Wedge product is zero

Question

Suppose $C$ is a scalar, $\mathcal{Y}$ is one-form and there is the equation

$$ dC\wedge \mathcal{Y}=0 \quad (1)$$

What is the most general solution of this equation?

Using the geometrical interpretation of the wedge product I would say that there is a solution $$ \mathcal{Y}=f(C)dC,\quad (2)$$ but is this the most general one? Probably the answer is no, but I would like to know other opinions.

In the tensorial form the equation (1) is $$\partial_s C Y_t - Y_s \partial_t C=0,\quad (3)$$ here $\mathcal{Y}=Y_adx^a$. Obviously (2) is a solution, but again it is not clear that (2) is the most general one.

Answer 1

There is the theorem (Theorem 3 in Wedge Product) saying that $v_1\wedge v_2\wedge \cdots \wedge v_k=0$ if and only if $$v_j=\sum_{i=1,i\ne j} a_i v_i$$ for some $j$, $j\le 1\le k$. Applying it to (1) we get the most general solution in the form $$\mathcal{Y}=f dC,$$ with some arbitrary function $f$. The solution (2) is correct ONLY if we require additionally $d\mathcal{Y}=0$, which gives $f=f(C)$.

Answer 2

For some 1-forms $X$ and $Y$ written as \begin{eqnarray} X &= \varepsilon_1 dx^1 + \varepsilon_2 dx^2 + \cdots + \varepsilon_n dx^n \\ Y &= \pi_1 dx^1 + \pi_2 dx^2 + \cdots + \pi_n dx^n \end{eqnarray}

Then the condition \begin{eqnarray} X \wedge Y &= 0 \\ \implies \sum_{1\le i<j \le n}(\varepsilon_i\pi_j-\pi_i\varepsilon_j)dx^i\wedge dx^j &=0 \end{eqnarray}

And the result you're looking for follows.