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Let $\{X_1,\dots ,X_m\}$ be $m$ independent vector fields on an $m$-dimensional smooth manifold $M$. Show that there are $m$ differential forms $\alpha_1, \dots, \alpha_m$ determined by the conditions that $\alpha_i(X_j) = 1$ if $i=j$ and $\alpha_i(X_j)$ if $i\neq j$ such that the form $\alpha_1 \wedge \dots \wedge \alpha_m$ is never zero.

Let $a\in M$, $U\subset M$ open such that $a\in U$ and $\mathrm{x} = (\mathrm{x}_1,\dots, \mathrm{x}_m):U\rightarrow \mathbb{R}^p$ a coordinate system. Now, since $\{X_1,\dots ,X_m\}$ are independent, $\{X_1,\dots ,X_m\}$ on $a$ form a basis of $T_aM$, so let $\alpha_i = \mathrm{dx_i}$ for $i= 1,\dots ,m$. Then $$\mathrm{dx_{i,a}}(X^{(j)}_a) = \frac{\partial \mathrm{x_i}}{\partial \mathrm{x}_j}(a) = 1$$ if $i=j$ and $0$ if $i\neq j$. So the conditions are satisfied. Furthermore $$\mathrm{dx}_1\wedge\dots\wedge\mathrm{dx_m}(X_1,\dots,X_m) =\mathrm{det}\ I_m = 1 \neq 0$$ and thus proving the result.

I think this proof is flawed. I started by fixing a point $a\in M$, but the coordinate system depends on $U$, and so the $\alpha_i$ are not unique. I feel like I wrote a bunch of symbols without really understanding them properly, so if someone could point out the mistakes and made and correct them explaining why, I'd be grateful.

EDIT I just realized, and this a wild guess, that maybe I could choose a differentiable partition of unity $\{\theta_i\}$ subordinate to the $U_i$ so defining $\omega = \sum_i \theta_i\alpha_i$ is then defined on any arbitrary point $a\in M$ regardless of the $U_i$.

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    The proof doesn't make sense: Since the basis $(X_i\vert_a)$ determines $(\alpha_i\vert_a)$, one cannot freely define $\alpha_i := dx_i$. (Indeed, unless all of the vector fields commute, this will not be true for *any* coordinates, let alone the ones chosen.) – 2017-02-08
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    Anyway, this observation shows that the claim is mostly linear-algebraic: For each point $a \in M$, $(\alpha_i\vert_a)$ is just the basis dual to $(X_i\vert_a)$, it follows from the definition that $(\alpha_1 \wedge \cdots \wedge \alpha_m)(X_1, \ldots, X_m)(a) > 0$. So, it remains just to show that the $\alpha_i$ are smooth. Smoothness is a local property, so it's enough to prove it in a local chart about an arbitrary point $a$, and we needn't worry about patching with a partition of unity. – 2017-02-08
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    @Travis Thank you! I just gathered together things without stop and thinking first if any of what I was doing make sense. I understand your approach, but I still don't understand why if I cover $M$ with coordinate charts $(\mathrm{x}_i, U_i)$, and if in each $U_i$ the differential form $\omega_i = \mathrm{dx}^{(i)}_1\wedge \dots \wedge \mathrm{dx}^{(i)}_m$ is never zero, then I can't use a partition of unity as in my edit to define $\omega$ in the whole $M$. – 2017-02-08
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    I'm not even sure what the purpose of that operation is, it doesn't seem to have much to do with the question, inasmuch as the $dx_k$ need not be nicely related to the $\alpha_k$. – 2017-02-09
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    @Travis I was trying to adapt the proof that orientability implies the existence of a non zero differential form of maximum degree without going through fixing an orientation first. I guess I was wrong. Thank you again. – 2017-02-09

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Hint: You can find a scalar product on $Vect(X_1,...,X_m)$ such that $\langle X_i,X_j\rangle =\delta_{ij}$ and extend it to a differentiable metric on $M$, write $\alpha_i=\langle X_i,.\rangle$.