4

Here is the problem 1-8 in Lee's introduction to smooth manifolds:

An angle function on a subset $U \subset S^1$ is a continuous function $ \theta: U\to \mathbb R$ such that $e^{i\theta(z)}=z$ for all $z \in U$. Show that there exists an angle function on an open subset $U\subset S^1$ if and only if $U \neq S^1.$ For any such angle function, show that $(U,\theta)$ is a smooth coordinate chart for $S^1$ with its standard smooth structure.

I wonder if $U\ne S^1$ is sufficient for the existence of such a continuous angle function(I feel that the statement should be "there exists an angle function on an open subset $U\subset S^1$ if and only if $U$ is not dense in $S^1.$").

Let's take a special case when $U=S^1-\{-1\}$. Suppose there is such a continuous angle function $ \theta: U\to \mathbb R$ such that $e^{i\theta(z)}=z$ for all $z \in U$. Then we must have $\theta(z)=arg(z)+2k\pi$. However, It seems that there will always be a "jump" somewhere on the circle making $\theta$ discontinuous. For example, if we define $\theta(z)=\arg(z)$, then when $z$ approaches $-1$ from the first quadrant and the four quadrant separately, $\theta(z)$ approaches to $\pi$ or $-\pi$.(This argument turns out to be wrong, see the answers and comments given below by repliers)

Since professor Lee doesn't make any corrections to this problem, it should be correct as given. But how to define such a continuous angle function for arbitrary proper open subset of $S^1$? Thanks in advance!

No One
  • 7,939

2 Answers2

4

Let $f: \mathbb{R} \rightarrow S^1; \quad x \mapsto e^{ix}$.

If there exists such an angle function, then $U$ can't be $S^1$: if it were, we would have that $f \circ \theta =Id_{S^1},$ hence $\theta$ is an injection. Since it is a continuous injection in a compact set, it is a homeomorphism with its image (which is a closed interval by preservation of connectedness and compactness), but this can't happen (why?).

If $U$ is not $S^1$, there exists a point which is not in $U$, let it be $p$. Consider $f^{-1}(\{p\})$. This is a set of equidistant points, with distance $2\pi$. Take a point $a \in f^{-1}(p)$, and consider $f|_{(a,a+2\pi)}$ . $f$ is easily verified to be a homeomorphism* with $S^{1} \backslash \{p\}$. Take now $\theta:=f|_{(a,a+2\pi)}^{-1}$. We have that $\theta|_U$ satisfies what you want.

*One can argue directly, by computing what is $f$ and $f^{-1}$, but also can argue indirectly as follows: $f|_{(a,a+2\pi)}$ is continuous, since it is the restriction of a continuous function. By considering an exhaustion of $(a,a+2\pi)$ by compact sets, it is clear that the restriction of $f$ to any of those compact sets is a homeomorphism. Since the images of those compact sets will also create an exhaustion of the image, $f|_{(a,a+2\pi)}$ must be a homeomorphism.

OBS: By exhaustion, I mean a sequence of compact sets such that $K_n \subset \text{int} K_{n+1}$.

Proving the lemma:

Let $X,Y$ be topological spaces, where $Y$ is Hausdorff, such that $X$ has an exhaustion by compact sets $K_n$, and let $F:X\rightarrow Y$ be a continuous bijective map such that $F(K_n)$'s exhaust $Y$. Then $F$ is a homeomorphism.

Proof:

$F$ is by hypothesis continuous, and being injective on compact sets (the codomain is Hausdorff), each $F|_{K_n}$ is a homeomorphism with its image. We will prove that $G:=F^{-1}$ is continuous.

Take $y \in Y$, together with a open neighbourhood $W$ of $G(y)$. Since $K'_n:=F(K_n)$ exhaust $Y$, there exists a $K'_N$ such that $y \in K'_N$. By hypothesis, $y \in \text{int}K'_{N+1}$. Due to our previous observation, $G|_{K'_{N+1}}$ is a homeomorphism. Therefore, $G|_{\text{int}K'_{N+1}}$ is continuous (in fact, a homeomorphism). Hence, there exists an open neighbourhood $V$ of $y$ such that $G(V) \subset W$. But a neighbourhood in the induced topology of an open set must be an open set in the topology itself. Hence, we found an open neighbourhood of $y$ which is taken inside $W$. This proves continuity of $G$.

Aloizio Macedo
  • 34,292
  • And by your definition, $\theta(e^{ix}):=f|_{(0,0+2\pi)}^{-1}(e^{ix})=x$. But $\theta(e^{i\frac{1}{n}})\to \frac{1}{n}$ while $\theta(e^{i (2\pi - \frac{1}{n})})\to 2\pi- \frac{1}{n}$. We need $\theta(z)$ to be continuous. – No One Mar 26 '16 at 03:19
  • By exhaustion, I mean that $K_n \subset \text{int}K_{n+1}$. Note that your counterexample does not satisfy this. – Aloizio Macedo Mar 26 '16 at 03:29
  • And in your second comment, the sequence $e^{i \frac{1}{n}}$ does not converge, nor does the other one, since $(1,0)$ is not on the domain (I assume you are talking about this case). So, there is no issue. – Aloizio Macedo Mar 26 '16 at 03:34
  • Sorry, could you be more explicit? I didn't understand what you are saying. – Aloizio Macedo Mar 26 '16 at 03:47
  • Note the part "Since the images of those compact sets will also create an exhaustion of the image". The problem in your case is on the image. – Aloizio Macedo Mar 26 '16 at 03:58
  • You mean $F(K_n)={ 0,1,1/2,\dots, 1/n }$ doesn't exhaust the image ${0,1,1/2,\dots }$? – No One Mar 26 '16 at 04:03
  • Yes. Note that $0 \notin \text{int}F(K_1)$, but $0 \in F(K_0)$. – Aloizio Macedo Mar 26 '16 at 04:08
  • It seems that you are using such a lemma in the last paragraph: Let$X,Y$ be topological spaces such that $X$ has a exhaustion by compact sets $K_n$, and let $F:X \to Y$ be a continuous map such that $F(K_n)$'s exhaust $Y$, if each $F|K_n$ is a homeomorphism, then $F$ is a homeomorphism. Could you prove it or show me some reference for this in your solution? Thank you! – No One Mar 26 '16 at 04:23
  • Let me restate my example here. I wonder if that argument works even when we assume that $K_n \subset \text{int}K_{n+1}$. Let $F:{0,1,2,\dots} \to {0,1,1/2,\dots}$ such that $F(0)=0, F(n)=1/n$. Let $K_n={ 0,1,2,\dots, n }$, then $K_n \subset \text{int}K_{n+1}$ because ${ 0,1,2,\dots }$ is discrete. The restriction of $F$ to $K_n$ is a homeomophism from $K_n={ 0,1,2,\dots, n }$ to ${ 0,1,1/2,\dots, 1/n }$, but ${ 0,1,2,\dots }$ and ${ 0,1,1/2,\dots }$ are not homeomorphic(The former one is noncompact). – No One Mar 26 '16 at 04:32
  • I've edited and added the lemma with some differences from your enunciation. Please tell me if any issues arise. – Aloizio Macedo Mar 26 '16 at 04:36
  • What would be an explicit expression of $f^{-1}$? Would we have to resort to working with $\operatorname{atan2}$ (or something comparable), or is there a nicer way? – Sha Vuklia Oct 06 '19 at 22:39
1

Your reasoning that $\arg$ isn't a continuous function on $S_1\setminus \{-1\}$ is not correct. Your argument is that it is not because $\lim_{z\rightarrow -1}\theta(z)$ does not exist, however, for something to be continuous on $S_1\setminus \{-1\}$, it is not necessary that this limit exist. It is merely necessary that, for any $y\in S_1\setminus\{-1\}$ we have that $\theta(y)=\lim_{z\rightarrow y}\theta(y)$ - and this does hold for the argument function.

Note that once you've found that $S_1\setminus \{-1\}$ admits an angle function, it is not hard to find an angle function for any set of the form $S_1\setminus \{z\}$ for any $z\in S_1$ just by conjugating your previous function with a rotation taking $z\mapsto -1$, and any proper open subset of $S_1$ is a subset of some $S_1\setminus \{z\}$.

Milo Brandt
  • 60,888