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!

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    $\begingroup$ (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. $\endgroup$ – Jack Lee Jan 24 '17 at 20:59
  • $\begingroup$ @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? $\endgroup$ – user143144 Jan 24 '17 at 21:07
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    $\begingroup$ Hint: It might help to put a Riemannian metric on $M$. $\endgroup$ – Jack Lee Jan 24 '17 at 21:09
  • $\begingroup$ math.stackexchange.com/questions/1107682/… $\endgroup$ – Moishe Kohan Jan 25 '17 at 11:45
  • $\begingroup$ @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. $\endgroup$ – user143144 Jan 25 '17 at 16:18

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