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Consider a compact interval $[a,b]$. If $[a,b]$ had global coordinates, then there would be an homeomorphism $f:[a,b]\to U$ where $U$ is an open subset of $\mathbb{R}$ or an open subset of $[0,\infty)$.

But $U$ cannot be open in $\mathbb{R}$ because $U$ is compact, so $U$ must be open in $[0,\infty)$. But also in this case I think U must be a closed interval, precisely $U=$[min$f$,max$f]$. But this set cannot be open in $[0,\infty)$. So i deduce that $[a,b]$ does not have global coordinates.

But then in John Lee's book Introduction to smooth manifolds i read a statement of the form "Let $t$ denote the stadard coordinate on $[a,b]$".

So how should I interpret the statement above?

Minato
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When I refer to the "standard coordinate" on a subset of $\mathbb R$ such as $[a,b]$, I just mean the standard coordinate of $\mathbb R$ restricted to the interval. You're right that it's not a "coordinate chart" in the sense that the term is defined for manifolds with boundary. Perhaps it's infelicitous to use the same term with two different meanings, but it would be far from the only time this happens in differential geometry!

Jack Lee
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    Thank you very much! So when you say $\omega$ is a smooth covector field on $[a,b]$, and then you write $w=f;dt$ with "$t$ standard coordinate on $[a,b]$", I have to think at $\omega$ as a smooth covector field defined on an open subset (or maybe interval) of $\mathbb{R}$, say $J$, containing $[a,b]$ and at $t$ as the standard coordinate on $J$? So in this way I have $\omega:J\to T^J$ smooth covector field on $J$ and I can write $\omega=f;dt$ with $f:J\to \mathbb{R}$ smooth, beacuse in this case I have that $t$ is a "true"* coordinate chart on $J$. Am I correct? – Minato Mar 01 '19 at 09:37
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    Beacuse if we think at $\omega$ as "literally" a smooth covector field on $[a,b]$, in the sense $\omega:[a,b]\to T^[a,b]$, then since we don't have smooth global coordinates (in the sense of a smooth global chart) on $[a,b]$ we cannot* write $w=f; dt$ on all $[a,b]$. But I should write, for example, $\omega=f; dt$ on $(a,b]$ with $t$ smooth chart on $(a,b]$ and $f:(a,b]\to \mathbb{R}$ smooth ; and $\omega=g;ds$ on $[a,b)$ with $s$ smooth chart on $[a,b)$ and $g:[a,b)\to \mathbb{R}$ smooth. Am I correct? – Minato Mar 01 '19 at 09:38
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    Because if we have $\omega:M\to T^M$ smooth covecor field on a smooth manifold $M$ with or without boundary, and $(U,\phi=(x^i))$ is a chart on $M$, then we have $\omega=\omega_i;dx^i$ only* in the domain of the chart, that is $U$.

    Please excuse me if I'm wandering.

     – Minato Mar 01 '19 at 09:39
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    Your first comment is essentially correct. The fact that $\omega$ is a smooth covector field on $[a,b]$ implies that it has an extension to a smooth covector field on an open subset of $\mathbb R$, which can be written in standard coordinates as $f,dt$. The original form $\omega$ can therefore be written as $f,dt$ on $[a,b]$. Here you can just consider $t$ as a smooth function on $[a,b]$, and $dt$ is a smooth covector field there. – Jack Lee Mar 02 '19 at 19:52
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    Please allow me one last question: now I have realized in what sense we intend the sentence " $\omega$ is a smooth covector field on $[a,b]$"! It is in the sense of Lemma 10.12, right? So we have $\pi: T^\mathbb{R}\to \mathbb{R}$ and $\omega:[a,b]\to T^\mathbb{R}$ is our smooth covector field in the sense that it admits an extension to a smooth local section of $\pi$ in a neighborhood of each point, and thus by Extension Lemma (Lemma 10.12) we have that $\omega$ is the restriction to $[a,b]$ of a smooth convector field on all of $\mathbb{R}$, right? – Minato Mar 06 '19 at 16:33
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    Yes, that's a good way to think about it. – Jack Lee Mar 06 '19 at 16:44
  • @JackLee Can you answer your competitor's example on $[a,b]$ without using the classification theorem in your book? –  Mar 11 '19 at 11:44