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Let M be a smooth manifold. Let $ J=[0,1] $ If $ \gamma : J \to M $ is a smooth curve, then for each $ t \in J $, the velocity vector $ \gamma '(t) \in T_{\gamma(t)} M$

I am self studying smooth manifolds and came across this statement. I am aware that each tangent vector in the tangent space of a manifold can be identified with a derivation; namely the directional derivative at a particular $ p \in M $

I cannot see how $ \gamma ' (t) $ can be written using the basis vectors of the tangent space

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  • $\begingroup$ What is you definition for $\gamma'(t)$? $\endgroup$
    – qualcuno
    Commented Sep 29, 2021 at 0:22
  • $\begingroup$ Try to cook up a derivation using gamma. Maybe by sending f to the derivative of $f\circ \gamma$ $\endgroup$ Commented Sep 29, 2021 at 0:36
  • $\begingroup$ I think you are looking at it the wrong way. The tangent space exists independently of your knowledge of its basis vectors. You can use $\gamma'(t)$ itself as your first basis vector, and there you have it. $\endgroup$
    – David K
    Commented Sep 29, 2021 at 4:13
  • $\begingroup$ On the other hand, to get a directional derivative you need to identify the vector first. So that is not a way to find the vector $\gamma'(t).$ $\endgroup$
    – David K
    Commented Sep 29, 2021 at 4:16

1 Answer 1

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One possible definition for $\gamma'$ is $\gamma = {d\gamma} \colon T(a,b) \to TM$, $d\gamma(v)(f) = v(f\gamma)$.

However, we may identify $T(a,b)$ with $(a,b)$ as $t \mapsto \frac{d}{dx}|_t$ and define $\gamma'(t) = d\gamma\left(\frac{d}{dx}|_{x=t}\right)$.

By definition, this is a derivation on $T_{\gamma(t)}M$ that sends a smooth function $f \colon M \to \mathbb{R}$ to

$$ \frac{d}{dx}|_{x=t}(f\gamma) = (f\gamma)'(t). $$

In particular, if $\varphi$ is a chart of $M$ around $\gamma(t)$, there are unique $b_1,\ldots,b_n \in \mathbb{R}$ such that $\gamma(t)' = \sum_i b_i \frac{\partial}{\partial \varphi^i}|_{\gamma(t)}$.

Moreover, evaluating at each coordinate function $\varphi^i \colon M \to \mathbb R$ we see that

$$ b_i = \gamma'(t)(b_i) = (\varphi^i \gamma)'(t) $$

and thus

$$ \gamma'(t) = \sum_{i=1}^n (\varphi^i \gamma)'(t) \frac{\partial}{\partial \varphi^i}\Big|_{\gamma(t)}. $$

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