# Lee's Smooth Manifolds Problem 1-8 - Angle function is a smooth coordinate chart.

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.

• 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 😄 Aug 9 '20 at 23:19
• @MaximilianJanisch its the one given by the stereographic projection :) Aug 11 '20 at 14:06

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?
• 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 $\theta$? Aug 10 '20 at 12:56
• Also, how do we know that $\theta$ is an homeomorphism? Aug 10 '20 at 13:23
• @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. Aug 10 '20 at 17:17
• The first equation in your comment shouldn't be $\theta(\theta^{-1}(t) + 2\pi i) = t$? Aug 11 '20 at 14:07
• 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? Aug 11 '20 at 14:10