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I am proving $TS^1$ is diffeomorphic to $S^1\times\mathbb{R}$. The following is my proof and I think it is wrong, because I only use the fact that $S^1$ is 1-dimentional. However, I do not know how to correct my proof. ($S^1$ is the unit circle).

For any $p\in S^1$, we can choose a chart $(U,\varphi)$ around it. Therefore, every $p\in S^1$ is associated with a vector $v_p^0=\frac{\partial}{\partial x}|_p$ if a chart is given.

Now, a function $F$ from $S^1\times\mathbb{R}$ to $TS^1$ is defined by $$F(p,\lambda)=(p,\lambda v_p^0)$$

I want to show that $F$ is a diffeomorphism.

Clearly, $F$ is injective. For any $(p,v)\in TS^1$, we have $$v=v^1\frac{\partial}{\partial x}|_p$$ $$v_p^0=v_p^{0,1}\frac{\partial}{\partial x}|_p$$ under some chart around $p$.

Therefore, we choose $\lambda=v^1/v_p^{0,1}$. $\lambda$ should be independet of choice of charts. Therefore, $F$ is also surjective.

Now choose two charts $(U\times\mathbb{R},\varphi\times i)$ and $(\pi^{-1}(U),\tilde{\varphi})$ for $S^1\times\mathbb{R}$ and $TS^1$, respectively. The expression of $F$ is \begin{align*} \hat{F}(q,x)&=\tilde{\varphi}\circ F\circ(\varphi\times i)^{-1}(q,x)\\ &=\tilde{\varphi}\circ F(p,x)\\ &=\tilde{\varphi}(p,xv_p^0)\\ &=(q,xv_p^{0,1}) \end{align*} $\hat{F}$ is smooth, since $v_p^{0,1}$ is smooth with respect to $p$.

For $F^{-1}$, my proof to show that it is smooth is similar. Therefore, $F$ is diffeomorphism.

However, I do not use any specific property of $S^1$ except that $S^1$ is 1-dimentional. Is my proof correct? If not, how to correct it?

Thanks!

YYF
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  • What is $v^{0,1}$? I mean, $\mathbb{S}^1$ is the unique compact one-dimensional connected smooth manifold--so if you used connectivity and compactness but you didn't. The proof of this should follow something along the lines of my proof here (http://math.stackexchange.com/questions/308691/let-g-be-a-lie-group-show-that-there-is-a-diffeomorphism-tg-cong-g-times-t/308798#308798) that the tangent bundle of every Lie group is trivial. – Alex Youcis Mar 26 '13 at 02:34
  • @AlexYoucis $v_p^{0,1}$ is just the component of $v_p^0$ under some chart. – YYF Mar 26 '13 at 02:43

2 Answers2

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Your function $F$ isn't well-defined. Since $S^1$ isn't diffeomorphic to $\mathbb{R}$, your coordinate patch $U$ can't contain every point of $S^1$. So, if $p\in S^1\setminus U$ and $\lambda$ is any real number, what is $F(p,\lambda)$?

edit: To answer your question below, consider the coordinate chart $\phi(x,y):=x$ defined on the open set $U:=\{(x,y)\in S^1: y>0\}$, where I'm viewing $S^1$ as a subset of $\mathbb{R}^2$. Let $x$ be the coordinate corresponding to this chart. Then for $p\in U$, your $v_p^0$ is $\frac{\partial}{\partial x}\Bigr|_p$. Now, consider the chart $\psi(x,y):=y$ defined on the open set $V:=\{(x,y)\in S^1: x>0\}$. Letting $y$ be the corresponding coordinate, your $v_p^0$ (which to emphasize the different coordinate chart I'll write as $w_p^0$) is $\frac{\partial}{\partial y}\Bigr|_p$. Are these the same? Well, the transition function from the $x$ coordinate chart to the $y$ coordinate chart is $$\psi\circ \phi^{-1}(x)=\psi\left(x,\sqrt{1-x^2}\right) = \sqrt{1-x^2},$$ which has derivative $$\frac{-x}{\sqrt{1-x^2}}=\frac{-\sqrt{1-y^2}}{y}.$$ So, for $p\in U\cap V$, $$v^0_p=\frac{\partial}{\partial x}\Biggr|_p = \frac{-\sqrt{1-y^2}}{y} \frac{\partial}{\partial y}\Biggr|_p\neq \frac{\partial}{\partial y}\Biggr|_p=w_p^0.$$ Thus, $v_p^0$ isn't well-defined.

Avi Steiner
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  • I think my $v_p^0$ is well-defined. It is defined using some coordinate. But, once it is defined, its value will not depend on the choice of coordinate. Therefore, for every $p$, $v_p^0$'s are well-defined. Therefore, the value of $F(p,\lambda)$ is just $(p,\lambda v_p^0)$, although $(p,\lambda v_p^0)$ has different expression under different coordiantes. Am I right? – YYF Mar 26 '13 at 04:33
  • Thank you for your detailed explanation, but I still do not understand why $v_p^0$ is not well-defined. When I defined $v_p^0$, I did not mean its component under every coordinate is 1. I just pick a coordinate, and define $v_p^0$ as $\frac{\partial}{\partial x}|_p$. Then, $v_p^0$ is defined. If the coordinate is changed, say $y$, $v_p^0$ should be $-\frac{\sqrt{1-y^2}}{y}\frac{\partial}{\partial y}|_p$, since $v_p^0$ is already defined in the coordinate $x$, which is picked at first. So, I am still confused, but really appreciate your explanation! – YYF Mar 26 '13 at 14:57
  • But it's not defined everywhere. – Avi Steiner Mar 26 '13 at 18:06
  • I am not sure that I understand what your mean. For every $p\in S^1$, a chart will be picked to define $v_p^0$. Therefore, definitely, $v_p^0\in T_pS^1$ for every $p\in S^1$. In other words, I think I am trying to define a vector field using a bunch of coordiante patches which are picked at the first place. Am I right? – YYF Mar 26 '13 at 20:27
  • In that case, how can you guarantee F is smooth? – Avi Steiner Mar 27 '13 at 21:04
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What you need to do is find the right atlas on $S^1$, say, the atlas with two charts given by polar coordinates on $S^1 \setminus \{-1\}$ and polar coordinates on $S^1 \setminus \{1\}$, so that $\tfrac{d}{dx}$ patches together to define a nowhere-vanishing vector field $\xi$ on $S^1$. Given such a vector field $\xi$, you can then define your $F : S^1 \times \mathbb{R} \to TS^1$ in a completely coordinate-independent way by $F(p,\lambda) := [(p,\lambda \xi_p)]$, and now use your concrete atlas on $S^1$ and your concrete nowhere-vanishing vector field $\xi$ to show that $F$ is indeed a diffeomorphism.

Branimir Ćaćić
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