One form wedge Product with $n - 1$ form. Let $a$ be an arbitrary one-form on $\mathbb{R}^{n}$. Then prove that there exists a $n - 1$ form $b$ on $\mathbb{R}^{n} - \{\mathbf{0}\}$ such that
$$
a \wedge b = dx^1 \wedge \cdots \wedge dx^n.
$$
I started to solve the question with taking a concrete example. I tried this on $\mathbb{R}^{3}$ and got $a_1 b_{23} - a_2 b_{13} + a_3 b_{12} = 1$ which satisfies the required conditions. But I can't figure out a way to generalise this.
 A: Using $\star dx_i$ as a basis for $n{-}1$ forms, let $b = b^i (\star dx_i)$. This is a nice basis because we have for all $i$ that $dx^i \wedge \star dx_i =dx^1 \wedge \cdots \wedge dx^n $ (no sum over $i$ here). $\star$ is the Hodge star operation.
Then $$a\wedge b = (a_i b^i ) dx^1 \wedge \cdots \wedge dx^n.$$ The assertion is then equivalent to the condition that for all $a_i(x)$ the equation $$\sum_{i=1}^n a_i(x) b^i (x) = 1$$ has at least one solution in the variables $b^i(x)$ for all $x$.
As Jack Lee points out in the comments, this is not true. If $a$ vanishes at at least one point then there exists $x$ such that $a_i(x) =0$ for all $i$. In this case the equation reduces to $0=1$ at $x$, which is inconsistent and has no solution for $b^i(x)$. Hence the assertion is false.
To produce an assertion which is true, assume that $a$ does not vanish anywhere on $\mathbb{R}^{n}$. Then for all $x$ there is at least one $i$ such that $a_i(x) \neq 0$. At each point $x$ the equation $\sum_{i=1}^n a_i(x) b^i (x) = 1$ is a single linear equation for the $n$ variables $b^i(x)$, which is an underdetermined linear system. This system is consistent given our assumption on $a$, and any consistent underdetermined linear system always has infinitely many solutions. Hence the assertion from the original post is true, provided that this additional assumption holds.
Edit;
Arctic Char points out that an instructive choice for the solution to the system could be $$b = \frac{ \star a}{\langle a,a\rangle},$$ where $\langle x,y\rangle$ is the inner product with the metric. Note that $a\wedge \star a = \langle a,a\rangle \sqrt{|\det g|} dx^1 \wedge \cdots \wedge dx^n$, where $g$ is the metric.
On $\mathbb{R}^{n}$ with Euclidean metric, we find that $b^i = \delta^{ij}\frac{ a_j}{a_1^2+...+a_n^2} =\frac{ a_i}{a_1^2+...+a_n^2}.$ It could be an instructive calculation to verify that $a\wedge b = dx^1 \wedge \cdots \wedge dx^n$ with this $b_i$.
