For $\omega$ and $\eta$ k-forms exist a $C^{1}$ function $f: \mathbb{R}^{3} \to \mathbb{R}$ such that $\eta = f\omega$.

Question

Let's consider a $k$- form $\omega$,

$$\omega = \sum_{i_{1} < ... < i_{k}} \omega_{i_{1}, ..., i_{k}} dx^{i_{1}} \wedge ... \wedge dx^{i_{k}}$$

$\omega$ is $C^{r}$ if $\omega_{i_{1},... ,i_{k}}$ is $C^{r}$. Consider $\Omega^k(U)$ the set of $C^{\infty}$ $k$-forms in $U$.

My question is the following:

Let $\omega,\ \eta\ \in \Omega^{1}(\mathbb{R}^{3})$. If $\omega(x) \neq 0$ for all $x \in \mathbb{R}^{3}$ and $\omega \wedge \eta = 0$, then exist a $C^{1}$ function $f: \mathbb{R}^{3} \to \mathbb{R}$ such that $\eta = f\omega$.

My attempt:

consider $\omega = \omega_{1}dx + \omega_{2}dy + \omega_{3}dz$ and $\eta = \eta_{1}dx + \eta_{2}dy + \eta_{3}dz$, thus

$$\omega \wedge \eta = (\omega_{1}\eta_{2} - \omega_{2}\eta_{1}) dx \wedge dy + (\omega_{1}\eta_{3} - \omega_{3}\eta_{1}) dx \wedge dz +\\ (\omega_{2}\eta_{3} - \omega_{3}\eta_{2}) dy \wedge dz$$

Using that $\omega \wedge \eta = 0$ and $\{dx \wedge dy, dx \wedge dz, dy \wedge dz \}$ is L.I. we have that

1. $\omega_{1}\eta_{2} = \omega_{2}\eta_{1}$;
2. $\omega_{1}\eta_{3} = \omega_{3}\eta_{1}$;
3. $\omega_{2}\eta_{3} = \omega_{3}\eta_{2}$.

Remember that we want a function $f$ such that $f\omega_{i} = \eta_{i}$. Suppose that exist $p \in \mathbb{R}^{3}$ such that $\omega_{3}(p) = 0$, then $\omega_{1}\eta_{3} = 0$ and $\omega_{2}\eta_{3} = 0$ and so $\omega_{1}(p) = 0$ or $\omega_{2}(p) = 0$ or $\eta_{3}(p) = 0$.

If $\eta_{3}(p) \neq 0$, then $\omega(p) = 0$, contradiction. So, we conclude that $\eta_{3}(p) = 0$. Hence, we can define $f$ like $$f(p) = \begin{cases} \frac{\eta_{3}}{\omega_{3}}(p), &\text{ if }\omega_{3}(p) \neq 0, \\ 0, & \text{ if }\omega_{3}(p) = 0.\end{cases}$$

My problem is prove that $f$ is a $C^{1}$ function. Someone can help me?

Answer 1

Unforunately your $f$ so defined might not be $C^1$: for example, if $\eta = \omega$, then from your construction you would get $f=1$ when $\omega_3(p) \neq 0$. So extending it by $0$ actually makes it discontinuous.

To deal with this, we consider the following: Let $U_i$, $i=1, 2, 3$ be the open sets

$$ U_i = \{ p\in \mathbb R^3: \omega_i (p) \neq 0\}.$$

Then since $\omega$ is nonzero everywhere, we have $U_1\cup U_2\cup U_3 = \mathbb R^3$.

On each $U_i$, define $f_i = \frac{\eta_i}{\omega_i}$. Note that they are well defined and $C^1$ (Indeed, $C^\infty$). Note that we can show $f_i = f_j$ on the intersection $U_i \cap U_j$: for example, using $$\omega_1 \eta_2 = \omega_2 \eta_1 $$ we have $$ f_2 =\frac{\eta_2}{\omega_2} = \frac{\eta_1}{\omega_1} = f_1$$ on $U_1\cap U_2$. Thus the function

$$f(x) = \begin{cases} f_1(x) & x\in U_1 \\f_2(x) & x\in U_2 \\f_3(x) & x\in U_3 \end{cases}$$ is a well-defined $C^1$ functions on $\mathbb R^3$ and $\eta = f\omega$.