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I am beginning to study multidimensional calculus using Spivak's Calculus on Manifolds, and so far I understand the purpose of considering the classes $C^n$ to be twofold. First, being in $C^n$ is a handy sufficient condition for the equality of mixed partials of degree $n$. Second, being in $C^1$ is a handy sufficient condition for the existence of the total derivative. I have two questions. (1) What are some other important reasons to be interested in the classes $C^n$? Is being in one of these classes necessary for anything or just always a handy sufficient condition? (2) How can the classes $C^n$ be defined in a coordinate-free way? The Wikipedia article on smooth functions claims that a coordinate-free definition is possible and that this can be seen by using a Banach space approach. I think I sort of understand what a Fréchet derivative, but I still do not understand this coordinate-free approach and how it is equivalent to the standard approach. It is not really explained in Wikipedia.

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Your question is not really a duplicate, but see this related one. –  user31373 Aug 1 '12 at 16:25

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The spaces $C^n$ – assuming its elements take on values in a vector space – are (infinite-dimensional) vector spaces themselves with a lot of structure (they admit a complete norm, making the spaces Banach spaces if you are a bit careful when choosing the domain and boundary behaviour). They and a zoo of additional spaces are a basis for several approaches to prove existence results for (ordinary and partial) differential equations. For partial differential equations they are not that useful but form a basis for other, more suitable spaces (spaces with Hölder continuous derivatives). Also for other spaces (Sobolev spaces) which are useful in PDE existence theory one has inclusion results into $C^n$ and can prove using differentialbility of abstract existence results using this (quite advanced) fact.

Interestingly, in differential geometry they are often not considered, but only $C^\infty$ is used.

A coordinate free way of saying that $f$ is of class $C^1(U, \mathbb{R})$ is by saying that there exists a continuous linear map $df:U\rightarrow L(\mathbb{R}^n, \mathbb{R})$ (the space of linear functions on $\mathbb{R}^n$, which is just its dual space) such that $df$ approximates $f$ in the following sense: $$\lim_{x\rightarrow y}\frac{f(x)-f(y)-df(y)(x-y)}{||x-y||} = 0$$ This reduces to the coordinate-based definition easily by introducing a coordinate system around $y$ and then checking that, e.g., $$ \frac{\partial f(y)}{\partial x_i} = df(y) e_i\;,$$ $e_i$ being the $i$-th standard basis vector, dual to the linear map $x^i$ which maps $v=\sum v^i e_i$ to $x^i(v)=v^i$. To get a feeling for this definition you need to work with it for a while. Higher-order coordinate-free definitions are then given recursively, i.e. $f\in C^2(U)$ iff $df \in C^1$. Writing down the target space soon becomes a bit clumsy, though.

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