# A way to compute the tangent bundle of the 1-sphere

I'm working on the following problem from Lee's Introduction to Smooth Manifolds:

Show that $T\mathbb{S}^1$ is diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$

I am trying to apply a procedure that I saw e.g. in the computation of the Lie Algebra of $SO(3)$:

For some point $p\in\mathbb{S}^1$ take a curve $\gamma(t)\in\mathbb{S}^1$ with $\gamma(0)=p$ and $\gamma'(0)=v$ where $v$ will be the tangent vector. Differentiating the defining property of the sphere ($x^2+y^2=1$) leads to $2x(0)\cdot x'(0)+2y(0)\cdot y'(0)=0 \Leftrightarrow p^T v=0$

Hence, $T_p\mathbb{S}^1=\left\{v\in\mathbb{R}^2:p^Tv = 0\right\}$. Since $p$ is not zero, this is a 1-dimensional subspace of $\mathbb{R}^2$ which is clearly diffeomorphic to $\mathbb{R}$. Then, $T\mathbb{S}^1$ is the union over all $T_p\mathbb{S}^1$ and thus diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$ (for each point $p\in\mathbb{S}^1$ we have a tangent space at p diffeomorphic to $\mathbb{R}$).

Is my reasoning correct? What would be the easiest way to show this?

• Your "argument" applies to all manifolds... Commented Mar 26, 2017 at 21:48
• @MarianoSuárez-Álvarez Thanks for the reply! It just seemed to be a bit too easy, that's why I wasn't sure... But $T\mathbb{S}^2$ is not diffeomorphic to $\mathbb{S}^2\times\mathbb{R}^2$ so where does it fail in that case? Commented Mar 26, 2017 at 22:33

## 1 Answer

... $T\mathbb{S}^1$ is the union over all $T_p\mathbb{S}^1$ and thus diffeomorphic to $\mathbb{S}^1\times\mathbb{R}$

As pointed out in the comment, your argument is not correct. And the mistake comes from the "thus" in the quoted comment. All you know in general is that the tangent bundle is locally a product:

$$TM|_{U} \cong U \times \mathbb R^n$$

for all small enough open set in a manifold $M$. There are lots of manifolds with nontrivial tangent bundle, as pointed out by yourself.

So in order to show $T\mathbb S^1 \cong \mathbb S^1 \times \mathbb R$, you need more inputs: for each $p\in \mathbb S^1$, from your description, there is a canonical choice of basis of $T_p\mathbb S^1$: for example, $\sqrt{-1} p$ is such a choice. So every vector $v\in T_p\mathbb S^1$ is given by $v = t \sqrt{-1} p$ for some $t\in \mathbb R$. Try to use this $t$ to write down an explicit isomorphism

$$T\mathbb S^1 \to \mathbb S^1 \times \mathbb R, \ \ \ (p, v)\mapsto ??.$$

• How did you come up with that basis? So we're identifying $\mathbb{R}^2$ with $\mathbb{C}$ now? In that case we could write p as $p=e^{i\phi}$ for some $\phi$ and could solve the equation $v=itp$ for t. This would lead to the following map and its inverse: $\mathbb{S}^1\times\mathbb{R}\rightarrow T\mathbb{S}^1,\>\> (p,t)\mapsto (p,itp)\> ,\>\>\>$ $T\mathbb{S}^1\rightarrow\mathbb{S}^1\times\mathbb{R},\>\> (p,v)\mapsto (p,-ip^*v)$ Commented Mar 27, 2017 at 10:35