Help deriving a vorticity equation I am reading Majda & Bertozzi (Vorticity and Incompressible Flow). In page 12 the following equation appears:
$$\frac{D \Omega}{Dt} + \Omega \mathcal{D} + \mathcal{D} \Omega = \nu \Delta \Omega$$
where $\frac{D}{Dt}$ is the convective/lagrangian/material derivative. $\Omega$ and $\mathcal{D}$ are $3$ by $3$ matrices, the first antisymmetric and the second symmetric, and $\nu$ is a scalar. Using that $\Omega$ is defined by $\Omega h = \frac{1}{2} \omega \times h $, where $\omega$ is a vector function representing vorticity, I should be able to get the following vorticity equation (which apparently plays a crucial role in the rest of the book):
$$ \frac{D \omega}{Dt} = \mathcal{D} \omega + \nu \Delta \omega.   $$
Any idea how?
Here is a link to the book
 A: Re-write the main given equation in index notation (following the Einstein summation convention)
$$ D_t \Omega_{ij} + \Omega_{ik}\mathcal{D}_{kj} + \mathcal{D}_{ik}\Omega_{kj} = \nu\triangle \Omega_{ij} \tag{1}$$
Small $\omega$ is defined by 
$$ \Omega_{ik}h^k = \frac12 \epsilon_{ijk}\omega_j h^k \tag{2}$$
which is the cross product definition. The $\epsilon_{ijk}$ is the Levi-Civita symbol (or fully antisymmetric tensor with $\epsilon_{123} = 1$). 
Plugging in (2) (which implies that $\Omega_{ij} = \frac12 \epsilon_{ikj}\omega_k$) into (1) we have that
$$ \epsilon_{ilj} D_t\omega_l + \epsilon_{ilk}\mathcal{D}_{kj}\omega_l + \mathcal{D}_{ik}\epsilon_{klj}\omega_l = \nu \epsilon_{ilj}\triangle \omega_l \tag{3}$$
Next we use the property of the Levi-Civita tensor,
$$ \epsilon_{jik}\epsilon_{jlk} = 2 \delta_{jl} \tag{4}$$
which means that multiplying (3) by $\epsilon_{imj}$ gives
$$ 2D_t\omega_m + \left(\epsilon_{ilk}\epsilon_{imj}\mathcal{D}_{kj} + \epsilon_{klj}\epsilon_{imj}\mathcal{D}_{ik}\right) \omega_l = \nu \triangle \omega_m \tag{5}$$
The antisymmetry properties of the Levi-Civita tensor, as well as the symmetry of the tensor $\mathcal{D}$ can be used to show that
$$ \epsilon_{ilk}\epsilon_{imj}\mathcal{D}_{kj} = \epsilon_{klj}\epsilon_{imj}\mathcal{D}_{ik} $$
So by another property of the Levi-Civita tensor,
$$ \epsilon_{ijk}\epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} \tag{6}$$
we conclude that (5) is equivalent to 
$$ D_t\omega_m + \omega_m (\delta_{kj}\mathcal{D}_{kj} - \mathcal{D}_{jm}) = \nu \triangle \Omega_m ~.$$
Which shows that you in fact omitted one necessary condition for your equation to hold, which is that $\mathcal{D}$, in addition to being symmetric, is also trace-free. 

If you have learned about differential forms, one should treat $\Omega$ as a differential two form on $\mathbb{R}^3$ and $\omega$ as a differential one form on $\mathbb{R}^3$ related by the Hodge star operator $\Omega = *\omega$. From this point of view the equation you want (the one for $\omega$) is merely obtained by taking the Hodge dual of the equation you are given (the one for $\Omega$) plus a little bit of multilinear algebra. 
A: $\newcommand{\D}{\mathcal D}\DeclareMathOperator{Tr}{Tr}$Adding an  approach. First, recall that $$\text{ if $ v\times h = w\times h$ for all $h$, then $v=w$. }\label{*}\tag{*}$$ (Indeed, the only vector perpendicular to all basis vectors is the zero vector.) Matrix-multiplying by $h$ and using $\Omega h= \omega\times h$, we already get (as $D/Dt$ is a scalar operator)
$$\frac{D}{Dt} \omega\times h + (\Omega \D + \D\Omega) h= \nu \Delta \omega \times h .$$
The difficulty, in view of \eqref{*}, is therefore in computing the following identity
$$ (\Omega \D + \D\Omega)h = -(\D \omega)\times h. \label{!}\tag{!}$$
Recall that an antisymmetric matrix is in one-to-one correspondence with cross products with vectors in $\mathbb R^3$:
$$ w=\begin{bmatrix}w_1\\w_2\\w_3\end{bmatrix}\in\mathbb R^3 \qquad \leftrightarrow \qquad [w]_\times=
\begin{bmatrix} 
0 &-w_3& w_2\\
w_3& 0 &-w_1 \\
-w_2& w_1& 0
 \end{bmatrix}\in \operatorname{Antisym}(\mathbb R,3)$$
(In particular $\Omega = [\omega]_\times$.) So since $\Omega \D + \D\Omega$ is antisymmetric, it represents the cross product with something. We just need to see that that something is $-\D\omega$, i.e.
$\Omega \D + \D\Omega = [-\D\omega]_\times$, i.e. if we set
$$\D = \left[d_1 \middle| d_2 \middle| d_3 \Rule{0em}{1em}{0em} \right] = \begin{bmatrix} d_1^T \\\hline d_2^T \\\hline d_3^T\end{bmatrix} =\begin{bmatrix} d_{11} & d_{12} & d_{13} \\d_{12} & d_{22} & d_{23} \\ d_{13} & d_{23} & d_{33} \\\end{bmatrix}, \quad d_{ij}=d_{ji}$$
then we need to show
$$ \Omega \D + \D\Omega =\begin{bmatrix} 
0 &d_3 \cdot \omega & -d_2\cdot \omega \\
-*& 0 &d_1\cdot \omega  \\
*& -*& 0
 \end{bmatrix}.$$
As $(\Omega\D)^T=\D^T\Omega^T  = -\D\Omega$, we just need to compute $\Omega\D$ to compute $\D\Omega$. Remaining computations;
\begin{align} \Omega\D  &= \Omega \left[d_1 \middle| d_2 \middle| d_3 \Rule{0em}{1em}{0em} \right] = \left[\omega\times d_1  \middle| \omega\times d_2 \middle| \omega \times d_3 \Rule{0em}{1em}{0em} \right]\\
&= \begin{bmatrix}
\omega_2d_{13} -\omega_3d_{12} 
& \omega_2d_{23} -\omega_3d_{22} 
& \omega_2d_{33} -\omega_3d_{32}
\\
\omega_3d_{11} -\omega_1d_{13} 
& \omega_3d_{21} -\omega_1d_{23} 
& \omega_3d_{31} -\omega_1d_{33}
\\
\omega_1d_{12} -\omega_2d_{11} 
& \omega_1d_{22} -\omega_2d_{21} 
& \omega_1d_{32} -\omega_2d_{31}
\end{bmatrix},\\
  \Omega \D + \D\Omega 
&= \Omega \D - (\Omega \D)^T 
\\
&=\begin{bmatrix} 
0 &\omega_2d_{23} -\omega_3d_{22} - (\omega_3d_{11} -\omega_1d_{13}) &\omega_2d_{33} -\omega_3d_{32} - (\omega_1d_{12} -\omega_2d_{11}) \\
-*& 0 &\omega_3d_{31}-\omega_1d_{33} - (\omega_1d_{22} -\omega_2d_{21})  \\
*& -*& 0
 \end{bmatrix} \\
&= \begin{bmatrix} 
0 &d_3 \cdot \omega -\omega_3\Tr\D& -d_2\cdot \omega + \omega_2 \Tr\D \\
-*& 0 &d_1\cdot \omega -\omega_1 \Tr \D \\
*& -*& 0
 \end{bmatrix}.\end{align}
So in fact, for general $\Tr\D$ , we have
$$ \Omega \D + \D\Omega = [(\Tr \D) \omega - \D\omega]_\times.$$
In the relevant case $\Tr\D=0$ the result \eqref{!} follows.
PS If there is a way to finish without "looking into the elements of the matrices", I would like to know.
