Show that the tangent bundle is trivial

differential-geometrysmooth-manifoldsspherestangent-bundle

So I was studying for a test and found this practice problem:

Let $S^1=\{ (x,y)\in \mathbb{R}^2, x^2+y^2=1 \}$.

i) Show that $S^1$ is a smooth manifold in $\mathbb{R}^2$ (Done)

ii) Show that the tangent bundle $TS^1$ is trivial.

Now, for ii) I know that you can show the tangent bundle is trivial if you set a fiber-preserving diffeomorphism (however I am not so sure how to do so) or if you show that $TS^1$ is isomorphic to $S^1\times \mathbb{R}$.

Are these the only two options? What are some common tools one can use to show that a tangent bundle is trivial? What is the intuition behind it?

Best Answer

One useful equivalent condition is that an $n$-manifold $M$ has trivial tangent bundle iff there exists a global frame, i.e. $n$ vector fields $E_1,\cdots,E_n$ which are everywhere linearly independent (in the sense that $\forall p\in M$, $E_1(p),\cdots,E_n(p)$ form a basis of $T_pM$). This is equivalent to your definition since, given such a global frame, there is an isomorphism $\psi:M\times\mathbb{R}^n\to TM$ given by $\psi(p,v^1,\cdots,v^n)=\sum_iv^iE_i(p)$.

For the circle, this means that $TS$ is trivial iff it admits a nonvanishing tangent vector field.