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Problem: Show that $T\mathbb{S}^1 \cong \mathbb{S}^1 \times \mathbb{R}$.

An attempt of a solution: The task is to find a diffeomorphism between the two given spaces.

Consider $F:T\mathbb{S}^1\rightarrow\mathbb{S}^1\times\mathbb{R}$ defined locally as

$F\big|_{TU_i^\pm}:TU_i^\pm\rightarrow U_i^\pm \times \mathbb{R} \quad\Big| \quad F\big|_{TU_i^\pm} (p, \alpha \ \partial_i^\pm) = (p,\alpha) $

Where $ U_i^+ = \{(x_1,x_2): x_i > 0 \}| $ and $ U_i^- = \{(x_1,x_2): x_i < 0 \} $ are the half circles, equipped with the charts $\varphi_i^\pm: U_i^\pm \rightarrow \mathbb{R}$ defined by $ \varphi_i^\pm(x_1,x_2) = x_i $. Note that $(\varphi_1^\pm)^{-1}(\alpha) = (\sqrt{1-\alpha^2}, \alpha)$ and $(\varphi_2^\pm)^{-1}(\alpha) = (\alpha, \sqrt{1-\alpha^2})$. It simply takes $ v_p \in T_pM $ and sends it to $\mathbb{R}$ through its coordinate chart.

$F$ is clearly a local diffeomorphism, so all that is left is showing that it is bijective, i.e., $F\big|_{TU_i^\pm} (p,v) = F\big|_{TU_j^\pm}(p,v) \iff F\big|_{TU_i^\pm} \circ F^{-1}\big|_{TU_j^\pm} = id_{U_i^\pm \cap U_j^\pm}$ if $p \in U_i^\pm \cap U_j^\pm$.

The inverse of $F\big|_{TU_i^\pm}$ is clearly

$F\big|_{TU_i^\pm}^{-1}: U_i^\pm \times \mathbb{R} \rightarrow TU_i^\pm \quad\Big| \quad F\big|_{TU_i^\pm} (p, \alpha) = (p,\alpha \partial_i^\pm) $

so that

$F\big|_{TU_i^\pm} \circ F^{-1}\big|_{TU_j^\pm} (p,\alpha) = (p,\alpha \partial_j^\pm)$

the change of coordinates $\partial_j \rightarrow \partial_i$ is given by the chain rule

$v_pf = \alpha \ \partial_j f = \alpha \ \partial_i \circ (\partial^{-1}_i \circ \partial_j)(f)$

This is where we got stuck. We're clearly not sure how one should proceed with the change of coordinates. I'm aware there are solutions to this problem here at StackExchange, but I found none taking this approach.

Any tip with formatting is welcome.

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  • $\begingroup$ Hint: First construct a nonvanishing section of this tangent bundle, then use this section to prove triviality of the bundle. $\endgroup$ – Moishe Kohan Jan 26 '18 at 22:59
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In fact $S^1$ is the first example of a Lie group, namely a manifold which is a group. This is the group of complex number of modulus 1.

The tangent space at the point $g$ is obtained by translating the tangent space at the point $1$ by the action of $g$. Here, the tangent space at the point $u$ is $u.L$ where $L$ is the vertical line $Re(z)=1$, or $u. (1+ ix)$ the trivialization is $F(u,x)= u+ iu x$.

This argument works for any Lie group.

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  • $\begingroup$ I absolutely could not answer this back in the day. Your comment answers the question. $\endgroup$ – big-lion Apr 19 '18 at 2:17
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    $\begingroup$ Nitpick: the first example of a Lie group is the trivial group. $\endgroup$ – Najib Idrissi May 15 '18 at 14:17

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