The question requires some definitions that I have listed below for your convenience. They can be found in Chapter VI, pages 297-298 of Bredon's Introduction to Compact Transformation Groups.

On a topological space $X$ a functional structure $F_{X}$ is an assingment defined on the collection of open sets $U\subset X$ taking $U\mapsto F_{X}(U)$ such that:

  1. $F_{X}(U)$ is a subalgebra of the algebra of continuous real valued functions on $U$ and contains all constant functions.

  2. If $V$ is open, $V\subset U$ and $f\in F_{X}(U)$ then $f|_{V}\in F_{X}(V)$.

  3. If $\{U_{i}\}$ is a collection of open sets and $f|_{U_{i}}\in F_{X}(U_{i})$ for all $i$ then $f\in F_{X}(\bigcup_{i}U_{i})$.

The pair $(X,F_{X})$ is called a functionally structured space. E.g. $(\mathbb{R}^{n},C^{\infty})$, where $\mathbb{R^{n}}$ has the usual topology and $C^{\infty}(U)=\text{all $C^{\infty}$ functions on $U$.}$

If $(X,F_{X})$ is a functionally structured space and $U\subset X$ is open then for $V\subset U$ open we define $F_{U}(V)=F_{X}(V)$ so that $(U,F_{U})$ is a functionally structured space.

A morphism of functionally structured spaces $(X, F_{X})$ and $(Y,F_{Y})$ is a continuous map $\varphi:X\rightarrow Y$ such that for any open set $V\subset Y$ and $f\in F_{Y}(V)$ we have $f\circ \varphi\in F_{X}(\varphi^{-1}(V))$. A morphism $\varphi$ is an isomorphism if $\varphi^{-1}$ exists as a morphism. We then say that $(X,F_{X})\simeq (Y,F_{Y})$.

An $n$-dimensional differentiable manifold is a second countable functionally structured Hausdorff space $(M,F)$ with the property that each point in $M$ has an open neighborhood $U$ such that $(U,F_{U})\simeq(V,C^{\infty}_{V})$ for some open set $V\subset \mathbb{R}^{n}$.

Question: Let $(X,F_{X})$ be a functionally structured space with the following property. Each point in $X$ has a neighborhood $U$ such that there are functions $f_{1},\ldots ,f_{n}\in F_{X}(U)$ verifying: A real valued function $g$ on $U$ is in $F_{X}(U)$ iff there exits a smooth real valued function $h$ of $n$ real variables such that $g(p)=h(f_{1}(p),\ldots,f_{n}(p))$ for all $p\in U$.

Does it follow that $\mathbf{(X,F_{X})}$ is an $\mathbf{n}$-dimensional differentiable manifold? (In the sense of the definition given above).

Edit: Sheaves are new to me, but after reading the comments I wanted to point out this and this MSE posts, and the wikipedia page and subwiki page where sheaves are used to define differentiable manifolds.

Edit: Sheaves my be used to define differentiable manifolds: Roughtly speaking, a differentiable manifold is a second countable Hausdorff topological space equipped with a subsheaf of the sheaf of continuous real valued function that is locally isomorphic to the sheaf of smooth real valued functions on some $\mathbb{R}^{n}$.

  • 3
    $\begingroup$ What you call a "functional structure" is more commonly known as a "subsheaf of the sheaf of continuous real-valued functions". Your question is local and can be phrased without recourse to such complications, though. I think you should probably add some other hypotheses: for example, if $X$ is a one-point space then it automatically satisfies your condition. Perhaps you should ask that the smooth function $g$ be unique. $\endgroup$ – Zhen Lin Aug 8 '13 at 18:47
  • $\begingroup$ I guess you mean that smooth function $h$ be unique, as $g$ is not assumed smooth. $\endgroup$ – John Aug 8 '13 at 18:52
  • $\begingroup$ @ZhenLin It looks slightly more restrictive than a general sheaf in that we ask that all constant functions are their. $\endgroup$ – Baby Dragon Aug 8 '13 at 19:33
  • $\begingroup$ Well, to be more precise, it has to be a subsheaf considered as a sheaf of $\mathbb{R}$-algebras, and as you say it has to be non-trivial if the space is non-empty. $\endgroup$ – Zhen Lin Aug 8 '13 at 20:09

The answer is negative:

Consider $X=\mathbb{R}$ with the usual topology and define a functional structure $F_X$ on $X$ as follows: For any open interval $I\subset\mathbb{R}$ let $F_{X}(I)$ consist only of constant functions. Now, an arbitrary open set $U\subset\mathbb{R}$ can be uniquely written (i.e. up to a permutation of the intervals) as the disjoint union of countably many open intervals. Therefore, we can define $F_{X}(U)$ to be the set of all real-valued functions which are constant on open intervals building up $U$.

Note that $(X,F_{X})$ as defined satisfies the hypothesis of the question but it is not a smooth manifold.

Additional hypotheses are then needed for an affirmative answer. E.g. requiring that the function $(f_{1},\ldots, f_{n})$ is locally invertible.

  • 2
    $\begingroup$ In other words you consider the sheaf of locally constant functions. $\endgroup$ – Martin Brandenburg Aug 8 '13 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.