I guess my question is about interpretation. My (very rough) understanding of the way we do calculus on a smooth manifold is as follows...

A smooth manifold is a topological space without necessarily any metrical structure- therefore, if we speak of "calculus" being done on smooth manifolds, then it is a calculus on spaces which are not generally metric. How then do we define a "derivative" for such a calculus? (In analysis we learn that a derivative is the (limit of) the ratio of distance functions evaluated on elements of two (functionally related) metric spaces.) And how then do we define the derivative's inverse- the integral? Now, I know that the objects which are differentiated and integrated on smooth manifolds are not functions, but differential forms. The theory is developed in such a way that this calculus of forms reduces - via coordinates - to good old calculus of functions on R^n endowed with the Euclidean metric. So a metric does sneak its way into this calculus on manifolds. What's going on here? Would it be fair to call calculus on manifolds a non-metrical calculus or not?

Perhaps my question is ill-defined, but here's my attempt to answer it...

The Euclidean metric only comes into play to allow us to do calculations (via the pullback). The existence of many different (diffeomorphic) coordinate spaces in which to do these calculations means that the Euclidean metric in any one coordinate space cannot be interpreted as carrying metrical information about the underlying manifold. Forms are the objects of calculus on a manifold precisely because the calculations we do with them (e.g. integration, exterior differentiation,...) can be defined without reference to coordinates (R^n) and are therefore invariant under coordinate transformations. So, yes, the calculus on a manifold - which apparently is the calculus of forms - is a non-metrical calculus.

I'm on my own here so any guidance is very much appreciated.

  • $\begingroup$ There's not really a question here that is answerable. $\endgroup$ – Thompson Aug 1 '16 at 23:11
  • $\begingroup$ I think this question could spark some interesting discussion. $\endgroup$ – ncmathsadist Aug 1 '16 at 23:39
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    $\begingroup$ @Thompson Is calculus on smooth manifolds an example of a calculus on spaces which aren't generally metric? $\endgroup$ – luis Aug 2 '16 at 0:30
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    $\begingroup$ "Do you need a metric to define concepts in calculus?" No. // On the other hand, every paracompact smooth manifold is metrizable, and also admit Riemannian metrics, so it is not even true that you are working on a space "without a metric (in either sense of the word)". You don't have to use it (see first part of my comment), but there is one (in fact many). $\endgroup$ – Willie Wong Aug 2 '16 at 2:39
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    $\begingroup$ @luis: no. It is the best linear approximation. Period. You can do calculus using any norm on $\mathbb{R}^n$, not necessarily one that comes from an inner product. You can do calculus even without a norm, just the linear structure. // Don't get me wrong, despite several misconceptions present in your question statement, the question is a great one. I might start writing something now, but I won't have the time to finish it until after at least a few days. $\endgroup$ – Willie Wong Aug 2 '16 at 3:33

Tangency revisited

Before talking about differential calculus, it may be best to conceptually understand what calculus is trying to tell us. One way to look at it is this: calculus is trying to tell us that fundamentally the functions $f$ with $f'(0) \neq 0$ is different from the functions $g$ with $g'(0) = 0$.

To emphasize this, let's go back to the notion of continuity.

Definition Given $(X,\tau)$ and $(Y,\sigma)$ two topological spaces, a function $f:X\to Y$ is said to be continuous at $x$ if for every open neighborhood $V$ of $f(x)$ there exists some open neighborhood $U$ of $x$ such that $f(U) \subseteq V$.

On the level of "continuity", we cannot tell apart $f$ and $g$ from above, since continuity lacks a quantitative expression to allow us to say that "as the neighborhood of $x$ shrinks, the corresponding neighborhood of $f(x)$ shrinks much faster". The basic idea of calculus is this:

Idea We say that $f:X\to Y$ is tangent to a constant at $x$ if for every open neighborhood $V$ of $f(x)$ there exists some open neighborhood $U$ of $x$, and a function $\gamma:(0,1] \to (0,1]$ satisfying $\lim_{s\to 0} \gamma(s)s^{-1} = 0$, such that for every $s$, $f(sU) \subseteq \gamma(s) V$.

Here, the undefined notation $sU$ means the intuitive notion of "shrinking $U$ by a factor of $s$", and similarly for $\gamma(s) V$. This notion can be made precise when $X$ is a topological vector space: the scaling $sU$ of $U$ around the point $x$ is $$ sU = \{ x + s(y-x): y\in U\}.$$ An indeed this gives rise to the standard definition of tangency in topological vector spaces (which may or may not have a norm).

Note that if you suitably define the scaling operations, you can generalize the notion of "tangency to a constant" to much more general settings.

In the case where $X$ and $Y$ are topological vector spaces, linearity allows us to add and subtract functions. This allows us to say that

Definition Two functions $f,g:X\to Y$ between topological vector spaces are said to be tangent at the point $x$ if $f(x) = g(x)$ and the function $f-g$ is tangent to 0 at $x$.

Definition A function $f:X\to Y$ is said to be differentiable at $x$ if there exists a continuous linear mapping $L:X\to Y$ such that $f$ is tangent to the function $x'\mapsto f(x) + L(x')$ at $x$. The linear mapping $L$ is said to be the derivative of $f$.

Exercise Prove the chain rule for topological vector spaces. That is to say, suppose $X,Y,Z$ are topological vector spaces, and suppose $f:X\to Y$ and $g:Y\to Z$ are function such that $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)\in Y$, then $g\circ f$ is differentiable at $x$, and the derivative is the continuous linear map $M\circ L$ where $L$ is the derivative of $f$ at $x$ and $M$ is the derivative of $g$ at $f(x)$.

(... to be continued ...)


Willie Wong has already said most of what needs to be said (though if you'd like a reference, see Lang's Differential and Riemannian Manifolds). But for a nice summary of those ideas in the context of the most general setting in which one can do calculus, you might also like to read the introduction of the book Calculus in Vector Spaces without Norm, by Frölicher and Bucher.

Let me also say that you seem to have an impoverished view of the notion of manifold. You are thinking of smooth manifolds modeled on Euclidean spaces. But manifolds can be modeled on more general spaces, such as Banach spaces and even Fréchet spaces. The resulting manifolds are called -- drumroll -- Banach manifolds and Fréchet manifolds. Since differentiation (and, in particular, the chain rule) makes sense in Banach spaces and in Fréchet spaces, we can lift the idea of differentiation to manifolds modeled on such spaces. Of course, given the greater generality, this project encounters some difficulties: it's not true that all the nice features of high school differential calculus carry over. But some semblance of the core notion of derivative does.

  • $\begingroup$ Aww, awesome nickname! $\endgroup$ – lisyarus Aug 7 '16 at 8:15
  • $\begingroup$ Although manifolds can be modeled, as you are saying, on more general spaces than $R^n $, and differentiation itself can then be defined and applied on the manifold because it is well defined on these spaces ( Banach space, Fréchet spaces ), I am wondering though if it is conceptually possible to define differentiation intrinsically on a manifold, i.e. without any reference to the homeomorphic spaces needed for the definition of the manifold ( $R^n$, or other more general spaces like Banach spaces, Fréchet spaces, etc.) ? Can you elaborate a little on this ? Many thanks. $\endgroup$ – user249018 Feb 11 '18 at 0:20

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