# Tangent bundle of $S^1$ is diffeomorphic to the cylinder $S^1\times\Bbb{R}$

How do I construct an explicit diffeomorphism between $TS^1$ and $S^1\times\Bbb{R}$?

It will be something like $\phi:TS^1\to S^1\times\Bbb{R}, (x,v)\to(x,...)$.

Also we know that for $x=(x_1,x_2)$ and $v=(v_1,v_2)$, $x_1^2+x_2^2=1$ and $v_1x_2+v_2x_1=0$. From these two equations we have \begin{align}v_1x_2&=-v_2x_1 \\v_1^2x_2^2&=v_2^2x_1^2\\v_1^2x_2^2+v_1^2x_1^2&=v_2^2x_1^2+v_1^2x_1^2\\v_1^2&=x_1^2(v_2^2+v_1^2)\end{align} And similarly $v_2^2=x_2^2(v_2^2+v_1^2)$. But I don't know where should I send $v$.

We can parametrise the circle $\mathbb S^1 \subset \mathbb R^2$ by $\theta \mapsto (\cos\theta,\sin\theta)$.

Any vector tangent to the circle at a point must be perpendicular to the position vector of that point. (This is a classical theorem: the radius is perpendicular to the tangent line.)

The tangent vectors are of the form $\lambda \mapsto (-\lambda\sin\theta,\lambda\cos\theta)$, for some $\lambda \in \mathbb R$.

The Tangent bundle to the circle can be parametrised by $$(\theta,\lambda) \longmapsto (\cos\theta,\sin\theta,-\lambda\sin\theta,\lambda\cos\theta).$$

If you like to be formal then you can write this as $$\mathrm f : (\theta,\lambda) \longmapsto (\cos\theta,\sin\theta;[-\lambda\sin\theta]\partial_x+[\lambda\cos\theta]\partial_y)$$

where $[-\lambda\sin\theta]\partial_x+[\lambda\cos\theta]\partial_y$ is a member of $T_p\mathbb R^2$, and $p = (\cos\theta,\sin\theta)$.

The cylinder appears because $\mathrm f$ is $2\pi$-periodic in $\theta$.

• Can you explain why you wrote tangent vector as a $((-\lambda \sin \theta) \delta_{x}+ (-\lambda \cos \theta) \delta _{y}$, I know that $/drlta_{X}$ and $\delta_{y}$ are a basis of tangent space at a point but how did you make sure that the scalars were what you wrote? – Mojojojo Sep 7 '18 at 11:58
• If the circle is parametrised by $\gamma(\theta) = (\cos\theta,\sin\theta)$, then a tangent vector is given by $\gamma'(\theta) = (-\sin\theta,\cos\theta)$. All tangent vectors are parallel to this one, and so are a scalar multiple: $(-\lambda\sin\theta,\lambda\cos\theta)$. – Fly by Night Sep 11 '18 at 15:06

A tangent vector to a circle is not really $(v_1,v_2)$. Rather, it is of the form $c\frac{d}{d\theta}$ where $c\in\mathbb{R}$ is a real number and $\frac{d}{d\theta}$ is the standard vector field globally defined on the whole circle. Therefore you can send $(x,v)$ to $(x,c)\in S^1\times\mathbb{R}$.

• The problem is that we haven't covered vector fields yet. – bluebox Apr 6 '14 at 13:57
• You don't need to know what a vector field is. If you know what a tangent bundle is, you just need to know that $\frac{d}{d\theta}$ gives you a nonzero vector at each point of the circle. – Mikhail Katz Apr 6 '14 at 14:08

As you know $(x_2,-x_1)\in T_xS^1$. So let $\phi:T_xS^1\to S_1\times\mathbb{R}$ given by $$\phi(x_1,x_2,v_1,v_2)=(x_1,x_2,<(x_2,-x_1),(v_1,v_2)>)=(x_1,x_2,v_1x_2-v_2x_1).$$ So $\phi^{-1}(x_1,x_2,t)=(x_1,x_2,t(x_2,-x_1))=(x_1,x_2,tx_2,-tx_1)$.

For a conceptual understanding as well, imagine you have a circle in the xy-plane centered at the origin. At each point on the circle, the tangent space "looks like" a line in the xy-plane tangent to that point. Now rotate this line until it becomes perpendicular to the xy-plane. Do this for all points on the circle. What we get is a cylinder!

• For someone studying the subject of manifolds for the first time, that was very helpful, thanks! – Mojojojo Sep 7 '18 at 11:38