I'm reading reading about differential forms and I keep finding alternative notions and I'm quite fed up with the inconsistency. The way I first learned about the is that a $k$-form $\omega$ is a smooth section of the $k$th exterior power of the cotangent bundle i.e. $\omega \in \Gamma \left(\bigwedge^k(T^\ast M)\right)$. This makes sense since $\omega : M \to \bigwedge^k(T^\ast M)$ gives us $\omega_p \in \bigwedge^k(T^\ast M)=\bigcup_{p \in M} \bigwedge^k(T^\ast_pM)$ which further means that $\omega_p$ is a multilinear map $T_pM \times \dots \times T_pM \to \mathbb{R}$. And since $\bigwedge^k(T^\ast M)$ is a vector space with sufficiently nice basis we can express $\omega$ as $$\omega = \sum_{I} a_I dx_I.$$
I'm now reading an example where they consider $\mathbb{R}^n -\{0\}$ and $\omega:\mathbb{R}^n-\{0\} \to \bigwedge^k(\mathbb{R}^n-\{0\})$ as $$\omega_p(v_1,\dots,v_{n-1})=dx_1 \wedge \dots \wedge dx_n \left(\frac{p}{|p|},v_1,\dots, v_{n-1} \right)$$ and this is supposed to be an $(n-1)$-form.
The issue here is that no general formula of $\omega$ is being given. It's only given that the evaluation at $p$ has that formula. How can I recover $\omega$ from this if I only now the evaluation at $p$? I have no information about $a_I(p)$ except that it's apparently $1$ all the time so is it safe to assume that $a_I =1$ since it evaluates to $1$ for all inputs?
Best Answer
What is meant is that you take the constant form $dx_1\wedge\dots\wedge dx_n$, then plug in $\tfrac{p}{|p|}$ as the first entry and leave the other $n-1$ entries free to obtain an $(n-1)$-Form. You can easily express this as a linear combination of the possible wedge products of $(n-1)$ of the $dx_i$ by plugging into the definition of the wedge product. The result should essentially be $$ \omega_x=\sum_{i=1}^n(-1)^{i-1}\frac{x_i}{|x|}dx_1\wedge\dots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\dots\wedge dx_n. $$