Is there a 'geometric' version of this derivation of the vorticity equation?

Question

Recall that for each vector $omegainmathbb R^3$, there is an anti-symmetric matrix $ [omega]_timesinmathbb R^{3times 3}$ (and vice-versa) such that $$[omega]_times h= omegatimes h.$$ 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
$ omegatimes(mathcal Dh) + mathcal D(omegatimes h) = (-mathcal Domega)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.

Funktorality · Accepted Answer

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 $$M=omega_timesmathcal D+mathcal Domega_times+(mathcal Domega)_times.$$ By the antisymmetry of the cross product, we easily have $$Momega=0.$$ So we can think of $M$ as an antisymmetric form with $omega$ in its kernel. It follows that $M$ takes the form $$M=comega_times.$$ (One way to see this is that the space of 3d antisymmetric 2-forms is $binom32=3$ dimensional, and contains the 3d space ${omega_times:omegainmathbb R^3}$.) Moreover, clearly $M$ is also linear in $mathcal D$. Therefore there’s a linear functional on $3times3$ 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 $$Mmathcal Domega=omega_timesmathcal Domega+mathcal Domega_timesmathcal Domega=L(mathcal D)omega_xmathcal Domegaquadquad(1)$$ which implies after taking $mathcal D$ again $$2mathcal Domega_xmathcal Domega=L(mathcal D)mathcal Domega_xmathcal Domega.quadquad(2)$$ 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) $mathcal Domega_timesmathcal Domegaequiv0$ But $omega_timesmathcal Domeganotequiv0$ so (1) implies $L(mathcal D)=1$. If $mathcal D$ projects to a 2d space, then $mathcal Domega_timesmathcal Domeganotequiv0$ 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 $$omega_timesmathcal D+mathcal Domega_times+(mathcal Domega)_times=(trmathcal D)omega_times$$ which of course implies the one in the question if $mathcal D$ is traceless.

Is there a 'geometric' version of this derivation of the vorticity equation?

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