I am a beginner at Manifolds. I tried the following problem, but not sure whether I got it correct not:
Find a differential 1-form on $S^1$
My attempt:
Define $f: S^1\to \mathbb{R}$ given by $f(x,y)=x$. Then definfe $df:S^1 \to \cup T_p^{*}S^1$ as, for any $p\in S^1$, $(df)_p:T_p S^1 \to \mathbb{R}$. Since $T_pS^1$ is a one dimensional vector space so take basis for it as $\{\frac{\partial}{\partial x}|_p \}$. Now finally,
$(df)_p(\frac{\partial}{\partial x})=\frac{\partial}{\partial x}f=\frac{\partial}{\partial x}x=1$.
Hence the differential one form on $S^1$ is $df$. Am I making sense or do my arguments have some errors?
Thanks for any insights!!
You have to be careful with local coordinates: they are indeed local so the expression you will get for $df$ depends on the given chart. I think some of your confusion stems from the fact that you are trying to use the ambient coordinates $(x,y)$ on $\mathbb R^2$ as local coordinates on $S^1$, as opposed to the usual approach of introducing an angular coordinate $\theta$ which reduces the amount of charts needed. Specifically, in order to make sense of your approach we need four charts
$U_x^+ = \{(x,y) \in S^1 : x > 0\}$, $U_x^- = \{(x,y) \in S^1 : x < 0\}$, $U_y^+ = \{(x,y) \in S^1 : y > 0\}$, $U_y^- = \{(x,y) \in S^1 : y < 0\}$,
as opposed to needing only two charts if we were working with an angular coordinate. For concreteness suppose that we look at the chart $\phi:U_y^+ \to \mathbb{R}$ which flattens down the northern hemisphere, $\phi(x,y) = x$, as well as the chart $\psi:U^+_x \to \mathbb R$ which flattens the right side of the circle, $\psi(x,y) = y$. The transition maps look like $\phi \psi^{-1}(y) = x = \sqrt{1-y^2}$. So your $df$, which is just $dx$ in the first chart, takes the form $d(\sqrt{1-y^2}) = \frac{-y}{\sqrt{1-y^2}} dy$ in the second chart.
In a similar vein you should be careful to note that the tangent vectors to $S^1$ denoted by $\frac{\partial}{\partial x}\rvert_p$ in these coordinates are not the usual horizontal tangent vectors in $\mathbb R^2$. If we denote by $\iota:S^1 \to \mathbb R^2$ the inclusion then for instance $\iota \phi^{-1}(x) = (x, \sqrt{1-x^2})$ so that $\iota_*(\frac{\partial}{\partial x}\rvert_p) = \left(1,\frac{-x}{\sqrt{1-x^2}}\right)$ (which is an actual tangent vector to the unit sphere).