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Say $(\mathcal E, V)$ and $(\mathcal F, W)$ are Euclidean spaces. A map $F : \mathcal E \to \mathcal F$ is said to be differentiable at $p_0 \in \mathcal E$ if there exists a linear map $L : V \to W$ such that $$ \lim_{p \to p_0} \frac{\lVert w - Lv \rVert}{\lVert v \rVert} = 0, $$ where $v = p - p_0$ is the unique vector in $V$ satisfying $p = p_0+v$ and $w = f(p)-f(p_0)$ is the unique vector in $W$ satisfying $f(p) = f(p_0) + w$. Such a map linear map $L$ can then be shown to be unique, and thus may be referred to as the differential of $F$ at $p$. Reading about differentiability classes, it seems to me that higher differentiability is defined only in terms of partial derivatives, which would require coordinates.

With $v$ and $w$ defined as before, can't, say, the second derivative of $F$ at $p_0$ be defined as the unique bilinear map $L_2 : V \times V \to W$ satisfying $$ \lim_{p \to p_0} \frac{\lVert w - L_1 v - \frac{1}{2} L_2[v,v] \rVert}{\lVert v \rVert} = 0 $$ where $L_1$ is the differential of $F$ at $p_0$? And how about generalizing to arbitrary smoothness of class $C^k$, defining it locally for a point $p_0$ via for each $i = 1, \dots k$ the existence of an $i$-linear map $L_i : V \times \cdots \times V \to W$ such that $$ \lim_{p \to p_0} \frac{\lVert v - P_k v \rVert}{\lVert v \rVert}, $$ where $$ P_k v = \sum_{i=1}^k \frac{L_i[v, \dots, v]}{i!}? $$ Is this sensible? I'm doing a writeup on differential geometry covering some background material, and want to keep it as coordinate-agnostic as possible. Any insight would be greatly appreciated.

markusas
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    For a coordinate-free approach (in the setting of topological vector spaces), see Lang's book "Differential manifolds." – Moishe Kohan Jun 03 '24 at 15:35
  • @MoisheKohan Thanks, just found a copy of the book "Differential and Riemannian Manifolds" by the same author, which I suppose is more or less the same? – markusas Jun 03 '24 at 16:04
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    If $F$ is class $C^1$, then $DF : \mathcal{E} \to \mathcal{L}(W)$ is continuous. You can define $C^2$ to mean $DDF : \mathcal{E} \to \mathcal{L}(\mathcal{L}(W))$ exists and is continuous. For finite-dimensional $V, W$ this is equivalent to second order partials existing and being continuous, but it is coordinate free and makes sense for Banach spaces $V$, $W$. Similarly, you can define $C^k$ to mean $D^k F$ exists and is continuous. – Mason Jun 03 '24 at 16:09
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    Btw, no to get the usual regularity of higher differentiability, you wouldn’t define things in this sort of manner of “having a Taylor polynomial”, because the naive formulation wouldn’t give the right answers even in one dimension; see this answer (to get correct formulations you’d need suitable continuity hypotheses on the remainder). Anyway as for defining higher order differentiability, it’s just done inductively; see this answer of mine. – peek-a-boo Jun 03 '24 at 16:34
  • @Kakashi And how do you define $DF$ and $DDF$ in this case? – markusas Jun 03 '24 at 16:43
  • @markusas It's the Frechet derivative just as you wrote. The Frechet derivative is defined for maps between Banach spaces $V, W$. In your language, $DF(p)$ is the differential of $F$ at $p$. Then since $DF : \mathcal{E} \to \mathcal{L}(V, W)$ and $\mathcal{L}(V, W)$ is a Banach space, $DDF(p)$ is the differential of $DF$ at $p$. So $DDF : \mathcal{E} \to \mathcal{L}(V, \mathcal{L}(V, W))$. – Mason Jun 03 '24 at 17:20
  • @MoisheKohan, Lang simply uses a Banach space instead of $\mathbb{R}^n$, mostly to allow for infinite dimensional manifolds. But there is really no essential difference. You just need a topological vector space, but that's still essentially a space of coordinates. – Deane Jun 03 '24 at 20:14
  • Concerning the many versions of Lang's book, see https://math.stackexchange.com/questions/3801180/updates-of-serge-lang-differential-manifolds. – KCd Jun 03 '24 at 21:11

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$\newcommand\A{\mathbb{A}}\newcommand\T{\mathbb{T}}$I do like to follow Lang's approach but slightly revised:

First, let $\T$ be a finite dimensional real vector space (viewed as a space of arrows). Then you can define the associated affine space $\A$ (viewed as a space of points) that is assumed to have operations \begin{align*} \A \times \T &\rightarrow \A\\ (a, t) &\mapsto a+t \end{align*} and \begin{align*} \A \times \A &\rightarrow \T\\ (a_1,a_2) &\mapsto a_2-a_1 \end{align*} that satisfy a certain set of axioms. You also need a topology on $\T$ and $\A$. Here, I don't know how to do this without defining a norm, but it is easy to show that the topology is independent of the norm chosen. The norm itself is not needed to do anything below.

You can then view $\A$ as the prototypical manifold and show that there is a natural isomorphism $$ T_a\A \simeq \T$$ for each $a \in \A$. Using the affine structure of $\A$, it is easy to define what smooth functions, maps, and their differentials are on any open subset of $\A$. It is also easy to show that the space of smooth functions is preserved under diffeomorphisms.

The next step is to show that any open subset $O \subset\A$ has the structure of a manifold, where there is a natural isomorphism $$T_pO \simeq \T.$$ I sometimes call $O$ a local manifold. This of course is simply a manifold that has an atlas with only one coordinate map. Since you don't need to worry about transition maps, it is much easier to introduce all of the fundamental objects associated with a manifold and their properties without using coordinates. It is straightfoward to define the tangent and cotangent bundles, the pushforward of a tangent vector, and the pullback of a cotangent vector.

Finally, you introduce the definition of a manifold. Everything done with local manifolds now automatically imply that everything you did with local manifolds extend naturally in a coordinate-free manner to manifolds.

I consider this approach as being analogous to teaching abstract vector spaces in linear algebra. I also liked it, because it makes it clear how many facts about manifolds follow directly from abstract linear algebra. I think this gets lost in most expositions of differential geometry.

But as I comment above, I also think that all I'm doing is hiding the coordinates and they're still really there. I just think that repackaging everything like this highlights the essential properties of the category of manifolds and maps between manifolds. On the other hand, when you need to prove nontrivial statements about manifolds or do calculations, you end up letting $\A = \mathbb{R}^n$ anyway.

Deane
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  • Thanks for the nice answer, I really like that approach. I suppose you obtain a norm on $\mathbb A$ by identifying it with $\mathbb T$ via a bijection $v \mapsto p + v$ for some point $p \in \mathbb A$ and further identifying $\mathbb T$ with $\mathbb R^{\text{dim} \mathbb T}$, and $\mathbb A$ then inherits the standard Euclidean dot product and thus the induced norm? – markusas Jun 07 '24 at 00:53
  • No, there's no norm on $\mathbb{A}$. To keep things abstract, you assume that $\mathbb{T}$ has an inner product. Then this is a flat Riemannian metric on the manifold $\mathbb{A}$. The inner product and induced norm are on $\mathbb{T}$ only, not $\mathbb{A}$. You want to think of $\mathbb{A}$ as a space of points. A point does not have a norm, and two points do not have a dot product. What the inner product on $\mathbb{T}$ does do is define the distance between two points in $\mathbb{A}$ and the angle between two line segments with a common endpoint. – Deane Jun 07 '24 at 03:12