Rotation of a curl? Given an arbitrary constant 3-D rotation matrix $\mathbf{R}$ and a 3-D vector field $\mathbf{A}$,
how can I find the vector field $\mathbf{B}$ such that
$$\nabla\times\mathbf{B}=\mathbf{R}\left(\nabla\times\mathbf{A}\right)$$ ?
I followed two ways, without really finding a way out


*

*I wrote the LHS and RHS explicitly. If I write $\mathbf{R}$ as the matrix
$$\mathbf{R} = \begin{pmatrix}a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$$
(which is obviously an overkill, since the actual degrees of freedom of an arbitrary rotation are just three), then my equation looks like the nasty
\begin{align*}
\partial_{y}B_{z}-\partial_{z}B_{y} & =a\left(\partial_{y}A_{z}-\partial_{z}A_{y}\right)+b\left(\partial_{z}A_{x}-\partial_{x}A_{z}\right)+c\left(\partial_{x}A_{y}-\partial_{y}A_{x}\right)\\
\partial_{z}B_{x}-\partial_{x}B_{z} & =d\left(\partial_{y}A_{z}-\partial_{z}A_{y}\right)+e\left(\partial_{z}A_{x}-\partial_{x}A_{z}\right)+f\left(\partial_{x}A_{y}-\partial_{y}A_{x}\right)\\
\partial_{x}B_{y}-\partial_{y}B_{x} & =g\left(\partial_{y}A_{z}-\partial_{z}A_{y}\right)+h\left(\partial_{z}A_{x}-\partial_{x}A_{z}\right)+i\left(\partial_{x}A_{y}-\partial_{y}A_{x}\right)
\end{align*}
which intimidates me quite a bit... Rearranging it acquires an interesting structure,
\begin{align*}
\partial_{y}B_{z}-\partial_{z}B_{y} & =\partial_{x}\left(cA_{y}-bA_{z}\right)+\partial_{y}\left(aA_{z}-cA_{x}\right)+\partial_{z}\left(bA_{x}-aA_{y}\right)\\
\partial_{z}B_{x}-\partial_{x}B_{z} & =\partial_{x}\left(fA_{y}-eA_{z}\right)+\partial_{y}\left(dA_{z}-fA_{x}\right)+\partial_{z}\left(eA_{x}-dA_{y}\right)\\
\partial_{x}B_{y}-\partial_{y}B_{x} & =\partial_{x}\left(iA_{y}-hA_{z}\right)+\partial_{y}\left(gA_{z}-iA_{x}\right)+\partial_{z}\left(hA_{x}-gA_{y}\right)
\end{align*}
but I don't really know how to continue.. I don't think that replacing the actual rotation matrix elements would even help..

*Then I tried a different, let's say geometric, way. I Fourier transformed both sides of my equation and, as (1) the Fourier transform of the $\nabla$ operator is $i\mathbf{k}$ and (2) the Fourier transform commutes with a rotation matrix, I obtained
$$i\mathbf{k}\times\tilde{\mathbf{B}}=\mathbf{R}\left(i\mathbf{k}\times\tilde{\mathbf{A}}\right)$$
which is a nice expression involving just vectors and not nasty partial derivatives. Now, I can rewrite this as
$$\mathbf{k}\times\tilde{\mathbf{B}}=\mathbf{R}\left(\mathbf{k}\right)\times\mathbf{R}\left(\tilde{\mathbf{A}}\right)$$
but this expression, although simple and nice, does not really help me find $\mathbf{B}$...
Perhaps I am just rusted and I am missing something simple....
 A: I think the reason you're having trouble is that there may be no such vector field $B$. Let's take an example: 
$$A(x, y, z) = \pmatrix{z\sin x + y \cos x\\ 0\\ 0}
$$
Then 
$$
curl ~ A(x, y, z) = \pmatrix{0 \\ \sin x \\ -\cos x}
$$
if I've done the computation right. And notice that the divergence of that is $0$, because the divergence of the curl of any vector field on a contractible set like $\Bbb R^3$ is always zero (at least if everything is $C^2$, which it certainly is in this case). 
Now let 
$$
R = \pmatrix{0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 1}.
$$
Then let's apply $R$ to the curl of $A$ and give it a name:
$$
Q = R(\nabla \times A) = \pmatrix{0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 1}\pmatrix{0 \\ \sin x \\ -\cos x} = \pmatrix{\sin x \\ 0 \\ -\cos x}.
$$
Now the divergence of $Q$ is 
$$
\nabla \cdot (R(\nabla \times A)) = \cos x + 0 + 0 = \cos x,
$$
which is not zero. That means that $Q$ cannot be the curl of any vector field. And that's why you couldn't find one. :) 
A: Let me add an interesting fact to the already existing answer: even though $x \mapsto R \: \text{rot}(F)(x)$ need not be a rotation field again (as John's answer demonstrates), the vector field $x \mapsto R \: \text{rot}(F)(R^{-1}x)$ is again a rotation field. In fact one has
\begin{equation}
R \: \text{rot}(F)(R^{-1}x) = \text{rot}(R F(R^{-1}x))
\end{equation}
I don't know of a nice way to show this though, except by direct computation.
