0
$\begingroup$

In the book by Frankel The geometry of physics on page 25 there is a definition of coordinate curve as follows: $x^i$(t)=constant for $i\neq \alpha$, and $x^\alpha(t)=t$. I do not understand why both, the argument and value of $x^i$, $t$, lie on $R^1$. I.e. one of them (either the argument of $x^i$ or the value) should lie in an open set $U$ of the manifold $M^n$! Can someone describe the precise what we are "really" dealing with? If you don't have the book, just type

"coordinate curve through a point,the curve being parameterized by"

into google,and follow the first link.

$\endgroup$
0
$\begingroup$

Remember that the coordinates themselves are functions $x^i : U \to \mathbb R$ smoothly parametrizing some open set $U$ of the manifold; i.e. so that taken together they produce a bijection from $U$ to some open subset of $\mathbb R^n$. Thus we can always talk about any local object on the manifold in terms of what it looks like from the point of view of the coordinates just by composing with $x^i$, which will bring us back to something defined on a familiar chunk of $\mathbb R^n$.

When we're talking about a curve $\gamma : \mathbb R \to U$, this idea leads us to deal with $\gamma^i = x^i \circ \gamma : \mathbb R \to \mathbb R^n$, which is a curve in Euclidean space (something we know how to deal with very well!) representing a curve on the manifold. In the case of coordinate curves, the representatives $x^i \circ \gamma$ are just the straight lines parallel to the standard axes, much like you would draw a grid on a map. The actual coordinate curves thus show you how the map $x^i$ wraps this grid on to the manifold.

$\endgroup$
  • $\begingroup$ Oh, could you be more elementary with your explanation then you were? My English is not good enough to understand the last 3 lines of your explanation.Could you be more verbose? Moreover what is wrong exactly with saying that the curve is given just by the inverse to $x^i$ i.e. $(x^i)^{-1}:\mathbb R\to U$? $\endgroup$ – user122424 Oct 11 '17 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.