# Why velocity vector belongs to the tangent space of a smooth manifold

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

• What is you definition for $\gamma'(t)$? Commented Sep 29, 2021 at 0:22
• Try to cook up a derivation using gamma. Maybe by sending f to the derivative of $f\circ \gamma$ Commented Sep 29, 2021 at 0:36
• 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. Commented Sep 29, 2021 at 4:13
• 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 $\gamma'(t).$ Commented Sep 29, 2021 at 4:16

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

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