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The local ring of C-infinity functions

Stemwede, 2022-07-09

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Setting

Definition. The local ring of $C^\infty$ functions around $0 \in \IR^n$ is defined as

$$ \AINF^n = \lim_\ra { \CINF(U,\IR) ,|, 0 \in U \subset \IR^n }. $$

The inductive limit is taken along the restriction maps $\rho_V: \CINF(U,\IR) \ra \CINF(V,\IR)$ for $U \subset V$.

More explicitly, an element $a \in \AINF$ is hence represented by a pair $(a,U), a \in \CINF(U,\IR)$. Two pairs $(a,U), (b,V)$ represent the same element in $\AINF^n$ if there is a $W \subset U \cap V, 0 \in W$ so that $\rho_W(a) = \rho_W(b)$.

Naturality. Any ($C^\infty$-)differentiable map $f: \IR^n \ra \IR^m$ with $f(0) = 0$ gives rise to a map $f^*: \AINF^m \ra \AINF^n$ (“pullback”), which sends $(b,V)$ to $(b \circ f, f^{-1}(V))$.

Evaluation. For $a \in \AINF^n$, we denote by $e[a] := ev_0[a] := a(0) \in \IR$, the evaluation of $a$ at $0 \in \IR^n$. This is operatin can be identified with the pullback along the injection $0: \IR^0 \ra \IR^n$, $e = 0^*$.

Constants. For $\lambda \in \IR$ the constant function $\IR^n \ra \IR, x \mapsto \lambda$ by the same letter $\lambda \in \AINF^n$. If we want to stress the function character of $\lambda$ we may use the symbol $\underline{\lambda}$. The resulting linear map $c: \IR \ra \AINF^n$ can be identified with $c = \pi^*$, where $\pi:\IR^n \ra \IR^0$ is the projection to a point.

Elementary Operations. Let $i \in {1,\dots,n}$. We have the following operations on $\AINF^n$.

  • Coordinate fuctions $x^i \in \AINF^n$ with $x^i(x_1,\dots,x_n) = x_i$ on any $0 \in U \subset \IR^n$,
  • Partial derivatives $ \del_i = \del_i^x = \frac{\del}{\del x^i}$ those are linear operators $\AINF^n \ra \AINF^n $.
  • Axis injections $x_i: \IR \ra \IR^n$, giving rise to maps $x_i^*: \AINF^n \ra \AINF^1$ via pullback.

For a multi-index $I \in \IN^n, I = (i_1, \dots, i_n)$, we define:

  • The monomial $x^I = (x^1)^{i_1} \cdot \dots \cdot (x^n)^{i_n} \in \AINF^n$
  • The differential operator $\del^I = \del_1^{i_1} \cdot \dots \cdot \del_n^{i_n}$

Elementary Relations.

  • Product rule. $\del_i[ a \cdot b ] = \del_i[a] \cdot b + a \cdot \del_i[b]$.
  • Iterated Product Rule. $$ \del^I[a \cdot b] = \sum_{J,K \in \IN^n, J+K=I} \frac{I!}{J! K!} \del^J[a] \cdot \del^K[b]. $$
  • Chain Rule. $\del_j f^*[b] = \del_j[ b \circ f ] = \sum_k \del_k[b] \circ f \cdot \del_i[f^k] = \sum_k f^{*}\del_k[b] \cdot \del_i[f^k]$, where $f^k$ are the components $f^k = x^k \circ f$.

Taylor Series

Definition. The taylor series of $a \in \AINF^n$ is the formal power series $$ T[a] = \sum_{I \in \IN^n} e \del^I [a] \cdot \frac{x^I}{I!} = \sum_{I \in \IN^n} \frac{\del^I a}{\del x^I}(0) \frac{x^I}{I!} \in \IR \dbrackets{ x^1, \dots, x^n }, $$ where $I! = i_1! \cdot \dots \cdot i_n!$. The degree-k taylor polyonmial is the degree-k truncation of the taylor series: $$ T_k[a] = \sum_{I, |I| \leq k} e \del^I [a] \cdot \frac{x^I}{I!} \in \IR[x^1, \dots, x^n]. $$

Properties.

  • For a polynomial $p \in \IR[x^1, \dots, x^n]$, we have $T_k[p] = p$ if $k \geq deg(p)$.
  • For $a,b \in \AINF^n$, we have $$ T(a \cdot b) = T(a) \cdot T(b), \quad\text{and}\quad T_k(a) \cdot T_k(b). $$

Proof The first property follows from the observation that $e \del^I[x^J] = I!$ if $I = J$ and zero otherwise. The second and third property follow from the iterated product rule.

Theorem (Borel) For every power series $p \in \IR\dbrackets{ x^1, \dots, x^n }$ there exists a $C^\infty$ function $a \in \AINF^n$ with $T(a) = p$. In other words, the map $T : \AINF^n \ra \IR\dbrackets{ x^1, \dots, x^n }$ is surjective, and hence $ \IR\dbrackets{ x^1, \dots, x^n } \isom \AINF^n / \ker(T)$.

For a proof see Ieke Moerdijk, Gonzalo E. Reyes: Models for Smooth Infinitesimal Analysis, p13. or ncatlab.

Commutative Algebra

$\IR$-Algebra Structure. The ring $\AINF^n$ inherits the structure of a unital $\IR$ algebra form $\CINF(U,\IR)$. In other words elements $a,b \in \AINF^n$ can be added, multiplied and multiplied by scalars $\lambda \in \IR$, so that the usual associativity and distributivity relations hold. The unit element $1 \in \AINF^n$ is the constant function $1$.

  • The structure maps $\rho_0: \CINF(U,\IR) \ra \AINF^n$ are morphisms of $\IR$-algebras.
  • The pullback maps $f^*$ are morphisms of $\IR$-algebras.
  • The “constants map” $c$ is a morphism of $\IR$-algebras.
  • The evaluation map $e$ is a morphism of $\IR$-algebras.

Denote the kernel of the evaluation map by $\fm = \ker(e) = \{ a \in \AINF^n | a(0) = 0 \} \subset \AINF^n$. It is a maximal ideal in $\AINF^n$, since the co-domain of $e$ is a field $\IR$.

Hadamard Lemma. The maxiamal $\fm$ ideal is generated by the coordinate fuctions $x^i$, i.e. every element $a \in \fm$ can be written in the form $a = \sum_{i=1}^n x^i \cdot a_i$ with $a_i \in \AINF^n$.

Proof. Let $a \in \fm$, and set $a_i(x) = \int_0^1 (\del_i a)(t \cdot x) dt$. Now $\frac{\del}{\del t} a(t \cdot x) = \sum_{i=1}^n (\del_i a)(tx) \cdot x^i(x)$, hence

$$ \sum_{i=1}^n x^i \cdot a_i = \int_0^1 \sum_{i=1}^n (\del_i a)(tx) \cdot x^i(x) dt = \int_0^1 \frac{\del}{\del t} a(t \cdot x) dt = a(x) - a(0) = a(x). qed. $$

Corollary.

  • If $a = \sum_{i=1}^n x^i \cdot a_i$ is a Hadamard representation of $a \in \fm$, then $a_i(0) = e \del_i[a]$
  • The powers $\fm^k$ are generated by the monomials $x^I, I \in \IN^n, |I| = k$.
  • The intersection $\fm^\inf := \cap_{k \geq 0} \fm^k$ is equal to $\ker(T) = \set{a}{ \del^I[a] = 0 \stext{for all} I}$.
  • The map $\IR[x^1, \dots, x^n] \ra \AINF^n$, mapping a polyomial to the associated $C^\infty$-function-germ is injective, and induces an isomorphism: $$ \IR[x^1, \dots, x^n] / (x^1, \dots, x^n)^k \lra \AINF^n / \fm^k. $$ In particular, the completions are isomorphic: $\IR\dbrackets{ x^1, \dots, x^n} \isom \hat \AINF^n$.

Lemma.

  • The evaluation map is the only $\IR$-alebgra morphism $\AINF^n \ra \IR$.

Lemma For an open set $0 \in U \subset \IR^n$, denote the restriction map by $\rho: C^\infty(U, \IR) \ra \AINF^n$, let $\fm_0 \subset C^\infty(U, \IR)$ be the kernel of the evaluation map at $0$. Then,

  • $\rho$ is surjective and induces and isomorphism $ C^\infty(U, \IR) / \ker(\rho) \isom \AINF^n$.
  • $\rho$ induces an isomophism $ C^\infty(U, \IR)_{\fm_0} \isom \AINF^n$.

Proof. To prove the first claim choose a representative $(V, a) \in \AINF^n$, with $V \subset U$. Choose $\eps > 0$ small enough so that the ball $B_{2\eps}$ lies in $V$. Choose a bump function $\delta$ with $\delta = 1$ on $B_{\eps}$ and $\delta = 0$ outside of $B_{2\eps}$. Then $a \cdot \delta $ is defined on $V$ and equal to zero outside of $B_{2\eps}$, and hence extends to U.

To prove the second claim we first note that $b(0) \neq 0$ implies that $b(x) \neq 0$ for $x$ in an open subset of $0$, hence $b$ is invertible in $\AINF^n$. This shows that $\rho$ induces a map from the localization. The map is surjective by the claim we just proved. To show the map is injective, assume that the formal quotient $(a,b)$ maps to $a/b = 0 \in \AINF^n$. This means that $a = 0$ in an open neightbourhood $V$ of $0$. Let $\delta$ be a bump function with $\delta(1) = 1$ and $\delta(x) = 0$ outside of $V$. Then $\delta \notin \fm_0$ but $\delta a = 0 \in C^\infty(U, \IR)$, showing that $(a,b) = 0 \in C^\infty(U, \IR)_{\fm_0}$.

Invariant Theory

Group Action. Let $\mathrm{Diff}_0^n$ be the group of local diffeomorphisms of $\IR^n$. Elements $\varphi \in \mathrm{Diff}_0^n$ are represented by diffeomorophisms $\varphi: U \ra V$, with $0 \in U, 0 \in V$. Two representatives $\varphi: U \ra V, \varphi': U' \ra V'$ are equivalent if they agree on $U \cap U'$.

Evenry element $\varphi \in \mathrm{Diff}_0^n$ gives rise to an liear isomorphism $ \varphi^* $ of $\AINF^n$ via the pullback operation. In this way we optain a group action $\mathrm{Diff}_0^n \ra GL(\AINF^n), \varphi \mapsto \varphi^*$.

Question: For any any natural “structure” $F$ on $\AINF^n$, classify elements of $F$ up to diffeomorphism.

Examples:

  • $F = id$: Classify local functions up to diffeomorphism. In the case $n=1$ we claim that $\AINF^1 / \mathrm{Diff}_0^n = c(\IR) \union \{ x^1 \}$. For $n > 2$
  • $F = T$ tangent space: Classify tangent vectors up to diffeopmorphis. We have $T(\AINF^n) / \mathrm{Diff}_0^n = \{ 0, \del_1 \}$, reflecting the fact that two non-zero tangent vector can be transformed into each other via a (linear) diffeomorphism.