Curl of a linear operator applied to a vector field

Question

Let $A$ be a $3\times 3$ matrix such that $A_{ij}$ is a smooth real valued function on $\mathbb{R}^3$ and let $v$ be a smooth vector field in $\mathbb{R}^3$. My problem is to simplify the expression: $$ \text{curl}(Av)\, . $$ Here is my initial attempt. Using the Einstein summation convention, we have $(Av)_i = A_{ij}v_j$ and $(\text{curl}(B))_i = \epsilon_{ijk}\partial_jB_k$. Then, \begin{align*} (\text{curl}(Av))_i &= \epsilon_{ijk}\partial_j(A_{kl}v_l)\\ &=\epsilon_{ijk}v_l\partial_jA_{kl}+ \epsilon_{ijk}A_{kl}\partial_jv_l\, . \end{align*} How can I simplify this further to get an expression involving the standard vector differential operators?

Answer 1

The "standard vector differential operators" operate on scalar fields, vector fields and their various products. Their outputs are again scalar or vector fields. But these operators cannot deal with "matricial objects". We also don't have, e.g., the gradient of a vector field, because this would be a "matricial object".

I don't know if the following could be of help: If $f$ is a scalar field and $\vec a$ a vector field then $${\rm curl}(f\,\vec a)=f\,{\rm curl}(a)+\nabla f\times\vec a\ .\tag{1}$$ Denote the $j^{\rm\, th}$ column of your matrix $A$ by $\vec A_j$. Then you can write the field $Av$ whose ${\rm curl}$ has to be calculated in the form $$Av=\sum_{j=1}^3 v_j\,\vec A_j\ ,\tag{2}$$ where now the components of the field $v$ appear as scalar functions. Applying $(1)$ to the three summands in $(2)$ individually we obtain $${\rm curl}(Av)=\sum_{j=1}^3\bigl(v_j\,{\rm curl}(\vec A_j)+\nabla v_j\times \vec A_j\bigr)\ .$$ But I don't see a way to encode each of the sums $$\sum_{j=1}^3 v_j\,{\rm curl}(\vec A_j)\>,\qquad\sum_{j=1}^3\nabla v_j\times \vec A_j$$ in a coordinate free way as "vectorial objects".