# [Math] tangent bundle and normal bundle

differential-geometrymanifolds

I have a problem about tangent bundle. It is known that the tangent bundle of most manifolds is not trivial: for example, the tangent bundle for $S^2$ is not $S^2\times \mathbb{R}^2$. However, for a surface embedded in $\mathbb{R}^n$, the normal bundle is simply a direct product (for example, in Novikov et al. book "Modern Geometry").

So how to visualize tangent bundle and normal bundle? When I try to visualize them, I just imagine a manifold with a tangent (normal) plane attached at each point. Why is this not a product space? And what is the difference between tangent bundle and normal bundle from this point?

Thanks a lot!

In general it's pretty difficult to visualise tangent bundles given that they have dimension twice the dimension of the base manifold; for example, $TS^2$ is $4$-dimensional. However, one can prove indirectly that this bundle is non-trivial (i.e. not a product). If it were a product, then it would have a nowhere-zero section. In other words, one could construct a continuous vector field on the sphere which doesn't vanish anywhere (explicitly, if $TS^2 \cong S^2 \times \mathbb{R}^2$ then we could take our vector field to be constant $(1, 0)$ in the second factor). And this is impossible, by the hairy ball theorem.
An example of a non-trivial normal bundle is easier. Here's one way. If $M$ is the total space of some vector bundle, and $N$ is the zero section, then the normal bundle of $N$ in $M$ is simply the bundle $M$ itself. So if $M$ is non-trivial (say $M$ is the Mobius bundle, i.e. a cylinder with a 'half twist'; this is non-trivial for the 'no nowhere-zero sections' reason above) then $N$ has non-trivial normal bundle in $M$.
As for why the tangent and normal bundles need not be the same, in general they don't even have the same dimension. If $N$ is an $n$-dimensional submanifold of an $m$-manifold $M$ then $TN$ is $2n$-dimensional whilst the normal bundle is $m$-dimensional.