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Consider a topological space $X$. Lee in Introduction to Smooth Manifolds wrote that it is impossible to introduce a smooth structure on the topological manifold based only on topology (i.e. preservable my homeomorphisms) since the square and the circle are homeomorphic, but the square we don't want to be smooth.

Anyway, when introducing topological/smooth manifolds one makes a strict connection with $\mathbb R^n$. It seems that we extremely need this space to introduce the notion of smoothness on other spaces, am I right? Namely, can we formulate the smoothness without talking about $\mathbb R^n$ at all? If yes, what is necessary to have: finite dimension? metrics? topology?

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Differentiability roughly means approximable by a linear map. How are you going to say "linear map" without using vector spaces? –  Qiaochu Yuan Aug 1 '11 at 19:04
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This question is (sort of) related. –  t.b. Aug 1 '11 at 19:12
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finite dimension is not necessary to have. For example, there are such a things as differentiable Hilbert manifolds. But like Qiaochu said, what we want is the linear structure of the tangent space. In the finite dimensional case, all (real) vector spaces are identical to $\mathbb{R}^n$. –  Willie Wong Aug 1 '11 at 20:28
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Another example commonly used is that of f(x):$\mathbb R\rightarrow \mathbb R$; f(x)=$x^{1/3}$ vs id:$\mathbb R\rightarrow\mathbb R$, so that, e.g., the identity is differentiable (smooth, actually) re the id chart, but not so re the f(x) chart(specifically at x=0) –  gary Aug 1 '11 at 21:11
    
Hey: don't I get some extra points for finally actually reading the actual question being asked and making a (reasonably) relevant post, after repeatedly answering questions that exist only in my head? –  gary Aug 1 '11 at 21:21

2 Answers 2

up vote 12 down vote accepted

Instead of introducing a smooth structure on the manifold $X$ by specifying a smooth structure using a "smooth" atlas, you could try instead to specify the set of "smooth" functions $X\to\mathbb{R}$. For example, the set of "smooth" functions for a compact submanifold is the restriction of the smooth functions of the embedding manifold to the submanifold. The problem is that this also works for non-smooth manifolds like the square, so you now need criteria how to distinguish non-smooth manifolds like the square from smooth manifolds like the circle, based on this set of "smooth" functions $X\to\mathbb{R}$. And even before that, you need criteria that such an "arbitrary" set of smooth functions must satisfy.

It might be fun trying to get such an approach to work. But at least for smooth manifolds, analytic manifolds and complex manifolds, the approach with the atlas is preferable. For differentiable manifolds, there might be questions where approaches like the one sketched above add value.

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You can require that the "smooth" functions have the structure of a $C^{\infty}$-ring (but you still need Euclidean space to talk about this): ncatlab.org/nlab/show/smooth+algebra –  Qiaochu Yuan Aug 1 '11 at 21:49
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I think such approaches have been studied in the past, maybe still today, You want a certain sheaf of smooth funtions with special properties. –  Olivier Bégassat Aug 1 '11 at 21:53
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The definition of smooth manifold that Thomas hints at is: a smooth manifold is a second countable Hausdorff topological space equipped with a subsheaf of the sheaf of continuous function that is locally isomorphic to the sheaf of smooth functions on some R^n. (Of course, this uses R^n.) You can find this definition in several textbooks (where there is usually an exercise asking you to prove it is equivalent to the atlas style of definition). –  Omar Antolín-Camarena Aug 2 '11 at 2:30
    
@Omar I actually like the definition of a manifold as a second countable Hausdorff topological space locally isomorphic to $\mathbb{R}^n$. I just wanted to specify the smooth structure on the manifold by something less arbitrary than an atlas. I hope that it's possible to derive the sheaf of smooth functions from the set of smooth functions via exhaustion by compact sets. I'm assuming finite dimension anyway, so I guess it should work. –  Thomas Klimpel Aug 2 '11 at 21:41

The basic idea of differential geometry is that you can differentiate functions in the spaces that you consider. "Smoothness" is simply the concept that there are functions that can be differentiated infinitely often.

The simplest functions are curves aka paths, i.e. functions from the reals (or an interval) to the model space $E$ you consider: $$ \gamma: [a, b] \to E $$ In order to define a differential quotent for such a curve you need to be able to write down $$ \lim_{h \to 0} \frac{1}{h} (\gamma(t+h) - \gamma(t)) $$ Define a path to be differentiable if all these limits exist.

In order for this to make sense, you need to have addition and scalar multiplication on $E$ and a topology so that the algebraic operations are continuous. This means that $E$ needs to be a topological vector space. Manifolds can then be defined to have locally homeomorphisms to your model space $E$.

If you use Hilbert spaces, you get Hilbert manifolds, the same holds true for Banach manifolds and Frechet manifolds. In finite dimensions, every topological vector space is algebraically isomorph to $\mathbb{R}^n$ and isomorph as topological vector space to $\mathbb{R}^n$ with its canonical topology, so there is not much choice.

Next, you can define smooth paths to be paths that are differentiable infinitely often, and define smooth maps of manifolds to be maps that map smooth paths to smooth paths.

This results in a definition of smooth/differential geometry in infinite dimensions.

For more details, see the book:

  • Krigl, Michor: "The Convenient Setting of Global Analysis"

In this setting we characterize smooth maps of manifolds via the images of smooth paths.

It is possible to generalize this to more general functions than smooth paths and get, for example, diffeological spaces.

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