TITLE: Is $(\nabla \times \nabla)$ an operator?
QUESTION [3 upvotes]: Is $(\nabla \times \nabla)$ an operator?
I am wondering if its possible to compute
\begin{align}
\vec{f} \cdot (\nabla \times \nabla)
\end{align}
Where $f$ is a vector. I would be interested in both cartesian and polar expressions.

REPLY [3 votes]: By equality of mixed partials, that is the fact that:
$$
\begin{align}
\frac{\partial^2g}{\partial y \partial z}&=\frac{\partial^2g}{\partial z \partial y}\\
\frac{\partial^2g}{\partial z \partial x}&=\frac{\partial^2g}{\partial x \partial z}\\
\frac{\partial^2g}{\partial x \partial y}&=\frac{\partial^2g}{\partial y \partial x},
\end{align}
$$
where $g$ is some function of $x$, $y$, and $z$, one has the following result for $C^2$ functions:
$$
\begin{align}
\vec{f}\cdot\left(\nabla \times \nabla\right) g
&=\vec{f}\cdot\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z}\\ 
\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \end{vmatrix} g\\
&= \vec{f}\cdot\left(\hat{x} \left(\frac{\partial^2g}{\partial y \partial z}-\frac{\partial^2g}{\partial z \partial y}\right) +\hat{y} \left(\frac{\partial^2g}{\partial z \partial x}-\frac{\partial^2g}{\partial x \partial z}\right) +\hat{z} \left(\frac{\partial^2g}{\partial x \partial y}-\frac{\partial^2g}{\partial y \partial x}\right)\right)\\
&=\vec{f}\cdot\vec{0}\\
&=0
\end{align}
$$
So, this is, as user297767 said, merely a fancy way of writing the zero operator.

If this is too similar to user297767's answer I apologize, but being as my edit of their post was rejected for "deviating from the intent of the post", I feel as though it is reasonable for me to post this as an answer.
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