- #1
kiuhnm
- 66
- 1
Let ##M## be an ##n##-dimensional (smooth) manifold and ##(U,\phi)## a chart for it. Then ##\phi## is a function from an open of ##M## to an open of ##\mathbb{R}^n##. The book I'm reading claims that coordinates, say, ##x^1,\ldots,x^n## are not really functions from ##U## to ##\mathbb{R}##, but from ##\mathbb{R}^n## to ##\mathbb{R}##. In other words, we should write $$
\phi(p) = (x^1(\phi(p)), \ldots, x^n(\phi(p)))
$$ instead of $$
\phi(p) = (x^1(p), \ldots, x^n(p)).
$$ That doesn't sound right to me. ##\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}##, which means that an element of ##\mathbb{R}^n## is already an ##n##-tuple, so why should we define an extra coordinate system on it? If we want a different coordinate system we just pick a different chart and not the same chart with a different coordinate system for ##\mathbb{R}^n##.
Do you agree or am I missing something?
\phi(p) = (x^1(\phi(p)), \ldots, x^n(\phi(p)))
$$ instead of $$
\phi(p) = (x^1(p), \ldots, x^n(p)).
$$ That doesn't sound right to me. ##\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}##, which means that an element of ##\mathbb{R}^n## is already an ##n##-tuple, so why should we define an extra coordinate system on it? If we want a different coordinate system we just pick a different chart and not the same chart with a different coordinate system for ##\mathbb{R}^n##.
Do you agree or am I missing something?