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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?

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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.

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  • $\begingroup$ Exactly what I was looking for, sharp answer. Thx $\endgroup$ – Marsl Jan 26 at 0:21
  • $\begingroup$ Glad to help! $$\phantom{}$$ $\endgroup$ – Ivo Terek Jan 26 at 0:58

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