$\textbf{Exercise 15.}$ Let $v$ be a differentiable vector field in $\mathbb{R}^3$. Prove that:
a) If div $v = 0$, then there exists a vector field $u \in \mathbb{R}^3$ such that rot $u = v$.
b) If rot $v = 0$, then there exists a function $f$ in $\mathbb{R}^3$ such tht grad $f = v$.
I will introduce some definitions and results that appear in some exercises of this same book that I think is necessary to prove this exercise.
We have the following definition in the exercise $10$ of the chapter $1$:
$\textbf{Definition of The Hodge star operation.}$ Given a $k$-form $\omega$ in $\mathbb{R}^n$ we will define an $(n-k)$-form $* \omega$ by setting
$$* (dx_{i_1} \wedge \cdots \wedge dx_{i_k}) = (-1)^{\sigma} (dx_{j_1} \wedge \cdots \wedge dx_{j_{n-k}})$$
and extending it linearly, where $i_1 < \cdots < i_k$, $j_1 < \cdots < j_{n-k}$, $(i_1, \cdots, i_k, j_1, \cdots, j_{n-k})$ is a permutation of $(1,2,\cdots,n)$ and $\sigma$ is $0$ or $1$ according to the permutation is even or odd, respectively.
We have that if $\omega$ is a $k$-form defined in $\mathbb{R}^n$, then $** \omega = (-1)^{k(n-k)} \omega$ by the item $c$ of the exercise $10$ of the chapter $1$.
The item $b$ of exercise $11$ of chapter $1$ states that if $v: \mathbb{R}^n \longrightarrow \mathbb{R}$ is a differentiable vector field and $\omega$ is the differential $1$-form obtained from $v$ by the canonical isomorphism induced by the inner product $\langle , \rangle$ and $\nu$ is the volume element of $\mathbb{R}^n$ (Exercise $9$), the divergence can be obtained as follow:
$$v \mapsto \omega \mapsto * \omega \mapsto d(* \omega) = \text{div} \ v \ \nu$$
$\textbf{Definition of the rotational.}$ Let be $v: \mathbb{R}^n \longrightarrow \mathbb{R}$ is a differentiable vector field. The rotational form $\text{rot} \ v$ is the $(n-2)$-form defined by
$$v \mapsto \omega \mapsto d\omega \mapsto *(d\omega) = \text{rot} \ v,$$
where $v \mapsto \omega$ is the correspondence between $1$-forms and vector fields induce by the inner product.
Keep in mind these definition and the item $b$ of the exercise $11$ of chapter $1$, I tried prove the exercise $15$.
$\textbf{My attempt:}$
a)
I tried another approach to do the item $a$:
By the question $14.a$ in chapter $4$ of this do Carmo's book, I know that
Using this and the question $14.b2$ in chapter $4$ of this do Carmo's book (which I attached previously on this post), I know that if $v := A e_1 + B e_2 + C e_3$ and $\text{div} \ v = 0$, then exist differentiable real functions $P,Q,R$ on $\mathbb{R}^3$ which satisfy the system given in question $14$ of chapter $4$, but the left hand side of the equations of the system is precisely the rotational of a differentiable vector field $u: \mathbb{R}^3 \longrightarrow \mathbb{R}$ defined by $u(x) := P(x) e_1 + Q(x) e_2 + R(x) e_3$ by the question $14.b2$. Thus, $\text{rot} \ u = v$.
b)
If $\text{rot} \ v = 0$, then $*(d\omega) = 0$. By the item $c$ of the exercise $10$ of the chapter $1$,
$$**(d\omega) = (-1)^{2(3-2)} d\omega = d\omega$$
and $**(d\omega) = 0$ by the definition of the Hodge star operation. Thus, $d\omega = 0$. Since $\mathbb{R}^3$ is contractible to a point for every point of $\mathbb{R}^3$, it follows from Poincaré's lemma that exists a $0$-form $f$ such that $\omega = df$. By the correspondence between vector fields and $1$-forms, we conclude that $\text{grad} \ f = v$.
I would like to know if my attempts for the items $a$ and $b$ are correct.
Thanks in advance!