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I want to understand something about defining tangent space with classes of equivalences of curves. I have the following definition:

A differentiable curve passing through point $x \in M$ (where $M$ is a manifold) is a function $\gamma : (-\epsilon, \epsilon) \to M$ with $\gamma(0) = x$ which is differentiable.

Now, let $\mathcal{C}(x)$ be the set of all curves passing through a point $x \in M$.

Let $\mathcal{F}_x$ to be the set of real-valued functions that are differentiable on open neighbourhoods of $x$. So, if $f_1,f_2:U_i\to \mathbb{R}$ are 2 functions from $\mathcal{F}_x$, then we can define $$f_1+f_2 = f_1\vert_{U_1\cap U_2} + f_2\vert_{U_1 \cap U_2}$$ $$f_1 \cdot f_2=f_1\vert_{U_1 \cap U_2} \cdot f_2\vert_{U_1 \cap U_2}$$

Then, we say that, if $\gamma_1, \gamma_2 \in \mathcal{C}(x)$, then $\gamma_1 \sim \gamma_2 \iff \frac{d(f \circ \gamma_1)}{dt}\vert_{t=0} = \frac{d(f \circ \gamma_2)}{dt}\vert_{t=0}$ for each $f \in \mathcal{F}(x)$.

But I don't understand why functions from $\mathcal{F}(x)$ needs to be real-valued for this to work. What is $f$ and what is the intuition behind this definition. It can't be $\mathbb{R}^n$ instead of $\mathbb{R}$? Thanks!

ProofSeeker
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  • not necessary. the point is that it suffices to ‘test against’ real-valued functions. you get the same equivalence relation if you replace the target space with a more general real Banach space $V$. – peek-a-boo Oct 05 '24 at 19:39
  • Don't you normally think of a manifold as embedded in some (high dimensional) $\Bbb{R}^n$? Shouldn't the tangents at a point be some (real) affine subspace of that $\Bbb{R}^n$? (There is a notion of complex manifold for which embedding into $\Bbb{C}^n$ is rare and for which the set of holomorphic functions (instead of real functions) is the sensible component of the definition of tangent space.) – Eric Towers Oct 05 '24 at 19:42
  • And what's the interpretation of this equivalence relation? I read is some books that it's equivalent to vectors having same velocity. but velocity of vector would be some sort of $\frac{d\gamma}{dt}\vert_{t=0}$ not $\frac{d(f \circ \gamma)}{dt}\vert_{t=0}$ – ProofSeeker Oct 05 '24 at 19:56
  • This is equivalent to having for one (or equivalently for every) chart $(U,\alpha)$ around the point $x$, $(\alpha\circ\gamma_i)’(0)\in\Bbb{R}^n$ are equal. So, this equivalence relation is saying in every chart, the velocity vectors are equal. See this question of mine and the corresponding section of Loomis and Sternberg for slightly more elaboration. – peek-a-boo Oct 05 '24 at 21:08

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