# Differential map of velocity vector

This is a very basic differential geometry question (please be patient, I am learning) I am given the definition of the differential map of $$\phi:M \to N$$ as

$$d\phi_p(v)(g)=v(g\circ\phi)$$

where $$v\in T_pM$$ is some vector and $$g$$ is a smooth function on $$N$$ and I also know about the chain rule for these. (I suppose, I need to use it but cannot see how just yet)

Now, we have defined tangent vectors as derivations and only later I learned that a velocity vector to a curve at a particular point of the manifold lies in the tangent space to that point. Let the curve be $$\alpha(t)$$ and its tangent vector $$\alpha '(t)$$. Now, its just stated that when pushing forward the velocity vector, I should do it like this

$$d \phi (\alpha'(t))=(\phi\circ\alpha)'(t)$$

But I cannot derive, why this is the right way and in particular I am also buzzed by the fact that I do not have to care about a particular point anymore. This pushforward seems to work fine for every $$t\in \mathbb{R}$$ whereas above I had to specify the point $$p$$ and my reference says explicitly: The differential map of $$\phi: M \to N$$ moves individual tangent vectors from $$M$$ to $$N$$ , but in general provides no way to move vector fields from $$M$$ to $$N$$ (or the reverse).

So why does it work for the entire tangents to the curve?

If $$\alpha:I \to M$$ is a curve defined in an open interval, then $$I$$ itself is a smooth manifold with a global chart $$t\colon I \to \Bbb R$$, and so by definition we have $$\alpha'(t) \doteq {\rm d}\alpha_t\left(\frac{\partial}{\partial t}\bigg|_t\right) \in T_{\alpha(t)}M.$$Here, under the identification $$T_tI \cong \Bbb R$$, the coordinate vector $$\partial/\partial t|_t$$ corresponds to the number $$1$$. This being understood, if $$\phi\colon M \to N$$ is smooth, we have that $$\phi\circ\alpha\colon I \to N$$ is a curve, and the above definition applied this time for $$\phi\circ \alpha$$ gives $$(\phi\circ\alpha)'(t) = {\rm d}(\phi\circ\alpha)_t\left(\frac{\partial}{\partial t}\bigg|_t\right) = {\rm d}\phi_{\alpha(t)} \circ {\rm d}\alpha_t\left(\frac{\partial}{\partial t}\bigg|_t\right) = {\rm d}\phi_{\alpha(t)}(\alpha'(t))\in T_{\phi(\alpha(t))}N,$$where in the second equal sign above we use the chain rule, and in the last one the definition of $$\alpha'(t)$$ again. If what bothers you is writing $${\rm d}\phi(\alpha'(t))$$ without indicating the base point $$\alpha'(t)$$, the reason for this is that it is not actually necessary to write it, because despite being a mild abuse of notation, there is no other possibility for base point since you know that $$\alpha'(t)\in T_{\alpha(t)}M$$ and that distinct tangent spaces are disjoint.
• Glad to help! $$\phantom{}$$ – Ivo Terek Jan 26 at 0:58