I want to know if it is possible to express the operation


$$
\nabla \phi \times (\nabla \times \mathbf A) 
$$


as the divergence of second order tensor field $T$. Here $ \phi$ is a scalar field and $\mathbf A$ is a solenoidal vector field ($\nabla \cdot \mathbf A=0$)


I have used all possible identities and finally I can only get 


$$ 
\nabla \phi \times (\nabla \times \mathbf A) = \nabla \phi \cdot(\nabla \mathbf A - \nabla \mathbf A^T) 
$$


Is it possible to find something like $ \nabla \phi \times (\nabla \times \mathbf A) = \nabla \cdot(\text{tensor}) $ ?




---


EDIT: I could derive an expression that may be equivalent to Carlo's answer. 


$$
\nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A-\nabla \mathbf A^T) \cdot \nabla \phi
$$
Now, from the identity $\nabla \cdot (\phi \mathbf T)=\phi\nabla\cdot \mathbf T+ \mathbf T^T \nabla \phi$ we can write


$$
(\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi = -\nabla \cdot(\phi(\nabla \mathbf A - \nabla \mathbf A^T))+\phi\nabla\cdot(\nabla \mathbf A - \nabla \mathbf A^T)
$$
Considering that $\nabla \cdot(\nabla \mathbf A)=\nabla^2 \mathbf A$ and $\nabla \cdot(\nabla \mathbf A^T)=\nabla (\nabla \cdot \mathbf A)=0 $:


$$
\nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi = -\nabla\cdot(\phi(\nabla \mathbf A-\nabla \mathbf A^T)+\phi\nabla^2 \mathbf A
$$ 
This form is convenient since the LHS of the equation already contains $\nabla^2 \mathbf A$.



Curl as a divergence... Is it possible?
Let me expand my comment that I think this is only possible if the Laplacian of $\phi$ vanishes.  

The object we are considering is
$$\mathbf V\equiv\nabla\phi\times(\nabla\times \mathbf A)=(\nabla \mathbf A)·\nabla \phi - (\nabla\phi)·\nabla \mathbf A.$$
Writing this in components,
$$V\_i=\sum\_j(\partial\_i A\_j) (\partial\_j\phi)-(\partial\_j\phi)(\partial\_j A\_i),$$
using the chain rule and $\nabla\cdot\mathbf A=0$,
$$V\_i=\sum\_j\partial\_j(\phi\partial\_i A\_j)-\partial\_j(A\_i\partial\_j\phi)+A\_i(\partial\_j\partial\_j\phi),$$
hence we arrive at
$$\mathbf V=\nabla\cdot\mathbf M^T+\mathbf A(\Delta\phi),$$
with $\Delta$ the Laplacian and the tensor
$$\mathbf M=\phi\nabla\mathbf A-\mathbf A\nabla\phi.$$
This has the form required in the OP if the Laplacian of $\phi$ vanishes.


