $\mathcal{U}\setminus\mathcal{U}(Z)$, and of a finite number of moments, $l_{k}(f)=\int_{\mathcal{U}}h_{k}(x)\cdot f(x)d^{n}x$, for some functions $\{h_{k}\}$ integrable on $\mathcal{U}$. Then $\cap Ker(l_{k})\twoheadrightarrow{\widehat{R}}$. Hence $\{(l_{k},c_{k})\}$ are $\widehat{f}$-resolvable iff $\{(l_{k},c_{k})\}$ are resolvable.

Remark 6.7**.**

One cannot impose an infinite collection of moments without non-trivial assumptions on $\{c_{k}\}$. Even if the conditions $\int_{\mathcal{U}}h_{k}\cdot f\cdot d^{n}x=c_{k}$ are resolvable on $R$, the solution can be unique. (This can happen e.g. for Hamburger moment problem.) Then no $\widehat{f}$-solvability of $\{(l_{k},c_{k})\}$ can be achieved.