# Curves and tangent vectors in a manifold setting

Consider the following definition: ($M$ denotes a manifold structure, $U$ are subsets of the manifold and $\phi$ the transition functions)

Def: A smooth curve in $M$ is a map $\gamma: I \rightarrow M,$ where $I \subset \mathbb{R}$ is an open interval, such that for any chart $(U,\phi)$, the map $\phi \circ \gamma : I \rightarrow \mathbb{R}^n$ is smooth.

My first question is, why do we define a smooth curve in this way? In particular, why is the map $\phi \circ \gamma$ a good object to consider? The only thing that comes to mind is that now we have a function defined from $\mathbb{R} \rightarrow \mathbb{R}^n$ so differentiation is well defined and thus one may introduce the concept of a tangent vector (as below).

Now let $f: M \rightarrow \mathbb{R}$ be a smooth function on $M$ and $\gamma: I \rightarrow M$, smooth curve as before. Then $f \circ \gamma : I \rightarrow \mathbb{R}$ is smooth. Hence we take a derivative to find the rate of change of $f$ along the curve $\gamma$: $$\frac{d}{dt}f(\gamma(t)) = [(f \circ \phi^{-1}) \circ (\phi \circ \gamma)]'(t) = \sum_{i=1}^n \left(\frac{\partial (f \circ \phi^{-1})}{\partial x^i}\right)_{\phi(\gamma(t))} \frac{d}{dt} x^i(\gamma(t))$$

My next question is to simply understand how this equation comes about. I can see it is some application of the chain rule but I am struggling with the precise details of the equation, mostly in how the final equality comes about and the subscript on the $\partial (f \circ \phi^{-1})/\partial x^i$ term. Many thanks!

On your first question, we want that differentiable manifolds serve as a generalization of surfaces of Euclidean space and of the Euclidean space itself. The diffeomorphism $\phi$ is a differentiable function from $U\subset M$ to $\mathbb{R}^n$ which possesses a differentiable inverse. Roughly speaking, it means that $M$ looks locally the same that $\mathbb{R}^n$. This way, we can define objects on manifolds throughout the Euclidean definition. In this sense, when we make the composition of the diffeomorphism $\phi$ with the curve on the manifold $\gamma$, we obtain a curve on the Euclidean space $\phi\circ\gamma$. Hence, it's natural to define $\gamma$ to be smooth when its associate Euclidean curve is smooth.
• Many thanks for your reply! I think I understand but let me apply it to a simpler case and then see if my thinking is right. Suppose we have some function $Y = f(\mathbf{x}(t))$ which I can write as $Y = f \circ \mathbf{x} \circ t$. Then $$\frac{df(\mathbf{x}(t))}{dt} = \sum_{i=1}^n \left(\frac{\partial f}{\partial x^i}\right)|_{\mathbf{x}(t)} \frac{d x^i(t)}{dt} = \sum_{i=1}^n \left(\frac{\partial f(\mathbf{x}(t))}{\partial x^i(t)}\right) \frac{d x^i(t)}{dt}?$$ – CAF Mar 17 '15 at 18:10
• Exactly, but the $x^i(t)$ notation in the partial derivative is uncommon (frankly, I've never seen it), just $x^i$ is okay. – Mr. K Mar 17 '15 at 18:25