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I asked this question in https://mathoverflow.net/, but was advised to ask it here. So here it is.
I just started a self-study of differential geometry and topology. And in several text I came accross the question, asking to show that the global coordinates cannot be defined on a circle $S_{1}$. It seems like quite easy question, but I cannot work the proof. I have a hunch that it has something to do with a Jacobian being zero at some points.
Suppose $S_{1}= [(x,y)\in R^{2} | x^{2}+y^{2}=1] $ and there exist a global coordinate system $u=f(x,y)$, $x=g(x,y)$. The Jacobian then is as follows: $$ J=\begin{vmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{vmatrix}$$ Since $y=\pm \sqrt{1-x^{2}}$ $$\frac{\partial f}{\partial y}=0$$ $$\frac{\partial g}{\partial y}=0$$ So I have that Jacobian is equal to zero. But this also feels not quite right. I'm I missing something?

Comment: the global coordinates means that the mapping is smooth bijective and has a non-zero Jacobian everywhere.

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Tomas
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    Dear Tomas, Since $S^1$ is a $1$-manifold, coordinates on $S^1$ consist of a single function, not a pair of functions. So you are trying to show that if $u: S^1 \to \mathbb R$ is a smooth function, it must have a critical point somewhere. Regards, – Matt E Dec 11 '12 at 18:37
  • I have a few questions. Let $f\colon S^1 \rightarrow R^n$ be your "global coordinates". $n$ is $2$? You mean $f$ is injective. Right? What is your definition of the smoothness of $f$? – Makoto Kato Dec 11 '12 at 19:56
  • @MakotoKato Yes, $n=2$. f is injective and f is smooth in terms of infinite differentiability. – Tomas Dec 11 '12 at 20:01
  • @Tomas I know. I meant, what is the definition of differentiability of $f$? $S^1$ is not an open subest of $R^2$. So the usual definition of the differentiability of a multivariate function cannot be applied to $f$. – Makoto Kato Dec 11 '12 at 20:23
  • @MakotoKato Made me think. There was no different definition of differentiability in my book, so I assume, that it is defined as usual, on open sets. I think this is the answer I was looking for: the global coordinate system does not exsist for whole circle, since it is a closed subset of $R^{2}$? – Tomas Dec 11 '12 at 20:43
  • @Tomas I think there is the definition of a differentiable manifold in your book. Did you read it? – Makoto Kato Dec 11 '12 at 21:16
  • @MakotoKato The fact is, that manifolds (differentiable manifolds) are defined in subsequent chapters. But This chapter is devoted purely to coordinate systems (or curvilinear coordinates), so I am supposed to answer this question without refering to manifolds. – Tomas Dec 11 '12 at 21:24
  • @Tomas OK. What is the definition of a coordinate system on $S^1$ in your book? – Makoto Kato Dec 11 '12 at 21:37
  • @Tomas: Could you write the full definition of a coordinate system that your book gives? – Jesse Madnick Dec 11 '12 at 21:40
  • Here is the definition of regular coordinate system (coordinate system definition is the same, just without the non-zero Jacobian condition): A regular coordinate system in a domain $C$ of Euclidean space $R^{n}$ is a system of smooth functions $ x^{1}(y^{1},...,y^{n}),..., x^{n}(y^{1},...,y^{n}) $ which map bijectively the domain $C$ onto domain $A$ in $R^{n}$ and are such the the Jacobian is not zero at all points of $C$. – Tomas Dec 11 '12 at 21:52
  • @Tomas What about the definition of a coordinate system on $S^1$? – Makoto Kato Dec 11 '12 at 21:59
  • @MakotoKato Well, my original question was how to show, that there is no global coordinate system on $S^{1}$. – Tomas Dec 11 '12 at 22:11
  • @Tomas That is an exercise of the book, right? Then it must have the definition of a coordinate system on $S^1$. Otherwise nobody cannot solve the problem. – Makoto Kato Dec 11 '12 at 22:13
  • @MakotoKato Since it is defined as a set $S^{1}=[(x,y) \in R^{2}|x^{2}+y^{2}=1]$ I assume it is initially defined Cartesian coordinates. – Tomas Dec 12 '12 at 08:44
  • @Tomas I'm asking the definition of your book. – Makoto Kato Dec 12 '12 at 10:34

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If there exists global coordinates on $S^1$, $S^1$ is homeomorphic to $R^n$ for some $n \ge 1$. But $S^1$ is compact while $R^n$ is not.

Makoto Kato
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  • Thank you for your response, but I found this question at the beginning of the book, i.e. just after the definition of global/local coordinate systems and definition of Jacobian for those coordinate systems. The notion of compact sets just to be introduced in the following chapters, so I'm supposed to answer this question just by definition of global coordinates (like in my original post), i.e. smoothness, bijectivity and zero/non-zero Jacobian. – Tomas Dec 11 '12 at 18:27
  • @Tomas Do you know that a bounded sequence in $R^n$ has a convergent subsequence? – Makoto Kato Dec 11 '12 at 18:32
  • yes, I do know this. It is one of the theorems in topology course. – Tomas Dec 11 '12 at 18:35
  • why the sequence $b_{m}=f^{-1}(a_{m})$ converge? Is it because the circle is a closed curve and $b_{m} \in S_{1}$? – Tomas Dec 11 '12 at 18:50
  • @Tomas Let $f\colon S^1 \rightarrow R^n$ be the global coordinates. Let $a_m = (m, 0,...,0) \in R^n$ for every integer $m \ge 1$. Let $b_m = f^{-1}(a_m)$. Since the sequence $(b_m)$ is bounded and $S^1$ is closed, it has a convergent subsequence whose limit lies in $S^1$. Hence $(a_m)$ has a convergent subsequence because $f$ is continuous. But this is impossible. – Makoto Kato Dec 11 '12 at 18:53
  • But in order to have a sequence $b_{m}=f^{-1}(a_{m})$, we need the sequence $a_{m}$ which is a subset of the image $f(S^{1})$. But, as I understand, we cannot garuantee that $a_{m} \in Image(f)$? – Tomas Dec 11 '12 at 19:03
  • @Tomas Perhaps I misunderstood your problem. – Makoto Kato Dec 11 '12 at 19:21
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The notion of "global coordinate system" is somewhat soft. As $S^1$ is a one-dimensional manifold, any coordinate system, local or global, on $S^1$ has one coordinate variable, and not two, as you suggest using $(x,y)$, or $(u,v)$.

I'm interpreting your question in the following way: Prove that there is no map $f:\ ]a,b[\ \to S^1$ which is a diffeomorphism. As $S^1$ is locally diffeomorphically parametrized by the polar angle $\phi$, such a map, if it existed, would have the form $$f:\quad ]a,b[\ \to S^1,\quad t\mapsto e^{i\phi(t)}\ ,$$ and would install on $S^1$ a global coordinate system, namely the coordinate $t$ ranging in the interval $\ ]a,b[\ $ of ${\mathbb R}$.

Assume that $f$ has the required properties. In particular, the function $\phi(\cdot)$ is continuous and injective, therefore monotone, say, monotonically increasing. It follows that $$\alpha:=\lim_{t\to a+}\phi(t)=\inf_{t\in\ ]a,b[}\phi(t)$$ and $$\beta:=\lim_{t\to b-} \phi(t)=\sup_{t\in\ ]a,b[}\phi(t)$$ exist, and $\alpha<\beta\leq\alpha+2\pi$; otherwise $f$ would not be injective. But from $\beta\leq \alpha+2\pi$ and the fact that $\ ]a,b[\ $ has neither a minimal nor a maximal element it follows that the value $\alpha+2k\pi$ is not taken by $\phi$ for any $k\in{\mathbb Z}$, whence $e^{i\alpha}\in S^1$ is not taken by $f$. Therefore $f$ would not be surjective.

It follows that such an $f$ cannot exist.

Christian Blatter
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  • Could you please explain, what does the function $\phi (t)$ means in $t\mapsto e^{i\phi(t)}$ and where it came from? since I do not see the connections. – Tomas Dec 11 '12 at 20:06