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The following is Problem 1-8 in Lee's Introduction to Smooth Manifolds, 2nd Edition:

By identifying $\mathbb R^2$ with $\mathbb C$, we can think of the unit circle $\mathbb S^1$ as a subset of the complex plane. An angle function on a subset $U \subset \mathbb S^1$ is a continuous function $\theta: U \to \mathbb R$ such that $e^{i \theta(z)} = z$. Show that there exists an angle function $\theta$ on an open subset $U$ of $\mathbb S^1$ if and only if $U \neq \mathbb S^1$ For any such angle function, show that $(U, \theta)$ is a smooth coordinate chart for $\mathbb S^1$ with its standard smooth structure.

My question regards the last part: showing that $(U, \theta)$ is a coordinate chart compatible with the standard smooth structure of $\mathbb S^1$.

I have no clue where to start. Any hints will be the most appreciated.

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    $\begingroup$ Do you know if the standard smooth structure given by the stereographic projection(s) or is it given abstractly with this ? Great book by the way 😄 $\endgroup$ Aug 9 '20 at 23:19
  • $\begingroup$ @MaximilianJanisch its the one given by the stereographic projection :) $\endgroup$ Aug 11 '20 at 14:06
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Ok, so Lee defines the coordinate charts as follows (this is in Example 1.2). First, let $U_i^+ = \{(x^1,x^2)\in \mathbb{S}^1\mid x^i > 0\}$, and similarly $U_i^- = \{(x^1,x^2)\in \mathbb{S}^1\mid x^i < 0\}$. Now define $\varphi_i^+:U_i^+\to\mathbb{R}$ as $\varphi_i(x^1,x^2) = x^i$ and $\varphi_i^-:U_i^-\to\mathbb{R}$ as $\varphi_i(x^1,x^2) = x^i$. Our atlas is therefore $\{U_i^\pm, \varphi^\pm\}$. Again, all of this is from Example 1.2.

Now, we are asked to show that for $U$ an open subset of $\mathbb{S}$ and $(U,\theta:U\to \mathbb{R})$ is also a smooth coordinate chart. This is also a chart if it is a homeomorphism, and if it is compatible with the atlas we already have. In otherwords, we want to show that $\varphi_i^\pm\circ\theta^{-1}$ and $\theta\circ (\varphi_i^\pm)^{-1}$ are smooth.

To do this, try writing out what $\theta$ and $\theta^{-1}$ are in terms of coordinates. Try composing them with $\varphi_i^\pm$ or its inverse. Also remember that $\theta\circ (\varphi_i^\pm)^{-1} = (\varphi_i^\pm\circ\theta^{-1})^{-1}$, so it's enough to show that one of them is smooth, and has a smooth inverse. Does this help?

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  • $\begingroup$ Thank you for the answer. This is what I had in mind, but my difficulty is that we don't know how $\theta$ looks like, do we? In the first part we indeed construct one $\theta$ as $\theta(e^{it}) = t$, but is this the only $\theta$? $\endgroup$ Aug 10 '20 at 12:56
  • $\begingroup$ Also, how do we know that $\theta$ is an homeomorphism? $\endgroup$ Aug 10 '20 at 13:23
  • $\begingroup$ @DaniloGregorin, you are correct that $\theta$ isn't uniquely defined as $\theta^{-1}(\theta(t)+2\pi i) = \theta(t)$. However, adding a multiple of $2\pi$ doesn't affect whether or not the map is a homeomorphism or smooth. $\theta$ is a continuous as the preimage of an interval is an open arc. It is open as an open arc maps to an open interval, and $\theta$ is a bijection, so it is a homeomorphism. Lastly, $\theta^{-1}(t) = (\cos(t),\sin(t))$. Try using that if you are having trouble looking at $\theta$ directly. $\endgroup$
    – memerson
    Aug 10 '20 at 17:17
  • $\begingroup$ The first equation in your comment shouldn't be $\theta(\theta^{-1}(t) + 2\pi i) = t$? $\endgroup$ Aug 11 '20 at 14:07
  • $\begingroup$ But suppose we are constrained to only "one turn" of $\mathbb S^1$, in such a way that we can't add multiples of $2\pi$. Then $\theta(e^{it}) = t$ is the only angle function? Is this what you are saying? $\endgroup$ Aug 11 '20 at 14:10

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