What is a nice form of curl(v)^2? Can $(\nabla \times v)\cdot(\nabla \times v)$ be written in a neat way?
Especially if $\nabla \cdot v = 0$...
 A: In geometric algebra, you will easily obtain a very simple expression. Write $\nabla\times v=-\nabla\wedge vI=-I\nabla\wedge v$ where $I$ is pseudoscalar and $I^2=II=-1$ in 3 dimension.
$$\begin{align}(\nabla\times v)\cdot(\nabla\times v)&=(-\nabla\wedge vI)\cdot(-I\nabla\wedge v)\\
&=\nabla\wedge v(II)\nabla\wedge v\\
&=-(\nabla v-\nabla\cdot v)(\nabla v-\nabla\cdot v)\\
&=-(\nabla v)^2+2(\nabla\cdot v)\nabla v-(\nabla\cdot v)^2\end{align}$$
Particularly, if $\nabla\cdot v=0$, the expression reduces to $-(\nabla v)^2$.
A: Using Einstein's notation we have (in 3 dimensions)
$$(\nabla \times v)_i=\epsilon_{ijk}\partial_j v_k,$$
for al $i=1,2,3$, denoting by $\epsilon_{ijk}$ the Levi Civita symbol.
Then
$$(\nabla \times v)\cdot (\nabla \times v)=(\nabla \times v)_i(\nabla \times v)_i=
\epsilon_{ijk}\partial_j v_k\epsilon_{irs}\partial_r v_s.$$
Using 
$$\epsilon_{ijk}\epsilon_{irs}=\delta_{jr}\delta_{ks}-\delta_{js}\delta_{kr}$$
we arrive at 
$$(\nabla \times v)\cdot (\nabla \times v)=(\delta_{jr}\delta_{ks}-\delta_{js}\delta_{kr})
\partial_j v_k\partial_r v_s=\partial_j v_k\partial_j v_k-\partial_j v_k\partial_k v_j=\sum_{j\neq k} \left[(\partial_j v_k)^2-\partial_j v_k\partial_kv_j\right].
$$
A: Let $\mathbf{B}=\nabla\times\mathbf{A}$ in the product rule, $\nabla\cdot(\mathbf{A}\times\mathbf{B}) = \mathbf{B}\cdot(\nabla\times\mathbf{A}) - \mathbf{A}\cdot(\nabla\times\mathbf{B})$:
$$\nabla\cdot(\mathbf{A}\times(\nabla\times\mathbf{A})) = (\nabla\times\mathbf{A})\cdot(\nabla\times\mathbf{A}) - \mathbf{A}\cdot(\nabla\times(\nabla\times\mathbf{A})),$$
or rearranging,
$$ (\nabla\times\mathbf{A})\cdot(\nabla\times\mathbf{A}) = \nabla\cdot(\mathbf{A}\times(\nabla\times\mathbf{A})) + \mathbf{A}\cdot(\nabla\times(\nabla\times\mathbf{A})) .$$
We can further use the common identity for second derivatives, $\nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A}, $ so that
$$(\nabla\times\mathbf{A})\cdot(\nabla\times\mathbf{A}) = \nabla\cdot(\mathbf{A}\times(\nabla\times\mathbf{A})) + \mathbf{A}\cdot\nabla(\nabla\cdot\mathbf{A}) - \mathbf{A}\cdot\nabla^2\mathbf{A}.$$
In the special case where $\nabla\cdot\mathbf{A} = 0$, this of course reduces to,
$$ (\nabla\times\mathbf{A})\cdot(\nabla\times\mathbf{A}) = \nabla\cdot(\mathbf{A}\times(\nabla\times\mathbf{A})) - \mathbf{A}\cdot\nabla^2\mathbf{A}.$$
