2

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.

Calvin Khor
  • 34,903
  • So, it is false that $\omega D+D\omega$ is antisymmetric – Vincenzo Tibullo Sep 18 '20 at 12:12
  • @enzotib thank you very much for the comment, i forgot a very important condition that $\mathcal D$ is symmetric. In application (see linked question) $\mathcal D$ is the symmetric part of the gradient of some vector field $u$, i.e. $\mathcal D = (\nabla u +\nabla u^T)/2$. This has already been edited into the question – Calvin Khor Sep 18 '20 at 12:13
  • I would be very amused and thankful if someone could give a good reason for the recent downvote on the question. Still amused either way – Calvin Khor Oct 22 '21 at 12:33

1 Answers1

2

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.

Funktorality
  • 3,221
  • Wow. I have never seen that trace characterisation! I also just bothered to type up my scribbles detailing the elementary matrix calculations as an answer in the question I linked. I'll check this more carefully soon (ps typo, a projection should satisfy $\mathcal D^2 = \mathcal D$) – Calvin Khor Sep 29 '20 at 07:46
  • @CalvinKhor Fixed, thanks. It’s a surprising characterization, but I couldn’t think of another clean way to introduce the trace. Would be interested in other approaches. – Funktorality Sep 29 '20 at 08:05
  • You say "Suppose $\mathcal{D}$ is a projection". What if it is not? – Vincenzo Tibullo Sep 29 '20 at 08:14
  • @enzotib It’s an interesting fact (see the MO link) that the trace is the only linear operator that has this action on projections. So it actually follows for all D. – Funktorality Sep 29 '20 at 08:18