$TS^1$ is Diffeomorphic to $S^1\times \mathbf R$.

Question

I know this is a very basic question. But I am unable to get every detail right.

I need to show that $TS^1$ is diffeomorphic to $S^1\times \mathbf R$.

(I am using the concept of derivations to define the tangent spaces.)

I asked this of my friends and this is what, in essence, I have learned from them:

Define $f:TS^1\to S^1\times \mathbf R$ as $$f \left( p;\ \lambda\left(p_2 \left.\frac{\partial}{\partial x_1}\right|_{p}-p_1\left.\frac{\partial}{\partial x_2}\right|_p \right) \right) = (p,\ \lambda) $$ for all $p=(p_1,p_2)\in S^1$ and all $\lambda\in \mathbf R$, and show that this is a diffeomorphism.

My problem with this is that:

$\left(p_2 \left.\frac{\partial}{\partial x_1}\right|_{p}-p_1\left.\frac{\partial}{\partial x_2}\right|_p \right)$ is not really a member of $T_pS_1$.

It lies in $di_p(T_pS^1)\subseteq T_p\mathbf R^2$, where $i:S^1\to \mathbf R^2$ is the inclusion map.

(Reason: The manifold structure of $S_1$ is governed by the fact that $i:S^1\to \mathbf R^2$ is a smooth embedding, and thus, a tangent vector $X_p\in T_p\mathbf R^2$ is in $di_p(T_pS^1)$ if an only if $X_p\xi=0$ for all $\xi\in \mathcal C^\infty(\mathbf R^2)$ with $\xi|S^1\equiv 0$.)

Keeping this in mind, I attempted the following:

Define $F:T\mathbf R^2\to \mathbf R^2\times \mathbf R^2$ as $$ F\left(p,\ a_1 \left. \frac{\partial}{\partial x_1} \right|_p+a_2\left.\frac{\partial}{\partial x_2}\right|_p\right) = (p,\ a_1, a_2) $$ for all $p\in \mathbf R^2$ and $a_1, a_2\in \mathbf R$. We note that $F$ is a diffeomorphism.

Now define $G:\mathbf R^2\times \mathbf R^2\to \mathbf R^2\times \mathbf R$ as $G(p, a_1, a_2)=(p, \sqrt{a_1^2+a_2^2})$. We note that $G$ is smooth.

Finally, since $i:S^1\to T\mathbf R^2$ is smooth (it is more than that), we have $di:TS^1\to T\mathbf R^2$ is also smooth.

Now $G\circ F\circ di: TS^1\to S^1\times \mathbf R$ is thus a smooth map, since composition of smooth maps is smooth.

But I am unable to show that this map is the required diffeomorphism.

Can somebody help?

Answer 1

The right construction is the one you start with, the vector is indeed tangent to $S^1$. The tangent to $S^1$ is the line perpendicular to $p=(p_1,p_2)$, which is generated by $(p_2, -p_1)=p_2(1,0)-p_1(0,1)$ (although the opposite vector is more natural, by orientation reasons). Then, the partial derivatives are exactly: $$ \tfrac{\partial}{\partial x_1}\big|_p=(1,0),\quad \tfrac{\partial}{\partial x_2}\big|_p=(0,1) $$ and we get the tangent vector under consideration. This said, $f$ is indeed a diffeo (the inverse is clear).

Thus, let us look at the identification of $T_pS^1\subset T_p\mathbb R^2$ in terms of derivations. To that end, we parametrize $S^1$ in the simplest standard way: $\varphi(t)=(\cos t,\sin t)$, and say $p=\varphi(\theta)$. Then the derivation $\frac{\partial}{\partial t}$ generates $T_pS^1$. It acts as follows: $$ \tfrac{\partial f}{\partial t}\big|_p=\tfrac{d}{dt}f(\cos t,\sin t)\big|_\theta. $$ Now $T_p\mathbb R^2$ is of course generated by the usual partial derivatives $\tfrac{\partial }{\partial x_1}\big|_p, \tfrac{\partial }{\partial x_2}\big|_p$, and we want to express $D=\tfrac{\partial }{\partial t}\big|_p$ in terms of those two derivations. That is, we look for coefficients $\alpha_1,\alpha_2$ such that $$ D=\alpha_1\tfrac{\partial }{\partial x_1}\big|_p+\alpha_2\tfrac{\partial }{\partial x_2}\big|_p. $$ Since $\tfrac{\partial x_i}{\partial x_j}=0$ or $1$ according to $i\ne j$ or $i=j$, we see that $\alpha_i=D(x_i)$, and so: $$ \begin{cases} \alpha_1=\tfrac{\partial x_1}{\partial t}\big|_p=\tfrac{d}{dt}\cos t\big|_\theta=-\sin\theta=-p_2,\\ \alpha_2=\tfrac{\partial x_2}{\partial t}\big|_p=\tfrac{d}{dt}\sin t\big|_\theta=\cos\theta=p_1, \end{cases} $$ that is: $$ \tfrac{\partial }{\partial t}\big|_p=-p_2\tfrac{\partial }{\partial x_1}\big|_p+p_1\tfrac{\partial }{\partial x_2}\big|_p. $$ This is the opposite of the one you have (as I had commented), the different sign coming from the way the parametrization $\varphi$ turns around the origin.

Concerning the $G\circ F\circ di$ proposal, the map we have is in fact $$ TS^1\to S^1\times\mathbb R:(x,u)\mapsto (x,\|u\|), $$ which is not injective: $(x,\pm u)\mapsto (x,\|u\|)$. Thus one cannot get a diffeo there.

Answer 2

A homeomorphism $TS^1 \to S^1 \times \mathbb{R}$ is just a trivialization, which is equivalent to giving a nowhere-vanishing section $s\in \Gamma(TS^1)$. To promote it to a diffeomorphism, we just need $s$ to be smooth. The space $S^1$ is a Lie group, so we can take $s(g) = DL_g(v)$ for some fixed $v\in T_{\operatorname{id}} S^1$, where $L_g : S^1 \to S^1$ is left-multiplication (which is also right-multiplication in this case) by $g$. Unwinding the definitions here gives the explicit map in the first line of your post.

Answer 3

Consider the standard coordinates on $\mathbb{S}^{1}$ with coordinate maps $\varphi_{1}, \varphi_{2}, \varphi_{3}$ and $\varphi_{4}$ where, for instance, $\varphi_{1}: (-1, 1)\rightarrow\mathbb{S}^{1}$ is defined by $\varphi(t) = (t, \sqrt{1-t^2})$. Then, for $(p_{1}, p_{2})\in\mathbb{S}^{1}$ with $p_{2} > 0$ (as an example) we can show that $$T_{p}\mathbb{S}^{1} = \text{span}\Big\{-p_{2}\frac{\partial}{\partial x_{1}}\Big|_{p}+p_{1}\frac{\partial}{\partial x_{2}}\Big|_{p}\Big\}$$ as a subspace of $T_{p}\mathbb{R}^{2}$ by using the fact that $T_{p}\mathbb{S^{1}} = \text{Im}(\text{d}(\varphi_{1})_{\varphi_{1}^{-1}(p)})$. Now define $F:T\mathbb{S}^{1}\rightarrow\mathbb{S}^{1}\times\mathbb{R}$ by $$F\Big(p_{1}, p_{2}, \lambda\Big(-p_{2}\frac{\partial}{\partial x_{1}}\Big|_{p}+p_{1}\frac{\partial}{\partial x_{2}}\Big|_{p}\Big)\Big) = (p_{1}, p_{2}, \lambda).$$ This is smooth as a map from $T\mathbb{S}^{1}\rightarrow\mathbb{R}^{2}\times\mathbb{R}$ since the coordinate representations look like (here we're using $\varphi_{1}$) $$(t, v)\mapsto \Big(t, \sqrt{1-t^{2}}, v\Big(\frac{\partial}{\partial x_{1}}\Big|_{\varphi_{1}(t)}-\frac{t}{\sqrt{1-t^2}}\frac{\partial}{\partial x_{2}}\Big|_{\varphi_{1}(t)}\Big)\Big)\mapsto\Big(t, \sqrt{1-t^2}, -\frac{v}{\sqrt{1-t^2}}\Big).$$ $F$ is also smooth as a map into $\mathbb{S}^{1}\times\mathbb{R}$ since it is an embedded submanifold of $\mathbb{R}^{2}\times\mathbb{R}$, and it is clearly bijective. The inverse can be shown to be smooth by regarding it as the restriction to $\mathbb{S}^{1}\times\mathbb{R}$ of a smooth map from $\mathbb{R}^{2}\times\mathbb{R}$ to $T\mathbb{R}^{2}$. It is also smooth as a map into $T\mathbb{S}^{1}$ since it is an embedded submanifold of $T\mathbb{R}^{2}$.