I found this in the fluid mechanics: $F(u)=\frac{\partial u_x}{\partial x}\frac{\partial u_y}{\partial y}+\frac{\partial u_x}{\partial x}\frac{\partial u_z}{\partial z}+\frac{\partial u_y}{\partial y}\frac{\partial u_z}{\partial z}-\frac{\partial u_x}{\partial z}\frac{\partial u_z}{\partial x}-\frac{\partial u_x}{\partial y}\frac{\partial u_y}{\partial x}-\frac{\partial u_y}{\partial z}\frac{\partial u_z}{\partial y}$
where $u=[u_{x},u_{y},u_{z}]$ is a vector field, I just wounder that how to indicate $F(u)$ with nabla operator $\nabla$?
Writing this expression in Einstein summation notation:
$$F(u)=\frac{1}{2}\frac{\partial u_i}{\partial x_i}\frac{\partial u_j}{\partial x_j} - \frac{1}{2}\frac{\partial u_i}{\partial x_j}\frac{\partial u_j}{\partial x_i}$$
Note that the outer product of $u$ and $\nabla$ is precisely $(\nabla\otimes u)_{ij}=\frac{\partial u_i}{\partial x_j}$, and so:
$$F(u)=\frac{1}{2}(\nabla\otimes u)_{ii}(\nabla\otimes u)_{jj} - \frac{1}{2}(\nabla\otimes u)_{ij}(\nabla\otimes u)_{ji}$$
Now note that, summing over repeated indices, $A_{ii}=\mathrm{tr}(A)$ for a matrix $A$, and $(A^2)_{ij}=A_{ik}A_{kj}$. Using these two facts, we have immediately that
$$F(u)=\frac{1}{2}(\mathrm{tr}(\nabla\otimes u))^2-\frac{1}{2}\mathrm{tr}((\nabla\otimes u)^2)$$
This is a nicely symmetric result. However, there is another helpful identity: the trace of the outer product is equivalent to the inner product. As such, we can write the first term as a divergence. Unfortunately, I'm not aware of a simple expression for the second term, so this is an alternative (not necessarily simpler or better) formulation:
$$F(u)=\frac{1}{2}(\nabla\cdot u)^2-\frac{1}{2}\mathrm{tr}((\nabla\otimes u)^2)$$