Where is the grad trace operator ($\nabla \operatorname{trace} A$ for matrix field $A$) used? I recently stumbled upon the linear partial differential operator $$\mathcal{L}A\colon=\nabla \mathrm{tr} A\quad\text{for matrix-valued fields } A\colon\mathbb{R}^d\rightarrow\mathbb{R}^{d\times d}.$$
In coordinates, this is $(\mathcal{L}A)_i=\partial_i(\sum_j A_{jj})$. In fact, the operator that arose was $$\nabla \mathrm{tr}-\mathrm{div},$$where the divergence is taken row wise, i.e., $(\mathrm{div}A)_i=\sum_j \partial_j A_{ij}$; also, my matrix fields are symmetric.
I could swear I've seen a fluid mechanics reference where the second operator appeared, but I cannot remember what it was. I could, of course, be wrong.
Could you please let me know if you've encountered any of these operators in applications or in any place in the literature?
Thank you!
 A: I suspect this is the same thing as the other answer but I lack the fluency in differential geometry to verify. Note that if you apply trace to a jacobian, you end up with div. So in the context of fluid mechanics, I bet it arose from the double curl formula
$$ \nabla\times(\nabla\times u) = [\nabla,\text{div}]u = \nabla \operatorname{div} u - \Delta u=  \nabla\operatorname{tr} (\nabla u) - \operatorname{div} (\nabla u)$$
For divergence free fields one can invert the Laplacian to recover the "velocity" $u$ from the "vorticity" $\omega=\nabla \times u$ (this is called the "Biot-Savart law")
A: $\def\tr{\mathrm{tr}}$
$\def\div{\mathrm{div}}$
$\def\grad{\mathrm{grad}}$
$\def\d{\mathrm{d}}$
$\def\R{\mathbb{R}}$
I will suppose your divergence operator is
$$(\div A)_i = \sum_{j=1}^d \partial_j A^j{}_{i}.$$
Admittedly, this answer is rather convoluted, but you can interpret your matrix field $A$ as a vector-valued differential $1$-form, since at any point $x\in\R^d$ you have an operator $A(x)$ which acts linearly on a vector $v\in\R^d$ and returns a vector $A(x)(v)\in\R^d$.
Then the components of the exterior derivative of $A$ are
$$(\d A)^k{}_{ij}=2\partial_{[i} A^{k}{}_{j]}
= \partial_iA^k{}_j-\partial_jA^k{}_i$$
where, the brackets are the antysimmetrization operator. Then your operator $\grad\circ\tr-\div$ is the exterior derivative followed by the trace (taken on the second index):
$$\begin{aligned}
(\tr_2(\d A))_i
&= \sum_{j=1}^d(\d A)^j{}_{ij} \\
&= \sum_{j=1}^d\partial_iA^j{}_j - \sum_{j=1}^d\partial_jA^j{}_i \\
&= \partial_i(\tr A) - (\div A)_i.
\end{aligned}$$
Hence,
$$\tr_2(\d A) = \grad(\tr A) - \div A.$$
I seriously doubt this is what you want, though?
