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I am currently reading through the section in Loomis and Sternberg's Advanced Calculus (Revised Edition) on tangent spaces, but I'm having trouble justifying one step of the argument (on pp. 373-374, shown below).

Here are the definitions and notations used by them. Let $M$ be a differentiable manifold (here they model their manifolds on a Banach space $V$ rather than some $\mathbb{R}^n$). Let $x \in M$, and let $\varphi : I \to M$ be a differentiable map where $I$ is an interval in $\mathbb{R}$ containing $0$, and $\varphi(0) = x.$ Then, they define an operator $D_{\varphi}: C^{\infty}(M) \to \mathbb{R}$ by $D_{\varphi}(f) = (f \circ \varphi)'(0)$. Next, they define an equivalence relation on all the curves passing through $x$ by $\varphi \sim \psi$ if and only if $D_{\varphi} = D_{\psi}$, and call an equivalence class of curves, $\xi$ to be a tangent vector at $x$.

So far so good. Next, they go to a chart $(W, \alpha)$, and the underlined section below is what I don't fully understand. I get that $\varphi \sim \psi$ if and only if for every $f \in C^{\infty}(M)$, $d(f \circ \alpha^{-1})_{\alpha(x)}((\alpha \circ \varphi)'(0))$ $= d(f \circ \alpha^{-1})_{\alpha(x)}((\alpha \circ \psi)'(0))$. But I don't see how to conclude from here that the two vectors $(\alpha \circ \varphi)'(0)$ and $(\alpha \circ \psi)'(0)$ are equal.

I'm guessing it has something to do with the fact that the two derivatives are equal for every $f$; if we somehow choose an $f$ such that the differential $d(f \circ \alpha^{-1})_{\alpha(x)} : V \to \mathbb{R}$ is injective, then its kernel is $\{0\}$, and thus equality follows. However I doubt this is always possible, since in general $\dim(V) > 1$, so a linear map from $V$ to $\mathbb{R}$ cannot be injective.

Any help justifying this step is much appreciated.

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Alp Uzman
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peek-a-boo
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2 Answers2

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The differential does not have to be injective. Note that for any vector space $W$, $\phi(v) = 0$ for any linear functional $\phi$ on $W$ implies that $v = 0$ necessarily. Therefore, since $\mathrm{d}F$ is always a linear functional regardless of the choice of $f$, you simply need to show that you can obtain any linear functional on $V$ with an appropriate choice of $f$.

Kurtland Chua
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For the sake of closure, here are some further details (though probably by now this question is resolved).

First note that by the definition of the $C^\infty$ (hence $C^1$) structure on $M$ the map $C^1(M;\mathbb{R})\to V^\ast$, $f\mapsto (f\circ\alpha^{-1})'(\alpha(x))$ is onto, where $V$ is the Banach space that is the local model for $M$. (Of course the map is not injective, as functions that have "contact of order $1$" at $x$ will produce the same topolinear functionals). As a consequence of the Hahn-Banach theorem $V^\ast$ separates points of $V$ (see e.g. When does $X^\star$ separate the points of $X$?), from which the statement follows.

It is interesting to note two points:

  • The same argument allows one to construct the tangent bundle of a $C^1$ manifold with a locally convex TVS local model.
  • This geometric method of considering velocities of curves is robust with low regularity. On the other hand for instance the definition of the tangent spaces through derivations does not work the way it is expected to work with low regularity; see the discussion in $C^{k}$-manifolds: how and why?.
Alp Uzman
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