# Coordinate curve, manifold, technical detail

I am reading "The Geometry of Physics: An Introduction" by Frankel, and am confused by the following construction in section 1.3a (perhaps available on Google Books using the Google query "from now on, we shall make no distinction between a vector"):

the $\alpha$-th coordinate curve is defined by $x^i(t)$=constant for $i\neq \alpha$ and $x^\alpha(t)=t$.

I do not understand this definition formally since the coordinates $x^\alpha$ are not functions of $t$!

Rather they are homeomorphisms like this

$x_U(U\cap V)\xrightarrow{{x_U^{-1}}}M^n\xrightarrow{{x_V}}\mathbb{R}$

for some open sets $U$ and $V$ in the manifold $M$ and $(x_U,U)$ is a chart.. How can I make $x_U$ to be a function of a real parameter $t$?

I'm not sure what you mean by $M^n$ or $x_U$/$x_V$: a curve is just a (sufficiently smooth) function $\gamma(t):[a,b]\to M$.
If $U\subset \mathbb{R}^n$ and $\phi:U\to M$ is a chart of $M$, with $p$ in the interior of $\operatorname{im} \phi$, then you can define the following curve:
$$\gamma(t) = \phi\left[\phi^{-1}(p) + t \mathbf{v}\right]$$ where the components of the vector $\mathbf{v}\in\mathbb{R}^n$ are $$\mathbf{v}^i = \begin{cases} 0, & i\neq \alpha\\ 1, & i=\alpha.\end{cases}$$ This curve is well-defined on a sufficiently small interval containing $t=0.$
• And how do I compute the derivative of this function $\gamma$:$d \gamma(t)/dt$ ?With what formulas or rules? The target of $\gamma$ is not $\mathbb{R}^n$ – user122424 Oct 22 '17 at 14:38