Difference between $\nabla(\nabla . F)$ and $\nabla^2F$, where $F$ is a vector function

Question

I recently started vector calculus and saw that following identity

$\nabla\times(\nabla\times F)=\nabla(\nabla.F)-\nabla^2F$

But aren't $\nabla(\nabla.F)$ and $\nabla^2F$ the same or I am missing something.

In a scalar function $f$ we define $\nabla^2f = \frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}$, is there a similar definition for $\nabla^2F$, where $F$ is a vector function

Answer 1

No. Typically $\nabla^2$, the Laplacian, is $\nabla\cdot\nabla$, is an operator which maps scalar functions to scalar functions. Clearly $$\nabla \nabla \cdot \neq \nabla \cdot \nabla$$ the first is the gradient of a divergence, the second is the divergence of the gradient.

Wikipedia defines the vector Laplacian $\nabla^2 F$, for vectorial $F$, by the formula you give. In Cartesian coordinates, the vector Laplacian is the scalar Laplacian applied component-wise, $$\nabla^2 [f_x,f_y,f_z] = [\nabla^2 f_x,\nabla^2 f_y,\nabla^2 f_z]$$

Answer 2

In fact, $\nabla^2 F$ is defined to be the divergence of the gradient, i.e. $\nabla\cdot \nabla F$. On the other hand, $\nabla(\nabla \cdot F)$ is the gradient of the divergence.