Difference between the tangent bundle $TM$ with $M\times \mathbb{R}^{\dim M}$

Question

The book I'm using is Lee's Introduction to Smooth Manifolds. I just had my first encounter with the tangent bundle and I'm asked to show that $T\mathbb{S}^1$ is diffeomorphic to $\mathbb{S}^1\times \mathbb{R}$. What is difficult for me is that I have no idea what exactly I should show because I'm not able to even tell the difference between them, or more generally, the difference between $TM$ with $M\times \mathbb{R}^{\dim M}$.

So far what I've learned is that, as a set, elements in $TM$ can be identified canonically with $M\times\mathbb{R}^{\dim M}$. Under this identification, the topology on $TM$ is given by declaring that for any smooth chart $(U,\varphi)$ of $M$ the subspace $U\times \mathbb{R}^{\dim M}\subset TM$ is open and is homeomorphic to that of the product manifold $M\times \mathbb{R}^{\dim M}$, so that the map $\varphi\times 1_{\mathbb{R}^{\dim M}}$ gives a chart on $TM$, if I'm getting it right.

A paragraph on Lee says that in general $TM$ is not even homeomorphic to the product $M\times \mathbb{R}^{\dim M}$, but I cannot figure out a reason. An open set $\mathcal{O}$ in $M\times\mathbb{R}^{\dim M}$ can be written as a union of its intersection with $U\times \mathbb{R}^{\dim M}$ where $U$ varies in a smooth atlas of $M$. Each of the intersection $\mathcal{O}\cap U\times \mathbb{R}^{\dim M}$ is open in $U\times \mathbb{R}^{\dim M}$, hence it is also open in $TM$, so their union is again open, concludes that $\mathcal{O}$ is open in $TM$. Similarly every open set in $TM$ is also open in $M\times\mathbb{R}^{\dim M}$, so the identification $TM\cong M\times\mathbb{R}^{\dim M}$ turns out to be a homeomorphism.

I cannot see what's wrong with the argument above. Is there any simple example that is helpful in understanding this issue? Or did I misunderstand the definition?

Thanks in advance.

Answer 1

It's true that the tangent bundle looks like $U \times \mathbb R^{{\rm dim}(M)}$ on a local neighbourhood $U$ (where $U$ is homeomorphic to an open disk). But when you fit these local neighbourhoods together, there could be some "twists" in the way that the $\mathbb R^{{\rm dim}(M)}$'s are glued together.

Let's take the $TS^2$ example, as mentioned by Alvin Jin. It's true that $TS^2$ is constructed by sticking a 2-plane of tangent vectors at each point in $S^2$. But geometrically (viewing the $S^2$ as being embedded in 3-dimensional Euclidean space), the 2-plane of tangent vectors at the North Pole is not parallel to the 2-plane of tangent vectors at a point on the equator. There is some twisting in the way that the various 2-planes of tangent vectors at the various points get glued together to form the entire tangent space $TS^2$.

To understand the twisting without viewing the $S^2$ as being embedded in a higher-dimensional Euclidean space, note that as Alvin points out, $TS^2$ has no global smooth non-vanishing vector fields (the Hairy Ball theorem). Whereas with the trivial bundle on $S^2$ (this is what you call $S^2 \times \mathbb R^2$), you can find a pair of smooth non-vanishing vector fields that span the tangent space at every point on the $S^2$.

(If you like, you can think of the trivial bundle $S^2 \times \mathbb R^2$ as taking an $S^2$ and sticking a 2-plane at each point in the $S^2$, but instead of orienting each 2-plane to be tangential to the $S^2$ at the point that it is attached to, all of the 2-planes are oriented parallel to the equatorial plane.)

Would you like another example? Why not think about the tangent bundle for the Mobius strip? That has an obvious twist in the way that the tangent vectors are glued together. Maybe you can decompose the Mobius strip into the union of two open neighbourhoods $U$ and $V$ (each being homeomorphic to the unit rectangle). Over $U$, the tangent bundle is $U \times \mathbb R^2$, and over $V$, the tangent bundle is $V \times \mathbb R^2$. But look how the tangent vectors glue together between $U$ and $V$!