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Let $M$ be a $3$ dimensional smooth manifold such that there exists two non vanishing independent vector fields $X_1, X_2 \in \mathfrak{X}(M)$. Given that $M$ is orientable, is $M$ also parallelizable?

Since $M$ is orientable, we can fix an orientation $\xi_M = \{ \xi_x : x\in M\}$, where $$\xi_x = \left[ \frac{\partial}{\partial \mathrm{x_1}}\bigg\vert_x,\dots, \frac{\partial}{\partial \mathrm{x}_3}\bigg\vert_x \right]$$ and $\mathrm{x} : U \to \mathbb{R}^m$ is some coordinate system.

Now, parallelizable means that we can choose as many independent vector fields as the dimension of the manifold, so we should be able to find $X_3 \in \mathfrak{X}(M)$ independent of the other two. But since the partial derivatives $\frac{\partial}{\partial \mathrm{x}_i}\big\vert_x$ form a basis for $T_xM$ for each $x\in M$, we can take the $\frac{\partial}{\partial \mathrm{x}_i}$ such that it is independent of $X_1$ and $X_2$, thus proving that $M$ is also parallelizable.

This has to be wrong, since the argument could be extended not only for $3$ dimensional orientable smooth manifolds, but for orientable smooth manifolds of arbitrary dimension. However, I can't see where the mistake is, so I'd like if anyone could point it out for me. Thank you!

user143144
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    (1) The word you want is "parallelizable," not "parallel." (2) Because your construction uses coordinates, it is only local (that is, it only works in the domain of a particular coordinate chart). To show that a manifold is parallelizable, you need to prove that there are as many independent global vector fields as the dimension of the manifold. – Jack Lee Jan 24 '17 at 20:59
  • @JackLee Thank you! I've edited the question accordingly. And I see, so then it is true that any smooth manifold is locally parallelizable, right? Could you give me any hint on how to find the other global vector I need? – user143144 Jan 24 '17 at 21:07
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    Hint: It might help to put a Riemannian metric on $M$. – Jack Lee Jan 24 '17 at 21:09
  • http://math.stackexchange.com/questions/1107682/elementary-proof-of-the-fact-that-any-orientable-3-manifold-is-parallelizable?rq=1 – Moishe Kohan Jan 25 '17 at 11:45
  • @MoisheCohen Thanks for the reference! However, I'm afraid that most (if not all) of the contents in the answers to that question are out of my reach. – user143144 Jan 25 '17 at 16:18
  • Then just follow Jack Lee suggestion: It works. – Moishe Kohan Jan 25 '17 at 16:23
  • @MoisheCohen I will, I just need some knowledge about Riemannian manifolds first. Roughly, I have to choose metric $g$ on $M$, and then get an independent vector field $X_3$ in a way kind of similar like the cross product of vectors in $\mathbb{R}^3$? – user143144 Jan 25 '17 at 16:50
  • Exactly! Now, how do you define the cross product geometrically? – Moishe Kohan Jan 25 '17 at 16:51
  • @MoisheCohen In $\mathbb{R}^3$? Well if the two given vectors are $u$ and $v$, then the norm of the cross product is the area of the parallelogram with sides $u$, $v$. – user143144 Jan 25 '17 at 16:59
  • Yes, what else do you need to determine this vector uniquely? – Moishe Kohan Jan 25 '17 at 17:39

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