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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?


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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 (…) that the tangent bundle of every Lie group is trivial. – Alex Youcis Mar 26 '13 at 2:34
@AlexYoucis $v_p^{0,1}$ is just the component of $v_p^0$ under some chart. – Y. Fan Mar 26 '13 at 2:43

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.

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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? – Y. Fan Mar 26 '13 at 4:33
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! – Y. Fan 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? – Y. Fan Mar 26 '13 at 20:27
In that case, how can you guarantee F is smooth? – Avi Steiner Mar 27 '13 at 21:04

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.

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