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For an $n$-dim differentiable manifold, the tangent space is identified with derivations.

Is this still true for infinitely dimensional manifolds?

On $\mathbb{R}^n$, the proof for the proposition that map $\tau:D_v\to \partial_v$ from directional derivative to a derivation is isomorphic relies on Taylor theorem. It seems, if in a Banach space $X$, we can forget the norm or topology for now, and consider its Hamel basis $B$, so that every $x\in X$ is a finite sum $x=\sum_i c_ie_i$ for $e_i\in B$. Then the Taylor theorem approach seems to still work. Then, once we get the conclusion that map $\tau$ is linearly isomorphic, a norm/topology on the space of derivations (tangent space) can be induced. Is that correct?

user760
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    The answer lies in availability of some form of Taylor's theorem, which probably depends on what is your model infinite-dimensional space is. You should include your definition of an infinite-dimensional manifold in your question. – Moishe Kohan Oct 20 '25 at 14:29
  • @MoisheKohan I just realized with a Hamel basis the coefficients might not be continuous, so does that mean we should at least restrict to those Banach spaces with Schauder basis? – user760 Oct 21 '25 at 03:58
  • This is your question, you decide on how to formulate it. Personally, I do not see why any basis is relevant here. There are several books on analysis on infinite-dimensional Banach spaces and topological vector spaces. Did you consult any of these textbooks regarding Taylor's theorem? – Moishe Kohan Oct 21 '25 at 08:56
  • Even on MSE, Taylor's formula for Banach spaces (with a description of the remainder) was discussed a number of times, for instance, here. – Moishe Kohan Oct 21 '25 at 09:56
  • @MoisheKohan The problem is, we can have $f(x)=f(p)+g(x)(p-x)$, where $g(x):=\int_0^1f'[p+t(p-x)]dt$ is smooth. But a derivation $\partial$ is a linear map from $C^\infty(U)$ to $\mathbb{R}$, and so $\partial (f(x))=0+\partial\left(g(x)(p-x)\right)$ cannot be further simplified, unless $g(x)(p-x)$ can be decomposed into sum of coefficients. Because neither $g$ nor $p-x$ is real valued. That's why I thought of basis. – user760 Oct 21 '25 at 13:51
  • @MoisheKohan For infinite dimensional manifolds, I've been reading Lang's Fundamentals of Differential Geometry, where he defines tangent vectors as equivalence classes of triples $(U, \varphi, v)$, not as derivations. Hence my question. – user760 Oct 21 '25 at 13:51
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    At the very least you will need some continuity assumption on derivations, as well as reflexivity assumption on your model Banach space (otherwise there will be counter-examples). – Moishe Kohan Oct 21 '25 at 18:29

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