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I'm quite new with manifold. Let $M$ a manifold of dimension $2$ and let $\Sigma\subset M$ a curve (i.e. a sub-manifold of dimension $1$). They say in my lecture : We choose a coordinate system s.t. $\Sigma=\{(x,y)\in \mathbb R^2\mid x=0 \}$.

Q1) I'm not sure how this is possible. How can I prove that such a coordinate system exist ? Alors, as written suggest that $\Sigma\subset \mathbb R^2$, but since $M$ is not $\mathbb R^2$ a priori, it may have a mistakes, no ?

Q2) If indeed there is a mistakes, suppose $M=\mathbb R^2$. Then, how is it possible that $\Sigma=\{(x,y)\mid x=0\}$ since we don't know the curve a priori ?

John
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  • I guess that you mean that you can choose a coordinate system such that it is locally of this form. In general this will not be true. – Severin Schraven Oct 06 '19 at 09:59
  • @SeverinSchraven: Isn't what I said ? 3rd line : "They say in my lecture : We choose a coordinate system s.t. Σ={(,)∈ℝ2∣=0}". Even if it's not true in general, it should be true when $M=\mathbb R^2$, no ? – John Oct 06 '19 at 10:03
  • I don't quite understand your last comment. Consider $$\Sigma = {(x,y)\in \mathbb{R}^2 \ : \ x^2+ y^2 =1}$$ this is submanifold of dimension 1. It is not possible to find a change of coordinates to make it globally of the form you wrote. It is just possible to pick charts such that in every chart is of that form. – Severin Schraven Oct 06 '19 at 10:30
  • @SeverinSchraven: Good point (sorry, I'm quite novice in manifold). And if $\Sigma={h(x)\mid x\in\mathbb R^2}$ ? – John Oct 06 '19 at 10:33
  • What would you assume on $h$? – Severin Schraven Oct 06 '19 at 10:46
  • Let say continuity @SeverinSchraven – John Oct 06 '19 at 10:49
  • I guess the thing you want to consider is $$ \Sigma = { (x, h(x)) } $$ ie the graph of a continuous function. This admits a global chart of your form. And hence every submanifold is locally of this form. – Severin Schraven Oct 06 '19 at 10:56
  • Yes, thank you. Could you please tell me why $\Sigma={(x,h(x))\mid x\in \mathbb R^2}$ has coordinate s.t. $\Sigma={(x,y)\mid x=0}$ ? (when $h$ is continuous) – John Oct 06 '19 at 10:57
  • Because you can consider the map $$ \mathbb{R} \rightarrow \Sigma, \ x \mapsto (x, h(x)) $$ this is a global chart. It is continuous and bijective. The inverse map is $\Sigma \rightarrow \mathbb{R), \ (x,y) \mapsto x $ is continuous as well. – Severin Schraven Oct 06 '19 at 11:07

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