Is there a 'geometric' version of this derivation of the vorticity equation? Recall that for each vector $\omega\in\mathbb R^3$, there is an anti-symmetric matrix $ [\omega]_\times\in\mathbb R^{3\times 3}$ (and vice-versa) such that $$[\omega]_\times h= \omega\times h.$$ Matrix product on the left, cross product of vectors on the right. Let $\mathcal D$ be a symmetric and trace-free  matrix (i.e. $\operatorname{tr}\mathcal D=\mathcal D_{11}+\mathcal D_{22}+\mathcal D_{33} = 0$). Then it is easy to check that
$$ [\omega]_\times \mathcal D + \mathcal D[\omega]_\times$$
is also anti-symmetric.
My question:
Is there a way (different from not-dupe) to show that in fact,
$$  [\omega]_\times \mathcal D + \mathcal D[\omega]_\times = [-\mathcal D \omega]_\times?$$Or alternatively, that
$ \omega\times(\mathcal Dh) + \mathcal D(\omega\times h) = (-\mathcal D\omega)\times h$ for all vectors $h$?
I am hoping perhaps for a proof that uses identities involving trace-free/symmetric/antisymmetric matrices, without "directly computing the components" like in the link above.
The calculation in the link above is straightforward, and since there are only 9 components in a matrix, you don't even need the Einstein summation notation. But I think it would be nice to see.
 A: I’m not sure this is the easiest way, so I invite anyone to try to improve it. I will actually prove a more general formula.
Let us define the matrix $$M=\omega_\times\mathcal D+\mathcal D\omega_\times+(\mathcal D\omega)_\times.$$ By the antisymmetry of the cross product, we easily have $$M\omega=0.$$ So we can think of $M$ as an antisymmetric form with $\omega$ in its kernel. It follows that $M$ takes the form $$M=c\omega_\times.$$ (One way to see this is that the space of 3d antisymmetric 2-forms is $\binom32=3$ dimensional, and contains the 3d space $\{\omega_\times:\omega\in\mathbb R^3\}$.) Moreover, clearly $M$ is also linear in $\mathcal D$. Therefore there’s a linear functional on $3\times3$ matrices $L$ such that $$M=L(\mathcal D)\omega_\times.$$
(I’m not sure what follows is the most efficient way to find $L$.) Suppose $\mathcal D$ is a projection (ie. $\mathcal D^2=\mathcal D$). Then $$M\mathcal D\omega=\omega_\times\mathcal D\omega+\mathcal D\omega_\times\mathcal D\omega=L(\mathcal D)\omega_x\mathcal D\omega\quad\quad(1)$$ which implies after taking $\mathcal D$ again $$2\mathcal D\omega_x\mathcal D\omega=L(\mathcal D)\mathcal D\omega_x\mathcal D\omega.\quad\quad(2)$$ We consider several cases depending on the dimensionality of the projection. If $\mathcal D$ projects to a 0d space, then $L(\mathcal D)=0$ (trivial). If $\mathcal D$ projects to a 1d space, then (trivial geometric argument) $\mathcal D\omega_\times\mathcal D\omega\equiv0$ But $\omega_\times\mathcal D\omega\not\equiv0$ so (1) implies $L(\mathcal D)=1$. If $\mathcal D$ projects to a 2d space, then $\mathcal D\omega_\times\mathcal D\omega\not\equiv0$ So (2) implies $L(\mathcal D)=2$. Finally, if $\mathcal D$ projects to 3d space, then it is the identity and trivially we have $L(D)=3$. In conclusion, $L$ is linear and maps projections to the dimension of the projection’s range. These properties characterize $L$ as the trace (see eg. Terry Tao’s answer here). So we actually have the beautiful formula $$\omega_\times\mathcal D+\mathcal D\omega_\times+(\mathcal D\omega)_\times=(tr\mathcal D)\omega_\times$$ which of course implies the one in the question if $\mathcal D$ is traceless.
