2
$\begingroup$

This is an exercise from Loring Tu's Introduction to Manifolds that I am stuck at. I know that a tangent bundle is trivial if it is isomorphic to the product bundle $M \times \mathbb{R}^{n}$. Here $n$ is the dimension of the (smooth of course) manifold $M$.

I think I have to construct a smooth frame from the the isomorphism and vice versa; but I cannot find a way to do so...Could anyone please help me?

$\endgroup$

2 Answers 2

4
$\begingroup$

On $\mathbb{R}^n$ there is the obvious basis $e_1,e_2,\dots,e_n$. Since $TM\cong M\times\mathbb{R}^n$ you have the $p\mapsto e_i$ for all $p$ defines a smooth section $X_i$ of $TM$. The $X_1,\dots,X_n$ defines a frame at every point.

Conversely, if you have global frame field $X_1,\dots,X_n$, then for $v\in T_pM$ we have $v=v^iX_i(p)$, and you send this $v$ to $(p,(v^i))\in M\times\mathbb{R}^n$. Check it works.

$\endgroup$
2
$\begingroup$

If there is a smooth frame $X_1,\dots,X_n$, then each tangent vector $V$ is uniquely written as $$V=v^1X_1+\dots+v^nX^n.$$

What can you say about the map $(x,V)\mapsto(x,v^1,\dots,v^n)$?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .