# Manifold , smooth curve, differentiation

Suppose $M^n$ is a smooth $n$-dimensional manifold, and $\gamma(t):[0,1]\to M^n$ is a smooth curve on it.How do I find the differentiation (derivation) $d\gamma(t)/dt$ so that in particular if $\gamma(t)=x^\alpha(t)$ is the $\alpha$th-coordinate function then $d\gamma(t)/dt=1$? What I do not understand is the fact that $\gamma$ leads to $M^n$ and not to $\mathbb{R}^n$, and I have problems with the precise,technical applications of charts $(U,\phi_U)$ where $U$ is an open set in the manifold $M^n$ and $\phi:U\to \phi(U)\subseteq\mathbb{R}^n$ is a homeomorphism.

• The expression $d\gamma(t)/dt=1$ does not make sense since $d\gamma(t)/dt \in T_{\gamma(t)}M$. Do you mean $dx^\alpha(t)/dt=1$ instead? – Jan Bohr Oct 22 '17 at 17:44
• Yes, I mean $dx^\alpha(t)/dt$ but $\gamma(t)=x^\alpha(t)$ by the second line. Maybe this precisely is my source of confusion? – user122424 Oct 22 '17 at 17:53
• I wish to consider the $\alpha$-th coordinate curve as a curve on the manifold $M$ and take the derivation of it to see that it will be $\partial/\partial x^\alpha$ – user122424 Oct 22 '17 at 17:59

I think you can do that locally in a chart easily by first make such curve in $\mathbb{R}^n$ and then define it to a curve in $M$ by compose the curve with chart map. For example we want to define coordinate curve of $x^1$ of some chart $(U,\varphi)$. Here is the construction :

Choose a chart $(U,\varphi)$ for $M$. Define a curve in $\varphi(U)=\hat{U} \subset \mathbb{R}^n$, $\alpha : I \rightarrow \hat{U}$ by $\alpha(t)=(t,0,\dots,0)$. Assume that $\alpha(I) \subset \hat{U}$. And then define $\gamma : I \rightarrow M$ by $$\gamma := \varphi^{-1} \circ \alpha$$ Therefore you'll have the curve that you want. In this case, the representation of $\gamma$ in this chart is $\hat{\gamma}(t) = (\varphi \circ \gamma )(t)= \alpha (t) = (t,0,\dots, 0)$. And therefore

$$\dot{\gamma}(t) = \frac{\partial}{\partial x^1}\Bigg|_{\gamma(t)}$$

It must be noted that the representation of this coordinate curve is not guaranteed to be the same in another chart $(V,\psi)$.

$\textbf{EDIT}$ :

When we have a curve on manifold $\gamma : I \rightarrow M$, the velocity vector $\dot{\gamma}(t)$ to this curve at some $t_0 \in I$ is $$\dot{\gamma}(t_0) = \frac{d\gamma^i}{dt}(t_0) \frac{\partial}{\partial x^i}\Bigg|_{\gamma(t_0)}$$ with $\gamma^i$ is just the coordinates of the curve in some local chart. That is $$\hat{\gamma} (t) = (\varphi \circ \gamma) (t) = (\gamma^1(t),\cdots, \gamma^n(t))$$ In your case above, $\hat{\gamma} (t) = (\gamma^1(t),\cdots, \gamma^n(t)) =\alpha(t) = (t,0,\cdots,0)$. So any other components of the velocity vector vanish except $\frac{d\gamma^1}{dt} = 1$. Surely you can modified the $\alpha : I \rightarrow \mathbb{R}^n$ such that the curve $\gamma$ is the any $i^{th}$-coordinate curve as originally you want. Look for example, John Lee's Smooth Manifold Ch.3 for the derivation of the velocity vector.

• How follows your $$\dot{\gamma}(t) = \frac{\partial}{\partial x^1}\Bigg|_{\gamma(t)}$$ I.e. how do we know that $\dot{\gamma}(t)$ is the first coordinate field? Only some intuition is needed. – user122424 Oct 23 '17 at 11:46
• @user122424 I'll add the details to the answer. – Sou Oct 23 '17 at 11:48
• Thank you very much. This is of importance for me. – user122424 Oct 23 '17 at 11:50
• Your $$\dot{\gamma}(t_0) = \frac{d\gamma^i}{dt}(t_0) \frac{\partial}{\partial x^i}\Bigg|_{\gamma(t_0)}$$ is exactly the chain rule, is that right? – user122424 Oct 23 '17 at 12:13
• @user122424 Yes${}{}{}$ – Sou Oct 23 '17 at 12:14