Bulding a base with additional property Lets say we have two smooth vector fields $v,w:\mathbb{R}^n\to\mathbb{R}^n$ such that $\|v(x)\|=\|w(x)\|=1,  \langle v(x),w(x)\rangle>0$ for $x\in\mathbb{R}^n$. How to construct a smooth vector fields $e_1,...,e_n :\mathbb{R}^n\to\mathbb{R}^n$, $\langle e_i(x),e_j(x)\rangle=\delta_{ij}, x\in\mathbb{R}^n$, such that
$$
\langle e_i(x),v(x)\rangle=v_i(x)>0 \text{ and} \langle e_i(x),w(x)\rangle=w_i(x)>0,\text{ for } x\in \mathbb{R}^n.
$$
Geometrically, it seems trivial but i have a problem when i have to do this formally.
 A: Edit: this is wrong, see comments.
My intuition is to build the basis to be "aimed at" $v+w$, i.e. so that $v+w$ is parallel to $u=\sum_i e_i$. This is obviously possible at a single point, and it's fairly simple to see that it satisfies the requirement: the angle between any $e_i$ and $u$ is at most $\pi/4$, and the angle between $u$ and either $v$ or $w$ is strictly less than $\pi/4$; so the angle between $e_i$ and $v$ or $w$ is less than $\pi/2$ as desired.
Thus we just need to show that we can choose such a basis in a smooth way. Letting $\delta_i$ denote the standard basis of $\mathbb R^n$, we see that this is equivalent to choosing a smooth field of rotations $R: \mathbb R^n \to SO(n)$ such that $R(x)(\delta_1 + \ldots +\delta_n)$ is parallel to $v(x)+w(x)$. Now it should be very intuitive that we can always do this locally - a proof of this should be attained easily using the implicit function theorem. To get a global solution, though, we need to stick our heads up in the clouds of topology for a little while.
Since all vectors involved are nonvanishing, we can normalize this to turn it into a more standard abstract problem: we have a smooth $X : \mathbb R^n \to S^{n-1}$ and we want to find a smooth $R : \mathbb R^n \to SO(n)$ such that $$R(x)p = X(x),\tag{1}$$ for some fixed $p \in S^{n-1}$. Since the map $\pi : SO(n) \to S^{n-1} : R \mapsto Rp$ is just the quotient map of $S^{n-1}=SO(n)/SO(n-1)$, in particular we know that $\pi$ is a smooth fiber bundle, and $(1)$ is just asking that $\pi \circ R = X$, i.e. that $R$ is a global section of the pullback bundle $X^* \pi$ over $\mathbb R^n$. Since every bundle over a contractible space is trivial, such a section exists.
This feels like overkill - perhaps if we relax the "aim" there is an easier proof. This was just my first idea.
