Pull back of a vector representing a 2-form in $\mathbb R^3$ Let $\omega$ be a $2$-form defined in $\mathbb R^3$. I know that it can be represented by a vector field $\xi$ in such a way:
$$
\omega_x (v,w) = \xi(x)\cdot (v\times w)
$$
($x,v,w \in \mathbb R^3$ and $\times$ is the vector product). And I know that the differential becomes the divergence:
$$
d \omega = (\nabla\cdot \xi)\ dx_1\wedge dx_2\wedge dx_3.
$$
Now if $T\colon \mathbb R^3 \to \mathbb R^3$ is a change of variables, I know that
$$
d(T^*\omega) = T^* (d\omega).
$$
I would like to prove the corresponding equality for vector fields (without passing to forms). I was able to make the corresponding computation for $1$-forms (and it wasn't trivial) but I'm stuck with $2$-forms.
What I've found is that if $\omega$ corresponds to $\xi$ as above, then 
$$
T^* \omega(v,w) = \omega(DTv,DTw) = \xi \cdot(DTv\times DTw) = (\det DT) (DT)^{-1}\xi\cdot(v\times w)
$$
hence the vector-field representing $T^*\omega$ is
$$
\xi^* = (\det DT)(DT)^{-1}\xi = \mathrm{adj}(DT) \xi.
$$
Now I would like to prove that
$$
\nabla \cdot \xi^* = (\det DT)\  \nabla \cdot \xi
$$
which should be the correct change of variables. But I have no clue on how to work on this...
 A: Divergence of a vector field $Y_p$ shows by how much the volume form $\mu$ (or infinitesimal volume elements) changes along the flow:
$$
(\operatorname{div} Y)\mu=\mathcal{L}_Y(\mu)=d(Y\lrcorner\mu),
$$
 where I also used the Cartan's magic formula. The interior product is the $(n-1)$-form $$[Y\lrcorner\mu](Y_1,Y_2,\ldots,Y_{n-1})=\mu(Y, Y_1,Y_2,\ldots,Y_{n-1}).
$$
The two expressions for divergence form will help us pull it back along a given map, or coordinate change, $\phi$. We now see the vector (field) Y as a result of pushforward by the map derivative $Y_{\phi(p)}=\phi_*(X_p)$. The pullback of the volume form shows the expected determinant of the Jacobian matrix, $f=det(\phi_*)$,
$$
\phi^*\mu=f\nu
$$
-- this is another factor that may cause the volume form $\nu$ to change.
The pull back of the interior product form is
$$
\phi^*(Y\lrcorner\mu)=X\lrcorner \phi^*\mu=X\lrcorner (f\nu).
$$
The pullback commutes with exterior derivative, and so
$$
\phi^*d(Y\lrcorner\mu)=d(\phi^*(Y\lrcorner\mu))=d(X\lrcorner f\nu)=\mathcal{L}_X(f\nu),
$$
where we've again used the Cartan's formula.
We will now use properties of Lie derivative and the atiderivation of the inner product to show that
$$
\mathcal{L}_X(f\nu)=(\mathcal{L}_X\;f)\nu +f\mathcal{L}_X\nu=\\
(X\lrcorner df)\nu+\left(\mathcal{L}_{fX}(\nu)-df\wedge(X\lrcorner\nu) \right)=\\
\mathcal{L}_{fX}(\nu)+\left((X\lrcorner df)\nu -df\wedge(X\lrcorner\nu) \right)=\\
=\mathcal{L}_{fX}(\nu)+X\lrcorner(df\wedge\nu)=\\
\mathcal{L}_{fX}(\nu)+0=\operatorname{div}(fX)\nu.
$$
Therefore,
$$\phi^*((\operatorname{div} Y)\mu)=\operatorname{div}(fX)\nu,$$
which suggests that the original volume form $\nu$ may change due to both divergence of the field along which we observe it and the mapping with the determinant scalar field $f$.
