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It is important in differential geometry that the space of derivations at a point $p\in V$ (where $V$ is a finite-dimensional real vector space) is isomorphic to the space of partial derivatives (or velocity vectors) at $p$. This can be proven using a very weak version of Taylor's theorem, which is the approach taken in Tu's "Intro to Manifolds" and described here.

Meanwhile it has also been described in these posts how Taylor's theorem can be stated in a completely coordinate-free form. Therefore, it is natural to suspect that Tu's proof can also be rewritten in a coordinate-free form. However, I can't see how to actually carry this out.

Tu's proof uses the following version of Taylor's theorem.

$$f(x) = f(p) + \sum (x^i - p^i)g_i(x) \qquad g_i(p) = \frac{\partial f}{\partial x^i}(p)$$

Meanwhile the coordinate-free version of the same theorem would look something like

$$f(x) = f(p) + \tilde{g}(x - p), \qquad \tilde{g}(x - p) = o(\|x - p\|).$$

Tu's approach seems to rely explicitly on the ability to express $\tilde{g}$ as a sum of products $(x^i - p^i)g_i(x),$ and I can't figure out whether this idea (and the remainder of the proof) can be expressed without coordinates.

WillG
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    Since the isomorphism between a vector space and its dual space is non-canonical, i.e. basis-dependent / coordinate-dependent, this is probably not possible. – Smiley1000 Oct 25 '24 at 06:22
  • You can look at the discussion on pp. 58-59 of Lee's Introduction to Smooth Manifolds. – Karthik Kannan Oct 25 '24 at 13:10

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The two versions are not equivalent. In the first version, the first order term is a linear function of $x-p$ and in the second, this is not necessarily so.

You can reformulate the second version to be more similar as follows: Given $p$, there exists a linear function $\ell$ $$ f(x) = f(p) + L(x-p) + o(|x-p|). $$

Also, you need a definition of a manifold that does not use coordinates. As @MoisheKohan has observed, you can simply define a manifold as being locally diffeomorphic to an open subset of a vector space. Here is a coordinate-free proof:

Let $p = 0$ and $O$ a neighborhood of $p$. For each $1$-form on $O$, let $$ Q_\theta(x) = \langle x,\theta(x)\rangle. $$ For each $f \in C^\infty(O)$, observe tht $$ D(Q_{f\theta}) = f(0)D(Q_\theta). $$ It follows that the function $$ v: \theta \mapsto D(Q_\theta) \in \mathbb R $$ is $C^\infty(O)$-linear and therefore $v \in (T^*_0)^* = T_0$. In particular, $$ D(Q_\theta) = \langle v,\theta(0)\rangle. $$

Any $f \in C^\infty(O)$ can be written as $$ f(x) = f(0) + \langle x,\theta(x)\rangle, $$ where $\theta$ is a $1$-form such that $\theta(0) = df(0)$. Then $$ D(f) = D(Q_\theta) = \langle v,\theta(0)\rangle = \langle v,df(0)\rangle. $$

Deane
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    You can define a smooth manifold as a locally compact space modeled on a topological vector space (with smooth transition maps). This is essentially Lang's viewpoint. – Moishe Kohan Oct 29 '24 at 17:59
  • @MoisheKohan, thanks. If I understand correctly, the key step is defining smooth functions, and a topological vector space is a good setting for that. – Deane Oct 29 '24 at 18:02
  • Right. It is all in Lang's book. – Moishe Kohan Oct 29 '24 at 18:05
  • I have it somewhere. – Deane Oct 29 '24 at 18:14
  • @MoisheKohan, then it really is worth writing out a coordinate-free proof. Any chance you want to provide that? – Deane Oct 29 '24 at 18:23
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    Similar arguments can also be found in Abraham and Marsden’s book (chapter 4). – peek-a-boo Oct 29 '24 at 19:13
  • I haven't gone through the second part of this answer yet, but regarding the first part, let me just say that I'm fine with explicitly restricting to finite-dimensional manifolds/vector spaces. My goal is not to extend this theorem to the infinite dimensional setting, but rather to see what it looks like without coordinates. – WillG Oct 29 '24 at 19:35
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    @WillG, yes. I understood that. My point was that it's not straightforward to do it in infinite dimensions so if you want only a finite-dimensional proof, you have to use the assumption that the manifold is finite-dimensional somewhere in the proof. The question is how to do that. When you learn about manifolds, the finite-dimensionality of manifolds is a consequence of the fact that there are only finitely many coordinates. – Deane Oct 29 '24 at 20:36
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    Sorry to ask a follow-up question after such a long delay, but I'm interested in the claim "Any $f \in C^\infty(O)$ can be written as $f(x) = f(0) + \langle x,\theta(x)\rangle,$ where...". I think this is the key fact that singles out finite dimension, and is typically justified using Taylor's theorem to explicitly construct such a 1-form in coordinates. Do you know if there's a slick way to justify this claim without appeal to Taylor's theorem, or at least minimizing reference to coordinates where possible? – WillG Jan 05 '25 at 02:01
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    The standard proof is: \begin{align} f(x) - f(0)& = \int_{t=0}^{t=1} \frac{d}{dt}f(tx),dt\&= \int_{t=0}^{t=1}\langle x,df(tx),dt\&= \langle x,\theta(x)\rangle, \end{align} where $$\theta(x) = \int_{t=0}^{t=1}df(tx),dt. $$ – Deane Jan 05 '25 at 02:55
  • By the way, this is the first step to a proof of Taylor’s theorem. I’ve never seen Taylor’s theorem used to prove this formula. – Deane Jan 05 '25 at 06:31
  • Yes, I was essentially referring to this computation as Taylor's theorem, though this is the first time I have seen it coordinate-free! Regarding the earlier (now deleted) discussion of whether this extends to infinite dimension—does it? I'm afraid I'm still confused about that. – WillG Jan 05 '25 at 18:43
  • In particular, we need a way to define 1-form-valued integration used in $\theta(x) = \int df(tx)dt$, and to show that it is smooth, and to show that the third "=" is justified. In finite dimension, we can do this with coordinates; in infinite dimension, I wonder whether it's still possible (e.g., by defining $\int df(tx)dt$ to be the one-form $\int df(tx)(\cdot)dt$). – WillG Jan 05 '25 at 18:43
  • First, note that it is straightforward to define $$ \int_0^1 F(t),dt$$ if $F: [0,1]\rightarrow V$ is continuous and $V$ is a Banach space. The standard definition of a Riemann integral works. Here, if the domain of $f$ is $V$, then $df: V \rightarrow V^$, where $V^$ is the space of bounded linear functionals. If $df$ is continuous, then so is $df\circ c$, where $c: [0,1]\rightarrow V$ is a continuous curve. – Deane Jan 05 '25 at 18:50
  • Ah, that looks nice—I have to think about this more. I guess I'm so unfamiliar with coordinate-free analysis that I don't have a strong intuition yet for where coordinates are necessary. – WillG Jan 05 '25 at 18:59
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    These links might be helpful: https://math.stackexchange.com/questions/2333438/reference-request-calculus-on-banach-spaces and https://math.stackexchange.com/questions/1057490/riemann-integrals-of-abstract-functions-into-banach-spaces – Deane Jan 05 '25 at 19:02